Edge Coloring Regular Graphs of High Degree
Edge Coloring Regular Graphs of High Degree L. Perkovic School of Computer Science CMU Pittsburgh USA
B. Reedy CNRS Paris France
Abstract We discuss the following conjecture: If G = (V; E ) is a -regular simple graph with an even number of vertices at most 2 then G is edge colorable. In this paper we show that the conjecture is true for large graphs if jV j < (2 ? ). We discuss related results.
1 Introduction The following conjecture has been proposed:
Conjecture 1 If
G = (V; E ) is a -regular simple graph with an even number of vertices at most 2 then G is 1-factorizable ( edge colorable).
This conjecture appeared in Chetwynd and Hilton [3] but may go back to G. A. Dirac in the early 1950's (Chetwynd and Hilton [3]). An interesting consequence is that for any regular graph G either G or its complement has a 1-factorization. This conjecture along with variants and related results can be found in the new book Graph Coloring Problems by Jensen and Toft [8]. E-mail: [email protected] y E-mail: breed@mu .cs.mcgill.ca
1
In order to describe related conjectures and results we need to de ne some terminology. A graph G = (V; E ) is simple if it contains no multiple edges and no loops. G is r-regular if the degree (i.e. the number of edges incident to a vertex) of every vertex is r. G is k edge colorable if the edge set E of G can be partitioned into k disjoint matchings. The chromatic index of G, denoted 0 (G), is the least k for which G has a k edge-colouring. We use (G) or simply to denote the maximum vertex degree in G. It is easy to see that 0 (G) . In 1964, Vizing [12] proved that for every simple graph G either 0 (G) = or 0 (G) = + 1. We note that Fournier [6], [1] designed a procedure to + 1 color a graph of maximum degree in polynomial time. A subgraph H of G is called overfull if jV (H )j is odd and jE (H )j > jV (H2)j?1 , where E (H ) is the set of edges in E induced by vertices in H . If a simple graph contains an overfull subgraph H then 0 (G) = + 1. This follows from the fact that in any edge-coloring of G at most jV (H2)j?1 edges in E (H ) can have the same color. It is not too hard to see that Conjecture 1 is a special case of Hilton's Overfull Subgraph Conjecture (Hilton and Johnson [7], Chetwynd and Hilton [4]):
Conjecture 2 (Hilton) If G is simple, > jV3 j and G contains no overfull
subgraph then G is edge colorable.
There are some partial results regarding p Conjecture 1. Chetwynd and Hilton ( [5] have proved the conjecture if 72?1) jV j. Furthermore Chetwynd and Hilton [3] note that R. Haggkvist has announced that for any > 0 there exists N > 0 so that every -regular graph G is 1-factorizable if G has an even number of vertices greater than N and ( 12 + )jV j. In this paper we prove the following theorem:
Theorem 1 There exists 0 s.t. for all 0 and for all -regular graphs G with an even number of vertices at most 2, if G contains no: (i) Bipartite subgraph H such that for all v in V (H ) : dH (v) ? 39=40 (ii) Subgraph H where jV j ? jV (H )j ? 39=40 such that for all v in V (H ) : dH (v ) ? 39=40 2
then G is edge colorable.
The result Haggkvist announced is a corollary of this theorem. Proof: Given , ( 21 + )jV j let G be a -regular graph with an even number of vertices greater than N = maxf20; ?40 g. Then G cannot contain a subgraph H satisfying (i) because jH j would be greater than 2 ? 239=40 jV j + 2jV j ? 2jV j39=40 > jV j which is a contradiction. Similarly G cannot contain a subgraph H satisfying (ii) of theorem 1. So G is -edge colorable. 2 More importantly, the methods we use in the proof of theorem 1 are very general and have been used to get similar partial results for Hilton's conjecture (Perkovic and Reed [10]). These partial results were enough to devise a polynomial expected time algorithm for edge coloring simple graphs (Perkovic and Reed [11]). But the proofs of these results are long and technical, while the proof of theorem 1 is relatively simple. We turn now to the proof of theorem 1. We present algorithm COLOUR that given a -regular graph G so that 0 and jV j 2 is even edge colors G with colors or returns a subgraph H satisfying conditions (i) or (ii) of theorem 1. The algorithm along with the proof of its correctness proves theorem 1. We assume 0 is large enough to satisfy a number of lower bounds appearing throughout the paper. In order to give the main idea of algorithm we need the concept of a reduction.
De nition 1
is a reduction of G if (G0) = (G) ? l and there exists a set of matchings fM1 ; M2; :::; Mlg such that G0 = G ? M1 ? M2 ? ::: ? Ml . G0
We remark that if G0 is a reduction of G and G0 has a (G0) edge-coloring then G has a (G) edge-coloring. We remark further that given disjoint matchings M1 ; :::; Ml in a graph G = G0 and setting Gi = Gi?1 ? Mi we have that each Gi is a reduction of G if and only if for each i, every vertex of maximum degree in Gi?1 is the endpoint of some edge of Mi . The main idea of the algorithm is that given a graph G we attempt to nd a bipartite reduction G0 of G. Since 0 (G0) = (G0) (by Konig's theorem [9]) we will have 0 (G) = (G). 3
To nd a bipartite reduction, we rst partition the vertices of G into sets H1 and H2 of equal size so that a number of conditions hold. We describe this in the following section. In section 3 we describe the algorithm COLOUR that constructs perfect matchings consisting mostly of edges within H1 and within H2 . Once we color all the edges in H1 and H2, the uncolored edges form a bipartite reduction of G. We introduce some notation we will need in our arguments. Given a subset H of V let NH (v ) = fu 2 H : (u; v ) 2 E g, dH (v ) = jNH (v )j; if H = G we simply write N (v ) or d(v ). If X is a subset of H then NH (X ) = [v2X N (v ).
2 The Split In the following section we present an algorithm COLOUR which edge colors a -regular graph G = (V; E ) (with jV j even and at most 2) or returns a subgraph H of G satisfying (i) or (ii) of theorem 1. However COLOUR needs to be given a special partition (H1; H2) of the vertices of G as an input. This partition has the property that certain sets of vertices split about evenly between H1 and H2. Let us de ne a split formally:
De nition 2 Let H1 and11=H20 2 be a partition of V . A set S V splits if jjS \ H1j ? jS \ H2jj < 2 . A partition (H1; H2) of satis ed:
V
is called a split partition if the following are
1. jH1j = jH2j (which implies jE (H1)j = jE (H2)j) 2. 8v 2 V; 8X; Y V of size less than log the following sets split: N (v ); N (X ); N (v ) \ N (X ); fw 2 N (X ) : dN (Y )(w) > ? 519=20g To prove that the desired partition exists we will show that a suitably de ned random split satis es the conditions with positive probability. We then apply COLOUR to some such partition obtaining either a edge coloring of G or a subgraph H of G satisfying (i) of (ii) of theorem 1. 4
2.1
The Partitioning Procedure
The following procedure constructs with positive probability a split partition (H1; H2) of V : We arbitrarily order the vertices in V . For each successive ordered pair of vertices switch the order of the pair with probability 1=2 and put the rst vertex in set H1 and the second in the set H2 . Let E (H1) and E (H2) be the edges induced by H1 and H2 respectively. It is easy to see that the resulting partition (H1; H2) of V satis es the rst condition of a split partition. We show that with positive probability the other condition in the de nition of a split partition is also satis ed in the following claim:
Claim 1 The probability that there is any v in V , X and Y subsets of V with jX j; jY j < log for which one of the following sets fails to split is less than 21 :
( ) ( ) ( ) \ N (X ); fw 2 N (X ) : dN (Y ) (w) > ? 519=20g
N v ;N X ;N v
Proof: Let jV j = n. There are at most log n(n +
n
!
n
!
log + n log +
!
n
! 3 log n log ) < n n
log
sets that we want to split. Let S = fv1 ; v2; v3; :::; vkg be one of them. We de ne for each v 2 V the random variable Xv : ( ?1 if v 2 H2 Xv = 1 if v 2 H 1
P Then jjS \ H1 j ? jS \ H2 jj j v2S Xv j and 11=20 X ?22=20 Pr(j Xv j > ) < 2e 8n 2 v2S which is less than (2n23log n )?1 for large enough. 5
2
Claim 1 implies that with probability at least 12 the partitioning procedure constructs a split partition. Before describing algorithm COLOUR on such a partition we describe a procedure that constructs a subgraph H of G satisfying (i) or (ii) of theorem 1 if COLOUR fails to edge color G. 2.2
The Subgraph Construction Procedure
The second property of a split partition is that certain sets of vertices split about evenly between H1 and H2. The reason we want these sets to split is that it enables us to construct a subgraph of G that satis es (i) or (ii) of the theorem 1 if we nd two sets X and Y such that X H1 , jH1j ? jX j > ? 9=10, Y H2 , jY j < jX j + 9=10 and 2
8v 2 X : dY (v) > 2 ? 9=10 Algorithm COLOUR fails to edge color G only if such sets X and Y are found. In this case we use the Subgraph Construction Procedure to nd a subgraph of G that violates a condition of theorem 1:
Input: A -regular graph G with an even number of vertices at most 2 a split partition (H1 ; H2) of V and sets X and Y such that X H1 , 9=10 9=10 and Y H2, jH1j ? jX j > 2 ? , jY j < jX j + 8v 2 X : dY (v) > 2 ? 9=10
Output: EITHER a bipartite subgraph H of G such that for all v in V (H ) : dH (v ) ? 39=40 OR a subgraph H of G where jV j ? jV (H )j ? 39=40 such that for all v in V (H ) : dH (v ) ? 39=40 Steps: 1. We construct the set Z = fv 2 Y : dX (v) 2 ? 219=20g. Since jY ? Z j < 32 19=20 we have then that for all v in X : dZ (v ) 2 ?219=20 and that for all v in Z : dX (v ) 2 ?219=20. 2. We construct sets X 0 X and Z 0 Z such that jX 0j and jZ 0j are less than log , Z N (X 0) and X N (Z 0 ) using the procedure given below (the construction of Z 0 is symmetric): 6
we construct X 0 iteratively; in each iteration we pick x 2 X ? X 0 so that jNZ (x) ? NZ (X 0)j is the largest; we stop once Z is contained in N (X 0). Obviously we have j(N (X 0) \ H2) ? Z j < 319=20 log and j(N (Z 0) \ H1) ? X j < 319=20 log . 3. We construct sets KX = fv 2 N (X 0) : dN (Z0)(v) > ? 519=20g and KZ = fv 2 N (Z 0) : dN (X 0) (v ) > ? 519=20g. If KX \ KZ = ; then KX [ KZ is a bipartite subgraph H of G such that for all v in V (H ) : dH (v ) ? 39=40. If KX \ KZ 6= ; then KX \ KZ is a subgraph H of G where jV j?jV (H )j ? 39=40 such that for all v in V (H ) : dH (v) ? 39=40.
Claim 2 In step 1. of the procedure we have jY ? Z j < 12 19=20. Proof: The number of edges between vertices in X and Y is jE (X; Y )j jX j( 2 ? 9=10) > > >
(jY j ? 9=10)( 2 ? 9=10) jY j( 2 ? 9=10) ? 9=10jY j jY j( 2 ? 29=10)
If jY ? Z j 32 19=20 we would have a contradiction with the edge count between X and Y : jE (X; Y )j jZ j( 2 + 11=20) + jY ? Z j( 2 ? 29=10) jY j( 2 ? 29=10)
2
Claim 3 In step 2. of the procedure j(N (X 0) ? Z ) \ H2j < (3 log )19=20. 7
X
0 is of size at most log and
Proof: Each vertex in Z is adjacent to at least 2 ? 219=20 vertices in X and jX j < 2 + 9=10. It follows that each vertex in Z ? NZ (X 0) is adjacent to at least =2 ? 219=20 vertices in X ? X 0 and jX ? X 0j < + 9=10. Thus we know that Z ? NZ (X 0) is reduced by at least a half 2
at each iteration (for > 1220). So there will be at most log2 iterations. j(N (X 0) ? Z ) \ H2j < 3 log 19=20 follows because every x 2 X 0 is adjacent to at most 219=20 + 11=20 vertices in H2 ? Z . 2
Claim 4 In step 3. of the procedure if KX \ KZ = ; then KX [ KZ is a bipartite subgraph H of G such that for all v in V (H ) : dH (v ) ? 39=40. If KX \ KZ = 6 ; then KX \ KZ is a subgraph H of G where jV j ? jV (H )j 39 = 40 ? such that for all v in V (H ) : dH (v ) ? 39=40. Proof: We rst show that X KZ \ H1 (the symmetric argument shows Z KX \ H2): Let v be in X . It follows that v is in N (Z 0), dY (v ) > 2 ? 9=10 and moreover dZ (v ) > 2 ? 219=20. Since Z is a subset of N (X 0) \ H2 we have jN (v) \ N (X 0) \ H2j > 2 ? 219=20 but since N (v) \ N (X 0) splits (by claim 1) we get jN (v ) \ N (X 0)j > ? 519=20. So v is in KZ \ H1 . It follows that (jN (X 0) ? KX ) \ H2 j < 319=20 log and jjN (Z 0) ? KZ ) \ H1 j < 319=20 log . Also since N (X 0), N (Z 0), KX and KZ split (by claim 1) and KX N (X 0) and KZ N (Z 0) we have (jN (X 0) ? KX ) \ H1 j) < 319=20 log + 11=20 (jN (Z 0) ? KZ ) \ H2j) < 319=20 log + 11=20 and it follows that
jN (X 0) ? KX j < 719=20 log jN (Z 0) ? KZ j < 719=20 log so jN (X 0) \ N (Z 0) ? KX \ KZ j < 1419=20 log . If KX \ KZ 6= ; then KX \ KZ is a subgraph of G that satis es (ii) of theorem 1: jKX \ KZ j < jN (X 0) \ N (Z 0)j < jX j + jZ j + 619=20 log < 8
jV (G)j ? + 39=40 and for all v 2 KX \ KZ : dN (X 0)(v) > ? 519=20 and dN (Z 0) (v ) > ? 519=20 which implies dN (X 0)\N (Z 0) (v ) > ? 1019=20 and nally dKX \KZ (v ) > ? 1019=20 ? 1419=20 log > ? 39=40 for
large enough. If KX \ KZ = ; then KX [ KZ de nes a subgraph of G that satis es condition (i) of theorem 1: let the bipartite graph be de ned by the partition (KX ; KZ ) where all the edges within KX and KZ are deleted. Then for all v 2 KX : dKZ (v ) > ? 519=20 ? 719=20 log > ? 39=40 and vice versa for large enough. 2
3 The Algorithm We now present the algorithm COLOU R that, given a -regular graph G with an even number of vertices at most 2 and a split partition (H1; H2) of V , edge colors G with colors or returns a subgraph H of G satisfying (i) or (ii) of theorem 1. Let B be the bipartite graph whose vertex set is de ned by the partition (H1; H2) and edge set by f(u; v ) 2 E : u 2 H1 ; v 2 H2 g. The goal of algorithm COLOUR is to nd a bipartite reduction G0 of G that is a subgraph of B . We attempt to do this in three steps. First (in part 3.1) we color the edges in E (H1) and E (H2) and obtain the set M of about 2 matchings that cover E (H1) and E (H2). Second (in part 3.2) we augment each matching in M to obtain a set of perfect matchings M0 (i.e. every matching in M0 hits every vertex in G) that we remove from G. This will leave a few edges in H1 and H2 uncolored. Finally, in part 3.3, we color all the edges left over in H1 and H2 , using a third set of matchings M00. We add edges from B to each matching in M00 to obtain a set of perfect matchings that we remove from G. We are then left with a bipartite reduction G0 of G. If we fail at any step we will show the existence of (and construct) sets X and Y such that X H1, Y H2, jH1j?jX j > 2 ? 9=10, jY j < jX j + 9=10 and
8v 2 X : dY (v) > 2 ? 9=10 Using the procedure from section 2.2 we can then nd a subgraph H of G 9
which satis es condition (i) or (ii) of theorem 1. 3.1
Coloring
E (H1) ands E (H2)
We now color the edges of H1 and H2. Let 3 = d 2 e + d 41 11=20e. We construct a 3 coloring of E (H1) [ E (H2), M = fM1 ; M2; :::; M3 g, which satis es the following properties:
1. For every two matchings Mi and Mj in M: jMi \ E (H1)j and jMj \ E (H1)j di er by at most one. jMi \ E (H2)j and jMj \ E (H2)j di er by at most one. 2. For every matching Mi in M: jMi \ E (H1)j = jMi \ E (H2)j A set of matchings satisfying these properties is called a balanced set of matchings.
Input: A -regular graph G with an even number of vertices at most 2
and a split partition (H1; H2) of V . Output: The graph G along with a split partition (H1; H2) of V and a balanced set of matchings M = fM1; M2 ; :::; M3 g covering E (H1) and E (H2). Step: We color E (H1) using 3 colors to get the set of disjoint matchings 0 using Fournier's algorithm; similarly we get a set of M10 ; M20 ; :::; M 3 disjoint matchings M100; M200; :::; M00 3 covering E (H2). We now modify these matchings so that property 1. in the de nition of a balanced set of matchings is satis ed: To satisfy it for H1 (the case H2 is the same) we will apply recursively the following procedure to two matchings Mj0 and Mi0 with the smallest and largest number of edges respectively: If the di erence between jMi0 j and jMj0 j is at most 1, we're done. If the di erence between jMi0 j and jMj0 j is equal to c > 1 then, in the subgraph of G de ned by all edge s in Mi0 and Mj0 , there will be at 10
least c components that are alternating paths that contain one more edge of Mi0 than Mj0 . We pick any b 2c c such components and switch the color of each edge on each component. The di erence between jMi0j and jMj0 j is at most 1 after this switch. When done, we order the matchings M10 ; M20 ; :::; M0 3 and 00 by non-decreasing number of edges and set Mi = M100; M200; :::; M 3 Mi0 [ Mi00. This satis es the second property of a balanced set of matchings.
Claim 5 Each matching Mi in M misses at most 11=20 vertices in H1 (H2).
Proof: By the de nition of a split partition no vertex will be missed by more than 21 11=20 matchings in M. Since all matchings in M miss the
same number of vertices, within 2, and H1 contains at most vertices, each matching misses at most 11=20
2 (3)?1 + 2 11=20 vertices in H1. 3.2
2
The Patching
We now augment each matching Mi in M so that it hits every missed vertex of G. That is, we produce a set of disjoint perfect matchings M0 = fM10 ; M20 ; :::; M0 3g so that Mi0 Mi [ B. We use a special kind of alternating path to augment a matching Mi . A patch with respect to a matching Mi between vertices x and y not incident to any edge in Mi is a path from x to y whose edges alternate between edges in the bipartition and edges in the matching Mi . We say that a vertex v is patched in Mi0 if v is in some edge of Mi ? Mi0. When we augment Mi along a patch some edges in E (H1) and E (H2) will get uncolored (those that belong to Mi ? Mi0 ). These edges will form reject graphs R1 and R2, respectively. We insure that the maximum degrees of R1 and R2 are small by limiting the number of times any vertex is patched. 11
Input A -regular graph G with an even number of vertices at most 2,
a split partition (H1; H2) of V (G) and a balanced set of matchings
M = fM1; M2; :::; M3g covering E (H1) and E (H2). Output EITHER a set of 3 disjoint matchings fM10 ; :::; M0 3g such that 0 is a ? 3 -regular reduction F of G with F = G ? M10 ? ::: ? M 3
the sets H1 and H2 inducing reject subgraphs R1 and R2 respectively such that jE (R1)j = jE (R2)j and for all v in Hi: dRi (v ) < 41 4=5 OR sets X and Y such that X H1, Y H2 , jH1j?jX j 2 ? 9=10, jY j jX j + 9=10 and
8v 2 X : dY (v) 2 ? 9=10
Steps Recursively for i = 1; 2; :::; 3 we construct a perfect matching Mi0 by augmenting along patches in F = G ? M10 ? M20 ? ::: ? Mi0?1 between
pairs of vertices missed by Mi . We then remove Mi0 and proceed to Mi0+1 . To construct a patch in F between a pair of vertices x in H1 and y in H2 missed by Mi we do the following: If there is an edge in F between x and y , add (x; y ) to Mi0 , and we're done. Assume then that there is no edge between x and y . We de ne a vertex v to be available if it was not patched in the last 41=5 matchings, was not used in a patch in this (i-th) iteration and is not matched in Mi with such a vertex. A vertex will be called unavailable if it is not available. We construct the sets Y 1 and Y 2 where Y 1 is the set of available neighbors of x in H2 that are hit by Mi , and Y 2 is the set of vertices that are matched with vertices in Y 1 in Mi . Similarly we construct sets X 1 and X 2 where X 1 is the set of available neighbors of y in H1 that are hit by Mi , and X 2 is the set of vertices that are matched with vertices in X 1 in Mi . If there is an edge between a vertex in X 2 and a vertex in Y 2 we augment Mi along a patch from x to y containing this edge. Add to R1 and R2 the edges left uncolored in H1 and H2 after this augmentation. If there is no such edge then set X = X 2 and Y = H2 ? Y 2 .
Claim 6 For i = 1; 2; :::; 3, after successfully constructing the perfect matching Mi0 and deleting it we have in F = G ? M10 ? ::: ? Mi0 : 12
(i) maxf(R1); (R2)g < 41 4=5 (ii) jE (R1)j = jE (R2)j (iii) (F ) = ? i Proof: (i) A vertex can have an incident edge rejected at most 413=5 times which
is less than 44=5 . (ii) In every augmentation over a patch the same number of edges is added to E (R1) and E (R2). (iii) Every vertex is hit by every perfect matching M10 ; :::; Mi0.
2 In claim 6 we show that if our Patching procedure succeeds in constructing the perfect matchings M10 ; :::; M0 3 then we get a ? 3-regular reduction F of G with the sets H1 and H2 inducing reject subgraphs R1 and R2 respectively such that jE (R1)j = jE (R2)j and for all v in Hi : dRi (v ) < 44=5 . We then use this reduction of G to nd a bipartite reduction of G in part 3.3. The following claim shows that if our Patching procedure was not succesful the sets X and Y satisfy the input conditions of the Subgraph Construction Procedure.
Claim 7 Assume that for some matching Mi and for some vertices x and y
missed by Mi we haven't found a patch using the patching procedure. Then the sets X and Y constructed in the procedure satisfy: jH1j?jX j 2 +9=10, jY j jX j + 9=10 and
8v 2 X : dY (v) 2 ? 9=10
Proof: Let's rst count the number of unavailable vertices in Hr (for r = 1 or r = 2). At most 2 vertices in Hr are patched in every patch. Since there can be at most 11=20 vertices in Hr missed by any matching (by claim 5) 13
there can be at most that many patches per matching. So the number of unavailable vertices is at most 9=10
2(4b1=5c + 1)2b11=20c < 2
for > 420. By claim 6 we must have that jX 1j, jX 2j, jY 1 j and jY 2 j are all greater than ? 3 9=10. Since jH1j and jH2j are at most we get jH1 j?jX j +11=20 2 4 2 and jY j jX j + 9=10. Again by claim 6 we get
8v 2 X : dY (v) 2 ? 4=5 2 3.3
The Reject Graph
In this part we deal with the reduction we obtained after the patching step. We will color all the edges in E (R1) and E (R2) to obtain a set of matchings M00 each of small size. We then augment each Mi00 in M00 by adding edges in B to it until we obtain a perfect matching.
Input A ? 3-regular reduction F of G with the sets H1 and H2 inducing reject subgraphs R1 and R2 respectively such that jE (R1)j = jE (R2)j
and for all v in Hi : dRi (v ) < 41 4=5 Output EITHER a set of d disjoint perfect matchings M00 such that F ? M100 ? ::: ? Md00 is a bipartite reduction of G OR sets X and Y such that X H1, Y H2 , jH1j?jX j 2 ? 9=10, jY j jX j + 9=10 and
8v 2 X : dY (v) 2 ? 9=10
Steps 1. Let 4 = maxf(R1); (R2)g + 1 d 14 4=5e. We color E (R1) and E (R2) (using Fournier's algorithm) with 4 colors to get a 14
set of matchings N = fN1; :::; N4 g. We modify our matchings in N so that it is a balanced set of matchings. We can do this using the same procedure as in part 3.1 when coloring the edges of H1 and H2. 2. We split each Ni into matchings of size at most 9=10 evenly split between E (H1) and E (H2) to get a set of matchings M00 = fM100; M200; :::; Md00g where d 92=10 . 3. For i = 1; :::; d we repeat the following in G0 = F ? M100 ? ::: ? Mi00?1 : Let U1 in H1 and U2 in H2 be the sets of vertices hit by Mi00. We extend Mi00 by nding a perfect matching in the bipartite subgraph P of G0 induced by H1 ? U1 and H2 ? U2 . If we're unable to nd such a matching let X be the subset in H1 ? U1 such that jX j > jNP (X )j. We set Y = NP (X ) [ U2.
Claim 8 In the step 3. of the procedure the sets X and Y satisfy: jH1j ? jX j 2 ? 9=10, jY j jX j + 9=10 and 8v 2 X : dY (v) 2 ? 9=10
Proof: It is clear that jY j < jX j + 9=10 and 8v 2 X : dY (v) 2 ? 9=10 Finally since jX j > jNP (X )j there must be a vertex v in H2 ? U2 with no neighbors in X in P . But every vertex in P \ H2 has at least 2 ? 92=10 ? 4=5 neighbors in H1 . 2
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[3] A. G. Chetwynd and A. J. W. Hilton. Regular graphs of high degree are 1-factorizable. Proc. London Math. Soc. Vol.50, pp. 193-206, 1985 [4] A. G. Chetwynd and A. J. W. Hilton. Star multigraphs with three vertices of maximum degree. Math. Proc. Cambridge Philos. Soc. Vol 100, pp. 303-317, 1986. [5] A. G. Chetwynd and A. J. W. Hilton. 1-factorizing regular graphs of high degree: an improved bound. Discrete Math. Vol. 75, pp. 103-112, 1989. [6] J. Fournier. Coloration des aretes d'un graphe. Cahiers Centre Etudes Rech. Oper. Vol. 15, pp. 311-314, 1973. [7] A. J. W. Hilton, P. D. Johnson, Graphs which are vertex-critical with respect to the edge-chromatic number. Math. Proc. Camb. Phil. Soc. Vol. 102, pp. 211-221, 1987. [8] T. R. Jensen and B. Toft. Graph coloring problems. Wiley-Interscience, 1995. [9] D. Konig. Uber graphen und ihre andwendung auf determinantentheorie und mengenlehre. Math. Ann. Vol. 77, pp. 453-465, 1916. [10] L. Perkovic and B. Reed. manuscript, 1995. [11] L. Perkovic and B. Reed. manuscript, 1995. [12] V. Vizing. On the estimate of the chromatic class of a p-graph. Diskret. Analiz. Vol. 3, pp. 25-30, 1964.
16
B. Reedy CNRS Paris France
Abstract We discuss the following conjecture: If G = (V; E ) is a -regular simple graph with an even number of vertices at most 2 then G is edge colorable. In this paper we show that the conjecture is true for large graphs if jV j < (2 ? ). We discuss related results.
1 Introduction The following conjecture has been proposed:
Conjecture 1 If
G = (V; E ) is a -regular simple graph with an even number of vertices at most 2 then G is 1-factorizable ( edge colorable).
This conjecture appeared in Chetwynd and Hilton [3] but may go back to G. A. Dirac in the early 1950's (Chetwynd and Hilton [3]). An interesting consequence is that for any regular graph G either G or its complement has a 1-factorization. This conjecture along with variants and related results can be found in the new book Graph Coloring Problems by Jensen and Toft [8]. E-mail: [email protected] y E-mail: breed@mu .cs.mcgill.ca
1
In order to describe related conjectures and results we need to de ne some terminology. A graph G = (V; E ) is simple if it contains no multiple edges and no loops. G is r-regular if the degree (i.e. the number of edges incident to a vertex) of every vertex is r. G is k edge colorable if the edge set E of G can be partitioned into k disjoint matchings. The chromatic index of G, denoted 0 (G), is the least k for which G has a k edge-colouring. We use (G) or simply to denote the maximum vertex degree in G. It is easy to see that 0 (G) . In 1964, Vizing [12] proved that for every simple graph G either 0 (G) = or 0 (G) = + 1. We note that Fournier [6], [1] designed a procedure to + 1 color a graph of maximum degree in polynomial time. A subgraph H of G is called overfull if jV (H )j is odd and jE (H )j > jV (H2)j?1 , where E (H ) is the set of edges in E induced by vertices in H . If a simple graph contains an overfull subgraph H then 0 (G) = + 1. This follows from the fact that in any edge-coloring of G at most jV (H2)j?1 edges in E (H ) can have the same color. It is not too hard to see that Conjecture 1 is a special case of Hilton's Overfull Subgraph Conjecture (Hilton and Johnson [7], Chetwynd and Hilton [4]):
Conjecture 2 (Hilton) If G is simple, > jV3 j and G contains no overfull
subgraph then G is edge colorable.
There are some partial results regarding p Conjecture 1. Chetwynd and Hilton ( [5] have proved the conjecture if 72?1) jV j. Furthermore Chetwynd and Hilton [3] note that R. Haggkvist has announced that for any > 0 there exists N > 0 so that every -regular graph G is 1-factorizable if G has an even number of vertices greater than N and ( 12 + )jV j. In this paper we prove the following theorem:
Theorem 1 There exists 0 s.t. for all 0 and for all -regular graphs G with an even number of vertices at most 2, if G contains no: (i) Bipartite subgraph H such that for all v in V (H ) : dH (v) ? 39=40 (ii) Subgraph H where jV j ? jV (H )j ? 39=40 such that for all v in V (H ) : dH (v ) ? 39=40 2
then G is edge colorable.
The result Haggkvist announced is a corollary of this theorem. Proof: Given , ( 21 + )jV j let G be a -regular graph with an even number of vertices greater than N = maxf20; ?40 g. Then G cannot contain a subgraph H satisfying (i) because jH j would be greater than 2 ? 239=40 jV j + 2jV j ? 2jV j39=40 > jV j which is a contradiction. Similarly G cannot contain a subgraph H satisfying (ii) of theorem 1. So G is -edge colorable. 2 More importantly, the methods we use in the proof of theorem 1 are very general and have been used to get similar partial results for Hilton's conjecture (Perkovic and Reed [10]). These partial results were enough to devise a polynomial expected time algorithm for edge coloring simple graphs (Perkovic and Reed [11]). But the proofs of these results are long and technical, while the proof of theorem 1 is relatively simple. We turn now to the proof of theorem 1. We present algorithm COLOUR that given a -regular graph G so that 0 and jV j 2 is even edge colors G with colors or returns a subgraph H satisfying conditions (i) or (ii) of theorem 1. The algorithm along with the proof of its correctness proves theorem 1. We assume 0 is large enough to satisfy a number of lower bounds appearing throughout the paper. In order to give the main idea of algorithm we need the concept of a reduction.
De nition 1
is a reduction of G if (G0) = (G) ? l and there exists a set of matchings fM1 ; M2; :::; Mlg such that G0 = G ? M1 ? M2 ? ::: ? Ml . G0
We remark that if G0 is a reduction of G and G0 has a (G0) edge-coloring then G has a (G) edge-coloring. We remark further that given disjoint matchings M1 ; :::; Ml in a graph G = G0 and setting Gi = Gi?1 ? Mi we have that each Gi is a reduction of G if and only if for each i, every vertex of maximum degree in Gi?1 is the endpoint of some edge of Mi . The main idea of the algorithm is that given a graph G we attempt to nd a bipartite reduction G0 of G. Since 0 (G0) = (G0) (by Konig's theorem [9]) we will have 0 (G) = (G). 3
To nd a bipartite reduction, we rst partition the vertices of G into sets H1 and H2 of equal size so that a number of conditions hold. We describe this in the following section. In section 3 we describe the algorithm COLOUR that constructs perfect matchings consisting mostly of edges within H1 and within H2 . Once we color all the edges in H1 and H2, the uncolored edges form a bipartite reduction of G. We introduce some notation we will need in our arguments. Given a subset H of V let NH (v ) = fu 2 H : (u; v ) 2 E g, dH (v ) = jNH (v )j; if H = G we simply write N (v ) or d(v ). If X is a subset of H then NH (X ) = [v2X N (v ).
2 The Split In the following section we present an algorithm COLOUR which edge colors a -regular graph G = (V; E ) (with jV j even and at most 2) or returns a subgraph H of G satisfying (i) or (ii) of theorem 1. However COLOUR needs to be given a special partition (H1; H2) of the vertices of G as an input. This partition has the property that certain sets of vertices split about evenly between H1 and H2. Let us de ne a split formally:
De nition 2 Let H1 and11=H20 2 be a partition of V . A set S V splits if jjS \ H1j ? jS \ H2jj < 2 . A partition (H1; H2) of satis ed:
V
is called a split partition if the following are
1. jH1j = jH2j (which implies jE (H1)j = jE (H2)j) 2. 8v 2 V; 8X; Y V of size less than log the following sets split: N (v ); N (X ); N (v ) \ N (X ); fw 2 N (X ) : dN (Y )(w) > ? 519=20g To prove that the desired partition exists we will show that a suitably de ned random split satis es the conditions with positive probability. We then apply COLOUR to some such partition obtaining either a edge coloring of G or a subgraph H of G satisfying (i) of (ii) of theorem 1. 4
2.1
The Partitioning Procedure
The following procedure constructs with positive probability a split partition (H1; H2) of V : We arbitrarily order the vertices in V . For each successive ordered pair of vertices switch the order of the pair with probability 1=2 and put the rst vertex in set H1 and the second in the set H2 . Let E (H1) and E (H2) be the edges induced by H1 and H2 respectively. It is easy to see that the resulting partition (H1; H2) of V satis es the rst condition of a split partition. We show that with positive probability the other condition in the de nition of a split partition is also satis ed in the following claim:
Claim 1 The probability that there is any v in V , X and Y subsets of V with jX j; jY j < log for which one of the following sets fails to split is less than 21 :
( ) ( ) ( ) \ N (X ); fw 2 N (X ) : dN (Y ) (w) > ? 519=20g
N v ;N X ;N v
Proof: Let jV j = n. There are at most log n(n +
n
!
n
!
log + n log +
!
n
! 3 log n log ) < n n
log
sets that we want to split. Let S = fv1 ; v2; v3; :::; vkg be one of them. We de ne for each v 2 V the random variable Xv : ( ?1 if v 2 H2 Xv = 1 if v 2 H 1
P Then jjS \ H1 j ? jS \ H2 jj j v2S Xv j and 11=20 X ?22=20 Pr(j Xv j > ) < 2e 8n 2 v2S which is less than (2n23log n )?1 for large enough. 5
2
Claim 1 implies that with probability at least 12 the partitioning procedure constructs a split partition. Before describing algorithm COLOUR on such a partition we describe a procedure that constructs a subgraph H of G satisfying (i) or (ii) of theorem 1 if COLOUR fails to edge color G. 2.2
The Subgraph Construction Procedure
The second property of a split partition is that certain sets of vertices split about evenly between H1 and H2. The reason we want these sets to split is that it enables us to construct a subgraph of G that satis es (i) or (ii) of the theorem 1 if we nd two sets X and Y such that X H1 , jH1j ? jX j > ? 9=10, Y H2 , jY j < jX j + 9=10 and 2
8v 2 X : dY (v) > 2 ? 9=10 Algorithm COLOUR fails to edge color G only if such sets X and Y are found. In this case we use the Subgraph Construction Procedure to nd a subgraph of G that violates a condition of theorem 1:
Input: A -regular graph G with an even number of vertices at most 2 a split partition (H1 ; H2) of V and sets X and Y such that X H1 , 9=10 9=10 and Y H2, jH1j ? jX j > 2 ? , jY j < jX j + 8v 2 X : dY (v) > 2 ? 9=10
Output: EITHER a bipartite subgraph H of G such that for all v in V (H ) : dH (v ) ? 39=40 OR a subgraph H of G where jV j ? jV (H )j ? 39=40 such that for all v in V (H ) : dH (v ) ? 39=40 Steps: 1. We construct the set Z = fv 2 Y : dX (v) 2 ? 219=20g. Since jY ? Z j < 32 19=20 we have then that for all v in X : dZ (v ) 2 ?219=20 and that for all v in Z : dX (v ) 2 ?219=20. 2. We construct sets X 0 X and Z 0 Z such that jX 0j and jZ 0j are less than log , Z N (X 0) and X N (Z 0 ) using the procedure given below (the construction of Z 0 is symmetric): 6
we construct X 0 iteratively; in each iteration we pick x 2 X ? X 0 so that jNZ (x) ? NZ (X 0)j is the largest; we stop once Z is contained in N (X 0). Obviously we have j(N (X 0) \ H2) ? Z j < 319=20 log and j(N (Z 0) \ H1) ? X j < 319=20 log . 3. We construct sets KX = fv 2 N (X 0) : dN (Z0)(v) > ? 519=20g and KZ = fv 2 N (Z 0) : dN (X 0) (v ) > ? 519=20g. If KX \ KZ = ; then KX [ KZ is a bipartite subgraph H of G such that for all v in V (H ) : dH (v ) ? 39=40. If KX \ KZ 6= ; then KX \ KZ is a subgraph H of G where jV j?jV (H )j ? 39=40 such that for all v in V (H ) : dH (v) ? 39=40.
Claim 2 In step 1. of the procedure we have jY ? Z j < 12 19=20. Proof: The number of edges between vertices in X and Y is jE (X; Y )j jX j( 2 ? 9=10) > > >
(jY j ? 9=10)( 2 ? 9=10) jY j( 2 ? 9=10) ? 9=10jY j jY j( 2 ? 29=10)
If jY ? Z j 32 19=20 we would have a contradiction with the edge count between X and Y : jE (X; Y )j jZ j( 2 + 11=20) + jY ? Z j( 2 ? 29=10) jY j( 2 ? 29=10)
2
Claim 3 In step 2. of the procedure j(N (X 0) ? Z ) \ H2j < (3 log )19=20. 7
X
0 is of size at most log and
Proof: Each vertex in Z is adjacent to at least 2 ? 219=20 vertices in X and jX j < 2 + 9=10. It follows that each vertex in Z ? NZ (X 0) is adjacent to at least =2 ? 219=20 vertices in X ? X 0 and jX ? X 0j < + 9=10. Thus we know that Z ? NZ (X 0) is reduced by at least a half 2
at each iteration (for > 1220). So there will be at most log2 iterations. j(N (X 0) ? Z ) \ H2j < 3 log 19=20 follows because every x 2 X 0 is adjacent to at most 219=20 + 11=20 vertices in H2 ? Z . 2
Claim 4 In step 3. of the procedure if KX \ KZ = ; then KX [ KZ is a bipartite subgraph H of G such that for all v in V (H ) : dH (v ) ? 39=40. If KX \ KZ = 6 ; then KX \ KZ is a subgraph H of G where jV j ? jV (H )j 39 = 40 ? such that for all v in V (H ) : dH (v ) ? 39=40. Proof: We rst show that X KZ \ H1 (the symmetric argument shows Z KX \ H2): Let v be in X . It follows that v is in N (Z 0), dY (v ) > 2 ? 9=10 and moreover dZ (v ) > 2 ? 219=20. Since Z is a subset of N (X 0) \ H2 we have jN (v) \ N (X 0) \ H2j > 2 ? 219=20 but since N (v) \ N (X 0) splits (by claim 1) we get jN (v ) \ N (X 0)j > ? 519=20. So v is in KZ \ H1 . It follows that (jN (X 0) ? KX ) \ H2 j < 319=20 log and jjN (Z 0) ? KZ ) \ H1 j < 319=20 log . Also since N (X 0), N (Z 0), KX and KZ split (by claim 1) and KX N (X 0) and KZ N (Z 0) we have (jN (X 0) ? KX ) \ H1 j) < 319=20 log + 11=20 (jN (Z 0) ? KZ ) \ H2j) < 319=20 log + 11=20 and it follows that
jN (X 0) ? KX j < 719=20 log jN (Z 0) ? KZ j < 719=20 log so jN (X 0) \ N (Z 0) ? KX \ KZ j < 1419=20 log . If KX \ KZ 6= ; then KX \ KZ is a subgraph of G that satis es (ii) of theorem 1: jKX \ KZ j < jN (X 0) \ N (Z 0)j < jX j + jZ j + 619=20 log < 8
jV (G)j ? + 39=40 and for all v 2 KX \ KZ : dN (X 0)(v) > ? 519=20 and dN (Z 0) (v ) > ? 519=20 which implies dN (X 0)\N (Z 0) (v ) > ? 1019=20 and nally dKX \KZ (v ) > ? 1019=20 ? 1419=20 log > ? 39=40 for
large enough. If KX \ KZ = ; then KX [ KZ de nes a subgraph of G that satis es condition (i) of theorem 1: let the bipartite graph be de ned by the partition (KX ; KZ ) where all the edges within KX and KZ are deleted. Then for all v 2 KX : dKZ (v ) > ? 519=20 ? 719=20 log > ? 39=40 and vice versa for large enough. 2
3 The Algorithm We now present the algorithm COLOU R that, given a -regular graph G with an even number of vertices at most 2 and a split partition (H1; H2) of V , edge colors G with colors or returns a subgraph H of G satisfying (i) or (ii) of theorem 1. Let B be the bipartite graph whose vertex set is de ned by the partition (H1; H2) and edge set by f(u; v ) 2 E : u 2 H1 ; v 2 H2 g. The goal of algorithm COLOUR is to nd a bipartite reduction G0 of G that is a subgraph of B . We attempt to do this in three steps. First (in part 3.1) we color the edges in E (H1) and E (H2) and obtain the set M of about 2 matchings that cover E (H1) and E (H2). Second (in part 3.2) we augment each matching in M to obtain a set of perfect matchings M0 (i.e. every matching in M0 hits every vertex in G) that we remove from G. This will leave a few edges in H1 and H2 uncolored. Finally, in part 3.3, we color all the edges left over in H1 and H2 , using a third set of matchings M00. We add edges from B to each matching in M00 to obtain a set of perfect matchings that we remove from G. We are then left with a bipartite reduction G0 of G. If we fail at any step we will show the existence of (and construct) sets X and Y such that X H1, Y H2, jH1j?jX j > 2 ? 9=10, jY j < jX j + 9=10 and
8v 2 X : dY (v) > 2 ? 9=10 Using the procedure from section 2.2 we can then nd a subgraph H of G 9
which satis es condition (i) or (ii) of theorem 1. 3.1
Coloring
E (H1) ands E (H2)
We now color the edges of H1 and H2. Let 3 = d 2 e + d 41 11=20e. We construct a 3 coloring of E (H1) [ E (H2), M = fM1 ; M2; :::; M3 g, which satis es the following properties:
1. For every two matchings Mi and Mj in M: jMi \ E (H1)j and jMj \ E (H1)j di er by at most one. jMi \ E (H2)j and jMj \ E (H2)j di er by at most one. 2. For every matching Mi in M: jMi \ E (H1)j = jMi \ E (H2)j A set of matchings satisfying these properties is called a balanced set of matchings.
Input: A -regular graph G with an even number of vertices at most 2
and a split partition (H1; H2) of V . Output: The graph G along with a split partition (H1; H2) of V and a balanced set of matchings M = fM1; M2 ; :::; M3 g covering E (H1) and E (H2). Step: We color E (H1) using 3 colors to get the set of disjoint matchings 0 using Fournier's algorithm; similarly we get a set of M10 ; M20 ; :::; M 3 disjoint matchings M100; M200; :::; M00 3 covering E (H2). We now modify these matchings so that property 1. in the de nition of a balanced set of matchings is satis ed: To satisfy it for H1 (the case H2 is the same) we will apply recursively the following procedure to two matchings Mj0 and Mi0 with the smallest and largest number of edges respectively: If the di erence between jMi0 j and jMj0 j is at most 1, we're done. If the di erence between jMi0 j and jMj0 j is equal to c > 1 then, in the subgraph of G de ned by all edge s in Mi0 and Mj0 , there will be at 10
least c components that are alternating paths that contain one more edge of Mi0 than Mj0 . We pick any b 2c c such components and switch the color of each edge on each component. The di erence between jMi0j and jMj0 j is at most 1 after this switch. When done, we order the matchings M10 ; M20 ; :::; M0 3 and 00 by non-decreasing number of edges and set Mi = M100; M200; :::; M 3 Mi0 [ Mi00. This satis es the second property of a balanced set of matchings.
Claim 5 Each matching Mi in M misses at most 11=20 vertices in H1 (H2).
Proof: By the de nition of a split partition no vertex will be missed by more than 21 11=20 matchings in M. Since all matchings in M miss the
same number of vertices, within 2, and H1 contains at most vertices, each matching misses at most 11=20
2 (3)?1 + 2 11=20 vertices in H1. 3.2
2
The Patching
We now augment each matching Mi in M so that it hits every missed vertex of G. That is, we produce a set of disjoint perfect matchings M0 = fM10 ; M20 ; :::; M0 3g so that Mi0 Mi [ B. We use a special kind of alternating path to augment a matching Mi . A patch with respect to a matching Mi between vertices x and y not incident to any edge in Mi is a path from x to y whose edges alternate between edges in the bipartition and edges in the matching Mi . We say that a vertex v is patched in Mi0 if v is in some edge of Mi ? Mi0. When we augment Mi along a patch some edges in E (H1) and E (H2) will get uncolored (those that belong to Mi ? Mi0 ). These edges will form reject graphs R1 and R2, respectively. We insure that the maximum degrees of R1 and R2 are small by limiting the number of times any vertex is patched. 11
Input A -regular graph G with an even number of vertices at most 2,
a split partition (H1; H2) of V (G) and a balanced set of matchings
M = fM1; M2; :::; M3g covering E (H1) and E (H2). Output EITHER a set of 3 disjoint matchings fM10 ; :::; M0 3g such that 0 is a ? 3 -regular reduction F of G with F = G ? M10 ? ::: ? M 3
the sets H1 and H2 inducing reject subgraphs R1 and R2 respectively such that jE (R1)j = jE (R2)j and for all v in Hi: dRi (v ) < 41 4=5 OR sets X and Y such that X H1, Y H2 , jH1j?jX j 2 ? 9=10, jY j jX j + 9=10 and
8v 2 X : dY (v) 2 ? 9=10
Steps Recursively for i = 1; 2; :::; 3 we construct a perfect matching Mi0 by augmenting along patches in F = G ? M10 ? M20 ? ::: ? Mi0?1 between
pairs of vertices missed by Mi . We then remove Mi0 and proceed to Mi0+1 . To construct a patch in F between a pair of vertices x in H1 and y in H2 missed by Mi we do the following: If there is an edge in F between x and y , add (x; y ) to Mi0 , and we're done. Assume then that there is no edge between x and y . We de ne a vertex v to be available if it was not patched in the last 41=5 matchings, was not used in a patch in this (i-th) iteration and is not matched in Mi with such a vertex. A vertex will be called unavailable if it is not available. We construct the sets Y 1 and Y 2 where Y 1 is the set of available neighbors of x in H2 that are hit by Mi , and Y 2 is the set of vertices that are matched with vertices in Y 1 in Mi . Similarly we construct sets X 1 and X 2 where X 1 is the set of available neighbors of y in H1 that are hit by Mi , and X 2 is the set of vertices that are matched with vertices in X 1 in Mi . If there is an edge between a vertex in X 2 and a vertex in Y 2 we augment Mi along a patch from x to y containing this edge. Add to R1 and R2 the edges left uncolored in H1 and H2 after this augmentation. If there is no such edge then set X = X 2 and Y = H2 ? Y 2 .
Claim 6 For i = 1; 2; :::; 3, after successfully constructing the perfect matching Mi0 and deleting it we have in F = G ? M10 ? ::: ? Mi0 : 12
(i) maxf(R1); (R2)g < 41 4=5 (ii) jE (R1)j = jE (R2)j (iii) (F ) = ? i Proof: (i) A vertex can have an incident edge rejected at most 413=5 times which
is less than 44=5 . (ii) In every augmentation over a patch the same number of edges is added to E (R1) and E (R2). (iii) Every vertex is hit by every perfect matching M10 ; :::; Mi0.
2 In claim 6 we show that if our Patching procedure succeeds in constructing the perfect matchings M10 ; :::; M0 3 then we get a ? 3-regular reduction F of G with the sets H1 and H2 inducing reject subgraphs R1 and R2 respectively such that jE (R1)j = jE (R2)j and for all v in Hi : dRi (v ) < 44=5 . We then use this reduction of G to nd a bipartite reduction of G in part 3.3. The following claim shows that if our Patching procedure was not succesful the sets X and Y satisfy the input conditions of the Subgraph Construction Procedure.
Claim 7 Assume that for some matching Mi and for some vertices x and y
missed by Mi we haven't found a patch using the patching procedure. Then the sets X and Y constructed in the procedure satisfy: jH1j?jX j 2 +9=10, jY j jX j + 9=10 and
8v 2 X : dY (v) 2 ? 9=10
Proof: Let's rst count the number of unavailable vertices in Hr (for r = 1 or r = 2). At most 2 vertices in Hr are patched in every patch. Since there can be at most 11=20 vertices in Hr missed by any matching (by claim 5) 13
there can be at most that many patches per matching. So the number of unavailable vertices is at most 9=10
2(4b1=5c + 1)2b11=20c < 2
for > 420. By claim 6 we must have that jX 1j, jX 2j, jY 1 j and jY 2 j are all greater than ? 3 9=10. Since jH1j and jH2j are at most we get jH1 j?jX j +11=20 2 4 2 and jY j jX j + 9=10. Again by claim 6 we get
8v 2 X : dY (v) 2 ? 4=5 2 3.3
The Reject Graph
In this part we deal with the reduction we obtained after the patching step. We will color all the edges in E (R1) and E (R2) to obtain a set of matchings M00 each of small size. We then augment each Mi00 in M00 by adding edges in B to it until we obtain a perfect matching.
Input A ? 3-regular reduction F of G with the sets H1 and H2 inducing reject subgraphs R1 and R2 respectively such that jE (R1)j = jE (R2)j
and for all v in Hi : dRi (v ) < 41 4=5 Output EITHER a set of d disjoint perfect matchings M00 such that F ? M100 ? ::: ? Md00 is a bipartite reduction of G OR sets X and Y such that X H1, Y H2 , jH1j?jX j 2 ? 9=10, jY j jX j + 9=10 and
8v 2 X : dY (v) 2 ? 9=10
Steps 1. Let 4 = maxf(R1); (R2)g + 1 d 14 4=5e. We color E (R1) and E (R2) (using Fournier's algorithm) with 4 colors to get a 14
set of matchings N = fN1; :::; N4 g. We modify our matchings in N so that it is a balanced set of matchings. We can do this using the same procedure as in part 3.1 when coloring the edges of H1 and H2. 2. We split each Ni into matchings of size at most 9=10 evenly split between E (H1) and E (H2) to get a set of matchings M00 = fM100; M200; :::; Md00g where d 92=10 . 3. For i = 1; :::; d we repeat the following in G0 = F ? M100 ? ::: ? Mi00?1 : Let U1 in H1 and U2 in H2 be the sets of vertices hit by Mi00. We extend Mi00 by nding a perfect matching in the bipartite subgraph P of G0 induced by H1 ? U1 and H2 ? U2 . If we're unable to nd such a matching let X be the subset in H1 ? U1 such that jX j > jNP (X )j. We set Y = NP (X ) [ U2.
Claim 8 In the step 3. of the procedure the sets X and Y satisfy: jH1j ? jX j 2 ? 9=10, jY j jX j + 9=10 and 8v 2 X : dY (v) 2 ? 9=10
Proof: It is clear that jY j < jX j + 9=10 and 8v 2 X : dY (v) 2 ? 9=10 Finally since jX j > jNP (X )j there must be a vertex v in H2 ? U2 with no neighbors in X in P . But every vertex in P \ H2 has at least 2 ? 92=10 ? 4=5 neighbors in H1 . 2
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