Factors in graphs with multiple degree constraints - Thompson Rivers ...
Factors With Multiple Degree Constraints in Graphs Richard C. Brewster, Sean McGuinness, and Morten Hegner Nielsen Thompson Rivers University Kamloops, BC V2C0C8 Canada August 9, 2013
Abstract For a graph G and for each vertex v ∈ V (G), let ΛG (v) = {EG (v, 1), EG (v, 2), . . . , EG (v, kv )} be a partition of the edges incident with v. Let ΛG = {ΛG (v) v ∈ V (G)}. We call the pair (G, ΛG ) a partitioned graph. Let k = maxv kv 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) ≤ dG (v, i) i = 1, . . . , kv (ii) u(v) ≤ t(v) ≤ dG (v) Pkv Pkv (iii) u(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) ≤ dH (v, i) ≤ f (v, i), i = 1, . . . , kv and (b) u(v) ≤ dH (v) ≤ t(v) where dH (v, i) = |E(H) ∩ EG (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 the existence of (0, f, t, u)-factors in a partitioned graph when u(v) < f (v, i) for all v and i. As a special case, we obtain Lov´asz’s (g, f )-factor theorem.
1
1 1.1
Introduction Notation
Let G be a graph. For U ⊆ V (G), let U = V (G)\U . For v ∈ V (G), let NG (v) denote the set of neighbours of v in G, let EG (v) be the set of edges incident with v, and let dG (v) = |EG (v)|. For U ⊆ V (G), we let EG (U ) = ∪v∈U EG (v) and dG (U ) = |EG (U )|. For sets U, W ⊆ V , EG (U, W ) shall denote the set of edges joining a vertex in U to a vertex in W and dG (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 P function defined on V (G), let φ(S) = s∈S φ(s). If ψ is another real-valued ψ(v) < φ(v)} and Vψ=φ = {v ∈ we define two sets V = {v ∈ V (G) ψ dG (v, i) − b g(v, i) − 1 = |(Y (v, i) ∪ Z(v, Si)) \ {x}|. Thus γ(G, 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 eb ∈ M. Now consider any pair v, i. We shall show that defH (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 defH (v, i) = max{g(v, i) − dH (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 defH (v, i) = max{g(v, i) − dH (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 defH (v, i) = µ(v, i) for all v, i and hence δ(G, g, f, t) ≤ b w). γ(G, b w) ≤ δ(G, g, f, t). To complete the proof, we need only show that γ(G, Suppose H is a minimum-deficiency (0, f, t, 0)-subgraph of G and let M b consisting of exactly those edges eb ∈ E(G) b where be the matching in G
9
e ∈ E(H). Suppose defH (v, i) > 0. Then defH (v, i) = g(v, i) − dH (v, i) and hence dG (v, i) = |X(v, i)| = dH (v, i) + defH (v, i) + dG (v, i) − g(v, i) = dH (v, i) + defH (v, i) + |Y (v, i) ∪ Z(v, i)|. 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 defH (v, i) vertices unmatched in X(v, i). Now suppose defH (v, i) = 0. Then |X(v, i)| − dH (v, i) satisfies |Y (v, i)| = dG (v, i) − f (v, i) ≤ |X(v, i)| − dH (v, i) ≤ dG (v, i) − g(v, i) = |Y (v, i) ∪ Z(v, i)|. Denote by M (v, i) a perfect matching between the |X(v, i)| − dH (v, i) M unsaturated vertices of X(v, i) and the set Y (v, i)∪Z 0 (v, i), for some Z 0 (v, i) ⊆ Z(v, i) where |Z 0 (v, i)| = |X(v, i)| − dH (v, i) − |Y (v, i)| = f (v, i) − dH (v, i). Then the extended matching M ∪ M (v, i) leaves exactly |Z(v, i) \ Z 0 (v, i)| = dH (v, i) − g(v, i) vertices unmatched in Z(v, i). It now follows that, for any v ∈ V (G), X (dH (v, i) − g(v, i)) ≤ t(v) = |W (v)|. i : defH (v,i)=0
Therefore, forSeach vertex v ∈ V (G) in turn, we may S further extend the matching M ∪ v,i M (v, i) by a perfect matching between i : defH (v,i)=0 (Z(v, i)\ Z 0 (v, i)) and some subset of W (v). The resulting matching has the same b w) ≤ δ(G, g, f, t). deficiency as H, and we conclude that γ(G, 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 matchb w); thus we have the following corollary (where digraphs may ing in (G, 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− − → (v), 0 ≤ g (v) ≤ f (v) ≤ d− → (v), G G → − 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 10
→ − + − + → (v) ≤ t(v) for all v ∈ V ( G ); (i) d− − → (v) ≤ f (v), d− → (v) ≤ f (v), and d− H H
H
P − (v) − d− (v), 0} + max{g + (v) − d+ (v), 0} → (ii) the deficiency v∈V (− max{g − → − → G) H H → − → − of H is minimum over all spanning subdigraphs of G that satisfy (i).
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); we let K(G\S) (respectively, Ko (G\S)) denote the set of components (respectively, odd components) of G\S. For each component K ∈ K(G\S), we say that K is odd with respect to w if K ∈ Ko (G\S) and all vertices of K have weight one. If K is not odd with respect to w, it is o (G\S) be the set of odd components of G\S with respect to even. We let Kw o (G\S)|. Let ξ (S) = max{o (G\S) − |S|, 0} and w and let ow (G\S) = |K w w w let ξw (G) = max{ξw (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) = ξw (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 ξw (S) > 0. Moreover, S ∗ is a maximum barrier for (G, w) if (I) S ∗ is a barrier for which ξw (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: o (G \ S ∗ ). (i) Each x ∈ S ∗ is adjacent to at least one odd component of Kw
11
(ii) Each x ∈ S ∗ with w(x) = 1 is adjacent to at least two odd components o (G \ S ∗ ); furthermore, if x is adjacent to no components in of Kw 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 ∗ ). (iv) For each pair x1 , x2 of non-adjacent vertices of G where w(xi ) = 1 and |NG (xi ) \ NG (x3−i )| ≤ 1, i = 1, 2, either {x1 , x2 } ⊆ S ∗ or {x1 , x2 } ⊆ S∗. o (G \ S ∗ ), then the Proof. If x ∈ S ∗ is adjacent to no odd components of Kw ∗ ∗ ∗ set T = S \ {x} satisfies ξw (T ) ≥ ξw (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 o (G \ S ∗ ). Let T = S ∗ \ {x}. If x is adjacent to only one component of Kw 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 ξw (T ) ≥ ξw (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 o (G \ S ∗ ). Kw 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 ∗ ) to which it is adjacent become one o (G \ T ); again, it follows that ξ (T ) ≥ ξ (S ∗ ) and odd component in Kw w w ∗ |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 ξw (T ) > ξw (S ∗ ), if x is o (G \ S ∗ ), and ξ (T ) = ξ (S ∗ ), if not adjacent to any odd component of Kw w w o (G \ S ∗ ). Since |T | < |S ∗ |, this x is adjacent to an odd component of Kw contradicts the minimality of S ∗ . This proves (iii) Suppose x1 , x2 are as stated in (iv) and x1 ∈ S ∗ and x2 ∈ S ∗ . Then x2 belongs to a component K in K(G \ S ∗ ). Suppose that x1 is adjacent to at least three components in K(G \ S ∗ ); such components contain vertices of NG (x1 ). Since |NG (x1 ) \ NG (x2 )| ≤ 1, at least two such components also contain vertices of NG (x2 ). Given that x2 ∈ V (K), both components must be contained in K, a contradiction. So x1 is adjacent to at most two components of K(G \ S ∗ ). The first part of (ii) implies that x1 is adjacent to
12
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 1.5.
5.1
e w) The weighted graph (G,
Let (G, ΛG ) 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 EG (v, i) by defining EG (v, i) = Ø, whenever i > kv . e w) as follows: for each v ∈ We shall construct a weighted graph (G, V (G), let X(v, 1), X(v, 2), . . . , X(v, kv ), Y (v, 1), Y (v, 2), . . . , Y (v, kv ) be disjoint sets with cardinalities |X(v, i)| = dG (v, i) and |Y (v, i)| = dG (v, i) − f (v, i), where i = 1, 2, . . . , kv . 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, . . . , kv }, 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 Ve (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 Ve (U ) = ∪v∈U Ve (v). For v ∈ V (G), i ∈ {1, . . . , kv }, and I ⊆ {1, . . . , kv }, let Ve (v, i) = X(v, i) ∪ Y (v, i)∪W (v)∪Z(v) and Ve (v, I) = ∪i∈I Ve (v, i). Given that |X(v)| = dG (v), we may assign to each edge e ∈ EG (v) a distinct vertex ve ∈ X(v). For each e as follows: if e has endvertices edge e ∈ E(G), we shall assign an edge ee in G e u and v, then let ee be the edge joining the vertices ue and ve in G. e Finally, the weight function w : V (G) → {0, 1} shall be defined as follows: S 0, if v ∈ u Z(u); w(v) = 1, otherwise. We have the following lemma whose proof is left as an easy exercise. 5.1 Lemma The partitioned graph (G, ΛG ) has an (0, f, t, u)-subgraph if and only if e w) has a matching which saturates all vertices of weight 1. (G,
13
5.2
Proof of the (0, f, t, u)-factor theorem
e defined above. For each We shall make one modification to the graph G v ∈ Vu=t , we may assume that kv = 1 (since t(v) = u(v) < f (v, i), for all i = 1, . . . , kv ); furthermore, we may assume that Y (v), W (v), and Z(v) are sets such that |W (v)| = dG (v) − u(v) and Y (v) = Z(v) = Ø. Lemma e Suppose that H is 5.1 is seen to hold even with this modification to G. a (0, f, t, u)-subgraph of (G, ΛG ). Let H = G\E(H). Let A1 , . . . , A2k be disjoint (possibly empty) subsets of V (G) and let A = A1 ∪ A2 ∪ · · · A2k . Let K0 be the set of components of G0 = G\A. We shall show that (2) holds. We first observe that k
2 X
k
k
2 X
2 X
dG (A1 ; Ai , Ii ) = dH (A1 ; Ai , Ii ) + dH (A1 ; Ai , Ii ). i=2 i=2 i=2 K0 where K is odd with respect to A1 , . . . , A2k . Since V (K)
(3)
Let K ∈ ⊆ Vu=t , it (v) = u(v) = t(v) for all v ∈ V (K). Thus u(V (K)) = P follows that dHP u(v) = v∈V (K) v∈V (K) dH (v) ≡ dH (V (K), V (K)) (mod 2). Given that K is odd, we have that dH (V (K), V (K))+dG (A1 , V (K)) ≡ 1 (mod 2). Since dG (A1 , V (K)) = dH (A1 , V (K)) + dH (A1 , V (K)) and dH (V (K), V (K)) = dH (A1 , V (K))+dH (A\A1 , V (K)), we see that dH (A1 , V (K))+dH (A\A1 , V (K)) ≡ 1 (mod 2). Since K is odd with respect to A1 , A2 , . . . , A2k we have that dG (V (K); Ai , I i ) = 0, i = 1, . . . , 2k . Thus we have dH (A\A1 , V (K)) = P2k P2k i=2 dH (Ai , Ii ; V (K)). It follows that dH (A1 , V (K))+ i=2 dH (Ai , Ii ; V (K)) ≡ 1 (mod 2). Thus we have k
q(A1 , . . . , A2k ) ≤ dH (A1 , A) +
2 X
dH (Ai , Ii ; A).
(4)
i=2
Thus k
2 X
k
dG (A1 ; Ai , Ii ) + q(A1 , . . . , A2k ) =
i=2
2 X
dH (A1 ; Ai , Ii )
i=2 k
+
2 X
dH (A1 ; Ai , Ii ) + q(A1 , . . . , A2k )
i=2 k
≤
2 X
k
dH (A1 ; Ai , Ii ) +
i=2
2 X
dH (A1 ; Ai , Ii )
i=2 k
+ dH (A1 , A) +
2 X i=2
14
dH (Ai , Ii ; A).
(5)
We also have k
2 X
dH (A1 ; Ai , Ii ) + dH (A1 , A) ≤ dH (A1 )
i=2
≤
X
(dG (v) − u(v))
v∈A1
= dG (A1 ) − u(A1 ),
(6)
and k
2 X
k
dH (A1 ; Ai , Ii ) +
i=2
2 X
dH (Ai , Ii ; A)
i=2
≤
k −1 2X
f (Ai , Ii ) + dH (A1 , A2k ) + dH (A2k , A)
i=2
≤
k −1 2X
f (Ai , Ii ) + t(A2k ).
(7)
i=2
It follows from (5), (6), and (7) that (2) holds. Suppose now that (G, ΛG ) has no (0, f, t, u)-subgraph. We shall find pairwise disjoint subsets of vertices A1 , A2 , . . . , A2k for which (2) does not hold. e w) Given that there is no (0, f, t, u)-subgraph, Lemma 5.1 implies that (G, e has no matching which saturates all vertices of weight 1. Thus ξw (G) > 0; let non−sing e e Denote by osing e ∗ S ∗ be a maximum barrier for G. (G\S ∗ ) w (G\S ) and ow the number of odd singleton and odd non-singleton components, respece ∗ with respect to w. tively, of G\S 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, . . . , kv . (iii) Y (v, i) ⊆ S ∗ or Y (v, i) ⊆ S ∗ , for all v ∈ V (G) and i = 1, . . . , kv . 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 ∈ Vu=t , then kv = 1 and Y (v) = Z(v) = Ø. Thus each vertex of W (v) is only adjacent to vertices in 15
X(v) = X(v, 1). It follows by (ii) and Lemma 4.2 (ii) that X(v) ⊆ S ∗ . Thus we may assume that v ∈ Vu ξw (S ∗ ), where the last inequality follows from the fact that f (v, i) > u(v). Thus ξw (T ∗ ) > ξw (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, . . . , kv }. Then X(v, i) ∪ Y (v, i) ⊆ V (K) for some e ∗ . Let T ∗ = S ∗ \Ve (v). Then all the vertices of component K of G\S e e ∗ . Furthermore, we see that V (v) belong to a single component of G\T X ξw (T ∗ ) ≥ ξw (S ∗ ) + |W (v)| + |Z(v)| − 1 − f (v, j) j6=i
= ξw (S ∗ ) + f (v) − u(v) − 1 −
X
f (v, j)
j6=i
= ξw (S ∗ ) + f (v, i) − u(v) − 1 ≥ ξw (S ∗ ). Thus ξw (T ∗ ) ≥ ξw (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, . . . , kv we have either X(v, i) ⊆ S ∗ and Y (v, i) ⊆ S ∗ or Ve (v, i) ⊆ V (K), for e ∗. some component K of G\S 16
Proof. Let v ∈ V (G) and suppose that W (v) ∪ Z(v) ⊆ S ∗ . Let i ∈ {1, . . . , kv }. 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) ∪ e ∗ . Now Lemma 4.2 W (v)∪Z(v) ⊆ V (K), for some component K of G\S ∗ (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). Recall that the sets I1 , I2 , . . . , I2k denote the subsets of {1, 2, . . . , k} where I1 = Ø and I2k = {1, 2, . . . , k}. For each v ∈ V (G), let Iv = {i ∈ W (v) ∪ Z(v) ⊆ S ∗ }. {1, . . . , k} X(v, i) ⊆ S ∗ }. Let A1 = {v ∈ V (G) ∗ For i = 2, . . . , 2k − 1, let Ai = {v ∈∗ V (G) Iv = Ii ; X(v) 6⊆ S }. Let A2k = {v ∈ V (G) X(v) ⊆ S } and A = A1 ∪ · · · ∪ A2k . By (iv), it follows that Iv = I1 = Ø for all v ∈ A1 . Let v ∈ Vu=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 ∈ A2k , or v ∈ A1 , or v ∈ A, according to the choices above. Thus Vu=t ⊆ A1 ∪ A2k ∪ A and hence A\(A1 ∪ A2k ) ⊆ Vu
Abstract For a graph G and for each vertex v ∈ V (G), let ΛG (v) = {EG (v, 1), EG (v, 2), . . . , EG (v, kv )} be a partition of the edges incident with v. Let ΛG = {ΛG (v) v ∈ V (G)}. We call the pair (G, ΛG ) a partitioned graph. Let k = maxv kv 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) ≤ dG (v, i) i = 1, . . . , kv (ii) u(v) ≤ t(v) ≤ dG (v) Pkv Pkv (iii) u(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) ≤ dH (v, i) ≤ f (v, i), i = 1, . . . , kv and (b) u(v) ≤ dH (v) ≤ t(v) where dH (v, i) = |E(H) ∩ EG (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 the existence of (0, f, t, u)-factors in a partitioned graph when u(v) < f (v, i) for all v and i. As a special case, we obtain Lov´asz’s (g, f )-factor theorem.
1
1 1.1
Introduction Notation
Let G be a graph. For U ⊆ V (G), let U = V (G)\U . For v ∈ V (G), let NG (v) denote the set of neighbours of v in G, let EG (v) be the set of edges incident with v, and let dG (v) = |EG (v)|. For U ⊆ V (G), we let EG (U ) = ∪v∈U EG (v) and dG (U ) = |EG (U )|. For sets U, W ⊆ V , EG (U, W ) shall denote the set of edges joining a vertex in U to a vertex in W and dG (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 P function defined on V (G), let φ(S) = s∈S φ(s). If ψ is another real-valued ψ(v) < φ(v)} and Vψ=φ = {v ∈ we define two sets V = {v ∈ V (G) ψ dG (v, i) − b g(v, i) − 1 = |(Y (v, i) ∪ Z(v, Si)) \ {x}|. Thus γ(G, 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 eb ∈ M. Now consider any pair v, i. We shall show that defH (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 defH (v, i) = max{g(v, i) − dH (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 defH (v, i) = max{g(v, i) − dH (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 defH (v, i) = µ(v, i) for all v, i and hence δ(G, g, f, t) ≤ b w). γ(G, b w) ≤ δ(G, g, f, t). To complete the proof, we need only show that γ(G, Suppose H is a minimum-deficiency (0, f, t, 0)-subgraph of G and let M b consisting of exactly those edges eb ∈ E(G) b where be the matching in G
9
e ∈ E(H). Suppose defH (v, i) > 0. Then defH (v, i) = g(v, i) − dH (v, i) and hence dG (v, i) = |X(v, i)| = dH (v, i) + defH (v, i) + dG (v, i) − g(v, i) = dH (v, i) + defH (v, i) + |Y (v, i) ∪ Z(v, i)|. 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 defH (v, i) vertices unmatched in X(v, i). Now suppose defH (v, i) = 0. Then |X(v, i)| − dH (v, i) satisfies |Y (v, i)| = dG (v, i) − f (v, i) ≤ |X(v, i)| − dH (v, i) ≤ dG (v, i) − g(v, i) = |Y (v, i) ∪ Z(v, i)|. Denote by M (v, i) a perfect matching between the |X(v, i)| − dH (v, i) M unsaturated vertices of X(v, i) and the set Y (v, i)∪Z 0 (v, i), for some Z 0 (v, i) ⊆ Z(v, i) where |Z 0 (v, i)| = |X(v, i)| − dH (v, i) − |Y (v, i)| = f (v, i) − dH (v, i). Then the extended matching M ∪ M (v, i) leaves exactly |Z(v, i) \ Z 0 (v, i)| = dH (v, i) − g(v, i) vertices unmatched in Z(v, i). It now follows that, for any v ∈ V (G), X (dH (v, i) − g(v, i)) ≤ t(v) = |W (v)|. i : defH (v,i)=0
Therefore, forSeach vertex v ∈ V (G) in turn, we may S further extend the matching M ∪ v,i M (v, i) by a perfect matching between i : defH (v,i)=0 (Z(v, i)\ Z 0 (v, i)) and some subset of W (v). The resulting matching has the same b w) ≤ δ(G, g, f, t). deficiency as H, and we conclude that γ(G, 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 matchb w); thus we have the following corollary (where digraphs may ing in (G, 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− − → (v), 0 ≤ g (v) ≤ f (v) ≤ d− → (v), G G → − 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 10
→ − + − + → (v) ≤ t(v) for all v ∈ V ( G ); (i) d− − → (v) ≤ f (v), d− → (v) ≤ f (v), and d− H H
H
P − (v) − d− (v), 0} + max{g + (v) − d+ (v), 0} → (ii) the deficiency v∈V (− max{g − → − → G) H H → − → − of H is minimum over all spanning subdigraphs of G that satisfy (i).
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); we let K(G\S) (respectively, Ko (G\S)) denote the set of components (respectively, odd components) of G\S. For each component K ∈ K(G\S), we say that K is odd with respect to w if K ∈ Ko (G\S) and all vertices of K have weight one. If K is not odd with respect to w, it is o (G\S) be the set of odd components of G\S with respect to even. We let Kw o (G\S)|. Let ξ (S) = max{o (G\S) − |S|, 0} and w and let ow (G\S) = |K w w w let ξw (G) = max{ξw (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) = ξw (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 ξw (S) > 0. Moreover, S ∗ is a maximum barrier for (G, w) if (I) S ∗ is a barrier for which ξw (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: o (G \ S ∗ ). (i) Each x ∈ S ∗ is adjacent to at least one odd component of Kw
11
(ii) Each x ∈ S ∗ with w(x) = 1 is adjacent to at least two odd components o (G \ S ∗ ); furthermore, if x is adjacent to no components in of Kw 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 ∗ ). (iv) For each pair x1 , x2 of non-adjacent vertices of G where w(xi ) = 1 and |NG (xi ) \ NG (x3−i )| ≤ 1, i = 1, 2, either {x1 , x2 } ⊆ S ∗ or {x1 , x2 } ⊆ S∗. o (G \ S ∗ ), then the Proof. If x ∈ S ∗ is adjacent to no odd components of Kw ∗ ∗ ∗ set T = S \ {x} satisfies ξw (T ) ≥ ξw (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 o (G \ S ∗ ). Let T = S ∗ \ {x}. If x is adjacent to only one component of Kw 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 ξw (T ) ≥ ξw (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 o (G \ S ∗ ). Kw 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 ∗ ) to which it is adjacent become one o (G \ T ); again, it follows that ξ (T ) ≥ ξ (S ∗ ) and odd component in Kw w w ∗ |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 ξw (T ) > ξw (S ∗ ), if x is o (G \ S ∗ ), and ξ (T ) = ξ (S ∗ ), if not adjacent to any odd component of Kw w w o (G \ S ∗ ). Since |T | < |S ∗ |, this x is adjacent to an odd component of Kw contradicts the minimality of S ∗ . This proves (iii) Suppose x1 , x2 are as stated in (iv) and x1 ∈ S ∗ and x2 ∈ S ∗ . Then x2 belongs to a component K in K(G \ S ∗ ). Suppose that x1 is adjacent to at least three components in K(G \ S ∗ ); such components contain vertices of NG (x1 ). Since |NG (x1 ) \ NG (x2 )| ≤ 1, at least two such components also contain vertices of NG (x2 ). Given that x2 ∈ V (K), both components must be contained in K, a contradiction. So x1 is adjacent to at most two components of K(G \ S ∗ ). The first part of (ii) implies that x1 is adjacent to
12
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 1.5.
5.1
e w) The weighted graph (G,
Let (G, ΛG ) 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 EG (v, i) by defining EG (v, i) = Ø, whenever i > kv . e w) as follows: for each v ∈ We shall construct a weighted graph (G, V (G), let X(v, 1), X(v, 2), . . . , X(v, kv ), Y (v, 1), Y (v, 2), . . . , Y (v, kv ) be disjoint sets with cardinalities |X(v, i)| = dG (v, i) and |Y (v, i)| = dG (v, i) − f (v, i), where i = 1, 2, . . . , kv . 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, . . . , kv }, 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 Ve (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 Ve (U ) = ∪v∈U Ve (v). For v ∈ V (G), i ∈ {1, . . . , kv }, and I ⊆ {1, . . . , kv }, let Ve (v, i) = X(v, i) ∪ Y (v, i)∪W (v)∪Z(v) and Ve (v, I) = ∪i∈I Ve (v, i). Given that |X(v)| = dG (v), we may assign to each edge e ∈ EG (v) a distinct vertex ve ∈ X(v). For each e as follows: if e has endvertices edge e ∈ E(G), we shall assign an edge ee in G e u and v, then let ee be the edge joining the vertices ue and ve in G. e Finally, the weight function w : V (G) → {0, 1} shall be defined as follows: S 0, if v ∈ u Z(u); w(v) = 1, otherwise. We have the following lemma whose proof is left as an easy exercise. 5.1 Lemma The partitioned graph (G, ΛG ) has an (0, f, t, u)-subgraph if and only if e w) has a matching which saturates all vertices of weight 1. (G,
13
5.2
Proof of the (0, f, t, u)-factor theorem
e defined above. For each We shall make one modification to the graph G v ∈ Vu=t , we may assume that kv = 1 (since t(v) = u(v) < f (v, i), for all i = 1, . . . , kv ); furthermore, we may assume that Y (v), W (v), and Z(v) are sets such that |W (v)| = dG (v) − u(v) and Y (v) = Z(v) = Ø. Lemma e Suppose that H is 5.1 is seen to hold even with this modification to G. a (0, f, t, u)-subgraph of (G, ΛG ). Let H = G\E(H). Let A1 , . . . , A2k be disjoint (possibly empty) subsets of V (G) and let A = A1 ∪ A2 ∪ · · · A2k . Let K0 be the set of components of G0 = G\A. We shall show that (2) holds. We first observe that k
2 X
k
k
2 X
2 X
dG (A1 ; Ai , Ii ) = dH (A1 ; Ai , Ii ) + dH (A1 ; Ai , Ii ). i=2 i=2 i=2 K0 where K is odd with respect to A1 , . . . , A2k . Since V (K)
(3)
Let K ∈ ⊆ Vu=t , it (v) = u(v) = t(v) for all v ∈ V (K). Thus u(V (K)) = P follows that dHP u(v) = v∈V (K) v∈V (K) dH (v) ≡ dH (V (K), V (K)) (mod 2). Given that K is odd, we have that dH (V (K), V (K))+dG (A1 , V (K)) ≡ 1 (mod 2). Since dG (A1 , V (K)) = dH (A1 , V (K)) + dH (A1 , V (K)) and dH (V (K), V (K)) = dH (A1 , V (K))+dH (A\A1 , V (K)), we see that dH (A1 , V (K))+dH (A\A1 , V (K)) ≡ 1 (mod 2). Since K is odd with respect to A1 , A2 , . . . , A2k we have that dG (V (K); Ai , I i ) = 0, i = 1, . . . , 2k . Thus we have dH (A\A1 , V (K)) = P2k P2k i=2 dH (Ai , Ii ; V (K)). It follows that dH (A1 , V (K))+ i=2 dH (Ai , Ii ; V (K)) ≡ 1 (mod 2). Thus we have k
q(A1 , . . . , A2k ) ≤ dH (A1 , A) +
2 X
dH (Ai , Ii ; A).
(4)
i=2
Thus k
2 X
k
dG (A1 ; Ai , Ii ) + q(A1 , . . . , A2k ) =
i=2
2 X
dH (A1 ; Ai , Ii )
i=2 k
+
2 X
dH (A1 ; Ai , Ii ) + q(A1 , . . . , A2k )
i=2 k
≤
2 X
k
dH (A1 ; Ai , Ii ) +
i=2
2 X
dH (A1 ; Ai , Ii )
i=2 k
+ dH (A1 , A) +
2 X i=2
14
dH (Ai , Ii ; A).
(5)
We also have k
2 X
dH (A1 ; Ai , Ii ) + dH (A1 , A) ≤ dH (A1 )
i=2
≤
X
(dG (v) − u(v))
v∈A1
= dG (A1 ) − u(A1 ),
(6)
and k
2 X
k
dH (A1 ; Ai , Ii ) +
i=2
2 X
dH (Ai , Ii ; A)
i=2
≤
k −1 2X
f (Ai , Ii ) + dH (A1 , A2k ) + dH (A2k , A)
i=2
≤
k −1 2X
f (Ai , Ii ) + t(A2k ).
(7)
i=2
It follows from (5), (6), and (7) that (2) holds. Suppose now that (G, ΛG ) has no (0, f, t, u)-subgraph. We shall find pairwise disjoint subsets of vertices A1 , A2 , . . . , A2k for which (2) does not hold. e w) Given that there is no (0, f, t, u)-subgraph, Lemma 5.1 implies that (G, e has no matching which saturates all vertices of weight 1. Thus ξw (G) > 0; let non−sing e e Denote by osing e ∗ S ∗ be a maximum barrier for G. (G\S ∗ ) w (G\S ) and ow the number of odd singleton and odd non-singleton components, respece ∗ with respect to w. tively, of G\S 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, . . . , kv . (iii) Y (v, i) ⊆ S ∗ or Y (v, i) ⊆ S ∗ , for all v ∈ V (G) and i = 1, . . . , kv . 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 ∈ Vu=t , then kv = 1 and Y (v) = Z(v) = Ø. Thus each vertex of W (v) is only adjacent to vertices in 15
X(v) = X(v, 1). It follows by (ii) and Lemma 4.2 (ii) that X(v) ⊆ S ∗ . Thus we may assume that v ∈ Vu ξw (S ∗ ), where the last inequality follows from the fact that f (v, i) > u(v). Thus ξw (T ∗ ) > ξw (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, . . . , kv }. Then X(v, i) ∪ Y (v, i) ⊆ V (K) for some e ∗ . Let T ∗ = S ∗ \Ve (v). Then all the vertices of component K of G\S e e ∗ . Furthermore, we see that V (v) belong to a single component of G\T X ξw (T ∗ ) ≥ ξw (S ∗ ) + |W (v)| + |Z(v)| − 1 − f (v, j) j6=i
= ξw (S ∗ ) + f (v) − u(v) − 1 −
X
f (v, j)
j6=i
= ξw (S ∗ ) + f (v, i) − u(v) − 1 ≥ ξw (S ∗ ). Thus ξw (T ∗ ) ≥ ξw (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, . . . , kv we have either X(v, i) ⊆ S ∗ and Y (v, i) ⊆ S ∗ or Ve (v, i) ⊆ V (K), for e ∗. some component K of G\S 16
Proof. Let v ∈ V (G) and suppose that W (v) ∪ Z(v) ⊆ S ∗ . Let i ∈ {1, . . . , kv }. 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) ∪ e ∗ . Now Lemma 4.2 W (v)∪Z(v) ⊆ V (K), for some component K of G\S ∗ (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). Recall that the sets I1 , I2 , . . . , I2k denote the subsets of {1, 2, . . . , k} where I1 = Ø and I2k = {1, 2, . . . , k}. For each v ∈ V (G), let Iv = {i ∈ W (v) ∪ Z(v) ⊆ S ∗ }. {1, . . . , k} X(v, i) ⊆ S ∗ }. Let A1 = {v ∈ V (G) ∗ For i = 2, . . . , 2k − 1, let Ai = {v ∈∗ V (G) Iv = Ii ; X(v) 6⊆ S }. Let A2k = {v ∈ V (G) X(v) ⊆ S } and A = A1 ∪ · · · ∪ A2k . By (iv), it follows that Iv = I1 = Ø for all v ∈ A1 . Let v ∈ Vu=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 ∈ A2k , or v ∈ A1 , or v ∈ A, according to the choices above. Thus Vu=t ⊆ A1 ∪ A2k ∪ A and hence A\(A1 ∪ A2k ) ⊆ Vu
