ASYMPTOTIC ENUMERATION BY DEGREE SEQUENCE OF GRAPHS WITH DEGREES o(n 1/2 ) Brendan D. McKay. Nicholas C. Wormald

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1 ASYPTOTIC ENUERATION BY DEGREE SEQUENCE OF GRAPHS WITH DEGREES on 1/ ) Brendan D ckay Computer Science Department, Australian National University, GPO Box 4, ACT 601, Australia Nicholas C Wormald Department of athematics, University of elbourne, Parkville, Vic 05, Australia Abstract We determine the asymptotic number of labelled graphs with a given degree sequence for the case where the imum degree is o EG) 1/ ) The previously best enumeration, by the first author, required imum degree o EG) 1/4 ) In particular, if k = on 1/ ), the number of regular graphs of degree k and order n is asymptotically nk)! nk/)! nk/ k!) exp k 1 ) k n 4 1n + Ok /n) Under slightly stronger conditions, we also determine the asymptotic number of unlabelled graphs with a given degree sequence The method used is a switching argument recently used by us to uniformly generate random graphs with given degree sequences 1 Introduction Where it is suitable, we will use the notation of [1] or [4] For any integer y 0, define [x] y = xx 1) x y +1) Let k = kn) =k 1,k,,k n ) be a sequence of nonnegative integers with even sum Define k = n i=1 k i and k =k 1 + k + + k n )/n Forr 0, further define r = r k) = i [k i] r and ν r = ν r k) = i kr i / k r n) It is easy to see that 1 = ν 0 = ν 1 ν ν, with the inequalities being equalities if and only if k 1 = k = = k n For simplicity, write = 1 Let Gk) be the set of all labelled simple graphs with degree sequence k, and define Gk) = Gk) We are concerned with the asymptotic value of Gk) asn any authors have obtained results by restricting the growth of the imum degree Work prior to [1] can be found summarised there ore recently, a completely different approach [] has born fruit for high degrees Interestingly, the result in both these extreme cases can be cast in a common form Theorem 11 Define λ = k/n 1) and γ = n k ν 1)/n 1) Suppose that either of the following is true: i) 1 k = o 1/4 ), 1

2 ii) k i k = On 1/+ɛ ) and min{ k,n k 1} >cn/log n for sufficiently small ɛ>0 and any c> Then Gk) λ λ 1 λ) 1 λ) n ) n ) n 1 1 exp 4 γ ) 4λ 1 λ) i=1 k i In this paper we will determine the asymptotic value of Gk) whenk = o 1/ ) The result will match Theorem 11 if some extra restrictions are imposed on the amount of variation amongst the degrees The method used closely resembles that of [1], the major improvement being the use of switching operations which lend themselves to easier analysis In Section 6, we will consider the case of unlabelled graphs The model Consider a set of points arranged in cells v 1,v,,v n of size k 1,k,,k n, respectively Take a partition P called a pairing) ofthe points into / parts called pairs)ofsize The degree of cell i is k i The multigraph GP ) associated with P has vertices v 1,v,,v n TheedgesofGP)are in correspondence with the pairs of P ; the pair x, y) corresponds to an edge v i,v j )ifx v i and y v j A loop of P is a pair whose two points lie in the same vertex, while a link is one involving two distinct vertices Two pairs are parallel if they involve the same cells The multiplicity of a pair is the number of pairs including itself) parallel to it A single pair is a pair of multiplicity one A double pair is a set of two parallel pairs of multiplicity two, whilst a triple pair is a set of three parallel pairs of multiplicity three If p is a point, then vp) is the cell containing that point For l, d, t 0, define C l,d,t = C l,d,t k) be the set of all pairings with degrees k, and exactly l loops, d double pairs, and t triple pairs, but no loops of multiplicity greater than one nor pairs of multiplictity greater than three We will make use of the following three operations on a pairing: the first two were introduced in [4] I l-switching: Take a loop {p 1,p 1} and two links {p,p } and {p,p }, such that five distinct cells are involved Replace these three pairs by {p 1,p }, {p 1,p } and {p,p } It is required that all of the pairs created or destroyed be single See Figure 1)

3 p 1 p 1 p 1 p 1 p p p p p p p p Figure 1 An l-switching II d-switching: Take a double link {{p 1,p 1 }, {p,p }}, wherevp 1 )=vp ), and two links {p,p } and {p 4,p 4}, such that six distinct cells are involved Replace these four pairs by {p 1,p }, {p,p 4 }, {p 1,p } and {p,p 4} Other than the original double link, all of the pairs created or destroyed must be single See Figure ) p 1 p p 1 p p 1 p p 1 p p 4 p 4 p 4 p 4 p p p p Figure A d-switching III t-switching: Take a triple link {{p 1,p 1 }, {p,p }, {p,p }}, where vp 1 ) = vp )= vp ), and three other links {p 4,p 4}, {p 5,p 5} and {p 6,p 6}, such that eight distinct cells are involved Replace these six pairs by {p 1,p 4 }, {p,p 5 }, {p,p 6 }, {p 1,p 4}, {p,p 5} and {p,p 6} Other than the original triple link, all of the pairs created or destroyed must be single See Figure ) The inverse of an l-switching will be termed an inverse l-switching, and similarly with the other switching types Note that an l-switching reduces by one the number of loops, without affecting the number of double or triple pairs We will use this fact to estimate the relative cardinalities of C l,d,t and C l 1,d,t The numbers of double and triple links are similarly affected by d-switchings and t-switchings, respectively

4 p 1 p 1 p 1 p 1 p p p p p p p p p 6 p 6 p 6 p 6 p 5 p 5 p 5 p 5 p 4 p 4 p 4 p 4 Figure A t-switching Preliminary results Let P be a random pairing with degrees k 1,k,,k n,where1 k = o 1/ ) We will begin with some elementary bounds on the probability that P has certain substructures The first follows from a simple count Lemma 1 The probability of t given pairs occurring in P is [/] t t [] t t) t Define P k) to be the probability that P contains no loops, and no links of multiplicity greater than one Since each labelled simple graph with degree sequence k 1,k,,k n corresponds to exactly k 1! k! k n! pairings, we have Gk) =! P k) 1) /)! / k 1! k n! Our task is thus reduced to computing P k) We first show that we can ignore pairs of high multiplicity, and bound the number of loops and other non-single links Define N 1 = log, 4 / ) N = log, / ) and N = log, / ) In the following lemma, and for the remainder of the paper, the notations O) and o ) refer to the passage of to infinity within the constraint that k = o) The implied constants will be uniform over all free variables unless otherwise stated 4

5 Lemma 1 P k) = N1 k / N N l=0 d=0 t=0 C l,d,t C 0,0,0 Proof By considering all the possibilities and applying Lemma 1, we find that the probability that P contains a loop of multiplicity greater than one is Ok/ ) Similarly, the probability of a link of multiplicity greater than three is Ok/ 6 )=ok/ ) Consider the probability that there are more than N 1 single loops By Lemma 1, the expected number of sets of l = N single loops is O / l /l! ) = o e/8) ) log = o1/ ) Similarly, the probability that there are more than N double links or more than N triple links is o1/ ) The lemma follows We will estimate C l,d,t / C 0,0,0 via estimates on the terms of the expansion C l,d,t C 0,0,0 = C l,d,t C l,d,t 1 C l,d,1 C l,d,0 C l,d,0 C l 1,d,0 C 1,d,0 C 0,d,0 C 0,d,0 C 0,d 1,0 C 0,1,0 C 0,0,0 Each of these terms can be estimated by means of one of the three switchings In the case of d-switchings, the analysis will be considerably more exacting, and we will need the following result adapted from [1] If K is a multigraph, let ek) denote its number of edges including loops, counting multiplicities), and let e 1 K) denote its number of loops If xx is an edge of K, thenµ K xx ) denotes its multiplicity, ie, the number of edges parallel to xx including itself If K and K are multigraphs with the same vertex set, then K +K is the multigraph with the same vertex set such that µ K+K xx )=µ K xx )+µ K xx ) for all {x, x } Similarly, K means K + K and K + xx is K with the multiplicity of xx increased by one Let L be a graph on n verticeswhichissimpleapartfromalooponeachvertex,and let H be a multigraph on the same vertex set with the restriction that if any edge xx has µ H xx ) 1, then xx is an edge of L Let l denote the imum degree of L Define CL, H) =CL, H; k) to be the set of all pairings P with degrees k such that the following are true for all {x, x }: a) If xx is an edge of L, thenµ GP ) xx )=µ H xx ) b) If xx is not an edge of L, thenµ GP ) xx ) 1 In other words, GP ) must be simple outside L and match H inside L Lemma Suppose that L is as defined above, and H and H + J satisfy the requirements given above for H Leth 1,h,,h n be the degrees of H, andletj 1,j,,j n be the degrees of J Then,ifk k + l )ej) =o), eh) =o), andcl, H), we have CL, H + J) CL, H) n i=1 = [k i h i ] j i e 1J)+eJ) [/ eh)] ej) {x,x } [µ H+Jxx )] µj xx ) 5 k k + l )ej)

6 Proof This is a special case of the combination of Theorems and 6 of [1] We will use Lemma to analyse the structure of C 0,d,0 For a pairing P C 0,d,0,let DP ) be the simple graph with vertices v 1,v,,v n and just those edges which correspond in position to the d double links of P Lemma 4 Let D = DP ) for some P C 0,d,0,where0 d N Let S be a simple graph on v 1,v,,v n which is edge-disjoint from D Letd 1,d,,d n be the degrees of D, s 1,s,,s n be the degrees of S, and suppose that es)k = o) Then the probability that S GP ) when P is chosen at random from those P C 0,d,0 such that DP )=D is n i=1 [k i d i ] si es)k exp O es) [/] es) + d) Proof The lemma is clearly true if s i >k i d i for any i, so suppose that that is not the case Define the graph L which has the edges of D and S as well as a loop on each vertex Then, for any J S, Lemma tells us that CL, D + J) CL, D) = fj), where fj) = n i=1 [k i d i ] ji k exp O ej) [/ d] ej) ej) and j 1,j,,j n are the degrees of J Now, the required probability can be written as fs) J S fj), and since the denominator is 1 + O es)k / ), the lemma follows In the following, we will find it convenient to write k v in place of k i if v = v i Lemma 5 Suppose that 0 d N and Choose v {v 1,v,,v n } and r 0 Then, if P is chosen at random from C 0,d,0,cellv is incident with exactly r double links with probability Q v r)/ k v / i 0 Q v i), where Q v i) = i [d] i [k v ] i i! i, ik exp O + i k + idk Proof Suppose that D = DP )forsomep C 0,d,0,andletw be a neighbour of v in D Let x and x be distinct vertices other than v, such that xx is not an edge of D Let L be the graph with the edges of D, plusxx, plus a loop on each vertex Let R = D vw, 0 α and 0 β, and, for vertex u, letr u denote the degree of u in R Then CL, R + αvw + βxx ) CL, R) = f R α, β; v, w, x, x ) k α+β α! β![/ d +] α+β 6,

7 where, by Lemma, [k v r v ] α [k w r w ] α [k x r x ] β [k x r x ] β, if w {x, x }, f R α, β; v, w, x, x )= [k v r v ] α [k w r w ] α+β [k x r x ] β, if w = x, and [k v r v ] α [k w r w ] α+β [k x r x ] β, if w = x For any simple graph X, letn[x] denote the number of pairings P C 0,d,0 such that DP )=X Then N[R + vw] = CL, R +vw) CL, R +vw + xx ), and similarly for N[R + xx ] Thus, when N[R + vw] 0, N[R + xx ] N[R + vw] = f R 0, ; v, w, x, x ) k f R, 0; v, w, x, x ) = [k x r x ] [k x r x ] [k v r v ] [k w r w ] k, ) since the terms involving f R 1, ; v, w, x, x )andf R, 1; v, w, x, x ) are small enough to be incorporated into the error term Suppose 1 i k v / Define Ri) to be the set of all simple graphs on V = {v 1,v,,v n } with exactly d 1 edges, of which exactly i 1 are incident with v For R Ri), let X R) denote the set of all distinct pairs {x, x } such that x v, x v and xx is not an edge of R Similarly, let WR) denotethesetofallw V such that w v and vw is not an edge of R If n i denotes the number of pairings P C 0,d,0 such that exactly i double links are incident with v, then 1 n i 1 = N[R + xx ] d i +1 and n i = 1 i R Ri) w WR) R Ri) xx X R) N[R + vw] From ) we find that, for any w and R Ri) for which the denominator is non-zero, xx X R) N[R + xx ] k = N[R + vw] [k v r v ] [k w r w ] + dk + k To see this, express the numerator as a sum over all ordered pairs xx V V and subtract those not in X R We can sum over w in a similar way to obtain, for any R Ri) for which the denominator is non-zero, xx X R) N[R + xx ] w WR) N[R + vw] = k [k v i 1)] + ik + dk Note that for sufficiently large, both the numerator and denominator must be non-zero since d = o / Since this is uniform over R, we conclude that n i = d i +1)[k v i 1)] k n i 1 i + ik + dk Identifying n r /n 0 as the quantity Q v r), we now obtain the lemma on taking the product over i 7

8 4 Analysis of the switchings We now analyse each of the switching types in turn, under the assumptions of Section Lemma 41 Suppose 0 l N 1, 0 d N and 1 t N Then,if > 0, C l,d,t C l,d,t 1 = k 1t k + l + d + t) Proof To simplify the consideration of equivalences, we will consider each of the points involved in the t-switching to be separately labelled with the labels p i and p i used above in the definition of a t-switching) Choose an arbitrary P C l,d,t,andletn = NP ) be the number of t-switchings which can be applied to P We can choose a triple link and its labelling in 1t ways, and choose three distinct labelled single links {p 4,p 4 }, {p 5,p 5 } and {p 6,p 6 } in [ l 4d 6t] ways Of these choices, some are not satisfactory Unwanted coincidences like vp 1 )=vp 4 )account for Otk ) choices, while those like vp 4 )=vp 5 ) account for Ot ) The forbidden cases where, for example, P already has a pair involving vp 1 )andvp 4 ) account for Otk ) Overall, we find that k N =1t + l + d + t Now choose an arbitrary P C l,d,t 1,andletN = N P )bethenumberofinverse t-switchings which can be applied to it We can choose two distinct -stars in [ ] ways Of these choices, we must eliminate those not permitted Unwanted coincidences, like vp 1 )= vp 1 ), vp 4)=vp 5 )orvp 4)=vp 1 ) account for Ok ) choices Cases where P already has a pair involving vp 1 )andvp 1)orvp 4 )andvp 4), for example, account for Ok 4 ) choices Finally, cases where either of the -stars include loops or non-single pairs account for O k l + d + t) ) choices Overall, we find that N = k k + l + d + t) The error term for N dominates that for N, so the lemma follows on considering the ratio N /N Lemma 4 Suppose 1 l N 1 and 0 d N Then C l,d,0 C l 1,d,0 = k l + k d + k l Proof We use the same method as for Lemma 41 Let P be an arbitrary member of C l,d,0,andletn = NP )bethenumberofl-switchings which can be applied to it We can choose and label the loop in l ways, then choose two 8

9 single labelled links in [ l 4d] ways This overcount needs to be corrected for unwanted coincidences like vp 1 )=vp )Olk ) choices), and unwanted adjacencies like a pair involving vp 1 )andvp )Olk Thus, N =l k + l + d Conversely, let P be an arbitrary member of C l 1,d,0,andletN = N P )bethenumber of inverse l-switchings which can be applied to it We can choose the -star in and {p,p } in ways This overcount needs to be corrected for unwanted coincidences like vp ) = vp )Ok ) choices), unwanted adjacencies like a pair involving vp )and vp )Olk + k ) choices), and unwanted involvement of loops or non-single pairs Ol + d)k ) choices) Thus k N = + k d + kl The lemma now follows on comparing N to N Lemma 4 Suppose 1 d N Then,if > 0, C 0,d,0 C 0,d 1,0 = k k 4d + d) Proof This can be proved by precisely the same method used for the previous two lemmas Details can be found in [4] Whilst we will use Lemma 4 in one special case, it is not sufficiently accurate for us in general The reason is that the number of double links in a random pairing is in general much higher than the numbers of loops or triple links However, Lemma 4 is the best that can be done using uniform counts over arbitrary members of C 0,d,0 and C 0,d 1,0 In order to do better, we need to consider averages over C 0,d,0 and C 0,d 1,0 Lemma 44 Suppose 1 d N and Then C 0,d,0 C 0,d 1,0 = 4d d 16d k + d + O Proof Define N to be the average number of possible d-switchings, where the average is over all P C 0,d,0 We can choose {p 1,p 1,p,p } in 4d ways and then {p,p,p 4,p 4} in at most [ 4] ways This gives us the initial overcount N N =4d[ 4] =4d 1/ ) ) However, some of these choices are not legal We can divide the set of illegal choices into three families: X 1 : These are choices involving too few vertices, for example if vp 1 )=vp )orvp )=vp 4) X : These are the choices for which the pairing already has a link involving vp 1 )andvp ) or the three other similar cases However, we exclude any choice which belongs to X 1 9

10 X : These are choices for which either {p,p } or {p 4,p 4} has multiplicity two However, we exclude any choice which belongs to X 1 In each of these three cases, we will consider the probability that randomly choosing one of the N possibilities described above gives that case, where the probability is taken over random P We will also bound the probability that X and X occur together Case X 1 : The probabilty of landing in X 1 is easily seen to be at most Ok / ), by just counting the cases Case X : Let P i i =1, ) be the probability that there is a pair {x, x } of multiplicity i such that vx) =vp 1 )andvx )=vp ) Note that our conditions on d, k, and imply that dk / = o1) From Lemma 5, the expected number of pairs of adjacent double links is Od 4 / )=Od k / ) Allowing k for the choices of p and for the choice of {p 4,p 4}, we find that P = O dk/ ) ) = Ok / ) P 1 is more involved For any choice of DP ), p 1, p 1, p and p, there are on average k k v r) + dk choices for p, p, p 4 and p 4,wherev = vp 1)andr is the number of double links incident with v This follows from Lemma 4 on summing over all the possibilities If K denotes the expected number of configurations included in the value of P 1, then by Lemma 5, where k K = k =4d S v i) = i [d 1] i [k v ] i i! i + dk + dk rk v r) v r 1 i 0 [k v ] S v i) i 0 Q v i), v Q v r) i 0 Q v i) ik exp O + i k + idk Since i 0 Q v i) 1, i 0 S vi) i 0 Q vi) = Q v i) S v i ) i 0 Using the inequality e x 1 < x e x, we find that Q v i) S v i) = i [d 1] i 1 [k v ] i i! i O e Oz) ikv + dikv + dkvz) ), where z = ik / +i k +idk )/ Hence the terms of the series i 1 Qv i) S v i) ) are boundedin magnitude by those of a geometric series with ratio o1), and thus by a constant multiple of the bound on the first term, that is by O k k + d)/ ) Overall, we find that P 1 = + O k ) + dk 10

11 Any two of the eight events counted in X single or double link in any of four positions) occur together with probability Ok / ), so altogether we find that X occurs with probability 4 + O k ) Case X : With the help of Lemmas 4 and 5, a routine calculation gives the probability of this case as 8d/ + Ok / ) Events X and X occur together with probability Ok / ), by similar reasoning Thus we have altogether that N =4d 1 4 8d + O k Conversely, define N to be the average number of possible inverse d-switchings, where the average is over all P C 0,d 1,0 For each choice of v = vp 1 ), there are at most [k v ] ways to choose p 1 and p A similar bound holds for vp 1), and so we an initial overcount N However, some of these choices are not legal and, as before, we divide these into a number of cases: Y 1 : These are choices involving too few vertices, for example vp 1 )=vp 1)orvp )=vp ) Y : These are choices where there is already a link involving vp 1 )andvp 1), excluding anything in Case Y 1 Y : These are choices where there is already a link involving vp )andvp ), or vp 4)and vp 4) Again, we exclude anything in case Y 1 Y 4 : These are choices for which one or more of the pairs chosen have multiplicity two, except any choice in case Y 1 These four cases can be analysed using the same method used for X 1 X For cases Y and Y, we can simply sum over all the possibilities using Lemma 4 For Case Y 4, we need Lemma 5 We will merely state the probability in each case, leaving the details to the reader k ) Case Y 1 : O k ) Case Y : + O k ) Case Y : + O 16d Case Y 4 : d ) + O The conjunction of any two of these cases gives no new error terms, so overall we have N = 1 The lemma now follows on comparing N to N 16d k + O + d 11

12 5 Consolidation With the aid of the lemmas in Section 4, we can now apply Lemma to estimate P k) As before, we assume that 1 k = o) Lemma 51 P k) =exp k 6 + O Proof Let 0 l N 1,0 d N and 0 t N If = 0, then clearly C l,d,t =0ift>0 Suppose instead that > 0 By Lemma 41, Summing over t, weobtain C l,d,t C l,d,0 = t tk 1 t t! t exp O k + l + d)+t k N t=0 which just happens to be true also for =0 Similarly, from Lemma 4, we have C l,d,t C l,d,0 =exp kk 1 + O + l + d), ) C l,d,0 C 0,d,0 = l lk l l! exp O l + lk d + kl Combining this with ) and summing over l, weobtain N 1 N l=0 t=0 C l,d,t C 0,d,0 =exp O k k + d) 4) Now suppose that From Lemma 44, we have that C 0,d,0 C 0,0,0 = d 4 d d! d exp 4d + 4d d d 8d dk + d) + O Combining this with 4) and summing over d with the help of the approximation d d /4 ), we obtain N N 1 N d=0 l=0 t=0 C l,d,t C 0,0,0 =exp O k 5) Inthecasewhere0< <, Lemma 4 gives the same result to within the same error In the trivial case = 0 which implies = 0) Equation 5) again holds The desired estimate now follows from Lemma We now have the result we have been seeking 1

13 Theorem 5 If 1 k = o 1/ ),then Gk) = =! /)! / k 1! k n! exp k O! /)! / k 1! k n! exp k ν 1 k 6ν ν ν4 ν ) + O 4 1n uniformly as Proof The first expression follows from Lemma 51 and Equation 1), and the second from the first using the simple bound kν = Ok ) Corollary 5 If 1 k = on 1/ ), the number of labelled regular graphs of degree k and order n is asymptotically as n nk)! nk/)! nk/ k!) exp k 1 ) k n 4 1n + Ok /n) k, Corollary 54 Define k min =min n i=1 k i Then Theorem 11 holds in the additional case iii) 1 k = o 1/ ) and k k min = o minn 1/8 k5/8,n 1/6 k1/ ) ) Proof For k = o 1/ ), the estimate of Gk) in Theorem 11 can be expanded as! /)! / k 1! k n! exp k ν 1 k 5 + ν +6ν 1ν ) k + O 4 1n Since ν =1+σ and ν =1+σ + σ,whereσ r = 1 n n i=1 k i/ k 1) r, we find that Theorem 11 holds provided k 6σ +6σ σ σ4 σ ) n = o1) Since σ r k k min )/ k) r, the claim follows after a routine calculation Corollary 54 adds additional support to the following conjecture, which first appeared in [] In fact, for k = o 1/ )wecantakeɛ =1/8 Conjecture There is some absolute constant ɛ > 0 such that the conclusion of Theorem 11 holds whenever 0 < k < n 1, k k min = o n ɛ min{ k,n k 1} 1/) and min{, n ) } as n 6 Unlabelled graphs Under some additional constraints, Theorem 5 can be applied to estimate the number of unlabelled graphs with a given degree sequence 1

14 Theorem 61 Let k be a graphical degree sequence with no entries of value 0, n 1 entries of value 1, and n entries of value Then, under any one of the following conditions, the number of unlabelled simple graphs with degree sequence k is asymptotically Gk)/n! i) n 1 = On 1/ ), n = On / ),andk 1 log n/ log log n; ii) k min 4 and k = o k 1/ n 1/1 ); iii) k min 5 and k = o k 1/ n /15 ); iv) k min 6 and k = o k 1/ n 1/4 1/kmin) ) Proof Under each of the conditions given, a slight weakening of Theorem 4 and Corollary 4 of [] shows that the expected number of non-trivial automorphisms of a random member of Gk) iso1), which implies the theorem Note that Theorem 61 covers the case of regular graphs of degree k for all k = kn) such that k = on 1/ ) Acknowledgement We wish to thank Wang Xiaoji for checking all the calculations References [1] B D ckay, Asymptotics for symmetric 0-1 matrices with prescribed row sums, Ars Combinatoria, 19A 1985) 15 5 [] B D ckay and N C Wormald, Automorphisms of random graphs with specified degrees, Combinatorica, ) 5-8 [] B D ckay and N C Wormald, Asymptotic enumeration by degree sequence of graphs of high degree, European J Combinatorics, ) [4] B D ckay and N C Wormald, Uniform generation of random regular graphs of moderate degree, J Algorithms, )