A GENERALIZATION OF THE RANDOM ASSIGNMENT PROBLEM

arXiv:math/0006146v1 [math.CO] 20 Jun 2000

A GENERALIZATION OF THE RANDOM ASSIGNMENT PROBLEM ¨ SVANTE LINUSSON AND JOHAN WASTLUND Abstract. We give a conjecture for the expected value of the optimal kassignment in an m × n-matrix, where the entries are all exp(1)-distributed random variables or zeros. We prove this conjecture in the case there is a zerocost k − 1-assignment. Assuming our conjecture, we determine some limits, as k = m = n → ∞, of the expected cost of an optimal n-assignment in an n by n matrix with zeros in some region. If we take the region outside a quarter-circle inscribed in the square matrix, this limit is thus conjectured to be π 2 /24. We give a computer-generated verification of a conjecture of Parisi for k = m = n = 7 and of a conjecture of Coppersmith and Sorkin for k ≤ 5. We have used the same computer program to verify this conjecture also for k = 6.

1. Introduction Suppose we are given an m by n array of nonnegative real numbers. A kassignment is a set of k entries, no two of which are in the same row or the same column (we “assign” rows to columns, or vice versa). The cost of the assignment is the sum of the entries. A k-assignment is optimal if its cost is minimal among all k-assignments. We let Fk (m, n) denote the expected cost of the optimal k-assignment in an m by n array of independent exponentially distributed random variables with mean 1. Aldous [A92] has shown the existence of limn→∞ Fn (n, n) = c. According to a conjecture of M´ezard and Parisi [MP85], c = π 2 /6. Moreover, Parisi [P98] has conjectured that Fn (n, n) = 1 + 1/4 + 1/9 + · · · + 1/n2 . This conjecture has been verified for n ≤ 7. For n ≤ 6 this can be found in [AS99], and the n = 7 case follows from the calculations in the appendix to this paper. In [CS98], a more precise conjecture was formulated. Conjecture 1.1. Fk (m, n) =

X

i,j≥0 i+j n · (1 − x − y + δ).

If we let i = ⌊n · x⌋ and j = ⌊n · y⌋, we certainly have

(1 − x) · iα + (1 − y) · jα > n − i − j + 2δ1 · n,

for some δ1 slightly smaller than δ/2. Let α′ be a random set constructed by letting each row belong to α′ with probability x, and each column with probability y. Since with probability at least 1/4, α′ will contain at least i rows and j columns, in order to prove that pn (x, y) tends to 0, it will suffice to show that the probability that one can cover Zn by adding n − i − j − 1 rows and columns to α′ tends to 0. Let B(p) denote the random variable that is 1 with probability p and 0 with probability 1 − p. The number of rows in α \ α′ is a sum of iα independent B(1 − x)distributed random variables, and similarly, the number of columns in α \ α′ is a sum of jα independent B(1 − y)-variables. In order that the size of α \ α′ be at most n − i − j − 1, one of these sums must deviate by at least n · δ1 from its mean value. By standard estimates in probability theory, this is at most e−c·n , for some constant c depending only on δ1 . Hence the probability that one can obtain a covering of Zn by adding at most n − i − j − 1 rows and columns to α′ is at most T (n) · e−c·n , which, by assumption, tends to 0 as n → ∞. Corollary 6.3. If K is compact, and the number of minimal coverings of Zn grows slower than (1 + ǫ)n for every ǫ, as n → ∞, then Z dx dy , lim Fn = n→∞ (1 − x)(1 − y) D

where D is the region in which M1−x,1−y (K) < 1 − x − y.

¨ SVANTE LINUSSON AND JOHAN WASTLUND

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6.1. An example: Zeros outside a quarter of a circle. Let K be the region x2 + y 2 ≥ 1, 0 ≤ x, y ≤ 1. To find Ma,b (K), for given a and b, we have to find the point where the slope of the curve x2 + y 2 = 1 is −a/b. The part of the curve which is above this point should be covered with a horizontal strip, and the part which is to the right of this point should be covered with a vertical strip. This point is   a b √ . ,√ a2 + b 2 a2 + b 2 Hence     a b Ma,b (K) = a 1 − √ +b 1− √ , a2 + b 2 a2 + b 2 which simplifies to p a + b − a2 + b 2 . The number T (n) of minimal coverings of Zn is at most n + 1, since any minimal covering consists of, for some i, 0 ≤ i ≤ n, columns i, . . . , n, and all rows that contain some element of Zn not covered by the columns. The region D is given by the inequality M (K1−x,1−y ) < 1−x−y, which becomes (1 − x)2 + (1 − y)2 > 1.

In other words, D is the part of the unit square which is outside the circle of radius 1 centered in the point (1, 1). By Corollary 6.3, we have Z π2 dx dy = . lim Fn = n→∞ 24 D (1 − x)(1 − y) 6.2. Generalizing the example. Fix p > 1. Let K be the region given by xp + y p ≥ 1, 0 ≤ x, y ≤ 1. We wish to compute Ma,b (K). To find the point on the curve xp + y p = 1 where the slope is −a/b, we compute:

dy = −xp−1 (1 − xp )1/p−1 = −(x/y)p−1 . dx It is not difficult to see that the point is   1

 

We have 

 Ma,b (K) = a 1 − 

1

a p−1

a

p p−1

+b

a a

p p−1

p p−1

1 p−1

+b

p p−1

p p−1

1/p ,  

b p−1

a

p p−1



+b

 1/p  .

p p−1

  1/p  + b 1 − 

b a

p p−1

1 p−1

+b

p p−1

p p−1



 1/p  =

a b =a−  1/p + b −  p 1/p = p p p p−1 p−1 p−1 p−1 a a +b +b

 p  p−1 p p . (21) = a + b − a p−1 + b p−1

Hence the inequality M1−x,1−y (K) < 1 − x − y becomes   p−1 p p p (1 − x) p−1 + (1 − y) p−1 > 1.

THE RANDOM ASSIGNMENT PROBLEM

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Again the number of minimal coverings of Zn is at most n + 1. To find the limit of Fn , we now by Corollary 6.3 wish to compute the integral Z 1 Z 1 dx dy , xy u 1/u x=0 y=(1−x ) where u = p/(p − 1). We eliminate the inner integral, and obtain
Z Z 1 − log (1 − xu )1/u 1 1 log(1 − xu ) dx = − dx. x u 0 x 0 We make the substitution t = xu , which gives dx = uxdt u−1 . We get  Z 1 Z 1 1 −t − t2 /2 − t3 /3 − . . . 1 log(1 − t) − 2 dt = − 2 dt = u 0 t u 0 t   1 1 1 1 π2 = 2 1 + + + ... = 2 · . (22) u 4 9 u 6 Substituting back p/(p − 1) for u,

2 2  π 1 · . lim Fn = 1 − n→∞ p 6 7. Remarks

Looking at the triangles of bi,j ’s one discover certain patterns. Define the property acyclic recursively by first letting the empty set be acyclic. Second, let Z be an acyclic set that has an optimal covering α with only rows or only columns. Let (i, j) be any position not covered by α. Then we define Z ∪ (i, j) to also be acyclic.
Corollary 7.1. Assume Conjecture 1.2. If Z is acyclic, then b0,0 = (−1)|Z| k−1 |Z| . This corollary can be proved by first specializing (18) to b0,0 and then study the set of bad sets and find the proper bijections which proves it by induction on |Z|. The proof we have found is much in the spirit of the proof in Section 5 of consistency for bi,j , i + j = k − 1. There are some highly non-trivial details and we omit the proof. Given a set of zeros Z, let Zrow (i), be the number of zeros in row i and Zcol (j) the number of zeros in column j. Ordering the nonzero Zrow (i), 1 ≤ i ≤ m by size, we get an integer partition of |Z|, call it λrow (Z). Similarly we define λcol (Z). P bi,j Conjecture 7.2. Given k and Z, write Fk,Z (m, n) = i+j