On the chromatic number of a random 5-regular graph

On the chromatic number of a random 5-regular graph J. Díaz A.C. Kapori G.D. Kemke L.M. Kiroui X. Pérez N. Wormald Abtract It wa only recently hown by Shi and Wormald, uing the differential equation method to analye an appropriate algorithm, that a random 5-regular graph aymptotically almot urely ha chromatic number at mot 4. Here, we how that the chromatic number of a random 5-regular graph i aymptotically almot urely equal to 3, provided a certain four-variable function ha a unique maximum at a given point in a bounded domain. We alo decribe extenive numerical evidence which trongly ugget that the latter condition hold. The proof applie the mall ubgraph conditioning method to the number of locally rainbow balanced 3-colouring, where a colouring i balanced if the number of vertice of each colour i equal, and locally rainbow if every vertex i adjacent to at leat one vertex of each of the other colour. 1 Introduction The chromatic number of random regular graph ha attracted much interet in recent year. For the uniform model G n,d of d-regular graph on n labelled vertice, recent work ha focued on the chromatic number for fixed d. All neceary background with repect to random regular graph a well a the pioneering reult about their chromatic number and other parameter can be found in the comprehenive review paper by the lat author [17]. A preliminary verion of ome of the reult of thi paper appeared in [6]. Author JD, LMK, and XP are partially upported by Future and Emerging Technologie programme of the EU under contract 001907 Dynamically Evolving, Large-Scale Information Sytem DELIS). JD wa partially upported by the Ditinció de la Generalitat de Catalunya per a la promoció de la recerca, 2002. ACK and LMK are partially upported by European Social Fund ESF), Operational Program for Educational and Vacational Training II EPEAEK II), and particularly Pythagora. Part of the reearch of LMK wa conducted while viiting on a abbatical the Departament de Llenguatge i Siteme Informàtic of the Univeritat Politècnica de Catalunya. GDK and NW are partially upported by NSERC. NW i alo upported by the Canada Reearch Chair program. Univeritat Politècnica de Catalunya, Departament de Llenguatge i Siteme Informàtic, Campu Nord Ed. Omega, 240, Jordi Girona Salgado, 1 3, E-08034 Barcelona, Catalunya, {diaz,xperez}@li.upc.edu) Univerity of Patra, Department of Computer Engineering and Informatic, GR-265 04 Patra, Greece, {kapori,kiroui}@ceid.upatra.gr) Department of Combinatoric and Optimization, Univerity of Waterloo, Waterloo, ON N2L3G1, Canada, {gdkemke, nwormald}@uwaterloo.ca). Reearch Academic Computer Technology Intitute, P.O. Box 1122, GR-261 10 Patra, Greece. 1

It i widely known that for d 2 a random d-regular graph i a.a.. not bipartite, and thu ha chromatic number at leat 3. We ay an event hold aymptotically almot urely a.a..) if it hold with probability tending to 1 a n. In thee aymptotic for G n,d we aume nd i alway even for feaibility.) Molloy and Reed ee [10]) gave a lower bound on the chromatic number for general d, and in particular they howed that for a random 6-regular graph it i a.a.. at leat 4. The baic ingredient of the proof wa the firt moment method: howing that the expected number of 3-colouring of a random regular graph converge to zero. Achliopta and Moore [2] proved that random 4-regular graph have chromatic number 3 w.p.p. i.e. with probability bounded away from 0 for large n, which they refer to a with poitive probability ). The proof wa algorithmic in the ene that it ued a backtracking-free algorithm baed on Brelaz heuritic. Subequently, Achliopta and Moore [3] howed that the chromatic number of a d-regular graph d 3) i a.a.. k or k + 1 or k + 2, where k i the mallet integer uch that d < 2k ln k. They alo howed that if furthermore d > 2k 1) ln k, then a.a.. the chromatic number i either k + 1 or k + 2. They alo obtained an upper bound on the probability that it i k + 2, which howed that 5-regular graph can be 4-coloured w.p.p. in the multigraph model that i commonly ued to analye random regular graph. Thi wa ubequently improved by Shi and the lat author [14, 15], who howed that the chromatic number of a random d-regular graph i a.a.. 3 for d = 4, 4 for d = 6, and either 3 or 4 for d = 5. In addition, they howed that a.a.. the chromatic number of a d-regular graph, for all other d up to 10, i retricted to a range of two integer.) Their proof were algorithmic. The above reult leave the main outtanding open quetion on the chromatic number of low-degree random regular graph a follow: i the chromatic number of a 5-regular graph a.a.. 3? Previou attempt of ome of the author of the preent paper to anwer the above quetion in the negative, uing refinement of the firt moment method, failed. Thee attempt computed the expected number of ucceively more retricted type of 3-colouring uch that whenever a generic 3-colouring exit, at leat one of the retricted type exit a well), and aimed at proving that it i a.a.. equal to zero. All attempt however gave expected value that tend to. Thee failure alo led to variou innovative attempt to deign an algorithm that would be amenable to rigorou mathematical analyi and that would at leat w.p.p. produce a 3-colouring for 5-regular graph. Thee attempt alo failed. Both the above failure were given a well founded empirical explanation by work in phyic. Building on a tatitical mechanic analyi of the pace of truth aignment of the 3-SAT problem, which ha not been hown yet to be mathematically rigorou, and on the Survey Propagation SP) algorithm for 3-SAT inpired by thi analyi ee e.g. [9] and the reference therein), Krz aka la et al. [8] provided trong evidence that 5-regular graph are a.a.. 3-colourable by an SP algorithm. They alo howed that the pace of aignment of three colour to the vertice legal or not, i.e. with no two adjacent vertice with the ame colour or not) conit of cluter of legal colour aignment inide which one can move from point to point by tep of mall Hamming ditance. However, to go from one cluter to another by 2

uch mall tep, it i neceary to go through aignment of colour that groly violate the requirement of legality high-energy colour aignment). Alo, the number of cluter that contain point with energy that i a local, but not global, minimum i exponentially large. A a reult, local earch algorithm are eaily trapped into uch local minima metatable tate). In thi article we reduce the problem of proving that random 5-regular graph are a.a.. 3-colourable to a problem of a totally different nature, involving imply howing that the maximum of a given function on a given bounded domain occur at a given location. In addition we decribe extenive calculation which trongly upport the hypothei that the maximum doe occur at the given location.) To achieve thi, we tudy locally rainbow balanced colouring of a 5-regular graph, where a colouring i balanced if the number of vertice of each colour i equal, and locally rainbow if every vertex i adjacent to vertice of all the other colour. We compute the expectation EY and variance σ 2 of the number Y of uch colouring aymptotically. Auming a hypothei tated below, we prove that σ 2 i aymptotically a contant time EY ) 2. A tandard econd moment inequality tate that Y i nonzero with probability at leat EY ) 2 /EY 2 ), which i hence bounded away from 0. Intead of thi reult, we obtain the tronger reult, that Y i a.a.. nonzero, by uing the mall ubgraph conditioning method ee [17]). Previou application of thi method have almot all been to cae where the random variable Y count ubgraph of ome type, uually regular panning ubgraph. Jut a few cae have applied it to other random variable, beginning with number of independent et [4]. The application in the preent paper ha a more ignificant conequence. The reaon behind the choice of thi particular ubet of colouring balanced and locally rainbow) i that our approach doe not work when applied to the full et of colouring. In fact, the econd moment of the number of ordinary 3-colouring grow large exponentially fater than the quare of it expectation ee the lat ection in [3]). Retricting the analyi to locally rainbow colouring make the expectation maller but fortunately the econd moment i decreaed even further and the requirement of our method are met. The extra condition that colouring are balanced jut make the computation impler. For our calculation, we ue the well-known pairing or configuration model P n,d which wa firt introduced by Bollobá [5]. A pairing in P n,d i a perfect matching on a et of dn point which are grouped into n cell of d point each. A random pairing naturally correpond in an obviou way to a random d-regular multigraph poibly containing loop or multiple edge), in which each cell become a vertex. Colouring of the multigraph then correpond to aignment of colour to the cell of the model. The reader hould refer to [17] for apect of the pairing model not explained here. The application of the mall ubgraph conditioning method call for the computation of joint moment of the number of locally rainbow balanced 3-colouring and hort cycle. It alo require an upper bound on the econd moment of Y, the number of locally rainbow balanced 3-colouring of the random 5-regular pairing P n,5. The etimation of the econd moment amount eentially to counting the number of pair of uch colouring on 5-regular graph. To give an exact expreion for EY 2 ) we had to um over a large number of variable 9 36). Thee variable expre the number of vertice that have a given pair of colour 3

out of the nine poible pair) and alo have a given ditribution of their five edge with repect to the pair of colour on the other endpoint of thee edge. A we will ee there are 36 poible ditribution.) The computation of the aymptotic value of thi expreion even within a polynomial factor) entail the computation of the global maximum of a function of 9 36 variable. In Section 5 we how how to reduce thi computation to the computation of the maximum of a four-variable continuou function F defined over a cloed and bounded convex domain. A the definition of F and it domain are technically involved, we potpone preenting them until Section 5, at which point the motivation behind the technicalitie become clearer. For the ake of eay reference, we repeat thee definition, and alo give an equivalent definition of F, in Section 7. Regarding the maximization of F, we how that the boundary of it domain contain no local maximizer and that the point 1/9, 1/9, 1/9, 1/9) in the interior of it domain i a local maximizer by howing that the Heian of ln F i negative definite at thi point). Although the definition of F involve another function with hundred of variable, we are able to obtain information on it value by a rather roundabout route. By numerically computing it value at a huge number of location over a fine grid of it domain, we obtain trong numerical evidence for the following. Hypothei 1.1 Maximum Hypothei) The four-variable function F n) ha a unique global maximum over it domain at the point 1/9, 1/9, 1/9, 1/9). We point out that for the cae of the ordinary not balanced locally rainbow) colouring there exit an analogue to function F which alo ha a local maximum at the point 1/9, 1/9, 1/9, 1/9) but unfortunately thi i not the global maximum. Provided that the Maximum Hypothei hold, we can etablih the chromatic number of the random 5-regular graph a.a.. Theorem 1.1 Under the Maximum Hypothei, the chromatic number of G n,5 i a.a.. 3. Thu, we have reduced the problem of proving that G n,5 a.a.. ha chromatic number 3, to howing that the maximum of a mooth function in a bounded domain occur at the very place that numerical calculation ugget. In the ret of thi article we prove Theorem 1.1 and decribe why we are convinced that the function ha it maximum at the required location. In Section 2 we explain the mall ubgraph conditioning method and how how it i ued to prove Theorem 1.1 in the cae that n i diviible by 6, uing the relevant reult from [17]. Thi aume the reult of certain calculation that are performed in Section 3 to 5. In Section 3 we compute joint moment of the number of locally rainbow balanced 3-colouring and hort cycle. We develop an exact expreion for the econd moment EY 2 ) in Section 4 and determine it aymptotic value, under the Maximum Hypothei, in Section 5. The argument for n not diviible by 6 i upplied in Section 6. Finally, in Section 7 we preent the empirical validation of the Maximum Hypothei. 2 Small ubgraph conditioning The mall ubgraph conditioning method wa introduced by Robinon and the lat author [12, 13]. See [7, Chapter 9] and [17] for a full expoition. 4

The etting for the method i a follow. A random variable, Y, count occurrence of ome tructure, and depend on a parameter n which tend to. The expectation EY tend to infinity, and we want to how that PY > 0) 1. The mall ubgraph conditioning method applie when the variance of Y i of the ame order a EY ) 2. The main computation required i the aymptotic value of ome joint moment of the number of certain mall ubgraph and the random variable Y. The reult which the method depend on can be tated a follow a conequence of [17, Corollary 4.2]). We ue [x] m := xx 1) x m + 1) to denote falling factorial.) Theorem 2.1 Let λ k > 0 and δ k 1 be real number for k = 1, 2,... and uppoe that for each n there are random variable X k = X k n), k = 1, 2,... and Y = Y n), all defined on the ame probability pace G = G n uch that X k i nonnegative integer valued, Y i nonnegative and EY > 0 for n ufficiently large). Suppoe furthermore that i) For each j 1, the variable X 1,..., X j are aymptotically independent Poion random variable with EX k λ k, ii) if µ k = λ k 1 + δ k ), then EY [X 1 ] m1 [X j ] mj ) EY j k=1 for every finite equence m 1,..., m j of nonnegative integer, iii) k λ k δ k 2 <, iv) EY 2 )/EY ) 2 exp k λ k δ k 2 ) + o1) a n. Then PY > 0 E) 1, where E i the event δk = 1{X k = 0}. µ m k k 2.1) Proof of Theorem 1.1 for n diviible by 6) For the application in the preent article we ue the probability pace G n = P n,5 with Y counting the number of locally rainbow balanced 3-colouring and X k counting the number of k-cycle for fixed k 1. We aume from now until the very end of thi proof that n i diviible by 6. We next dicu how the four condition of Theorem 2.1 are verified in thi etting. It i well-known e.g., ee [17]) that condition i) i atified by λ k = 4 k /2k). In 3.1) and 3.2) we will ee that condition ii) hold for the function δ k = 15 k + 2 5) k + 2 3) k. 2.2) Subtituting thi function into condition iii) and iv), we ee that the um i λ k δk 2 = k k = k 5 1) k 15 k + 2 5) k + 2 3) k) 2 2k ) k ) k 1 4 4 + 4 + 4 2k 225 45 ) k 4 + 4 75 ) k ) k ) ) k 4 4 4 + 8 + 4. 9 15 25 5

Uing the identity 1 k 2k xk = 1 ln1 x), thi um become 2 221 λ k δk 2 = 1 ) ) 4 ) 4 49 79 5 2 ln 225 45 75 9 k 3 13 5 13 = ln 7 6 11 4 79 2 13 17 ) 4 11 15 ) 8 ) ) 4 21 25 ). 2.3) To verify condition iv), we will need the aymptotic value of the firt and econd moment of Y. Later in thi article we will prove that ) 2 EY 2 3 6 5 3 n 1 25 2.4) 11 3 2πn) 2 24 and, under the Maximum Hypothei, We compute the ratio EY 2 ) 2 2 3 19 5 16 1 7 6 11 7 79 2 13 17 2πn) 2 EY 2 ) EY ) 2 3 13 5 13 7 6 11 4 79 2 13 17, ) n 25. 2.5) 24 which matche 2.3), etablihing condition iv). Having verified the four condition, we may apply the mall ubgraph conditioning method to conclude PY > 0 E) 1, where E i the event δk = 1{X k = 0}. To interpret the event E in the concluion, we note that δ 1 = 1 and for k 2 we have δ k 15 2 + 25) 2 + 23) 2 < 1. So the concluion read P Y > 0 X 1 = 0) 1. Becaue P X 2 = 0) i bounded away from 0 for large n, it follow that Y > 0 a.a.. for the imple graph G n,5. Thi prove Theorem 1.1 provided that n i diviible by 6. The cae that n i 2 or 4 mod 6 are treated in Section 6. Thi i done by modifying certain part of the argument for n 0 mod 6) to handle a mall number of precoloured vertice, and then applying an aymptotic equivalence between random graph pace. 3 Joint moment The goal of thi ection i to compute aymptotic value of ome joint moment for the random variable which count locally rainbow balanced 3-colouring and hort cycle in random regular graph. On the pace P n,5, let Y be the random variable counting the number of locally rainbow balanced 3-colouring. We begin by computing the aymptotic value of EY. 6

Lemma 3.1 where ) n 5n/6)! 3 EY An) n/3, n/3, n/3 P n,5 3 ) 3 2 An) = 30 n/3. 11πn Proof. To compute thi expected value we mut count, for each of the n n/3,n/3,n/3) way to aign vertice to equal-ized colour clae, the number of pairing which make the colouring locally rainbow and balanced. All thee aignment are equivalent, o fix one of them. Becaue the three colour clae have equal ize, the number of edge between any two colour clae mut be 5n/6. In our dicuion, the point in a vertex inherit the colour of that vertex. To count the pairing which make the colouring locally rainbow and balanced we proceed in two tep. Firt, at each vertex v, we chooe for each point in v the colour of the point it i paired with. Thi mut be done carefully to enure that each vertex will be adjacent to at leat two colour and that the number of edge between the colour clae will be 5n/6 a required. Then, for each pair of colour clae, we pair up the appropriate point between thee clae in one of 5n/6)! way. Thu, the econd tep give u a factor of 5n/6)! 3. To determine the number of choice in the firt tep, we oberve that each colour cla produce an equivalent contribution. We fix one colour cla, ay colour 1, and contruct the ordinary generating function which count the number of way of chooing the colour of the neighbour of each point within the cla, with the indeterminate x marking one of the two poible colour. At each vertex, each of the 5 point can be aigned a mate i.e. the other point in it pair) of either one of the two colour, provided that not all of the point are aigned to the ame colour. Thu the contribution of each vertex to the generating function i x + 1) 5 x 5 1, giving u the generating function x + 1) 5 x 5 1 ) n/3. Exactly 5n/6 of thee choice mut be for the colour marked by x, o the total number of choice for the firt tep i letting quare bracket denote extraction of a coefficient) N = [x 5n/6 ] x + 1) 5 x 5 1 ) n/3 for each colour cla. Combining thee reult, we have ) n 5n/6)! 3 EY = N 3. n/3, n/3, n/3 P n,5 Uing the addle-point method ee e.g. Section 12.1 in [11]) we will etimate N uing a contour integral along the path z = 1. We begin by ubtituting z = expiθ) and expanding 7

in θ. N = 1 2πi = 1 2π z =1 π π = 1 2π 25 2) n/3 = 1 2π 30n/3 z + 1) 5 z 5 1) n/3 dz z 5n/6 e iθ5n/6 e iθ + 1) 5 e iθ5 1 ) n/3 dθ π ) exp 45 + 45) 5 + 1)2 1+5 nθ 2 + Onθ 3 ) dθ π 242 5 2) 2 π exp 11 ) 72 nθ2 + Onθ 3 ) dθ. π For θ n 2/5, the contribution to the integral i aymptotically I = 1 2π 30n/3 exp 11 ) 72 nθ2 dθ = 1 72π 2π 30n/3 11n = 3 2 30 n/3. 11πn For θ > n 2/5, the abolute value of e iθ + 1) 5 e iθ5 1 ) n/3 i 5 1 5 n/3 e iθ )e n/3 iθj j j=1 2 5 4 + e iθ + 1 ) n/3 2 5 4 + 2 + 2 co n 2/5 ) ) n/3 = 30 14 n 4/5 + O n 8/5)) n/3 n = 30 n/3 exp 3 ln 1 1 120 n 4/5 + O n 8/5) )) = 30 n/3 exp 1 360 n1/5 + O n 3/5) ), which i oi). Therefore the expreion for I give the correct aymptotic etimate for N, which i N 3 2 30 n/3. 11πn Combining thi with our above reult, we get Lemma 3.1. 8

From Lemma 3.1 it i eay to deduce the aymptotic value of EY a tated in 2.4). Simply ubtitute P n,5 = 5n)!/2 5n/2 5n/2)!) and apply Stirling formula. We omit the calculation. We now move cloer to our goal of computing joint moment for locally rainbow balanced 3-colouring and hort cycle. For fixed k 1, let the random variable X k count the number of k-cycle in P n,5. We will actually work with rooted oriented cycle, which introduce a factor of 2k into the counting. It will be helpful to have the following definition. For a rooted oriented cycle in a coloured graph, define it colour type to be the equence of colour on it vertice. To calculate the expected value of Y X k, we will count, for each locally rainbow balanced 3-colouring and each rooted oriented k-cycle, the number of pairing which contain thi cycle and repect thi colouring. A before, there are n n/3,n/3,n/3) way to chooe the balanced 3-colouring. All are equivalent, o fix one. To enumerate the cycle and pairing which repect thi colouring, we will um over all colour type T. Once a colour type ha been choen, each vertex of the cycle can be placed in the pairing model by chooing a vertex of the correct colour and an ordered pair of point in that vertex to be ued by the cycle. Hence, in total, there are aymptotically 5 4 n/3) k way to place the rooted oriented cycle in the pairing model. We now have EY X k ) 1 2k n n/3, n/3, n/3 ) 20n 3 ) k 1 P n,5 ft ), where ft ) i the number of pairing which repect a fixed colouring and fixed rooted oriented cycle of colour type T and make the colouring locally rainbow. To etimate the function ft ), we will again fix one colour cla j and contruct an ordinary generating function. The generating function will count the number of way of chooing the colour of the neighbour of each point within the cla, with the indeterminate x marking one of the two poible colour. For j = 1, 2, 3, let α j T ) count the number of j-coloured vertice in colour type T whoe two neighbour in the cycle have different colour. Let α jt ) count the number of j-coloured vertice in colour type T whoe two neighbour in the cycle both have the colour marked by x. Let α j T ) count the remaining j-coloured vertice in T. We alo define β j T ) = α jt )+α j T ). For any vertex through which the cycle doe not pa, the contribution to the generating function i, a before, x + 1) 5 x 5 1. For a cycle vertex whoe neighbour in the cycle have different colour, we can aign the neighbour colour for the remaining point in any way, giving u x + 1) 3. But for a cycle vertex whoe neighbour in the cycle have the ame colour, we mut enure that thi vertex get at leat one neighbour of a different colour o that the colouring i locally rainbow. Thi give u x + 1) 3 x 3 if the neighbour have the colour marked by x, and x + 1) 3 1 otherwie. Combining thee function, the number of way of chooing the neighbour of each point within colour cla j i given by the coefficient of x 5n/6 in the expreion x + 1) 5 x 5 1 ) n/3 α j T ) α j T ) α j T ) ) x + 1) 3 α j T ) x + 1) 3 x 3) α j T ) x + 1) 3 1 ) α j T ). Earlier in thi ection we ued the addle-point method to etimate a imilar coefficient. A imple comparion with that previou application make it eay to ee that the current 9 T

coefficient i aymptotically 30 n/3 α jt ) β j T ) 8 α jt ) 7 β jt ) 3 2 11πn. After the colour of the neighbour of each point ha been choen, it remain to pair up the point between each two colour clae. Since the k pair in the cycle have already been choen, the number of way to do thi i aymptotically 5n/6)! 3 5n/6) k. Putting αt ) = α 1 T ) + α 2 T ) + α 3 T ) and βt ) = β 1 T ) + β 2 T ) + β 3 T ), we conclude that ft ) 30n αt ) βt ) 8 αt ) 7 βt ) 3 3 2 3 ) 3 5n/6)!3 11πn 5n/6) k An) 5n/6)!3 5n/6) k ) αt ) ) βt ) 8 7. 30 30 Letting c α = 8/30 and c β = 7/30, it remain to etimate S = T c αt ) α c βt ) β where the um i taken over all colour type T. In other word, we need to enumerate the colour type, introducing a factor of c α for each cycle vertex whoe neighbour have different colour, and a factor of c β for each of the remaining cycle vertice. It i helpful to view each colour type a a equence of ordered pair of colour: the colour at the endpoint of each edge, taken in the order induced by the orientation of cycle. One could conider each poible pair to be a tate in a Markov chain. Number the tate a follow. tate pair of colour 1 12 2 21 3 31 4 13 5 23 6 32 Each colour type on k vertice then correpond to a equence of k + 1 tate where the firt tate equal the lat tate. For example, conider the colour type with colour equence 1, 2, 3, 2. It correpond to the tate equence 1, 5, 6, 2, 1. The tranition from tate 1 to tate 5 repreent to a vertex of colour 2) whoe neighbour in the cycle have different colour 1 and 3); hence it hould introduce a factor of c α. Thu, in the matrix below, the entry at poition 1, 5) i c α. In thi way we can contruct a matrix which account for all poible 10

tranition, and ue it to obtain the deired enumeration. The above um S equal TrM k ), where Tr denote the trace, and M i the tranition matrix 0 c β 0 0 c α 0 c β 0 0 c α 0 0 c α 0 0 c β 0 0 0 0 c β 0 0 c α. 0 0 c α 0 0 c β 0 c α 0 0 c β 0 1 2 The eigenvalue of thi matrix are c β + c α, c β + c α, 1c 2 α + 1 2 3c 2 α + 4c 2 β. The lat two eigenvalue have multiplicity 2. Thu S = c β + c α ) k + c β + c α ) k + 2 1 2 c α + 1 2 ) k 3c 2 α + 4c 2 β + 2 1 2 c α 1 k 3c 2 2 α + 4cβ) 2. 3c 2 α + 4c 2 β, and 1 2 c α Since c β + c α = 7/30 + 8/30 = 1/2, we may write where which i 2.2). We conclude that EY X k ) 1 2k S = 1 2 k 1 + δ k) δ k = 15 k + 2 5) k + 2 3) k 3.1) n n/3, n/3, n/3 ) 20n and hence, combining thi reult with the previou lemma, 3 ) k 1 P n,5 An)5n/6)!3 5n/6) k S, EY X k ) EY 1 2k 8k S 4k 2k 1 + δ k). The above argument i eaily extended to work for higher moment, by counting the pairing that contain a given locally rainbow balanced 3-colouring and et of oriented cycle of the appropriate length. The contribution from cae where the cycle interect turn out to be negligible, for the following reaon. Suppoe that the cycle form a ubgraph H with ν 11

vertice and µ edge, and the total length of cycle i ν 0. Then in the cae of dijoint cycle, ν = µ = ν 0. A factor of Θn ν 0 ν ) i lot if there i a reduction in the number of vertice of H, compared with the dijoint cae, becaue of the reduced number of way of placing the cycle on the coloured vertice. Similarly, a factor Θn ν 0 µ ) i gained in the function f for the reduction in the number of edge of H, becaue of the correponding increae in the number of point to be paired up at the end. Thu, the contribution from uch an arrangement of cycle to the quantity being etimated i of the order of n ν µ time that of the contribution from dijoint cycle. In all non-dijoint cae, H ha more edge than vertice, ince it minimum degree i at leat 2, and it ha at leat one vertex of degree at leat 3. There are only finitely many iomorphim type of H to conider, o the contribution from the cae of dijoint cycle i of the order of n time the ret. The ignificant term in thi cae decompoe into a product of the factor correponding to the individual cycle. Conequently, we obtain the following reult, a required for 2.1) in accordance with 2.2): EY [X 1 ] m1 [X j ] mj ) EY j ) 4 k mk 2k 1 + δ k). 3.2) k=1 4 Exact Expreion for the Second Moment Given a pairing P P n,5, let R P be the cla of locally rainbow balanced 3-colouring of P. Let Y be the random variable that count the number of locally rainbow balanced 3-colouring in P n,5. Then, it i eaily hown that EY 2 ) = {P, C 1, C 2 ) P P n,5, C 1, C 2 R P }. 4.1) P n,5 Below we aume we are given a pairing P and two locally rainbow balanced 3-colouring C 1 and C 2 on P. Recall that a pairing i a perfect matching on 5n point which are organized into n cell of 5 point each. For i, j = 0, 1, 2, let V be the et of cell coloured with i and j with repect to colouring C 1 and C 2, repectively. Let n = V /n and let E be the et of point in cell of V. Since C 1 and C 2 are balanced, we have n = 1/3, j, n = 1/3, i, 4.2) i and therefore n = 1. Alo, for r, t { 1, 1}, let E be the et of point in E which are matched with point in E i+r,j+t. Here and throughout the article, the arithmetic in the indice i modulo 3.) Let m = E /n. For fixed i and j, it i convenient to think of the four variable m ) { 1,1} a the entrie of a 2 2 matrix. The row and column are indexed by -1 and 1, with -1 for the firt row or column. We have that m = 5n, and therefore, m = 5. And, ince matching et of point hould have equal cardinalitie, we alo have that j m = m i+r,j+t r, t. 4.3) 12

Let v be a cell in V. The pectrum of cell v i a 2 2 nonnegative integer matrix. The row and column are indexed by -1 and 1, with -1 for the firt row or column. Cell v i aid to have pectrum if out of it five point, r, t { 1, 1}, are matched to point in cell of V i+r,j+t. The um of the entrie of i 5 becaue of the 5-regularity of the graph. Each row and column um i at leat 1 becaue both C 1 and C 2 are locally rainbow. We let S denote the et of poible pectra. One can check that S = 36. For each i, j Z 3 and pectrum S, we denote by d the caled with repect to n) number of cell which belong to V and have pectrum. We have m = S d, 4.4) n = S d, 4.5) and therefore, d = 1. Throughout thi paper we refer to the et of the nine number n a the et of the overlap variable. We alo refer to the et of the thirty-ix number m a the et of the matching variable. We refer to the 9 36 number d a the pectral variable. We conider the polytope { D = d ) Z3, S R 324 : d 0 i, j,, j, d = 1 i, 3 i, d and the dicrete ubet = 1 3 j, d I = D = r, td i+r,j+t ) 1 n Z324. } i, j, r, t, In view of 4.2) 4.5), note that I contain the et of equence d ) Z3, S that correpond to ome pair of locally rainbow balanced 3-colouring. Given a fixed equence d ) I, let u denote by ) n d n) the multinomial coefficient that count the number of way to ditribute the n vertice into clae of cardinality d n for all poible value of i, j and. Define m by 4.4). Alo let ) 5 tand for 5!/!. Let N = {P, C 1, C 2 ) P P n,5, C 1, C 2 R P }. By counting the way to aign pectra to cell, and then colour to point in cell given their pectra, and finally the number of matching between colour clae, we have N = { ) n ) ) d 5 n m d n) n)! ) )} 1/2. 4.6) I,, Dividing thi by P n,5 = 5n)!/2 5n/2 5n/2)!), we obtain { EY 2 ) = 25n/2 5n/2)! ) n ) ) d 5 n 5n)! d n) I,, m n)! ) )} 1/2. 4.7) 13

5 Aymptotic Value of EY 2 ) In thi ection, we complete the proof of the main theorem given in Section 2 by howing that equation 2.5) hold, auming the Maximum Hypothei. For ake of implicity, we will often write d to denote the tuple d ) Z3, S. Let u conider the function ˆF d) = 5 )) d ) ) 1 m 2 m,, d defined in D, where m denote d convention that 0 0 = 1 and 0 ln 0 = 0. We define fd) = ln ˆF d) =, gd) =, d ) m 1/4 Lemma 5.1 The econd moment atifie, a before. Throughout thi article we oberve the ) ) 5 ln ln d + 1 2 m ln m,,, d ) 1/2, hn) = 2 1/2 2πn) 305/2 5 5n/2. 5.1) EY 2 ) = hn) d I qn, d)e fd)n 5.2) where, a n and uniformly over all d, qn, d) = On 162 ) and qn, d) gd) provided all d and m are bounded away from 0. Proof. We apply Stirling formula and perform imple manipulation to 4.7) to obtain: EY 2 ) = 25n/2 5n/2)!n! 5 d n ) m 5n)! d n)! ) ) 1/2 d I, n)!, πn 5 5n/2 n/e) n 5 d n ) m n/e) 5n/2 d n)! ) ) 1/2 d I, n)!, = hn) 2πn) 153 n/e) n 5 d n ) m n/e) 5n/2 d n)! ) ) 1/2 d I, n)!. 5.3), Now we need to uniformly approximate the factorial of everal number not necearily growing large with n. Stirling formula alo implie k! = 2πηk)k/e) k for all k 0, where ηk) k if k, and ηk) = Θk + 1) for all k 0. In particular, η i nonzero.) So we have 5 d n ) d, n)!, m n)! ) 1/2 ) 14

= = =, 5 d n ) 2πηd n) d n e ) d n, n/e) 5n/2, ηnm ) 1/4 n 1/4 ) 2πn) 153 n/e) n, ηnd ) 1/2 n 1/2 ) n/e) 5n/2 qn, 2πn) 153 d)efd)n n/e) n 2πηm n), 5 ) d m n e ) m ) d n, n 1/2 ) ) m m n/2 for a function q of the type in the tatement of the lemma. Combining thi with 5.3) yield the tatement of the lemma. Notice that the number of term in 5.2) i at mot n + 1) 324, ince each coordinate of any d I mut be a rational in 1 Z between 0 and 1. We conider the maximum bae of the n exponential part of the term in 5.2), taken over all point in the polytope D: { M = max 5 5/2 e fd)}. d D Thi i well defined, due to the compactne of the domain and the continuity of the expreion. Note that the exponential behavior of the econd moment i governed by M ince the number of term in the um in 5.2) i polynomial with repect to n. In the next ubection we determine the value of M under the Maximum Hypothei. In the following ubection, baed on that fact and uing a Laplace-type integration argument, we compute the ub-exponential factor in the aymptotic expreion of the econd moment, and obtain 2.5). 5.1 Computing M We will maximize ˆF in two phae. In the firt one, we will maximize ˆF auming the matching variable m are fixed contant. Thee contant mut be compatible with the polytope D over which ˆF i defined, o we define M to be the et of vector m of 2 2 matrice m ) Z3 uch that 4.4) hold for ome d D. We will often conider variable d and m for fixed i, j Z 3. To implify notation, we delete the indice i and j when they are fixed throughout the formula. We alo define, for any 0 < c R, D c) = {d ) S R 36 : d 0, d = c}, and let M c) be the et of 2 2 matrice m uch that 4.4) hold for ome d ) S D c) after deleting upercript i and j). We will ue d to denote both point in D and D c). The meaning will be clear from the context. In order to give an alternative characterization of the matching variable m, we conider 15

the following equation for all ordered pair i, j), i, j Z 3, and all r, t { 1, 1}: m 0, m + m r, t 4m + m r, t), m + m 4m r, t + m r, t). 5.4) That i, for all uch i and j, the entrie of m are nonnegative, neither row um i greater than 4 time the other, and neither column um i greater than 4 time the other. Lemma 5.2 Let c > 0 R. The et M c) can be alternatively decribed a the polytope containing all matrice m uch that m = 5c, 5.5) i, and the contraint in 5.4) hold. Similarly, M i the polytope containing all vector m of matrice m uch that m = 5/3, m = 5/3, 5.6) and the contraint in 4.3), 5.4) hold. Proof. Let A be the et of matrice m atifying 5.4) and 5.5). M c) A : Let m be a matrix in M c). Then, for ome d D c), we have m = d = d = 5d = 5c, and 5.5) i atified. Moreover, we oberve that for any pectrum, we have 0, + r, t 4 + r, t ) and + 4 r, t + r, t ). Then m mut atify the contraint in 5.4), ince it i a poitive linear combination of pectra, and m A. A M c) : A i a polytope and o it i the convex hull of it vertice: [ [ ] 0 c c 0 [ c 0 0 4c ], c 3c ], [ 0 c 4c 0 ], 3c c ], j, [ 0 4c c 0 ], [ c 3c 0 c ] [ 4c 0, 0 c ] [ 3c c, c 0 Each of thee vertice v ha the hape of ome pectrum time c. By making d = c and d = 0 for, we how that v M c). Moreover, we oberve that M c) i a convex et, ince it i the image of D c) under a linear mapping. Then M c) mut contain the convex hull of the vertice of A, and thu A. The econd tatement in the lemma follow eaily from thi and from the definition of M. 16

For any fixed m M, let F m) be the maximum of ˆF retricted to d D uch that 4.4) hold. To expre F m) in term of m, we will ue the matrix function [ ] x y Φ = ) 5 x 1, 1 y 1,1 z 1, 1 w 1,1 z w S = x + y + z + w) 5 x + y) 5 x + z) 5 y + w) 5 z + w) 5 + x 5 + y 5 + z 5 + w 5 5.7) and, for each of the nine poible pair i, j), i, j Z 3, conider the 4 4 ytem µ Φµ µ = m, r, t = 1, 1, 5.8) in the matrix variable µ. Lemma 5.3 For any m in the interior of M, each of the nine ytem in 5.8) ha a unique poitive olution. Moreover, in term of the olution of thee ytem, F m) =, m ) 1 2 µ ) m, and the equation remain valid for m on the boundary of M if the expreion on the right i extended by continuity. Proof. We aume that m i a fixed vector in the interior of M. In order to compute F m), it i ufficient to maximize the function ˆF d) for nonnegative d ubject to 4.4), ince the other contraint are trivially atified. We oberve that the factor ) m 1 2 m i contant, and that variable d with different pair of indice i, j) appear in different factor of ˆF and alo in different contraint. Thu, it i ufficient to maximize, eparately for each i, j Z 3, the function G = S 5 ) d ) d, 5.9) over nonnegative d ubject to the matrix contraint 4.4). From now on in thi proof, we fix i and j and thu omit upercript a dicued above. Let R be the polytope containing all d = d ) S uch that d i nonnegative for all S, and atifying 4.4). The fact that m i in the interior of M implie that R contain point with all the d trictly poitive. In fact, the interior of R conit of all thoe point in R with thi property. For any point d 0 on the boundary of R we elect a egment joining d 0 with ome interior point. We oberve that, in moving along the egment from the interior of R toward d 0, the directional derivative of ln G contain the um of ome bounded term plu ome term of the type ln d with poitive coefficient, which become large a we approach d 0. Hence, G doe not maximize at the boundary of R. 17

We temporarily relax the contraint 4.4) and oberve that the Heian of ln G i negative definite for any tuple of poitive d. Hence ln G i trictly concave in that domain and alo in the interior of R, ince linear contraint do not affect concavity. Thu, the maximum of G i unique and occur in the only tationary point of ln G in the interior of R. We are now in a good poition to apply the Lagrange multiplier method to look for tationary point of ln G. We conider ln G = ) ) 5 d ln ln d, 5.10) for poitive d ubject to the four contraint: L = d m = 0, r, t { 1, 1}. 5.11) For each one of the four contraint L in 5.11) a Lagrange multiplier i λ introduced. Then we obtain the following equation: ) 5 ln 1 ln d = λ, S 5.12) which, together with the contraint 5.11) have a unique olution when d i the only tationary point of ln G. Let u define µ = exp λ 1/5). After exponentiating 5.12), and noting that the um of the i 5, we have ) 5 d = µ ), S, 5.13) and combining thi with 5.11) give m = ) 5 5 ) = µ µ r,t µ r,t )r,t ) r,t µ r,t )r,t, r, t { 1, 1} ), r, t { 1, 1}. 5.14) By contruction, thi ytem ha a unique poitive olution, and 5.13) give the maximizer of G in term of thi olution. From 5.7), we oberve that 5.14) i exactly the ame ytem a the one in 5.8). Now the maximum of G can be obtained by plugging 5.13) into 5.9), reulting in max d R Gd) = 1 µ ) m, 5.15) and the required expreion for F m) follow by elementary computation. 18

Let u now define for any d D 1/9) the auxiliary function Ĝd) = ) 5 d ) ) m ) 1 2 m, 5.16) d where m = d. Recall that 0 0 = 1.) Lemma 5.4 The function Ĝ take it maximum on D 1/9) in the interior of D 1/9). Proof. It i eay to ee that the boundary of D 1/9) comprie the point where for at leat one, d = 0 and d = 1/9. We oberve that it i ufficient to prove the tatement for ln Ĝ. The continuity of ln Ĝ at the boundary point of D 1/9) follow from the fact that lim x 0 xx = 1. After proving ln Ĝ i continuou at the boundary of D 1/9), take any d on the boundary. Here d 0 = 0 for ome 0. Then d 1 > 0 for ome 1 ince the um of entrie of d i 1/9. At any point d uch that d > 0, ln Ĝ d = ln ) 5 1 ln d + 5 2 + 1 2 ln m. 5.17) Note: if d > 0 then all the m correponding to a nonzero are alo necearily nonzero.) A a firt cae, uppoe none of the m i zero at d. Then at a point d + ɛe 0 ɛe 1 here E denote the vector with 1 in it coordinate and zero elewhere) ln Ĝ d 0 ln Ĝ d 1 a ɛ 0. Since the firt partial goe to and the econd i bounded.) Hence there i no maximum at d. Next uppoe preciely one m i zero at d fix uch value of r and t). Pick an uch that = 1. Then d = 0 at d. So rename a 0 and ue the above argument, chooing again any 1 with d 1 > 0. Now the unbounded term in ln Ĝ d 0 are ln d 0 + 1 2 0) ln m and we have m d 0 becaue 0 ) = 1. It follow that there i no maximum at d. For two different m equal to zero at d, pick the pectrum 0 to have 1 in one of the correponding poition, and zero in the other. Then the ame argument a above give the reult. So no local maximum occur on the boundary. The reult follow. Lemma 5.5 The function Ĝ ha a unique maximum in D 1/9) at the point where all the d are equal to 5 ) /8100. The function value at the maximum i 5 5/2 25/24) 1/9. Proof. We note that 4.4) map the interior of D 1/9) into the interior of M 1/9). A a reult and in view of Lemma 5.4, the maximum of Ĝ, under mapping 4.4), doe not occur on the boundary of M 1/9). 19

Aume that m i a fixed matrix in the interior of M 1/9). We firt maximize Ĝ in D 1/9) ubject to the matrix contraint 4.4). Denote thi maximum by Gm). By arguing a in the proof of Lemma 5.3, we have Gm) = ) m ) 1 m 2, µ where the µ are the unique poitive olution of the ytem in 5.8) after deleting upercript i and j. Moreover, the maximizer i given in term of thi olution by 5.13). We now maximize G in the interior of M 1/9), by applying the Lagrange multiplier method to ln Gm) = ) 1 m 2 ln m ln µ, ubject to m = 5/9. We need ome preliminary computation. By adding the four equation in 5.8) and taking into account 5.7), we have 5Φµ) = m. In view of thi, we have for all r, t { 1, 1} r,t { 1,1} Thi allow u to compute m r,t ln µ r,t m = r,t { 1,1} m r,t µ r,t = µ r,t m r,t { 1,1} Φµ) µ µ r,t r,t m = Φµ) m = 1 5. 5.18) ln Gm) m = 1 2 ln m + 1 2 ln µ r,t m r,t ln µ r,t m = 1 2 ln m ln µ + 3 10, 5.19) and obtain the equation 1 2 ln m ln µ + 3 10 = λ, r, t { 1, 1}, 5.20) where λ i the Lagrange multiplier introduced by the ingle contraint. After exponentiating 5.20), and defining λ = expλ 3/10), we can write m µ = λ, r, t { 1, 1}. 5.21) 20

We relabel the entrie of the matrice m and µ a [ ] [ m1 m 2 µ1 µ, 2 m 3 m 4 µ 3 µ 4 ]. Combining 5.21) and 5.8) and after ome manipulation, we get We can factorize the following equation: µ i Φ µ j µ j Φ µ i = 0, i, j {1,..., 4}. µ 1 Φ µ 4 µ 4 Φ µ 1 = 0, and get where µ 1 µ 4 ) P = 0, P = 120 µ 1 µ 2 µ 3 µ 4 + 20 µ 1 3 µ 2 + 20 µ 1 3 µ 3 + 25 µ 1 3 µ 4 + 30 µ 1 2 µ 2 2 +30 µ 1 2 µ 3 2 + 35 µ 1 2 µ 4 2 + 5 µ 4 4 + 20 µ 1 µ 2 3 + 20 µ 1 µ 3 3 +25 µ 1 µ 4 3 + 5 µ 1 4 + 60 µ 1 2 µ 2 µ 3 + 80 µ 1 2 µ 2 µ 4 + 80 µ 1 2 µ 3 µ 4 +60 µ 2 2 µ 3 µ 4 + 60 µ 1 µ 2 2 µ 3 + 90 µ 1 µ 2 2 µ 4 + 60 µ 1 µ 2 µ 3 2 +80 µ 1 µ 2 µ 4 2 + 90 µ 1 µ 3 2 µ 4 + 80 µ 1 µ 3 µ 4 2 + 60 µ 2 µ 3 2 µ 4 +60 µ 2 µ 3 µ 4 2 + 20 µ 2 3 µ 3 + 20 µ 2 3 µ 4 + 30 µ 2 2 µ 3 2 + 30 µ 2 2 µ 4 2 +20 µ 2 µ 3 3 + 20 µ 2 µ 4 3 + 20 µ 3 3 µ 4 + 30 µ 3 2 µ 4 2 + 20 µ 3 µ 4 3, which i trictly poitive, o µ 1 = µ 4. Similarly, we can factorize µ 2 Φ µ 3 µ 3 Φ µ 2 = 0, and get where µ 2 µ 3 ) Q = 0, Q = 120 µ 1 µ 2 µ 3 µ 4 + 20 µ 1 3 µ 2 + 20 µ 1 3 µ 3 + 20 µ 1 3 µ 4 + 30 µ 1 2 µ 2 2 +30 µ 1 2 µ 3 2 + 30 µ 1 2 µ 4 2 + 5 µ 3 4 + 5 µ 2 4 + 20 µ 1 µ 2 3 + 20 µ 1 µ 3 3 +20 µ 1 µ 4 3 + 90 µ 1 2 µ 2 µ 3 + 60 µ 1 2 µ 2 µ 4 + 60 µ 1 2 µ 3 µ 4 +80 µ 2 2 µ 3 µ 4 + 80 µ 1 µ 2 2 µ 3 + 60 µ 1 µ 2 2 µ 4 + 80 µ 1 µ 2 µ 3 2 +60 µ 1 µ 2 µ 4 2 + 60 µ 1 µ 3 2 µ 4 + 60 µ 1 µ 3 µ 4 2 + 80 µ 2 µ 3 2 µ 4 +90 µ 2 µ 3 µ 4 2 + 25 µ 2 3 µ 3 + 20 µ 2 3 µ 4 + 35 µ 2 2 µ 3 2 + 30 µ 2 2 µ 4 2 +25 µ 2 µ 3 3 + 20 µ 2 µ 4 3 + 20 µ 3 3 µ 4 + 30 µ 3 2 µ 4 2 + 20 µ 3 µ 4 3, 21

which i alo tricly poitive, o µ 2 = µ 3. Finally, we ubtitute µ 4 by µ 1 and µ 3 by µ 2 in and then factorize it to obtain where µ 1 Φ µ 2 µ 2 Φ µ 1 = 0, µ 1 µ 2 ) R = 0, R = 70 µ 1 4 + 275 µ 1 3 µ 2 + 415 µ 1 2 µ 2 2 + 275 µ 1 µ 2 3 + 70 µ 2 4, which i again trictly poitive, o µ 1 = µ 2. Hence, all the µ i are equal and all the m i are equal). Since the m i are equal and um to 5/9, each mut equal 5/36. Subtituting thi value into any of equation 5.8), and remembering that the µ i are equal, give each µ i = 2 2/5 3 4/5 5 2/5. Thi how that the Lagrange multiplier problem ha a unique olution. Thi olution mut correpond to the unique tationary point of G in the interior of M 1/9), which mut then be a maximum. Finally, 5.13) give the maximizer of Ĝ in D 1/9) when the m and the µ ) are fixed to be equal. The maximum value of Ĝ i computed from it definition. Now we recall the definition of the nine overlap variable from Section 4. We oberve that 4.5) map D into a polytope of dimenion 4. The vector n ) in thi polytope can be expreed in term of four variable by n 0,2 = 1/3 n 0,0 n 0,1, n 1,2 = 1/3 n 1,0 n 1,1, n 2,0 = 1/3 n 0,0 n 1,0, n 2,1 = 1/3 n 0,1 n 1,1, n 2,2 = n 0,0 + n 0,1 + n 1,0 + n 1,1 1/3, 5.22) where the variable n 0,0, n 0,1, n 1,0 and n 1,0 take arbitrary nonnegative real value uch that n 0,0 +n 0,1 1 3, n1,0 +n 1,1 1 3, n0,0 +n 1,0 1 3, n0,1 +n 1,1 1 3, n0,0 +n 0,1 +n 1,0 +n 1,1 1 3. 5.23) We are now in a good poition to define the function F ued in the tatement of the Maximum Hypothei. We firt define the domain of F. Thi i the et of all nonnegative real vector n = n 0,0, n 0,1, n 1,0, n 1,1 ) atifying 5.23). For each n in the domain of F, we compute the nine overlap variable from 5.22) and define F n) to be the maximum of ˆF d) over D ubject to the contraint in 4.5). Thi definition of F i repeated in Section 7, which alo contain an alternative equivalent definition. Let b = b ) S, Z3 be the point in D where b = 5 ) for all i, j,. Now we return to 8100 our main function f, which wa defined in 5.1). Lemma 5.6 Under the Maximum Hypothei, the function f ha a unique maximizer in D at b. Moreover, M := max d D { 5 5/2 e fd)} = 25/24. Proof. Recall that f = ln ˆF. The Maximum Hypothei implie that any maximizer of ˆF on D mut atify S d = 1/9, for all i, j Z 3. Let u momentarily relax the contraint 22

in 4.3), and maximize each factor Ĝ d) = 5 ) d ) d m ) 1 2 m ), eparately in D 1/9). In view of Lemma 5.5, b i the unique maximizer and the maximum value of each factor i 5 5/2 25/24) 1/9. We oberve that the contraint in 4.3) are alo atified by b. Therefore b i the unique maximizer of ˆF and the maximum function value i 5 5/2 25/24) 1/9) 9 = 5 5/2 25/24. 5.2 Subexponential Factor Here we complete the computation of the aymptotic expreion of EY 2 ) under the Maximum Hypothei by uing a tandard Laplace-type integration technique. Firt we need the following reult, whoe proof we omit Lemma 5.7 The following ytem of 24 equation in the variable d ha rank 23: t 1d t 1d i+1,j+t = 0 i, j, t, d = 1/3 i, d = 1/3 j. Moreover, after relabelling the variable a d 1,..., d 324, the olution can be expreed by d 1,..., d 301 free, d k = L k d 1,..., d 301, 1/6), k = 302,..., 324, where L k are linear function with coefficient in Z. Hereinafter, we relabel d a d 1,..., d 324 in the ene of Lemma 5.7. The b are alo relabelled a b 1,..., b 324 accordingly. Recall that b wa defined a 5 ) /8100.) For a point d = d 1,..., d 324 ) D, the firt 301 coordinate will be often denoted by d = d 1,..., d 301 ) for implicity. Let ɛ > 0 be fixed but mall enough. We conider the cube of ide 2ɛ centered on b j, i, Q = {d 1,..., d 301 ) R 301 : d k [b k ɛ, b k + ɛ], k} and the dicrete ubet ) J = Q 1 n Z301. Let u define their extenion to higher dimenion: Q = { d 1,..., d 324 ) R 324 : d 1,..., d 301 ) Q, d k = L k d 1,..., d 301, 1/6), k = 302,..., 324}, 23

where the L k are a in Lemma 5.7, and J = Q ) 1 n Z324. Note that b i an interior point of D, and that for each k the function L k, 1/6) i continuou. Then, if ɛ i choen mall enough, we can enure that for ome δ > 0 d Q, d k > δ and d k b k < δ, k = 1,..., 324, 5.24) and hence Q D. Moreover, ince n i alway diviible by 6, for each k the function L k, 1/6) map point from 1 n Z301 into 1 Z, and o J I. n Now recalling the definition of f, g and h in 5.1), we define for any d 1,..., d 301 ) Q fd 1,..., d 301 ) = fd 1,..., d 324 ) gd 1,..., d 301 ) = gd 1,..., d 324 ), where d k = L k d 1,..., d 301, 1/6), k = 302,..., 324. From Lemma 5.6 and by traightforward computation we obtain the following: Lemma 5.8 The following tatement hold: Under the Maximum Hypothei, f ha a unique maximum in D at b. Under the Maximum Hypothei, f ha a unique maximum in Q at b, with e fb) = e f b) = 25 24 55/2 58.2309. The Heian H of f at b i negative definite, and det H = 2 175 3 1078 5 310 7 12 11 14 13 17 79 4. g b) = 2 90 3 558 5 171 0. Both f and g are of cla C in Q. We compute the contribution to EY 2 ) of the term around b and get the following. Lemma 5.9 Under the Maximum Hypothei, d J qn, d)e fd)n 2πn)301/2 det H g b)e n f b) = 23 3 19 5 16 2πn) 301/2 7 6 11 7 79 2 2 13 17 ) n 25 5 5n/2. 24 Proof. From 5.24), we ee that for all d J Q we mut have d k > δ k. Thu, by their definition, all the m are bounded away from 0, qn, d) gd) and we can write qn, d)e fd)n gd)e nfd) = g d)e n f d). 5.25) d J J J 24

We note that both f and g and it partial derivative up to any fixed order are uniformly bounded in the compact et Q. Then, by repeated application of the Euler-Maclaurin ummation formula ee [1], p. 806), we have aymptotically a n grow large g d)e n f d) n 301 g x)e n f x) d x. 5.26) J We oberve from Lemma 5.8 that we are in good condition to apply Laplace method a developed in the multivariate cae by Wong [16, Theorem IX.5.3]. We obtain Q g x)e n f x) d x The reult follow from 5.25), 5.26), 5.27) and Lemma 5.8. Now we deal with the remaining term of the um. Q ) 301/2 1 2π det H g b)e n f b). 5.27) n Lemma 5.10 Under the Maximum Hypothei, there exit ome poitive real α < e fb).t. qn, d)e fd)n = o α n ). I\J Proof. Let B be the topological cloure of D\Q. We recall from Lemma 5.8 that f ha a unique maximum in D at point b / B. Then, ince B i a compact et and f i continuou, there mut be ome real β < fb) uch that f x) β x B. Now we oberve that all term in the um I\J qn, d)efd)n can be uniformly bounded by Cn 162 e βn, for ome fixed contant C. Note furthermore that there i a polynomial number of term at mot n + 1) 324 ) in the um. Hence, the reult hold by taking for intance α = e β + e fb) )/2. From Lemmata 5.9 and 5.10, I gd)e fd)n 23 3 19 5 16 2πn) 301/2 7 6 11 7 79 2 2 13 17 ) n 25 5 5n/2 24 and finally, from thi and Lemma 5.1, we conclude the following. Theorem 5.1 Under the maximum hypothei, EY 2 ) 2 2 3 19 5 16 1 7 6 11 7 79 2 13 17 2πn) 2 ) n 25, 24 which i 2.5). 25

6... and for n not diviible by 6 Since 5-regular graph have an even number of vertice, we only need to conider n 2 or 4 mod 6). One poibility i to rework the whole argument of thi paper but with lightly unbalanced colouring. Intead, the aymmetry in the argument can be omewhat reduced by uing an argument relating different model of random regular graph. We firt treat the cae n 0 mod 6) in more depth, and prove the following. Theorem 6.1 Fix nonnegative integer j 12, j 23 and j 13 and et j = j 12 + j 23 + j 13. Conider the 5-regular graph with n 0 mod 6) vertice and a ditinguihed ordered et of j edge, no two being incident with the ame vertex. Let G be choen uniformly at random from uch tructure. Under the Maximum Hypothei, G a.a.. ha a 3-colouring in which the firt j 12 ditinguihed edge have end vertice coloured 1 and 2, the next j 23 have end vertice coloured 2 and 3, and the ret have end vertice coloured 1 and 3. Proof. Conider the probability pace Ω n with uniform probability ditribution, and whoe underlying et conit of pairing in P n,5 with an ordered et J of j ditinguihed pair of point, uch that no two pair in J are incident with the ame vertex. Let Ŷ denote the number of locally rainbow balanced 3-colouring of a pairing containing J, in which the ditinguihed pair join vertice of the preaigned colour. We will how that EŶ 3 j EY, 6.1) that 3.2) hold with Y replaced by Ŷ and no other adjutment), and that under the Maximum Hypothei, EŶ 2 ) 9 j EY 2 ). 6.2) The theorem then follow immediately by the argument in the lat few entence of the proof of Theorem 1.1. To how 6.1), we apply the ame method a in the proof of Theorem 1.1. Firt, rework the proof of Lemma 3.1, but after aigning vertice to colour clae, elect which pair are in J. Thi can be done in aymptotically 5n/3) 2j way, ince each ditinguihed pair mut join point belonging to vertice of two given colour clae, and if uch pair of vertice are randomly choen, a.a.. no vertex i repeated. Then, for thoe vertice containing a point in a pair in J, the generating function x+1) 5 x 5 1 i adjuted to either x+1) 4 x 4 or x+1) 4 1 ince the choice of colour for the mate of one of the point i already determined. Compare thi with the imilar adjutment made in the derivation of 3.2).) Finally, the remaining part of the) matching between colour clae are choen a before, o between two colour clae where there are j 0 ditinguihed edge, the number of matching i 5n/6 j 0 )!. On the other hand, the total number of choice of the pairing with the ordered et J ditinguihed i aymptotically 5n/2) j P n,d, ince we can chooe firt the pairing and next the ditinguihed edge at random. Thee will atify the nonadjacency condition a.a.. Comparing with the computation of EY, thi produce EŶ 5n/3) 2j 15 2j 30 2j 5n/6) j 5n/2) j EY = 3 j EY, 26