k-connectivity of uniform s-intersection graphs

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1 k-connectivity of uniform -interection graph Mindauga Blozneli, Katarzyna Rybarczyk Faculty of Mathematic and Informatic, Vilniu Univerity, Vilniu, Lithuania, Faculty of Mathematic and Computer cience, Adam Mickiewicz Univerity, Poznań, Poland, Abtract Let W 1,..., W n be independent random ubet of [m] = {1,..., m}. Auming that each W i i uniformly ditributed in the cla of d-ubet of [m] we tudy the uniform random interection graph G n, m, d on the vertex et {W 1,... W n }, defined by the adjacency relation: W i W j whenever W i W j. We how that a n, m the edge denity threhold for the property that each vertex of G n, m, d ha at leat k neighbour i aymptotically the ame a that for G n, m, d being k-connected. 1 Introduction Let H = Hn, m, d be a random bipartite graph with bipartition V, W, where V = n, W = m, and where each vertex from V chooe d neighbour in W uniformly at random and independently of the other vertice of V. Given a natural number, 1 < d, the uniform random interection graph G = G n, m, d i defined a the graph on the vertex et V, where u, v V are adjacent denoted by u v if they hare at leat common neighbour in H. We refer to the vertice of V a enor, and the vertice of W we call key. Thi random graph model ha been widely tudied in the literature mainly a a model of ecure wirele enor network that ue random preditribution of key ee [1], [7], [10], [11], [12], [17]. Our tudy i motivated by fact that k connectivity of G i an important characteritic of the reliability of the enor network a well a it reilience againt attack by an adverary controlling a certain number of enor ee [6]. Key word and phrae: random interection graph, k-connectivity, wirele enor network. Mathematic Subject Claification 2010: primary 05C80, 05C40, econdary 05C07. 1

2 We tudy the threhold for the property C k that G i k-connected, i.e., that G i connected and that the removal of any et of at mot k 1 vertice doe not diconnect the graph. Here k = 1, 2,... i arbitrary, but fixed. For thi purpoe we conider a equence of random graph {G n, m, d, n = 1, 2,... }, where m = mn a n, and the number = n and d = dn may depend on n. In particular, they may tend to infinity a n, but at a low rate, ee 1 below. We aume that n < dn. By δg we denote the minimum degree of a graph G. We denote by p = pn, m, d, the edge-probability in G n, m, d. We alway aume that expreion o, O refer to the cae where n, and all inequalitie are aumed to hold for n which i large enough. A neceary condition for a graph to be k-connected i that it ha no vertex of degree le than k. Our firt reult how that the threhold for the propertie C k and δ G n, m, d k coincide. Theorem 1. Let k 1 be an integer. Let γ 0, 1. Let m, n +. Aume that + 2 d 5ln 3d /3d ln n 2 ln n 1 γ 1 and for ome θ > 1 we have Then ln n ln 1/2 n np θ ln n. 2 P G n, m, d C k P δg n, m, d k n γ+o1 d +2 d 4. 3 Our next theorem give the threhold for the property δ G n, m, d k. Theorem 2. Let k 1 be an integer. Let m, n +. Aume that d 2 = om ln 1 n, 4 1 d 1 ln n 1 + k 1d 1 ln ln n +. 5 Then for we have np = ln n + k 1 ln ln n + x n with x n = oln n 6 lim P δg n, m, d k = n 0 if x n, e e x k 1! if x n x, 1 if x n +. 7 Combining 3 and 7 we obtain the threhold for the property C k. Theorem 3. Let k 1 be an integer. Let m, n +. Suppoe that for ome γ 0, 1 condition 1 i atified. Aume that 4, 5 hold. Then for p atifying 6 we have 0 if x n, lim P G n, m, d C k = e e x k 1! if x n n x, 8 1 if x n +. We remark that in the tatement of Theorem 2 and 3 the edge-probability p can be replaced by the expreion only involving d, and m ˆp = d2!m, 2

3 where a b = aa 1 a b+1 for any poitive integer b. Indeed, a we hall ee in Lemma 1 below, condition 4 implie that p = ˆp1 oln 1 n. Therefore, for nˆp = ln n + k 1 ln ln n + x n, with x n = Oln n, we have np = nˆp + o1. In particular, Theorem 2 and 3 remain true with p replaced by ˆp. In the following corollary of Theorem 2 condition 4 and 5 are replaced by a impler, but more tringent condition d = oln 1/2 n. Corollary 1. Let k 1 be an integer. Let m, n +. Aume that d = oln 1/2 n. Suppoe that nˆp = ln n + k 1 ln ln n + x n, with x n = oln n. Then 7 hold. In the particular cae where 1 i contant we have the following reult. Corollary 2. Let k 1 be an integer. Let 0 < α < 0.2. Let m, n +. Aume that 1 and d = Oln α n. Suppoe that nˆp = ln n + k 1 ln ln n + x n. Then 8 hold. We note that the condition x n = oln n doe not appear in Corollary 2. Theorem 3 and Corollary 2 ay that the edge denity threhold for the property that G n, m, d i k-connected i the ame a that of the binomial random graph Gn, p, where edge are inerted independently, ee [5], [9], [13]. Our reult are obtained under the aumption that < d. In the cae where = d the random graph G n, m, d i a union of dijoint clique. It i connected alo k-connected whenever all enor have choen the ame collection of key. Thi happen with probability m n 1, d which doe not depend on k. Related work. For k = 1 the edge denity threhold for the property δg n, m, d 1 ha been hown in [12]. For 1 the connectivity and k-connectivity of G 1 n, m, d ha been tudied in [1], [7], [17], [20], [21]. For > 1 the connectivity threhold of G n, m, d ha been hown in [3]. Our proof of Theorem 1 differ from thoe of [1], [7], [17], [20], [21]. It relie on an expanion property of G n, m, d etablihed in [3]. 2 Proof Before the proof we introduce ome notation and formulate an auxiliary lemma. For a et Ω and a natural number t, we denote by Ω t the collection of t-element ubet of Ω. The et of key adjacent to a enor v V in H i denoted W v. We ay that a enor v cover a et of key B if B W v. Subet of W of ize are called joint. Lemma 1. ee, e.g., Lemma 6 of [4] Given integer 1 d m, let W 1, W 2 be independent random ubet of the et W = {1,..., m} uch that W 1 and W 2 are uniformly ditributed in the cla of ubet of W of ize d. Then 1 d 2 m + 1 d ˆp P W 1 W 2 = P W 1 W 2 ˆp. Proof of Theorem 3. The reult follow by Theorem 1 and Theorem 2. The fact that 1 indeed implie that the quantity in the right-hand ide of 3 tend to 0 i hown in [3] ee the proof of Theorem 1 in [3]. Proof of Corollary 1. We hall how that condition 4, 5, 6 of Theorem 2 are atified. The bound nˆp = Oln n implie that ˆp 1 cn ln 1 n, for ome contant c > 0. Furthermore, the inequality d >! implie m > ˆp 1. Hence, we have m cn ln 1 n. The later inequality implie 4, ince m e 1 1+o1 ln n e ln1/2 n > d 2 ln 2 n, for < d = oln 1/2 n. 3

4 Let u how 5. For 2 1 d we have 1 d The quantity on the left ide of 5 i bounded from below by ln n k ln ln n. Hence, it tend to +, ince < d = oln 1/2 n. For > 2 1 d we write 1 d 1 d 1 1 d 1 > d 2. Now the quantity on the left ide of 5 i bounded from below by d 2 ln n k + 1d 1 ln ln n. It tend to +, ince d = oln 1/2 n. Finally, 4 and Lemma 1 imply p = ˆp1 Od 2 /m = ˆp on 1. nˆp = ln n + k 1 ln ln n + x n, with x n = oln n, implie 6. Hence, the relation Proof of Corollary 2. Firt we conider the cae where x n = oln n. In thi cae we have nˆp = Oln n and we derive 4, 5, 6 from the bound d = oln α n a in the proof of Corollary 1 above. The bound d = oln α n alo implie 1. Hence, condition of Theorem 3 are atified and we obtain 8. Uing a coupling argument we extend the reult to the cae where the condition x n = oln n i violated. We note that except for ome particular cae, we do not know how to contruct a proper coupling of random interection graph. One exception i the cae = 1, where uch a coupling i available. In [1] ee alo the proof of Corollary 1 in [3] it i hown that if m = hm for ome integer h then there i a common probability pace on which G 1 n, m, d G 1 n, m, d with probability 1. In particular, we have PG 1 n, m, d C k PG 1 n, m, d C k. If, in addition, m and m are uch that the firt probability tend to 1 the econd probability tend to 0, then the econd probability tend to 1 the firt probability tend to 0 a well. Therefore it i enough to et m = m m = m and m m uch that the edge probability in G 1 n, m, d G 1 n, m, d follow 6 with x n x n. Proof of Theorem 1. We ue the ame notation a in [3]. Conider an Hn, m, d uch that 1 and 2 hold. The et of key adjacent to a enor v V in H i denoted by W v. Given, let H = H n, m, d be a bipartite graph with bipartition V, W, where v V and B W are adjacent whenever v cover B. Hence, Hn, m, d define H n, m, d and H n, m, d define G = G n, m, d. We note that every enor cover d joint and the probability that a given enor cover a joint choen uniformly at random i d m 1. We denote r = d and p = r m 1. A joint B W i called thin if the number v V I {B W v} of enor that cover B i le than k = r 2 ln ln n 1 ln n ; otherwie, B i fat. A enor v V i tiny if every B W v i thin and it i heavy if every B Wv i fat. Otherwie, v i mall. A ubet S V i heavy if all it member are heavy. We remark that our choice of k enure that any et of heavy enor ha a large neighbourhood in G, ee the property A 5 below. We fix k and conider the following propertie of a graph H cf. [3]. A 2 : no two tiny enor are within ditance 8 from each other 8 hop in graph H ; A 3 : every fat joint i covered by at mot + 1r mall enor; A 4 : there are fewer than 2p 1 k 1 mall enor; A 5 : for any heavy et of enor S V of ize S 2n/3 we have NS min { + 1r 2 + r + 1 S, 2p 1 }. Here NS = { u V \ S : u v for ome v S } denote the neighbourhood of S in G. 4

5 Let A denote the event that the random graph H atifie all the propertie A 2, A 3, A 4, A 5. In [3] it wa hown that PA = 1 o1. More preciely, we have, ee Lemma 4 and 5 in [3], 1 PA i 1 + o1 n γ d +2r 4 + n r o1, i = 2, 3, 4, 5. 9 We remark that although our definition of the property A 4 differ from that of [3], where only the cae k = 1 i conidered, the argument of the proof of the upper bound for 1 PA 4 in Lemma 4 in [3] applie to an arbitrary, but fixed k. Hence 9 hold. Now we derive 3. For thi purpoe we how that the event A {δg k} implie S V with 1 S n + 1/2 we have NS k implie the k-connectivity property of G. In order to how that A {δg k} implie 10 we partition V = V T V S V H, where V T, V S and V H denote the et of tiny, mall and heavy enor repectively. For S V T 10 follow from δg k and the property A 2. For S V T V S with S V S we find a fat joint covered by a mall enor, ay v, from S. By A 3, thi fat joint i covered by at leat k + 1r > k heavy enor which are neighbour of v from outide S. Here, the latter inequality follow from 1. Now conider a et S uch that S H := S V H i nonempty. In the cae where H := S H i le than k, we fix a fat joint of a heavy vertex v S H and in view of A 3 we find at leat k +1r heavy enor that cover thi joint. Among thee heavy enor at leat k + 1r H k + 1r k > k are from outide S, where the latter inequality follow from 1. Hence NS k. Now aume that H k. Heavy vertice of S H all together contain at mot H r fat joint and thee can be covered by at mot H r + 1r mall enor, by property A 3. In the cae where + 1r 2 + r + 1 S H < 2p 1, the property A 5 yield that the et NS H ha at leat +1r 2 +r +1 H enor and we know that there are at mot + 1r 2 H mall enor among them. Hence NS H contain at leat r + 1 H r + 1k > k heavy enor and, obviouly, thee are from outide of S. Finally, in the cae where + 1r 2 + r + 1 H 2p 1, the inequality NS H 2p 1 implie that NS H contain at leat k heavy enor, becaue by A 4 the total number of mall enor the graph G i le than 2p 1 k. Proof of Theorem 2. Denote λ n = e xn /k 1! and λ = e x /k 1!. Let X n denote the number of vertice of G n, m, d of degree at mot k 1. In view of the identity PδG n, m, d k = PX n = 0 it uffice to how 7 with PδG n, m, d k replaced by PX n = 0. For thi purpoe we prove that, for t = 1, 2,..., lim n + λ t n EX n t = Let u how that 11 implie 7. For x n +, 11 implie EX n = o1 and we obtain 1 PX n = 0 = PX n 1 EX n = o1, by Markov inequality. For x n, 11 implie EX n 2 /EX 2 n = 1 o1 and we obtain 1 PX n = 0 = PX n 1 = 1 o1 uing the Paley-Zygmund inequality PX n 1 EX 2 n/ex 2 n. Finally, for x n x, 11 implie EX n t = λ t 1 + o1, for every t = 1, 2,.... By the method of moment, we obtain that X n converge in ditribution to the Poion ditribution with mean λ. Hence, PX n = 0 e λ. Let u prove 11. Given t, the number X n t count t-ubet of the et of vertice having degree at mot k 1, thu X n t = t! V V, V =t I B V, where I BV i the indicator of the event B V := {all vertice from V have degree at mot k 1}. It follow now, by ymmetry, that EX n t = t! n t PB, where B := BV, V := {v 1,..., v t } 5

6 Thu in order to how 11 we are left with proving PB = 1 + o1 λ n /n t. 12 In the proof of 12 we approximate PB by the probability that W v1,..., W vt are dijoint, all vertice from V are of degree exactly k 1 and their neighbourhood are dijoint. Therefore we conider the following event. C 0 : each vertex from V ha degree k 1, W v1,..., W vt are dijoint and every v V \ V ha at mot one neighbour in V and if uch a neighbour exit, it hare with v exactly key, while any other member of V ha no common key with v; C 1 : the et of vertice from V \ V, having at leat one neighbour in V, can be divided into dijoint ubet V 1,..., V t V \ V uch that for every i = 1,..., t we have V i k 1 and all member of V i are neighbour of v i we note that any vertex from V i i allowed to be a neighbour of v j V for j i. We have C 0 B C 1, i.e., event C 0 implie event B, and event B implie event C 1. Hence, Thu in order to prove 12 it i enough to how that PC 0 PB PC PC 0 = 1 + o1 λ n /n t, PC o1 λ n /n t. 14 The proof of 14 i technical. In order to avoid cumberome formulae we introduce the notation T = nˆptk 1 k 1! t e tnˆp, τ = d2 m. We oberve, that ˆp τ /! τ. Let u how that p = ˆp on 1. We note that 4 implie 1 > d2 d2 m. Hence, m > d2 d+1 m d+1. Next, the inequalitie d2 d + 1 > d 1 2 d 2 imply d 2 m > d 2 d 2 m d+1. Combining thi inequality with the inequalitie ˆp p ˆp1 m+1 d ee Lemma 1 we obtain ˆp p ˆp1 d 2 /m. Hence, p = ˆp1 Oτ. Now, the bound τ = oln 1 n, ee 4, implie ˆp = Op, and the bound p = On 1 ln n, ee 6, implie the deired bound p = ˆp Oˆpτ = ˆp Opτ = ˆp on 1. In particular, we have nˆp = ln n+k 1 ln ln n+x n +o1. The latter relation implie T = λ n /n t 1 + o1. 15 Evaluation of PC 0. We have PC 0 = N 1 N 2 N 3 p k 1t 1 1 p 2 n t k 1t, 16 where N 2 = m t d count all poible collection Wv1,..., W vt of the et of key that can be m! aigned to v 1,..., v t, while N 1 = count the collection of non-interecting et. n t! k 1! t n t k 1t! d! t m td! Furthermore, N 3 = count the number of way to aigning neighbourhood d m td each of ize k 1 to the vertice v 1,..., v t, and p 1 = d m i the conditional probability d that, given non interecting et W v1,... W vt, the vertex v V \V and the vertex v i V hare exactly key, while any other member of V ha no common key with v. Finally, p 2 denote 6

7 the conditional probability that given non interecting et W v1,... W vt, the vertex u V \ V i adjacent to ome vertex v j from V. Let u we evaluate 16. A direct calculation how that N 1 N 3 p k 1t 1 = m td n t k 1t d 2 k 1t m td d N 2 m d t k 1! t! m d = e Oτ n k 1t d 2 m td d k 1t k 1! t!m d e Oτ 1 + o1 = nˆpk 1t 1 + o1. 17 k 1! t In the lat tep we ued the fact that d 2 = om implie τ = o1 and e Oτ = 1 + o1. Now we etimate the value of 1 p 2 n t k 1t. Let u V \ V be fixed and W v1 W v2 =. By incluion-excluion, tp u v 1 W v1 t 2 P u v1, u v 2 W v1, W v2 p2 tp u v 1 W v1. Note that Pu v 1, u v 2 W v1, W v2 m 2 d 2 d 2 m d and o Pu v 1, u v 2 W v1, W v2 ˆp 2. Here d 2 count pair B1, B 2 of joint B 1 W v1, B 2 W v2 and m 2 m 1 d 2 d i the probability that Wu cover a given pair B 1, B 2 of dijoint joint. On the other hand Pu v 1 W v1 = p = ˆp1 + Oτ implie p 2 = tˆp1 + Oτ + ˆp = tˆp1 + Oτ. Therefore we have 1 p 2 n t k 1t = exp{n t k 1t ln1 p 2 } = exp{ tnˆp}1 + o1. 18 In the econd identity of 18 we expanded the logarithm in power of p 2 and ued np 2 2 = Onˆp 2 = Oˆp ln n and ˆp ln n τ ln n = o1, ee 4. Finally, we ubtitute 17 and 18 to 16. Then, by 15, we get the firt relation of 14. Upper bound for PC 1. We firt collect ome auxiliary reult. We define random variable Z 1 = W v1 W v2, Z 2 = W v1 W v2 W v3,, Z t 1 = W v1 W vt 1 W vt and the random vector Z = Z 1,..., Z t 1. We note that d W v1 W vi td, for i t. Thu for i < t and for any integer 0 z d we have PZ i = z W v1,..., W vi td m d m 1 td z m d d z z d z d = d z. 19 z! m d Now we evaluate m d d z /m d = m z e Oτ and bound, for z d, td z z! d z tz d z z! d d z eet z 2 z/. 20! In the lat tep we ued the inequality d z e z+1 d z, which follow by Stirling approximation, and the imple inequalitie d z d z and! z z!. From 19 and 20 we obtain d PZ i = z W v1,..., W vi eet z 2 z/!m e Oτ = et z ˆp z/ e + o1 21 7

8 uniformly in 1 i t 1, W v1,... W vi and z d. Given an arbitrary vector z = z 1,..., z t 1 with coordinate from {0, 1,..., d}, let z = I {z1 }z 1,..., I {zt 1 }z t 1. The et of indice of non-zero coordinate of z i denoted J z = {i 1, i 2,..., i r }. Here we aume that i 1 < < i r. For any given z, we have P Z = z r PZ ij = z ij Z ih = z ih, 1 h < j j=1 i J z et z i ˆp zi/ e + o1. 22 In the econd inequality we ued the fact that upper bound 21 obviouly extend to conditional probabilitie PZ ij = z Z ih = z ih, 1 h < j. Denote S z := i J z z i. From 22 we obtain P Z = z exp{s z 1 ln ˆp + ln t + 2}1 + o1. 23 We call a joint B occupied, if it i covered by ome W vi, for 1 i t. We oberve that if the event { Z = z} hold then the number of occupied joint i at leat N z = t d t 1 zj j=1. In particular, we have N z = t d in the cae where J z =. For J z we have N z = t i J z z i d d t d 1 S z d. 24 Now fix u V \ V and, auming that the realized et W v1,..., W vt atify the condition Z = z, conider the conditional probability denoted p z of the event that u i adjacent to ome v j V, given W v1,..., W vt. Oberving that p z i the probability that a random ubet of W of ize d cover an occupied joint we obtain m td m 1 p z N z t d 1 S z ˆpe Oτ. 25 d d In the econd inequality of 25 we applied 24 and the relation d m td m 1 d d = ˆpe Oτ. Now we are ready to how an upper bound for PC 1. By the law of total probability, PC 1 = z PC 1 Z = zp Z = z = I 1 + I 2, 26 where I 1 = z:j z= PC 1 Z = zp Z = z and I 2 = z:j z PC 1 Z = zp Z = z. We firt etimate the conditional probabilitie PC 1 Z = z. Recall that C 1 occur when the et of neighbour of V = {v 1,..., v t } can be divided into dijoint ubet V 1,..., V t uch that V i k 1 and element of V i are neighbour of the i-th vertex of V, 1 i t. Denote l i = V i, 1 i t, and l := l l t. For any z we have, by the union bound, PC 1 Z = z tk 1 l=0 T l z, T l z := l 1 + +l t=l, 0 l 1,...,l t k 1 n t l l 1! l t! ˆpl 1 p z n l t. 27 In the definition of T l z, n t l l 1! l t! i the number of way we may chooe et V 1,..., V t, given their cardinalitie. Moreover, given V 1,..., V t and v j V, and u V 1 V t, the number ˆp i an upper bound for the probability that u cover a joint belonging to v j ee Lemma 1, and 1 p z i an upper bound on the probability that given u V \ V V 1 V t doe not cover any joint of any vertex from V. 8

9 Let u contruct an upper bound for I 1. For thi purpoe we etimate every T l z, with J z =. The latter relation implie S z = 0 and we have p z tˆpe Oτ = tˆp + on 1, ee 25, 4, and 1 p z tk 1 = 1 o1. We obtain T tk 1 z = n t tk 1 k 1! t ˆp tk 1 1 p z n kt 1 + o1t. For l = tk 1 j, where j 1, we have for ome contant C t,k,j depending only on t, k, j T l z C t,k,j T nˆp j. Combining thi inequality and the relation nˆp = 1+o1 ln n we obtain from 27 that PC 1 Z = z T 1 + o1. We note that the latter inequality hold uniformly in z atifying J z =. Hence, we have I 1 T 1 + o1. 28 Now we contruct an upper bound for I 1. Given z with J z = {i 1,..., i r } we etimate T l z. From 25 combined with the relation e Oτ = 1 + Oτ = 1 + oln 1 n we obtain 1 p z n l t 1 p z n kt exp{ n ktp z } exp{ nˆpt d 1 S z + o1}. The latter inequalitie imply T tk 1 z 1 + o1t e nˆpt d 1 S z, 29 T l z C t,k,j T nˆp j e nˆpt d 1 S z, l = tk 1 j, j Combining 27, 29, 30 we obtain the inequality PC 1 Z = z 1 + o1t e nˆpt d 1 S z. Oberving that S z in the right hand ide depend only on z we conclude that I o1t z 0 e nˆpt d 1 S z P Z = z. 31 Here the um run over all z that are not equal to 0 = 0,..., 0. Next, we invoke 23 to obtain e nˆpt d 1 S z P Z = z e S zξ, ξ := d 1 nˆp + 1 ln ˆp + ln t + 2. We remark that 5 implie ξ. Hence i 1 eiξ = o1. Now, given ξ with i 1 eiξ < 1, define independent random variable Y 1,..., Y t 1 with the common ditribution PY j = i = e iξ, i = 1, 2,..., and PY j = 0 = 1 i 1 eiξ. The inequalitie e S zξ P max Y j t 1PY j t 1 imply z 0 z 0 e S zξ t 1 z i e iξ = o1. We conclude that I 2 = ot. Combining thi bound with 28 and 26 we obtain the econd inequality of 14. Acknowledgement. Author thank referee for their remark. M. Blozneli acknowledge upport of Lithuanian Reearch Council grant MIP-067/2013. K. Rybarczyk i partially upported by the National Science Centre grant - DEC- 2011/01/B/ST1/

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