Adaptive Rumor Spreading

Transcription

1 José Correa 1 Marcos Kiwi 1 Neil Olver 2 Alberto Vera VU Amsterdam and CWI July 27, /21

2 The situation 2/21

3 The situation 2/21

4 The situation 2/21

5 The situation 2/21

6 Introduction Rumors in social networks: contents, updates, new technology, etc. 3/21

7 Introduction Rumors in social networks: contents, updates, new technology, etc. In viral marketing campaigns, the selection of vertices is crucial. Domingos and Richardson (2001) 3/21

8 Introduction Rumors in social networks: contents, updates, new technology, etc. In viral marketing campaigns, the selection of vertices is crucial. Domingos and Richardson (2001) An agent (service provider) wants to efficiently speed up the communication process. 3/21

9 Rumor spreading Models differ in time and communication protocol. Demers et al. (1987) and Boyd et al. (2006) 4/21

10 Rumor spreading Models differ in time and communication protocol. Demers et al. (1987) and Boyd et al. (2006) In simple cases, the time to activate all the network is mostly understood. 4/21

11 Rumor spreading Models differ in time and communication protocol. Demers et al. (1987) and Boyd et al. (2006) In simple cases, the time to activate all the network is mostly understood. Even in random networks the estimates are logarithmic in the number of nodes. Doerr et al. (2012) and Chierichetti et al. (2011) 4/21

12 Opportunistic networks We have an overload problem, an option is to exploit opportunistic communications. 5/21

13 Opportunistic networks We have an overload problem, an option is to exploit opportunistic communications. A fixed deadline scenario has been studied heuristically along with real large-scale data. Whitbeck et al. (2011) 5/21

14 Opportunistic networks We have an overload problem, an option is to exploit opportunistic communications. A fixed deadline scenario has been studied heuristically along with real large-scale data. Whitbeck et al. (2011) Control theory based algorithms greatly outperform static ones. Sciancalepore et al. (2014) 5/21

15 The model Bob communicates and shares information. 6/21

16 The model Bob communicates and shares information. Bob meets Alice according to a Poisson process of rate λ/n. λ/n 6/21

17 The model Bob communicates and shares information. Bob meets Alice according to a Poisson process of rate λ/n. Every pair of nodes can meet and gossip. λ/n 6/21

18 The problem There is a unit cost for pushing the rumor. Opportunistic communications have no cost. At time τ all of the graph must be active. 7/21

19 The problem There is a unit cost for pushing the rumor. Opportunistic communications have no cost. At time τ all of the graph must be active. We want a strategy that minimizes the overall number of pushes. 7/21

20 Adaptive and non-adaptive A non-adaptive strategy pushes only at times t = 0 and t = τ. 8/21

21 Adaptive and non-adaptive A non-adaptive strategy pushes only at times t = 0 and t = τ. An adaptive strategy may push at any time, with the full knowledge of the process evolution. 8/21

22 Introduction Model Proofs Other results and open questions Adaptive and non-adaptive A non-adaptive strategy pushes only at times t = 0 and t = τ. An adaptive strategy may push at any time, with the full knowledge of the process evolution. Number of active nodes Number of active nodes Push τ t 0 t3 τ t 8/21

23 Main result Define the adaptivity gap as the ratio between the expected costs of non-adaptive and adaptive. Theorem In the complete graph the adaptivity gap is constant. 9/21

24 Adaptive can be arbitrarily better r v1 v2 v3 vk 10/21

25 Adaptive can be arbitrarily better With a small deadline, non-adaptive activates all of the v i s. Adaptive activates only the root, then at some time t pushes to the inactive v i s. r v1 v2 v3 vk 10/21

26 Adaptive can be arbitrarily better With a small deadline, non-adaptive activates all of the v i s. Adaptive activates only the root, then at some time t pushes to the inactive v i s. An adaptivity gap of log k log log k is easy to prove. r v1 v2 v3 vk 10/21

27 Non-adaptive λ k k N n k N λ = 1. λ k := k(n k) n. 1 n/2 n k 11/21

28 Non-adaptive Optimal non-adaptive pays almost the same at t = 0 and at t = τ. λ k k N n k N λ = 1. λ k := k(n k) n. 1 n/2 n k 11/21

29 Non-adaptive Optimal non-adaptive pays almost the same at t = 0 and at t = τ. - A 2-approximation is easy to see. λ k k N n k N λ = 1. λ k := k(n k) n. 1 n/2 n k 11/21

30 Non-adaptive Optimal non-adaptive pays almost the same at t = 0 and at t = τ. - A 2-approximation is easy to see. Non-adaptive does not push more than n/2 rumors. Therefore, neither adaptive. λ k k N n k N λ = 1. λ k := k(n k) n. 1 n/2 n k 11/21

31 Big deadline: τ (2 + δ) log n Starting from a single active node, the time until everyone is active is 2 log n + O(1). The time is exponentially concentrated. Jason (1999) Just starting with one node has cost 1 + ε, therefore adaptivity does not help. 12/21

32 Small deadline: τ 2 log log n A Poisson process of unit rate gives the randomness. t 13/21

33 Small deadline: τ 2 log log n A Poisson process of unit rate gives the randomness. Given the points S i and S i+1, the rescaling S i+1 S i λ i inter-arrival time. is the k λ k λ k+1 λ i t 13/21

34 Small deadline: τ 2 log log n A Poisson process of unit rate gives the randomness. Given the points S i and S i+1, the rescaling S i+1 S i λ i inter-arrival time. A push can be seen as adding a point. is the k λ k λ k+1 λ i λ i+1 t 13/21

35 Small deadline (cont.): τ 2 log log n A clairvoyant strategy knows the realization, therefore outperforms adaptive. k λ k λ k+1 λ i λ i+1 t 14/21

36 Small deadline (cont.): τ 2 log log n A clairvoyant strategy knows the realization, therefore outperforms adaptive. We show that clairvoyant adds points only at the beginning. k λ k λ k+1 λ i λ i+1 t 14/21

37 Small deadline (cont.): τ 2 log log n A clairvoyant strategy knows the realization, therefore outperforms adaptive. We show that clairvoyant adds points only at the beginning. Clairvoyant chooses the best number of initial pushes, given the realization. k λ k λ k+1 λ i λ i+1 t 14/21

38 Small deadline (cont.): τ 2 log log n Say we start with k initial pushes. 15/21

39 Small deadline (cont.): τ 2 log log n Say we start with k initial pushes. We know the inter-arrival distributions. We know the non-adaptive cost; it pays Ω( n log n ). 15/21

40 Small deadline (cont.): τ 2 log log n Say we start with k initial pushes. We know the inter-arrival distributions. We know the non-adaptive cost; it pays Ω( n log n ). Lemma Clairvoyant is considerably better than non-adaptive with probability at most 1 n 2. 15/21

41 Small deadline (cont.): τ 2 log log n Say we start with k initial pushes. We know the inter-arrival distributions. We know the non-adaptive cost; it pays Ω( n log n ). Lemma Clairvoyant is considerably better than non-adaptive with probability at most 1 n 2. In this case we can prove the gap to be 1 + o(1). 15/21

42 Other deadlines Insight: adaptive interferes when the cost of pushing is less than or equal to that of not pushing, i.e., 1 + cost(k + 1 active nodes) cost(k active nodes). ( ) 16/21

43 Other deadlines Insight: adaptive interferes when the cost of pushing is less than or equal to that of not pushing, i.e., 1 + cost(k + 1 active nodes) cost(k active nodes). ( ) The expected cost remains the same, it is a martingale, thus the condition should be met only a few times. 16/21

44 Other deadlines Insight: adaptive interferes when the cost of pushing is less than or equal to that of not pushing, i.e., 1 + cost(k + 1 active nodes) cost(k active nodes). ( ) The expected cost remains the same, it is a martingale, thus the condition should be met only a few times. A relaxed strategy pushes for free, but with certain conditions. 16/21

45 Other deadlines Insight: adaptive interferes when the cost of pushing is less than or equal to that of not pushing, i.e., 1 + cost(k + 1 active nodes) cost(k active nodes). ( ) The expected cost remains the same, it is a martingale, thus the condition should be met only a few times. A relaxed strategy pushes for free, but with certain conditions. - Pushes only when ( ) holds. - Does not push after n/2. 16/21

46 Other deadlines Insight: adaptive interferes when the cost of pushing is less than or equal to that of not pushing, i.e., 1 + cost(k + 1 active nodes) cost(k active nodes). ( ) The expected cost remains the same, it is a martingale, thus the condition should be met only a few times. A relaxed strategy pushes for free, but with certain conditions. - Pushes only when ( ) holds. - Does not push after n/2. Relaxed outperforms adaptive. 16/21

47 Other deadlines (cont.) Relaxed adaptive can be described by thresholds φ k. 17/21

48 Introduction Model Proofs Other results and open questions Other deadlines (cont.) Relaxed adaptive can be described by thresholds φ k. Let K(t) be the number of active nodes at time t. cost( K(t)) L(t ) L( tk+1) φk+2 φk+1 φk cost( K(0)) t tk+1 t H(0) H(L(t)) 17/21

49 Introduction Model Proofs Other results and open questions Other deadlines (cont.) Relaxed adaptive can be described by thresholds φ k. Let K(t) be the number of active nodes at time t. We transform the process: H (L(t)) := λ K(t) log cost( K(t)) φ K(t). cost( K(t)) L(t ) L( tk+1) φk+2 φk+1 φk cost( K(0)) t tk+1 t H(0) H(L(t)) 17/21

50 Introduction Model Proofs Other results and open questions Other deadlines (cont.) Relaxed adaptive can be described by thresholds φ k. Let K(t) be the number of active nodes at time t. We transform the process: H (L(t)) := λ K(t) log cost( K(t)) φ K(t). cost( K(t)) L(t ) ) L( tk+1) L(t) φk+2 φk+1 φk cost( K(0)) t t tk+1 H(0) tt H(L(t)) 17/21

51 Introduction Model Proofs Other results and open questions Other deadlines (cont.) We show that each time H touches zero, relaxed wins exactly 1 compared to non-adaptive. L(t ) L( t k+1) L(t) H(0) t k+1 t H(L(t)) 18/21

52 Introduction Model Proofs Other results and open questions Other deadlines (cont.) We show that each time H touches zero, relaxed wins exactly 1 compared to non-adaptive. Essentially H (s) is dominated by s 2 Poiss(s). L(t ) L( t k+1) L(t) H(0) t k+1 t H(L(t)) 18/21

53 Introduction Model Proofs Other results and open questions Other deadlines (cont.) We show that each time H touches zero, relaxed wins exactly 1 compared to non-adaptive. Essentially H (s) is dominated by s 2 Poiss(s). The number of times H (s) touches zero is constant. L(t ) L( t k+1) L(t) H(0) t k+1 t H(L(t)) 18/21

54 Additional results The target set version has a constant adaptivity gap. 19/21

55 Additional results The target set version has a constant adaptivity gap. The maximization problem has a 1 + o(1) adaptivity gap. 19/21

56 General model /21

57 General model 2 1 λ 1, /21

58 General model λ 1,2 = 1 2 λ 1,3 3 λ 4,5 = /21

59 General model We need to keep track of the set of active nodes. λ 1,2 = 1 2 λ 1,3 3 λ 4,5 = /21

60 General model We need to keep track of the set of active nodes. Even the non-adaptive problem is difficult in this setting! λ 1,2 = 1 2 λ 1,3 3 λ 4,5 = /21

61 Conjectures and open problems Is there a broader class of graphs maintaining the constant gap result? - High conductance/connectivity. - Metric induced rates. 21/21

62 Conjectures and open problems Is there a broader class of graphs maintaining the constant gap result? - High conductance/connectivity. - Metric induced rates. Additive gap for the complete graph is constant, i.e., cost NA cost A = O(1). 21/21