Edge Disjoint Hamilton Cycles

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1 Edge Disjoit Hamilto Cycles April 26, Itroductio l +l l +c I the late 70s, it was show by Komlós ad Szemerédi ([7]) that for p =, the limit probability for G(, p) to cotai a Hamilto cycle equals the limit probability for G(, p) to have miimum degree at least 2. A few years later, Ajtai, Komlós ad Szemerédi ([1]) have show a hittig time versio of this i the G(, m) model. Say 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. Is it true that for every 0 p 1 the radom graph G(, p) has property H with high probability? This is clear wheever δ(g) = 0. I the early 80s, Bollobás ad Frieze ([3]) have proved that cojecture for δ(g) = O(1). I this talk I pla to prove the result for p() (1 + o(1)) l /. This is a result of Frieze ad Krivelevich from 08 ([4]). Remark 1. The cojecture is owadays kow to be true for every p. It was proved for the rage l 50 / p 1 l 9 / 1/4 by Kox, Küh ad Osthus i 13 ([6]), i a rather techically complicated paper. Later, Krivelevich ad Samotij ([8]) have closed the gap for the sparse case, ad Küh ad Osthus ([9]) have closed the gap for the dese case. This is the mai result we ited to prove: Theorem 2. Let p = p() (1 + o(1)) l /. The whp G(, p) has property H. Remark 3. I this talk I will ot cosider the extra ear perfect matchig, expected i the case where δ(g) is odd. This adds some techicality, but othig really differet. 2 Prelimiaries 2.1 Probability Theorem 4 (Cheroff bouds, [5], Theorem 2.1). Let X Bi (, p), µ = p, a 0. The the followig iequality holds: ) P (X µ a) exp ( a2. 2µ 0 Git reviso 94a

2 Defiitio 5. A mootoe icreasig graph property P is a set of graphs which is closed upwards; that is, if G P ad G H the H P. Similarly, a mootoe decreasig graph property Q is a set of graphs which is closed dowwards; that is, if G Q ad H G the H Q. Theorem 6 (The FKG iequality for mootoe graph properties, [2], Theorem 6.3.3). Let P 1, P 2 be two mootoe icreasig graph properties ad Q 1, Q 2 be two mootoe decreasig graph properties. Let G G(, p). The: P (G P 1 P 2 ) P (G P 1 ) P (G P 2 ), P (G Q 1 Q 2 ) P (G Q 1 ) P (G Q 2 ), P (G P 1 Q 1 ) P (G P 1 ) P (G Q 1 ). 2.2 Sprikle sprikle I the proof, we will use several stadard techiques/tricks. The first trick is the trick of spriklig radom edges. Formally, we d like to preset G as a uio of G 0, which is very similar to G, ad some radom leftovers, R. This ca be achieved easily by takig p 0 ad ρ so that 1 p = (1 p 0 )(1 ρ) ad lettig ρ = o(1/), thus decomposig G G(, p) to G = G 0 R where G 0 G(, p 0 ) ad R G(, ρ). I which sese are G ad G 0 similar? I the followig: Claim 7. Fix G 0, let R G(, ρ) ad G = G 0 R; the whp δ(g 0 ) = δ(g). Proof. Clearly, δ(g 0 ) δ(g), as G cotais all the edges of G 0 ad more. Now, let v G 0 with d G0 (v) = δ(g 0 ). As ρ = o(1/), d R (v) = 0 whp (a stadard first momet argumet), implyig δ(g) d G (v) = d G0 (v) = δ(g 0 ). From ow o, write δ 0 = δ(g 0 ). It follows that it is eough to prove that G cotais (whp) δ 0 /2 edge disjoit Hamilto cycles ad a edge disjoit ear perfect matchig if δ 0 is odd. We also assume that p = (1 + o(1)) l /, as otherwise δ 0 1 ad there s othig ew to prove. We also ote that from this assumptio it follows that δ(g) = o(l ); this will follow from the followig claims. Let D k be the radom variable coutig the umber of vertices i G(, p) with degree exactly k. Clearly, D k = v [] D k(v), where D k (v) is the idicator of the evet that v is of degree k. Note that E (D k ) = ( ) 1 E (D k (v)) = P (d(v) = k) = p k (1 p) 1 k. k v [] Thus, lettig k = δ l ad p = (1 + ε) l / for ε = o(1), ( ) 1 E (D k ) = p δ l 1 δ l (1 p) δ l ( ) ( 1)p δ l e p δ l ( ) 1 + ε δ l ε δ l(1/δ) ε = ω(1), δ 2

3 if we take δ = δ() to be large eough, say, δ = ε. Claim 8. For k = O(l ), if E (D k ) = ω(1) the Var (D k ) = o ( E 2 (D k ) ). Proof. Let u v be two vertices. Note that ( k 1 ( k ) p k 1 ) p k = k k (1 + o(1)) = (1 + o(1)) = O(1), p l thus Cov (D k (u), D k (v)) = P (d(u) = k = d(v) u v) P (u v) +P (d(u) = k = d(v) u v) P (u v) P 2 (d(u) = k) (( ) ) 2 2 = p k 1 (1 p) 1 k p k 1 (( ) ) 2 2 (( ) p k (1 p) 2 k (1 p) )p k (1 p) 1 k k k = O ( E 2 (D k )p 2) ( ( )) 1 + O E 2 (D k ) 2 1 p 1 = o ( E 2 (D k ) 2). It follows that Var (D k ) E (D k ) + u v Cov (D k (u), D k (v)) = o ( E 2 (D k ) ). For techical reasos, we ll defie a very particular ρ so that ρ = o(1/) will hold. Set d 0 = mi {k E (D k ) 1}. Claim 9. d 0 = o(l ). Proof. As we ve see, for k = δ l, δ = o(1), E (D k ), ad d 0 < k, so d 0 = o(l ). Note that d 0 approximates δ(g); formally, Claim 10. whp, δ(g) d 0 2. Proof. Note that E (D k+1 ) E (D k ) = ( ) 1 k+1 p k+1 (1 p) 2 k ( ) 1 = k p k (1 p) 1 k ( 1 k)p (k + 1)(1 p). As we ve see, d 0 = o(l ). Thus it follows that for b 1, E (D d0 b 1) = E (D d0 b) (d 0 b)(1 p) ( 1 (d 0 b 1))p < d p = ε = o(1), 3

4 ad by a Markov s iequality ad the uio boud, d 0 1 P ( b 1, D d0 b 1 > 0) (ε ) b ε 1 ε = o(1). b=1 I additio, 1 ( 1 d 0 )p E (D d0 +1) = E (D d0 ) (d 0 + 1)(1 p) 2 p ad by Chebyshev s iequality ad the previous claim, d 0 = ω(1), P (D d0 +1 = 0) P ( D d0 +1 E (D d0 +1) 1) Var (D d 0 +1) E 2 (D d0 +1) = o(1). Therefore, whp there is o vertex with degree at most d 0 2 ad there is a vertex with degree d 0 +1, thus δ(g) d 0 2. Corollary 11. whp, δ(g) = o (l ). We the defie ρ = 2001(d 0 + l l ), l ad observe that ρ = o(1/) (agai, sice d 0 = o(l )), ad that p 0 = p(1 + o(1)). 2.3 Properties of radom graphs I this sectio we give a list of properties, each occurig whp, i the radom graph G 0 G(, p 0 ). Defie the set SMALL: SMALL = { v V (G) d G0 (v) 0.1 l }. Lemma 12. The radom graph G 0 G(, p 0 ) with p 0 defied earlier, has whp the followig properties: (P1) There is o o-empty path of legth at most 4 i G 0 such that both of its (possibly idetical) edpoits lie i SMALL. (P2) Every set U V (G) with U 100/ l spas at most U (l ) 1/2 edges i G 0. (P3) For every two disjoit sets U, W V (G) with U 100/ l, W U l /10000, E G0 (U, W ) < 0.09 U l. (P4) For every two disjoit sets U, W V (G) with U 100/ l, W /4, E G0 (U, W ) 0.1 U l. 4

5 Proof of (P1). Fix a vertex v. Note that P (v SMALL) = 0.1 l k=0 P (Bi ( 1, p 0 ) = k) ( ) l 0.1 l p0 (1 p 0 ) 0.1 l ( ) 10ep 0.1 l 0.1 l e p 0( l ) l l e (1 o(1)) l < l Now fix u v. The probability that u, 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 (choosig l 1 ier vertices ad for each such choice requirig l edges). Furthermore, as there s exactly oe edge of K coectig u with v, coditioig o the evet u SMALL caot icrease the probability of v SMALL by too much: P (u, v SMALL) P (v SMALL u SMALL) P (u SMALL) P (v SMALL {u, v} / E) P (u SMALL) 1 P (v SMALL) P (u SMALL) 1 p. Note also that u, v SMALL is a mootoe decreasig evet ad d(u, v) 4 is a mootoe icreasig evet. Thus, accordig to the FKG iequality, Therefore, P (u, v SMALL d(u, v) 4) P (u, v SMALL) P (d(u, v) 4). P (u, v SMALL d(u, v) 4) P (u, v SMALL) P (d(u, v) 4) P (v SMALL) P (u SMALL) P (d(u, v) 4) (1 + o(1)) l4 (1 + o(1)) < 2.1. Applyig the uio boud over all possible pairs of u, v we establish (P1). Proof of (P2). For a give U [] with U = u 100/ l, let A U E(U) u l 1/2. By the uio boud, P ( U, U = u 100/ l, A U ) 5 100/ l 100/ l 100/ l ( )( ( u ) 2) u u l 1/2 p u l1/2 ( ( e eup ) ) l 1/2 u u 2 l 1/2 e u ( 2u l 1/2 be the evet by which ) l 1/2 u.

6 We ow separate the sum ito two: ( l e u ) l 2u l 1/2 1/2 ( u l e ) l 2 l 3/2 1/2 = o(1), ad 100/ l u=l e u ( ) l 2u l 1/2 1/2 u ( ( u l e ) 1/2 1 ( ) ) l 2 l 1/2 1/2 u ( e ( ) ) 1 l 1/2 1 ( ) u l 2 l 1/2 1/2 = o(1). l Proof of (P3). For a give U [] with U = u 100/ l ad W [] with W u = u l 10000, let A U,W be the evet by which E(U, W ) 0.09u l. By the uio boud, P ( U, W, A U,W ) 100/ l 100/ l 100/ l 100/ l 100/ l 100/ l u l /10000 w=1 ( u )( )( ) uw p w 0.09u l ( )( )( ) uu u u u p 0.09u l u ( (e u u ( e 0.09u l 0.09u l ) ( e ) ( l /10000 eu ) ) p 0.09 l u u 0.09 l l /10000 ( e 0.09 ( ( u 2 ) ) 0.08 l u u u ( /u ) u = o(1), ) 0.09 l ( u ) l l ) u as 0.08/u 8. Proof of (P4). Fix U, W, U 100/ l, W /4. Note that the umber of edges betwee U, W i G 0 is biomially distributed with U W trials ad success probability p 0, hece E ( E G0 (U, W ) ) (1 + o(1)) U l /4. 6

7 By Cheroff bouds (Theorem 4), P ( E G0 (U, W ) 0.1 U l ) P ( E G0 (U, W ) 0.25 U l 0.15 U l ) ) (0.15 U l )2 exp ( U l < exp ( U l ) < exp ( 4). Now, the umber of pairs U, W is at most 4, uio boud gives that the probability that such a pair exists is at most 4 e 4 = o(1). 2.4 Expaders, rotatios ad boosters Oe of the key cocepts i may coectivity ad Hamiltoicity related problems is that of a expader. Defiitio 13. For every c > 0 ad every positive iteger R we say that a graph G = (V, E) is a (R, c)-expader if every subset of vertices U V of cardiality U R satisfies N G (U) c U. Claim 14. Let G be a (k, 2)-expader o vertices, with k > 4. The, G is coected. Proof. Sice every set of cardiality at most /4 expads, every coected compoet must be of cardiality at least 3/4, ad there s room for oly 1 such compoet. Our approach will cosist of that cocept, budled with the so-called rotatio-extesio techique, itroduced by Pósa i 76 ([10]). Here we will cover the techique, icludig a key lemma. Give a path P = (v 0,..., v m ), we ca exted it by addig v m+1 which is ot part of the path but is a eighbour of v m, or we ca rotate it by fidig a eighbour v i of v m iside the path, addig the edge {v m, v i } ad erasig the edge {v i, v i+1 } (1 i < m). v m+1 v 0 P v i v i+1 v m Figure 1: Pósa extesio e v 0 P v i v i+1 v m Figure 2: Pósa rotatio Lemma 15. Let G be a graph, P a path of maximal legth i G, P the set of all (rooted) paths obtaied by P be a sequece of rotatios, U the set of edpoits of these paths, N ad N + the sets of vertices immediately preceedig ad followig the vertices of U alog P, respectively. The, N(U) N N +. 7

8 Proof. Deote P = (v 0,..., v m ). Let u U, v / (U N N + ), ad let P u be a rotatio of P edig at u. If v / P the {u, v} / E, otherwise we could have exteded P u ad get a loger path, cotradictig our assumptio. Thus, v P. Let v, v + be its two possible eighbours i P. Suppose {u, v} E. The, we ca rotate P u to get P w edig at w, where w is a eighbour of v. If w is v or v +, we get a cotradictio, as this puts v i N or N +. Thus, oe of the edges i P betwee v ad v, v + broke durig a rotatio. Let s look at whe it has happeed; the, if {v, v} broke, that rotatio has made v a ed vertex, ad if {v, v + } broke, that rotatio has put v N, N +. Thus, {u, v} / E. Corollary 16. N(U) N N + 2 U 1. Corollary 17. Let G be a coected o-hamiltoia (k, 2)-expader. The G cotais a path of (edge) legth 3k 1. Proof. Let P be a path of maximal legth m (coutig i edges) i G. Recall that N(U) 2 U 1, that is, U does ot expad, hece U > k. Let U U with U = k. Sice P is maximal, N(U ) V (P ), thus V (P ) 3k, hece P is of legth at least 3k 1. I order to utilize that lemma for our eeds, we itroduce the otio of a booster: Defiitio 18. Give a graph G, a o-edge e = {u, v} of G is called a booster if addig e to G creates a graph G, which is either Hamiltoia or whose maximum path is loger tha that of G. Note that techically every o-edge of a Hamiltoia graph G is a booster by defiitio. Boosters advace a graph towards Hamiltoicity whe added; addig sequetially boosters clearly brigs ay graph o vertices to Hamiltoicity. Corollary 19. Let G be a coected o-hamiltoia (k, 2)-expader. The G has at least (k+1)2 2 boosters. Proof. Let P be a path of maximal legth m (coutig i edges) i G. Agai, U > k. We ow seek of (k+1)2 2 o-edges which, whe added, create a cycle of legth m + 1. Fix a set u 1,..., u k+1 of ed vertices. For each, let P i be the rotatio of P edig at u i. For such i, fix u i as a startig vertex, ad let P i be the set of rotatios of P i. Let U i be the set of edpoits retrieved that way. As before, U i > k. Let u (i) 1,..., u(i) k+1 be a set of such ed vertices. Note that for every i, j [k + 1], u i, u (i) j are ot coected, as if they were, we would have a cycle of legth m + 1, ad either ed up with a Hamilto cycle, or, if m + 1 <, sice G is coected, get a loger path. As each o-edge was couted at most twice that way, we have at least (k + 1) 2 /2 such o-edges, each is a booster. 8

9 3 The proof The outlie of the proof is as follows: we split the graph R ito δ 0 /2 idetically distributed radom graphs R i. We start with G 0, fidig eough boosters i R 1 to get a Hamilto cycle, deletig its edges ad ed up i G 1, ad cotiuig so: give G i 1 (1 i δ 0 /2 ), we fid boosters i R i to get a Hamilto cycle H i, ad by deletig it we get G i. Durig the process, we ll keep the followig attributes of G i : (I1) δ(g i ) 2 (I2) G i is a (/3 c/ l, 2)-expader (that will follow from (P1)-(P4)) (I3) G i is coected (I4) G i has a path of legth at least c/ l (I5) G i has quadratic umber of boosters. If δ 0 is odd, we ll eed a fial stage to create a ear perfect matchig. 3.1 Formal argumet We may assume that δ 0 2, otherwise there s othig ew to prove. For 1 i δ 0 /2 defie ρ i by 1 ρ = (1 ρ i ) δ 0/2. Observe that ad thus ρ i 1 ρ = (1 ρ i ) δ 0/2 1 ρ i δ 0 /2, ρ δ 0 /2 = 2001(d 0 + l l ) 4000 δ 0 /2 l l. Now let R = δ 0 /2 i=1 where R i G (, ρ i ), ad let G i be the graph obtaied from G 0 i j=1 R i after havig deleted the first i Hamilto cycles, assumig that the previous i 1 stages were ideed successful. Let i < δ 0 /2. To see (I1), ote that every vertex had its degree i G 0 reduced by at most 2i i G i. Thus, R i, δ(g i ) δ 0 2i δ 0 2( δ 0 /2 1) 2. To see (I2), we ow show that G i is a (k, 2)-expader for k = /3 100/(3 l ). For that, let X be a vertex set with t vertices. Cosider the followig two cases: 9

10 X S X L N Gi (X S ) X N Gi (X L ) (X S N G0 (X S )) Figure 3: Case I Case 1 (t 100/ l ): Deote X S = X SMALL ad X L = X X S. Deote t S = X S ad t L = X L. Observe that N Gi (X S, V X) 2t S t L ; ideed, by property (P1), every vertex i X S has at least 2 uique eighbours i V X S, ad at most t L of these 2t S eighbours lie i X L. By property (P2), X L spas at most t L (l ) 1/2 edges i G 0. Recall that the miimal degree of X L i G 0 is at least 0.1 l ; thus, at least 0.09t L l edges leave X L i G 0. But the by (P3), the set of eighbours of X L must be of cardiality greater tha t L l / By (P1) agai, at most t L of these fall ito X S N G0 (X S ). As i G i each vertex has lost at most δ 0 eighbours comparig to what it had i G 0, we have that i G i, the set of eighbours of X L outside X S N G0 (X S ) is at least t L (l / δ 0 ) t L l / Altogether, as claimed. N Gi (X) 2t S t L + t L l / t S + 2t L = 2t, Case 2 (100/ l t /3 100/(3 l )): Assume to the cotrary that N Gi (X) < 2t. I that case we ca fid a vertex set Y disjoit to X N Gi (X) of cardiality 3t. Thus, i G 0 there were at most 2 δ 0 /2 mi {t, 3t} edges betwee X ad Y. If t /4 the 3t /4 ad by (P4) we should have had E G0 (X, Y ) 0.1t l δ 0 t. If t /4 the 3t 3 (/3 100/(3 l )) = 100/ l, ad agai by (P4) we should have had E G0 (Y, X) 0.1 Y l δ 0 Y. The proof of Theorem 2 will follow from: Lemma 20. 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, P (G R is ot Hamiltoia) < exp ( Ω(k)). Proof. Note that the followig properties hold for G: (I3) G is coected (due to Claim 14) 10

11 (I4) G has a path of legth at least 3k 1 (due to Corollary 17) (I5) If a supergraph of G is o-hamiltoia it has at least 2 /20 boosters (due the Corollary 19, ad sice (/3 k + 1) 2 /2 > 2 /20). We split the radom graph R ito 6k idetically distributed graphs R = 6k R i where R i G(, p i ) ad p i p/(6k) = 20/ 2. Set G 0 = G ad for i [6k] let i=1 G i = G i R j. At stage i we add to G i 1 the ext radom graph R i. We call a stage successful if the maximal legth of a path i G i is loger tha that of G i 1, or if G i is Hamiltoia. Clearly, if at least 3k + 1 stages are successful the the fial graph G 6k is Hamiltoia (due to (I5)). 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 boosters of G i 1. Thus, stage i is usuccessful with probability at most (1 p i ) 2 /20 < 1/e. Thus, the umber of successful stages S is a radom variable which stochastically domaiates Bi (6k, 1 1/e). Therefore, puttig c = 1 1/e ad usig Cheroff bouds (Theorem 4), P (G R is ot Hamiltoia) P (S 3k) exp ( (6c 3)2 k 2 ) < exp ( Ω(k)). 2 6ck j=1 Proof of Theorem 2. Suppose we have G i 1 for 1 i δ 0 /2. We have show that G i 1 is a (/3 k, 2)-expader for k = 100/(3 l ) = o(). R i is a radom graph with probability ρ i 4000/( l ) = 120k/ 2. Therefore by the above lemma, G i 1 R i is ot Hamiltoia with probability at most exp ( Ω(k)). Uio boud over all δ 0 /2 steps yields the desired result. Refereces [1] Miklós Ajtai, Jáos Komlós, ad E Szeraerédi, First occurrece of hamilto cycles i radom graphs, North-Hollad Mathematics Studies 115 (1985), [2] Noga Alo ad Joel H. Specer, The probabilistic method, 3rd ed., Wiley Series i Discrete Mathematics ad Optimizatio, Joh Wiley & Sos, [3] Béla Bollobás ad Ala M Frieze, O matchigs ad hamiltoia cycles i radom graphs, North-Hollad Mathematics Studies 118 (1985), [4] Ala Frieze ad Michael Krivelevich, O two hamilto cycle problems i radom graphs, Israel Joural of Mathematics 166 (2008), o. 1, [5] Svate Jaso, Tomasz Luczak, ad Adrzej Ruciński, Radom graphs, Wiley Series i Discrete Mathematics ad Optimizatio, Joh Wiley & Sos, [6] Fiachra Kox, Daiela Küh, ad Deryk Osthus, Edge-disjoit hamilto cycles i radom graphs, Radom Structures & Algorithms (2013). 11

12 [7] Jáos Komlós ad Edre Szemerédi, Limit distributio for the existece of hamiltoia cycles i a radom graph, Discrete Mathematics 43 (1983), o. 1, [8] Michael Krivelevich ad Wojciech Samotij, Optimal packigs of hamilto cycles i sparse radom graphs, SIAM Joural o Discrete Mathematics 26 (2012), o. 3, [9] Daiela Kuh ad Deryk Osthus, Hamilto decompositios of regular expaders (2013). [10] Lajos Pósa, Hamiltoia circuits i radom graphs, Discrete Mathematics 14 (1976), o. 4,