高级算法 (Fall 2022)/Problem Set 1

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Problem 1

Modify the Karger's Contraction algorithm so that it works for the weighted min-cut problem. Prove that the modified algorithm returns a weighted minimum cut with probability at least [math]\displaystyle{ \frac{2}{n(n-1)} }[/math]. The weighted min-cut problem is defined as follows.

  • Input: an undirected weighted graph [math]\displaystyle{ G(V, E) }[/math], where every edge [math]\displaystyle{ e \in E }[/math] is associated with a positive real weight [math]\displaystyle{ w_e }[/math];
  • Output: a cut [math]\displaystyle{ C }[/math] in [math]\displaystyle{ G }[/math] such that [math]\displaystyle{ \sum_{e \in C} w_e }[/math] is minimized.

Problem 2

Consider the function [math]\displaystyle{ f:\mathbb{R}^n\to\mathbb{R} }[/math] defined as

[math]\displaystyle{ f(\vec x)=f(x_1,x_2,\dots,x_n)=\prod_{i=1}^{n}(a_ix_i-b_i) }[/math],

where [math]\displaystyle{ \{a_i\}_{1\le i\le n} }[/math] and [math]\displaystyle{ \{b_i\}_{1\le i\le n} }[/math] are unknown coefficients satisfy that [math]\displaystyle{ a_i, b_i\in \mathbb{Z} }[/math] and [math]\displaystyle{ 0\le a_i, b_i \le n }[/math] for all [math]\displaystyle{ 1\le i\le n }[/math].

Let [math]\displaystyle{ p\gt n }[/math] be the smallest prime strictly greater than [math]\displaystyle{ n }[/math]. The function [math]\displaystyle{ g:\mathbb{Z}_p^n\to\mathbb{Z}_p }[/math] is defined as

[math]\displaystyle{ g(\vec x)=g(x_1,x_2,\dots,x_n)=\prod_{i=1}^{n}(a_ix_i-b_i) }[/math],

where [math]\displaystyle{ + }[/math] and [math]\displaystyle{ \cdot }[/math] are defined over the finite field [math]\displaystyle{ \mathbb{Z}_p }[/math].

By the properties of finite field, for any value [math]\displaystyle{ \vec r\in\mathbb{Z}_p^n }[/math], it holds that [math]\displaystyle{ g(\vec r)=f(\vec r)\bmod p }[/math].

Since the coefficients [math]\displaystyle{ \{a_i\}_{1\le i\le n} }[/math] and [math]\displaystyle{ \{b_i\}_{1\le i\le n} }[/math] are unknown, you can't calculate [math]\displaystyle{ f(\vec x) }[/math] directly. However, there exists an oracle [math]\displaystyle{ O }[/math], each time [math]\displaystyle{ O }[/math] gets an input [math]\displaystyle{ \vec x }[/math], it immediately outputs the value of [math]\displaystyle{ g(\vec x) }[/math].

1. Prove that [math]\displaystyle{ f\not\equiv 0 \Rightarrow g\not\equiv 0 }[/math].

2. Use the oracle [math]\displaystyle{ O }[/math] to design an algorithm to determine whether [math]\displaystyle{ f\equiv 0 }[/math], with error probability at most [math]\displaystyle{ \epsilon }[/math], where [math]\displaystyle{ \epsilon\in (0,1) }[/math] is a constant.

Problem 3

In Balls-and-Bins model, we throw [math]\displaystyle{ m }[/math] balls independently and uniformly at random into [math]\displaystyle{ n }[/math] bins. We know that the maximum load is [math]\displaystyle{ \Theta\left(\frac{\log n}{\log\log n}\right) }[/math] with high probability when [math]\displaystyle{ m=\Theta(n) }[/math]. The two-choice paradigm is another way to throw [math]\displaystyle{ m }[/math] balls into [math]\displaystyle{ n }[/math] bins: each ball is thrown into the least loaded of two bins chosen independently and uniformly at random(it could be the case that the two chosen bins are exactly the same, and then the ball will be thrown into that bin), and breaks the tie arbitrarily. When [math]\displaystyle{ m=\Theta(n) }[/math], the maximum load of two-choice paradigm is known to be [math]\displaystyle{ \Theta(\log\log n) }[/math] with high probability, which is exponentially less than the maxim load when there is only one random choice. This phenomenon is called the power of two choices.

Here are the questions:

  • Consider the following paradigm: we throw [math]\displaystyle{ n }[/math] balls into [math]\displaystyle{ n }[/math] bins. The first [math]\displaystyle{ \frac{n}{2} }[/math] balls are thrown into bins independently and uniformly at random. The remaining [math]\displaystyle{ \frac{n}{2} }[/math] balls are thrown into bins using the two-choice paradigm. What is the maximum load with high probability? You need to give an asymptotically tight bound (in the form of [math]\displaystyle{ \Theta(\cdot) }[/math]).
  • Replace the above paradigm to the following: the first [math]\displaystyle{ \frac{n}{2} }[/math] balls are thrown into bins using the two-choice paradigm while the remaining [math]\displaystyle{ \frac{n}{2} }[/math] balls are thrown into bins independently and uniformly at random. What is the maximum load with high probability in this case? You need to give an asymptotically tight bound.
  • Replace the above paradigm to the following: assume all [math]\displaystyle{ n }[/math] balls are thrown in a sequence. For every [math]\displaystyle{ 1\le i\le n }[/math], if [math]\displaystyle{ i }[/math] is odd, we throw [math]\displaystyle{ i }[/math]-th ball into bins independently and uniformly at random, otherwise, we throw it into bins using the two-choice paradigm. What is the maximum load with high probability in this case? You need to give an asymptotically tight bound.

Problem 4

In class, we saw how to estimate the number of distinct elements in a data stream using the Flajolet-Martin algorithm. Consider the following alternative formulation of the distinct elements problem: given an [math]\displaystyle{ N }[/math] dimensional vector [math]\displaystyle{ x }[/math], we want to process a stream of arbitrary increments to entries in [math]\displaystyle{ x }[/math]. In other words, if we see a number [math]\displaystyle{ i\in 1,\dots,N }[/math] in the stream, we update entry [math]\displaystyle{ x_i\gets x_i + 1 }[/math]. Our goal is to estimate [math]\displaystyle{ \left \|x\right \|_0 }[/math], which measures the number of non-zero entries in [math]\displaystyle{ x }[/math]. With [math]\displaystyle{ x }[/math] viewed as a histogram that maintains counts for [math]\displaystyle{ N }[/math] potential elements, [math]\displaystyle{ \left \|x\right \|_0 }[/math] is exactly the number of distinct elements processed. In this problem we will develop an alternative algorithm for estimating [math]\displaystyle{ \left \|x\right \|_0 }[/math] that can also handle decrements to entries in x. Specifically, instead of the stream containing just indices [math]\displaystyle{ i }[/math], it contains pairs [math]\displaystyle{ (i, +) }[/math] and [math]\displaystyle{ (i, −) }[/math]. On receiving [math]\displaystyle{ (i, +) }[/math], [math]\displaystyle{ x }[/math] should update so that [math]\displaystyle{ x_i\gets x_i + 1 }[/math] and on receiving [math]\displaystyle{ (i, −) }[/math], [math]\displaystyle{ x }[/math] should update so that [math]\displaystyle{ x_i\gets x_i - 1 }[/math]. For this problem we will assume that, at the end of our stream, each [math]\displaystyle{ x_i \ge 0 }[/math] (i.e. for a specific index we can’t receive more decrements than increments).

  • Consider a simpler problem. For a given value [math]\displaystyle{ T }[/math], let’s design an algorithm that succeeds with probability [math]\displaystyle{ (1 − \delta) }[/math], outputing LOW if [math]\displaystyle{ T \lt \frac{1}{2}\left \|x\right \|_0 }[/math] and HIGH if [math]\displaystyle{ T \gt 2\left \|x\right \|_0 }[/math]:
    • Assume we have access to a completely random hash function [math]\displaystyle{ h(\cdot) }[/math] that maps each [math]\displaystyle{ i }[/math] to a random point in [math]\displaystyle{ [0, 1] }[/math]. We maintain the estimator [math]\displaystyle{ s=\sum_{i:h(i)\lt \frac{1}{2T}}x_i }[/math] as we receive increment and decrement updates. Show that, at the end of our stream,
      (i) If [math]\displaystyle{ T \lt \frac{1}{2}\left \|x\right \|_0 }[/math], [math]\displaystyle{ \Pr_h[s=0]\lt 1/e\approx 0.37 }[/math];
      (ii) If [math]\displaystyle{ T \gt 2\left \|x\right \|_0 }[/math], [math]\displaystyle{ \Pr_h[s=0]\gt 0.5 }[/math].
    • Using this fact, show how to use [math]\displaystyle{ k=O(\log 1/\delta) }[/math] independent random hash functions, and corresponding individual estimators [math]\displaystyle{ s_1, s_2, . . . , s_k }[/math], to output LOW if [math]\displaystyle{ T \lt \frac{1}{2}\left \|x\right \|_0 }[/math] and HIGH if [math]\displaystyle{ T \gt 2\left \|x\right \|_0 }[/math]. If neither event occurs you can output either LOW or HIGH. Your algorithm should succeed with probability [math]\displaystyle{ (1 − \delta) }[/math].
  • Using [math]\displaystyle{ O(\log N) }[/math] repetitions of your algorithm for the above decision problem (with [math]\displaystyle{ \delta }[/math] set appropriately), show how to obtain an estimate [math]\displaystyle{ F }[/math] for [math]\displaystyle{ \left \|x\right \|_0 }[/math] such that [math]\displaystyle{ \frac{1}{4}\left \|x\right \|_0\le F\le 4\left \|x\right \|_0 }[/math] w.h.p.(with probability [math]\displaystyle{ 1-O(1/N) }[/math]).