Randomized Algorithms (Spring 2010)/Fingerprinting

Fingerprinting

Checking identities

Example: Checking matrix multiplication

Consider the following problem:

• Given as the input three [math]\displaystyle{ n\times n }[/math] matrices [math]\displaystyle{ A,B }[/math] and [math]\displaystyle{ C }[/math],
• check whether [math]\displaystyle{ C=AB }[/math].

Algorithm (Freivalds)

pick a vector [math]\displaystyle{ r \in\{0, 1\}^n }[/math] uniformly at random; if [math]\displaystyle{ A(Br) = Cr }[/math] then return "yes" else return "no";

If [math]\displaystyle{ AB=C }[/math] then [math]\displaystyle{ A(Br) = Cr }[/math] for any [math]\displaystyle{ r \in\{0, 1\}^n }[/math], thus the algorithm always returns "yes".

Lemma

If [math]\displaystyle{ AB\neq C }[/math] then for a uniformly random [math]\displaystyle{ r \in\{0, 1\}^n }[/math], [math]\displaystyle{ \Pr[A(Br) = Cr]\le \frac{1}{2} }[/math].

Example: Checking polynomial identities

Consider the following problem:

• Given as the input two multivariate polynomials [math]\displaystyle{ P_1(x_1,\ldots,x_n) }[/math] and [math]\displaystyle{ P_2(x_1,\ldots,x_n) }[/math],
• check whether [math]\displaystyle{ P_1\equiv P_2 }[/math].

Algorithm (Schwartz-Zippel)

pick [math]\displaystyle{ r_1, \ldots , r_n }[/math] independently and uniformly at random from a set [math]\displaystyle{ S }[/math]; if [math]\displaystyle{ P_1(r_1, \ldots , r_n) = P_2(r_1, \ldots , r_n) }[/math] then return “yes” else return “no”;

Theorem (Schwartz-Zippel)

Let [math]\displaystyle{ Q(x_1,\ldots,x_n) }[/math] be a multivariate polynomial of degree [math]\displaystyle{ d }[/math] defined over a field [math]\displaystyle{ \mathbb{F} }[/math]. Fix any finite set [math]\displaystyle{ S\subset\mathbb{F} }[/math], and let [math]\displaystyle{ r_1,\ldots,r_n }[/math] be chosen independently and uniformly at random from [math]\displaystyle{ S }[/math]. Then [math]\displaystyle{ \Pr[Q(r_1,\ldots,r_n)=0\mid Q\not\equiv 0]\le\frac{d}{|S|}. }[/math]

Fingerprinting

Suppose we want to compare two items [math]\displaystyle{ Z_1 }[/math] and [math]\displaystyle{ Z_2 }[/math]. Instead of comparing them directly, we compute random fingerprints [math]\displaystyle{ \mathrm{FING}(Z_1) }[/math] and [math]\displaystyle{ \mathrm{FING}(Z_2) }[/math] and compare these. The fingerprints has the following properties:

• If [math]\displaystyle{ Z_1\neq Z_2 }[/math] then [math]\displaystyle{ \Pr[\mathrm{FING}(Z_1)=\mathrm{FING}(Z_2)] }[/math] is small.
• It is much more to compute and compare the fingerprints than to compare [math]\displaystyle{ Z_1 }[/math] and [math]\displaystyle{ Z_2 }[/math] directly.

For Freivald's algorithm, the items to compare are two [math]\displaystyle{ n\times n }[/math] matrices [math]\displaystyle{ AB }[/math] and [math]\displaystyle{ C }[/math], and given an [math]\displaystyle{ n\times n }[/math] matrix [math]\displaystyle{ M }[/math], its random fingerprint is computed as [math]\displaystyle{ \mathrm{FING}(M)=Mr }[/math] for a uniformly random [math]\displaystyle{ r\in\{0,1\}^n }[/math].

For the Schwartz-Zippel algorithm, the items to compare are two polynomials [math]\displaystyle{ P_1(x_1,\ldots,x_n) }[/math] and [math]\displaystyle{ P_2(x_1,\ldots,x_n) }[/math], and given a polynomial [math]\displaystyle{ Q(x_1,\ldots,x_n) }[/math], its random fingerprint is computed as [math]\displaystyle{ \mathrm{FING}(Q)=Q(r_1,\ldots,r_n) }[/math] for [math]\displaystyle{ r_i }[/math] chosen independently and uniformly at random from some fixed set [math]\displaystyle{ S }[/math].

Communication complexity

Alice and Bob are two entities. Alice has a private input [math]\displaystyle{ x }[/math] and Bob has a private input [math]\displaystyle{ y }[/math]. Together they want to compute a function [math]\displaystyle{ f(x,y) }[/math] by communicating with each other. This is the model of communication complexity due to Yao.

In the communication complexity model, the local computational costs are ignored. The complexity of algorithms (also called communication protocols here) are measured by the number of bits communicated between Alice and Bob.

One of the basic function is IDENTITY, defined as

[math]\displaystyle{ \mathrm{IDENTITY}(x,y)= \begin{cases} 1& \mbox{if } x=y,\\ 0& \mbox{otherwise.} \end{cases} }[/math]

A trivial way to solve IDENTITY is to let Bob send [math]\displaystyle{ y }[/math] to Alice. Supposed that [math]\displaystyle{ x,y\in\{0,1\}^n }[/math], this costs [math]\displaystyle{ n }[/math] bits of communications.