Edge- onne tivity augmentation of graphs andhypergraphsAttila Bernáth

Ph.D. thesis submitted to Eötvös Loránd University,Fa ulty of Natural S ien es, Institute of Mathemati sDo toral S hool: Mathemati sDire tor: Miklós La zkovi hDo toral Program: Applied Mathemati sDire tor: György Mi haletzkySupervisors:Béla Vizvári, Dr. habil., asso iate professorTamás Király, Ph.D., resear h fellowThe dissertation was prepared at the Operations Resear h Department of Eötvös LorándUniversity and at the Egerváry Resear h Group on Combinatorial Optimization(MTA-ELTE)O tober 2009

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Contents1 Introdu tion 51.1 A knowledgements - Köszönetnyilvánítás . . . . . . . . . . . . . . . . . . . 51.2 Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.3 Notations and preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . 101.3.1 Graphs and hypergraphs . . . . . . . . . . . . . . . . . . . . . . . . 101.3.2 Set fun tions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141.4 Ora les . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171.5 G-polymatroids and their interse tions . . . . . . . . . . . . . . . . . . . . 192 Edge- onne tivity augmentation by adding (hyper)edges 232.1 Examples: edge- onne tivity requirement fun tions . . . . . . . . . . . . . 242.1.1 Global edge- onne tivity augmentation . . . . . . . . . . . . . . . . 242.1.2 Global ar - onne tivity augmentation of mixed hypergraphs . . . . 252.1.3 Node-to-area edge- onne tivity augmentation . . . . . . . . . . . . 262.1.4 Lo al edge- onne tivity augmentation . . . . . . . . . . . . . . . . . 282.2 Polyhedra and various obje tive fun tions of the augmentation . . . . . . . 292.3 Covering with graph edges: the splitting-o� te hnique . . . . . . . . . . . . 312.3.1 Contra tion of tight sets . . . . . . . . . . . . . . . . . . . . . . . . 343 Covering skew-supermodular fun tions with hyperedges 373.1 Merging hyperedges . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 383.2 Weak overing of positively skew-supermodular fun tions . . . . . . . . . . 413.3 Covering two positively skew-supermodular fun tions . . . . . . . . . . . . 453.4 Appli ations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 473.4.1 Lo al edge- onne tivity augmentation of hypergraphs . . . . . . . . 473.4.2 The node-to-area onne tivity augmentation problem in hypergraphs 483.4.3 Augmenting the global ar - onne tivity of mixed hypergraphs . . . 493

4 Contents4 Covering skew-supermodular fun tions with graph edges 514.1 Previous results - brief history . . . . . . . . . . . . . . . . . . . . . . . . . 524.2 A simple lemma and algorithm . . . . . . . . . . . . . . . . . . . . . . . . 534.2.1 General observations on the stu k situation . . . . . . . . . . . . . . 554.2.2 Simple proofs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 574.3 Stu k situation for spe ial skew-supermodular fun tions . . . . . . . . . . . 594.3.1 Crossing supermodular fun tions . . . . . . . . . . . . . . . . . . . 604.3.2 Crossing negamodular fun tions . . . . . . . . . . . . . . . . . . . . 634.4 Further appli ations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 664.4.1 Lo al edge- onne tivity augmentation of hypergraphs . . . . . . . . 664.4.2 Global ar - onne tivity augmentation of mixed hypergraphs . . . . 684.4.3 Rank-respe ting augmentation of hypergraphs with negamodular on-straints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 705 Covering symmetri rossing supermodular fun tions 755.1 Previous results - brief history . . . . . . . . . . . . . . . . . . . . . . . . . 765.1.1 Partition onstrained global edge- onne tivity augmentation of a graph 785.2 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 815.3 Proof of the theorem of Ben zúr and Frank . . . . . . . . . . . . . . . . . . 855.4 Partition onstrained overing problem . . . . . . . . . . . . . . . . . . . . 905.4.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 905.4.2 The degree-spe i�ed problem . . . . . . . . . . . . . . . . . . . . . 975.4.3 The minimum version . . . . . . . . . . . . . . . . . . . . . . . . . 1125.4.4 Appli ation: Partition onstrained global edge- onne tivity augmen-tation of a hypergraph . . . . . . . . . . . . . . . . . . . . . . . . . 1186 Sour e lo ation in hypergraphs 1216.1 Problem formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1226.2 Compatible requirement- and weight fun tion . . . . . . . . . . . . . . . . 1236.3 Implementation of Step (*) . . . . . . . . . . . . . . . . . . . . . . . . . . . 1246.4 Improving the running time for uniform requirements . . . . . . . . . . . . 1267 Open problems 129

Chapter 1Introdu tion1.1 A knowledgements - KöszönetnyilvánításKöszönetet szeretnék mondani mindenekel®tt a saládomnak. Köszönöm a szüleimnek asok bíztatást és lehet®séget: megtanultam t®lük a munka szeretetét és azt, hogy a legjobbmotivá ió a dí séret. Köszönöm feleségemnek, hogy a nehéz pillanatokban se hagyottel süggedni. Köszönöm a gyermekeimnek, hogy megtanítottak értékelni és kihasználni aztaz id®t amit együtt tölthetünk, és nemkülömben azt az id®t, amit külön töltünk.Köszönöm pedagógusaimnak, tanáraimnak, akik az általános iskolától, s®t az óvodátólkezdve bontogatták az érdekl®désemet a világ, azon belül is a matematika iránt. Haddemlítsem nevükön a matematika tanáraimat. Köszönöm Molnár Éva 1-2. osztályos tanítónénimnek, Tóth János 3-4. osztályos tanító bá simnak, Németh Ágnes fels® tagozatos tanárnénimnek az általános iskolából, Bíróné Mihály� Erzsébet tanárn®nek a gimnáziumból.Az egyetem matematika-�zika, kés®bb matematikus szakán már sok matematika tanárnevét kellene megemlítenem, ezért nem teszek kísérletet egy teljes felsorolásra. Kiemel-ném viszont Szamuely Tamás algebra gyakorlatvezet®met, aki el®ször bíztatott arra, hogymatematikus szakra jöjjek. Köszönöm továbbá Frank Andrásnak, hogy lebilin sel® el®adá-saival olthatatlan tudásszomjat ébresztett bennem a kombinatorikus optimalizálás és agráfelmélet iránt, és hogy rengeteg szakmai segítséget nyújtott és anyagi lehet®séget biz-tosított arra, hogy kutassam ezeket a területeket.Köszönöm témavezet®mnek, Vizvári Bélának, hogy elvállalta a témavezetést. Köszönömtovábbá kollégáimnak a sok együttm¶ködést és segítséget. Közülük is el®sorban konzu-lensemnek, Király Tamásnak volt meghatározó szerepe abban, hogy ez a dolgozat egyál-talán megszületett, mert mindig hajlandó volt végighallgatni az elképzeléseimet, mindigszámítani lehetett rá, hogy észreveszi (és általában ki is javítja) a hibákat a gondolatmenetem-ben. Köszönöm a társszerz®imnek, hogy megpróbáltak velem együtt gondolkodni, megosz-5

6 Chapter 1. Introdu tiontották velem részeredményeiket és ötleteiket és meghallgatták az enyémeket. Hadd em-lítsem ®ket ismét név szerint, még ha néhányukkal a közös publiká ió nem is szerepelebben a disszertá ióban: Henning Bruhn, Gerbner Dániel, Roland Grappe, Satoru Iwata,Gwenaël Joret, Király Zoltán, Szigeti Zoltán, és természetesen Király Tamás. Köszönömtovábbá az EGRES soport minden tagjának a sok segítséget, hogy elviseltek a disszertá ióel®készítésének és megírásának minden (különösen a legvégs®) periódusában, ellen®riztéka kézirataimat és meghallgatták a bizonyításvázlataimat.1.2 OverviewThis thesis is devoted to edge- onne tivity augmentation whi h is the following: we aregiven a network (modelled by a graph or hypergraph) that we somehow want to make morerobust against the failure (deletion) of its onne tions (edges or hyperedges). The operationthat we an do is to introdu e new onne tions between already existing nodes.Most of this thesis is about this notion of edge- onne tivity augmentation (apart from thelast hapter about sour e lo ation where we onsider a di�erent augmentation te hnique �more details later). We emphasize that though we sometimes allow dire ted stru tures asthe basi network to be augmented (dire ted graphs and hypergraphs, or more generallymixed graphs and hypergraphs), the new onne tions to be added will always beundire ted. This is be ause our te hniques (e.g. polyhedral onsiderations) do not applywhen we augment with dire ted ar s or hyperar s.The obje tive fun tion of the augmentation is also ommon to the problems we investi-gate: we want to minimize the total size of the hyperedges to be added. When weare only allowed to add graph edges then this is two times the number of the new edges.A little bit more general obje tive fun tion is also onsidered in the so alled minimumnode- ost augmentation problems (exa t de�nition to be given later). The reason we onsider these obje tive fun tions is simple: the polyhedral relations show that these aretra table, while if a ost of a new hyperedge does not only depend on the ost of its verti esthen the simplest augmentation problems already be ome intra table.The starting point of our investigations is Menger's theorem (with its many di�erentversions), whi h allows us to reformulate the augmentation problem into a overingproblem. By overing problem we mean a problem of the following form: given a setfun tion p : 2V → Z ∪ {−∞} over the �nite ground set V , �nd a graph (or hypergraph)G = (V, E) overing the fun tion p, meaning that dG(X) ≥ p(X) has to hold for anysubset X ⊆ V (here dG(X) denotes the number of (hyper)edges of G interse ting both Xand V − X). A whole hapter (Chapter 2) is devoted to showing how di�erent versions

Se tion 1.2. Overview 7of the edge- onne tivity augmentation problem an be formulated as a overing problemwith a suitably hosen requirement (or de� ien y) fun tion p.The above notion of edge- onne tivity augmentation is quite well studied and has a hugeliterature. For two re ent surveys we refer the reader to [43℄ and [21℄. Let us �rst (in thisand the next paragraph) restri t ourselves to augmentation with graph edges. The�rst result is due to Watanabe and Nakamura [45℄ on global edge- onne tivity augmentationof graphs. An important milestone is due to Cai and Sun [13℄ and Frank [19℄ who realizedthat the splitting-o� te hnique introdu ed by Lovász [33℄ is a very useful tool in solvingedge- onne tivity augmentation problems with graph edges. This method will also be usedin this thesis. Frank observed the importan e of skew-supermodular fun tions and provedin [19℄ the deep result that is behind the onne tion between degree-spe i�ed augmentationand augmentation with a minimum number of edges. With this method and the splitting-o� theorem due to Mader [34℄, Frank [19℄ solved the lo al edge- onne tivity augmentation ofgraphs. Another result due to Bang-Jensen, Frank and Ja kson [2℄ uses a similar approa hto in rease the global ar - onne tivity of mixed graphs with undire ted edges. A re ent ap-proa h is the node-to-area onne tivity augmentation introdu ed by Japanese resear hers.Here the requirement is not between node-pairs, but between nodes and nodesets. Thoughthe basi problem is NP-hard, a surprising observation due to Ishii and Hagiwara [26℄ saysthat if we restri t ourselves to the (most interesting) ase when the requirements are all atleast 2, then the problem be omes tra table. Bang-Jensen and Ja kson [4℄ generalized theresult of Watanabe and Nakamura in an other dire tion and solved the problem of aug-menting the global edge- onne tivity of a hypergraph with graph edges. An abstra t versionof this result, the problem of overing a symmetri rossing supermodular fun tion withgraph edges was solved by Ben zúr and Frank [5℄. Yet another generalization of Watanabeand Nakamura's result, the partition onstrained global edge- onne tivity augmentation ofgraphs was solved by Bang-Jensen, Gabow, Jordán and Szigeti [3℄.What an we add to the widely studied area of edge- onne tivity augmentation (withgraph edges) in this thesis? We have found a new and onvenient approa h to the splitting-o� te hnique whi h simpli�es the dis ussion to a great extent. The main breakthrough isa simple lemma (Lemma 4.6) that, instead of only looking at dangerous sets (that are theimportant obje ts in splitting-o� theorems) as in previous proofs, onsiders the interrelationbetween dangerous sets and sets with maximum de� ien y and thus simpi�es the furtherdis ussion. This gives a simpli� ation of many of the known proofs and enables us to provenew results, too. Furthermore, we have given an uni�ed dis ussion of the results mentionedabove based on the properties of the de� ien y fun tions of the problems. The most general lass of set fun tions is skew-supermodular fun tions that are hard to handle in general (the

8 Chapter 1. Introdu tionsymmetry of our fun tion an be assumed by the symmetrizing operation). However,in many ases the fun tion to be handled is from a more restri ted lass, for exampleit is the symmetrized of a rossing supermodular fun tion as in global ar - onne tivityaugmentation of mixed hypergraphs, or it is the symmetrized of a rossing negamodularfun tion as in the node-to-area onne tivity augmentation problem.If we are also allowed to use hyperedges of arbitrary size then the problems be omesimpler. Our starting point is Theorem 3.2 of Szigeti on overing skew-supermodular fun -tions with hyperedges. The proof of this theorem was onsiderably simpli�ed by TamásKirály, who used the hyperedge merging te hnique, whi h might be onsidered as a gener-alization of splitting-o�. This proof is very fas inating, and with some further observationsand some polyhedral results we (Tamás Király and the author) managed to generalizeSzigeti's theorem in many dire tions. For example we have shown that the hypergraph overing our fun tion an be hosen nearly uniform, or in some restri ted ases we aneven solve the simultaneous overing of two fun tions optimally. Appli ations in- lude the lo al edge- onne tivity augmentation of hypergraphs, the global ar - onn etivityof mixed hypergraphs with undire ted hyperedges, and the node-to-area onne tivity aug-mentation in hypergraphs.In Chapter 6 we onsider versions of the sour e lo ation problem, whi h an beinterpreted as a di�erent notion of edge- onne tivity augmentation. Here, instead of addingnew (hyper)edges onne ting existsing nodes, we are allowed to ontra t a suitably hosenset of nodes ( alled sour e set). The motivation of this problem omes from the followingappli ation: given a network onsisiting of omputers, say, and some onne tions betweenthem, we want to de ide whi h of these omputers to use as servers for a servi e that wewant to provide for our users (who are also lo ated at the omputers of this networks).The requirement is that the users must have good edge- onne tivity to the set of servers,and the ost of hoosing a node as a server may vary from one node to the other. We onsider hypergraphi generalizations of ertain sour e lo ation problems in Chapter 6.In this thesis we will only on entrate on polynomial solvability of the problems. Wewill often expli itly des ribe algorithms, too, and we show that they have a polynomialrunning time, but usually we do not aim to a hieve the best running times. Instead, wetry to �nd a on eptually simple algorithm for our problem.The stru ture of the thesis is as follows. In the rest of this hapter we introdu e themost important notations and de�nitions that we will need, we introdu e the appropriateora les needed in our abstra t set overing algorithms, and we re all the important resultsabout g-polymatroids.Chapter 2 is still an introdu tory hapter. Here we �rst of all show how the edge-

Se tion 1.2. Overview 9 onne tivity augmentation problems that we plan to speak about an be reformulated to overing problems. This means the introdu tion of de� ien y fun tions and showing theirmain properties. This an be found in Se tion 2.1. In Se tion 2.2 we show the tra tableobje tive fun tions of our optimization problems, and we show the onne tion betweendegree-spe i�ed overing and the minimum version. Lastly, in Se tion 2.3 we re all thebasi fa ts about the splitting-o� te hnique needed later.Chapter 3 is devoted to overing a skew-supermodular fun tion with hyperedges. First wegive the result on merging hyperedges originally due to Tamás Király, but in a generalizedform. In Se tion 3.2 we introdu e the notion of weak overing of a set fun tion andwe show how this relates to g-polymatroids. Furthermore we show that if we use thesmallest number of hyperedges then weak overing implies overing. These are the mainbuilding blo ks of our results on overing a skew-supermodular fun tion with nearly uniformhyperedges and overing two skew-supermodular fun tions simultaneously. In Se tion 3.4we state the appli ations of these results.In Chapter 4 we turn to overing skew-supermodular fun tions with graph edges. Aftera brief review of previous results we give our key lemma, Lemma 4.6. In fa t this lemmawas also observed by Nutov in 2005, but he did not publish a omplete proof of it (aspe ial ase of this lemma also appears in the Ph.D. thesis of Ben Cosh [15℄, but the proofis rather umbersome). We have found a very simple proof for this lemma. We look atthe onsequen es of this lemma in the subsequent se tions: we show how a greedy typesplitting-o� algorithm would get stu k in general (Se tion 4.2.1), and why it will not getstu k in many well known spe ial ases (Se tion 4.2.2 on simple proofs of known results).Then in Se tion 4.3 we des ribe (as far as we an) this stu k situation for spe ial skew-supermodular fun tions. In Se tion 4.4 we give appli ations of our results of the previousse tions.Chapter 5 is devoted to overing symmetri rossing supermodular fun tions with graphedges. This is a spe ial ase of the problem of the previous hapter, however we thoughtthat this order moving �from the general to the spe ial� is more onvenient in our dis ussion.After reviewing the known results and giving some preliminaries, we give a relatively simpleproof of the theorem of Ben zúr and Frank that solves the problem of overing a symmetri rossing supermodular fun tion with graph edges. This proof enables us to generalize theresult further by posing extra onditions on the graph edges allowed in the overing. Theproblem onsidered is the partition onstrained version of overing a symmetri rossingsupermodular fun tion. We give a solution of this problem in Se tion 5.4. In Se tion 5.4.4we des ribe the appli ation of global edge- onne tivity augmentation of a hypergraph witha multipartite graph.

10 Chapter 1. Introdu tionChapter 6 is devoted to hypergraphi versions of the sour e lo ation problem. In Se tion6.1 we introdu e the problems that we want to solve: in parti ular we introdu e an abstra tversion of the sour e lo ation problem. In Se tion 6.2 we give a greedy type algorithm thatsolves the tra table version (that is, ompatible requirement- and weight fun tion) of ourproblem in general. However, it is not lear how to implement a step of it, sin e it isnot known how to minimize an interse ting posimodular fun tion in polynomial time. InSe tion 6.3 we show how to implement this step of the previous algorithm if the fun tionis also submodular, whi h is true for the hypergraphi sour e lo ation. Finally, in Se tion6.4 we show a leverer and faster algorithm for the ase when the requirement fun tion isuniform.In Chapter 7 we enumerate some open problems that, if solved, would have �t into thisthesis.1.3 Notations and preliminariesIn the following se tions we introdu e the notions and notations that will be used in thethesis.1.3.1 Graphs and hypergraphsIn the whole thesis V will be a �nite ground set and we will usually use n = |V |. Forsubsets X and Y of V , we use the notation X − Y := {v ∈ X : v /∈ Y }, X + Y = X ∪ Y ,and X = V − X (if the ground set is lear from the ontext). Two subsets X, Y ⊆ V are alled o-disjoint if X ∪ Y = V .A partition of the set V is a olle tion X1, X2, . . . , Xt of nonempty disjoint subsetsof V with V = ∪t1Xi (in some ases we allow that some members of a partition an beempty, but we will always state this expli itly). A subpartition of V is a partition of asubset of V . Let S(V ) denote the set of all subpartitions of the set V . If a set X has onlyone element x then we will all it a singleton and we will often write x instead of {x}(for example if d : 2V → R is a set fun tion then d(x) means d({x})). The hara teristi fun tion of a set X will be denoted by χX : V → {0; 1}, i.e. χX(v) = 1 if v ∈ X and

χX(v) = 0 otherwise. For s, t ∈ V , a set X ⊆ V is an st-set if s /∈ X and t ∈ X.In the thesis Z+ (Q+, R+) will denote the set of non-negative integer (rational, real)numbers.Sets X, Y ⊆ V are properly interse ting if X∩Y, X−Y and Y −X are all nonempty:this terminology is due to András Frank and we de ided to use it in order to larify thedistin tion between interse ting (i.e. not disjoint) set pairs and properly interse ting set

Se tion 1.3. Notations and preliminaries 11pairs. If furthermore X ∪ Y 6= V then we say that the two sets are rossing. For afamily F ⊆ 2V let co(F) = {X ⊆ V : X ∈ F}. We say that F is a ring family ifX ∩ Y, X ∪ Y ∈ F holds for any X, Y ∈ F . We say that F is an interse ting family( rossing family) if X ∩ Y, X ∪ Y ∈ F holds for any interse ting (respe tively rossing)pair X, Y ∈ F .An undire ted hypergraph (or shortly hypergraph) H = (V, E) is a pair of a �niteset V and a family E of subsets of V (repetitions are allowed). The set V is alled thenode set of the hypergraph, the family E is alled the edge set of the hypergraph. Anelement e of E will be alled a hyperedge. The ardinality of e is denoted by |e|.If the ardinality of every hyperedge is 2 in a hypergraph then it is alled a graph. Agraph is usually denoted by G = (V, E), the elements of E are alled edges. A graph an ontain parallel edges and uv or (u, v) will denote an arbitrary edge between the nodesu, v ∈ V . Note however that in some ases we will allow loops in graphs, so stri tlyspeaking a graph (with loops) is not a spe ial ase of a hypergraph, but this small abuseof notation will not ause any misunderstanding.In a hypergraph H , a path between nodes s and t is an alternating sequen e of distin tnodes and hyperedges s = v0, e1, v1, e2, . . . , ek, vk = t, su h that vi−1, vi ∈ ei for all i between1 and k. H is onne ted if there is a path between any two distin t nodes.For a hypergraph H = (V, E) and a set X ⊆ V we say that a hyperedge e ∈ E enters Xif neither e ∩ X nor e ∩ (V − X) is empty, and we de�ne ∆H(X) = {e ∈ E : e enters X}(the set of hyperedges entering X) and dH(X) = ∆H(X) (the degree of X in H). The setfun tion dH : 2V → Z+ is a symmetri submodular fun tion (see the de�nition in Se tion1.3.2). For graphs we will sometimes have to ount loop edges in the degree of singetons,therefore we introdu e the fun tion d+

G : V → Z for a graph G = (V, E) to mean thatd+

G(v) is dG(v) plus two times the number of loops in ident to v (for a hypergraph H letd+

H(v) be dH(v) plus the number of singleton hyperedges {v}). For a graph G = (V, E) andX, Y ⊆ V we introdu e the notation dG(X, Y ) to be equal to the number of edges of Gwith one endpoint in X−Y and the other in Y −X. Let dG(X, Y ) = dG(X, Y ) = dG(Y, X).We will use that the following equalities hold for any graph G = (V, E) and X, Y ⊆ V .

dG(X) + dG(Y ) = dG(X ∩ Y ) + dG(X ∪ Y ) + 2dG(X, Y ), (1.1)dG(X) + dG(Y ) = dG(X − Y ) + dG(Y − X) + 2dG(X, Y ). (1.2)More generally, one an prove the following equalities for any hypergraph H = (V, E) and

X, Y ⊆ V .dH(X) + dH(Y ) = dH(X ∪ Y ) + dH(X ∩ Y ) + 2d1(X, Y ) + d2(X, Y ), (1.3)dH(X) + dH(Y ) = dH(Y − X) + dH(X − Y ) + 2d1(X, Y ) + d2(X, Y ), (1.4)

12 Chapter 1. Introdu tionwhere d1(X, Y ) is the number of hyperedges interse ting only X−Y and Y −X and neitherX ∩ Y nor V − (X ∪ Y ), d2(X, Y ) is the number of hyperedges interse ting X − Y andY −X and exa tly one of X∩Y and V −(X∪Y ), and d1(X, Y ) = d1(X, Y ) = d1(X, Y ) andd2(X, Y ) = d2(X, Y ) = d2(X, Y ) (i.e. d1(X, Y ) is the number of hyperedges interse tingonly X ∩Y and V − (X ∪Y ), d2(X, Y ) the number of hyperedges interse ting X ∩Y andV − (X ∪ Y ) and exa tly one of X − Y and Y − X).We say that the hypergraph H overs a set fun tion p if dH(X) ≥ p(X) for anyX ⊆ V (shortly dH ≥ p). The total size of the hypergraph is the sum of the ardinalitiesof the hyperedges: if our hypergraph is a graph then this is two times the number of theedges of this graph. The rank of a hypergraph is the size of the largest hyperedge init. A hypergraph is said to be uniform if the size of every hyperedge is the same. Wesay that the hypergraph is nearly uniform, if the largest hyperedge is at most one biggerthan the smallest one.Sin e the thesis is about edge- onne tivity augmentation, the following de�nition is ofbasi importan e.De�nition 1.1. Given a hypergraph H = (V, E) and sets X, Y ⊆ V , the edge- onne tivitybetween X and Y , denoted by λH(X, Y ), is the maximum number of edge-disjoint pathsstarting in X and ending in Y (we say that λH(X, Y ) = ∞ if X ∩ Y 6= ∅). The subs riptH may be omitted if no onfusion an arise.It is well known that Menger's theorem an be generalized for hypergraphs:Theorem 1.2. Let H = (V, E) be a hypergraph, and S, T ⊆ V . Then

λH(S, T ) = min{dH(X) : T ⊆ X ⊆ V − S}.For nodes u, v ∈ V we will also say that λH(u, v) is the lo al edge- onne tivitybetween u and v. A hypergraph H = (V, E) is k-edge- onne ted if λH(u, v) ≥ k for anyu, v ∈ V (where k is a positive integer). By Menger's theorem this is equivalent to sayingthat dH(X) ≥ k for any nonempty X ( V . More generally, for a fun tion r : V × V → Z+we say that H is r-edge- onne ted if λH(u, v) ≥ r(u, v) for any u, v ∈ V . Observe thatwe an assume that r is symmetri , i.e. r(u, v) = r(v, u) for any u, v ∈ V . This follows fromthe observation that H is r-edge- onne ted if and only if H is r′-edge- onne ted, wherer′(u, v) = max{r(u, v), r(v, u)} for any u, v ∈ V .In some ases it will be useful to allow in undire ted hypergraphs that nodes havemultipli ities in the hyperedges. So let us introdu e the notion of multihypergraph asa pair H = (V, E) of a �nite set V and a family E = {e : V → Z+} of fun tions, alledmultihyperedges (sometimes simply hyperedges). Multihyperedges are onsidered to be

Se tion 1.3. Notations and preliminaries 13multisets identi�ed by their hara teristi fun tions, i.e. e(v) equals the multipli ity of thenode v in the hyperedge e. This is just a natural generalization of loop edges in graphs.A hypergraph is learly a spe ial ase of a multihypergraph, if we identify the hyperedgeswith their hara teristi ve tors. For a multihypergraph we an de�ne the underlyinghypergraph that simply means that we forget about the multipli ities of the nodes in thehyperedges (i.e. if H = (V, E) is a multihypergraph then the underlying hypergraph of His (V, {max(e, 1) : e ∈ E})). Most of the de�nitions for multihypergraphs do not use themultipli ities of the nodes in the multihyperedges: if H = (V, E) is a multihypergraph thena path between two nodes, the lo al edge- onne tivity in H , the degree of a setX ⊆ V is de�ned through the underlying hypergraph of H . The most importantdi�eren e is in the de�nition of the fun tion d+

H : for a multihypergraph H = (V, E) andnode v ∈ V , d+H(v) =

∑

e∈E e(v). Also, the ardinality of a multihyperedge e is de�nedwith |e| =∑

v∈V e(v). A fun tion m : v → Z+ will often be alled a degree spe i� ation.We say that the multihypergraph H satis�es the degree-spe i� ation m : v → Z+ ifd+

H(v) = m(v) for every v ∈ V . We note that the notion of multihypergraphs is not veryimportant in this thesis: we only introdu e them be ause some theorems have a simplerform if we formulate them for multihypergraphs instead of hypergraphs, and we will try toshow these simpli� ations. In fa t we only speak about multihypergraphs in Chapter 3.Another generalization of a hypergraph is a mixed hypergraph. A mixed hypergraphM = (V,A) is a pair of a �nite set V and a family A ontaining nonempty ordered pairsof subsets of V (the same pair an o ur more than on e). The elements of A are alledhyperar s. For a hyperar a = (Ta, Ha) ∈ A, the set Ta is alled the tail set of a, whileHa is alled the head set of a. More intuitively we an think of a hyperar a as thesubset Ta ∪Ha of V , in whi h every node is either a head node, a tail node or even both(head-tail node), su h that every hyperar ontains at least one head and at least one tail.Hopefully it will not ause any onfusion if we sometimes refer to a hyperar a = (Ta, Ha)as a single set Ta ∪ Ha: for example this way we de�ne the size of a as |a| = |Ta ∪ Ha|,or the rank of a mixed hypergraph as the maximum size of its hyperar s. When wesay that v is a tail node of a hyperar a then we also allow that it is a head-tail node (andsimilarly for head nodes). An undire ted hypergraph an be onsidered (for our purposes)as a spe ial mixed hypergraph where every node in a hyperar is a head-tail node of thishyperar . For a mixed hypergraph M = (V,A) and some X ⊆ V , the restri tion of M toX is a mixed hypergraph de�ned as M [X] = (X, {a ∈ A : a ⊆ X}).A mixed graph is the spe ial ase of a mixed hypergraph where every hyperar is of ardinality two. For a mixed graph G and sets X, Y ⊆ V let dG(X, Y ) denote the numberof (undire ted or dire ted) edges of G with one endpoint in X −Y and the other in Y −X.

14 Chapter 1. Introdu tionIn a mixed hypergraph M , a path between nodes s and t is an alternating sequen e ofdistin t nodes and hyperar s s = v0, a1, v1, a2, . . . , ak, vk = t, su h that vi−1 is a tail nodeof ai and vi is a head node of ai for all i between 1 and k. A hyperar a enters a set X ifthere is a head node of a in X and there is a tail node of a in V −X. For a set X we de�ne̺M(X) = |{a ∈ A : a enters X}| (the in-degree of X) and δM (X) = ̺M(V − X) (theout-degree of X). It is easy to he k that both ̺ and δ are submodular (again, see thede�nition later). Given a mixed hypergraph M = (V,A) and sets S, T ⊆ V , let λM(S, T )denote the maximum number of ar -disjoint paths starting in S and ending in T (we saythat λM(S, T ) = ∞ if S ∩ T 6= ∅). The following version of Menger's theorem holds formixed hypergraphs:Theorem 1.3. Let M = (V,A) be a mixed hypergraph, and S, T ⊆ V . Then

λM(S, T ) = min{̺M(X) : T ⊆ X ⊆ V − S}.De�nition 1.4. If M = (V,A) is a mixed hypergraph, r ∈ V is a designated root node,and k, l are nonnegative integers, then we say that M is (k, l)-ar - onne ted from r ifλM(r, v) ≥ k and λM(v, r) ≥ l for any v ∈ V .1.3.2 Set fun tionsBy set fun tions we will usually mean fun tions of the form p : 2V → Z ∪ {∞,−∞}. Letus give the most important de�nitions for set fun tions needed in the thesis.Let p be a set fun tion. We will say that a set X ⊆ V is p-positive if p(X) > 0. We willwidely use the notation Mp = max{p(X) : X ⊆ V }. Any fun tion m : V → R also indu esa set fun tion (that will also be denoted by m) with the de�nition m(X) =

∑

v∈X m(v)for any X ⊆ V . This notation together with the simpli� ation p({x}) = p(x) introdu edearlier for a set fun tion p an again ause some onfusion, but we hope that after thisremark it will not (i.e. if p is a set fun tion then p(X) is not ne essarily equal to ∑

v∈X p(v)).A set fun tion p : 2V → Z∪{−∞} is alled skew-supermodular if at least one of thefollowing two inequalities holds for every X, Y ⊆ V :p(X) + p(Y ) ≤ p(X ∩ Y ) + p(X ∪ Y ), (∩∪)

p(X) + p(Y ) ≤ p(X − Y ) + p(Y − X). (−)A set fun tion is symmetri if p(X) = p(V − X) for every X ⊆ V . Symmetri skew-supermodular fun tions will be very important in this thesis. If p : 2V → Z ∪ {−∞} is aset fun tion and (∩∪) holds for some sets X, Y ⊆ V then we say that X and Y satisfy(∩∪) , or shortly that X(∩∪) Y : if we don't expli itly say whi h fun tion is meant then we

Se tion 1.3. Notations and preliminaries 15always mean p. The same notation is used for (−) . Note that X(∩∪) Y is equivalent withX(−) Y .A set fun tion p : 2V → Z ∪ {−∞} is alled supermodular if it satis�es (∩∪) forany pair X and Y . A set fun tion b is alled submodular if −b is supermodular. Themost important example of a symmetri submodular fun tion in this thesis is the degreefun tion of a hypergraph. A set fun tion p : 2V → Z ∪ {−∞} is alled negamodularif it satis�es (−) for any pair X and Y . A set fun tion b is alled posimodular if −b isnegamodular. Sometimes the supermodular inequality (∩∪) does not hold for any pair ofsets X, Y , only for spe ial pairs. If p satis�es (∩∪) for any properly interse ting pair X, Y ,then we say that it is interse ting supermodular. If p satis�es (∩∪) only for rossingpairs X, Y , then we say that p is rossing supermodular. Interse ting or rossingsubmodular (negamodular, posimodular) fun tions are de�ned analogously.Let us weaken further the notions introdu ed above. We say that a fun tion p : 2V →

Z ∪ {−∞} is positively skew-supermodular if at least one of (∩∪) or (−) holds forany X, Y ⊆ V with p(X), p(Y ) > 0. Note that we do not require that su h a fun tionis nonnegative, unlike it is usually assumed in the literature. For a set fun tion p letp+(X) = max(p(X), 0) for any X ⊆ V : if p is skew-supermodular then p+ is positivelyskew-supermodular. We an also generalize the notion of rossing supermodular fun tions:a set fun tion p : 2V → Z ∪ {−∞} is alled positively rossing supermodular if itsatis�es (∩∪) for any rossing pair X and Y with p(X), p(Y ) > 0. However positively rossing (or skew-) supermodular fun tions annot be handled algorithmi ally: this will bedetailed later. Fortunately, in our appli ations we will only meet rossing supermodularand skew-supermodular fun tions. We an de�ne positively rossing negamodularfun tions analogously.A symmetri , (positively) rossing supermodular fun tion p will satisfy both (∩∪) and(−) for an arbitrary (p-positive) rossing pair X, Y ⊆ V . An important observation is thefollowing: if p is (positively) skew-supermodular and H is a hypergraph, then so is p− dH .Similar statement holds if p is (positively) rossing supermodular or (positively) rossingnegamodular et .Let q : 2V → Z∪{−∞} be a set fun tion. De�ne the omplement of q as q(X) = q(X)and the symmetrized of q by qs(X) = max{q(X), q(X)} for any X ⊆ V : note that qsis a symmetri fun tion. Observe that a hypergraph H overs q if and only if H overs qs.Note that a rossing supermodular (or rossing negamodular) fun tion is not ne essarilyskew-supermodular: we annot ensure any of (∩∪) and (−) for o-disjoint sets (i.e. whenX ∪ Y = V ). However the symmetrized of su h fun tions is already skew-supermodular.The following laim is easy to prove, the proof is left to the reader.

16 Chapter 1. Introdu tionClaim 1.5. The symmetrized of a rossing supermodular, rossing negamodular or skew-supermodular fun tion is (symmetri and) skew-supermodular.Similar statement holds for the symmetrized of positively rossing supermodular, ross-ing negamodular or skew-supermodular fun tions: their symmetrized will be positivelyskew-supermodular. We will often use the following strengthening of (∩∪) and (−) im-plied by (1.1) and (1.2). Assume that p is of form p = p0 − dG with some positivelyskew-supermodular fun tion p0 and graph G = (V, E). For two p0-positive subsets X andY of V , at least one of the following two inequalities holds.

p(X) + p(Y ) ≤ p(X ∩ Y ) + p(X ∪ Y ) − 2dG(X, Y ), (1.5)p(X) + p(Y ) ≤ p(X − Y ) + p(Y − X) − 2dG(X, Y ). (1.6)Again, if p0 is symmetri and positively rossing supermodular and X, Y are rossingand p0-positive, then the reader an see that both (1.5) and (1.6) holds. The followingsimilar statement is implied by (1.3) and (1.4).Lemma 1.6. Assume that p is of form p = p0 − dH with some symmetri positively skew-supermodular fun tion p0 and hypergraph H = (V, E). Let X, Y ⊆ V .(i) If X and Y satisfy (∩∪) for p0, but there is a hyperedge e ∈ E su h that e ∩ (X −

Y ) 6= ∅, e ∩ (Y − X) 6= ∅, and e interse ts at most one of X ∩ Y and X ∪ Y , thenp(X) + p(Y ) < p(X ∩ Y ) + p(X ∪ Y ).(ii) If X and Y satisfy (−) for p0, but there is a hyperedge e ∈ E su h that e ∩ (X ∩

Y ) 6= ∅, e ∩ (X ∪ Y ) 6= ∅, and e interse ts at most one of X − Y and Y − X, thenp(X) + p(Y ) < p(X − Y ) + p(Y − X).The following property of a positively rossing supermodular fun tion will be very useful.Claim 1.7. If q is a positively rossing supermodular fun tion, and X, Y are q-positive rossing sets with q(Y ) ≥ q(X ∩ Y ) then q(X) ≤ q(X ∪ Y ).The following laim generalizes the pre eding one, it an be proved by a simple indu tion.Claim 1.8. Let q be a positively rossing supermodular fun tion and X1, . . . , Xk be q-positive subsets of V so that Xj rosses ⋃j−1

i=1 Xi and q(Xj ∩ (⋃j−1

i=1 Xi)) ≤ q(Xj) for anyj = 2, . . . , k. Then q(

⋃j−11 Xi) ≤ q(

⋃j1 Xi) for any j = 2, . . . , k. Consequently q(X1) ≤

q(⋃j

1 Xi) ≤ q(⋃k

1 Xi) for any j = 1, 2, . . . , k.Claim 1.8 will be usually applied under mu h simpler ir umstan es in the followingform (in many ases ij will just be 1 for every j when we apply Claim 1.9).

Se tion 1.4. Ora les 17Claim 1.9. Let {W1, . . . , Wk} be a subpartition of V so that ⋃k1 Wi 6= V , and for every

j = 2, . . . , k there exists an ij su h that 1 ≤ ij < j, q(Wij) = 1, and q(Wij ∪ Wj) ≥ 1.Then q(⋃k

1 Wi) ≥ 1.Proof. Apply Claim 1.8 for the sets Wij ∪ Wj .On the other hand, if q is positively rossing negamodular then an important observationis the following.If X, Y ⊆ V are rossing, q(X) = q(Y ) = Mq > 0, then q(X∩Y ) = Mq and q(X∪Y ) = Mq.(1.7)The following laim generalizes this statement in one dire tion. It is proved by a simpleindu tion.Claim 1.10. Let q : 2V → Z be rossing negamodular and assume that X0, X1, X2, . . . , Xtare subsets of V (where t ≥ 0) su h that q(X0) = q(X1) = q(X2) = · · · = q(Xt) = Mq > 0and Xi rosses X0 ∪⋃

j<i Xj for any i = 1, 2, . . . , t. Then q(X0 ∪⋃

j≤t Xj) = Mq.1.4 Ora lesIn our abstra t algorithms we will usually deal with some set fun tion p : 2V → Z∪{−∞}.However it is usually not allowed to enumerate all the fun tion values for every X ⊆ V .So we will think of the fun tion as something available through an ora le. But whatkind of questions an the ora le answer for us? The most straightforward idea is a fun tionevaluation ora le: we pass it a subset X ⊆ V and it tells us the value p(X). However thiswill usually not be su� ient for us: with this we annot even de ide whether the fun tiontakes �nite values at all. There are other problems with this ora le, too, therefore we willneed a more lever ora le. One an think of many types of ora les: we will introdu ehere two onvenient ones. The �rst one is the simple fun tion evaluation ora le, and these ond is a tri ky one, whi h will always be su� ient for our purposes. Then we will showthe relationship between these two ora les for di�erent types of fun tions. We say that ahypergraph H = (V, E) is given expli itly if it is en oded by enumerating the (di�erent)hyperedges as ve tors in {0, 1}V and for every su h hyperedge its multipli ity is given bya binary integer.De�nition 1.11. An evaluation ora le for a set fun tion p : 2V → Z ∪ {−∞} takes asinput a subset X of V and it returns the fun tion value p(X).

18 Chapter 1. Introdu tionDe�nition 1.12. A maximizing ora le for a set fun tion p : 2V → Z ∪ {−∞} takes asinput a hypergraph H (given expli itly) and a ve tor x ∈ RV+ and it returns max{p(Z) −

dH(Z) − x(Z) : Z ⊆ V } and it also gives a set Z ⊆ V that maximizes this quantity.This se ond ora le appears also in [5℄ and [29℄. Our algorithms will usually use a max-imizing ora le for ps: note that if we have a maximizing ora le for p and one for p, thenwe an implement a maximizing ora le for ps using these two ora les. Let us give somemotivation why the maximizing ora le is of the form given above. We will often want tode ide whether a given x ∈ ZV+ satis�es (1.8) or not, be ause it is a ne essary ondition onthe existen e of a hypergraph overing p with degree-sequen e x: this an be done with amaximizing ora le for ps by alling it with x and the empty hypergraph. Similarly, in many ases we have already hosen some hyperedges that we want to in lude in the hypergraph overing some fun tion p0: if we denote this hypergraph by H then a degree-spe i� ation

x for the remaining hyperedges has to satisfy (1.8) with p = p0 − dH and this again an bede ided with a maximizing ora le for ps0.

x(Z) ≥ ps(Z) for every Z ⊆ V. (1.8)Note that usually we annot test the properties of the fun tion p laimed by the ora le(e.g. supermodularity, skew-supermodularity, symmetry et .), sin e that would need toomany ora le alls. Therefore in this model our algorithms will only give orre t answers ifthey an a ess the appropriate ora le and the fun tion has the laimed properties.Note that the maximizing ora le an learly provide us Mp = max{p(X) : X ⊆ V }.However it is not lear whether we an implement the evaluation ora le with help of amaximizing ora le: if we want to determine p(X) for some X ⊆ V then we need to �ndan x ∈ RV+ and a hypergraph H su h that X be omes the only maximizer of max{p(Z) −

dH(Z) − x(Z) : Z ⊆ V }. However if p(X) = −∞ then this never happens.Claim 1.13. If p(X) ≥ 0 then we an evaluate p(X) with a maximizing ora le for p. If theset fun tion p has only �nite values and we are given some lower bound L ≤ min{p(Z) :

Z ⊆ V }, then an evaluation ora le an be implemented with help of a maximizing ora le.Proof. Let M = Mp + 1 for the �rst statement and M = Mp + 1 − L for the se ond.Let H be the disjoint union of an arbitrary tree with node set X and one with node setV − X and let the multipli ity of every edge of H be M . Let furthermore x(v) = M forevery v ∈ V − X and x(v) = 0 for every v ∈ X. Call the maximizing ora le to maximizep − dH − x. One an see that p(Z) − dH(Z) − x(Z) ≤ p(Z) − M ≤ Mp − M , if Z ⊆ Vis di�erent from X, whi h is in both statements stri tly less than p(X). Thus the uniquemaximizer is X and the maximum is p(X).

Se tion 1.5. G-polymatroids and their interse tions 19It is more important to hara terize the fun tion lasses in whi h we an (or we annot)implement the maximizing ora le with polynomially many alls to the evaluation ora le.First we show that for positively skew-supermodular fun tions the maximizing ora le annot be redu ed to the evaluation ora le. This is true even for symmetri positively rossing supermodular fun tions. Consider the example p(X) = 1 if X = X0 or X = V −X0for some �xed X0 (and 0 otherwise): this fun tion is learly symmetri and positively rossing supermodular, but we need exponentially many alls to the evaluation ora le tode ide whether Mp > 0 or not.We have similar problems if the fun tion an take −∞ as value, too. The same ex-ample (but with −∞ instead of zero values) shows that the maximizing ora le annot beimplemented with polynomially many alls to the evaluation ora le even if p is symmetri and rossing supermodular (but an take the value −∞, too). We note that maximizing a( rossing) supermodular fun tion having −∞ among the fun tion values (and thus imple-menting the maximizing ora le for su h fun tions) an be solved if we have a little strongerora le than the fun tion evaluation ora le given above: this an be found in [22℄.On the other hand, for a �nitely valued rossing supermodular fun tion p, the maximiz-ing ora le an be implemented through general submodular fun tion minimizationte hniques (see e.g. [27℄ or [40℄). This is true sin e the fun tion p−dH −x is still rossingsupermodular (if H is an arbitrary hypergraph and x ∈ RV+) and the maximization of a rossing supermodular fun tion an be redu ed to the maximization of a (fully) supermod-ular fun tion.It is a hallenging open problem to de ide whether one an implement the maximizingora le for a �nitely valued skew-supermodular fun tion. This is essentially the sameproblem as the problem of maximizing su h a fun tion. The simplest open question iswhether we an maximize a �nitely valued interse ting negamodular fun tion (withpolynomially many alls to the fun tion evaluation ora le).In Se tion 2.1 we will show how to implement the maximizing ora le for the edge- onne tivity requirement fun tions appearing in our appli ations.1.5 G-polymatroids and their interse tionsIn this se tion we give some ba kground on (integer) g-polymatroids, of ourse withoutaiming for ompleteness. For a omplete introdu tion we refer the reader to [22℄. In the restof Se tion 1.5 we assume that the set fun tions p : 2V → Z∪{−∞} and b : 2V → Z∪{+∞}satisfy that p(∅) = b(∅) = 0.We say that the fun tions p, b form a strong pair, if p is supermodular, b is submodular

20 Chapter 1. Introdu tionand they satisfy the following ross-inequality for any pair X, Y ⊆ V :b(X) − p(Y ) ≥ b(X − Y ) − p(Y − X). (1.9)If p, b form a strong pair then the following polyhedron is alled a generalized polyma-troid, or shortly a g-polymatroid:

Q(p, b) = {x ∈ RV : p(Z) ≤ x(Z) ≤ b(Z) for any Z ⊆ V }. (1.10)It is onvenient to onsider the empty set as a g-polymatroid, too, though it annot bede�ned with a strong pair. The notion of g-polymatroids was introdu ed by Frank in[18℄ as a onvenient ommon generalization of polymatroids, ontrapolymatroids, basepolyhedra and submodular polyhedra. Note that the system (1.10) has an exponentialsize in |V |. Therefore we will think of it as a system that is given only impli itly: we an a ess the fun tions p and b through some ora le. A basi question is whether we an test membership in Q(p, b) and whether we an optimize a linear obje tive fun tionover Q(p, b) in polynomial time. We note that by the ellipsoid method due to Gröts hel,Lovász and S hrijver [25℄, these two problems (optimization and separation) are equivalent,therefore we will usually just show that we an solve one of them. Frank proved severalni e properties of g-polymatroids. We will need the following.Theorem 1.14. (A. Frank, [18℄) A g-polymatroid is an integer polyhedron. If the fun -tions p and b are given with an evaluation ora le and (p, b) is a strong pair then one anoptimize a linear obje tive fun tion over Q(p, b) in polynomial time.It is true that the formula (1.10) de�nes a g-polymatroid under more general ir um-stan es, too. We will need the following result. The fun tions p, b form a weak pair ifp : 2V → Z ∪ {−∞} is interse ting supermodular, b : 2V → Z ∪ {+∞} is interse tingsubmodular and they satisfy the ross-inequality for properly interse ting set pairs.Theorem 1.15. (A. Frank, [18℄) The polyhedron de�ned by (1.10) is a (possibly empty)g-polymatroid even is p and b form a weak pair.Sin e we will de�ne g-polymatroids with even weaker set fun tions we will need thestronger maximizing ora le in order to be able to �nd elements of these g-polymatroids.The following proposition is a orollary of Theorem 1.15.Theorem 1.16. For an integer g-polymatroid Q ⊆ RV , fun tions l : V → Z ∪ {−∞} andu : V → Z ∪ {∞}, and numbers α ∈ Z ∪ {−∞}, β ∈ Z ∪ {∞}, the polyhedron

Q ∩ {x ∈ RV : l(v) ≤ x(v) ≤ u(v) for every v ∈ V, α ≤ x(V ) ≤ β}is a (possibly empty) integer g-polymatroid.

Se tion 1.5. G-polymatroids and their interse tions 21An important spe ial ase of a g-polymatroid in this thesis is the ontrapolymatroid,whi h is the following polyhedron for a monotone supermodular fun tion p : 2V → Z ∪

{−∞}:C(p) = {x ∈ RV : x(Z) ≥ p(Z) ∀Z ⊆ V, x ≥ 0}. (1.11)Note that the monotoni ity of p together with p(∅) = 0 implies that p ≥ 0. We pointout that di�erent monotone supermodular fun tions de�ne di�erent ontrapolymatroids,sin e the polyhedron C(p) determines its de�ning monotone supermodular fun tion p bythe following relation:

p(Z) = min{x(Z) : x ∈ C(p)}. (1.12)Again, C(p) de�ned by (1.11) is a (nonempty) ontrapolymatroid under more general ir umstan es, too, by the following result.Theorem 1.17. (A. Frank, [19℄, [2℄) If p : 2V → Z+ is positively skew-supermodularthen C(p) = C(p′) with the (uniquely de�ned) monotone supermodular fun tionp′(X) = max{

t∑

i=1

p(Xi) : X1, X2, . . . , Xt is a subpartition of X}. (1.13)The interse tion of two g-polymatroids is again an integer polyhedron, in fa t it isa spe ial submodular �ow polyhedron. We omit the de�nition of submodular �owpolyhedra sin e we will not need them in this thesis, we will only use the following result.Theorem 1.18. (A. Frank, [18℄) The interse tion of two g-polymatroids is an integerpolyhedron.

22 Chapter 1. Introdu tion

Chapter 2Edge- onne tivity augmentation byadding (hyper)edgesThe most natural approa h to edge- onne tivity augmentation is the following: we are givena graph or hypergraph and we want to add edges or hyperedges to it in order to meetsome edge- onne tivity requirements. One possible obje tive fun tion is to minimizethe total size of the hyperedges needed. The main on ern of this thesis is indeed thisnotion of edge- onne tivity augmentation (note however that we will use a di�erent notionof edge- onne tivity augmentation in Chapter 6). Sometimes we will augment dire tedstru tures (i.e. mixed graphs or mixed hypergraphs), though we emphasize that we willalways want to add undire ted edges or hyperedges to our initial (possibly dire ted)graph or hypergraph. This problem an be formulated as the problem of overing somespe ial set fun tion with a graph or hypergraph as we will see shortly. In this thesisthe most general lass of set fun tions that we onsider will be positively skew-supermodularfun tions. Be ause of its importan e let us state this overing problem expli itly.Problem 2.1 (Covering a symmetri , positively skew-supermodular fun tion with (hy-per)edges). Given a symmetri , positively skew-supermodular fun tion p : 2V → Z∪{−∞}with a maximizing ora le, �nd a (hyper)graph H of minimum total size overing p.The (hyper)graph H to be found will also be alled the augmenting (hyper)graph. Bythe remarks on the symmetrized of a set fun tion, symmetry matters only in the followingsense: if p is not symmetri then we need to have a ess to a maximizing ora le for ps.Note that the problem has already two versions, depending on whether we allow arbitraryhyperedges or only graph edges in overing our fun tion. The fun tion p will also be alledthe requirement fun tion, or the de� ien y fun tion.In the subsequent se tions �rst we give the examples that motivate Problem 2.1 by show-23

24 Chapter 2. Edge- onne tivity augmentation by adding (hyper)edgesing that it is indeed a general framework for edge- onne tivity augmentation. Inthe following examples the obje tive is always to minimize the total size of the augmenting(hyper)graph H (if H has to be a graph then this is the same as minimizing the numberof edges in H). We will dis uss di�erent obje tives in the Se tion 2.2.2.1 Examples: edge- onne tivity requirement fun tionsIn the following se tions we will show appli ations where the augmentation problem anbe formulated as overing a skew-supermodular set fun tion. There are always di�erentversions of the problems depending on further onstraints: we lay the emphasis on thede� ien y fun tion here only. First we give a onstru tion that will be used in the proofsbelow.De�nition 2.2. Given a mixed hypergraph M = (V,A), an undire ted hypergraph H =

(V, E) and a fun tion x : V → R+, we de�ne the mixed hypergraph M ′ = M ′(M, H, x) asfollows. Introdu e a new node z /∈ V and let the vertex set of M ′ be V + z. Conne t zwith every v ∈ V with a graph edge having apa ity x(v): the ar set of M ′ onsists of the(disjoint) union of E and A and the ( apa itated) graph edges in ident to z.Note that if M itself is an undire ted hypergraph in the above de�nition, too, then so isM ′.2.1.1 Global edge- onne tivity augmentationThe most natural question in edge- onne tivity augmentation is the following.Problem 2.3 (Global edge- onne tivity augmentation problem). Given a graph orhypergraph H0 = (V, E0) and a positive integer k, �nd a (hyper)graph H = (V, E) su h thatH0 + H is k-edge- onne ted.Note that the problem has many versions depending on whether H0 and H is a graphor a hypergraph. By Menger's theorem (Theorem 1.2) this is equivalent to the problem of overing the following set fun tion:

p(X) :=

k − dH0(X) for any nonempty X ( V ,

0 for X = ∅ and X = V .(2.1)This fun tion p has very ni e properties: it is symmetri and rossing supermodular. Thusthe global edge- onne tivity augmentation problem is indeed a spe ial ase ofProblem 2.1 with the symmetri rossing supermodular fun tion de�ned in (2.1).

Se tion 2.1. Examples: edge- onne tivity requirement fun tions 25This problem has already many versions. The �rst version is when H0 is a graph and Hhas to be a graph, too: this is the global edge- onne tivity augmentation problemof graphs. This problem was solved by Watanabe and Nakamura in [45℄. The se ondversion is when H0 is a hypergraph, but we want to augment it only with graph edges: thisproblem is the global edge- onne tivity augmentation problem of hypergraphswith graph edges. This problem was solved by Bang-Jensen and Ja kson in [4℄. Our ontribution to these topi s is simpli�ed proofs that enable us to handle generalizations,too. See Se tions 4.2.2 and Chapter 5 for more details. The version when H may ontainhyperedges an be solved under more general ir umstan es: see Se tion 2.1.4.2.1.2 Global ar - onne tivity augmentation of mixed hypergraphsA bit more di� ult problem arises when we want to augment a mixed (hyper)graph, insteadof an undire ted one (however we only want to add undire ted (hyper)edges to it!).Problem 2.4 (Global ar - onne tivity augmentation of mixed hypergraphs). LetM = (V,A) be a mixed hypergraph, r ∈ V be a designated root node, and k, l be nonnegativeintegers. Find a (hyper)graph H = (V, E) of minimum total size su h that λM+H(r, v) ≥ kand λM+H(v, r) ≥ l for any v ∈ V (i.e. M + H is (k, l)-ar - onne ted from r).Again, the problem has many versions. Let us de�ne the set fun tion q = qM,r,k,l by

qM,r,k,l(X) :=

k − ̺M(X) for any X 6= ∅ with r /∈ X,

l − ̺M(X) for any X 6= V with r ∈ X

0 for X = ∅ and X = V .

(2.2)The set fun tion q is rossing supermodular, but it is not symmetri . For an undire tedhypergraph H one an he k using the dire ted version of Menger's theorem (Theorem1.3) that M + H is (k, l)-ar - onne ted from r if and only if dH overs q or equivalentlyp = qs, whi h is a spe ial symmetri skew-supermodular set fun tion. Thus the globalar - onne tivity augmentation of mixed hypergraphs is indeed a spe ial ase ofProblem 2.1 where p is the symmetrized of a rossing supermodular fun tion qde�ned in (2.2).Lemma 2.5. The maximizing ora le an be implemented for the fun tion p = qs

M,r,k,lintrodu ed in this se tion.Proof. Assume that we want to maximize p− dH − x for some hypergraph H and x ∈ RV+.Constru t the mixed hypergraph M ′ = M ′(M, H, x) as shown in De�nition 2.2 and observe

26 Chapter 2. Edge- onne tivity augmentation by adding (hyper)edgesthat λM ′(r, v) and λM ′(v, r) an be determined for any v ∈ V with standard network �owte hniques. The maximum of p − dH − x is equal to maxv∈V {k − λM ′(r, v), l − λM ′(v, r)}and the maximizer an be found similarly.If M is a mixed graph and we want to add graph edges to it then this problem wassolved by Bang-Jensen, Frank and Ja kson [2℄: for a simpli�ed proof see Se tion 4.2.2. IfM is a mixed hypergraph and we only allow the addition of graph edges then the problemis unsolved. If we also allow hyperedges then the problem an be solved even with morerestri tions, for example we an require that the augmenting hypergraph has to be nearlyuniform. See details of this version in Se tion 3.4.3.2.1.3 Node-to-area edge- onne tivity augmentationA di�erent set fun tion arises when we want to solve a node-to-area onne tivity aug-mentation problem. The problem is the following.Problem 2.6 (Node-to-area onne tivity augmentation problem). Given a (hy-per)graph H0 = (V, E0), a olle tion of subsets W of V and a fun tion r : W → Z+, ouraim is to �nd a (hyper)graph H of minimum total size su h that

λH0+H(x, W ) ≥ r(W ) for any W ∈ W and x ∈ V. (2.3)Some versions of this problem will be detailed below. Note that the global edge- onne tivity augmentation problem is a spe ial ase where the areas are all the singletonsand the requirement of every area is the same.Let us show why this problem is a spe ial ase of Problem 2.1. De�neRN2A(X) = max{r(W ) : W ∈ W, W ∩ X = ∅} for any ∅ 6= X ⊆ V and RN2A(∅) = 0.(2.4)This is a monotone de reasing fun tion: by that we mean that RN2A(X) ≥ RN2A(Y ) forany ∅ 6= X ⊆ Y . This easily implies that RN2A is rossing negamodular (in fa t it is eveninterse ting negamodular), and by Menger's theorem (Theorem 1.2) a (hyper)graph H isa feasible solution to the node-to-area onne tivity augmentation problem if and only if

dH overs q = RN2A − dH0or equivalently p = qs. Thus the node-to-area onne tivityaugmentation problem is indeed a spe ial ase of Problem 2.1 where p is thesymmetrized of a rossing negamodular fun tion q. In fa t the rossing negamodularfun tion q is of spe ial form, sin e it is the di�eren e of a monotone de reasing fun tionand a symmetri submodular fun tion.The node-to-area onne tivity augmentation problem in graphs is the spe ial ase of this problem when H0 is a graph and H has to be a graph, too. This problem was

Se tion 2.1. Examples: edge- onne tivity requirement fun tions 27introdu ed by Ishii and Hagiwara in [26℄. It is in general NP- omplete (even if H0 is theempty graph and r(W ) = 1 for every W ∈ W): be ause of the beauty and simpli ity ofthe proof given by Zoltán Király we present the proof of this.Theorem 2.7. The (general) node-to-area onne tivity augmentation problem in graphs isNP - omplete, even if the graph to be augmented is the empty graph and the requirement is1 for every area.Proof (Z. Király). We redu e the three dimensional mat hing problem (Problem SP1 in[23℄) to our problem. Let H = (V, E) be an instan e of the three dimensional mat hingproblem (that is, H is a 3-uniform hypergraph, where 3 divides |V | and the question iswhether V an be overed with disjoint hyperedges of H). Let W = {V − X ⊆ V : |X| ∈

{1, 2, 3} and X /∈ E} and let r : W → Z+ be de�ned by r(W ) = 1 for ea h W ∈ W.One an he k that the optimal solution of the node-to-area onne tivity augmentationproblem in graphs de�ned by W, r and the empty hypergraph H0 ontains exa tly 2|V |/3edges (its omponents are paths of length 2) if and only if H ontains a three dimensionalmat hing.Sin e the node-to-area onne tivity augmentation problem in graphs is NP - omplete,the authors of [26℄ assume that r ≥ 2 and surprisingly the problem be omes tra table:they give a polynomial time algorithm that solves it. Their proof was simpli�ed by Szigetiand Grappe in [24℄. We further simplify a part of this proof: we show that a greedyalgorithm fails only slightly for this problem in Se tion 4.3.2. In the thesis the node-to-area onne tivity augmentation problem in graphs will mean the ase when r ≥ 2 todistinguish from the general node-to-area onne tivity augmentation problem in graphs,whi h is NP - omplete.The node-to-area onne tivity augmentation problem in hypergraphs is the ase when H0 and H an be arbitrary hypergraphs. We will dis uss this problem inSe tion 3.4.2. This version an be solved in polynomial time even without the assumptionthat r ≥ 2.We will onsider a third version of this problem: assume that H0 is a hypergraph ofrank at most γ and we an only add hyperedges of size at most γ to it: this is the rank-respe ting node-to-area onne tivity augmentation problem in hypergraphs.As the γ = 2 ase shows, this is again NP - omplete in general, so we need the r(W ) 6= 1requirement, but under this assumption we solve the problem in Se tion 4.4.3.A similar proof to that of Lemma 2.5 justi�es the following lemma.Lemma 2.8. The maximizing ora le an be implemented for the fun tion p = (RN2A−dH0)sintrodu ed in this se tion.

28 Chapter 2. Edge- onne tivity augmentation by adding (hyper)edgesProof. Assume that we want to maximize p− dH − x for some hypergraph H and x ∈ RV+.Constru t the hypergraph M ′ = M ′(H0, H, x) as shown in De�nition 2.2. The maximumof p − dH − x is equal to max{r(W ) − λM ′(W, v) : W ∈ W, v ∈ V − W} whi h (togetherwith a maximizer) an be determined with standard network �ow te hniques.2.1.4 Lo al edge- onne tivity augmentationIf we onsider the lo al edge- onne tivity of graphs or hypergraphs then we obtain a sym-metri skew-supermodular de� ien y fun tion that is not the symmetrized of a rossingsupermodular or rossing negamodular fun tion. Let us introdu e the problem.Problem 2.9 (Lo al edge- onne tivity augmentation problem). Let H0 be a (hy-per)graph and let r : V ×V → Z+ be a symmetri edge- onne tivity requirement (i.e. r(u, v) =

r(v, u) for every u, v ∈ V ). Our aim is to �nd a (hyper)graph H of minimum total sizesu h that λH0+H(u, v) ≥ r(u, v) for every pair of nodes u, v.This lass of problems is alled lo al edge- onne tivity augmentation problems, sin e therequirement is lo ally de�ned for the pairs of nodes. Note that the global edge- onne tivityaugmentation is the spe ial ase when r(u, v) is the same for any pair u, v ∈ V . The lo aledge- onne tivity augmentation of graphs is the problem when H0 is a graph and Hhas to be a graph, too. The lo al edge- onne tivity augmentation of hypergraphsis the ase when H0 is a hypergraph and H an be an arbitrary hypergraph, too. We willalso onsider other versions to be de�ned later.Let us de�ne the set fun tion Rloc as Rloc(∅) = Rloc(V ) = 0 andRloc(X) = max{r(u, v) : u ∈ X, v /∈ X} for any nonempty X ( V. (2.5)By Menger's theorem λH0+H(u, v) ≥ r(u, v) for every pair of nodes u, v if and only if

dH0+H(X) ≥ Rloc(X) for any X ⊆ V , in other words if and only if H overs p = Rloc−dH0.An important observation is the following.Theorem 2.10 (A. Frank [19℄). The fun tion Rloc de�ned by (2.5) is skew-supermodular.Sin e the fun tion Rloc, and therefore Rloc − dH0

is symmetri and skew-supermodular,the lo al edge- onne tivity augmentation problem is indeed a spe ial ase ofProblem 2.1 where p is the symmetri skew-supermodular fun tion of formRloc − dH0

and Rloc is de�ned in (2.5). We mention that the possibility that r(u, v) = 1for ertain pairs u, v again auses di� ulties in some ases, so we will often assume thatr(u, v) 6= 1 for any pair u, v, but here this is only a te hni al assumption and the omplexityof the problem does not depend on it. The following lemma an again be proved similarlyto Lemma 2.5.

Se tion 2.2. Polyhedra and various obje tive fun tions of the augmentation 29Lemma 2.11. The maximizing ora le an be implemented for the fun tion p = Rloc − dH0introdu ed in this se tion.Proof. Assume that we want to maximize p− dH − x for some hypergraph H and x ∈ RV+.Constru t the hypergraph M ′ = M ′(H0, H, x) as shown in De�nition 2.2. The maximumof p(Z) − dH(Z) − x(Z) is equal to max{r(u, v) − λM ′(u, v) : u, v ∈ V } whi h (togetherwith a maximizer) an be determined with standard network �ow te hniques.The lo al edge- onne tivity augmentation of graphs was solved by Frank [19℄ using thesplitting lemma of Mader [34℄. Our ontribution is a simple proof of this result. If H0 is ahypergraph and we want to augment it with graph edges to meet lo al edge- onne tivityrequirements, then we obtain an NP - omplete problem, as was shown in [16℄. The asewhen H may ontain arbitrarily large hyperedges was solved by Szigeti [42℄. In Se tion3.4.1 we generalize his results and show, for example, that one an �nd an optimal solutionthat is nearly uniform. If H an ontain hyperedges but its rank annot be bigger thanthat of H0 then the problem was solved by Ben Cosh [15℄: we present a simpler proof ofthis result in Se tion 4.4.1.2.2 Polyhedra and various obje tive fun tions of theaugmentationAs we have seen, the edge- onne tivity augmentation problems that we onsiderhere an be formulated as the problem of �nding a (hyper)graph overing a positivelyskew-supermodular set fun tion, i.e. a overing problem. Let us look at this problem interms of the obje tive fun tion.The most simple obje tive fun tion of edge- onne tivity augmentation would be to min-imize the number of edges of the augmenting graph (or the total size of the augment-ing hypergraph): let us all this version of the problem the minimum augmentationproblem. However it is more omfortable to speak about another problem, the so- alleddegree spe i�ed augmentation problem. This is the following: given a symmetri positively skew-supermodular set fun tion p and a degree spe i� ation m : V → Z+,the question is whether a graph (or hypergraph) G overing p exists that satis�es thedegree-spe i� ation meaning that d+

G(v) = m(v) has to hold for every v ∈ V (so this isnot an optimization problem, but a de ision problem). A natural ne essary onditionof the existen e of the graph G is that m(X) =∑

v∈X m(v) ≥ p(X) for any X ⊆ V ,sin e dG(X) ≤∑

v∈X dG(v) ≤∑

v∈X d+G(v) =

∑

v∈X m(v) for any X ⊆ V . This ne essary ondition and the ni e properties of the fun tion p allow us to redu e the minimum version

30 Chapter 2. Edge- onne tivity augmentation by adding (hyper)edges(and even more general versions) of the augmentation problem to the degree-spe i�ed ver-sion. We say that the degree-spe i� ation m ∈ ZV+ is admissible if m(X) ≥ p(X) for any

X ⊆ V . If m is not admissible then a set X with m(X) < p(X) will be alled de� ient.For a set fun tion p : 2V → Z ∪ {−∞} we have introdu ed the polyhedronC(p) = {x ∈ RV : x(Z) ≥ p(Z) ∀Z ⊆ V, x ≥ 0}. (2.6)As it was mentioned in Se tion 1.5, this is an (integer) ontrapolymatroid for a positivelyskew-supermodular fun tion p. We note that, by the properties of a ontrapolymatroid, apolynomial algorithm to the degree spe i�ed overing problem will give rise to a solution tothe minimum version of the problem, and to more general versions, su h as the minimumnode- ost problem. By the properties of g-polymatroids (see Se tion 1.5), min{x(V ) : x ∈

C(p)} = min{1 · x : x ∈ C(p)} = max{∑

X∈X p(X) : X is a subpartition of V }. De�nethe Subpartition Lower Bound by SLB(p) = max{∑

X∈X p(X) : X is a subpartition ofV } = min{1 · x : x ∈ C(p)}: this is obviously a lower bound on the minimum total size ofany hypergraph overing p. We say that m ∈ C(p) ∩ ZV is minimal if m′ ∈ C(p) ∩ ZV ,m′ ≤ m implies that m′ = m. By Edmonds' greedy algorithm and Theorem 1.17 we havethe following.Corollary 2.12. For a ve tor m ∈ C(p), m is minimal if and only if m(V ) = SLB(p).By the properties of a ontrapolymatroid we an handle the following,minimum node- ost overing problem, too. This is the following: assume that we are given a ostfun tion c : V → R+ and the task is to �nd a (hyper)graph H overing p that minimizes∑

v∈V c(v)dH(v). The minimum version of the overing problem orresponds to the asewhen c(v) = 1 for every v ∈ V . Note that we an assume about any optimal solution H ofsu h a minimum node- ost overing problem that it does not ontain singleton hyperedges,so dH(v) = d+H(v) for any v ∈ V .The reader an ask why we onsider exa tly these obje tive fun tions. The answer issimple: these are the tra table ones. For example it is easy to see that the followingsimplest minimum ost edge- onne tivity augmentation problem in graphs is already NP - omplete: given a graph G0 and a requirement k ∈ Z+, �nd a set of edges F of minimum ost su h that G + F is k-edge- onne ted, where the ost of hoosing the edge uv is givenfor any u, v ∈ V . If k = 2 and G0 is the empty graph then this problem is equivalent tothe Symmetri Travelling Salesman Problem, as one an he k. Let us de�ne formally the onsidered versions of the problems in terms of the obje tive fun tion.De�nition 2.13. Let us be given the problem of overing a set fun tion p : 2V → R∪{−∞}with a hypergraph H (where we might have further onstraints on the hypergraph H). The

Se tion 2.3. Covering with graph edges: the splitting-o� te hnique 31minimum version of the problem is to �nd a hypergraph H overing p of minimumtotal size. The minimum node- ost version of the problem is to �nd a (hyper)graphH overing p that minimizes ∑

v∈V c(v)dH(v) for a given ost fun tion c : V → R+. Thedegree-spe i�ed version is to �nd a hypergraph H overing p that also satis�es a givendegree-spe i� ation m ∈ ZV+.For sake of larity let us outline Edmonds' greedy algorithm to �nd a ve tor m ∈ C(p)that minimizes {∑v∈V c(v)x(v) : x ∈ C(p)} for a positively skew-supermodular fun tion

p : 2V → Z ∪ {−∞} (given by a maximizing ora le) and an arbitrary c : V → R+ (i.e. thegreedy algorithm for a ontrapolymatroid). Determine Mp = max{p(X) : X ⊆ V } andset initially m(v) = Mp for every v ∈ V . Order the elements of V su h that c(v1) ≥

c(v2) ≥ · · · ≥ c(vn). Starting with i = 1 su essively de rease m(vi) by a maximumpossible value that maintains m ∈ C(p): note that this maximum value an be determinedby a logarithmi sear h. By the well known results of Edmonds [17℄, the resulting mwill be an integer minimizer of {∑v∈V c(v)x(v) : x ∈ C(p)}. This algorithm an learlybe implemented to run in polynomial time if a maximizing ora le is available for p andit shows the ideas of the redu tion of the minimum node- ost overing problem to thedegree-spe i�ed overing problem.Noti e that using Edmonds' greedy algorithm we an implement the evaluation ora le forthe monotone supermodular fun tion p′ de�ned in (1.13) that de�nes the same ontrapoly-matroid C(p), if the positively skew-supermodular fun tion p is given with a maximizingora le. This is true sin e p′(Z) = min{x(Z) : x ∈ C(p)} by (1.12), therefore to evaluatep′(Z) we only need to minimize χZx over C(p).Be ause of the arguments given above we will always speak about the degree-spe i�ed overing problem and we will usually only state the solution of the minimum version as onsequen es.2.3 Covering with graph edges: the splitting-o� te h-niqueWhen we want to over a set fun tion by a degree spe i�ed graph then the usualte hnique is splitting-o�. The splitting-o� operation was originally introdu ed by Lovász[33℄ and subsequently developed further by Mader [34℄ and others (we mention that adire ted version of splitting-o� an be de�ned, too, but we will only use undire ted splitting-o�). In the rest of Se tion 2.3 let p : 2V → Z ∪ {−∞} be a symmetri , positively skew-supermodular fun tion that satis�es p(∅) ≤ 0 and m : V → Z+ a nonnegative fun tion

32 Chapter 2. Edge- onne tivity augmentation by adding (hyper)edgessatisfying m(X) ≥ p(X) for any X ⊆ V (i.e. an integer element of C(p), or with otherwords an admissible degree-spe i� ation). We would like to de ide whether there is a graphG overing p that satis�es d+

G(v) = m(v) for every v ∈ V .For a node v ∈ V we say that v is positive if m(v) > 0, and neutral otherwise. Theset of positive nodes will be denoted by V +. Assume u, v ∈ V + are two positive nodes(possibly u = v, but then m(u) ≥ 2 is assumed). The operation splitting-o� (or shortlysplitting) at u and v onsists of repla ing m by m′ and p by p′ wherem′ = m − χ{u} − χ{v} and p′(X) =

{

p(X) − 1 if |X ∩ {u, v}| = 1,

p(X) otherwise. (2.7)(Note that p′ = p− d(V,{uv}) where d(V,{uv}) denotes the ut fun tion of the graph (V, {uv})having only one edge between u and v, therefore p′ is positively skew-supermodular, too.)The edge uv will be alled a split edge. One an observe that this is indeed the usualnotion of splitting-o�: if we introdu e a graph K = (V +s, E) with every edge of E in identto s and dK(s, v) = m(v) for any v ∈ V then we are ba k at the well known splitting-o�operation.If m′(X) ≥ p′(X) for any X ⊆ V then we say that the splitting o� is admissible (orthe pair u, v is admissible). A set X is dangerous if m(X)−p(X) ≤ 1 and it is alledtight if m(X) − p(X) = 0. The following laim an be easily veri�ed.Claim 2.14. If m ∈ C(p)∩ZV and u, v ∈ V + then the splitting-o� at u and v is admissibleif and only if there is no dangerous set X ontaining both u and v.We will also say that su h a dangerous set X blo ks the splitting at u and v, orsimply that X blo ks u and v.We will sometimes need to do the inverse of a splitting-o�. So assume that e is a splitedge. The unsplitting operation of e is simply the reverse of the splitting-o� operation:me = m+χ{u}+χ{v} and pe = p+d(V,{(uv)}). Of ourse, this operation is always admissible,that is me ∈ C(pe), if m was admissible (sin e m(X)− p(X) ≤ me(X)− pe(X) for any setX).An admissible splitting-o� sequen e is a sequen e of splitting-o�s, where ea h stepis admissible (where note that the fun tions p and m are always updated after ea h step).It is important to observe that the order of su h a sequen e an be arbitrarily hanged: ifthe admissible splitting-o� at u, v is followed by the admissible splitting-o� at x, y, then�rst performing the splitting-o� at x, y and then at u, v is also an admissible splitting-o�sequen e (this is be ause the unsplitting is always admissible). A omplete admissiblesplitting-o� is an admissible splitting-o� sequen e whi h de reases m(V ) to zero. Note

Se tion 2.3. Covering with graph edges: the splitting-o� te hnique 33that there exists a graph overing the fun tion p and satisfying the degree-spe i� ation mif and only if there exists a omplete admissible splitting-o�.Now we give some lemmas on splitting-o� that will be needed later.Claim 2.15. Let p : 2V → Z ∪ {−∞} be a symmetri fun tion and m ∈ C(p). If T ⊆ Vis tight and D ⊆ V is dangerous satisfying T ∪ D = V then m(T ∩ D) = 0.Proof. We an assume that m(V ) > 0 and T ∩D 6= ∅, otherwise the laim is trivially true.Assuming that m(T ∩ D) > 0 and using that p(D) = p(T − D) and p(T ) = p(D − T ) bythe symmetry of p, we get a ontradi tion from the following inequalities: p(T ) + p(D) =

p(T −D)+p(D−T ) ≤ m(T −D)+m(D−T ) ≤ m(T )−1+m(D)−1 ≤ p(T )−1+p(D).Lemma 2.16. Assume that p : 2V → Z∪{−∞} is a symmetri , positively skew-supermodularfun tion and m ∈ C(p) is minimal. If the pair u, v ∈ V + is admissible, then splitting-o�this pair the subpartition lower bound de reases by two, that is SLB(p′) = SLB(p) − 2 forthe modi�ed fun tion p′ = p − d(V,{uv}).Proof. Sin e m′ = m−χ{u}−χ{v} ∈ C(p′), ertainly SLB(p′) ≤ SLB(p)− 2 = m(V )− 2 =

m′(V ). On the other hand, if SLB(p) =∑

X∈X p(X) for some subpartition X of V , then uand v must be in two di�erent members of X by the minimality of m and the admissibilityof u, v, therefore SLB(p′) ≥∑

X∈X p′(X) = SLB(p) − 2.We mention that the minimality is indeed ne essary in Lemma 2.16. As an example, on-sider the lo al edge- onne tivity augmentation of a C5 = ({x1, x2, x3, x4, x5}, {x1x2, x2x3,

x3x4, x4x5, x5x1}) with onne tivity requirements r(xi, xj) = 3 for any i, j ∈ {2, 3, 4, 5}and r(x1, xj) = 0 for any j ∈ {2, 3, 4, 5}. Choosing the degree-spe i� ation m = 1, thesplitting-o� at x2, x5 is admissible and SLB(p′) = SLB(p) − 1 = 3.Let Mp = max{p(X) : X ⊆ V }. A set X with p(X) = Mp will be alled p-maximal.Clearly, if Mp ≤ 0 then any splitting-o� is admissible, so we will usually assume thatMp > 0. In this ase if X and Y are two p-maximal sets then either both of X ∩ Y andX∪Y are p-maximal (if X and Y satisfy (∩∪) ) or both of X−Y and Y −X are p-maximal(if X and Y satisfy (−) ) by the positively skew-supermodularity of p.We mention the following simple observation about loop edges. When we onsiderthe degree-spe i�ed version of a overing problem, then usually we allow loop edges inthe graph to be found. Therefore we usually use the terminology �a graph G satis�es thedegree-spe i� ation m� to mean that d+

G(v) = m(v) for every v ∈ V (note that d+G(v) also ounts the loop edges). However we ould avoid this by the following transformation: if Gsatis�es the degree-spe i� ation m and ontains a loop edge in ident to some node u, butthere exist another edge vw (possibly another loop but) not in ident to u then deleting

34 Chapter 2. Edge- onne tivity augmentation by adding (hyper)edgesthese two edges and introdu ing the edges uv and uw the obtained graph G′ still satis�esthe degree-spe i� ation m and dG′(X) ≥ dG(X) for any X ⊆ V , i.e. if G overs somefun tion p then G′ also overs p. Repeating this operation we an get rid of the loop edges,unless there exists a node v ∈ V that has m(v) > m(V − v). Therefore if we also made theassumptionm(v) ≤ m(V − v) for every v ∈ V (2.8)in our theorems, then we ould avoid dealing with loop edges. However we will followan approa h that allows loop edges. Note that in the partition onstrained problems inChapter 5 the ondition (2.8) is impli itly present among the others (and loop edges arenot allowed in partition onstrained problems).2.3.1 Contra tion of tight setsA widely used ingredient of splitting-o� algorithms is ontra tion of tight sets. This anbe done sin e it does not hange the admissibility stru ture (the set of admissible pairsdoes not hange) and most of the proofs be ome simpler if tight sets are ontra ted. Inthis se tion we give the ba kground of this simpli� ation, however we will always try toavoid the use of ontra tion wherever the dis ussion does not be ome unne essarily di� ultwithout it.If T ⊆ V then ontra ting T roughly means that from now on we onsider it to bea singleton. Formally this means that we de�ne V/T = V − T + vT where vT is a newnode not in V . For any set fun tion p : 2V → Z ∪ {−∞} we de�ne p/T : 2V/T →

Z ∪ {−∞} by p/T (X) = p(X) if vT 6∈ X and p/T (X) = p(X − vT + T ) if vT ∈ X. Form : V → R de�ne m/T : V/T → R with m/T (v) = m(v) if v 6= vT and m/T (vT ) = m(T ):observe that regarding m to be a set fun tion would give the same de�nition. In this ontra ted problem a splitting-o� is admissible if it is admissible with respe t to p/T .Note that p/T will inherit the interesting properties of p investigated in this paper (e.g.symmetry, (positively) rossing supermodularity, (positively) skew-supermodularity et .).Contra tion of a hypergraph H = (V, E) is understood in the obvious way as H/T =

(V/T, {e ∈ E : T ∩ e = ∅} ∪ {e − T + vT : e ∈ E , T ∩ e 6= ∅}), so we will usually not onsider hyperedges with multipli ities (multihyperedges) after a ontra tion. However, forthe graph of the edges split so far we must ount the multipli ity in the loop edges obtainedthis way in order to satisfy the degree spe i� ation: this will not ause any onfusion. One an he k that dH/T = dH/T . The ontra ted image v/T of a node v ∈ V is de�nedas v/T = vT , if v ∈ T , and v/T = v otherwise. The main observation about ontra tion isthe following.

Se tion 2.3. Covering with graph edges: the splitting-o� te hnique 35Lemma 2.17. Let p : 2V → Z ∪ {−∞} be a symmetri , positively skew-supermodularfun tion, m ∈ C(p), and u, v ∈ V with m(u), m(v) > 0. If we ontra t a p-positive tightset T then the splitting at u/T and v/T is admissible with respe t to p/T if and only if thesplitting at u and v is admissible with respe t to pProof. By the de�nition of p/T , if the splitting-o� at u and v was admissible then it learly stays admissible. Let us prove the other dire tion. Assume that u/T, v/T be omesadmissible while u, v was not admissible, i.e. there was a set X ⊆ V with p(X) ≥ m(X)−1with u, v ∈ X (a dangerous set with respe t to p: note that the inequality p(X) ≥ m(X)−1implies that X is p-positive). Clearly, neither T ⊆ X nor X ∩ T = ∅ an hold. If (∩∪)holds for X and T then X ∪ T is also dangerous, a ontradi tion. So (−) must hold forthem, meaning X − T is also dangerous and u, v ∈ X − T , a ontradi tion again.This lemma allows us to simplify some of the proofs by assuming that every tight set is asingleton. Note that if p is not only positively skew-supermodular but skew-supermodular,then we an also ontra t a tight set T with p(T ) = 0.

36 Chapter 2. Edge- onne tivity augmentation by adding (hyper)edges

Chapter 3Covering skew-supermodular fun tionswith hyperedgesThis hapter is devoted to overing symmetri , positively skew-supermodular fun tionswith hyperedges. The starting result is Theorem 3.2 of Szigeti. The appli ation he had inmind was lo al edge- onne tivity augmentation of hypergraphs by hyperedges of minimumtotal size. Here we generalize this theorem in many dire tions and we will onsider otherappli ations, too. The results of this se tion were found together with Tamás Király andthey appeared in [12℄.Re all that the hypergraph H = (V, E) is said to over the fun tion p : 2V → Z∪{−∞}if dH(X) ≥ p(X) for every X ⊆ V . We will also use a di�erent notion of overing: H issaid to weakly over the fun tion p : 2V → Z∪{−∞} if bH(X) ≥ p(X) for every X ⊆ V ,wherebH(X) = |{e ∈ E : e ∩ X 6= ∅}|.Note that bH is submodular, monotone (but not symmetri ), and

bH(X) + bH(Y ) ≥ bH(X − Y ) + bH(Y − X) + |{e ∈ E : ∅ 6= e ∩ Y ⊆ X ∩ Y }|.Noti e that bH(v) = d+H(v) for a hypergraph H and a node v. The entral problem ofthis hapter is the following.Problem 3.1 (Covering a positively skew-supermodular fun tion with hyperedges). Letus be given a symmetri , positively skew-supermodular fun tion p : 2V → Z ∪ {−∞} witha maximizing ora le. The minimum version of the problem is to �nd a hypergraph H overing p of minimum total size. The degree-spe i�ed version is to �nd a hypergraph

H overing p that also satis�es a given degree-spe i� ation m ∈ ZV+.37

38 Chapter 3. Covering skew-supermodular fun tions with hyperedgesIn [42℄, Szigeti proved the following result, whi h solves the degree-spe i�ed version ofthe problem given above. It states that if the degree-spe i� ation is admissible then therequired hypergraph exists.Theorem 3.2 (Z. Szigeti, [42℄). Let p : 2V → Z ∪ {−∞} be a symmetri , positivelyskew-supermodular set fun tion, and m : V → Z+ a degree spe i� ation.(i) There exists a hypergraph H s.t. d+H(v) = m(v) for every v ∈ V and dH(X) ≥ p(X)for every X ⊆ V if and only if

∑

v∈X

m(v) ≥ p(X) for every X ⊆ V . (3.1)(ii) Furthermore, if m(v) ≤ Mp = max{p(X) : X ⊆ V } for any v ∈ V , then H an be hosen so that it onsists of exa tly Mp hyperedges.In this hapter we give di�erent kinds of extensions of this theorem. The stru ture of this hapter is the following. First we show a simple proof of Szigeti's theorem with a di�erentmethod ompared to his: we use the operation of merging hyperedges. This already allowsus to generalize his result slightly in one dire tion: in Theorem 3.3 we an tell when itis possible to obtain a hypergraph H∗ overing the fun tion p by merging hyperedges ofa hypergraph H given in advan e. In the next se tion �rst we show a generalization ofS hrijver's supermodular olouring theorem, that is losely related to hypergraphs weakly overing a positively skew-supermodular fun tion p. We show this relation in Theorem3.9. Then we observe that under ertain ir umstan es weak overing already implies overing. This together with the polyhedral observations on hypergraphs weakly over-ing the set fun tion give extensions of Theorem 3.2: for example the optimal hypergraph an always be hosen to be nearly uniform. Furthermore, we obtain that under some ir umstan es we an solve the problem of overing two skew-supermodular fun tions si-multaneously. In Se tion 3.4 we give appli ations of our results. We note that algorithmi aspe ts are not adressed in this hapter, we simply give minmax results here. The algo-rithm that we suggest �nds the hyperedges of the augmenting hypergraph one by one, thusit is not polynomial. However, probably a similar argument as the one given in [29℄ wouldgive polynomial algorithms for the problems dis ussed in this hapter.3.1 Merging hyperedgesLet H = (V, E) be a hypergraph. By merging two disjoint hyperedges of H we meanthe operation of repla ing them by their union. �Merging some hyperedges of H� means

Se tion 3.1. Merging hyperedges 39repeating this operation a few times. We note that part (i) of the following theorem dueto Tamás Király already appeared in [30℄ and in Hungarian in [31℄.Theorem 3.3. Let H = (V, E) be a hypergraph, and let p : 2V → Z∪{−∞} be a symmetri ,positively skew-supermodular set fun tion with k = Mp = max{p(X) : X ⊆ V } ≥ 0, forwhi hbH(X) ≥ p(X) for every X ⊆ V . (3.2)(i) Then by merging some hyperedges of H we an obtain a hypergraph H∗ = (V, E∗) that overs p.(ii) Furthermore, if there are k hyperedges f 1, f 2, . . . , fk in H su h that every hyperedgein H − {f 1, . . . , fk} is a singleton and bH(v) ≤ k for any v ∈ V , then the mergingoperations an be organized in a way that H∗ = (V, {f 1

∗ , f 2∗ . . . , fk

∗ }) where f i ⊆ f i∗ forevery i = 1, . . . , k.Proof. We prove (i) by indu tion on the number of hyperedges of H (it is learly true if

E = ∅). A set X ⊆ V is alled tight if bH(X) = p(X). By the properties of bH and p, ifX and Y are p-positive and tight, then either X ∩ Y and X ∪ Y are tight, or X − Y andY −X are tight. Furthermore, if X and Y are p-positive and tight and there is a hyperedgee su h that ∅ 6= e ∩ Y ⊆ X ∩ Y , then X ∩ Y and X ∪ Y are tight.Let e0 be an arbitrary hyperedge of H . If there is no tight set X su h that e0 ⊆ X,then let H ′ := H − e0 and p′ = p − dH0

where H0 = (V, {e0}). The set fun tion p′ issymmetri and positively skew-supermodular, and bH′(X) ≥ p′(X) for every X ⊆ V , soby indu tion there is a hypergraph H ′∗, obtained by merging some hyperedges of H ′, su hthat dH′

∗(X) ≥ p′(X) for every X ⊆ V . It follows that H∗ := H ′

∗ + e0 overs p. We anthus assume that there is a tight set X0 su h that e0 ⊆ X0: let X0 be a maximal tight set ontaining e0.Suppose that there is no hyperedge e ∈ E su h that e ∩ X0 = ∅. Then p(V − X0) =

p(X0) = bH(X0) > bH(V −X0) sin e e0 ⊆ X0, ontradi ting (3.2). Thus there is a hyperedgee1 ∈ E su h that e1 ∩X0 = ∅. Consider the hypergraph H ′ := (V, E − {e0, e1}+ (e0 ∪ e1)),i.e. the hypergraph obtained by merging e0 and e1. If bH′(Y0) < p(Y0) for some Y0 ⊆ V ,then e0 ∩ Y0 6= ∅, e1 ∩ Y0 6= ∅, and Y0 was tight. Sin e ∅ 6= e0 ∩ Y0 ⊆ X0 ∩ Y0, X0 ∪ Y0 isalso tight, whi h ontradi ts the maximality of X0.We proved that H ′ and p satisfy (3.2), so by indu tion there is a hypergraph H∗, obtainedby merging some hyperedges of H ′ (hen e obtained by merging some hyperedges of H),that overs p.The proof of (ii) is similar to the proof of Theorem 3.2 by Szigeti. We will use thefollowing observation. Let X, Y ⊆ V su h that X is tight and p(Y ) = k. If (∩∪) applies

40 Chapter 3. Covering skew-supermodular fun tions with hyperedgesfor X and Y then p(X ∪ Y ) ≤ p(Y ) = k implies that p(X ∩ Y ) ≥ p(X) = bH(X) ≥

bH(X ∩ Y ) ≥ p(X ∩ Y ), so every inequality is satis�ed with equality here (in ludingp(X ∪ Y ) = k). On the other hand, if (−) applies for X and Y then p(Y −X) ≤ p(Y ) = kimplies that p(X − Y ) ≥ p(X) = bH(X) ≥ bH(X − Y ) ≥ p(X − Y ), so every inequality issatis�ed with equality here (in luding p(Y − X) = k).We will prove the statement indire tly: suppose that H , p and f 1, . . . , fk form a oun-terexample with k as small as possible and, subje t to that, |V − fk| as small as possible.Trivially, k > 0. Suppose that there is a set Y with p(Y ) = k that is disjoint from fk. Sin ebH(Y ) ≥ p(Y ) = k there must be a hyperedge e ∈ E − {f 1, . . . , fk−1, fk} that interse tsY : sin e these hyperedges are singletons, in fa t e ⊆ Y . Let H ′ be obtained from H bymerging fk and e into a hyperedge f ′k. We laim that H ′ does not violate (3.2): if it doesthen there was a tight set X su h that fk ∩ X 6= ∅, e ∩ X 6= ∅ (e ⊆ X in fa t). ButbH(X) > b(X ∩ Y ) be ause of the edge fk, implying that (∩∪) annot apply for X and Y .On the other hand, bH(X) > bH(X − Y ) be ause of the hyperedge e, so (−) annot applyfor X and Y either, a ontradi tion. By the minimal hoi e of H , the statement is true forH ′, but then also for H , a ontradi tion. So in our minimal ounterexample fk interse tsevery set Y with p(Y ) = k.Similarly we laim that in this minimal ounterexample fk must over every vertex vwith bH(v) = k. Assume that this is not the ase and v is su h a vertex not overed by fk.Then there must be a hyperedge e ∈ E − {f 1, . . . , fk−1, fk} that overs v and if we mergefk with e then the hypergraph H ′ obtained will not violate (3.2), sin e every set Y thatinterse ts both e and fk has bH(Y ) ≥ k +1, so it annot be tight. Therefore the statementis true for H ′, and then also for H , a ontradi tion.We laim that there is no tight set X satisfying fk ⊆ X. Assume that this is not trueand let X be su h a tight set. Let Y be an arbitrary set with p(Y ) = k (su h a setexists by the de�nition of k). If (∩∪) applies for X and Y then p(X ∪ Y ) = k, but thenp(V − (X ∪ Y )) = k, too, but this set is not overed by fk. However, (−) annot apply forX and Y either, be ause then Y − X would be a set with p(Y − X) = k not overed byfk. So we really obtained that there is no tight set ontaining fk.Let H ′ = H − fk and p′ = p − dHk , where Hk = (V, {fk}). Then max{p′(X) : X ⊆

V } = k − 1, bH′(v) ≤ k − 1 for every v ∈ V , and bH′ ≥ p′, thus H ′, p′ and f 1, . . . , fk−1must satisfy the statement to be proved (otherwise H was not a minimal ounterexample),so there exists a hypergraph H ′∗ = (V, {f 1

∗ , . . . , fk−1∗ }) that overs p′ and satis�es f i ⊆ f i

∗for any i between 1 and k − 1. But then one an easily he k that H∗ = H ′∗ + fk satis�esour requirements, so H was not a ounterexample.Theorem 3.2 orresponds to the ase when H onsists of hyperedges of size 1, and m(v)

Se tion 3.2. Weak overing of positively skew-supermodular fun tions 41is the multipli ity of {v} in H . We mention that the assumptions of Theorem 3.3 (ii) arereally needed: if a hypergraph H weakly overs a positively skew-supermodular fun tion pand H has more than k = Mp nonsingleton hyperedges then we annot ne essarily mergehyperedges of H with maintaining the inequality bH ≥ p (but by Theorem 3.3 (i), only ifH in fa t overs p). A simple example above the 3 element ground set V = {v1, v2, v3}is the following: let p(X) = 2 for any nonempty X ( V and p(∅) = p(V ) = 0 and letH = (V, {v1v2, v2v3, v3v1}).3.2 Weak overing of positively skew-supermodular fun -tionsIn this se tion we �rst show a generalization of S hrijver's supermodular olouring theorem.We want to des ribe this in the ontext of hypergraphs instead of olourings: let us showthe onne tion. A k- olouring is a partition X1, . . . , Xk of V (where some of these lassesmay even be empty). If a set fun tion p : 2V → Z∪ {−∞} is also given then a k- olouringis good (for p) if |{i : Xi ∩ X 6= ∅}| ≥ p(X) for any X ⊆ V . Observe that the olouringX1, . . . , Xk is good if and only if the hypergraph with edge set X1, . . . , Xk weakly oversthe set fun tion p.Let us state the important results on supermodular olourings. The following theoremis alled the supermodular olouring theorem.Theorem 3.4 (S hrijver, [39℄). Let p1, p2 : 2V → Z ∪ {−∞} be interse ting supermodularfun tions and let k be a positive integer. There exists a olouring X1, X2, . . . , Xk of V thatis good for both p1 and p2 if and only if

pi(X) ≤ min{k, |X|} for any X ⊆ V and i = 1, 2.A ni e and simple proof of this theorem was given by Éva Tardos (a further simpli�edform of this proof an be found in S hrijver's book [41℄, pages 849-851). The key of herproof is the following lemma.Lemma 3.5 (Tardos, [44℄). Let p : 2V → Z ∪ {−∞} be an interse ting supermodularfun tion and let k be a positive integer. Assume that p(X) ≤ min{k, |X|} for any X ⊆ V .Then the polyhedronQ = {x ∈ RV : x(Z) ≥ 1 if p(Z) = k, x(Z) ≤ |Z| − p(Z) + 1 ∀Z ⊆ V, 0 ≤ x ≤ 1}is a nonempty integer g-polymatroid.

42 Chapter 3. Covering skew-supermodular fun tions with hyperedgesTo prove Theorem 3.4 it is enough to show that the integer elements of the polyhedronde�ned in Lemma 3.5 orrespond to the hara teristi ve tors of the possible olour lassesof good olourings: we will see exa t details in a more general setting later. Szigeti showedthat Theorem 3.2 implies a spe ial ase of the supermodular olouring theorem. In this se -tion we develop tools to prove a reverse impli ation. First we show that the supermodular olouring theorem is true in a more general form, namely for positively skew-supermodularfun tions instead of interse ting supermodular fun tions � this was observed by TamásKirály in [28℄. Then we show that for symmetri , positively skew-supermodular setfun tions weak overing implies overing if the number of hyperedges equalsthe maximum value of the set fun tion. This enables us to derive generalizations ofSzigeti's theorem.Let p : 2V → Z ∪ {−∞} be an arbitrary positively skew-supermodular fun tion and letC = C(p) be the ontrapolymatroid de�ned in (1.11). Let k ≥ max{p(X) : X ⊆ V } andlet y ≤ (k, k, . . . , k) be an integer ve tor in C (note that the ve tor (k, k, . . . , k) is in C,so su h a y exists). Our aim is to �nd a hypergraph of k hyperedges satisfying the degree-spe i� ation y that weakly overs p. Note that in the following theorems the symmetry ofp is not ne essarily assumed.Let us de�ne the polyhedron

Q = Q(p, k, y) = {x ∈ RV : x(Z) ≥ 1 if p(Z) = k,

x(Z) ≤ y(Z) − p(Z) + 1 ∀Z ⊆ V, 0 ≤ x ≤ y}.One an observe that y/k ∈ Q, so Q is non-empty.The proof of the following lemma essentially follows the line of the proof of Lemma 3.5that appears in S hrijver's book [41℄.Lemma 3.6. Q is a (nonempty, integer) g-polymatroid.Proof. We will show that Q′ = y−Q is a g-polymatroid, from whi h the statement follows.One an see thatQ′ = {x ∈ RV : x(Z) ≥ p(Z) − 1 ∀Z ⊆ V, x(T ) ≤ y(T ) − 1 if p(T ) = k, 0 ≤ x ≤ y}.Let

D = {T ⊆ V : p(T ) = k but p(Y ) < k for any Y ( T},andC = {X ⊆ V : X ⊆ T for some T ∈ D or X ∩ T = ∅ ∀T ∈ D}.

Se tion 3.2. Weak overing of positively skew-supermodular fun tions 43The positively skew-supermodularity of p implies that D is a subpartition. It is also learthatQ′ = {x ∈ RV : x(Z) ≥ p(Z) − 1 ∀Z ⊆ V, x(T ) ≤ y(T ) − 1 ∀T ∈ D, 0 ≤ x ≤ y}.We laim that Q′ is a tually equal toQ′′ = {x ∈ RV : x(Z) ≥ p(Z) − 1 ∀Z ∈ C, x(T ) ≤ y(T ) − 1 ∀T ∈ D, 0 ≤ x ≤ y}.We only need to show that Q′′ ⊆ Q′. Let x ∈ Q′′ and Z ⊆ V be arbitrary: we have toshow that x(Z) ≥ p(Z) − 1. To prove this assume that Z interse ts t > 0 members of Dand prove by indu tion on t. Let T ∈ D be one of the t members of D interse ted by Z.Assume T and Z satisfy (∩∪) . This implies p(Z) ≤ p(Z ∩ T ) (sin e p(T ) is maximum).Then x(Z) ≥ x(Z ∩ T ) ≥ p(Z ∩ T )− 1 ≥ p(Z)− 1, as laimed. Otherwise T and Z satisfy

(−) , implying p(Z) ≤ p(Z − T ). Then x(Z) ≥ x(Z − T ) ≥ p(Z − T )− 1 ≥ p(Z)− 1, sin eZ − T interse ts t − 1 members of D, so we an use indu tion.Let f(X) = max{0,

∑ti=1(p(Xi) − 1): X1, . . . , Xt is a subpartition of X} for any X ∈ Cand −∞ otherwise, and let g(X) = y(X) − 1 for any X ∈ D and ∞ otherwise. The setfun tion g is interse ting submodular be ause it has �nite values on a subpartition, and it learly satis�es the ross inequality (1.9) with f for any properly interse ting pair X, Y .To see that f is interse ting supermodular, let U be a maximal member of C, and let

fU and p′U be the set fun tions f and p(X) − 1 restri ted to the subsets of U . Then p′Uis skew-supermodular and fU is the supermodular fun tion de�ning C(p′U) (see Theorem1.17). Sin e the maximal members of C form a partition, f is interse ting supermodularon V .We an on lude that f and g form a weak pair, so Q(f, g) is a g-polymatroid byTheorem 1.15. Furthermore, by Theorem 1.16, Q′ = Q′′ = Q(f, g) ∩ {x : 0 ≤ x ≤ y} isalso a g-polymatroid, sin e it is the interse tion of a g-polymatroid with a box.The integer points of Q are losely related to multihyperedges of a multihypergraph thatweakly overs p. More pre isely, the following statement an be proved with a similar proofto that of Theorem 3.9. Let Q∗ = Q∩ {x ∈ RV : x(v) = 1 if y(v) = k}: by Theorem 1.16,Q∗ is an integer g-polymatroid, too (note that d+

H(v) is not ne essarily equal to bH(v) formultihypergraphs).Theorem 3.7. An integer ve tor is in Q∗ if and only if it is a multihyperedge of a mul-tihypergraph H = (V, E) ontaining k multihyperedges, whi h weakly overs p and satis�esd+

H(v) = y(v) for any v ∈ V .

44 Chapter 3. Covering skew-supermodular fun tions with hyperedgesHowever, we would like to speak of hypergraphs instead of multihypergraphs. Thuswe would like to prove a version of Theorem 3.7 that applies to hypergraphs instead ofmultihypergraphs. We an do this, moreover, we an prove a similar statement aboutnearly uniform hypergraphs. To this end, let us modify Q slightly. By the basi propertiesof g-polymatroids we have the following.Corollary 3.8. The polyhedronQ1 = Q1(p, k, y) = Q∗ ∩

{

x ∈ RV : 0 ≤ x ≤ 1,

⌊

y(V )

k

⌋

≤ x(V ) ≤

⌈

y(V )

k

⌉} (3.3)is a non-empty integer g-polymatroid.Proof. The statement is implied by Theorem 1.16. The non-emptiness follows from thefa t that yk∈ Q1 sin e y(v) ≤ k for every v ∈ V .Now we show that Q1 is exa tly the onvex hull of the hara teristi ve tors of hyperedgesthat an appear in the type of hypergraphs we are looking for. To avoid some trivialdegenerate ases we assume that y(V ) ≥ k.Theorem 3.9. An integer ve tor x ∈ ZV is in Q1 if and only if it is the hara teristi ve tor of a hyperedge of a nearly uniform hypergraph H = (V, E) ontaining k hyperedges,whi h weakly overs p and satis�es bH(v) = y(v) for any v ∈ V .Proof. If H = (V, E) is a nearly uniform hypergraph ontaining k hyperedges that weakly overs p and satis�es bH(v) = y(v) for every v ∈ V , then learly χe ∈ Q1 ∩ ZV for any

e ∈ E .Let x ∈ Q1 ∩ ZV . We need to prove that there is a hypergraph H with the desiredproperties. We prove by indu tion on k. Let Hk = (V, {ek}) where χek= x. If k = 1,then Hk satis�es the onditions. Otherwise, let p∗ = p − bHk

, and let y∗ = y − x. Theset fun tion p∗ is positively skew-supermodular. By the hoi e of x we have max{p∗(X) :

X ⊆ V } ≤ k − 1, y∗ ∈ C(p∗), y∗(v) ≤ k − 1 for every v ∈ V , and y∗(V ) ≥ k − 1. LetQ∗

1 = Q1(p∗, k − 1, y∗). By Corollary 3.8, Q∗

1 is also a non-empty g-polymatroid; let x∗ bean arbitrary integer element of Q∗1. By indu tion there is a nearly uniform hypergraph H∗with k − 1 hyperedges (one of them with hara teristi ve tor x∗) satisfying bH∗ ≥ p∗ and

bH∗(v) = y∗(v) for every v ∈ V . Thus H = H∗ + Hk weakly overs p and bH(v) = y(v)for every v ∈ V . In addition, sin e H∗ is nearly uniform and ⌊y(V )/k⌋ ≤ y∗(V )/(k − 1) ≤

⌈y(V )/k⌉ H , it follows that H is nearly uniform, too.The above theorem of ourse implies the following.

Se tion 3.3. Covering two positively skew-supermodular fun tions 45Corollary 3.10. Let p : 2V → Z ∪ {−∞} be a positively skew-supermodular fun tion, letk ≥ max{p(X) : X ⊆ V } be an integer, and let y ∈ C(p) be an integer ve tor with y(V ) ≥ kand y(v) ≤ k for every v ∈ V . Then there is a nearly uniform hypergraph ontaining khyperedges that weakly overs p and satis�es bH(v) = y(v) for every v ∈ V .Next, we prove that for symmetri set fun tions we an strengthen the above result byrepla ing �weakly overs� by � overs�, provided that k = Mp = max{p(X) : X ⊆ V }.Lemma 3.11. If p : 2V → Z ∪ {−∞} is a symmetri positively skew-supermodularfun tion, k = max{p(X) : X ⊆ V }, and H is a hypergraph ontaining exa tly k hyperedges,then bH ≥ p implies that H overs p.Proof. The simplest way of proving this at this point is just to say that it obviously followsfrom Theorem 3.3 (i). However we give a dire t proof, too. Suppose that H does not over p, so there is a set X with bH(X) ≥ p(X) > dH(X) = bH(X)− iH(X), where iH(X)denotes the number of hyperedges of H indu ed by X. By the assumptions there is a setT with p(T ) = k. Sin e H ontains exa tly k hyperedges, p(X ∪ T ) = p(V − (X ∪ T )) ≤

k − iH(X) and p(T − X) ≤ k − iH(X) also follows. If (∩∪) applies for X and T thenp(X ∩ T ) ≥ p(X) + iH(X) > bH(X) ≥ bH(X ∩ T ), and if (−) applies for X and Tthen p(X − T ) ≥ p(X) + iH(X) > bH(X) ≥ bH(X − T ), both of whi h ontradi ts ourassumptions.This immediately implies the following generalization of Theorem 3.2.Theorem 3.12. Let p : 2V → Z ∪ {−∞} be a symmetri , positively skew-supermodularfun tion, let k = max{p(X) : X ⊆ V }, and let y ∈ C(p) be an integer ve tor with y(v) ≤ kfor every v ∈ V . Then there is a nearly uniform hypergraph ontaining k hyperedges that overs p and satis�es bH(v) = y(v) for every v ∈ V .Note that if we want to �nd a hypergraph of minimum total size that overs p, it su� esto �nd a ve tor y ∈ C(p) with y(V ) minimal. Su h a ve tor automati ally satis�es y(v) ≤ kfor every v ∈ V be ause the ve tor de�ned by y′(v) = min{k, y(v)} is also in C(p). We aneven go a bit further: instead of minimizing the total size of H , we may want to minimize∑

v∈V c(v) · bH(v), where c is a non-negative ost fun tion on the nodes. In this ase weshould �nd a ve tor y ∈ C(p) with ∑

v∈V c(v)y(v) minimal.3.3 Covering two positively skew-supermodular fun tionsAs in the ase of S hrijver's supermodular olouring theorem, these results an be gen-eralized using the fa t that the interse tion of two integer g-polymatroids is an integerpolyhedron (a spe ial type of submodular �ow polyhedron).

46 Chapter 3. Covering skew-supermodular fun tions with hyperedgesLet p1, p2 : 2V → Z ∪ {−∞} be two positively skew-supermodular fun tions, and letk ≥ max{pi(X) : X ⊆ V, i = 1, 2}. Let y ∈ C(p1) ∩ C(p2) ∩ ZV su h that y(v) ≤ k forevery v ∈ V and y(V ) ≥ k (for example (k, k, . . . , k) is su h a ve tor). Let

R = R(p1, p2, k, y) := Q1(p1, k, y) ∩ Q1(p2, k, y),where Q1 is de�ned as in (3.3). Then y/k ∈ R, and R is the interse tion of two integerg-polymatroids, so R is a non-empty integer polyhedron.Theorem 3.13. An integer ve tor x ∈ ZV is in R if and only if it is the hara teristi ve tor of a hyperedge of a nearly uniform hypergraph H = (V, E) ontaining k hyperedgeswhi h weakly overs both p1 and p2 and satis�es bH(v) = y(v) for every v ∈ V .Proof. The proof is analogous to the proof of Theorem 3.9; the only di�eren e is that wehave to de�ne p∗1 and p∗2, and use indu tion on R(p∗1, p∗2, k − 1, y∗).The results above imply the following �skew-supermodular olouring theorem� whi h isan extension of S hrijver's supermodular olouring theorem. It follows from Theorem 3.13by hoosing y = (1, 1, . . . , 1)T ∈ RV .Theorem 3.14 (T. Király, [28℄). Let p1, p2 : 2V → Z ∪ {−∞} be two positively skew-supermodular fun tions and k ≥ 1 an integer. Then there is a k- olouring that is good forboth p1 and p2 if and only if

pi(X) ≤ min{k, |X|} for any X ⊆ V and i = 1, 2.Using Lemma 3.11 we obtain the following theorem.Theorem 3.15. Let p1 : 2V → Z ∪ {−∞} and p2 : 2V → Z ∪ {−∞} be two symmetri ,positively skew-supermodular set fun tions su h thatmax{p1(X) : X ⊆ V } = max{p2(X) : X ⊆ V } = k, (3.4)and let y ∈ C(p1) ∩ C(p2) be an integer ve tor with y(v) ≤ k for every v ∈ V . Thenthere is a nearly uniform hypergraph ontaining k hyperedges that overs both p1 and p2and satis�es bH(v) = y(v) for every v ∈ V .This theorem an be used to �nd a hypergraph of minimum total size that overs both

p1 and p2 provided that (3.4) holds. By the argument presented at the end of the previousse tion, it su� es to �nd a ve tor y ∈ C(p1)∩C(p2) with y(V ) minimal, sin e su h a ve torautomati ally satis�es y(v) ≤ k for every v ∈ V . We an also minimize ∑

v∈V c(v) · bH(v),where c is a non-negative ost fun tion on the nodes.We will show in Se tion 3.4.1 that without the assumption that the maximum values ofthe two set fun tions are the same, the problem of minimizing the total size of H be omesNP-hard.

Se tion 3.4. Appli ations 473.4 Appli ationsIn this se tion we present appli ations of our results. Besides the lo al edge- onne tivityaugmentation of hypergraphs dis ussed by Szigeti in [42℄, we show two other appli ations.3.4.1 Lo al edge- onne tivity augmentation of hypergraphsThe lo al edge- onne tivity augmentation of hypergraphs with hyperedges ofminimum total size (solved by Szigeti in [42℄) is the following problem.Problem 3.16. Given a hypergraph H0 = (V, E0) and a symmetri edge- onne tivity re-quirement r : V × V → Z+, �nd a hypergraph H of minimum total size su h that H0 + His r-edge- onne ted, meaning thatλH0+H(u, v) ≥ r(u, v) for every u, v ∈ V. (3.5)Let us de�ne the set fun tion Rloc with (2.5). As we have seen in Se tion 2.1.4, ahypergraph H satis�es (3.5) if and only if H overs p = Rloc − dH0

. Sin e p is a posi-tively skew-supermodular fun tion, applying Theorem 3.12 gives the following extension ofSzigeti's result.Theorem 3.17. The optimal solution of the lo al edge- onne tivity augmentation of hy-pergraphs with hyperedges of minimum total size an be hosen to be nearly uniform.Now we use Theorem 3.15 to get an even more general result. Consider the followingproblem, the simultaneous lo al edge- onne tivity augmentation problem. Giventwo hypergraphs H1, H2 on the same ground set V , two symmetri requirement fun tionsr1, r2 : V × V → Z, and a nonnegative ost fun tion c : V → R+, we want to �nd ahypergraph H of minimum total ost su h that Hi + H is ri-edge- onne ted for i = 1, 2(the ost of H is ∑

v∈V c(v)bH(v)).Theorem 3.18. Assume that max{r1(u, v) − λH1(u, v) : u, v ∈ V } = max{r2(u, v) −

λH2(u, v) : u, v ∈ V }, in other words Mp1

= Mp2, where p1 and p2 are the dei�en yfun tions of the two instan es. If y ∈ C(p1)∩C(p2)∩ZV then there exists a nearly uniformhypergraph H = (V, E) satisfying the degree-spe i� ation y that solves the simultaneouslo al edge- onne tivity augmentation problem (i.e. overs both p1 and p2).The following theorem shows that without the assumption on the maximum de� ien iesthe problem be omes NP - omplete: the redu tion is similar to that used in [16℄.Theorem 3.19. The simultaneous lo al edge- onne tivity augmentation problem of twohypergraphs is in general NP - omplete, even if the ost fun tion is onstant.

48 Chapter 3. Covering skew-supermodular fun tions with hyperedgesProof. The problem is learly in NP . To show its ompleteness onsider the Spe ialBin-Pa king Problem (SBP). An instan e of this problem onsists of a set of positiveintegers W = {w1, w2, . . . , wn} (weights), a set of positive integers B = {b1, b2, . . . , bm}(bins) su h that γ =∑n

i=1 wi =∑m

j=1 bj . The SBP problem asks whether there exists apartition W1, W2 . . . , Wm of W su h that ∑

w∈Wjw = bj for every j = 1, 2, . . . , m. Thisproblem is shown to be strongly NP - omplete in [15℄, i.e. it remains NP - omplete even ifthe weights and bins are unary en oded. We will redu e the unary-en oded SBP problemto our problem. For ea h weight wi ∈ W onsider a set Xi su h that |Xi| = wi andsimilarly, for ea h bin bj ∈ B let Yj be su h that |Yj| = bj . The sets Xi (i = 1, 2, . . . , n) and

Yj (j = 1, 2, . . . , m) are assumed to be pairwise disjoint. Let X = ∪ni=1Xi and Y = ∪m

j=1Yj.The ground set of the two hypergraphs is V = X ∪ Y . The edge-set of H1 is the unionof a omplete graph on X and a omplete graph on Y . The requirement fun tion r1 isuniformly γ. The edge-set of H2 onsists of hyperedges Yj for every j = 1, 2, . . . , m andthe requirement is r2(u, v) = 1 if u, v ∈ Xi for some i, and 0 otherwise. One an he k thatthere is a hypergraph H of total size at most 2γ su h that Hi + H is ri-edge- onne ted fori = 1, 2 if and only if the SBP problem is solvable: the hypergraph H must be in fa t agraph, more pre isely a omplete mat hing between X and Y (note that in H1 the degreeof any node is γ − 1). The details are left to the reader.3.4.2 The node-to-area onne tivity augmentation problem in hy-pergraphsThe node-to-area onne tivity augmentation problem for graphs was solved by Ishii andHagiwara [26℄. Here we onsider a hypergraphi version of the problem. Note that this isnot a generalization of the node-to-area onne tivity augmentation of graphs.Problem 3.20 (Node-to-area onne tivity augmentation problem in hypergraphs). Givena hypergraph H0 = (V, E0), a olle tion of subsets W of V and a fun tion r : W → Z+, ouraim is to �nd a hypergraph H of minimum total size su h that

λH0+H(x, W ) ≥ r(W ) for any W ∈ W and x ∈ V. (3.6)De�ne the set fun tion RN2A by (2.4) and let q = RN2A − dH0. The fun tion q isinterse ting negamodular, so p = qs is positively skew-supermodular. From Theorems 3.12and 3.15 we get the following.Theorem 3.21. The optimal solution of the node-to-area onne tivity augmentation prob-lem in hypergraphs an be hosen to be nearly uniform. If we have two instan es of the

Se tion 3.4. Appli ations 49problem (on the same node set) su h that for the orresponding requirement fun tions p1and p2 we have max{p1(X) : X ⊆ V } = max{p2(X) : X ⊆ V } and y ∈ C(p1)∩C(p2)∩ZVthen there exists a nearly uniform hypergraph H = (V, E) satisfying the degree-spe i� ationy that solves the simultaneous node-to-area edge- onne tivity augmentation problem (i.e. overs both p1 and p2).Note that the node-to-area onne tivity augmentation problem for graphs was solved in[26℄ for the ase when the set fun tion RN2A does not take the value 1; it was shown thatwithout this assumption the problem is NP-hard. By the above theorem, this assumptionis not needed in the hypergraphi ase.3.4.3 Augmenting the global ar - onne tivity of mixed hypergraphsThis problem is a hypergraphi version of the global ar - onne tivity augmentation ofmixed graphs with undire ted edges solved by Bang-Jensen, Ja kson and Frank [2℄. Theproblem is the following.Problem 3.22 (Augmenting the global ar - onne tivity of mixed hypergraphs). Let M =

(V,A) be a mixed hypergraph, r ∈ V is a designated root node, and k, l be nonnegativeintegers. Find a hypergraph H of minimum total size su h that M+H is (k, l)-ar - onne tedfrom root r.Sin e this problem is a spe ial ase of the problem of overing a positively skew-supermodularfun tion with hyperedges, it was also solved by Szigeti. Our results imply the followinggeneralizations.Theorem 3.23. If M = (V,A) is a mixed hypergraph, r ∈ V is a designated root node, k, lare nonnegative integers, and y ∈ ZV is a degree-spe i� ation then there exists a hypergraphH su h that M +H is (k, l)-ar - onne ted from r and d+

H(v) = y(v) for every v ∈ V if andonly if y ∈ C(p) with p = qsM,r,k,l. Furthermore, H an be hosen to be nearly uniform.Theorem 3.24 (Simultaneous global ar - onne tivity augmentation of mixed hypergraphs).If we have two instan es of Problem 3.22 given by (M1, r1, k1, l1) and (M2, r2, k2, l2) on a ommon node set V , and for the set fun tions q1 = qM1,r1,k1,l1, q2 = qM2,r2,k2,l2 we have

max{q1(X) : X ⊆ V } = max{q2(X) : X ⊆ V },then there exists a nearly uniform hypergraph H = (V, E) satisfying the degree-spe i� ationy that solves the simultaneous lo al edge- onne tivity augmentation problem (i.e. oversboth q1 and q2) if and only if y ∈ C(qs

1)∩C(qs2)∩ZV . Furthermore, H an be hosen to benearly uniform.

50 Chapter 3. Covering skew-supermodular fun tions with hyperedges

Chapter 4Covering skew-supermodular fun tionswith graph edges: a new approa h tosplitting-o�In this hapter we mainly onsider the problem of overing our fun tion only with graphedges. However our approa h gives rise to a di�erent question, too, whi h is the following:augment the edge- onne tivity of a (mixed or undire ted) hypergraph without in reasingits rank. This question naturally arose from our approa h to splitting-o�, as it will be learshortly. We will all it the rank respe ting augmentation problem: exa t de�nitionswill be given later. Nevertheless, the main on ern of this hapter is the following problem.Problem 4.1 (Covering a positively skew-supermodular fun tion with graph edges). Letus be given a symmetri , positively skew-supermodular fun tion p : 2V → Z∪{−∞} with amaximizing ora le. The minimum version of the problem is to �nd a graph G overingp with a minimum number of edges. The degree-spe i�ed version is to �nd a a graph G overing p that also satis�es a given degree-spe i� ation m ∈ ZV

+.We remind the reader that this problem is in general NP - omplete, even for spe ialskew-supermodular fun tions. Let us mention two ases that are already NP - omplete.It was shown in [16℄ that the lo al edge- onne tivity augmentation of a hypergraph witha minimum number of graph edges (introdu ed in Se tion 2.1.4) is NP - omplete. Theproof given there shows that the lo al edge- onne tivity augmentation of a hypergraph witha degree-spe i�ed graph is also NP - omplete. The other NP - omplete spe ial ase is thegeneral node-to-area onne tivity augmentation problem in graphs (see Theorem 2.7).The results presented in this hapter appeared in our joint paper with Tamás Király[11℄, ex ept for those in Se tion 4.4.3 whi h appeared in [7℄.51

52 Chapter 4. Covering skew-supermodular fun tions with graph edges4.1 Previous results - brief historyAs we have mentioned in the previous se tion, the problem of overing a symmetri skew-supermodular fun tion with a minimum number of graph edges is in general NP - omplete.However, many spe ial ases an be solved in polynomial time: let us review some of these ases.Global edge- onne tivity augmentation of graphs was solved by Watanabe and Nakamura[45℄. However, their algorithm is not strongly polynomial. Frank [19℄ gave the �rst stronglypolynomial algorithm for this problem. These results will be stated in Chapter 5, sin ethey fall in the lass of overing a symmetri rossing supermodular fun tion.Frank also proved in [19℄ that the lo al edge- onne tivity augmentation of graphs an alsobe solved in strongly polynomial time. His proof uses the splitting-o� te hnique to solvethe degree spe i�ed problem. The solution of the lo al edge- onne tivity augmentationproblem relies on the lassi al splitting lemma of Mader:Lemma 4.2 (Mader's lemma). Let G = (V + s, E) be su h that there is no ut edgein ident to s and dG(s) > 3. Then there exists a splitting-o� at s that preserves the lo aledge- onne tivities in V .Frank dis overed the relationship between the degree-spe i�ed and the minimum versionsof augmentation problems that we sket hed in Se tion 2.2. He also observed that Mader'slemma ited above is essentially equivalent to the degree-spe i�ed lo al edge- onne tivityaugmentation of graphs. Thus he gave the solution of the problem that is the following.Theorem 4.3 (A. Frank, [19℄). Let G0 be a graph and let r : V ×V → Z+ be a symmetri edge- onne tivity requirement su h that r(u, v) 6= 1 for any pair u, v ∈ V . There exists agraph G satisfying the degree spe i� ation m ∈ ZV+ su h that G0 + G is r-edge- onne ted ifand only if m(V ) is even and

m(X) ≥ R(X) − dG0(X) for any X ⊆ V,where Rloc is de�ned with (2.5). The minimum number of graph edges that make G0 r-edge- onne ted is equal to

max{⌈1/2∑

X∈X

(Rloc(X) − dG0(X))⌉ : X is a subpartition of V }.Note that Frank a tually solved the more general problem without the assumption that

r(u, v) 6= 1, but the answer is a little more te hni al if this an also happen.The next result is about augmenting the ar - onne tivity of a mixed graph with undi-re ted edges. For simpli ity, we only state the minimum version.

Se tion 4.2. A simple lemma and algorithm 53Theorem 4.4 (J. Bang-Jensen, A. Frank, and B. Ja kson, [2℄). Let M0 = (V, E0) be amixed graph and k ≥ 2 integer. The minimum number of (undire ted) graph edges thatmake M0 k-ar - onne ted is equal tomax{⌈1/2

∑

X∈X

k − βM0(X))⌉ : X is a subpartition of V },where βM0

(X) = min{̺M0(X), δM0

(X)} for any X ⊆ V .Note that the above theorems state that solution of the minimum version only dependson the bound SLB, and the solution of the degree-spe i�ed version exists if and only if thedegree spe i� ation m is admissible (and m(V ) is even).In the following problem this is not exa tly the ase: the optimal solution of the min-imum version an be one bigger than the bound ⌈12SLB⌉ obtained from subpartitions.This problem is the node-to-area onne tivity augmentation problem solved by Ishii andHagiwara.Theorem 4.5 (Ishii and Hagiwara [26℄). Let a node-to-area onne tivity augmentationproblem be given by the graph G0 = (V, E0), a olle tion of subsets W of V and a fun tion

r : W → Z+ \ {1}. The minimum number of graph edges |F | su h that G0 +F satis�es thearea requirements is equal to ⌈12SLB(p)⌉, unless there is a ertain on�guration alled W- on�guration present, in whi h ase the optimum is ⌈1

2SLB(p)⌉+ 1, where p = Rs

N2A − dG0and RN2A is de�ned by (2.3).The de�nition of a W- on�guration is de�ned in [26℄ where it is alled P-property.In Se tion 4.3.2 we will give a simple proof of the fa t that the optimum is at most⌈1

2SLB(p)⌉ + 1.4.2 A simple lemma and algorithmIn the results below we will use the splitting-o� te hnique. The basi s of this te hnique aredes ribed in Se tion 2.3. The starting point of our results is Lemma 4.6. This lemma wasalso found by Nutov who sket hed a proof in [36℄. For the spe ial ase when p is obtainedfrom lo al edge- onne tivity augmentation requirements in a hypergraph, this lemma wasimpli itly also shown by Ben Cosh in [15℄. However the proof presented here is simplerthan the previous ones and its onstru tiveness has further appli ations, too.Lemma 4.6. Let p : 2V → Z ∪ {−∞} be a symmetri , positively skew-supermodularfun tion and m ∈ C(p) ∩ ZV . If Mp = max{p(X) : X ⊆ V } > 1 then there is anadmissible splitting-o�.

54 Chapter 4. Covering skew-supermodular fun tions with graph edgesProof. Let Y be a minimal set satisfying p(Y ) = Mp. By symmetry, p(V − Y ) = Mp,too, so we an hoose a minimal set Z ⊆ V − Y satisfying p(Z) = Mp. Sin e Mp ≥ 1we an hoose y ∈ Y, z ∈ Z with m(y), m(z) > 0. We laim that the splitting at yand z is admissible. Assume that it is not and onsider a dangerous set X ontainingy and z (note that p(X) > 0). Sin e m(X − Y ) ≤ m(X) − m(y) ≤ m(X) − 1 andp(Y − X) < Mp by the minimality of Y , X and Y annot satisfy (−) , sin e that wouldmean m(X) − 1 + Mp ≤ p(X) + p(Y ) ≤ p(X − Y ) + p(Y − X) < m(X − Y ) + Mp ≤

m(X) − 1 + Mp, a ontradi tion. So X and Y must satisfy (∩∪) , whi h implies (usingm(X ∩ Y ) = m(X) − m(X − Y ) ≤ m(X) − m(z) ≤ m(X) − 1) that p(X ∪ Y ) = Mpand m(X − Y ) = 1, sin e m(X) − 1 + Mp ≤ p(X) + p(Y ) ≤ p(X ∩ Y ) + p(X ∪ Y ) ≤

m(X ∩Y ) +Mp ≤ m(X)− 1 + Mp. Now X ∪Y and Z annot satisfy (−) sin e this wouldgive p(Z − (X ∪ Y )) = Mp, ontradi ting the minimality of Z. Therefore X ∪ Y and Zsatisfy (∩∪) implying that p(Z ∩ (X ∪ Y )) = Mp, whi h is only possible if Z ⊆ X ∪ Y .But 2 ≤ Mp = p(Z) ≤ m(Z) ≤ m(X − Y ) = 1 gives a ontradi tion.Let us mention an important onsequen e of this lemma. If there is no admissiblesplitting-o�, then p ≤ 1 and every pair u, v ∈ V + is in a dangerous set X: this meansthat p(X) = 1 and m(X) = 2, hen e m ≤ 1. The following orollary is immediate: itgeneralizes a Theorem 3.2 of Szigeti and in a spe ial ase it was also observed by Ben Cosh(see [15℄).Corollary 4.7. If p is a symmetri , positively skew-supermodular fun tion and m ∈

C(p) ∩ ZV then there is a hypergraph H overing p satisfying the degree-spe i� ation mthat ontains at most one hyperedge of size greater than 2.Proof. If there is an admissible splitting-o� then the statement follows by indu tion onm(V ). Otherwise p ≤ 1 by Lemma 4.6: observe that a hypergraph ontaining the onlyhyperedge V + overs p in this ase by the symmetry of p.Consider the following greedy algorithm for the degree-spe i�ed overing problem (notethat this algorithm an be implemented to run in polynomial time only with a maximizingora le for p).Algorithm GREEDYCOVERbegin INPUT A symmetri , positively skew-supermodular fun tion p : 2V → Z∪{−∞}(given with a maximizing ora le) and m ∈ C(p) ∩ ZV .OUTPUT A graph G = (V, E) and a hyperedge e su h that the hypergraph G + e overs p and d+

G+e(v) = m(v) for every v ∈ V .

Se tion 4.2. A simple lemma and algorithm 551.1. Initialize G = (V, ∅).1.2. While there exists an admissible pair u, v do1.3. Let m = m − χ(u) − χ(v) and p = p − d(V,{(u,v)}) and G = G + (uv).1.4. Output G and e where χe = m.endClearly, if one an test membership in C(p − dG) in polynomial time for any graph G(whi h an be done using a maximizing ora le) then this algorithm an be implemented torun in polynomial time using the following observations. If a pair u, v ∈ V + is admissible atsome point, then by a binary sear h we an determine the maximum number of admissiblesplittings at u and v. Furthermore, if a pair of positive nodes u, v is not admissible at somepoint, then it will not be ome admissible later, sin e dangerous sets remain dangerous.We say that the algorithm got stu k if the hyperedge e in the output is of size greaterthan 0. Using the symmetry of p we an show that the algorithm annot get stu k withm(V ) = 2.4.2.1 General observations on the stu k situationWe an read out many things about the situation when the algorithm GREEDYCOVER getsstu k from Lemma 4.6. In the following lines we will do this. So for the rest of Se tion4.2 let p : 2V → Z ∪ {−∞} be symmetri and positively skew-supermodular, m ∈ C(p)and we assume that there is no further admissible splitting o�. By Lemma 4.6 and theremarks following it, every pair u, v ∈ V + is in a dangerous set X: sin e p ≤ 1 this meansthat p(X) = 1 and m(X) = 2, hen e m ≤ 1. The interesting ase for us will be the asewhen the splitting pro edure gets stu k with m(V ) ≥ 4: we either want to �nd a graph,when m(V ) has to be even, or we want to solve a rank-respe ting problem when gettingstu k with m(V ) = 3 is satisfa tory. In the rest of Se tion 4.2 we assume that we are atthis stu k situation with m(V ) ≥ 4.Observe that the algorithm GREEDYCOVER an be modi�ed in an obvious way if G + eis not a feasible output (for example e is too big): we an repla e e with any onne tedhypergraph on V +. For example if we are only allowed to use graph edges then we annoti e that with m(V ) − 1 graph edges we an �nish the pro edure: any spanning treeon V + will over p. However, we ould possibly over p with less edges, as the examplep(X) = 1 if |X| ∈ {1, 2, |V | − 2, |V | − 1} (and p(X) = 0 otherwise) shows. Though thereis a lower bound: one needs at least ⌈2(m(V ) − 1)/3⌉ edges to �nish the pro edure. Thespe ial ase of the following lemma when m is minimal was also proved by Nutov in [36℄,who used it to devise approximation algorithms for the problem of overing a symmetri ,

56 Chapter 4. Covering skew-supermodular fun tions with graph edgespositively skew-supermodular fun tion with a minimum number of graph edges.Lemma 4.8. Let p : 2V → Z ∪ {−∞} be a symmetri , positively skew-supermodularfun tion and m ∈ C(p) ∩ ZV . Assume that there is no admissible splitting-o�. Then any(in lusionwise) minimal graph K overing p has at least ⌈2(m(V ) − 1)/3⌉ edges.Proof. We laim that we an assume that the edges of F onne t positive nodes: onsiderany edge e = (xy) ∈ E(K), where at least one of x and y is not positive. Sin e K − e doesnot over p, the fun tion p′ = p − dK−e has positive values, however obviously p′ ≤ p ≤ 1.We laim that the family F = {X ⊆ V : x ∈ X, y /∈ X, p′(X) = 1} is losed underinterse tion and union. Let X, Y ∈ F : then p′ annot satisfy (−) for X and Y , sin ethen K would not over p. So (∩∪) holds for X and Y and fun tion p′, whi h impliesthat X ∩ Y, X ∪ Y ∈ F , as laimed. Sin e m ∈ C(p) ∩ ZV , there must be a positive nodex0 ∈ ∩F and a positive node y0 ∈ V − ∪F , so K ′ = K − e + (x0y0) also overs p anditerating this we arrive at a graph (with the same number of edges) that has only edgesbetween positive nodes.It is lear that K[V +] annot ontain a omponent of ardinality 2, sin e any pair ofpositive nodes is ontained in a dangerous set, therefore an edge of K must leave this set.If m is not minimal then it is possible that there exists a v ∈ V + that is not ontainedin a tight set, but in this ase any other u ∈ V + must be ontained in a tight set, sin eif u1, u2 ∈ V + then Xu1,v and Xu2,v annot satisfy (∩∪) (sin e their interse tion is nottight), so they satisfy (−) , implying that u1 and u2 are both ontained in a tight set, as laimed. Therefore every omponent of K[V +] must be of ardinality at least 3, ex ept forat most one singleton omponent (however, if m is minimal, then there is no su h singleton omponent). So if C denotes the set of these omponents then |C| ≤ (m(V ) − 1)/3 + 1.Using this we have

|E(K)| ≥∑

C∈C

(|V (C)| − 1) ≥ m(V ) − (m(V ) + 2)/3 = 2(m(V ) − 1)/3.Let us give a lemma that will be useful later. Assume x0, x1, x2 ∈ V are three di�erentpositive nodes and X0, X1, X2 are three dangerous sets blo king them with xi ∈ Xj ∩ Xkfor any {i, j, k} = {0, 1, 2}. (Sin e we assume that m(V ) ≥ 4 the three sets X0, X1, X2 arepairwise rossing here.) We will say that X0, X1 and X2 form a blo king-triangle. Wesay that X0 is slim if X0 ∩ X1 ∩ X2 = ∅ and X0 − (X1 ∪ X2) = ∅.Lemma 4.9 (Slimming Lemma). Assume that X1 and X2 satisfy (∩∪) . Then (X0−(X1∩

X2)) ∩ (X1 ∪ X2) is also dangerous and blo ks x1, x2.Proof. Sin e X1 and X2 satisfy (∩∪) , p(X1 ∩ X2) = p(X1 ∪ X2) = 1. Now X1 ∩ X2and X0 annot satisfy (∩∪) , sin e that would imply that p(X0 ∩ X1 ∩ X2) = 1, but

Se tion 4.2. A simple lemma and algorithm 57m(X0 ∩ X1 ∩ X2) = 0. This implies that p(X ′

0) = 1 where X ′0 = X0 − (X1 ∩ X2). Now X ′

0and X1∪X2 annot satisfy (−) , sin e that would give p(X ′0− (X1∪X2)) = 1 ontradi ting

m(X ′0 − (X1 ∪ X2)) = 0. So we obtain from (∩∪) that p(X ′

0 ∩ (X1 ∪ X2)) = 1 and learlyx1, x2 ∈ X ′

0 ∩ (X1 ∪ X2).We note that the family of sets blo king a �xed pair of nodes u, v ∈ V + is losed underunion and interse tion. Let us denote the unique minimal member of this family by Xuv.Observe, that for 4 di�erent nodes u, v, x, y ∈ V + we have Xuv ∩ Xxy = ∅: they annotsatisfy (∩∪) sin e m(Xuv ∩ Xxy) = 0, so they satisfy (−) , and then by minimality theymust be disjoint.4.2.2 Simple proofsIn this subse tion we give simple proofs of lassi al results in order to demonstrate thesimpli ity of our approa h. First we give a simple proof of a spe ial ase of a theorem ofBen zúr and Frank. They proved in [5℄ that the problem of overing a symmetri , pos-itively rossing supermodular set fun tion by a minimum number of graph edges an besolved in polynomial time (if the fun tion is given with a maximizing ora le): a ompleteproof of their theorem will be given in Se tion 5.3. In a spe ial ase this problem an besolved greedily. Many proofs below onsider the situation when the Algorithm GREEDY-COVER gets stu k. In most of the ases we an assume that this is already the ase in thebeginning, sin e after some steps we are again at an instan e of our starting problem: anexample of this is Theorem 4.10. Note that a symmetri positively rossing supermodularfun tion is also positively skew-supermodular (whi h is not ne essarily the ase without thesymmetry). Furthermore, a symmetri , positively rossing supermodular fun tion satis�esboth (∩∪) and (−) if X and Y are rossing p-positive sets.Theorem 4.10. Let p0 : 2V → Z∪{−∞} be a symmetri , positively rossing supermodularfun tion that does not take 1 as value, and G = (V, E) be an arbitrary graph. Thenthe Algorithm GREEDYCOVER does not get stu k with input p = p0 − dG and arbitrarym ∈ C(p) ∩ ZV if m(V ) is even.Proof. Assume that the algorithm GREEDYCOVER gets stu k (at start). Then m(V ) ≥ 4must hold. Consider a blo king triangle X, Y, Z. By Lemma 4.6 and the observationsabove, any pair of this three sets must satisfy (∩∪) and (−) for p with equality. Usingthe Slimming Lemma (Lemma 4.9) we an assume that X, Y and Z are all slim. However,p(X ∩ Y ) = p0(X ∩ Y )− dG(X ∩ Y ) = 1 and p0(X ∩ Y ) 6= 1 implies that there must be anedge in G leaving X ∩ Y . But in presen e of su h an edge we are able to �nd two sets outof X, Y, Z that annot satisfy (−) or (∩∪) with equality by (1.5) and (1.6).

58 Chapter 4. Covering skew-supermodular fun tions with graph edgesThe above theorem in ludes the lassi al splitting theorem of Lovász that an be usedfor global edge- onne tivity augmentation of graphs.Lemma 4.11 (Lovász' lemma). Let G = (V + s, E) be k-edge- onne ted in V , wherek ≥ 2. Assume dG(s) is even. Then there exists a splitting-o� at s that preserves k-edge- onne tivity in V .Proof. Let G′ = G[V ] and p : 2V → Z de�ned by p(X) = k − dG′(X) for any ∅ 6= X 6= Vand p(∅) = p(V ) = 0. Let m(v) = dG(s, v) for any v ∈ V . With these notations the lemmafollows from Theorem 4.10.Next we give a simple proof of Mader's lassi al splitting lemma.Proof of Lemma 4.2. If there is no ut edge in ident to s then λG(u, v) ≥ 2 for any pairof s-neighbours u, v. Let us de�ne R(X) = max{λG(x, y) : x ∈ X, y ∈ V − X} for any Xwith ∅ 6= X 6= V and R(∅) = R(V ) = 0 and p(X) = R(X)− dG[V ](X) for any X ⊆ V . Letm(v) = dG(s, v) for any v ∈ V . By Theorem 2.10, the fun tion R (and onsequently p)is a symmetri and skew-supermodular fun tion. By assumption, m(X) ≥ p(X) holds forevery X ⊆ V . Assume that there is no splitting-o� and take a blo king triangle X, Y, Z onsisting of maximal dangerous sets. Consider the following two ases.Case I.: Assume that two of these three sets (wlog. X and Y ) satisfy (∩∪) . Then, usingthe Slimming Lemma, substitute Z by Z ′ = (Z−(X∩Y ))∩(X ∪Y ). Let R(Z ′) = λG(z, v)with z ∈ Z ′ and v ∈ V − Z ′ and assume wlog. that z ∈ X ∩ Z ′ implying R(Z ′) ≤

R(X ∩ Z ′). Sin e there is no ut edge in ident to s, dG(Y ∩ Z ′) ≥ R(Y ∩ Z ′) ≥ 2. ThendG(Z ′)−1 ≤ R(Z ′) ≤ R(X∩Z ′) ≤ dG(X∩Z ′) = dG(Z ′)−dG(Y ∩Z ′)+dG(X∩Z ′, Y ∩Z ′) ≤

dG(Z ′) − 2 + dG(X ∩ Z ′, Y ∩ Z ′) implies that dG(X ∩ Z ′, Y ∩ Z ′) > 0, but then X and Y annot satisfy (∩∪) with equality by (1.5).Case II.: Assume that X, Y and Z pairwise satisfy (−) . This implies that p(X −Y ) = 1, onsequently Z and X − Y annot satisfy (−) , sin e m((X − Y ) − Z) = 0. Thus theysatisfy (∩∪) whi h implies by the maximality of Z that X − (Y ∪ Z) = ∅. Similarly we an prove that Y − (Z ∪ X) = Z − (X ∪ Y ) = ∅. Using that there is a neighbour of s notin X ∪ Y ∪Z we an dedu e that R(X ∪Y ∪Z) ≥ 2. However, sin e X, Y and Z pairwisesatisfy (−) with equality, there must not be an edge of G[V ] leaving X ∪ Y ∪ Z by (1.6).But this would imply that p(X ∪ Y ∪ Z) ≥ 2, ontradi ting Lemma 4.6.Finally we will give a simple proof of a theorem of Bang-Jensen, Frank and Ja kson [2℄on undire ted splitting-o� in mixed graphs: the k = l ase is a spe ial ase of Theorem3.2 of [2℄, so we also manage to extend slightly this spe ial ase. We mention that theauthors of [2℄ proved a more general theorem on the augmentation of mixed graphs that

Se tion 4.3. Stu k situation for spe ial skew-supermodular fun tions 59also ontained Mader's Lemma as a spe ial ase, but we only on entrate on a spe ial aseof their theorem, whi h is identi al to the k = l ase of our theorem below. Interestingly,if either k = 1 or l = 1 in this problem then the answer is di�erent and this is not overedby the following theorem.Theorem 4.12 (Bang-Jensen, Frank, Ja kson). Let M = (V + s, E) be a mixed graphand assume that s is only in ident with undire ted edges. Let r ∈ V and k, l ∈ N − {1}integers and assume that λM(r, v) ≥ k and λM(v, r) ≥ l for any v ∈ V . Then there existsa splitting-o� at s preserving this property, provided that dM(s) > 3.Proof. We an assume that M − s is a digraph (by substituting undire ted edges byoppositely dire ted pairs of ar s): let us denote this digraph by D = (V, A) and letm(v) = dM(s, v) for any v ∈ V . Let the fun tion q be de�ned by q(∅) = q(V ) = 0,q(X) = k − ̺D(X) for any nonempty X ⊆ V − r and q(X) = l − ̺D(X) for any X ( Vwith r ∈ X (i.e. q = qD,r,k,l as de�ned in (2.2)). Then one an he k that q is rossingsupermodular and thus p = qs is skew-supermodular. Sin e M is (k, l)-ar - onne ted fromr (apart from s), m(X) ≥ p(X) for any X ⊆ V . Assume that there is no splitting-o�.Consider a blo king triangle X, Y, Z. We an assume without loss of generality that eitherq(X) = q(Y ) = 1 or q(X) = q(Y ) = 1 so X and Y must satisfy (∩∪) with equality, imply-ing that dD(X, Y ) = 0. By Lemma 4.9 we an assume that Z is slim. If r /∈ Z then either̺D(Z) = k − 1 or δD(Z) = l − 1: assume the former, the other ase being analogous. But̺D(Z ∩X) ≥ k−1 and ̺D(Z ∩Y ) ≥ k−1 together with k ≥ 2 implies that dD(X, Y ) > 0,a ontradi tion. If r ∈ Z (wlog. r ∈ Z ∩ X) then either ̺D(Z) = l − 1 or δD(Z) = k − 1:assume the former, and observe that ̺D(Z ∩X) ≥ l − 1 and ̺D(Z ∩ Y ) ≥ k − 1 > 0 againimply that dD(X, Y ) > 0, thus yield a ontradi tion.4.3 Stu k situation for spe ial skew-supermodular fun -tionsIn this se tion we want to hara terize the stu k situation if the symmetri , positivelyskew-supermodular fun tion p is of form qs with some spe ial fun tion q. Throughout inSe tion 4.3 we will assume (4.1) by Lemma 2.17 on ontra tion:

p-positive tight sets are singletons. (4.1)Re all that for a pair u, v ∈ V + the unique minimal set blo king them is denoted by Xuv.Observe that for four nodes x, y, u, v ∈ V +

Xxy(−) Xyu and Xyu(−) Xuv ⇒ |Xxy| = |Xyu| = |Xuv| = 2. (4.2)

60 Chapter 4. Covering skew-supermodular fun tions with graph edgesThis is true sin e p-positive tight sets are singletons and Xxy∩Xuv = ∅ as we noted in theend of Se tion 4.2.1. For the subsequent two subse tions let us introdu e some notations. Ifp is the symmetrized of a fun tion q then for any set X either p(X) = q(X) or p(X) = q(X)(possibly both). In the former ase we say that X is of q-type, in the latter we say thatX is of q-type (so X an be of both types). We introdu e two (undire ted, simple) graphson the set of positive nodes: the edge set of the q-graph (q-graph) onsists of the pairsu, v of positive nodes having q(Xuv) = 1 (q(Xuv) = 1, respe tively). Sin e there is noadmissible splitting, the union of these two graphs is the omplete graph (on the set ofpositive nodes), and an edge may belong to both graphs. We will all this 2-edge- oloured omplete graph the qq-graph.4.3.1 Crossing supermodular fun tionsIn this subse tion we hara terize the stu k situation if p is the symmetrized of a positively rossing supermodular fun tion q. Re all that a set fun tion q : 2V → Z ∪ {−∞} is alledpositively rossing supermodular if it satis�es (∩∪) whenever X and Y are rossing andq(X), q(Y ) are both positive. One an he k that the omplement of a positively rossingsupermodular fun tion is also positively rossing supermodular, and the symmetrized of apositively rossing supermodular fun tion is positively skew-supermodular.If two p-positive rossing sets X and Y are of the same type then they will satisfy (∩∪) .If furthermore p(X) = p(Y ) = 1 then their interse tion and union is also of the same typeas X and Y and p(X ∪ Y ) = p(X ∩ Y ) = 1 (here we use that p ≤ 1). On the other hand ifX and Y are of di�erent types then p(X − Y ) = p(Y − X) = 1. Also note that from anythree sets there are two of the same type.If p is symmetri and positively rossing supermodular, then it is easy to he k thatevery node is positive by (4.1) (one an �nd examples showing that this does not holdin general, if only the positively skew-supermodularity of p is assumed). However we willprove this statement for our more general ase, when p is the symmetrized of a positively rossing supermodular fun tion q. First it is useful to prove the following lemma.Lemma 4.13. If p is the symmetrized of a positively rossing supermodular fun tion qthen |Xuv| = 2 for any u, v ∈ V +.Proof. Assume that there are nodes x, z ∈ V + su h that |Xxz| > 2. By possibly omple-menting q we an assume that Xxz is of q-type. Let y ∈ V −Xxz be another positive node.We laim that Xxy must be of q-type, too. If not, then Xxz−Xxy = z, Xxy−Xxz = y, sin ethey are tight. But then Xyz annot be of q-type (sin e this would imply Xyz ∩ Xxz = zand Xxy − Xyz = x, a ontradi tion), neither of q-type (for a similar reason). So we have

Se tion 4.3. Stu k situation for spe ial skew-supermodular fun tions 61proved that for any y ∈ V + − {x, z} the set Xxy is of q-type. So the union of these setsY =

⋃

y∈V +−{x,z} Xxy is also of q-type, and has p(Y ) = q(Y ) = 1. However this impliesthat 1 = p(V − Y ) = m(z) = m(V − Y ), so it is tight, whi h ontradi ts |Xxz| > 2 (notethat Y ∩ Xxz = x).The lemma implies that the edge set of the q-graph (q-graph) onsists of the pairs u, vof positive nodes having q({u, v}) = 1 (q({u, v}) = 1, respe tively). Observe that a non-singleton onne ted omponent X 6= V of the q-graph is also of q-type and has q(X) = 1by Claim 1.9 (and similarly for the q-graph). This immediately implies the result promisedbefore.Lemma 4.14. If p is the symmetrized of a positively rossing supermodular fun tion qthen every node is positive.Proof. Suppose not, then the set of positive nodes V + 6= V must be onne ted in atleast one of the two graphs (sin e the union of two dis onne ted graphs annot be the omplete graph), so p(V +) = 1. But then p(V − V +) = 1 by the symmetry, ontradi tingm(V − V +) = 0.What is more, this implies the following surprising observation.Lemma 4.15. If p is the symmetrized of a supermodular fun tion q, then p(X) = 1 forany X with ∅ 6= X 6= V (i.e. q(X) = 1 or q(V − X) = 1 for every su h set).Proof. By the pre eding argument, any non-singleton X ( V must be onne ted in atleast one of the two graphs, so has p(X) = 1 (it is also easy to see for singletons, usingm(V ) ≥ 4).Consequently we have a rossing family F ontaining all sets with q value 1, and thefamily of the omplements of this family co(F) (these are the sets with q value 1), and theunion of these two families is 2V − {∅, V }. In the following theorem we will hara terizesu h families (for sake of brevity we will also add ∅ and V to the family: these sets analways be added to or removed from a rossing family).Let x ∈ V and let X1, . . . , Xt be t ≥ 1 pairwise disjoint subsets of V − x (possibly t = 1and X1 = ∅). We introdu e the following family:

Fx,X1,...,Xt= {X ⊆ V : x ∈ X or X ⊆ Xi for some i ∈ 1, . . . , t}.Theorem 4.16. Let F ⊆ 2V be a rossing family with ∅, V ∈ F that satis�es F ∪ co(F) =

2V . Then either V has exa tly four elements and F = 2V \ {{y, z}} for some y 6= z,y, z ∈ V or there exists a node x ∈ V and X1, . . . , Xt pairwise disjoint subsets of V −x forsome t ≥ 1 su h that either F or co(F) is equal to Fx,X1,...Xt

or Fx,X1,...Xt∪ {V − x}.

62 Chapter 4. Covering skew-supermodular fun tions with graph edgesProof. We an learly assume that V has at least 3 elements. We introdu e 2 (simpleundire ted) graphs on V : for sake of simpli ity we all them blue and red. The bluegraph is B = (V, {(u, v) : {u, v} ∈ F}), and the red is R = (V, {(u, v) : {u, v} ∈ co(F)})(so some edges might belong to both graphs). It might seem that these graphs don't haveevery information on F , but it turns out that they almost do. Again, we have that theunion of these two graphs is the omplete graph, and a non-singleton onne ted omponentX 6= V of B is in F (so V −X is in co(F)). This implies that if B[V −{u, v}] is onne tedfor nodes u 6= v, then (u, v) ∈ R, and vi e versa. If (u, v) ∈ B then we will say that thisedge is blue, if (u, v) /∈ R then we will say that this edge is pure blue.Claim 4.17. There is a node x ∈ V su h that either B or R ontains the edges (x, v) forevery v ∈ V − x.Proof. Assume indire tly that every node v ∈ V is entered by a pure red edge and by a pureblue edge, too. Consider an edge (u, v) that is pure blue: this means that B[V − {u, v}] isdis onne ted, so there is a bipartition X, Y of V − {u, v} su h that every edge is pure redbetween X and Y . Assume wlog. that the pure red neighbour x of v is in X and onsidertwo ases.CASE I. |X| ≥ 2. Sin e R[V −{v, x}] must be dis onne ted, every edge of the form (u, y)must be pure blue for any y ∈ V − {u, v, x}. So the pure red edge entered by u must bethe edge (u, x). Now onsider any x′ ∈ X − x: sin e B[V − {x, x′}] is onne ted, this edgeis red, but then x is not entered by a pure blue edge, a ontradi tion.CASE II. X = {x}. Then there is a bipartition Y1, Y2 of Y + u su h that every edgebetween Y1 and Y2 is pure blue. Assume that u ∈ Y1 and onsider any y ∈ Y1 − u: sin eR[V −{u, y}] is onne ted, this edge is blue. Then the only possibility for a pure red edgein ident to u is ne essarily the edge (u, x) whi h again means that there is no pure blueedge leaving x, �nishing the proof of the laim. �So onsider the vertex x given by this laim and assume wlog. that (x, v) is blue for anyv ∈ V − x. We distinguish again two ases.CASE I. There exist two properly interse ting sets Y, Z ∈ F su h that Y ∪ Z = V − x.We laim that this Y and Z an be hosen su h that their symmetri di�eren e is of ardinality two. Indeed, for any y ∈ Y − Z the set V − {x, y} also belongs to F sin eZ and Y − y = (Y − y + x) ∩ Y both belong to F , they are rossing and this is theirunion. So Z an be substituted by Z ′ = V − {x, y} and similarly Y an be substitutedby Y ′ = V − {x, z} for some z ∈ Z − Y . Now if |V | > 4 then this implies that F = 2V .To prove this onsider an arbitrary A ⊆ V − x: if A ⊆ Y ′ then learly A ∈ F , sin e Ais the interse tion of the rossing sets A + x ∈ F and Y ′ ∈ F . Similarly, A ⊆ Z ′ implies

Se tion 4.3. Stu k situation for spe ial skew-supermodular fun tions 63that A ∈ F . Therefore we assume that y, z ∈ A. If A 6= {y, z} then A ∈ F follows sin eA is the union of A − y and A − z that are rossing sets in F . The only remaining aseis A = {y, z}. Let v ∈ V − {x, y, z} arbitrary, then A is the interse tion of {v, y, z} ∈ Fand {x, y, z} ∈ F , that are rossing sets if |V | > 4, implying that A ∈ F , as laimed. If|V | = 4 then F an also be 2V \ {{y, z}}.CASE II. There does not exist two properly interse ting sets Y, Z ∈ F su h that Y ∪Z =

V −x. Let the maximal sets of F properly ontained in V −x be X1, X2, . . . , Xt: these arepairwise disjoint and sin e ∅ ∈ F , we have t ≥ 1. One an simply he k that F is eitherFx,X1,...Xt

or Fx,X1,...Xt∪ {V − x}, as laimed above. �A simple orollary that is worth mentioning is the following.Theorem 4.18. Let F ⊆ 2V be a ring family with ∅, V ∈ F that satis�es F ∪ co(F) = 2V .Then there exists a node x ∈ V and a (possibly empty) set X1 ⊆ V − x su h that either For co(F) is equal to Fx,X1

.4.3.2 Crossing negamodular fun tionsRe all that a set fun tion q : 2V → Z∪{−∞} is alled positively rossing negamodular if itsatis�es (−) whenever X and Y are q-positive rossing subsets. Note that the symmetrizedof a positively rossing negamodular fun tion is positively skew-supermodular, but the omplement of a positively rossing negamodular fun tion is not ne essarily positively rossing negamodular. An important spe ial ase is amonotone de reasing fun tion: bythat we mean a fun tion q that satis�es q(∅) ≤ 0 but q(X) ≥ q(Y ) for any ∅ ( X ⊆ Y ⊆ V .In this se tion we want to hara terize the stu k situation if p = qs with a positively rossing negamodular fun tion q. Re all that for a pair u, v ∈ V + the unique minimalset blo king them is denoted by Xuv. If Xxy, Xyu and Xuv are of the same type for fourdi�erent positive nodes x, y, u, v then they all must be of ardinality two by (4.2).As the NP - ompleteness of the general node-to-area onne tivity augmentation problemin graphs shows, we annot expe t a ni e hara terization of the stu k situation of theAlgorithm GREEDYCOVER in general, if p = qs with an arbitrary rossing negamodularfun tion q. Therefore we restri t ourselves to some spe ial positively rossing negamodularfun tions in this se tion: we assume thatq = R − dH0

with a positively rossing negamodular fun tion Rsatisfying R(X) 6= 1 for any X ⊆ V and an arbitrary hypergraph H0.(4.3)The following lemma hara terizes the stu k situation of the Algorithm GREEDYCOVERwith the input p = qs and m ∈ C(p)∩ZV . A hyperedge of H0 is alled an m-large hyperedgeif it ontains at least two nodes of V +.

64 Chapter 4. Covering skew-supermodular fun tions with graph edgesLemma 4.19. If p = qs for a fun tion q given with (4.3), m ∈ C(p) ∩ ZV , there is noadmissible splitting-o� and m(V ) ≥ 5 then there exists an m-large hyperedge. Furthermore,the number of positive nodes that are avoided by an m-large hyperedge is at most one.Proof. Assume that there is no m-large hyperedge. Apply (−) for Xab and Xbc and p (withan arbitrary a, b, c ∈ V +) to get that every a ∈ V + is ontained in a non-singleton hyperedgee. We laim that neither the q-graph nor the q-graph an ontain a path onsisting of 3edges. Assume indire tly that for some four nodes x, y, u, v ∈ V + the sets Xxy, Xyu, Xuv areall of the same type: then (4.2) gives that they all are of ardinality 2. But then Xxy andXyu annot satisfy (−) with equality by the nonsingleton hyperedge ontaining y, provingour laim. One an he k that the edge set of a omplete graph on at least 5 nodes annotbe de omposed into 2 sets su h that neither of them ontains a path of 3 edges, so theremust be an m-large hyperedge.Assume that there is an m-large hyperedge e that avoids x ∈ V +. Sin e e is m-large, thereexist u, v ∈ V + ∩ e. If the sets Xxu and Xxv satis�ed (∩∪) then they would have to satisfyit with equality, but their interse tion would be x by (4.1) and then they annot satisfy(∩∪) with equality by the presen e of the hyperedge e (see Lemma 1.6 (i)). Therefore Xxuand Xxv must be of the same type. If e avoids another positive node y then Xxu and Xyu annot be of the same type for similar reasons. This implies that e annot avoid a thirdpositive node, so it ontains at least 3 positive nodes, sin e m(V ) ≥ 5. Then the typeof Xuv and Xux must be di�erent, sin e they annot satisfy (−) with equality be ause ofthe edge e that is not ontained in Xuv. But then the type of Xuv and Xuy would be thesame, whi h annot hold for the same reason, so e annot avoid the se ond positive nodey. Furthermore, these observations on the qq-graph show that x an be the only positivenode that is avoided by an m-large hyperedge.We mention that if m(V ) = 4 then we don't ne essarily have m-large hyperedges: anexample an be found in [26℄. The following example shows that even if there are m-largehyperedges, they might ontain 2 positive nodes if m(V ) is only 4. Let H0 be the graph onnode set {x0, x1, x2, x3, y} and with edge set {x0y, x0y, x1x2, x2x3, x3x1} (that is, a triangleand two parallel edges). The areas are W = {{y, x1}, {y, x2}, {y, x3}} and they all haverequirement 3. The only minimal degree-spe i� ation is m = χV −y, there is no admissiblesplitting-o�, and the edges of the triangle are m-large (hyper)edges.As an appli ation of this lemma onsider the following generalization of the node-to-area onne tivity augmentation problem.Problem 4.20 (Rank-respe ting node-to-area onne tivity augmentation problem in hy-pergraphs). Given a hypergraph H0 = (V, E0) of rank at most γ, a olle tion of subsets W

Se tion 4.3. Stu k situation for spe ial skew-supermodular fun tions 65of V and a fun tion r : W → Z+ satisfying r ≥ 2, �nd a hypergraph H of minimum totalsize su h that λH0+H(x, W ) ≥ r(W ) for any W ∈ W and x ∈ V and the rank of H is atmost γ.If we de�ne RN2A with (2.4) and set q = RN2A − dH0then it is lear that H0 + Hsatis�es the area requirements if and only H overs q. Sin e RN2A does not take 1 asvalue, we an apply Lemma 4.19 and obtain that the Algorithm GREEDYCOVER fails onlyslightly for this problem. Sin e the maximizing ora le an be implemented for this spe ialfun tion RN2A by Lemma 2.8, the Algorithm GREEDYCOVER an be implemented to runin polynomial time for this ase.Theorem 4.21. Let an instan e of the minimum total size node-to-area onne tivity aug-mentation problem in hypergraphs be given by the hypergraph H0 = (V, E0) of rank at most

γ ≥ 2, W ⊆ 2V and r : W → Z+ with r ≥ 2. Then the Algorithm GREEDYCOVER gives asolution that ontains only graph edges and one hyperedge of size at most γ + 1, if γ > 2and γ + 2 if γ = 2.We mention that, though our proof does not rely on this, after ontra tion of a set Tthe fun tion RN2A/T an be de�ned with a node-to-area requirement fun tion as follows:if RN2A was de�ned with W and r then let W ′ = {W ∈ W : T ∩W = ∅} ∪ {W − T + vT :

T ∩ W 6= ∅} and let r′(W ) = r(W ) if vT /∈ W and r′(W ′) = r(W ) if W ′ = W − T + vT .One an he k that W ′ and r′ de�ne RN2A/T .Note that for any γ there are examples where the Algorithm GREEDYCOVER wouldoutput a hyperedge of size greater than γ. For γ = 2 an example an be found in [26℄, forbigger values onsider the following example. Let V ontain γ + 2 nodes x0, x1, . . . , xγ, yand the hypergraph H0 ontain two hyperedges {x0, y} and {x1, . . . , xγ}. The areas areof the form W = {{y, xi} : i = 1, 2, . . . , γ} and r(W ) = 2 for any W ∈ W. One an he k that the (only) minimal degree-spe i� ation is m = χV −y and there is no admissiblesplitting-o�. Also note that the SLB annot be a hieved in this example without in reasingthe rank.Figure 4.3.2 is an illustration with γ = 4. The (hyper)edges are drawn with a solid line,the areas are illustrated with dashed lines. The empty ball is the neutral node y.If we spe ialize our results for graphs (γ = 2) we obtain that a greedy algorithm (theobvious modi� ation of GREEDYCOVER) uses at most one more edge (i.e. at most twomore total size) than ne essary. Our results do not hara terize the ases when the SLB inthe node-to-area augmentation problem in graphs an be a hieved, this an be found in[26℄ and [24℄, but they imply that a greedy algorithm an only fail by at most one (edge)for this problem.

66 Chapter 4. Covering skew-supermodular fun tions with graph edges

Figure 4.1: An instan e of Problem 4.20 where the SLB annot be a hievedA more areful analysis of the stu k situation shows that a slight modi� ation of theAlgorithm GREEDYCOVER will solve the rank respe ting node-to-area onne tivity aug-mentation problem in hypergraphs (Problem 4.20) optimally for γ ≥ 3. In fa t, we willsolve the abstra t version Problem 4.25 for any γ ≥ 3. This an be found in Se tion 4.4.3.4.4 Further appli ations4.4.1 Lo al edge- onne tivity augmentation of hypergraphsIn this se tion we onsider the lo al edge- onne tivity augmentation of hypergraphswithout in reasing the rank. Let H0 = (V, E0) be a hypergraph of rank at most γ, andlet r : V ×V → Z+ \ {1} be a symmetri edge- onne tivity requirement that does not take1 as value. Let us de�ne the set fun tion Rloc with (2.5). Our aim is to �nd a hypergraphH of minimum total size su h that H0 + H overs Rloc, that is, λH0+H(u, v) ≥ r(u, v)for every pair of nodes u, v. Sin e Rloc is a skew supermodular fun tion, the AlgorithmGREEDYCOVER gives a solution that ontains graph edges and at most one hyperedge.The question we want to answer is whether the size of this hyperedge is at most γ. One ase when this is obviously not true is when γ = 2 and the SLB is odd: then the size of thehyperedge will be 3. The following theorem shows that this is the only ex eptional ase.Note that this theorem generalizes the theorem of Frank [19℄ on lo al edge- onne tivityaugmentation of graphs. After having proved this theorem we have been informed thatBen Cosh had also proved it in his PhD thesis [15℄. We have noti ed that ombining ourideas with those in [15℄ the following simple proof an be given.

Se tion 4.4. Further appli ations 67Theorem 4.22. Let an instan e of the minimum total size lo al edge- onne tivity aug-mentation problem be given by a hypergraph H0 = (V, E0) of rank at most γ ≥ 2, and thesymmetri edge- onne tivity requirement r : V × V → Z+ \ {1}. Then the hyperedge in theoutput of the Algorithm GREEDYCOVER is of size at most γ, if γ > 2, and it is of size atmost 3, if γ = 2.Proof. We will prove that the hyperedge in the output of the Algorithm GREEDYCOVERis of size at most γ for any minimal input m ∈ C(p) ∩ ZV : observe that this ontainsmore general augmentation problems, e.g. the minimum node- ost version, too. We anassume that the Algorithm GREEDYCOVER is stu k already at the beginning and assumeindire tly that m(V ) > max{3, γ}. This proof is easier told if we think of the lassi aldes ription of splitting-o�: instead of the degree spe i� ation m, add a new node s tothe hypergraph and introdu e m(v) parallel edges between s and any v ∈ V (of ourse,from the results above m(v) ≤ 1 for any v, so there are no parallel edges in ident to s:the positive nodes be ome the neighbours of s) and denote the hypergraph obtained thisway with H ′ (i.e. H ′ has node set V + s and it ontains the hyperedges of H0 and onlygraph edges in ident to s). By the assumptions, H ′ is r-edge- onne ted in V . We anfurther assume that r(u, v) = λH′(u, v) for any u, v ∈ V , sin e if we in rease r(u, v) thenwe do not reate new admissible splittings (note, that possibly new sets be ome tight andr(u, v) > 1 if u and v are neighbours of s). So we assume that Rloc is de�ned by (2.5)with λH′ substituted in the pla e of r and p = Rloc − dH0

as before. We further assumethat tight sets are singletons, whi h implies that λH′(u, v) = min(dH′(u), dH′(v)) for anyu, v ∈ V . Let t be a neighbour of s su h that dH′(t) = min{dH′(v) : v is a neighbour ofs}. Let u and v be neighbours of s (distin t from ea h other and from t). The following laim was already used in [20℄ (Claim 4.1): for ompleteness, we in lude a proof.Claim 4.23. Xtu and Xtv satisfy (−) (with fun tion Rloc and thus with p, too).Proof. Assume that they do not satisfy (−) with Rloc: then they must satisfy (∩∪) . Thustheir interse tion is tight, so it is t. On the other hand, sin e dH′(t) ≤ dH′(u), Rloc(Xtu) ≤

Rloc(Xtu−t), and similarly Rloc(Xtv) ≤ Rloc(Xtv−t), implying that Rloc(Xtu)+Rloc(Xtv) ≤

Rloc(Xtu − t) + Rloc(Xtv − t) = Rloc(Xtu −Xtv) + Rloc(Xtv −Xtu), so they satisfy (−) afterall, a ontradi tion. �So there exists a set X ⊆ V ontaining t, su h that Xtu = X + u for any neighbour u ofs (using that tight sets are singletons). Sin e p(X) ≤ 1 (by Lemma 4.6) and Rloc(X) ≥ 2(using that t ∈ X), there must be a hyperedge e in H0 entering X. We laim thatthis hyperedge must ontain every neighbour of s (ex ept possibly t), ontradi ting thehypothesis that the rank of H0 is at most γ. Assume that it ex ludes a neighbour u of

68 Chapter 4. Covering skew-supermodular fun tions with graph edgess and �x an arbitrary other neighbour v of s (distin t from u and t). Sin e Xtu and Xtvmust satisfy (−) with equality for p, this implies that e ⊆ Xtv. But then Xtu and Xtw willnot satisfy (−) with equality for a fourth neighbour w of s. �We mention that the minimality of m is ru ial in the proof above: if m is not minimalthen a simple example shows that the greedy algorithm an fail and produ e a hyperedgeof size γ + 1. As an example, onsider the following hypergraph H0 = (V, {V − x}), whereV is an arbitrary �nite set and x ∈ V is a �xed node (i.e. H0 has only one hyperedgeV − x). Let r(u, v) = 2 between any u, v ∈ V − x and r(x, v) = 0 between x and anarbitrary v ∈ V − x. If we set m(a) = 1 for every a ∈ V , whi h is an admissible but notminimal degree-spe i� ation, then there does not exist an admissible splitting-o�, as one an he k. Thus the Algorithm GREEDYCOVER outputs the hyperedge V that has rankone greater than that of H0.4.4.2 Global ar - onne tivity augmentation of mixed hypergraphsIn this se tion we give an appli ation of the results of Se tion 4.3.1 about global edge- onne tivity augmentation of mixed hypergraphs. If M = (V,A) is a mixed hypergraphand X ⊆ V then ontra ting X yields the mixed hypergraph M/X = (V/X,A/X) thefollowing way: for every a = (Ta, Ha) ∈ A let T ′

a = Ta if Ta∩X = ∅ and let T ′a = Ta−X+vXotherwise, similarly let H ′

a = Ha if Ha ∩X = ∅ and let H ′a = Ha−X +vX otherwise. Then

A/X = {a′ = (T ′a, H

′a) : a ∈ A}. Observe that ̺M/X = ̺M/X . If the root node r is in Xthen the ontra ted node vX will be ome the new root node. This shows that ontra tinga set de�nes a ontra ted problem the natural way.Let M = (V,A) be a mixed hypergraph and let k, l be nonnegative integers su h that

k 6= 1 and l 6= 1. We assume that M is of rank at most γ. We want to make M (k, l)-ar - onne ted by adding an undire ted, degree spe i�ed hypergraph that also has rank atmost γ. Is it true that the Algorithm GREEDYCOVER will output su h a hypergraph?The answer is �almost yes�: sometimes this an only be done by adding a hyperedge of ardinality γ + 1 (even for k = l = 2). As an example, onsider the following mixedhypergraph M = (V,A): let |V | ≥ 3 and x, y ∈ V be two nodes. There are 3 hyperar s inA: one is a digraph ar (x, y), the se ond is (y, V − x− y) and the third is (V − x− y, x).Finally let k = l = 2 and γ = |V | − 1. It is easy to see that the SLB is |V | and the onlyway to a hieve it is to add the hyperedge V .Figure 4.4.2 is an illustration: the digraph ar and one of the other two hyperar sare drawn with solid lines, the third hyperar is drawn with dashed lines. We use the onvention that the tails of a hyperar are denoted by an �o� and heads by an �x� (ex ept

Se tion 4.4. Further appli ations 69for the digraph ar , whi h is denoted by an arrow).

x yFigure 4.2: A mixed hypergraph that annot be made 2-ar - onne ted with a hypergraphmeeting the SLB without in reasing the rankHowever, we an prove the following result.Theorem 4.24. If M is of rank at most γ ≥ 2 and k, l ∈ N − {1} are integers, then we an make M (k, l)-ar - onne ted greedily by the addition of graph edges and one hyperedgeof size at most γ + 1.Proof. Let q = qM,r,k,l as de�ned in (2.2) and p = qs and let m ∈ C(p)∩ZV . We an assumethat the Algorithm GREEDYCOVER is stu k already at start. We have to prove that m(V )is at most γ + 1. We an also assume that tight sets are singletons (and delete singletonhyperedges, sin e they are irrelevant for onne tivity), so by the observations in Se tion4.3.1, every node is positive. By Theorem 4.16, there is an x ∈ V su h that (by possiblyreversing every hyperar of M and swit hing the role of k and l) every set X 6= V withx ∈ X has q(X) = 1 (observe that this onsequen e is also true for the sporadi exampleon 4 nodes). First we laim that V − x annot ontain hyperar s. Assume that it does ontain a hyperar a, let v be an arbitrary head node of a, and let X = Ta ∪ Ha − v + xand Y = {v, x}. These sets are rossing (sin e |a| < |V − x| by the assumption) and ofq-type, but (∩∪) annot hold with equality for them by the presen e of the hyperar a,a ontradi tion. So every hyperar of M ontains x. We laim that if v 6= x is a tail ofa hyperar a = (Ta, Ha) satisfying |a| ≥ 3, then x ∈ Ha and Ta − v − x = ∅. To see this onsider the rossing sets X = a − v and Y = {v, x}. Then q(X) = q(Y ) = 1 but one

70 Chapter 4. Covering skew-supermodular fun tions with graph edges an he k that, by the presen e of the hyperar a, (∩∪) annot hold with equality for Xand Y , a ontradi tion. So the hyperar s leaving any v ∈ V − x all enter x and su h ahyperar annot leave two su h nodes. This implies that ̺(x) =∑

v∈V −x δ(v). If x = rthen l − 1 = ̺(x) =∑

v∈V −x δ(v) = |V − x|(l − 1) ontradi ting that |V | > 2 and l > 1.On the other hand, if x 6= r then k − 1 = ̺(x) =∑

v∈V −x δ(v) = (|V | − 2)(l− 1) + (k − 1),again ontradi ting that |V | > 2 and l > 1.4.4.3 Rank-respe ting augmentation of hypergraphs with negamod-ular onstraintsIn this se tion we will give appli ations of the results of Se tion 4.3.2: we will solve therank-respe ting node-to-area onne tivity augmentation problem (Problem 4.20) for anyγ ≥ 3. Furthermore, we will solve a generalization of that problem, alled the rank-respe ting augmentation of hypergraphs with negamodular onstraints whi h is de�ned asfollows.Problem 4.25 (Rank-respe ting augmentation of a hypergraph with negamodular on-straints). Assume that R : 2V → Z∪{−∞} is positively rossing negamodular, R does nottake 1 as value, and let q = R − dH0

with a hypergraph H0 = (V, E0) of rank at most γ.Find a hypergraph H of rank at most γ overing q = R − dH0and having smallest possibletotal size.In this se tion we will solve this problem for any γ ≥ 3 (we don't solve it for γ = 2). Tothis end we will slightly modify the Algorithm GREEDYCOVER and we use the observationsof Se tion 4.3.2, where we have shown that this algorithm only fails slightly for this problem.The result of this se tion appeared in [7℄.Assume that the Algorithm GREEDYCOVER (with input p = (R − dH0

)s and a minimalm ∈ C(p) ∩ ZV ) did not output a feasible hypergraph for Problem 4.25. Let the outputof the algorithm be G + e (where G is a graph, e is a hyperedge of size γ + 1). Our ideais the following: if the graph G does not ontain edges at all, then it is easy to see thatthe SLB annot be a hieved (see Lemma 4.30), but one more is already enough (and it issimple to see how to rea h it: any onne ted hypergraph on V + will be a good solution).Otherwise, if G ontains an edge ab, then an appropriate node c ∈ e an be deletedfrom e and added to ab (thus reating a hyperedge {a, b, c} of size 3) and the hypergraphH ′ = (V, E(G) − {ab} + {a, b, c} + e′) of total size meeting the Subpartition Lower Boundis a feasible solution, where e′ = e − c. In what follows we show that almost any hoi e ofc will be good. In the rest of Se tion 4.4.3 we assume that we are at the stu k situation ofthe Algorithm GREEDYCOVER (so the notations p, m, V + are meant for this ase). We will

Se tion 4.4. Further appli ations 71denote the output of the algorithm by G + e (where G is a graph and e is the hyperedgewith χe = m) and V + will just be another synonym for e. The following lemma tells usthe ondition that we have to satisfy when hoosing node c.Lemma 4.26. Let p = p0 − dG with a symmetri , positively skew-supermodular fun tionp0 and a graph G, and let m ∈ C(p). Assume that there does not exist an admissiblesplitting-o�. Let ab ∈ E(G) and c ∈ V + be arbitrary. Let the hypergraph G′ be obtainedfrom G by deleting the edge ab and adding the 3-hyperedge {a, b, c}, let p′ = p0 − dG′ andlet m′ = m − χ{c}. Then m′ ∈ C(p′) if and only if there is no set X ⊆ V satisfyingp(X) = m(X) = 1, c ∈ X and {a, b} ∩ X 6= ∅.Proof. If m′ /∈ C(p′) then there must be a set X ⊆ V su h that m′(X) < p′(X). Sin ep′ ≤ p ≤ 1, we get that p′(X) = p(X) = 1. This together with m ∈ C(p) gives thatm(X) ≥ p(X) = p′(X) = 1 > m′(X), implying that c ∈ X and {a, b} ∩ X 6= ∅, as laimed.The following orollary of Lemma 4.19 an be read out from its proof.Corollary 4.27. If there is no admissible splitting-o�, tight sets are singletons and m(V ) >

γ ≥ 4 then there exists an m-large hyperedge f and a node x ∈ V + su h that f = V + − x.In this ase either the q-graph or the q-graph is the omplete graph on V +−x and the othergraph is the omplement (i.e. a star entered at x).As we have already mentioned, if m(V ) = 4 then we don't ne essarily have m-largehyperedges: an example an be found in [26℄. The example in page 64 showed that evenif there are m-large hyperedges, they might ontain 2 positive nodes if m(V ) is only 4.However, the se ond statement of Corollary 4.27 still holds.Lemma 4.28. If there is no admissible splitting-o�, tight sets are singletons and m(V ) = 4and γ ≤ 3 then there exists a spe ial node x ∈ V + su h that either the q-graph or the q-graph is the omplete graph on V + − x and the other graph is the omplement (i.e. a star entered at x).Proof. We have to prove that neither the q-graph, nor the q-graph ontains a path of 3edges. Assume that for the four nodes v1, v2, v3, v4 ∈ V + the sets Xv1v2, Xv2v3

and Xv3v4are all of the same type. By (4.2), |Xv1v2| = |Xv2v3

| = |Xv3v4| = 2. Sin e p(Xv2v3

) = 1,a hyperedge h of H0 leaves the set Xv2v3. Assume wlog. that h ontains v2: sin e Xv2v3and Xv1v2

satis�es (−) for p with equality, by Lemma 1.6.ii the hyperedge h must ontainv1 and h ⊆ {v1, v2, v3}, too (note that |h| ≤ 3). Sin e Xv2v3

and Xv3v4satis�es (−) withequality, by Lemma ii h annot ontain v3 (in fa t we have proved that h = v1v2). Be ause

72 Chapter 4. Covering skew-supermodular fun tions with graph edgesof the edge h, the type of Xv2v3and Xv1v3

must be the same. Sin e p(v3) = 1, there mustbe an edge g of H0 leaving v3: this edge annot leave {v2, v3, v4}, sin e (−) must hold withequality for Xv2v3and Xv3v4

. In any ase, either Xv2v3and Xv1v3

, or Xv1v3and Xv3v4

willnot satisfy (−) with equality.In Corollary 4.27 and Lemma 4.28 above we have shown that there exists a uniquepositive node x su h that either the q-graph or the q-graph is the omplete graph on V +−xand the other graph is the omplement (i.e. a star). This translated to the situation before ontra tion means, that for any u ∈ V + there exists a tight set X(u) that was ontra ted(so these sets are disjoint).In the γ ≥ 4 ase X(u)∪X(v) is dangerous for any u, v ∈ V + − x. Furthermore thereexists at least one m-large hyperedge f ∈ H0 that has size γ and satis�es that |f∩X(u)| = 1for any u ∈ V + −x. Let Y = V −∪u∈V +−xX(u), so X(x) ( Y (it must be a proper subset,sin e p(Y ) 6= 1, sin e no hyperedge leaves Y ). In this ase it is not hard to see that everyset Y ∪X(u) (u ∈ V +−x) is of q-type: this will be proved in Lemma 4.30. It is lear, thatthe graph G does not have edges between the partition lasses {X(u) : u ∈ V +−x}∪{Y }.On the other hand, in the γ ≤ 3 ase there exists a set Z ⊆ V −∪u∈V +X(u) su h thatX(u)∪X(v)∪Z is dangerous for any u, v ∈ V + − x and these sets all have the same type(Z is empty in the γ > 3 ase). Let Y = V −Z −∪u∈V +−xX(u): again X(x) ( Y (it mustbe a proper subset, sin e p(Y ) 6= 1, sin e no hyperedge leaves Y ). In this ase the edges ofG are either indu ed in a member of the partition {X(u) : u ∈ V + − x} ∪ {Y, Z}, or they an even go between two lasses (but only between X(u) and X(v) or between X(u) andZ, where u, v ∈ V + − x). In both ases we an prove the following lemma.Lemma 4.29. The sets X(u) (u ∈ V + − x) are maximal tight sets.Proof. Assume that there is u ∈ V +−x and a tight set X ) X(u). Let v, w ∈ V + −{u, x}be arbitrary and observe that type of X(v)∪X(w)∪ Z and that of Y ∪X(v) ∪X(w)∪ Zis di�erent: this follows from (1.7) applied to X(v) ∪ X(w) ∪ Z and Y ∪ X(v). Thisimplies that X ∩X(v) = X ∩Z = ∅, sin e (∩∪) for X and either of X(v)∪X(w)∪Z andY ∪X(v)∪X(w)∪Z (and p) would give a ontradi tion. We only need to prove that X∩Yis empty. Assume that it is not and distinguish the following two ases. Clearly, the typeof X and Y ∪X(z) has to be the same for any z ∈ V +−x, otherwise p(X∩(X(z)∪Y )) = 1would follow from (∩∪) for X and X(z) ∪ Y , and it would give a ontradi tion.CASE I: Every set Y ∪X(z) (z ∈ V +−x) is of q-type. By (1.7) applied to X(v)∪X(w)∪Zand Y ∪X(v), the set Y ∪X(v)∪X(w)∪Z would be of q-type, implying that q(X∩Y ) = 1,a ontradi tion.

Se tion 4.4. Further appli ations 73CASE II: Every set Y ∪ X(z) (z ∈ V + − x) is of q-type. Apply Claim 1.10 for X0 = Xand the sets X(u) ∪ X(z) ∪ Z (z ∈ V + − x) to get that p(X ∪ Z ∪⋃

z∈V +−x X(z)) = 1.Let Y ′ = V − (X ∪ Z ∪⋃

z∈V +−x X(z)) = Y − X: it has p-value 1, so there must be ahyperedge h leaving it. This hyperedge annot leave Y , so it enters Y ∩ X. But in this ase the sets X ∪X(v)∪Z and X ∪X(w)∪Z ould not satisfy (−) with equality, thoughboth of them have q-value 1. This ontradi tion �nishes the proof of this lemma.This Lemma shows that our idea an be implemented the following way: if G ontainsan edge ab, and c ∈ V + − x is su h that X(c) ∩ {a, b} = ∅, then the hypergraph H ′ =

(V, E(G) − {ab} + {a, b, c} + e′) of total size meeting the Subpartition Lower Bound is afeasible solution, where e′ = V + − c. Sin e an arbitrary edge ab of G an interse t at mosttwo of the sets {X(v) : v ∈ V + − x}, su h a c an always be found.Next we show that if G does not ontain edges at all then the SLB annot be a hieved.Lemma 4.30. For any nonempty U ( V + − x the set Y ∪⋃

u∈U X(u) has p value 1.Proof. The lemma is obvious if m(V ) = 4 by (1.7). Assume that m(V ) ≥ 5: in this asewe show that every set Y ∪ X(u) (u ∈ V + − x) is of q-type. Assume not, then every su hset is of q-type by Corollary 4.27, onsequently X(v)∪X(w) is of q-type for v, w ∈ V +−x.By Claim 1.10 X(v)∪X(w) is also of q-type be ause its omplement an be built up as theunion of a q-type set and some q-type sets. But this also implies that ∪u∈V +−xX(u) hasp-value 1, and so has its omplement Y , but this annot be the ase, sin e no hyperedgeleaves Y . Thus we have shown that every set Y ∪X(u) (u ∈ V + − x) is of q-type and thisimplies the lemma by Claim 1.10.This orollary implies that if the Algorithm GREEDYCOVER applied for Problem 4.25outputs G+e where G has no edges at all, and e is too big, then the SLB annot be a hieved.Let us give the pseudo ode of the modi�ed version of the Algorithm GREEDYCOVER thatwe have suggested.Algorithm NEGAMODULAR_COVERbegin INPUT A rossing negamodular fun tion R : 2V → Z (given with an ora le) thatsatis�es R(X) 6= 1 for any X ⊆ V , and a hypergraph H0 = (V, E0) of rank at most γ(where γ ≥ 3)OUTPUT A hypergraph H = (V, E) overing R − dH0

having smallest total size andrank ≤ γ1.1. Let q = R − dH0and p = qs and �nd a minimal m ∈ C(p) ∩ ZV1.2. Initialize H = (V, ∅)

74 Chapter 4. Covering skew-supermodular fun tions with graph edges1.3. While there exists an admissible pair u, v do1.4. Let m = m − χ(u) − χ(v) and p = p − d(V,{uv}) and H = H + uv1.5. EndWhile1.6. If m(V ) ≤ γ then let H = H + e where χe = m1.7. Else (i.e. m(V ) = γ + 1)1.8. If E(H) = ∅ then let E(H) = {ab, V + − b} with arbitrary a, b ∈ V +1.9. Else1.10. Let ab ∈ E(H) be arbitrary and c ∈ V + − x su h that X(c)∩ {a, b} = ∅ (wherex ∈ V + is the spe ial node given by Corollary 4.27 and Lemma 4.28)1.11. Let E(H) = E(H) − ab + {{a, b, c}, {V + − c}}1.12. EndIf1.13. EndIf1.14. Output H and STOPend

Chapter 5Covering symmetri rossingsupermodular fun tions with graphedgesA symmetri (positively) rossing supermodular fun tion is a spe ial ase of a (positively)skew-supermodular fun tion. Re all that set fun tion p : 2V → Z∪{−∞} is alled positively rossing supermodular if it satis�es the following inequality for every rossing pair X, Y ⊆ Vwith p(X), p(Y ) > 0:p(X) + p(Y ) ≤ p(X ∩ Y ) + p(X ∪ Y ). (∩∪)Observe that (∩∪) trivially holds if X ⊆ Y or Y ⊆ X. If furthermore p is symmetri (i.e.

p(X) = p(V −X) for any X ⊆ V ) then it will also satisfy the following inequality for every rossing pair X, Y ⊆ V with p(X), p(Y ) > 0:p(X) + p(Y ) ≤ p(X − Y ) + p(Y − X). (−)Again, (−) will always hold if X ∩ Y = ∅ or X ∪ Y = V (the latter by symmetry). Weemphasize that we do not assume the nonnegativity of our fun tion p.The �rst half of this hapter is about overing a symmetri (positively) rossingsupermodular fun tion with graph edges. This problem is related to global edge- onne tivity augmentation. Let us give the spe ial ases that motivate this problem. Theglobal edge- onne tivity augmentation problem of graphs was de�ned in Se tion2.1.1. It was solved by Watanabe and Nakamura in [45℄. A bit more di� ult problem,the global edge- onne tivity augmentation problem of hypergraphs with graphedges was solved by Bang-Jensen and Ja kson. The abstra t problem of overing a sym-metri (positively) rossing supermodular fun tion with graph edges was solved by Ben zúr75

76 Chapter 5. Covering symmetri rossing supermodular fun tionsand Frank in [5℄. In Se tion 5.3 we give a relatively simple proof of their result. The proof ontains some further simpli� ations due to Zoltán Szigeti.The se ond part of the hapter is devoted to overing a symmetri (positively) rossing supermodular fun tion with partition onstraints. A spe ial ase of thisproblem, the partition onstrained global edge- onne tivity augmentation of agraph was solved by Bang-Jensen, Gabow, Jordán and Szigeti in [3℄. This is motivated bythe edge- onne tivity augmentation of a bipartite graph while maintaining bipartiteness.The more general global edge- onne tivity augmentation of a hypergraph with abipartite graph was solved by Ben Cosh in [15℄. In Se tion 5.4 we solve the abstra tproblem of overing a symmetri , positively rossing supermodular fun tion with partition onstraints and we spe ialize our results for the problem of global edge- onne tivityaugmentation of a hypergraph with a multipartite graph.The stru ture of the hapter is the following. In Se tion 5.1 we brie�y review the previ-ous results on overing symmetri , positively rossing supermodular fun tions with graphedges. In Se tion 5.3 we show an algorithmi proof of Theorem 5.5 of Ben zúr and Frank:this proof appeared in [8℄. In Se tion 5.4 we solve the problem of overing symmetri ,positively rossing supermodular fun tions with partition onstraints: the results of thisse tion appeared in [10℄ and [9℄.5.1 Previous results - brief historyGlobal edge- onne tivity augmentation of graphs was solved by Watanabe and Nakamura[45℄. They proved the following hara terization for the minimum version of the problem:Theorem 5.1 (Watanabe, Nakamura [45℄). Let G0 = (V, E0) be a graph, and k ≥ 2 aninteger. G0 an be made k-edge- onne ted by adding at most γ new edges if and only if∑

Z∈F

(k − dG0(Z)) ≤ 2γ for every subpartition F of V .Note that the theorem does not hold for k = 1, the answer is di�erent in this ase, itdepends on the number of omponents of G0 (although this ase is very simple, in gen-eralizations of the above theorem this ase will ause most of the di� ulties). Watanabeand Nakamura showed that a minimum ardinality augmentation an be obtained in poly-nomial time by repeatedly in reasing the edge- onne tivity of the graph by one using theminimum number of edges. However, this algorithm is not strongly polynomial.Frank [19℄ gave the �rst strongly polynomial algorithm for this problem. The algorithmrelies on the following result on erning the degree spe i�ed augmentation:

Se tion 5.1. Previous results - brief history 77Theorem 5.2. Let G0 = (V, E0) be a graph, k ≥ 2 an integer, and m : V → Z+ a degreespe i� ation su h that m(V ) is even. There is a graph G su h that d+G(v) = m(v) for every

v ∈ V and G0 + G is k-edge- onne ted if and only ifm(X) ≥ k − dG0

(X) for every ∅ 6= X ⊂ V . (5.1)In [14℄, Cheng gave a formula on the minimum number of graph edges that an be addedto an initial (k − 1)-edge- onne ted hypergraph su h that the resulting hypergraph is k-edge- onne ted. Bang-Jensen and Ja kson [4℄ extended this result to the ase when theinitial hypergraph an be arbitrary. Let c(H) denote the number of onne ted omponentsof the hypergraph H . The min-max theorem is the following:Theorem 5.3 (Bang-Jensen, Ja kson [4℄). Let H0 = (V, E0) be a hypergraph, and k apositive integer. There is a graph G with γ edges su h that H0 + G is k-edge- onne ted ifand only if the following hold:2γ ≥

∑

Z∈F

(k − dH0(Z)) for every sub-partition F of V, (5.2)

γ ≥ c(H0 − E ′0) − 1 for every E ′

0 ⊆ E0 for whi h |E ′0| = k − 1. (5.3)Bang-Jensen and Ja kson used a sophisti ated splitting-o� te hnique to prove this result,their method gives rise to a polynomial-time algorithm. Note that the k = 1 ase is notex luded in Theorem 5.3.The general problem of overing a symmetri (positively) rossing supermodular fun tionwith a minimum number of graph edges was solved by Ben zúr and Frank. The solutionof the degree spe i�ed version is the following. One natural ne essary ondition of theexisten e of a graph G satisfying the degree spe i� ation m and overing the symmetri positively rossing supermodular fun tion p is that m ∈ C(p) (of ourse we assume that

m(V ) is even). However this is not su� ient, as the problem of making a graph onne tedshows. Let us introdu e another ne essary ondition. A partition X = {X1, X2, . . . , Xt}of V is alled p-full if p(∪i∈IXi) > 0 for any nonempty I ( {1, 2, . . . , t}. The maximum ardinality of a p-full partition is the dimension of p and is denoted by dim(p). It is easyto see that any graph overing p must have at least dim(p) − 1 edges: any graph overingp has to onne t the members of any p-full partition.Theorem 5.4 (Ben zúr and Frank [5℄). Let p0 : 2V → Z∪{−∞} be a symmetri , positively rossing supermodular set fun tion and m0 ∈ C(p0) ∩ ZV with m0(V ) even. There existsa graph G overing p0 with d+

G(v) = m0(v) for any v ∈ V if and only if m0(V )/2 ≥

dim(p0) − 1.

78 Chapter 5. Covering symmetri rossing supermodular fun tionsBy the methods seen so far Theorem 5.4 implies the following theorem on the minimumnumber of graph edges overing a symmetri , positively rossing supermodular fun tion.Theorem 5.5 (Ben zúr and Frank [5℄). Let p : 2V → Z∪{−∞} be a symmetri , positively rossing supermodular set fun tion. The minimum number of graph edges overing p isequal to the maximum of the following two quantities:max{⌈

1

2

∑

X∈X

p(X)⌉ : X is a subpartition of V }, (5.4)dim(p) − 1. (5.5)5.1.1 Partition onstrained global edge- onne tivity augmentationof a graphBang-Jensen et al. [3℄ were motivated by the following problem: given a bipartite graphand a positive integer k, augment this graph with a minimum number of new edges to makeit k-edge- onne ted, while maintaining the bipartiteness. As a onvenient generalizationthey introdu ed the following problem (note that the previous problem is a spe ial ase ofProblem 5.6 only if the starting bipartite graph is onne ted, otherwise the two- olouringis not unique).Problem 5.6 (Partition onstrained global edge- onne tivity augmentation of a graph).We are given a graph G0 = (V, E0), a partition P = {P1, P2, . . . , Pr} of the set V and apositive integer k. The problem is to �nd a graph G su h that G0 + G is k-edge- onne ted,and G ontains only edges between the lasses of P. In the minimum version thenumber of edges of G is to be minimized. In the degree-spe i�ed version G has to satisfya given degree-spe i� ation m ∈ ZV

+.This problem was solved by Bang-Jensen, Gabow, Jordán and Szigeti in [3℄ for anyk ≥ 2 (the problem is easy for k = 1). In the solution of the minimum version the valueΦ = max{αG, β1

G, ..., βrG} is a straightforward lower bounds on the minimum number ofedges, where

αG := max{⌈12

∑

X∈F k − dG(X)⌉ : F is a subpartition of V },βi

G := max{∑

X∈F k − dG(X) : F is a subpartition of Pi} for every i = 1, 2, . . . , r.The �rst value αG is related to the Subpartition Lower Bound. The value βiG is a lowerbound sin e we annot add an edge indu ed by Pi. However there are examples where thislower bound fails (by one). One example is when we want to make a C4 3-edge- onne tedwhile maintaining its bipartiteness: the above bound is only 2, but we learly need at least

Se tion 5.1. Previous results - brief history 793 edges. Another example is when we want to make a C6 3-edge- onne ted, but we arenot allowed to onne t opposite nodes, then the above lower bound 3 annot be a hieved.These examples an be generalized with the following de�nitions. Re all that S(X) denotesthe set of all subpartitions for a set X.De�nition 5.7. A partition A = {A1, A2, A3, A4} of V is alled a C4- on�guration (forG,P, k) if(i) k ≥ 2 is odd,(ii) k > dG(Ai) for every i = 1, 2, 3, 4,(iii) dG(Ai, Ai+2) = 0 for every i = 1, 2,(iv) k − dG(Ai) + k − dG(Ai+2) = Φ for both i = 1, 2,(v) there is a olour lass P ∈ P and an index i ∈ {1, 2} su h that for j = i and j = i+2there exist subpartitions Xj ∈ S(Aj) satisfying ∑

X∈Xjk − dG(X) = k − dG(Aj) and

Xi ∪ Xi+2 ∈ S(P ).De�nition 5.8. A partition A = {A1, A2, A3, A4, A5, A6} of V is alled a C6- on�gura-tion (for G,P, k) if(i) k ≥ 2 is odd,(ii) dG(Ai) = k − 1 for every i = 1, 2, 3, 4, 5, 6,(iii) dG(Ai, Ai+1) = (k − 1)/2 for every i = 1, 2, 3, 4, 5, 6,(iv) there exist three distin t olour lasses P1, P2, P3 ∈ P and subsets A′i ⊆ Ai satisfying

dG(′Ai) = k − 1 for every i = 1, 2, 3, 4, 5, 6 su h that A′j ∪ A′

j+3 ⊆ Pj for everyj = 1, 2, 3.With these de�nitions the solution of the minimum version of Problem 5.6 is the follow-ing.Theorem 5.9 (Bang-Jensen, Gabow, Jordán and Szigeti [3℄). Let us be given a graph

G = (V, E), a partition P = {P1, P2, . . . , Pr} of the set V , and an integer k ≥ 2. Theminimum number of graph edges between the lasses of P and making G k-edge- onne tedis Φ, unless a C4 or a C6 on�guration exists, in whi h ase it is Φ + 1. Su h an edge set an be found in polynomial time.

80 Chapter 5. Covering symmetri rossing supermodular fun tionsThe solution of the degree-spe i�ed problem is the following. We an formulate thefollowing natural ne essary onditions of the existen e of a solution with degree fun tionm if k ≥ 2:

m(X) ≥ k − dG0(X) holds for any nonempty X ( V, (5.6)

m(V ) is even, (5.7)m(Pi) ≤ m(V − Pi) for every i = 1, 2, . . . , r. (5.8)However, similarly to the minimum version some ex eptional ases show that these onditions are not su� ient. These ex eptional ases are alled obsta les: they are relatedto the on�gurations de�ned above, though there is an important di�eren e. This is thefollowing: it is possible that the optimum of the minimum version is Φ, but there is adegree-spe i� ation m satisfying (5.6)-(5.8) and yielding m(V ) = 2Φ, for whi h the degree-spe i�ed version is not solvable (be ause of an obsta le). This is illustrated in Figure 5.1.

1 1

11

0 0

Figure 5.1: A C4-obsta le. The minimum number of edges between bla k and white nodesmaking this graph 3-edge- onne ted is two, but the degree-spe i� ation given here doesnot yield a solution with 2 edges.Let us give the de�nition of the obsta les. In the following 2 de�nitions we assume thatm satis�es (5.6)-(5.8).De�nition 5.10. A partition A = {A1, A2, A3, A4} of V is alled a C4-obsta le (forG,P, k, m) if(i) k ≥ 2 is odd,(ii) m(Ai) = k − dG(Ai) for every i = 1, 2, 3, 4,(iii) dG(Ai, Ai+2) = 0 for every i = 1, 2,(iv) m(A1 ∪ A3) = m(V )/2 and the positive nodes of A1 ∪ A3 all have the same olour.De�nition 5.11. A partition A = {A1, A2, A3, A4, A5, A6} of V is alled a C6-obsta le(for G,P, k, m) if

Se tion 5.2. Preliminaries 81(i) k ≥ 2 is odd,(ii) m(Ai) = 1 and dG(Ai) = k − 1 for every i = 1, 2, 3, 4, 5, 6,(iii) dG(Ai, Ai+1) = (k − 1)/2 for every i = 1, 2, 3, 4, 5, 6,(iv) the positive node in Ai and in Ai+3 have the same olour for every i = 1, 2, 3.The authors of [3℄ have shown that these are the only obsta les of the existen e of asolution with degree fun tion m.Theorem 5.12 (Bang-Jensen, Gabow, Jordán and Szigeti [3℄). Let us be given a graphG0 = (V, E0), a partition P = {P1, P2, . . . , Pr} of the set V , and an integer k ≥ 2. Assumethat the degree-spe i� ation m ∈ ZV

+ satis�es (5.6)-(5.8). Then there exists a graph Gsatisfying the degree-spe i� ation m and the partition onstraints P su h that G0 + G isk-edge- onne ted, unless k is odd and G0 ontains a C4- or a C6-obsta le.The main idea in the solution of the minimum version of Problem 5.6 is to �nd a degree-spe i� ation m satisfying (5.6)-(5.8) with m(V ) as small as possible, and trying to avoid reating a C4- and a C6-obsta le. The authors of [3℄ show that this is not possible if andonly if the problem instan e ontains a on�guration.5.2 PreliminariesIn this se tion we give preliminaries about the methods that we will use in proving Theorem5.4 and in solving the partition onstrained version overing problem. We start with twosimple observations about tight and dangerous sets. In many ases however we omit thereferen e to these lemmas, sin e the situation will be mu h simpler: for example p ≤ 1 willhold.Lemma 5.13. Assume that p : 2V → Z ∪ {−∞} is a symmetri , positively rossingsupermodular fun tion and m ∈ C(p). If D is a p-positive dangerous set and T is a tightset with m(D ∩ T ) > 0 then one of D and T ontains the other.Proof. Certainly, T is also p-positive. They annot satisfy (−) , sin e that would implym(D) − 1 + m(T ) ≤ p(D) + p(T ) ≤ p(D − T ) + p(T − D) ≤ m(D − T ) + m(T − D) ≤

m(D) − 1 + m(T ) − 1, a ontradi tion. This implies the lemma, sin e two p-positive setsalways satisfy (−) unless one ontains the other.Lemma 5.14. Assume that p : 2V → Z ∪ {−∞} is a symmetri , positively rossingsupermodular fun tion and m ∈ C(p). If a p-positive dangerous set D and a p-positivetight set T ross ea h other then D ∪ T and D ∩ T is also dangerous.

82 Chapter 5. Covering symmetri rossing supermodular fun tionsProof. Assume that one of D∪T and D∩T is not dangerous and apply (∩∪) for D and Tto get m(D)−1+m(T ) ≤ p(D)+p(T ) ≤ p(D∩T )+p(D∪T ) ≤ m(D∩T )+m(D∪T )−2,a ontradi tion.The following simple lemma due to Ben zúr and Frank follows easily from Claim 1.9.Lemma 5.15. If p : 2V → Z ∪ {−∞} is a symmetri , positively rossing supermodularfun tion and {X1, X2, . . . , Xt} is a partition of V satisfying p(X1) = 1 and p(X1 ∪Xi) > 0for any i = 1, 2, . . . , t, then this partition is p-full.However we will need the following, slightly more ompli ated lemma.Lemma 5.16. Let p : 2V → Z ∪ {−∞} be a symmetri , positively rossing supermodularfun tion and {V1, V2, . . . , Vk} be a partition of V satisfying p(Vi) = 1 for any i = 1, 2, . . . , k(where k ≥ 4). Let furthermore U1i , U2

i , . . . , U tii be a partition of Vi (where ti ≥ 1 is aninteger) for any i = 1, 2, . . . , k su h that p(Vi ∪ U l

j) > 0 for any possible i, j, l. Assumefurthermore that p(U11 ) = 1. Then the partition U = {U j

i : i = 1, 2, . . . , k and j =

1, 2, . . . , ti} is p-full.Proof. Let i ∈ {1, 2, . . . , k} be arbitrary and U ′ ⊆ U su h that Vi ∪ (⋃

U ′) 6= V . By Claim1.9, p(Vi ∪ (⋃

U ′)) > 0. Applying this and the symmetry of p we get that p(U11 ∪ U l

j) =

p(V − (U11 ∪ U l

j)) > 0 for any possible j, l ( hoose an arbitrary Vi ⊆ V − (U11 ∪ U l

j) and anappropriate U ′ ⊆ U). But then we an apply Lemma 5.15 in order to �nish this proof.Our approa h in the algorithms to be given in this hapter is the following: we per-form some splitting-o� steps, and sometimes we undo one splitting-o� in order to make aprogress. Therefore we introdu e the following operations. Let p0 : 2V → Z ∪ {−∞} besymmetri , positively rossing supermodular and m0 ∈ C(p0) ∩ ZV with m0(V ) even. As-sume that we have performed some admissible splitting-o�s and let G be the graph of thesplit edges. Let furthermore p = p0 − dG and m(v) = m0(v)− dG(v) for every v ∈ V . Pi kan edge uv = e ∈ G. Re all that the unsplitting operation of e is simply the reverse of thesplitting-o� operation: me = m+χ{u}+χ{v}, Ge = G−e and pe = p+d(V,{(uv)}) = p0−dGe .Of ourse, this is always admissible, that is me ∈ C(pe).In the sequel, we will use the following operation. Let uv ∈ E(G) and x ∈ V +. The edge-swit h operation (at x, u, v) is the following: let G′ = G−uv+xu, let m′ = m−χ{x}+χ{v}and p′ = p0 − dG′ (i.e. unsplit uv and split-o� at x, u: the edge xu will also be alled asplit edge). If m′ ∈ C(p′) then the edge-swit h is admissible. If there is a set X ⊆ Vwith 0 = p(X) = m(X) − 1, x, u ∈ X but v /∈ X then the edge-swit h at x, u, v is learly not admissible: we will say that su h a set is a (x, u, v)-swit hblo king set. An

Se tion 5.2. Preliminaries 83(x, uv)-swit hblo king set is a set whi h is either an (x, u, v)-swit hblo king set, or an(x, v, u)-swit hblo king set.If x and y are two (distin t) positive nodes and uv is an edge of G then a one- hange(at x, u, v, y) is the following operation: m′ = m− χ{x} − χ{y}, G′ = G − e + ux + vy andp′ = p0 − dG′ (i.e. unsplit uv and split-o� at x, u and at y, v: the edges xu and yv will alsobe alled split edges). It will be an admissible one- hange if m′ ∈ C(p′). Note that if theedge-swit h at x, u, v or the one at y, v, u is not admissible (for example be ause there existsan (x, u, v)-swit hblo king set, or an (y, v, u)-swit hblo king set), then the one- hange atx, u, v, y is not admissible.A partition of V into p-positive tight sets will be alled a p-tight partition. In thearguments below after performing one of the above operations we will usually (unlessstated otherwise) repla e the fun tions m and p by the modi�ed fun tions. The followingstatements will be useful in the proofs of this hapter.Claim 5.17. Let p : 2V → Z ∪ {−∞} be a symmetri , positively rossing supermodularfun tion and m ∈ C(p) ∩ ZV . Assume that p ≤ 1. Then a set X ⊆ V having p(X) = 1 annot ross a p-positive tight set Y .Proof. Assume that X rosses Y . Note that p(Y ) = m(Y ) = 1. By possibly omplementingX we an assume that m(Y ∩X) = 0. But (∩∪) for X and Y implies that p(Y ∩X) = 1,a ontradi tion with m ∈ C(p).Lemma 5.18. Let p0 : 2V → Z ∪ {−∞} be a symmetri , positively rossing supermodularfun tion, G be a graph and p = p0 − dG and m ∈ C(p)∩ZV . Let x and y be two (distin t)positive nodes and uv be an edge of G. The edge-swit h at x, u, v is not admissible if andonly if there exists a set X ontaining x and u but ex luding v whi h is dangerous withrespe t to p and m. If the edge-swit h at x, u, v and the one at y, v, u are both admissible thenthe one- hange at x, u, v, y is not admissible if and only if there exists a set X ontainingx, u, v, y whi h is dangerous with respe t to p and m.Proof. The �rst statement is lear: sin e the unsplitting at u and v is always admissible,the edge-swit h x, u, v is not admissible if and only if there exists a set X that is dangerous ontaining x and u after this unsplitting (i.e. with respe t to puv and muv). If this setalso ontains v then m(X)− p(X) does not hange during the edge-swit h, so X does notbe ome de� ient. Therefore X does not ontain v, therefore m(X) − p(X) will de reaseby two after the edge-swit h, so it be omes de� ient if and only if it was dangerous withrespe t to p and m.To prove the se ond statement observe that the one- hange at x, u, v, y an be onsideredas the sequen e of a one- hange at x, u, v followed by a splitting-o� at y, v. Let m′ and p′

84 Chapter 5. Covering symmetri rossing supermodular fun tionsbe the fun tions after the edge-swit h at x, u, v. If the splitting is not admissible after thisedge-swit h then there is an X ontaining y and v whi h is dangerous with respe t to p′and m′. Note that this X is not dangerous with respe t to puv and muv (this is be ause theedge-swit h y, v, u was also admissible originally), muv(X)−puv(X) > m′(X)−p′(X) musthold, in other words X must ontain x and u. But then m(X) − p(X) = m′(X) − p′(X),therefore X was also dangerous with respe t to p and m.We will only apply the operations introdu ed above under spe ial ir umstan es. Thenext lemma des ribes this. Note that if p ≤ 1 then a dangerous set ontaining a positivenode will have p-value 0 or 1.Lemma 5.19. Let p0 : 2V → Z ∪ {−∞} be a symmetri , positively rossing supermodularfun tion, G be a graph and p = p0 − dG and m ∈ C(p) ∩ ZV . Assume that p ≤ 1 and{V1, V2, . . . , Vk} is a p-tight partition of V (i.e. m(Vj) = p(Vj) = 1 for every j = 1, 2, . . . , k),where k ≥ 4, and let vj ∈ Vj ∩ V + for every j ∈ {1, 2, . . . , k}. Then1. the tight partition {V1, V2, . . . , Vk} is unique,2. if uv ∈ E(G) is indu ed in Vi for some i ∈ {1, 2, . . . , k} and vr, vs ∈ V + − vi then(a) if X is a (vr, u, v)-swit hblo king set, then X−Vi = Vr and there is no (vr, v, u)-swit hblo king set,(b) if X is a (vr, u, v)-swit hblo king set, and Y is a (vs, u, v)-swit hblo king set then

X − Y = Vr and Y − X = Vs,( ) the edge-swit h at vr, u, v is not admissible if and only if there exists a (vr, u, v)-swit hblo king set,(d) the one- hange at vr, u, v, vs is not admissible if and only if there exists a (vr, u, v)-swit hblo king set, or a (vs, v, u)-swit hblo king set,3. if uv ∈ E(G) is su h that u ∈ V1 and v ∈ V2 then(a) the edge-swit h at v1, v, u is always admissible,(b) the edge-swit h at v3, u, v is not admissible if and only if p(V1 ∪ V3) = 1,( ) the one- hange at v3, u, v, v1 is not admissible if and only if p(V1 ∪ V3) = 1.Proof. To prove the �rst statement, suppose that the partition {V ′1 , V

′2 , . . . , V

′k} also satis�esthat m(V ′

i ) = p(V ′i ) = 1 for every i = 1, 2, . . . , k (assume that these sets are indexed su hthat m(Vi∩V ′

i ) = 1 for every i = 1, 2, . . . , k). Sin e V ′i annot ross Vi for all i, and Vi ( V ′

i

Se tion 5.3. Proof of the theorem of Ben zúr and Frank 85would imply that V ′i rosses another Vj (sin e k > 2), this means that Vi = V ′

i for everyi = 1, 2, . . . , k.To prove (2a) apply (∩∪) for X and Vi to get that p(X ∪ Vi) = 1, implying thatX − Vi = Vr by Claim 5.17. If X is a (vr, u, v)-swit hblo king set and Y is a (vr, v, u)-swit hblo king set then (1.5) for X and Y gives that p(X ∪ Y ) = 1, ontradi ting Claim5.17.To show (2b) apply (1.6) for X and Y .To prove (2 ) observe that if the edge-swit h at vr, u, v is not admissible then by Lemma5.18 there is a dangerous set X with vr, u ∈ X and v /∈ X. Sin e m(V ) ≥ 4, this set rossesVi, therefore it is a (vr, u, v)-swit hblo king set, as laimed.The proof of (2d) is similar: if the edge-swit h at vr, u, v or the one at vs, v, u is notadmissible then this is be ause of a swit hblo king set by (2 ). If both of these edge-swit hoperations are admissible but the one- hange at vr, u, v, vs is not admissible then by Lemma5.18 there exists a dangerous set X ontaining vr, u, v, vs. Sin e m(X) ≥ 2 by the nodesvr, vs ∈ X, this is only possible if m(X) = 2 = p(X) + 1. But this X then rosses Vi, ontradi ting Claim 5.17.To prove (3a) assume that the edge-swit h at (v1, v, u) is not admissible. Use Lemma5.18 and note that the dangerous set X will ross V1, so it annot have p-value 1, i.e. itis a (v1, v, u)-swit hblo king set. Apply (∩∪) for X and V2 to get that p(X ∪ V2) = 1, ontradi ting Claim 5.17.To prove (3b) assume that the edge-swit h at v3, u, v is not admissible. By Lemma 5.18there is a dangerous set X ontaining v3, u and not ontaining v. Assume that p(X) =

0. Then X rosses V1 (sin e v1 /∈ X), therefore applying (1.6) to X and V1 gives a ontradi tion. Therefore p(X) = 1, and then by Claim 5.17 X = V1 ∪ V3.To prove (3 ) assume that the one- hange at v3, u, v, v1 is not admissible but p(V1∪V3) <

1, i.e. the edge-swit h at (v3, u, v) is admissible. By Lemma 5.18 there exists a dangerousset X ontaining v3, u, v, v1, thus m(X) = 2 = p(X)− 1 and X rosses V2, sin e v2 /∈ X, a ontradi tion by Claim 5.17.5.3 Proof of the theorem of Ben zúr and FrankIn this se tion we will give an algorithmi proof of Theorem 5.4. This proof appeared in [8℄.The idea is the following: �rst we perform an arbitrary sequen e of admissible splitting-o�sas long as possible. When there is no more admissible splitting-o�, then we try to unsplitone single edge su h that we obtain two admissible splitting-o�s (this operation was alled an admissible one- hange). Surprisingly, this will be enough: we prove that this

86 Chapter 5. Covering symmetri rossing supermodular fun tionsalgorithm �nds the required degree-spe i�ed graph, if the onditions of Theorem 5.4 hold.Proof of Theorem 5.4. The ne essity of the onditions is lear. The proof of su� ien yuses the algorithm sket hed above. In fa t we will prove a little more, as in this proof thealgorithm takes as input the fun tion p0 and an arbitrary m0 ∈ C(p0) ∩ ZV with m0(V )even (that is, we don't assume about the input that m0(V )/2 ≥ dim(p0)−1 holds). We willprove that if this algorithm does not terminate with �nding the required degree-spe i�edgraph then the ondition m0(V )/2 ≥ dim(p0) − 1 did not hold. In this ase we will also�nd a p0-full partition of size dim(p0). In fa t we will prove that the algorithm �nds alongest possible admissible splitting sequen e. Let us formally des ribe the algorithm.First Step of the Algorithm: Perform an arbitrary sequen e of admissible splitting steps aslong as there exists one.Before ontinuing the des ription of the algorithm we des ribe the situation when thereis no further admissible splitting-o�. As usual, let the graph of the edges split so far bedenoted by G, p = p0−dG and m(v) = m0(v)−d+G(v) for any v ∈ V . By Lemma 4.6, p ≤ 1and any pair u, v ∈ V + is in a dangerous set X: this means that p(X) = 1 and m(X) = 2,hen e m ≤ 1. We an further assume that m(V ) ≥ 4 (if m(V ) = 0 then the theorem isproved and learly m(V ) = 2 annot be the ase). Let the positive nodes be v1, v2, . . . , vk(where k = m(V ) is of ourse even).Lemma 5.20. Under the assumptions made above(i) for any i ∈ {1, 2, . . . , k} there exists a unique maximal tight set Vi ontaining vi,(ii) the set blo king vi and vj is Vi ∪ Vj for any i, j ∈ {1, 2, . . . , k},(iii) p(∪i∈IVi) = 1 for any nonempty I ( {1, 2, . . . , k},(iv) the sets V1, V2, . . . , Vk form a partition of V .Proof. The sets that we onsider will always have positive p value, so we an use (∩∪) and

(−) if two of them ross. A set X blo king a pair u, v ∈ V + and another set Y blo kinga pair w, v ∈ V + ross ea h other, meaning that p(X ∩ Y ) = 1, so v ∈ V + is in a tightset. If T1 and T2 are two tight sets ontaining the positive node x ∈ V + then T1 and T2 annot ross ea h other, sin e then (−) would imply that p(T1−T2) = 1 > m(T1−T2) = 0,a ontradi tion. Thus one of T1 and T2 must ontain the other, so indeed there exists aunique maximal tight set Vi ontaining vi for every i.To prove (ii) let i, j be two di�erent indi es between 1 and k. It is straightforward thatVi and Vj have to be disjoint (otherwise p(Vi ∩Vj) = 1 would follow from (∩∪) ). Similarly,

Se tion 5.3. Proof of the theorem of Ben zúr and Frank 87a set X blo king vi and vj must ontain Vi (and Vj), otherwise p(Vi −X) = 1 would followfrom (−) . On the other hand, if l ∈ {1, 2, . . . , k} is di�erent from i and j and Y is a setblo king vi and vl then (∩∪) implies that X ∩ Y = Vi (sin e it is tight) and (−) impliesthat X − Y = Vj (sin e it is tight again). This �nishes the proof of (ii), and then (iii)follows from Claim 1.9.The only thing to be proved to get (iv) is that ∪ki=1Vi = V : but if this was not the asethen Claim 1.9 would also imply that p(∪k

i=1Vi) = 1, whi h would give a ontradi tion,sin e m(V − ∪ki=1Vi) = 0 and p(V − ∪k

i=1Vi) = p(∪ki=1Vi) = 1.One simple observation shows that G does not have edges between two lasses Vi and

Vj of this partition: if it had, then hoosing a third index l ∈ {1, 2, . . . , k} and using thatX = Vi ∪ Vl and Y = Vj ∪ Vl has to satisfy (∩∪) with equality would give a ontradi tionby (1.1).The next step of the algorithm tries to �nd admissible one- hanges as long as possible.That is, our aim is to �nd an edge uv = e of G (spanned by Vi, say) and two positivenodes vr and vs su h that an admissible one- hange an be performed. Note that r = i isnot a good hoi e, sin e p(Vi ∪ Vs) = 1; s = i is not good, either, so vr and vs are bothdistin t from vi. By Lemma 5.19 (2d), the obsta le of the admissibility of the one- hange atvr, u, v, vs is a swit hblo king set. The following lemma tells us that if a one- hange is notadmissible at vr, u, v, vs for some edge uv ∈ E(G), then there is no admissible one- hangeat this edge at all.Claim 5.21. If X is a (vr, uv)-swit hblo king set (where uv ∈ E(G) is indu ed by Vi forsome i and vr ∈ V + − vi), then e = ∆G(X) and Vj ∪ (X ∩ Vi) is the unique (vj, uv)-swit hblo king set for any j 6= i.Proof. Without loss of generality assume that X is a (vr, u, v)-swit hblo king set. We willuse that m(V ) ≥ 4: let i, r be the indi es of the statement and let q, s ∈ {1, 2, . . . , k}−{i, r}be distin t indi es. Apply (∩∪) to X and Y = Vr ∪Vq to get that p(X ∪Y ) ≥ 0, but sin eit annot be 1 by Claim 5.17, it must be 0. Now apply (−) to X ∪ Y and Vr ∪ Vs to obtainthat p(Vq∪(X∩Vi)) ≥ 0, but again by Claim 5.17 it annot be one, so Vq∪(X∩Vi) is indeeda (vq, uv)-swit hblo king set for any q 6= i. Apply (1.6) for X and Vq ∪ (X ∩Vi) to get thate = ∆G(X). Finally Lemma 5.19 (2b) gives the uniqueness of (vj , u, v)-swit hblo king setfor any j 6= i.After these preliminaries the des ription of the algorithm is ontinued.Se ond Step of the Algorithm: Perform an arbitrary sequen e of admissible one- hangeoperations as long as there exists one. If m(V ) de reases to 2 then �nish the pro edure

88 Chapter 5. Covering symmetri rossing supermodular fun tionswith a splitting-o� (whi h is of ourse admissible) and terminate.Observe that an admissible one- hange will not reate an admissible splitting-o� byLemma 5.20 (iii), unless m(V ) be omes 2.Let us now des ribe the situation when there is no further admissible one- hange.For simpli ity let us denote the remaining degree spe i� ation again by m, the obtainedgraph by G and p = p0 − dG. Again, if m(V ) ≤ 3 then we are done, so assume thatm(V ) ≥ 4. We will now show how to obtain a p0-full partition of size greater thanm0(V )/2 + 1, whi h �nishes the proof of the theorem. Furthermore this partition will beof size m(V )+ |E(G)|, whi h shows that the dimension of p0 is not greater than this, sin eadding a spanning tree on V + to G overs p0 and has size m(V ) + |E(G)| − 1. To this endlet us des ribe the stru ture of the obsta les of further admissible one- hanges. By Claim5.21 for every split edge e indu ed in Vi, say, there exists a uniquely de�ned set Xe ⊆ Vi−visu h that Xe ∪ Vj is the unique (vj , e)-swit hblo king set for any j 6= i. Sin e dG(Xe) = 1for any edge, this implies that G does not ontain a y le.Claim 5.22. Let e, f be two (distin t) edges of G spanned by Vi. Then Xe and Xf annot ross ea h other.Proof. Assume the ontrary and onsider two positive nodes vr and vs (distin t from vi).Apply (−) for the rossing sets X = Vr ∪ Xe and Y = Vs ∪ Xf to get that p(X − Y ) =

p(Y −X) = 0. Sin e at least one of X − Y and Y −X is entered by at least one of e andf , this ontradi ts the uniqueness of the sets Xe and Xf .This laim shows that the sets {Xe : e ∈ G} form a laminar family. Visually, if weintrodu e an orientation ~G of G su h that an edge e ∈ G is oriented to enter Xe, then we getthat the undire ted omponents of ~G are arbores en es and the laminar family {Xe : e ∈ G} an naturally be related to these arbores en es (it is known that an arbores en e naturallyde�nes a laminar family), though we don't need this observation below.Consider the laminar family U = {V1, V2, . . . , Vk}

⋃

{Xe : e ∈ G} and let U∗ = U −⋃

W∈U ,W(U W for any U ∈ U . The following laim �nishes the proof of the theorem, sin ethe family {U∗ : U ∈ U} has m(V ) + |E(G)| members.Claim 5.23. The set U∗ is not empty for any U ∈ U . The partition {U∗ : U ∈ U} isp0-full.Proof. Observe that U∗ is never empty, sin e if U = Vi for some i then vi ∈ U∗, and ifU = Xe then one endpoint of e is in U∗. Now we want to show that p0(Vi ∪ U∗) > 0for any i ∈ {1, 2, . . . , k} and U ∈ U . We an assume that U ⊆ Vj where j 6= i andlet l ∈ {1, 2, . . . , k} − {i, j}. Apply Claim 1.9 for Vl and the maximal sets of the family

Se tion 5.3. Proof of the theorem of Ben zúr and Frank 89{W ∈ U : W ( U} to dedu e that p0(Vl ∪X) > 0 where X =

⋃

W∈U ,W(U W . Now (−) forVi∪U and Vl∪X gives that p0(Vi∪U∗) ≥ p0(Vi∪U)+p0(Vl∪X)−p0(Vl) ≥ 1+1−1 ≥ 1, as laimed. Now Lemma 5.16 �nishes the proof of this laim, sin e if U is a minimal memberof the family U then p0(U) = p0(U

∗) = 1 an be easily veri�ed.�Let us give the pseudo ode of the algorithm that we suggest (for simpli ity we do notread out the p-full partition of size dim(p) as in the proof above). We note that by thearguments given after Algorithm GREEDYCOVER, this algorithm an also be implementedto run in polynomial time.Algorithm SYMCROS_COVERbegin INPUT: A symmetri , positively rossing supermodular fun tion p : 2V → Z ∪

{−∞} (given with a maximizing ora le) and an admissible degree spe i� ation m : V →

Z+ (with m(V ) even).OUTPUT: A graph G = (V, E) overing p and satisfying d+G(v) = m(v) for every

v ∈ V , or the statement that the ondition m(V )/2 ≥ dim(p) − 1 did not hold.1.1. Initialize G = (V, ∅).1.2. While there exists an admissible splitting-o� at u, v, perform it, i.e.1.3. Let m = m − χ(u) − χ(v) and p = p − d(V,{(u,v)}) and G = G + (uv).1.4. EndWhile1.5. While there exists an admissible one- hange at x, u, v, y, perform it, i.e.1.6. Let m = m−χ(x)−χ(y), G = G−uv +xu+vy and p = p+d(V,{uv})−d(V,{xu,vy}).If m(V ) = 2 then �nish with an admissible splitting-o�, i.e.1.7. Let m = m − χ(u) − χ(v) (where V + = {u, v}) and p = p − d(V,{(u,v)}) andG = G + (uv).1.8. EndWhile1.9. If m(V ) = 0 then output G.1.10. Otherwise return �The ondition m(V )/2 ≥ dim(p) − 1 did not hold!�.endIn fa t our proof implies the following de� ient form of Theorem 5.4 (where an admissiblesplitting sequen e means an arbitrary sequen e of admissible splitting-o�s).Theorem 5.24. Let p0 : 2V → Z ∪ {−∞} be a symmetri , positively rossing supermod-ular set fun tion and m0 ∈ C(p0) ∩ ZV with m0(V ) even. If m0(V )/2 < dim(p0) − 1then the longest admissible splitting sequen e onsists of m0(V )− dim(p0) splitting-o�s. If

m0(V )/2 ≥ dim(p0) − 1 then there exists a omplete admissible splitting-o�.

90 Chapter 5. Covering symmetri rossing supermodular fun tionsProof. Consider an arbitrary running of the algorithm sket hed above. If it gets stu k withremaining degree spe i� ation m and graph G, then we have seen that m(V ) + |E(G)| =

dim(p0). Sin e m(V ) + 2|E(G)| = m0(V ), this shows that (after arbitrary hoi es in thealgorithm) |E(G)| = m0(V ) − dim(p0). Sin e a longest admissible splitting sequen e is learly a valid running of the algorithm (there annot exist an admissible one- hange aftera longest splitting sequen e), this �nishes the proof.5.4 Partition onstrained overing problemIn this se tion we will also onsider partition (or olour) onstraints. That is we arealso given a partition P = {P1, P2, . . . , Pr} of the set V ( olour lasses) and we are onlyallowed to introdu e edges between two members of this partition. If v ∈ Pi for somev ∈ V and i then we will also use the notation c(v) = i (that is, v has olour i). Weformulate the problems that we want to solve in this se tion.Problem 5.25 (Partition onstrained overing of a symmetri , positively rossing super-modular fun tion). Let us be given a symmetri , positively rossing supermodular fun tionp : 2V → Z ∪ {−∞} with a maximizing ora le and a partition P = {P1, P2, . . . , Pr} of theset V . The problem is to �nd a graph G overing p that ontains only edges between the lasses of P. In the minimum version the number of edges of G is to be minimized. Inthe degree-spe i�ed version G has to satisfy a given degree-spe i� ation m ∈ ZV

+.Note that loop edges in G are prohibited by the partition onstraints, so d+G(v) = dG(v)for any v ∈ V .This se tion ontains joint results with Roland Grappe and Zoltán Szigeti. These resultsare not published yet. They have been a epted for presentation at the ACM-SIAM Sym-posium on Dis rete Algorithms (SODA10) whi h will be held in Austin, Texas in January2010: see [10℄.5.4.1 PreliminariesLower bounds and ne essary onditionsIn this subse tion we give natural lower bounds for the minimum version of Problem 5.25and natural ne essary onditions for the degree-spe i�ed version of that problem. We startwith the latter.Assume that the degree-spe i�ed version of Problem 5.25 is given. We have seen that anatural ne essary ondition of the existen e of a solution is that the degree-spe i� ation is

Se tion 5.4. Partition onstrained overing problem 91admissible, that ism(X) ≥ p(X) for any X ⊆ V. (5.9)It is easy to see that the following onditions are also ne essary.

m(V ) is even, (5.10)m(Pi) ≤ m(V − Pi) for every i = 1, 2, . . . , r. (5.11)We will say that a degree-spe i� ation m satisfying (5.9)-(5.11) is allowed. A furtherne essary ondition of the existen e of a solution with degree fun tion m is that

m(V )/2 ≥ dim(p) − 1. (5.12)We emphasize that an allowed degree-spe i� ation does not ne essarily satisfy the dimen-sion ondition (5.12).However the onditions (5.9)-(5.12) are still not su� ient, there are some ex eptional ases when this graph does not exist as this was already seen at the problem of parti-tion onstrained global edge- onne tivity augmentation. These ex eptions will be alledobsta les here, too. We will give the de�nition of these obsta les in Se tion 5.4.2.Turning to the minimum version of the problem: by the previous arguments the valueφ = max{αp, β

1p , . . . , β

rp, dim(p)−1} is a lower bound on the number of edges of the desiredgraph, where

αp = ⌈SLB(p)/2⌉ = max{⌈

∑

X∈X p(X)

2⌉ : X is a subpartition of V }

βip = max{

∑

Y ∈F

p(Y ),F subpartition of Pi}, for i = 1, . . . r.Again, as seen at the problem of partition onstrained global edge- onne tivity augmen-tation, in some ases this lower bound annot be a hieved. These ex eptional ases willbe alled on�gurations, their des ription will be given in Se tion 5.4.3. Sin e obsta lesand on�gurations are losely related to ea h other, they have a ommon root, we give thedes ription of this ommon root in the next se tion. We all them onstru tions: theyexpress ertain properties of the set fun tion p, i.e. the partition P does not play a role inthe de�nition of onstru tions.Constru tionsWe will use the following onvention. If A1, A2 . . . , Al are some subsets of V , then Al+1will denote A1, Al+2 will be A2 and so on (and similarly A0 will be Al). Furthermore,

92 Chapter 5. Covering symmetri rossing supermodular fun tionsa onstru tion (an obsta le, or a on�guration, see later) will always be a partition A =

{X1, X2, . . . , Xt} of V : though the order of the partition lasses does matter in somesense in these de�nitions, we de ided to use the set-notation {X1, X2, . . . , Xt} insteadof the sequen e-notation (X1, X2, . . . , Xt), sin e some reordering will always be possible.Hopefully after this remark this will not ause a onfusion. Re all that m ∈ C(p) is saidto be minimal if m(V ) = SLB(p) = max{∑

X∈X p(X) : X is a subpartition of V }.De�nition 5.26. A partition A = {A1, A2, A3, A4} of V is alled a C∗

4- onstru tion for

p if1. p(Ai) + p(Ai+1) − p(Ai ∪ Ai+1) is odd for every i = 1, 2, 3, 4,2. p(Ai−1 ∪ Ai) + p(Ai ∪ Ai+1) = p(Ai−1) + p(Ai+1) for every i = 1, 2, 3, 4,3. p(A1 ∪ A3) ≤ 0 and p(A2 ∪ A4) ≤ 0,4. p(A1) + p(A3) = p(A2) + p(A4) = 12SLB(p).Note that the order of the partition lasses in a C∗

4 - onstru tion matters only in the fol-lowing sense: if {A1, A2, A3, A4} is a C∗4 - onstru tion, then so is {A2, A3, A4, A1} ( y li alreindexing) and {A4, A3, A2, A1} (reversing).The following stru ture arises only for our abstra t form of the problem, it does not existin the framework of graphs or hypergraphs.De�nition 5.27. A partition A = {A1, A2, A3, A4, B1, . . . , Bt} (where t ≥ 1) of V is alleda C∗

5- onstru tion for p if1. p(Ai) = 1 for every i = 1, 2, 3, 4,2. p(Bj) = 2 for every j = 1, . . . , t,3. p(Ai ∪ Bj) = p(Ai ∪ Ai+1) = 1 for every i = 1, 2, 3, 4 and j ∈ {1, . . . , t},4. p(A1 ∪ A3) ≤ 0 and p(A2 ∪ A4) ≤ 0,5. SLB(p) =

∑

X∈A p(X) = 2t + 4.Observe that we an y li ally reindex the lasses Ai, or reverse the order of them in aC∗

5 - onstru tion, while the lasses Bj an ome in any order.De�nition 5.28. A partition A = {A1, A2, A3, A4, A5, A6} of V is alled a C∗

6- onstru tionfor p if1. p(Ai) = 1 for every i = 1, 2, 3, 4, 5, 6,

Se tion 5.4. Partition onstrained overing problem 932. p(Ai ∪ Ai+1) = 1 for every i = 1, 2, 3, 4, 5, 6,3. p(Ai ∪ Aj) ≤ 0 for every 1 ≤ i ≤ 4 and i + 2 ≤ j ≤ i + 4,4. SLB(p) =∑6

i=1 p(Ai) = 6.Again note that we an y li ally reindex the lasses, or reverse the order of the lassesin a C∗6 - onstru tion.We say that a partition A is a onstru tion if it is either a C∗

4 -, a C∗5 - or a C∗

6 - onstru tion. Note, that if A is a onstru tion and m ∈ C(p), then the ondition m(V ) =SLB(p) (minimality of m) is equivalent to saying that every X ∈ A is tight. Also note thatif A is a C∗4 - onstru tion and m ∈ C(p) is minimal then De�nition 5.26.1 requires that

m(Ai ∪ Ai+1) − p(Ai ∪ Ai+1) has to be odd for every i = 1, 2, 3, 4.Claim 5.29. If A = {A1, A2, A3, A4} is a C∗4 - onstru tion for p then p(Ai) > 0 for ev-ery i = 1, 2, 3, 4. Consequently, a set ontaining at least three members of A annot bedangerous.Proof. Assume for example that p(A2) = 0 (it annot be negative sin e that would implySLB(p) =

∑

i p(Ai) <∑

i6=2 p(Ai) ≤ SLB(p)) and let m ∈ C(p) be minimal. De�nition5.26.2 applied to i = 2, the tightness of A1, A2 and A3, and the admissibility of m gives thatp(A1 ∪A2)+ p(A2 ∪A3) = p(A1)+ p(A3) = m(A1)+m(A3) = m(A1 ∪A2)+m(A2 ∪A3) ≥

p(A1 ∪A2) + p(A2 ∪A3), implying that A1 ∪A2 and A2 ∪A3 are both tight, ontradi tingDe�nition 5.26.1.To get the orollary assume that a set X ontaining A1 ∪ A2 ∪ A3 is dangerous. Sin ep(X) = p(V − X), this implies that SLB(p) ≥ p(X) + p(V − X) = 2p(X) ≥ 2(m(X) −

1) ≥ 2(SLB(p)/2 + 1 − 1), i.e. p(X) = SLB(p)/2, but m(V − X) ≤ SLB(p)/2 − 1, a ontradi tion.Claim 5.30. Let A = {A1, A2, A3, A4} be a C∗4 - onstru tion, m ∈ C(p) be minimal and

i ∈ {1, 2, 3, 4} arbitrary. If Ai ∪ Ai+1 is not dangerous then a pair x ∈ Ai ∩ V + andy ∈ Ai+1 ∩ V + is admissible. If we split-o� this pair then A be omes a C∗

4 - onstru tion forthe modi�ed fun tion p′ = p − d(V,{xy}).Proof. Without loss of generality let i = 1. Assume that the pair x, y is not admissible andlet X be a dangerous set blo king it. We an assume that X is a maximal dangerous set,thus by Lemma 5.13 A1 ∪ A2 ⊆ X. Similarly, if for example A3 ∩ X 6= ∅ then A3 and Xwould ross ea h other (sin e A4 6⊆ X by Claim 5.29), and then Lemma 5.14 would implythat A3 ⊆ X, ontradi ting Claim 5.29. Therefore X = A1 ∪ A2, whi h is not dangerousby the assumption. The se ond statement is easy to he k (to get De�nition 5.26.4 useLemma 2.16).

94 Chapter 5. Covering symmetri rossing supermodular fun tionsThe following properties of a C∗5 - onstru tion an be proved.Claim 5.31. If A = {A1, A2, A3, A4, B1, . . . , Bt} is a C∗

5 - onstru tion and m ∈ C(p) isarbitrary with m(V ) = SLB(p) then(i) p(Ai∪⋃

j∈J Bj) = p(Ai∪Ai+1∪⋃

j∈J Bj) = 1 for all J ⊆ {1, . . . , t} and i ∈ {1, 2, 3, 4},(ii) p(∪j∈JBj) = 2 for all nonempty J ⊆ {1, . . . , t},(iii) p(Ai ∪ Ai+2 ∪⋃

j∈J Bj) = 0 for all J ⊆ {1, . . . , t} and i ∈ {1, 2}(iv) If X rosses a member Y of the partition A then p(X) ≤ p(X ∪ Y ) and p(X) ≤

p(X − Y ). If furthermore m(X ∩ Y ) < m(Y ) then p(X) < p(X ∪ Y ), and similarlyif m(Y − X) < m(Y ) then p(X) < p(X − Y ).(v) If X rosses a member of the partition A then p(X) ≤ 1 ( onsequently p ≤ 2),(vi) the splitting at u and v is admissible for any positive pair u ∈ Ai ∪ Bj1 and v ∈

Ai+2 ∪ Bj2 (where i ∈ {1, 2, 3, 4} and j1, j2 ∈ {1, 2, . . . , t} are distin t),(vii) Splitting-o� at a ∈ A1∩V + and b ∈ B1∩V + the partition {A1∪B1, A2, A3, A4, B2, . . . , Bt}be omes a C∗5 - onstru tion for the modi�ed fun tion p′, if t > 1, and it be omes a C∗

4 - onstru tion for the modi�ed fun tion p′, if t = 1 (the analogous statement holds foran arbitrary a ∈ Ai ∩ V + and b ∈ Bj ∩ V +, too).Proof. Let us �rst prove 1 = p(A1 ∪ A2 ∪⋃

j∈J Bj) of (i). Claim 1.8 applied to A1 ∪ A2and A1 ∪ Bj (j = 1, 2, . . . , t) gives that 1 = p(A1 ∪ A2) ≤ p(A1 ∪ A2 ∪⋃

j∈J Bj) ≤

p(A1 ∪ A2 ∪⋃t

j=1 Bj) = p(A3 ∪ A4) = 1 for all J ⊆ {1, 2, . . . , t}, what was to be proved.The same proof gives that 1 = p(A1 ∪ A2 ∪ A3 ∪⋃

j∈J Bj) for all J ⊆ {1, 2, . . . , t}, whi his the se ond statement of (i).Now we prove (ii). By (i), p(A1∪A2 ∪A3) = p(A3∪A4∪A1) = 1. Let us apply (∩∪) forthese two sets to get that p(A1 ∪ A2 ∪ A3 ∪ A4) ≥ 2. Choose an arbitrary i ∈ {1, 2, . . . , t}and apply Claim 1.8 for sets A1∪A2∪A3∪A4 and A1∪Bj (j ∈ {1, 2, . . . , t}− i) to get that2 ≤ p(A1 ∪A2 ∪A3 ∪A4 ∪

⋃

j∈J Bj) ≤ p(A1 ∪A2 ∪A3 ∪A4 ∪⋃

j∈{1,2,...,t}−i Bj) = p(Bi) = 2for all J ( {1, 2, . . . , t}, proving (ii).Let us prove (iii): �rst note that by (∩∪) applied to A1 ∪⋃

j∈J Bj and A3 ∪⋃

j∈J Bjwe get that p(A1 ∪ A3 ∪⋃

j∈J Bj) ≥ 0 for any J ⊆ {1, . . . , t}. Assume that p(A1 ∪ A3 ∪⋃

j∈J Bj) > 0 for some J ⊆ {1, . . . , t}. Apply Claim 1.8 to A1 ∪A3 ∪⋃

j∈J Bj and A1 ∪Bj

(j ∈ {1, 2, . . . , t} − J) to get that 1 ≤ p(A1 ∪ A3 ∪⋃

j∈J Bj) ≤ p(A1 ∪ A3 ∪⋃t

j=1 Bj) =

p(A2 ∪ A4) ≤ 0, a ontradi tion.

Se tion 5.4. Partition onstrained overing problem 95To prove (iv) observe that m(X∩Y ) ≤ m(Y ) and m(Y −X) ≤ m(Y ), so the tightness ofY implies that p(X ∩Y ) ≤ p(Y ) and p(Y −X) ≤ p(Y ). Thus, applying (∩∪) for X and Ygives that p(X) ≤ p(X ∪ Y ), and applying (−) for X and Y gives that p(X) ≤ p(X − Y ).Similarly, m(X ∩ Y ) < m(Y ) implies p(X) < p(X ∪ Y ), and m(Y − X) < m(Y ) impliesp(X) < p(X − Y ).Assume that (v) does not hold and hoose an X violating (v) with maximum p-value.Let Y be a member of A rossing X: sin e m(Y ) > 0, at least one of m(X ∩ Y ) < m(Y )and m(Y − X) < m(Y ) holds, so (iv) gives that p(X ∪ Y ) or p(X − Y ) is stri tly greaterthan p(X). Let X ′ be either X ∪ Y or X − Y su h that p(X ′) > p(X), so X ′ annot rossother sets of the partition A and p(X ′) ≥ 3, ontradi ting the previous results (note thatif X ′ ∪ Y ′ = V for some Y ′ ∈ A then p(X ′) = p(V − X ′) ≤ m(V − X ′) ≤ 2 also follows).To prove (vi) assume that a dangerous set X ontains the positive nodes u ∈ A1 ∪ Bj1and v ∈ A3 ∪ Bj2 . Note that p(X) = 2 an only hold for X = Bj1 ∪ Bj2 , whi h is notdangerous, implying that p(X) = 1 = m(X) − 1 must be the ase. Furthermore we anassume that X ∩ Y = ∅ for any Y ∈ A satisfying m(Y ∩X) = 0, sin e we ould substituteX with X − Y by (iv). Similarly, Y ⊆ X an be assumed about any Y ∈ A satisfyingm(Y − X) = 0, sin e we ould substitute X with X ∪ Y by (iv). So u ∈ A1 and v ∈ A3would imply that X = A1 ∪ A3, whi h is not dangerous. On the other hand, if u ∈ Bj1then m(Bj1 −X) < m(Bj1) implies by (iv) that p(X −Bj1) ≥ 2, whi h annot be the asesin e m(X − Bj1) = 1.The proof of (vii) is simple using the properties we have proved so far.Claim 5.32. If A = {A1, A2, A3, A4, A5, A6} is a C∗

6 - onstru tion for p, then p(X) ≤ 1for any X ⊆ V .Proof. Assume that p(X) > 1 for some X ⊆ V and assume that we have hosen a maximalX with this property. Note that X ⊆ Ai (or V − X ⊆ Ai) annot hold for some i byDe�nition 5.28.4. Thus a set Ai annot ross X sin e then Claim 1.7 for X and Ai wouldgive that p(X ∪Ai) ≥ p(X), ontradi ting the maximality of X. Thus, by De�nition 5.28,X has to be the union of exa tly 3 members of A. Apply again Claim 1.7 for X andAi ∪ Ai+1 with some Ai ⊆ X satisfying Ai+1 ⊆ V − X to get a ontradi tion.Claim 5.33. If A = {A1, A2, A3, A4, A5, A6} is a C∗

6 - onstru tion for p and m ∈ C(p)is minimal then the pair ai ∈ Ai ∩ V + and ai+2 ∈ Ai+2 ∩ V + is admissible for any i =

1, 2, 3, 4, 5, 6. Splitting-o� su h a pair we get C∗4 - onstru tion (for the modi�ed fun tion

p′ = p − d(V,{aiai+2})).Proof. Without loss of generality let i = 1. The �rst statement follows from Claim 5.32:if the pair a1, a3 was not admissible then the set blo king X it would have p(X) = 1, and

96 Chapter 5. Covering symmetri rossing supermodular fun tionsX = A1 ∪ A3 would hold by Claim 5.17, ontradi ting De�nition 5.28.3. For the se ondstatement, observe that (∩∪) for A1 ∪ A2 and A2 ∪ A3 gives that p(A1 ∪ A2 ∪ A3) = 1.Using this one an verify that {A1 ∪ A2 ∪ A3, A4, A5, A6} be omes a C∗

4 - onstru tion.Lemma 5.34. If A is a onstru tion for p and m ∈ C(p) is minimal then there exists a omplete admissible splitting-o�. Consequently, dim(p) − 1 ≤ SLB(p)/2, if a onstru tionexists for p.Proof. Sin e in a basi C∗4 - onstru tion the splitting-o� at nodes a ∈ A1 ∩ V + and b ∈

A3 ∩ V + is admissible, Claim 5.30, 5.31 and 5.33 give the statement.Claim 5.35. If A is a onstru tion and m ∈ C(p) is minimal, then every X ∈ A is amaximal tight set.Proof. Assume that X ′ is tight for some X ′ ) X ∈ A. For any Y ∈ A, if X ′ ∩ Y 6= ∅ thenY ⊆ X ′ must hold (sin e otherwise X ′ and Y would be properly interse ting, and then byClaim 2.15 they would be rossing, onsequently X ′ ∪ Y would also be tight). Thus X ′ isthe union of some members of A.1. If A is a C∗

4 - onstru tion then Ai ∪ Ai+1 is not tight by De�nition 5.26.1, Ai ∪ Ai+2is not tight by De�nition 5.26.3 and Ai ∪ Ai+1 ∪ Ai+2 is not tight by Claim 5.29.2. If A is a C∗5 - onstru tion then the union of some members of A annot be tight byClaim 5.31.3. If A is a C∗6 - onstru tion then the union of some members of A annot be tight byClaim 5.32.

The following lemma shows that if a onstru tion exists then it is unique.Lemma 5.36. If A is a partition of V then it satis�es at most one of De�nition 5.26,De�nition 5.27, and De�nition 5.28. If A and A′ are two di�erent partitions of V then atmost one of them is a onstru tion.Proof. The �rst statement is straightforward: the three de�nitions require di�erent numberof partition lasses or di�erent fun tion values of the lasses. The se ond statement is dueto the fa t that hoosing an arbitrary minimal m ∈ C(p), every set of a onstru tion is amaximal tight set.

Se tion 5.4. Partition onstrained overing problem 975.4.2 The degree-spe i�ed problemLet us be given a symmetri , positively rossing supermodular fun tion p : 2V → Z∪{−∞}with a maximizing ora le, a partition P = {P1, P2, . . . , Pr} of the set V and a degree-spe i� ation m ∈ ZV+. In this se tion we want to solve the degree-spe i�ed version ofProblem 5.25.In Se tion 5.4.2 we will assume that the lasses of P are indexed su h that

m(P1) ≥ m(Pi) for any i = 1, 2, . . . , r. (5.13)We will also say that the olour of P1 is red, and any other olour is non-red. Note thatunder the assumption (5.13) the ondition (5.11) is equivalent to requiring that m(P1) ≤

m(V − P1).Before giving the des ription of the obsta les we �rst des ribe the building blo ks of ouralgorithm for solving the problem at hand. The algorithm will be similar to AlgorithmSYMCROS_COVER: we perform splitting-o�, one- hange and edge-swit h operations whileonly taking are of the allowedness of the degree spe i� ation. Therefore we introdu e thefollowing operations.Assume that we are given an allowed degree-spe i� ation m (we emphasize that thisdoes not require the dimension ondition (5.12)). An admissible splitting-o� at u andv is alled allowed if c(u) 6= c(v) and the degree-spe i� ation after the splitting-o� isagain allowed. This last ondition is equivalent to requiring that if m(P1) = m(V )/2 thenan allowed splitting-o� uses exa tly one red node. After an allowed splitting-o� we willalways reindex the lasses of the partition P in order to maintain that m(P1) is maximal. A omplete allowed splitting-o� is a sequen e of allowed splitting-o� steps that de reasesm(V ) to zero. Observe that a solution to the degree-spe i�ed problem exists if and only ifthere is a omplete allowed splitting-o�.Let uv ∈ E(G) be a split edge and x ∈ V + and onsider the edge-swit h operation atx, u, v. If m was allowed, c(u) 6= c(x), and m′ is allowed, then the edge-swit h operation isallowed. Pi k an edge uv = e of G and (distin t) positive nodes x and y. The admissibleone- hange at x, u, v, y is an allowed one- hange (at x, u, v, y) if c(u) 6= c(x), c(v) 6= c(y)and the degree-spe i� ation after the one- hange is still allowed.After performing any of the above operations we will always repla e the fun tions m andp with the modi�ed fun tions m′ and p′. We will also reindex the lasses of the partitionP in order to maintain that m(P1) is maximal.

98 Chapter 5. Covering symmetri rossing supermodular fun tionsObsta lesIn this se tion we give the des ription of the instan es of the degree-spe i�ed Problem 5.25that satisfy the ne essary onditions (5.9)-(5.11), but still not solvable. It will be useful togive these de�nitions again in two levels, therefore �rst we give some preliminary de�nitions.Note that in the following de�nitions only the allowedness of the degree-spe i� ation m isassumed, the dimension ondition (5.12) will be a onsequen e. The following de�nitionsassume that we are given an instan e of the degree-spe i�ed version of Problem 5.25.De�nition 5.37. A C∗4 - onstru tion A = {A1, A2, A3, A4} for p is alled a C∗

4-semiobsta lefor (p, P, m) if (5.9)-(5.11) holds and m(V ) = SLB(p).De�nition 5.38. A C∗

5 - onstru tion A = {A1, A2, A3, A4, B1, . . . , Bt} for p is alled aC∗

5-semiobsta le for (p, P, m) if (5.9)-(5.11) holds, and m(V ) = SLB(p).De�nition 5.39. A C∗

6 - onstru tion A = {A1, A2, A3, A4, A5, A6} for p is alled a C∗

6-semiobsta le for (p, P, m) if (5.9)-(5.11) holds, and m(V ) = SLB(p).In other words, a semiobsta le is nothing else but a onstru tion A and an alloweddegree-spe i� ation m a hieving the lower bound m(V ) = SLB(p). Note that su h adegree-spe i� ation does not ne essarily exist (i.e. the existen e of a onstru tion does notimply the existen e of a semiobsta le).De�nition 5.40. A C∗

4 -semiobsta le A = {A1, A2, A3, A4} for (p,P, m) is alled a C∗

4-obsta le for (p, P, m) if there exists an i0 ∈ {1, 2} su h that the positive nodes of

Ai0 ∪ Ai0+2 all have olour 1.Observe that m(P1) = m(V − P1) = m(V )/2 in a C∗4 -obsta le. A C∗

4 -obsta le will be alled basi , if m(V ) = 4.De�nition 5.41. A C∗5 -semiobsta le A = {A1, A2, A3, A4, B1, . . . , Bt} for (p,P, m) is alled a C∗

5-obsta le for (p, P, m) if1. Either m(P1) = m(V − P1) and there is an i0 ∈ {1, 2} su h that m(P1 ∩ Ai0) =

m(P1 ∩ Ai0+2) = m(P1 ∩ Bj) = 1 for every j = 1, 2, . . . , t,2. Or m(P1) = m(P2) = m(V )/2−1, m(P1∩A1) = m(P1∩A3) = m(P2∩A2) = m(P2∩

A4) = 1, and there exists a j0 ∈ {1, 2, . . . , t} su h that m(P1 ∩Bj) = m(P2 ∩Bj) = 1for any j ∈ {1, 2, . . . , t} − j0, ( onsequently m((P1 ∪ P2) ∩ Bj0) = 0).De�nition 5.42. A C∗6 -semiobsta le A = {A1, A2, A3, A4, A5, A6} for (p,P, m) is alled a

C∗

6-obsta le for (p, P, m) if the positive node in Ai and in Ai+3 have the same olourfor every i = 1, 2, 3.

Se tion 5.4. Partition onstrained overing problem 99Observe that the C∗6 -obsta le is quite restri ted: it only exists with m(V ) = 6 and ifthere are at least 3 olours (i.e. |P| ≥ 3). This last statement is be ause the positive nodesin A1, A2 and A3 must have di�erent olours. Note that the de�nition of the obsta les (andthat of semiobsta les) also allows y li al reindexing and reversing the order of the sets Ai.Lemma 5.43. If a C∗

5 -semiobsta le A = {A1, A2, A3, A4, B1, . . . , Bt} is not a C∗5 -obsta le,then there exists a omplete allowed splitting-o�.Proof. Assume that A is a semiobsta le for (p,P, m) that is not a C∗

5 -obsta le. Let {ai} =

Ai ∩ V + for every i = 1, 2, 3, 4 and {bj , b′j} = Bj ∩ V + for every j = 1, 2, . . . , t (where

bj = b′j might even hold for some values of j, and V + = {v ∈ V : m(v) > 0}). We use theobservation that if t > 1 then the admissible splitting-o� at a ∈ Ai and b ∈ Bj for somei ∈ {1, 2, 3, 4} and j ∈ {1, 2, . . . , t} gives a new C∗

5 -semiobsta leA−{Ai, Bj}+{Ai∪Bj} forthe modi�ed p and m. Similarly, if t = 1 then this way we obtain a basi C∗4 -semiobsta le.Our approa h is the following: we perform an arbitrary allowed splitting-o� at some

a ∈ Ai and b ∈ Bj and we he k that the semiobsta le after this step is an obsta le ornot. If it is then we show how to modify the step in order to avoid the trouble: in most ofthe ases this modi� ation will be an allowed edge-swit h. This way we get an indu tiveproof on t and it is easy to see that the t = 0 base ase is true: there is a omplete allowedsplitting-o� in a basi C∗4 -semiobsta le whi h is not a C∗

4 -obsta le.Assume that after the allowed splitting-o� at a1 and b1, say, the partition A′ = {A1 ∪

B1, A2, A3, A4, B2, . . . , Bt} is a (C∗5 - or a basi C∗

4 -) obsta le for (p′,P, m′), where p′ andm′ are the modi�ed fun tions. In this proof we will assume that m′(P1) ≥ m′(P2) ≥ · · · ≥

m′(Pr). As usual, we will say that the olour of P1 is red. Distinguish the following ases.1. De�nition 5.41.1 (�rst option of a C∗5 -obsta le) or De�nition 5.40 (basi C∗

4 -obsta le)holds for A′ (with p′, m′ of ourse). Assume that the olour of bj is red for everyj = 2, . . . , t. There are two sub ases.(a) a2, a4 are both red. Then b′1 is not red (sin e the splitting we have performedwas allowed) and b1 is not red either (otherwise A would have been a C∗

5 -obsta le for (p,P, m) satisfying De�nition 5.41.1). Perform the edge-swit hat a2, b1, a1, i.e. repla e the edge a1b1 with the edge a2b1. Let the modi�edfun tions be m′′ and p′′: note that m′′(V ) = m′(V ) and m′′(P1) ≤ m′(P1). Thisedge-swit h is allowed, sin e c(a2) 6= c(b1) as we have seen, m′′(P1) ≤ m′(P1)and c(a1) = c(a3) = c(b′1) = c(b′2) = · · · = c(b′t) annot hold, sin e then Awould be a C∗5 -obsta le for (p,P, m) satisfying De�nition 5.41.1 with olour lass

P2. Moreover A′′ = {A1, A2 ∪ B1, A3, A4, B2, . . . , Bt} annot be an obsta le for

100 Chapter 5. Covering symmetri rossing supermodular fun tions(p′′,P, m′′), sin e that would mean that there is a olour lass P ∈ P − P1su h that a1, a3 ∈ P and b′j ∈ P for every j = 2, . . . , t (note that De�nition5.41.2 annot hold for A′′ sin e c(b′1) 6= c(a4)). But this implies that A was aC∗

5 -obsta le for (p,P, m): if m(P ∩ B1) = 1 then it satis�es De�nition 5.41.1with lass P and if m(P ∩B1) = 0 then it satis�es De�nition 5.41.2 with lassesP1 and P (m(P ∩ B1) = 2 annot be the ase sin e then m(P ) > m(V )/2).(b) b′1, a3 are both red. Here we also assume that m′(P2) < m′(V )/2, sin e them′(P1) = m′(P2) = m′(V )/2 ase was handled just before (swit h P1 and P2).Sin e c(a1) 6= c(b1), at most one of a1 and b1 is red. Moreover a1 annot be red,sin e then A would have been a C∗

5 -obsta le for (p,P, m) satisfying De�nition5.41.1 with lass P1. If b1 is red then repla e the edge a1b1 with the edge a2b′1:sin e the number of red positive nodes does not hange, this way we destroyedthe obsta le, sin e c(b′1) 6= c(a4). If b1 is not red, either, then repla e the edge

a1b1 with the edge a1b′1 (i.e. apply the edge-swit h at b′1, a1, b1). Let the modi�edfun tions be m′′ and p′′. Sin e c(a2) = c(a4) = c(b′2) = · · · = c(b′t) annot holdby m′(P2) < m′(V )/2, this edge-swit h is allowed. For the same reason, thisway we get rid of the obsta le: neither De�nition 5.41.1 nor De�nition 5.40 anhold for A′ and p′′,P, m′′ sin e c(a2) = c(a4) = c(b′2) = · · · = c(b′t) is not true,and De�nition 5.41.2 annot hold sin e c(b1) 6= c(a3).2. De�nition 5.41.2 holds for A′ and p′,P, m′. Let P1 be the olour of b′1 and a3, andlet P2 be the olour of a2 and a4. Repla e the edge a1b1 with the edge a2b

′1 (notethat this is not an edge-swit h). Let the modi�ed fun tions be m′′ and p′′. Sin e

m′′(P1) ≤ m′(P1), m′′(P2) ≤ m′(P2) and c(a1) 6= c(b1), we have m′′(Pi) ≤ m′′(V )/2for every i = 1, 2, . . . , r (that is, m′′ is allowed). Furthermore, De�nition 5.41.2 annot hold any more, sin e c(b1) = c(a4) and c(a1) = c(a3) would imply that A wasa C∗5 -obsta le for (p,P, m) satisfying De�nition 5.41.2 with olour lasses P1, P2.The following result motivates the de�nition of the obsta les.Lemma 5.44. If there is an obsta le A for (p,P, m), then there is no omplete allowedsplitting-o� (though there is a omplete admissible splitting o�).Proof. In this proof, ai will denote a node of Ai ∩V + and bi one of Bi∩V +. We �rst provethe result when there exists a C∗

4 -obsta le. The two other ases redu e to it.1. If A is a C∗4 -obsta le, we may assume that a maximum olour lass is in ident to

A1 ∪ A3. Ea h allowed splitting-o� has to ontain one positive node of A1 ∪ A3 and

Se tion 5.4. Partition onstrained overing problem 101one of A2 ∪ A4, therefore m(Ai ∪ Ai+1) − p(Ai ∪ Ai+1) may only de rease with aneven number after any sequen e of allowed splitting-o�. Then De�nition 5.26.1 andDe�nition 5.26.2 imply that m(Ai) may never de rease to zero for any i. Thereforethere is no omplete allowed splitting-o�.2. If A is a C∗5 -obsta le, then there are two ases. Sin e every set of {B1, . . . Bt} istight, an allowed splitting-o� may not ontain two positive nodes of the same set of

{B1, . . . Bt}.(a) De�nition 5.41.1 holds. Sin e the splitting-o� at ai, ai+1 is not admissible forany positive nodes (ai, ai+1) ∈ Ai × Ai+1, ea h allowed splitting-o� involves apositive node of a set Bi. Perform an allowed splitting-o� (either at ai, bj orat bi, bj, j 6= i). Then we get a C∗5 -obsta le, either A ∪ (Ai ∪ Bj) − Ai − Bj or

A∪ (Bi ∪Bj)−Bi −Bj . Note that the number of sets of the obsta le de reasedby one. Repeat until the obsta le onsists of four sets, then by Claim 5.31 it isa C∗4 -obsta le and we may apply 1.(b) De�nition 5.41.2 holds. If there exist a omplete allowed splitting-o�, then oneof its allowed splitting-o�s is at u, v with u ∈ Bj0. Perform this splitting-o�,then we get a C∗

5 -obsta le where 5.41.1 holds, a ontradi tion.3. If A is a C∗6 -obsta le, sin e Ai ∪ Ai+1 is dangerous and c(ai) = c(ai+3), the onlyallowed splitting-o� are at ai, ai+2, for i = 1, 2, 3. By (∩∪) applied to Ai ∪ Ai+1 and

Ai+1 ∪ Ai+2, we have p(Ai ∪ Ai+1 ∪ Ai+2) = m(Ai ∪ Ai+1 ∪ Ai+2) + 2. Perform anallowed splitting o� at ai, ai+2 for some i ∈ {1, 2, 3}, then Ai ∪ Ai+1 ∪ Ai+2 is tightand we are ba k to Case 1 with the C∗4 -obsta le {Ai ∪Ai+1 ∪Ai+2, Ai+3, Ai+4, Ai+5}.

Note that if A is an obsta le then φ = αp =∑

A∈A p(A).The splitting-o� theoremTheorem 5.45. Let p0 : 2V → Z∪{−∞} be a symmetri , positively rossing supermodularset fun tion, P a partition of V and m0 ∈ ZV+ an allowed degree-spe i� ation so that

12m0(V ) ≥ dim(p0)− 1. Then there exists a solution of Problem 5.25 satisfying the degree-spe i� ation m0, unless there exists a C∗

4 -, C∗5 -, or a C∗

6 -obsta le for (p0,P, m0).Proof. In the proof of this theorem we will give an algorithm.

102 Chapter 5. Covering symmetri rossing supermodular fun tionsFirst Step of the Algorithm: Perform an arbitrary sequen e of allowed splitting-o� operationsas long as there exists one.Let G be the graph of the edges split so far, p = p0 − dG and m(v) = m0(v) − dG(v) forall v ∈ V . Let us hara terize the situation when there is no further allowed splitting-o�.For symmetri , positively rossing supermodular fun tions we an extend Lemma 4.6 evenfor this partition onstrained ase.Lemma 5.46. Let p : 2V → Z ∪ {−∞} be a symmetri , positively rossing supermodularfun tion and m be an allowed degree spe i� ation. If p(X) > 1 for some X ⊆ V then thereis an allowed splitting-o�.Proof. Assume that there is no allowed splitting-o� and let Mp = max{p(X) : X ⊆ V },whi h is by assumption at least 2. Let Y be a minimal set satisfying p(Y ) = Mp. Bysymmetry, p(V − Y ) = Mp, too, so we an hoose a minimal set Z ⊆ V − Y satisfyingp(Z) = Mp. We know from the proof of Lemma 4.6 that any pair of positive nodesy ∈ Y, z ∈ Z is admissible, so there is an allowed splitting o� if we an hoose su h ared-non-red pair y ∈ Y, z ∈ Z. Assume that either all positive nodes in Y ∪ Z are of olour 1, or none of them are of olour 1. Choose x ∈ V + − (Y ∪ Z) of olour di�erentfrom 1 in the �rst ase, and of olour 1 in the se ond ase, and let y ∈ Y and z ∈ Zbe arbitrary positive nodes. If the splitting at x and y is not admissible then there is adangerous set X ontaining x and y. Sin e m(X − Y ) ≤ m(X) − m(y) ≤ m(X) − 1 andp(Y −X) < Mp by the minimality of Y , X and Y annot satisfy (−) , sin e that would meanm(X)−1+Mp ≤ p(X)+p(Y ) ≤ p(X−Y )+p(Y −X) < m(X−Y )+Mp ≤ m(X)−1+Mp, a ontradi tion. So Y ⊆ X must hold. But then X an only be dangerous if p(X) = Mp andm(X) = m(Y )+1 = Mp+1. Similarly, if the splitting at x and z is not admissible then thereexists an X ′ ontaining x and Z and satisfying p(X ′) = Mp and m(X ′) = m(Z)+1 = Mp+1.Sin e X and X ′ must ross (be ause even V + − (X ∪ X ′) is not empty), this implies thatp(X ∩ X ′) = Mp, but this ontradi ts the fa t that m(X ∩ X ′) = m(x) = 1. �By Lemma 5.46, sin e any pair v ∈ V + ∩ P1 and u ∈ V + − P1 is in a dangerous setX, we have 1 ≥ p(X) ≥ m(X) − 1 ≥ m({u, v}) − 1 ≥ 2 − 1 thus m ≤ 1. Most of thestatements of Lemma 5.20 hold here, too, the proofs are only a little more ompli ated(they are presented here for the reader's onvenien e).Lemma 5.47. If there is no allowed splitting-o�, let V + = {v1, v2, . . . , vk}. Then thefollowing hold.(i) for all i ∈ {1, 2, . . . , k} there exists a unique maximal tight set Vi ontaining vi,

Se tion 5.4. Partition onstrained overing problem 103(ii) if the splitting at vi and vj is not admissible for some i, j ∈ {1, 2, . . . , k} then the setblo king vi and vj is Vi ∪ Vj,(iii) the sets V1, V2, . . . , Vk form a partition of V .Proof. The sets that we onsider will always have positive p value, so we an use (∩∪)and (−) if two of them ross. Observe that sin e m(V ) ≥ 4, a set X blo king a pairv ∈ V + ∩ P1, u ∈ V + − P1 and another set Y blo king a pair v ∈ V + ∩ P1, w ∈ V + − P1 ross ea h other, implying that p(X∩Y ) = p(X−Y ) = 1, so every x ∈ V + is in a tight set.Similarly, if T1 and T2 are two tight sets ontaining the positive node x ∈ V + then T1 and T2 annot ross ea h other, sin e then (−) would imply that p(T1−T2) = 1 > m(T1−T2) = 0,a ontradi tion. Thus one of T1 and T2 must ontain the other, so indeed there exists aunique maximal tight set Vi ontaining vi for every i.Let i, j be two di�erent indi es between 1 and k su h that the splitting-o� at vi and vjis not admissible. It is straightforward that Vi and Vj have to be disjoint by Claim 5.17.Similarly, a set X blo king vi and vj must ontain Vi (and Vj) by Claim 5.17. On the otherhand, if l ∈ {1, 2, . . . , k} is di�erent from i and j for whi h vi and vl is not admissible (notethat su h an l exists), and Y is a set blo king vi and vl then (∩∪) implies that X ∩Y = Vi(sin e it is tight) and (−) implies that X − Y = Vj (sin e it is tight again). This �nishesthe proof of (ii).The only thing to be proved to get (iv) is that ∪k

i=1Vi = V : but if this was not the asethen Claim 1.9 would also imply that p(∪ki=1Vi) = 1, whi h would give a ontradi tion,sin e m(V − ∪k

i=1Vi) = 0 and p(V − ∪ki=1Vi) = p(∪k

i=1Vi) = 1. �If m(V ) = 4 then there may exist an admissible but not allowed splitting-o�. Howeverthis annot happen if m(V ) ≥ 6.Lemma 5.48. If m(V ) ≥ 6 and there is no allowed splitting-o�, then there is no admissiblesplitting-o�, either.Proof. Let us introdu e the nonadmissibility graph N on node set V + that ontains anedge between two nodes x, y ∈ V + (sharing possibly the same olour) if and only if thesplitting-o� at x and y is not admissible. By the arguments above, if there is no allowedsplitting-o� then any pair v ∈ V + ∩P1, u ∈ V + −P1 is onne ted with an edge in N . Notethat for any nonempty subset Z ( V + satisfying that N [Z] is onne ted we have thatp(∪vi∈ZVi) = 1 by Claim 1.9. This implies that if vi, vj ∈ V + is any pair and N − {vi, vj}is onne ted, then vivj is also an edge of N , sin e p(Vi ∪ Vj) = 1 by the symmetry of p.We have to prove that N is a lique in this ase. Let us �rst onsider a pair x, y ∈ V +−P1:sin e V +−P1−{x, y} is not empty by our assumptions (sin e m is allowed and m(V ) ≥ 6),

104 Chapter 5. Covering symmetri rossing supermodular fun tionsit is easy to he k that N − {x, y} is onne ted, so xy ∈ E(N). Finally, if x, y ∈ V + ∩ P1then by the pre eding argument N − {x, y} is also onne ted, so again xy ∈ E(N). �Se ond Step of the Algorithm, First Main Case: If there does not exist an admissible splitting-o�, then perform an arbitrary sequen e of allowed one- hange operations, until m(V )de reases to 2, when the pro edure an be �nished with a last allowed splitting-o�. Outputthe graph found and terminate.Let us prove the orre tness of this part of the algorithm. We an use the observationsmade after the First Step of Algorithm SYMCROS_COVER. One su h observation is thatthere are no split edges between the lasses of our tight partition. Also, by Lemma 5.20(iii), an allowed one- hange operation will not reate an admissible splitting-o�, unlessm(V ) be omes 2. The key observation is given in the following Lemma.Lemma 5.49. Assume that there is no admissible splitting-o� and there is no allowedone- hange. Then there does not exist an admissible one- hange, either.Proof. Let uv ∈ E(G): sin e there is no admissible splitting-o�, uv ⊆ Vi for some i. We willshow that there is no admissible one- hange using this edge. Let vr ∈ V +∩P1, vs ∈ V +−P1be di�erent from vi. By possibly ex hanging u and v we an a hieve that u /∈ P1 and the olour of v and vs is di�erent. Sin e the one- hange at vr, u, v, vs is not allowed, there mustexist a swit hblo king set X, without loss of generality we assume that it is (the unique)(vr, u, v)-swit hblo king set. By Claim 5.21 there is no admissible one- hange at this edge,sin e p(Vj ∪ (X ∩ Vi)) = 0 for all j 6= i, what was to be proved. �This lemma shows that we an again refer to the arguments given in the previous se tion,in parti ular we an thus prove the orre tness of this ase of the algorithm. We only needto prove that this ase of the algorithm annot get stu k, we will always �nd an allowedone- hange, unless m(V ) de reases to 2. But if this was not the ase, then the steps thatwe have done so far an be onsidered as a running of the Algorithm SYMCROS_COVER,and by the previous lemma the Algorithm SYMCROS_COVER would get stu k after thesesteps, too. But the Algorithm SYMCROS_COVER annot get stu k with su h an input asours, a ontradi tion.Now we an ontinue with the des ription of the Se ond Main Case of the se ond stepof the algorithm.Se ond Step of the Algorithm, Se ond Main Case: If there exists an admissible splitting-o�(in whi h ase m(V ) = 4 by Lemma 5.48), then try to �nd an allowed one- hange. If this

Se tion 5.4. Partition onstrained overing problem 105is found then �nish the pro edure with a last allowed splitting-o�, output the graph foundand terminate.Before ontinuing the des ription of the algorithm �rst we prove some lemmas about the ase when there is no allowed one- hange. Let the graph of the edges split so far be denotedby G, p = p0 − dG and m(v) = m0(v) − dG(v) for every v ∈ V . Let V + = {v1, v2, v3, v4}indexed in the way su h that the splitting-o� at v1 and v3 is admissible (and then of oursethe one at v2, v4 is admissible, too, but either c(v1) = c(v3), or c(v2) = c(v4)), and letVi be the maximal tight set ontaining vi for all i. Sin e the splitting-o� at v1 and v3 isadmissible (but not allowed), G might ontain some edges between the lasses Vi, but onlybetween onse utive ones (i.e. between Vi and Vi+1 for some i ∈ {1, 2, 3, 4}).By Lemma 5.19 (2d) the obsta le of the admissibility of the one- hange along an edgeuv ∈ E(G) indu ed in one of the sets Vi and two positive nodes distin t from vi is aswit hblo king set. Furthermore, we an show the following.Lemma 5.50. Assume that e = uv is an edge of G indu ed in V1, say. Then there is no(v3, e)-swit hblo king set, onsequently either the one- hange at v3, u, v, v2, or the one atv3, v, u, v2 is admissible. (The same is true with v4 instead of v2 here!)Proof. Assume indire tly that X is su h a swit hblo king set: but sin e X and V1 ross ea hother this implies that p(X ∪V1) = 1, ontradi ting the admissibility of the splitting-o� atv1 and v3. �Lemma 5.51. Assume that uv is an edge of G su h that u ∈ V1 and v ∈ V2. Then eitherc(u) = c(v1) = c(v3) or c(v) = c(v2) = c(v4).Proof. Otherwise there would be an allowed one- hange. Without loss of generality we anassume that c(v1) = c(v3) = 1. Assume that u /∈ P1. If c(v) 6= c(v4) then the one- hangeat v3, u, v, v4 would be allowed, so c(v) = c(v4). If c(v2) 6= c(v4) then the one- hange atv3, u, v, v1 is allowed by Lemma 5.19 (3 ). �The next lemma is about a ase when there is an edge indu ed in V1.Lemma 5.52. Assume that the one- hange at v2, u, v, v4 is admissible where the edge uvis indu ed in V1. Then there exists a (v2, v, u)-swit hblo king set X2, and a (v4, u, v)-swit hblo king set X4. These sets are disjoint, p(V1 − X2) = p(V1 − X4) = 0 and p(V1 −

(X2 ∪ X4)) = 1. Furthermore c(v2) = c(u) and c(v) = c(v4) both must hold, onsequentlythe olour of v2 and that of v4 must be di�erent.

106 Chapter 5. Covering symmetri rossing supermodular fun tionsProof. The existen e of X2 follows from Lemma 5.50: if it does not exist then both the one- hanges at v3, u, v, v2 and at v3, v, u, v2 are admissible and one of them would be allowed.Similarly for X4. If X2 ∩ X4 6= ∅ then (1.5) for these two sets give that p(X2 ∪ X4) = 1,whi h annot be the ase by Claim 5.17. Let Yi = V1 − Xi for both i = 2, 4: (−) forV1 and Xi gives that p(Yi) ≥ 0 for both i = 2, 4. Now (1.5) for Y2 and Y4 gives thatp(Y2) = p(Y4) = 0 and p(Y2 ∩ Y4) = 1, as laimed, sin e Y2 ∩ Y4 = V1 − (X2 ∪ X4).To prove the last statement assume that for example c(v2) 6= c(u): then the one- hangeat v2, u, v, v3 would be allowed, unless c(v) = c(v3), but in this ase the one- hange atv3, u, v, v4 would be allowed, a ontradi tion. �End of the Algorithm: Let V1, V2, V3, V4 be the partition into maximal tight sets and vi ∈ Vibe the positive nodes, s.t. v1 and v3 is admissible (note that we ontinue the des ription ofthe Se ond Step, Se ond Main Case).CASE A There is an edge e ∈ E(G) indu ed in a lass Vi and another edge f ∈ E(G)between onse utive lasses: �nd a omplete allowed splitting-o� by unsplitting these twoedges.CASE B Every edge of G goes between onse utive lasses Vj and Vj+1 (in luding the ase E(G) = ∅): if either c(v1) = c(v3) = c(e ∩ (V1 ∪ V3)) for every edge e ∈ G, orc(v2) = c(v4) = c(e ∩ (V2 ∪ V4)) for every edge e ∈ G, then {V1, V2, V3, V4} was a C∗

4 -obsta le, otherwise a omplete allowed splitting-o� an be found (by unsplitting only 2edges of G).CASE C Every edge of G is indu ed in a lass of the partition {V1, V2, V3, V4}.Sub ase (i) There is an edge uv indu ed in V1, say, for whi h the one- hange atv2, u, v, v4 is admissible. If uv is the only edge of G then there was a C∗

6 -obsta le,otherwise a omplete allowed splitting-o� an be found (by unsplitting 2 edges).Sub ase (ii) For any edge uv of G indu ed in some Vi, the one- hange neither atvi−1, u, v, vi+1 nor at vi−1, v, u, vi+1 is admissible. Then there exists a partition satisfyingC∗

5 -semiobsta le in the input: if it is not a C∗5 -obsta le (whi h is easy to he k) then �nda omplete allowed splitting-o�.Let us �rst justify CASE A.Lemma 5.53. If G ontains an edge indu ed in some set Vi, and another edge betweensome Vj and Vj+1 then we an �nd a omplete allowed splitting-o� after unsplitting thesetwo edges.Proof. Assume that e = uv is an edge of G indu ed in V1, say, and G also ontains an edge

f between two lasses, too. We �rst prove the following laim.

Se tion 5.4. Partition onstrained overing problem 107Claim 5.54. It is not possible that (vi, u, v)-swit hblo king sets exist for both i = 2 andi = 4. (Consequently, either the one- hange at v2, u, v, v4, or the one at v2, v, u, v4 isadmissible.)Proof. Assume indire tly that Xi is a (vi, u, v)-swit hblo king set for both i = 2 and 4 and onsider two ases. In both ases we an assume that one endpoint of f is in V2.In the �rst ase the edge f is between V2 and V3: apply (∩∪) for X2 and X4 to get thatp(X2 ∪ X4) = 0, then apply (∩∪) for X2 ∪ X4 and V3 ∪ V4: using the edge f you get a ontradi tion. In the other ase the edge f is between V1 and V2: apply (∩∪) for X2 andV1 to get that f must be indu ed in X2 and then apply (−) for X4 and V1 to see that f annot enter X4. This ontradi ts Lemma 5.19 (2b).By possibly ex hanging the role of u and v we an assume that the one- hange atv2, u, v, v4 is admissible. Let f = xy su h that y ∈ V2: by Lemma 5.51 and Lemma 5.52,x must be red in this ase. We have again two ases: either x ∈ V1 or x ∈ V3. In both ases G − e − f + ux + vv3 + v2v4 + v1y is a graph that satis�es our requirements. This an be justi�ed the following way. By Lemma 5.52, there exists a (v2, v, u)-swit hblo kingset X2, and a (v4, u, v)-swit hblo king set X4. Applying (∩∪) for X2 and V2 ∪ V3 givesthat p(V3 ∪ X2) = 0. Similarly, (∩∪) for X4 and V1 gives that p(X4 ∩ V1) = 0. Nowapply the edge-swit h operation at v3, v, u, whi h is learly allowed by Lemma 5.50 andlet p′ and m′ be the modi�ed fun tions. By the pre eding observations, the partitionV1 − (X2 ∪ X4), X4 ∩ V1, V3 ∪ X2, V4 is a p′-tight partition ( learly p′ ≤ 1). Now we laimthat the one- hange at v1, y, x, u be ame admissible. By Lemma 5.19 (2d) we only have toshow that there is no (v1, y, x)-swit hblo king set, and there is no (u, x, y)-swit hblo kingset, either (with respe t to p′ and m′, of ourse). If X was a (v1, y, x)-swit hblo king setthen (1.6) for X and V2 and p′ would give a ontradi tion (note that p′(V2) = p(V2) = 1).On the other hand if X was a (u, x, y)-swit hblo king set then (∩∪) for X and V3 ∪ X2would give that p′(V3 ∪ X2 ∪ X) = 1 = p(V3 ∪ X2 ∪ X), ontradi ting Claim 5.17 (appliedto p), sin e this set rosses V1. Figure 5.4.2 illustrates the proof. �We are left with two ases: either all edges of G go between ( onse utive) lasses ofV1, V2, V3, V4, or all of them are indu ed in the lasses of V1, V2, V3, V4. Let us �rst des ribeCASE B, when every edge of G goes between ( onse utive) lasses of the partitionV1, V2, V3, V4.Lemma 5.55. Assume that G ontains only edges between the lasses Vi. If either c(v1) =

c(v3) = c(e ∩ (V1 ∪ V3)) for every edge e ∈ G, or c(v2) = c(v4) = c(e ∩ (V2 ∪ V4)) forevery edge e ∈ G then {V1, V2, V3, V4} was a C∗4 -obsta le, otherwise a omplete allowedsplitting-o� an be found by unsplitting two edges of G.

108 Chapter 5. Covering symmetri rossing supermodular fun tions

u

v

yx

v1

v2

v3v4

V1 V2

V3V4

X2

X4

Figure 5.2: Illustration for the proof of Lemma 5.53.Proof. Assume that there is an edge e su h that c(v1) 6= c(e∩ (V1∪V3)) and an edge f su hthat c(v2) 6= c(f∩(V2∪V4)). These imply by Lemma 5.51 that c(v2) = c(v4) = c(e∩(V2∪V4))and c(v1) = c(v3) = c(f ∩ (V1 ∪ V3)). Assume without loss of generality that e = uv whereu ∈ V1 and v ∈ V2.Apply the edge-swit h operation at v1, v, u whi h is allowed by Lemma 5.19 (3a) andnote that the fun tion value of the sets Vi and Vi∪Vi+1 (i = 1, 2, 3, 4) does not hange afterthis edge-swit h. Furthermore, sin e the splitting at v3 and u is admissible after this edge-swit h by Lemma 5.19 (3 ) (sin e the one- hange at v3, u, v, v1 was admissible originally),this means that we are again at the Se ond Main Case of our algorithm (with the sametight partition V1, V2, V3, V4), and by Lemma 5.51 we reated an allowed one- hange usingedge f . �Now we an des ribe CASE C, when every edge of G is indu ed in the lassesof the partition V1, V2, V3, V4. The �rst sub ase of this ase is when there is an edge uvindu ed in V1, say, for whi h the one- hange at v2, u, v, v4 is admissible. The next lemmadeals with this ase.Lemma 5.56. Assume that G ontains and edge uv indu ed in V1 and the one- hange atv2, u, v, v4 is admissible. If uv is the only edge of G then there was a C∗

6 -obsta le, otherwise

Se tion 5.4. Partition onstrained overing problem 109a omplete allowed splitting-o� an be found by unsplitting uv and an arbitrary other edgeof G.Proof. By Lemma 5.52 there exists a (v2, v, u)-swit hblo king set X2, and a (v4, u, v)-swit hblo king set X4. Let Yi = V1 − Xi for i = 2, 4 and let A1 = Y2 ∩ Y4, A2 = Y4 − Y2 =

V1∩X2, A3 = V2, A4 = V3, A5 = V4 and A6 = Y2−Y4 = V1∩X4. By Lemma 5.52 p(A1) = 1,and similarly p(A2) = p(A6) = 0 follows from (−) for Y1 and Y2. These together with theprevious observations give that by unsplitting e the sets A1, A2, . . . , A6 form a C∗6 -obsta lefor p′,P, m′ (where p′ = pe and m′ = me are the modi�ed fun tions after the unsplitting).Thus if e is the only edge of G then the proof is ompleted.Assume that G ontains another edge xy. Observe that m′(P1) = m′(P2) = m′(P3) = 2and p′ ≤ 1 again holds (if p′(X) = 2 for some set X then X ⊆ V1 and e enters X, say

v ∈ X and u /∈ X, but then (∩∪) for X and X2 and p gives a ontradi tion). Let ai ∈ Aibe the positive node in Ai (i.e. a1 = v1, a2 = v, a3 = v2, a4 = v3, a5 = v4, a6 = u). Reindex( y li ally) the sets su h that x ∈ A1. Consider the following two ases.CASE I: If y /∈ A1, then we an assume that y ∈ A2. Perform the allowed splitting-o� ata3 and a5 and observe that we arrive in the situation given in Lemma 5.53, so there is a omplete allowed splitting-o� sequen e (note that it is found by unsplitting two edges: uvand xy).CASE II: If y ∈ A1, too. Let the olour of a1 be red. Sin e A1, A2, . . . , A6 forms a C∗

6 -obsta le for p′,P, m′, there is no (ai, xy)-swit hblo king set for i ∈ {3, 4, 5} (if X was su ha set then (∩∪) for X and A1 would give that p′(X ∪A1) = p(Ai ∪A1) = 1, ontradi tingDe�nition 5.28.3). This implies by Lemma 5.19 (2d) that hoosing two of a3, a4 and a5(say ai and aj) the one- hange at ai, x, y, aj is admissible. There are two sub ases: if oneof x and y is red (say x) then the one- hange at a3, x, y, a4 and the one at a5, x, y, a4 areboth allowed. In the other sub ase none of x and y is red: then we an assume thatc(x) 6= c(a5) and c(y) 6= c(a3) (if any of these does not hold then swit h x and y), so boththe one- hanges at a5, x, y, a4 and the one at a4, x, y, a3 are allowed. It remains to he kthat in both sub ases one of the two allowed one- hanges will result that the splitting ata2 and a6 be omes admissible, thus a omplete allowed splitting-o� sequen e is found. �The only remaining ase is when, for any edge uv of G indu ed in (say) V1 theone- hange neither at v2, u, v, v4 nor at v2, v, u, v4 is admissible. In other words, for everysu h edge there are sets X2, X4, su h that (after possibly ex hanging u and v) Xi is a(vi, u, v)-swit hblo king set for both i = 2 and 4. By Lemma 5.19 (2b), X2 ∩V1 = X4 ∩ V1,so let Xuv = X2 ∩ V1. Furthermore Xi = Xuv ∪ Vi for both i = 2, 4. The set Xe an bede�ned the same way for any e indu ed in some other Vi. In what follows we will restri t

110 Chapter 5. Covering symmetri rossing supermodular fun tionsourselves to the edges of G indu ed in V1. Observe that p(V2 ∪ V4 ∪ Xe) = 0 for an edgee ⊆ V1.Lemma 5.57. If e and f are two edges of G indu ed in V1 then one of Xe and Xf ontainsthe other.Proof. They annot ross ea h other by the same proof as that of Claim 5.22 ( hoosevr = v2 and vs = v4 in that proof). Suppose that they are disjoint. Observe that p(V2 ∪

V4 ∪ Xe ∪ Xf) = 0 (apply (∩∪) for the rossing sets V2 ∪ V4 ∪ Xe and V2 ∪ V4 ∪ Xf ) andp(V3 ∪ V4 ∪Xe) = 0 (apply (∩∪) for the rossing sets V3 ∪ V4 and V4 ∪Xe). Apply (−) forthese two sets to get that p(V2 ∪ Xf) = 1, a ontradi tion. �So the edges of G indu ed in V1 an be �ordered�: E(G[V1]) = {e1, e2 . . . , el} su h that1 ≤ i < j ≤ l implies Xei

( Xej. For all i ∈ {1, 2, . . . , l} let Bei

= Xei+1− Xei

(whereXel+1

= V1 here!). The sets Be an be similarly de�ned for edges e indu ed in other setsVi (i ≥ 2). For all i = 1, 2, 3, 4 let us de�ne Ai = Vi −

⋃

e⊆ViBe (e.g. A1 = Xe1

). Figure5.4.2 is an illustration.e2

e3

el

e1

e

V3V4

Be2

Be

Xe

Xe2

V1 V2v1 v2

v4v3Figure 5.3: An illustration for the se ond sub ase of CASE C

Se tion 5.4. Partition onstrained overing problem 111The next lemma �nishes the proof of the theorem.Lemma 5.58. The partition {A1, A2, A3, A4} ∪ {Be : e ∈ E(G)} is a C∗5 -semiobsta le.Proof. We need to he k that the onditions of De�nition 5.27 (with, of ourse set fun tion

p0 and degree-spe i� ation m0) hold. First note that p0(Vi−1 ∪Ai) = p0(Vi+1 ∪Ai) = 1 forall i = 1, 2, 3, 4. Furthermore we will use the following observations.Claim 5.59. For every edge e ⊆ Vi we have p0(Vi−1 ∪ Be) ≥ 1 and p0(Vi+1 ∪ Be) ≥ 1.Proof. Assume that i = 1 and e = ej for some j ∈ {1, 2, . . . , l}. Apply (−) for V4 ∪ Xej+1and V2 ∪ Xejto get the statement.It is easy to see that De�nition 5.27.1 holds: for example apply (∩∪) for V2 ∪ A1 and

V4∪A1 to get that p0(A1) ≥ 1, and use that p0(A1) ≤ m0(A1) = 1. The proof of De�nition5.27.2 will be given later.Let us prove 3 of De�nition 5.27. Again, without loss of generality it su� es to show thatp0(A1 ∪A2) = p0(A1 ∪Be) = 1 for all e ∈ E(G). We will show that p0(V − (A1 ∪Be)) = 1(the proof of p0(A1 ∪ A2) = 1 is analogous), so let X = V − (A1 ∪ Be). Observe that itis enough to show that p0(X) ≥ 1, by (∩∪) for X and Y = Vj ∪ Be (where Vj ⊆ X andj ∈ {2, 4}) give that p0(X) ≤ p0(X ∪ Be) = p0(A1) = 1. In order to prove that p0(X) ≥ 1we will use Claim 1.9: assume that e ⊆ Vi (where i might even be 1) and apply Claim1.9 for the subpartition Q, if i = 1, and for the subpartition Q ∪ {Ai}, if i 6= 1, whereQ = {Vj : j ∈ {2, 3, 4} − i} ∪ {Bf : f ⊆ Vi ∪ V1, f 6= e}. By he king the ases a ordingto the value of i one an see that Claim 1.9 an indeed be applied.To prove De�nition 5.27.4 assume indire tly that p0(A1 ∪ A3) ≥ 1 and apply Claim 1.8for the fun tion p0 and the sets A1 ∪ A3, {A1 ∪ Be : e ⊆ V1} and {A3 ∪ Be : e ⊆ V3}: bythe previous part we get that p0(V1 ∪ V3) = p(V1 ∪ V3) ≥ 1, ontradi ting the admissibilityof v1 and v3.Finally we prove De�nition 5.27.2. Note that p0(A1∪A2∪A3) ≥ 1 follows from (∩∪) forA1∪A2 and A2∪A3, and similarly p0(A3∪A4∪A1) ≥ 1. Apply (∩∪) for these two sets to getthat 1 + 1 ≤ p0(A1 ∪A2 ∪A3) + p0(A3 ∪A4 ∪A1) ≤ p0(A1 ∪A3) + p0(A1 ∪A2 ∪A3 ∪A4) ≤

0 + p0(A1 ∪ A2 ∪ A3 ∪ A4), by De�nition 5.27.4. Choose an arbitrary e ∈ E(G) andapply Claim 1.8 for sets A1 ∪ A2 ∪ A3 ∪ A4 and A1 ∪ Bf (f ∈ E(G) − e) to get that2 ≤ p(A1 ∪ A2 ∪ A3 ∪ A4 ∪

⋃

f∈E(G)−e Bf) = p(Be). Sin e p(Be) ≤ m(Be) = 2, De�nition5.27.2 is proved. �Now it is easy to he k whether the partition found is indeed a C∗5 -obsta le by he kingthe olours: if it is then there is no omplete allowed splitting-o�, otherwise one an �nda omplete allowed splitting-o� by Lemma 5.43.

112 Chapter 5. Covering symmetri rossing supermodular fun tions�5.4.3 The minimum versionIn this se tion, we are given a symmetri , positively rossing supermodular fun tion on

V and a partition P = {P1, . . . Pr} of V . We show how to �nd a graph having its edgesbetween di�erent members of P and overing p with a minimum number of edges. Themain on ern here is to �nd an allowed degree spe i� ation m minimizing m(V ) su h thatm(V )/2 ≥ dim(p) − 1 also holds, and to avoid getting an obsta le.First, we explain how to �nd an allowed degree spe i� ation m a hieving the lowerbound m(V ) = 2φ: su h a degree-spe i� ation will be alled an optimal extension. Then,we des ribe the instan es for whi h the lower bound φ may not be a hieved in the minimumversion of Problem 5.25, that is the on�gurations. Finally, we prove the theorem solvingthis problem, see Theorem 5.66.ExtensionFor u ∈ V let Xu be a minimal tight set ontaining u (if u is not ontained in a tight setthen Xu := V ). If the dependen e on the degree spe i� ation m has to be emphasized,then we write Xm

u . The following lemma will be often used in this se tion.Lemma 5.60. Let m be an admissible degree-spe i� ation. Let u be a positive node andu′ ∈ Xm

u . Then m′ := m − χ{u} + χ{u′} is again admissible, and Xm′

u′ = Xmu .Proof. If u is not in a tight set then there is nothing to prove. Otherwise, we know that

Xmu is the unique minimal tight set ontaining u, proving the �rst statement of the lemma.To prove the se ond statement observe that Xm

u is also m′-tight, thus Xm′

u′ ⊆ Xmu . Sin e aset X with u /∈ X ∋ u′ annot be m′-tight by the admissibility of m and the de�nition of

m′, this �nishes the proof.Note that the lemma also implies that if u, v are positive nodes with Xu ∩ Xv = ∅ andu′ ∈ Xu, v′ ∈ Xv, then m − χ{u} + χ{u′} − χ{v} + χ{v′} is also admissible.An allowed degree-spe i� ation m satisfying the dimension ondition (5.12) and a hiev-ing the lower bound m(V ) = 2φ is alled an optimal extension for (p, P). Below, wedes ribe how to �nd an optimal extension. The algorithm is formulated in a way su h thatany optimal extension an be its output.Algorithm OPT_EXTbegin

Se tion 5.4. Partition onstrained overing problem 113INPUT: A symmetri , positively rossing supermodular fun tion p : 2V → Z ∪

{−∞} (given with a maximizing ora le) and a partition P = {P1, P2, . . . , Pr} of V .OUTPUT: A degree-spe i� ation m : V → Z+ satisfying (5.9)-(5.12) with m(V ) = 2φ.1.1. Pi k m ∈ C(p) ∩ ZV su h that m(V ) is minimum.1.2. If m(V ) is odd, let m := m + χ{u} for some u ∈ V . Then m(V ) = 2αp.1.3. If m(V ) < 2(dim(p)− 1) (whi h an be he ked using Algorithm SYMCROS_COVER)then let m = m+m′ with an arbitrary m′ ∈ ZV+ satisfying m′(V ) = 2(dim(p)−1)−m(V ).1.4. While m(P1) > ⌈m(V )

2⌉ (where (5.13) is assumed) do1.5. If Xu * P1 for some u ∈ P1 ∩ V + then let m := m − χ{u} + χ{u′}, for some

u′ ∈ Xu − P1,1.6. Otherwise let m = m + m′ with an arbitrary m′ ∈ ZV+ satisfying m′(V ) = 2m(P1)−

m(V ) and m′(P1) = 0.endIt is lear that the algorithm above outputs an optimal extension. Note that if either ofStep 1.2, 1.3 or 1.6 holds then m(V ) > SLB(p), therefore there is no obsta le for (p,P, m).Con�gurationsIt is possible that Algorithm OPT_EXT has found an optimal extension but we annot �nda omplete allowed splitting-o�, sin e there is an obsta le. In this se tion we show how wetry to modify the optimal extension in order to getting rid of this obsta le, and we des ribethe stru tures ( alled on�gurations) where this problem annot be avoided sin e anyoptimal extension ontains an obsta le. First we give the de�nition of the on�gurations.In order to simplify the de�nitions below, we introdu e the following de�nitions. A pair(X1, X2) of disjoint sets of V is alled a P -pair if P ∈ P and there exist a subpartitionFi of Xi su h that ∑

X∈Fip(X) = p(Xi) for i = 1, 2 and F1 ∪F2 is a subpartition of P . Asubpartition X of V is alled P -subpartition if P ∈ P and there exist a set X ′ ⊆ X forevery X ∈ X su h that p(X ′) = 1 and ⋃

X∈X X ′ ⊆ P .De�nition 5.61. A C∗4 - onstru tion A = {A1, A2, A3, A4} for p is alled a C∗

4- on�gu-ration for (p, P) if βi

p ≤ αp for every i = 1, 2, . . . , r and there exist P ∈ P and i0 ∈ {1, 2}su h that (Ai0 , Ai0+2) is a P -pair.De�nition 5.62. A C∗5 - onstru tion A = {A1, A2, A3, A4, B1, . . . Bt} for p is a C∗

5- on�g-uration for (p, P) if βi

p ≤ αp for every i = 1, 2, . . . , r and1. Either there exist P ∈ P and i0 ∈ {1, 2} su h that (Ai0 , Ai0+2) is a P -pair and{B1, . . . Bt} is a P -subpartition,

114 Chapter 5. Covering symmetri rossing supermodular fun tions2. Or there exist P ′, P ′′ ∈ P and j0 ∈ {1, . . . t} su h that (A1, A3) is a P ′-pair, (A2, A4)is a P ′′-pair, and {Bj , j 6= j0} is both a P ′- and a P ′′-subpartition.De�nition 5.63. A C∗6 - onstru tion A = {A1, A2, A3, A4, A5, A6} for p is a C∗

6- on�gu-ration for (p, P) if there exist 3 distin t olour lasses P1, P2, P3 su h that (Ai, Ai+3) isa Pi-pair for i = 1, 2, 3.Note that βi

p ≤ 2 holds for every i = 1, 2, . . . , r in a C∗6 - on�guration, sin e otherwiseSLB(p) ≥ 7 would follow from De�nition 5.63. There is a strong relation between on�gu-rations and obsta les, whi h is shown in the following two lemmas.Lemma 5.64. If a on�guration exists for (p,P), then for every optimal extension m,there exists an obsta le for (p,P, m).Proof. Let A be a on�guration for (p,P) and m an optimal extension for (p,P). Sin e

∑

A∈A p(A) ≤∑

A∈A m(A) = m(V ) = 2φ =∑

A∈A p(A), we have m(A) = p(A) for allA ∈ A. This implies the lemma.We mention that if A is a C∗

5 - on�guration satisfying De�nition 5.62.2 then this doesnot ne essarily imply that an optimal extension will satisfy De�nition 5.41.2: it is evenpossible that |P| = 2!Lemma 5.65. If no on�guration exists for (p,P), then there is an optimal extension msu h that no obsta le exists for (p,P, m).Proof. Let m be an optimal extension for (p,P) and suppose that A is an obsta le for(p,P, m). Throughout this proof, we will very often repla e m by m − χ{u} + χ{u′} foru′ ∈ Xu, using impli itly Lemma 5.60. In every ase below we will show how to modify theoptimal extension m this way in order to destroy the obsta le A. Sin e A is the unique onstru tion for p by Lemma 5.36, this way we make sure that no obsta le exists any more.Note that Xu ⊆ A for every A ∈ A and u ∈ A ∩ V +.In this proof we will assume that m(P1) ≥ m(P2 ≥ · · · ≥ m(Pr) and we will use thatthe tie an be broken arbitrarily. We will say that P1 is red and P2 is blue. Note that iffor example A is a C∗

4 -obsta le but not a C∗4 - on�guration then there exists a red positivenode u su h that Xu * P1. As there are three di�erent possible obsta les, there are three ases.1. If A is a C∗

4 -obsta le, then De�nition 5.61 does not hold and for every maximum olour lass Pi there exists a positive node u ∈ Pi su h that Xu * Pi. There are two ases.

Se tion 5.4. Partition onstrained overing problem 115(a) If m(P1) > m(P2), then repla e m by m − χ{u} + χ{u′}, for some u′ ∈ Xu − P1.(b) If m(P1) = m(P2), then pi k ui ∈ Pi su h that Xui* Pi, for i = 1, 2. Repla e

m by m − χ{u1} + χ{u′

1} − χ{u2} + χ{u′

2} for some u′

i ∈ Xui− Pi, for i = 1, 2.After these modi� ations De�nition 5.40 does not hold any more, therefore A is nota C∗

4 -obsta le anymore.2. A is a C∗5 -obsta le. Note that if m(P1) = m(V )

2in a C∗

5 -obsta le, then two oppositesets Ai, say A1 and A3, and every Bi have exa tly one red positive node. There arefour ases, depending on m(P1) and m(P2). In ea h ase, we will just show that wemay destroy both De�nition 5.41.1 and De�nition 5.41.2. Visually, De�nition 5.41.1 orresponds to A1, A3 and every Bi has exa tly one red positive node, and De�nition5.41.2 to Bj0 has no red and no blue positive node, A1, A3 have a red positive node,A2 and A4 have a blue positive node and every Bj, j 6= j0 has exa tly one red andone blue positive node.(a) m(P1) = m(V )

2and m(P2) < m(V )

2−1. Sin e De�nition 5.62.1 does not hold, thereexists a red positive node u su h that Xu−P1 6= ∅. Repla e m by m−χ{u}+χ{u′}with u′ ∈ Xu − P1. Then no olour lass P may satisfy m(P ) = m(V )

2. Supposethat now De�nition 5.41.2 holds, then ne essarily u′ ∈ P2 and m(P2) = m(V )

2−1.But either u was a positive node of Ai for some i ∈ {1, 3} and now the positivenode of Ai is blue whereas the positive node of Ai+2 is still red. Thus De�nition5.41.2 is not satis�ed. Or u belongs to some Bi and now Bi has no red positivenode. Therefore De�nition 5.41.2 does not hold.(b) m(P1) = m(P2) = m(V )

2− 1, then every Bj 6= Bj0 has exa tly one red andone blue positive node, and Bj0 has no red and no blue positive node. Sin eDe�nition 5.62.2 does not hold, we may assume that there exists a red positivenode u su h that Xu − P1 6= ∅. Repla e m − χ{u} + χ{u′} with u′ ∈ Xu − P1.Then De�nition 5.41.2 does not hold anymore, and of ourse 5.41.1 annot holdeither.( ) m(P1) = m(V )

2and m(P2) = m(V )

2− 1. Sin e De�nition 5.62.1 does not hold,there exists a red positive node u su h that Xu −P1 6= ∅. If there is su h a nodewith Xu−P1−P2 6= ∅, then repla e m by m−χ{u}+χ{u′} with u′ ∈ Xu−P1−P2.Then De�nition 5.41.1 does not hold, and either De�nition 5.41.2 holds and thisis Case (b), or De�nition 5.41.2 does not hold.Now, suppose for every red positive node, we have Xu ⊂ P1 ∪ P2. Re all thatat least one red positive node u satis�es Xu ∩ P2 6= ∅.

116 Chapter 5. Covering symmetri rossing supermodular fun tionsIf su h a node is in A1 or A3, then repla e m by m−χ{u}+χ{u′} with u′ ∈ Xu∩P2.Now P2 is the only olour lass satisfying m(P2) = m(V )2

but one of A1, A3 hasa blue positive node and the other a red one, therefore neither De�nition 5.41.1nor 5.41.2 may hold.Otherwise, let u be a red positive node in Bi su h that Xu ∩P2 6= ∅ and repla em by m− χ{u} + χ{u′} with u′ ∈ Xu ∩ P2. Then De�nition 5.41.1 does not holdfor red. If afterwards De�nition 5.41.1 holds for blue, then A2 and A4 have ablue positive node and every Bj 6= Bi has exa tly one red and one blue positivenode. Ne essarily, the positive node in Bi beside u′ is neither blue nor red. Butnow sin e De�nition 5.62.2 does not hold, there exists a positive node v /∈ Bisu h that Xv − P 6= ∅ for some P ∈ {P1, P2}. If v belongs to some Ai then v isblue and it is identi al with the ase above for red.Now v ∈ Bj 6= Bi. If v is red, then the modi� ation we have performed was nota good one, therefore we will undo it and perform an other one: repla e m bym− χ{u′} + χ{u} − χ{v} + χ{v′} with v′ ∈ Xv −P1. Then Bi has no blue positivenode and Bj has no red one, hen e none of De�nition 5.41.1 and De�nition5.41.2 may hold. Otherwise v is blue, then repla e m by m − χ{v} + χ{v′} withv′ ∈ Xv − P2. Now Bi has no red positive node and Bj has no blue ones, hen eneither De�nition 5.41.1 nor 5.41.2 holds.(d) m(P1) = m(P2) = m(V )

2, then every Bi has exa tly one red and one blue positivenode. Sin e De�nition 5.62.1 does not hold for P1, there exists a red positivenode u su h that Xu − P1 6= ∅. If Xu − P1 − P2 6= ∅, then repla e m by

m − χ{u} + χ{u′} with u′ ∈ Xu − P1 − P2 and this is Case ( ). Otherwise, letu be a red positive node su h that Xu ∩ P2 6= ∅. Sin e De�nition 5.62.1 doesnot hold for P2, there exists a blue positive node v su h that Xv − P2 6= ∅. IfXv−P2−P1 6= ∅, we dealt with this ase just before (for P1). Otherwise, repla em by m− χ{u} + χ{u′} − χ{v} + χ{v′} with u ∈ Xu ∩P2 and v′ ∈ Xv ∩ P1. Thereare three ases.i. If at least one of u and v is in some Ai then we destroyed De�nition 5.41.1,therefore A is not a C∗

5 -obsta le any more.ii. If the ase above does not happen, then we may assume that u ∈ Bi 6= Bj ∋

v be ause De�nition 5.62.2 does not hold. Then Bi has two blue positivenodes and Bj has two red ones, hen e neither De�nition 5.41.1 nor 5.41.2may hold.3. If A is a C∗6 -obsta le, then there exists a olour lass P ∈ P and a positive node

Se tion 5.4. Partition onstrained overing problem 117u ∈ P su h that Xu * P , then repla e m by m− χ{u} + χ{u′} for some u′ ∈ Xu − P .Then A is not a C∗

6 -obsta le any more.We emphasize that if there exists an obsta le for (p,P, m), then the proof of Lemma5.65 provides an algorithm to de ide if there exists a on�guration for (p,P).Theorem for the minimum versionBy exploiting the relations between on�gurations and obsta les and by applying oursplitting-o� result, we may now prove the following theorem solving the minimum ver-sion of Problem 5.25. It states that the lower bound φ may always be a hieved unless thereexists a on�guration.Theorem 5.66. Let p : 2V → Z+ be a symmetri , positively rossing supermodular setfun tion and P = {P1, . . . Pr} a partition of V . Then the minimum number of edges betweendi�erent members of P resulting in a graph that overs p is φ unless a on�guration exist,in whi h ase it is φ + 1.Proof. Let OPT (p,P) be the minimum number of edges between di�erent members of Presulting in a graph that overs p. The following Lemmas prove the theorem.Lemma 5.67. OPT (p,P) ≥ φ. If there exists a on�guration for p, then the inequality isstri t.Proof. We have seen that OPT (p,P) ≥ α, OPT (p,P) ≥ dim(p) − 1 and OPT (p,P) ≥ βifor i = 1, . . . r.Suppose there exists a on�guration for (p,P) and the inequality is not stri t. Let F bea minimum set of edges su h that (V, F ) overs p and it satis�es the partition onstraints,and let m be the degree spe i� ation de�ned by m(v) := dF (v) for every v ∈ V . By theminimality of F , m is an optimal extension for (p,P). Sin e there is a on�guration for(p,P), by Lemma 5.64, there is an obsta le for (p,P, m). But this ontradi ts Lemma5.44.Lemma 5.68. OPT (p,P) ≤ φ + 1. If there exists no on�guration for (p,P), then theinequality is stri t.Proof. If there exists no on�guration for (p,P), then by Lemma 5.65 there exists anoptimal extension m for (p,P) whi h ontains no obsta le. Hen e by Theorem 5.45 thereexists a omplete admissible splitting-o� and the stri t inequality follows. If there exists a

118 Chapter 5. Covering symmetri rossing supermodular fun tions on�guration for (p,P), let m be an optimal extension for (p,P). By Lemma 5.64, thereexists an obsta le for (p,P, m). Repla e m by m′ := m + χ{u} + χ{v} for some u, v withoutviolating m(P ) ≤ m(V )2

for every P ∈ P. Now, the sets ontaining u and v annot be tight,therefore there is no obsta le for (p,P, m′). By Theorem 5.45, there exists a ompleteallowed splitting-o� and the inequality follows.�Algorithm for the minimum versionIn this se tion, we des ribe the algorithm that, given a symmetri positively rossingsupermodular set fun tion p : 2V → Z ∪ {−∞} with a maximizing ora le, and a partition

P = {P1, . . . Pr} of V , �nds a minimum number of edges between di�erent members ofP resulting in a graph that overs p. It onsists of three major steps, extension, thensplitting-o�, and �nally determining if a on�guration exists.1. Find an optimal extension m for (p,P) by applying the algorithm of Se tion 5.4.3.Re all that m(V ) = 2φ.2. Apply the algorithm des ribed in the proof of Theorem 5.45 to (p,P, m).(a) If there is a omplete allowed splitting-o�, then we have found the desired graphhaving m(V )

2= φ edges.(b) Otherwise, we have found an obsta le A for (p,P, m).3. Apply the algorithm des ribed in the proof of Lemma 5.65 to A.(a) If it �nds another optimal extension m′ for (p,P) su h that no obsta le existsfor (p,P, m′), then Theorem 5.45 provides a omplete allowed splitting-o� andthereby the desired graph with φ edges.(b) Otherwise A is a on�guration for (p,P). The algorithmi proof of Lemma 5.68provides the desired graph with φ + 1 edges.5.4.4 Appli ation: Partition onstrained global edge- onne tivityaugmentation of a hypergraphIn this subse tion we spe ialize the results of this se tion for the problem of global edge- onne tivity augmentation of a hypergraph with a multipartite graph. Theresults of this subse tion were presented at the European Conferen e on Combinatori s,

Se tion 5.4. Partition onstrained overing problem 119Graph Theory and Appli ations (Euro omb 2009) whi h was held in Bordeaux, Fran e inSeptember 2009: see [9℄.Problem 5.69 (Partition onstrained global edge- onne tivity augmentation of a hyper-graph). Assume that we are given a hypergraph H0 = (V, E0), a partition P = {P1, P2, . . . , Pr}of the set V and a positive integer k. The problem is to �nd a graph G su h that H0 + Gis k-edge- onne ted, and G ontains only edges between the lasses of P. In the min-imum version the number of edges of G is to be minimized. In the degree-spe i�edversion G has to satisfy a given degree-spe i� ation m ∈ ZV+.This problem was solved by Ben Cosh [15℄ in the bipartition onstrained ase, i.e. when

r = 2. Here we give a omplete solution of this problem. Let p be the fun tion de�ned by(2.1). Let Φ be the maximum of the following values.α = max{⌈

1

2

∑

X∈X

(k − dH0(X))⌉ : X ∈ S(V )},

β = max{max{∑

Y ∈Y

(k − dH0(Y )) : Y ∈ S(P )} : P ∈ P},

ω(H0) = max{# omponent(H0 −F) − 1 : F ⊆ E , |F| = k − 1}.One an formulate the de�nition of onstru tions, obsta les and on�gurations by spe- ializing the abstra t de�nitions given above. Interestingly, there is no C∗5 - onstru tion forthis fun tion p used here.Lemma 5.70. There is no C∗

5 - onstru tion for the fun tion de�ned by (2.1).Proof. Assume that A = {A1, A2, A3, A4, B1, . . . , Bt} is a C∗5 - onstru tion. By Claim 5.31,

p(B) = 2, p(Ai ∪ B) = 1, p(A1 ∪ A2 ∪ B) = 1, p(A1 ∪ A3 ∪ B) ≤ 0. Apply (1.3) and (1.4)to get that, for any rossing pair X, Y ⊆ V

p(X) + p(Y ) = p(X ∩ Y ) + p(X ∪ Y ) − 2d1(X, Y ) − d2(X, Y ), (5.14)p(X) + p(Y ) = p(X − Y ) + p(Y − X) − 2d1(X, V − Y ) − d2(X, V − Y ). (5.15)(The de�nition of d1(X, Y ) and d2(X, Y ) an be found after Equation (1.3).) Equation(5.14) applied to X = A1∪B and Y = A2∪B gives that there exists exa tly one hyperedge

e in the hypergraph H0 that enters A1, A2 and exa tly one of B and A3 ∪A4. If e enters Bthen apply (5.15) to X = A1∪B and Y = A3∪B to get a ontradi tion. Otherwise, withoutloss of generality assume that e enters A3: now (5.14) applied to the pair X = A1 ∪B andY = A3 ∪ B gives a ontradi tion.

120 Chapter 5. Covering symmetri rossing supermodular fun tionsWe spe ialize the de�nition of the onstru tions below, from whi h the de�nition ofobsta les and on�gurations is easy to devise. To larify the distin tion from the abstra tproblem solved in Se tion 5.4, we will use the notation C4- and C6- onstru tion. Similarly,we will use C4- and C6-obsta le/ on�guration for this spe ial ase.De�nition 5.71. A partition {A1, . . . , A4} of V is a C4- onstru tion for (H0, k) if1. There exists a set A of hyperedges su h that k − |A| is odd and A = ∆H0(A1) ∩

∆H0(A3) = ∆H0

(A2) ∩ ∆H0(A4),2. α = k − dH0

(A1) + k − dH0(A3) = k − dH0

(A2) + k − dH0(A4).De�nition 5.72. A partition {A1, . . . , A6} of V is a C6- onstru tion for (H0, k) if1. k − dH0

(Ai) = 1, for all 1 ≤ i ≤ 6,2. k − dH0(Ai ∪ Ai+1) = 1, for all 1 ≤ i ≤ 6, (A7 = A1)3. α = 3.Thus the solution of the degree-spe i�ed version of Problem 5.69 is the following.Theorem 5.73. We are given a hypergraph H0 = (V, E0), a partition P = {P1, P2, . . . , Pr}of the set V and a positive integer k. Assume that the degree-spe i� ation m : V → Zsatis�es that m(V ) is even, m(Pi) ≤ m(V − Pi) for every i = 1, 2, . . . , r, and m(X) ≥

k − dH0(X) for any nonempty X ( V . Then there exist a graph G satisfying the degree-spe i� ation m su h that H0 +G is k-edge- onne ted, and G ontains only edges betweenthe lasses of P, unless there exists a C4- or a C6-obsta le for H0, k,P, m.The solution of the minimum version of Problem 5.69 is the following.Theorem 5.74. Let H0 = (V, E0) be a hypergraph, P a partition of V and k an integer.Then the minimum number of new edges between di�erent members of P to be added to H0in order to make it k-edge- onne ted is Φ + 1 if H0 ontains a C4- or a C6- on�guration,and Φ otherwise.

Chapter 6Sour e lo ation in hypergraphs �another way of edge- onne tivityaugmentationIn this hapter we give a di�erent approa h to edge- onne tivity augmentation. This isthe following: given a (hyper)graph, �nd a nonempty subset S of the nodes ( alled sour eset) su h that ontra ting the set S will result in a (hyper)graph satisfying the edge- onne tivity requirements. The obje tive is to minimize the total weight of the sour eset to be found (where nodes have nonnegative weights). This lass of problems is alledsour e lo ation problems, be ause it omes from the following appli ation: given a(hyper)graph H = (V, E), de ide where to lo ate the set of sour es S ⊆ V su h thatnodes have good edge- onne tivity to this sour e set, and lo ation osts are minimized.Sour e lo ation problems have re ently been intensively studied, not only in undire tedstru tures but also in dire ted ones (e.g. digraphs). In this thesis we will on entrate ongeneralizations of the following problem.Problem 6.1 (Sour e Lo ation in Graphs). Let us be given a graph G = (V, E), a weightfun tion w : V → R+ and a requirement fun tion r : V → R+ on the nodes. Find anonempty subset of nodes S su h that λG(S, v) ≥ r(v) for every node v ∈ V − S andw(S) :=

∑

s∈S w(s) is minimum.This problem was raised in [1℄ where they show the NP - ompleteness of this problem.However, it is shown in [1℄ that the spe ial ase when w is onstant an be solved inO(|V |M) time, where M is the time needed for one maximum �ow omputation in ournetwork. A minor observation shows that the same algorithm solves another spe ial ase,namely when r is onstant, but the authors of [1℄ give another algorithm for this ase that121

122 Chapter 6. Sour e lo ation in hypergraphsimproves the running time to O(|V |(|E| + |V | log |V |)).In this se tion we introdu e a generalization of Problem 6.1 for (undire ted) hypergraphs.Then we generalize the problem further and formulate an abstra t sour e lo ation problem.Of ourse we an not solve this problem in general (it ontains an NP - omplete problem),but we observe that it an be solved with an adaptation of the greedy algorithm introdu edin [1℄ in the spe ial ase when the requirement and the weight fun tions are ompatible(see the de�nition in Se tion 6.2). This ontains the ase when either of these fun tionsis uniform ( onstant, in other words), but in the subsequent subse tion we give anotheralgorithm for uniform requirement fun tion that gives a better running time. In every ase we also spe ialize our results for the hypergraphi sour e lo ation problem, too. Theresults presented in this hapter appeared in [6℄.6.1 Problem formulationA straightforward generalization of Problem 6.1 introdu ed above is the following:Problem 6.2 (Sour e Lo ation in Hypergraphs). Let us be given an undire ted hypergraphH = (V, E), a weight fun tion w : V → R+ and a requirement fun tion r : V → R+ onthe nodes. Find a nonempty subset of nodes S su h that λH(S, v) ≥ r(v) for every nodev ∈ V − S, and w(S) :=

∑

s∈S w(s) is minimum.By Theorem 1.2 we an reformulate the problem in the following way: �nd a subset ofnodes S su h that dH(X) ≥ max{r(v) : v ∈ X} for every ∅ 6= X ⊆ V − S and w(S)is minimum. However, sin e Problem 6.1 is a spe ial ase of this problem, this is anNP - omplete problem. In what follows we will on entrate on the two spe ial ases ofthis problem that were solvable for graphs; namely the onstant weight and the onstantrequirement ase, or more pre isely a ommon generalization of these two ases whenthese two fun tions are ompatible. We will show that the algorithms given in [1℄ anbe modi�ed appropriately to work in this more general setting, too. What is more, wewill show that they work in an even more abstra t environment, too. To this end let usintrodu e the following, abstra t form of Problem 6.2. Re all that a set fun tion d : 2V → Ris posimodular if it satis�es d(X) + d(Y ) ≥ d(X − Y ) + d(Y − X) for any X, Y ⊆ V .Problem 6.3 (Abstra t Sour e Lo ation). We are given a fun tion d : 2V → R+ that isposimodular and submodular, a weight fun tion w : V → R+ and a requirement fun tionr : V → R+ on the nodes. Find a subset of nodes S with w(S) minimum subje t to

d(X) ≥ max{r(v) : v ∈ X} for every X ⊆ V − S. (6.1)

Se tion 6.2. Compatible requirement- and weight fun tion 123For brevity, a set S ⊆ V satisfying (6.1) will be alled sour e set. As a remark we men-tion that the nonnegativity of the fun tion d is not really a restri tion: an arbitrary posi-modular fun tion d′ attains its minimum at the empty set, so de�ning d(X) = d′(X)−d′(∅)gives a nonnegative posimodular fun tion and we an modify the requirement fun tion sim-ilarly to be able to redu e the more general problem without the nonnegativity onstraintsto our problem. On the other hand, the nonnegativity of the weight fun tion is a naturalrequirement, sin e any superset of a valid sour e set is again a valid sour e set. Also, we ould have required the nonemptyness of a sour e set: that would not have hanged theproblem signi� antly.In the following se tions we will show that the spe ial ase of this abstra t problem whenthe requirement and the weight fun tions are ompatible an be solved using appropriateadaptations of the algorithms given in [1℄.Let us introdu e some terminology. A set X ⊆ V is alled de� ient if d(X) < max{r(v) :

v ∈ X}. It is obvious that a set S is a valid sour e set if and only if it meets every de� ientset, whi h is equivalent to requiring that S has to meet every minimal de� ient set.6.2 Compatible requirement- and weight fun tionIn this se tion we will give an algorithm to solve Problem 6.3 in the spe ial ase whenthe requirement- and weight fun tions are ompatible. This is an adaptation of the greedyalgorithm given in [1℄. Let us de�ne �rst what we mean by these fun tions being ompatible.De�nition 6.4. Two fun tions r, w : V → R are ompatible if there is an orderingv1, v2, . . . , vn of V su h that r(v1) ≤ r(v2) ≤ · · · ≤ r(vn) and w(v1) ≥ w(v2) ≥ · · · ≥ w(vn).Observe that ompatibility of two fun tions an be easily he ked in O(n log n) time andit is indeed a symmetri relation. Now we an give the greedy algorithm to solve Problem6.3: in this algorithm we only assume the posimodularity of the fun tion d but we leaveone question open that will only be answered when d is also submodular!Algorithm GREEDYbegin INPUT A posimodular fun tion d : 2V → R+ given with a fun tion evaluation ora le, arequirement fun tion r : V → R+ and a weight fun tion w : V → R+ on the �nite set

V . Assume that r and w are ompatible fun tions.OUTPUT A minimum weight sour e set S.1.1. Let S = V . Order V s.t. r(v1) ≤ r(v2) ≤ · · · ≤ r(vn) and w(v1) ≥ w(v2) ≥ · · · ≥

w(vn).

124 Chapter 6. Sour e lo ation in hypergraphs1.2. For i = 1 to n do1.3. (*) If there is no de� ient set X with S ∩ X = {vi} then S := S − vi.1.4. Output S.endOne thing remains to make lear: how to implement Step (*); to be able to do this wewill also assume the submodularity of the fun tion d. Let us speak about this later and �rst he k the orre tness of the algorithm: the following argument was taught to the author byAndrás Frank. Let us denote the urrent set S in the ith iteration before exe uting step (*)by Si (so S1 = V ) and let the output of the algorithm be Sn+1 = {vi1 , vi2 , . . . , vit} for somet ≥ 0. A simple indu tive argument shows that Si is a valid sour e set for any i between 1and n + 1. We only need to show that Sn+1 has minimum weight among the sour e sets.If, in the ith step of the for loop, Si − vi is not a valid sour e set then there must be ade� ient set Xi with Si∩Xi = {vi}. We an assume that this Xi is minimal de� ient: thereis a minimal de� ient set in Xi but it must ontain vi otherwise Si was not a valid sour eset. In parti ular Xi ⊆ {v1, v2, . . . , vi} whi h also implies that max{r(v) : v ∈ Xi} = r(vi).We will show that these sets Xi1, Xi2 , . . . , Xit are pairwise disjoint minimal de� ient sets.Assume that this is not the ase and suppose ij < ik are su h that Xij ∩ Xik 6= ∅. It is lear that vik /∈ Xij be ause Xij ⊆ {v1, v2, . . . , vij}. But vij ∈ Xik an not hold either sin ein the ikth iteration Xik is only overed by vik from among the nodes of Sik whi h also ontains vij . But then we have a ontradi tion from

r(vij ) + r(vik) > d(Xij) + d(Xik) ≥ d(Xij − Xik) + d(Xik − Xij ) ≥ r(vij) + r(vik).(The �rst inequality follows from the de� ien y of sets Xij and Xik and the last is be- ause they are minimal de� ient sets.) From these observations the optimality of thealgorithm follows: we have given an optimal overing of the disjoint minimal de� ient setsXi1 , Xi2, . . . , Xit .6.3 Implementation of Step (*)The only thing to show is the subroutine that implements Step (*). We note that asubroutine he king whether a given set S is a valid sour e set or not would also su� e butthe one we gave is easier to implement in our ase (and also gives a better running time).In the ith iteration we de�ne the set fun tion di : 2V −Si → R with di(X) := d(X + vi) forany X ⊆ V − Si. Clearly, Si − vi is a valid sour e set if and only if min{di(X) : X ⊆

V − Si} ≥ r(vi). In order to be able to de ide this we should be able to minimize theposimodular fun tion d over the nonempty subsets, or in other words we should be able to

Se tion 6.3. Implementation of Step (*) 125minimize an interse ting posimodular fun tion. It is an open question already mentionedin [38℄ whether this an be done for an arbitrary posimodular set fun tion d. Therefore wealso assume that d (and thus di for every i) is submodular in the rest of this se tion. Soany algorithm for submodular fun tion minimization will solve this problem. This yieldsthe following result:Theorem 6.5. Algorithm GREEDY solves the Abstra t Sour e Lo ation Problem with om-patible requirement and weight fun tion in O(n SFM(n, γ)) time.Here SFM(n, γ) denotes the time needed to de ide whether a submodular fun tion onan n element ground set takes on a value below a given bound, where the fun tion is givenby an ora le and γ denotes the time of a all to this ora le.We mention that the dual solution (disjoint de� ient sets) an be obtained, sin e al-gorithms for submodular fun tion minimization an produ e these, too. We also pointout that de iding whether a submodular fun tion takes a value less than a given boundB an be formulated as Problem 6.3 with uniform weight fun tion (following the ideas ofNarayanan [35℄, for example), so any algorithm solving Problem 6.3 with uniform weightfun tion will need essentially at least SFM(n, γ) time.The following orollary is a dire t extension of Theorems 1 and 2 of [1℄ and it improvesthe result (Theorem 24) in [38℄ by a fa tor of n:Corollary 6.6. Problem 6.3 with a uniform weight fun tion (and arbitrary requirements) an be solved in O(n SFM(n, γ)) time.Another orollary of Theorem 6.5 is that Problem 6.3 with a uniform requirement fun -tion (and arbitrary weights) an be solved in O(n SFM(n, γ)) time. However, in Se tion6.4 we will improve the running time for this spe ial ase.We an spe ialize our algorithm to the hypergraphi sour e lo ation problem: in this ase Step (*) an be regarded as a minimum (S − vi, vi) ut problem in our hypergraphH = (V, E) that an be translated to a maximum-�ow problem in a graph that has O(n+|E|)nodes and O(||E||) edges, where ||E|| denotes the sum of the ardinalities of the hyperedges.This yields the following result:Theorem 6.7. Algorithm GREEDY solves the Sour e Lo ation Problem in Hypergraphs with ompatible requirement and weight fun tion in O(nM(n+ |E|, ||E||)) time, where M(n′, m′)denotes the time needed for a maximum-�ow omputation in a graph with n′ nodes and m′edges.

126 Chapter 6. Sour e lo ation in hypergraphs6.4 Improving the running time for uniform require-mentsIn this se tion we will give an algorithm to solve Problem 6.3 in the spe ial ase when therequirement fun tion is onstant, that is r(v) = k, ∀v ∈ V , for some k ∈ R+. This is anadaptation of the algorithm solving a spe ial ase of this problem that was given in [1℄.Let us make one simple observation: minimal de� ient sets are disjoint in this ase. Thisis proved by the following argument. Suppose X and Y are two su h sets with X ∩Y 6= ∅.Sin e they an not ontain ea h other, ∅ 6= X − Y ( X and ∅ 6= Y − X ( Y whi h givesa ontradi tion together withk + k > d(X) + d(Y ) ≥ d(X − Y ) + d(Y − X) ≥ k + k.(The �rst inequality follows from de� ien y and the last from minimality.)In [37℄ Queyranne introdu es the notion of a pendent pair and gives an algorithm to�nd a pendent pair. The algorithm omputes an ordering of the nodes (that will be alledMax-Adj ordering here) and �nds the pendent pair using this ordering. This Max-Adjordering is nothing else but the appropriate generalization of the ordering used in thealgorithm by Nagamo hi and Ibaraki to �nd a minimum ut in a graph. Let us give thede�nitions and results.De�nition 6.8. If h : 2V → R is a symmetri submodular fun tion then an ordering

(v1, v2, . . . , vn) is a Max-Adj ordering for this fun tion if it satis�es the following:h({vi}) − h({v1, . . . , vi}) ≥ h({vj}) − h({v1, . . . , vi−1, vj}) ∀2 ≤ i ≤ j ≤ n.We note that the �rst element v1 of su h an ordering an be spe i�ed arbitrarily.De�nition 6.9. If h : 2V → R is a symmetri submodular fun tion then an ordered pairof nodes (s, t) is alled a pendent pair if

h({t}) = min{h(X) : X ⊆ V, s /∈ X, t ∈ X}.Lemma 6.10 ([37℄). If (v1, v2, . . . , vn) is a Max-Adj ordering for the fun tion h then(vn−1, vn) is a pendent pair. Su h an ordering an be al ulated with O(n2) alls to theh-value ora le and O(n2) other operations.After these preliminaries we an present the algorithm that solves Problem 6.3 when therequirement fun tion is uniform. Let us �rst give a sket h of this algorithm.We want to al ulate a minimum weight sour e set S and we start with S = ∅. Themain idea is the following: the minimal de� ient sets form a subpartition of V . In ea h

Se tion 6.4. Improving the running time for uniform requirements 127step of the algorithm we �nd a suitable pair of nodes that is not separated by a (yetun overed) minimal de� ient set. Then we ontra t this pair of nodes, and repeat thisuntil one minimal de� ient set be omes a singleton. Then we noti e this and in lude anelement of this set in S. But what shall we do with the minimal de� ient sets that arealready overed? We should somehow in rease the value of fun tion d on sets that ontainsu h a set: this is best des ribed using an auxiliary graph and its ut fun tion.After a ertain number of su h ontra tions every node in the resulting set is the imageof a ontra ted set in the original ground set. For every su h ontra ted set we rememberthe � heapest� element in it, i.e. the one with the smallest weight. This is done with theancestor fun tion in the algorithm below. This way, when we ontra t a minimal de� ientset into a singleton we an pi k the heapest element from it to in lude in our set S. The orre t formulation of the algorithm is des ribed by the following pseudo ode.Algorithm CONSTANT_DEMANDbegin INPUT A posimodular and submodular fun tion d : 2V → R+, a requirement k ∈ R+and a weight fun tion w : V → R+ on the �nite set V .OUTPUT A minimum weight sour e set S.1.1. Let S = ∅, V ′ = V + s with a new node s and G = (V ′, E) an auxiliary graph: in thebeginning E is empty but later we add edges in E (one endpoint of these will always be

s). De�ne the symmetri submodular fun tion h : 2V ′

→ R by h(X) = d(X) + kdG(X)for any X ⊆ V and h(X) = h(V ′ − X) if s ∈ X ⊆ V ′.1.2. For ea h v ∈ V1.3. ancestor(v) := v.1.4. If d({v}) < k then S := S + v and add an edge between s and v to G (so thefun tion h hanges here!).1.5. End for.1.6. While |V ′| > 2 and ∃v ∈ V ′ − s with (s, v) /∈ E1.7. Constru t the Max-Adj ordering of V ′ for the fun tion h starting with s and takethe last two elements in this ordering (a pendent pair) (vn′−1, vn′).1.8. Contra t vn′−1 and vn′ into a node v′ and let the an estor of v′ be the heaper of thean estor of vn′−1 and that of vn′ (so V ′ gets smaller: the fun tion h and the graphG follow these hanges the obvious way).1.9. If h({v′}) < k then add an edge between s and v′ to G and S := S + ancestor(v′)(fun tion h hanges, too!).1.10. End while.1.11. Output S.

128 Chapter 6. Sour e lo ation in hypergraphsendLet us prove the orre tness of the above algorithm. In the argumentation below we willthink of the set V as the original ground set, while the set V ′ always means the urrentground set after some ontra tions (note that s is never ontra ted with any other node).Also, for a set X ⊆ V we will use the notation X ′ for the image of this set after the ontra tions made so far: this only makes sense if there is no set Y ⊆ V with X ∩ Y 6= ∅and (V − X) ∩ Y 6= ∅ that was ontra ted.Lemma 6.11. In Step 1.7 of the above algorithm(i) if X ⊆ V is a minimal de� ient set that is not overed yet (i.e. S ∩ X = ∅ with the urrent S) then the image X ′ of X makes sense, and(ii) there is no (yet un overed) minimal de� ient set X ⊆ V for whi h |X ′∩{vn′−1, vn′}| =

1.Proof. Initially (i) is true. At any stage, if (i) is true, then the statement of (ii) makessense: assume it is not true and there is a minimal de� ient set X ⊆ V for whi h X ′separates vn′−1 and vn′. But then d(X ′) = h(X ′) sin e X is not overed yet by S andk > d(X) = d(X ′) = h(X ′) ≥ h(vn′) ≥ k,gives a ontradi tion (the se ond inequality follows from the properties of the Max-Adjordering; for the last inequality observe that we always assure in the algorithm that h(v) ≥

k for singletons v ∈ V ′−s). Finally, if (ii) is true then (i) will be true in the next iteration,after ontra ting vn′−1 and vn′ , too.Lemma 6.12. Algorithm CONSTANT_DEMAND solves Problem 6.3 with uniform re-quirement fun tion in O(n3γ + n3) time (where γ denotes the time of a all to the d-valueora le).Proof. The orre tness of the algorithm follows from Lemma 6.11 and the de�nition of theancestor fun tion. It is straightforward to he k that Step 1.7 dominates the algorithmwhi h gives the running time indi ated above.If we spe ialize our algorithm to hypergraphs, that is to Problem 6.2, then we ansubstitute Step 1.7 of our algorithm with a similar subroutine des ribed in [32℄. Thisyields the following running time:Lemma 6.13. Algorithm CONSTANT_DEMAND solves Problem 6.2 with uniform re-quirement fun tion in O(n2 log(n) + n||E||) time (||E|| denotes the sum of the ardinalitiesof the hyperedges).

Chapter 7Open problemsI don't think that anyone ever has �nished a thesis (ar any mathemati al writeup) withoutleaving some of the questions he or she wanted to answer open. Let me enumerate theproblems that I was wondering about during writing this thesis.1. Maximizing skew-supermodular fun tions, maximizing rossing (interse t-ing) negamodular fun tions. In Se tion 1.4 we have observed that a maximizingora le an be implemented for a rossing supermodular fun tion p : 2V → R givenwith an evaluation ora le by using standard submodular fun tion minimization te h-niques. However this is unknown if the fun tion p is only skew-supermodular. Thesimplest open problem is whether a maximizing ora le an be implemented for aninterse ting negamodular fun tion p : 2V → R given with an evaluation ora le. Notethat these problems were almost always present in our abstra t algorithms, even inSe tion 6.2, where we wanted to minimize an interse ting posimodular fun tion.2. Rank-respe ting augmentation of a hypergraph with negamodular on-straints for γ = 2. This problem is a slight generalization of the rank-respe tingnode-to-area onne tivity augmentation problem in hypergraphs. In Se tion 4.4.3 wehave solved this problem for any γ ≥ 3 but we did not bother about the γ = 2 ase, as the interesting spe ial ase (the node-to-area onne tivity augmentation ingraphs) has already been solved by Ishii and Hagiwara.3. Rank-respe ting global ar - onne tivity augmentation of mixed hyper-graphs. In Se tion 4.4.2 we have shown that the Algorithm GREEDYCOVER appliedto the global ar - onne tivity augmentation of a mixed hypergraph M �nds graphedges plus a hyperedge that is at most one bigger than the rank of M . Probably,similarly to the results in Se tion 4.4.3 we ould solve the rank-respe ting version ofthis problem, too. 129

130 Chapter 7. Open problems4. Covering a (positively) skew-supermodular fun tion of form p0 − dG withgraph edges, where p0 does not take the value 1. In the proofs of the NP - ompleteness of Problem 4.1 one seems to need that the skew-supermodular fun tionoften takes the value 1. Possibly this is a key point, as was the ase with the node-to-area onne tivity augmentation in graphs, and we possibly obtain a polynomiallysolvable problem if we restri t ourselves to fun tions of the above form.5. Covering a rossing supermodular fun tion with graph edges. Ben zúr andFrank [5℄ has solved the problem of overing a symmetri rossing supermodular fun -tion with graph edges. Probably this ould be generalized to not ne essarily symmetri fun tions, too. An appli ation would be the global ar - onne tivity augmentation ofmixed hypergraphs with graph edges.6. Minimum node- ost partition onstrained overing of a a symmetri posi-tively rossing supermodular fun tion (with graph edges). In Se tion 5.4 wehave solved the minimum- and the degree-spe i�ed version of Problem 5.25. The min-imum node- ost version does not follow straightforwardly from the degree-spe i�edversion as usually, be ause of the partition onstraints. However, probably this ispolynomially solvable, too.

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Összefoglaló

A dolgozat nagyobb részében azzal a hagyományos élösszefügg®ség-növelési feladattal

foglalkozunk, amikor új élek illetve hiperélek hozzáadásával kell egy gráf illetve hiper-

gráf élösszefügg®ségét megnövelni. Hangsúlyozzuk, hogy míg a növelend® hipergráf tar-

talmazhat irányított hiperéleket is, addig a növel® (hiper)gráf mindig irányítatlan ebben

a dolgozatban. A legfontosabb célfüggvény mindig a hozzáadandó új (hiper)élek összmé-

retének minimalizálása. Ezeket a feladatokat mindig egy ferdén szupermoduláris halmaz-

függvény gráfélekkel avagy hiperélekkel való fedésének absztrakt feladatává fogalmazzuk

át. Alkalmazásként azonban a következ® élösszefügg®ség-növelési feladatok variánsai le-

begnek a szemünk el®tt: hipergráfok globális illetve lokális élösszefügg®ségének növelése,

vegyes hipergráfok globális élösszefügg®ségének növelése, illetve az úgynevezett node-to-

area élösszefügg®ség-növelési feladat. A variánsok f®leg abban különböznek, hogy meg-

engedjük e tetsz®legesen nagy hiperélek hozzáadását, avagy megpróbáljuk ezek méretét

korlátozni, extrém esetben csak gráféleket engedünk meg.

A megfelel® bevezetések után a dolgozat 3. fejezetében azt az esetet vizsgáljuk, amikor

tetsz®leges méret¶ hiperélek hozzáadása megengedett, és Szigeti Zoltán ide vonatkozó téte-

lét általánosítjuk több irányban is. A 4. fejezetben ezzel éppen ellentétben azt vizsgáljuk,

hogy mi történik, ha az ember csak gráfélek hozzáadásával próbálkozik. Ez a megköze-

lítés megválaszolja azt az esetet, amikor az új hiperélekt®l azt várjuk, hogy a rangot ne

(nagyon) növeljék meg. Végezetül az 5. fejezetben azt vizsgáljuk, hogy mi van, ha ragasz-

kodunk ahhoz, hogy csak gráféleket engedünk meg, de cserébe csak a hipergráfok globális

élösszefügg®ségének növelésére korlátozódunk. A feladat absztrakt alakjának megoldását

Benczúr és Frank tétele szolgáltatja: erre adunk egy viszonylag egyszer¶ bizonyítást, ami

lehet®vé teszi, hogy a feladat partíciókorlátos változatát is megoldjuk.

A dolgozat 6. fejezetében egy másik élösszefügg®ség-növelési fogalmat tekintünk, az úgy-

nevezett forrástelepítési feladatkört. Itt a (hiper)gráf élösszefügg®ségét egy alkalmasan

választott ponthalmaz összehúzásával akarjuk megnövelni. Általánosítjuk Arata és társai

gráfokra vonatkozó eredményeit egy absztrakt formában, ami a hipergra�kus változatot is

magában foglalja.

Summary

The main part of this thesis deals with the traditional notion of edge-connectivity aug-

mentation, when the edge-connectivity of a graph or hypergraph has to be augmented

by introducing new edges or hyperedges. We emphasize that though the hypergraph to

be augmented can even contain directed hyperedges, the new (hyper)edges are always

undirected in this thesis. The most important objective function is always to minimize

the total size of the new (hyper)edges. The problems are always treated in the abst-

ract framework of covering a skew-supermodular function with graph edges or hyperedges.

However, as applications we will consider variants of the following edge-connectivity aug-

mentation problems: global or local edge-connectivity augmentation of hypergraphs, global

arc-connectivity augmentation of mixed hypergraphs, and the so called node-to-area edge-

connectivity augmentation problem. The variants mainly di�er on whether we allow the

addition of arbitrarily large hyperedges, or we restrict the size of these, in the extremal

case we only allow graph edges.

After the appropriate introduction, Chapter 3 deals with the case when we allow hype-

redges of arbitrary size: we generalize the theorem about this case due to Zoltán Szigeti

in many directions. On the other hand in the 4th chapter we try to answer what happens

if one wants to use only graph edges. This approach answers the problem when we want

that the rank of the hypergraph should not increase (too much). Finally, in Chapter 5 we

insist on using only graph edges, but we restrict ourselves to the global edge-connectivity

augmentation of undirected hypergraphs. The abstract form of this problem was solved

by Benczúr and Frank: we give a relatively simple proof of their result which enables us

to generalize it to the partition constrained case, too.

In Chapter 6 we investigate a di�erent edge-connectivity augmentation technique, the

class of the so called source location problems. Here the edge-connectivity of the (hy-

per)graph is to be augmented by contracting a suitable set of nodes. We generalize the

results of Arata et al on graphs to an abstract framework which also includes the hyper-

graphic version of the problem.