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Disjoint Spanning Trees 1 Forest Partitioning Disjoint Spanning Trees Luis Goddyn 1 Forest Partitioning A forest in a graph G is a set of edges which induces an acyclic subgraph of G. We say that a set J of Sk edges of a graph G is k-partitionable if J = i=1 Ji is the union of k forests Ji in G. (We may assume the forests to be disjoint, where convenient.) We consider the following question. Forest Partitioning Problem Given a graph G = (V; E), find a k-partitionable set J ⊆ E of maximum size, and provide a certificate that proves jJj is indeed maximum. For any edge set A ⊆ E of a graph G = (V; E), we define the rank of A, written r(A) to be the maximum size of a subforest of A. Thus r(A) is the number of vertices minus the number of connected components in the induced subgraph G[A] := (V; A). We now describe a barrier which prevents G from having a large k-partitionable set. Suppose that G contains a dense subgraph. In particular, suppose that A is a set of edges for which jAj is large but r(A) is Sk small. Then for any k-partitionable J = i=1 Ji, each set Ji \A is a subforest of A, whence jJi \Aj ≤ r(A). It follows that k X jJj = jJi \ Aj + jJ \ (E − A)j ≤ kr(A) + jE − Aj: (1) i=1 The surprising fact is that some such inequality is always tight. Theorem 1 For any graph G and positive integer k, maxfjJj : J ⊆ E is k-partitionableg = minfjE − Aj + kr(A): A ⊆ Eg: Before the proof, we give two applications. By setting J = E and solving for k we solve a forest covering problem. Corollary 2 The minimum number of forests needed to cover the edges of a graph G is maxfdjAj=r(A)e : A ⊆ Eg: Informally, the `forest covering number' of a graph is determined by its `densest' subgraph. Similarly, if G is connected we may equate jJj with kr(E) = k(jV j − 1) to solve a tree packing problem. Corollary 3 The maximum number of edge disjoint spanning trees which can be packed into a connected graph G is minfbjE − Aj=(jV j − 1 − r(A))c : A ⊆ Eg: This expression is best interpreted in terms of the graph G=A obtained by contracting the edges in A. It is easy to see that r(G=A) = jV (G=A)j − 1 = jV j − 1 − r(A), so the expression we are minimizing is bjE(G=A)j=r(G=A)c. Informally, the `tree packing number' is determined by the `least dense contraction'. In graphs the relevant subsets A arise from vertex partitions. For any partition π ` V , let θ(π)(δ(π)) denote the set of edges with its ends in the same (resp. different) parts. A set A ⊆ E is closed if every e 2 E belonging to some circuit C ⊆ A [ feg also belongs to A. Thus closed sets are precisely those of the form θ(π) for some π ` V . It is not hard to see that any set A achieving the minimum in Theorem 1 must be closed. For a given closed set A ⊆ E there is a unique most refined partition π = (πi) ` V satisfying θ(π) = A; namely π is the partition for which each induced subgraph G[πi] is connected. For such π we observe that r(θ(π)) = jV j − jπj and r(δ(π)) = jπj − 1. Thus we may replace the formulae in the above three results by the following expressions, respectively. kjV j + minfjδ(π)j − kjπj : π ` V g: maxfdjθ(π)j=(jV j − jπj)e : π ` V g: minfbjδ(π)j=(jπj − 1)c : π ` V g: 1 2 The Algorithm Proving Theorem 1 The proof of Theorem 1 consists of an algorithm (which happens to be polytime) which, when given G and a k-partition (Ji;:::;Jk) of J ⊆ E, produces either a k-partition of J + e for some e 2 E − J, or a set A ⊆ E satisfying jJj = jE − Aj + kr(A). Thus for a given G and k we may find a largest partitionable set and a suitable A ⊆ E by iterating this algorithm starting with Ji = ;, i = 1::k. Given G, k and (Ji) we set J = [Ji and construct an auxiliary arc-coloured digraph D = D(G; k; (Ji)) with vertex set fs; tg [ E(G) and three types of arcs. For every e; f 2 E and i 2 f1; : : : ; kg, D has 1. an arc (s; e) (having no colour), if e 2 E − J; 2. an arc (e; t) of colour i, if Ji + e is acyclic. 3. an arc (e; f) of colour i, if Ji + e − f is acyclic; Now we seek a shortest path P from s to t in D. If P does not exist, then we output the set A of vertices in V (D) − fsg which are reachable from s by some dipath in D. Since there are no arcs in δ+(fsg [ A) of types 1,2 or 3, we deduce the following facts respectively. 1. E − A ⊆ J 2. For any i 2 f1; : : : ; kg and e 2 A, Ji [ feg has a cycle. 3. For any i 2 f1; : : : ; kg and e 2 A and f 2 A, if f belongs to the cycle in Ji [ feg, then f 2 A. In other words, Ji \ A is a spanning forest of A whence jJi \ Aj = r(A). From facts 1. and 3., it we find that the inequality (1) is tight for the pair (A; (Ji)), so A proves that jJj is indeed maximum. If a shortest s; t-path P = (s; e1; e2; : : : ; es; t) does exist, then by the construction of D, we have e1 2 E − J. We shall the edges of P and their colours to guide an `augmenting sequence', which modify the sets (Ji) in such a way as to produce a k-partition of set J [ fe1g. Let i be the colour of the arc (e1; e2) (we allow here that e2 = t). We modify the forest Ji by adding e1 to it. If e2 = t, then (e1; t) is of type 2, and we output the resulting partition of J + e. Otherwise, 0 (e1; e2) is an arc of type 3. so Ji + e1 has a fundamental cycle which contains e2. Let i be the colour of (e2; e3) (again, we permit e3 = t). Then we then we modify both Ji and Ji0 by transfering e2 from Ji to Ji0 . This restores Ji to being acyclic, but now either we are done (if e3 = t) or Ji0 + e3 has a fundamental cycle containing e4. We continue modifying (Ji) in this manner until we have traversed P . Clearly J is now larger by one element, but it remains to prove that each modification to a forest Ji leaves it acyclic. This is fact not trivial, but is easier to prove by relying on the fact P was chosen to be a shortest path in D. Suppose for example exactly two arcs (ea; ea+1), (eb; eb+1) in P have label i, where a < b. The modification in step a leaves Ji acyclic by construction of D. However at the start of step b, it may no longer be true that Ji + eb − eb+1 is acyclic, since Ji has been changed in step a. This trouble can only have occurred if eb+1 also belonged to the fundamental cycle Ji + ea at the start of step a. This would imply that (ea; eb+1) is an arc of D, contradicting that P is a shortest path. Equivalently, we have argued that the set sp(J) of edges of G which have both endpoints in the same connected component of G[J] is equal to sp(J + eb − eb+1). If Ji has been modified more than twice we must argue more carefully. Suppose that label i occurs in steps a1; a2; : : : at in that order along P . For j = 0; : : : ; t − 1 we define (j) J = Ji + eat−j − eat−j +1 + eat−j+1 − eat−j+1+1 + ··· + eat − eat+1 Here we are adding and subtracting edges from J in `reverse order' to the way it was done in the (t) (0) algorithm, so J = Ji input at the start of the algorithm and J is the set after all modifications have been made. We now argue by induction on j that the set sp(J (j)) of edges spanned by each connected component of J (j) is the same as sp(J (0)). We leave out the details of this argument. In the end, we have (0) (t) (0) that sp(J ) = sp(J ) = sp(Ji), which implies J is acyclic, since it has the same cardinality as Ji. 2 3 Generalization to Matroids The above algorithm in fact works in a more general framework. Let M1, M2,...,Mk be a list of matroids all having the same set E for its elements. We say that J ⊆ E is partitionable (with respect to the list) S There exist subsets Ji, i = 1; =dots; k such that Each Ji is an independent set of Mi and J = Ji. An obvious modification of the above algorithm, and its proof of correctness solves the following problem. Matroid Partitioning Problem Given matroids M1,..., Mk on E, find a k-partitionable set J ⊆ E of maximum size, and provide a certificate that proves jJj is indeed maximum.
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