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SIAM JOURNAL ON COMPUTING, 24(6):1278–1304, 1995.

IDENTIFYING THE MINIMAL TRANSVERSALS OF A HYPERGRAPH ANDRELATED PROBLEMS

THOMAS EITERy AND GEORG GOTTLOByAbstract. The paper considers two decision problems on hypergraphs, hypergraph saturation and recognition

of the transversal hypergraph, and discusses their significance for several search problems in applied computerscience. Hypergraph saturation, i.e., given a hypergraphH, decide if every subset of vertices is contained in orcontains some edge ofH, is shown to be co-NP-complete. A certain subproblem of hypergraph saturation,thesaturation of simple hypergraphs (i.e. Sperner families),is shown to be under polynomial transformation equivalentto transversal hypergraph recognition, i.e., given two hypergraphsH1;H2, decide if the sets inH2 are all theminimal transversals ofH1. The complexity of the search problem related to the recognition of the transversalhypergraph, the computation of the transversal hypergraph, is an open problem. This task needs time exponentialin the input size; it is unknown whether an output-polynomial algorithm exists. For several important subcases,for instance if an upper or lower bound is imposed on the edge size or for acyclic hypergraphs, output-polynomialalgorithms are presented. Computing or recognizing the minimal transversals of a hypergraph is a frequent problemin practice, which is pointed out by identifying important applications in database theory, Boolean switching theory,logic, and AI, particularly in model-based diagnosis.

Key words. hypergraphs, hypergraph transversals, hitting sets, independent sets, graph algorithms, NP-complete,polynomial time, output-polynomial time, database design, dependency inference, key computation, distributed data-bases, coteries, model-based diagnosis, satisfiability, prime implicants

AMS subject classifications.05C65, 05C85, 05C90, 06E30, 68P15, 68Q20, 68Q25, 68R10, 68T30, 94C10

1. Introduction. Hypergraph theory [7] is an important subfield of discrete mathemat-ics with many relevant applications in both theoretical andapplied computer science. In thispaper, we study complexity issues of relevant computational problems on hypergraphs.

A hypergraphH is a set of subsets (edges) of a finite set of vertices. A hypergraph issimple if none of its edges is contained in any other of its edges. We say that a hypergraph issaturated if every subset of the vertex set is contained in anedge or contains an edge of thehypergraph.

The first problem we consider is the test whether a given hypergraphH is saturated. Thisdecision problem will be referred to asHYPERGRAPH SATURATION(H-SAT). We will alsodeal with a restricted version of this problem whereH is supposed to be a simple hypergraph.This subproblem is calledSIMPLE HYPERGRAPH SATURATION(SIMPLE-H-SAT).

The second main problem – and probably the more important onefrom the applicationviewpoint – concerns hypergraph transversals. A transversal (or hitting set) of a hypergraphH is a subset of the vertices ofH which intersects each edge ofH. A transversal is minimal ifit does not properly contain any other transversal. The setTr(H) of all minimal transversalsof a hypergraphH is itself a hypergraph called the transversal hypergraph ofH.

Our second problem can thus be formulated as follows: Given two hypergraphsG andH, decide whetherG is the transversal hypergraph ofH. This decision problem, referred toasTRANSVERSAL HYPERGRAPH(TRANS-HYP), is closely related to the search problem ofcomputingTr(H) for a given hypergraphH.

The complexity ofTRANS-HYP is to date an open issue. The problem is in co-NP,but there is no proof of co-NP-completeness. On the other hand, while the problem ispolynomial-time solvable ifG orH is a graph, no polynomial algorithm for the solution of thegeneral version ofTRANS-HYP is currently known. Similarly, it is an open problem, whetheryChristian Doppler Laboratory for Expert Systems, Information Systems Department,Technical University of Vienna, Paniglgasse 16, A-1040 Wien, Austria.Email: (eiterjgottlob)@vexpert.dbai.tuwien.ac.at

1278

MINIMAL TRANSVERSALS OF A HYPERGRAPH 1279Tr(H) can be computed in output-polynomial total time (i.e., in time polynomial in the com-bined sizes of the input and the output). This complexity problem was posed independently byseveral researchers [41, 14, 29]. Note that the existence ofan output-polynomial algorithm forcomputingTr(H) would imply the polynomial-time solvability ofTRANS-HYP; vice versa,if TRANS-HYPis co-NP-complete, then no output-polynomial algorithm for the computationof Tr(H) is likely to exist.

This paper presents an extensive study of the above-mentioned complexity and computa-tion problems. We derive some new complexity results and exhibit relationships to otherrelevant problems such as satisfiability or hypergraph 2-colorability. Particular attentionis paid to restricted versions of the general problem classes which are either polynomial-time solvable or�pm-equivalent (i.e., equivalent under polynomial-time transformations) toTRANS-HYP. The considered problems have important applications, e.g., in database theory,switching theory, and artificial intelligence, from which it will become clear that computingthe minimal transversals of a hypergraph is a frequently encountered problem of computerscience. In the rest of this section, we outline the structure of the paper and highlight themost important results.

Our study of hypergraph problems starts in Section 2 with thedefinitions of basic hyp-ergraph concepts and a brief overview of results known so far. In Section 3 we show thatH-SAT is co-NP-complete and relate this problem to the well-known problemof hypergraph2-colorability. SIMPLE-H-SAT turns out to be�pm-equivalent to the problem of 2-coloring ahypergraph whose edges are mutually intersecting. In Section 4 it is shown thatSIMPLE-H-SAT is�pm-equivalent toTRANS-HYP. A restricted version of the latter problem is the ques-tion whether a given hypergraph is equal to its own transversal hypergraph, i.e.,Tr(H) = H.This problem, which we callSELFTRANSVERSALITY, has attracted much interest by math-ematicians [36, 6, 7] but no complexity results have been derived so far. We show thatSELFTRANSVERSALITYis�pm-equivalent toTRANS-HYP.

Section 5 is dedicated to the investigation of polynomial subcases of the main problems.The most interesting results are briefly summarized as follows:

1. H-SAT becomes solvable in polynomial time if the cardinalities ofthe edges of theinput hypergraphH differ by at most a constantk. In other words, if all edges ofH haveapproximately the same size.

2. TRANS-HYP is decidable in polynomial time if the cardinalities of the edges ofone of the input hypergraphs, sayH, are bounded by a constant. We present an output-polynomial algorithm for computingTr(H) in this case. Note that this result immediatelyleads to a relevant generalization of well-known output-polynomial methods for computingthe maximal independent sets of graphs [52, 34, 29]. (Indeed, the maximal independent setsof a graph or hypergraph are exactly the complements of its minimal transversals.)

3. TRANS-HYP is decidable in polynomial time if one of the input hypergraphs isacyclic. We present an output-polynomial algorithm for computingTr(H) for acyclic hy-pergraphsH. The type of acyclicity we consider is�-acyclicity as defined by Fagin in [16].Note that�-acyclicity is among the weakest types of acyclicity; our results and methods alsohold if we replace�-acyclicity by any of the several stronger types of acyclicity from theliterature, cf. [16, 7].

Several applications where recognizing or computing the transversal hypergraph playsa fundamental role are described in Section 6. Our results can be profitably used in thoseapplications. Section 7 concludes the paper; we state some open problems and give direc-tions for further research. For convenient readability, a cross-reference list of definitions isprovided in the appendix.

1280 T. EITER and G. GOTTLOB

2. Preliminaries and previous results.

DEFINITION 2.1. A hypergraph is a pair(V; E) of a finite setV = fv1; : : : ; vng and asetE of subsets ofV . The elements ofV are called vertices, the elements ofE edges.

Note that some authors, e.g. [7], state that the edge-set as well as each edge must benonempty and that the union of all edges yields the vertex set.

For notational convenience, we will identify a hypergraph with its edge-set and vice versaif there is no danger of ambiguity. Thus for hypergraphsH = (V; E) andH0 = (V 0; E 0),we writeE 2 H for E 2 E , H [ H0 for (V [ V 0; E [ E 0), min(H) for (V;min(E)) etc.Moreover, if not stated otherwise, it is presupposed that hypergraphs have the same set ofvertices which is denoted byV , thatn denotes the number of vertices, and thatm denotes thenumber of edges.

A hypergraphH is calledsimpleif for each pairH;H 0 of distinct edges ofH it holdsthatH 6� H 0. Simple hypergraphs are also calledSperner familiesaccording to [49], whereit is proved that the cardinalityjHj of a simple hypergraphH is bounded by

� n[n=2]�; which is

asymptotic to� 2� �1=2 2nn�1=2 [3]. For example, the set of minimal models for a finite set of

propositional clauses or the minimal keys of a database relation form a Sperner system.LetH be a hypergraph. The setVe(H) = SH2HH are theessential verticesofH. The

rank r(H) of H is defined byr(H) = maxf jEj : E 2 Hg, and itsanti-rank ar(H) byar(H) = minfjEj : E 2 Hg. For example, every graph is a hypergraph of rank 2.H iscalledintersectingif no pair of its edges is disjoint, i.e., for allE1; E2 2 H, E1 \ E2 6= ;:Singleton edges in a hypergraph are referred to asloops.EXAMPLE: Consider the hypergraphH = ff1g; f2; 3g; f1; 3; 4gg on V = f1; 2; 3; 4g. Wenote thatVe(H) = V , i.e., all vertices are essential, and thatH is not simple. Moreover,ar(H) = 1, r(H) = 3, andH is not intersecting.

DEFINITION 2.2. LetH be a hypergraph. Thenmin(H) (resp.max(H) ) denotes the setof minimal (resp. maximal) edges ofH with respect to set inclusion, i.e.,min(H) = fE 2 H :there exists noE0 2 H with E0 � Eg (resp.max(H) = fE 2 H : there exists noE0 2 HwithE0 � Eg).

Clearly,min(H) andmax(H) are simple hypergraphs onV . For the hypergraphH =ff1g; f2; 3g; f1; 3; 4ggof the last example, we havemin(H) = ff1g; f2; 3ggandmax(H) =ff2; 3g; f1; 3; 4gg.DEFINITION 2.3 (Transversal).LetH be a hypergraph. A setT � V is a transversal ofH if for eachE 2 H, T \E 6= ;. A transversalT is minimal if no proper subsetT 0 of T is a

transversal.EXAMPLES: Consider the hypergraphH = ff1g; f2; 3g; f1; 3; 4gg on V = f1; 2; 3; 4gagain.H has two minimal transversals:f1; 2g andf1; 3g. Note that; is a minimal transversalof H = ;, since everyT � V vacuously fulfills the transversal criterion, and thatH = f;ghas no transversal, since noT � V has nonempty intersection with;.

Transversals are also calledhitting setsand in case of a graphvertex covers. The min-imal vertex covers of a graph as well as the maximal independent sets are output-efficientlycomputable. Anindependent setof a graphG is a vertex subsetS � V that contains no edgeof G, and it is maximal if no proper superset of it has this property. Clearly,S is a maximalindependent set ofG iff V � S is a minimal vertex cover ofG. Finding the lexicographicallyfirst maximal independent set is LOGSPACE-hard [11], while finding the lexicographicallylast maximal independent set isNP-hard [29]. (Given thatV is ordered,V1 � V comes lex-icographically beforeV2 � V if the first element at whichV1 andV2 disagree is inV1 [29].)The maximal independent sets of a graph can be output with polynomial-time delay [52, 34],even if they have to be in lexicographic order [29]. Moreover, fast parallel algorithms are

MINIMAL TRANSVERSALS OF A HYPERGRAPH 1281

known, cf. [31].Notice that transversals are closely related to edge covers. An edge coverof a hypergraphH is an edge subsetH0 � H such that

SE2H0 E yieldsV , and is minimal if no proper subsetis an edge cover. The computation of minimal transversals and minimal edge covers is easilytransformable into each other, cf. [35], thus results on oneof these problems apply to theother.

Finding a minimal transversal of a hypergraphH is efficiently possible: IfH = ;, then; is the only minimal transversal ofH. If H 6= ;, thenV = fv1; : : : ; vng is a transversal ofH. If we defineT0 = V; and for alli, 1 � i � n,Ti = � Ti�1; if Ti�1 � fvig is not a transversal;Ti�1 � fvig; if Ti�1 � fvig is a transversal,

thenTn is a minimal transversal ofH. Thus a minimal transversal ofH can be found in timeO(m�n). Finding a minimum cardinality transversal, however, isNP-hard, cf. [21].DEFINITION 2.4. LetH be a hypergraph onV . The transversal hypergraphTr(H) ofH is the hypergraph onV whose edges are the minimal transversals ofH.The following propositions capture important relations between a hypergraphH andTr(H), cf. [7].PROPOSITION2.1. For every hypergraphH, Tr(H) is a simple hypergraph andTr(H)

= Tr(min(H)).PROPOSITION2.2. LetG andH be simple hypergraphs. Then,

(i) G = Tr(H) if and only ifH = Tr(G);(ii) Tr(G) = Tr(H) iff G = H;(iii) Tr(Tr(H)) = H.

The following simple algorithm to determineTr(H) is given in [7] (cf. also [38, 35]).LetH = fE1; : : : ; Emg, m � 0. Note thatTr(fEig) = ffeg : e 2 Eig. Define a sequenceT 0; : : : ; T m byT 0 = f;g; and for alli, 1 � i � m, T i = minfT [ feg : T 2 T i�1 ande 2 Eig.Then,T j = Tr(fE1; : : : ; Ejg), 0 � j � m; and thusT m = Tr(H).

It is easy to see that the computation of the transversal hypergraph is inherently expo-nential. For example, consider the hypergraphFn = ff2i � 1; 2ig : 1 � i � ng onVn = f1; : : : ; 2ng. Then,Tr(Fn) = ffx1; : : : ; xng : xi 2 f2i; 2i� 1g; 1 � i � ng. SinceTr(Fn) has2n edges, the computation ofTr(Fn) needs space (hence also time) exponentialin the input size.

Since an algorithm for computingTr(H) that runs in time polynomial in the input sizeIS is not possible, an algorithm is desirable which works at least in polynomial time if thenumber of minimal transversals (resp. the output sizeOS) is taken into account. Thereare various possibilities to define appealing notions of output-polynomiality, e.g. output-polynomial total time, incremental-polynomial time, polynomial-time delay computability[29], or P-enumerability [54].

The most general of these concepts isoutput-polynomial total time, which requires thatan algorithm works in time bounded by a polynomial inIS andOS. Note that an output-polynomial total time algorithm may run for exponentially many steps inIS without produ-cing any output. The same holds for the notion ofP-enumerability, where a search algorithmcomputes all solutions of a problem in timep(IS)N , wherep is a polynomial andN is thenumber of solutions [54]. To consider incremental computations, an algorithm meets the


incremental-polynomial timecriterion if it outputs all solutionss1; : : : ; sN to the respectiveproblem one by one as follows. The time until the first output and the time between output ofsolutionssi; si+1, 1 � i < N , is bounded by a polynomial in the combined sizes of the inputand all solutionss1; : : : ; si, and the algorithm stops in polynomial time after the outputof thelast solution. A more restrictive criterion is output withpolynomial-time delay. An algorithmwith this property generates the solutions one by one such that the time until the first solutionis output and between the output of consecutive solutions isbounded by a polynomial inIS.Clearly, any such algorithm P-enumerates the solutions.

It is easy to show that the above algorithm does not stop in output-polynomial total time.Consider the complete graphKn on verticesf1; : : : ; ng, i.e.,Kn = f fi; jg : 1 � i < j �n g, wheren � 4 is even. Let the edgesE1; : : : ; Em, m = jKnj, ofKn be ordered such thatE1 = f1; 2g; E2 = f3; 4g; : : : ; En=2 = fn� 1; ng. ComputingTr(Kn) by T 0; : : : ; T m, wefind thatjT n=2j = 2n=2; becauseT n=2 = Tr(fE1; : : : ; En=2g) andfE1; : : : ; En=2g is justthe hypergraphFn=2 from above. AsTr(Kn) is given byT m = ff1; : : : ; ng � fig : 1 �i � ng andm = n(n� 1)=2, we haven = jTr(Kn)j = �(jKnj1=2). RelatingjT n=2j to thej-th power,j arbitrarily fixed, ofjKnj = maxf jKnj; jTr(Kn)j g we getjT n=2j = 2�(jKnj1=2); thus jT n=2j > jKnjjfor sufficiently largen. Hence, the size ofT n=2 is not bounded by any polynomial in theinput and output size. Consequently, the above algorithm isnot efficient with respect tooutput-polynomiality.

Though there are several algorithms which involve, in a moreor less obvious way, thecomputation of transversal hypergraphs, (e.g., [38, 35, 46, 14, 40, 42, 45]), unfortunately nooutput-polynomial transversal hypergraph algorithm is known today; moreover, it is even anopen question whether such an algorithm is likely to exist atall, cf. [14, 29, 43]. We willshow in Section 4 that the complexity of the closely related decision problemTRANS-HYP,which is, given two hypergraphsG andH, to decide ifG = Tr(H), is of crucial importanceto answer this question.

The second kind of problem in our study is the complexity of hypergraph saturation, acovering problem for the powersetP(S) of a finite setS considered in [51, 10]. We introducesome notation first.

DEFINITION 2.5. Let S be a finite set, and letX � S: ThenCovS(X) = fY � S :Y � X or Y � Xg. For any hypergraphH onV , Cov(H) = SH2H CovV (H).In words,Cov(H) is the set of setsX � V that arecoveredby an edgeE ofH as subset

(X � E) or as superset(X � E).DEFINITION 2.6. A hypergraphH on V is called saturated if and only ifCov(H) =P(V ).According to the definition, a hypergraphH is saturated iff every setV 0 � V is contained

in at least one of edge ofH or contains at least one edge ofH.EXAMPLE: Consider the hypergraphH = ff1g; f2; 3g; f1; 3; 4gg onV = f1; 2; 3; 4g again.Then,Cov(H) = P(V )� f2; 4g, i.e.,H covers all subsets ofV butf2; 4g. Hence,H is notsaturated. If we addf2; 4g toH, the resulting hypergraphH0 = ff1g; f2; 3g; f1; 3; 4g; f2; 4ggis saturated. Note that saturation of a hypergraphG does not imply saturation of eithermin(G)or max(G). As an example considerH0. Neithermin(H0) = ff1g; f2; 3g; f2; 4gg normax(H0) = ff2; 3g; f1; 3; 4g; f2; 4gg is saturated, becausef3; 4g andf1; 2g respectivelyare not covered.

The complexity of checking if a hypergraph is saturated has,to our best knowledge, notbeen studied yet in the literature.


We assume that the reader is familiar with the basic conceptsof the theory ofNP-completeness (cf. [21]). co-� denotes the problem complementary to decision problem�;�pm denotes polynomial-time transformability. Decision problems� and�0 are called�pm-equivalentiff � �pm �0 and�0 �pm �. Problems that are solvable in polynomial time arereferred to astractable, and those which are most likely not solvable in polynomial time,amongst them allNP-hard problems, asintractable.

3. Hypergraph saturation. Imagine you are a friend of Toni who manages a pizzarestaurant that is famous for its rich variety of pizze.1

The pizze differ in their topping, and accordingly every pizza has its special name. Forexample, on aPizza Margheritathere are tomatoes, onions, and some cheese, on aPizza alTonnotuna fish and onions, and on aPizza Provincialetomatoes, ham, corn, onions, mush-rooms, salami, pepperonies, and cheese. Now Toni wants to enlarge his pizza offer by a newpizza, but the new pizza should be neither the “Grande” version of another pizza, i.e., have allthe food of another pizza on it and some additional, nor the “Mini” version, i.e., all the foodon it is also on some other pizza. Toni has been testing quite anumber of food collections forthe new pizza yet, but each of his compositions turned out to be a Grande or a Mini version ofone of the many pizze. This makes Toni wondering if there is any food collection for a newpizza at all. Since you are an expert in computer science, he asks you to write a computerprogram for answering this question. Under the principle ofquot capita tot gustus, whichmeans that every collection of food is appropriate for a pizza, you find that Toni’s problem isjust the hypergraph saturation problem if the food available is considered as setV of verticesand the pizze already offered, described by food collections, are considered as edges of ahypergraph. In a more precise formulation,

Problem: HYPERGRAPH SATURATION (H-SAT)Instance: A hypergraphH = (V; E) on verticesV = fv1; : : : ; vng:Question: IsH saturated ?A simple brute force algorithm would be to test all possible2n vertex sets subsequently

until some collection is found that is not covered byH; this collection would be a new pizzawhich satisfies the requirements. However, ifH is saturated, the algorithm recognizes thisonly after2n covering checks – too many if the number of food items on Toni’s pizze is takeninto account, as then the algorithm would take weeks to run. Thus a more subtle one thanthe naive exponential algorithm is needed to solve the problem efficiently. However, chancesare very low that there is some algorithm substantially faster than the brute force algorithm,because the problem is co-NP-complete; this is shown in this section. First, we need someadditional definitions.

DEFINITION 3.1. Let V be a set and letV 0 � V . ThenV 0 = V � V 0. For everyhypergraphH onV , the complemented hypergraph ofH is the hypergraphH = fH : H 2Hg onV .

The following relationships betweenH andH are straightforward, but nevertheless im-portant:

PROPOSITION 3.1. Let H be a hypergraph. ThenH = H, andH is simple iffH issimple.

DEFINITION 3.2. A hypergraphH is called self-complemented if and only ifH = H.Notice that if a hypergraphH is self-complemented, then for eachX � V we haveX 2 Cov(H) iff X 2 Cov(H).2-colorability of a hypergraph, also known as set splitting, is defined as follows.1Problems on pizze are studied elsewhere, cf. [24, p. 4].


DEFINITION 3.3. LetH be a hypergraph. A partitioning(A;B) of V , i.e.,A [ B = VandA \ B = ;, is a 2-coloring ofH iff for everyE 2 H, it holds thatE 6� A andE 6� B.H is 2-colorable iff there exists a 2-coloring forH.

In words, a hypergraph is2-colorable if and only if there is an assignment of one of2“colors” to each vertex such that each edge has two colors. Deciding whether a hypergraphH is 2-colorable (HYPERGRAPH 2-COLORABILITY, HP2C) is anNP-complete problem [36,21]. We prove that this problem remainsNP-complete for the following restriction:

THEOREM 3.2. HP2C remainsNP-complete even if the hypergraphH is self-comple-mented.

Proof. Consider the hypergraphH0 =(V 0; E 0), whereV 0 = V [fe; fg for somee; f 62 V;andE 0 = H [ fE [ fe; fg : E 2 Hg: It is clear thatH0 can be constructed fromH inpolynomial time; moreover,H0 is self-complemented. We claim thatH is 2-colorable if andonly if H0 is 2-colorable. To prove theonly if direction, let(A;A) be any 2-coloring ofH.If e,f are colored differently, only an edge ofH0 that is also an edge ofH can prevent a 2-coloring ofH0 . Hence(A [ feg; (V �A) [ ffg) is a 2-coloring ofH0. For theif direction,if (B;B) is any 2-coloring ofH0, it does not color any edge fromH monochromaticallybecauseH � H0, and since no edge ofH containse or f , clearly(B � fe; fg; B � fe; fg)is a 2-coloring ofH.

Using the relationship between 2-colorable hypergraphs and saturated hypergraphs, weshow now the intractability ofH-SAT. We refer to the following lemma.

LEMMA 3.3. LetH be a self-complemented hypergraph. ThenH is not 2-colorable iffH is saturated.Proof. If H is not 2-colorable, for every partitioning(A;A) of V there existsH 2 H

such thatH � A or H � A. As self-complementarity ofH implies thatA 2 Cov(H) iffA 2 Cov(H), it follows thatA;A 2 Cov(H); consequently,H is saturated, which provestheonly if direction. For theif direction, assumeH is 2-colored by(A;A). We show thatfor everyH 2 H, it holds thatA 6� H andA 6� H ; hence,A 62 Cov(H), which means thatH is not saturated. ConsiderH 2 H. Since(A;A) is a 2-coloring ofH, we haveA 6� H .If A � H , thenA � H, which by self-complementarity ofH contradicts that(A;A) is a2-coloring ofH. ThusA 6� H . This proves theif direction.

THEOREM 3.4. H-SAT is co-NP-complete, even ifH is self-complemented.Proof. Membership in co-NP is easy to show: Guess a subsetV 0 � V and check ifV 0 2Cov(H) holds. This can clearly be done in time polynomial inn andm. co-NP-hardness

under the asserted restriction follows immediately from Theorem 3.2 and Lemma 3.3.This result can be strengthened as follows.THEOREM 3.5. H-SAT remains co-NP-complete even ifH is self-complemented and all

edges have sizek or n� k, for fixedk � 3.Proof. HP2CremainsNP-complete even if no edge ofH contains more than three vertices

[21]. Applying the transformation in the proof of Theorem 3.2 on such a hypergraphH, weobtain a self-complemented hypergraphH0 = (V 0; E 0) such that everyE 2 H0 contains atmost3 vertices or at leastn0 � 3 vertices, wheren0 = jV 0j = n + 2, with the property thatH0 is saturated iffH is not 2-colorable. Hence,H-SAT remains co-NP-complete ifH hasthe form ofH0. Therefore, we assume in the following without loss of generality thatH isself-complemented and that every edge contains at most3 vertices or at leastn� 3 vertices.

We show first the result fork = 3. Without loss of generality assume thatn > 2k = 6.One can test in polynomial time if for eachX � V with jX j � 3, X 2 Cov(H) holds. IfH does not cover all the setsX , thenH is not saturated. Otherwise, replace inH every edgeE with jEj < 3 by all setsE0 of size3 such thatE � E0 � V , and every edgeF withjF j > n� 3 by all subsetsF 0 � F of sizen� 3. LetH00 be the resulting hypergraph. Notice


thatH00 is self-complemented and contains only edges of size 3 orn � 3; furthermore, itfollows from the construction ofH00 thatX 2 Cov(H00) for eachX � V such thatjX j � 3.

We claim thatH00 is saturated iffH saturated. To show this, assume thatH is not satur-ated. Then, sinceH is self-complemented, there existsX � V such that3 < jX j < n � 3andX =2 Cov(H). By construction ofH00, it follows thatX =2 Cov(H00). Thus,H00 is notsaturated. The proof of the converse direction is analogous.

SinceH00 can be constructed in polynomial time,H-SAT remains co-NP-complete underthe asserted restriction fork = 3.

For k > 3, the proof is analogous, where in each place “3” is replaced by “k”. Noticesincek is a constant, it can be checked in polynomial time if for eachX � V with jX j � k,it holds thatX 2 Cov(H) (there are fewer thanknk suchX), andH00 can be constructed inpolynomial time.

The following result is an immediate consequence of Theorem3.5 and Lemma 3.3.COROLLARY 3.6. HP2C is NP-complete even ifH is self-complemented and all edges

have sizek or n� k, for fixedk � 3.The casek = 3 in Theorem 3.5 marks the intractability frontier, as the case k = 2 is

polynomial.THEOREM 3.7. H-SAT is polynomial ifH is self-complemented and all edges have size2 or n� 2.Proof. Let G be the graph onV with edgesfE 2 H : jEj = 2g. We reduce deciding

saturation ofH to testing certain properties ofG, which can be done in polynomial time. Inwhat follows, we assume without loss of generality thatG is not empty (i.e.,H is not empty)and thatn � 4.

CLAIM A: Letv 2 V be arbitrary. Ifv belongs to all edges ofG, thenH is not saturatediff Ve(G) � V .

To prove theif direction of this claim, assume thatVe(G) � V . Let v0 2 V � Ve(G)be arbitrary. It is readily checked that the partitioning(fv; v0g; V � fv; v0g) is a 2-coloringof H. Thus, by Lemma 3.3,H is not saturated. This proves theif direction. For theonly ifdirection, assume thatH is not saturated. Then, by Lemma 3.3 there must exist a 2-coloringof H; let (B;B) be an arbitrary such 2-coloring. SinceG � H, (B;B) is a 2-coloring ofG.Assume thatVe(G) = V . Then, it follows thatB = fvg orB = fvg; assume without loss ofgenerality thatB = fvg. LetE 2 G be arbitrary. Sincev 2 E, it followsE � V �fvg = B.SinceH is self-complemented, we haveE 2 H; this implies that(B;B) is not a 2-coloringof H, which is a contradiction. Thus, it follows thatVe(G) 6= V , and henceVe(G) � V .

CLAIM B: If no v 2 V belongs to all edges ofG, thenH is not saturated iffG is 2-colorable.

Theonly if direction is immediate from Lemma 3.3. For theif direction, let(B;B) beany 2-coloring ofG. Since nov 2 V belongs to all edges ofG, the 2-coloring must satisfyjB \ Ve(G)j � 2 andjB \ Ve(G)j � 2. It follows thatB =2 Cov(G) andB =2 Cov(G). Sincethe latter is equivalent toB =2 Cov(G) andH = G [ G, we haveB =2 Cov(H). This meansthatH is not saturated, and theif direction is shown.

In summary, by Claims A and B, the problem of deciding whetherH is saturated isefficiently reducible to checking whetherVe(G) 6� V or to deciding whetherG is not 2-colorable. The former is trivially polynomial, and the latter is polynomial since checking2-colorability of a graph is polynomial [22].

It follows thatH-SAT is polynomial under the asserted restriction.

Simple hypergraph saturation. Toni has a brother Luigi, who also manages a bigpizzeria. Luigi is proud that each of his pizze is “originale” which means that in his res-taurant there are no Grande or Mini versions of pizzas. As he hears from the plans of his


brother, Luigi also intends to enlarge his pizza offer by a new pizza, which has, of course,to be a pizza originale. It is clear that Luigi’s problem is a restricted version of Toni’s prob-lem. Since all pizze are “originali”, in the corresponding hypergraph saturation problem thehypergraph of pizze is simple. Thus in terms of hypergraphs,we have

Problem: SIMPLE HYPERGRAPH SATURATION (SIMPLE-H-SAT)Instance: A simple hypergraphH = (V; E) on verticesV = fv1; : : : ; vng.Question: IsH saturated ?The question is if this restriction on the input makesH-SAT tractable. ThoughSIMPLE-

H-SAT is clearly in co-NP, a precise classification is still open.A saturation test for two simple hypergraphs is easily reducible to a test for one hyper-

graph.DEFINITION 3.4. LetH be a hypergraph onV and letv be a (possibly new) vertex.

The hypergraphHv is defined byHv = (V [ fvg; fH [ fvg : H 2 Hg). LetG andH besimple hypergraphs onV , and lete; f be two (possibly new) vertices. ThenCe;f (G;H) =(V [ fe; fg;Ge [ Hf [ fV; fe; fgg):

PROPOSITION3.8. LetG andH be simple hypergraphs which do not contain; or V asedges, and lete; f 62 V be distinct. ThenCe;f (G;H) is simple, andCe;f (G;H) is saturatediff G andH are saturated.

Proof. It is readily verified thatCe;f (G;H) is simple. To show the other property, assumefirst thatCe;f (G;H) is not saturated. That is, there existsX � V [ fe; fg such thatX =2Cov(Ce;f (G;H)). SinceV 2 Ce;f (G;H) andfe; fg 2 Ce;f (G;H), X must contain exactlyone ofe andf . If e 2 X , it follows thatX � feg =2 Cov(G); hence,G is not saturated.Analogously, iff 2 X , it follows thatH is not saturated. Thus theif direction holds. For theonly if direction, assume without loss of generality thatG is not saturated. LetX � V suchthatX 62 Cov(G). Then, we have thatX [ feg =2 Cov(Ce;f (G;H)), i.e.,Ce;f (G;H) is notsaturated.

We obtain thatSIMPLE-H-SATremains for the following subcase as hard as in the generalcase:

PROPOSITION 3.9. SIMPLE-H-SAT is �pm-equivalent to the subcase whereH is self-complemented.

Proof. Notice that a hypergraphH is saturated if and only ifH is saturated: Indeed, foreveryX � V , we haveX 2 Cov(H) iff X 2 Cov(H). Since the hypergraphCe;f (G;H)is self-complemented forH = G, the result follows from Proposition 3.8. In terms ofcolorability, the following result can be formulated as an immediate consequence of Propos-ition 3.9 and Lemma 3.3. Observe thatHP2C for simple hypergraphs is stillNP-hard, as ahypergraphH is 2-colorable iffmin(H) is 2-colorable.

COROLLARY 3.10. HP2Cfor simple and self-complementedH is�pm-equivalent to co-SIMPLE-H-SAT.

4. Relating hypergraph transversals and saturation.Problems involving hypergraphtransversals appear in many practical applications (cf. Section 6), so their computational com-plexity is of major interest. Moreover, several computational problems on hypergraphs canbe reformulated in terms of transversals; for example, finding a 2-coloring for a hypergraphH is equivalent to finding a transversalT of H such thatT is also a transversal.

In this section we give first a characterization of saturatedhypergraphs in terms of atransversal relation and then we will consider the problem of transversal recognition, wherewe will show that this problem is�pm-equivalent toSIMPLE-H-SAT. Moreover, we will invest-igate into necessary and sufficient criterions for simple hypergraph saturation and transversalhypergraph recognition, and we will show how the latter problem reduces to a very specificsubcase.


4.1. Definitions and basic facts.We need some additional definitions first.DEFINITION 4.1 (cf. [7]). LetH andG be two hypergraphs. ThenH � G iff for everyH 2 H there existsG 2 G such thatH � G, 2 andH � G iff for everyH 2 H there existsG 2 G such thatH � G.For every hypergraphH, ; � H, andf;g � H iff ; 2 H. Note thatH � G (resp.H �G) does not implyG � H (resp.G � H); for example, letH = ff1gg andG = ff1g; f2gg.� is in some sense dual to� with respect to complementation. For every hypergraphH,H � fV g, andfV g � H iff V 2 H. Notice that� and� are reflexive, transitive, and define

partial orders on the simple hypergraphs onV . We observe the following properties:PROPOSITION4.1. LetG andH be hypergraphs. Then,G � H () G � min(H) () min(G) � min(H);G � H () G � max(H) () max(G) � max(H):We have the following characterization of saturated hypergraphs.THEOREM 4.2. LetH 6= ; be a hypergraph. Then the following statements(i)–(iii) are

equivalent:(i) H is saturated. (ii) Tr(max(H)) � min(H) (iii) Tr(H) � HProof. (i) ) (ii). AssumeH is saturated. ConsiderT 2 Tr(max(H)). We haveT 2 Cov(H), but sinceT is a transversal ofmax(H), for everyE 2 H it holds thatT 6� E.

But thenE0 � T for someE0 2 H, which impliesT � E00 for someE00 2 min(H). Itfollows thatTr(max(H)) � min(H).(ii) ) (iii). Check thatmax(H) = min(H). Hence,Tr(max(H)) = Tr(min(H)) =Tr(H). SinceTr(max(H)) � min(H), it follows from Proposition 4.1 thatTr(H) � H.(iii) ) (i). AssumeH is not saturated, i.e., there existsX � V such thatX 62 Cov(H).SinceX 6� E for everyE 2 H, X is a transversal ofH. Let X 0 � X be a minimaltransversal ofH. Now consider anyE 2 H; sinceE 6� X , we haveE 6� X 0. Consequently,Tr(H) 6� H.

COROLLARY 4.3. To test ifTr(H) � H holds for a hypergraphH is co-NP-complete,even ifH is self-complemented and contains only edges of sizek andn� k, for fixedk � 3.

Proof. Membership in co-NP follows since a guess forT 2 Tr(H) such that for everyE 2 H, T 6� E can be verified in polynomial time. co-NP-hardness under the assertedrestriction follows from Theorems 4.2 and 3.5.

We remark that the equivalence of(i) and (iii) in Theorem 4.2, restricted to simplehypergraphs, is implicit in a result of [51], which investigates Sperner families in the con-text of relational database theory. Notice that this criterion does not describe the transversalhypergraphexplicitly. We will provide such a description later in this section.

4.2. The main problem. BesidesSIMPLE-H-SATthe recognition of the transversal hyp-ergraph is the second issue of central concern in our study. We will show now that the com-plexity of this decision problem is of importance for computing the transversal hypergraph,and then we will demonstrate�pm-equivalence ofSIMPLE-H-SATand transversal hypergraphrecognition. Let us start with a precise problem statement:

Problem: TRANSVERSAL HYPERGRAPH (TRANS-HYP)Instance: Two hypergraphsG = (V; E1) and H = (V; E2) on verticesV =fv1; : : : ; vng.Question: DoesH = Tr(G) hold ?2On [7, p. 45]� is denoted by�.


The problem is in co-NP, since a guess forT 2 Tr(G) such thatT =2 H can be verifiedin polynomial time. However, there is no trivial evidence that the problem is inNP, even lessthat it is inP. We observe thatTRANS-HYPefficiently reduces to the following subproblem:

PROPOSITION4.4. TRANS-HYP is�pm-equivalent to the subcase in whichG andH aresimple.

Proof. Note thatH = Tr(G) implies thatH is simple. Furthermore,Tr(G) = Tr(min(G)).Sincemin(G) is a simple hypergraph and computable in polynomial time, the result follows.

The significance ofTRANS-HYPon the related search problem, namely the computationof the transversal hypergraph, is captured by the followingtheorem.

THEOREM 4.5. If TRANS-HYP is not in P, then no output-polynomial total time al-gorithm for computing the transversal hypergraph exists.

Proof. Assume that an algorithmA computesTr(F) fromF in output-polynomial totaltime. Let its runtime be bounded by a polynomialp(IS(F); N); whereIS denotes the inputsize andN = jTr(F)j. To solveTRANS-HYP, construct an algorithmA0 which works asfollows: A0 simulatesA applied onG, but does this for at mostp(IS(G); jHj) steps. IfAdid not terminate withinp(IS(G); jHj) steps, thenH 6= Tr(G). Otherwise,A0 compares thehypergraph output byA toH; these hypergraphs are identical iffH = Tr(G). It is easy to seethat the runtime ofA0 can be bounded by a polynomialq(IS(G); jHj). (The book-keepingbetween simulation steps can be done in polynomial time.) Hence,A0 is polynomial withrespect to the input size. The result follows.

4.3. Characterizations of the transversal hypergraph.Since one can test in polyno-mial time for everyX � V whetherX is a minimal transversal for a given hypergraphH, wewonder if we can infer thatG = Tr(H) if all edges ofG are minimal transversals ofH andvice versa, since this can be checked in polynomial time. Theanswer is no:

PROPOSITION4.6. Let G andH be simple hypergraphs. ThenH � Tr(G) andG �Tr(H) does not implyH = Tr(G).Proof. We give a counterexample. LetV = fa; : : : ; gg. Consider the hypergraphsH = f fa; b; cg; fa; d; gg; fa; e; fg; fb; d; fg; fb; e; gg; fc; d; eg; fc; f; gg;fd; e; f; gg; fb; c; e; fg; fb; c; d; gg; fa; c; e; gg; fa; c; d; fg; fa; b; f; gg g;G = f fa; b; c; dg; fa; b; c; eg; fa; b; c; fg; fa; b; c; gg; fa; b; d; gg; fa; c; d; gg;fa; d; e; gg; fa; d; f; gg; fa; b; e; fg; fa; c; e; fg; fa; d; e; fg; fa; e; f; gg;fa; b; d; fg; fb; c; d; fg; fb; d; e; fg; fb; d; f; gg; fa; b; e; gg; fb; c; e; gg;fb; d; e; gg; fb; e; f; gg; fa; c; d; eg; fb; c; d; eg; fc; d; e; fg; fc; d; e; gg g:

It holds thatH � Tr(G) andG � Tr(H), but G 6= Tr(H) asfc; f; gg 2 Tr(H) andfc; f; gg =2 G. 3For further characterizations of the transversal hypergraph of a simple hypergraph we

introduce a special delete operator which removes from every edge of a hypergraph a vertexin all possible ways:

DEFINITION 4.2. Let V be a finite set, and letHyp(V ) = 22V denote the set ofall hypergraphs onV . The mapping� : Hyp(V ) ! Hyp(V ) is defined by�(H) =(V; fH � fvg : H 2 H; v 2 Hg).3H = P7 [ P7 � ffa; b; d; eg g andG = fE [ fvg : E 2 P7 � ffc; f; ggg; v 2 Eg for the hypergraphP7 = ffa; b; cg; fc; d; eg; fb; d; fg; fa; e; fg; fa; d; gg; fb; e; gg; fc; f; ggg on V = fa; : : : ; gg from [7,p.47,Fig.1].


EXAMPLES: For the hypergraphH = ff1g,f2,3g,f1,3,4gg on V = f1; 2; 3; 4g, we have�(H) = f;; f2g; f3g; f1; 3g; f1; 4g; f3; 4gg. Note in particular that�(;) = ; and that�(f;g) = ;.We observe the following relationships.PROPOSITION 4.7. Let G andH be simple hypergraphs such thatVe(G) 6= ; andVe(H) 6= ;. Then

(i) H � Tr(G) impliesH � �(G), but the converse does not hold.(ii) �(H) � Tr(H), i.e., every edge of�(H) is a transversal ofH.(iii) Tr(H) � �(H), i.e., every minimal transversal ofH is contained in some edge of�(H).

Proof.(i): ConsiderT 2 H. SinceT 2 Tr(G), it follows that for everyv 2 T there existsG 2 Gsuch that(T�fvg)\G = ;; hence,T�fvg � G: This implies thatT � G[fvg = G� fvg,from which it follows immediately thatH � �(G).

The following example shows that the converse does not hold.ConsiderH = ff1gg andG = ff1g; f2gg onV = f1; 2g. Then,�(G) = ff1; 2gg andH � �(G), butH 6� Tr(G) =ff1; 2gg:(ii): ConsiderH 2 H. H is not a transversal ofH, but sinceH is simple, it is a transversalof the hypergraphH � fHg. Thus for everyv 2 H , H [ fvg is a transversal ofH; hence,H [ fvg = H � fvg contains a minimal transversal ofH. The claim follows immediatelyfrom the definition of�.(iii): For every minimal transversalT of H and every vertexv 2 T there exists an edgeHofH such thatT \H = fvg; hence,T � H [fvg; butH [fvg = H � fvg 2 �(H).

Note that in the above proposition(i) states a necessary, but not sufficient conditionfor TRANS-HYP. With the �-operator we have the following important characterization ofsaturated simple hypergraphs:

THEOREM 4.8. Let H be a simple hypergraph. ThenH is saturated if and only ifTr(H) = min(�(H)).Proof. We prove theonly if direction first. AssumeH is saturated, butTr(H) 6=min(�(H)). Note that this assumption impliesH 6= ; andH 6= f;g, henceVe(H) 6= ;.CLAIM A: Tr(H) 6� min(�(H)). Proof: Suppose to the contrary thatTr(H) �min(�(H)); hence, sinceTr(H) 6= min(�(H)), there exists an edgeE 2 min(�(H)) �Tr(H). From Proposition 4.7(ii), it follows thatE is a transversal ofH; thus,E con-

tains someT 2 Tr(H). Consider any suchT . Sincemin(�(H)) is simple, it follows thatT =2 min(�(H)). Hence,T gives rise to a contradiction ofTr(H) � min(�(H)). This provesthe claim.

From Claim A and the assumptionTr(H) 6= min(�(H)), it follows that there existsT 2 Tr(H) such thatT 2 Tr(H)�min(�(H)).CLAIM B: H \ T 6= ;. Proof: Assume to the contrary thatH \ T = ;, i.e. thatH � T .

Clearly,H is not a transversal ofH; however, sinceH is simple, for everyv 2 H it holds thatH[fvg is a transversal ofH. From this andH � T , T 2 Tr(H) we infer thatT = H[fv0gfor somev0 2 H . Hence,T 2 �(H), which means thatT is a superset of some edge inmin(�(H)). SinceT 2 Tr(H), we infer from Proposition 4.7(ii) thatT 2 min(�(H)). Wereached a contradiction. ThusH 6� T , i.e.H \ T 6= ;.

Since for eachH 2 H, a vertexe exists such thatE = H [ feg 2 �(H), it follows fromClaim B thatT is a transversal ofH. SinceH is saturated, we have by Theorem 4.2(iii) thatTr(H) � H. Thus there existsH 2 H such thatT � H , which is equivalent toT � H . Butthis means thatT is not a transversal ofH; this is a contradiction to the existence ofT whichfollowed from Claim A. Thus, the assumptionTr(H) 6= min(�(H)), from which we derived


Claim A, is not consistent. This proves theonly if direction.To prove theif direction, assume thatTr(H) = min(�(H)), butH is not saturated.

Notice thatH 6= ;. SinceH is not saturated, there exists a setT � V such thatT 62 Cov(H).In particular, we have for everyH 2 H thatT 6� H , and henceT \H 6= ;. This means thatT is a transversal ofH. From the assumptionTr(H) = min(�(H)), we infer thatT � E forsomeE 2 min(�(H)). By the definition of�(H), we have thatE = H [ fvg for someHandv such thatH 2 H andv 2 H . Hence,T � H [fvg, and thusT � H . The latter meansT � H , which impliesT 2 Cov(H). Thus we reached a contradiction to the existence ofT .This proves theif direction.

Given a simple hypergraphH, one can easily computeH0 = min(�(H)) in polynomialtime. Since checking whether each edge ofH is a minimal transversal ofH0 and vice versais possible in polynomial time, we wonder if this necessary condition for saturation ofH isalso sufficient. Unfortunately, this does not hold:

PROPOSITION 4.9. Let H be a simple hypergraph, and letH0 = min(�(H)). ThenH � Tr(H0) andH0 � Tr(H) does not imply thatH is saturated. The same holds ifH isself-complemented.

A counterexample for unrestrictedH is not hard to find (e.g., in the proof of Proposi-tion 4.6,G = min(�(H)) ). Imposing self-complementarity onH makes this more challen-ging, in particular to find a counterexample with as few edgesor vertices as possible; we canoffer an instance ofH with 308 edges on22 vertices.

It follows immediately from Theorem 4.8 thatSIMPLE-H-SAT is�pm-transformable intoTRANS-HYP. The next theorem implies the converse, which establishes thatSIMPLE-H-SATandTRANS-HYPare�pm-equivalent. We use the following simple, but helpful lemma:

LEMMA 4.10. If H is simple, then for eachE 2 Tr(H), there exists noE0 2 H suchthatE0 � E.

Proof. Indeed, ifE 2 Tr(H), thenE \E0 6= ; for all E0 2 H; hence,E0 �E 6= ;, i.e.,E0 6� E.THEOREM 4.11. LetG1 andG2 be two simple hypergraphs onV , and lete; f 62 V be

distinct. Define a hypergraphH = (V 0;F1 [ F2), whereV 0 = V [ fe; fg, F1 = G1, andF2 = fE [ fe; fg : E 2 G2g. ThenG1 = Tr(G2) iff H is simple and saturated.Proof. For theonly if direction, letG1 = Tr(G2) hold. We first show thatH is simple.

SupposeH is not simple. SinceF1;F2 constitute simple hypergraphs, there must existF1 2 F1; F2 2 F2 such thatF1 � F2 or F2 � F1. By Lemma 4.10 and the constructionof F1 andF2 it follows thatF1 � F2 is impossible to hold. However,F2 � F1 is alsoimpossible becausee 2 F2 but e 62 F1. Thus we have reached a contradiction, andH issimple. Now we show thatH must also be saturated. Assume this is not the case, i.e., thereexistsX � V 0; X 62 Cov(H). Since for allF 2 F2 we haveX 6� F , it follows thatX is a

transversal ofF2 = G2 = G2. SinceF1 constitutesTr(G2), this implies thatE � X for someE 2 F1, which means thatX 2 Cov(H), a contradiction. ThusH is saturated; theonly ifdirection is proved.

To prove theif direction, assume thatH is simple and saturated, butG1 6= Tr(G2).We show thatG1 6� Tr(G2). Assume thatG1 � Tr(G2). Hence, there exists a setE 2G1 � Tr(G2). SinceG1 � Tr(G2) andG1 is simple, it follows thatE is not a transversalof G2. SinceE is not a transversal ofG2, there exists someE0 2 G2 such thatE \ E0 = ;;hence,E � E0[fe; fg. By construction ofH, we have thatE;E0[fe; fg 2 H. This meansthatH is not simple, which is a contradiction. Thus it follows thatG1 6� Tr(G2).

SinceG1 6= Tr(G2) andG1 6� Tr(G2), there exists aT 2 Tr(G2) � G1. ConsiderT [ feg. BecauseT is a transversal ofG2 =F2, it is clear that there is noF 2 F2 suchthatT [ feg � F . There is noF 2 F2 such thatT [ feg � F sincef 2 F; f 62 T , and


there is noF 2 F1 with the propertyT [ feg � F becausee 62 F . SinceH is saturated,this implies that there is someF 2 F1 such thatF � T [ feg, from which clearlyF � Tfollows. BecauseH is simple andF2 � H, it holds for everyG 2 F2 thatF 6� G; thus,F isa transversal ofF2 = G2. SinceT 2 Tr(G2), it follows thatF = T . Hence,T 2 G1, whichis a contradiction toT 2 Tr(G2)� G1. This proves theif direction.

Now the above stated relationship betweenSIMPLE-H-SATandTRANS-HYP is easy toestablish:

THEOREM 4.12.SIMPLE-H-SATis�pm-equivalent toTRANS-HYP.Proof. Theorems 4.8 and 4.11 imply thatSIMPLE-H-SAT�pm-reduces toTRANS-HYP

and vice versa.

4.4. Selftransversal hypergraphs.As announced above, we show that recognizing thetransversal hypergraph�pm-reduces to theSELFTRANSVERSALITYproblem, which is as fol-lows.

Problem: SELFTRANSVERSALITYInstance: A hypergraphH = (V; E) on verticesV = fv1; : : : ; vng.Question: DoesH = Tr(H) hold ?Selftransversal hypergraphs have attracted much interestby mathematicians, cf. [36, 6,

7]. The simplest example of such a hypergraph isH = (fxg; ffxgg). Another exampleis the hypergraphH = ffx; yg; fx; yg, fy; zg g. For a thorough study of selftransversalhypergraphs, the reader is referred to [7, chapter 2]. We have the following result.

THEOREM 4.13.To decide if a hypergraphH satisfiesTr(H) = H is�pm-equivalent toSIMPLE-H-SAT.

Proof. We show thatTRANS-HYP is �pm-reducible toSELFTRANSVERSALITY, fromwhich by Theorem 4.12 the result follows. Let without loss ofgenerality beG andH twosimple, nonempty hypergraphs onV and lete; f =2 V be two new distinct vertices. Define ahypergraphD onV 0 = V [ fe; fg byD = Ge [ Hf [ ffe; fgg. We claim thatG = Tr(H)iff D = Tr(D).

To prove this, considerTr(D). We have thatfe; fg 2 Tr(D) and alsofe; fg 2 D.Every T 2 Tr(D) � ffe; fgg distinct from fe; fg contains eithere or f , but not both.Assume thatT containse, but notf . Since no edge inHf containse and since all other edgesin D � Hf containe, we infer thatT � feg must be a transversal ofH. Conversely, ifT 0is a transversal ofH, thenT 0 [ feg is easily shown to be a transversal ofD. This impliesthat the minimal transversals ofD which containe but notf are given byT e = fE [ feg :E 2 Tr(H)g = Tr(H)e. From this it is immediately verified thatG = Tr(H) if and only ifGe = T e. In the same way one can show that the setT f of minimal transversals ofD whichcontainf but note satisfiesT f = Hf if and only ifH = Tr(G). SinceG andH are simple,by Proposition 2.2G = Tr(H) iff H = Tr(G). Hence, sinceTr(D) = T e [ T f [ ffe; fgg,we have thatG = Tr(H) iff D = Tr(D); the claim is proved.

From this result, we obtain by the close relation between 2-colorable intersecting hyper-graphs and selftransversal hypergraphs [7] the following result on deciding 2-colorability ofintersecting hypergraphs.

COROLLARY 4.14.HP2Cfor intersecting hypergraphs is�pm-equivalent to co-SIMPLE-H-SAT.

Proof. Notice thatH is intersecting iffmin(H) is intersecting, and thatH is 2-colorableiff min(H) is 2-colorable. Thus without loss of generality we may assume thatH is simpleand intersecting, and in addition, thatH contains no loops, i.e. singleton edges. A simple in-tersecting hypergraphH without loops satisfiesH = Tr(H) if and only if it is not 2-colorable[7]. Thus, ifH is intersecting, 2-colorability ofH is�pm-reducible to co-SELFTRANSVERSALITY,which by Theorem 4.13 is�pm-equivalent to co-SIMPLE-H-SAT.


5. Polynomial cases.Although we do not know whetherSIMPLE-H-SATandTRANS-HYP are intractable in their general problem statement, in practical occurrences there are(natural) restrictions on the instances of a problem such that the problem reduces to a sub-problem of the general problem. Since a subproblem might be computationally easier, thereis hope to get efficient algorithms for it even if the more general problem is intractable. Forexample, in the pizza baker’s problem it seems quite reasonable to assume that all pizze haveapproximately the same number of food items; in this case, however, the problem becomespolynomial-time solvable for both Toni and Luigi. Narrowing the “frontier” between the in-tractable general problem and tractable subcases [21], we identify some important subcasesof SIMPLE-H-SATandTRANS-HYPwhich are inP, among them the recognition of the min-imal transversals of a hypergraph with bounded edge size andhypergraph saturation if thedifference between rank and antirank is bounded. In particular, we present algorithms foroutput-efficient computation of the transversal hypergraph, if the edge-size is small (prob-ably the most important of the restrictions treated for practice) and if it is large, or if thehypergraph is acyclic.

5.1. Restrictions on the edge size.Let us consider restrictions on the size of the edgesof a hypergraph. A straightforward restriction of this kindwhich is important in practice is aconstant upper bound. Fortunately, the transversal hypergraph is output-efficiently comput-able in this case, which implies that the minimal transversals of such a hypergraph can berecognized in polynomial time and that also saturation of a simple hypergraph of this form isefficiently decidable.

DEFINITION 5.1. LetH be a hypergraph on verticesV , and letx 2 V . The degreed(x;H) of vertexx in hypergraphH is defined asd(x;H) = jfE 2 H : x 2 Egj.In the proof we apply a result from combinatorial studies by Berge and Duchet:LEMMA 5.1 ([7, p.58, Corollary 1]).LetG be a simple hypergraph, and letk � 2 be an

integer. Thenr(Tr(G)) � k iff for all G0 � G with jG0j = k + 1 there exists anE 2 G s.t.E � fv 2 V : d(v;G 0) > 1g:THEOREM 5.2. LetH be a hypergraph. If the size of the edges ofH is bounded by some

constantk, i.e.,r(H) � k, thenTr(H) is computable in incremental-polynomial time.Proof. We describe an algorithm which has the desired property. The following char-

acterization ofTr(H) for a simple, nonempty hypergraphH is obtained from Lemma 5.1.Let r = r(H), and for any hypergraphF and integeri � 0 denote byFji the hypergraphF i = fE 2 F : jEj � ig. Then,G = Tr(H) () (G � Tr(H)) ^ P ^ Q ;(1)

where P � Tr(G)jr � H;Q � there exists noG0 � G with jG0j = r + 1 such that for allE 2 G,E 6� fx 2 V : d(x;G 0) > 1g.It is readily checked that (1) holds ifr � 1. We show that (1) also holds ifr � 2. If G=Tr(H) andH is simple, thenH = Tr(G), and theonly if direction follows by Lemma 5.1.For theif direction, ifH is simple,G � Tr(H), andP holds, thenH � Tr(G). To show this,considerH 2 H. SinceG � Tr(H), H is a transversal ofG. LetH 0 � H be an arbitraryminimal transversal ofG. SincejH 0j � r, it follows from P thatH 0 2 H; asH is simple, itfollows thatH 0 = H . Hence,H � Tr(G).

Since by validity ofQ and Lemma 5.1r(Tr(G)) � r, it follows from P thatH = Tr(G).We derive now from (1) a method for incrementally computingTr(H), whereH is

simple and nonempty. IfG � Tr(H) andG 6= Tr(H), thenP or Q is false. In this case,someT1 2 Tr(H) such thatT1 =2 G can be found as follows.

MINIMAL TRANSVERSALS OF A HYPERGRAPH 1293(i) Assume thatP is false. Hence, there existsT 2 Tr(G)jr �H.CLAIM A: T is a transversal ofH. Proof: Notice thatT is not a transversal ofH iff there

exists anH 2 H such thatT \H = ;, which is equivalent toH � T . SinceG � Tr(H),His a transversal ofG. AsT 2 Tr(G)jr �H, T is a minimal transversal ofG. SinceH � T , itfollows thatH = T ; hence,T 2 H. This is a contradiction. Consequently,T is a transversalof H.

CLAIM B: T =2 Cov(G). Proof: Assume to the contrary thatT 2 Cov(G). Assumefurthermore that there existsG 2 G such thatG � T . This impliesG \ T = ;, which meansthatT is not a transversal ofG. However, by definitionT 2 Tr(G)jr � H, and thusT is atransversal ofG. We reached a contradiction, and thusG does not exist. Assume otherwisethat there existsG 2 G such thatT � G. This implies thatG =2 Tr(H), which contradictsG � Tr(H). This shows thatT =2 Cov(G).

By Claims A and B,T is a transversal ofH such thatT =2 Cov(G). ThusT containssomeT1 2 Tr(H) such thatT1 =2 G; in fact, everyT1 2 Tr(H) such thatT1 � T has thisproperty.(ii) Assume thatQ is false. Hence, there existsG0 � G with jG 0j = r + 1 such that foreveryE 2 G we haveE 6� fx 2 V : d(x;G 0) > 1g. Note that this is only possible ifjGj > r(H), which implies thatr(H) � 2. By Lemma 5.1, it follows thatr(Tr(G)) > r,thusH 6= Tr(G).

CLAIM C: T = fx 2 V : d(x;G 0) > 1g is a transversal ofH andT =2 Cov(G). Proof:Note that ifT is a transversal ofH, thenT =2 Cov(G), as no edge ofG is contained inT ,and, on the other hand, for noE 2 G it holds thatT � E asE 2 Tr(H). Thus to prove ourclaim, it remains to show thatT is a transversal ofH. Assume thatT is not a transversal ofH. Hence, there exists someE 2 H such thatE \ T = ;, i.e.,E � T . However, this isimpossible: SinceT contains no edge ofG, T is a transversal ofG, and thus also a transversalof G0. From the definition ofT , each vertex ofT appears in at most one edge ofG0. SinceG0 hasr + 1 edges, it follows that no transversalT 0 of G0 with T 0 � T can have fewer than

Algorithm Trans (H)input : simple, nonempty hypergraphH onV .output: Tr(H) edge by edge, i.e. incrementally all minimal transversals of H.r r(H); G ;;

loop = � G � Tr(H) � =if (there exists aT 2 Tr(G)jr �H) then

minimizeT to a minimal transversalT1 ofH;else= � H � Tr(G);G � Tr(H) � =

if (there existsG0 � G with jG0j = r + 1 such thatfor all E 2 G, E 6� fx 2 V : d(x;G0) > 1g)

thenminimizefx 2 V : d(x;G0) > 1g to a minimal transversalT1 ofH;

else= � G = Tr(H) � =exit loop;

fi;fi;output(T1); G G [ fT1g;

endloop;TABLE 1

Algorithm computingTr(H) in incremental-polynomial time ifr(H) � k, for constantk � 0.r + 1 vertices. Check that everyH 2 H is a transversal ofG 0; sincejH j � r, it follows thatE 6� T , which is a contradiction. ThusT is a transversal ofH, and the claim is proved.


SinceT =2 Cov(G), T contains someT1 2 Tr(H) such thatT1 =2 Cov(G); in fact, everyT1 2 Tr(H) such thatT1 � T has this property.

Utilizing this result, the algorithmTrans in Table 1 incrementally outputs the minimaltransversals of a simple, nonempty hypergraphH. The correctness ofTrans can be easilyproved from (1) and the described additional transversal determination; note thatG � Tr(H)is a loop invariant.

To prove the theorem, assume thatr(H) � k. If H is not simple, computemin(H) inpolynomial time and proceed with that hypergraph. Sincek is a constant, the conditions ofbothif statements can be checked in time polynomial in the size ofG,H andV . nk is an upperbound on the number of vertex sets to check in the outerif , andjGjk+1 is an upper bound forthe number of subhypergraphsG 0 � G (each of them computable in polynomial time) to testin the innerif . In both cases, all tests are clearly polynomial. Furthermore,T1 is computablein polynomial time (cf. Section 2). Thus an execution of the loop body needs time polynomialin the size ofG;H andn. Every pass of the loop but the last yields an additional minimaltransversalT1 of H for output. The result follows.

The following result is an immediate consequence of the lasttheorem.COROLLARY 5.3. TRANS-HYPis polynomial ifr(H) � k or r(G) � k, for fixedk � 0.Note that Theorem 5.2 generalizes the well-known result that the maximal independent

sets of a graph are output-efficiently computable to hypergraphs with edge-size bounded bya constant. (Recall that a maximal independent set is the complement of a minimal transver-sal). Thus the question in [29], whether there is an output-polynomial total time algorithmfor generating all maximal independent sets of a hypergraph, is affirmatively answered forhypergraphs of bounded edge-size.

Let us now consider a constant upper bound on the edge size. Inthis case the computationof the transversal hypergraph is efficiently possible even in the input size.

THEOREM 5.4. If ar(H) � n � k, for fixedk � 0, thenTr(H) is computable ininput-polynomial time.

Proof. Without loss of generality we assume thatH is simple. We know from Propos-ition 4.7(iii) thatTr(H) � �(H). Sincer(�(H)) � k + 1, only the setsV 0 � V of up tok + 1 elements are candidates for minimal transversals, and there are no more than roughlynk+1 of them.

From Theorems 5.4 and 4.8, we immediately have the followingresult.COROLLARY 5.5. TRANS-HYPandSIMPLE-H-SATare polynomial ifar(H) � n � k,

for fixedk � 0.As the third restriction, let us consider hypergraphs wherethe edges differ in their size

by at most some given constant. We note a simple, but important lemma.LEMMA 5.6. LetH be a hypergraph onV , anda = ar(H); r = r(H). ThenH is not

saturated if and only if there exists a setV 0 � V , a � jV 0j � r, such thatV 0 62 Cov(H).Proof. Assume there existsV 0 � V; such thatV 0 62 Cov(H). If jV 0j < a, we may add

any verticesv1; : : : ; va�jV 0j from V � V 0 to V 0, andV 0 [ fv1; : : : ; va�jV 0jg 62 Cov(H) willhold. If jV 0j > r, we may remove any verticesv1; : : : ; vjV 0j�r from V 0 without establishingV 0�fv1; : : : ; vjV 0j�rg 2 Cov(H): Thus theonly if direction holds. Theif direction is trivial.

THEOREM 5.7.H-SAT is polynomial ifr(H)� ar(H) � k, for fixedk � 0.Proof. Let a = ar(H), andr = r(H). There is a simple saturation algorithm which

is polynomial for this restriction: Check for eachX � V , jX j = a; a + 1; : : : ; r, whetherX =2 Cov(H), and output “no” if the first suchX is found and “yes” if none is found. ByLemma 5.6, this algorithm is correct.


Let us analyze the complexity of this algorithm. ConsiderY 2 H. The exact number ofthe setsX covered byY is given byC(Y; a; r) = r�jY jXi=0 �n� jY ji �+ jY j�aXi=0 �jY ji �� 1;(2)

where the first (resp. second) term on the right hand side of the equation equals the numberof the setsX covered as supersets (resp. subsets). The numberC(Y; a; r) is bounded by thesum of the number of setsX � V with jX j � r that a fixedY 0 � V with jY 0j = a coversas superset and the number of setsX � V with jX j � a that a fixedY 0 � V with jY 0j = rcovers as subset. Letc = r � a+ 1. Thus,C(Y; a; r) � r�aXi=0 ��n� ai �+�ri�� 2c �max��n� ai �;�ri� : 0 � i � c� 1�� 2cnc�1:The last inequation uses thatu � v implies

�uw� � �vw� and that�uw� � uw. As c � k+1, we

haveC(Y; a; r) � s(n) for some polynomials(n). Hence,H covers at mostm�s(n) of thesetsX .

Note that the subsets ofV of sizei can be systematically output with polynomial-timedelayp(n) (cf. [12] for a suitable order). Moreover, testingX 2 Cov(H) can be done inpolynomial timeq(m;n).

Hence ifH is not saturated, the algorithm finds the firstX such thatX =2 Cov(H) intimeO(m�s(n)(p(n) + q(m;n))), i.e., in polynomial time. IfH is saturated, then all setsX are checked. Letf(n; a; r) = �na� + � na+1� + � � � + �nr� denote their number. SinceH issaturated, it covers each of the setsX . On the other hand,H covers at mostm �s(n) of thesetsX ; consequently,f(n; a; r) � m�s(n). This ensures that the algorithm again terminatesin timeO(m�s(n)(p(n) + q(m;n))).

COROLLARY 5.8 (to Theorem 5.7).SIMPLE-H-SATis polynomial ifr(H)�ar(H) � k,for fixedk � 0.

5.2. Acyclic hypergraphs. Like in graph theory, the notion of acyclicity is appealing inhypergraph theory from a theoretical as well as a practical point of view. In some contexts,acyclic hypergraphs gain special attention. E.g., in relational database theory acyclic hyper-graphs (also calledtree schemes) were introduced in [8] and later used by many authors (cf.[53, 30]), especially in relational database design, cf. [2, 4, 5, 17].

SeveralNP-complete problems on hypergraphs become polynomial for acyclic hyper-graphs [57]. Since it is not straightforward to carry over the definition of a cycle from graphsto hypergraphs, there are many notions of acyclicity in a hypergraph, cf. [16, 18, 7]. We referto �-, �-, -, andBerge-acyclicity as stated in [16], where the proper inclusion hierarchyBerge-acyclic) -acyclic) �-acyclic) �-acyclic is proved.

The notion of�-acyclicity came up in the context of relational database theory [23, 58, 5,16]. A hypergraphH is �-acyclic iff H = ; orH is by theGYO-reduction (due to Graham[23] and (independently) to Yu and Ozsoyoglu [58]), that is by repeated application of one ofthe two rules:

(1) If vertexv occurs in only one edgeE, removev fromE;(2) If distinct edgesE;E0 satisfyE0 � E, removeE0,


f de

c

b

a

FIG. 1.�-acyclic, but�-cyclic hypergraphH = ffa; b; cg; fc; d; eg; fa; e; fg; fa; c; egg.reducible to the hypergraphf;g.

For example, the hypergraph in Figure 1 is�-acyclic. Notice that�-acyclicity of ahypergraphH can be checked with an algorithm of Tarjan and Yannakakis [50] in timeO(n+t), wheren is the number of vertices andt is the total size of the edges ofH.

Note that rule 2 implies that a hypergraphH is �-acyclic if and only ifmax(H) is �-acyclic. Consequently, an�-acyclic hypergraphH may contain an�-cyclic subhypergraphH0 � H. For example, let in Figure 1 beH0 = ffa; b; cg; fc; d; eg; fa; e; fgg. However,this is not what one expects from acyclicity, as it is surprising that a cycle disappears byadding edges. Fagin points out this anomaly [16] and introduces the more natural concept of�-acyclicity.

DEFINITION 5.2. A hypergraphH is �-acyclic iff every subhypergraphH0 � H is�-acyclic.The hypergraph in Figure 1 is�-cyclic, i.e., not�-acyclic. Fagin presents in [16] various

equivalent definitions of�-acyclicity, among them a related acyclicity criterion by Graham[23], and he also gives a polynomial-time algorithm to test�-acyclicity.

Our next aim is to show that the transversal hypergraph of a�-acyclic hypergraph isefficiently computable with respect to the input and the output size.

DEFINITION 5.3. LetH be a hypergraph onV . For everyV 0 � V; the partial hyper-graph ofH generated byV 0 isHV 0 = fE \ V 0jE 2 Hg:

For convenience, let us call any vertex anear nodeof the hypergraphH if it occursin exactly one edge ofH (cf. [53, p.698]). We observe the following rule for transversalcomputation.

LEMMA 5.9. LetH be a simple hypergraph onV with an ear nodev that occurs in edgeE. Thenffvg [ T : T 2 Tr(HE � f;g)g [ Tr(HV�fvg) is a partitioning ofTr(H).Proof. If H is simple, then every essential vertexv 2 V occurs in at least one minimal

transversal ofH. (This is readily shown from Proposition 2.2.) However,Tr(HE � f;g)is never empty. Now considerTr(H). The minimal transversals can be partitioned intoTr(H) = T 1 [ T 2, whereT 1 contains all minimal transversals that containv andT 2 allothers. SinceTr(Tr(H)) = H if H is simple (Proposition 2.2),E is a minimal transversalof Tr(H). It follows thatE \ T = fvg for everyT 2 Tr(H) that containsv. Indeed, iffvg � E \T would hold, then, sinceE is the only edge ofH that containsv, T �fvg wouldbe a transversal ofH, which contradictsT 2 Tr(H). Thus,E \ T = fvg. This impliesT � fvg � E: From this it is clear thatT 1 = ffvg [ T : T 2 Tr(HE � f;g)g. It is

immediate thatT 2 = Tr(HV�fvg):THEOREM 5.10. The minimal transversals of a�-acyclic hypergraphH are P-enum-

erable.Proof. Recall that a problem is P-enumerable if there is some algorithm which computes

all solutions to the problem in timep(IS)N , wherep is some polynomial in the input sizeISandN is the number of solutions. To include the case in which the problem hasno solution,


we slightly modify this convention fromp(IS)N to p(IS)(N + 1); this modification is notsubstantial.

To prove the theorem, we proceed as follows: first we note somefacts regarding�-acyclic hypergraphs, and then we give an algorithm for transversal computation. We mayassume without loss of generality thatH has no inessential vertices and thatH is simple,sincemin(H) is polynomial-time computable,Tr(H) = Tr(min(H)), andmin(H) is �-acyclic ifH is �-acyclic.

There is a simple observation on ear nodes of�-acyclic hypergraphs.FACT 1: Every simple�-acyclic hypergraphH that contains a nonempty edge has an

ear node.Indeed, ifH is �-acyclic, then it is also�-acyclic, and theGYO-reduction mustsucceed. However, asH is simple, rule 2 is not applicable, and asH 6= f;g, rule 1 mustapply. HenceH has an ear node.

FACT 2: LetH be a�-acyclic hypergraph and letV 0 � V . ThenHV 0 is �-acyclic.(See[16, p. 530].)

Algorithm BetaTr (H)input : Simple�-acyclic hypergraphH.output: Tr(H).

if H = ; then return(f;g)else

if H = f;g then return(;)elsev an ear node inE 2 H;H1 min(HE � f;g); H2 min(HV�fvg);T 1 BetaTr(H1); T 2 BetaTr(H2);

return (fT [ fvg : T 2 T 1g [ T 2)fi;

fi;TABLE 2

Output-polynomial algorithm for computingTr(H) of a simple,�-acyclic hypergraphH.

Now consider the procedureBetaTr in Table 2, where the input hypergraphH is simple.Let for every hypergraphG denoteve(G) = jVe(G)j in what follows.

The correctness ofBetaTr is shown by induction onve(H). If ve(H) = 0, thenH = ;orH = f;g, and the result is correct. Now considerve(H) > 0 and assume the hypothesis iscorrect for simple hypergraphs with fewer essential nodes thanH. AsH is simple,�-acyclic,andve(H) > 0, by Fact 1H has an ear node; hence, a vertexv in edgeE will be found.From Lemma 5.9, we haveTr(H) = ffvg [ T : T 2 Tr(HE � f;g)g [ Tr(HV�fvg);which is equivalent toTr(H) = ffvg [ T : T 2 Tr(min(HE � f;g))g [ Tr(min(HV�fvg)):Since a hypergraph is�-acyclic iff every subhypergraph of it is�-acyclic, in conjunctionwith Fact 2 it follows thatH1 andH2 are simple,�-acyclic hypergraphs. BecauseH1 andH2 have fewer essential vertices thanH, by the induction hypothesis we haveTr(Hi) =BetaTr(Hi) = T i, for i = 1; 2. ThusTr(H) = ffvg[T : T 2 Tr(HE � f;g)g[Tr(HV�fvg) = ffvg[T : T 2 T 1g[T 2:


This is exactly whatBetaTr(H) returns ifVe(H) 6= ;; hence, the claimed statement holdsfor H.

It remains to prove the claim on the complexity ofBetaTr. Denote bycalls(H) the totalnumber of recursive calls toBetaTr in computingBetaTr(H). We show by induction onve(H) that calls(H) � 2ve(H)�jTr(H)j:Forve(H) = 0, we have eitherH = ; orH = f;g; the statement clearly holds. Now considerthe caseve(H) > 0 and assume the statement is correct for hypergraphs with fewer essentialvertices. We have that calls(H) = calls(H1) + calls(H2) + 2:By the induction hypothesis, we obtain thatcalls(H1) � 2ve(H1)�jTr(H1)j � 2(ve(H)� 1)�jTr(H1)j � 2ve(H)�jTr(H1)j � 2;the last inequation holds becauseH1 6= f;g, and thereforejTr(H1)j > 0. Similarly, weobtain by the induction hypothesis thatcalls(H2) � 2ve(H2)�jTr(H2)j � 2ve(H)�jTr(H2)j:Thus calls(H) = calls(H1) + calls(H2) + 2� 2ve(H)�jTr(H1)j � 2 + 2ve(H)�jTr(H2)j+ 2= 2ve(H)�jTr(H1)j+ 2ve(H)�jTr(H2)j:Now, sincejTr(H)j = jTr(H1)j + jTr(H2)j, it follows thatcalls(H) � 2ve(H)�jTr(H)j,i.e., the claimed statement holds.

The body ofBetaTr can be clearly executed in polynomial time. Thus, the total runtimeof BetaTr(H) is bounded byp(m;n) �(calls(H) + 1) for some polynomialp in m andn(i.e., the number of edges and vertices ofH). Sincep(m;n)�(calls(H) + 1) � p(m;n)�(2ve(H)�jTr(H)j+ 1);it follows that the runtime ofBetaTr(H) is bounded byq(m;n) �(jTr(H)j + 1) for somepolynomialq. Thus,BetaTr(H) P-enumerates the minimal transversals ofH. The resultfollows.

We remark that it is not difficult to modifyBetaTr such that the minimal transversals areoutput with polynomial-time delay; we leave this to the reader.

From Theorem 5.10, we obtain together with Theorems 4.5 and 4.12 immediately thefollowing polynomial cases of the transversal problem and the saturation problem.

COROLLARY 5.11. TRANS-HYP and SIMPLE-H-SATare polynomial for�-acyclic hy-pergraphs.

6. Overview of related problems and applications.In this section we give a shortoverview of some related problems and applications in different fields of computer science.For space reasons the exposition is rather succinct. In particular, we omit several formaldefinitions and all proofs. A full discussion, exact definitions, all proofs, and more materialcan be found in the extended report [15]; other interesting problems related toTRANS-HYPhave recently been studied in [32].


6.1. Clause satisfiability. We have identified the following restrictions of the well-knownSATISFIABILITY problem which are�pm-equivalent toTRANS-HYP(resp. co-TRANS-HYP).

Problem: INTERSECTING MONOTONE SAT (IMSAT)Instance: A setC of clauses such that each clause is either positive (i.e., consists

entirely of positive literals) or negative (i.e., consistsentirely of negativeliterals) and for each positive clauseC1 and negative clauseC2 of C,there exists an atomu such thatu 2 C1 and:u 2 C2 .

Question: Is C satisfiable ?

Problem: SYMMETRIC INTERSECTING MONOTONE SAT (SIMSAT)Instance: Restriction ofIMSAT to instancesC where the negative clauses are pre-

cisely all clausesC� such thatC� = f:u : u 2 C+g for some positiveclauseC+ 2 C. (By this restriction, nonempty positive clauses ofC aremutually intersecting.)

Question: Is C satisfiable ?Both problems are�pm-equivalent to co-SIMPLE-H-SAT. Interestingly, unlikeSATISFIA-

BILITY or most known other restrictions ofSATISFIABILITY , these problems become poly-nomial-time decidable as soon as the cardinality of the clauses in the problem instances isbounded by a constant.

6.2. Design of relational databases.For the basic concepts of relational database the-ory consult [39] or [53]. IfF is a set of functional dependencies (FDs) on a set of attributesU , thenF+ denotes the closure ofF , i.e., all those dependencies onU that follow fromF .Let RS = hU; F i be a relation schema, whereU is a finite set of attributes andF is a setof functional dependencies. A relation instanceR overU is anArmstrong relationfor RSiff the functional dependencies that hold inR are preciselyF+. It was recently advocatedthat Armstrong relations can be used as a very profitable toolin database design [44, 40]. Inthis context it is important to compute an Armstrong relation from a given set of functionaldependencies and vice-versa. The following related decision problem is�pm-equivalent toSIMPLE-H-SAT:

Problem: FD-RELATION EQUIVALENCEInstance: A relation instanceR and a setF of FDs in BCNF, both on a set of

attributesU .Question: IsR an Armstrong relation forF ?A relation schemeRS = hU; F i is in Boyce-Codd Normal Form(BCNF) if for everyX ! Y 2 F+ such thatY 6� X ,X is a superkey forRS, i.e.,X ! U .By using Corollary 5.3 it can be shown that the problemFD-RELATION EQUIVALENCE

becomes polynomial ifF is in BCNF and for eachX ! Y of F , X contains fewer thankattributes for some constantk. Moreover, we can show that, under the same restrictions, theproblem ofgeneratingan Armstrong relation forF can be done in output-polynomial time.

A relation instanceR onU is in BCNF if the schemahU; FRi is in BCNF, whereFR isthe set of all FDs that hold onR. The following problem can be shown to be�pm-equivalentto co-SIMPLE-H-SAT:

Problem: ADDITIONAL KEY for relation instancesInstance: A relation instanceR on attributesU , a setK of minimal keys forR.Question: Is there a minimal key for the schemeRS = hU; FRi not contained inK?Note that the additional key problem for relationschemesis polynomial-time solvable by

an algorithm of Lucchesi and Osborn [37].


6.3. Updates in distributed databases.In distributed databases, mutual exclusion ofgroups of sites, which is necessary for executing critical operations, can be realized by de-fining a priori a set of groups (quorums) that intersect each other [33]. A group of sitescan perform the critical operation only if it contains a quorum of this set. Clearly, it is nat-ural to consider only minimal quorums, i.e., no quorum should properly contain any otherquorum. In terms of hypergraph theory, this means that the specified quorums should con-stitute a simple, intersecting hypergraph on the set of sites, termed acoterie in [20]. Forexample, the hypergraphC = f fa; b; cg; fa; b; dg; fa; c; dg; fb; c; dg g is a coterie on ver-ticesfa; b; c; dg. Coteries appear to be a fruitful concept for access controlin distributeddatabases, cf. [20, 25, 19].

A coterieA dominatesa coterieB if and only ifA 6= B and for eachB 2 B there existsanA 2 A such thatB � A. A coterieA is nondominated(ND) if there is no coterieBthat dominatesB. In practice, ND coteries are desired for decision agreement groups withrespect to reliability considerations. Unfortunately, noefficient test for nondomination of acoterie is known; in [20] merely an algorithm which has worstcase run time exponential inthe number of vertices is described. Interestingly, this problem turns out to be an instance oftheSELFTRANSVERSALITYproblem.

THEOREM 6.1. A coterieC is ND if and only ifC = Tr(C).Since every selftransversal hypergraph is a coterie, we getfrom Theorems 6.1 and 4.13

the following.COROLLARY 6.2. Checking if a coterie is ND(ND COTERIE) is �pm-equivalent to

SIMPLE-H-SAT.Note thatND COTERIEis by Theorem 6.1 and Corollary 5.3 polynomial for small quor-

ums.In the extended report, we also show that for a recent generalization of this approach

to bi- and semicoteries, which model read and write operations by read quorums and writequorums [19, 25], the complexity of testing nondomination is not increased and remains�pm-equivalent toTRANS-HYP.

6.4. Boolean switching theory.We assume that the reader is familiar with the basicconcepts of switching circuits and Boolean functions (BFs); for details we refer to the stand-ard literature (e.g., [56]). The design of efficient algorithms for the computation of primeimplicants from a function table or a logical expression hasbeen topic of research over dec-ades.

It is well-known that every monotone Boolean function has a uniqueminimal conjunctivenormal formconsisting of a conjunction of disjunctions of atoms where no conjunct subsumesany other.

In [15] we show the following results. Given a monotone Boolean expressionE anda setP of prime implicants ofE, it is NP-complete to determine whether there exists anadditional prime implicant ofE. However, ifE is in minimal conjunctive normal form,then the same problem is�pm-equivalent to co-SIMPLE-H-SAT. In particular, in the lattercase, the prime implicants ofE are precisely the minimal transversals of the conjuncts ofE.Thus, the polynomial subcases for computing all minimal hypergraph transversals describedin Section 5 carry over to the problem of computing the prime implicants of a monotoneBoolean function in conjunctive normal form. This problem is also investigated in [9], whichgives some new results.

6.5. Model-based diagnosis.Basic techniques for model-based diagnosis have beendeveloped within AI by de Kleer and Williams [13], and by Reiter [45]. To introduce thenecessary concepts briefly, asystemis a pair(SD;COMP ), whereSD, thesystem descrip-tion, is a set of usually first-order sentences andCOMP is a set of constants which model the


systemcomponents. This general system description is used together with a setOBS of first-order sentences, which are particularobservationson the system behavior, to diagnose faults.The system descriptionSD makes use of a distinguished predicateAB(c) which interprets“componentc operates in abnormal mode”. Now adiagnosisfor (SD;COMP;OBS) is aminimal set� � COMP of components such thatT = SD [OBS [ fAB(c) j c 2 �g [ f:AB(c) j c 2 COMP ��gis consistent.4 Of course, if there is no fault,� = ;must be a diagnosis, otherwiseSD is nota sound system description.

In [45] a characterization of diagnoses in terms of so calledconflict sets is given. Aconflict setis a setC � COMP such thatSD[OBS[f:AB(c) jc 2 Cg is inconsistent. Aconflict setC is minimal if and only if no proper subset ofC is a conflict set. The fundamentaltheorem on conflict sets and diagnoses in [45] put into hypergraph terminology is

THEOREM6.3 ([45]).� � COMP is adiagnosisfor (SD;COMP;OBS) if and onlyif � 2 Tr(C), whereC denotes the set of all minimal conflict sets of(SD;COMP;OBS).

From Theorems 6.3 and 4.12 we immediately have the followingtheorem:THEOREM 6.4. LetC be the set of minimal conflict sets of(SD;COMP;OBS), andD

a set of diagnoses for(SD;COMP;OBS). GivenC andD for input, deciding if there is anadditional diagnosis not contained inD is�pm-equivalent to co-SIMPLE-H-SAT.

The determination of diagnoses from the given minimal conflict sets is essential in pop-ular algorithms for model-based diagnosis [45, 13]. Deciding if an already computed set ofdiagnoses is complete with respect to a given set of minimal conflict sets, i.e., consists of alldiagnoses, is an important subproblem if diagnoses are computed incrementally. Therefore,the complexity of the additional diagnosis problem is of crucial interest.

7. Conclusion. We have studied two computational problems on hypergraphs in this pa-per, namely deciding saturation of a hypergraph (H-SAT) and the recognition of the transver-sal hypergraph (TRANS-HYP). The latter problem is closely related to computing all minimaltransversals of a hypergraph.

The complexity of computing (resp. recognizing) all minimal transversals of a hyper-graph is an open problem [20, 41, 14, 29]. We showed that checking saturation for a simplehypergraph (SIMPLE-H-SAT) is computationally equivalent toTRANS-HYP; without the re-striction to simple hypergraphs,H-SAT was proved co-NP-complete in its general version aswell as for various subcases.

In the study ofSIMPLE-H-SAT and TRANS-HYP, we investigated the relationships ofthese problems to well-studied problems such as satisfiability and hypergraph 2-colorability.Several�pm-equivalent subproblems were exhibited, the most important among themSELF-TRANSVERSALITY. Moreover, narrowing the open problems “frontier”, we showed thatsome generalizations of these two problems are intractableand that several important sub-cases are polynomial. Some of those results use algorithms that compute under certain restric-tions all minimal transversals of a hypergraph in output-polynomial total time; the probablymost important restriction is a constant upper bound on the edge size.

Our results apply to various problems in database theory, switching theory, logic, andartificial intelligence which are all closely related toSIMPLE-H-SATandTRANS-HYP.

For future research, we present the following open questions:1. What is the complexity ofSIMPLE-H-SATandTRANS-HYP? In particular, are these

problems co-NP-complete, or can they be solved in polynomial time (or, lessrestrictive, in4It is presupposed thatT is always in a decidable subclass of first-order predicate calculus.


nondeterministic polynomial time) ? In connection with this, are the minimal transversals ofa hypergraphH computable in output-polynomial total time ?

2. How do SIMPLE-H-SAT and TRANS-HYP relate to other open problems inNP-completeness (cf. [28]) ? The most famous such problem,GRAPH ISOMORPHISM, hasbeen studied extensively in the literature. It is known thatif GRAPH ISOMORPHISMis NP-complete andP 6= NP, then the polynomial hierarchy collapses at level two [48],whichis not expected by the experts [27]. However, there seems to be no trivial relation betweenGRAPH ISOMORPHISMon the one hand andSIMPLE-H-SATandTRANS-HYPon the other.

3. What is the gain by using randomized algorithms, probabilistic algorithms, or in-teractive proof systems (cf. [26, 27]) forSIMPLE-H-SAT andTRANS-HYP, and what aboutweaker forms of reductions than polynomial transformability (especially, randomized reduc-tions, cf. [1, 47, 55])? Schoning [47] defines withinNP a Low HierarchyL0; L1; : : :, whereL0 = P; L1 = NP\co-NP, and a High HierarchyH0; H1; : : : ;whereH0 =NP-complete,and the other classesHi; i > 0, correspond to weakened versions ofNP-completeness suchas -completeness [1] (fully contained inH1) etc. which are believed still strong enough thatno problem inP satisfies one of them. Classifying co-SIMPLE-H-SAT into the Low or HighHierarchy would give strong evidence that the problem is notNP-complete or intractable,respectively.

Acknowledgments.The authors thank L. Lovasz and M. Yannakakis for interesting dis-cussions of this work and valuable comments. They would further like to thank the refereesof previous versions of this paper, as well as H. Mannila and D. Plaisted, for their commentsand helpful suggestions.

Note added in proof.(1) In a recent paper, Fredman and Khachiyan showed thatTr(H) can be recog-nized in timeO(nlog n) (“On the Complexity of Dualization of Monotone DisjunctiveNormal Forms,”Technical Report LCS-TR-225, Dept. of Computer Science, Rutgers University, 1994). This impliesthat all problems that are�pm-equivalent to TRANS-HYP can be recognized within the same timebound. Moreover, it strongly suggests that all these problems are not co-NP-complete and providespartial evidence to the thesis that TRANSHYP is close to the border between polynomiality and co-NP-hardness.

(2) As pointed out by V. Gurvich,�pm-equivalence between TRANSHYP and SELFTRANSVER-ALITY (Theorem 4.13) was earlier proved by P.D. Seymour in the context of Boolean functions inhis masters thesis (see p. 309 of “On the two-coloring of hypergraphs,” Quart. J. Math. Oxford (3) 25,(1974), pp. 303–312).

REFERENCES

[1] L. Adleman and K. Manders,Reducibility, randomness, and intractability, in Proceedings of the9th ACMSymposium on the Theory of Computing, 1977, pp. 151–163.

[2] C. Batini, A. D’Atri, and M. Moscarini,Formal tools for top-down and bottom-up generation of acyclicrelational database schemata, in Proceedings7th Int. Conf. on Graph-Theoretic Concepts in ComputerScience, Linz, Austria 1981.

[3] C. Beeri, M. Down, R. Fagin, and R. Statman,On the structure of Armstrong relations for functional depend-encies, Journal of the ACM, 31 (1984), pp. 30–46.

[4] C. Beeri, R. Fagin, D. Maier, A. Mendelzon, J. Ullman, andM. Yannakakis,Properties of acyclic data-base schemata, in Proceedings13th ACM Symposium on the Theory of Computing, May 1981 1981,pp. 355–362.

[5] C. Beeri, R. Fagin, D. Maier, and M. Yannakakis,On the desirability of acyclic database schemes, Journal ofthe ACM, 30 (1983), pp. 479–513.

[6] C. Benzaken,Critical hypergraphs for the weak chromatic number, Journal of Combinatorial Theory, SeriesB, 29 (1980), pp. 328–338.

[7] C. Berge,Hypergraphs, vol. 45 of North Holland Mathematical Library, Elsevier Science Publishers B.V.(North-Holland), 1989.


[8] P. A. Bernstein and D. W. Chiu,Using semijoins to solve relational queries, Journal of the ACM, 28 (1981),pp. 25–40.

[9] C. Bioch and T. Ibaraki,Complexity of dualization and identification of positive boolean functions, tech.report, Erasmus University Rotterdam, 1994.

[10] G. Burosch, J. Demetrovics, and G. O. H. Katona,The poset of closures as a model of changing databases,Order, 4 (1987), pp. 127–142.

[11] S. Cook,A taxonomy of problems with fast parallel algorithms, Information and Control, 64 (1985), pp. 2–22.[12] D. E. Daykin, J. Godfrey, and A. J. W. Hilton,Existence theorems for Sperner families, Journal of Combinat-

orial Theory, Series A, 17 (1974), pp. 245–251.[13] J. de Kleer and B. C. Williams,Diagnosing multiple faults, Artificial Intelligence, 32 (1987), pp. 97–130.[14] J. Demetrovics and V. D. Thi,Keys, antikeys and prime attributes, Annales Univ. Sci. Budapest, Sect. Comp.,

8 (1987), pp. 35–52.[15] T. Eiter and G. Gottlob,Identifying the minimal transversals of a hypergraph and related problems, Tech.

Report CD-TR 91/16, Christian Doppler Laboratory for Expert Systems, TU Vienna, Austria, January1991.

[16] R. Fagin,Degrees of acyclicity for hypergraphs and relational database schemes, Journal of the ACM, 30(1983), pp. 514–550.

[17] R. Fagin, A. Mendelzon, and J. Ullman,A simplified universal relation assumption and its properties, ACMTrans. on Database Syst., 7 (1982), pp. 343–360.

[18] J.-C. Fournier and M. L. Vergnas,A class of bichromatic hypergraphs, Annals of Discrete Mathematics, 21(1984), pp. 21–27.

[19] A. Fu, Enhancing Concurrency and Availability for Database Systems, PhD thesis, Simon Fraser University,Burnaby, B.C., Canada V5A 1S6, 1990.

[20] H. Garcia-Molina and D. Barbara,How to assign votes in a distributed system, Journal of the ACM, 32 (1985),pp. 841–860.

[21] M. Garey and D. S. Johnson,Computers and Intractability – A Guide to the Theory of NP-Completeness, W.H. Freeman, New York, 1979.

[22] M. R. Garey, D. S. Johnson, and L. Stockmeyer,Some simplified NP-complete graph problems, TheoreticalComputer Science, 1 (1976), pp. 237–267.

[23] M. Graham,On the universal relation, tech. report, Univ. of Toronto, Toronto, Ont., Can., September 1979.[24] R. Graham, D. E. Knuth, and O. Patashnik,Concrete Mathematics, Addison Wesley, 1989.[25] T. Ibaraki and T. Kameda,Theory of coteries, Tech. Report CSS/LCCR TR90–09, Simon Fraser University,

Burnaby, B.C., Canada V5A 1S6, August 1990. To appear in IEEETransactions on Parallel and Distrib-uted Systems.

[26] D. S. Johnson,The NP-completeness column – an ongoing guide, Journal of Algorithms, 5 (1984), pp. 147–160. Column 10: Updates, Updates, Updates.

[27] , The NP-completeness column – an ongoing guide, Journal of Algorithms, 9 (1988), pp. 426–444.Column 21: Interactive Proof Systems for Fun and Profit.

[28] , The NP-completeness column – an ongoing guide, Journal of Algorithms, 11 (1989), pp. 144–151.Column 22: The Story So Far.

[29] D. S. Johnson, M. Yannakakis, and C. H. Papadimitriou,On generating all maximal independent sets, Inform-ation Processing Letters, 27 (1988), pp. 119–123.

[30] P. Kanellakis,Elements of Relational Database Theory, in Handbook of Theoretical Computer Science, J. vanLeeuwen, ed., vol. B, Elsevier Science Publishers B.V. (North-Holland), 1990, ch. 17.

[31] R. Karp and A. Wigderson,A fast parallel algorithm for the maximal independent set problem, Journal of theACM, 32 (1986), pp. 762–773.

[32] D. Kavvadias, C. Papadimitriou, and M. Sideri,On Horn envelopes and hypergraph transversals, in Proceed-ings 4th International Symposium on Algorithms and Computation (ISAAC-93), W. Ng, ed., LNCS 762,Hong Kong, December 1993, Springer.

[33] L. Lamport, The implementation of reliable distributed multiprocess systems, Comp. Networks, 2 (1978),pp. 95–114.

[34] E. Lawler, J. Lenstra, and A. Rinnooy Kan,Generating all maximal independent sets: NP-hardness andpolynomial-time algorithms, SIAM J. Comp., 9 (1980), pp. 558–565.

[35] E. L. Lawler,Covering problems: duality relations and a new method of solution, SIAM J. Appl. Math., 14(1966), pp. 1115–1132.

[36] L. Lovasz,Coverings and colorings of hypergraphs, in Proceedings of the 4th Conference on Combinatorics,Graph Theory, and Computing, 1973, pp. 3–12.

[37] C. L. Lucchesi and S. Osborn,Candidate keys for relations, Journal of Computer and System Sciences, 17(1978), pp. 270–279.

[38] K. Maghout,Sur la determination des nombres de stabilite et du nombrechromatique d’un graphe, in ComptesRendus des Seances de l’Academie des Sciences, vol. 248, Paris, Gauthier-Villars, 1959, pp. 3522–3523.

[39] D. Maier,The Theory of Relational Databases, Computer Science Press, Rockville, Md., 1983.


[40] H. Mannila and K.-J. Raiha,Design by example: an application of Armstrong relations, Journal of Computerand System Sciences, 22 (1986), pp. 126–141.

[41] , Dependency inference, in Proceedings of the13th VLDB, 1987, pp. 155–158.[42] , Algorithms for inferring functional dependencies, Tech. Report A-1988-3, University of Tampere,

Dept. of Comp. Sc., Series of Publ. A, April 1988.[43] , Generating Armstrong databases for sets of functional and inclusion dependencies. manuscript,

August 1988.[44] , Practical algorithms for finding prime attributes and testing normal forms, in Proceedings of the 8th

ACM SIGACT-SIGMOD-SIGART Symposium on Principles of Database Systems, 1989, pp. 128–133.[45] R. Reiter,A theory of diagnosis from first principles, Artificial Intelligence, 32 (1987), pp. 57–95.[46] B. Roy,Algebre moderne et Theorie des graphes, vol. II, Dunod, Paris, 1970.[47] U. Schoning,A low and a high hierarchy within NP, Journal of Computer and System Sciences, 27 (1983),

pp. 14–28.[48] , Graph isomorphism is in the Low Hierarchy, Journal of Computer and System Sciences, 37 (1988),

pp. 312–323.[49] E. Sperner,Ein Satz uber Untermengen einer endlichen Menge, Mathematische Zeitschrift, XXVII (1928),

pp. 544–548.[50] R. E. Tarjan and M. Yannakakis,Simple linear time algorithms to test chordality of graphs,test acyclicity of

hypergraphs, and selectively reduce acyclic hypergraphs, SIAM J. Comp., 13 (1984), pp. 566–579.[51] V. D. Thi, Minimal keys and antikeys, Acta Cybernetica, 7 (1986), pp. 361–371.[52] S. Tsukiyama, M. Ide, H. Ariyoshi, and I. Shirakawa,A new algorithm for generating all maximal independent

sets, SIAM J. Comp., 6 (1977), pp. 505–517.[53] J. D. Ullman,Principles of Database and Knowledge Base Systems, vol. 2, Computer Science Press, 1988.[54] L. G. Valiant,The complexity of computing the permanent, Theoretical Computer Science, 8 (1979), pp. 189–

201.[55] L. G. Valiant and V. V. Vazirani,NP is as easy as detecting unique solutions, Theoretical Computer Science,

47 (1986), pp. 85–93.[56] I. Wegener,The Complexity of Boolean Functions, Wiley - Teubner series in computer science, Wiley & Sons

Ltd and Teubner, Stuttgart, 1987.[57] M. Yannakakis,Algorithms for acyclic database schemes, in Proceedings7th Conf. on Very Large Database

Systems, Cannes, France 1982, pp. 82–94.[58] C. T. Yu and M. Ozsoyoglu,An algorithm for tree-query membership of a distributed query, in Proceedings

IEEE COMPSAC, 1979, pp. 306–312.

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