Approximating The Permanent

Approximating The Permanent

Amit Kagan

Seminar in Complexity 04/06/2001

Topics

• Description of the Markov chain

• Analysis of its mixing time

Definitions

• Let G = (V₁, V₂, E) be a bipartite graph on n+n vertices.

• Let  denote the set of perfect matchings in G.

• Let (y, z) denote the set of near-perfect matchings with holes only at y and z.

•      _y,_z  ⁽ y^, z⁾

|(u,v)|/|| Exponentially Large

It has only one perfect matching...

u v

Observe the following bipartite graph:

|(u,v)|/|| Exponentially Large

But two near-perfect matchings with holes at u and v.

u v

|(u,v)|/|| Exponentially Large

• Concatenating another hexagon,

– adds a constant number of vertices,

– but doubles the number of near-perfect matchings, – while the number of perfect matchings remains 1.

. . .

Thus we can force the ratio |(u,v)|/|| to be exponentially large.

The Breakthrough

• Jerrum, Sinclair, and Vigoda [2000]

introduced an additional weight factor.

• Any hole pattern (including that with no holes) is equally likely in the stationary distribution π.

• π will assign Ω(1/n²) weight to perfect matchings.

Edge Weights

• For each edge (y, z)  E, we introduce a positive weight (y, z).

• For a matching M, (M) = _{(i, j)M}(i, j).

• For a set of matchings S, (S) = _MS(M).

• We will work with the complete graph on n+n vertices:

 (e) = 1 for all e  E

 (e) = ξ ≈ 0 for all e  E

1

1 ξ

The Stationary Distribution

• The desired distribution π over Ω is

(M)  (M), where







 

 

 M

M

v u v

u M

v u w M M

if )

(

, some for

) , ( if

) , ( ) ) (

( 



w : V₁ × V₂  ⁺ is the weight function, to be specified shortly

The Markov Chain

1. Choose an edge e=(u,v) uniformly at random.

2. (i) If M   and e  M, let M’ = M\{e}, (ii) if M  (u,v), let M’ = M{e},

(iii) if M  (u,z) where z  v, and (y,v)  M, let M’ = M{e}\{(y,v)},

(iv) if M  (y,v) where y  u, and (u,z)  M, let M’ = M{e}\{(u,z)}.Metropolis rule

3. With probability min{1,(M’)/(M)} go to M’; otherwise, stay at M.

The Markov Chain (cont.)

• Finally, we add a self-loop probability of ½ to every state.

• This insures the MC is aperiodic.

• We also have irreducibility.

Detailed Balance

• Consider two adjacent matchings M and M’ with

(M) ≤ (M’).

 (M)P(M, M’) = (M’)P(M’, M) P(M,M’) > 0

=: Q(M,M’) )

(M (M) M m

M P

M m M

P

' π

π 1

2 ) 1

,' (

1 2

) 1 ' ,

(











• The transition probabilities between M and M’

may be written

m

M M

 2

)) ' (

), (

min( 

The Ideal Weight

• Recall that (M)  (M), where

• Ideally, we would take w = w*, where







 

 

 M

M

v u v

u M

v u w M M

if )

(

, some for

) , ( if

) , ( ) ) (

( 



)) ,

( (

) ) (

, (

* u v u v

w 





 

((u,v))







) , ( vu M 

(M) ( ( , ))

)) , ( (

)

( u v

v

u 



 



 _

λ(M)w(u,v)







) , (

) (

) , (

v u M

M v

u w



 = λ() = ()

The Concession

• We will content ourselves with weights w satisfying

) , ( 2

) , 2 (

) ,

( _*

*

z y w

z y z w

y

w  

• This perturbation will reduce the relative

weight of perfect and near-perfect matchings by at most a constant factor (4).

The Mixing Time Theorem

Assuming the weight function w satisfies the above inequality for all (y,z)  V₁ × V₂ , then the mixing time of the MC is bounded above by () = O(m⁶n⁸(n logn + log^-1)), provided the initial state is a perfect matching of

maximum activity.

Edge Weights Revisited

• We will work with the complete graph on n+n vetices.

• Think of non-edges e  E as having a very small activity of 1/n!.

• The combined weight of all invalid matchings is at most 1.

• We begin with activities  whose ideal

weights w* are easy to compute, and progress towards our target activities.

 ≡ 1

*(e) = 1^/n! for all e  E

*(e) = 1/n! for all e  E

Step I

• We assume at the beginning of the phase w(u,v) approximates w*(u,v) within ratio 2 for all (u,v).

• Before updating an activity, we will find for each

(u,v) a better approximation, one that is within ratio c for some 1 < c < 2.

• For this purpose we use the identity

) (

π

)) ,

( (

π )

, (

*

) , (



 u v

v u w

v u

w 

) (

) , ( )) ,

( (







 u v w u v

Step I (cont.)

• The mixing time theorem allows us to sample, in polynomial time, from a

distribution ’ that is within variation distance  of π.

• We choose  = c₁/n², take O(n² log ^-1)

samples from ’, and use sample averages.

• Using a few Chernoff bounds, we have,

with probability 1- (n²+1), approximation within ratio c to all of w*(u,v).

c₁ > 0 is a sufficiently small constant

Step I (conclusion)

Taking c = 6/5 and using O(n² log ^-1)

samples, we obtain refined estimates w(u,v) satisfying

5w*(u,v)/6 ≤ w(u,v) ≤ 6w*(u,v)/5

Step II

• We update the activity of an edge e

 (e) ← (e) * exp(-1/2)

• The ideal weight function w* changes by at most a factor of exp(1/2).

• Since 6exp(1/2)/5 < 2, our estimates w after step I approximate w* within ratio 2 for the new activities.

≈ 1.978

Step II (cont.)

• We use the above procedure repeatedly to reduce the initial activities to the target

activities.

 ≡ 1

*(e) = 1/n! for all e  E

*(e) = 1/n! for all e  E

• This requires O(n² · n log n) phases.

• Each phase requires O(n² log ^-1) samples.

• Each sample requires O(n²¹ log n)

simulation steps (mixing time theorem).

 Overall time - O(n²⁶ log² n log ^-1)

The  Error

• We need to set  so that the overall failure probability is strictly less than , say /2.

• The probability that any phase fails is at most O(n³ log n · n²).

• We will take  = c₂ / n⁵ log n .

Time Complexity

)) log

log (

( ⁿ

^{n n}  

O

• Running time of generating a sample:

)) log

(log log

( ⁿ

^{n n}  

O

• Running time of the initialization:

Conductance

• The conductance of a reversible MC is defined as

=min_{S}(S), where

• Theorem:

For an ergodic, reversible Markov chain with self- loops probabilities P(y,y)  ½ for all states x,

) ( ) (

) , ( )

( ) (

) , ) (

( S S

y x Q S

S

S S

S Q ^{x S y S}



 



 





 

) ln

) ( 2 (ln

)

( ₂  ^1  ^1

 



_x x

Canonical Paths

• We define canonical paths γ_I,F from all I  Ω to all F 

.

• Denote Γ = { γ_I,F : (I, F)  Ω × }.

• Certain transitions on a canonical path will be deemed chargeable.

• For each transition t denote cp(t) = {(I, F) : γ_I,F contains t as a chargeable transition}

I  F

• If I  , then I  F consists of a collection of alternating cycles.

• If I  (y,z), then I  F consists of a

collection of alternating cycles together with a single alternating path from y to z.

y

z

Type A Path

• Assume I  .

• A cycle v₀  v₁  …  v_2k = v₀ is unwound by:

We assume w.l.g. that the edge (v₀, v₁) belongs to I

(i) removing the edge (v₀, v₁),

(ii) successively, for each 1 ≤ i ≤ k – 1,

exchanging the edge (v_2i, v_2i+1) with (v_2i-1, v_2i), (iii) adding the edge (v_2k-1, v_2k).

• All these transitions are deemed chargeable.

Type A Path Illustrated

v₀ v₁ v₁

v₂

v₃

v v₆

v₀ v₇

Type B Path

• Assume I  (y,z).

• The alternating path y = v₀  …  v_2k+1 = z is unwound by:

(i) successively, for each 1 ≤ i ≤ k, exchanging the edge (v_2i-1, v_2i) with (v_2i-2, v_2i-1), and

(ii) adding the edge (v_2k, v_2k+1).

• Here, only the above transitions are deemed chargeable.

Type B Path Illustrated

y z

Congestion

• We define a notion of congestion of Γ:

• Lemma I

Assuming the weight w approximates w*

within ratio 2, then τ(Γ) ≤ 16m.











 





 

) ( cp ) , (

) (

) ) (

( max 1

: ) (

t F

T I

t I F

t

Q  



Lemma II

• Let u,y  V₁, v,z  V₂. Then,

(i) λ(u,v)λ((u,v)) ≤ λ(), for all vertices u,v with u  v.

(ii) λ(u,v)λ((u,z))λ((y,v)) ≤ λ()λ((y,z)), for all distinct vertices u,v,y,z with u  v.

• Observe that M_u,z  M_y,v  {(u,v)} decomposes into a collection of cycles together with an odd- length path O joining y and z.

Corollary III

Let u,y  V₁, v,z  V₂. Then,

(i) w*(u,v) ≥ λ(u,v), for all vertices u,v with u  v.

(ii) w*(u,z)w*(y,v) ≥ λ(u,v)w*(y,z), for all distinct vertices u,v,y,z with u  v.

(iii) w*(u,z)w*(y,v) ≥ λ(u,v) λ(y,z), for all distinct vertices u,v,y,z with u  v and y  z.

Proof of Lemma I

• For any transition t = (M,M’) and any pair of states I, F  cp(t), we will define an encoding η_t(I,F)  Ω such that η_t : cp(t) → Ω is an injection, and

π(I)π(F) ≤ 8 min{π(M), π(M’)}π(η_t(I,F)) = 16m Q(t)π(η_t(I,F))

• Summing over I,F  cp(t), we get

m F

I m

F t I

Q I F t

t t

F I

16 ))

, ( ( 16

) ( ) ) (

( 1

) ( cp ) , ( )

( cp ) , (

















The Injection η

• For a transition t = (M,M’) which is

involved in stage (ii) of unwinding a cycle, the encoding is

η_t(I,F) = I  F  (M  M’) \ {(v₀, v₁)}.

• Otherwise, the encoding is

η_t(I,F) = I  F  (M  M’).

From Congestion to Conductance

• Corollary IV

Assuming the weight function w approximates w* within ratio 2 for all (y,z)  V₁ × V₂ , then

 ≥ 1/100τ³n⁴ ≥ 1/10⁶m³n⁴.

• Proof

• Set α = 1/10τ^{n2 .}

• Let (S,Ŝ) be a partition of the state-space.

Case I

• π(S  ) / π(S) ≥ α and π(Ŝ  ) / π(Ŝ) ≥ α.

• Just looking at canonical paths of type A we have a total flow of π(S  )π(Ŝ  ) ≥

α²π(S)π(Ŝ) across the cut.

• Thus, τQ(S,^{Ŝ) ≥ α2}π(S)π(Ŝ), and,

 (S) = Q(S,Ŝ)/π(S)π(Ŝ) ≥ α² /τ ^{= 1/100}τ^3n4.

1/10τn²

Case II

• Otherwise, π(S  ) / π(S) < α .

• Note the following estimates:

π() ≥ 1/4(n²+1) ≥ 1/5n² π(S  ) < απ(S) < α

π(S \ ) = π(S) – π(S  ) > (1 – α)π(S) Q(S \ , S  ) ≤ π(S  ) < απ()

Case II (cont.)

• Consider the cut (S \ , Ŝ  ).

• The weight of canonical paths (all chargeable as they cross the cut) is π(S \ )π() ≥ (1 – α)π(S)/5n² ≥ π(S)/6n².

1/10τn²

• Hence, τQ(S \ ,Ŝ  ) ≥ π(S)/6n².

• Q(S,Ŝ) ≥ … ≥ π(S)π(Ŝ)/15τn².

 (S) = Q(S,Ŝ)/π(S)π(Ŝ) ≥ 1/15τn².

Summing It Up

• Starting from an initial state X₀ of maximum activity guarantees π(X₀) ≥ 1/n!, and hence, log(π(X₀)^-1) = O(n log n).

• We showed (S) ≥ 1/100τ³ⁿ⁴, and hence,

(S)^-1 = O(τ^{3n4) = O(}m^3n4).

• Thus, according to the conductance theorem,

_x0() = O(m⁶n⁸(n logn + log^-1)).