Phase transitions in discrete structures
Amin Coja-Oghlan
Goethe University Frankfurt
Overview
1
The physics approach.
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The second moment method.
“Quiet planting”.
A physics-inspired rigorous approach.
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Basics.
Replica symmetry (“Belief Propagation”).
The 1RSB cavity method.
“Classical” rigorous work.
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3
[following Mézard, Montanari ’09]
The Kauzmann transition.
The free entropy in the 1RSB phase.
Random k-SAT.
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A rigorous Belief Propagation-based approach.
Amin Coja-Oghlan (Frankfurt)
Random k-SAT
2 / 40
The Boltzmann distribution
Let X be a finite set of spins and let N be a “large” integer.
Consider a probability distribution on X N defined by
M
1 Y
µ(x) =
ψa (x),
Z
with Z =
a=1
M
X Y
ψa (x).
x∈X N a=1
Assume that ψa depends only on the components ∂a ⊂ [N] of x.
Thus, ψa (x) = ψa (x∂a ).
We call µ the Boltzmann distribution and Z the partition function.
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The factor graph
Suppose we are given the Boltzmann distribution
M
1 Y
ψa (x∂a ).
µ(x) =
Z
a=1
We set up the bipartite factor graph.
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Square vertices: the factors ψa , a = 1, . . . , M.
Round vertices: the variables x1 , . . . , xM .
Conncect each factor ψa with all variables xi , i ∈ ∂a.
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The factor graph
Example: the 1-dimensional Ising model
There are N “sites” with spins in X = {−1, 1}.
For any a ∈ {1, . . . , N − 1} we define
ψa (x) = exp(βxa xa+1 )
(β > 0).
Hence, ∂a = {a, a + 1}.
h P
i
Q
N−1
.
This yields µ(x) ∝ M
exp(βx
x
)
=
exp
β
x
x
a
a+1
a
a+1
a=1
a=1
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The factor graph
Example: the Potts antiferromagnet
Let X = [K ] = {1, 2, . . . , K }.
Let G = (V , E ) be a graph on V = [N].
For an edge e joining vertices i, j in G let
ψe (x) = exp −β · 1xi =xj .
This gives rise to
Y
µ(x) =
ψe (x) = exp [−β · #monochromatic edges] .
e∈E
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The factor graph
Example: the Potts antiferromagnet
Let X = [K ] = {1, 2, . . . , K }.
Let G = (V , E ) be a graph on V = [N].
For an edge e joining vertices i, j in G let
ψe (x) = exp −β · 1xi =xj .
This gives rise to
Y
µ(x) =
ψe (x) = exp [−β · #monochromatic edges] .
e∈E
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The free entropy
Generally, the partition function Z scales exponentially with N.
Therefore, we are interested in the free entropy
1
N
ln Z .
In fact, we care mostly about the theormdynamic limit N → ∞, i.e.,
lim
N→∞
1
ln Z ,
N
the free entropy density.
Amin Coja-Oghlan (Frankfurt)
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The free entropy
Generally, the partition function Z scales exponentially with N.
Therefore, we are interested in the free entropy
1
N
ln Z .
In fact, we care mostly about the theormdynamic limit N → ∞, i.e.,
lim
N→∞
1
ln Z ,
N
the free entropy density.
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Does this limit exist?
If so, can we compute/approximate it?
Is the limit an analytic function, or are there phase transitions?
Amin Coja-Oghlan (Frankfurt)
Random k-SAT
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The free entropy
Generally, the partition function Z scales exponentially with N.
Therefore, we are interested in the free entropy
1
N
ln Z .
In fact, we care mostly about the theormdynamic limit N → ∞, i.e.,
lim
N→∞
1
ln Z ,
N
the free entropy density.
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Does this limit exist?
If so, can we compute/approximate it?
Is the limit an analytic function, or are there phase transitions?
Example: the Ising model
1-dim: the free entropy density is analytic, i.e., no phase transition.
2-dim: there is a phase transition.
Amin Coja-Oghlan (Frankfurt)
Random k-SAT
[Onsager 1949]
7 / 40
Disordered systems
Two levels of randomness
The Boltzmann distribution itself is random.
Hence, the free entropy density becomes
1
ln Z .
lim E
N→∞
N
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Disordered systems
Two levels of randomness
The Boltzmann distribution itself is random.
Hence, the free entropy density becomes
1
ln Z .
lim E
N→∞
N
Example: the Edwards-Anderson model
With x as in the 1-dim Ising model, we let
" N−1
#
X
µ(x) ∝ exp β
Ji xi xi+1 .
i=1
The Ji are independent Gaussian random variables.
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Disordered systems
Example: the diluted mean-filed Potts antiferromagnet
Let X = {1, . . . , K } for some “small” K ≥ 2.
Consider a random graph G (N, M) on N vertices with M edges.
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Disordered systems
Example: the diluted mean-filed Potts antiferromagnet
Let X = {1, . . . , K } for some “small” K ≥ 2.
Consider a random graph G (N, M) on N vertices with M edges.
For any edge e of G (N, M) joining vertices i, j, define
ψe (x) = exp −β · 1xi =xj .
Amin Coja-Oghlan (Frankfurt)
Random k-SAT
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Disordered systems
Example: the diluted mean-filed Potts antiferromagnet
Let X = {1, . . . , K } for some “small” K ≥ 2.
Consider a random graph G (N, M) on N vertices with M edges.
For any edge e of G (N, M) joining vertices i, j, define
ψe (x) = exp −β · 1xi =xj .
The Boltzmann distribution
Y
µ(x) ∝
ψe (x) = exp [−β · #monochromatic edges]
e
is random as it depends on the graph G (N, M).
Amin Coja-Oghlan (Frankfurt)
Random k-SAT
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Disordered systems
Example: the diluted mean-filed Potts antiferromagnet
Let X = {1, . . . , K } for some “small” K ≥ 2.
Consider a random graph G (N, M) on N vertices with M edges.
For any edge e of G (N, M) joining vertices i, j, define
ψe (x) = exp −β · 1xi =xj .
The Boltzmann distribution
Y
µ(x) ∝
ψe (x) = exp [−β · #monochromatic edges]
e
is random as it depends on the graph G (N, M).
New parameter: the denisty α = M/N.
Amin Coja-Oghlan (Frankfurt)
Random k-SAT
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Disordered systems
Example: coloring random graphs
Let X = {1, . . . , K } for some “small” K ≥ 3.
For any edge e of G (N, M) joining vertices i, j, define
ψe (x) = exp −∞ · 1xi =xj = 1xi 6=xj [“zero temperature”].
Then
µ(x) ∝
Y
ψe (x) = 1no edge is monochromatic .
e
is the uniform distribution over proper K -colorings of G (N, M).
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Disordered systems
Question
[Erdős, Rényi 1960]
Is there a phase transition in this model (in terms of α = M/N)?
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Random k-SAT
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Disordered systems
Example: random k-SAT
Let X = {0, 1} and fix some “small” K ≥ 3.
Think of x1 , . . . , xN as Boolean variables.
Let Φ be an expression of the form
(x1 ∨ x̄17 ∨ · · · ∨ x29 ) ∧ (x̄11 ∨ x2 ∨ · · · ∨ x̄1 ) ∧ · · ·
|
{z
} |
{z
}
k-clause Φ1
k-clause Φ2
with M “clauses”, chosen uniformly at random.
For i = 1, . . . , M let
ψi (x) = 1clause Φi
Then µ(x) =
QM
Amin Coja-Oghlan (Frankfurt)
i=1 ψi (x)
evaluates to true .
is uniform over satisfying assignments.
Random k-SAT
11 / 40
Belief Propagation
Goals
To compute the free entropy on trees.
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the Belief Propagation equations.
the Bethe free entropy.
The replica symmetric ansatz.
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Belief Propagation as a distributional fixed point equation.
application to “diluted mean-field models”.
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Random k-SAT
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Example: the 1-dimensional Ising model
Let X = {±1}, x ∈ X N and
"
µ(x) ∝ exp β
N−1
X
xi xi+1 + βB
i=1
N
X
#
xi .
i=1
Goal: to compute the marginal µ(xi ) of xi .
Amin Coja-Oghlan (Frankfurt)
Random k-SAT
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Example: the 1-dimensional Ising model
For j ∈ [N] define two distributions ν̂→j , ν̂j→ on X by
"
ν̂→j (xj ) ∝
X
exp β
x1 ,...,xj−1 ∈X
ν̂j← (xj ) ∝
exp β
xj+1 ,...,xN ∈X
Amin Coja-Oghlan (Frankfurt)
xi xi+1 + βB
j−1
X
i=1

X
j−1
X
Random k-SAT
N−1
X
i=j
#
xi ,
i=1
xi xi+1 + βB
N
X

xi  .
i=j+1
14 / 40
Example: the 1-dimensional Ising model
For j ∈ [N] define two distributions ν̂→j , ν̂j→ on X by
"
ν̂→j (xj ) ∝
X
exp β
x1 ,...,xj−1 ∈X
ν̂j← (xj ) ∝
xi xi+1 + βB
j−1
X
i=1

X
j−1
X
exp β
xj+1 ,...,xN ∈X
N−1
X
#
xi ,
i=1
xi xi+1 + βB
i=j
N
X

xi  .
i=j+1
Then
µ(xj ) ∝ ν̂→j (xj ) · exp(βBxj ) · νj← (xj ).
Amin Coja-Oghlan (Frankfurt)
Random k-SAT
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Example: the 1-dimensional Ising model
For j ∈ [N] define two distributions ν̂→j , ν̂j→ on X by
"
ν̂→j (xj ) ∝
X
exp β
x1 ,...,xj−1 ∈X
ν̂j← (xj ) ∝
xi xi+1 + βB
j−1
X
i=1

X
j−1
X
exp β
xj+1 ,...,xN ∈X
N−1
X
#
xi ,
i=1
xi xi+1 + βB
i=j
N
X

xi  .
i=j+1
Then
µ(xj ) ∝ ν̂→j (xj ) · exp(βBxj ) · νj← (xj ).
Moreover, we have the recurrence
X
ν̂→j+1 (xj+1 ) =
ν̂→j (xj ) exp [βxj xj+1 + βBxj ] .
xj ∈X
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Random k-SAT
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Example: the 1-dimensional Ising model
We have ν̂→j+1 (xj+1 ) =
P
xj ∈X
ν→j (xj ) exp [βxj xj+1 + βBxj ] .
Setting
u→j =
ν̂→j (1)
1
ln
,
2β ν̂→j (−1)
we can rephrase the above as
u→j+1 = f (u→j + B) with f (x) = β −1 atanh(tanh(β) tanh(βx)).
The function f has a unique fixed point u ∗ .
Hence, for j sufficiently far from the boundary, we find
µ(xj ) ∼ tanh(β(2u ∗ + B)).
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Belief Propagation on trees
Suppose that the factor graph of µ(x) is a tree.
Generalising the above example, we define messages recursively by
Y
(t+1)
(t)
νj→a (xj ) ∝
ν̂b→j (xj ),
b∈∂j\{a}
(t)
ν̂b→j (xj )
∝
X
ψb (x∂b )
x∂b\j
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Y
(t)
νi→b (xi ).
i∈∂b\j
t counts the number of recursive steps,
∂j is the neighborhood of j,
the sum ranges over X ∂b\j , with xj is given upfront.
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Belief Propagation on trees
Suppose that the factor graph of µ(x) is a tree.
Generalising the above example, we define messages recursively by
Y
(t+1)
(t)
νj→a (xj ) ∝
ν̂b→j (xj ),
b∈∂j\{a}
(t)
ν̂b→j (xj )
∝
X
ψb (x∂b )
x∂b\j
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Y
(t)
νi→b (xi ).
i∈∂b\j
t counts the number of recursive steps,
∂j is the neighborhood of j,
the sum ranges over X ∂b\j , with xj is given upfront.
Define the Belief Propagation marginal of variable xj as
ν (t) (xj ) ∝
Y
(t−1)
ν̂b→j (xj ).
b∈∂j
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Belief Propagation on trees
Theorem
Assume that the factor graph is a tree of diameter T .
1
The BP equations have a unique fixed point ν ∗ .
2
For t > T we have ν (t) = ν ∗ , regardless of the initialisation.
3
For all variables xj we have
ν ∗ (xj ) = µ(xj ),
i.e., the BP marginals coincide with the Boltzmann marginals.
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Belief Propagation on trees
Theorem (the Bethe free entropy)
Given the messages ν, define
F (ν) =
M
X
Fa (ν) +
a=1
Fa (ν) = ln
X
Fi (ν) −
i=1
ψa (x∂a )
x∂a
Fi (ν) = ln
N
X
Y
X
Fia (ν), where
(i,a)
νi→a (xi ),
i∈∂a
XY
ν̂b→i (xi ),
xi b∈∂i
Fia (ν) = ln
X
νi→a (xi )ν̂a→i (xi ).
xi
If the factor graph is a tree, then ln Z = F (ν ∗ ).
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Belief Propagation on trees
On trees, BP converges to its unique fixed point.
This fixed point yields
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the correct marginals,
the free entropy.
Example: the 1-dim Ising model.
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Belief Propagation on infinite trees
Question
Does this formalism survive the thermodynamic limit?
That is, does it extend to infinite trees?
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Belief Propagation on infinite trees
Question
Does this formalism survive the thermodynamic limit?
That is, does it extend to infinite trees?
Examples of infinite trees
The infinite regular tree.
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Each variable/factor node has the same number of outgoing edges.
The factors ψa are identical.
Infinite random trees.
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The numbers of outgoing edges is are independent Poisson variables.
The factors ψa might have some randomness, too.
Generally, we assume that sub-trees are identically distributed.
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Random k-SAT
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Belief Propagation on infinite trees
The distributional BP equations
We interpret the Belief Propagation equations
Y
νj→a (xj ) ∝
ν̂b→j (xj ),
b∈∂j\{a}
ν̂b→j (xj ) ∝
X
ψb (x∂b )
x∂b\j
Y
νi→b (xi ).
i∈∂b\j
as distributional fixed point equations.
That is, if
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ψb , |∂b|, |∂j| are i.i.d. from the distribution defining the tree,
the νi→b are i.i.d. from a distribution ν∗ over distributions,
then νj→a has distribution ν∗ .
The Bethe free entropy is determined by ν∗ .
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Belief Propagation on infinite trees
Example: coloring regular trees
Consider the graph K -coloring problem on the d-ary tree.
By symmetry, all the marginals are identical, i.e.,
νj→a (xj ) = 1/K
for all j, xj .
Thus, ν∗ is the measure concentrated on the uniform distribution.
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Belief Propagation on infinite trees
Example: random K -SAT
Each variable node has a Po(α) number of children.
The factors ψa are i.i.d. random clauses of length K .
The fixed point distribution ν∗
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is not known to exist,
but is conjectured to be highly non-trivial and not discrete,
based on numerical evidence (“population dynamics”).
This mirrors the asymmetric combinatorics of the problem.
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The replica symmetric ansatz
Diluted mean-field models
Think of the factor graphs in models such as
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random graph K -coloring,
random K -SAT.
The factor graph is an Erdős-Rényi-like sparse random graph.
Thus, roughly speaking,
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the vertices are bounded degree,
the shortest cycle has length Ω(ln N).
In other words, locally the graph looks like the infinite random tree!
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The replica symmetric ansatz
The replica symmetric “solution”
Let ν∗ be the BP fixed point on the ∞ tree.
Let φ∗ be the resulting Bethe free entropy.
Then the replica symmetric prediction is that
E[ln Z ]
,
N→∞
N
φ∗ = lim
with Z the partition function in, e.g., random K -SAT.
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The replica symmetric ansatz
The replica symmetric “solution”
Let ν∗ be the BP fixed point on the ∞ tree.
Let φ∗ be the resulting Bethe free entropy.
Then the replica symmetric prediction is that
E[ln Z ]
,
N→∞
N
φ∗ = lim
with Z the partition function in, e.g., random K -SAT.
Hypotheses
The fixed point ν∗ exists and is only one “relevant” fixed point.
There is correlation decay.
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The replica symmetric ansatz
Example: ferromagnetic Ising on sparse random graphs
[DM ’08]
The replica symmetric solution is correct.
Paramagnetic/ferromagnetic phase transition.
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The replica symmetric ansatz
Example: coloring random graphs
Let K = #colors, α = M/N density of the graph.
Then Z = #K -colorings of G (N, M).
According to the replica symmetric solution,
1
1
E[ln Z ] ∼ ln E[Z ] = ln K + α ln(1 − 1/K ).
N
N
The col/uncol transition occurs at α = (K − 21 ) ln K + oK (1).
Rigorous work:
[ACO, Vilenchik ’13]
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The r.s. solution is correct only for
1
α < (K − ) ln K − ln 2 + oK (1).
2
I
The col/uncol transition occurs at α ≤ (K − 12 ) ln K −
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1
2
+ oK (1).
27 / 40
Multiple Belief Propagation fixed points
Random factor graphs
In random graph K -coloring, the factor graph is similar to G (N, M).
Thus, it’s far from being a tree.
In effect, there can be multiple BP fixed points.
These mirror long range correlations.
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Multiple Belief Propagation fixed points
xi
w
The reconstruction problem
Suppose µ is defined by a factor graph G .
Pick a variable xi and let G ω be the depth-ω neighborhood.
What is the impact of the boundary on xi ?
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Multiple Belief Propagation fixed points
Correlation decay: Gibbs uniqueness
Let y ω be a boundary configuration.
The worst-case influence of the boundary can be cast as
E[sup |µ (xi |y ω ) − µ(xi )|].
yω
Here the expectation is over the choice of µ.
If
lim
lim E[sup |µ (xi |y ω ) − µ(xi )|] = 0,
ω→∞ N→∞
yω
we say there is Gibbs uniqueness.
Observe that this is a purely local property.
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Random k-SAT
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Multiple Belief Propagation fixed points
Correlation decay: non-reconstruction
The average influence of the boundary configuration can be cast as
X
E
µ(y ω )|µ (xi |y ω ) − µ(xi )|.
yω
If
lim
lim E
ω→∞ N→∞
X
µ(y ω )|µ (xi |y ω ) − µ(xi )| = 0,
yω
we say there is non-reconstruction.
Otherwise reconstruction is possible.
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Random k-SAT
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Multiple Belief Propagation fixed points
Theorem
[Achlioptas, ACO ’08]
If the density satisfies
α = M/N >
1
+ ε K ln K ,
2
then in random graph K -coloring reconstruction is possible.
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Multiple Belief Propagation fixed points
Theorem
[Achlioptas, ACO ’08]
If the density satisfies
α = M/N >
1
+ ε K ln K ,
2
then in random graph K -coloring reconstruction is possible.
In fact, suppose we initialise BP with one random K -coloring.
Then for Ω(n) variables xi the BP marginals satisfy ν ∗ (xi ) = 1.
Thus, these variables are “locally frozen”.
Interpretation: µ decomposes into a large number of clusters.
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The 1RSB cavity method
1-step replica symmetry breaking
In the above scenario, µ is a convex combination
µ=
N
X
wi µ i
i=1
of probability measures µi (corresponding to “clusters”).
Here wi ∝ exp(Fi · N), with Fi the free entropy of µi .
Key idea: perform Belief Propagation on the level of the µi !
The parameter β is replaced by the Parisi parameter y .
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The 1RSB cavity method
Idea
A Boltzmann distribution on the µi .
This distribution is defined by
Py (µi ) ∝ wiy .
In terms of the complexity function
Σ(φ) =
1
ln # {i ∈ [N ] : Fi ∼ φ} ,
N
the partition function Ξ(y ) of ξy satisfies
Ξ(y ) = y φ + Σ(φ), with φ s.t.
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Random k-SAT
∂Σ
= −y .
∂φ
34 / 40
The 1RSB cavity method
Belief Propagation one level up
What is the factor graph of Py ?
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The “old” messages νi→a , ν̂a→i are the variables.
There are factor nodes enforcing the BP equations.
Additional factor nodes implement the Parisi parameter y .
(This is possible due to Bethe’s formula.)
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Random k-SAT
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The 1RSB cavity method
Belief Propagation one level up
What is the factor graph of Py ?
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I
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The “old” messages νi→a , ν̂a→i are the variables.
There are factor nodes enforcing the BP equations.
Additional factor nodes implement the Parisi parameter y .
(This is possible due to Bethe’s formula.)
On this factor graph, write the Belief Propagation equations.
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1RSB cavity equations.
Complication: the spins are continuous.
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Random k-SAT
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The 1RSB cavity method
Belief Propagation one level up
What is the factor graph of Py ?
I
I
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I
The “old” messages νi→a , ν̂a→i are the variables.
There are factor nodes enforcing the BP equations.
Additional factor nodes implement the Parisi parameter y .
(This is possible due to Bethe’s formula.)
On this factor graph, write the Belief Propagation equations.
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1RSB cavity equations.
Complication: the spins are continuous.
The BP fixed point for Py yields Σ(φ).
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Analogous to the Bethe free entropy.
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Random k-SAT
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The 1RSB cavity method
Belief Propagation one level up
What is the factor graph of Py ?
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The “old” messages νi→a , ν̂a→i are the variables.
There are factor nodes enforcing the BP equations.
Additional factor nodes implement the Parisi parameter y .
(This is possible due to Bethe’s formula.)
On this factor graph, write the Belief Propagation equations.
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1RSB cavity equations.
Complication: the spins are continuous.
The BP fixed point for Py yields Σ(φ).
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Analogous to the Bethe free entropy.
The 1RSB equations can be viewed as distributional equations.
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the thermodynamic limit.
Setting y = 0 yields the Survey Propagation equations.
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The 1RSB cavity method
mical, conand q-COL
lc
10
14
19
ls [14]
10
15
20
αd,+
αd
αc
αs
The geometry of
the Boltzmann
distribution of the different phase transiti
Fig.
2. Pictorial representation
The
replica solutions
symmetric
s known
that
of aphase.
rCSP. At αd,+ some clusters appear, but for αd,+
comprise
only
an exponentially small fraction of solutions. For αd
Dynamic 1RSB.
s put forward
are split among about eNΣ∗ clusters of size eNs∗ . I
sented Static
in Ta- 1RSBsolutions
(aka condensation, Kauzmann transition).
the set of solutions is dominated by a few large clusters (with str
large k [11].
weights), and above αs the problem does not admit solutions an
) confirming
or q-coloring
[14], the valcorresponding to two distinct intuitions. According to
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The 1RSB cavity method
Example: for what d is the random d-regular graph K -colorable?
Setting y = 0, we consider the distribution
P0 (µi ) =
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1
,
N
N = #clusters.
Random k-SAT
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The 1RSB cavity method
Example: for what d is the random d-regular graph K -colorable?
Setting y = 0, we consider the distribution
P0 (µi ) =
1
,
N
N = #clusters.
Suppose that each µi is characterized by frozen variables.
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Encode µi as a map ζi : [N] → {1, 2, . . . , K , ∗}.
∗ = unfrozen (“joker color”).
Amin Coja-Oghlan (Frankfurt)
Random k-SAT
37 / 40
The 1RSB cavity method
Example: for what d is the random d-regular graph K -colorable?
Setting y = 0, we consider the distribution
P0 (µi ) =
1
,
N
N = #clusters.
Suppose that each µi is characterized by frozen variables.
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Encode µi as a map ζi : [N] → {1, 2, . . . , K , ∗}.
∗ = unfrozen (“joker color”).
This gives rise to SP messages:
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Qi→j (xi ) =P0 -probability that w/out j, i is frozen to color xi .
Qi→j (xi ) =P0 -probability that w/out j, i is unfrozen.
Amin Coja-Oghlan (Frankfurt)
Random k-SAT
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The 1RSB cavity method
Example: for what d is the random d-regular graph K -colorable?
We have the fixed point equations
P
Q
Qi→j (xi ) =
x
P∂i\j
∈N (xi )
x∂i\j ∈D
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Q
k∈∂i\j
k∈∂i\j
Qk→i (xk )
Qk→i (xk )
(xi ∈ [K ]).
D contains all (c1 , . . . , cd−1 ) such that [K ] 6⊂ {c1 , . . . , cd−1 }.
N (x) contains all (c1 , . . . , cd−1 ) ∈ D with [K ] \ {x} ⊂ {c1 , . . . , cd−1 }.
Let Qi→j (∗) = 1 −
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P
xi
Qi→j (xi ).
Random k-SAT
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The 1RSB cavity method
Example: for what d is the random d-regular graph K -colorable?
We have the fixed point equations
P
Q
Qi→j (xi ) =
x
P∂i\j
∈N (xi )
x∂i\j ∈D
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Q
k∈∂i\j
k∈∂i\j
Qk→i (xk )
Qk→i (xk )
(xi ∈ [K ]).
D contains all (c1 , . . . , cd−1 ) such that [K ] 6⊂ {c1 , . . . , cd−1 }.
N (x) contains all (c1 , . . . , cd−1 ) ∈ D with [K ] \ {x} ⊂ {c1 , . . . , cd−1 }.
P
Let Qi→j (∗) = 1 − xi Qi→j (xi ).
Simplification: it might seem reasonable to assume that
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Qi→j (xi ) = a for some fixed 0 ≤ a ≤ 1/q and all xi ∈ [K ].
Qi→j (∗) = 1 − Ka.
The above equation becomes
Pq−1
r K −1
(1 − (r + 1)a)d−1
r =0 (−1)
r
a = Pq−1
.
r q
d−1
r =0 (−1) r +1 (1 − (r + 1)a)
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Random k-SAT
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The 1RSB cavity method
Example: for what d is the random d-regular graph K -colorable?
We obtain the complexity ( N1 ln #clusters)
Σ = ln
"q−1
X
r =0
(−1)r
#
q
d
(1 − (r + 1)a)d − ln(1 − Ka2 )
r +1
2
with a the solution to
Pq−1
r
=0 (−1)
a = Prq−1
r
r =0 (−1)
K −1
(1 − (r + 1)a)d−1
r
.
q
d−1
r +1 (1 − (r + 1)a)
Now, K -colorability should occur iff Σ > 0.
Asymptotically, this yields a threshold of
dK −col = (2K − 1) ln K − 1 + oK (1).
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Conclusion
The cavity method as a non-rigorous formalism.
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free entropy,
phase transitions.
There are two variants:
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the replica symmetric version (“Belief Propagation”).
the 1RSB version (“Survey Propagation”).
Can we develop a rigorous foundation?
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Random k-SAT
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