Adaptive Rumor Spreading - Cornell UniversityI In viral marketing campaigns, the selection of vertices is crucial. Domingos and Richardson (2001) I An agent (service provider) wants

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Introduction Model Proofs Other results and open questions

Adaptive Rumor Spreading

Jose Correa 1 Marcos Kiwi 1

Neil Olver 2 Alberto Vera 1

1Universidad de Chile

2VU Amsterdam and CWI

July 27, 2015

Adaptive Rumor Spreading Universidad de Chile

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The situation


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The situation


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The situation


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The situation


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Introduction

I Rumors in social networks: contents, updates, newtechnology, etc.

I In viral marketing campaigns, the selection of vertices iscrucial. Domingos and Richardson (2001)

I An agent (service provider) wants to efficiently speed upthe communication process.


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Introduction





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Introduction





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Rumor spreading

I Models differ in time and communication protocol. Demers etal. (1987) and Boyd et al. (2006)

I In simple cases, the time to activate all the network ismostly understood.

I Even in random networks the estimates are logarithmic inthe number of nodes. Doerr et al. (2012) and Chierichetti et al. (2011)


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Rumor spreading





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Rumor spreading





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Opportunistic networksI We have an overload problem, an option is to exploit

opportunistic communications.

I A fixed deadline scenario has been studied heuristicallyalong with real large-scale data. Whitbeck et al. (2011)

I Control theory based algorithms greatly outperform staticones. Sciancalepore et al. (2014)


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opportunistic communications.I A fixed deadline scenario has been studied heuristically

along with real large-scale data. Whitbeck et al. (2011)



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opportunistic communications.I A fixed deadline scenario has been studied heuristically

along with real large-scale data. Whitbeck et al. (2011)



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The model

I Bob communicates and shares information.

I Bob meets Alice according to a Poisson process of rate λ/n.I Every pair of nodes can meet and gossip.


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The model

I Bob communicates and shares information.I Bob meets Alice according to a Poisson process of rate λ/n.

I Every pair of nodes can meet and gossip.

λ/n


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The model

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

λ/n


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The problem

I There is a unit cost for pushing the rumor.I Opportunistic communications have no cost.I At time τ all of the graph must be active.

We want a strategy that minimizes the overall number ofpushes.


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The problem

I There is a unit cost for pushing the rumor.I Opportunistic communications have no cost.I At time τ all of the graph must be active.

We want a strategy that minimizes the overall number ofpushes.


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Adaptive and non-adaptiveI A non-adaptive strategy pushes only at times t = 0 and

t = τ .

I An adaptive strategy may push at any time, with the fullknowledge of the process’ evolution.

0

1

2

3

4

5

t

Number of active nodes

τ

b

b

bc

bc

0

1

2

3

4

5

t


τ

b

b

bc

bc

bc

bcb

b

Push

t3


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t = τ .I An adaptive strategy may push at any time, with the full

knowledge of the process’ evolution.

0

1

2

3

4

5

t


τ

b

b

bc

bc

0

1

2

3

4

5

t


τ

b

b

bc

bc

bc

bcb

b

Push

t3


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t = τ .I An adaptive strategy may push at any time, with the full

knowledge of the process’ evolution.

0

1

2

3

4

5

t


τ

b

b

bc

bc

0

1

2

3

4

5

t


τ

b

b

bc

bc

bc

bcb

b

Push

t3


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Main result

Define the adaptivity gap as the ratio between the expectedcosts of non-adaptive and adaptive.

TheoremIn the complete graph the adaptivity gap is constant.


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Adaptive can be arbitrarily better

I With a small deadline, non-adaptive activates all of the vi ’s.I Adaptive activates only the root, then at some time t ′

pushes to the inactive vi ’s.I An adaptivity gap of log k

log log k is easy to prove.

r

v1 v2 v3 vk


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pushes to the inactive vi ’s.

I An adaptivity gap of log klog log k is easy to prove.

r

v1 v2 v3 vk


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pushes to the inactive vi ’s.I An adaptivity gap of log k

log log k is easy to prove.

r

v1 v2 v3 vk


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Non-adaptive

I Optimal non-adaptive pays almost the same at t = 0 andat t = τ .

- A 2-approximation is easy to see.

I Non-adaptive does not push more than n/2 rumors.Therefore, neither adaptive.

1 nk

λk

n/2

kN n− kN

λ = 1.λk := k(n−k)

n .


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Non-adaptiveI Optimal non-adaptive pays almost the same at t = 0 and

at t = τ .

- A 2-approximation is easy to see.I Non-adaptive does not push more than n/2 rumors.

Therefore, neither adaptive.

1 nk

λk

n/2

kN n− kN

λ = 1.λk := k(n−k)

n .


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at t = τ .- A 2-approximation is easy to see.


1 nk

λk

n/2

kN n− kN

λ = 1.λk := k(n−k)

n .


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at t = τ .- A 2-approximation is easy to see.


1 nk

λk

n/2

kN n− kN

λ = 1.λk := k(n−k)

n .


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Big deadline: τ ≥ (2 + δ) log n

I Starting from a single active node, the time until everyoneis active is 2 log n +O(1).

I The time is exponentially concentrated. Jason (1999)

I Just starting with one node has cost 1 + ε, thereforeadaptivity does not help.


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Small deadline: τ ≤ 2 log log n

I A Poisson process of unit rate gives the randomness.

I Given the points Si and Si+1, the rescaling Si+1−Siλi

is theinter-arrival time.

I A push can be seen as adding a point.

tb b b b b


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I A Poisson process of unit rate gives the randomness.I Given the points Si and Si+1, the rescaling Si+1−Si

λiis the

inter-arrival time.

I A push can be seen as adding a point.

tb b b b b

b

b

b

bk

λk λk+1 λi


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I A Poisson process of unit rate gives the randomness.I Given the points Si and Si+1, the rescaling Si+1−Si

λiis the

inter-arrival time.I A push can be seen as adding a point.

tb b b b bbc

b

b

b

bk

λk λk+1 λi λi+1


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Small deadline (cont.): τ ≤ 2 log log n

I A clairvoyant strategy knows the realization, thereforeoutperforms adaptive.

I We show that clairvoyant adds points only at thebeginning.

I Clairvoyant chooses the best number of initial pushes,given the realization.

tb b b b bbc

b

b

b

bk

λk λk+1 λi λi+1


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tb b b b bbc

b

b

b

bk

λk λk+1 λi λi+1


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tb b b b bbc

b

b

b

bk

λk λk+1 λi λi+1


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Say we start with k initial pushes.

I We know the inter-arrival distributions.I We know the non-adaptive cost; it pays Ω( n

log n ).

LemmaClairvoyant is considerably better than non-adaptive withprobability at most 1

n2 .


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Say we start with k initial pushes.I We know the inter-arrival distributions.I We know the non-adaptive cost; it pays Ω( n

log n ).


n2 .


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log n ).


n2 .


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log n ).


n2 .

In this case we can prove the gap to be 1 + o(1).


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Other deadlines

Insight: adaptive interferes when the cost of pushing is lessthan 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 thecondition should be met only a few times.

I A relaxed strategy pushes for free, but with certainconditions.

- Pushes only when (?) holds.- Does not push after n/2.

I Relaxed outperforms adaptive.


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Other deadlines








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Other deadlines








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Other deadlines








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Other deadlines








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Other deadlines (cont.)

I Relaxed adaptive can be described by thresholds φk .

I Let K (t) be the number of active nodes at time t.I We transform the process:

H (L(t)) := λK(t) log cost(K (t))φK(t)

.


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Other deadlines (cont.)I Relaxed adaptive can be described by thresholds φk .I Let K (t) be the number of active nodes at time t.

I We transform the process:


.

bc

t

cost(K(t))

φk

φk+1

φk+2

cost(K(0))

t′ tk+1

bc

bc

b

b

b

b

L(t)L(t′) L(tk+1)

bb

H(L(t))

bc

bc

H(0)


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Other deadlines (cont.)I Relaxed adaptive can be described by thresholds φk .I Let K (t) be the number of active nodes at time t.I We transform the process:


.

bc

t

cost(K(t))

φk

φk+1

φk+2

cost(K(0))

t′ tk+1

bc

bc

b

b

b

b

L(t)L(t′) L(tk+1)

bb

H(L(t))

bc

bc

H(0)


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Other deadlines (cont.)I Relaxed adaptive can be described by thresholds φk .I Let K (t) be the number of active nodes at time t.I We transform the process:


.

bc

t

cost(K(t))

φk

φk+1

φk+2

cost(K(0))

t′ tk+1

bc

bc

b

b

b

b

L(t)L(t′) L(tk+1)

bb

H(L(t))

bc

bc

H(0)

bc

t

cost(K(t))

φk

φk+1

φk+2

cost(K(0))

t′ tk+1

bc

bc

b

b

b

b

L(t)L(t′) L(tk+1)

bb

H(L(t))

bc

bc

H(0)


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Other deadlines (cont.)I We show that each time H touches zero, relaxed wins

exactly 1 compared to non-adaptive.

I Essentially H (s) is dominated by s − 2 Poiss(s).I The number of times H (s) touches zero is constant.

bc

t

cost(K(t))

φk

φk+1

φk+2

cost(K(0))

t′ tk+1

bc

bc

b

b

b

b

L(t)L(t′) L(tk+1)

bb

H(L(t))

bc

bc

H(0)


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exactly 1 compared to non-adaptive.I Essentially H (s) is dominated by s − 2 Poiss(s).

I The number of times H (s) touches zero is constant.

bc

t

cost(K(t))

φk

φk+1

φk+2

cost(K(0))

t′ tk+1

bc

bc

b

b

b

b

L(t)L(t′) L(tk+1)

bb

H(L(t))

bc

bc

H(0)


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exactly 1 compared to non-adaptive.I Essentially H (s) is dominated by s − 2 Poiss(s).I The number of times H (s) touches zero is constant.

bc

t

cost(K(t))

φk

φk+1

φk+2

cost(K(0))

t′ tk+1

bc

bc

b

b

b

b

L(t)L(t′) L(tk+1)

bb

H(L(t))

bc

bc

H(0)


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Additional results

I The target set version has a constant adaptivity gap.

I The maximization problem has a 1 + o(1) adaptivity gap.


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Additional results

I The target set version has a constant adaptivity gap.I The maximization problem has a 1 + o(1) adaptivity gap.


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General model

2

1

5 4

3


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General model

2

1

5 4

3λ1,3


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General model

2

1

5 4

3λ1,3

λ1,2 =∞

λ4,5 = 0


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General modelWe need to keep track of the set of active nodes.

2

1

5 4

3λ1,3

λ1,2 =∞

λ4,5 = 0


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General modelWe need to keep track of the set of active nodes.Even the non-adaptive problem is difficult in this setting!

2

1

5 4

3λ1,3

λ1,2 =∞

λ4,5 = 0


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Conjectures and open problems

I Is there a broader class of graphs maintaining the constantgap result?

- High conductance/connectivity.- Metric induced rates.

I Additive gap for the complete graph is constant, i.e.,costNA − costA = O(1).


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Conjectures and open problems

I Is there a broader class of graphs maintaining the constantgap result?

- High conductance/connectivity.- Metric induced rates.

I Additive gap for the complete graph is constant, i.e.,costNA − costA = O(1).


Documents

Adaptive Rumor Spreading - Cornell UniversityI In viral marketing campaigns, the selection of vertices is crucial. Domingos and Richardson (2001) I An agent (service provider) wants