Factors in Graphs With Multiple Degree Constraints

Transcription

1 Factors in Graphs With Multiple Degree Constraints Richard C. Brewster, Morten Hegner Nielsen, and Sean McGuinness Thompson Rivers University Kamloops, BC V2C0C8 Canada October 4, 2011 Abstract For a graph G and for each vertex v V (G), let Λ G (v) = {E G (v, 1), E G (v, 2),..., E G (v, k v )} be a partition of the edges incident with v. Let Λ G = {Λ G (v) v V (G)}. We call the pair (G, Λ G ) a partitioned graph. Let k = max v k v and let g, f : V (G) {1,..., k} N and t, u : V (G) N be functions where for all vertices v V (G) (i) g(v, i) f(v, i) d G (v, i) i = 1,..., k v (ii) u(v) t(v) d G (v) (iii) u(v) k v i=1 f(v, i) and i=1 g(v, i) t(v). A subgraph H of the partitioned graph is said to be a (g, f, u, t)- factor if all vertices v V (G) satisfy (a) g(v, i) d H (v, i) f(v, i), i = 1,..., k v (b) u(v) d H (v) t(v) and where d H (v, i) = E(H) E G (v, i). In this paper, we shall show via a reduction to a matching problem, that there is a good algorithm for determining whether a partitioned graph has a (g, f, u, t)-factor. Secondly, we shall also prove a theorem which characterizes when (0, f, t, u)-factors exist in a partitioned graph when u(v) < f(v, i) for all v and i. As a special case, we obtain Lovász s (g, f)-factor theorem. 1

2 1 Introduction 1.1 Notation Let G be a graph. For U V (G), let U = V (G)\U. For v V (G), let N G (v) denote the set of neighbours of v in G, let E G (v) be the set of edges incident with v, and let d G (v) = E G (v). For U V (G), we let E G (U) = v U E G (v) and d G (U) = E G (U). For sets U, W V, E G (U, W ) shall denote the set of edges joining a vertex in U to a vertex in W and d G (U, W ) shall denote the cardinality of this set. For a real-valued function φ defined on V (G) and a set S V (G), we let φ(s) = s S φ(s). If ψ is another real-valued function defined on V (G), we define two sets V ψ<φ = {v V (G) ψ(v) < φ(v)} and V ψ=φ = {v V (G) ψ(v) = φ(v)}. For a digraph G and sets X, Y V ( G) we let d + G (X, Y ) (respectively, d G (X, Y )) denote the number of arcs directed from X to Y (respectively, Y to X). We let d G (X, Y ) = d + G (X, Y ) + d G (X, Y ). Let (G, Λ) be a partitioned graph (as defined in the abstract). For U, V V (G) and I, J {1, 2,..., k}, let E G (U, I) = v U,i I E G (v, i), E G (U; V, J) = E G (U) E G (V, J) and E G (U, I; V, J) = E G (U, I) E G (V, J). Let d G (U), d G (U, I), d G (U; V, I), and d G (U, I; V, J) denote the respective cardinalities of these sets. In addition, we let E G (U, I; V ) = E G (V ; U, I) and d G (U, I; V ) = d G (V ; U, I). Let φ : V (G) {1, 2..., k} N. For v V (G), I {1,..., k}, and V V (G), let φ(v) = k i=1 φ(v, i), φ(v, I) = i I φ(v, i), and φ(v, I) = v V φ(v, I). 1.2 Factors in Graphs Suppose that f is a nonnegative integer-valued function defined on the vertices of a graph G where 0 f(v) d G (v) for all vertices v V (G). A subgraph H of G is called an f-factor of G if d H (v) = f(v) for all v V (G). If f(v) = k for all v V (G) then an f-factor is usually referred to as a k-factor. The theory of graph factors dates back at least to 1947 with Tutte s 1-Factor Theorem [8], which characterizes the graphs having a perfect matching. Later developments include Tutte s f-factor Theorem [9] which we shall describe below. For disjoint sets of vertices S, T V (G), 2

3 we let qg (S, T ) denote the number of components K of G\S T such that d G (V (K), T ) + f(v (K)) is odd. 1.1 Theorem ( Tutte ) A graph G has an f-factor iff every pair of disjoint subsets S, T V (G) satisfies d G (T ) f(t ) + f(s) d G (S, T ) q G(S, T ) 0. In [5], Lovász extended Tutte s Theorem above. Suppose g, f are nonnegative integer-valued functions defined on the vertices of a graph G where g(v) f(v) d G (v) for all v V (G). A subgraph H is called a (g, f)-factor if g(v) d H (v) f(v) for all v V (G). Below, we state Lovász s (g, f)- Factor Theorem which characterizes when a graph has a (g, f)-factor. For a short proof of this theorem, we refer the reader to [6]. 1.2 Theorem ( Lovász ) A graph G has a (g, f)-factor iff every pair of disjoint subsets S, T V (G) satisfies d G (T ) g(t ) + f(s) d G (S, T ) q G (S, T ) 0. (1) Here q(s, T ) is the number of components K in G\S T where V (K) V f=g and f(v (K)) + d G (S, V (K)) is odd. While numerous variations of this theorem have appeared (see [4]), there are still several other possible avenues for further research. For example, consider the following problem for digraphs. Suppose we are given a digraph G and functions f, f +, u, t defined on V ( G) where f (v) d G (v), f + (v) d + G (v), and u(v) t(v) d G (v) for all vertices v. Is there a theorem describing necessary and sufficient conditions (in a similar manner to the above theorem) for the existence of a subdigraph H for which d H (v) f (v), d + H (v) f + (v), and u(v) d H (v) t(v) for all vertices v? We shall show that this is possible given certain extra conditions. Observe that a directed graph is a special example of a partitioned graph; at every vertex, the out-directed and in-directed edges at a vertex form a partition of its incident edges. In this paper we shall extend the above factor theorems to partitioned graphs. More precisely, we shall address two general questions: 1.3 Question Is finding a (g, f, t, u)-factor in a partitioned graph a polynomial problem? 3

4 1.4 Question Is there a (g, f, t, u)-factor theorem for partitioned graphs? We shall demonstrate that the answer to the first question is yes, by showing that the problem is equivalent to finding a certain matching in an expanded graph, described in Section 2. The answer to the second question seems quite complicated in general and for this reason we shall focus on the case where g = 0. Building on the themes of Tutte s f-factor and Lovász s (g, f)-factor theorems, we shall give a theorem which characterizes when (0, f, t, u)-factors exist in partitioned graphs, in the case when u(v) < f(v, i) for all v and i. We shall describe the theorem below, but first we need some notation. Let (G, Λ) be a partitioned graph and let g, f, t, u be functions as described in the abstract. Since for any (g, f, t, u)-subgraph H we have that d H (v, i) f(v, i) for all v and i, it follows that d H (v) f(v) for all v. Because of this, we may always assume that t(v) f(v) for all v. Suppose g = 0. Then H is a (0, f, t, u)-subgraph of (G, Λ G ) if for all vertices v V (G) we have d H (v, i) f(v, i) for all i {1,..., k v } and u(v) d H (v) t(v). Let I 1, I 2,..., I 2 k denote the subsets of {1, 2,..., k} where we shall assume that I 1 = Ø and I 2 k = {1, 2,..., k}. For a set I {1, 2,..., k} let I = {1, 2,..., k}\i. For pairwise disjoint (possibly empty) subsets A 1, A 2,..., A 2 k of V (G), let K be the set of components of G = G\{A 1 A 2 A 2 k}). We shall say that a component K K is odd with respect to A 1,..., A 2 k if V (K) V u=t ; u(v (K)) + d G (A 1, V (K)) 1(mod 2); d G (V (K); A i, I i ) = 0, i = 2, 3,..., 2 k. Let q(a 1,..., A 2 k) denote the number of such odd components. In Section 5, we shall prove the following theorem: 1.5 Theorem Suppose u(v) < f(v, i) for all v V (G) and i = 1,..., k v. Then the partitioned graph (G, Λ G ) has a (0, f, t, u)- subgraph if and only if all choices of 2 k pairwise disjoint (possibly empty) sets A 1,..., A 2 k V (G) satisfy 2 k 1 d G (A 1 ; A i, I i )+q(a 1,..., A 2 k) d G (A 1 ) u(a 1 )+t(a 2 k)+ f(a i, I i ). (2) 4

5 The case k = 1 in the above theorem is seen to be just the (g, f)-factor theorem (Theorem 1.2). The above theorem can also be applied to digraphs. Suppose we are given a digraph G and functions f, f +, u, t defined on V ( G) where f (v) d G (v), f + (v) d + G (v), and u(v) t(v) d G(v) for all vertices v. Then a subdigraph H is an (0, f, t, u)-factor if d H (v) f (v), d + H (v) f + (v), and u(v) d H (v) t(v) for all vertices v. Applying Theorem 1.5 in the case k = 2 for digraphs, one obtains the following theorem: 1.6 Theorem Suppose u(v) < f (v) and u(v) < f + (v) for all vertices v V ( G). Then G has a (0, f, t, u)-factor if and only if for all choices of four disjoint (possibly empty) sets A i V ( G), i = 1, 2, 3, 4 we have d + G (A 1, A 2 ) + d G (A 1, A 3 ) + d G (A 1, A 4 ) + q(a 1, A 2, A 3, A 4 ) d G (A 1 ) u(a 1 ) + t(a 4 ) + f (A 2 ) + f + (A 3 ). We mention that in the case u = 1, the problem of finding a (0, f, t, u)-factor in a directed graph amounts to determining whether there exist certain oriented stars which partition the vertices of the graph; that is, whether a star-packing exists or not. The authors in [2] provide necessary and sufficient conditions for such star-packings to exist in a digraph. While their theorem is simpler, it applies only to the case where u = 1. 2 The Expanded Graph Ĝ Suppose (G, Λ G ) is a partitioned graph where Λ G (v) = {E G (v, 1),..., E G (v, k v )} for all vertices v V (G). Let g, f, t, u be nonnegative integer-valued functions satisfying conditions (i) and (ii) as defined in the abstract. A key concept in this paper is the expanded graph Ĝ which we associate with the partitioned graph (G, Λ G ) and the functions g, f, t, u. We construct such a graph as follows. First, to each vertex v in G, we associate the following subgraph in Ĝ: For each set E G(v, i), create three disjoint sets X(v, i), Y (v, i), and Z(v, i), containing d G (v, i), d G (v, i) f(v, i), and f(v, i) g(v, i) vertices, respectively. Join each vertex of X(v, i) to each vertex of Y (v, i) Z(v, i). Next, add a set W (v) consisting of t(v) k v i=1 g(v, i) vertices and join each 5

6 v t(v) = 8 d(v, 1) = 4 f(v, 1) = 2 g(v, 1) = 1 d(v, 2) = 6 f(v, 2) = 5 g(v, 2) = 2 d(v, 3) = 3 f(v, 3) = 1 g(v, 3) = 0 W (v) = t(v) g(v, 1) g(v, 2) g(v, 3) Y (v, 1) = d(v, 1) f(v, 1) Z(v, 1) = f(v, 1) g(v, 1) Y (v, 2) Z(v, 2) Y (v, 3) Z(v, 3) X(v, 1) X(v, 2) X(v, 3) Figure 1: Expanding a vertex v in Ĝ where k v = 3. of them to each vertex of k v i=1 Z(v, i). Finally, for each edge e = uv E(G) we shall associate an edge ê E(Ĝ) where, given that e E G(u, i) E G (v, j), the edge ê will join a vertex of X(u, i) to a vertex in X(v, j). We may (and shall) choose the edges ê so that they form a matching in Ĝ; that is, for any two distinct edges e, f E(G), the edges ê and f are nonincident. In Figure 1, we illustrate the subgraph corresponding to a vertex v,when k v = 3. We call the resulting graph the (G, Λ G, g, f, t, u)-expanded graph and denote it by Ĝ(Λ G, g, f, t, u). 3 Reduction to a Matching Problem In this section, we shall show that the problem of finding a (g, f, t, u)-factor in a partitioned graph reduces to a matching problem on the expanded graph. Let (G, Λ G ) be a partitioned graph and let g, f, t, u be nonnegative 6

7 integer-valued functions as described in the abstract. Let Ĝ = Ĝ(G, Λ G, g, f, t, u) be the expanded graph. For convenience, we let X(v) = k v i=1 X(v, i), Y (v) = kv i=1 Y (v, i), and Z(v) = k v i=1 Z(v, i). 3.1 Theorem For all v V (G), let W 0 (v) W (v) where W 0 (v) = t(v) max{u(v), g(v)}. Then (G, Λ G ) has a (g, f, t, u)-factor iff Ĝ has a matching which saturates all vertices of V (Ĝ)\ v V (G) W 0(v). Proof. Suppose M is a matching in Ĝ which saturates all vertices of V (Ĝ)\ v V (G) W 0(v). Let H be the subgraph of G induced by the edges e E(G) where ê M. Then d H (v, i) is the number of edges of M matching vertices of X(v, i) to vertices of u V (G)\{v} X(u). Since all the vertices of Y (v, i) must be matched to X(v, i), it follows that X(v, i) d H (v, i) Y (v, i) and hence d H (v, i) X(v, i) Y (v, i) = f(v, i) for all v and i. Also, we see that X(v, i) d H (v, i) vertices in X(v, i) are matched to vertices in Y (v, i) Z(v, i). Thus d G (v, i) d H (v, i) = X(v, i) d H (v, i) Y (v, i) Z(v, i) = d G (v, i) g(v, i). Consequently, we have g(v, i) d H (v, i) for all v and i. Moreover, there are exactly d H (v, i) g(v, i) vertices of Z(v, i) which are not matched to vertices of X(v, i) and hence these vertices are matched to vertices in W (v). Thus d H (v) g(v) = k v i=1 (d H(v, i) g(v, i)) W (v) = t(v) g(v) and it follows that d H (v) t(v) for all v. However, since all the vertices of W (v)\w 0 (v) are matched to vertices of Z(v), we see that W (v)\w 0 (v) = max{u(v), g(v)} g(v) d H (v) g(v). Thus u(v) d H (v) for all v. We conclude that H is a (g, f, t, u)-subgraph of (G, Λ G ). Suppose, on the other hand, that H is a (g, f, t, u)-subgraph of (G, Λ G ). Let M = {ê e E(H)}. Then M is seen to be a matching in Ĝ. We shall show that M can be extended to a matching M of Ĝ which saturates all vertices in V (Ĝ)\ v V (G) W 0(v). We first observe that there are exactly X(v, i) d H (v, i) = d G (v, i) d H (v, i) M -unsaturated vertices in X(v, i). For all v V (G) and i = 1,..., k v do the following: match the d G (v, i) d H (v, i) M -unsaturated vertices of X(v, i) to vertices in Y (v, i) Z(v, i) so that all vertices of Y (v, i) are matched. This is possible since Y (v, i) = d G (v, i) f(v, i) d G (v, i) d H (v, i) Y (v, i) Z(v, i) = d G (v, i) g(v, i). This leaves d H (v, i) g(v, i) vertices of Z(v, i) unmatched. Let Z (v, i) denote this unmatched set of vertices and let Z (v) = k v i=1 Z (v, i). Now we 7

8 can match the vertices of Z (v) to vertices of W (v) since Z (v) = k v i=1 (d H (v, i) g(v, i)) = d H (v) g(v) W (v) = t(v) g(v). Furthermore, we can match all the vertices of W (v)\w 0 (v) in such a matching since W (v)\w 0 (v) = max{u(v), g(v)} g(v) d H (v) g(v) = Z (v). Let M be the resulting matching. Then M saturates all vertices of V (Ĝ)\ v V (G) W 0(v). This completes the proof. We shall now look at the problem when u = 0. Suppose no (g, f, t, 0)- subgraph of (G, Λ) exists. Extending an analogous concept for (g, f)-factors (see [3]), one can measure the extent to which a (0, f, t, 0)-factor falls short of being a (g, f, t, 0)-factor by the notion of g-difficiency. For an (0, f, t, 0)- subgraph H, we define the (g, v, i)-deficiency of H to be def H (v, i) = max{g(v, i) d H (v, i), 0} and the (g, v)-deficiency of H to be def H (v) = kv i=1 def H(v, i). Finally, we define the g-deficiency of H to be def H = v V def H(v) and let δ(g, g, f, t) = min{def H H is an (0, f, t, 0)-subgraph of (G, Λ G )}. The Minimum Deficiency Problem (MDP) for (G, Λ G ) is defined as follows: Given: (G, Λ G ) and the functions g, f, and t. Find: a (0, f, t, 0)-subgraph H of G for which def H = δ(g, g, f, t). In the next theorem, we shall show that there is a good (polynomial) algorithm to MDP by reducing it to a weighted matching problem on the expanded graph Ĝ. For a weighting w : V (Ĝ) N, if M is a matching of Ĝ, then we let w(m) be the sum of the weights of vertices covered by M. We define the deficiency of M with respect to w to be defĝ,w (M) = w(v (Ĝ)) w(m) and let γ(ĝ, w) = min{def (M) M a matching of Ĝ}. Ĝ,w 8

9 3.2 Theorem Let w : V (Ĝ) N be the weight function w(v) = Then δ(g, g, f, t) = γ(ĝ, w). 0 if v u V (G) W (u); 1 if v u V (G) X(u); 2 if v u V (G)(Y (u) Z(u)). Proof. Let M be a maximum-weight matching in Ĝ. Notice that every vertex in v (Y (v) Z(v)) is M-saturated for, if x were an M-unsaturated vertex belonging to some set Y (v, i) Z(v, i), then the maximality of M would imply that the vertices of X(v, i) were all matched to (Y (v, i) Z(v, i)) \ {x}, contradicting the fact that X(v, i) = d G (v, i) > d G (v, i) g(v, i) 1 = (Y (v, i) Z(v, i)) \ {x}. Thus γ(ĝ, w) equals the number of M-unsaturated vertices of v X(v). Let µ(v, i) (for all v, i) denote the number of M-unsaturated vertices of X(v, i) and let H be the spanning subgraph of G consisting of the edges e E(G) where ê M. Now consider any pair v, i. We shall show that def H (v, i) = µ(v, i) for all v, i. Suppose µ(v, i) 1. By maximality of M, every vertex of Y (v, i) Z(v, i) is matched to a vertex in X(v, i) and so def H (v, i) = max{g(v, i) d H (v, i), 0} = max{g(v, i) ( X(v, i) Y (v, i) Z(v, i) µ(v, i)), 0} = µ(v, i). Suppose µ(v, i) = 0. Let β denote the number of edges of M between X(v, i) and Z(v, i); note that β Z(v, i) = f(v, i) g(v, i). Then def H (v, i) = max{g(v, i) d H (v, i), 0} = max{g(v, i) ( X(v, i) Y (v, i) β), 0} = max{g(v, i) f(v, i) + β, 0} = 0 = µ(v, i). It follows that def H (v, i) = µ(v, i) for all v, i and hence δ(g, g, f, t) γ(ĝ, w). To complete the proof, we need only show that γ(ĝ, w) δ(g, g, f, t). Suppose H is a minimum-deficiency (0, f, t, 0)-subgraph of G and let M be the matching in Ĝ consisting of exactly those edges ê E(Ĝ) where e E(H). Suppose def H (v, i) > 0. Then def H (v, i) = g(v, i) d H (v, i) and hence d G (v, i) = X(v, i) = d H (v, i) + def H (v, i) + d G (v, i) g(v, i) = d H (v, i) + def H (v, i) + Y (v, i) Z(v, i). 9

10 Denote by M(v, i) a perfect matching between Y (v, i) Z(v, i) and any Y (v, i) Z(v, i) M-unsaturated vertices in X(v, i); then the extended matching M M(v, i) leaves exactly def H (v, i) vertices unmatched in X(v, i). Now suppose def H (v, i) = 0. Then X(v, i) d H (v, i) satisfies Y (v, i) = d G (v, i) f(v, i) X(v, i) d H (v, i) d G (v, i) g(v, i) = Y (v, i) Z(v, i). Denote by M(v, i) a perfect matching between the X(v, i) d H (v, i) M- unsaturated vertices of X(v, i) and the set Y (v, i) Z (v, i), for some Z (v, i) Z(v, i) where Z (v, i) = X(v, i) d H (v, i) Y (v, i) = f(v, i) d H (v, i). Then the extended matching M M(v, i) leaves exactly Z(v, i) \ Z (v, i) = d H (v, i) g(v, i) vertices unmatched in Z(v, i). It now follows that, for any v V (G), i : def H (v,i)=0 (d H (v, i) g(v, i)) k v i=1 (d H (v, i) g(v, i)) = t(v) g(v) = W (v). Therefore, for each vertex v V (G) in turn, we may further extend the matching M v,i M(v, i) by a perfect matching between i : def H (v,i)=0(z(v, i)\ Z (v, i)) and some subset of W (v). The resulting matching has the same deficiency as H, and we conclude that γ(ĝ, w) δ(g, g, f, t). It is well known that a maximum-weight matching in a (vertex- or edge-) weighted graph can be found in polynomial time (see [7]). Note also that the proof of Theorem 3.2 shows how to construct a minimumdeficiency (0, f, t, 0)-subgraph of a graph G from a maximum-weight matching in (Ĝ, w); thus we have the following corollary (where digraphs may contain parallel arcs and 2-cycles). 3.3 Corollary Let G be a digraph with associated functions g, g +, f, f +, t : V ( G) N satisfying that 0 g (v) f (v) d G (v), 0 g + (v) f + (v) d + G (v), and g (v) + g + (v) t(v) f (v) + f + (v) for every v V ( G). Then there is a polynomial time algorithm for constructing a spanning subdigraph H of G satisfying (i) d (v) f (v), d + (v) f + (v), and d H H H (v) t(v) for all v V ( G); (ii) the deficiency ( ) v V ( max{g (v) d G) (v), 0} + max{g + (v) d + (v), 0} H H of H is minimum over all spanning subdigraphs of G that satisfy (i). 10

11 An immediate consequence of Corollary 3.3 is that a cycle factor (i.e., a spanning collection of pairwise disjoint directed cycles) in a digraph can be found in polynomial time, if one exists. 4 The Tutte-Berge Formula and Barriers We shall make use of the classical Tutte-Berge formula (see [1], Corollary 16.12). Suppose w : V (G) {0, 1} is a weighting of the vertices of a graph G. Let S V (G); for each component K K(G\S), we say that K is odd with respect to w if K K o (G\S) and all vertices of K have weight one. If K is not odd with respect to w, it is even. We let K o w(g\s) be the set of odd components of G\S with respect to w and let o w (G\S) = K o w(g\s). Let ξ w (S) = max{o w (G\S) S, 0} and let ξ w (G) = max{ξ w (S) S V (G)}. We then have the following theorem which can be derived from the classic Tutte-Berge formula (where w(v) = 1 for all vertices v). 4.1 Theorem ( Weighted Tutte-Berge ) γ(g, w) = ξ w (G) Let G be a graph and w : V (G) {0, 1} be a weighting of the vertices of G. We say that a set S V (G) is a barrier for the weighted graph (G, w) if ξ w (S) > 0. Moreover, S is a maximum barrier for (G, w) if (I) S is a barrier for which ξ w (S ) is maximum; (II) S is minimum among all barriers that satisfy (I). The next lemma describes some useful properties pertaining to S. 4.2 Lemma Suppose S is a maximum barrier for the weighted graph (G, w). Then: (i) Each x S is adjacent to at least one odd component of K o w(g \ S ). (ii) Each x S with w(x) = 1 is adjacent to at least two odd components of K o w(g \ S ); furthermore, if x is adjacent to no components in K(G \ S ) having a vertex of weight zero, then x is adjacent to at least three odd components. (iii) Each x S with w(x) = 0 is adjacent to at least two components of K(G \ S ). 11

12 (iv) For each pair x 1, x 2 of non-adjacent vertices of G where w(x i ) = 1 and N G (x i ) \ N G (x 3 i ) 1, i = 1, 2, either {x 1, x 2 } S or {x 1, x 2 } S. Proof. If x S is adjacent to no odd components of Kw(G o \ S ), then the set T = S \ {x} satisfies ξ w (T ) ξ w (S ) and T < S, contradicting the choice of S. This proves (i). Let x S where w(x) = 1. By (i), x is adjacent to at least one odd component of Kw(G o \ S ). Let T = S \ {x}. If x is adjacent to only one odd component, then x together with all the components of K(G \ S ) that are adjacent to x become one even component in K(G \ T ); this means that ξ w (T ) ξ w (S ) and so T is a barrier for (G, w) with T < S, contradicting the choice of S. So x is adjacent to at least two odd components of Kw(G o \ S ). If x is adjacent to no components K K(G\S ) having vertices of weight zero, and x is adjacent to exactly two odd components, then x together with all the components in K(G \ S ) which it is adjacent to become one odd component in Kw(G o \ T ); again, it follows that ξ w (T ) ξ w (S ) and T < S, contradicting the choice of S. This proves (ii). Suppose x S has weight w(x) = 0 and is adjacent to at most one component of K(G \ S ). Let T = S \{x}. Then ξ w (T ) > ξ w (S ), if x is not adjacent to any odd component of Kw(G o \ S ), and ξ w (T ) = ξ w (S ), if x is adjacent to an odd component of Kw(G o \ S ). Since T < S, this contradicts the minimality of S. This proves (iii) Suppose x 1, x 2 are as stated in (iv) and x 1 S and x 2 S. Then x 2 belongs to a component K in K(G \ S ). Suppose that x 1 is adjacent to at least three components in K(G \ S ); such components contain vertices of N G (x 1 ). Since N G (x 1 ) \ N G (x 2 ) 1, at least two such components also contain vertices of N G (x 2 ). Given that x 2 V (K), both components must be contained in K, a contradiction. So x 1 is adjacent to at most two components of K(G \ S ). The first part of (ii) implies that x 1 is adjacent to exactly two components, both of which are odd; however, this contradicts the last part of (ii). This proves (iv). 5 A theorem on (0, f, t, u)-factors In this section, we shall prove Theorem

13 5.1 The weighted graph ( G, w) Let (G, Λ G ) be a partitioned graph, and let g, f, u, t be functions satisfying conditions (i),(ii), and (iii) given in the abstract where we shall assume that g = 0. For convenience, we shall extend the notation E G (v, i) by defining E G (v, i) = Ø, whenever i > k v. We shall construct a weighted graph ( G, w) as follows: for each v V (G), let X(v, 1), X(v, 2),..., X(v, k v ), Y (v, 1), Y (v, 2),..., Y (v, k v ) be disjoint sets with cardinalities X(v, i) = d G (v, i) and Y (v, i) = d G (v, i) f(v, i), where i = 1, 2,..., k v. Let W (v) and Z(v) be disjoint sets, where W (v) = f(v) t(v) and Z(v) = t(v) u(v). For all v V (G) and all i {1,..., k v }, each vertex of X(v, i) shall be joined to each vertex of Y (v, i) W (v) Z(v). For v V (G), let Ṽ (v) = X(v) Y (v) W (v) Z(v). For a subset U V (G), let X(U) = v U X(v), Y (U) = v U Y (v), and Ṽ (U) = v UṼ (v). For v V (G), i {1,..., k v }, and I {1,..., k v }, let Ṽ (v, i) = X(v, i) Y (v, i) W (v) Z(v) and Ṽ (v, I) = i IṼ (v, i). Given that X(v) = d G(v), we may assign to each edge e E G (v) a distinct vertex v e X(v). For each edge e E(G), we shall assign an edge ẽ in G as follows: if e has endvertices u and v, then let ẽ be the edge joining the vertices u e and v e in G. Finally, the weight function w : V ( G) {0, 1} shall be defined as follows: { 0, if v w(v) = u Z(u); 1, otherwise. We have the following lemma whose proof is left as an easy exercise. 5.1 Lemma The partitioned graph (G, Λ G ) has an (0, f, t, u)-subgraph if and only if ( G, w) has a matching which saturates all vertices of weight Proof of the (0, f, t, u)-factor theorem We shall make one modification to the graph G defined above. For each v V u=t, we may assume that k v = 1 (since t(v) = u(v) < f(v, i), for all i = 1,..., k v ); furthermore, we may assume that Y (v), W (v), and Z(v) are sets such that W (v) = d G (v) u(v) and Y (v) = Z(v) = Ø. Lemma 5.1 is seen to hold even with this modification to Ĝ. Suppose that H is a (0, f, t, u)-subgraph of (G, Λ G ). Let H = G\E(H). Let A 1,..., A 2 k be disjoint (possibly empty) subsets of V (G) and let A = A 1 A 2 A 2 k. Let 13

14 K be the set of components of G = G\A. We shall show that (2) holds. We first observe that d G (A 1 ; A i, I i ) = d H (A 1 ; A i, I i ) + d H (A 1 ; A i, I i ). (3) Let K K where K is odd with respect to A 1,..., A 2 k. Since V (K) V u=t, it follows that d H (v) = u(v) = t(v) for all v V (K). Thus u(v (K)) = v V (K) u(v) = v V (K) d H(v) d H (V (K), V (K)) (mod 2). Given that K is odd, we have that d H (V (K), V (K))+d G (A 1, V (K)) 1 (mod 2). Since d G (A 1, V (K)) = d H (A 1, V (K)) + d H (A 1, V (K)) and d H (V (K), V (K)) = d H (A 1, V (K))+d H (A\A 1, V (K)), we see that d H (A 1, V (K))+d H (A\A 1, V (K)) 1 (mod 2). Since K is odd with respect to A 1, A 2,..., A 2 k we have that d G (V (K); A i, I i ) = 0, i = 1,..., 2 k. Thus we have d H (A\A 1, V (K)) = 2 k d H(A i, I i ; V (K)). It follows that d H (A 1, V (K))+ 2 k d H(A i, I i ; V (K)) 1 (mod 2). Thus we have Thus q(a 1,..., A 2 k) d H (A 1, A) + d G (A 1 ; A i, I i ) + q(a 1,..., A 2 k) = We also have + d H (A i, I i ; A). (4) d H (A 1 ; A i, I i ) d H (A 1 ; A i, I i ) + q(a 1,..., A 2 k) d H (A 1 ; A i, I i ) + + d H (A 1, A) + d H (A 1 ; A i, I i ) + d H (A 1, A) d H (A 1 ) 14 d H (A 1 ; A i, I i ) d H (A i, I i ; A). (5) (d G (v) u(v)) v A 1 = d G (A 1 ) u(a 1 ), (6)

15 and d H (A 1 ; A i, I i ) + 2 k 1 2 k 1 d H (A i, I i ; A) f(a i, I i ) + d H (A 1, A 2 k) + d H (A 2 k, A) f(a i, I i ) + t(a 2 k). (7) It follows from (5), (6), and (7) that (2) holds. Suppose now that (G, Λ G ) has no (0, f, t, u)-subgraph. We shall find pairwise disjoint subsets of vertices A 1, A 2,..., A 2 k for which (2) does not hold. Given that there is no (0, f, t, u)-subgraph, Lemma 5.1 implies that ( G, w) has no matching which saturates all vertices of weight 1. Thus ξ w ( G) > 0; let S be a maximum barrier for G. Denote by o sing w ( G\S ) and ow non sing ( G\S ) the number of odd singleton and odd non-singleton components, respectively, of G\S with respect to w. Using Lemma 4.2, one can show the following: (i) W (v) Z(v) S or W (v) Z(v) S, for all v V (G). (ii) X(v, i) S or X(v, i) S, for all v V (G) and i = 1,..., k v. (iii) Y (v, i) S or Y (v, i) S, for all v V (G) and i = 1,..., k v. In addition, we have the following properties (iv) (vii): (iv) For all v V (G), if W (v) Z(v) S, then X(v) S and Y (v) S. Proof. Suppose W (v) Z(v) S. If v V u=t, then k v = 1 and Y (v) = Z(v) = Ø. Thus each vertex of W (Z) is only adjacent to vertices in X(v) = X(v, 1). It follows by (ii) and Lemma 4.2 (ii) that X(v) S. Thus we may assume that v V u<t and hence Z(v) Ø. Suppose X(v, i) S for some i {1,..., k v }. Let I = {j X(v, j) S }. Clearly, I Ø, since each vertex of W (v) Z(v) is adjacent to at least one odd component of G\S. Let T = S \(W (v) Z(v) j I Y (v, j)). 15

16 Then we see that ξ w (T ) ξ w (S ) + W (v) + Z(v) j I ξ w (S ) + k v f(v, j) u(v) j=1 j I ξ w (S ) + f(v, i) u(v) > ξ w (S ), f(v, j) f(v, j). where the last inequality follows from the fact that f(v, i) > u(v). Thus ξ w (T ) > ξ w (S ) and this contradicts the fact that S is a maximum barrier. It follows that X(v) S. We shall show that Y (v) S. Suppose to the contrary that Y (v, i) S for some i {1,..., k v }. Then X(v, i) Y (v, i) V (K) for some component K of G\S. Let T = S \Ṽ (v). Then all the vertices of Ṽ (v) belong to a single component of G\T. Furthermore, we see that ξ w (T ) ξ w (S ) + W (v) + Z(v) 1 j i = ξ w (S ) + f(v) u(v) 1 j i = ξ w (S ) + f(v, i) u(v) 1 ξ w (S ). f(v, j) f(v, j) Thus ξ w (T ) ξ w (S ); however, T < S and this contradicts the fact that S is a maximum barrier. Therefore, Y (v) S. This completes the proof of (iv). (v) For all vertices v V (G), if W (v) Z(v) S, then for i = 1,..., k v we have either X(v, i) S and Y (v, i) S or Ṽ (v, i) V (K), for some component K of G\S. Proof. Let v V (G) and suppose that W (v) Z(v) S. Let i {1,..., k v }. If X(v, i) S, then (iii) and Lemma 4.2 (i) imply that Y (v, i) S. We suppose therefore that X(v, i) S. Then X(v, i) W (v) Z(v) V (K), for some component K of G\S. Now Lemma 4.2 (ii) implies that Y (v, i) S. In particular, this means that Y (v, i) V (K) as well. This completes the proof of (v). 16

17 Recall that the sets I 1, I 2,..., I 2 k denote the subsets of {1, 2,..., k} where I 1 = Ø and I 2 k = {1, 2,..., k}. For each v V (G), let I v = {i {1,..., k} X(v, i) S }. Let A 1 = {v V (G) W (v) Z(v) S }. For i = 2,..., 2 k 1, let A i = {v V (G) I v = I i ; X(v) S }. Let A 2 k = {v V (G) X(v) S } and A = A 1 A 2 k. By (iv), it follows that I v = I 1 = Ø for all v A 1. Let v V u=t. Then X(v) = X(v, 1) and Y (v) = Z(v) = Ø. There are three possibilities for v, namely: either X(v) S and W (v) S, or X(v) S and W (v) S, or X(v) W (v) S. Thus either v A 2 k, or v A 1, or v A, according to the choices above. Thus V u=t A 1 A 2 k A and hence A\(A 1 A 2 k) V u<t. (vi) For all v A\(A 1 A 2 k) we have Ṽ (v, I v) V (K), for some even component K of G\S with respect to w. Proof. Suppose v A i, for some i {2,..., 2 k 1}. Clearly, W (v) Z(v) S, since v A 1. Thus, by (i), it follows that W (v) Z(v) S. Now (v) implies that Ṽ (v, I v) V (K), for some component K of G\S. However, since v V u<t, we have Z(v) Ø. Consequently, K must be an even component. (vii) Ṽ (v) S, for all v A. Proof. By definition of A i, i = 1, 2,..., 2 k we have that X(v) S and W (v) Z(v) S for all v A. Let v A. Then the vertices of X(v) W (v) Z(v) are contained in one component of G\S. Thus by Lemma 4.2 (ii) we see that Y (v) S and we have Ṽ (v) S. This completes the proof of (vii). ( G\S ) = q(a 1, A 2,..., A 2 k). Let K be a non- If Ṽ (v) V (K) Ø, then X(v) V (K) Ø since K is a non-singleton component. Thus V (K) X(A 2 k) = Ø and furthermore, V (K) X(A\A 1 A 2 k) = Ø by (vi). If V (K) X(A) = Ø, then V (K) X(A 1 ) Ø and K is seen to be a single edge and an even component, contradicting the fact that K is odd. It follows that K contains vertices of X(A). Thus, by (vii), there is a component C of G \ A, where Ṽ (V (C)) V (K). By (vi), it follows that d G(V (C); A i, I i ) = 0, i = 2,..., 2 k. Thus we see that V (K) = Ṽ (V (C)) + d G(V (C), A 1 ). We shall show that o non sing w singleton odd component of G\S with respect to w. Given that K is odd with respect to w, it must be that Z(V (C)) = Ø. Consequently, V (C) V u=t. Noting that Ṽ (V (C)) u(v (C)) (mod 2), 17

18 we infer from the above that V (K) u(v (C)) + d G (V (C), A 1 ) (mod 2). In other words, K is a non-singleton odd component of G\S (with respect to w) if and only if C is a component of G\A which is odd with respect to A 1,..., A 2 k. Thus o non sing w ( G\S ) = q(a 1, A 2,..., A 2 k). Let u V ( G) where u is a singleton odd component of G\S with respect to w. We see that u Ṽ (v) for some v A. Suppose first that v A i, for some i {2,..., 2 k 1}. Then W (v) Z(v) S and furthermore, since X(v) S, the vertices of W (v) Z(v) together with some vertices of X(v) belong to a non-singleton component of G\S. Now (v) implies that the number of singleton odd components contained in Ṽ (v) is just Y (v, I i ). Thus the total number of singleton odd components contained in Ṽ (A 2 A 2 k 1) is 2 k 1 Y (A i, I i ). If v A 1, then (iv) implies that the number of singleton odd components in Ṽ (v) equals the number of vertices of X(v) adjacent to vertices in X(A\A 1 ) S. Thus the total number of these odd components is exactly 2 k d G(A 1 ; A i, I i ). If v A 2 k, then X(v) S and hence Y (v) W (v) Z(v) S (by Lemma 4.2 (i)). Thus the number of singleton odd components in Ṽ (v) is Y (v) + W (v) = d G (v) f(v) + f(v) t(v) = d G (v) t(v). Thus the total number of singleton odd components contained in Ṽ (A 2k) is d G (A 2 k) t(a 2 k). From the above, we see that o sing 2 k 1 Y (A i, I i ) + d G (A 2 k) t(a 2 k). So We also observe that o w ( G\S ) = o sing w ( G\S ) + ow non sing ( G\S ) = w 2 k 1 d G (A 1 ; A i, I i ) + Y (A i, I i ) ( G\S ) = 2 k d G(A 1 ; A i, I i )+ + d G (A 2 k) t(a 2 k) + q(a 1,..., A 2 k). (8) S Y (A 1 ) + W (A 1 ) + Z(A 1 ) + = d G (A 1 ) u(a 1 ) + X(A i, I i ) X(A i, I i ) 2 k 1 = d G (A 1 ) u(a 1 ) + X(A i, I i ) + d G (A 2 k). (9) 18

19 Since ξ w (S ) > 0, it follows that S < o w ( G\S ). Thus, by equations (8) and (9), we obtain that d G (A 1 ) u(a 1 ) + Thus d G (A 1 ; A i, I i ) + 2 k 1 2 k 1 X(A i, I i ) + d G (A 2 k) < Y (A i, I i ) + d G (A 2 k) t(a 2 k) + q(a 1,..., A 2 k). d G (A 1 ) u(a 1 ) + < 2 k 1 ( X(A i, I i ) Y (A i, I i ) ) + t(a 2 k) d G (A 1 ; A i, I i ) + q(a 1,..., A 2 k). Observing that X(A i, I i ) Y (A i, I i ) = f(a i, I i ), i = 2,..., 2 k 1, we conclude from the above that 2 k 1 d G (A 1 ) u(a 1 )+ f(a i, I i )+t(a 2 k) < d G (A 1 ; A i, I i )+q(a 1,..., A 2 k). Therefore, (2) does not hold for the sets A 1,..., A 2 k. This completes the proof of the theorem. References [1] J.A. Bondy and U.S.R. Murty, Graph Theory, GTM 244, Springer, [2] R.C. Brewster, P. Hell, and R. Rizzi, Oriented star packings, J. Combin. Theory Series B 98 (2008) [3] P. Hell and D.G. Kirkpatrick, Algorithms for degree constrained graph factors of minimum deficiency, J. Algorithms 14, (1993) [4] Q. R. Yu and G. Liu, Graph Factors and Matching Extensions, Higher Ed. Press, Beijing, [5] L. Lovász, Subgraphs with prescribed valencies, J. Combin. Theory 8 (1970)

20 [6] H. Lu and Q. Yu, Constructive proof of deficiency theorem of (g, f)- factor, Sci. China Math. (2010), doi: /s [7] C. Papadimitriou and K. Steiglitz, Combinatorial Optimization, Dover, New York, [8] W. Tutte, The factorization of linear graphs, J. London Math. Soc. 22 (1947) [9] W. Tutte, A short proof of the factor theorem for finite graphs, Canad. J. Math. 6 (1954) [10] W. Tutte, Graph Factors, Combinatorica 1 (1981)