minimum congestion spanning trees in graphsfacpub.stjohns.edu/ostrovsm/2009 new york graph...

Minimum Congestion Spanning Trees in Graphs

Mikhail OstrovskiiDepartment of Mathematics and Computer Science

St. John’s UniversityWeb page: http://facpub.stjohns.edu/ostrovsm/

e-mail: [email protected]

Graph Theory Day, November 2009

Mikhail Ostrovskii, St. John’s University Minimum Congestion Spanning Trees in Graphs

Main definitions

I Let G be a graph and let T be a spanning tree in G . For eachedge e of T let Ae and Be be the vertex sets of thecomponents of T − e. By eG (Ae , Be) we denote the numberof edges in G with one end vertex in Ae and the other endvertex in Be .

I In the graph below a spanning tree is shown using fat edges.Consider an edge e colored red. The correspondingcomponents Ae and Be are shown in blue and black,respectively. In this case eG (Ae , Be) = 5. In addition to ethere are 4 green edges joining Ae and Be .

Main definitions

I Let G be a graph and let T be a spanning tree in G . For eachedge e of T let Ae and Be be the vertex sets of thecomponents of T − e. By eG (Ae , Be) we denote the numberof edges in G with one end vertex in Ae and the other endvertex in Be .

I In the graph below a spanning tree is shown using fat edges.Consider an edge e colored red. The correspondingcomponents Ae and Be are shown in blue and black,respectively. In this case eG (Ae , Be) = 5. In addition to ethere are 4 green edges joining Ae and Be .

Main definitions

I We define the edge congestion of G in T by

ec(G : T ) = maxe∈E(T )

eG (Ae , Be).

I The name comes from the following analogy. Imagine thatedges of G are roads, and edges of T are those roads whichare cleaned from snow after snowstorms. If we assume thatfor each edge in G there is a flow of traffic between its ends;these flows are the same for each edge, and that after asnowstorm each driver takes the corresponding (unique)detour in T , then ec(G : T ) describes the traffic congestionat the most congested road of T .

I We are interested in a spanning tree which minimizes thecongestion.

Main definitions

eG (Ae , Be).

Main definitions

eG (Ae , Be).

Basic definitions and some references

I We define the spanning tree congestion of G by

s(G ) = min{ec(G : T ) : T is a spanning tree of G}. (1)

Each spanning tree T in G satisfying ec(G : T ) = s(G ) iscalled a minimum congestion spanning tree.

I At present there exist two approaches to estimates of thespanning tree congestion from below.

I One of them is based on Jordan’s notion of a centroid and onisoperimetric estimates. This method was suggested in[Ostrovskii 2004] and developed in [Castejon-Ostrovskii 2009],[Kozawa-Otachi-Yamazaki 2009], and [Law 2009].

I The second known approach to estimates of the spanning treecongestion from below is based on the notion of a dual treeand (at present) is applicable to planar graphs only. Themethod was suggested in [Hruska 2008] and developed in[Ostrovskii 2009].

Estimates from below

I The purpose of this talk is to describe the mentionedapproaches.

I I would like to mention that in many cases the estimatesobtained on these lines do not give us exact values of thespanning tree congestion. So for many collections of graphsthere is still some work to be done.

I The purpose of this talk is to describe the mentionedapproaches.

I I would like to mention that in many cases the estimatesobtained on these lines do not give us exact values of thespanning tree congestion. So for many collections of graphsthere is still some work to be done.

I Now we describe results on trees needed for estimates frombelow. They go back to [C. Jordan 1869].

I Let u be a vertex of a tree T . If we delete all edges incidentto u from T we get a graph with several connectedcomponents (k + 1 components if k edges were deleted). Themaximal number of vertices in components of the obtainedgraph is called the weight of T at u. A vertex v of T is calleda centroid vertex if the weight of T at v is minimal possible.The weight of a centroid vertex is called the weight of T andis denoted by w(T ).

I Theorem 1. A tree T has one or two centroid vertices. Twocentroid vertices are present only in the case when T has anedge whose removal splits T into two components with thesame number of vertices. In particular, in such a case|V (T )| = 2w(T ), where |V (T )| is the number of elements inV (T ).

Applying Jordan’s theorem to estimates from below

I Denote by ∆G the maximum degree of the graph G . Let Tbe an optimal spanning tree in G , that is, ec(G : T ) = s(G ).Let u be a centroid of T . Since T is a subgraph of G , thereare at most ∆G edges incident with u in T .

I Hence at least one of the components, we denote its vertexset by A, in the graph T − u has at least

⌈|V (G )| − 1

⌉vertices, and at most

|V (G )|2

vertices.

(The notation dxe is used for the rounded up real number x .)

I The edge connecting u with A is used in eG (A, V (G )− A)detours. Therefore any inequality estimating from below thenumber of edges in G joining a set A containing at least OG

and at most |V (G )|/2 vertices with V (G )− A provides anestimate of s(G ) from below.

⌈|V (G )| − 1

|V (G )|2

vertices.

⌈|V (G )| − 1

|V (G )|2

vertices.

Isoperimetry

I Estimates of the type mentioned at the end of the previousslide have been systematically studied, they are calledisoperimetric.

I The general edge-isoperimetric problem is to determine, for agraph G , the minimum edge-boundary over subsets of verticesof a fixed order.

I In more detail. Let 0 ≤ c ≤ |V |, for S ⊂ V (G ) by S wedenote the complement of S in V (G ). The purpose is to find

mG (c) = min{eG (S , S) : S ⊂ V (G ), |S | = c}.

I Many estimates of this type are known, see surveys [Bezrukov1999] and [Leader 1991]. We are going to consider a simplecase in which we are able to prove the correspondingisoperimetric inequality rather than to refer to a known result.

Isoperimetry

Square grids

I We are interested in the spanning tree congestion of theso-called grid graphs. We use the following notation andterminology. We use the notation [k] for the set{0, 1, 2, . . . , k − 1}. The grid graph is the graph with vertexset [k]n = {0, 1, 2, 3, . . . , k − 1}n in which x = (xi )

n1 is joined

to y = (yi )n1 if and only if |xi − yi | = 1 for some i and xj = yj

for all j 6= i .

I We restrict our attention to the two-dimensional case, that is,the case when n = 2. We call the corresponding graphs squaregrids. The enclosed figure shows the square grid for k = 4.

• • • •

Square grids

I We are interested in the spanning tree congestion of theso-called grid graphs. We use the following notation andterminology. We use the notation [k] for the set{0, 1, 2, . . . , k − 1}. The grid graph is the graph with vertexset [k]n = {0, 1, 2, 3, . . . , k − 1}n in which x = (xi )

n1 is joined

to y = (yi )n1 if and only if |xi − yi | = 1 for some i and xj = yj

for all j 6= i .

I We restrict our attention to the two-dimensional case, that is,the case when n = 2. We call the corresponding graphs squaregrids. The enclosed figure shows the square grid for k = 4.

• • • •

Spanning tree congestion of square grids

I Our purpose is to find the exact value of the spanning treecongestion for square grids.

I Theorem 2 [Hruska 2008, Castejon-Ostrovskii 2009].s([k]2) = k for k ≥ 2.

I Proof. Observe that for [k]2 the maximum degree is 4,

therefore O[k]2 =

⌈k2 − 1

⌉. We need to estimate from below

the number of edges connecting a set A ⊂ V ([k]2) withV ([k]2)− A, where A is a set containing between O[k]2 and

k2/2 vertices. Our purpose is to show that this number is ≥ k .

therefore O[k]2 =

⌈k2 − 1

therefore O[k]2 =

⌈k2 − 1

Estimate from below for square grids

I Let t be the number of horizontal levels containing verticesfrom A and c be the number of vertical levels containingvertices from A.

I Obviously there are three possibilities:

1. Both t < k and c < k.2. Exactly one of the numbers t and c is equal to k.3. t = c = k.

I Let t be the number of horizontal levels containing verticesfrom A and c be the number of vertical levels containingvertices from A.

I Obviously there are three possibilities:

1. Both t < k and c < k .2. Exactly one of the numbers t and c is equal to k.3. t = c = k.

I Case 1. In this case each vertical level intersecting A as wellas each horizontal level intersecting A contains an edge whichjoins A and V ([k]2)− A. Hence the number of such edges is≥ t + c. On the other hand, |A| ≤ tc .

I Recall the well-known inequality t + c ≥ 2√

tc . We get

t + c ≥ 2√|A| ≥ 2

√⌈k2 − 1

⌉. It is easy to check that this

implies t + c ≥ k.

I Case 2. First we consider the case t = k , c < k. Then eachhorizontal level contains edges joining A with V ([k]2)− A.Since the number of horizontal levels is k, we are done in thiscase.

I The case c = k , t < k is completely analogous.

tc . We get

t + c ≥ 2√|A| ≥ 2

√⌈k2 − 1

tc . We get

t + c ≥ 2√|A| ≥ 2

√⌈k2 − 1

tc . We get

t + c ≥ 2√|A| ≥ 2

√⌈k2 − 1

I Case 3. If t = c = k , let is denote by r the set of thosevertical levels, for which all vertices are in A and by s the setof those horizontal levels, for which all vertices are in A.

I The number of edges joining A and V ([k]2)− A is at least(k − r) + (k − s), whereas the number of vertices in A is atleastk(r + s)− rs ≥ k(r + s)−

)2= (r + s)(k − ((r + s)/4)).

Hence k2

2 ≥ (r + s)(k − ((r + s)/4)). It is a simple Calculusproblem to show that this implies (r + s) ≤ k . Now theestimate of the number of edges implies that the number ofedges joining A and V ([k]2)− A is at least k.

I Thus, we have proved s([k]2) ≥ k .

)2= (r + s)(k − ((r + s)/4)).

Hence k2

)2= (r + s)(k − ((r + s)/4)).

Hence k2

Estimates from above

I It remains to show that there exist trees for which thesebounds are attained. Now we construct optimal spanningtrees Tk in [k]2. We construct them somewhat differently forodd and even k .

I First let k = 2r + 1, r ∈ N. In this case the edge set of Tk

consists of

I (A) ‘Vertical’ edges joining the vertices (r , y) and (r , y + 1),y ∈ {0, . . . , k − 2};

I (B) ‘Horizontal’ edges joining the vertices (x , y) and(x + 1, y), x ∈ {0, . . . , k − 2}, y ∈ {0, . . . , k − 1} (theenclosed figure shows the result in the case k = 3).

• • •

I The verification of the equality ec([k]2 : Tk) = k isstraightforward.

consists of

• • •

consists of

• • •

consists of

• • •

I Now let k = 2r , r ∈ N. In this case the edge set of Tk

consists of

I (A) Two sets of ‘vertical’ edges:(1) edges joining the vertices (r , y) and (r , y + 1),

y ∈ {0, . . . , k − 2};(2) edges joining the vertices (r − 1, y) and (r − 1, y + 1),

y ∈ {0, . . . , k − 2};I (B)

(1) ‘Horizontal’ edges joining the vertices (x , y) and (x + 1, y),x ∈ {0, . . . , k − 2}\{r − 1}, y ∈ {0, . . . , k − 1};

(2) A horizontal edge joining (r − 1, r) and (r , r). (The enclosedfigure shows the resulting graph for k = 4.)

• • • •

consists ofI (A) Two sets of ‘vertical’ edges:

(1) edges joining the vertices (r , y) and (r , y + 1),y ∈ {0, . . . , k − 2};

(2) edges joining the vertices (r − 1, y) and (r − 1, y + 1),y ∈ {0, . . . , k − 2};

I (B)(1) ‘Horizontal’ edges joining the vertices (x , y) and (x + 1, y),

x ∈ {0, . . . , k − 2}\{r − 1}, y ∈ {0, . . . , k − 1};(2) A horizontal edge joining (r − 1, r) and (r , r). (The enclosed

figure shows the resulting graph for k = 4.)

• • • •

consists ofI (A) Two sets of ‘vertical’ edges:

(1) edges joining the vertices (r , y) and (r , y + 1),y ∈ {0, . . . , k − 2};

(2) edges joining the vertices (r − 1, y) and (r − 1, y + 1),y ∈ {0, . . . , k − 2};

I (B)(1) ‘Horizontal’ edges joining the vertices (x , y) and (x + 1, y),

x ∈ {0, . . . , k − 2}\{r − 1}, y ∈ {0, . . . , k − 1};(2) A horizontal edge joining (r − 1, r) and (r , r). (The enclosed

figure shows the resulting graph for k = 4.)

• • • •

End of the proof

I The verification of the equality ec([k]2 : Tk) = k in this caseis also straightforward.

I An exact isoperimetric inequality is known forhigh-dimensional grids [Bollobas-Leader 1991]. Applying theargument with centroids, we get from such estimates anestimate for the spanning tree congestion from below. But fordimensions ≥ 3 matching estimates from above are notknown. For this reason we do not know exact values of thespanning tree congestion for grids [k]n of dimension n = 3 andhigher (with the exception of few small values of k and n).

End of the proof

I The verification of the equality ec([k]2 : Tk) = k in this caseis also straightforward.

I An exact isoperimetric inequality is known forhigh-dimensional grids [Bollobas-Leader 1991]. Applying theargument with centroids, we get from such estimates anestimate for the spanning tree congestion from below. But fordimensions ≥ 3 matching estimates from above are notknown. For this reason we do not know exact values of thespanning tree congestion for grids [k]n of dimension n = 3 andhigher (with the exception of few small values of k and n).

minimum congestion spanning trees in graphsfacpub.stjohns.edu/ostrovsm/2009 new york graph...

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