On random multi-dimensional assignment problems

TL;DR

This paper investigates the structure and approximation algorithms for random multi-dimensional assignment problems, especially focusing on three dimensions with costs decomposed into sums of independent exponential variables, showing a simple greedy algorithm achieves a near-optimal approximation.

Contribution

It introduces a new model for 3D assignment problems with costs as sums of independent exponentials and proves a simple greedy algorithm is a (3+o(1))-approximation, improving known bounds.

Findings

Greedy algorithm achieves (3+o(1))-approximation in 3D case.

Cost structure as sums of independent exponentials impacts approximation quality.

Contrast with previous results where costs are fully independent exponential variables.

Abstract

We study random multidimensional assignment problems where the costs decompose into the sum of independent random variables. In particular, in three dimensions, we assume that the costs $W_{i,j,k}$ satisfy $W_{i,j,k}=a_{i,j}+b_{i,k}+c_{j,k}$ where the $a_{i,j},b_{i,k},c_{j,k}$ are independent exponential rate 1 random variables. Our objective is to minimize the total cost and we show that w.h.p. a simple greedy algorithm is a $(3+o(1))$-approximation. This is in contrast to the case where the $W_{i,j,k}$ are independent exponential rate 1 random variables. Here all that is known is an $n^{o(1)}$-approximation, due to Frieze and Sorkin.

Equations

$$Minimize\left\{\sum _{i=1}^n{W}_{i,\sigma (i),\tau (i)}:\sigma ,\tau \text{ are permutations of }[n]\right\}.$$

$$W_{i,j,k}=a_{i,j}+b_{i,k}+c_{j,k},1\le i,j,k\le n,$$

$$\text{mbox}\mathbf{Pr}(|S-N\mu |\ge \epsilon N\mu )\le 2e^{-{\epsilon }^{2}N\mu /3}.$$

$$W_n=\min \left\{a_i+b_j+c_{i,j}:\text{ai,bj,ci,j,i,j∈[n] are independent uniform [0,1] random variables}\right\}.$$

$$L=\log n,I=\left\{i:a_i\le \frac{L}{n^{2/3}}\right\},J=\left\{j:b_j\le \frac{L}{n^{2/3}}\right\},X=\left\{(i,j)\in I\times J:c_{i,j}\le \frac{L}{n^{2/3}}\right\}.$$

$$|I,|J|\in \left[\frac{1}{2}L{n}^{1/3},\frac{3}{2}L{n}^{1/3}\right]q.s.$$

$$|X|\in \left[\frac{L^3}{8},10L^3\right]q.s.$$

$$\text{mbox}\mathbf{E}(W_n\mid {\mathcal{E}}_M)\text{mbox}\mathbf{Pr}({\mathcal{E}}_M)\le \text{mbox}\mathbf{E}(W_n)\le \text{mbox}\mathbf{E}(W_n\mid {\mathcal{E}}_M)+3\text{mbox}\mathbf{Pr}(\neg {\mathcal{E}}_M).$$

$$\text{mbox}\mathbf{Pr}(\neg {\mathcal{E}}_M)=O\left(\frac{L^{15}}{n}\right).$$

$$p=\frac{L}{n^{2/3}}.$$

$$\text{mbox}\mathbf{Pr}(\neg (i))=O\left(\left(\genfrac{}{}{0ex}{}{K}{5}\right)p^4\right)=O\left(\frac{L^9{n}^{5/3}}{n^{8/3}}\right)=O\left(\frac{L^9}{n}\right).$$

$$\text{mbox}\mathbf{Pr}(\neg (ii))=O\left({\left(\left(\genfrac{}{}{0ex}{}{K}{4}\right)p^3\right)}^2\right)=O\left(\frac{L^{14}{n}^{8/3}}{n^4}\right)=O\left(\frac{L^{14}}{n^{4/3}}\right).$$

$$\text{mbox}\mathbf{Pr}(\neg (iii))=O\left({\left(\left(\genfrac{}{}{0ex}{}{K}{3}\right)p^2\right)}^3\right)=O\left(\frac{L^{15}{n}^3}{n^4}\right)=O\left(\frac{L^{15}}{n}\right).$$

$$\varphi (u)=\text{mbox}\mathbf{Pr}(U\ge pu)=1-\frac{1}{6}\sum _{k=0}^{\lfloor u\rfloor }(-1{)}^k(\genfrac{}{}{0ex}{}{3}{k})(u-k{)}^3.$$

$$\text{mbox}\mathbf{E}({\hat{W}}_n\mid {\mathcal{E}}_L,{\mathcal{E}}_M,|M|)$$

$$=p{\int }_{u=0}^3\varphi (u{)}^{|M|}du$$

$$=p(I_1+I_2+I_3),$$

$$I_1$$

$$={\int }_{u=0}^{1/L^{2/3}}{\left(1-\frac{u^3}{6}\right)}^{|M|}du+{\int }_{u=1/L^{2/3}}^1{\left(1-\frac{u^3}{6}\right)}^{|M|}du$$

$$={\int }_{u=0}^{1/|L^{2/3}}\exp \left\{-|M|u^3/6+O(|M|u^6)\right\}du+O(e^{-\Omega (|M|/L^2)})$$

$$=\left(1+O\left(\frac{1}{L^2}\right)\right){\int }_{u=0}^{1/|L^{2/3}}{e}^{-|M|u^3/6}du+O(e^{-\Omega (|M|/L^2)})$$

$$=\left(1+O\left(\frac{1}{L^2}\right)\right){\int }_{u=0}^{\infty }{e}^{-|M|u^3/6}du$$

$$=\frac{\left(1+O(L^{-2}\right))}{|M{|}^{1/3}}{\int }_{x=0}^{\infty }{e}^{-x^3/6}dx$$

$$=\frac{\left(6^{1/3}\Gamma (4/3)+O(L^{-2})\right)}{|M{|}^{1/3}}.$$

$$I_2={\int }_{u=1}^2\varphi (u{)}^{|M|}du\le {\left(\frac{5}{6}\right)}^{|M|}\text{ and }{I}_3={\int }_{u=2}^3\varphi (u{)}^{|M|}du\le {\left(\frac{5}{6}\right)}^{|M|}.$$

$$\text{mbox}\mathbf{E}({\hat{W}}_n\mid {\mathcal{E}}_L,{\mathcal{E}}_M,|M|)=\frac{\left(6^{1/3}\Gamma (4/3)+O(L^{-2})\right)}{|M{|}^{1/3}}p.$$

$$\text{mbox}\mathbf{E}({\hat{W}}_n\mid {\mathcal{E}}_L,{\mathcal{E}}_M)=\left(6^{1/3}\Gamma (4/3)+O(L^{-2})\right)\times \text{mbox}\mathbf{E}((Bin(|I|\cdot |J|,p)-O(1){)}^{-1/3})\times p.$$

$$\text{mbox}\mathbf{E}((Bin(N,q)-O(1){)}^{-1/3})$$

$$=\sum _{k=(1-\epsilon )Nq}^{(1+\epsilon )Nq}\left(\genfrac{}{}{0ex}{}{n}{k}\right)q^k(1-q{)}^{N-k}(k-O(1){)}^{-1/3}+2e^{-{\epsilon }^{2}Nq/3}$$

$$=\frac{1+O(\epsilon )}{(Nq{)}^{1/3}}+O(e^{-{\epsilon }^{2}Nq/3}),$$

$$E({\hat{W}}_n\mid {\mathcal{E}}_L,{\mathcal{E}}_M)\approx \frac{6^{1/3}\Gamma (4/3)p}{(|I|\cdot |J|\cdot p{)}^{1/3}}$$

$$E({\hat{W}}_n\mid {\mathcal{E}}_M)\approx \frac{6^{1/3}\Gamma (4/3)}{n^{2/3}}.$$

$$\text{mbox}\mathbf{E}({\hat{W}}_n\mid \mathcal{A})$$

$$=\frac{1}{\text{mbox}\mathbf{Pr}({\hat{W}}_n\le p{L}^{-1/2})}{\int }_{u=0}^{p{L}^{-1/2}}\text{mbox}\mathbf{Pr}({\hat{W}}_n\ge u)du-p{L}^{-1/2}\frac{\text{mbox}\mathbf{Pr}({\hat{W}}_n>p{L}^{-1/2})}{\text{mbox}\mathbf{Pr}({\hat{W}}_n\le p{L}^{-1/2})}.$$

$$\text{mbox}\mathbf{Pr}({\hat{W}}_n>p{L}^{-1/2})={\left(1-\frac{(L^{-1/2}{)}^3}{6}\right)}^{|M|}\le e^{-|M|L^{-1/6}/6}.$$

$$\text{mbox}\mathbf{Pr}({\hat{W}}_n\ge u)\ge {\left(1-\frac{u^3}{6}\right)}^{|M|}$$

Full text

1 Introduction

The (planar) three dimensional assignment problem is a natural generalisation of the classical assignment ptoblem. As an optimization problem it can be expressed as follows: we are given real values $W_{i,j,k}$ for $i,j,k\in[n]$ and we are asked to

[TABLE]

This is an NP-hard problem and occurs for example as a practical problem [2]. In this paper we study the following simple greedy heuristic:

Several authors have considered the average case where the $W_{i,j,k}$ are random variables. Kravtsov [3] considered the case where the $W_{i,j,k}$ are chosen randomly from $[1,M]$ where $M=n^{\alpha}$ for some $\alpha<1$. Here the minimum is at least $n$ and it is not difficult to show (see Section 4 that with the choice of $m=n-\log n$ that w.h.p. (i) greedy($m$) runs in polynomial time and (ii) it outputs a solution of value $n+o(n)$. In this case Step 6 can be completed via the choice of an arbitrary completion.

It is more difficult to analyse the case where $L\gg n$ and the case where the $W_{i,j,k}$ are independent exponential rate 1 random variables is (essentially) a scaled version of such a case. This case was considered by Frieze and Sorkin [1] and they proved the following theorem.

Theorem 1** (Frieze and Sorkin).**

Suppose that the $W_{i,j,k}$ are independent EXP(1) random variables and that $Z_{n}$ denote the value of the optimum. Then (a) $\frac{1}{n}\leq\mbox{{\bf E}}(Z_{n})=O\left(\frac{\log n}{n}\right)$ and (b) there is a polynomial time algorithm that w.h.p. finds a solution of value $\frac{1}{n^{1-o(1)}}$.

This is where the problem stands for such $W_{i,j,k}$ and here we consider the case where

[TABLE]

where the $a_{i,j},b_{i,k},c_{j,k}$ are independent uniform $[0,1]$ random variables.

We note that the problem considered in [2] was of the form given in (1). We will prove the following theorem.

Theorem 2**.**

There exist constants $c_{1},c_{2}$ such that (a) $\mbox{{\bf E}}(Z_{n})\geq c_{1}n^{1/3}$ and (b) greedy($n-n^{1/4}$) finds a solution of expected value at most $c_{2}n^{1/3}$. In this case Step 6 can be completed by choosing an arbitrary completion.

Before giving a proper proof, we give a heuristic argument for (a). Fix $i$ and consider $W_{i,j,k}$. For $W_{i,j,k}$ to be of order $n^{-\alpha}$ say we need each of 3 uniform $[0,1]$ varables to be of order $n^{-\alpha}$. This happens with probability $O(n^{-3\alpha})$ and there are $n^{2}$ choices and $3\alpha=2$ gives the largest value for $\alpha$. Summing over $i$ gives (a).

We discuss the rigorous proof of Theorem 2 in Section 2 and in Section 3 we consider the extension to higher dimensions.

1.1 Preliminaries

We sometimes refer to the Hoeffding bounds for the $S=S_{1}+S_{2}+\ldots+S_{N}$ where $S_{1},S_{2},\ldots,S_{N}\in[0,1]$ are independent and $\mbox{{\bf E}}(S_{1})+\mbox{{\bf E}}(S_{2})+\cdots+\mbox{{\bf E}}(S_{N})=N\mu$:

[TABLE]

We say that a sequence of events ${\cal E}_{n}$ occur quite surely if $\mbox{{\bf Pr}}(\neg{\cal E}_{n})=O(n^{-K})$ for any constant $K>0$.

2 Proof of Theorem 2

We begin by analysing the distribution of the smallest weight element in $\Pi_{i}=\left\{i\right\}\times[n]^{2}$.

2.1 Weights in a fixed plane

Let

[TABLE]

Lemma 3**.**

$\mbox{{\bf E}}(W_{n})\approx c_{1}n^{-2/3}$, where $c_{1}=6^{1/3}\Gamma(3/4)$, where $\Gamma$ denotes Euler’s Gamma function.

Proof.

Let

[TABLE]

It follows from (2) that

[TABLE]

Conditional on the sizes of $I,J$ we have $|X|$ is distributed as $B(|I|\cdot|J|,L/n^{2/3})$. It follows from (2) that

[TABLE]

Thus let ${\cal E}_{L}$ denote the even that $|X|\in\left[\frac{L^{3}}{8},10L^{3}\right]$.

Let ${\cal E}_{M}$ denote the event that the edges in $X$ almost form a matching. By this we mean that the graph induced by $X$ consists of a matching $M$ plus at most 4 extra edges $Y$. Then,

[TABLE]

We first deal with $\mbox{{\bf Pr}}(\neg{\cal E}_{M})$ by showing that.

[TABLE]

Let

[TABLE]

Condition on $I,J$ satisfying (4). Let $\Gamma_{X}$ be the graph induced by $X$ and note that it is distributed as the binomial random graph $G_{|I|,|J|,p}$.

Claim 1**.**

The following holds with probability $1-O(L^{15}/n)$: (i) $\Gamma_{X}$ has no component with 4 or more edges and (ii) $\Gamma_{X}$ has at most one component with 3 edges and (iii) $\Gamma_{X}$ has at most 2 components with 2 edges.

Proof of claim: Let $K=|I|+|J|$.

[TABLE]

[TABLE]

[TABLE]

End of proof of claim.

Now given ${\cal E}_{M}$ we let $\widehat{W}_{n}$ denote the minimum weight in $M$ and we see that $\widehat{W}_{n}$ is the minimum of $|M|$ independent copies of $U=(U_{1}+U_{2}+U_{3})p$ where $U_{1},U_{2},U_{3}$ are independent uniform $[0,1]$.

Thus

[TABLE]

It follows that

[TABLE]

where

[TABLE]

Now because $\phi(u)$ dereases monotonically with $u$ we have

[TABLE]

Thus,

[TABLE]

Integrating $|M|$ from (11) we obtain

[TABLE]

Given ${\cal E}_{L}$ we see that the binomial is q.s. much greater than 4. Now, for $Nq$ large, we have, from (2), that for $\varepsilon>0$,

[TABLE]

provided $\varepsilon^{2}Nq\gg\log Nq$.

It then follows from (12) that

[TABLE]

Arguing as for (13) and using the independence and concentration of $|I|,|J|$ around $Ln^{1/3}$, we see that

[TABLE]

We now have to deal with the at most 4 edges in $Y$, since $W_{n}=\min\left\{\widehat{W}_{n},Z\right\}$ where $Z$ is the minimum of at most 4 copies of $(U_{1}+U_{2}+U_{3})p$, where $U_{1},U_{2},U_{3}$ are i.i.d. $U[0,1]$. Clearly $\mbox{{\bf E}}(W_{n})\leq\mbox{{\bf E}}(\widehat{W}_{n})$ and we need to argue that it is not much smaller. So, let ${\cal A}=\left\{\widehat{W}_{n}\leq pL^{-1/2}\leq Z\right\}$. Now we have $\mbox{{\bf Pr}}({\cal A})=1-O(L^{-1/2})$ and $\mbox{{\bf E}}(W_{n})\geq\mbox{{\bf E}}(\widehat{W}_{n}\mid{\cal A})\mbox{{\bf Pr}}({\cal A})$ and so we only have to verify now that $\mbox{{\bf E}}(\widehat{W}_{n}\mid{\cal A})$ is asymptotically equal to $\mbox{{\bf E}}(\widehat{W}_{n})$. Now because $\widehat{W}_{n}$ and $Z$ are independent, we have, given $|M|$,

[TABLE]

Now

[TABLE]

Furthermore,

[TABLE]

and so integral in the first term of (16) is at least

[TABLE]

Thus

[TABLE]

and we can proceed as for our estimation of $I_{1}$.

The lemma now follows after applying (6) and (7). ∎

This proves Part (a) of Theorem 2, since clearly, $\mbox{{\bf E}}(Z_{n})\geq n\mbox{{\bf E}}(W_{n})$.

2.2 Analysis of Greedy

Let now $W_{m}$ denote the the weight of the triple $(i,j,k)$ added in the $m$th round of greedy.

Lemma 4**.**

If $m\leq n-n^{1/4}$ then

[TABLE]

Proof.

We let $I_{m},J_{m}$ be as defined in (3), where we replace $n$ in the definition by $\nu_{m}=n-m+1$. We keep $L$ as $\log n$ though and replace $p$ by $p_{m}=\frac{L}{\nu_{m}}$. The values $a_{m,j},b_{m,k}$ are independent of the first $m-1$ rounds of greedy. Now $|I|_{m},|J_{m}|$ are distributed as $Bin(\nu_{m},L\nu_{m}^{-2/3})$ and equation (2) implies that (4) holds q.s. with $n$ replaced by $\nu_{m}$. Next define $X_{m}$ iteratively via $X_{0}=\emptyset$ and

[TABLE]

We will show below that

[TABLE]

Observe that $c_{i,j}$ for $(i,j)\in X_{m}$ is unconditioned by the history of greedy to this point. Indeed, we will not have needed to expose its value in order to compute the sequence $W_{1},W_{2},\ldots,$ $W_{m-1}$. But if (18) holds then the analysis of Section 2.1 implies that

[TABLE]

Indeed, going back to (12) we

[TABLE]

and continue as before.

It remains to verify (18). Thus let $Y_{m}=(I_{m}\times J_{m})\cap\bigcup_{l<m}X_{l}$ and $Z=|Y_{m}|$. Now the sequence of choices $I_{\ell},J_{\ell},\ell\leq m$ are independent and then for $(x,y)\in I_{m}\times J_{m}$ and $\ell<m$ we have

[TABLE]

It follows (using (4)) that

[TABLE]

Unfortunately, this is not good enough to prove (18). Instead, suppose that $S=\left\{(x_{i},y_{i}),i\in[s]\right\}\subseteq I_{m}\times J_{m}$ where $s=O(1)$ and $S$ is a matching. Then,

[TABLE]

Thus,

[TABLE]

if $s=40$. Finally observe that if the maximum size of $S=s\leq 40$ and $|I_{m}|,|J_{m}|\leq 10L\nu_{m}^{1/3}$ then $|Y_{m}|\leq s(|I_{m}|+|J_{m}|)\leq 10sL\nu_{m}^{1/3}$ and the condition in (18) holds. ∎

Given Lemma 4 we see that the expected cost of the assignment produced by greedy is at most

[TABLE]

The final $n-n^{1/4}$ steps cost at most $3$ per step and this completes the proof of Theorem 2.

3 Higher Dimensions

Consider for example 4 dimensions. Here we have two reasonable options.

$W_{i,j,k,l}=a_{i,j}+b_{i,k}+c_{i,l}+d_{j,k}+e_{j,l}+f_{k,l}$. 2. 2.

$W_{i,j,k,l}=a_{i,j,k}+b_{i,j,l}+c_{i,k,l}+d_{j,k,l}$.

We have not considered the first option. The second option is a strightforward generalisation of what we have done so far. Here we will sketch a proof as a series of bullet points that the optimum and the greedy solution for the $d$-dimensional problem grow at rate $n^{1/d}$ in expectation. By the $d$-dimensional problem we mean

[TABLE]

where

[TABLE]

We claim that Theorem 2 can be generalised to

Theorem 5**.**

Suppose that $d\geq 3$. Then there exist constants $c_{d},C_{d}$ such that (a) $\mbox{{\bf E}}(Z_{n})\geq c_{d}n^{1/d}$ and (b) greedy($n^{1/(d+1)}$) finds a solution of expected value at most $C_{d}n^{1/d}$. In this case Step 6 can be completed by choosing an arbitrary completion.

**Proof Sketch:

**We can follow the argument in Lemma 3 essentially replacing $n^{1/3}$ by $n^{1/d}$ and $n^{2/3}$ by $n^{(d-1)/d}$. In effect, we make the following replacements:

1. (a):

$p$ becomes $L/n^{(d-1)/d}$. 2. (b):

$I,J$ will be replaced by $I_{1},\ldots,I_{d-1}$ of expected size $np$. 3. (c):

In which case $X$ becomes $\left\{{\bf j}\in(I_{1}\times\cdots\times I_{d-1}):W_{i,{\bf j}}\leq p\right\}$. 4. (d):

(5) becomes $|X|=\Theta(L^{d})$. 5. (e):

A matching now means a matching in a $(d-1)$-uniform hypergraph $H$ induced by $I_{1}\times\cdots\times I_{d-1}$. In the proof of Claim 1, we now let $K=|I_{1}|+\cdots+|I_{d-2}|$. We now claim that with probability $\frac{1}{n^{1-o(1)}}$ there are at most $\frac{d}{\ell-1}$ components of $H$ with $\ell\leq d+1$ edges and no components with $d+2$ or more edges. Indeed, the probability that there are $a$ components of $H$ with $\ell$ edges can be bounded by

[TABLE]

This verifies the claim and shows that if ${\cal E}_{M}$ is the event that $X$ defines a matching plus $O(1)$ edges, then $\neg{\cal E}_{M}$ is unlikely enough so that we can use (6). 6. (f):

The sum $p(I_{1}+I_{2}+I_{3})$ becomes $p(I_{1}+\cdots+I_{d})$ which is dominated by $pI_{1}$ where

[TABLE] 7. (g):

After this we find that (14) becomes

[TABLE] 8. (h):

Because the $|I_{j}|$ are strongly concentrated about their means, this results in replacing (15) by

[TABLE]

Multiplying by $n$ gives us part (a) of Theorem 5 with $c_{d}=(d!)^{1/d}\Gamma(1+1/d)$. 9. (i):

The essential part of (b) is the inequality (21). For this, where $S=\left\{x_{\bf i_{l}}:l=1,2,\ldots,s\right\}$ is a matching in $H$ and $m\geq n-n^{1/(d+1)}$, we use

[TABLE]

for $s\geq 3d(d+1)/(d-1)^{2}$.

We deduce from this that we can replace (22) by

[TABLE]

The final $n-n^{1/(d+1)}$ steps cost at most $d$ per step and this completes our sketch proof of Theorem 5.

4 Greedy for small $L$

When $W_{i,j,k}$ is chosen uniformly from $[1,M=n^{\alpha}],0<\alpha<1$ we

1. (a):

Let $Z_{m}$ denote the cost of the $m$th triple. Then for $1\leq m\leq n$ and $a\geq 1$,

[TABLE]

if

[TABLE]

Putting $m_{0}=n-(3M\log n)^{1/2}$ we see that $a$ satisfies (23) for

[TABLE]

It follows that w.h.p. and in expectation that if $m_{1}=n-\log n$, then

[TABLE]

5 Greedy versus Greedy

There is another version of the greedy algorithm where at each step we choose the “tple” of minimum weight that can be added to the current choice. Let $E(\lambda)$ denote the exponential rate $k$ random variable i.e. $\mbox{{\bf Pr}}(E(\lambda)\geq u)=e^{-\lambda u}$. We consider the $d$-dimensional case and argue next that if the weights $W_{i_{1},\ldots,i_{d}}$ are independent $E(1)$ then the value of the solution given by the two algorithms is the same in distribution. So let $G_{n,1}$ be the value returned by the algorithm described above and let $G_{n,2}$ be the value returned by algorithm described in this section. We claim that $G_{n,1}$ and $G_{n,2}$ have the same distribution.

The distribution of $G_{n,1}$ is $E(n^{d-1})+G_{n-1,1}$ and the distribution of $G_{n,2}$ is $E(n^{d})(1+(n-1))+G_{n-1,2}$. The term $E(n^{d})(n-1)$ is a result of the fact that conditioning an exponential to be greater than $x$ is equivalent to adding $x$ to a copy of that variable. Then observe that $E(n^{d-1})=nE(n^{d})$. The claim follows by induction.

Note that coincidentally, when $d=3$, $\mbox{{\bf E}}(G_{n,1})$ is equal to the expected optimum value for the $d=2$ case, see [4] and [5]. This does not generalise.

6 Final Comments

We have analysed a random multi-dimensional assignment problem with a particular form of objective fucntion. We have shown that w.h.p. there is a simple greedy algorithm that is a $(3+o(1)$-approximation to the minimum. It is possible to replace the 3 here by $3-\varepsilon$, by arguing that w.h.p. the optimum solution must use the (at least) second smallest $j,k$ (when $d=3$) for $\Omega(n)$ values of $i$. We omit the details as the real aim is to replace 3 by 1.