RC 21133 (94490) 18 March 1998
Mathematics
IBM Research Report Constructive Bounds and Exact Expectations For the Random Assignment Problem Don Coppersmith, Gregory B. Sorkin IBM T.J. Watson Research Center P.O. Box 218 Yorktown Heights, New York 10598
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CONSTRUCTIVE BOUNDS AND EXACT EXPECTATIONS FOR THE RANDOM ASSIGNMENT PROBLEM DON COPPERSMITH AND GREGORY B. SORKIN
Abstract
The random assignment problem is to choose a minimum-cost perfect matching in a complete n n bipartite graph, whose edge weights are chosen randomly from some distribution such as the exponential distribution with mean 1. In this case it is known that the expectation does not grow unboundedly with n, but approaches some limiting value c? between 1.51 and 2. The limit is conjectured P tonbe 2 =6, while a recent conjecture has it that for nite n, the expected cost is i=1 1=i2 . This paper contains two principal results. First, by de ning and analyzing a constructive algorithm, we show that the limiting expectation is c? < 1:94. Second, we extend the nite-n conjecture to partial assignments on complete m n bipartite graphs, and prove it in some limited cases. 1. Introduction The linear assignment problem is that of nding a minimum-weight (or minimum-cost) matching in a complete bipartite graph with weighted edges. Alternatively the \left" and \right" vertices of the graph may be regarded as rows and columns of a matrix, whose entries are the weights of the corresponding edges; then Pn the minimum assignment problem is to nd the permutation minimizing i=1 ai;(i) | the selection of n entries, all from di erent rows and di erent columns, whose sum is as small as possible. The assignment problem has been considered since 1923, with publication dating back at least as far as 1946 [Eas46]. An n n instance of the assignment problem can be expressed as a linear program (LP) (see Figure 1), and therefore solved in polynomial time. To solve the assignment problem the LP variables must be 0 or 1, not fractional, but that is always true of the optimal \basic solutions" to this LP, and in fact for the more general \transportation problem" LP [Dan63]. There are also ecient algorithms for the minimum-weight matching problem, running in time O(n5=2). Random instances of the assignment problem have been considered since at least 1962 [Kur62]. Here, if an n n instance A has a minimal solution of cost A? , Date : 18 March 1998. Key words and phrases. Statistical physics; weighted matching; bipartite matching. 1
2
COPPERSMITH AND SORKIN
Primal: minimize
n n X X i=1 j =1
Dual: aij xij
subject to xij > 0 (i = 1 : : :n; j = 1 : : :n) X xij = 1 (i = 1 : : :n) j
X
i
maximize
n X i=1
ui +
n X j =1
vj
subject to aij ? ui ? vj > 0 (i = 1 : : :n; j = 1 : : :n)
xij = 1 (j = 1 : : :n) Figure 1. The (primal) linear program describing the assignment problem for a matrix A, and its dual LP. The variables xij in basic optimal solutions to the primal LP will have values of 0 or 1 (not fractional), xij = 1 indicating assignment of row i to column j. The dual variables ui are called row duals, and the vj column duals.
what is of interest is F(n) = EA? . Random instances are de ned by choosing independent, identically distributed (i.i.d.) edge weights. The uniform distribution over the interval (0; 1) is the most commonly used, but the exponential distribution with parameter 1 (and distribution function P(x) = exp(?x) and mean 1) is a better choice for several reasons that will become clear. For brevity, when we say a random variable (r.v.) is \exponential" or has exponential distribution, parameter 1 is presumed unless explicitly stated otherwise. It is known that, remarkably, as n ! 1 , E A? has a well-de ned, nite limit c? . Intuitively, the greater number of assignment edges needed for larger n is o set by the increased choice available, in particular the availability of some very cheap edges. The exact value of c? is not known. An upper bound of 3 was published in 1979 [Wal79], and a lower bound of 1 + 1=e was established in the same year [Laz79, Laz93]. The existence of a limit is shown in [Ald92], which also asserts that the uniform and exponential distributions give the same limiting expectation: intuitively, only very small values appear in the minimum-cost assignment, and near 0 both distributions have density 1, and are indistinguishable. It is conjectured that c? = 2 =6 [MP87, MP85]. 2. Outline In Section 3 we use an (unoriginal) analysis of a natural greedy algorithm to introduce some properties of the exponential distribution, and a conjecture of Parisi's on the expected assignment cost F(n) for a nite-size instance. In Section 5 we review the known upper bounds for c? . In Section 6 we review the known lower bounds; these also set the stage for our new upper bounds, in Sections 7{ 11. Completing the paper's structural symmetry, in Section 12 we introduce a new
RANDOM ASSIGNMENT
3
conjecture, generalizing Parisi's, for the expectation F(k; m; n) of the minimumcost k-assignment (k-edge matching) in a complete m n bipartite graph; the conjecture is proved in some special cases. This section can be read independently of the others.
Sections 3{6 review primary sources, and have substantial overlap with the excellent review in [Ste97, Chapter 4]. However, we take a new perspective on the problem and the role of the exponential distribution; we include recent results (notably [Oli92] and [Par98]); and we introduce new techniques that we will call upon in later sections. 3. Greedy algorithm It is impossible to think about the random assignment problem without considering the natural greedy algorithm:
Greedy Algorithm
Repeat n times in all: Choose the smallest element in matrix A. Exclude the row and column of the chosen element. To calculate the expected cost of the solution produced we capitalize on a property of the exponential distribution: Property 1. The 1minimum of a collection of n i.i.d. exponential r.v.'s with pa1 ). rameter (mean ) is an exponential r.v. with parameter n (mean n Relabeling rows and columns for convenience, take the rst element chosen by the algorithm to be a11 . As the minimum of all n2 elements, a11 is exponentially distributed with mean 1=n2 . The second element chosen, a22 , is the minimumof the (n ? 1) (n ? 1) elements in the other rows and columns; it is tempting to suppose that they are exponentially distributed and that E a22 = 1=(n ? 1)2 . Continuing this line of reasoning, we would conclude that E A? = 1=n2 +1=(n ? 1)2 + +1=12 , and in particular (exploiting a well-known identity) that limn!1 F(n) = 2 =6. However, there is a critical error in this analysis. After selection of a11 , the remaining elements of A are exponentially distributed conditioned upon being at least as large as a11 . A corrected analysis uses the memoryless property of the exponential distribution: Property 2. For an exponential r.v. X with parameter (density P(x) = 1 e?x ), PrfX > s + t j X > tg = PrfX > sg for s; t > 0 . Equivalently, conditional upon X > t, the distribution of X ? t is again exponential, with the same parameter.
4
COPPERSMITH AND SORKIN
In the greedy algorithm, revealing the smallest value in A to be a11 conditions the remaining values in A: they are a11 plus i.i.d. exponential r.v.'s with parameter 1. Thus E a22 = E a11 + 1=(n ? 1)2 = 1=n2 + 1=(n ? 1)2 . Continuing, E a11 = 1=n2 E a22 = 1=n2 + 1=(n ? 1)2 .. . E ann = 1=n2 + 1=(n ? 1)2 + + 1=12: Adding columnwise shows that the expected cost of the greedy solution is exactly 1=n + 1=(n ? 1) + + 1=1 = H(n), the n'th harmonic number. Since F(n) is bounded, and H(n) ! 1 as n ! 1 , greedy is a bad algorithm.1 4. Conjectured exact expected costs Strikingly, the incorrect analysis of the greedy algorithm had it that the soluP tion cost was ni=1 1=i2 , approaching 2 =6 as n ! 1 . This result matches the prediction of calculations based on the \replica method" of statistical physics: Conjecture 3 ([MP87, MP85]). Given an n n array of i.i.d., uniform (0; 1)
random variables, the minimum assignment has an expected cost which approaches 2 =6 as n ! 1 .
The replica method is known to give correct answers in many cases, but is not mathematically rigorous. For the assignment problem, in brief, it goes as follows. Given an instance A and an \inverse temperature" parameter , de ne the \partition function" Z( ) as a sum over all assignments , X Z( ) = e? cost() ?! e? A? ; !1
? ln Z = E lim !1 ?1 lim Z " ? 1 : = E lim !1 "!0 " When " is a positive integer rather than a limiting value, the last expression can be computed in closed form, and taken to 0. Substituting " = 0 into this expression (\analytic continuation") completes the calculation. To make it rigorous would require justifying the analytic continuation, and | worse | the exchange of the two limits. E A?
Simulations give supporting evidence to Conjecture 3. In 1969, building and solving pseudo-random systems of size n = 100, Donath estimated a limiting expected cost of about 1.6 [Don69]. Simulations of that size can now be done in 1 But note how the exponential distribution streamlines the analysis; in the uniform (0; 1) distribution, by contrast, the distribution of X conditioned upon X > t is uniform but not over (0; 1), the minimum of n uniform variables has a distribution that is not uniform, and the combination of the two is messier still.
RANDOM ASSIGNMENT
5
seconds on a laptop computer, and simulations for n = 10; 000 remain consistent with c? = 2 =6 [PR93]. The third and best reason to believe that c? = 2 =6 is provided by a remarkable new conjecture by Parisi, which (bizarrely) is also in agreement with the incorrect analysis of the greedy algorithm: Conjecture 4 ([Par98]). Given an n n array of i.i.d., exponentially distributed random variables, the minimum-cost assignment has an expected value of precisely Pn 2. 1=i i=1
Note that Conjecture 4 requires the exponential distribution; for uniform (0; 1) it would already be false at n = 1. However, since the exponential and uniform distributions are equivalent as n ! 1 , proof of Conjecture 4 would immediately validate Conjecture 3 as well. Because of the strong conjecture's elegance, and because we have proved it through n = 4 (see section 12), we believe it must certainly be true. 5. Upper bounds The rst upper bound, c? 6 3, was established in [Wal79]. The idea is to nd a perfect matching among the very cheapest edges of the bipartite graph. On two sets of n vertices, de ne a random 2-out bipartite graph by randomly choosing 2 out of the n possible edges on each vertex. The key fact is: Lemma 5 (Walkup). A random 2-out bipartite graph has a perfect matching with probability at least 1 ? 5=n . Theorem 6 (Walkup). c? 6 3. Proof. Lemma 5 cannot be applied directly to the graph de ned by the two cheapest edges on each vertex of the input graph A. In that graph, if row i has (i; j) as one of its two cheapest edges, edge (i; j) stands a good chance of being one of the two cheapest edges for column j as well: the selections are not independent.
The resolution (see Figure 2) is to generate the random weights for the graph A as aij = minfbij ; cij g ; if B and C have edge weights i.i.d. according to the exponential distribution with parameter 2 (mean 1=2), then A has edge weights i.i.d. according to the exponential distribution with parameter 1, as desired.2 De ne a random 2-out subgraph in A corresponding to the 2 cheapest edges out of each row vertex bi in B and the 2 cheapest edges into each column vertex cj in C. By Lemma 5, with high probability this random 2-out graph contains a perfect matching, which is an assignment for A. 2 Walkup worked with the uniform (0; 1) distribution, requiring the distributions for b and c ij ij to be carefully constructed to make aij = minfbij ;cij g uniform, and making it a bit more work to compute the expected rst- and second-smallest values among n such r.v.'s. The exponential is easier to work with and more elegant.
6
COPPERSMITH AND SORKIN rows
cols
b c
ij
2/n
ij 4/n
a
ij
Figure 2. At left, a random 2-out bigraph is de ned by having
each \row" vertex randomly choose two \columns" and vice-versa. At center, if each undirected edge weight aij is the smaller of two edge weights bij and cij , a random 2-out bigraph is generated by the two cheapest edges from each \row" in B, and from each \column" in C. At right, if edge weights of B and C are exponentially distributed with parameter 2, a \ rst" row-vertex edge (generated by B) has expected cost E aij 6 Ebij = 2=n, and the second has expected cost 2=n+2=(n ? 1) 4=n; similarly for edges generated by C; so the 2-out graph's edges have mean expectation . 3=n.
The selection of the matching edges in the 2-out subgraph is made without regard to cost, so the expected cost of the matching is just n times the cost of a typical edge in the 2-out graph. The cheapest of the n i.i.d. mean-2 exponential edges out of a row bi has expected cost 2=n (by Property 1), and the second cheapest has expectation 2=n + 2=(n ? 1) 4=n (by Properties 1 and 2). Since a corresponding edge in A has cost aij = minfbij ; cij g 6 bij , if a perfect matching =n = 3. exists, asymptotically, its cost satis es E[cost] 6 n 2=n+4 2 Walkup's proof is non-constructive in the sense that it builds a matching for an edge-weighted complete bipartite graph A only when given the random graphs B and C from which A was derived. Karp, Kan, and Vohra [KKV94] circumvent this problem by showing how to construct appropriate, independent matrices B and C from A: Toss a fair coin, and either let bij = aij and cij = aij + X (where X is a mean-2 exponential r.v.), or else swap the roles of bij and cij .3 By symmetry, bij and cij have identical distributions; that this distribution is mean-1 exponential and that they are independent can be shown without any calculation. The smaller of the two is equal to aij and therefore has mean-1 exponential distribution (as would the smaller of two mean-2 exponential r.v.'s), and the larger is the smaller plus a mean-2 exponential r.v.(as would be the larger of two mean-2 exponential r.v.'s); thus the joint distribution of (aij ; smaller; larger) is just as for the process described in the proof of Theorem 6, leading to the same conclusion for the joint distribution of (aij ; bij ; cij ).4 3 This probabilistic construction is a tiny part of [KKV94], whose primary contribution is an ecient heuristic for nding a perfect matching in the 2-out graph; it has expected running time O(n), and O(n ln n) in the worst case. 4 Again, the presentation in [KKV94] is based on the uniform distribution rather than the exponential. Validation of the distributions and independence of bij and cij is described as calculational, and not explicitly presented | the exponential simpli es matters.
RANDOM ASSIGNMENT
7
The upper bound was reduced by Karp [Kar84, Kar87] a decade ago. Theorem 7 (Karp). c? 6 2. The proof is non-constructive, and is a subtle application of LP duality. The theorem was generalized to other linear programs with randomized cost functions in [DFM86]. An attempt to explain the intuition behind the more general theorem, and indeed a good explication of the random assignment problem generally, is given by Steele [Ste97]. In Sections 7{11, we shall present an assignment algorithm whose solution's expected cost is less than 2. 6. Lower bounds Feasible (not necessarily optimal) solutions to an assignment problem's dual LP (see Figure 1) underlie all the lower bounds on the expected assignment cost. The key is the transformation of an instance A: Lemma 8. For real nP-vectors uPand v , if A ?? u1T ? 1vT = A0 (that is, aij ? ui ? n n 0 ? vj = aij ), then A = i=1 ui + j =1 vj + A0 . Proof. Since any assignment selects preciselyPone value from each row and each P
column, cost(A; ) = cost(A0 ; ) + ni=1 ui + nj=1 vj . Thus the minimum-cost P assignments for A and A0 are achieved by the same , and A? = A0 ? + ni=1 ui + P n v . j =1 j Corollary 9.PIf A ? u1T ? 1vT = A0 , and A0 is elementwise non-negative, then P ? A > ui + vj . Lemma 10. EA? > 1. Proof. Let u be the vector of row minima, and v = 0 . By Property 1, E ui? = ? 0
1=n. Applying Lemma 8 and Corollary 9, and taking expectations, EA = E A + P E ui > n 1=n = 1. Theorem 11 (Lazarus). c? > 1 + 1=e. Proof. As above, subtract the row minima ui in A to give A0 and conclude that
= 1 + E A0 ? . By Property 2, subtracting ui from values aij other than the row-i minimum produces a collection of i.i.d. exponential r.v.'s a0ij = aij ? ui .5 Thus A0 is (like A) an i.i.d. exponential matrix, except that in each row it has a single zero, in a random column.
EA?
5 This argument, based on the exponential distribution, is easier than Lazarus's for the uniform distribution. The memoryless property of the exponential means that when row and/or column minima are subtracted from an i.i.d. exponential matrix, the result is again i.i.d. exponential (outside of the zeros); this \self-reducibility" is most useful. (The matrix A , and its \residue" after subtraction of the row and column minima, are far from independent of one another, but that is not a concern.)
8
COPPERSMITH AND SORKIN
In the large-n limit, the probability that a column contains no zeros is 1=e. Subtract the column minima vj in A0 to give A00 . If column j contained a zero, vj = 0; otherwise (by Property 1) vj has exponential distribution with mean 1=n. P So in the large-n limit, E nj=1 vj = n 1=e 1=n = 1=e, and by Lemma 8, X X E A? = E A00? + E u i + E vj (1) > 1 + 1=e: Improved lower bounds from Olin [Oli92] take o from inequality (1) and replace > 0 with something stronger. In A00 , many rows still contain only one zero, not having acquired more zeros when column minima were subtracted. When k such rows | a k-group | all have their zeros in the same column, they con ict: for at least k ? 1 of them, the element selected for the minimum-cost assignment must come from a di erent column. Estimation of the fraction of rows forming a k-group, and of the cost of selecting nonzero elements for at least k??1 1 rows in each P ?e ) (k ? H(k)), group, shows that in the large-n limit, EA00 ? > 1 k=1 Poisk (e ? 6 giving c > 1:471. A similar approach is taken by Goemans and Kodialam [GK93], P1 1 Pi?1 ? achieving a bound of c > 3 ? e + i=1 ii! j =0 (i ? j) Poisj (i=e) > 1:441. Olin takes her calculations a step further, deriving the current best bound of about 1.51. That she does it all with the uniform distribution rather than the exponential is an unfortunate but impressive tour de force.
EA00 ?
7. Outline of Sections 8{11 There is at least the impression that the stream of lower bounds for c? | 1.368, 1.441, 1.471, 1.51 | is never-ending; that the dual constructions underlying them could be improved inde nitely towards 2 =6 1:645. By contrast, the upper bound has rested at exactly 2 for over a decade, and is purely non-constructive. In the following sections, we derive a smaller upper bound by de ning and analyzing an assignment algorithm which, applied to a large i.i.d. exponential matrix A, produces an assignment with expected cost E A? 6 1:94. There are three key points. 1. We begin with a partial assignment of size about :81n implicit in Lazarus's lower-bound construction. The lower-bound construction is meant to \give away" as little as possible, which is why it also makes a good starting point for an algorithmic solution and an upper bound. 2. To complete even a partial assignment of size n ? 1 to a complete assignment in the naive manner would increase the assignment cost by 1, which when we are gunning for 2=6 is intolerably huge. However, we show how an (n ? 1)assignment can be completed to an n-assignment quite simply, with additional
We have simpli ed Olin's notation (and that of Goemans and Kodialam), replacing Pki=1 with H (k) (the k 'th harmonic number), and replacing e? k =k! with Poisk () (the density of the Poisson distribution with parameter at k ). In fact the Poisson process has a very natural role (see Section 8), but perhaps it was masked in the uniform-distribution framework. 6
RANDOM ASSIGNMENT .49
.36
9 .15
0 0 0 .49
0 0 0 0 0
.16
0 0 x 0 0 0 0 0
.35
0 0 0
Figure 3. The zero structure of the matrix resulting from Algo-
rithm 0: subtraction of row and column minima, and replacement of selected zeros (such as in the spot marked with an \x") by fresh random variables.
cost o(1). Dogged extension of the technique allows us to complete an initial assignment of size about :81n to an n-assignment with additional cost of only about 0:56 altogether. 3. In both the initial construction and the successive augmentations, the \working" matrix will consist entirely of zeros and i.i.d. exponential values. 8. Initial assignment Begin with an i.i.d. exponential matrix A, and | following Lazarus's lead | form a \reduced" matrix A0 according to the following algorithm:
Algorithm 0 Input: An n n matrix A of i.i.d. exponential elements. For i = 1 : : :n: Subtract from row i its minimum value ui . For j = 1 : : :n: Subtract from column j its minimum value vj . For i = 1 : : :n: For j = 1 : : :n: If element (i; j) is zero, has other zeros in its row and in its column, and is not the last zero remaining in its column, replace it with an exponential r.v.. A typical reduced matrix is depicted in Figure 3, and is described by the following theorem:
10
COPPERSMITH AND SORKIN
Theorem 12. For n asymptotically large, Algorithm 0 reduces an n n matrix A of i.i.d. exponential r.v.'s to a matrix A0 in which
there is at least one zero in each row and in each column; a zero may share either its row or its column with other zeros, but not both; the collection of nonzero elements are i.i.d. exponential random variables; and, with probability 1 ? o(1) ,
the fraction of columns having k > 1 zeros is asymptotically equal to
col(k) = Poisk (e?e?1 ); the fraction of rows having k > 1 zeros is asymptotically equal to row(k) = (1 ? p?) Poisk (1=e) + (p? ) Poisk?1(1=e); ?1
where p? = (e?e?e ? e?1)=(1 ? e?e?1 ) ; and the fraction of zeros unique in their row and column is asymptotically equal to
row(1) = col(1)
8 < ?1 = e?e?1 :e?1 + e?e?e
?1 e?1 e?1 ? e?e?e + 1 ? e?e?1
9 = ;
:4914:
Proof. The rst two assertions hold by construction: A zero is generated in every
row and every column, and zeros are removed when they share their column and their row with others. The third assertion follows from Property 1.
When row minima are subtracted, the number of zeros contained in any column j is the sum of n Bernoulli(1=n) random variables; this has binomial distribution B(n; 1=n), which (as n ! 1 ) converges in distribution to Pois(1). Immediately, the expected number of columns having j zeros is n Poisj (1). The actual number is tightly concentrated about this expectation, by Azuma's inequality [McD89]: the location of the zero in any row changes the number Y (j) of columns with j zeros by at most 2, so Pr[ jY (j) ? E Y (j)j > t] 6 2 exp(?2t2 =4n). Thus almost surely, Y (j) = Poisj (1) (n + o(n)) (for example o(n) = n2=3 will do). In the remainder of this proof, all statements will be \almost surely" and \almost exactly" by the same logic; we shall omit the repetitive phrasing. From the above, n Pois0 (1) = n=e columns contain no row-induced zeros, so subtracting column minima introduces n=e new zeros. They fall in random rows, with a density of 1=e new zeros per row, so the number of column-induced zeros in each row is distributed as Pois(1=e). Consider a column containing j row-induced zeros. If the rst of their j rows acquired k > 0 additional column-induced zeros | which occurs with net probability 1 ? Pois0(1=e) = 1 ? e?e?1 | then the row's original zero is replaced, leaving the row with only the k new zeros. The same is true of all the rst j ? 1 zero-bearing
RANDOM ASSIGNMENT
11
rows of this column. But if the rst j ? ?11 rows do all acquire one or more additional zeros | which occurs w.p.(1 ? e?e )j ?1 | then the j'th row gets to keep its original zero plus any new column-induced zeros. Thus the number of rows with Pn Pn ?1 j ?1 ? e k > 1 zeros is n j =1 Poisj (1)(1 ? e ) Poisk?1(1=e) + n j =1 Poisj (1)(j ? (1 ? e?e?1 )j ?1) Poisk (1=e). Letting 1
X ?1 p? = Poisj (1)(1 ? e?e )j ?1 j =1 ?1 = (e?e?e ? e?1 )=(1 ? e?e?1 )
:4308;
the fraction of rows having k > 1 zeros is row(k) = p? Poisk?1(1=e) + (1 ? p? ) Poisk (1=e): Note that this does not properly count the number of rows with just 1 zero because rows keeping only their original zero are not counted. The number of columns with k > 1 zeros is more easily counted. A column only acquires multiple zeros in the row phase; a zero is removed if any of the n=e column zeros falls into the same row and if the zero is not the last remaining in its column. Forgetting this exceptional \and " only results in miscounting columns with no or one zeros. Then the probability that a zero is allowed to remain,?1is the probability that no column zero falls into its row, which is Pois0 (1=e) = e?e . Thus a column may be modeled as having n slots, into each of which a zero is placed w.p.1=n,?1and allowed to remain w.p.e?e?1 ; this is equivalent to ?just placing zeros w.p.e?e =n, 1 ? e which for n large is the Poisson process Pois(e ). So the fraction of columns containing k > 1 zeros is col(k) = Poisk (e?e?1 ):
The only thing we have not yet counted is rows which neither have 2 or more zeros, nor are associated with a column having 2 or more zeros. With = e?e?1 ,
12
COPPERSMITH AND SORKIN
these are counted by row(1) = 1 ?
1 X
k=2
= 1? "
k col(k) ?
1 X
k=2
1 X k=2
row(k)
(2)
e? k?1=(k ? 1)! 1
1
#
X X ? p? Poisk (e?1 ) + (1 ? p?) Poisk (e?1 ) k=2 k=1 h i ?1 ?1 ?1 ? ? ? e = 1 ? (1 ? e ) ? p (1 ? e ) + (1 ? p? )(1 ? e?e ? e?e e?1) 8
?1 < ?e?1 = e?e e?1 + e?e :
:4914
?1 e?1 e?1 ? e?e?e + 1 ? e?e?1
9 = ;
Such rows are in natural association with similar columns | those neither having 2 or more zeros, nor associated with a row having 2 or more zeros. It is therefore a check on the veracity of our calculations that row(1) = col(1), the latter de ned P1 P by col(1) = 1 ? 1 k=2 k row(k) ? k=2 col(k). Numerical values for row(k) (the fraction of rows having zero-cost edges to k columns), and col(k) (the fraction of columns for which k rows are in competition), are: k=1 2 3 4 5 row(k) .4914 .1364 .0234 .0028 .0002 col(k) .4914 .1199 .0277 .0048 .0007
Corollary 13. For asymptotically large graphs, the size of a maximum assignment contained in the bipartite graph B1 formed by edges which have minimum cost?1for either?their row or their column is almost surely almost exactly n times 2 ? e?e ? 1 ? e (approximately :8073n ). e?e Proof. Given a matrix A, subtract o row and column minima to form A00 ; B1 is
the graph of zero-cost edges of A00 . The reduced matrix A0 output by Algorithm 0 is the same as A00 , except that some of its zeros are removed. If a zero in row i was removed, there was at least one other zero in row i, in a column j of its own. Since column j cannot be assigned to any other row, there is no harm in assigning row i to column j, at which point the replaced zero in row i is of no use. Thus the size of a maximum zero-cost assignment in A0 is the same as the size of a maximum zero-cost assignment in A00 , which is the size of a maximum assignment in B1 . The size of a maximum zero-cost assignment in A0 is: the number of rows with a unique zero lying in a column also having a unique zero (the same zero), plus the number of rows with several zero-column options (choose any one), plus the number of columns with several zero-row options (choose any one). By Lemma 12,
RANDOM ASSIGNMENT
13
0
0
s cheapest
a in
0
0
0
0
.
.
.
.
.
.
0
0 a nn
a ni cheapest
Figure 4. At left, the naive completion of an (n ? 1)-assignment
adds expected cost Eann = 1. At center, adding p the cheapest pair ain + ani and deleting a adds expected cost O( 1=n). At right, ii p with s = b 2nc , nding the s smallest elements in column n (depicted without loss of generality as elements 1 : : :s), then choosing the smallest corresponding element |^ of row n, gives a pair with p E (a|^n + an|^) 2=n. P P1 this is almost surely almost exactly n times row(1) + 1 j =2 row(j) + j =2 col(j). Reduction to closed form parallels equation (2) and following.
Pittel and Weishaar [PW98] also compute the size of a maximum assignment in B1 , as a warm-up towards analyzing the size of a maximum assignment in the graph B2 formed of the rst- and second-cheapest edges out of each row column, and similarly for B3 . They use di erent methods, and end with a de nite integral whose numerical value agrees with our closed-form expression. Corollary 13 means that at a cost of 1+1=e (in row and column dual variables), Lazarus's construction produces an assignment of size about :81n consisting entirely of zero-weight edges, with i.i.d. exponential edge weights elsewhere. How cheaply can this assignment be completed? 9. Completing an (n ? 1) -assignment It is illustrative rst to consider a simpler question. Suppose an n n matrix A has n ? 1 zeros along its diagonal (a zero-cost assignment of size n ? 1), and other entries are i.i.d. exponential. How cheaply can this assignment be completed? The simplest way of completing it is of course to assign row n to column n. This adds 1 to the expected cost, since E ann = 1; as our goal is to nd an assignment of cost less than 2 (ideally, 2=6), an additional cost of 1 is far too much. A better way to complete the assignment (see Figure 4) is to remove from the matching some edge aii , and add the two edges ain and ani ; this is augmentation by the alternating path ani , aii , ain . There are n ? 1 ways of doing this (choosing
14
COPPERSMITH AND SORKIN
i 2 1;p: : : ; n ? 1), and the cheapest will have expected cost E mini (ain + ani ) = O(1= n). The same principle can be applied more generally, and implemented a little di erently. With reference again to Figure 4, we de ne an algorithm as follows:
Algorithm 1 Input: An n n matrix A whose elements are i.i.d. exponential except on an all-zero m-assignment, faii = 0gmi=1 . p Let s = b 2mc In column m + 1, choose the cheapest s elements ai;m+1 , for i 2 1 : : :m. Let S be the set of row indices i chosen. Let |^ = argmini2S fam+1;i g be the column index of the cheapest corresponding element in row m + 1. Let u = 0 , except for um+1 = am+1;|^. Let v = 0 , except for vm+1 = a|^;m+1 . Let A0 = A ? u1T ? 1vT . \Recondition" A0 : Replace any negative entries of A0 with i.i.d. exponential r.v.'s. Replace a|0^^| (which was zero) with an exponential r.v.
Algorithm 1 augments an m-assignment with a 3-edge alternating path: adding the edge am+1;|^, removing the edge a|^;|^, and adding the edge a|^;m+1 . Lemma 14. Given an n n matrix A with elements aii = 0 for i = 1 : : :m, and i.i.d. exponential random elsewhere. Then P Algorithmp1 produces u , v , and A0 P 0 T T such that: A 6 A + u1 + 1v ; E i ui + E j vj 6 2=m + O(1=m) ; and the elements of A0 are i.i.d. exponential except on an (m + 1) -assignment all of whose edge weights are zero.
Proof. The rst claim, that A 6 A0 + u1T + 1vT , is immediate by construction. For the second claim, as the smallest of s elements, E am+1;|^ = 1=s. Furthermore, the smallest element occurs in random position, i.e., |^ 2 S is chosen uniformly at random. For i 2 S, the elements ai;m+1 have expectations 1=m; : : : ; s=m, p so s +1 among themPthe (random) element a|^;m+1phas expectation 2m . With s = b 2mc , P E ui + E vj = Eam+1;|^ + E a|^;m+1 = 2=m + O(1=m).
A0 has an all-zero (m + 1)-matching, consisting of the edges a|^;m+1 and am+1;|^, along with the original assignment edges except for a|^|^. To see that all entries of A0 are still i.i.d. exponential, note that only A's m+1'st row and column have been changed, or even looked at. When the chosen a|^;m+1 is subtracted from column m+1, entries ai;m+1 that were larger than a|^;m+1 become i.i.d. exponential r.v.'s, by the memoryless property of the exponential. Entries that were originally smaller than a|^;m+1 become negative, and are replaced with fresh, independent exponential random variables. Similar logic applies to row m+1. The
RANDOM ASSIGNMENT
15
now-extraneous zero in a|0^;|^ is also replaced with a fresh exponential, rendering A0 i.i.d. exponential outside of the (m + 1)-assignment, as claimed. Corollary 15. An nn matrix whose edge weights are i.i.d. exponential, except on an m -assignment where all edge p weights are zero, has a complete, n -assignment of expected cost E A? 6 (n ? m)( 2=m + O(1=m)) . In particular, if n ? m = o(n1=2) , EA? = o(1) . Proof. Let A(m) = A, and for k = m : : :n ? 1, apply Algorithm 1 to A(k), gener-
ating output matrix A(k+1) in which the k'th row is also p assigned with a zero-cost edge. By Lemmas 8 and 14, E A(k)? 6 E A(k + 1)? + 2=k + O(1=k). Summing over k, p E A? = (n ? m) ( 2=m + O(1=m)): 10. An upper bound less than 3 For a cost O(1), Algorithm 1 can only expand an assignment by o(n1=2) edges. A more careful construction will allow us to take this to (n) edges, and thus to complete the :81n matching of Section 8 (the Lazarus construction) at nite cost. As before we will be augmenting the assignment via a 3-edge alternating path, but where before the path starting at row m + 1 was constrained to end at column m + 1, now it will be allowed to end at any column from m + 1 to n. Before introducing the algorithm, let us establish a lemma. Lemma 16. Let v be an n-vector of i.i.d. exponential elements hidden from a
player who performs a sequence of operations, each operation consisting of (1) masking an element of v that has not yet been masked or revealed, (2) revealing the value and location of the smallest non-masked element of v , or (3) stopping. Let v0 be the vector obtained by subtracting the last revealed value r^ = v^{ from v , and replacing all negative entries with i.i.d. exponential r.v.'s. Then v0 has a 0 in location ^{ (not necessarily random), and is elsewhere i.i.d. exponential (though not independent of v ). Proof. In fact, outside of the 0 element, the values of v0 are independent even when conditioned upon the sequence of masked locations, revealed locations, and revealed
values. De ne ri to be the value of the last element revealed before element vi was masked, ri = 0 if vi was masked before any element was revealed, and ri = r^ (the last revealed value) if vi was never masked. Clearly v^{0 = v^{ ? r^ = 0. All other revealed values are vi < r^, so vi ? r^ < 0, and the vi0 's are fresh i.i.d. exponential variables. Now consider the restriction of all vectors to the set of non-revealed locations. At the end of the process, their values are i.i.d. exponential variables conditioned by vi > ri , so vi ? ri is an i.i.d. exponential vector independent of r; in turn positive entries in vi ? r^ = (vi ? ri) ? (^r ? ri ) form i.i.d. exponential components of v0 , while negative entries are replaced by fresh i.i.d. exponential entries.
16
COPPERSMITH AND SORKIN
For m=n > 1=2, the following algorithm augments an m-assignment by one.
Algorithm 2
Input: matrix A; m-assignment of zero-weight edges, with m > n=2; positive integer s 6 m=(n ? m). For each unmatched column k = m + 1 : : :n in turn: Choose the s cheapest elements aik among rows i 2 1 : : :m. For each row i chosen: De ne parent(i) = k. Eliminate row i from future choices. Let S be the set of row indices i chosen. ( jS j = s(n ? m).) Let |^ = argminj 2S fam+1;j g , the index of the cheapest of the s(n ? m) elements in row m + 1 and columns S. Add edges am+1;|^ and a|^;parent(^|) to the assignment, and remove edge a|^;|^. Let u = 0 , except for um+1 = am+1;|^. Let v = 0 , except for vparent(^|) = a|^;parent(^|) . Let A0 = A ? u1T ? 1vT . Replace any negative elements a0ij with fresh exponential r.v.'s. Replace element a|0^;|^ with a fresh exponential r.v. Lemma 17. Given an n n matrix A with elements aii = 0 for i = 1 : : :m, and i.i.d. exponential random elsewhere, with n ? m 0 . Then Algorithm 2 produces A0 , u , and v such that: A 6 A0 + u1T + 1vT ; X X m s + 1 ln E ui + E vj 6 s(n ?1 m) + 2s(n ? m) m ? s(n ? m) ; i
j
and the elements of A0 are i.i.d. exponential except on an (m + 1) -assignment all of whose edge weights are zero. Proof. The rst claim, that A 6 A0 + u1T + 1vT , is immediate by construction.
For the second claim,
X
X
ui + E vj = am+1;|^ + a|^;parent(^|) : As the smallest of s(n ? m) elements, E am+1;|^ = 1=s(n ? m). Furthermore, the smallest element occurs in random position: |^ 2 S is chosen uniformly at random, and the expectation of the chosen column element a|^;parent(^|) is merely the expected mean of all the elements aj;parent(j ) generated by the algorithm. The rst s of these elements chosen, from amongst rows 1 : : :m, have expectations 1=m; : : : ; s=m; the next s chosen (from amongst s fewer rows) have expectations 1=(m ? s); : : : ; s=(m ? s); and so forth, the last column's elements having expectations 1=(m ? (n ? m ? 1)s); : : : ; s=(m ? (n ? m ? 1)s). The mean is n?X m?1 1 1 s + 1 E a|^;parent(^|) = 2 n ? m k=0 m ? ks Z n?m 1 dx: s + 1 6 2(n ? m) m ? xs 0 E
RANDOM ASSIGNMENT
17
Thus
E
X
ui + E
X
m s + 1 ln vj 6 s(n ?1 m) + 2s(n ? m) m ? s(n ? m) :
Finally, we must show that apart from the m + 1 zeros, the elements of A0 are i.i.d. exponential r.v.'s. Before am+1;|^ is chosen, row m + 1 was unrevealed. Its revelation can be simulated by: choosing |^ 2 S uniformly at random; setting am+1;j equal to an exponential random variable with mean 1=jS j ; for the remaining j 2 S ? |^, setting am+1;j = am+1;|^ +xj , where the xj are independent exponentials with mean 1; and lling in the other elements as am+1;j = xj , again with xj i.i.d. exponential r.v.'s. Subtracting am+1;|^ from each am+1;j , and replacing negative results with fresh exponentials, results in an i.i.d. exponential n-vector. The above simulation begins by choosing |^, which may be done by specifying in advance, at random, the time | from 1 to s(n ? m) | at which |^ was added to S; this gives us the ability to stop generating the set S once |^ is produced, say while choosing from column k^ . It is crucial that all columns were equally represented, with s choices each, for it is this that ensures that k^ is chosen uniformly at random from m+1; m+2; : : : ; n, and in turn that the columns k 6= k^ not chosen retain their unbiased i.i.d. exponential distribution. (Otherwise, for example, high-probability selection of a column k^ with particularly small entries would bias the untouched columns to have large entries.) In column k^ , rows i chosen for columns k < k^ are masked, then minimal values aj;k^ are chosen from the remaining rows, ending in the generation of a|^;k^ . This mask-and-minimum process obeys the rules of Lemma 16, thus subtraction of a|^;k^ from the column (and replacement of negative entries with zeros) produces a vector whose entries are exponential r.v.'s independent of one another, and of the rest of the matrix.
Corollary 18. An n n i.i.d. exponential matrix A has an assignment whose expected cost is less than 2.92.
Proof. Given a matrix which initially has an all-zero assignment of size m, apply Algorithm 2 repeatedly (n ? m times), producing vectors u(m + 1) and v(m + 1), u(m + 2) and v(m + 2), : : : , up to u(n) and v(n). By Lemma 17, and choosing s at each iteration to minimize the cost bound added, the total weight of these
18
COPPERSMITH AND SORKIN
vectors satis es n X m0 =m+1
E
X
ui (m0 ) +
n X
Z
X
vj (m0 )
m0 s + 1 ln + min 6 + s(n ? m0 ) 2(n ? m0 )s m0 ? s(n ? m0 ) m0 =m+1 s2 Z n 0 1 m s + 1 0 smin ? m0 ) + 2(n ? m0 )s ln m0 ? s(n ? m0 ) dm m 2 + s(n Z 0 1 + s + 1 ln 1 ? x min = 1 ? x ? sx d(?xn): 1?m=n s2 + xns 2xns Z 1?m=n 1 + s + 1 ln 1 ? x = min+ xs 2xs 1 ? x ? sx dx: s2 x=0
Z
1
Z Z
?1
?e?1 ? e?e?e :81, numerical integration yields P b1 =0 2 ? Pe PnFor m=n = 0 0 m0 =m+1 E ( ui(m ) + vj (m )) 1:51. That is, a matrix A which is i.i.d. exponential except for an all-zero assignment of size (1 + o(1))b1n, has assignment cost E A0 ? 6 1:51 + o(1). Applying Algorithm 0 to an i.i.d. exponential matrix A produces a matrix which includes an all-zero assignment of size (1 + o(1))b1n, some additional zeros, and i.i.d. exponential r.v.'s. Replacing its non-assignment zeros ?with new exponential r.v.'s produces a matrix A0 as required, and with E A? 6 EA0 + 1 + 1=e. Thus for a random i.i.d. exponential matrix A, EA? 6 1 + 1=e + 1:51 = 2:92.
To our knowledge, this is the rst proof of c? < 3 that is based on algorithmically constructing an assignment of such cost. 11. An upper bound less than 2 To reduce the upper bound further we will use similar methods, but augmenting an m-assignment via an alternating path of more than 3 edges, typically from row m+1 to one of the unassigned, or \free", columns. Roughly speaking, for each free column j we nd an assigned row i with small edge weight aij ; then we know that we can reassign row i to column j at low cost, i.e., adding (i; j) to the assignment and deleting (i; i) from it would not increase the assignment cost too much. Thus we can treat column i as being \free" too, expanding the pool of free columns. Once the pool is suciently large, we nd the free column to which row m + 1 can be assigned most cheaply, and carry out the implied chain of reassignments. (In the above example, if row m+1 got assigned to column i, we would use the alternating path m + 1; i; j.)
RANDOM ASSIGNMENT
19
Algorithm 3A (k -doubling algorithm) Input: matrix A; m-assignment of zero-weight edges; a list C0 of \free" columns; a set R0 of \working" rows. Set R = R0 . For ` = 1 : : :k: Set C` = C`?1 . For each j 2 C`?1 , in order: Set ^{ = argmini2R fai;j g . De ne parent(^{) = j. Set R = R n ^{ . Set C` = C` ;^{ . Let u = v = 0 . Set |^ = argminj 2Ck fam+1;j g . Add the edge (m + 1; |^) to the assignment, and set um+1 = am+1;|^. Set j = |^. Until j 2 C0 , repeat: Delete edge (j; j) from the assignment. Add the edge (j; parent(j)) to the assignment and set vparent(j ) = aj;parent(j ) . Set A0 = A ? u1T ? 1vT , and replace any negative entries of A0 with fresh i.i.d. exponential r.v.'s. Lemma 19. Given an n n matrix A with an assignment of zero-weight edges, aii = 0 for i = 1 : : :m ; a set of working rows R [m] ; and a set of free columns C [n] n [m] ; and let the entries ai;j for i 2 R and i 6= j be i.i.d. exponential. Then, with r = jR0j and c = jC0j , Algorithm 3A produces A0 , u , and v in which: A 6 A0 + u1T + 1vT ; the elements of A0 are i.i.d. exponential except on an (m+1) assignment all of whose edge weights are zero; E E
X
i X
j
ui 6 1c 2?k ; and
k X vj 6 ? 1c 2?` ln(1 ? (2` ? 1)c=r): `=1
(3) (4)
Proof. Paralleling the proof of Lemma 17, the rst assertion is immediate by con-
struction, as is the existence of an all-zero (m + 1)-assignment in A0 .
Selection of |^ may be simulated by choosing, in advance and uniformly at random, an index t between 1 and 2k c, and letting |^ be the t'th element of the list Ck . Values in the row m+1 can be simulated by choosing am+1;|^ from the exponential distribution with mean 1=jCkj = 1c 2?k , and, for j 6= |^, setting am+1;j = am+1;|^+Xj for j 2 Ck , and am+1;j = Xj otherwise, where X is a vector of i.i.d. exponential-1 r.v.'s. Subtraction of am+1;|^ from this row yields i.i.d. exponential entries, a single zero, and negative entries replaced by i.i.d. exponentials. On the vertex set consisting of all columns, Algorithm 3A generates a forest of directed hypercubes, each hypercube rooted at a column in C0 , with directed
20
COPPERSMITH AND SORKIN
15
16
13
14 7
8
5
6 11
12
9
10 3 1
4 2
Figure 5. Illustration showing the result after 3-doubling, start-
ing with two unmatched columns, 1 and 2. Edges % produce 1st-generation children, " 2nd-generation, and - 3rd-generation.
edges running from parent(j) to j. (See Figure 5.) The index t xed in advance corresponds to a node in the doubling forest; excluding the node t itself, the path from t back to its root is the set of columns reassigned by Algorithm 3A. The proof that the residual matrix A0 is i.i.d. exponential (outside of its zeros) will be by induction on t: from t to t0 , where parent(t0 ) = t in the doubling forest. Referring to Figure 5, the induction might take us from the column j(t) corresponding to t = 3, to the column j(t0 ) corresponding to its second child, t0 = 11. We assert that at the time it was selected, no entries in column j(t) had yet been revealed. This is certainly true for the rst c columns (depth 1). When column j(t) is rst selected, certain rows (those assigned to columns previously selected) are already hidden. Then, within column j(t), a sequence of operations is performed, each consisting of revealing the minimumnon-hidden element, or hiding an element. If t0 is the `'th child of t, the sequence of operations is halted after the `'th \minimum" operation. The minimumlies in row j(t0 ) (by de nition) and (since each column has only one parent), no elements of the corresponding column j(t0 ) have been revealed (proving that part of the inductive hypothesis). Meanwhile, the sequence of mask-and-minimum operations performed on column j(t) is in accordance with Lemma 16; therefore when the element aj (t0);j (t) is subtracted from the column, and negative entries are replaced with fresh exponentials, the result is a vector of exponential r.v.'s independent of one another and of the rest of the matrix A0 . It only remains to estimate the values of the selected elements. For u , only um+1 is nonzero; chosen as the minimum of jCk j = 2k c entries in A, its expectation is Eum+1 = 1=(2k c): For v , the nonzero entries are: vparent(^|) = a|^;parent(^|) ; vparent(parent(^|)) = aparent(^|);parent(parent(^|)) ; and so forth. If we associate cost aj;parent(j ) with each
RANDOM ASSIGNMENT
21
7 1/(r−6)
6 1/(r−5)
3
2
5
1/(r−2) 1/(r−2)+ 1/(r−4)
4 1/r+1/(r−1) +1/(r−3)
1/r + 1/(r−1)
1 root
1/r
Figure 6. A 3-doubling forest (k = 3) with a single tree (c = 1). Edges are generated in the order labeled by their terminal vertices; its expected cost is shown along each edge. Note for example that the cost 1=r of edge 1 contributes to the costs of the paths back from its descendants 3, 5, and 7, but also (through biasing the distribution on later edges from the root) to the cost of edges 2 and 4, and consequently to the cost of the path back from 6 through 2. In all, edge 1 contributes to the path cost for all 23 ? 1 = 7 nodes excepting the root itself.
directed edge from parent(j) to j in the doubling forest, the expected total of elements of v is the expected cost of a path from a random node of the forest back to its root. The cost of each edge in the forest may be expressed as the cost of its preceding sibling edge, plus a \new" cost distributed exponentially depending on the number of working rows from which the edge is chosen. For example, in Figure 6, the expected cost of the 4th edge is 1=r + 1=(r ? 1) (coming from its siblings), plus a \new" term of 1=(r ? 3). Because of this, not only does each edge's \new" cost contribute to the path costs from all nodes \downstream" of it, but also to the cost of all its later sibling edges, and the paths from all their downstream nodes. In all, a new edge added in generation ` of the doubling contributes to the costs of paths back from 2k?`+1 ? 1 nodes. The i'th edge in the rst tree of the forest has expected \new" cost 1=(r ?(i?1)c), while the corresponding edge in the j'th tree has cost 1=(r ? (i ? 1)c ? j + 1). The total \new" cost of these edges is c?1 X j =0
1
Z
c
1 r ? (i ? 1)c ? x dx 0 = ln(r ? (i ? 1)c) ? ln(r ? ic):
r ? (i ? 1)c ? j 6
22
COPPERSMITH AND SORKIN
An edge i added in the `'th generation of the doubling contributes to 2k?`+1 ? 1 paths, so, summing over all generations `: E[cost
of all paths] = = =
k X
` ?1 2X
`=1 k X
i=2`?1
(2k?`+1 ? 1)
[ln(r ? (i ? 1)c) ? ln(r ? ic)]
(2k?`+1 ? 1)[ln(r ? (2`?1 ? 1)c) ? ln(r ? (2` ? 1)c)]
`=1 kX ?1 `=1
(2k?` ? 2k?`+1) ln(r ? (2` ? 1)c)
+ (2k ? 1) ln r ? ln(r ? (2k ? 1)c) = (2k ? 1) lnr ? Dividing by the E
c2k n X j =1
k X `=1
2k?` ln(r ? (2` ? 1)c):
nodes, the expected path cost from a random node is
k X vj = 1c (1 ? 2?k ) lnr ? 1c 2?` ln(r ? (2` ? 1)c):
= ? 1c
k X `=1
`=1
2?` ln(1 ? (2` ? 1)c=r):
Remark 20. In Lemma 19, E P i ui + E P j vj is minimized by k? (r=c) = blog2(1 + (1 ? 1=e) r=c)c; for f ? (r; c) =
(r=c) 1 2?k? ? 1 k X ?` ` c c `=1 2 ln(1 ? (2 ? 1)c=r): ?
Proof. Disregarding the uniform factor of 1=c, increasing from k ? 1 to k increases
the summed expectation (equations (3)+(4)) by (2?k ? 2?(k?1)) ? 2?k ln(r ? (2k ? 1)c). This is less than or equal to 0 when k 6 log2 (1 + (1 ? 1=e)(r=c)).
Algorithm 3B Input: Matrix A; m-assignment of zero-weight edges; additional entry am+1;m = 0; a list C0 63 m of \free" columns; a set R0 63 m of \working" rows. Like Algorithm 3A except: Set |^ = argminj 2Ck fminfam;j ; am+1;j gg . If |^ was a selection from row m, swap rows m and m + 1, so we may assume that |^ was a selection from row m + 1. Proceed with Algorithm 3A, leaving row m assigned to column m.
RANDOM ASSIGNMENT
23
Lemma 21. Under the conditions of Lemma 19, but with am+1;m = 0, the same
conclusions hold, except that row m of A0 is tainted (its elements are not i.i.d. P exponential r.v.'s), and E i ui = 21c 2?k .
Proof. Exactly paralleling the proof of Lemma 19. Recall that row m+1 is unbiased
because its minimum revealed element um+1 = am+1;|^ is subtracted from it; some elements of row m are biased because they are known to be at least um+1 , but um+1 is not subtracted from them. The expectation of um+1 is cut by half because it is chosen from two rows rather than one. Remark 22. In Lemma 21, E P i ui + E P j vj is minimized by ky (r=c) = blog2 (1 ? (1 + e?1=2 ) r=c)c; for kyX (r=c) y 1 1 y ? k f (r; c) = 2c 2 ? c 2?` ln(1 ? (2` ? 1)c=r): `=1
Proof. As for Remark 20.
Algorithm 3 Input: n n matrix A of i.i.d. exponential elements; value " > 0.
Execute Algorithm 0 (row/column/recondition). Assign all rows with unique zeros to their corresponding columns. For each column with several zeros, assign the rst corresponding row. Let C be the set of unassigned columns. Let R be the set of assigned rows. For columns with several zeros, re-sort the rows to move their corresponding \second" rows to the end (leaving 3rd and further rows earlier). For each \third and subsequent" row m: Assign row m using Algorithm 3A, with alternating path ending at column j 2 C . Set R = R [ m, and C = C n j. If column j's zero was in row i, and row i now has a unique zero, in column j 0 , assign i to j 0 , and set R = R; i, and C = C n j 0 . For each \second" row m: Assign row m using Algorithm 3B, with alternating path ending at column j 2 C . Set R = R [ m, and C = C n j. Set aside row m's biased twin m0 (whose zero was in the same column): R = R n m0 . If column j's zero was in row i, and row i now has a unique zero, in column j 0 , assign i to j 0 , and set R = R; i, and C = C n j 0 . When fewer than "n rows remain to be assigned, nish up with Algorithm 2. Theorem 23. In the limits as " ! 0 and (then) as n ! 1, for an n n matrix of
i.i.d. exponential r.v.'s, Algorithm 3 produces a complete assignment with expected cost less than 1.94. (Thus c? < 1:94 .) Proof. Treat as a group the columns of R whose zeros lie in a single row. By
Theorem 12, the number of columns in a row's bunch is initially described by a
24
COPPERSMITH AND SORKIN
random variable X = Y + Z, with Y Bernoulli(p? ) and Z Pois(1=e). (Values X < 2 correspond to rows with no such bunches.) As rows are assigned, random columns from C are consumed, and the resulting \depopulated" distribution is therefore described by X 0 = Y 0 +Z 0 , with Y 0 Bernoulli(p? ) and Z 0 Pois(=e), where describes the degree of depopulation and the distribution is again valid restricted to X > 2. (It will not be an issue that singleton columns, consumed at once by Algorithm 3, are not accurately described by this model.) For any constant 0 < < 1 (i.e., not a vanishingly small function of n), the expected total number of bunched columns is given by nc(), where c() = p?
1 X
k Poisk?1(=e) + (1 ? p? )
1 X
k Poisk (=e)
k=2 k=2 = e?=e (2 p? =e) + (1 ? e?=e)(p? + =e):
(5) Moreover, as n ! 1 , by the law of large numbers the actual count is within a factor 1 + o(1) of the expectation, with probability 1 ? o(1). At the same time, the number of rows still to be added is equal to the sum, over all bunches of size 2 or more, of the bunch size minus 1: one column in the bunch will go to the row responsible for the bunch, while the others will get assigned to the rows through which Algorithm 3 iterates. Thus the expected number of rows to be assigned is nt(), where t() = p?
1 X
(k ? 1) Poisk?1(=e) + (1 ? p? )
k=2 = e?=e (1 ? p? ) + (p? + =e ? 1);
1 X
(k ? 1) Poisk (=e)
k=2
and again the exact number is asymptotically close.
(6)
As ranges from 1 to 0, the number of rows to be assigned is asymptotically nt(), where t() ranges from about .1927 to 0, and the number of free columns is jCj = nc(), where c() ranges from about .3556 to 0. The number of rows falling into the algorithm's phase B | the number of \second" rows per column with multiple zeros | is just n times the fraction ty of multiple-row columns: ty =
1 X
k=2
col(k) =
1 X
k=2
Poisk (e?e?1 ) = 1 ? e?e
?e?1
(1 + e?e?1 ) :1531:
In Phase A, the sets R and C are always complementary, so the number of working rows is jRj = n(1 ? c()). In Phase B, opposing the addition of rows to R there is also the removal of one \biased twin" row per iteration. Thus the number of working rows is n(1 ? c() ? [ty ? t()]). Observe that for f = f ? or f = f y , f(nr; nc) = n1 f(r; c). De ne (
? if :1927 > t() > ty ; f() = f y (1 ? c(); c()); f (1 ? c() ? (ty ? t()); c()); if ty > t() > 0:
RANDOM ASSIGNMENT
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Figure 7. Graph of f() vs t(). The area under the curve | the rst integral in expression (7) | is the key component in the assignment cost bound c? < 1:94. Reading the graph from right to left, as Algorithm 3 progresses, ranges from 1 to 0; t() decreases from about :19 to 0; and the incremental cost f() of adding an edge starts at 2:53 and | for the most part | increases. The sudden drop in f() at t :15 represents the transition from Algorithm 3A to 3B; there is also an invisible cup in f , with minimum occurring around t = :188.
The total expected assignment cost from Algorithm 3 is Z " 1 f()d(n t()) + Z " min 1 + s + 1 ln 1 ? x 1 + 1=e + 1 ? x ? sx dx: (7) 0 s2 + xs 2xs =1 n p For " small (therefore x small), the last term's integrand is minimized by s p p 2=x, R p where its value is also 2=x. Thus its contribution is 0" 2=xdx = 2 2", which vanishes asymptotically.R7 Thus asymptotically, Algorithm 3 gives an assignment of expected cost 1 + 1=e + 0=1 f()dt() < 1:94; the value is arrived at by numerical integration with care to over-estimate.
Z
12. Exact optima Parisi's conjecture [Par98] (shown above as Conjecture 4) seems to o er the best \handle" on the assignment problem. To that end, we have veri ed it to the extent we could, and in the process generalized it further: 7 This term is due to Algorithm 2, which is invoked only to allow analysis when = o(1), and equations (5) and (6) are no longer guaranteed to estimate c and t to within a factor 1 + o(1).
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COPPERSMITH AND SORKIN
Conjecture 24. Given an m n array of i.i.d., exponentially distributed random variables aij with mean 1, select k 6 min(m; n) elements, no two in the same row or column, such as to minimize the sum of those variables. (A k -assignment.) The expected value of this sum is
1 (m ? i)(n ? j) i;j 0; i+j
