View source for 高级算法 (Fall 2024)/Problem Set 1

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== Problem 1 (Min-cut/Max-cut) ==
For any '''<math>\alpha \ge 1</math>''', a cut is called an '''<math>\alpha</math>'''-approximate min-cut in a multigraph '''<math>G</math>''' if the number of edges in it is at most '''<math>\alpha</math>''' times that of the min-cut.  Prove that the number of '''<math>\alpha</math>'''-approximate min-cuts in a multigraph '''<math>G</math>''' is at most '''<math>n^{2\alpha} / 2</math>'''.  ('''''Hint''''': Run Karger's algorithm until it has '''<math>\lceil 2\alpha \rceil</math>''' supernodes. What is the chance that a particular '''<math>\alpha</math>'''-approximate min-cut is still available? How many possible cuts does this collapsed graph have?)

== Problem 2 (Fingerprinting) ==
Two rooted trees <math>T_1</math> and <math>T_2</math> are said to be isomorphic if there exists a one to one mapping <math>f</math> from the nodes of <math>T_1</math> to those of <math>T_2</math> satisfying the following condition: <math>v</math> is a child of <math>w</math> in <math>T_1</math> if and only if <math>f(v)</math> is a child of <math>f(w)</math> in <math>T_2</math>. Observe that no ordering is assumed on the children of any vertex. Devise an efficient randomized algorithm for testing the isomorphism of rooted trees and analyze its performance. '''''Hint:''''' Recursively associate a polynomial <math>P_v</math> with each vertex <math>v</math> in a tree <math>T</math>.

== Problem 3 (Hashing and Sketching) ==
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>N</math> dimensional vector <math>x</math>, we want to process a stream of arbitrary increments to entries in <math>x</math>. In other words, if we see a number <math>i\in 1,\dots,N</math> in the stream, we update entry <math>x_i\gets x_i + 1</math>. Our goal is to estimate <math>\left \|x\right \|_0</math>, which measures the number of non-zero entries in <math>x</math>. With <math>x</math> viewed as a histogram that maintains counts for <math>N</math> potential elements, <math>\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>\left \|x\right \|_0</math> that can also handle '''decrements''' to entries in <math>x</math>. Specifically, instead of the stream containing just indices <math>i</math>, it contains pairs <math>(i, +)</math> and <math>(i, −)</math>. On receiving <math>(i, +)</math>, <math>x</math> should update so that <math>x_i\gets x_i + 1</math> and on receiving <math>(i, −)</math>, <math>x</math> should update so that <math>x_i\gets x_i - 1</math>. For this problem we will assume that, at the end of our stream, each <math>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>T</math>, let’s design an algorithm that succeeds with probability <math>(1 − \delta)</math>, outputing '''LOW''' if <math>T < \frac{1}{2}\left \|x\right \|_0</math> and '''HIGH''' if <math>T > 2\left \|x\right \|_0</math>:
#* Assume we have access to a completely random hash function <math>h(\cdot)</math> that maps each <math>i</math> to a random point in <math>[0, 1]</math>. We maintain the estimator <math>s=\sum_{i:h(i)<\frac{1}{2T}}x_i</math> as we receive increment and decrement updates. Show that, at the end of our stream,    (i)  If <math>T < \frac{1}{2}\left \|x\right \|_0</math>, <math>\Pr_h[s=0]<1/e\approx 0.37</math> and (ii)  If <math>T > 2\left \|x\right \|_0</math>, <math>\Pr_h[s=0]>0.5</math>.
#* Using this fact, show how to use <math>k=O(\log 1/\delta)</math> independent random hash functions, and corresponding individual estimators <math>s_1, s_2, . . . , s_k</math>, to output '''LOW''' if <math>T < \frac{1}{2}\left \|x\right \|_0</math> and '''HIGH''' if <math>T > 2\left \|x\right \|_0</math>. If neither event occurs you can output either '''LOW''' or '''HIGH'''. Your algorithm should succeed with probability <math>(1 − \delta)</math>.
# Using <math>O(\log N)</math> repetitions of your algorithm for the above decision problem (with <math>\delta</math> set appropriately), show how to obtain an estimate <math>F</math> for <math>\left \|x\right \|_0</math> such that <math>\frac{1}{4}\left \|x\right \|_0\le F\le 4\left \|x\right \|_0</math> w.h.p.(with probability <math>1-O(1/N)</math>).

== Problem 4 (Concentration of measure) ==
Consider the [[wikipedia:Erdős–Rényi_model#Definition|Erdős–Rényi random graph]] <math>G(n, p)</math> where every two vertices in the graph are connected randomly and independently with probability <math>p</math>. We denote <math>G \sim G(n, p)</math> if <math>G</math> is generated in this way. Recall that <math>\chi(G)</math> is the chromatic number of the graph <math>G</math>.

'''(a.)''' For <math>0 < p_1 < p_2 < 1</math>, let <math>G_1 \sim G(n, p_1)</math> and let <math>G_2 \sim G(n, p_2)</math>. Compare <math>\mathbf{E}[\chi(G_1)]</math> and <math>\mathbf{E}[\chi(G_2)]</math> and prove it.

'''(b.)''' For <math>G \sim G(n, n^{-\alpha})</math> with <math>\alpha > 5/6</math> and constant <math>C > 0</math>, prove that every subgraph of <math>G</math> on <math>C\sqrt{n \log n}</math> vertices is <math>3</math>-colorable with probability <math>1 - o(1)</math> when <math>n</math> is large enough. ('''''Hint''''': <math>\binom{n}{k} \leq (en/k)^k</math>.)

'''(c.)''' For <math>G \sim G(n, n^{-\alpha})</math> with <math>\alpha > 5/6</math>, show that <math>\chi(G)</math> is concentrated on four values with probability <math>1 - o(1)</math> when <math>n</math> is large enough. To be more exact, show that there exists an integer <math>u</math> such that <math>u \leq \chi(G) \leq u+3</math> with probability <math>1 - o(1)</math> when <math>n</math> is large enough.

== Problem 5 (Dimension reduction) ==
['''Inner product'''] Fix parameters <math>d>0, \delta,\epsilon\in(0,1)</math>. Let <math>A\in \mathbb{R}^{k\times d}</math> be a random matrix with <math>k = O(\log(1/\delta)/\epsilon^2)</math> rows, and entries of <math>A</math> are chosen i.i.d. from Gaussian distribution with mean <math>0</math> and variance <math>1/k</math>. Prove that for any <math>x,y\in \mathbb{R}^d</math>: <math>|x^\top y - (Ax)^\top(Ay)|\leq \epsilon(\|x\|_2^2 + \|y\|_2^2)</math> with probability <math>\geq 1-\delta</math>.

['''Linear separability'''] In machine learning, the goal of many classification methods is to seperate data into classes using a hyperplane. A hyperplane in <math>\mathbb{R}^d</math> is characterized by a unit vector <math>a\in \mathbb{R}^d (\|a\|_2 = 1)</math> and <math>c\in \mathbb{R}</math>. It contains all <math>z\in \mathbb{R}^d</math> such that <math>a^\top z = c</math>. Suppose our dataset consists of <math>n</math> '''unit''' vectors in <math>\mathbb{R}^d</math>. These points can be separated into two linearly separable sets <math>X,Y</math> where <math>|X|+|Y| = n</math>. That is, for all <math>x\in X</math>, <math>a^\top x>c</math> and for all <math>y\in Y</math>, <math>a^\top y<c</math> (or vice versa). Furthermore, suppose that the <math>\ell_2</math> distance of each point in <math>X</math> and <math>Y</math> to this separating hyperplane is at least <math>\epsilon</math>. When this is the case, the hyperplane is said to have margin <math>\epsilon</math>.
# Show that <math>X,Y</math> can be separated by the hyperplane characterized by <math>a\in \mathbb{R}^d (\|a\|_2 = 1)</math> and  <math>c\in \mathbb{R}</math> with margin <math>\epsilon</math> is equivalent to the following condition: for all <math>x\in X</math>, <math>a^\top x \geq c+\epsilon</math> and for all <math>y\in Y</math>, <math>a^\top y \leq c-\epsilon</math> (or vice versa).
# Show that if we use a Johnson-Lindenstrauss map <math>A\in \mathbb{R}^{k\times d}</math> (the scaled Gaussian matrix given in the lecture) to reduce our data points to <math>O(\log n/\epsilon^2)</math> dimensions, then with probability at least <math>9/10</math>, the dimension reduced data can still be separated by a hyperplane with margin <math>\epsilon/4</math>. ('''''Hint''''': use the fact that JLT preserves inner product.)