Combinatorial optimization and the mean field modelJohan WästlundChalmers University of TechnologySweden

Random instances of optimization problems

Random instances of optimization problems

Random instances of optimization problems Typical distance between nearby points is of

order n-1/2

Random instances of optimization problems A tour consists of n links, therefore we

expect the total length of the minimum tour to scale like n1/2

Beardwood-Halton-Hammersley (1959):

73.02/1 pN

nL

92.062.0

Mean field model of distance

Distances Xij chosen as i.i.d. variables Given n and the distribution of distances,

study the random variable Ln If the distribution models distances in d

dimensions, we expect Ln to scale like n1-1/d

In particular, pseudo-dimension 1 means Ln is asymptotically independent of n

Mean field model of distance

The edges of a complete graph on n vertices are given i. i. d. nonnegative costs

Exponential(1) distribution.

Mean field model of distance

We are interested in the cost of the minimum matching, minimum traveling salesman tour etc, for large n.

Mean field model of distance

Convergence in probability to a constant?

Matching Set of edges that gives a pairing of all points

Statistical Physics / C-S Spin configuration Hamiltonian Ground state energy

Temperature Gibbs measure Thermodynamic limit

Feasible solution Cost of solution Cost of minimal

solution Artificial parameter T Gibbs measure n→∞

Statistical physics

Replica-cavity method of statistical mechanics has given spectacular predictions for random optimization problems

M. Mézard, G. Parisi 1980’s Limit of /12 for minimum matching on the

complete graph (Aldous 2000) Limit 2.0415… for the TSP (Wästlund 2006)

A. Frieze (2004): “Up to now there has been almost no progress analysing this random model of the travelling salesman problem.”

N. J. Cerf et al (1997): “Researchers outside physics remain largely unaware of the analytical progress made on the random link TSP.”

Non-rigorous derivation of the /12 limit Matching problem on Kn for large n. In principle, this requires even n, but we shall

consider a relaxation Let the edges be exponential of mean n, so

that the sequence of ordered edge costs from a given vertex is approximately a Poisson process of rate 1.

Non-rigorous derivation of the /12 limit The total cost of the minimum matching is of

order n. Introduce a punishment c>0 for not using a

particular vertex. This makes the problem well-defined also for

odd n. For fixed c, let n tend to infinity. As c tends to infinity, we expect to recover

the behavior of the original problem.

Non-rigorous derivation of the /12 limit For large n, suppose that the problem

behaves in the same way for n-1 vertices. Choose an arbitrary vertex to be the root What does the graph look like locally around

the root? When only edges of cost <2c are considered,

the graph becomes locally tree-like

Non-rigorous derivation of the /12 limit Non-rigorous replica-cavity method Aldous derived equivalent equations with the

Poisson-Weighted Infinite Tree (PWIT)

Non-rigorous derivation of the /12 limit Let X be the difference in cost between the

original problem and that with the root removed.

If the root is not matched, then X = c. Otherwise X = i – Xi, where Xi is distributed like X, and i is the cost of the i:th edge from the root.

The Xi’s are assumed to be independent.

Non-rigorous derivation of the /12 limitIt remains to do some calculations.

We have

where Xi is distributed like X

),,min( ii XcX

Non-rigorous derivation of the /12 limit Let )(exp)(exp)()( ufdttFuXPuF

u

X

-u

Non-rigorous derivation of the /12 limit Then if u>-c,

)()()(' ufeuFuf

)()(' ufeuf

Non-rigorous derivation of the /12 limit

)()()()( )(')(')(')(' ufufufuf edudeufe

dudeufufuf

Hence )()( ufuf ee is constant

Non-rigorous derivation of the /12 limit

The constant depends on c and holds when

–c<u<c

f(-u)

f(u)

Non-rigorous derivation of the /12 limit From definition, exp(-f(c)) = P(X=c) =

proportion of vertices that are not matched, and exp(-f(-c)) = exp(0) = 1

e-f(u) + e-f(-u) = 2 – proportion of vertices that are matched = 1 when c = infinity.

Non-rigorous derivation of the /12 limit

1 ee yx

6

2Area

Non-rigorous derivation of the /12 limit What about the cost of the minimum

matching?

Non-rigorous derivation of the /12 limit

Non-rigorous derivation of the /12 limit

Non-rigorous derivation of the /12 limit Hence J = area under the curve when f(u) is

plotted against f(-u)! Expected cost = n/2 times this area In the original setting = ½ times the area = /12.

The equation has the explicit solution

This gives the cost

)1log()( ueuf

121)1log(

21 2

dueeu

u

The exponential bipartite assignment problem

n

The exponential bipartite assignment problem Exact formula conjectured by Parisi (1998)

Suggests proof by induction Researchers in discrete math, combinatorics and

graph theory became interested Generalizations…

2

191

411)(

nCE n

Generalizations

by Coppersmith & Sorkin to incomplete matchings

Remarkable paper by M. Buck, C. Chan & D. Robbins (2000)

Introduces weighted vertices Extremely close to proving Parisi’s

conjecture!

Incomplete matchings

nm

Weighted assignment problems Weights 1,…,m, 1,…, n on vertices Edge cost exponential of rate ij

Conjectured formula for the expected cost of minimum assignment

Formula for the probability that a vertex participates in solution (trivial for less general setting!)

The Buck-Chan-Robbins urn process Balls are drawn with probabilities proportional

to weight

1 2

3

Proofs of the conjectures

Two independent proofs of the Parisi and Coppersmith-Sorkin conjectures in 2003 (Nair, Prabhakar, Sharma and Linusson, Wästlund)

Rigorous method

Relax by introducing an extra vertex Let the weight of the extra vertex go to zero Example: Assignment problem with 1=…=m=1, 1=…=n=1, and m+1 =

p = P(extra vertex participates) p/n = P(edge (m+1,n) participates)

Rigorous method

p/n = P(edge (m+1,n) participates) When →0, this is

Hence

By Buck-Chan-Robbins urn theorem,

1,,1,,1,,1,,,1 nmknmknmknmknm CCCClP

n

pCCE nmknmk 01,,1,, lim

11

kmmmp

Rigorous method

Hence

Inductively this establishes the Coppersmith-Sorkin formula

nkmnmmn

CECE nmknmk )1(1

)1(11

1,,1,,

)1(

1)1(

1)1(

1)1(

11,,

knmnkmnmnmmn

CE nmk

Rigorous results

Much simpler proofs of Parisi, Coppersmith-Sorkin, Buck-Chan-Robbins formulas

Exact results for higher moments Exact results and limits for optimization

problems on the complete graph

The 2-dimensional urn process

2-dimensional time until k balls have been drawn

Limit shape as n→∞

Matching:

TSP/2-factor:

1 ee yx

12

12

1

ee yx yx

041548.2)(21

0

dxxy

6)(

2

0

dxxy

Mean field TSP

If the edge costs are i.i.d and satisfy P(l<t)/t→1 as t→0 (pseudodimension 1), then as n →∞,

...041548.2* LLp

n

For the TSP, the replica-cavity approach gives

)())(1()(' ufeufuf

)())(1()(' ufeufuf

It follows that

is constant, and = 1 by boundary conditions Replica-cavity prediction agrees with the

rigorous result (Parisi 2006)

)()(

2)(1

2)(1 ufuf eufeuf

Further exact formulas

nnn

n kn

kkC1

32

1

2

1

4 /114/12/15var

2

1)3(4)2(4n

On

LP-relaxation of matching in the complete graph Kn

121

91

411)(

2

2

nCE n

Future work

Explain why the cavity method gives the same equation as the limit shape in the urn process

Reprove results of one method with the other Find the variance with the replica method Find rigorously the distribution of edge costs

participating in the solution (there is an exact conjecture)