My slide for DISC19.

1. Phase Transitions of Best-of-Two and
Best-of-Three on Stochastic Block Models
Nobutaka Shimizu and Takeharu Shiraga
(Chuo University)(The University of Tokyo)
DISC19

2. /182
Sync. voting processes on a graph, Best-of-two (Bo2)
and Best-of-three (Bo3). Each vertex supports one of
two distinct opinions.
‣In each round, every node updates his opinion according
to some rule.
u
• (Bo2): Sample two random neighbors and . Then, take
the majority among opinions of and .
x y
x, y u
• (Bo3): Sample three random neighbors and take their majority.
‣Markov Chain on . Each state is denoted by .
‣ = consensus time starting from (r.v.).
2V
A ⊆ V
T(A) A
• c.f. In the well-known Voter process, adopts the opinion of a random neighbor.u
u y
x
u y
x

3. /18
Bo2 on with the half-and-half initial state.K500
3

4. /18
Known results (Bo2 & Bo3)
4
‣ For Bo2 and Bo3 on , w.h.p for any [DGMSS11].Kn T(A) = O(log n) A
‣ opinions setting: Bo2 and Bo3 on satisfies
for any if is moderate [BCNPT18], [GL18],
[BCNPST14].
k > 2 Kn
T(A) = O(k log n) A k
‣ For Bo2 and Bo3 on expander graphs, under
some initial bias assumption on [CER14], [CERRS15], [CRRS17].
T(A) = O(log n)
A
• core-periphery graph [CNNS18]
• regular stochastic block model [CNS19]
‣ Bo2 on various graphs are investigated recently
c.f. In Voter, it is known that [CEOR13].E[T(A)] = O(n)

5. /18
Known results (Bo2 & Bo3)
5
‣ w.h.p for any on [DGMSS11]. Much faster
than Voter.
T(A) = O(log n) A Kn
‣ If there are opinions, for any on if
is moderate [BCNPT18], [GL18], [BCNPST14].
k > 2 T(A) = O(k log n) A Kn
k
‣ Bo2 and Bo3 reaches consensus assumng an initial bias
on expander graphs [CER14], [CERRS15], [CRRS17]
O(log n)
• core-periphery graph [CNNS18]
• regular stochastic block model [CNS19]
‣ Bo2 on various graphs are investigated recently
c.f. In Voter, it is known that [CEOR13].E[T(A)] = O(n)
How graph structure affect the Bo2 and Bo3 ?

6. /18
This Work
‣Bo2 and Bo3 on Stochastic Block Model (SBM)
‣SBM is a random graph consists of two
vertex groups each of size , and two vertices
form an edge with probability
G(2n, p, q)
n
u, v
G(300, 0.1, 0.01)
6
if and are in the same group.u v
{
p
q o.w.
‣Generate and then set initial
opinion state and observe the process.
G(2n, p, q)

7. /187
Question: Where is the boundary?🤔?
q
p
= 0.4
q
p
= 0.1

8. /18
Our result (phase transition)
Consider Bo2 on such that and are constants.G(2n, p, q) p q
Theorem 1
• If , w.h.p. for any .
• If , w.h.p. for some .
q/p > 5 − 2 T(A) = O(log n) A ⊆ V
q/p < 5 − 2 T(A) ≥ exp(Ω(n)) A ⊆ V
8
Remarks:
1. Formally, we show that, w.h.p. exhibits a “nice”
structure such that the statements above hold.
G(2n, p, q)
2. Here, w.h.p. means “with probability for some const ”.1 − n−c
c > 0
3. Bo3 is more likely to reach consensus (since ).1/7 < 5 − 2
Consider Bo3 on such that and are constants.G(2n, p, q) p q
Theorem 2
• If , w.h.p. for any .
• If , w.h.p. for some .
q/p > 1/7 T(A) = O(log n) A ⊆ V
q/p < 1/7 T(A) ≥ exp(Ω(n)) A ⊆ V

9. /18
Two components of our proof
9
1. Structual property of .G(2n, p, q)
Tool: Janson’s ineq & Kim-Vu’s ineq
2. Bo2 and Bo3 on “nice” graphs
Tools: Jacobi matrix, competitive dynamical system.

10. /1810
Idea behind the proof (Bo3)
The case of (i.e., ) is previously known.
We expect that the same technique works for the case
of contsants .
p = q = 1 K2n
p, q > 0
‣Bo3 is a M.C. on in general graphs. But on in .
‣Let and be the set of opinion 1 and that in the next round.
Then
2V
{0,…, n} Kn
A A′
|A′| =
∑
v∈V
1[v∈A′]
is the sum of independent random variables, which concentrates
on its expectation.
E[|A′|] = ?

11. /19
α′ ≈ E[α′] =
E[|A′|]
n
= g(α) + O(n−1
) .
E[|A′| |A] =
∑
v∈A
Pr[v ∈ A′] =
∑
v∈V
g
(
degA(v)
deg(v) )
.
11
‣Evaluate . Indeed, letting
, we have
E[|A′|]
g(x) =
(
3
2)
x2
(1 − x) +
(
3
3)
x3
0
1
0.5 α g(α)
degA(v) = 3
v
‣Let On , we have
and thus
α = |A|/n . Kn
degA(v)
deg(v)
=
|A|
n
+ O(n−1
) = α + O(n−1
)
‣The orbit of is approximated by a discrete-time
dynamical system on .
α
[0,1]
deg(v) = 5

12. /18
Key lemma (concentration of for Bo3 on )E[α′|A] G(2n, p, p)
‣The lemma says “the behavior of can be approximated by a
discrete-time dynamical system.”
‣We can evaluate , which leads to the symmetry breaking.
‣The result can be extended to .
‣The proof relies on concentration inequalities.
α
Var[α′|A]
G(2n, p, q)
12
Suppose that . Then satisfies the
following w.h.p.:
p = ω(log n/n) G(2n, p, p)
∀A ⊆ V, E[α′|A] = g(α) +O ((np)−0.5
),
where g(x) := 3x2
− 2x3
.
‣This observation holds for . We prove it for !!Kn G(2n, p, q)

13. /18
On G(2n,p,q)
‣ Let .Ai := A ∩ Vi (i = 1,2)
A1
A2
13
‣ Fix the time step .
‣ : configuration at the current step .
‣ : that of the next step .
t
A t
A′ t + 1

14. /18
2D dynamical system
‣Consider for each .E[α′i] = E[|A′i |/n] i = 1,2
• This yields a 2D dynamical system on .[0,1]2
14
α′1 = 3 (
α1 + rα2
1 + r )
2
− 2 (
α1 + rα2
1 + r )
3
± O((np)−1/2
),
α′2 = 3 (
rα1 + α2
1 + r )
2
− 2 (
rα1 + α2
1 + r )
3
± O((np)−1/2
) .
‣For Bo3, our key lemma implies
where r = q/p .

15. /18
thresholdr <
15
thresholdr >
Vertical axis = Horisontal axis = .α1 α2

16. /1816
threshold (absorbing)r < threshold (saddle).r >

17. /18
Intuitive Explanation (p≠q)
‣If , two absorbing points appear:q/p < 1/7
It takes exponential time to escape from the ε-
neighborhood of the absorbing points.
17
‣We used competitive dynamical system theory to prove
the orbit convergence for any initial point .(α1, α2)
‣We do not explore the case where .r = 1/7
‣If , these absorbing points disappear:q/p > 1/7
Escape from the fixed point in time.O(log n)

18. /18
• Approximate orbit by a dynamical system.
Summary
‣ Technical tools:
• Local behavior around fixed points.
• Global behavior (orbit convergence)
‣ Bo2 and Bo3 on SBM
• phase transition on q/p.
• consensus time: log vs. exponential.
18
• Community detection is a potential application.
• The case threshold is left as future task.
• We obtained similar results for sparse SBM under initial bias
assumption.
q/p =
Concentration inequalities to analyze random graphs.
Essentially known in [Doerr et al. 2011].
Competitive dynamical system theory.