组合数学 (Spring 2013)/Problem Set 2

Set Balancing

Supposed that we have an [math]\displaystyle{ n\times m }[/math] matrix [math]\displaystyle{ A }[/math] with 0-1 entries. We are looking for a [math]\displaystyle{ b\in\{-1,+1\}^m }[/math] that minimizes [math]\displaystyle{ \|Ab\|_\infty }[/math].

Recall that [math]\displaystyle{ \|\cdot\|_\infty }[/math] is the infinity norm (also called [math]\displaystyle{ L_\infty }[/math] norm) of a vector, and for the vector [math]\displaystyle{ c=Ab }[/math],

[math]\displaystyle{ \|Ab\|_\infty=\max_{i=1,2,\ldots,n}|c_i| }[/math].

We can also describe this problem as an optimization:

[math]\displaystyle{ \begin{align} \mbox{minimize } &\quad \|Ab\|_\infty\\ \mbox{subject to: } &\quad b\in\{-1,+1\}^m. \end{align} }[/math]

This problem is called set balancing for a reason.

The problem arises in designing statistical experiments. Suppose that we have [math]\displaystyle{ m }[/math] subjects, each of which may have up to [math]\displaystyle{ n }[/math] features. This gives us an [math]\displaystyle{ n\times m }[/math] matrix [math]\displaystyle{ A }[/math]: [math]\displaystyle{ \begin{array}{c} \mbox{feature 1:}\\ \mbox{feature 2:}\\ \vdots\\ \mbox{feature n:}\\ \end{array} \left[ \begin{array}{cccc} a_{11} & a_{12} & \cdots & a_{1m}\\ a_{21} & a_{22} & \cdots & a_{2m}\\ \vdots & \vdots & \ddots & \vdots\\ a_{n1} & a_{n2} & \cdots & a_{nm}\\ \end{array} \right], }[/math] where each column represents a subject and each row represent a feature. An entry [math]\displaystyle{ a_{ij}\in\{0,1\} }[/math] indicates whether subject [math]\displaystyle{ j }[/math] has feature [math]\displaystyle{ i }[/math]. By multiplying a vector [math]\displaystyle{ b\in\{-1,+1\}^m }[/math] [math]\displaystyle{ \left[ \begin{array}{cccc} a_{11} & a_{12} & \cdots & a_{1m}\\ a_{21} & a_{22} & \cdots & a_{2m}\\ \vdots & \vdots & \ddots & \vdots\\ a_{n1} & a_{n2} & \cdots & a_{nm}\\ \end{array} \right] \left[ \begin{array}{c} b_{1}\\ b_{2}\\ \vdots\\ b_{m}\\ \end{array} \right] = \left[ \begin{array}{c} c_{1}\\ c_{2}\\ \vdots\\ c_{n}\\ \end{array} \right], }[/math] the subjects are partitioned into two disjoint groups: one for -1 and other other for +1. Each [math]\displaystyle{ c_i }[/math] gives the difference between the numbers of subjects with feature [math]\displaystyle{ i }[/math] in the two groups. By minimizing [math]\displaystyle{ \|Ab\|_\infty=\|c\|_\infty }[/math], we ask for an optimal partition so that each feature is roughly as balanced as possible between the two groups. In a scientific experiment, one of the group serves as a control group (对照组). Ideally, we want the two groups are statistically identical, which is usually impossible to achieve in practice. The requirement of minimizing [math]\displaystyle{ \|Ab\|_\infty }[/math] actually means the statistical difference between the two groups are minimized.

We propose an extremely simple "randomized algorithm" for computing a [math]\displaystyle{ b\in\{-1,+1\}^m }[/math]: for each [math]\displaystyle{ i=1,2,\ldots, m }[/math], let [math]\displaystyle{ b_i }[/math] be independently chosen from [math]\displaystyle{ \{-1,+1\} }[/math], such that

[math]\displaystyle{ b_i= \begin{cases} -1 & \mbox{with probability }\frac{1}{2}\\ +1 &\mbox{with probability }\frac{1}{2} \end{cases}. }[/math]

This procedure can hardly be called as an "algorithm", because its decision is made disregard of the input [math]\displaystyle{ A }[/math]. We then show that despite of this obliviousness, the algorithm chooses a good enough [math]\displaystyle{ b }[/math], such that for any [math]\displaystyle{ A }[/math], [math]\displaystyle{ \|Ab\|_\infty=O(\sqrt{m\ln n}) }[/math] with high probability.

Theorem

Let [math]\displaystyle{ A }[/math] be an [math]\displaystyle{ n\times m }[/math] matrix with 0-1 entries. For a random vector [math]\displaystyle{ b }[/math] with [math]\displaystyle{ m }[/math] entries chosen independently and with equal probability from [math]\displaystyle{ \{-1,+1\} }[/math], [math]\displaystyle{ \Pr[\|Ab\|_\infty\gt 2\sqrt{2m\ln n}]\le\frac{2}{n} }[/math].

Proof. Consider particularly the [math]\displaystyle{ i }[/math]-th row of [math]\displaystyle{ A }[/math]. The entry of [math]\displaystyle{ Ab }[/math] contributed by row [math]\displaystyle{ i }[/math] is [math]\displaystyle{ c_i=\sum_{j=1}^m a_{ij}b_j }[/math]. Let [math]\displaystyle{ k }[/math] be the non-zero entries in the row. If [math]\displaystyle{ k\le2\sqrt{2m\ln n} }[/math], then clearly [math]\displaystyle{ |c_i| }[/math] is no greater than [math]\displaystyle{ 2\sqrt{2m\ln n} }[/math]. On the other hand if [math]\displaystyle{ k\gt 2\sqrt{2m\ln n} }[/math] then the [math]\displaystyle{ k }[/math] nonzero terms in the sum [math]\displaystyle{ c_i=\sum_{j=1}^m a_{ij}b_j }[/math] are independent, each with probability 1/2 of being either +1 or -1. Thus, for these [math]\displaystyle{ k }[/math] nonzero terms, each [math]\displaystyle{ b_i }[/math] is either positive or negative independently with equal probability. There are expectedly [math]\displaystyle{ \mu=\frac{k}{2} }[/math] positive [math]\displaystyle{ b_i }[/math]'s among these [math]\displaystyle{ k }[/math] terms, and [math]\displaystyle{ c_i\lt -2\sqrt{2m\ln n} }[/math] only occurs when there are less than [math]\displaystyle{ \frac{k}{2}-\sqrt{2m\ln n}=\left(1-\delta\right)\mu }[/math] positive [math]\displaystyle{ b_i }[/math]'s, where [math]\displaystyle{ \delta=\frac{2\sqrt{2m\ln n}}{k} }[/math]. Applying Chernoff bound, this event occurs with probability at most [math]\displaystyle{ \begin{align} \exp\left(-\frac{\mu\delta^2}{2}\right) &= \exp\left(-\frac{k}{2}\cdot\frac{8m\ln n}{2k^2}\right)\\ &= \exp\left(-\frac{2m\ln n}{k}\right)\\ &\le \exp\left(-\frac{2m\ln n}{m}\right)\\ &\le n^{-2}. \end{align} }[/math] The same argument can be applied to negative [math]\displaystyle{ b_i }[/math]'s, so that the probability that [math]\displaystyle{ c_i\gt 2\sqrt{2m\ln n} }[/math] is at most [math]\displaystyle{ n^{-2} }[/math]. Therefore, by the union bound, [math]\displaystyle{ \Pr[|c_i|\gt 2\sqrt{2m\ln n}]\le\frac{2}{n^2} }[/math]. Apply the union bound to all [math]\displaystyle{ n }[/math] rows. [math]\displaystyle{ \Pr[\|Ab\|_\infty\gt 2\sqrt{2m\ln n}]\le n\cdot\Pr[|c_i|\gt 2\sqrt{2m\ln n}]\le\frac{2}{n} }[/math]. [math]\displaystyle{ \square }[/math]

How good is this randomized algorithm? In fact when [math]\displaystyle{ m=n }[/math] there exists a matrix [math]\displaystyle{ A }[/math] such that [math]\displaystyle{ \|Ab\|_\infty=\Omega(\sqrt{n}) }[/math] for any choice of [math]\displaystyle{ b\in\{-1,+1\}^n }[/math].