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This is an elementary (yet important) fact in matrix analysis.

Statement

Let M be an n\times n complex Hermitian matrix which means M=M^* where * denote the conjugate transpose operation. Let \lambda_1, \neq \lambda_2 be two different  eigenvalues of M. Let x, y be the two eigenvectors of M corresponding to the two eigenvalues \lambda_1 and \lambda_2, respectively.

Then the following is true:

\boxed{\lambda_1 \neq \lambda_2 \iff \langle x, y \rangle = 0.}

Here \langle a,b\rangle denote the usual inner product of two vectors x,y, i.e.,

\langle x,y \rangle := y^*x.

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Informally speaking, Impagliazzo’s hardcore lemma says that if a boolean function is “hard to compute on average” by small circuits, then there exists a set of inputs on which the same function is “extremely hard to compute on average” by slightly smaller circuits.

In this post, I am going to explain how I understand the proof of the hardcore lemma presented in the Arora-Barak complexity book (here). Because the formal proof can be found in the book I intend to write in an informal way. I think some subtleties are involved in turning the context of the lemma into a suitable two-player zero-sum game. Doing so enables one to use von Neumann’s minimax theorem to effectively “exchange the quantifiers” in the contrapositive statement of the lemma. Although the Arora-Barak proof mentions these subtleties, I am going to explore these in more detail and in a more accessible way for a beginner like me.

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You must have known it already, but I can’t stop myself from showing off the interplay of const with pointers in C/C++.

Here goes the code. It’s simple.


void main() {

	int arr[]={1,2,3,4}; // data

	const int *p1 = &arr[0]; // non-const ptr to const data
	int const *p2 = &arr[0]; // non-const ptr to const data
	int* const p3 = &arr[0]; // const ptr to non-const data
	const int* const p4 = &arr[0]; // const ptr to const data

	p1++; // ok
	p1[0]++; // compile error: modifying const data

	p2++; // ok
	p2[0]++; // compile error: modifying const data

	p3++; // compile error: modifying const ptr
	p3[0]++; // ok 

	p4++; // compile error: modifying const ptr
	p4[0]++; // compile error: modifying const data

}

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We want to show that for 1\leq k\leq n, the following holds: {n\choose k} \leq \Big(\frac{n\cdot e}{k}\Big)^k.

First, recall the power series e^k=\sum_{i=0}^\infty{\frac{k^i}{i!}}. For k \geq 1, this sum is definitely larger than its kth term only. In other words, e^k \geq k^k/k!, which implies k! \geq (k/e)^k. Now,

\begin{matrix}{n\choose k}  &=& \frac{n!}{k! \cdot (n-k)!}\\  &=& \frac{n(n-1)(n-2)\cdots (n-(k-1))}{k!}\\  &\leq& \frac{n^k}{k!} \\  &\leq& \frac{n^k}{(k/e)^k} \\  &=& (\frac{en}{k})^k.\\  \end{matrix}

This proof is extracted from the StackExchange discussion here. For more inequalities/bounds for binomial coefficients, see Wikipedia.

We will show that the eigenvalues of the n\times n Laplacian matrix L of the complete graph K_n are \{0,1\} where  the eigenvalue 0 has (algebraic) multiplicity 1 and the eigenvalue n has multiplicity n-1.

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In this post, we provide a proof of Kirchhoff’s Matrix Tree theorem [1] which is quite beautiful in our biased opinion. This is a 160-year-old theorem which connects several fundamental concepts of matrix analysis and graph theory (e.g., eigenvalues, determinants, cofactor/minor, Laplacian matrix, incidence matrix, connectedness, graphs, and trees etc.). There are already many expository articles/proof sketches/proofs of this theorem (for example, see [4-6, 8,9]). Here we compile yet-another-expository-article and present arguments in a way which should be accessible to the beginners.

Statement

Suppose  L is the Laplacian of a connected simple graph G with n vertices. Then the number of spanning trees in G, denoted by t(G), is the following:

t(G)=\frac{1}{n} \times \{ the product of all non-zero eigenvalues of L\ \}.

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In this post I will discuss my own understanding of the spectral sparsification paper by Daniel Spielman and Nikhil Srivastava (STOC 2008). I will assume the following:

  1. The reader is a beginner, like me, and have already glanced through the Spielman-Srivastava paper (from now on, the SS paper).
  2. The reader has, like me, a basic understanding of spectral sparsification and associated concepts of matrix analysis. I will assume that she has read and understood the Section 2 (Preliminaries) of the SS paper.
  3. The reader holds a copy of the SS paper while reading my post.

First, I will mention the main theorems (actually, I will mention only what they “roughly” say).

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Introduction

This is a beautiful theorem which states that the difference between a positive integer a and a prime power of a, a^p-a, is divisible by p. It’s history can be found in Wikipedia: Fermat posited the theorem (and as usual, did not give the proof) while Euler first published a proof using mathematical induction. Below, we will state the theorem and provide a simple-to-understand proof using only modular arithmetic, followed by another simple proof using mathematical induction.

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Proofs from the Book

Proofs from the Book

This book, inspired from Paul Erdős‘ notion of the Book, presents some beautiful proofs to some intriguing math problems. Both the problems and proofs presented here are elegant, clear, and artistic.

Riemann zeta function is a rather simple-looking function. For any number s, the zeta function \zeta(s) is the sum of the reciprocals of all natural numbers raised to the s^\mathrm{th} power.

\begin{array}{ccl}\zeta(s)&=&\displaystyle\sum_{n=1}^{\infty}{\frac{1}{n^s}}\\&=&1+\frac{1}{2^s}+\frac{1}{3^s}+\frac{1}{4^s}+\frac{1}{5^s}+\frac{1}{6^s}+\cdots\end{array}

Although the variable s is a complex number, for the sake of simplicity, we will treat s as real.  (Real numbers are a subset of complex numbers.)

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