Information Processing Letters 112 (2012) 697–700

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Information Processing Letters

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On minimal arbitrarily partitionable graphs ✩

Olivier Baudon a,b, Jakub Przybyło c,∗,1, Mariusz Wozniak c,1

a Univ. Bordeaux, LaBRI, UMR 5800, 33400 Talence, Franceb CNRS, LaBRI, UMR 5800, 33400 Talence, Francec AGH University of Science and Technology, al. A. Mickiewicza 30, 30-059 Krakow, Poland

a r t i c l e i n f o a b s t r a c t

Article history:Received 19 January 2012Accepted 19 June 2012Available online 20 June 2012Communicated by Jinhui Xu

Keywords:Minimal arbitrarily partitionable graphArbitrary vertex decomposable graphInterconnection networks

A graph G = (V , E) of order n is called arbitrarily partitionable, or AP for short, if givenany sequence of positive integers n1, . . . ,nk summing up to n, we can always partitionV into subsets V 1, . . . , Vk of sizes n1, . . . ,nk , resp., inducing connected subgraphs in G . Ifadditionally G is minimal with respect to this property, i.e. it contains no AP spanningsubgraph, we call it a minimal AP-graph. It has been conjectured that such graphs aresparse, i.e., there exists an absolute constant C such that |E| � Cn for each of them. Weconstruct a family of minimal AP-graphs which prove that C � 1 + 1

30 (if such C exists).© 2012 Elsevier B.V. All rights reserved.

1. Introduction

Consider a simple graph G = (V , E) of order n. A se-quence τ = (n1, . . . ,nk) of positive integers is called ad-missible for G if it is a partition of n, i.e., n1 + · · · + nk = n.If additionally there exists a partition (V 1, . . . , Vk) of thevertex set V such that each V i induces a connected sub-graph of order ni in G , then we say that τ is realizablein G , while (V 1, . . . , Vk) is called a realization of τ in G . Ifevery admissible sequence is also realizable in G , then wesay that this graph is arbitrarily partitionable (or arbitraryvertex decomposable, see e.g. [7]) or we call it an AP-graphfor short.

The cornerstone of the study of arbitrarily partition-able graphs are the well-known results of Gyori [6]and Lovász [8], who showed independently that everyk-connected graph is arbitrarily partitionable into (at most)k parts. In fact, with additional restrictions, i.e., that one

✩ Research partially supported by the partnership Hubert Curien Polo-nium 22658VG.

* Corresponding author. Tel.: +48 12 617 46 38; fax: +48 12 617 31 65.E-mail address: przybylo@wms.mat.agh.edu.pl (J. Przybyło).

1 Research partially supported by the Polish Ministry of Science andHigher Education.

0020-0190/$ – see front matter © 2012 Elsevier B.V. All rights reserved.http://dx.doi.org/10.1016/j.ipl.2012.06.010

may fix any representative for each subset of an expectedvertex partition, they showed that the two properties areequivalent, see [6] or [8] for details. For the small valuesof k also the polynomial-time algorithms have been given,in particular, for k = 2, k = 3 [9,13], and for the case k = 4restricted to the planar graphs [10].

The problem of arbitrarily partitionable graphs wasthen rediscovered, already with the definition we use, byBarth et al. [1] in the following applicable setting. Supposeour graph is a model of a interconnection computer net-work, and we have k potential users each of whom wemust provide with a connected subnetwork consisting ofa given number of computing resources. Then if only ournetwork forms an AP-graph, we may always satisfy the re-quirements of any number of potential clients, providedonly that we have sufficiently many computers in the net-work. Note that every AP-graph must be connected, andmoreover the property of being AP is monotone, i.e., closedunder the operation of adding edges. Thus if only G con-tains a spanning tree which is AP, then so is G . Lots ofresearch on the AP-property has been devoted to treesthen. In particular in [2] Barth and Fournier proved thefollowing.

Theorem 1. The maximum degree of each AP-tree is at most 4.

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Fig. 1. Graph Ap .

Fig. 2. Graph J .

n

From the algorithmic point of view, it has been provedin [2] that deciding whether a given partition τ is realiz-able in a tree T is NP-complete, even for trees of degreeat most 3. A polynomial-time algorithm for tripods (treesconsisting of three paths sharing one common vertex) waspresented in [1]. Other results concerning the family andsubfamilies of tress and related complexity problems arecontained, e.g. in [3,5,7,11].

On the other hand there are known examples of AP-graphs which do not contain any spanning AP-trees, lotsof examples of such graphs are included in this paper.This fact made the problem of AP-graphs even more com-plex, and led to the following definition introduced byRavaux [12]. We say that G is a minimal AP-graph if G isAP and removing any of its edges makes it not AP. Obvi-ously each AP-tree is a minimal AP-graph. Conversely, if Gis a minimal AP-graph which is not a tree, then it containsno spanning AP-tree. An example of such a graph was in-cluded in [12]. Though such graphs exist, Ravaux suspectsthat they cannot be too complicated (too dense) after all,and formulated in [12] the following intriguing conjecture,which is still wide open.

Conjecture 2. If G = (V , E) is a minimal AP-graph of order n,then |E| = O (n).

In this paper we present an infinite family of mini-mal AP-graphs containing arbitrarily many cycles and withasymptotic density greater than one, hence “significantly”larger than in the case of trees, see Corollary 7.

2. The family of minimal AP-graphs

Definition 3. Let p be a prime number, p � 3, and letA p denote a graph with the vertex set V = {v1, . . . ,

v(p+1)(p+2)} and the edge set E = {vi vi+1: i = 1, . . . ,

p+2, p+4, . . . , (p+1)(p+2)−1}∪{v p+2 v p+4, v p+3 v2p+3}(hence |E| = |V | = (p + 1)(p + 2)). This is a unicyclicgraph which may also be constructed by adding to thepath v1 v2 . . . v p+2 v p+4 v p+5 . . . v(p+1)(p+2) a new vertexv p+3, together with the edges v p+2 v p+3 and v p+3 v2p+3(note that there are only p − 1 vertices on this pathbetween v p+2 and v2p+3), see Fig. 1. We shall call v1and v(p+1)(p+2) the first and the last, resp., and v p+3 thetop vertex of A p . Moreover, we say that the vertices

v p+2, . . . , v2p+3 form a block in A p , and analogously asabove, we shall call v p+2, v2p+3 and v p+3 the first, the lastand the top, resp., vertex of this block.

Definition 4. Assume n = l(p + 1)(p + 2), where l = sp + rfor some integers r, s with r ∈ {0, . . . , p − 1}, and letJn denote the graph with n vertices, v1, . . . , vn , and

n + l − s − 1 edges constructed from a path v1 v2 . . . vn

by removing the edges vi+1 vi+2 and adding the edgesvi vi+2, vi+1 vi+p+1 for each i = t(p + 1)(p + 2) + (p + 2)

with t �≡ p − 1 (mod p). (In other words, Jn is created bytaking l − s copies of A p and s copies of P (p+1)(p+2) , ar-ranging them into sequence so that every p-th graph ofthis sequence is a path, and then for each pair of consecu-tive graphs in this sequence, joining the last vertex of thefirst of them with the first vertex of the second one, seeFig. 2.)

Theorem 5. Let p be a prime integer, p � 3. For each n > 0divisible by (p + 1)(p + 2), Jn is a minimal AP-graph.

Proof. Let n = l(p + 1)(p + 2) > 0, where l = sp + r forsome integers r, s with r ∈ {0, . . . , p − 1}.

First we show that removing any edge e from Jn makesit not AP. Obviously this is the case if e does not belongto any cycle (because then removing it disconnects thegraph).

Assume then that e lies on a cycle vi vi+2 vi+3 . . .

vi+p+1 vi+1 vi for some i = t(p + 1)(p + 2) + (p + 2). Notethat n ≡ 0 (mod p + 1) and n ≡ 0 (mod p + 2), henceτp+1 = (p + 1, . . . , p + 1) and τp+2 = (p + 2, . . . , p + 2)

should be realizable in the obtained graph (if it is AP).The structure of our graph resembles a path, hence, sincei − 1 ≡ 0 (mod p + 1) and i ≡ 0 (mod p + 2), it isobvious that in a realization of τp+1 or τp+2 the ver-tices v1, . . . vi−1 or v1, . . . vi , resp., must be covered by(i − 1)/(p + 1) or i/(p + 2), resp., subsets of this re-alization. Consequently, if e = vi vi+1, then one of thesubsets of the realization of τp+1 must be a set V ′ ={vi, vi+2, vi+3, . . . , vi+p+1}, and after removing it, we areleft with an isolated vertex vi+1, see Fig. 3. If on the otherhand e = vi vi+2, we are analogously unable to cover vi+2while trying to find a realization of τp+1, see Fig. 4. Fi-nally, in all the remaining cases (for all the other edges

O. Baudon et al. / Information Processing Letters 112 (2012) 697–700 699

Fig. 3. Graph Jn with e = vi vi+1 removed.

Fig. 4. Graph Jn with e = vi vi+2 removed.

Fig. 5. Graph Jn with some other edge e from the block removed.

of our cycle), we immediately create an isolated path con-taining less than p vertices while trying to realize τp+2,see Fig. 5.

Second, we show that Jn itself is AP. Let us fix any ad-missible sequence τ for Jn . If there are any 1’s in thissequence, choose as many top vertices with the greatestindexes (the ones “from the right”) as their realizationsas possible. Then remove the used vertices from Jn andthe used 1’s from τ . Denote by J ′ the obtained graph, andthe obtained sequence by τ ′ . Note that if we had enough1’s, then J ′ would be a path, and we could easily findthe realization of the remaining elements of the sequence.Therefore, we may assume that there are still some blocksleft in J ′ (the ones on the left), and that there are no 1’sin τ ′ . We shall try to find a realization of the remaining el-ements “from left to right”, i.e., such that the realizationsof subsequent elements of τ ′ consist of consecutive ver-tices (with respect to their indexes), with an exception ofthe top vertices, where for every such vertex we must onlyrequire that it is in the same subset of the realization asthe first or the last vertex of the block to which it belongs.To do this, it is sufficient to find an ordering (n1, . . . ,nk) ofthe elements of τ ′ such that if S = {s1, . . . , sk} is the set ofthe sums of the form si = ∑

j�i n j , i = 1, . . . ,k, then:

(*) for no block of J ′ with the first vertex va and the lastvertex vb we have a,b − 1 ∈ S .

Then we will be able to realize our sequence τ ′ from leftto right according to this ordering in such a way that eachtop vertex is in the same subset of the realization as thefirst or the last vertex of its block (dependent on whethera /∈ S or b − 1 /∈ S , resp.).

Suppose then that there is no such ordering, and letus consider an ordering τ ′ = (n1, . . . ,nk) for which all,say f , elements p are at the beginning of τ ′ (i.e., τ ′ =(p, . . . , p,n f +1, . . . ,nk) with n j �= p for j > f ), and whichmaximizes the number of first consecutive blocks forwhich the condition from (*) holds. Let vq be the first ver-tex of the first block (counting from the left) for which ourcondition does not hold, hence there exist sc, sd ∈ S suchthat sc = q and sd = q + p. Note that sc, sd /∈ {s1, . . . , s f }(hence c > f ), since s1, . . . , s f ≡ 0 (mod p), while i �≡0 (mod p) for the first vertex vi of any block. Conse-quently, nc, . . . ,nk �= p. Moreover, sd − sc = nc+1 + · · · +nd = p and nc+1 + · · · + nk � (p + 1)(p + 2) − (p + 2) =(p + 1)2 > p, hence it is easy to reorder the elementsnc+1, . . . ,nk so that q + p do not belong to the obtainednew set S , what yields a contradiction with the assumed“maximality” of the ordering of τ ′ . To see that such re-ordering exists, it is enough to note that for any set ofintegers A (A = {nc+1, . . . ,nk}), each greater than 1 anddistinct from a prime number p, whose sum is greaterthan p, there always exist l1, . . . , lt ∈ A (t � 0) such thatl1 +· · ·+ lt−1 < p and l1 +· · ·+ lt > p. Indeed, it is obviousif all elements of A are the same, since p is a prime num-ber. Otherwise, suppose nc+1 � nc+2 � · · · � nk , nc+1 < nkand nc+1 +· · ·+n j = p for some j. Then nc+2 +· · ·+n j < pand nc+2 + · · · + n j + nk > p. �3. Final conclusions and remarks

Conclusion 6. There exist minimal AP-graphs with arbitrar-ily many arbitrarily long cycles.

This conclusion may also be derived from a different re-sult. Observe that if we have some AP-graph, then we may

700 O. Baudon et al. / Information Processing Letters 112 (2012) 697–700

make a minimal AP-graph of it by deleting its consecutiveedges till we reach the moment that deleting any otheredge makes the obtained graph not AP. Note also that byTheorem 1 neither AP-graph may contain a vertex v whoseremoval disconnects the graph into more than four compo-nents (since otherwise we might construct of it an AP-treewith maximum degree greater than 4 by replacing each ofthe obtained components with paths of the correspondingorders and joining them to the vertex v). A balloon is agraph obtained of some number of paths by joining oneend of each such path to a root vertex r1, and the otherend to a root vertex r2. In [4] it is proved that there ex-ist AP-balloons constructed of arbitrarily many paths (andthus containing arbitrarily many cycles). Then if we removemore than 6 edges from such a balloon so that it remainsconnected, then surely removing one of the two root ver-tices next disconnects the obtained graph into more than4 components. Consequently, we may construct minimalAP-graphs containing arbitrarily many cycles of the cor-responding balloons (constructed of more than 6 paths)by removing at most six edges from each of them. Un-fortunately, it is also proven in [4] that the lengths ofpaths making up such balloons must increase exponen-tially, hence the density of such graphs, ε(G) := ‖G‖/|G|,tends to 1 as their order grows (the same is in fact thecase for minimal AP-trees). The construction presented inthis paper implies something more. It provides the bestknown lower bound regarding the conjecture on the linear(in order) size of minimal AP-graphs.

Corollary 7. Let C be the smallest positive constant (if it ex-ists) such that for each minimal AP-graph with n vertices and medges, m � Cn. Then

C � 31

30.

Proof. We must show that for each ε > 0 there exists aminimal AP-graph G such that

‖G‖|G| >

31

30− ε.

Let p = 3 and n = (s3 + 2)(3 + 1)(3 + 2), where s � 0 isan integer. Then by Theorem 5, there exists a minimal AP-graph Jn with n + (s3 + 2) − s − 1 edges, and hence,

‖ Jn‖| Jn| = 1 + 2s + 1

60s + 40−−−→s→∞

31

30. �

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