The ores of random hypergraphs with a given degreesequen e. �Colin CooperyAugust 30, 2003Abstra tWe study random r-uniform n vertex hypergraphs with �xed degree sequen e d =(d1; :::; dn), maximum degree � = o(n1=24) and total degree �n where � < 1. We givethe whp size, number of edges and degree sequen e of the �- ores (� � 2) up to an errorof O(n2=3�4=3 log n). In the ase of graphs (r = 2) we give further stru tural details su has the number of tree omponents and, for the ase of smooth degree sequen es, the size ofthe mantle. We give various examples, su h as the ores of r-uniform hypergraphs with anear Poisson degree sequen e, and an improved upper bound for the �rst linear dependen eamong the olumns in the independent olumn model of random Boolean matri es.1 Introdu tionAn interesting question in the theory of random graphs on erns the size of thelargest omponent of su h a graph. In their formative paper [8℄, Erd}os and R�enyiproved a strong di hotomy for the size C1 of the largest omponent of the randomgraph Gn;m when m = n=2, onstant. Erd}os and R�enyi showed that, in Gn;m,if < 1 then whp1 C1 = O(logn) and that if > 1 then C1 � G( )n for somefun tion G( ) > 0. A omponent of order n is alled a giant omponent. When = 1 the situation is more ompli ated and mu h e�ort has gone into an analysisof this ase. See e.g. Bollob�as [3℄, Lu zak [12℄, Lu zak, Pittel and Wierman[13℄, Janson, Knuth, Lu zak and Pittel [10℄ and the books by Bollob�as [4℄ and byJanson, Lu zak and Ru i�nski [11℄.The �- ore of a hypergraph G is the largest vertex indu ed subgraph of minimumdegree �. It an be obtained by repeatedly removing edges in ident with verti es�Revised RSA version: File rhg ore 4.texyDepartment of Computer S ien e, King's College, London WC2R 2LS, UK. e-mail: ooper�d s.k l.a .uk1An event E o urs with high probability (whp) if Pr(E) tends to 1 as n!11

of degree less then � until no su h verti es remain. Quite possibly the �- ore isempty.Let V�(G) denote the number of verti es in the �- ore of G. Given a spa e ofrandom graphs, we an ask if a non-empty �- ore exists whp. A losely relatedquestion on erns the whp emergen e of a giant �- ore (V�(G) = �(n)) in Gn;mor Gn;p or other suitably parameterized sequen es of spa es. See Pittel, Spen erand Wormald [20℄ for a substantial treatment of this problem. A re ent paper byKim [14℄ gives results for �- ores of r-uniform hypergraphs Gn;p using an a Poisson lumping te hnique.Let n be �xed, let d = (d1; d2; : : : ; dn) and let � = (d1 + d2 + � � � + dn)=n. Letr � 2 be an integer dividing �n, so that �n = mr. Let G(d) be the set ofsimple r-uniform hypergraphs G = (V;E) with vertex set V = [n℄ and edge setE = fe1; :::; et; :::; emg, where jetj = r; t = 1; :::; m. For i 2 [n℄ the degree of vertexi is di. Let � = maxi di be the maximum degree of G. Let Li be the number ofverti es of degree i in G(d). Thus the total degree of G is �n = P iLi, and � isthe average degree. We assume L0 = 0 unless stated otherwise.This paper studies the whp properties of �- ores (� � 2) of simple random r-uniform n vertex hypergraphs G with degree sequen e d. The whp size of the�- ore (� � 2) depends primarily on a single parameter b� = b�(�), where b� is thesmallest solution for t 2 [0; �n=r℄ ofq(t) = ��1� rt�n� ; (1)and the fun tion q(t) is given byq(t) = Xi�� Lin iXj=� j�ij���1� rt�n�r�1r �j �1� �1� rt�n�r�1r �i�j : (2)We note that q(�n=r) = 0, so equation (1) always has a solution b�.Let a(t) = �(1 � rt=�n) � q(t) where q(t) is given by (2). Thus b� is the smallestnon-negative solution of a(t) = 0. The fun tion a(t) omes from the analysis ofthe algorithm whi h we use to generate a random multi-hypergraph with degreesequen e d. To simplify notation we measure all errors in the analysis of thealgorithm in terms of a single parameter � where� = n2=3�4=3 logn:We say that the degree sequen e d satis�es the Separation Property if the following onditions hold:S1: � = o(n1=24), � <1 and P i(i� 1)Li=(�n) <1.2

S2: If t < b� �K� then na(t) = (�), where K is a large positive onstant.S3: na0(b�) < 0 onstant, and a(b� + h) = ha0(b�)(1 + o(1)), for jhj = O(�).We give a brief informal dis ussion of our hoi e of these onditions. ConditionS1 is somewhat arbitrary, but ensures there is a positive probability that themulti-hypergraph we generate is simple. Conditions S2, S3 are mainly to avoidthresholds. We identify a threshold for a �- ore with a minimum of a(t) tou hingthe t axis at t < �n=r. Condition S3 requires a(t) to be learly de reasing at theroot b�. This is to avoid a near minimum, whi h we identify with being lose to athreshold. Also we need (2) to have a smooth Taylor expansion about b�. Given thefun tional form of q(t) it seems reasonable that this ould be proved dire tly, butwe have not done so. Condition S2 avoids degree sequen es very lose to a possiblese ondary �- ore threshold (earlier than b�). It also avoids degree sequen es whi hdo not �t the framework of our analysis. For example, a spa e of graphs onsistingof a entral vertex with O(�) paths of length O(n=�) atta hed. Looking for a 2- ore by removing verti es of degree 1 means that for O(n) steps only O(�) verti esof degree 1 are ever exposed.We aim to answer two questions. Do graphs in G(d) have a signi� ant �- orewhp and if so, what are its properties? Theorem 1 gives the order of magnitudeof the ore, and Theorem 2 gives the pre ise details if � logn � b� � �n=r�� logn.Theorem 1. Let G(d) be the spa e of simple r-uniform n vertex hypergraphs withdegree sequen e d satis�ng � = o(n1=24), � <1 and P i(i� 1)Li=(�n) <1. Letq(t) = Xi�� Lin iXj=� j�ij���1� rt�n� r�1r �j �1� �1� rt�n� r�1r �i�j ;and let a(t) = �(1� rt=�n)� q(t).Let � � 2, and let b�(�) be the smallest solution in [0; 1℄ to a(t) = 0. The followingresults hold whp(i) If b�(�) � �n=r � � logn and a(t) = (�) for t � �n=r � � logn,(a) if r = 2; � = 2 the number of edges in the �- ore is O(� logn),(b) if r = 2; � � 3 or r � 3; � � 2 the �- ore is empty.(ii) If � logn � b�(�) � �n=r� � logn and d satis�es S2, S3, the number of edgesin the �- ore is � (�n=r � b�).(iii) If �(�) < � logn and d satis�es S3, the �- ore ontains �n=r � O(� logn)edges.Let V�(G); E�(G) be the number of verti es and edges of the �- ore (respe tively).Let Li;j(G) be the number of verti es of degree i in G whi h have degree j the3

�- ore. In the ase of graphs (r = 2), let R(G) be the number of isolated tree omponents. The following theorem gives these stru tural details of the �- ore.Theorem 2. Let G(d) be the spa e of simple r-uniform hypergraphs with a degreesequen e d whi h satis�es the Separation Property.Let � � 2. Let � logn � b�(�) � �n=r� � logn, and let bx = (1� rb�=�n) r�1r . Thenwhp E�(G) = �nr � b� +O(�)V�(G) = Xi�� Li iXj=� �ij� bxj (1� bx)i�j +O(�)Li;j(G) = Li�ij� bxj(1� bx)i�j +O(�):Also, for graphs (r = 2), whpR(G) = Xi�0 Li(1� bx)i � �n2 (1� bx)2 +O(�):In Se tion 2 we de�ne an algorithm, onstru t to generate a random multi-hypergraph. In Se tion 3 we give the proof of Theorems 1, 2. In Se tion 4 we lookat results relating to 2- ores of graphs. We ombine our results with the work ofMolloy and Reed [16, 17℄ on giant omponents of smooth degree sequen es, to givethe size of the mantle of su h sequen es. In Se tion 5 we review results for �- oresof hypergraphs Gn;p; Gn;m and give an appli ation to upper-bounding the rank ofrandom Boolean matri es.The rest of this se tion ontains an informal dis ussion of results, and examples.Re all that the size of the �- ore depends on a single parameter b�, whi h is thesmallest solution for t 2 [0; �n=r℄ of q(t) = � (1� rt=�n). We de�ne the followingvariables li = Li=n; pi = ili=�; � = r�1r ; x = �1� rt�n�� : (3)Using this notation q(t) an we written as q(x(t)) whereq(x) = Xi�� liXj�� j�ij�xj(1� x)i�j; (4)and a(t) as a(x(t)) = �x1=� � q(x). We see that q(x) = �xf(x) wheref(x) = Xi�� pi Xj���1 �i�1j �xj(1� x)i�1�j: (5)4

Let bx be de�ned by bx = �1� rb��n��. Thus a(bx) = 0 and q(bx) satis�esq(bx) = �bx1=� = �bxf(bx): (6)The solutions to this are bx = 0 andbx = Xi�� pi Xj���1 �i�1j �bxj(1� bx)i�1�j!r�1 (7)= (f(bx))r�1:Informally, the probability bx that an edge in general position survives in the �- oreis the probability that ea h of its r�1 terminal verti es survives by having at leastj � � � 1 surviving des endants. This orresponden e with a bran hing pro esswas noted in [20℄.The results of this paper are for a �xed degree sequen e d. If the degree sequen ed = d( ) is parameterized by (e.g. li( ) � ie� =i!), then it may be possible touse the results of this paper to prove that the �- ore exhibits threshold behaviouras varies.We note that a(�n=r) = 0 always, and presume a(0) > 0. We say the parameter-ization of d( ) is riti al at � if, (i) for < �, a(t; ) > 0 for t < �n=r, (ii) at � a minimum of a(t; �) tou hes the t-axis at b�( �), (iii) for > �, a(t; ) rossesthe t-axis at some b�( ) < �n=r.Using q(x) = �xf(x), the riti al ondition (ii) that a(t) = a0(t) = 0 an be writtenas x 1r�1 = f(x) = (r � 1)xf 0(x); (8)whi h is a onvenient set of equations for the riti al parameters ( �; bx( �)).It seems natural to interpret the point � as the threshold for the �- ore. Assumingd( ) satis�es S1 and S2, to prove threshold behaviour at � the following approa h an be adopted.(i) Derive the riti al value � using (8).(ii) Find an �! 0 su h that if = �� � then b� > �n=r�� logn, and if = �+ �then � logn < b� < �n=r � � logn.(iii) Prove the degree sequen e d( ) satis�es S3 for � � + �.(iv) Apply Theorem 1. For � ��� the (possibly empty) ore is of size O(� logn)whp. For � � + �, the size of the ore as �(�n=r � b�) whp. The whpproperties of the ore for � � + � are given by Theorem 2.5

We next give two examples of thresholds in r-uniform hypergraphs obtained inthis way. The �rst example is the well known ase of random hypergraphs Gn; =nor Gn;m. We note that similar results have been given by several authors (see eg:[20℄). For these hypergraphs, the degree sequen e is approximately nPo( ). Thisapproximation is lose enough to produ e a urate results, as is shown in Se tion5. For the sequen e (li = ie� =i!; i � 0), thenq(x) = x�1� e� x�1 + x + � � �+ ( x)��2(�� 2)!�� : (9)Using q(x) = �xf(x) and applying (8) we �nd the following. Let y = x. The riti al values ( �; by) are given by the solutions of(r � 1)(�� 1)� 1 = y� + y2�(�+ 1) + � � �+ yj�(j) + � � � = y� (�� 2)!eyy��1(r � 1)�r�1 :The se ond example is the emergen e of (d+ 1)- ores in graphs with a mixture ofverti es of degree d and d+ 2 in the proportions 1� and respe tively2. For thesequen e (ld = 1� ; ld+2 = ), then � = d+ 2 andq(x) = (d+ 2) �(d+ 1)xd+1(1� x) + xd+2� :Using q(x) = �xf(x) and applying (8) gives the riti al values for graphs asbx = d2 � 1d2 � = d2d(d+ 1)(d+ 2)(d2 � 1)d�1 � 2d2d�1 :For example, if d = 1, the threshold for a 2- ore o urs when the proportion ofverti es of degree 3 is 1=4.2 Generating hypergraphs with a �xed degree sequen e.Given a hypergraph G, the obvious way to �nd the �- ore is to repeatedly deletean edge in ident with a vertex of degree less than � in the remaining subgraphH, until either H is empty or all verti es have degree at least �. We use analgorithm, onstru t, to generate uniformly at random (uar) a on�gurationmulti-hypergraph F with degree sequen e d. In the version of onstru t whi hwe analyse, the edges e1; e2; :::; e�n=r of F are generated in an order suitable fordeletion by the pro ess given above to obtain the �- ore.2This ni e example was suggested by Andrzej Ru inski6

We say a hypergraph G = (V;E) is simple if no edge e 2 E ontains a repeatedvertex, and there are no repeated edges in E (ie: E is not a multi-set). We remarkthat as far as the proofs in this paper are on erned there is nothing to prevent amore restri tive de�nition de�nition of simple, for example that no pair of edgesshare a ommon pair of verti es (the edges are of o-degree 1).Let G(d) be the set of hypergraphs with vertex set V = [n℄ and degree sequen ed. The standard method for generating uniformly distributed simple graphs (andhypergraphs) with a given degree sequen e is to use the on�guration model of Bol-lob�as [2℄ whi h is a probabilisti interpretation of the ounting formula of Benderand Can�eld [1℄.Let W = [�n℄. Let d = (d1; d2; :::; dn) and let P (d) be the ordered partitionW1;W2; :::;Wn of W into sets of size jWij = di. Thus (maxx 2 Wi) + 1 = min y 2Wi+1. De�ne �d : W ! [n℄ by �(w) = i if w 2 Wi. We refer to the elements ofW as ( on�guration) points, and the sets Wi i 2 [n℄ as verti es. Relative to thepartition P (d), whi h remains �xed in all future dis ussions, the degree sequen eof W is d.Let F be a partition of W into m = �n=r sets of size r. Let be the set of su hpartitions F of W . Thus jj = (�n)!=((�n=r)!(r!)�n=r). We refer to the elementsF of as on�gurations, the elements of F as edges.Let (F ) denote the multi-hypergraph with vertex set [n℄ and edge multiset EF =f�(e) = f�(x1); �(x2); :::; �(xr)g : e = fx1; x2; :::; xrg 2 Fg. Relative to the �xedpartition P (d) the degree sequen e of (F ) is d. Let M(d) = f (F ) : F 2g be the set of underlying multi-hypergraphs and G(d) the subset of simplehypergraphs. It is a standard result that under the mapping the subset G(d)has uniform measure if does.In the ase of graphs, Frieze and Lu zak [9℄ des ribe a simple algorithm for obtain-ing a u.a.r multi-graph on�guration F from . The algorithm whi h was usedin [7℄ under the name onstru t, generalizes naturally to the uar generation ofmulti-hypergraph on�gurations:Algorithm onstru tbeginW (0) := [�n℄F (0) := ;C(0) = (F (0);W (0))For s = 1 to �n=r dobeginChoose a point xs 2 W (s� 1) using a deterministi rule (R*)Choose r � 1 points ys = (ys;2; :::; ys;r)7

uniformly at random without repla ement from W (s� 1) n fxsgF (s) := F (s� 1) [ ffxs; ys;2; :::; ys;rggW (s) := W (s� 1) n fxs; ys;2; :::; ys;rgC(s) = (F (s);W (s))endF := F (�n=r)endThe set C(s) = (F (s);W (s)) des ribes the output of onstru t after step s. It onsists of a set of edges F (s) and a set of remaining on�guration points W (s).Let v be a vertex. At the end of step s, Wv(s) = Wv\W (s) is the set of remaining on�guration points of vertex v. Let d(v; s) = jWv(s)j. We all d(v; s) the residualdegree of vertex v at (the end of) step s.During step s, the algorithm generates the edge sequen e es = (xs; ys;2; :::; ys;r).Let �(es) = fu; v1; :::; vr�1g = f�(xs); �(ys;2); :::; �(ys;r)g denote the underlyingmulti-hypergraph edge generated. We refer to xs as the initial point and to ys;k; k =2; :::; r as the terminal points, and similarly to u = �(xs) as the initial vertex of theedge, and to v2 = �(ys;2); :::; vr = �(ys;r) as the terminal verti es. If the ontextis lear we suppress the �, thus statements su h as v 2 es mean z 2 es; v = �(z)et .Let C(s) = fC(s)g be the possible outputs of onstru t after step s. Be ause ofthe deterministi hoi e of xs at (R*) we have jC(s)j = jC(s�1)j(�n�r(s�1)�1)(r�1)=(r � 1)! and thus jC(s)j = (�n)(rs)=((�n=r)(s)(r!)s). Also, the probabilitythat C(s) is the output at step s is 1=jC(s)j.Algorithm onstru t generates on�guration multi-hypergraphs. The proofs inthis paper are based on the analysis of onstru t. In order for the proofs to bevalid for simple hypergraphs we require that the subset the on�gurations indu edby G(d) has onstant measure.In general, the expe ted number of loops (edges with repeated verti es) is O(�2).The expe ted number of repeated edges is O(�2r=nr�2). This is o(1=pn) for r � 3and � = o(n1=24). By requiring that � is onstant and P i(i � 1)Li=(�n) < 1( ondition S1 of the Separation Property) we an ensure thatG(d) is large enough.Lemma 3. [6℄ Let � be onstant, let P i(i� 1)Li=(�n) be �nite and let the max-imum degree � = o(n1=24), then Pr(G is simple ) > �(n) where �(n) > 0 onstantand limn!1 �(n) = � > 0.Finally, we remark that given the sequen e of edges e1; :::; et generated at the endof step t of the algorithm, the remaining on�guration is generated uar from the8

set W (t). Thus, onditional on the residual degree sequen e and the hypergraphbeing simple, the �- ore is a random hypergraph with the given degree sequen e.Properties of algorithm onstru tWe will use a version of onstru t whi h �rst generates the edges outside the(possibly empty) �- ore, and then generates the �- ore. We all this version onstru t(�). It hooses initial point xt+1 using the following rule:(R*):Sele t xt+1 from W (t) as follows.Æ(t) = minu2V fjWu(t)j : jWu(t)j > 0gU(t) = fu : jWu(t)j = Æ(t)gX(t) = [u2U(t)Wu(t)xt+1 = minfy : y 2 X(t)gInformally we an write:(R*): Sele t xt+1 from W (t) by hoosing the smallest point label fromthe verti es of minimum positive residual degree.A losely asso iated version of onstru t, alled onstru t-[v℄ uses the fol-lowing rule, whi h di�ers only in that vertex v is never sele ted as initial.(R*): Sele t xt+1 from W (t) � Wv(t) by hoosing the smallest pointlabel from the verti es of minimum positive residual degree other thanv, else stop.For the purpose of analysis, we say onstru t is in Phase I at step t, if for allsteps T � t the point xT hosen as initial omes from a vertex of degree less than� (ie: 8T�t d(�(xT ); T � 1) < �). If some vertex of degree at least � has been hosen as initial by the end of step t, we say the algorithm is in Phase II.Let C(t) = (F (t);W (t)) be some �xed output of onstru t after t steps, whereF (t) is the set of generated edges, and W (t) is the set of unused on�gurationpoints. The algorithm has no memory, so starting with C(t) the outputs C(t +� ;C(t)) are exa tly those obtained by running it for � steps with initial input(F (t);W (t)). In the dis ussions below we often treat C(t) as C(0) in this way.In what follows we al ulate the probability Pr(d(v; t) = j) as a fun tion of v; t; jand other variables, parti ularly d; n. Generally we have Pr(d(v; t) = j; d(v; s) =9

i;d(s);d; n;Alg ), where i � j, s � t, d(s) is the degree sequen e at step sand Alg is the version of onstru t. In its simplest form it is Pr(d(v; t) =j;d; n; onstru t(�)). We an �x on an output C(s) = (F (s);W (s)) withd(v; s) = i and residual degree sequen e d(s), and restart the algorithm with C(s)as input. In onstru t(�), the probability Pr(d(v; t) = j) is not a fun tion ofv, and any other vertex w su h that d(w; 0) = d(v; 0) would give the same value.For onstru t(�) let C�(j; t) be the subset of outputs C(t) = (F (t);W (t)) ofC�(t) su h that d(v; t) = j (ie: jWv(t)j = j). Let P� denote probability overoutputs of onstru t(�). For onstru t-[v℄, we similarly use the notationCv(t), and Pv respe tively.Let d(v; 0) = i, and for onstru t-[v℄ let �i;j(0; t; v) = Pv(d(v; t) = j). Thus�i;j(0; t; v) = jCv(j; t)jjCv(t)j :The notation �i;j(0; t) = �i;j(t) assumes we start at time 0 with d(v; 0) = i andW (0) = �n on�guration points. The notation �i;j(s; t) assumes d(v; s) = i andwe start at step s+1 with W (s) = �n�rs points. The values of these probabilitiesare given by the following lemma, whi h is proved in the Appendix.Lemma 4. Let � = (r � 1)=r and let x(s; t) = ((�n� rt)=(�n� rs))�. Let�i;j(s; t) = �ij�xj(1� x)i�j: (10)Let d(v; s) = i, then in onstru t-[v℄�i;j(s; t; v) = �i;j(s; t)(1 + �(s; t));where �(s; t) = O � �3�n�rt + �3t�s� : (11)Remark 1. We note thatlXi=j �l;i(0; s)�i;j(s; t) = �l;j(0; t): (12)ProofLet x = � �n�rs�n �� ; y = � �n�rt�n�rs�� then xy = � �n�rt�n ��. The result follows fromXi �li�xi(1� x)l�i�ij�yj(1� y)i�j = �lj�(xy)j(1� xy)l�j:10

2Let Lj(T ) = jfv : d(v; T ) = jgj ount the verti es of residual degree j in W (T ).Given an output C(T ), let L(T ) = (L0(T ); L1(T ); :::; L�(T )). Let A(t) = L1(t) +2L2(t) + � � �+ (�� 1)L��1(t), be the number of available points.We say C = C(t) is � -good if A(t) � r� . If C(t) is � -good and in Phase I thenthis guarantees that all outputs C(t+ � ;C) are in Phase I. The following lemmais pivotal in the proof.Lemma 5. If C(t) is � -good and i � j � � then for C(t + �) 2 C(t + � ;C) in onstru t(�) we haveP�(d(v; t+ �) = j ; d(v; t) = i; C(t) is � -good ) = �i;j(t; t + � ; v):Proof We laim thatjCv(t+ � ;C)j = jC�(t+ � ;C)j;Cv(j; t + � ;C) = C�(j; t + � ;C):The outputs obtained are equivalent to running the appropriate version of on-stru t for � steps, starting with C(t) (ie: jW (0)j = �n � rt). Any version of onstru t starting with W (0) = �n satis�esjC(�)j = (�n)(r�)=(�n=r)(�)(r!)� :Be ause A(t) � r� neither version of onstru t hooses an initial point xT fromany vertex of degree d(w; T ) � � at any step T = 1; :::; � .It is enough to give a one step indu tion. Let t < T � t + � . Given C(T � 1)su h that d(v; T � 1) � � the same hoi e of xT 62 Wv(T � 1) is made by rule(R*) in both algorithms. Thus the outputs C(T ;C(T � 1)) of onstru t-[v℄and onstru t(�) are the same.Any outputs of onstru t-[v℄ su h that d(v; t+�) = j � � have identi al outputsin onstru t(�). 2Lemma 6. If C = C(t) is � -good then for all j � �P� �jLj(t+ �)� ELj(t+ �)j >p2�(r � 1)2� logn ;C� = O(n��):The probability is over outputs C�(t + � ;C) of onstru t(�) given the inputC = (F (t);W (t)).Proof We use a standard martingale inequality. All outputs f 2 C�(t+� ;C)are equiprobable. Without loss of generality we establish a bije tion (f) = f 011

between outputs f = (e1; :::; e� ) and any other output f 0 = (e01; :::; e0� ) where thedi�eren e o urs on the �rst (ordered) edge e01.Let the digraph � have vertex set W = [Wv and edge set f ! (f). We anwrite f = (z1; :::; zr� ) so that f(j) = (z1; :::; zj). Thus for s = 0; :::; r� , �(s) hasedges f(s) ! (f(s)) and � = �(r�).Let e1 = (x1; y1;2; :::; y1;r) and let e01 = (x01; y01;2; :::; y01;r). Certainly x1 = x01 as it is hosen deterministi ally from W (t). Let yi;1 ! y0i;1; i = 2; :::; r.We now onsider ej; j � 2. Let xj ! x0j where x0j is hosen deterministi ally by(R*) given f 0(r(j�1)). Let yj;i ! yj;i if yj;i 62 f 0(r(j�1)+ i�1). If yj;i is alreadyused in f 0 then it has indegree 1 and out degree 0 in the digraph �(r(j�1)+ i�1).Hen e it is the terminal vertex of a path in that digraph with initial vertex u ofindegree 0 (ie: u 2 f; u 62 f 0). Let (yj;i) = u ompleting a y le in the digraph �.We now prove that jLj(f)�Lj( (f))j � 2(r� 1). Note there are at most 2(r� 1)verti es of degree j in f (or (f)) in ident with e1 or e01. We laim that if v 62e1; v 62 e01 and d(v; �; f) = j � � then d(v; �; (f)) = j.If d(v; �; f) � � then d(v; 0) � �. Let s � 2 be the �rst step, if any, that a point z =zr(s�1)+i of v is sele ted in f or f 0. z is not in xs ! x0s as, by guarantee, only pointsof degree less than � are sele ted as initial, and d(v; s� 1; f 0) = d(v; s� 1; f) � �.If z is sele ted for f then z ! z so the degree of v in f and f 0 remains the same.What if z is sele ted for f 0 but not f? Then w! z where w is already sele ted inf 0 and hen e z is the initial vertex of a path in �(r(s� 1) + i� 1) terminating atw. Thus z was previously sele ted in f but not f 0. This is a ontradi tion, as nopoints of v were sele ted previously (or, general indu tion: all previously sele tedpoints of v were mapped to themselves). 23 The stopping time of Phase IDuring Phase I of onstru t(�), the initial vertex v = �(xs) of the edge wegenerate (and remove) at ea h step s has residual degree at most �� 1. Let �+ 1be the �rst step at whi h the minimum residual degree is at least �. Thus �is the stopping time of Phase I. The remaining (ungenerated) multi-hypergraphH = G � fe1; :::; e�g is the �- ore of the multi-hypergraph G. At the end ofPhase I the output is C(�) = (F (�);W (�)). We have generated and removededges F (�) = fe1; :::; e�g. Phase II then generates a random �- ore on�guration onditional on the degree sequen e of W (�).Let Li(t) be the number of verti es of residual degree i at the end of step t of12

onstru t(�), and let Li(0) = Li. Let Q(t) be the total degree of verti es ofresidual degree at least �. ThusQ(t) = Xi�� iLi(t): (13)At the end of step t, A(t) the total degree of verti es of residual degree less than�, is related to Q(t) by A(t) +Q(t) + rt = �n: (14)The stopping time � for Phase I is de�ned as � = minfs : A(s) = 0g.Re all that we have de�ned q(t) byq(x(t)) = Xi�� li iXj=� j�ij�xj (1� x)i�j ;where li = Li=n and x(t) = (1� rt=�n)�. Alsoa(t) = � �1� rt�n�� q(t);and that b� is the smallest non-negative solution of a(t) = 0. We laim that whpb� is a good approximation to �.Let � = dn2=3�4=3 logne (15)�0 = b� �K�; �1 = b� +K�; (16)where K is a large positive (generi ) onstant. We prove that for t � �0 whp onstru t(�) is in Phase I (or �0 � 0) and for t � �1 whp onstru t(�) is inPhase II (or �1 > �n=r).To do this we exhibit a sequen e of steps tk = k�; k = 0; :::; d�0=�e su h that whpA(tk) � r� . Conditional on A(tk) � r� , we an guarantee that onstru t(�)pro eeds to step tk+1 using only initial points xt from verti es of degree less than�. Thus we are always removing a non- ore edge and the algorithm remains inPhase I.Lemma 7. Let �0; �1 be given by (16). Let b� be the smallest solution of a(t) = 0,and let � logn � b� � �n=r � � logn, then whp � 2 [�0; �1℄.Proof Lower Bound. Let S = d�0=�e and for k = 0; 1; :::; S let tk = k� .Let � = O(�3=�) and let � = O(p2(r � 1)2�� logn). Re all that � = o(n1=24).The proof is indu tive. The indu tion is from tk to tk+1, on the proposition thatfor � � j � � with probability 1� O(k�n��)Lj(tk) = k� + (1 + k�)Xi�j Li�i;j(0; k�);13

where �i;j(0; t) is given by (10). Given L(tk) satisfying this, from Lemma's 4, 5and 6, with probability 1�O(n��)Lj(tk+1) = ELj(tk+1) + �= � +Xi�j Li(tk)�i;j(tk; tk + �)= � +Xi�j �i;j(tk; tk + �) k� + (1 + k�)Xl�i Ll�l;i(0; k�)!= (k + 1)� + (1 + (k + 1)�)Xl�i LlXi�j �l;i(0; k�)�i;j(k�; (k + 1)�)= (k + 1)� + (1 + (k + 1)�)Xl�j Ll�l;j(0; (k + 1)�):Thus Q(tk) = Xj�� jLj(tk)= �2k� + (1 + k�)Xi�j LiXj�� j�i;j(0; tk)= �2k� + (1 + k�)nq(tk):Thus for k � S jQ(tk) � nq(tk)j = �2k� + k�nq(t) = o(�):Hen e jA(tk)� na(tk)j = o(�): (17)By ondition S2 na(t) � K� for t � �0. Thus A(tk) � r� .Upper Bound. The expe ted number of new points added to A(t) during step tof Phase I is (1 + o(1))(r � 1)(�� 1) �L�(t)�n� rt :We prove below that whp(r � 1)�(�� 1) L�(�0)�n� r�0 = 1� + o(1); (18)where 0 < � 1 onstant is given by = �na0(b�). We omplete the proof ofthe upper bound as follows: Let F (s) be the number of new points added to A(s)at step s = �0 + 1; :::; �0 + T , where T = K� . Thus PF (s) is sto hasti ally14

dominated by the sum of non-negative independent random variables boundedabove by (r � 1)(�� 1) and with expe tation(1 + o(1))(r � 1)�(�� 1)L�(�0) + (s� �0)(r � 1)�n� rs :If (1 � ) = o(1) then whp P�0+Ts=�0+1 F (s) = o(T ) and Phase I ends in A(�0) +O(T ) = O(K�) steps. If (1� ) > 0 onstant thenPr(Xs F (s) � (1 + �)T (1� )) � exp���2T (1� )23(r�)2 � ;and whp A(�0 + T ) � A(�0) + (1 + �)T (1� )� T:As (whp) A(�0) = O(�), Phase I halts after at most O(K�) steps.We next prove (18). Let I = [�0 � �; :::; �0 � 1℄. Note thatE(A(�0) j A(�0 � �)) =A(�0 � �)� � � (1 + o(1))(r � 1)Xt2I A(t)�n� rt+ (1 + o(1))(�� 1)(r � 1)Xt2I �L�(t)�n� rt :By Lemma 6 we have whpjA(�0)� E(A(�0) j A(�0 � �))j = o(�):Also jA(t+ 1)� A(t)j � max(r; (�� 1)(r � 1)� 1)jL�(t + 1)� L�(t)j � (r � 1):Thus for X = A;L� we haveXt2I X(t)�n� rt = �(X(�0) +O(�))�n� r�0 :From (17), whpA(�0)� A(�0 � �) = na(�0)� na(�0 � �) + o(�):For h = O(�) by ondition S3a(b� + h) = ha0(b�)(1 + o(1)):Putting this all together gives (18). 215

Let M(t) be the number of verti es of residual degree at least � and letm(t) = Xi�� li iXj=� �ij�xj (1� x)i�j : (19)Let Li;j(t) be the number of verti es v su h that d(v; 0) = i; d(v; t) = j � � andlet li;j(t) = li�ij�xj(1� x)i�j: (20)The following orollary summarizes results deriving from the previous lemmas.Lemma 8. With probability 1� o(1) for all � � j � i � � and for all t � �1jLi;j(t)� nli;j(t)j = O(�)jLi(t)� nli(t)j = O(�)jM(t) � nm(t)j = O(�)jA(t)� na(t)j = O(�)jQ(t)� nq(t)j = O(�):We next onsider the ase where b� > �n=r � � logn.Lemma 9. Let � = o(n1=24) and let r = 2; � � 3 or r � 3; � � 2, then whp a�- ore of size O(� logn) is empty.Proof We use a rude ounting argument in the on�guration model. LetC(s; t) be the probability there exists a �- ore of s verti es and t edges. Letd(v; S) denote the degree of vertex v in S � V and let(s; t) = fS : jSj = s; S = fv1; :::; vsg; d(vi; S) � �; Xvi2S d(vi; S) = rtg:Thus C(s; t) = O(D(s; t)) whereD(s; t) = X(s;t)�d(v1) + � � �+ d(vs)rt ���n=rt ���nrt� :Let 1 � s � � logn. Repla ing d(vi) by � we obtainD(s; t) � �ns�(�s)rtrt! ��n=rt ���nrt�� �nes �s��se�n �rt ��nert �t :16

Note that rt = �s+ j; j � 0, and thus for r � 2��se�n �r �nert = O ���sn �r�1! = O n1=24 � log2 nn17=72�r�1! = o(1):Thus Xs Xrt��sD(s; t) � Xs �nes ���se�n ����ners ��=r!s� Xs O��� � sn��(1� 1r)�1�s ;whi h is o(1) provided � � 3 when r = 2, or � � 2 when r � 3. 2The ase when b� < � logn follows from the upper bound proof in Lemma 7.This ompletes the proof of Theorem 1 and the main part of Theorem 2.4 Results on 2- ores of graphsWe estimate the number of small tree omponents, and in the ase of smoothsequen es the size of the mantle and number of edges in small omponents. Also,for 2- ores of graphs we establish simpler onditions whi h imply Theorems 1, 2.Let d be a degree sequen e satisfying Theorem 2. The number of verti es V2(G)in the 2- ore of an r-uniform hypergraph G 2 G(d) is whp V2(G) = �n + O(�)where � = 1� �bx �1� bx1=(r�1)��Xi�0 li(1� bx)i: (21)We now onsider isolated tree omponents in graphs (r = 2). Let R(0) ountisolated verti es of G, (ie. di = 0), and for T > 0 let R(T ) also ount verti eswhose last remaining on�guration point was sele ted as the terminal point yt ofthe edge et = (xt; yt) generated during steps t � T of Phase I. At the stoppingtime � of Phase I, R(G) = R(�) is the number of isolated tree omponents.Lemma 10. whp R(G) = n +O(�), where = Xi�0 li(1� bx)i � �2(1� bx)2: (22)17

Proof In the ase of 2- ores of graphs, A(t) is also the number of verti es ofdegree 1. The balan e equation for verti es isn = M(t) + A(t) + t +R(t);where M(t) is the number of verti es of residual degree at least 2 at the end ofstep t. At �, A(�) = 0 so that R(�) = n���M(�). The result follows from (21)whi h gives the value of � = m(b�). 2The stru ture of graphs with a smooth degree sequen eA sequen e (Li; i = 1; 2; :::), giving the number of verti es of degree i, is smooth inthe sense used by Molloy & Reed [16, 17℄ if limn!1 Li=n = �i where �i is onstantand li = �i + o(1). Thus �i = limn!1 li should su h a limit exist.For 2- ores, rearranging the de�nition of q(x) in (4) givesq(x) = �x� xXi�1 ili(1� x)i�1:Let = 1 � bx and qi�1 = pi = ili=�. The ondition a(b�) = 0 an be writtenq(bx) = �bx2. This is satis�ed if = 1 or if = Xi�0 qi i: (23)Equation (23) is the equation for the extin tion probability of an i.i.d. bran hingpro ess X with distribution of progeny fqi : i � 0g and expe ted value EX = d.Provided q0 > 0, su h a bran hing pro ess only has a solution 6= 1 if d > 1.The ondition d > 1 orresponds to the whp existen e of a non-trivial 2- ore (ie.bx > 0). This ondition an be written as Pi�0 i(i� 1)li > � or equivalently, using� = P ili as Xi�0 i(i� 2)li > 0:This is the Molloy-Reed ondition for the whp existen e of a giant omponent.Hen e, in the ase where (Li; i = 1; 2; :::) is smooth we an sket h in some stru -tural details on the size of the mantle and number of edges in small omponents.Theorem 1 of [16℄ gives the (whp) proportion3 � of verti es in the (unique) giant omponent as � = 1�Xi�0 �i i;3When we refer to the proportion � of verti es (resp. edges, omponents et ) with the stated property, it is tobe understood that the asso iated random variable X(G) = n� +O(�) whp.18

where = 1�bx is the solution of (23). But � = �+�, where � is the proportion ofverti es in the mantle of the giant (arbores en es rooted on the 2- ore), and � theproportion of verti es in the 2- ore, given by (21) is � = 1�� (1� )�Pi�0 �i i.Thus the proportional size of the mantle is � = � (1� ).The proportion of edges Æ in the giant is given by Æ = � + � where � is given by�=r � b�=n. Thus the proportion of edges on small omponents, �, is � = �2 � Æ =�2 2. The papers [16, 17℄ show that whp the number of verti es on small non-tree omponents is o(n1=4 log3 n).4.1 Conditions implying the Separation PropertyLemma 11. Let d = Pi�0(i � 1)ili=�. For � = 2; r = 2, provided � = o(n1=24),l1 > 0 and d > 1 then the onditions of Theorems 1, 2 hold.Proof We �rst prove that � logn � b� � �n=r � � logn. Let z = 1 � (1 �2t=�n)12 . Let g(z) = Pi�0((i + 1)li+1=�)zi. As d > 1 and l1 > 0 the smallestsolution of the equation z = g(z) is the unique solution in (0; 1).We next prove the Separation Property. We have thata(z) = �(1� z)(g(z) � z):Certainly g(z) > z for z < . The fun tion g(z) � z is monotone de reasing in[0; + �℄ for some � > 0. This follows be ause g0(z) is an in reasing fun tion ofz, and at z = , the slope of y = g(z) is less than the slope of y = z. Thus0 < g0( ) < 1. It follows from g0( ) < 1 and dz=dt = 1=((1� z)�n) thata0(b�) = �(1� )(g0( ) � 1)z0(b�) = (g0( )� 1)=n:That a(t) has a uniform Taylor expansion about b� is similarly proved. 25 Poisson degree sequen esAs a sanity he k, we reassure ourselves that the results of Theorems 1, 2 andLemma 8 for a Poisson Po( ) degree sequen e give the whp standard results forthe �- ore of the hypergraph models Gn;p; Gn;m (see eg: [4, 10, 19, 14℄). The degreesequen e (Li; i � 0) of random hypergraphs Gn;p; p = =n and Gn;m; m = d n=reis (whp) approximately the Poisson sequen e (n�i; i � 0) where �i = ( i=i!)e� .For i = �(logn), Li = 0, whp. For i = o(logn), the number of verti es of degree19

i is (whp) Li = n ii! e� +O(ipn logn):Moreover Pi��(log n) n�i = n�a for any a > 0.What kind of error do we introdu e by repla ing (Li) by (n�i)? Let b� be thestopping time of (Li) and let e� be the stopping time obtained from a(t) = 0 wherea(t) uses (�i). The error in b� and the properties X(G) of the ores of G (number ofverti es, edges, degree sequen e) is o(pn log3 n), whi h is within the o(�) boundsof this paper.5.1 2- ores of Poisson hypergraphsIn the ase of 2- ores of r-uniform Poisson hypergraphs the size of the 2- ore isdetermined by the largest solution bx in [0; 1℄ tobx = (1� e� bx)r�1: (24)The (limiting) proportion of edges, � and verti es, �, in the 2- ore is given by� = rbx rr�1 (25)� = bx 1r�1 � bx+ bx rr�1 : (26)We note that � an also be written as � = 1� e� bx(1 + bx).Some threshold values ( �; bx) for emerging 2- ores of r-uniform hypergraphs aretabulated below. r 3 4 5 6 7 � 2.45 3.1 3.5 3.8 4.0bx 0.52 0.62 0.67 0.70 0.72� 0.36 0.57 0.68 0.74 0.78� 0:31 0:41 0:42 0:41 0:39The relationship (24) and the results in the table are given in the paper [18℄.From (20), the proportion of verti es of the 2- ore with degree j � � (in the 2- ore) is Po( bx), and the proportion of verti es of the 2- ore with degree J � 0 inthe mantle is � ( (1�bx))JJ! e� (1�bx):For graphs, the detailed stru ture of the mantle is given in [19℄.20

5.2 Upper bounds for the rank of random Boolean matri esConsider a n�m matrix whose olumns onsist of the vertex-edge in iden e ve torsof a random r-uniform hypergraph on n verti es. The m olumns are independentand sampled with repla ement. All olumns have exa tly r entries whi h are 1,and n� r entries whi h are 0. The probability of any olumn is 1=�nr�.The rank of su h random matri es over GF(2) is sharply on entrated, but it isdiÆ ult to obtain the expe ted value of the rank as a fun tion of m. The followingquestion has been studied extensively. At what value of m does the �rst lineardependen e o ur among the olumns (whp)?A lower bound for the �rst linear dependen e an be obtained from the smallest msu h that the expe ted number of dependen ies among the olumns does not tendto zero (see e.g. [15℄ or [5℄). The number of entries per row is well approximatedby a Po( ), = rm=n degree sequen e. Let n0(m) is the number of rows whoseentries are all zero. A trivial upper bound an be found when the number of olumns m ex eeds the number of non-zero rows n� n0(m) � n(1� e� ).An improved upper bound is obtained by �nding the smallest value of m forwhi h the number of edges in the 2- ore ex eeds the number of verti es whp. Thesolution to the equation �=� = 1 is obtained from (24), (25), (26). The tablebelow gives the details.r (non-zero entries per olumn) 3 4 5 6m=n lower bound ([15℄, [5℄) 0.8895 0.9672 0.9892 0.9962m=n upper bound (from m = n� n0) 0.9404 0.9801 0.9930 0.9974m=n upper bound (from �=� = 1 in 2- ore) 0.9179 0.9768 0.9924 0.9974Referen es[1℄ E.A.Bender and E.R.Can�eld, The asymptoti number of labelled graphs withgiven degree sequen es, Journal of Combinatorial Theory, Series A 24 (1978)296-307.[2℄ B.Bollob�as, A probabilisti proof of an asymptoti formula for the number oflabelled regular graphs, European Journal on Combinatori s 1 (1980) 311-316.[3℄ B.Bollob�as, The evolution of random graphs, Transa tions of the Ameri anMathemati al So iety 286 (1984) 257-274.[4℄ B.Bollob�as, Random Graphs, Cambridge University Press (2nd edition 2001).[5℄ C. Cooper. Asymptoti s for dependent sums of random ve tors. RandomStru tures and Algorithms. 14 (1999) 267-292.21

[6℄ C. Cooper and A. Frieze, The size of the largest strongly onne ted omponentof a random digraph with a given degree sequen e, submitted (2002).[7℄ C. Cooper, A. Frieze and B. Reed, Hamilton y les in random regular graphsof non- onstant degree, Combinatori s, Probability and Computing, 11 (2001),249-262.[8℄ P. Erd}os and A. R�enyi, On the evolution of random graphs, Publ. Math. Inst.Hungar. A ad. S i. 5 (1960) 17-61.[9℄ A. Frieze and T. Lu zak. On the independen e and hromati numbers ofregular random graphs. Journal of Combinatorial Theory Series B, 54 (1992)123-132.[10℄ S. Janson, D.E. Knuth, T. Lu zak and B.G. Pittel, The birth of the giant omponent, Random Stru tures and Algorithms 4 (1993) 233-358.[11℄ S. Janson, T. Lu zak and A. Ru i�nski, Random Graphs, Wiley, (2000).[12℄ T. Lu zak, Components behavior near the riti al point of the random graphpro ess, Random Stru tures and Algorithms 1 (1990) 287-310.[13℄ T. Lu zak, B.G. Pittel and J.C. Wierman, The stru ture of a random graphnear the point of the phase transition, Transa tions of the Ameri an Mathe-mati al So iety 341 (1994) 721-748.[14℄ J. H. Kim, The Poisson loning model for random graphs, submitted (2002).[15℄ V.F. Kol hin. Random graphs and systems of linear equations in �nite �elds,Random Stru tures and Algorithms 5 (1994) 135-146.[16℄ M. Molloy and B.A. Reed, A Criti al Point for Random Graphs with a GivenDegree Sequen e, Random Stru tures and Algorithms 6 (1995) 161-180.[17℄ M. Molloy and B.A. Reed, The Size of the Largest Component of a RandomGraph on a Fixed Degree Sequen e, Combinatori s, Probability and Comput-ing 7 (1998) 295-306.[18℄ B. S. Majewski, N. C. Wormald, G. Havas and Z.J. Cze h, A family of perfe thashing methods, Computer Journal 39 (1996) 547-554.[19℄ B. Pittel, On tree ensus and the giant omponent in sparse random graphs,Random Stru tures and Algorithms 1.3 (1990) 311-342.[20℄ B. Pittel, J. Spen er and N. Wormald, Sudden emergen e of a giant k- orein a random graph, Journal of Combinatorial Theory Series B, 67, (1996)111-151.22

6 Appendix. Proof of Lemma 4: Distribution of residualdegreeLet 0 � s � t � (�n� d(v; 0))=r. Let � = (r � 1)=r, and let�k;k�j(s; t) = �kj�� �n� rt�n� rs��(k�j)�1� � �n� rt�n� rs���j :Lemma 12. �k;k(s; t; v) = �k;k(s; t) O �1 + rk3�n�rt + 1�n�rs� : (27)Proof When t = s we require that �k;k(s; s) = 1 and (27) satis�es this.Assume t > s. During step t of algorithm onstru t-[v℄, on�guration points ofvertex v an only be sele ted as terminal points yt;i; i = 2; :::; r. The hoi e of thepoint yt;i is made uniformly from a set of size �n� (r(t� 1) + (i� 1)), thusPr(�(yt;i) 6= v) = 1� k�n� (r(t� 1) + (i� 1)) :Thus �k;k(s; t) = t�1Y�=s r�1Yi=1 �1� k�n� (r� + i)�= exp �k(r � 1) t�1X�=s 1�n� r� +O� k2r(t�s)(�n�rs)(�n�rt)�!= exp�k� log �n� rt�n� rs +O� 1�n�rs�+O� k2r(t�s)(�n�rs)(�n�rt)��= � �n� rt�n� rs��k �1+O� rk2�n�rt�+O� 1�n�rs�� : (28)The last line follows be ause (t� s)=(�n� rs) � t=�n < 1 as rt � �n. 2Lemma 13. Let Æ(k; s; t) = rk3�n� rt + 1�n� rs; (29)�(j; s; t) = j2(t� s) ; (30) (j; s; t) = j3(t� s) ; (31)and let �(k; j; s; t) = O(Æ(k; t) + �(j; s; t) + (j; s; t)): (32)23

Let j � 0, then �k;k�j(s; t; v) = �k;k�j(s; t) (1 + �(j; k; s; t)) : (33)ProofLet (�l; il); l = 1; :::; j; i = 2; :::; r denote the (step, iteration) at whi h thetransition of vertex degree k � l + 1 ! k � l o urs. There are two possibilities:That all transitions o ur at distin t steps � (probability pk;k�j(s; t : 1)).That at least 2 transitions o ur during iterations i = 2; :::; r of some step �(probability pk;k�j(s; t : 2)).We prove below thatp(s; t : 1) = �k;k�j(s; t) (1 + �(j; s; t)) (1 + Æ(k; s; t)) (34)p(s; t : 2) � �k;k�j(s; t) (j; s; t); (35)whi h will omplete the proof of the lemma.The probability that a on�guration point of vertex v is hosen by onstru t-[v℄at iteration i of step � is given by l + 1�n� (r(� � 1) + (i� 1)) :Thus the probability of a = 1; :::; r� 1 transitions from l+ a to l o urring at � is�r � 1a � (l + a)(a)(�n� r�)a �1+O( ra�n�r� )� :Case pk;k�j(s; t : 1). Let �0 = s andT1 = f� = (�1; :::; �j) : s = �0 < �1 < �2 < � � � < �j � tgbe the set of possible sequen es of transition points. From (28) we have,p(s; t : 1) = X�2T1 "j�1Yi=0 ��n� r(�i+1 � 1)�n� r�i �(k�i)� (r � 1)(k � i)�n� r�i+1 �1+O� r(k�i)2�n�rt ��#�� �n� rt�n� r�j�(k�j)� �1+O� r(k�j)2�n�rt �� :Thus p(s; t : 1) = (k)(j)� �n� rt�n� rs�(k�j)� �1+O� rk2j�n�rt��� (r � 1)j(�n� rs)j� X�2T1 jYi=1 1(�n� r�i)1�� :24

For A > 0, 1=(1� x)A is monotone in reasing in x, sotXs+1 1(�n� r�)A = Z ts 1(�n� r�)Ad� +O� 1(�n� rt)A� :For i � 1 let fi(t) = tX�=s+1 1(�n� r�)i(1��) ;then f1(t) = 1r � 1 ((�n� rs)� � (�n� rt)�) +O� 1(�n�rt)1��� ;and, noting that � = (r � 1)=r � 1=2,f2(t) = O� 1(�n�rt)2�2��+8<: 12 (log(�n� 2s)� log(�n� 2t)) r = 21r�2 ((�n� rs)2��1 � (�n� rt)2��1) r � 3For j � 1 let Fj(t) = X�2T1 jYi=1 1(�n� r�i)1�� ;then F1(t) = f1(t) and for j � 21j! (f1(t))j � Fj(t) � 1j! (f1(t))j �1� �j2� f2(t)(f1(t))2� :For B > 0 onstant � �n� rt�n� rs�B = 1� C �Br(t� s)�n� rs � ; (36)for some onstant C > 0, and in parti ular if � r(t�s)�n�rs� = o(1) then C = (1 + o(1)).Thus f2(t)f1(t)2 = O(1)t� s:This ompletes the proof of (34).Case pk;k�j(s; t : 2). For r = 2 (graphs), p(s; t : 2) = 0 always. Let r � 3; j � 2,then pk;k�j(s; t : 2) < (k)(j)� �n� rt�n� rs��(k�j) (r � 1)j(�n� rs)j��j 1(j � 2)!(f1(t))j f2(t)(f1(t))2� �k;k�j(s; t) j3(t� s)25

The last line follows from (36). This ompletes the proof of (35). 2

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