Maximizing Polynomials Subject to Assignment Constraints ∗
Konstantin Makarychev
Microsoft Research
Maxim Sviridenko
University of Warwick
Abstract
We study the q-adic assignment problem. We first give an O(n(q−1)/2 )-approximation algorithm for the Koopmans–Beckman version of the problem improving upon the result of Barvinok.
Then, we introduce a new family of instances satisfying “tensor triangle inequalities” and give
a constant factor approximation algorithm for them. We show that many classical optimization
problems can be modeled by q-adic assignment problems from this family. Finally, we give
several integrality gap examples for the natural LP relaxations of the problem.
1
Introduction
In 1963, Lawler [13] proposed the following problem, which he called the q-adic assignment problem:
We are given a set [n] = {1, . . . , n} of indices and a 2q-dimensional array cu,v for u, v ∈ [n]q . The
goal is to find a permutation π : [n] → [n] optimizing (maximizing or minimizing) the expression
X
cu,π(u) ,
(1)
u∈[n]q
where π(u) = (π(u1 ), . . . , π(uq )) for the vector u = (u1 , . . . , uq ). The maximization and minimization problems are called the maximum and minimum q-adic assignment problems respectively.
This is a natural generalization of the quadratic assignment problem (QAP) originally defined by
Koopmans and Beckman [12] in 1957 for which q = 2 and cu,v = wu dv for all u, v ∈ [n]2 . Thus,
we call the problem with decomposable coefficients cu,v = wu dv the Koopmans–Beckman version
of the q-adic assignment problem throughout the paper.
Similarly, to the quadratic assignment problem, the general q-adic assignment problem has
many practical applications and can be used to model variety of optimization problems. For
example, Winter and Zimmerman [19] used the cubic assignment problem for scheduling. Burkard,
Cela and Klinz [7] suggested applications of the 4-adic (or quartic) assignment problem in VLSI
synthesis. The Koopmans-Beckman version of the problem can also be viewed as a variant of the
hypergraph isomorphism problem studied by Babai et al. [3] where we would like to maximize
the product of weights of hyperedges that are mapped into each other. (However, since techniques
used in hypergraph or graph isomorphism algorithms heavily use the fact that all hyperedges
must be mapped these techniques cannot be used for the q-adic assignment problem.) But despite
enormous amount of applications, the problem remains highly intractable both from theoretical [15]
and practical viewpoints even when q = 2. For larger q, e.g. for q = 4, the problem cannot be
solved exactly in practice even if n = 14 (see [6]).
∗
an extended abstract of this paper appeared in Proceedings of ICALP2011
1
1.1
Overview of the Results
In this paper, we study the maximization version of the problem with nonnegative coefficients.
There are no known approximation algorithms that have a proven performance guarantee for the
general case. The Koopmans–Beckman version of the maximum q-adic assignment problem was
considered by Barvinok [4] who designed an approximation algorithm with performance guarantee εnq/2 and running time nO(q/ε) for any ε > 0. The best known approximation algorithm
√
with performance guarantee O( n) for the special case of q = 2, i.e. the Maximum QAP, was
given in [17] and [15]. In this paper, we present a new Oq (n(q−1)/2 )-approximation algorithm
for the Koopmans–Beckman version of the maximum q-adic assignment problem thus improving
upon the result of Barvinok [4] and matching the performance guarantee of [15] for the Maximum
QAP. The Oq (·) notation hides terms dependent only on q. Note, that we cannot hope to get a
poly-logarithmic approximation ratio, since such possibility was ruled out (under very plausible
complexity assumptions) even for the maximum quadratic assignment problem in [15] (using the
same method as in [15], one can show that under slightly stronger assumptions (see [5]), there
exists δ > 0 and q such that the maximum q-adic assignment problem is NP-hard to approximate
within a factor of nδ ). Thus, one of our main goals is to identify special families of instances
that can be solved with a much better performance guarantee. Arkin, Hassin and Sviridenko [2]
and, then Nagarajan and Sviridenko [17] showed how to get a constant approximation ratio in the
Koopmans–Beckman version of the maximum quadratic assignment problem, if the coefficients dv
satisfy the triangle inequality. We give a very general analog of this result for the general (i.e., not
the Koopmans–Beckman version) maximum q-adic assignment problem. We further describe combinatorial optimization problems that can be expressed as special cases of this problem. Finally, we
give Oq (n(q−1) )-approximation algorithm for the general case of the maximum q-adic assignment
problem (improving on the trivial O(nq ) approximation
algorithm).
(q−1)/2 q−1 n
for the
We give an integrality gap examples of Ωq ln n for the general case, and Ωq n √ln n
Koopmans–Beckman case of the maximum q-adic assignment problem.
1.2
Tensor Triangle Inequality
Let K = {k1 , . . . , kq−1 } be a set of (q − 1) distinct numbers in [n], i.e. ka 6= kb for all a 6= b.
For each vector v ∈ [n]q and each coordinate s ∈ [q], let Vs (v, K) be the set of (q − 1)! vectors
obtained from v by replacing all coordinates except the coordinate s with indices k1 , . . . , kq−1 in
an arbitrary order. For example for q = 3, s = 1, vector v = (v1 , v2 , v3 ) and K = {k1 , k2 } ⊂ [n] we
get V1 (v, K) = {(v1 , k1 , k2 ), (v1 , k2 , k1 )}.
Definition 1.1. We say that an instance of the maximum q-adic assignment problem satisfies the
tensor triangle inequality if for every K = {k1 , . . . , kq−1 } and u, v ∈ [n]q ,
cu,v ≤
q
X
s=1
X
v′ ∈V
cuv′ .
(2)
s (v,K)
Note that the total number of terms on the right hand side of (2) is q! and the total number of
n
different sets K is q−1
. We now discuss various special cases of the tensor triangle inequality:
1. For q = 2 the 4-dimensional tensor (c(i,j),(p,q) ) for i, j, p, q ∈ [n] satisfies the tensor triangle
inequality if for any r ∈ [n] the following inequality holds
c(i,j),(p,q) ≤ c(i,j),(p,r) + c(i,j),(r,q) .
2
(3)
In the Koopmans–Beckman version of the problem, c(i,j),(p,q) = wij dpq and the condition (3)
just becomes the triangle inequality dpq ≤ dpr + drq . If in addition dpq is symmetric, i.e.
dpq = dqp we obtain the maximum quadratic assignment problem with the triangle inequality
for which constant factor approximation algorithms are known [2, 17].
If wij is the weight of the edge (i, j) in a directed graph, dpq = 1 if p < q and dpq = 0 if p ≥ q,
then the tensor triangle inequality is also satisfied, and the problem is exactly equivalent to
the Maximum Acyclic Subgraph Problem.
More generally, the condition (3) models the Koopmans–Beckman variant of the maximum
quadratic assignment problem in which matrix dpq is not symmetric but satisfies the triangle
inequality. There was no constant factor algorithm known for this problem before and the
techniques from [2, 17] do not seem to generalize to this case.
2. In the Betweenness Problem, we are given the set T of triples (i, j, k), we would like to find
a permutation π and maximize the number of triples such that either π(i) < π(j) < π(k) or
π(i) > π(j) > π(k). This problem has applications in computational biology and was studied
in [9, 14]. This problem is modeled by the q-adic assignment problem in the Koopmans–
Beckman form with q = 3. We define c(i,j,k),(p,q,r) = w(i,j,k)d(p,q,r) such that w(i,j,k) = 1 for
each (i, j, k) ∈ T , w(i,j,k) = 0, for each (i, j, k) 6∈ T and d(p,q,r) = 1 if p < q < r or p > q > r
and d(p,q,r) = 0, otherwise. The condition (3) is satisfied since if dpqr = 1 then at least one of
the term one right hand side of (3) must be one for any choice of k1 , k2 . We note however that
the random assignment gives a 3 approximation for the Betweenness Problem, our algorithm
gives a worse constant.
3. Finally, the condition (3) models the generalization of the Maximum Acyclic Subgraph Problem considered by Guruswami et al. [10]. In this problem for each customer we are given a
list of preferences in the form i1 > i2 > · · · > iq the goal is to find a permutation π that maximizes the total number of satisfied customers, where a customer is satisfied if his preferences
are satisfied, i.e. π(i1 ) > π(i2 ) > · · · > π(iq ). Guruswami et al. [10] show that assuming
the Unique Games Conjecture there is no approximation algorithm for this problem with
performance guarantee better than q! (note, that a bound of q! is trivial, because a random
assignment gives a q! approximation). Since this problem can be modeled using condition (3)
it shows that our e2 q! approximation for the maximum q-adic assignment problem is optimal
up to a factor e2 .
2
Linear Programming Relaxation
For any vector u ∈ [n]q , let us be the s-th coordinate of vector u. The maximum q-adic assignment
problem can be formulated as the problem of maximizing the value of degree q multilinear (or
tetrahedral) polynomial subject to assignment constraints:
max
X
u,v∈[n]
c
q u,v
q
Y
xus vs ,
(4)
s=1
X
i∈[n]
3
xij = 1,
j ∈ [n],
(5)
X
j∈[n]
xij = 1,
i ∈ [n],
xij ∈ {0, 1}.
(6)
(7)
We linearize the objective function (4) and relax the integrality constraint (7). Let Sq be the
set of all permutations on q elements. For each u ∈ [n]q and σ ∈ Sq , let uσ denote the
vectorv
1 2 3
with coordinates vs = uσ(s) i.e., σ permutes the coordinates of the vector u (e.g., if σ =
,
2 1 3
u = (5, 17, 10), then uσ = (17, 5, 10)). In our linear programming relaxation we keep assignment
variables xij for i, j ∈ [n] and we define new variables yu,v for u, v ∈ [n]q that have the value
Q
q
s=1 xus vs if the variables xij are integral.
Let R = {u ∈ [n]q : ui 6= uj , if i 6= j}. The fact that ourQobjective function (4) is a multilinear
polynomial and assignment constraints (5)-(6) imply that qs=1 xus vs = 0 if u ∈ [n]q \ R or v ∈
[n]q \ R. Therefore, we can restrict our attention to the q-tuples from the set R. We now formally
define our linear programming relaxation
X
max
cu,v yu,v ,
(8)
u,v∈R
X
yu,v ≤ xivs ,
v ∈ R, s ∈ [q], i ∈ [n],
(9)
u∈R:us =i
X
yu,v ≤ xus j ,
u ∈ R, s ∈ [q], j ∈ [n],
(10)
v∈R:vs =j
yu,v = yuσ ,vσ ,
X
xij = 1,
i∈[n]
X
xij = 1,
j∈[n]
yu,v ≥ 0,
xij ≥ 0,
u, v ∈ R, σ ∈ Sq ,
(11)
j ∈ [n],
(12)
i ∈ [n],
(13)
u, v ∈ R,
(14)
i, j ∈ [n].
(15)
The
integral solution of the maximum q-adic assignment problem
P valid forQan
Pconstraints (9) are
q
since u∈R:us =i yu,v = u∈R:us =i s′ =1 xus′ vs′ and we can iteratively apply the equations (12) for
each coordinate of vector u except coordinate s. Analogously, we can show that Q
constraints (10)
are valid. The validity of constraints (11) follows from the fact that the product qs=1 xus vs does
not depend on the order of factors.
In Section 6, we show that our linear programming relaxation has integrality gap Ω(nq−1 / log n)
for the general case of the maximum q-adic assignment problem. However, in the Koopmans–
Beckman form, i.e. when cu,v = wu dv for all u, v ∈ [n]q the integrality gap is O(n(q−1)/2
√). We
present an approximation algorithm in Section 5 and an almost matching Ω(n(q−1)/2 / log n)integrality gap example is presented in Section 7 of the paper.
3
General Case
In this section, we show how to get Oq (nq−1 ) approximation in the general case. Note, that to
get Oq (nq ) approximation we could simply let π : [n] → [n] to be a random permutation sampled
uniformly from the set of all permutations. To get an improvement, we map some of the indices
uniformly at random, and some according to the probability distribution given by the LP. We do
it in such a way that for many u ∈ R, one and only one index i ∈ u is mapped according to the
4
LP and the rest (q − 1) indices in u are mapped uniformly at random. We analyze this rounding
scheme term by term. We compare the expected profit we get from the term cu,v with the profit
of the LP solution (i.e., cu,v yu,v ). We show that we lose only a constant factor by mapping the
index i, and we lose a factor of n−(q−1) by mapping the rest of indices. We give the details of this
rounding scheme in the following lemma.
Lemma 3.1. Let (x∗ , y ∗ ) be an optimal solution of the linear program (8)–(15). There exists a
randomized polynomial time algorithm that finds an integral solution to the linear program (8)–(15)
with expected value at least
2(q−1)
!
−1 X
q
1 − 1q
X
n
x∗us vs .
cu,v
q!
q−1
s=1
u,v∈R
Proof. The matrix X ∗ = (x∗ij ) for i, j ∈ [n] is doubly stochastic since it satisfies the constraints
(12) and (13). Therefore, it can be represented as a convex combination of permutation matrices,
i.e.
X
X∗ =
λτ Π(τ )
(16)
τ ∈[N ]
PN
where N ≤ n2 , τ =1 λτ = 1, λτ ≥ 0 for all τ ∈ [N ] and each Π(τ ) is a permutation matrix. Our
randomized algorithm chooses one permutation matrix Π(τ ) with probability distribution given by
coefficients λτ . We define the first random permutation ϕ(i) = j if Πij (τ ) = 1. Then, by (16),
Pr[ϕ(i) = j] = x∗ij for any i, j ∈ [n]. We define the second random permutation ψ : [n] → [n] to be
a random permutation on [n] chosen uniformly among all permutations.
We now combine the mappings ϕ and ψ. Choose the set X ⊆ [n] at random such that each
index i ∈ [n] belongs to X independently with probability 1/q. We define the mapping π as follows:
if i ∈ X, then π(i) = ϕ(i); if i ∈
/ X and ψ(i) ∈
/ ϕ(X), then π(i) = ψ(i). All other elements (i ∈
/X
and ψ(i) ∈ ϕ(X)) are mapped one-to-one to elements in [n] in an arbitrary way so that π is a
permutation.
We estimate probability, that π(u) = v for u, v ∈ R. For every s ∈ [q], let As (u, v) be the
following event
^
As (u, v) = {ϕ(us ) = vs } {ψ(us′ ) = vs′ for all s′ ∈ [q], s′ 6= s}.
Since the permutations ϕ and ψ are defined independently, we have
−1
n
∗
.
Pr[As (u, v)] = xus ,vs · (q − 1)!
q−1
Let Bs (u, v) be the event
Bs (u, v) = {us ∈ X}
^
/ ϕ(X) for all s′ ∈ [q], s′ 6= s}.
/ X and vs′ ∈
{us′ ∈
Since each i ∈ [n] is chosen to the set X independently with probability 1/q and since | ∪s′ 6=s
{us′ , ϕ−1 (vs′ )}| ≤ 2(q − 1) (always, and thus independently of As (u, v)) we have
1
Pr [Bs (u, v) | As (u, v)] ≥
q
5
1
1−
q
2(q−1)
.
For every s ∈ [q], events As (u, v) and Bs (u, v) together imply the event {π(u) = v}, also events
Bs (u, v) are disjoint for different s ∈ [q], therefore
−1
q
q
X
X
x∗us ,vs
n
1
1 2(q−1)
Pr[As (u, v) ∧ Bs (u, v)] ≥
.
Pr[π(u) = v] ≥
×
1−
(q − 1)! q − 1
q
q
s=1
s=1
We are now ready to estimate the expected value of the solution.
#
"
X
X
=
cu,v Pr[π(u) = v]
cu,π(u)
E
u∈R
u,v∈R
−1
q
X
x∗us ,vs
n
1 2(q−1)
1
≥
cu,v ×
1−
×
(q − 1)! q − 1
q
q
s=1
u,v∈R
2(q−1)
!
−1 X
q
1 − 1q
X
n
∗
xus ,vs .
cu,v
=
q!
q−1
X
u,v∈R
s=1
The main result of this section, Theorem 3.2, easily follows from Lemma 3.1.
Theorem 3.2. Let LP ∗ be the optimal value of the linear program (8)–(15). There exists a randomized approximation algorithm for the maximum q-adic
assignment problem (1) that finds a solution
n
with expected value at least LP ∗ /(e2 (q − 1)! q−1
), i.e. the approximation ratio of the algorithm is
n
e2 (q − 1)! q−1
= Oq (nq−1 ).
Proof. The algorithm of Lemma 3.1 finds a solution with expected value
2(q−1)
!
−1 X
q
1 − 1q
X
n
∗
xus vs
≥
cu,v
q!
q−1
s=1
u,v∈R
2(q−1)
−1 X
−1
1 − 1q
n
n
e−2
∗
cu,v · q · yu,v ≥
LP ∗ .
q!
(q − 1)! q − 1
q−1
u,v∈R
4
Problems Satisfying the Tensor Triangle Inequality
The following lemma provides us with an upper bound on the value of the linear programming
relaxation for problems satisfying the tensor triangle inequality.
Lemma 4.1. Let (x∗ , y ∗ ) be an optimal solution of the linear programming relaxation (8)–(15). If
the input instance of the maximum q-adic assignment problem satisfies the tensor triangle inequality,
i.e. the inequality (2), then for the optimal value LP ∗ of the linear programming relaxation (8)–(15)
we have
!
−1 X
q
X
n
x∗us vs
cu,v
LP ∗ ≤
q−1
s=1
u,v∈R
6
P
∗
and
Proof. We apply the inequality (2) to each term in the expression LP ∗ = u,v∈[n]q cu,v yu,v
average over different choices of the set K. We obtain


q
X
X X
X n −1
∗

cuv′  yu,v
LP ∗ ≤
q−1
u,v∈R
K:|K|=q−1 s=1 v′ ∈Vs (v,K)


−1 X
q
X
X
n
∗ 
=
cu,v′ 
yu,v
q−1
s=1 v∈R:vs =vs′
u,v′ ∈R
!
−1 X
q
X
n
∗
xus ,vs ,
cu,v
≤
q−1
s=1
u,v∈R
where the last inequality follows from P
constraints (10). The first equality is derived by considering
∗ generate the term c ′ y ∗ after applying the
which terms in the expression LP ∗ = u,v∈R cu,v yu,v
u,v u,v
tensor triangle inequality.
Combining Lemmas 4.1 and 3.1 we obtain the following theorem.
Theorem 4.2. There exists a randomized approximation algorithm with approximation ratio e2 q!
for the maximum q-adic assignment problem satisfying the generalized triangle inequality (2).
5
Koopmans–Beckman Case: The Problem with Decomposable
Coefficients
We now give an algorithm for the Koopmans–Beckman variant of the problem (an extension of our
earlier work [15]). In this case, all weights cu,v have the form cu,v = wu dv . It is convenient to think
that vectors u, v ∈ R are hyperedges and wu and dv are weights of these hyperedges. We write
i ∈ u, if for some s, i = us . If u′ ∈ [n]q and u′′ ∈ [n]q do not have common vertices i ∈ [n], then we
say that u′ and u′′ are disjoint.
Theorem 5.1. There exists a polynomial-time randomized approximation algorithm for the Koopmans–
Beckman version of the q-adic assignment problem with the approximation ratio O(q 2 n(q−1)/2 ).
Proof. The input of the algorithm is a collection of weights {wu : u ∈ R} and {dv : v ∈ R}. The
output is a permutation π : [n] → [n]. Fix the parameter
'
& s
' & s
n!
n
1
1
=
(q − 1)!
.
M=
q−1
q
q (n − q + 1)!
Solve the linear programming relaxation (8)–(15). Denote the solution by (x∗ , y ∗ ) and the value of
the objective function by LP ∗ . Now, greedily find a set of disjoint “heavy” (according to the LP)
hyperedges E = {e ∈ R} using the following algorithm.
7
• Initialization: Let U and E be subsets of R. Initially, set U = R; E = ∅.
• while (U =
6 ∅)
– Find the most expensive edge e ∈ U with respect to the cost function
X
∗
lp-cost(e) =
we dv ye,v
.
(17)
v∈R
– Add e to the set E.
– For every vertex i ∈ e, we denote e(i) = e.
– Sequentially, for each i ∈ e and s ∈ {1, . . . , q}: we set Ai,s = {u ∈ U : us = i} and
then remove elements of Ai,s from the set U . Thus, we partition all hyperedges in U
that intersect with e into disjoint sets Ai,s (where i ∈ e, s ∈ {1, . . . , q}), such that for
every u ∈ Ai,s , us = i. Note also that by removing hyperedges of Ai,s from U we ensure
that each vertex i has at most one hyperedge e(i) associated with it. Moreover, each
hyperedge belongs to exactly one set Ai,s .
– Let Bi,s be the set of M hyperedges u from Ai,s with the maximum value of wu . If
|Ai,s | < M , then let Bi,s = Ai,s .
Note, that ∪i∈[n] ∪s∈{1,...,q} Ai,s = R and Ai,s ∩ Ai′ ,s′ = ∅ for (i, s) 6= (i′ , s′ ).
We let I = ∪e∈E e = {i ∈ [n] : ∃e ∈ E s.t. i ∈ e}. By convention, we define Ai,s = ∅ if i does not
belong to any edge e ∈ E. Note, that e(i) is defined for every i ∈ I, and Ai,s = ∅ if i ∈
/ I. We have
LP
∗
=
XX
∗
wu dv yu,v
=
X
X
X
∗
wu dv yu,v
i∈[n] s=1 u∈Ai,s \Bi,s v∈R
|
∗
wu dv yu,v
i∈[n] s=1 u∈Ai,s v∈R
u∈R v∈R
q
X
=
q
X X
XX
{z
+
q
X X
XX
∗
.
wu dv yu,v
i∈[n] s=1 u∈Bi,s v∈R
}
LPI∗
|
{z
∗
LPII
}
∗ . We now consider two cases. In
Denote the first term in the sum LPI∗ and the second term LPII
each case we run a separate approximation algorithm. Our final algorithm could be viewed as an
algorithm that runs these two approximation algorithms and chooses the best solution.
I. First, assume, that LPI∗ ≥ LP ∗ /2. Let mi,s = min{wu : u ∈ Bi,s }, if |Bi,s | = M ; and mi,s = 0, if
|Bi,s | < M . Note, that
max{wu : u ∈ Ai,s \ Bi,s } ≤ mi,s ≤
8
1 X
wu .
M
u∈Bi,s
(18)
We have
LPI∗
=
q
XX
X
X
∗
wu dv yu,v
i∈[n] s=1 u∈Ai,s \Bi,s v∈R
≤
X
q
X
X
X
∗
mi,s dv yu,v
i∈[n] s=1 u∈R:us =i v∈R
≤
≤
q
XX
X
X
∗
mi,s dv yu,v
i∈[n] s=1 u∈Ai,s \Bi,s v∈R
q X
XX
mi,s dv x∗i,vs .
i∈[n] s=1 v∈R
In the last inequality, we used LP constraint (9). Then, using upper bound (18) on mi,s , we get


q X
q X
XX
XX
X
 1
mi,s dv x∗i,vs ≤
LPI∗ ≤
wu  dv x∗i,vs
M
≤
Hence,
i∈[n] s=1 v∈R
q
XX
1
M
i∈[n] s=1 v∈R
X X
wu dv x∗i,vs =
i∈[n] s=1 u:us =i v∈R
q X
X
s=1 u,v∈R
1
M
u∈Bi,s
q X
X
X
wu dv x∗us ,vs .
s=1 u∈R v∈R
1
wu dv x∗us ,vs ≥ M · LP ∗ .
2
By Lemma 3.1, there exists a randomized polynomial-time algorithm that finds a solution of cost
!
−1 X
q
X
(1 − 1q )2(q−1)
n
∗
ALGI ≥
xus vs
wu dv
q!
q−1
s=1
u,v∈R
q
n
1
LP ∗
1 q (q − 1)! q−1
∗
LP
≥
≥
q−1 .
n
2e2
q! q−1
2e2 q 2 n 2
∗ ≥ LP ∗ /2. For every e ∈ E, select independently at random v ∈ R
II. Now, assume, that LPII
∗ , and set φ(e) = v. Then, pick elements e ∈ E in a random order and for each
with probability ye,v
s = 1, . . . , q set π(es ) = φ(e)s , if the vertex φ(e)s does not have a preimage yet; and discard the
vertex es , otherwise. In the end, map all discarded vertices in an arbitrary way.
∗ .
We need to estimate the probability that π(e) = v. The probability that φ(e) = v equals ye,v
′
′
Let N (v) be the number of e for which φ(e ) intersects v. We estimate E[N (v) | φ(e) = v] as
follows
X
X
X X
E[N (v) | φ(e) = v] = 1 +
ye∗′ v′ ≤ 1 +
ye∗′ v′
= 1+
e′ ∈E\{e} v′ :v′ ∩v6=∅
q X
q
XX
X
e′ ∈E s=1 s′ =1 v′ :v′ ′ =vs
s
e′ ∈E v′ :v′ ∩v6=∅
ye∗′ v′ ≤ 1 +
q X
q
XX
e′ ∈E s=1 s′ =1
x∗e′ ′ vs .
s
Recall, P
that all
hyperedges e′ in E are disjoint. Thus the expression above can be bounded above
n Pq
by 1 + i=1 s=1 x∗ivs = 1 + q. This bound implies that
∗
yev
1
| φ(e) = v ≥
Pr[π(e) = v] = Pr[φ(e) = v] · E
N (v)
(q + 1),
9
where the last inequality follows from the Jensen’s inequality for the convex function x 7→ 1/x.
Hence, the cost of the solution returned by the algorithm is at least
ALGII ≥
XX
we dv
e∈E v∈R
∗
yev
.
q+1
On the other hand using that lp-cost(u) ≤ lp-cost(e(i)) for u ∈ Bi,s ⊂ Ai,s (see (17)), that |Bi,s | ≤
M , and that Bi,s = ∅ for i ∈
/ I, we get
!
!
q
q
n X
X X
X
X X
XX
∗
∗
∗
wu dv yu,v
LPII
=
we(i) dv ye(i),v
≤
i=1 s=1 u∈Bi,s
=
q
XX
i∈I s=1
= qM
|Bi,s |
{z
}
lp-cost(u)
X
∗
we(i) dv ye(i),v
v∈R
XX X
e∈E i∈e
i∈I s=1 u∈Bi,s
v∈R
|
v∈R
∗
we dv ye,v
!
!
2
≤ qM
=q M
X X
i∈I
X X
e∈E
v∈R
|
{z
lp-cost(e(i))
∗
we(i) dv ye(i),v
v∈R
∗
we dv ye,v
v∈R
!
!
}
.
Combining the inequalities, we get
ALGII ≥
∗
LPII
LP ∗
≥
.
(q−1)
q 2 (q + 1)M
4n 2 q(q + 1)
Thus, we have showed that the approximation ratio of the algorithm is O(q 2 n
6
(q−1)
2
).
Integrality Gap for the General Case
We will use a variant of Bernstein inequality (see for example [16], Theorem 2.3 (b)).
Lemma 6.1 (see e.g., [16], Theorem 2.3P
(b)). Let the random variables X1 , . . . , Xm be independent,
with 0 ≤ Xk ≤ 1 for each k. Let Sm = m
k=1 Xk , let µ = E[Sm ]. Then for any ε > 0,
2
ε µ
− 2(1+ε/3)
Pr[Sm ≥ (1 + ε)µ] ≤ e
Pr[Sm ≤ (1 − ε)µ] ≤ e−ε
2 µ/2
Theorem
q−1 6.2. For each q ≥ 2, the integrality gap of the LP relaxation of the q-adic assignment is
Ωq nln n .
Proof. Consider the following random instance of the maximum q-adic assignment problem: For
each u, v ∈ R, let cu,v = 1 independently with probability
p=
n ln n
q! nq
and cu,v = 0 with probability 1 − p. Consider any fixed integral solution xij , i, j ∈ [n] of the
maximum q-adic assignment problem on that random instance. The total number of pairs u, v for
10
Q
u, v ∈ R such that s∈[q] xus ,vs = 1 is exactly q! nq . Therefore, the expected value of the fixed
integral solution on our random instance is exactly µ1 = pq! nq = n ln n. Let S be the value of our
integral solution. By Lemma 6.1 with ε = 2,
Pr[S > (1 + ε)µ1 ] ≤ e−6n ln n/5 .
Now, using that the total number of possible integral solutions is n! < nn = en ln n , we get by the
union bound that the value of all integral solutions for our random instance is at most (1 + ε)µ1 =
3n ln n with probability at least 1 − e−n ln n/5 .
Let P ⊆ R be the random set of pairs (u, v) such that cu,v = 1 or cuσ ,vσ = 1 for some permutation
σ : [q] → [q]. Define the following fractional solution of the linear programming relaxation (8)–(15):
xij = 1/n for all i, j ∈ [n]; yu,v = y = 10q!n1 ln n for all (u, v) ∈ P; and yu,v = 0 for all (u, v) ∈ R×R\P.
For convenience, we define an equivalence relation ∼ on pairs (u, v) ∈ R × R: Let (u′ , v ′ ) ∼ (u′′ , v ′′ ),
if for some σ ∈ Sq , (u′ , v ′ ) = (u′′σ , vσ′′ ). Then yu′ ,v′ = yu′′ ,v′′ if (u′ , v ′ ) ∼ (u′′ , v ′′ ); and yu′ ,v′ and yu′′ ,v′′
are independent otherwise.
We claim that this fractional solution is feasible with high probability. The constraints (12) and
(13) are always satisfied, because xij = 1/n. The constraint (11) is satisfied, because if (u, v) ∈ P,
then (uσ , vσ ) ∈ P for every σ ∈ Sk . So it remains to check the feasibility of constraints (9), (10).
Consider one of these constraints. E.g.,
X
yu,v ≤ xivs .
u∈R:us =i
The total number of yuv variables on the left hand side is (q − 1)! n−1
q−1 . These variables are
independent, since (u′ , v) ∼ (u′′ , v) only if u′ = u′′ . For each yuv , Pr[yuv = y] ≤ q!p since there are
q! pairs (u, v) ∈ R that could have cu,v = 1. Therefore the expected value of the left hand side is
at most
n−1
n−1
n ln n
1
1
(q − 1)!
· q!p · y = (q − 1)!
· n ·
=
.
q−1
q−1
10q!n ln n
10n
q
We apply Lemma 6.1 with ε = 9. We scale the values of random variables yu,v and the expectation by factor 1/y = 10q!n ln n, so that the values of random variables belong to {0, 1} and the
expectation equals µ2 = q! ln n. The probability that the value of the left hand side is more than
−
ε2 µ2
81
(1 + ε)/(10n) = 1/n, i.e. this constraint is violated, is at most e 2(1+ε/3) = e−q! ln n 8 ≤ n−10q! .
The total number of constraints (9) and (10) is 2qnq! nq ≤ n3q . Thus, by the union bound, all
constraints (9), (10) are satisfied with probability at least 1 − n−7q .
The expected value of this fractional solution is exactly




X
X
X
µ3 = E 
cu,v yu,v  = E y
cu,v  = y
Pr[cu,v = 1]
u,v∈R
u,v∈R
u,v∈R
2
n
n ln n
1 n
1
2
=
·
· q!
= Ωq (nq ),
= y · |R| · p =
q
10q!n ln n
10 q
q! nq
By Lemma 6.1 with ε = 1/2, the value of the fractional solution is less than µ3 /2 with probability
at most e−µ3 /8 .
11
By the union bound again for all bad events, we obtain that with probability at least
1 − n−7q − e−n ln n/5 − e−µ3 /8 > 0
the ratio between best fractional and best integral solutions for our random instance is at least
q−1 1 n
n
10 q
= Ωq
.
2n ln n
ln n
7
Integrality Gap for the Koopmans–Beckman Case
Theorem 7.1. The integrality gap
LP relaxation of the Koopmans–Beckman case of the
q of the
nq−1
q-adic assignment problem is Ωq
ln n .
Proof. In the Koopmans–Beckman variant of the maximum q-adic assignment problem we have
cu,v = wu dv for all u, v ∈ R. Consider the following random instance. For each u ∈ R we define
wu = 1 independently with probability
s
n ln n
p=
q! nq
and wu = 0 with probability 1 − p. Similarly, for each v ∈ R we define dv = 1 independently with
probability p and dv = 0 with probability 1 − p. The analysis of the integrality gap for this random
instance is very similar to the analysis in Section 6. The only difference is that now the coefficients
in the objective function are not chosen completely independently, as a result the bound on the
integrality gap decreases.
Q
Consider any fixed integral solution xij , i, j ∈ [n]. Let W = {(u, v) : u, v ∈ R, s∈[q] xus ,vs = 1}.
I.e., (u, v) ∈ W , if u is mapped to v. We have |W | = q! nq . Therefore, the expected value of the
integral solution on our random instance is exactly p2 q! nq = n ln n. We can also view the value of
this specific solution as a sum of q! nq independent boolean random variables Y(u,v) for (u, v) ∈ W
where Y(u,v) = wu dv . By Lemma 6.1 with ε = 2, the value of the integral solution is larger
than 3n ln n with probability at most e−6n ln n/5 . The total number of possible integral solutions
is n! < nn = en ln n . Thus, by the union bound, the value of all integral solutions for our random
instance is at most 3n ln n with probability at least 1 − e−n ln n/5 .
Let P ⊆ R be the random set of vectors u ∈ R such that wu = 1 or wuσ = 1 for some permutation
σ : [q] → [q]. Let Q ⊆ R be the random set of vectors v ∈ R such that dv = 1 or dvσ = 1 for
some permutation σ : [q] → [q]. Define the following fractional solution of the linear programming
relaxation (8)–(15): xij = 1/n for all i, j ∈ [n]; yu,v = y = 10q!np ln n for all (u, v) ∈ P × Q; and
yu,v = 0 for all (u, v) ∈ R × R \ P × Q. In other words, yuv = y · (maxσ wuσ ) · (maxσ dvσ ). We define
an equivalence relation ∼ on vectors u ∈ R: Let u′ ∼ u′′ , if for some σ ∈ Sq , u′ = u′′σ . We then pick
a subset R′ ⊂ R that contains exactly one representative from each equivalence class. For example,
we may pick R′ = {(u1 , . . . , uk ) ∈ R : u1 < u2 < · · · < uq }. Then for every u ∈ R, there exists one
and only one permutation σ such that uσ ∈ R′ . Observe, that for a fixed v ∈ R, yu′ ,v = yu′′ ,v if
u′ ∼ u′′ , and that the variables yu,v , where u ∈ R′ , are independent given maxσ dvσ .
12
We claim that this fractional solution is feasible with high probability. As in the previous section,
constraints (11), (12), and (13) are always satisfied. We now check the feasibility of constraints (9),
(10). Consider one of these constraints for some v ∈ R, s ∈ [q] and value us ∈ [n]:
X
yu,v ≤ xivs .
u∈R:us =i
The left hand side can be rewritten as follows (below, we write i ∈ u if ut = i for some t ∈ [q]):
X
X
X
yu,v =
yuσ ,v
′
u∈R:us =i
u∈R :i∈u
=
X
u∈R′ :i∈u
= (q − 1)!
X
σ∈Sq :uσ(s) =i
X
yu,v
σ∈Sq :uσ(s) =i
yu,v
X
u∈R′ :i∈u
= (q − 1)!y max dvσ ·
max wuσ
σ
u∈R′ :i∈u σ
X
≤ (q − 1)!y
max wuσ .
′
u∈R :i∈u
σ
′
The total number of random variables maxσ wuσ in the sum above is n−1
q−1 . For each u ∈ R ,
Pr[maxσ wuσ = 1] ≤ q!p. Therefore,
i
hX
n−1
−1 n − 1
q!p = p
E
max wuσ ≤
q!p2
u∈R′ :i∈u σ
q−1
q−1
n ln n
−1 n − 1
= p−1 q ln n.
= p
q!
q−1
q! nq
p
√
Note that for q ≥ 2 (for q = 1 the problem is trivial), p−1 q ln n ≥ 2 n(n − 1)/(n ln n) ln n > n.
We apply Lemma 6.1 with ε = 9:
hX
i
√
−1
− n
.
Pr
max
w
≥
10p
q
ln
n
≤
e
u
σ
′
u∈R :i∈u
σ
√
Hence, with probability 1 − e− n ,
X
X
yu,v ≤ (q − 1)!y
u∈R:us =i
u∈R′ :i∈u
max wuσ = (q − 1)!
σ
p
1
10p−1 q ln n = .
10q!n ln n
n
√
By the union bound, all 2qnq! nq constraints (9), (10) are satisfied with probability at least 1−e n/2 .
Finally, we compute the expected value of the fractional solution. Note that if wu = 1 and
dv = 1, then yu,v = y, hence the expected value equals




#
#
"
"
X
X
X
X
y wu dv  = yE
dv
yu,v wu dv  = E 
wu · E
E
u,v∈R
u,v∈R
u∈R
v∈R
2
n2
n ln n 3
1
n
2
2
= y pq!
=
(q!)
n
q
10q!n ln n q! q
q
s
n
q+1
1
1 n ln n q
= Ωq (n 2 (ln n) 2 ).
=
10
q!
13
P
P
Applying the Bernstein inequality (Lemma 6.1), we get that both sums u∈R wu and v∈R dv
are concentrated around their expectations with high probabily. Hence, the integrality gap of the
instance is
!
!
r
q+1
nq−1
n 2 (ln n)1/2
= Ωq
.
Ωq
n ln n
ln n
References
[1] W. P. Adams and T. A. Johnson, Improved Linear Programming-based Lower Bounds for
the Quadratic Assignment Problem, DIMACS Series in Discrete Mathematics and Theoretical
Computer Science, 16, 1994, 43-77.
[2] E. Arkin, R. Hassin and M. Sviridenko, Approximating the Maximum Quadratic Assignment
Problem, Information Processing Letters, 77, 2001, pp. 13-16.
[3] L. Babai, P. Codenotti, Isomorphism of Hypergraphs of Low Rank in Moderately Exponential
Time. In Proc. 39th Ann. IEEE Symp. on Theory of Computing (FOCS 2008), IEEE Comp.
Soc. Press 2008, pp. 667-676.
[4] A. Barvinok, Estimating L∞ norms by L2k norms for functions on orbits. Found. Comput.
Math. 2 (2002), no. 4, 393–412.
[5] M. Bellare, S. Goldwasser, C. Lund, and A. Russel, Efficient Probabilistically Checkable Proofs
and Applications to Approximation. Proceedings of the Twenty-Fifth Annual ACM Symposium on Theory of Computing (STOC 1993), pages 294–304.
[6] R.E. Burkard, E. Cela, Heuristics for biquadratic assignment problems and their computational
comparison, European Journal of Operations Research v. 83 (1995), pp. 283-300.
[7] R.Burkard, E. Cela abd B. Klinz, On the biquadratic assignment problem. Quadratic assignment and related problems (New Brunswick, NJ, 1993), 117–146, DIMACS Ser. Discrete Math.
Theoret. Comput. Sci., 16, Amer. Math. Soc., Providence, RI, 1994.
[8] E. Cela, The Quadratic Assignment Problem: Theory and Algorithms, Springer, 1998.
[9] B. Chor and M. Sudan, A Geometric Approach to Betweenness, SIAM J. Discrete Math. 11(4):
511-523 (1998).
[10] V. Guruswami, J. Håstad, R. Manokaran, P. Raghavendra and M. Charikar, Beating the
Random Ordering is Hard: Every ordering CSP is approximation resistant. SIAM J. Comput.
40(3) (2011), pp. 878–914.
[11] R. Hassin, A. Levin and M. Sviridenko, Approximating the minimum quadratic assignment
problems, ACM Transactions on Algorithms 6(1), (2009).
[12] T. C. Koopmans and M. Beckman, Assignment problems and the location of economic activities, Econometrica 25 (1957), pp. 5376.
14
[13] E. Lawler, The quadratic assignment problem. Management Science 9 1962/1963, pp. 586–599.
[14] Y. Makarychev, Simple Linear Time Approximation Algorithm for Betweenness, Microsoft
Research Technical Report MSR-TR-2009-74.
[15] K. Makarychev, R. Manokaran and M. Sviridenko, Maximum Quadratic Assignment Problem:
Reduction from Maximum Label Cover and LP-based Approximation Algorithm, ICALP 2010,
pp. 594–604.
[16] C. McDiarmid, Concentration. Probabilistic methods for algorithmic discrete mathematics,
195–248, Algorithms Combin., 16, Springer, Berlin, 1998.
[17] V. Nagarajan and M. Sviridenko, On the maximum quadratic assignment problem, Mathematics of Operations Research 34(4), pp. 859-868 (2009).
[18] M. Queyranne, Performance ratio of polynomial heuristics for triangle inequality quadratic
assignment problems, Operations Research Letters, 4, 1986, 231-234.
[19] T. Winter and U. Zimmermann, Real-time dispatch of trams in storage yards. Mathematics
of industrial systems, IV (Valparaiso, 1996). Ann. Oper. Res. 96 (2000), 287–315.
15