Ultra-Fast Asynchronous Rumor Spreading Ali Pourmiri - - PowerPoint PPT Presentation

▶
ultra fast asynchronous rumor spreading
SMART_READER_LITE
LIVE PREVIEW

Ultra-Fast Asynchronous Rumor Spreading Ali Pourmiri - - PowerPoint PPT Presentation

Feb 05, 2024 320 likes •1.17k views

Ultra-Fast Asynchronous Rumor Spreading Ali Pourmiri alipourmiri@gmail.com University of Isfahan 17 April 2019 IPMCCC19 Ali (Isfahan) Rumor Spreading 17 April 1 / 53 Push Protocol (Synchronous) Demers, Gealy, Greene, Hauser, Irish,

slide-1
SLIDE 1

Ultra-Fast Asynchronous Rumor Spreading

Ali Pourmiri

alipourmiri@gmail.com

University of Isfahan

17 April 2019 IPMCCC’19

Ali (Isfahan) Rumor Spreading 17 April 1 / 53

slide-2
SLIDE 2

Push Protocol (Synchronous)

Demers, Gealy, Greene, Hauser, Irish, Larson, Manning, Shenker, Sturgis, Swinehart, Terry, Woods’87

  • 1. The ground is a simple connected graph.
  • 2. At time 0, one vertex knows a rumour.
  • 3. At each time-step 1, 2, . . . , every informed vertex tells the rumour

to a random neighbour.

Ali (Isfahan) Rumor Spreading 17 April 2 / 53

slide-3
SLIDE 3

Push Protocol (Synchronous)

Demers, Gealy, Greene, Hauser, Irish, Larson, Manning, Shenker, Sturgis, Swinehart, Terry, Woods’87

  • 1. The ground is a simple connected graph.
  • 2. At time 0, one vertex knows a rumour.
  • 3. At each time-step 1, 2, . . . , every informed vertex tells the rumour

to a random neighbour. Remark 1. Informed vertex may call a neighbour in consecutive steps. Remark 2. If a vertex receives the rumour at time t, it starts passing it from time t + 1.

Ali (Isfahan) Rumor Spreading 17 April 2 / 53

slide-4
SLIDE 4

Push Protocol (Synchronous)

Demers, Gealy, Greene, Hauser, Irish, Larson, Manning, Shenker, Sturgis, Swinehart, Terry, Woods’87

  • 1. The ground is a simple connected graph.
  • 2. At time 0, one vertex knows a rumour.
  • 3. At each time-step 1, 2, . . . , every informed vertex tells the rumour

to a random neighbour. Remark 1. Informed vertex may call a neighbour in consecutive steps. Remark 2. If a vertex receives the rumour at time t, it starts passing it from time t + 1. Spread Time: the first time everyone knows the rumour.

Ali (Isfahan) Rumor Spreading 17 April 2 / 53

slide-5
SLIDE 5

Application: distributed computing

Ali (Isfahan) Rumor Spreading 17 April 3 / 53

slide-6
SLIDE 6

Application: distributed computing

Rumour spreading advantages: Simplicity, locality, no memory Scalability, reasonable link loads Robustness

Ali (Isfahan) Rumor Spreading 17 April 3 / 53

slide-7
SLIDE 7

Example: a path

Ali (Isfahan) Rumor Spreading 17 April 4 / 53

slide-8
SLIDE 8

Example: a path

informTime(0) = 0

Ali (Isfahan) Rumor Spreading 17 April 4 / 53

slide-9
SLIDE 9

Example: a path

informTime(0) = 0 informTime(1) = 1

Ali (Isfahan) Rumor Spreading 17 April 5 / 53

slide-10
SLIDE 10

Example: a path

informTime(0) = 0 informTime(1) = 1 informTime(2) = 1 + Geo(1/2)

Ali (Isfahan) Rumor Spreading 17 April 6 / 53

slide-11
SLIDE 11

Example: a path

informTime(0) = 0 informTime(1) = 1 informTime(2) = 1 + Geo(1/2) informTime(3) = 1 + Geo(1/2) + Geo(1/2)

Ali (Isfahan) Rumor Spreading 17 April 7 / 53

slide-12
SLIDE 12

Example: a path

informTime(0) = 0 informTime(1) = 1 informTime(2) = 1 + Geo(1/2) informTime(3) = 1 + Geo(1/2) + Geo(1/2) informTime(4) = 1 + Geo(1/2) + Geo(1/2) + Geo(1/2)

Ali (Isfahan) Rumor Spreading 17 April 8 / 53

slide-13
SLIDE 13

Example: a path

informTime(0) = 0 informTime(1) = 1 informTime(2) = 1 + Geo(1/2) informTime(3) = 1 + Geo(1/2) + Geo(1/2) informTime(4) = 1 + Geo(1/2) + Geo(1/2) + Geo(1/2) E[Spread Time] = 1 + 3 × 2 = 7

Ali (Isfahan) Rumor Spreading 17 April 9 / 53

slide-14
SLIDE 14

Example: a path

informTime(0) = 0 informTime(1) = 1 informTime(2) = 1 + Geo(1/2) informTime(3) = 1 + Geo(1/2) + Geo(1/2) informTime(4) = 1 + Geo(1/2) + Geo(1/2) + Geo(1/2) E[Spread Time] = 1 + 3 × 2 = 7 = 2n − 3

Ali (Isfahan) Rumor Spreading 17 April 10 / 53

slide-15
SLIDE 15

Example: a star

❧♥

Ali (Isfahan) Rumor Spreading 17 April 11 / 53

slide-16
SLIDE 16

Example: a star

When k + 1 vertices are informed and n − 1 − k uninformed, after E[Geo(n−k−1

n−1 )] = n−1 n−1−k more rounds a new vertex will be

informed. ❧♥

Ali (Isfahan) Rumor Spreading 17 April 11 / 53

slide-17
SLIDE 17

Example: a star

When k + 1 vertices are informed and n − 1 − k uninformed, after E[Geo(n−k−1

n−1 )] = n−1 n−1−k more rounds a new vertex will be

informed. E[Spread Time] = n − 1 n − 1 + n − 1 n − 2 + · · · + n − 1 2 + n − 1 1 ≈ n ❧♥ n

Ali (Isfahan) Rumor Spreading 17 April 11 / 53

slide-18
SLIDE 18

Improving the protocol Uninformed vertices ask the informed ones...

Ali (Isfahan) Rumor Spreading 17 April 12 / 53

slide-19
SLIDE 19

Push-Pull Protocol (Synchronous)

Demers, Gealy, Greene, Hauser, Irish, Larson, Manning, Shenker, Sturgis, Swinehart, Terry, Woods’87

  • 1. The ground is a simple connected graph.
  • 2. At time 0, one vertex knows a rumour.
  • 3. At each time-step 1, 2, . . . ,

every informed vertex sends the rumour to a random neighbour (PUSH); and every uninformed vertex queries a random neighbour about the rumour (PULL).

Ali (Isfahan) Rumor Spreading 17 April 13 / 53

slide-20
SLIDE 20

Push-Pull Protocol (Synchronous)

Demers, Gealy, Greene, Hauser, Irish, Larson, Manning, Shenker, Sturgis, Swinehart, Terry, Woods’87

  • 1. The ground is a simple connected graph.
  • 2. At time 0, one vertex knows a rumour.
  • 3. At each time-step 1, 2, . . . ,

every informed vertex sends the rumour to a random neighbour (PUSH); and every uninformed vertex queries a random neighbour about the rumour (PULL). Remark 1. Vertices may call the same neighbour in consecutive steps. Remark 2. If a vertex receives the rumour at time t, it starts passing it from time t + 1. Spread Time: the first time everyone knows the rumour.

Ali (Isfahan) Rumor Spreading 17 April 13 / 53

slide-21
SLIDE 21

Example: a star

push protocol: n ❧♥ n rounds push-pull protocol: 1 or 2 rounds

Ali (Isfahan) Rumor Spreading 17 April 14 / 53

slide-22
SLIDE 22

Example: a path

Ali (Isfahan) Rumor Spreading 17 April 15 / 53

slide-23
SLIDE 23

Example: a path

informTime(0) = 0

Ali (Isfahan) Rumor Spreading 17 April 15 / 53

slide-24
SLIDE 24

Example: a path

informTime(0) = 0 informTime(1) = 1

Ali (Isfahan) Rumor Spreading 17 April 16 / 53

slide-25
SLIDE 25

Example: a path

informTime(0) = 0 informTime(1) = 1 informTime(2) = 1 + ♠✐♥{Geo(1/2), Geo(1/2)} = 1 + Geo(3/4)

Ali (Isfahan) Rumor Spreading 17 April 17 / 53

slide-26
SLIDE 26

Example: a path

informTime(0) = 0 informTime(1) = 1 informTime(2) = 1 + Geo(3/4) informTime(3) = 1 + Geo(3/4) + Geo(3/4)

Ali (Isfahan) Rumor Spreading 17 April 18 / 53

slide-27
SLIDE 27

Example: a path

informTime(0) = 0 informTime(1) = 1 informTime(2) = 1 + Geo(3/4) informTime(3) = 1 + Geo(3/4) + Geo(3/4) informTime(4) = 1 + Geo(3/4) + Geo(3/4) + 1

Ali (Isfahan) Rumor Spreading 17 April 19 / 53

slide-28
SLIDE 28

Example: a path

informTime(0) = 0 informTime(1) = 1 informTime(2) = 1 + Geo(3/4) informTime(3) = 1 + Geo(3/4) + Geo(3/4) informTime(4) = 1 + Geo(3/4) + Geo(3/4) + 1 E[Spread Time] = 2 + 2 × 4/3 = 14/3

Ali (Isfahan) Rumor Spreading 17 April 20 / 53

slide-29
SLIDE 29

Example: a path

informTime(0) = 0 informTime(1) = 1 informTime(2) = 1 + Geo(3/4) informTime(3) = 1 + Geo(3/4) + Geo(3/4) informTime(4) = 1 + Geo(3/4) + Geo(3/4) + 1 E[Spread Time] = 2 + 2 × 4/3 = 14/3 = 4 3n − 2 (versus 2n − 3 for push)

Ali (Isfahan) Rumor Spreading 17 April 21 / 53

slide-30
SLIDE 30

Known results

  • n complete graph

push: ❧♦❣2 n + ❧♥ n + o(❧♦❣ n) push-pull: ❧♦❣3 n + o(❧♦❣ n) ❧♦❣ ❧♦❣ ♠❛① ❧♦❣ ❧♦❣

Ali (Isfahan) Rumor Spreading 17 April 22 / 53

slide-31
SLIDE 31

Known results

  • n complete graph

push: ❧♦❣2 n + ❧♥ n + o(❧♦❣ n) push-pull: ❧♦❣3 n + o(❧♦❣ n) Barabasi-Albert preferential attachment graph has Spread Time Θ(❧♦❣ n), PUSH alone has Spread Time poly(n). ❧♦❣ ♠❛① ❧♦❣ ❧♦❣

Ali (Isfahan) Rumor Spreading 17 April 22 / 53

slide-32
SLIDE 32

Known results

  • n complete graph

push: ❧♦❣2 n + ❧♥ n + o(❧♦❣ n) push-pull: ❧♦❣3 n + o(❧♦❣ n) Barabasi-Albert preferential attachment graph has Spread Time Θ(❧♦❣ n), PUSH alone has Spread Time poly(n). Random graphs with power-law expected degrees (a.k.a. the Chung-Lu model) with exponent ∈ (2, 3) has Spread Time Θ(❧♦❣ n). ♠❛① ❧♦❣ ❧♦❣

Ali (Isfahan) Rumor Spreading 17 April 22 / 53

slide-33
SLIDE 33

Known results

  • n complete graph

push: ❧♦❣2 n + ❧♥ n + o(❧♦❣ n) push-pull: ❧♦❣3 n + o(❧♦❣ n) Barabasi-Albert preferential attachment graph has Spread Time Θ(❧♦❣ n), PUSH alone has Spread Time poly(n). Random graphs with power-law expected degrees (a.k.a. the Chung-Lu model) with exponent ∈ (2, 3) has Spread Time Θ(❧♦❣ n). If Φ is Cheeger constant (conductance) and α is the vertex expansion (vertex isoperimetric number), Spread Time ≤ C ♠❛①{Φ−1 ❧♦❣ n, α−1 ❧♦❣2 n}.

Ali (Isfahan) Rumor Spreading 17 April 22 / 53

slide-34
SLIDE 34

Key Idea : rigorously analyze the size of informed nodes until time t, It.

slide-35
SLIDE 35

Key Idea : rigorously analyze the size of informed nodes until time t, It. Examples: complete graphs, G(n, p) It+1 ∼ (1 + c)It

slide-36
SLIDE 36

Key Idea : rigorously analyze the size of informed nodes until time t, It. Examples: complete graphs, G(n, p) It+1 ∼ (1 + c)It Key Idea: efficient connectors facilitate the communication between large degree nodes

slide-37
SLIDE 37

Key Idea : rigorously analyze the size of informed nodes until time t, It. Examples: complete graphs, G(n, p) It+1 ∼ (1 + c)It Key Idea: efficient connectors facilitate the communication between large degree nodes Example: Chung-Lu, preferential attachment,.. .

slide-38
SLIDE 38

Toward a more realistic model...

Ali (Isfahan) Rumor Spreading 17 April 24 / 53

slide-39
SLIDE 39

Asynchronous Rumor Spreading Protocols

Boyd, Ghosh, Prabhakar, Shah’06

  • 1. A simple connected graph and each node has a Poisson clock of

rate 1

  • 2. At the beginning, one vertex knows a rumor
  • 3. As soon as the Poisson clock of a vertex rings, he contacts a

random neighbor and pushes (or pulls) the rumor to (from) his neighbor

Ali (Isfahan) Rumor Spreading 17 April 25 / 53

slide-40
SLIDE 40

Asynchronous Rumor Spreading Protocols

Boyd, Ghosh, Prabhakar, Shah’06

  • 1. A simple connected graph and each node has a Poisson clock of

rate 1

  • 2. At the beginning, one vertex knows a rumor
  • 3. As soon as the Poisson clock of a vertex rings, he contacts a

random neighbor and pushes (or pulls) the rumor to (from) his neighbor Remark 1. The number of calls by a node has Poisson Dist. of rate 1 Remark 2. The time distribution between any two consecutive rings of a node has Exponential Dist. of rate 1

Ali (Isfahan) Rumor Spreading 17 April 25 / 53

slide-41
SLIDE 41

Asynchronous Rumor Spreading Protocols

Boyd, Ghosh, Prabhakar, Shah’06

  • 1. A simple connected graph and each node has a Poisson clock of

rate 1

  • 2. At the beginning, one vertex knows a rumor
  • 3. As soon as the Poisson clock of a vertex rings, he contacts a

random neighbor and pushes (or pulls) the rumor to (from) his neighbor Remark 1. The number of calls by a node has Poisson Dist. of rate 1 Remark 2. The time distribution between any two consecutive rings of a node has Exponential Dist. of rate 1 Spread Time: the first time everyone knows the rumour.

Ali (Isfahan) Rumor Spreading 17 April 25 / 53

slide-42
SLIDE 42

Example: Asynchronous push on a path

informTime(0) = 0 informTime(1) = Exp(1)

Ali (Isfahan) Rumor Spreading 17 April 26 / 53

slide-43
SLIDE 43

Example: Asynchronous push on a path

informTime(0) = 0 informTime(1) = Exp(1) informTime(2) = Exp(1) + Exp(1/2) informTime(3) = Exp(1) + Exp(1/2) + Exp(1/2)

Ali (Isfahan) Rumor Spreading 17 April 27 / 53

slide-44
SLIDE 44

Example: Asynchronous push on a path

informTime(0) = 0 informTime(1) = Exp(1) informTime(2) = Exp(1) + Exp(1/2) informTime(3) = Exp(1) + Exp(1/2) + Exp(1/2) informTime(4) = Exp(1) + Exp(1/2) + Exp(1/2) + Exp(1/2)

Ali (Isfahan) Rumor Spreading 17 April 28 / 53

slide-45
SLIDE 45

Example: Asynchronous push on a path

informTime(0) = 0 informTime(1) = Exp(1) informTime(2) = Exp(1) + Exp(1/2) informTime(3) = Exp(1) + Exp(1/2) + Exp(1/2) informTime(4) = Exp(1) + Exp(1/2) + Exp(1/2) + Exp(1/2) E[Spread Time] = 1 + 3 × 2 = 2(n − 2) + 1

Ali (Isfahan) Rumor Spreading 17 April 29 / 53

slide-46
SLIDE 46

Example: Asynchronous push-pull on a path

informTime(0) = 0 informTime(1) = ♠✐♥{Exp(1), Exp(1/2)} = Exp(3/2)

Ali (Isfahan) Rumor Spreading 17 April 30 / 53

slide-47
SLIDE 47

Example: Asynchronous push-pull on a path

informTime(0) = 0 informTime(1) = ♠✐♥{Exp(1), Exp(1/2)} = Exp(3/2) informTime(2) = Exp(3/2) + ♠✐♥{Exp(1/2), Exp(1/2)} = Exp(3/2) + Exp(1) informTime(3) = Exp(3/2) + Exp(1) + Exp(1)

Ali (Isfahan) Rumor Spreading 17 April 31 / 53

slide-48
SLIDE 48

Example: Asynchronous push-pull on a path

informTime(0) = 0 informTime(1) = Exp(3/2) informTime(2) = Exp(3/2) + Exp(1) informTime(3) = Exp(3/2) + Exp(1) + Exp(1) informTime(4) = Exp(3/2) + Exp(1) + Exp(1) + ♠✐♥{Exp(1/2), Exp(1)} = Exp(3/2) + Exp(1) + Exp(1) + Exp(3/2)

Ali (Isfahan) Rumor Spreading 17 April 32 / 53

slide-49
SLIDE 49

Example: Asynchronous push-pull on a path

informTime(0) = 0 informTime(1) = Exp(3/2) informTime(2) = Exp(3/2) + Exp(1) informTime(3) = Exp(3/2) + Exp(1) + Exp(1) informTime(4) = Exp(3/2) + Exp(1) + Exp(1) + Exp(3/2) E[Spread Time] = 2 + 4/3 = n − 3 + 4/3

Ali (Isfahan) Rumor Spreading 17 April 33 / 53

slide-50
SLIDE 50

Known results

  • n Gn,p, p ∈ (❧♦❣ n

n , 1]

push-pull: w.h.p. ST = ❧♦❣ n + O(1) ❧♦❣ ❧♦❣ ❧♦❣

Ali (Isfahan) Rumor Spreading 17 April 34 / 53

slide-51
SLIDE 51

Known results

  • n Gn,p, p ∈ (❧♦❣ n

n , 1]

push-pull: w.h.p. ST = ❧♦❣ n + O(1)

  • n Chung-lu random graph

push-pull: w.h.p. ST = O(❧♦❣ ❧♦❣ n) ❧♦❣

Ali (Isfahan) Rumor Spreading 17 April 34 / 53

slide-52
SLIDE 52

Known results

  • n Gn,p, p ∈ (❧♦❣ n

n , 1]

push-pull: w.h.p. ST = ❧♦❣ n + O(1)

  • n Chung-lu random graph

push-pull: w.h.p. ST = O(❧♦❣ ❧♦❣ n) For every graph G push-pull: w.h.p. STasynch = O(STsynch + ❧♦❣ n) STsynch STasynch ≤ √npolylog(n)

Ali (Isfahan) Rumor Spreading 17 April 34 / 53

slide-53
SLIDE 53

Toward a "bit" more realistic model...

Ali (Isfahan) Rumor Spreading 17 April 35 / 53

slide-54
SLIDE 54

Multiple-Rate Asynchronous Rumor Spreading

  • P. and Ramezani’19

Each node u has a Poisson clock of rate ru chosen from a given distribution At the beginning, one vertex knows a rumor As soon as the Poisson clock of a vertex rings, it pushes (pulls) the rumor to (from) a random neighbor contacts a random neighbor. Spread TimeST(ε): For every ε ∈ [0, 1), this is the first time when (1 − ε)fraction of nodes gets informed,

Ali (Isfahan) Rumor Spreading 17 April 36 / 53

slide-55
SLIDE 55

Our results

Theorem (P., Ramezani’19+) Suppose that R is a power law prob. dist. with exponent β ∈ (2, 3). Let us consider the push-pull protocol on an n-node complete

  • graph. Then, with constant probability, we have

ST(ε) = O(1 + ❧♦❣(1/ε))

Ali (Isfahan) Rumor Spreading 17 April 37 / 53

slide-56
SLIDE 56

Proof Sketch

Ali (Isfahan) Rumor Spreading 17 April 38 / 53

slide-57
SLIDE 57

Proof Sketch

Ali (Isfahan) Rumor Spreading 17 April 39 / 53

slide-58
SLIDE 58

Proof Sketch

Ali (Isfahan) Rumor Spreading 17 April 40 / 53

slide-59
SLIDE 59

Proof Sketch: Initial Phase

  • Each node u was assigned a random number ru from R ∝ k −β,

β ∈ (2, 3) ❧♦❣ ♠✐♥ ❧♦❣

Ali (Isfahan) Rumor Spreading 17 April 41 / 53

slide-60
SLIDE 60

Proof Sketch: Initial Phase

  • Each node u was assigned a random number ru from R ∝ k −β,

β ∈ (2, 3)

  • It: set of informed nodes until time t

❧♦❣ ♠✐♥ ❧♦❣

Ali (Isfahan) Rumor Spreading 17 April 41 / 53

slide-61
SLIDE 61

Proof Sketch: Initial Phase

  • Each node u was assigned a random number ru from R ∝ k −β,

β ∈ (2, 3)

  • It: set of informed nodes until time t
  • Recursively define w0 = 2

4(β−2) (3−β)2 +1, ..., w2k = (n/❧♦❣2 n) 1 β−1 , ..

wi :=        ♠✐♥

  • wi−1

2(i−1)/2

  • 1

β−2 , ⌊(n/❧♦❣2 n) 1 β−1 ⌋

  • if i is odd,

wi−1 2i/2

  • therwise.

(1)

Ali (Isfahan) Rumor Spreading 17 April 41 / 53

slide-62
SLIDE 62

Proof Sketch: Initial Phase

  • Each node u was assigned a random number ru from R ∝ k −β,

β ∈ (2, 3)

  • It: set of informed nodes until time t
  • Recursively define w0 = 2

4(β−2) (3−β)2 +1, ..., w2k = (n/❧♦❣2 n) 1 β−1 , ..

  • Corresponding to each wi, define random variable

Ti := ♠✐♥{t : |It| ≥ wi} if i is even, ♠✐♥{t : ∃u ∈ It with ru ≥ wi}

  • therwise.

Ali (Isfahan) Rumor Spreading 17 April 42 / 53

slide-63
SLIDE 63

Ti := ♠✐♥{t : |It| ≥ wi} if i is even, ♠✐♥{t : ∃u ∈ It with ru ≥ wi}

  • therwise.

Ali (Isfahan) Rumor Spreading 17 April 43 / 53

slide-64
SLIDE 64

Ti := ♠✐♥{t : |It| ≥ wi} if i is even, ♠✐♥{t : ∃u ∈ It with ru ≥ wi}

  • therwise.

Remark Random variable T2k stochastically dominates the time is required for the initial phase.

Ali (Isfahan) Rumor Spreading 17 April 43 / 53

slide-65
SLIDE 65

Lemma

  • 1. E[T0] = O(1)
  • 2. for some constant C, E[Ti+1 − Ti] ≤ C −i

Corollary E[T2k] = E[T2k − T0] + E[T0] =

2k−1

  • i=0

E[Ti+1 − Ti] + E[T0] =

2k−1

  • i=0

E[E[Ti+1 − Ti|Ti]] + E[T0] = O(1)

Ali (Isfahan) Rumor Spreading 17 April 44 / 53

slide-66
SLIDE 66

Lemma

  • 1. E[T0] = O(1)
  • 2. for some constant C, E[Ti+1 − Ti] ≤ C −i

Proof of 1. T0 = ♠✐♥{t : |It| ≥ w0 = 2

4(β−2) (3−β)2 +1}

⇒ E[T0] = O(1)

Ali (Isfahan) Rumor Spreading 17 April 45 / 53

slide-67
SLIDE 67

Proof of 2. Fix an even i, then Ti = ♠✐♥{t : |It| ≥ wi} Ti−1 = ♠✐♥{t : ∃u ∈ It with ru ≥ wi−1}

Ali (Isfahan) Rumor Spreading 17 April 46 / 53

slide-68
SLIDE 68

Proof of 2. Fix an even i, then Ti = ♠✐♥{t : |It| ≥ wi} Ti−1 = ♠✐♥{t : ∃u ∈ It with ru ≥ wi−1} Since |It| = o(n), for every t ∈ [Ti−1, Ti],

  • u∈It

ru(n − |It|) n

  • Poission Rate(push attempt)

≥ wi−1(1 − o(1)) > wi−1 2

Ali (Isfahan) Rumor Spreading 17 April 46 / 53

slide-69
SLIDE 69

Proof of 2. Fix an even i, then Ti = ♠✐♥{t : |It| ≥ wi} Ti−1 = ♠✐♥{t : ∃u ∈ It with ru ≥ wi−1} Since |It| = o(n), for every t ∈ [Ti−1, Ti],

  • u∈It

ru(n − |It|) n

  • Poission Rate(push attempt)

≥ wi−1(1 − o(1)) > wi−1 2 So for every t ∈ [Ti−1, Ti], a new node gets informed with an exponential dist. of rate

2 wi−1 .

Ali (Isfahan) Rumor Spreading 17 April 46 / 53

slide-70
SLIDE 70

Proof of 2. So for every t ∈ [Ti−1, Ti], a new node gets informed within an exponential dist. of rate

2 wi−1 .

Ali (Isfahan) Rumor Spreading 17 April 47 / 53

slide-71
SLIDE 71

Proof of 2. So for every t ∈ [Ti−1, Ti], a new node gets informed within an exponential dist. of rate

2 wi−1 .

Only consider the push protocol, to inform wi new nodes E[Ti+1 − Ti|Ti] ≤ 2wi wi−1 = 2−i/2 ⇒ E[Ti+1 − Ti] = E[E[Ti+1 − Ti|Ti]] ≤ C −i

Ali (Isfahan) Rumor Spreading 17 April 47 / 53

slide-72
SLIDE 72

Proof of 2. So for every t ∈ [Ti−1, Ti], a new node gets informed within an exponential dist. of rate

2 wi−1 .

Only consider the push protocol, to inform wi new nodes E[Ti+1 − Ti|Ti] ≤ 2wi wi−1 = 2−i/2 ⇒ E[Ti+1 − Ti] = E[E[Ti+1 − Ti|Ti]] ≤ C −i Similar technique works for odd i’s.

Ali (Isfahan) Rumor Spreading 17 April 47 / 53

slide-73
SLIDE 73

Our results

Theorem (P., Ramezani’19+) Suppose that R is a prob. dist. with mean µ = O(1) and bounded

  • variance. Let us consider the push protocol on an n-node complete
  • graph. Then,

E[ST(0)] = 2 ❧♦❣ n

µ

+ ω(1) w.h.p., we have ST(0) = 2 ❧♦❣ n µ + ω(

  • ❧♦❣ n)

Ali (Isfahan) Rumor Spreading 17 April 48 / 53

slide-74
SLIDE 74

E[ST(0)] = 2 ❧♦❣ n

µ

+ ω(1) ❧♦❣

slide-75
SLIDE 75

E[ST(0)] = 2 ❧♦❣ n

µ

+ ω(1) ❧♦❣

slide-76
SLIDE 76

E[ST(0)] = 2 ❧♦❣ n

µ

+ ω(1) tj : required time to inform (j + 1)-th node tj : exponentially distributed with rate n−j

n−1Sj

E[tj ] = E[E[tj |Sj ]] = E[ n − 1 (n − j )Sj ] = n − 1 n − j E[ 1 Sj ]

it needs more cal.

∼ n − 1 (n − j )µj ❧♦❣

slide-77
SLIDE 77

E[ST(0)] = 2 ❧♦❣ n

µ

+ ω(1) tj : required time to inform (j + 1)-th node tj : exponentially distributed with rate n−j

n−1Sj

E[tj ] = E[E[tj |Sj ]] = E[ n − 1 (n − j )Sj ] = n − 1 n − j E[ 1 Sj ]

it needs more cal.

∼ n − 1 (n − j )µj E[ST(0)] =

n−1

  • j =1

E[tj ] = 2 ❧♦❣ n µ ± O(1)

slide-78
SLIDE 78

Our results

Theorem (P., Ramezani’19+) Suppose that R is a power law prob. dist. with exponent β ∈ (2, 3). Let us consider the push-pull protocol on an n-node complete graph. Then, with constant probability, we have ST(ε) = O(❧♦❣(1/ε))

❧♦❣

❧♦❣ ❧♦❣

Ali (Isfahan) Rumor Spreading 17 April 50 / 53

slide-79
SLIDE 79

Our results

Theorem (P., Ramezani’19+) Suppose that R is a power law prob. dist. with exponent β ∈ (2, 3). Let us consider the push-pull protocol on an n-node complete graph. Then, with constant probability, we have ST(ε) = O(❧♦❣(1/ε)) Theorem (P., Ramezani’19+) Suppose that R is a prob. dist. with mean µ = O(1) and bounded

  • variance. Let us consider the push protocol on an n-node complete
  • graph. Then,

E[ST(0)] = 2 ❧♦❣ n

µ

+ ω(1) w.h.p., we have ST(0) = 2 ❧♦❣ n µ + ω(

  • ❧♦❣ n)

Ali (Isfahan) Rumor Spreading 17 April 50 / 53

slide-80
SLIDE 80

Multiple-Call Rumor Spreading (synch. version)

Panagiouto, P., Sauerwald’13

Each node u was assigned a random number ru chosen from a given distribution At the beginning, one vertex knows a rumor In each round, node u pushes (pulls) the rumor to (from) ru random neighbors. Spread Time: the first time when all nodes become informed.

Ali (Isfahan) Rumor Spreading 17 April 51 / 53

slide-81
SLIDE 81

Asynchronous vs Synchronous

algorithm distribution multiple-call multiple-rate push E[R] = µ < ∞, V[R] < ∞ ST(0) =

❧♦❣ n ❧♦❣(1+µ) + ❧♦❣ n µ

± o(❧♦❣ n) ST(0) = ❧♦❣ n

µ

± o(❧♦❣ n) push-pull R is a power law with β ∈ (2, 3) ST(ε) = Θ(❧♦❣ ❧♦❣ n) ST(ε) = Θ(1)

Ali (Isfahan) Rumor Spreading 17 April 52 / 53

slide-82
SLIDE 82

Asynchronous vs Synchronous

algorithm distribution multiple-call multiple-rate push E[R] = µ < ∞, V[R] < ∞ ST(0) =

❧♦❣ n ❧♦❣(1+µ) + ❧♦❣ n µ

± o(❧♦❣ n) ST(0) = ❧♦❣ n

µ

± o(❧♦❣ n) push-pull R is a power law with β ∈ (2, 3) ST(ε) = Θ(❧♦❣ ❧♦❣ n) ST(ε) = Θ(1)

The asynchronous model propagates the rumor much faster

Ali (Isfahan) Rumor Spreading 17 April 52 / 53

slide-83
SLIDE 83

Any Question? Thank you! alipourmiri@gmail.com