O two Hamilto cycle problems i radom graphs Ala Frieze Michael Krivelevich November 16, 2006 Abstract We study two problems related to the existece of Hamilto cycles i radom graphs. The first questio relates to the umber of edge disjoit Hamilto cycles that the radom graph G,p cotais. δ(g)/2 is a upper boud ad we show that if p (1 + o(1)) l/ the this upper boud is tight whp. The secod questio relates to how may edges ca be adversarially removed from G,p without destroyig Hamiltoicity. We show that if p K l/ the there exists a costat α > 0 such that whp G H is Hamiltoia for all choices of H as a -vertex graph with maximum degree (H) αk l. 1 Itroductio I this paper, we give results o two problems related to Hamilto cycles i radom graphs. 1.1 Edge Disjoit Hamilto Cycles It was show by Komlós ad Szemerédi [8] that if p = l +ll +c the, lim Pr(G,p is Hamiltoia) = lim Pr(δ(G,p ) 2). Bollobás [3], Ajtai, Komlós ad Szemerédi [1] proved a hittig time versio of this statemet, i.e., whp 1, as we add radom edges e 1,e 2,...,e m oe by oe to a empty graph, the graph G m = Departmet of Mathematical Scieces, Caregie Mello Uiversity, Pittsburgh PA 15213, U.S.A. Research supported i part by NSF grat CCR-0200945. Departmet of Mathematics, Raymod ad Beverly Sackler Faculty of Exact Scieces, Tel Aviv Uiversity, Tel Aviv 69978, Israel. E-mail: krivelev@post.tau.ac.il. Research supported i part by USA-Israel BSF Grat 2002-133 ad by grat 526/05 from the Israel Sciece Foudatio. 1 A sequece of evets E is said to occur with high probability (whp) if lim Pr(E ) = 1 1
([], {e 1,e 2,...,e m }) becomes Hamiltoia at exactly the poit whe the miimum degree reaches two. Let us say that a graph G has property H if it cotais δ(g)/2 edge disjoit Hamilto cycles plus a further edge disjoit (ear) perfect matchig i the case δ(g) is odd. (Here a (ear) perfect matchig is oe of size /2 ). Bollobás ad Frieze [5] showed that whp G m has property H as log as the miimum degree is O(1). It is reasoable to cojecture that whp G,p has property H for ay 0 p 1. Our first result is to show that this is true for p (1 + o(1)) l/ which stregthes the o-hittig time versio the result quoted from [5]. Theorem 1 Let p() (1 + o(1)) l/. The whp G,p has property H. We remark that Frieze ad Krivelevich [7] showed that if p is costat the whp G,p almost satisfies H i the sese that it cotais (1 o(1))δ(g,p )/2 edge disjoit Hamilto cycles. 1.2 Robustess of Hamiltoicity I recet times, there is icreasig iterest i graphs which are oly partially radom. For example, Bohma, Frieze ad Marti [2] cosidered graphs of the form G = H + R where H is arbitrary, but with high miimum degree ad R is radom. I this sectio we cosider graphs of the form G = R H where R is radom ad H is a arbitrary subset of R, subject to some restrictios. I particular R = G,p Sudakov ad Vu [10] have recetly show that if p > (l) 4 / ad if G = G,p the whp G H is Hamiltoia for all choices of H as a -vertex graph with maximum degree (H) (1/2 ε)p. Here ε > 0 is a arbitrarily small costat. Note that this boud o (H) is essetially best possible, otherwise R H could be a bipartite graph with a ueve partitio. I this ote we reduce p to O(l /) but ufortuately, we have to reduce the boud o (H) as well. Theorem 2 Let G = G,p where p K l / for some sufficietly large costat K > 0. There exists a costat α > 0 such that whp G H is Hamiltoia for all choices of H as a -vertex graph with maximum degree (H) αk l. 2 Proof of Theorem 1 2.1 Prelimiaries Observe first that the assumptio o the edge probability i this theorem ca be easily see to be essetially equivalet to the assumptio that the miimum degree δ(g) of G,p almost surely satisfies: δ(g) = o(log ). 2
Notatio: For a graph G = (V,E) ad two disjoit vertex subsets U,W we deote: N(U,W) := {w W : w has a eighbor i U} ; N(U) := N(U,V \ U) ; E(U,W) := {e E(G) : e U = 1, e W = 1} ; e(u,w) := E(U,W). Defiitio 1 A graph G = (V,E) is called a (k,c)-expader if N(U) c U for every subset U V (G) of cardiality U k. Set { d 0 = d 0 (,p) = mi k : Oe ca prove that whp δ(g,p ) satisfies (say): ( 1 k ) } p k (1 p) 1 k 1. δ(g) d 0 l l. Ideed, u k = ( ) 1 k p k (1 p) 1 k is the expected umber of vertices of degree k ad u k+1 /u k = ( 1 k)p. Sice d (k+1)(1 p) 0 = o(l) we see that u d0 l l = o(1). Furthermore, u d0 +l l ad we ca use the Chebyshev iequality to show that u d0 +l l 0 whp. Defie ρ = 2001(d 0 + l l), l observe that ρ = o(1/). Defie p 0 = p 0 () by 1 p = (1 p 0 )(1 ρ), (1) observe that p 0 = p ρ(1 o(1)). We ca thus decompose G G,p as G = G 0 R, where G 0 G,p0, R G,ρ. Notatio. δ 0 = δ(g 0 ). Claim 1 For a fixed G 0, almost surely over the choice of R G,ρ, δ(g 0 ) = δ(g 0 R). Proof Clearly, δ(g 0 ) δ(g 0 R). I the opposite directio, take a vertex v of miimum degree i G 0. Recall that ρ = o(1/), ad therefore the edges of R almost surely miss v, implyig δ(g 0 R) d G0 R(v) = d G0 (v) = δ(g 0 ). It thus follows that i order to prove Theorem 1 it is eough to prove that almost surely G 0 R cotais δ 0 /2 disjoit Hamilto cycles, plus a edge disjoit (ear) perfect matchig if δ 0 is odd.. Of course we may (ad will ideed) assume that p() = (1 + o(1)) l/, as otherwise whp δ 0 1 ad there is othig ew to prove. 3
2.2 Properties of G 0 = G,p0 Defie SMALL = {v V : d G0 (v) 0.1 l}. Lemma 3 The radom graph G 0 = G,p0, with p 0 defied by (1), has whp the followig properties: (P1) G 0 does ot cotai a path of at most four distict edges (with possibly idetical edpoits), both of whose edpoits lie i SMALL. (P2) Every vertex has at most oe eighbor i SMALL. (P3) Every set U V of size U 100/ l spas at most U (l ) 1/2 edges i G 0. (P4) For every two disjoit subsets U,W V satisfyig: U 100/ l, W 10 4 U l, e G0 (U,W) < 0.09 U l. (P5) For every two disjoit subsets U,W V satisfyig: U 100/ l, W /4, e G0 (U,W) 0.1 U l. Proof The above are rather stadard statemets about radom graphs, so we will be relatively brief i our argumets. We start with provig P1. Observe that for a vertex v V (G 0 ), the degree of v is biomially distributed with parameters 1 ad p 0. Therefore, ( ) 1 Pr[v SMALL] = Pr[B( 1,p 0 ) = k] 0.1 l p 0.1l (1 p) 1 0.1l 0.1 l k 0.1l ( ) 0.1l 10ep 0.1 l e p( 1 0.1l ) < 29 0.1l e (1 o(1))l l < 0.6. Also, for a fixed pair u v V (G 0 ) the probability that u ad v are coected by a path of legth l i G 0 is at most l 1 p l 0 = ((1 + o(1)) l) l 1. Therefore, usig the FKG iequality, Pr[(u,v SMALL)&(dist(u,v) 4)] Pr[u SMALL]Pr[v SMALL]Pr[dist(u,v) 4] 4 0.6 0.6 (1 + o(1)) l4 < 2.1. Applyig the uio boud over all possible pairs of distict vertices u,v (O( 2 ) of them), we establish P1. The case where u = v is treated similarly. 4
Property P2 follows directly from P1. Properties P3, P4 are straightforward first momet calculatios which we thus omit. We coclude with provig P5. Fix U,W. The the umber of edges betwee U ad W is distributed biomially with parameters U W ad p 0 ad has thus expectatio U W p 0 (1+o(1)) U l/4. Therefore by applyig stadard Cheroff-type bouds o the lower tail of the biomial distributio, it follows that } (0.25 U l 0.1 U l )2 Pr[e G0 (U,W) 0.1 U l] exp { = exp{ 2 0.15 2 U l } 2 0.25 U l < exp{ 4}. As the pair (U,W) ca be chose i at most 4 ways, P5 follows by applyig the uio boud. 2.3 Pósa s Lemma ad its cosequeces Defiitio 2 Let G = (V, E) be a o-hamiltoia graph with a logest path of legth l. A pair (u,v) E(G) is called a hole if addig (u,v) to G creates a graph G which is Hamiltoia or cotais a path loger tha l. I additio, if the maximum size of a matchig i G is m < /2 the (u,v) E(G) is called a hole if addig (u,v) to G creates a graph G which is cotais a matchig of size m + 1. Lemma 4 Let G be a o-hamiltoia coected (k, 2)-expader. The G has a path of legth at least 3k 1 ad at least k 2 /2 holes. Proof Let P = (v 0,...,v k ) be a logest path i graph G. A Pósa rotatio of P [9] with v 0 fixed gives aother logest path P = (v 0,...v i v k...v i+1 ) created by addig edge (v k,v i ) ad deletig edge (v i,v i+1 ). Let END G (v 0,P) be the set of edpoits obtaied by a sequece of Pósa rotatios startig with P, keepig v 0 fixed ad usig a edge (v k,v i ) of G. Each vertex v j END G (v 0,P) ca the be used as the iitial vertex of aother set of logest paths whose edpoit set is END G (v j,p), this time usig v j as the fixed vertex, but agai oly addig edges from G. Let END G (P) = {v 0 } END G (v 0,P). The Pósa coditio (see, e.g., [4], Ch.8.2) N(END G (v,p)) 2 END G (v,p) 1 for v END G (P) together with the fact that G is a (k, 2)-expader implies that END G (v,p) > k. The coectivity of G implies that closig a logest path to a cycle either creates a Hamilto cycle or creates a loger path. For every v END G (P) ad for every u END G (v,p), a pair (u,v) is a hole. This shows that the umber of holes is at least k 2 /2 (each hole is couted at most twice for both its edpoits). As all eighbors i G of a subset U END G (v,p) of size U = k belog 5
to P, due to the maximality of P, ad G is a (k, 2)-expader, it follows that the legth of P is at least 3k 1. The followig lemma is take from [5]. Lemma 5 Let G be a (k, 1)-expader which does ot cotai a matchig of size /2. The G has a matchig of size at least k ad at least k 2 /2 holes. Proof Let M deote the set of maximum size matchigs i G ad let M M. Fix v ucovered by M ad ow let S 0 be the set of vertices reachable from v by a eve legth alteratig path with respect to M. Clearly, every vertex of S 0 is either v or is covered by M. Let x N(S 0 ). The x is covered by M, as otherwise we ca get a larger matchig by usig a alteratig path from v to y S 0, ad the a edge (y,x). Let y 1 satisfy (x,y 1 ) M. We show that y 1 S 0. Now there exists y 2 S 0 such that (x,y 2 ) E(G). Let P be a eve legth alteratig path from v termiatig at y 2. If P cotais (x,y 1 ) we ca trucate it to termiate with (x,y 1 ), otherwise we ca exted it usig edges (y 2,x) ad (x,y 1 ). It follows that N(S 0 ) < S 0 (as v S 0, v is ot covered by M). Recallig that G is a (k, 1)- expader, we derive that S 0 > k. But the obviously the uio S 0 N(S 0 ) has at least 2k vertices ad thus has at least 2k 1 vertices from M. This implies: M k. Now we prove that G has at least k 2 /2 holes. Fix v ucovered by M ad ow let S be the other vertices ucovered by M. Let S 1 S be the set of vertices reachable from S by a eve legth alteratig path with respect to M. As before we ca prove that S 1 > k. For every u S 1 there is a eve legth alteratig path with respect to M edig at u. Replacig the edges alog this path belogig to M with those outside of M gives a maximum matchig M M ot coverig u. Thus (u,v) is a hole. Repeatig ow the above argumet with u,m istead of v,m, respectively, gives at least k holes touchig u. Sice S 1 k, ad each hole is couted at most twice, altogether we get at least S 1 k/2 k 2 /2 holes, as required. 2.4 Proof idea We split the radom graph R ito δ 0 /2 idetically distributed radom graphs R i. We the create δ 0 /2 Hamilto cycles H i (plus a matchig if eeded). We use the radom edges of R i to fill a hole. Oce H i is created its edges are deleted from the graph ad we proceed to the ext phase. At the i-th stage, by the defiitio of δ 0, the graph G i has miimum degree at least 2, moreover, most vertices i it have degree aroud l (as each vertex loses at most δ 0 = o(l) eighbors durig the process), ad therefore G i is coected, is a ( c/ l, 2)-expader by properties P1-P5 ad has a path P i of legth at least c/ l. We gradually augmet P i to a Hamilto path, ad the to a Hamilto cycle. At each substage of augmetig P i, the curret graph has a quadratic umber of holes, ad therefore a costat umber of radom edges are expected to augmet the curret path to a loger oe/close a Hamilto cycle. If δ 0 is odd, we eed a fial stage to create a (ear) perfect matchig. 6
2.5 Formal argumet We may assume that δ 0 2 as otherwise there is othig ew to prove. Defie ρ i by observe that We the represet where R i G(,ρ i ). ρ i 1 ρ = (1 ρ i ) δ 0/2 ρ δ 0 /2 = 2001(d 0 + l l) δ 0 /2 l R = δ 0 /2 i=1 R i, 4000 l. For i = 1,..., δ 0 /2, let G i be a graph obtaied from G 0 i 1 j=1 R j after havig deleted the first i 1 Hamilto cycles (assumig that the previous i 1 stages were successful, of course). Each vertex v has its degree i G 0 reduced by at most 2(i 1) i G i. Therefore if i δ 0 /2 the the miimum degree δ(g i ) satisfies δ(g i ) δ 0 2(i 1) 2. If δ 0 is odd, the δ(g δ0 /2 ) 1. We will ow show that if i δ 0 /2 the G i is a (k, 2)-expader for k = /3 100/(3 l). Let X V be a set of X = t vertices. Case 1: t 100/ l. Deote X 0 = X SMALL, X 0 = t 0, X 1 = X \X 0, X 1 = t 1. Observe first that N Gi (X 0,V \X) 2t 0 t 1. Ideed, i G i all edges touchig X 0 have their secod edpoit outside X 0, by Property P1. We curretly have at least two edges per each vertex i X 0. By Property P2 each vertex outside SMALL has at most oe eighbor i X 0 i the graph G i. Thus the other edpoits of the edges from G i touchig X 0 are distict, ad at most t 1 of them lad i X 1. Now, X 1 spas at most t 1 (l ) 1/2 edges i G 0, by Property P3. As the degrees i G 0 of all vertices i X 1 are at least 0.1 l, by the defiitio of SMALL, at least 0.09t 1 l edges leave X 1 i G 0. But the by Property P4 N G0 (X 1 ) 10 4 t 1 l. By Property P1 at most t 1 of those eighbors fall ito X 0 N G0 (X 0 ), implyig: N G0 (X 1,V \ X) N G0 (X 0,V \ X) 10 4 t 1 l t 1. As i G i every vertex lost at most δ 0 eighbors compared to G 0, we have N Gi (X 1,V \ X) N G0 (X 0,V \ X) 10 4 t 1 l t 1 δ 0 t 1 10 5 t 1 l. Altogether, as claimed. N Gi (X) 2t 0 t 1 + 10 5 t 1 l 2t, 7
Case 2: t 100/ l. Recall that t /3 100/(3 l). Assume to the cotrary that N Gi (X) < 2 X. The i G i there is a vertex subset Y disjoit from X such that Y = 3t, ad G i has o edges betwee X ad Y. But the there were at most 2 mi { δ 0 /2 X, δ 0 /2 Y } edges betwee X ad Y i G 0. If t /4, the 3t /4, ad we get a cotradictio to Property P5 with X,Y substituted for U,W, respectively. If /4 t /3 (100)/(3 l), the 3t 100/ l, agai cotradictig Property P5 with Y,X istead of U,W, respectively. We have proved that give properties P1-P5 of G 0, for each i the graph G i is determiistically a (/3 100/(3 l ), 2)-expader. A similar argumet, i the case where δ 0 is odd, shows that the graph G δ0 /2 100/(2 l ), 1)-expader. is a (/2 Recall that a radom graph R i added at the i-th stage is distributed accordig to G,ρi ρ i 4000, so ρ l i 120 100 ad ρ 2 3l i > 20 100. Theorem 1 will thus follow from: 2 2l Lemma 6 (a) Let G = (V,E) be a (/3 k, 2)-expader o vertices, where k = o(). Let R be a radom graph G,p with p() = 120k/ 2. The Pr[G R is ot Hamiltoia] < e Ω(k). (b) Let G = (V,E) be a (/2 k, 1)-expader o vertices, where k = o(). Let R be a radom graph G,p with p() = 20k/ 2. The Pr[G R does ot cotai a (ear) perfect matchig] < e Ω(k). Proof (a) Observe that by Pósa s Lemma ad its cosequeces (Lemma 4): G is coected (Due to expasio of G there is o room for two coected compoets); G has a path of legth at least 3k 1 (due to Lemma 4); If a supergraph of G i is o-hamiltoia it has at least (/3 k) 2 /2 > 2 /20 holes. We split the radom graph R ito 6k idepedet idetically distributed graphs R = 6k i=1 8 R i, with
where R i G,pi ad p i p/(6k) = 20/ 2. Set G 0 = G, ad for each i = 1,...6k defie G i = G i R j. At Stage i we add to G i 1 the ext radom graph R i. A stage i is called successful if a logest path i G i+1 is loger tha that of G i, or if G i+1 is already Hamiltoia. Clearly, if at least 3k + 1 stages are successful the the fial graph G 6k is Hamiltoia. Observe that for Stage i to be successful, if G i 1 is ot yet Hamiltoia, it is eough for the radom graph R i to hit oe of the holes of G i 1. Thus, Stage i is usuccessful with probability at most (1 p i ) 2 /20 < 1/e. Let X be the radom variable coutig the umber of successful stages. The X stochastically domiates Bi(6k, 1 1/e). Hece by stadard estimates o the tails of the biomial distributio, j=1 Pr[G R is ot Hamiltoia] Pr[X 3k] < e Ω(k), as claimed. The proof of (b) is similar. 3 Proof of Theorem 2 We will prove the result for G,m, m = 1 2 K l. This implies the result for the G,p model. This time we will use the colorig argumet of Feer ad Frieze [6]. Cosider the followig properties: (Q1) K l /2 δ(g) (G) 2K l. (Q2) S (Q3) implies E(S) 2 S. K 3 (l ) 2 S /(K l ) implies N(S) (K l/5) S. K 3 (l ) 2 (Q4) If S,T are disjoit sets of vertices ad S T /10 the e(s,t) (K l /20) T. Lemma 7 If K is sufficietly large, G = G,m satisfies Q1 Q4 whp. Proof We will prove that G,p has these properties where p = K l/. Iflatig error probabilities by O( 1/2 ) will show them for G,m. Q1, Q2 are simple first momet calculatios. We will check Q3, Q4. The size of N(S) is distributed as the biomial B( s, 1 (1 p) s ). Now 1 (1 p) s sp/2 if sp 1. Applyig a Cheroff boud we see that Pr( S failig Q3) /(K l ) s= K 3 (l ) 2 9 ( ) e ( s)sp/32 = o(1). s
Similarly, Pr( S,T failig Q4) s /10 t /10 ( s )( t ) e K l T /80 = o(1). I the followig we will asssume that K is sufficietly large ad α is sufficietly small so that our claimed iequalities hold. We do ot attempt to optimise, sice we are far from gettig α close to 1/2. Now let H be a graph with (H) αk l ad let X be ay βm subset of E(G H) satisfyig (X) 2βK l. Here we will be assumig 1 β α. Let Γ = G H X. Lemma 8 If Q1 Q4 hold the (a) Γ is a (/30, 2)-expader. (b) Γ is coected. Proof (a) (i) S. 3K 3 (l ) 2 By costructio, we have δ(γ) (1/2 α 2β)K l. So if N Γ (S) < 2 S we fid that N Γ (S) S cotais at least ((1/2 α 2β)K l ) S /2 edges, cotradictig Q2. (ii) S /(K l ). 3K 3 (l ) 2 It follows from Q3 that (iii) /(K l ) S /30. Choose S S of size /(K l ). The N Γ (S) ((1/5 α 2β)K l ) S 2 S. N Γ (S) N Γ (S ) S (1/5 α 2β) S 2 S. (b) It follows from (a) that if Γ is ot coected the each compoet is of size at least /10. But the Q4 implies that there are at least (1/20 α 2β)K T l edges betwee each compoet i Γ, cotradictio. We ow resort to our colorig argumet. Let G 1,G 2,...,G M, M = ( ( ) 2) m be a eumeratio of graphs with vertex set [] ad m edges. For each i let H i be a fixed sub-graph of G i with (H i ) αk l such that G i H i is o- Hamiltoia, if oe exists. Otherwise H i is a arbitrary sub-graph of G i with the same restrictios o the maximum degree. If graph G is o-hamiltoia, let λ(g) deote the legth of the logest 10
path i G ad let λ(g) = if G is Hamiltoia. Now for a graph G i, let X i,1,x i,2,..., be a eumeratio of all βm-subsets of E(G i H i ). Let Γ i,j = G i H i X i,j. The let (a) G i satisfies Q1 Q4 (b) λ(g i H i ) = λ(g i H i X i,j ) 1 a i,j = (c) G i H i is ot Hamiltoia (2) (d) (X) 2βK l 0 otherwise The otatio A B stads for A (1 o(1))b. Lemma 9 If G i satisfies (a) ad (c) of (2) the j a i,j ( ) (1 α)m βm. Proof H i has at most 1 αk l = αm edges ad to esure (b) all we have to do is avoid some 2 fixed logest path of Γ i,j. Furthermore, almost all choices of βm edges will iduce a sub-graph with maximum degree at most 2βK l. Lemma 10 Let N = ( 2). The, ( )( )( ) N m (1 β)m a i,j m (899/900) βm. m βm αm i,j Proof Let K i,j = G i X i,j ad for a fixed graph K with (1 β)m edges let us estimate the umber of (i,j) with K i,j = K ad a i,j = 1. For each sub-graph H K with (H) αk l, we let θ(k,h) deote the umber of choices of βm edges X such that (i) K + X satisfies Q1 Q4 ad (ii) λ(k H) = λ(k + X H). The a i,j θ(k,h). (3) K,H i,j This is because for each (i,j) with a i,j = 1 there is a correspodig K i,j = G i X i,j such that G i = K i,j + X i,j satisfies Q1 Q4 ad a H i such that λ(k i,j H i ) = λ(k i,j + X i,j H i ). Now if K + X satisfies Q1 Q4 the from Lemmas 4 ad 8 we see that to esure λ(k H) = λ(k + X H), X must avoid at least (/30) 2 /2 edges i.e. ( ) N (1 β)m θ(k,h) (899/900) βm. βm Cosequetly, αm ( )( )( N (1 β)m N (1 β)m θ(k, H) (1 β)m t βm K,H t=0 ( )( )( N (1 β)m N (1 β)m m (1 β)m αm βm ( )( )( ) N m (1 β)m = m (899/900) βm. m βm αm 11 ) (899/900) βm ) (899/900) βm
Let ν H deote the umber of i such that G i satisfies Q1 Q4 ad yet G i H i o-hamiltoia ad let M = ( N m). We must show that νh = o(m). It follows from Lemma 9 that ( ) (1 α)m a i,j ν H. βm O the other had, Lemma 10 implies ν H ( N ) m( m βm m ( m i,j )( (1 β)m αm ( (1 α)m βm ) (899/900) βm ) ) βm ( ) αm me (1 β)e (899/900) βm (1 α)m βm α = o(1), ad Theorem 2 follows. Refereces [1] M. Ajtai, J. Komlós ad E. Szemerédi, The first occurrece of Hamilto cycles i radom graphs, Aals of Discrete Mathematics 27 (1985) 173-178. [2] T. Bohma, A.M. Frieze ad R. Marti, How may radom edges make a dese graph Hamiltoia?, Radom Structures ad Algorithms 22 (2003) 33-42. [3] B. Bollobás, The evolutio of sparse graphs, Graph Theory ad Combiatorics. (Proc. Cambridge Combiatorics Coferece i Hoour of Paul Erdős (B. Bollobás; Ed.)) Academic Press (1984) 35-57. [4] B. Bollobás, Radom graphs, 2d Ed., Cambridge Uiv. Press, Cambridge, 2001. [5] B. Bollobás ad A.M. Frieze, O matchigs ad hamiltoia cycles i radom graphs Aals of Discrete Mathematics 28 (1985) 23-46. [6] T.I. Feer ad A.M. Frieze, O the existece of hamiltoia cycles i a class of radom graphs, Discrete Mathematics 45 (1983) 301-305. [7] A.M. Frieze ad M. Krivelevich, O packig Hamilto Cycles i ε-regular Graphs, Joural of Combiatorial Theory B, 94 (2005) 159-172. [8] J. Komlós ad E. Szemerédi, Limit distributios for the existece of Hamilto circuits i a radom graph, Discrete Mathematics 43 (1983) 55-63. 12
[9] L. Pósa, Hamiltoia circuits i radom graphs, Discrete Mathematics 14 (1976) 359-364. [10] B. Sudakov ad V. Vu, Resiliece of graphs, submitted. 13