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Information Processing Letters www.elsevier.com/locate/ipl

On minimal arbitrarily partitionable graphs ✩ c ,1 Olivier Baudon a,b , Jakub Przybyło c,∗,1 , Mariusz Wozniak ´ a b c

Univ. Bordeaux, LaBRI, UMR 5800, 33400 Talence, France CNRS, LaBRI, UMR 5800, 33400 Talence, France 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

Article history: Received 19 January 2012 Accepted 19 June 2012 Available online 20 June 2012 Communicated by Jinhui Xu Keywords: Minimal arbitrarily partitionable graph Arbitrary vertex decomposable graph Interconnection networks

a b s t r a c t A graph G = ( V , E ) of order n is called arbitrarily partitionable, or AP for short, if given any sequence of positive integers n1 , . . . , nk summing up to n, we can always partition V into subsets V 1 , . . . , V k of sizes n1 , . . . , nk , resp., inducing connected subgraphs in G. If additionally G is minimal with respect to this property, i.e. it contains no AP spanning subgraph, we call it a minimal AP-graph. It has been conjectured that such graphs are sparse, i.e., there exists an absolute constant C such that | E | Cn for each of them. We 1 construct a family of minimal AP-graphs which prove that C 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 sequence τ = (n1 , . . . , nk ) of positive integers is called admissible for G if it is a partition of n, i.e., n1 + · · · + nk = n. If additionally there exists a partition ( V 1 , . . . , V k ) of the vertex set V such that each V i induces a connected subgraph of order ni in G, then we say that τ is realizable in G, while ( V 1 , . . . , V k ) is called a realization of τ in G. If every admissible sequence is also realizable in G, then we say that this graph is arbitrarily partitionable (or arbitrary vertex decomposable, see e.g. [7]) or we call it an AP-graph for 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 every k-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 Polonium 22658VG. Corresponding author. Tel.: +48 12 617 46 38; fax: +48 12 617 31 65. E-mail address: [email protected] (J. Przybyło). 1 Research partially supported by the Polish Ministry of Science and Higher 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 expected vertex partition, they showed that the two properties are equivalent, see [6] or [8] for details. For the small values of k also the polynomial-time algorithms have been given, in particular, for k = 2, k = 3 [9,13], and for the case k = 4 restricted to the planar graphs [10]. The problem of arbitrarily partitionable graphs was then rediscovered, already with the deﬁnition we use, by Barth et al. [1] in the following applicable setting. Suppose our graph is a model of a interconnection computer network, and we have k potential users each of whom we must provide with a connected subnetwork consisting of a given number of computing resources. Then if only our network forms an AP-graph, we may always satisfy the requirements of any number of potential clients, provided only that we have suﬃciently many computers in the network. Note that every AP-graph must be connected, and moreover the property of being AP is monotone, i.e., closed under the operation of adding edges. Thus if only G contains a spanning tree which is AP, then so is G. Lots of research on the AP-property has been devoted to trees then. In particular in [2] Barth and Fournier proved the following. Theorem 1. The maximum degree of each AP-tree is at most 4.

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Fig. 1. Graph A p .

Fig. 2. Graph J n .

From the algorithmic point of view, it has been proved in [2] that deciding whether a given partition τ is realizable in a tree T is NP-complete, even for trees of degree at most 3. A polynomial-time algorithm for tripods (trees consisting of three paths sharing one common vertex) was presented in [1]. Other results concerning the family and subfamilies of tress and related complexity problems are contained, e.g. in [3,5,7,11]. On the other hand there are known examples of APgraphs which do not contain any spanning AP-trees, lots of examples of such graphs are included in this paper. This fact made the problem of AP-graphs even more complex, and led to the following deﬁnition introduced by Ravaux [12]. We say that G is a minimal AP-graph if G is AP and removing any of its edges makes it not AP. Obviously each AP-tree is a minimal AP-graph. Conversely, if G is a minimal AP-graph which is not a tree, then it contains no spanning AP-tree. An example of such a graph was included in [12]. Though such graphs exist, Ravaux suspects that 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 minimal AP-graphs containing arbitrarily many cycles and with asymptotic 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 let A p denote a graph with the vertex set V = { v 1 , . . . , v ( p +1)( p +2) } and the edge set E = { v i v i +1 : i = 1, . . . , p + 2, p + 4, . . . , ( p + 1)( p + 2)− 1}∪{ v p +2 v p +4 , v p +3 v 2p +3 } (hence | E | = | V | = ( p + 1)( p + 2)). This is a unicyclic graph which may also be constructed by adding to the path v 1 v 2 . . . v p +2 v p +4 v p +5 . . . v ( p +1)( p +2) a new vertex v p +3 , together with the edges v p +2 v p +3 and v p +3 v 2p +3 (note that there are only p − 1 vertices on this path between v p +2 and v 2p +3 ), see Fig. 1. We shall call v 1 and v ( p +1)( p +2) the ﬁrst and the last, resp., and v p +3 the top vertex of A p . Moreover, we say that the vertices

v p +2 , . . . , v 2p +3 form a block in A p , and analogously as above, we shall call v p +2 , v 2p +3 and v p +3 the ﬁrst, the last and the top, resp., vertex of this block. Deﬁnition 4. Assume n = l( p + 1)( p + 2), where l = sp + r for some integers r , s with r ∈ {0, . . . , p − 1}, and let J n denote the graph with n vertices, v 1 , . . . , v n , and n + l − s − 1 edges constructed from a path v 1 v 2 . . . v n by removing the edges v i +1 v i +2 and adding the edges v i v i +2 , v i +1 v i + p +1 for each i = t ( p + 1)( p + 2) + ( p + 2) with t ≡ p − 1 (mod p ). (In other words, J n is created by taking l − s copies of A p and s copies of P ( p +1)( p +2) , arranging them into sequence so that every p-th graph of this sequence is a path, and then for each pair of consecutive graphs in this sequence, joining the last vertex of the ﬁrst of them with the ﬁrst vertex of the second one, see Fig. 2.) Theorem 5. Let p be a prime integer, p 3. For each n > 0 divisible by ( p + 1)( p + 2), J n is a minimal AP-graph. Proof. Let n = l( p + 1)( p + 2) > 0, where l = sp + r for some integers r , s with r ∈ {0, . . . , p − 1}. First we show that removing any edge e from J n makes it not AP. Obviously this is the case if e does not belong to any cycle (because then removing it disconnects the graph). Assume then that e lies on a cycle v i v i +2 v i +3 . . . v i + p +1 v i +1 v i for some i = t ( p + 1)( p + 2) + ( p + 2). Note that n ≡ 0 (mod p + 1) and n ≡ 0 (mod p + 2), hence τ p+1 = ( p + 1, . . . , p + 1) and τ p+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, since i − 1 ≡ 0 (mod p + 1) and i ≡ 0 (mod p + 2), it is obvious that in a realization of τ p +1 or τ p +2 the vertices v 1 , . . . v i −1 or v 1 , . . . v i , resp., must be covered by (i − 1)/( p + 1) or i /( p + 2), resp., subsets of this realization. Consequently, if e = v i v i +1 , then one of the subsets of the realization of τ p +1 must be a set V = { v i , v i +2 , v i +3 , . . . , v i + p +1 }, and after removing it, we are left with an isolated vertex v i +1 , see Fig. 3. If on the other hand e = v i v i +2 , we are analogously unable to cover v i +2 while trying to ﬁnd a realization of τ p +1 , see Fig. 4. Finally, in all the remaining cases (for all the other edges

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Fig. 3. Graph J n with e = v i v i +1 removed.

Fig. 4. Graph J n with e = v i v i +2 removed.

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

of our cycle), we immediately create an isolated path containing less than p vertices while trying to realize τ p +2 , see Fig. 5. Second, we show that J n itself is AP. Let us ﬁx any admissible sequence τ for J n . If there are any 1’s in this sequence, choose as many top vertices with the greatest indexes (the ones “from the right”) as their realizations as possible. Then remove the used vertices from J n and the used 1’s from τ . Denote by J the obtained graph, and the obtained sequence by τ . Note that if we had enough 1’s, then J would be a path, and we could easily ﬁnd the realization of the remaining elements of the sequence. Therefore, we may assume that there are still some blocks left in J (the ones on the left), and that there are no 1’s in τ . We shall try to ﬁnd a realization of the remaining elements “from left to right”, i.e., such that the realizations of subsequent elements of τ consist of consecutive vertices (with respect to their indexes), with an exception of the top vertices, where for every such vertex we must only require that it is in the same subset of the realization as the ﬁrst or the last vertex of the block to which it belongs. To do this, it is suﬃcient to ﬁnd an ordering (n1 , . . . , nk ) of the elements of τ such that if S = {s1 , . . . , sk } is the set of the sums of the form si = j i n j , i = 1, . . . , k, then: (*) for no block of J with the ﬁrst vertex v a and the last vertex v b we have a, b − 1 ∈ S. Then we will be able to realize our sequence τ from left to right according to this ordering in such a way that each top vertex is in the same subset of the realization as the ﬁrst or the last vertex of its block (dependent on whether a∈ / S or b − 1 ∈ / S, resp.).

Suppose then that there is no such ordering, and let us 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 which maximizes the number of ﬁrst consecutive blocks for which the condition from (*) holds. Let v q be the ﬁrst vertex of the ﬁrst block (counting from the left) for which our condition does not hold, hence there exist sc , sd ∈ S such that 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 v i of any block. Consequently, 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 elements nc +1 , . . . , nk so that q + p do not belong to the obtained new set S, what yields a contradiction with the assumed “maximality” of the ordering of τ . To see that such reordering exists, it is enough to note that for any set of integers A ( A = {nc +1 , . . . , nk }), each greater than 1 and distinct from a prime number p, whose sum is greater than p, there always exist l1 , . . . , lt ∈ A (t 0) such that l1 + · · · + lt −1 < p and l1 + · · · + lt > p. Indeed, it is obvious if all elements of A are the same, since p is a prime number. Otherwise, suppose nc +1 nc +2 · · · nk , nc +1 < nk and nc +1 + · · · + n j = p for some j. Then nc +2 + · · · + n j < p and nc +2 + · · · + n j + nk > p. 2 3. Final conclusions and remarks Conclusion 6. There exist minimal AP-graphs with arbitrarily many arbitrarily long cycles. This conclusion may also be derived from a different result. Observe that if we have some AP-graph, then we may

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make a minimal AP-graph of it by deleting its consecutive edges till we reach the moment that deleting any other edge makes the obtained graph not AP. Note also that by Theorem 1 neither AP-graph may contain a vertex v whose removal disconnects the graph into more than four components (since otherwise we might construct of it an AP-tree with maximum degree greater than 4 by replacing each of the obtained components with paths of the corresponding orders and joining them to the vertex v). A balloon is a graph obtained of some number of paths by joining one end of each such path to a root vertex r1 , and the other end to a root vertex r2 . In [4] it is proved that there exist AP-balloons constructed of arbitrarily many paths (and thus containing arbitrarily many cycles). Then if we remove more than 6 edges from such a balloon so that it remains connected, then surely removing one of the two root vertices next disconnects the obtained graph into more than 4 components. Consequently, we may construct minimal AP-graphs containing arbitrarily many cycles of the corresponding balloons (constructed of more than 6 paths) by removing at most six edges from each of them. Unfortunately, it is also proven in [4] that the lengths of paths making up such balloons must increase exponentially, hence the density of such graphs, ε (G ) := G /|G |, tends to 1 as their order grows (the same is in fact the case for minimal AP-trees). The construction presented in this paper implies something more. It provides the best known 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 exists) such that for each minimal AP-graph with n vertices and m edges, m Cn. Then

C

31 30

.

Proof. We must show that for each minimal AP-graph G such that

> 0 there exists a

G 31 − . > |G | 30 Let p = 3 and n = (s3 + 2)(3 + 1)(3 + 2), where s 0 is an integer. Then by Theorem 5, there exists a minimal APgraph J n with n + (s3 + 2) − s − 1 edges, and hence,

Jn 2s + 1 31 − −→ . =1+ s− →∞ | Jn| 60s + 40 30

2

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