Random matching and traveling salesman problems

Johan Wästlund

Chalmers University of Technology

Sweden

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.

Matching

Set of edges giving a pairing of all points

Traveling salesman

Tour visiting all points

Walkup’s theorem

Theorem (Walkup 1979): The expected cost of the minimum matching is bounded

Bipartite model

n

RL

Walkup’s theorem

= cost of the minimum assignment. Modify the graph model: Multiple edges with costs

given by a Poisson process This obviously doesn’t change the minimum

assignment

nA

Walkup’s theorem

Give each edge a random direction

Choose the five cheapest edges from each vertex.

We show that whp this set contains a perfect matching

Hall’s criterion

An edge set contains a perfect matching iff for every subset S of L,

SS )(

Hall’s criterion

If Hall’s criterion holds, an incomplete matching can always be extended.

Hall’s criterion

If Hall’s criterion fails for S, then it also fails for

S

T

(S)

RST c )(

Hall’s criterion

Here we can take |S| + |T| = n+1 If Hall’s criterion fails, then it fails for some S (in

L or in R) with

2

2n

S

Walkup’s theorem

Walkup’s theorem

Walkup’s theorem

The directed edges from a given vertex have costs from a rate n/2 Poisson process

The 5 cheapest edges have expected costs 2/n, 4/n, 6/n, 8/n, 10/n.

The average cost in this set is 6/n, and there are n edges in a perfect matching

Walkup’s theorem

If Hall’s criterion holds, there is a perfect matching of expected cost at most 6.

What about the cases of failure?

Walkup’s theorem

Randomly color the edges Red p Blue 1-p Take the 5 cheapest blue edges from each

vertex. If Hall’s criterion holds, this gives a matching of cost 6/(1-p)

Otherwise the red edges 1-1, 2-2 etc give a matching of cost n/p.

Walkup’s theorem

Total expected cost

Take p = 1/n for instance. For large n, the expected cost is < 6 + o(1) This completes the proof.

p

n

nO

p

5

1

1

6

Walkup’s theorem

Actually

but we return to this…

2

1

9

1

4

11)(

nAE n

Walkup’s theorem

Walkup’s theorem obviously carries over to the complete graph (for even n)

The method also works for the TSP, minimum spanning tree, and other related problems

Natural conjecture: E(cost) converges in probability to some constant.

Statistical physics

The typical edge in the optimum solution has cost of order 1/n, and the number of edges in a solution is of order n.

Analogous to spin systems of statistical physics

Disordered Systems

Spin glasses AuFe random alloy Fe atoms interact

Statistical physics

Each particle essentially interacts only with its close neighbors

Macroscopic observables (magnetic field) arise as sums of many small terms, and are essentially independent of individual particles

Statistical physics

Convergence in probability to a constant?

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, W. Krauth, 1980’s Limit of /12 for minimum matching on the

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

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 limit

It 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 edu

deufe

du

deufufuf

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.

nK

L

nL

K

K-L matching

K-L matching

Similarly, the K-L matching problem leads to the equations:

)](min[d

iiYKX )](min[

d

iiXLY

• has rate K and has rate L• min[K] stands for K:th smallest

Shown by Parisi (2006) that this system has an essentially unique solution

The ground state energy is given by

where x and y satisfy an explicit equation

For K = L = 2 (equivalent to the TSP), this equation is

K-L matching

0

ydx

12

12

1

eeyx yx

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

1

9

1

4

11)(

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 were announced on March 17, 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(

111,,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

eeyx yx

041548.2)(2

1

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 →∞,

A. Frieze proved that whp a 2-factor can be patched to a tour at small cost

...041548.2* LLp

n

Further exact formulas

nnn

n kn

kkC1

3

2

1

2

1

4 /11

4/12/15var

2

1)3(4)2(4

nO

n

LP-relaxation of matching in the complete graph Kn

12

1

9

1

4

11)(

2

2

nCE n

Future work

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

Establish more detailed cavity predictions Use proof method of Nair-Prabhakar-Sharma

in more general settings

Thank you!