高级算法 (Fall 2022)/Problem Set 2
Problem 1
Suppose we want to estimate the value of [math]\displaystyle{ Z }[/math]. Let [math]\displaystyle{ \mathcal{A} }[/math] be an algorithm that outputs [math]\displaystyle{ \widehat{Z} }[/math] satisfying [math]\displaystyle{ \Pr[ (1-\epsilon)Z \leq \widehat{Z} \leq (1+\epsilon )Z] \geq \frac{3}{4} . }[/math]
We run [math]\displaystyle{ \mathcal{A} }[/math] independently for [math]\displaystyle{ s }[/math] times, and obtain the outputs [math]\displaystyle{ \widehat{Z}_1,\widehat{Z}_2,\ldots,\widehat{Z}_s }[/math].
Let [math]\displaystyle{ X }[/math] be the median (中位数) of [math]\displaystyle{ \widehat{Z}_1,\widehat{Z}_2,\ldots,\widehat{Z}_s }[/math]. Find the number [math]\displaystyle{ s }[/math] such that [math]\displaystyle{ \Pr[ (1-\epsilon)Z \leq X \leq (1+\epsilon )Z] \geq 1 - \delta . }[/math]
Express [math]\displaystyle{ s }[/math] as a function of [math]\displaystyle{ \delta }[/math]. Make [math]\displaystyle{ s }[/math] as small as possible.
Remark: in this problem, we boost the probability of success from [math]\displaystyle{ \frac{3}{4} }[/math] to [math]\displaystyle{ 1-\delta }[/math]. This method is called the median trick.
Hint: Chernoff bound.
Problem 2
Let [math]\displaystyle{ X }[/math] be a random variable with expectation [math]\displaystyle{ 0 }[/math] such that moment generating function [math]\displaystyle{ \mathbf{E}[\exp(t|X|)] }[/math] is finite for some [math]\displaystyle{ t \gt 0 }[/math]. We can use the following two kinds of tail inequalities for [math]\displaystyle{ X }[/math].
Chernoff Bound
- [math]\displaystyle{ \begin{align} \mathbf{Pr}[|X| \geq \delta] \leq \min_{t \geq 0} \frac{\mathbf{E}[e^{t|X|}]}{e^{t\delta}} \end{align} }[/math]
[math]\displaystyle{ k }[/math]th-Moment Bound
- [math]\displaystyle{ \begin{align} \mathbf{Pr}[|X| \geq \delta] \leq \frac{\mathbf{E}[|X|^k]}{\delta^k} \end{align} }[/math]
- Show that for each [math]\displaystyle{ \delta }[/math], there exists a choice of [math]\displaystyle{ k }[/math] such that the [math]\displaystyle{ k }[/math]th-moment bound is stronger than the Chernoff bound.
Hint: Consider the Taylor expansion of the moment generating function and apply the probabilistic method.
- Why would we still prefer the Chernoff bound to the (seemingly) stronger [math]\displaystyle{ k }[/math]-th moment bound?
Problem 3
Let [math]\displaystyle{ \Omega }[/math] be a finite set of size [math]\displaystyle{ m }[/math]. Let [math]\displaystyle{ \mathcal{A}=\{A_1,A_2,\dots,A_n\} }[/math] be a family of [math]\displaystyle{ n }[/math] subsets of [math]\displaystyle{ \Omega }[/math]. For any "coloring" [math]\displaystyle{ \chi:\Omega\rightarrow \{-1,+1\} }[/math] that maps each element of [math]\displaystyle{ \Omega }[/math] to [math]\displaystyle{ -1 }[/math] or [math]\displaystyle{ +1 }[/math], and any subset [math]\displaystyle{ A\subseteq \Omega }[/math], we define [math]\displaystyle{ \begin{align} \chi(A)\triangleq \sum\limits_{a\in A}\chi(a). \end{align} }[/math]
Define the discrepancy of the family of subsets [math]\displaystyle{ \mathcal{A} }[/math] with respect to the "coloring"[math]\displaystyle{ \chi }[/math] by
[math]\displaystyle{ \begin{align} \quad\textsf{disc}(\mathcal{A},\chi)\triangleq \max\limits_{A\in \mathcal{A}}|\chi(A)|, \end{align} }[/math]
and the discrepancy of [math]\displaystyle{ \mathcal{A} }[/math] by
[math]\displaystyle{ \begin{align} \quad\textsf{disc}(\mathcal{A})\triangleq \min\limits_{\chi:\Omega\rightarrow\{-1,+1\}} \textsf{disc}(\mathcal{A},\chi). \end{align} }[/math]
- Show that
[math]\displaystyle{ \quad\textsf{disc}(\mathcal{A})\leq \sqrt{2m\ln{(2n)}}. }[/math]
Hint: Apply the probabilistic method.
- Suppose [math]\displaystyle{ m }[/math] is even. Show further that
[math]\displaystyle{ \quad\textsf{disc}(\mathcal{A})\leq \sqrt{m\ln{(2n)}}. }[/math]
Hint: Set [math]\displaystyle{ \chi(i+\frac{m}{2})=-\chi(i) }[/math] for [math]\displaystyle{ 1\leq i\leq \frac{m}{2} }[/math]
Problem 4
Given an undirected graph [math]\displaystyle{ G(V, E) }[/math] with maximum degree [math]\displaystyle{ \Delta }[/math], where the vertex set [math]\displaystyle{ V }[/math] is partitioned into [math]\displaystyle{ r }[/math] disjoint subsets, i.e., [math]\displaystyle{ V = S_1 \cup S_2 \cup \cdots \cup S_r }[/math] with [math]\displaystyle{ S_i \cap S_j = \varnothing }[/math] for all [math]\displaystyle{ i \neq j }[/math]. We call a vertex set [math]\displaystyle{ T }[/math] is a transversal of the partition [math]\displaystyle{ \{S_1, S_2, \cdots, S_r\} }[/math], if [math]\displaystyle{ |T \cap S_i| = 1 }[/math] for all [math]\displaystyle{ 1 \leq i \leq r }[/math]. Assume that [math]\displaystyle{ |S_i| \geq 2e\Delta }[/math] for all [math]\displaystyle{ 1 \leq i \leq r }[/math].
1. Prove that there must be an independent transversal (a transversal which is also an independent set) of [math]\displaystyle{ \{S_1, S_2, \cdots, S_r\} }[/math].
Hint: Lovász Local Lemma.
2. Design a randomized algorithm that finds an independent transversal of [math]\displaystyle{ \{S_1, S_2, \cdots, S_r\} }[/math] in expected polynomial time.