A more rapidly mixing Markov chain for graph colorings

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A More Rapidly Mixing Markov Chain forGraph ColoringsU

Martin Dyer and Catherine GreenhillSchool of Computer Studies, University of Leeds, Leeds, UK, LS2 9JT; e-mail:

{ }dyer, csg @scs.leeds.ac.uk

Recei ed 27 August 1997; accepted 13 July 1998

Ž .ABSTRACT: We define a new Markov chain on proper k-colorings of graphs, and relateits convergence properties to the maximum degree D of the graph. The chain is shown tohave bounds on convergence time appreciably better than those for the well-knownJerrumrSalas]Sokal chain in most circumstances. For the case ks2D, we provide adramatic decrease in running time. We also show improvements whenever the graph isregular, or fewer than 3D colors are used. The results are established using the method ofpath coupling. We indicate that our analysis is tight by showing that the couplings used areoptimal in a sense which we define. Q 1998 John Wiley & Sons, Inc. Random Struct. Alg., 13,285]317, 1998

1. INTRODUCTION

Markov chains on the set of proper colorings of graphs have been studied inw x w xcomputer science 9 and statistical physics 13 . In both applications, the rapidity of

convergence of the chain is the main focus of interest, though for somewhatw xdifferent reasons. The papers 9, 13 introduced a simple Markov chain, which we

refer to as the JSS chain, and related its convergence properties to the maximumŽ .degree D of the graph. Specifically, they established independently that the chain

Žconverges in time polynomial in the size of the graph i.e., has the rapid mixing

Correspondence to: Martin Dyer* Supported by ESPRIT Working Group RAND2.Q 1998 John Wiley & Sons, Inc. CCC 1042-9832r98r030285-33

285

DYER AND GREENHILL286

. Žproperty , provided that the number of colors k exceeds 2D. Note that the JSS.chain is known to converge eventually provided only kGDq2. Jerrum proved this

using the coupling technique while Salas and Sokal used Dobrushin’s uniquenesscriterion.

wThe two results have different merits. Jerum’s result can be extended but withU Ž 2 . 1 xan V n increase in running time to the case ks2D, whereas Salas and

w xSokal’s extends to the more general situation of the Potts model 2 .The transition procedure of the JSS chain is simple: choose a vertex ¨ and a

color c both uniformly at random and recolor ¨ with c if this results in ¨ beingproperly colored. A slight variation of this procedure involves choosing the color cuniformly at random from the set of colors which could properly recolor ¨.

The contribution of this paper is to define and analyze a new Markov chain oncolorings. The transition procedure of the new chain can be thought of as anextension of the second version of the JSS chain. An edge, rather than a vertex, is

Ž .chosen uniformly at random and the endpoints of the edge are properly recoloreduniformly at random from the permissible color pairs.

We establish the following properties for the new chain. If ks2D then ourU Ž 2 .bound on convergence time for the new chain is V n faster than that for the

JSS chain. Further, we show that the new chain always requires fewer than half thenumber of steps of the JSS chain whenever 2DFk-3D, or the underlying graph isregular. This is somewhat surprising, since one step of the new chain is no morecomplex than two steps of the Jerrum chain. Our convergence results are obtained

w xusing the method of path coupling, introduced in 3 . We give a rather preciseanalysis of the chain, and provide some evidence that coupling techniques areunlikely to improve upon our analysis of this chain.

The reader may be tempted to think that our results simply imply that the JSSchain is actually faster than its known bound. This could be the case, but we stressthat there is at presently absolutely no evidence for this hypothesis. The improvedconvergence rate over the JSS chain may be an artifact of the analysis, or mayrepresent a genuinely faster convergence, since the new chain certainly has ‘‘morepower’’ than two steps of JSS. For example, we prove that the new chain convergeseventually for kGDq1.

Although we do not address the question here, it can be shown that the problemof counting proper k-colorings of arbitrary graphs of maximum degree D isaP-hard for any fixed k, D such that kG3, DG3. The proof takes as its startingpoint the known fact that counting proper k-colorings of arbitrary graphs is

Ž w x. w xaP-hard when kG3 see 8 . Proofs are presented in 6 .The plan of the paper is as follows. A review of the path coupling method is

presented in Section 2. The rapid mixing of the new chain is established in Section3 and a variation of the chain is defined which improves the mixing time fornonregular graphs. The mixing times of the JSS chain and the new chain are thencompared. Finally, in Section 4 it is shown that the couplings used to prove rapidmixing of the new chain are optimal couplings in a precise sense which we define.

1 U Ž . Ž .V ? is the notation which hides factors of log n .

MARKOV CHAIN FOR GRAPH COLORINGS 287

2. THE PATH COUPLING METHOD

In this section some notation is introduced and the path coupling method isreviewed. Let V be a finite set and let MM be a Markov chain with state space V,transition matrix P, and unique stationary distribution p . In order for a Markovchain to be useful for almost uniform sampling or approximate counting, it mustconverge quickly toward its stationary distribution p . We make this notion moreprecise below. If the initial state of the Markov chain is x then the distribution of

tŽ . tŽ .the chain at time t is given by P y sP x, y . The total äriation distance of thexMarkov chain from p at time t, with initial state x, is defined by

1t td P , p s P x , y yp y .Ž . Ž .Ž . ÝTV x 2ygV

w x Ž . Ž t .Following Aldous 1 , let t « denote the least value T such that d P , p F«x TV xŽ . Ž . � Ž .for all tGT. The mixing time of MM, denoted by t « , is defined by t « smax t « :x

4xgV . A Markov chain is said to be rapidly mixing if the mixing time is boundedŽ y1 .above by some polynomial in n and log « , where n is a measure of the size of

the elements of V. Throughout this paper all logarithms are to base e.There are relatively few methods available to prove that a Markov chain is

rapidly mixing. One such method is coupling. A coupling for MM is a stochasticŽ . Ž . Ž .process X , Y on V=V such that each of X , Y , considered marginally, is at t t t

Ž w x.faithful copy of MM. The Coupling Lemma see, for example, Aldous 1 states thatw xthe total variation distance of MM at time t is bounded above by Prob X /Y , thet t

probability that the process has not coupled. The following result is used to obtainan upper bound on this probability and hence an upper bound for the mixing time.

w xThe proof of the second statement follows that given in 12, Lemma 4 .

Ž .Theorem 2.1. Let X , Y be a coupling for the Marko chain MM and let d be anyt tinteger älued metric defined on V=V. Suppose that there exists bF1 such that

E d X , Y Fbd X , YŽ . Ž .tq1 tq1 t t

for all t. Let D be the maximum älue that d achie es on V=V. If b-1 then theŽ .mixing time t « of MM satisfies

log D«y1Ž .t « F .Ž .

1ybŽ .

If bs1 and there exists a)0 such that

Prob d X , Y /d X , Y GaŽ . Ž .tq1 tq1 t t

for all t, then

2eDy1t « F log « .Ž . Ž .

a


Proof. Clearly,

t tE d X , Y Fb d X , Y Fb D ,Ž . Ž .t t 0 0

and, since d is nonnegative and integer valued,

w xProb X /Y FE d X , Y .Ž .t t t t

Ž . Ž y1 . Ž y1 .If b-1 then the Coupling Lemma implies that t « F log D« rlog b . TheŽ y1 .first statement follows as log b )1yb. Suppose now that bs1. Then the

Ž . Ž . Ž Ž ..2process Z t defined by Z t s Dyd X , Y ya t is a submartingale, sincet t

2E d X , Y yd X , Y Ga .Ž . Ž .Ž .tq1 tq1 t t

Ž . Ž . x, yThe differences Z tq1 yZ t are bounded. Let T denote the first time thatX sY where X sx and Y sy. Then T x, y is a stopping time for Z. The optionalt t 0 0

Ž w x.stopping theorem for submartingales see, for example, 15, Theorem 10.10applies and allows us to conclude that

d x , y 2 Dyd x , y D2Ž . Ž .Ž .x , yw xE T F F .

a a

u 2 v x, yLet Ts eD ra . By Markov’s inequality, the probability that T GT is at mostey1. If we run s independent trials of length T then the probability that we havenot coupled by the end of the sT steps is at most eys. Let tssT. Now eys F« and

Ž y1 .only if sG log « . Since s is an integer it follows that

y1tGT log « .Ž .Ž .This gives the stated bound on the mixing time t « . B

In general, the problem of obtaining an estimate for b is quite difficult. Thew xpath coupling method, introduced in 3 , provides a simplification. It involves

Ž .defining a coupling X , Y by considering a path, or sequence,t t

X sZ , Z , . . . , Z sYt 0 1 r t

between X and Y where the Z satisfy certain conditions.t t i

Theorem 2.2. Let d be an integer älued metric defined on V=V which takes� 4 Ž .älues in 0, . . . , D . Let S be a subset of V=V such that for all X , Y gV=Vt t

there exists a path,

X sZ , Z , . . . , Z sYt 0 1 r t

Ž .between X and Y such that Z , Z gS for 0F l- r andt t l lq1

ry1

d Z , Z sd X , Y .Ž . Ž .Ý l lq1 t tls0

Ž . Ž X X. Ž .Define a coupling X, Y ¬ X , Y of the Marko chain MM on all pairs X, Y gS.Apply this coupling along the gi en sequence from X to Y to obtain the new sequencet t


X X X Ž . X XZ , Z , . . . , Z . Then X , Y is also a coupling for MM, where X sZ and Y sZ .0 1 r t t tq1 0 tq1 rw Ž X X.x Ž . Ž .Moreoër, if there exists bF1 such that E d X , Y Fbd X, Y for all X, Y gS

then

E d X , Y Fbd X , Y .Ž . Ž .tq1 tq1 t t

Ž . Ž .Proof. Clearly X , Y are faithful copies of MM, as they are constructed fromt tinductive application of a coupling of MM defined on elements of S. Now

ry1X XE d X , Y FE d Z , ZŽ . Ž .Ýtq1 tq1 l lq1

ls0

ry1X Xs E d Z , ZŽ .Ý l lq1

ls0

ry1

Fb d Z , ZŽ .Ý l lq1ls0

sbd X , Y ,Ž .t t

using linearity of expectation and the fact that d is a metric. This proves thetheorem. B

Remark 2.1. The path coupling technique can greatly simplify the argumentsrequired to prove rapid mixing by coupling. For example, the set S is often takento be

Ss X , Y gV=V : d X , Y s1 .� 4Ž . Ž .

Here one need only define and analyze a coupling on pairs which are at distance 1w xapart. Some applications of path coupling can be found in 3]5, 7 .

3. A NEW MARKOV CHAIN FOR GRAPH COLORINGS

Ž .Let Gs V, E be a graph and let D be the maximum degree of G. Let k be apositive integer and let C be a set of size k. A map from V to C is called a

Ž .k-coloring or simply a coloring when k is fixed . A vertex ¨ is said to be properlycolored in the coloring X if ¨ is colored differently from all of its neighbors. Acoloring X is called proper if every vertex is properly colored in X. A sufficientcondition for the existence of a proper k-coloring of every graph with maximum

Ž . Vdegree D is kGDq1. Let V G :C be the set of all proper k-colorings of G.kŽ .We will denote the color assigned to a vertex ¨ in the coloring X by X ¨ .

Ž . w xA simple Markov chain with state space V G was proposed by Jerrum 9 andkw x Ž .by Salas and Sokal 13 . We refer to this chain as the JSS chain. If the JSS chain is

in state X at time t then the state at time tq1 is determined using the followingtprocedure: choose a vertex ¨ uniformly at random from V and choose a color cuniformly at random from C. Let X X denote the coloring obtained from X byrecoloring vertex ¨ with the color c. If ¨ is properly colored in X X then let


X sX X, otherwise let X sX . In the transition procedure the color c istq1 tq1 tchosen uniformly at random from the set C of all colors. This is an example of a so

Ž w x.called Metropolis Markov chain see 10, p. 511 . There is a nonzero probabilitythat the state X X will be rejected. We can remove this probability by choosing thecolor c uniformly at random from the set

� 4L ¨ sC_ X u : u , ¨ gE . 3.1� 4Ž . Ž . Ž .X tt

This set contains just those colors which can be used to color the vertex ¨ properlyin X . We refer to this amended chain as the heat bath JSS chain and denote thist

Ž Ž ..chain by JJ V G . Both the original JSS chain and the heat bath JSS chain arekirreducible for kGDq2 and rapidly mixing for kG2D. A simple path coupling

Ž .argument can be used to establish Jerrum’s result, that the mixing time t « ofJŽ Ž ..JJ V G satisfiesk

kyDy1t « F n log n« 3.2Ž . Ž . Ž .J ky2D

Ž . 3 Ž y1 .when k)2D. Path coupling also establishes that t « F6 Dn log « whenJks2D.

Ž Ž ..In this section we define a new Markov chain MM V G with state space1 kŽ . Ž Ž ..V G . We show that MM V G is irreducible for kGDq1 and rapidly mixingk 1 k

for kG2D. The chain apparently mixes most rapidly when the graph G isD-regular. By introducing self-loop edges, the D-regular mixing time can beachieved for nonregular graphs if 2DFk-3D. At the end of this section, acomparison is made between the mixing times of the new chain and the mixingtime of the heat bath JSS chain. It is shown that the new chain is faster than theheat bath JSS chain if G is D-regular or if 2DFkF3Dy1. If ks2D then the

U Ž 2 .new chain is V n times faster than the JSS chain. We will use the metric knownŽ . Ž .as the Hamming distance. Given X, YgV G the Hamming distance H X, Yk

between X and Y counts the number of vertices colored differently in X and Y.Ž Ž ..We begin by defining the transitions of the Markov chain MM V G . A1 k

� 4transition from state X involves choosing an edge ¨ , w uniformly at random fromŽ . Ž .E, then choosing colors c ¨ , c w uniformly at random from those pairs such that

both ¨ and w are properly colored in the coloring obtained from X by recoloring ¨Ž . Ž .with c ¨ and w with c w . To make this more precise, we make some definitions.

� 4For each edge cs ¨ , w in E let S be the set of colors defined by¨ , w , X

� 4S s X u : u , ¨ gE, u/w .� 4Ž .¨ , w , X

Let C e denote the set of all maps which assign colors to both endpoints of e, that is

e � 4C s c : ¨ , w ªC .� 4e Ž . eWe will refer to elements of C as color maps. Let AA e denote the subset of CX

defined by

AA e s cgC e : c ¨ gC_S , c w gC_S , c ¨ /c w .� 4Ž . Ž . Ž . Ž . Ž .X ¨ , w , X w , ¨ , X

� 4 eGiven es ¨ , w gE and cgC , denote by X the coloring obtained from Xeª cŽ . Ž .by recoloring ¨ by c ¨ and w by c w . Then both ¨ and w are properly colored in


Ž . Ž Ž ..X if and only if cgAA e . Suppose that the Markov chain MM V G is ineª c X 1 kstate X at time t. Then the state at time tq1 is determined by the followingtprocedure:

( ) � 4i choose an edge es ¨ , w uniformly at random from E,( ) Ž .ii choose c uniformly at random from AA e and let X sX .X tq1 eª ct

Ž Ž .. Ž Ž ..The chain MM V G is clearly aperiodic. We prove below that MM V G is1 k 1 kŽ Ž ..irreducible for kGDq1. Therefore MM V G is ergodic when kGDq1. Let1 k

Ž . < Ž . < � 4N e s AA e for each es ¨ , w gE and write w;¨ if w is a neighbor of ¨ .X XŽ Ž ..The transition matrix P of MM V G has entries1 k

y1¡ < <E N e if XsY ,Ž .Ž .Ý XegE

y1< < � 4E N ¨ , w if X and Y differ only at some vertex ¨ ,Ž .Ž .Ý X~ w;¨P X , Y sŽ .y1< <E N e if X , Y differ only at both endpointsŽ .Ž .X

of some edge e,¢0 otherwise.

Ž .for all X, YgV G . Suppose that X and Y differ just at one or both endpoints ofk� 4 Ž . Ž .the edge es ¨ , w . Then S sS and S sS . Hence AA e sAA e¨ , w , X ¨ , w , Y w , ¨ , X w , ¨ , Y X Y

Ž . Ž .and N e sN e . This implies that the transition matrix P is symmetric.X YŽ Ž ..Therefore the stationary distribution of MM V G is the uniform distribution on1 k

Ž .V G .k� 4Given any coloring X and any edge ¨ , w , the sets S , S both have at¨ , w , X w , ¨ , X

Ž .most Dy1 elements. Since we may easily show that

< < < < < <N e s ky S ky S y ky S jS , 3.3Ž . Ž .Ž . Ž . Ž .X ¨ , w , X w , ¨ , X ¨ , w , X w , ¨ , X

Ž . Ž .Ž . Ž .it follows that N e G kyD kyDq1 for all edges egE. Therefore N e isX XŽ Ž ..always positive when kGDq1. The next result shows that MM V G is irre-1 k

ducible for kGDq1.

Theorem 3.1. Let G be a graph with maximum degree D. The Marko chainŽ Ž ..MM V G is irreducible for kGDq1.1 k

Proof. The new chain can perform any move of the JSS chain and the latter isknown to be irreducible for kGDq2. Therefore we need only show thatŽ Ž .. Ž .MM V G is irreducible when ksDq1. Let X and Y be two elements of V G1 k k

Ž . Ž .where H X, Y G1. We show that H X, Y can be reduced by performing aŽ Ž ..suitable series of transitions of MM V G with initial state X, relabeling the1 k

Ž .resulting state by X after each transition here Y remains fixed . Suppose that thecolorings disagree at vertex ¨. Let U be the set of neighbors u of ¨ such thatŽ . Ž . Ž .X u sY ¨ . Assume that U is nonempty, otherwise H X, Y can be decreased by

Ž .recoloring ¨ with Y ¨ . Clearly all vertices in U are nonadjacent and the coloringsX, Y disagree at each element of U. Any vertex ugU which has fewer than D


Ž .colors surrounding it can be colored with some color other than Y ¨ , decreasing< < Ž .U without increasing H X, Y . Therefore we may assume that each element of Uhas D differently colored neighbors. In particular ¨ is the only neighbor of u

Ž . < <colored X ¨ , for all ugU. If U )1 then ¨ can be legally recolored with someŽ . Ž . Ž .color c ¨ /X ¨ without changing H X, Y . Then all vertices in U may be

Ž . Ž .recolored with X ¨ without increasing H X, Y . Finally ¨ may be recolored withŽ . Ž . < <Y ¨ , decreasing H X, Y by 1. Otherwise U s1. If u is the unique element of U

� 4then the edge u, ¨ may be chosen and the colors of its endpoints may be switchedŽ Ž .. Ž .by one move of the chain MM V G . This move decreases H X, Y by at least 1.1 k

B

We now prove that the new chain is rapidly mixing when kG2D, using pathcoupling on pairs at Hamming distance 1 apart. The result is stated in terms of theminimum degree d of the graph G.

Theorem 3.2. Let G be a graph with minimum degree d and maximum degree D.Ž Ž .. Ž .The Marko chain MM V G is rapidly mixing for kG2D. The mixing time t « of1 k 1

Ž Ž ..MM V G satisfies1 k

kyD kyDq1Ž . Ž . y1< <t « F E log n« .Ž . Ž .1 22d k y 3Dy2 kq2 Dy1Ž . Ž .Ž .Ž . Ž Ž ..Let T « denote the mixing time of MM V G in the case that G is D-regular. Then1 k

kyD kyDq1Ž . Ž . y1T « F n log n« .Ž . Ž .222 k y 3Dy2 kq2 Dy1Ž . Ž .Ž .

Ž . Ž . Ž . Ž .Proof. Let X , Y be a pair of elements of V G =V G and let rsH X , Y .t t k k t tWe wish to define a path,

X sZ , Z , . . . , Z sY 3.4Ž .t 0 1 r t

Ž .between X and Y of length r such that H Z , Z s1 for 0F i- r. In generalt t i iq1we cannot guarantee that such a path contains only proper colorings. To see this,

� 4consider elements X , Y which differ only at both endpoints of a single edge ¨ , wt tsuch that

Y ¨ sX ¨ , Y w sX ¨ .Ž . Ž . Ž . Ž .

However, we can always construct such a path with elements Z in the set of alliV Ž Ž .. Vcolorings C . The Markov chain MM V G can be extended to the set C of all1 k

colorings, since it does not assume that the current state X is a proper coloring intŽ V . Ž .the transition procedure. Denote the extended chain by MM C . Since N e isX

positive for any edge e and any coloring X, a transition is always possible at e in Xand both the endpoints of the chosen edge are properly colored after the transi-tion. Moreover, no properly colored vertex can become improperly colored bysubsequent transitions. Therefore the improper colorings are transient states, and


Ž V .the stationary distribution p of MM C is given by

y1V G if XgV G ,Ž . Ž .k kp X sŽ . ½ 0 otherwise.

Ž Ž .. Ž V .The chain MM V G acts exactly as the chain MM C when the starting state of1 kŽ V . Ž Ž ..both chains is a proper coloring. Hence if MM C is rapidly mixing then MM V G1 k

Ž V .is rapidly mixing with no greater mixing time. We now prove that MM C is rapidlymixing using the path coupling method on pairs at Hamming distance 1 apart.

Let X and Y be elements of CV which differ just at the vertex ¨ . The sets� 4S , S are equal when one of u or w equal ¨ , or when w, ¨ fE andw , u, X w , u, Y

� 4u, ¨ fE. We drop the subscripts X, Y for these sets. Let d denote the degree ofthe vertex ¨ . The only edges which can affect the Hamming distance between X

� 4 � 4 � 4and Y are the d edges of the form ¨ , w and all edges w, u where w, ¨ gE and� 4 Ž . Ž .u/¨ . If the chosen edge is es ¨ , w then AA e sAA e . Moreover these sets areX Y

� 4nonempty. Therefore we couple at edges es ¨ , w by choosing the same color mapŽ . X Xc uniformly at random from AA e and letting X sX and Y sY . HereX e¬ c eª c

Ž X X.H X , Y s0 for each choice of c.� 4 � 4Now suppose that the chosen edge is es w, u where w, ¨ gE and u/¨ . We

proceed as follows:

( ) Ž . Ž .i choose c gAA e uniformly at random and let c gAA e be chosen asX X Y Ydescribed below.

( ) X Xii let X sX and let Y sY .eª c eª cX Y

Ž .For the remainder of the proof write AA for AA e as the edge e is fixed:X Xsimilarly write AA , N , N . For ease of notation a color map cgC e will be writtenY X Y

² Ž . Ž .:as c w , c u . Note that

0 if c sc ,¡ X YX X ~H X , Y y1sŽ . 2 if c w /c w and c u /c u ,Ž . Ž . Ž . Ž .X Y X Y¢

1 otherwise.

� 4 � 4There are two configurations, depending on whether u, ¨ gE. If u, ¨ fE then� 4 � 4 � 4call w, u a Configuration 1 edge, while if u, ¨ gE then call w, u a Configuration

2 edge. Let M be defined by

kyDq2Ms . 3.5Ž .

kyDq1y kyDŽ . Ž .In the case analysis that follows, we show that every Configuration 1 edge

w Ž X X. xcontributes less than M to E H X , Y y1 and every Configuration 2 edgeŽ .contributes less than 2 M. Let d w denote the degree of the vertex w and write

¨ ;w if w is a neighbor of ¨ . Then the total contribution of all Configuration 1and 2 edges is less than

d w y1 MFd Dy1 M . 3.6Ž . Ž . Ž .Ž .Ýw;¨

A Configuration 2 edge is allowed to contribute twice as much as a Configuration 1Ž .edge since it is counted twice in the summation given in 3.6 .


Before embarking upon the case analysis we must establish a useful inequality.� 4 � 4Let w, u be any edge such that w, ¨ gE and u/¨ . We claim that

< <ky S q1u , w , X FM . 3.7Ž .NX

< <Let as S . Then aFDy1. After some rearranging, this implies thatu, w , X

kyD kyaq1 F kyDq2 kyay1 . 3.8Ž . Ž . Ž . Ž . Ž .

< < < <Let bs S . Then bFDy1 and S jS Gb. It follows thatw , u, X w , u, X u, w , X

kyay1 kyDq1 F kyay1 kybŽ . Ž . Ž . Ž .< <F kya kyb y ky S jSŽ . Ž . Ž .w , u , X u , w , X

sN .X

Ž .Combining this with 3.8 , we see that

< <kyD ky S q1 kyDq1 F kyDq2 N ,Ž . Ž . Ž .Ž .u , w , X X

Ž . Ž .establishing 3.7 . Note that 3.7 also holds with the roles of w and u reversed, orafter replacing X by Y, or both. This will be used throughout the remainder of theproof. Now the different cases which arise for Configuration 1 and 2 edges areidentified. Then a coupling will be described for each subcase and the contribution

w Ž X X. xof the coupling to E H X , Y y1 will be calculated.� 4Suppose first that w, u is a Configuration 1 edge. Here S sS so weu, w , X u, w , Y

Ž . Ž .drop the subscripts X, Y. If X ¨ gS and Y ¨ gS then AA sAA . Herew , u, Y w , u, X X YŽ X X.we can always use the same color map c in both X and Y, so H X , Y s1 with

Ž .probability 1. Otherwise without loss of generality assume that Y ¨ fS .w , u, XŽ .There are two cases to consider: for Case 1 suppose that X ¨ fS and forw , u, Y

Ž .Case 2 suppose that X ¨ gS . Within each case there are four subcases:w , u, YŽ . Ž . Ž . Ž . Ž .Subcase i where X ¨ fS and Y ¨ fS ; Subcase ii where X ¨ fSu, w u, w u, w

Ž . Ž . Ž . Ž . Ž .and Y ¨ gS ; Subcase iii where X ¨ gS and Y ¨ fS ; Subcase ivu, w u, w u, w

Ž . Ž . Ž . Ž .where X ¨ gS and Y ¨ gS . Clearly Subcases 1 ii and 1 iii are equivalent,u, w u, w

Ž .by exchanging X and Y. The reader may verify that the situation for Subcase 2 iiiŽ .is exactly as described below for Subcase 2 i . Hence the coupling procedure

Ž . Ž .described below for Subcase 2 i works in Subcase 2 iii also. Similarly SubcaseŽ . Ž .2 iv may be handled identically to Subcase 2 ii . Therefore it suffices to consider

Ž . Ž . Ž . Ž . Ž .only Subcases 1 i , 1 ii , 1 iv , 2 i , and 2 ii for a Configuration edge 1.� 4 Ž . Ž .Now suppose that w, u is a Configuration 2 edge. If X ¨ gS , Y U gw , u, Y

Ž . Ž .S , X ¨ gS , and Y ¨ gS then AA sAA . Here we can always usew , u, X u, w , Y u, w , X X YŽ X X .the same color map c in both X and Y, so H X , Y s1 with probability 1.

Ž .Otherwise without loss of generality assume that Y ¨ fS . The same twow , u, X

Ž . Ž .cases arise: Case 1 where X ¨ fS and Case 2 where X ¨ gS . Again,w , u, Y w , u, Y

Ž . Ž .within each case there are four subcases: Subcase i where X ¨ fS andu, w , YŽ . Ž . Ž . Ž . Ž .Y ¨ fS ; Subcase ii where X ¨ fS and Y ¨ gS ; Subcase iiiu, w , X u, w , Y u, w , X

Ž . Ž . Ž . Ž .where X ¨ gS and Y ¨ fS ; Subcase iv where X ¨ gS andu, w , Y u, w , X u, w , Y


Ž . Ž .Y ¨ gS . In Subcase iv notice that S sS . Therefore the methodsu, w , X u, w , X u, w , YŽ . Ž .used in Configuration 1, Subcase iv may be used here. Again Subcases 1 ii and

Ž .1 iii are equivalent, by exchanging X and Y. Moreover by exchanging the roles ofŽ . Ž .u and w we see that Subcases 1 ii and 2 i are equivalent. Therefore it suffices to

Ž . Ž . Ž . Ž .consider Subcases 1 , 2 i , 2 ii , and 2 iii for a Configuration 2 edge.

( ) � 4 Ž .Configuration 1, Subcase 1 i . Here u, ¨ fE. Moreover X ¨ fS andw , u, YŽ . < < < < Ž . Ž .Y ¨ fS , so S s S . Also X ¨ fS and Y ¨ fS . It followsw , u, X w , u, Y w , u, X u, w u, w

that N sN . The set AA is obtained from AA by replacing the color mapX Y Y X² Ž . Ž .: ² Ž . Ž .:Y ¨ , X ¨ with the color map X ¨ , Y ¨ and replacing every other color map

² Ž . Ž .: ² Ž . Ž .:of the form Y ¨ , c u with X ¨ , c u . Here choose c gAA uniformly atX XŽ . Ž . Ž . Ž . Ž . Ž .random and let c sc unless c w sY ¨ . If c w sY ¨ and c u /X Ÿ X X X X

² Ž . Ž .: ² Ž . Ž .:then let c s X ¨ , c u . Finally if c s Y ¨ , X ¨ then let c sY X X Y² Ž . Ž .:X ¨ , Y ¨ . This defines a bijection between AA and AA so each element of AAX Y Y

Ž . Ž . Ž X X.is chosen with equal probability. If c w sY ¨ then H X , Y s2 unlessXŽ . Ž . Ž X X. Ž X X.c u sX ¨ , in which case H X , Y s3. In all other cases H X , Y s1. ThereX

< < Ž . Ž .are exactly ky S y1 elements c of AA such that c w sY ¨ . Thereforeu, w X X Xw Ž X X. xthe contribution to E H X , Y y1 made by this edge is

< < < <ky S y2 2 ky Su , w u , wq s -M .N N NX X X

( ) � 4 Ž .Configuration 1, Subcase 1 ii . Here u, ¨ fE. Moreover X ¨ fS andw , u, YŽ . < < < < Ž . Ž .Y ¨ fS , so S s S . Also X ¨ fS and Y ¨ gS . It followsw , u, X w , u, Y w , u, X u, w u, w

that N sN q1. The set AA can be obtained from AA by deleting the color mapX Y Y X² Ž . Ž .: ² Ž . Ž .:Y ¨ , X ¨ and replacing all other color maps of the form Y ¨ , c u by² Ž . Ž .:X ¨ , c u . Here choose c gAA uniformly at random and let c sc unlessX X Y X

Ž . Ž . Ž . Ž . Ž . Ž . ² Ž . Ž .:c w sY ¨ . If c w sY ¨ and c u /X ¨ then let c s X ¨ , c u .X X X Y X² Ž . Ž .:Finally if c s Y ¨ , X ¨ then choose c uniformly at random from AA . WeX Y Y

now check that each element of AA is equally likely to be the result of thisYŽ .procedure. Let c be the element of AA chosen uniformly at random in Step iX X

Ž . Ž .and let c gAA . Suppose first that c w /X ¨ . Then c is the result of theY Y Y YŽ . Ž . ² Ž . Ž .:coupling procedure if a c sc , or b c s Y ¨ , X ¨ and c is the elementX Y X Y

of AA chosen uniformly at random. Therefore the probability that c is the resultY Yof the coupling procedure is

y1 y1² :w xProb c sc qProb c s Y ¨ , X ¨ N sN . 3.9Ž . Ž . Ž .X Y X Y Y

Ž . Ž . Ž . ² Ž . Ž .:Now suppose that c w sX ¨ . Then c is chosen if a c s Y ¨ , c u , orY Y X YŽ . ² Ž . Ž .:b c s Y ¨ , X ¨ and c is the element of AA chosen uniformly at random.X Y Y

² Ž . Ž .: Ž .Replacing c with Y ¨ , c u in 3.9 we see that c is chosen with probabilityY Y YN y1 in this case also. Therefore each element of AA is chosen with equalY Y

w Ž X X . xprobability, as required. Now we calculate the contribution to E H X , Y y1 . IfŽ . Ž . Ž X X. Ž . Ž . Ž . Ž .c w / Y ¨ then H X , Y s 1. If c w s Y ¨ and c u / X ¨ thenX X XŽ X X. < <H X , Y s2. There are exactly ky S y1 such elements in AA . Finally ifu, w X

² Ž . Ž .: Ž X X. Ž < < . y1c s Y ¨ , X ¨ then H X , Y s2 with probability kq S y1 N , oth-X w , u, Y Y


Ž X X. w Ž X X. xerwise H X , Y s3. Therefore the contribution to E H X , Y y1 in this case is

< < < < < <ky S y1 2 N y ky S y1 ky S q1Ž .u , w Y w , u , Y u , wq - FM .N N N NX X Y X

( ) � 4 Ž .Configuration 1, Subcase 1 iv . Here u, ¨ fE. Moreover X ¨ fS andw , u, YŽ . < < < < Ž . Ž .Y ¨ fS , so S s S . Also X ¨ gS and Y ¨ gS . It followsw , u, X w , u, Y w , u, X u, w u, w

that N sN . The set AA is obtained from AA by replacing every color map of theX Y Y X² Ž . Ž .: ² Ž . Ž .:form Y ¨ , c u with X ¨ , c u . Here choose c gAA uniformly at randomX X

Ž . Ž . Ž . Ž . Ž .and let c s c unless c w s Y ¨ . If c w s Y ¨ then let c w sY X X X Y² Ž . Ž .:X ¨ , c u . This defines a bijection between AA and AA so each element of AAX X Y Y

Ž . Ž . Ž X X.is chosen with equal probability. If c w sY ¨ then H X , Y s2 and ifXŽ . Ž . Ž X X. < <c w /Y ¨ then H X , Y s1. There are exactly ky S elements c of AAX u, w X X

Ž . Ž . w Ž X X. xsuch that c w sY ¨ . Therefore the contribution to E H X , Y y1 made byXthis edge is

< <ky Su , w-M .

NX

( ) � 4 Ž .Configuration 1, Subcase 2 i . Here u, ¨ fE. Moreover, X ¨ gS andw , u, YŽ . < < < < Ž . Ž .Y ¨ fS , so S s S y1. Also X ¨ fS and Y ¨ fS . Itw , u, X w , u, X w , u, Y u, w u, w

Ž < < .follows that N sN q ky S y1 . The set AA may be obtained from AA byX Y u, w Y X< < Ž .deleting the ky S y1 color maps which assign the color Y ¨ to w. Before weu, w

can describe the coupling procedure we define some notation. The set C_S isw , u, Ythe disjoint union of the subsets C and C , where1 2

C sC_ S jS and C sS _S . 3.10Ž . Ž .1 u , w w , u , Y 2 u , w w , u , Y

< <If b is an element of C then there are ky S y1 elements c gAA such that1 u, w Y YŽ . < <c w sb, while if bgC then there are ky S such elements. Let n :Y 2 u, w Y

C _S ªR be defined byw , u, Y

< < y1ky S y1 N if bgC ,Ž .u , w Y 1n b s 3.11Ž . Ž .Y y1½ < <ky S N if bgC ,Ž .u , w Y 2

for all bgC_S . Then n is a probability measure on C_S . Thew , u, Y Y w , u, Ycoupling procedure can now be described. Choose c gAA uniformly at randomX X

Ž . Ž . Ž . Ž . Ž .and let c sc unless c w sY ¨ . If c w sY ¨ then choose c w gC_Y X X X YŽ . Ž .S according to the probability distribution n . If c w gC then let c u sw , u, Y Y Y 1 Y

Ž . Ž . Ž . Ž . Ž . Ž .c u unless c u sc w , in which case let c u sY ¨ . If c w gC then letX X Y Y Y 2

y1¡ < < < <c u with probability ky S y1 ky S ,Ž . Ž . Ž .X u , w u , w~c u sŽ .Y y1¢ < <Y ¨ with probability ky S .Ž . Ž .u , w

We now check that each element of AA is chosen with equal probability. LetYŽ .c gAA be the element chosen uniformly at random in step i and let c gAA .X X Y Y

Ž . Ž . Ž .Suppose first that c w gC and c u /Y ¨ . Then c is the result of theY 1 Y Y


Ž . Ž . ² Ž . Ž .: Ž .coupling procedure if and only if a c sc , or b c s Y ¨ , c u and c wX Y X Y Yis the element of C_S chosen. Therefore the probability that c results inw , u, Y Ythis case is

² :w xProb c sc qProb c s Y ¨ , c u ?n c wŽ . Ž . Ž .Ž .X Y X Y Y Y

y1 < < y1sN 1q ky S y1 NŽ .Ž .X u , w Y

sN y1 .Y

Ž . Ž . Ž .Suppose now that c w gC and c u sY ¨ . Then c is the result of theY 1 Y YŽ . Ž . ² Ž . Ž .: Ž .coupling procedure if and only if a c sc , or b c s Y ¨ , c w and c wX Y X Y Y

is the element of C_S chosen. The probability that c occurs is given byw , u, Y Y

y1² :w xProb c sc qProb c s Y ¨ , c w ?n c w sN .Ž . Ž . Ž .Ž .X Y X Y Y y Y

Ž . Ž . Ž .Now let c gAA be such that c w gC and c u /Y ¨ . Then c is the resultY Y Y 2 Y YŽ . Ž . ² Ž . Ž .:of the coupling procedure if and only if a c sc , or b c s Y ¨ , c u , theX Y X y

Ž . Ž . Ž .chosen element of C_S is c w and c u sc u . Therefore probabilityw , u, Y Y Y Xthat c is chosen is given byY

w xProb c scX Y

y1² : < < < <qProb c s Y ¨ , c u ?n c w ? ky S y1 ky SŽ . Ž . Ž . Ž . Ž .Ž .X Y Y y u , w u , w


sN y1 .Y

Ž . Ž . Ž .Finally suppose that c w gC and c u sY ¨ . Then c occurs if and onlyY 2 Y YŽ . Ž . Ž . Ž . Ž .if a c sc , or b c w sY ¨ , the element chosen from C_S is c wX Y X w , u, Y Y

Ž . Ž .and c u sY ¨ . Hence the probability that c is chosen in this case is given byY Y

y1< <w xProb c sc qProb c w sY ¨ ?n c w ? ky SŽ . Ž . Ž . Ž .Ž .X Y X Y Y u , w


sN y1 .Y

Therefore every element of AA is chosen with equal probability. Now weYw Ž X X. xcalculate the contribution which is made to E H X , Y y1 using this procedure.

Ž . Ž . Ž X X. Ž . Ž .If c w /Y ¨ then H X , Y s1. Suppose then that c w sY ¨ . Then weX XŽ . Ž Ž ..choose an element c w from C_S with probability n c w . Suppose firstY w , u, Y Y Y

Ž . < < Ž X X.that c w gC . Then ky S y2 elements of AA give H X , Y s2 and oneY 1 u, w XŽ X X. Ž . Ž X X.gives H X , Y s3. If however c w gC then H X , Y s2 with probabilityY 2

Ž < < .Ž < <.y1 Ž X X. Ž < <.y1ky S y1 ky S and H X , Y s3 with probability ky S .u, w u, w u, w


w Ž X X. xTherefore the contribution made to E H X , Y y1 in this case is

< < < < < < < < < <ky S y1 ky S ky S ky S y1 ky S q1u , w u , w u , w u , w u , w< < < <C ? q C ? ?1 2ž / ž /< <N N N N ky SY X Y X u , w

< < < <ky S y1 ky S 1u , w u , w< < < < < < < <s C ky S q C ky S yŽ . Ž .1 u , w 2 u , wž / ž /N N N N N NX Y X Y X Y

< < < <ky S S _Su , w u , w w , u , Ys yN N NX X Y

< <ky Su , ¨-

NX

-M .

( ) � 4 Ž .Configuration 1, Subcase 2 ii . Here u, ¨ fE. Moreover X ¨ gS andw , u, Y

Ž . < < < < Ž . Ž .Y ¨ fS , so S s S y1. Also X ¨ fS and Y ¨ gS . Itw , u, X w , u, X w , u, Y u, w u, w

Ž < <.follows that N sN q ky S . The set AA may be obtained from AA byX Y u, w Y X< < Ž .deleting the ky S color maps which assign the color Y ¨ to w. Before definingu, w

the coupling procedure we define the family of maps j : C_S ªR, whereq w , u, YqgC , as follows. Given qgC , let1 1

< < y1ky S N if b/q ,Ž .u , w Yj b sŽ .q y1½ < <S _S N if bsq ,Ž .w , u , Y u , w Y

for all bgC_S . Then j is a probability measure on C_S for allw , u, Y q w , u, YqgC , since1

< < < < < <ky S y1 ky S q S _S sN . 3.12Ž .Ž . Ž .w , u , Y u , w w , u , Y u , w Y

We may now state the coupling procedure. Choose c gAA uniformly at randomX XŽ . Ž . Ž . Ž .and let c sc unless c w sY ¨ . If c u gS _S then choose c w gY X X X w , u, Y u, w Y

Ž . Ž . Ž .C _S according to the distribution n and let c u sc u . If c u fSw , u, Y Y Y X X w , u, Y

Ž . Ž .then c u gC . In this case choose c w gC_S according to the probabil-X 1 Y w , u, Y

Ž . Ž . Ž . Ž . Ž .ity distribution j . Let c u sc u if c w /c u and choose c uc Žu. Y X Y X YXŽ . Ž .uniformly at random from S _S if c w sc u . Let us check that eachw , u, Y u, w Y X

element of AA is equally likely to be chosen by this procedure. Denote by c theY XŽ .element of AA chosen uniformly at random in step i and let c gAA . SupposeX Y Y

Ž . Ž .first that c w gC and c u gS _S . Then c is the result of theY 2 Y w , u, Y u, w YŽ . Ž . ² Ž . Ž .: Ž .coupling procedure if and only if a c sc , or b c s Y ¨ , c u and c wX Y X Y Y

is the element of C_S chosen. Therefore the probability that c is chosen isw , u, Y Ygiven by


y1 < < y1sN 1y ky S NŽ .Ž .X u , w Y

sN y1 .Y


Ž . Ž .Suppose next that c w gC and c u gS _S . Then c is chosen ifY 1 Y w , u, Y u, w Y

Ž . Ž . ² Ž . Ž .: Ž .and only if a c sc , or b c s Y ¨ , c u and c w is the element ofX Y X Y YŽ . ² Ž . Ž .:C _S chosen, or c c s Y ¨ , c w , the chosen element of C_S isw , u, Y X Y w , u, Y

Ž . Ž .c w and the chosen element of S _S is c u . Therefore the probabilityY w , u, Y u, w Ythat c is chosen is given byY


y1² : < <qProb c s Y ¨ , c w ?j c w ? S _SŽ . Ž . Ž .Ž .X Y c Žw . Y w , u , y u , wY

sN y1 1qn c w qN y1Ž .Ž .Ž .X Y Y Y

y1 < < y1sN 1q ky S NŽ .Ž .X u , w Y

sN y1 .Y

Ž . Ž .Finally suppose that c w gC_S and c u gC . Then c is chosen ifY w , u, Y Y 1 Y

Ž . Ž . ² Ž . Ž .: Ž .and only if a c sc , or b c s Y ¨ , c u and c w is the chosen elementX Y X Y Yof C_S . Hence the probability that c is chosen in this case is given byw , u, Y Y

² :w xProb c sc qProb c s Y ¨ , c u ?j c wŽ . Ž . Ž .Ž .X Y X Y c Žu. YY

y1 < < y1sN 1y ky S NŽ .Ž .X u , w Y

sN y1 .Y

Therefore every element of AA has equal probability of being chosen. NowYw Ž X X. x Ž X X.the contribution to E H X , Y y1 will be calculated. As usual H X , Y s1 if

Ž . Ž . Ž . Ž . Ž .c w /Y ¨ . Suppose then that c w sY ¨ . If c u gS _S thenX X X w , u, Y u, w

Ž X X. Ž . Ž X X. <H X , Y s 2. If c u g C then H X , Y s 3 with probability S _X 1 w , u, Y

< y1 Ž X X. w Ž X X. xS N , otherwise H X , Y s2. Therefore the contribution to E H X , Y y1u, w Yis

< < < < < < < < < < < <S _S C S _S ky S C S _Sw , u , Y u , w 1 w , u , Y u , w u , w 1 w , u , Y u , wq 1q s qž /N N N N N NX X Y X X Y

< < < < < <ky S C ky S y1Ž .u , w 1 u , w- q

N N NX X Y

< <ky Su , wFNX

FM .

( ) � 4 Ž .Configuration 2, Subcase 1 i . Here u, ¨ gE. Moreover X ¨ fS andw , u, YŽ . < < < < Ž . Ž .Y ¨ fS , so S s S . Also X ¨ fS and Y ¨ fS . Itw , u, X w , u, Y w , u, X u, w , Y u, w , X

follows that N sN . The set AA is obtained from AA by replacing each of theX Y Y X< < ² Ž . Ž .: ² Ž . Ž .:ky S color maps of the form c w , Y ¨ with c w , X ¨ , and replacingw , u, X

< < ² Ž . Ž .: ² Ž . Ž .:each of the ky S color maps of the form Y ¨ , c u with X ¨ , c u . Weu, w , X


now describe the coupling in this case. Choose c gAA uniformly at random andX Xlet

¡c if c w /Y ¨ and c u /Y ¨ ,Ž . Ž . Ž . Ž .X X X~² :c w , X ¨ if c u sY ¨ ,Ž . Ž . Ž . Ž .c s X XY ¢² :X ¨ , c u if c w sY ¨ .Ž . Ž . Ž . Ž .X X

This defines a bijection between AA and AA . Therefore each element of AA isX Y YŽ X X . Ž . Ž . Ž .equally likely to be chosen. Now H X , Y s2 whenever c w sY ¨ or c u sX X

Ž . Ž X X. w Ž X X. xY ¨ . Otherwise H X , Y s1. Therefore the contribution to E H X , Y y1 inthis case is

< < < < < < < <2 ky1 y S q S ky S y1 ky S y1Ž . Ž .w , u , X u , w , X w , u , X u , w , XF2 max ,½ 5N N NX X X

-2 M .( ) � 4 Ž .Configuration 2, Subcase 2 i . Here u, ¨ gE. Moreover X ¨ gS andw , u, Y

Ž . < < < < Ž . Ž .Y ¨ fS , so S s S y1. Also X ¨ fS and Y ¨ fS . Itw , u, X w , u, X w , u, Y u, w , Y u, w , XŽ < < .follows that N sN q ky S y1 . The set AA may be obtained from AA byX Y u, w , X Y X

< < Ž .deleting the ky S y1 color maps which assign the color Y ¨ to w andu, w , X< < ² Ž . Ž .:replacing each of the ky S y1 color maps of the form c w , Y ¨ withw , u, X

² Ž . Ž .: Ž .c w , X ¨ . Recall the sets C and C defined in 3.10 and the probability1 2Ž .distribution n defined in 3.11 . We use a coupling based on that used in SubcaseY

Ž . Ž .2 i for Configuration 1 edges. Choose c gAA uniformly at random. If c w /X X XŽ . Ž . Ž . ² Ž . Ž .:Y ¨ then let c sc unless c u sY ¨ , in which case let c s c w , X ¨ . IfY X X Y XŽ . Ž . Ž .c w sY ¨ then choose c w gC_S according to the probability distribu-X Y w , u, Y

Ž . Ž . Ž . Ž . Ž .tion n . If c w gC then let c u sc u unless c u sc w , in which caseY Y 1 Y X X YŽ . Ž . Ž .let c u sY ¨ . If c w gC then letY Y 2

y1¡ < < < <c u with probability ky S y1 ky S ,Ž . Ž . Ž .X u , w , X u , w , X~c u sŽ .Y y1¢ < <Y ¨ with probability ky S .Ž . Ž .u , w , X

Then each element of AA is equally likely to be the outcome of this procedure,YŽ .following the proof of Subcase 2 i for Configuration 1 edges. The contribution to

w Ž X X. x Ž . Ž . Ž . Ž .E H X , Y y1 will now be calculated. If c w /Y ¨ and c u /Y ¨ thenX XŽ X X. Ž . Ž . Ž X X.H X , Y s1. If c u sY ¨ then H X , Y s2. Finally the contribution toXw Ž X X. x Ž . Ž .E H X , Y y1 from elements c gAA such that c w sY ¨ is given byX X X

< < < <ky S S _Su , w , X u , w , Y w , u , Yy ,N N NX X Y

Ž .following the proof for Subcase 2 i , Configuration 1. Therefore the contribution tow Ž X X. xE H X , Y y1 in this case is given by

< < < < < <2ky1y S q S S _SŽ .w , u , X u , w , X u , w , Y w , u , YyN N NX X Y

< < < <ky S ky Sw , u , X u , w , X-2 max ,½ 5N NX X

-2 M .


( ) � 4 Ž .Configuration 2, Subcase 2 ii . Here u, ¨ gE. Moreover X ¨ gS andw , u, YŽ . < < < < Ž . Ž .Y ¨ fS , so S s S y1. Also X ¨ fS and Y ¨ gS . Itw , u, X w , u, X w , u, Y u, w , Y u, w , X

Ž < < < <.follows that N sN q S y S . The set AA may be obtained from AAX Y w , u, Y u, w , X Y X

< < Ž .by deleting the ky S color maps which assign the color Y ¨ to w and addingu, w , X< < Ž .the ky S color maps which assign the color X ¨ to u. First suppose thatw , u, Y

N sN . Then we couple as follows. Fix a bijection f : C_S ªC_S . LetX Y u, w , X w , u, Yc be chosen uniformly from AA and let c be defined byX X Y

c if c w /Y ¨ ,Ž . Ž .X Xc sY ½ ² :f c u , X ¨ if c w sY ¨ .Ž . Ž . Ž . Ž .Ž .X X

This defines a bijection between AA and AA . Hence each element of AA is equallyX Y YŽ X X . Ž . Ž . Ž X X. Ž .likely to result. Here H X , Y s1 if c w /Y ¨ and H X , Y s3 if c w sX X

Ž . w Ž X X . xY ¨ . Therefore the contribution to E H X , Y y1 in this situation is

< <2 ky SŽ .u , w , X-2 M .

NX

For the remainder of this subcase assume without loss of generality that N )N .X YBefore describing the coupling in this case, we must define two families of maps.

Ž .Recall the sets C and C defined in 3.10 . For qgC let h : AA ªR be de-1 2 1 q Yfined by

¡ y1< < < <S y S N if b u sq ,Ž .Ž .w , u , Y u , w , X Y

y1< <ky S N if b u sX ¨ , b w /q ,Ž . Ž . Ž .Ž .~ w , u , Y Yh b sŽ .qy1< < < <ky S q S _S y1 N if b u sX ¨ , b w sq ,Ž . Ž . Ž .Ž .w , u , Y w , u , Y u , w , Y Y¢

0 otherwise,

for all bgAA . Then h is a probability distribution on AA for all qgC , sinceY q Y 1

< < < < < < < < < <ky S y1 S y S q ky S y1 ky SŽ . Ž . Ž . Ž .w , u , Y w , u , Y u , w , X w , u , Y w , u , Y

< < < <q ky S q S _S y1Ž .w , u , Y w , u , Y u , w , Y

< < < <s ky S y1 ky S y1Ž . Ž .w , u , Y u , w , Y

< < < <q ky S q S _S y1Ž .w , u , Y w , u , Y u , w , Y

< < < < <sN q C y ky S q S _SŽ .Y 1 u , w , Y w , u , Y u , w , Y

sN . 3.13Ž .Y

Now for qgS _S define the map z : AA ªR byw , u, X u, w , X q Y

¡ y1< < < <S y S N if b u sq ,Ž .Ž .w , u , Y u , w , X Y

y1< <ky S N if b u sX ¨ , b w gC ,Ž . Ž . Ž .Ž .~ w , u , Y Y 1z b sŽ .qy1< <ky S q1 N if b u sX ¨ , b w gC ,Ž . Ž . Ž .Ž .w , u , Y Y 2¢

0 otherwise,


for all bgAA . Then z is a probability distribution on AA for all elements q ofY q YS _S , sincew ,u, X u, w , X

< < < < < < < < < < < < < <ky S S y S q C ky S q C ky S q1Ž . Ž . Ž . Ž .w , u , Y w , u , Y u , w , X 1 w , u , Y 2 w , u , Y

< < < < < <s ky S ky S y1 q S _SŽ . Ž .w , u , Y u , w , Y u , w , Y w , u , Y

< < < < < <sN q C y ky S q S _SŽ .Y 1 w , u , Y u , w , Y w , u , Y

sN . 3.14Ž .Y

We may now define the coupling when N )N . Choose c gAA uniformly atX Y X XŽ . Ž . Ž . Ž . Ž .random and let c sc unless c w sY ¨ . If c w sY ¨ and c u gCY X X X X 1

Ž .then choose c gAA according to the probability distribution h . If c u fCY Y c Žu. X 1XŽ .then c u gS _S . Here choose c gAA according to the probabilityX w , u, X u, w , X Y Y

distribution z . We must prove that each element of AA is equally likely to bec Žu. YX

chosen by this procedure. Denote by c the element of AA chosen uniformly atX XŽ . Ž . Ž .random in step i and let c gAA . Suppose first that c u /X ¨ . Then c isY Y Y Y

Ž . Ž .chosen by the coupling procedure if and only if a c sc , or b c sX Y X² Ž . Ž .: w Ž .Y ¨ , c u and c is the element selected from AA with probability h cY Y Y c Žu. YY

Ž . Ž . x Ž .if c u gC and with probability z c otherwise . Note that h c sY 1 c Žu. Y c Žu. YY Y

Ž < < < <. y1 Ž . Ž . Ž < < < <. y1S y S N if c u gC and z c s S y S N ifw , u, Y u, w , X Y Y 1 c Žu. Y w , u, Y u, w , X YY

Ž .c u gS _S , so these two probabilities are the same. Therefore theY w , u, X u, w , Xprobability that c is chosen in this case is given byY

y1² : < < < <w xProb c sc qProb c s Y ¨ , c u ? S y S NŽ . Ž . Ž .X Y X Y w , u , Y u , w , X Y

y1 < < < < y1sN 1q S y S NŽ .Ž .X w , u , Y u , w , X Y

sN y1 .Y

Ž . Ž . Ž .Suppose next that c u sX ¨ and c w gC . Then c is the result of theY Y 1 YŽ . Ž . Ž . Ž . � Ž .4coupling procedure if and only if a c w sY ¨ , c u gC _ c w and c isX X 1 Y Y

w Ž .x Ž . Ž . Ž .the element chosen from AA with probability h c , or b c w sY ¨ ,Y c Žu. Y XXŽ . Ž . w Ž .xc u sc w and c is the element chosen from AA with probability h c ,X Y Y Y c Žw . YY

Ž . Ž . Ž . Ž .or c c w sY ¨ , c u gS _S and c is the element chosen fromX X w , u, X u, w , X Yw Ž .xAA with probability z c . Therefore the probability that c is chosen in thisY c Žu. Y YX

case is given by

Prob c w sY ¨ , c u gC _ c w ?h c� 4Ž . Ž . Ž . Ž . Ž .X X 1 Y c Žu. YX

² :qProb c s Y ¨ , c w ?h cŽ . Ž . Ž .X Y c Žw . YY

qProb c w sY ¨ , c u gS _S ?z cŽ . Ž . Ž . Ž .X X w , u , X u , w , X c Žu. YX

y1 y1 < < < <sN N C y1 ky SŽ . Ž .X Y 1 w , u , Y

< < < < < < < <q ky S q S _S y1 q S _S ky SŽ . Ž .w , u , Y w , u , Y u , w , Y w , u , X u , w , X w , u , Y

y1 y1 < < < < < <sN N N q C y ky S q S _S y1Ž .X Y Y 1 w , u , Y w , u , Y u , w , Y

y1 y1 < < < <sN N N q S y SŽ .X Y Y w , u , Y u , w , X

sN y1 .Y


Ž . Ž . Ž .Finally suppose that c u sX ¨ and c w gC . Then c is chosen by theY Y 2 YŽ . Ž . Ž . Ž .coupling procedure if and only if a c w sY ¨ , c u gC and c is theX X 1 Y

w Ž .x Ž . Ž . Ž . Ž .element of AA chosen with probability h c , or b c w sY ¨ , c u gY c Žu. Y X XX w Ž .xS _S and c is the element of AA chosen with probability z c .w , u, X u, w , X Y Y c Žu. YX

Therefore the probability that c is chosen in this case is given byY

Prob c w sY ¨ , c u gC ?h cŽ . Ž . Ž . Ž .X X 1 c Žu. YX

qProb c w sY ¨ , c u gS _S ?z cŽ . Ž . Ž . Ž .X X w , u , X u , w , X c Žu. YX

y1 y1 < < < < < < < <sN N C ky S q S _S ky S q1Ž . Ž .X Y 1 w , u , Y w , u , X u , w , X w , u , Y

y1 y1 < < < < < <sN N N q C y ky S q S _SŽ .X Y Y 1 w , u , Y w , u , X u , w , X

y1 y1 < < < <sN N N q S y SŽ .X Y Y w , u , Y u , w , X

sN y1 .Y

Therefore every element of AA has equal probability of being the result of theYw Ž X X.procedure. We must now calculate the contribution of the coupling to E H X , Y

x Ž . Ž . Ž X X. Ž . Ž .y1 . If c w /Y ¨ then H X , Y s1. Suppose now that c w sY ¨ andX XŽ . Ž X X.c u gC . Then H X , Y s2 with probability,X 1

< < < < < < y1ky S y1 S y S N ,Ž . Ž .w , u , Y w , u , Y u , w , X Y

Ž X X. < <otherwise H X , Y s3. This follows as there are ky S y1 elements c gAAw , u, Y Y Y

Ž . Ž . Ž < < < <. y1with c u sc u , and each is chosen with probability S y S NY X w , u, Y u, w , X Y

Ž . Ž . Ž .using the probability distribution h . If c w sY ¨ and c u gS _Sc Žu. X X w , u, X u, w , XX

< < Ž . Ž .then there are ky S elements c gAA such that c u sc u . Each isw , u, Y Y Y Y X

Ž < < < <. y1selected with probability S y S N using the probability distributionw , u, Y u, w , X Y

Ž X X.z . Therefore H X , Y s2 with probability,c Žu.X

< < < < < < y1ky S S y S N ,Ž . Ž .w , u , Y w , u , Y u , w , X Y

Ž X X.and H X , Y s3 otherwise. Therefore the contribution of this edge to theŽ X X.expected value of H X , Y y1 is given by

< < < < < < < <C ky S y1 S y SŽ . Ž .1 w , u , Y w , u , Y u , w , X2yž /N NX Y

< < < < < < < <S _S ky S S y SŽ . Ž .w , u , X u , w , X w , u , Y w , u , Y u , w , Xq 2yž /NX

< < < < < < < <ky S ky S S y SŽ . Ž .u , w , X w , u , Y w , u , Y u , w , Xs 2yž /N NX Y

< < < < < <C S y SŽ .1 w , u , Y u , w , XqN NX Y


2< < < < < << < ky S S _S q ky SŽ . Ž .ky S u , w , X u , w , Y w , u , Y w , u , Yu , w , Xs qN N NX X Y

< < < < < <C S y SŽ .1 w , u , Y u , w , XqN NX Y

< < < < < < < < < <ky S ky S ky S ky S q1 y CŽ . Ž . Ž .u , w , X w , u , Y u , w , X w , u , Y 1s qN N NX X Y

< < < < < < < <ky S ky S N q S y SŽ . Ž .u , w , X w , u , Y Y w , u , Y u , w , Xs qN N NX X Y

< < < <ky S ky Su , w , X w , u , Ys qN NX Y

< < < <ky S ky Su , w , X w , u , YF2 max ,½ 5N NX Y

-2 M .

( ) � 4 Ž .Configuration 2, Subcase 2 iii . Here u, ¨ gE. Moreover X ¨ gS andw , u, YŽ . < < < < Ž . Ž .Y ¨ fS , so S s S y1. Also X ¨ gS and Y ¨ fS . Itw , u, X w , u, X w , u, Y u, w , Y u, w , X

Ž . Ž < < < <.follows that N sN q2 ky1 y S q S . The set AA may be ob-X Y w , u, X u, w , X Y< <tained from AA by deleting the ky S y1 color maps which assign the colorX u, w , X

Ž . < < Ž .Y ¨ to w and deleting the ky S y1 color maps which assign the color Y ¨w , u, XŽ .to u. Recall the sets C and C defined in 3.10 and the probability distribution n1 2 Y

Ž . X Xdefined in 3.11 . The set C and probability distribution n may be defined by2 Yexchanging the roles of w and u in these definitions. That is, let

C X sS _S . 3.15Ž .2 w , u , Y u , w , Y

The sets C and C X form a partition of C_S . The probability distribution nX:1 2 u, w , Y Y

C _S ªR is defined as follows:u, w , Y

< < y1ky S y1 N if bgC ,Ž .w , u , Y Y 1Xn b s 3.16Ž . Ž .Y Xy1½ < <ky S N if bgC ,Ž .w , u , Y Y 2

for all bgC_S . Before we can describe the coupling we must define twou, w , Y< <more families of maps. Fix a C -cycle s which permutes the elements of C . For1 1

Ž � 4.qgC let the map c : C_ S j q ªR be defined by1 q w , u, Y

¡ y1< <ky S q1 N if bgC ,Ž .u , w , Y Y 2

y1 y1~ < <c b s 3.17ky S N if bgC _ q , qs ,Ž . Ž .� 4Ž .q u , w , Y Y 1

y1 y1¢ < < < <ky2 S q S N if bsqs ,Ž .u , w , Y w , u , Y Y


Ž � 4. Ž � 4.for all bgC_ S j q . Then c is a probability measure on C_ S j qw , u, Y q w , u, Yfor all qgC , since1

< < < < < < < < < < < <C ky S q1 q C y2 ky S q ky2 S q SŽ . Ž . Ž . Ž .2 u , w , Y 1 u , w , Y u , w , Y w , u , Y

< < < < < < < <s ky S ky S q C y ky SŽ . Ž . Ž .w , u , Y u , w , Y 1 w , u , Y

< < < < < <sN q C q C y ky SŽ .Y 1 2 w , u , Y

sN . 3.18Ž .Y

X Ž � 4.We define c : C_ S j q ªR for all qgC by exchanging the roles of wq u, w , Y 1

and u, and replacing sy1 by s in the definition of c . That is,q

¡ y1 X< <ky S q1 N if bgC ,Ž .w , u , Y Y 2X y1~ < <c b s 3.19� 4ky S N if gC _ q , qs ,Ž . Ž .Ž .q w , u , Y Y 1

y1¢ < < < <ky2 S q S N if bsqs ,Ž .w , u , Y u , w , Y Y

Ž � 4. X Ž � 4.for all bgC_ S j q . Then c is a probability measure on C_ S j qu, w , Y q u, w , Y

for all qgC , by adapting the proof for c .1 qThe coupling may now be defined. Choose c gAA uniformly at random and letX X

Ž . Ž . Ž . Ž . Ž . Ž . Ž . Xc sc unless c w sY ¨ or c u sY ¨ . If c w sY ¨ and c u gCY X X X X X 2Ž . Ž .then choose c w gC_S according to the distribution n and let c u sY w , u, Y Y Y

Ž . Ž . Ž . Ž . Ž .c u . If c w gC and c u sY ¨ then choose c u gC_S accordingX X 2 X Y u, w , YX Ž . Ž . Ž . Ž .to the distribution n and let c w sc w . Now suppose that c w sY ¨ andY Y X X

Ž . Ž . Ž � Ž .4.c u gC . Choose c w gC_ S j c u according to the distributionX 1 Y w , u, Y XŽ . Ž . Ž . Ž . Ž .c and let c u sc u . Finally suppose that c w gC and c u sY ¨ .c Žu. Y X X 1 XX

Ž . Ž � Ž .4.Here choose c u gC_ S j c w according to the distribution cY u, w , Y X c Žv .9xŽ . Ž .and let c w sc w . Let us check that each element of AA is equally likely to beY X Y

the result of this coupling procedure.Ž .Denote by c the element of AA chosen uniformly at random in step i and letX X

Ž . Ž .c sAA . Now c is the result of the coupling if and only if a c sc , or bY Y Y X Y² Ž . Ž .: Ž . wc s c w , Y ¨ and c u is the element of C_S chosen with probabilityX Y Y u, w , Y

XŽ Ž .. Ž . XŽ Ž .. Ž . x Ž .n c u if c w gC , or probability c c u if c w gC , or c c sY Y Y 2 c Žw . Y Y 1 XY² Ž . Ž .: Ž . wY ¨ , c u and c w is the element of C_S chosen with probabilityY Y w , u, Y

Ž Ž .. Ž . X Ž Ž .. Ž . xn c w if c u gC , or with probability c c w if c u gC . SupposeY Y Y 2 c Žu. Y Y 1YŽ . Ž . Xfirst that c w gC and c u gC . Then the probability that c is the result ofY 2 Y 2 Y

the procedure is given by

X² :w xProb c sc qProb c s c w , Y ¨ ?n c uŽ . Ž . Ž .Ž .X Y X Y Y Y

² :qProb c s Y ¨ , c u ?n c wŽ . Ž . Ž .Ž .X Y Y Y

< < < <ky S ky Sw , u , Y u , w , Yy1sN 1q qX ž /N NY Y

sN y1 ,Y


Ž . Ž .as required. Suppose next that c w gC and c u gC . Then the probabilityY 2 Y 1that c is chosen is given byY

X² :w xProb c sc qProb c s c w , Y ¨ ?n c uŽ . Ž . Ž .Ž .X Y X Y Y Y

² :qProb c s Y ¨ , c u ?c c wŽ . Ž . Ž .Ž .X Y c Žu. YY

< < < <ky S y1 ky S q1w , u , Y u , w , Yy1sN 1q qX ž /N NY Y

sN y1 .Y

Ž . Ž . XIf c w gC and c u gC then essentially the same calculation as theY 1 Y 2previous one shows that c is chosen with probability N y1. Finally suppose thatY Y

Ž . Ž . Ž . Ž .c w gC and c u gC . If c u /c w s then the probability that c isY 1 Y 1 Y Y Ychosen is given by

X² :w xProb c sc qProb c s c w , Y ¨ ?c c uŽ . Ž . Ž .Ž .X Y X Y c Žw . YY

² :qProb c s Y ¨ , c u ?c c wŽ . Ž . Ž .Ž .X Y c Žu. YY

< < < <ky S ky Sw , u , Y u , w , Yy1sN 1y qX ž /N NY Y

sN y1 .Y

Ž . Ž .However if c u sc w s then the probability that c is chosen is given byY Y YX² :w xProb c sc qProb c s c w , Y ¨ ?c c w sŽ . Ž . Ž .Ž .X Y X Y c Žw . YY

y1² :qProb c q Y ¨ , c u ?c c u sŽ . Ž . Ž .Ž .X Y c Žu. YY

< < < < < < < <ky2 S q S ky2 S q Sw , u , Y u , w , Y u , w , Y w , u , Yy1sN 1q qX ž /N NY Y

sN y1 .Y

Therefore every element of AA is equally likely to be the result of the coupling.Yw Ž X X. xFinally the contribution of this coupling to E H X , Y y1 must be calculated.

Ž X X . Ž . Ž . Ž . Ž .This is not difficult, as H X , Y s2 whenever c w sY ¨ or c u sY ¨ , andX XŽ X X. w Ž X X.H X , Y s1 otherwise. Therefore the contribution of this edge to E H X , Y y

x1 is given by

< < < < < < < <2 ky1 y S q S ky S y1 ky S y1Ž . Ž .w , u , X u , w , X w , u , X u , w , XF2 max ,½ 5N N NX X X

-2 M .This completes the case analysis. We have established that the maximum

contribution of any Configuration 1 edge is less than M and the maximumŽ .contribution of any Configuration 2 edge is less than 2 M. It follows by 3.6 that

w Ž X X. xthe total contribution of all Configurations 1 and 2 edges to E H X , Y y1 isŽ .less than d Dy1 M. The only other edges which affect the Hamming distance are

� 4 Ž X X.the d edges of the form ¨ , w , where H X , Y s0 with probability 1. Therefore


Ž X X .the expected value of H X , Y after one step of the coupling may be boundedabove as follows,

dX XE H X , Y y1 - M Dy1 y1Ž . Ž .Ž .

< <E22d K y 3Dy2 kq2 Dy1Ž . Ž .Ž .

sy . 3.20Ž .< <E kyDq1 kyDŽ . Ž .

Ž .Now the right-hand side of 3.20 is a decreasing function of k which is negativewhen kG2D and positive when Dq1FkF2Dy1. Let b be defined by

22d k y 3Dy2 kq2 Dy2Ž . Ž .Ž .bs1y . 3.21Ž .

< <E kyD kyDq1Ž . Ž .

Ž . w Ž X X .xThen 3.20 implies that E H X , Y -b-1 whenever kG2D. Hence whenŽ V .kG2D the Markov chain MM C is rapidly mixing with mixing time

log n«y1 kyD kyDq1Ž . Ž . Ž . y1< <t « F s E log n«Ž . Ž .1 221yb d k y 3Dy2 kq2 Dy1Ž . Ž .Ž .Ž Ž ..by Theorems 2.1 and 2.2. It follows that MM V G is rapidly mixing for kG2D1 k

with the same mixing time. Finally suppose that G is D-regular. Then dsD andŽ .Eq. 3.21 becomes

222 k y 3Dy2 kq2 Dy1Ž . Ž .Ž .bs1y .

n kyD kyDq1Ž . Ž .

Ž Ž ..Therefore the mixing time of MM V G in this case satisfies1 k

kyD kyDq1Ž . Ž . y1T « F n log n« , 3.22Ž . Ž . Ž .222 k y 3Dy2 kq2 Dy1Ž . Ž .Ž .as stated. B

Remark 3.2. The proof given in Theorem 3.2 uses the path coupling method toŽ Ž ..show that the Markov chain MM V G is rapidly mixing. The fact that this proof1 k

is quite technical suggests that a proof of the same result by the standard couplingmethod might be impossibly complex.

� 4Remark 3.3. Let es w, u gE be an edge chosen by the transition procedure ofŽ Ž .. Ž .the Markov chain MM V G . The set AA e used in the transition procedure has1 k X

Ž 2 .O k elements. When k is large it may be too expensive to form the entire setŽ .AA e simply in order to choose an element uniformly at random from it. RecallX

Ž .the probability distribution n defined in 3.11 , Theorem 3.2. Consider theXfollowing procedure:

( ) Ž . Ž Ž ..i choose a color c w from C_S with probability n c w ,w , u, X X( ) Ž . Ž � Ž .4.ii choose a color c u from C_ S j c w uniformly at random.u, w , X


Ž . Ž . Ž .Now n b measures the proportion of elements cgAA e such that c w sb. ItX XŽ . Ž . Ž .follows from 3.3 and 3.11 that this procedure produces elements of AA eX

uniformly at random. Assuming that the set C is linearly ordered, the set differ-Ž .ence of two sets of size at most k can be formed in O k operations using

membership arrays. Moreover, linear algorithms exist for selecting elements fromŽ .sets of size O k according to a given probability distribution. See, for example,

w x14 . Therefore this is an efficient implementation of the transition procedure ofŽ Ž ..MM V G .1 k

Ž Ž ..Remark 3.4. In Theorem 3.2 the Markov chain MM V G was shown to be1 krapidly mixing for kG2D, where D is the maximum degree of the graph G. The

< < < <bound on the mixing time of the chain is proportional to E rd , where E is thenumber of edges and d is is the minimum degree of G. Therefore the chain maybe D times more rapidly mixing for a D-regular graph than for a graph withmaximum degree D and minimum degree 1. We can improve this situation byembedding the graph G in a D-regular multigraph GX with vertex set V and edgemultiset EX, constructed as follows. Suppose that ¨ is a vertex with degreeŽ . Ž . Ž Ž .. � 4d ¨ -D. If Dyd ¨ is even we add Dyd ¨ r2 self-loop edges ¨ , ¨ of weight

Ž . Ž Ž . .2, and if Dyd ¨ is odd then we add Dyd ¨ y1 r2 self-loop edges of weight 2Ž Ž X..and one of weight 1. Edges in E have weight 2. Let MM V G denote the2 k

Ž X.Markov chain with state space V G and with transitions governed by thekfollowing procedure: choose an edge egEX with probability proportional to its

Ž Ž ..weight. If egE then perform one transition of MM V G on the edge e. If1 k� 4es ¨ , ¨ is a self-loop edge then perform one step of the heat-bath JSS chain on

Ž Ž X.. Ž .the vertex ¨ . Then MM V G can be thought of as acting on V G . The proof of2 k kŽ Ž ..Theorem 3.2 may be extended to prove that MM V G is rapidly mixing for2 k

kG2D. If 2DFkF3Dy1 then the mixing time is bounded above by

kyD kyDq1Ž . Ž . y1n log n« .Ž .222 k y 3Dy2 kq2 Dy1Ž . Ž .Ž .

Ž . Ž Ž ..This is the same bound as that given in 3.22 for the mixing time of MM V G1 kŽ Ž ..when the graph G is D-regular. If kG3D then the mixing time of MM V G is2 k

Ž y1 .bounded above by n log n« . This may be reduced to

Dq1 kyD kyDq1Ž . Ž . Ž . y1n log n«Ž .222D k y 3Dy2 ky2 Dy1Ž . Ž .Ž .

by altering the transition procedure so that edges are chosen from EX uniformly atrandom. The details are omitted.

We conclude this section with a comparison of the mixing time of the new chainwith the mixing time of the heat bath JSS chain. We will assume that G isD-regular. Note that, by Remark 3.4, the comparison is also valid for nonregulargraphs with maximum degree D whenever 2DFkF3Dy1. By Theorem 3.2 the


Ž . Ž Ž ..mixing time t « of MM V G satisfies1 1 k

kyD kyDq1Ž . Ž . y1t « F n log n« .Ž . Ž .1 222 k y 3Dy2 kq2 Dy1Ž . Ž .Ž .

We shall not compare the new chain directly with the heat bath JSS chain asŽ Ž ..performing one transition of MM V G is more complicated than performing one1 k

Ž Ž ..transition of the heat bath JSS chain. However, a transition of MM V G is at1 kmost as complicated as two steps of the heat bath JSS chain. Suppose that ks2D.

Ž .By defining a suitable coupling one can show that, for any pair X, Y gV=V, theŽ .probability that the Hamming distance changes after one step is at least 1r nD

Ž .unless X, Y are a so-called stuck pair, one in which no changes in the Hammingdistance is possible. One may define a coupling on stuck pairs so as to guarantee

Ž .that they are not stuck in the next step. Let t « denote the mixing time of the2 Jtwo-step JSS chain. Then the probability that the Hamming distance changes in

Ž .two steps is at least 1r nD . Therefore by the second part of Theorem 2.2,

t « F3 Dn3 log «y1 ,Ž . Ž .2 J

Ž Ž ..when ks2D. By Theorem 3.2 the mixing time of MM V G satisfies1 2D

D Dq1Ž . y1t « F n log n« .Ž . Ž .1 4

The ratio r of these upper bounds is

12n2 log «y1Ž .rs .y1Dq1 log n«Ž . Ž .

U Ž 2 .Therefore the JSS chain is V n times slower than the new chain when ks2D.Suppose now that k)2D. Again we compare the two-step JSS chain with the newchain. Let r be the ratio of the upper bound on the mixing time of the JSS chain

Ž .with twice the upper bound on the mixing time of the new chain, as given in 3.2and Theorem 3.2. The ratio is

22k y 3Dy2 kq2 Dy1Ž . Ž .Ž .rs

ky2D kyDq1Ž . Ž .ky2Dq2

s1q ,ky2D kyDq1Ž . Ž .

that is, always greater than 1. For example, if Ds4 and ks2Dq1 then the JSSchain is 50% slower than the new chain eën allowing the JSS chain two steps foreach step of the new chain.


4. OPTIMALITY OF THE COUPLINGS

In this section we prove that the couplings used in Theorem 3.2 are optimal, in theŽ X X.sense that they minimize the expected value of H X , Y . Therefore we cannot

Ž Ž ..hope to improve the analysis of the Markov chain MM V G using the coupling1 kmethod applied to Hamming distance.

The optimality of the couplings used in Theorem 3.2 will be established asfollows: each coupling will be related to a solution of an associated transportation

w xproblem, which will be shown to be optimal. This idea is explored further in 5 ,where the techniques of this section are generalized. First we give a very brief

w xdefinition of a transportation problem. For more details see, for example, 11 . Letm and n be positive integers and let K be an m=n matrix of nonnegativeintegers. Let a be a vector of m positive integers and let b be a vector of npositive integers such that Ým a sÝn b sN. An m=n matrix Z of nonnega-is1 i js1 jtive numbers is a solution of the transportation problem defined by a, b, and K ifÝn Z sa for 1F iFm and Ým Z sb for 1F jFn. The cost of this solutionjs1 i, j i is1 i, j jis measured by Ým Ýn K Z . An optimal solution of this transportation prob-is1 js1 i, j i, jlem is a solution which minimizes the cost. The entries of an optimal solution areall integers. The elements of a are called row sums and the elements of b arecalled column sums. Efficient methods exist for solving transportation problems.

Ž w x.One such is the Hungarian method see, for example, 11, Chap. 20 .Ž Ž ..We now describe how the couplings defined for MM V G give rise to solution1 k

of a related transportation problem. Let X and Y be two colorings which differ� 4 � 4only at vertex ¨ and let es w, u be the chosen edge, where w, ¨ gE and u/¨.

Ž Ž ..A coupling for MM V G at the edge e defines a probability distribution f :1 kŽ . Ž . Ž . Ž .AA e =AA e ªR on the joint probability space AA e =AA e . As in Section 3X Y X Y

Ž .write AA for AA e , similarly write AA , N , and N . ThenX X Y X Y

X Xf c , c sProb X sX , Y sY .Ž .X Y eª c eª cX Y

Ž . y1 Ž .The map f satisfies Ý f c , c sN for all c gAA and Ý f c , cc g AA X Y X X X c g AA X YY Y X Xy1 � 4sN for all c gAA . Let h: AA =AA ª 0, 1, 2 be defined byY Y Y X Y

0 if c sc ,¡ X Y~h c , c sŽ . 2 if c w /c w and c u /c u ,Ž . Ž . Ž . Ž .X Y X Y X Y¢1 otherwise.

Ž X X.Then the expected value of H X , Y y1 which results from this coupling isgiven by

f c , c h c , c .Ž . Ž .Ý Ý X Y X Yc gAA c gAAX YX Y

Refer to this quantity as the cost of the coupling. Let a be the N -dimensionalXvector with each entry equal to N and let b be the N -dimensional vector withY Yeach entry equal to N . Let K be the matrix with N rows and N columns,X X Y

Ž .corresponding to the elements of AA and AA , respectively, such that the c , cX Y X YŽ . Ž .entry of K is h c , c . Let Z be the N =N matrix whose c , c entry equalsX Y X Y X Y

Ž .N N f c , c . Then Z is a solution of the transportation problem defined by a,X Y X Y


b, and K. An optimal solution of this transportation problem corresponds to an� 4optimal coupling at the edge w, u . The cost of the optimal solution equals N NX Y

times the cost of the optimal coupling.We may now prove that the couplings defined in the proof of Theorem 3.2 are

optimal. We do this by tracing the steps which would be performed by theHungarian method with input corresponding to the given couplings. We follow the

w xdescription of the Hungarian method given in 11, Chap. 20 . Note that it takes atmost two steps of the Hungarian method to establish optimality in each case.

Theorem 4.1. The couplings used in the proof of Theorem 3.2 are optimal.

Ž .Proof. Let X and Y be two elements of V G which differ just at the vertex ¨ .k� 4 � 4Let es w, u be an edge in G such that u/¨ and w, ¨ gE. In the proof of

� 4 � 4Theorem 3.2 we consider two types of edges: if u, ¨ fE then w, u is aConfiguration 1 edge, otherwise it is a Configuration 2 edge. Five subcases areidentified for Configuration 1 edges and four subcases are identified for Configura-tion 2 edges, and a coupling is described for each. Recall the notation established

Ž .in Theorem 3.2. Fix a linear ordering on the elements of C such that x ¨ andŽ .y ¨ appear last. Unless otherwise stated, the sets AA and AA are orderedX Y

lexicographically with respect to this order. This gives an ordering on the rows andŽ . Ž .columns of the N =N matrix of costs K, with c , c entry equal to h c , cX Y X Y X Y

Ž .for all c , c gAA =AA . We now show that each coupling is optimal by tracingX Y X Ythe steps which would be performed by the Hungarian method. Throughout the

Ž .proof the i, i entries of a matrix are called diagonal entries, even if the matrix isnot square.

( ) � 4 Ž . Ž .Configuration 1, Subcase 1 i . Here u, ¨ fE, X ¨ fS , Y ¨ fS ,w , u, Y w , u, XŽ . Ž .X ¨ fS , and Y ¨ fS . In this case N sN and the set AA is obtainedu, w u, w X Y Y

² Ž . Ž .:from the set AA by replacing the color map Y ¨ , X ¨ by the color mapX² Ž . Ž .: ² Ž . Ž .:X ¨ , Y ¨ and replacing all other color maps of the form Y ¨ , c u by² Ž . Ž .:X ¨ , c u . The square matrix K of costs has the following form: the first

Ž < < .N y ky S y1 rows have a zero element on the diagonal and all otherX u, w< <entries either 1 or 2. The next ky S y2 rows have a 1 on the diagonal, a 2 inu, w

the last column and every other entry is equal to 1 or 2. Each entry of the final rowis either 1 or 2, with a 2 in the last column. Let K Ž1. denote the matrix obtained

< <from K by first subtracting 1 from the last ky S y1 rows and then subtractingu, wŽ . Ž1.1 from the last column. The i, i entry of K is zero for 1F iFN . Make each ofX

these zero elements an independent zero with multiplicity N . This set of N 2X X

independent zeros in K Ž1. gives an optimal solution of the original transportationŽ .problem which corresponds exactly to the coupling described for Subcase 1 i ,

Configuration 1 in the proof of Theorem 3.2.

( ) � 4 Ž . Ž .Configuration 1, Subcase 1 ii . Here u, ¨ fE, X ¨ fS , Y ¨ fS ,w , u, Y w , u, XŽ . Ž .X ¨ fS , and Y ¨ gS . In this case N sN q1 and the set AA is obtainedu, w u, w X Y Y

² Ž . Ž .:from the set AA by deleting the color map Y ¨ , X ¨ and replacing all otherX² Ž . Ž .: ² Ž . Ž .:color maps of the form Y ¨ , c u by X ¨ , c u . The matrix K of costs has the

Ž < <.following form: the first N y ky S rows have a zero element on the diagonalX u, w< <and every other entry is either 1 or 2. The next ky S y1 rows have a 1 on theu, w


diagonal and every other entry is either 1 or 2. Every entry of the final row is either1 or 2. Let K Ž1. be the matrix obtained from K by subtracting 1 from the last

< < Ž1.ky S rows. There is a zero in every row and every column of K . In particular,u, wŽ .there is a zero in the i, i position for 1F iFN . Make each of these elements anY

independent zero with multiplicity N . Each entry of the last row of K Ž1. is eitherY0 or 1, and the entry in the column corresponding to c is zero if and only ifY

Ž . Ž .c u sX ¨ . Make each of these zero elements an independent zero with multi-Y< < Ž1.plicity 1. This puts ky S y1 independent zeros in the last row of K . A setw , u, Y

of covering lines for K Ž1. is formed by taking the all columns which contain a zeroentry in the final row, together with all rows which contain a zero entry in anyother column. No independent zero lies on the intersection of two covering lines,so the independent set of zeros is maximal for K Ž1.. The minimum uncoveredelement of K Ž1. is 1. Let K Ž2. be the matrix obtained from K Ž1. by subtracting 1from every uncovered element and adding 1 to every element which is coveredtwice. The last row of K Ž2. consists entirely of zeros. Therefore the set ofindependent zeros can be increased by adding the remaining elements of the lastrow as independent zeros with multiplicity 1. This set of N N independent zerosX Yin K Ž2. gives an optimal solution of the original transportation problem which

Ž .corresponds exactly to the coupling given for Subcase 1 ii , Configuration 1 in theproof of Theorem 3.2.

( ) � 4 Ž . Ž .Configuration 1, Subcase 1 iv . Here u, ¨ fE, X ¨ fS , Y ¨ fS ,w , u, Y w , u, XŽ . Ž .X ¨ gS , and Y ¨ gS . In this case N sN and the set AA is obtainedu, w u, w X Y Y

² Ž . Ž .:from the set AA by replacing all color maps of the form Y ¨ , c u byX² Ž . Ž .:X ¨ , c u . The square matrix K of costs has the following form: the first

Ž < <.N y ky S rows have a zero element on the diagonal and all other entriesX u, w< <either 1 or 2. The final ky S rows have a 1 on the diagonal and every otheru, w

entry is equal to 1 or 2. Let K Ž1. denote the matrix obtained from K by subtracting< < Ž . Ž1.1 from the last ky S rows. The i, i entry of K is zero for 1F iFN . Makeu, w X

each of these zero elements an independent zero with multiplicity N . This set ofXN 2 independent zeros in K Ž1. gives an optimal solution of the original transporta-X

Ž .tion problem which corresponds exactly to the coupling described for Subcase 1 iv ,Configuration 1 in the proof of Theorem 3.2.

( ) � 4 Ž . Ž .Configuration 1, Subcase 2 i . Here u, ¨ fE, X ¨ gS , Y ¨ fS ,w , u, Y w , u, XŽ . Ž . Ž < < .X ¨ fS , and Y ¨ fS . In this case N sN q ky S y1 and the setu, w u, w X Y u, w

Ž < < .AA is obtained from the set AA by deleting the ky S y1 color maps whichY X u, wŽ .assign the color Y ¨ to w. The matrix K of costs has the following form: the first

Ž < < .N y ky S y1 rows have a zero element on the diagonal and every otherX u, w< <entry is either 1 or 2. Every entry of the last ky S y1 rows is either 1 or 2,u, w

with every row containing at least one entry which equals 1. Let K Ž1. be the matrix< <which is obtained from K by subtracting 1 from each of the last ky S y1u, w

rows. Now K Ž1. has a zero in every row and every column. In particular, there is aŽ . Ž1.zero in the i, i position of K for 1F iFN . Make each of these zero elementsY

an independent zero with multiplicity N . Now consider the row corresponding toYŽ . Ž . Ž .c where c w sY ¨ . If c u gS then this row has a zero entry only inX X X w , u, Y

Ž . Ž . Ž .the columns associated with c where c u sc u and c w gC_S . IfY Y X Y w , u, YŽ . Ž .c u fS then c u gC and the row has zero entries only in the columnsX w , u, Y X 1


Ž . Ž . Ž . Ž � Ž .4.associated with c where c u sc u and c w gC_ S j c u . LetY Y X Y w , u, Y X< <each of these zero entries be independent zeros with multiplicity ky S y1.u, w

< < Ž .This places N y C independent zeros in the row if c u gS and placesY 2 X w , u, YŽ < < < < . Ž .N y C qky S y1 independent zeros in the row if c u gC . A set ofY 2 u, w X 1

covering lines for K Ž1. is formed by taking all columns with a zero entry in the final< <ky S y1 rows, together with all rows with a zero entry in any other column.u, w

No independent zero lies on the intersection of two covering lines, so the set ofindependent zeros is maximal for K Ž1.. The minimum uncovered element of K Ž1. is1. Let K Ž2. be the matrix formed by subtracting 1 from all uncovered elements ofK Ž1. and adding 1 to all elements of K Ž1. which are covered twice. The set ofindependent zeros can now be extended as all uncovered elements in the last

< < Ž2. Ž2.ky S y1 rows of K are 0. Consider the row of K associated with theu, wŽ . Ž .element c where c w sY ¨ . Make each of the following elements of this rowX X

an independent zero with multiplicity 1: those in a column associated with cYŽ . Ž . Ž . Ž .where c u sY ¨ and c w gC . If c u gC then also make the element inY Y 2 X 1

² Ž . Ž .:the column associated with c s c u , Y ¨ an independent zero with multiplic-Y X< <ity ky S y1. Then every row now has N independent zeros. This set ofu, w Y

N N independent zeros in K Ž2. gives an optimal solution of the original trans-X YŽ .portation problem which corresponds exactly to the coupling for Subcase 2 i ,

Configuration 1 given in Theorem 3.2.

( ) � 4 Ž . Ž .Configuration 1, Subcase 2 ii . Here u, ¨ fE, X ¨ gS , Y ¨ fS ,w , u, Y w , u, XŽ . Ž . Ž < <.X ¨ fS , and Y ¨ gS . In this case N sN q ky S and the set AA isu, w u, w X Y u, w Y

Ž < <.obtained from the set AA by deleting the ky S color maps which assign theX u, w

Ž .color Y ¨ to w. The matrix K of costs has the following form: the first N yXŽ < <.ky S rows have a zero element on the diagonal and every other entry isu, w

< <either 1 or 2. Every entry of the last ky S rows is either 1 or 2, with every rowu, wcontaining at least one entry which equals 1. Let K Ž1. be the matrix obtained from

< < Ž1.K by subtracting 1 from each of the last ky S rows. Now K has a zero inu, wŽ .every row and every column. In particular there is a zero in the i, i position for

1F iFN . Let each of these elements be an independent zero with multiplicity ofYŽ . Ž1.N . Recall the probability measure n defined in 3.11 . Consider the row of KY Y

Ž . Ž . Ž .associated with c where c w sY ¨ . If c u gS then every element ofX X X w , u, Ythe row is either 0 or 1, and the entry in the column associated with c is zero ifY

Ž . Ž .and only if c u sc u . Let these zero entries be independent zeros withY XŽ .multiplicity n c N . This places N independent zeros in the row correspondingY Y Y

Ž .to c . Now suppose that c u gC . Then every element in the row associatedX X 1with c is either 0 or 1 and the entry in the column associated with c is zero ifX Y

Ž . Ž . Ž . Ž � Ž .4.and only if c u sc u and c w gC_ S j c u . Let each of theseY X Y w , u, Y X< < Ž .zero elements be an independent zero with multiplicity ky S . By 3.12 , thisu, w

< <gives N y S _S independent zeros in the row associated with c . A set ofY w , u, Y u, w Xcovering lines for K Ž1. is formed by taking all columns associated with c gAAY Y

Ž . Ž . Ž � Ž .4.where c u gC and c w gC_ S j c u , together with all rows whichY 1 Y w , u, Y Ycontain a zero entry in any other column. No independent zero lies on theintersection of two covering lines, so the set of independent zeros is maximal forK Ž1.. The minimum uncovered element of K Ž1. is 1. Let K Ž2. be the matrix formedby subtracting 1 from all uncovered elements of K Ž1. and adding 1 to all elements


of K Ž1. which are covered twice. The set of independent rows can now be extended< < Ž2.as all uncovered elements in the last ky S rows of K are 0. Consider the rowu, w

Ž2. Ž . Ž . Ž .of K associated with the element c where c w sY ¨ and c u gC . MakeX X X 1each of the following elements of this row an independent zero with multiplicity 1:

Ž . Ž . Ž .those in a column associated with c where c w sc u and c u gY Y X YS _S . There are now N independent zeros in this row. This set of N Nw , u, Y u, w Y X Yindependent zeros in K Ž2. gives an optimal solution for the original transportation

Ž .problem which corresponds exactly to the coupling for Subcase 2 ii , Configura-tion 1 given in Theorem 3.2.

( ) � 4 Ž . Ž .Configuration 2, Subcase 1 i . Here u, ¨ gE, X ¨ fS , Y ¨ fS ,w , u, Y w , u, XŽ . Ž .X ¨ fS , and Y ¨ fS . In this case N sN and the set AA is obtainedu, w , Y u, w , X X Y Y

² Ž . Ž .:from AA by replacing each color map of the form c w , Y ¨ with the color mapX² Ž . Ž .: ² Ž . Ž .:c w , Y ¨ , and replacing every color map of the form Y ¨ , c u with² Ž . Ž .:X ¨ , c u . The matrix K of costs has the following form: the row correspondingto c has a 0 on the diagonal and every other entry equal to 1 or 2, unlessX

Ž . Ž . Ž . Ž .c w sY ¨ or c u sY ¨ . For these rows there is a 1 on the diagonal andX Xevery other entry is equal to 1 or 2. Let K Ž1. be the matrix obtained from K by

Ž . Ž .subtracting 1 from each of the rows associated with c where c w sY ¨ orX XŽ . Ž . Ž1.c u sY ¨ . Now K has a zero in every row and every column. In particular,X

Ž .there is a zero in the i, i position for 1F iFN . Let each of these elements be anXindependent zero with multiplicity N . This set of N 2 independent zeros in K Ž1.

X Xgives an optimal solution of the original transportation problem which corresponds

Ž .exactly to the coupling described for Subcase 1 i , Configuration 2 in the proof ofTheorem 3.2.

( ) � 4 Ž . Ž .Configuration 2, Subcase 2 i . Here u, ¨ gE, X ¨ gS , Y ¨ fS ,w , u, Y w , u, X

Ž . Ž . Ž < < .X ¨ fS , and Y ¨ fS . In this case N sN q ky S y1 and theu, w , Y u, w , X X Y u, w , X< <set AA may be obtained from AA by deleting the ky S y1 color maps whichY X u, w , X

Ž . < <assign the color Y ¨ to w and replacing the ky S y1 color maps of thew , u, X² Ž . Ž .: ² Ž . Ž .:form c w , Y ¨ with c w , X ¨ . The matrix K costs has the following form:

the row corresponding to c has a 0 on the diagonal and every other entry equal toXŽ . Ž . Ž . Ž . Ž . Ž .1 or 2, unless c w sY ¨ or c u sY ¨ . If c u sY ¨ then the row has a 1X X X

Ž . Ž .on the diagonal and every other entry is equal to 1 or 2. If c w sY ¨ then everyXentry in the row is either 1 or 2, with at least one entry equal to 1. Let K Ž1. be thematrix obtained from K by subtracting 1 from the row corresponding to cX

Ž . Ž . Ž . Ž . Ž1.whenever c w sY ¨ or c u sY ¨ . Then every row and column of KX XŽ . Ž1.contains a zero entry. In particular, the i, i entry of K is zero for 1F iFN .Y

Make each of these entries an independent zero with multiplicity N . TheYŽ .remainder of the proof follows just as in the proof of Subcase 2 i for Configuration

Ž .1 above, to show that the coupling described in Theorem 3.2 for Subcase 2 i ,Configuration 2 is optimal.

( ) � 4 Ž . Ž .Configuration 2, Subcase 2 ii . Here u, ¨ gE, X ¨ gS , Y ¨ fS ,w , u, Y w , u, XŽ . Ž . Ž < < < <.X ¨ fS , and Y ¨ gS . In this case N sN q S y S andu, w , Y u, w , X X Y w , u, Y u, w , X

< <the set AA may be obtained from AA by deleting the ky S maps whichY X u, w , XŽ . < <assign the color Y ¨ to w and adding the ky S color maps which assign thew , u, Y


Ž .color X ¨ to u. We order the elements of AA lexicographically but order theXŽ < <.elements of AA as follows. The first N y ky S elements of AA are theY Y w , u, Y Y

� Ž . Ž .4 < <elements c : c u /X ¨ , ordered lexicographically. The last ky S ele-Y Y w , u, Y� Ž . Ž .4 Ž .ments of AA are the elements c : c u sX ¨ ordered by c w . We do this soY Y Y Y

Ž < <.that the ith element of AA and AA agree, for 1F iFN y ky S . Then theX Y X u, w , XŽ < <.matrix K of costs has the following form: the first N y ky S rows have a 0X u, w , X

< <on the diagonal and every other entry is 1 or 2. Each entry of the last ky Su, w , Xrows is either 1 or 2, with each row having at least one entry equal to 1, and each

< < Ž1.entry in the last ky S columns equal to 2. Let K be the matrix obtainedw , u, Y< <from K by first subtracting 1 from the last ky S rows, and then subtracting 1u, w , X

< <from the last ky S columns. Then there is a zero entry in each row andw , u, Ycolumn of K Ž1..

First suppose that N sN . The diagonal entry of each row of K Ž1. equals zero.X YLet each of these be an independent zero with multiplicity N . This set of N 2

X Xindependent zeros gives an optimal solution of the original transportation problem

Ž .which corresponds exactly to the coupling described for Subcase 2 ii , Configura-Ž . wtion 2 when N sN in the proof of Theorem 3.2. Note that the bijection f :X Y

C _S ªC_S used in the proof of Theorem 3.2 can be defined as follows:u, w , X w , u, Y² Ž . Ž .:if the ith element of AA is Y ¨ , c u and the ith element of AA isX X Y

² Ž . Ž .: Ž Ž .. Ž . Ž < <. xc w , X ¨ then let f c u sc w , for N y ky S - iFN .Y X Y X u, w , X XNow suppose that N )N . The following zero elements of K Ž1. are taken asX Y

Ž .independent zeros with the stated multiplicities. Let the i, i entry be an indepen-Ž < <.dent zero with multiplicity N for 1F iFN y ky S . Now consider the rowY X u, w , X

² Ž . Ž .:associated with Y ¨ , c u . Let each entry in a column associated withX² Ž . Ž .: < < < <c w , c u be an independent zero with multiplicity S y S . FirstY X w , u, Y u, w , X

Ž .suppose that c u gC . Then let each entry in a column associated withX 1² Ž . Ž .: Ž . Ž .c w , X ¨ where c w /c u be an independent zero with multiplicityY Y X

< < ² Ž . Ž .:ky S , and let the entry in the column associated with c u , X ¨ be anw , u, Y XŽ < < < < . Ž .independent zero with multiplicity ky S q S _S y1 . By 3.13 ,w , u, Y w , u, Y u, w , Y

Ž .this places N independent zeros in that row. Finally suppose that c u gY X² Ž . Ž .:S _S . Then let each entry in a column associated with c w , X ¨ beu, w , X w , u, X Y

< < Ž .an independent zero. The multiplicity of this entry is ky S if c w gC andw , u, Y Y 1< < Ž .ky S q1 if c w gC . This ensures that there are N independent zeros inw , u, Y Y 2 Y

Ž . Ž1.the row, by 3.14 . This set of N N independent zeros in K gives an optimalX Ysolution of the original transportation problem which corresponds exactly to the

Ž . Ž .coupling described for Subcase 2 ii , Configuration 2 when N )N in Theo-X Yrem 3.2.

( ) � 4 Ž . Ž .Configuration 2, Subcase 2 iii . Here u, ¨ gE, X ¨ gS , Y ¨ fS ,w , u, Y w , u, X

Ž . Ž . Ž . Ž < <X ¨ gS , and Y ¨ fS . In this case N sN q2 ky1 y S qu, w , Y u, w , X X Y w , u, X< <. < <S and the set AA may be obtained from AA by deleting the ky S y1u, w , X Y X u, w , X

Ž . < <color maps which assign the color Y ¨ to w and deleting the ky S y1 colorw , u, XŽ .maps which assign the color Y ¨ to u. The matrix K of costs has the following

form: the row associated with c has a 0 on the diagonal and every other entryXŽ . Ž . Ž . Ž . Ž . Ž . Ž .equal to 1 or 2, unless c w sY ¨ or c u sY ¨ . If c w sY ¨ or c u sX X X X

Ž .Y ¨ then every entry of the row is 1 or 2, with at least on entry equal to 1 in eachrow. Let K Ž1. be the matrix obtained from K by subtracting 1 from each row


Ž . Ž . Ž . Ž .corresponding to c where c w sY ¨ or c u sY ¨ . There is a zero in everyX X Xrow and every column of K Ž1.. In particular, the diagonal element of the row

Ž . Ž . Ž . Ž .corresponding to c is zero whenever c w /Y ¨ and c u /Y ¨ . Make eachX X Xof these entries an independent zero with multiplicity N . We now describe how toYselect independent zeros in the remaining rows.

X X Ž . Ž .Recall the probability distributions n , n , c , and c defined in 3.11 , 3.16 ,Y Y q qŽ . Ž . ² Ž . Ž .:3.17 , and 3.19 , respectively. Consider the row associated with Y ¨ , c u . IfX

Ž . X ² Ž . Ž .:c u gC then let the entry in the column associated with c w , c u be anX 2 Y XŽ Ž .. Ž .independent zero with multiplicity h c w N for all c w gC_S . IfY Y Y Y w , u, Y

Ž . ² Ž . Ž .:c u gC then let the entry in the column associated with c w , c u be anX 1 Y XŽ Ž .. Ž . Žindependent zero with multiplicity c c w N for all c w gC_ S jc Žu. Y Y Y w , u, YX

� Ž .4. Ž .c u . This places N independent zeros in each of these rows, by 3.18 and theX Yfact that n is a probability distribution. Finally consider the row associated withY² Ž . Ž .: Ž .c w , Y ¨ . If c w gC then let the entry in the column associated withX X 2² Ž . Ž .: XŽ Ž ..c w , c u be an independent zero with multiplicity n c u N for allX Y Y Y Y

Ž . Ž .c u gC_S . If c w gC then let the entry in the column associated withY u, w , Y X 1² Ž . Ž .: XŽ Ž ..c w , c u be an independent zero with multiplicity c c u N for allX Y c Žw . Y YX

Ž . Ž � Ž .4.c u gC_ S j c w . This places N independent zeros in each of theseY u, w , Y X Y

rows, as both nX and c

X are probability distributions. This set of N N indepen-Y q X Ydent zeros in K Ž1. gives an optimal solution to the original transportation problem

Ž .which corresponds exactly to the coupling for Subcase 2 iii , Configuration 2 givenin Theorem 3.2. This completes the proof. B

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