Faster rumor spreading with multiple calls

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Introduction
Randomized rumor spreading is an important primitive for spreading information in networks.The goal is to spread a piece of information, the so-called rumor, from an arbitrary node to all the other nodes.Randomized rumor spreading protocols are based on the simple idea that every node picks a random neighbor and these two nodes are able to exchange information in that round.This paradigm ensures that the protocol is local, scalable, and robust against network failures (cf.[13,15]).Therefore, these protocols have been successfully applied in other contexts such as replicated databases [8], failure detection [30], resource discovery [23], load balancing [3], data aggregation [25], and analysis of the spread of computer viruses [2].
The most basic variant of randomized rumor spreading is the Push protocol.At the beginning, there is a single node who knows of some rumor.Then in each of the following rounds every informed node calls a random neighbor chosen independently and uniformly at random and informs it of the rumor.The Pull protocol is symmetric, here every uninformed node calls a random neighbor chosen independently and uniformly at random, and if that neighbor happens to be informed the node becomes informed.The Push-Pull protocol is simply the combination of both protocols.Most studies in randomized rumor spreading concern the runtime, which is the number of rounds required until the rumor reaches all other nodes, and the communication overhead, which is the total number of information exchanges, produced by these protocols (see e.g.[24]).
In one of the first papers in this area, Frieze and Grimmett [19] proved that if the underlying graph is a complete graph with n nodes, then the runtime of Push is log 2 n + log n ± o(log n) with high probability 1 , where log n denotes the natural logarithm of n.This result was later strengthened by Pittel [29].For the standard Push-Pull protocol, Karp et al. [24] proved a runtime bound of log 3 n + O(log log n).In order to overcome the large number of Θ(n log n) calls, Karp et al. also presented an extension of the Push-Pull protocol together with a termination mechanism that spreads a rumor in O(log n) rounds using only O(n log log n) messages.Doerr and Fouz [9] proposed a new protocol using only Push calls that achieves a runtime of (1 + o(1)) log 2 n using only O(n • f (n)) calls (and messages), where f (n) is an arbitrarily slowly growing function.
Besides the complete graph, randomized rumor spreading protocols have been shown to be efficient also on other topologies.In particular, their runtime is at most logarithmic in n for topologies ranging from basic networks, such as random graphs [15,14,16] and hypercubes [15], random regular graphs [1,17], graphs with constant conductance [27,6,20], constant weak conductance [4] or constant vertex expansion [22,21], to more complex structures including preferential attachment graphs modeling social networks [5].In particular, recent studies establishing a sub-logarithmic runtime on certain social network models [10,11,18] raise the question whether it is possible to achieve a sub-logarithmic runtime also on the complete graph.In addition to analyses on static graphs, there are also studies on mobile geometric graphs, e.g., [7,28].Since all aforementioned protocols require Θ(log n) rounds to spread the rumor on a complete graph, we equip nodes with the possibility of calling more than one node in each round.Specifically, we assume that the power of a node u, denoted by C u , is a random variable, which has the same distribution as a random variable R with support on the positive integers and which is independent of u.In order to keep the overall communication cost small, we focus on distributions R satisfying u∈V C u = O(n) with high probability -in particular, R has bounded mean.Our aim is to understand the impact of the distribution of R on the runtime of randomized rumor spreading.In particular, we seek for conditions on R which are necessary (and/or sufficient) for a sublogarithmic runtime.
Our first result concerns the Push protocol for the case where R has bounded mean and bounded variance, which is the most basic setting.Let T total be the first round in which all nodes are informed.
Theorem 1.1.Consider the Push protocol and assume that R is a distribution with Note that by putting R ≡ 1, we retain the classic result by Frieze and Grimmett.Our next result addresses the case where we drop the assumption on the variance, and it provides a lower bound of Ω(log n) on the number of rounds.Although this result is less precise than Theorem 1.1, it demonstrates that it is necessary to consider the Push-Pull protocol in order to achieve a sub-logarithmic runtime.
. Then with probability 1 − o(1), the Push protocol needs at least Ω(log n) rounds to inform all nodes.
We point out that the lower bound in Theorem 1.2 is tight up to constant factors, as the results in [19,29] for the standard Push protocol imply an upper bound of O(log n) rounds.We now consider the Push-Pull protocol and extend the lower bound of Ω(log n) from Theorem 1.1.Then for any constant > 0, with probability 1 − the Push-Pull protocol needs at least Ω(log n) rounds to inform all nodes.
Theorem 1.3 establishes that an unbounded variance is necessary to break the Ω(log n) lower bound.An important distribution with bounded mean but unbounded variance is the power law distribution with exponent β 3, i.e., there are constants 0 We are especially interested in power law distributions, because they are scale invariant and have been observed in a variety of settings.Our main result below shows that this natural distribution achieves a sublogarithmic runtime.
Theorem 1.4.Assume that R is a power law distribution with 2 < β < 3. Then the Push-Pull protocol informs all nodes in Θ(log log n) rounds with probability 1 − o(1).
the electronic journal of combinatorics 22(1) (2015), #P1.23 Notice that if R is a power law distribution with β > 3, then Theorem 1.3 applies because the variance of R is bounded.Hence our results reveal a dichotomy in terms of the exponent β: if 2 < β < 3, then the Push-Pull protocol finishes in O(log log n) rounds, whereas for β > 3 the Push-Pull protocol finishes in Θ(log n) rounds2 .While a very similar dichotomy was shown in [18] for random graphs with a power law degree distribution, our result here concerns the spread of the rumor from one to all nodes.
In the case β = 3 we show that the runtime is close to the one in the β > 3 case.
Theorem 1.5.Assume that R is a power law distribution with β = 3.Then the Push-Pull protocol informs all nodes in Θ log n log log n rounds with probability 1 − o(1).
Finally, we also argue that it is necessary that the C u 's are chosen once and for all at the beginning, and they are not updated in each round.Indeed, suppose we generate in the tth round a new variable C t u , which is the number of calls made by u in that round.Then we prove the following lower bound.
Theorem 1.6.Assume that R is any distribution with E [R] = O(1).Then with probability 1 − o(1), the Push-Pull protocol needs Ω(log n) rounds to inform all nodes.

Notations and Preliminaries
We introduce some notation that will be used throughout the paper without further reference.In our setting, the Push, Pull and Push-Pull protocols proceed like the classic ones except that in each round, every (un)informed node u calls C u node(s) chosen independently and uniformly at random and sends (requests) the rumor.For any of these protocols, we let I t be the set of informed nodes at the end of round t and U t the set of uninformed nodes.We write V = I t ∪ U t for the vertex set of the graph, and we assume |V| = n.The size of I t and U t is denoted by I t and U t .We indicate the set of newly informed nodes in round t + 1 by N t and its size is N t .Let S t be the number of Push calls in round t + 1, so S t = u∈It C u I t .Let us define N Pull .The size of every set divided by n will be denoted by the corresponding small letter, so i t , n t and s t are used to denote I t /n, N t /n, and S t /n, respectively.Further, let Moreover, let ∆ = max u∈V C u .
We will use extensively the following two concentration inequalities.The first one is a Chernoff-type bound.
In particular, The next inequality is known as the Bounded Difference inequality.

Theorem 2.2 ([26]
). Suppose that X 1 , X 2 . . ., X n are independent random variables and every X i , 1 i n, takes a value from a finite set A i .Let f : 1 i n A i → R be a real-valued function so that there exist c 1 , c 2 , . . ., c n with sup Then, for every λ > 0, 3 Some Useful Facts for Power Law Distributions Let R be a power law probability distribution with exponent β, i.e., there are constants 0 < c 1 < c 2 so that for every integer z 1, and Pr [C u 1] = 1.In this section we collect some basic properties of R. Proof.Since β > 3 Let C u , u ∈ V be independent, power-law distributed random variables with exponent β.Then, with probability the electronic journal of combinatorics 22(1) (2015), #P1.23 Proof.By definition, Applying the union bound yields the claim.
Proposition 3.3.Let β > 2. Let C u , u ∈ V be independent, power-law distributed random variables with exponent β.Then, for every z = O(n Proof.For u ∈ V let I u be the indicator random variable for the event C u z.Since the C u 's are independent and identically distributed, so are the I u 's.By linearity of expectation Applying Theorem 2.1 to the random variable X := u∈V I u yields that

Push Protocol
In this section we will show two general lemmas for the Push protocol that are valid for any R with support on the positive integers.They will be used when analyzing the Push and the Push-Pull protocols.
Lemma 4.1.Consider the Push protocol and suppose that S t log c n, where c > 0 is an arbitrary constant.Then with probability at least 1 − O(n −1 log 2c n) we have Proof.Recall that S t is the number of Push calls in round t + 1.By applying the union bound, the probability that an informed node receives a call in round t + 1 is bounded by StIt n .So, with probability at least 1 − StIt n , none of the calls are sent to a node in I t .Conditioning on this event, consider all calls one by one in an arbitrary order; then the probability that the i−th call informs a different node from the previous i − 1 calls is Ut .Therefore the conditional probability that S t calls inform S t different nodes is at least So the probability that S t calls inform S t different uninformed nodes is at least where the above equality holds because I t S t log c n and U t = n(1 − o(1)).The claim follows.
Lemma 4.2.Consider the Push protocol.Then with probability at least Proof.Since N t is always bounded by S t , n t s t .To see the lower bound, let for v ∈ U t Z v be the indicator random variable for the event v ∈ I t+1 .Then N t = v∈Ut Z v .Since the Z v 's are identically distributed random variables, to be the function counting the number of newly informed nodes in round t + 1.Then For each change in just one coordinate of f , the following statement holds: sup Therefore by applying Theorem 2.2, we obtain So with probability 1 − o(1/log n) we have Now we estimate . By the definition of Push Using that 1 − x e −x 1 − x + x 2 for any x 0 We now plug the value obtained by the above formula into (1) and normalize it.So we obtain where the last inequality comes from the fact that i t s t . Corollary Proof.By assumption we have for every 1 i k that S t+i = o(n).Applying Corollary 4.3 shows that with probability 1 − o( 1 log n ) Using an inductive argument and the union bound for k implies the statement.

Push Protocol with Bounded Mean and Bounded Variance
This section is devoted to the proof of Theorem 1.1.Recall that T total := min{t | I t = n}, i.e., the first round in which all nodes are informed.We claim that if and To prove this result, we study the protocol in three consecutive phases.In the following we give a brief overview of the proof.
• The Preliminary Phase.This phase starts with just one informed node and ends when I t log 5 n.Similar to the proof of the birthday paradox we show that in each round every Push call informs a different uninformed node and thus the number of informed nodes increases by S t I t .Hence after O(log log n) rounds there are at least log 5 n informed nodes.Further, since E [R] = O(1), after O(log log n) rounds we also have S t log O (1) n for all these rounds.
• The Middle Phase.This phase starts when log 5 n I t S t log O(1) n and ends when I t n/log log n.First we show that the number of Push calls S t increases by a factor of approximately 1 + E [R] as long as the number of informed nodes is o(n).Then we prove that the number of newly informed nodes in round t + 1 is roughly the same as S t .Therefore an inductive argument shows that it takes log 1+E[R] n ± o(log n) rounds to reach n/log log n informed nodes.
• The Final Phase.This phase starts when I t n log log n and ends when all nodes are informed with high probability.In this phase, we first prove that after o(log n) rounds the number of uninformed nodes decreases to n/log 5 n.Then we show the probability that an arbitrary uninformed node remains uninformed is e − E[R]±o(1/log n) .Finally, an inductive argument establishes that it takes log e E[R] n ± o(log n) rounds until every node is informed.
In the following we present the detailed proofs for these phases.Before that we show the following proposition.
Proposition 5.1.Let > 0 and let R be a random variable with support on the positive integers such that E [R] = O(1) and Var [R] = O(1).Let δ 0 be such that U t = n 1−δ , for some round t.Then, with probability Proof.For k ∈ N let us define a random variable

By linearity of expectation
Since C u has bounded variance the last sum is in O(1).Thus, the electronic journal of combinatorics 22(1) (2015), #P1.23

The Preliminary Phase
This phase starts with one informed node and ends when I t log 5 n and S t log O(1) n.Let T 0 be the first round in which the number of informed nodes exceeds log 5 n.
Lemma 5.2.For any t = O(log log n), with probability at least 1 − log −3 n, we have Proof.We will bound the expected number of calls in each round t as follows: where the last inequality comes from the fact that N t−1 S t−1 .Since the origin of the rumor is chosen before determining the C u 's we have Applying the law of total expectation yields By using Markov's inequality we have that and the claim follows for any t = O(log log n).
where the inequality comes from the fact that S t I t .So, with probability at least there exists a round T 0 = O(log log n) such that I T 0 log 5 n and S T 0 log O(1) n.

The Middle Phase
The phase starts when log 5 n I t S t log O(1) n and ends when I t n/log log n.Let T 1 be the first round so that I T 1 n/log log n.The main result of this subsection is that the electronic journal of combinatorics 22(1) (2015), #P1.23 Lemma 5.4.Suppose for a round t we have s t = Ω(n −1 • log 5 n) and s t = o(1).Then for Proof.Consider the random variable u∈Nt C u .Since N t is fixed and the random variables C u , u ∈ N t are independent we obtain that Chebychev's inequality implies that Using the above formula and the fact that N t S t we have Since S t is a non-decreasing function in t and log 5 n I t S t , with probability 1 − o(1/log n) An inductive argument and the union bound for all k events that violate the above inequality shows that for any In order to prove the left hand side of (2), we use Lemma 4.2 which states with probability Using the lower bound in (3) and the above formula implies that with probability 1 − o(1/log n), An inductive argument and the union bound for all k events that violate the above inequality show that for any integer k for which ( Inequality (4) yields that with probability 1 − o(k/log n), where a := 1+o(1).F (s t ) is a non-decreasing function in s t and hence for any k = O(log n) and 1 j k, Hence by combining the above inequality and (5), we conclude that for any integer k, where (1 the electronic journal of combinatorics 22(1) (2015), #P1.23 where d 1 and d 2 are constants which do not depend on i.Since (1 + E [R]) k s t = o(1) and s t = Ω( log 5 n n ), for any 1 i k, Lemma 5.5.Suppose that log 5 n n i t s t log O (1) n n . Then for any k = O(log n) with where f 1 > 0 and f 2 > 0 are constants.
Proof.It is easy to see that Applying Lemma 5.4 implies that for any integer k for which ( ) the following upper bound holds: where f 1 > 0 is a constant.On the other hand, Lemma 4.2 yields that with probability 1 − o( 1 log n ), Another application of Lemma 5.4 shows that with probability 1 Using these two inequalities, as long as (1 + E [R]) k s t = o(1), we have with probability the electronic journal of combinatorics 22(1) (2015), #P1.23 where f 2 > 0 and d > 0 are constants.Since log 5 n n i t s t , we obtain that By combining equations ( 7) and ( 6) we infer that with probability 1 − o( k log n ),

The Final Phase
This phase starts with at least n log log n informed nodes and ends when all nodes get informed.Let T 1 be the first round in which I T 1 n log log n and let T 2 be the first round in which all nodes are informed with probability 1−o(1).We will show that with probability Proof.We define the indicator random variable Z v for every v ∈ U t and any round t T 1 : Thus, where for simplicity we omit the conditioning of U t+1 on U t when dealing with the Z v 's.
Using the fact that 1 , we can approximate the value Pr [Z v = 1] as follows, Since for every u, v ∈ U t , we have that Therefore, Applying Chebychev's inequality implies that with probability 1 − o( 1 log n ), the electronic journal of combinatorics 22(1) (2015), #P1.23 Combining inequalities ( 8) and ( 9) yields that with probability 1 − o( 1 log n ), According to the value of U t , we consider two cases.
• Suppose that U t n log 5 n .Note that s t i t 1 log log n by the assumption of the lemma.Since s t is a non-decreasing value in t and U t < n the recursive formula (10) implies that with probability 1 − o( 1 log n ), Using an inductive argument shows that with probability 1 − o( k log n ), Hence after at most k 0 = 6 log log 2 n rounds with probability 1 − o(1) the number of uninformed nodes decreases to n log 6 n + O( n log 2 n), where c > 0 is a constant.
• Suppose that U t n log 5 n .If we set n δ = log 5 n, then applying Proposition 5.1 implies that for any t for which On the other hand, using Chebychev's inequality yields that with probability 1 Combining the above equality and equality (11) results into an approximation for s t which is not best possible but it suffices for our purpose.We know that So, Therefore, s t can be replaced by Inequality (10) where α = e − E[R]±o(1/ log n) .So as long as U t log 5 n with probability 1 − o( 1 log n ), In order to lower bound U t+k we apply the lower bound (12) inductively.So we have that with probability 1 − o( k log n ), Applying inequality (13) yields that with probability 1 − o( k log n ), Thus, where c > 0 is a constant and the last inequality holds because k−1 i=0 α ). Combining the inequalities ( 13) and ( 14) yields for any k satisfying Hence by taking k = log e E[R] n − o(log n), with probability 1 − o(1), the number of uninformed nodes after T 1 + k 0 + k rounds decreases to log 5 n, so we have at most log 5 n uninformed nodes.Using the fact that for every x 0, 1 − x e −x , the probability that a node does not get informed after k 1 additional rounds is bounded from above by We already know that Thus the union bound implies that the probability that every node in U t does not get informed is bounded by log . By choosing k 1 = Θ(log log n) we conclude that with probability 1 − o(1) all nodes get informed.So we have with probability

Push Protocol with Bounded Mean
This section is devoted to the proof of Theorem 1.2.
Proof.In the Push protocol, in round t + 1, at most S t randomly chosen uninformed nodes are informed.This implies that E [S t+1 | S t ] increases by at most E [R] • S t .Since the origin of the rumor is chosen without knowing C u , E [S 0 ] = E [R].Using the law of total expectation yields that By applying Markov's inequality, we conclude that Hence Ω(log n) rounds are necessary to inform all nodes with probability 1 − o(1).

Lower Bound for Push-Pull
Before we present our results about the Push-Pull protocol we show the following general lemma.Recall that S 0 = C u , where u is the single node that is aware of the rumor at the beginning of the protocol.Proof.We know the probability that an uninformed node u gets informed by Pull in round t + 1 is bounded by I t • C u /n.Therefore using this bound we have u∈Ut On the other hand the probability that a node u ∈ U t gets informed by Push in round t + 1 is at most S t /n.So we get that u∈Ut where the last inequality follows by C u C 2 u .Combining the above inequalities implies that Applying the law of total expectation yields that Using Markov's inequality implies that and the claim follows.u ] = O(n).Another application of Markov's inequality implies that with probability 1 − /2, u∈V C 2 u = O(n).Therefore using a union bound for failure probability of two mentioned events implies that for fixed > 0 with probability at least 1 − , {C u : u ∈ V} is a good sequence.Conditioning on the event that {C u : u ∈ V} is a good sequence, using Lemma 7.1 implies that with probability at least 1 − o(1) the Push-Pull protocol needs Ω(log n) rounds to inform n nodes and the result follows.
9 Push-Pull Protocol with Power Law Distribution 2 < β < 3 In this section we analyze the Push-Pull protocol where R is a power law distribution with 2 < β < 3 and show that it only takes Θ(log log n) rounds to inform all with probability 1 − o(1).To prove the upper bound of O(log log n), we study the protocol in three consecutive phases and show that each phase takes only O(log log n) rounds.After that we show the lower bound Ω(log log n).

Proof of the Upper Bound
The Preliminary Phase.This phase starts with just one informed node and ends when I t n 1 β−1 /(2 • log n).Let T 1 be the first round where I T 1 n 1 β−1 /(2 log n).We will show that T 1 = O(log log n).First we claim that O(log log n) rounds are sufficient to have log O (1) n informed nodes.Then we will show that in round t + 1 with probability 1 − e −Ω(log n) there exists a node u with C u I 1+γ t , γ := 3−β 2(β−2) > 0, which pulls the rumor and consequently S t+1 I 1+γ t .Then considering only Push calls it follows that with probability 1 − o( 1 log n ), So in every two rounds, I t is increased by a factor of 1 2 I γ t and hence after O(log log n) rounds the phase ends.For a complete proof see the following lemma.), Thus as long as S t log 2 3−β n, in each round the number of informed nodes is at least doubled.So we conclude that with probability 1 − o(1), O(log log n) rounds are sufficient to inform log . Now for any T 0 t T , we can apply Proposition 3.3 and conclude that with probability So, We will bound the probability that none of u ∈ L(I 1+γ t ) gets informed by Pull calls in round t + 1 as follows, Since for any t T 0 , I t log 2 3−β n, we have that with probability at least 1 − n −c 1 , at least one node in L(I 1+γ t ) gets informed by Pull in round t + 1. Hence we have that Let us now consider the Push calls in round t + 2. By applying Lemma 4.1 we know that as long as Thus, the electronic journal of combinatorics 22(1) (2015), #P1.23 An inductive argument shows that for any integer k 1 as long as I 1+γ , where C = 2 γ = O(1).So we conclude that after T 0 +2k rounds, where k = o(log 1+γ log n), there are two cases: either I T 0 +2k n 1 β−1 /(2 log n) which means T 1 T 0 +2k = O(log log n) and we are done, or In the latter case, we change the value γ to γ which satisfies I 1+γ T 0 +2k = n 1 β−1 / log n and a similar argument shows that

The Middle Phase.
This phase starts with at least n 1 β−1 /(2 log n) informed nodes and ends when I t n log n .Let T 2 be the first round in which n log n nodes are informed.We will show that T 2 − T 1 = O(log log n).In contrast to the Preliminary Phase where we focus only on an informed node with maximal C u , we now consider the number of informed nodes u with a C u above a certain threshold Z t+1 which is inversely proportional to I t .Lemma 9.2.Suppose that I t n 1 β−1 /(2 log n) for some round t.Then with probability where Z t+1 := n log log n It .
Proof.We consider two cases.If at least 1  4 of the nodes in L(Z t+1 ) are already informed (before round t+1), then the statement of the lemma is true.Otherwise |L(Z t+1 )∩U t+1 | > 3 4 L(Z t+1 ).In the latter case, we define Let X u be an indicator random variable for every u ∈ L (Z t+1 ) so that X u := 1 if u gets informed by Pull in round t + 1, 0 otherwise.
Then we define a random variable X to be X := We know that I t n 1 β−1 /(2 log n) and also I t is a non-decreasing function in t, so where the last inequality holds because β < 3. Now we can apply Proposition 3.3 (see appendix) to infer that with probability 1 Therefore, Then applying Theorem 2.1 results into So with probability 1 − o( 1 n ), we have that where the last inequality holds because the electronic journal of combinatorics 22(1) (2015), #P1.23 By applying Proposition 3.3, we conclude that with probability 1 . Therefore, with probability 1 − o( 1 n ), As long as S t+1 = o(n), we can apply Lemma 4.2 for the Push protocol to round t + 2 implying that with probability 1 − o( 1 log n ), Thus, .
By an inductive argument, we obtain that for any integer k 1 with S t+k = o(n), it holds with probability 1 − o( k log n ), Therefore there exists k = O(log

The Final Phase.
This phase starts with at least n log n informed nodes.Since the runtime of our Push-Pull protocol is stochastically smaller than the runtime of the standard Push-Pull protocol (i.e.C u = 1 for every u ∈ V ), we simply use the result by Karp et.al in [24,Theorem 2.1] for the standard Push-Pull protocol which states that once I t n log n , additional O(log log n) rounds are with probability 1 − o(1) sufficient to inform all n nodes.

Proof of the Lower Bound
Since increasing the number of informed nodes can only decrease the runtime of the protocol, we may assume that at the beginning there are log b n random informed nodes, where b := max{4, 2 + 3 (3−β)  β−2 }.Applying Markov's inequality to the random variable S 0 implies that with probability 1 − o( 1 log n ), log b n S 0 log 2+b n.In the following we lower bound the number of rounds to reach n 1 log log n informed nodes.We do this by keeping track of the largest value of C u among all informed nodes and show that this value does not exceed I We denote the size of M i with M i .By definition, for any 1 i i * , M i L(2 i−1 ).Applying Proposition 3.3 implies that with probability 1 − o( 1 n ) for any (1−β) .Let us define the indicator random variable Z i u for every u ∈ U t ∩ M i as follows: n .Let P i be the probability that at least one node in U t ∩ M i gets informed by Pull in round t + 1.Then, for any 1 i i * − 1, So as long as I t n 1 log log n , P i * = o( 1 log 3 n ).We define ∆ t := S 1 β−2 t log 3 β−2 n.Let 1 i t i * be the smallest integer so that 2 it ∆ t .Then for any i t i i * we have, Let E t be the event that no node with C u ∆ t gets informed by Pull in round t + 1.Then we have Let us define S (1) We know that E [R] = O(1), so Markov'e inequality implies that with probability 1 − O( 1 log n ), S 0 = O(log n).Conditioning on the event that for every u ∈ V, C u < n we get So applying Markov's inequality yields that with probability 1 − O( Proof.By considering the Push call we have that the size of newly informed nodes is bounded by S t .Since they are chosen randomly, we have that On the other hand we have that Pr [u gets informed by Pull in round t + 1] where the last inequality holds because we assume that It the electronic journal of combinatorics 22(1) (2015), #P1.23 Thus, Combining ( 21) and (23) implies that We know that I t+1 (z) = I 0 (z) + t i=1 N i (z).Using the linearity of expectation we have that where the last inequality follows from inequality (22).Therefore we infer that with probability 1 − o( 1 log n ), |U t (z) ∩ L * (z)| L * (z) 2 .
Lemma 10.4.Suppose that I t = e Ω( log n log log n ) and S t n log 6 n .Then with probability 1−o(1), the Push-Pull protocol needs O log n log log n rounds to inform at least e log n− log n log log n nodes.
Proof.Let X u be an indicator random variable for every u ∈ U t (z) ∩ L * (z) so that X u := 1 if u gets informed by Pull in round t + 1, 0 otherwise.
Then we define the random variable X t (z) := u∈Ut(z)∩L * (z) X u .Let us define z t = min{I where the last inequality holds because Pr [R = z] c z 3 .Since I t = e Ω( log n log log n ) and X u 's are independent, applying a Chernoff bound 2.1 implies that with probability 1 − o( 1 n ), Using the above inequality and inequality (24) shows that that with probability 1−o( An inductive argument shows that for any integer k as long as S t+2k = n log 6 n with probability 1 − o(1), Thus there is a k = O log n log log n so that after t + 2k rounds there are at least e log n− log n log log n informed nodes.
Recall that S t is the number of Push calls by informed nodes in round t + 1.Therefore, N Push t S t and Hence, By using the law of total expectation, we conclude that E [I t ] < (1 + 4 • E [R]) t .If we set T = c • log n, where c > 0 is a small constant, then So with probability 1 − o(1), we need at least c • log n rounds to inform all nodes.

t
and N Push t to be the set of newly informed nodes by Pull and Push calls in round t + 1, respectively.The size of N Pull t and N Push t are denoted by N Pull t and N Push t

Corollary 5 . 6 .
With probability 1 − o(1) we have |T 1 − log 1+E[R] n| = o(log n).Proof.Applying Corollary 5.3 shows that with probability 1 − o(1), T 0 = O(log log n), where T 0 is the first round in which log 5 n n i T 0 s T 0 log O(1) n n .Now we can apply Lemma 5.5 and set k = log 1+E[R] n − o(log n) such that with probability at least 1 − o(1) we have 1 log log n i T 0 +k A log log n , where A > 1 is a constant.Then we conclude that with probability 1 − o(1), |T 1 − log 1+E[R] n| = o(log n).

Lemma 7 . 1 .
Consider the Push-Pull protocol and {C u : u ∈ V} be a sequence of positive integers.Then with probability 1 − o(1), the Push-Pull protocol needs at least Ω log n − log S 0 log u∈V C 2 u /n rounds to inform all nodes.

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the electronic journal of combinatorics 22(1) (2015), #P1.Push-Pull Protocol with Bounded Mean and Bounded Variance This section is devoted to the proof of Theorem 1.3.Proof.{C u : u ∈ V} be a sequence of positive integers each of which is generated independently according to some distribution R with E [R] = O(1) and Var [R] = O(1).We call {C u : u ∈ V} a good sequence if u∈V C 2 u = O(n) and S 0 = O(1).Since the origin of the rumor is chosen without knowing C u , E [S 0 ] = E [R].Applying Markov's inequality implies that for any constant > 0 with probability at least 1 − /2, S 0 = O(1).Since R is a probability distribution with bounded variance, u∈V E [C 2

3 β− 2 n 1 β− 1 1 β− 1 /
with high probability.By Fact 3.2, with probability 1 − o(1log n ) we have max u∈V C u n log n.Let i * be the smallest positive integer so that 2 i * n log n.Then i * < log n.Let us define the set M e −a .Applying Fact 10.1 shows that for any z = O(n

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[I t+1 (z)|S i , I i , 0 i t] = I 0 (z) + t i=0 E [N i (z)|S i , I i ] I 0 (z) + Pr [R = z] • t i=0 (S i + 3 • I i • z) 1 + Pr [R = z] • (t + 1) • (S t + z • 3 • I t ),where the last inequality comes from the fact that S i and I i are non-decreasing function in t.By assumption z min{ n It•log 6 n , O(n )} and S t n log 6 n , for any round t = O(log n) we have thatE [I t+1 (z)|S i , I i , 1 i t] 2 • (t + 1) • (S t + 3 • I t • z) • Pr [R = z] n • Pr [R = z] log 4 n .Applying Markov's inequality shows that with probability 1− o( 1 log n ) for any round t = O(log n), I t+1 (z) log 2 n • E [I t+1 (z)|S i , I i , 0 i t] n • Pr [R = z] log 2 n L * (z) 2 ,

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I t = u∈Ut Pr [u gets informed by Pull in round t + 1] 4.3.Consider the Push protocol.Then with probability at least 1 − o(1/log n) for any round t in which S t n/8 we have that I t+1 I t + S t /2.Proof.If 1 S t log n, then Lemma 4.1 yields that with probability 1 − o(1/log n) we have N t = S t .If log n S t n/8, then 2s 2 t s t /4 and 2 s t n −1 log log n s t /4.Thus, Lemma 4.2 guarantees that with probability at least 1 − o(1/log n) n t s t − 2s 2 t − 2 s t log log n n s t 2 .Corollary 4.4.Consider the Push protocol.For any round t and positive integer k Therefore we have that with probability 1 − o(1), the sequence {C u : u ∈ V} is good.Before we present a proof for the upper bound we show following two lemmas.
1 4 )}.Then with probability 1 − o( 1 log n ), for any round t = O(log n) we have that For any positive integer k in which I t+k ∈ [e Ω( log n log log n ) , e log n− log n log log n ], we have that e Ω( log n log log n ) z t .Hence from the above inequality we conclude that here exists a constant C 1 so thatS t+1 C 1 • I t • log n log log n C 1 • I t • log n.Considering only Push transmission for S t = o(n) and applying Lemma 4.2 implies that with probability 1 − o( 1 log n ) t • C • log z t .