Freivalds’ Algorithm

Freivalds, Rusins. “Probabilistic Machines Can Use Less Running Time.” IFIP congress. Vol. 839. 1977.

Problem Statement

Suppose we are given as input two $n\times n$ matrices $A$ and $B$ over ${\mathbb{F}}_p$, where $p>n^2$ is a prime number. The fastest known algorithm for accomplishing this task in time roughly $\mathcal{O}(n^{2.37286})$.

Suppose someone hands us a matrix $C$. Our goal is to check whether or not the product matrix $A\cdot B=C$ in $\mathcal{O}(n^2)$ time.

Algorithm

First, choose a random $r\in {\mathbb{F}}_p$, and let $x=(1,r,r^2,...,r^{n-1})$. Then compute $y=Cx$ and $z=A\cdot Bx$, outputting $1$ if $y=z$ and $0$ otherwise.

Time Cost

1. vector $x=(1,r,r^2,...,r^{n-1})$ can be done with $\mathcal{O}(n)$ total multiplication operations.
2. Multiply an $n\times n$ matrix by an n-dimensional vector can be done in $\mathcal{O}(n^2)$ time.
   2.1. $y=Cx$ : $\mathcal{O}(n^2)$ time
   2.2. $w=Bx$ : $\mathcal{O}(n^2)$ time
   2.3. $z=Aw$ : $\mathcal{O}(n^2)$ time

Explaination

Recall the Reed-Solomon Fingerprinting: Encode vector $a$ and $b$ with Reed-Solomon Encoding: $p_a(x)={\sum }_{i=1}^n{a}_i{r}^{i-1}$,$p_b(x)={\sum }_{i=1}^n{b}_i{r}^{i-1}$. If $a_i=b_i$ for all $i=1,...,n$, then $p_a(r)=p_b(r)$ for every possible choice of $r$. Otherwise, if there is even one $i$ such that $a_i\ne b_i$, $p_a(r)=p_b(r)$ with probability at least $1-(n-1)/p$. (It can be proved by Schwartz-Zippel Lemma.)

The encoding is distance-amplifying: if $a$ and $b$ differ on even a single coordinate, then their encodings will differ on a $1-(n-1)/p$ fraction of coordinates. Due to the distanceamplifying nature of the code, checking equality of two vectors a and b was reduced to checking equality of a single randomly chosen entry of the encodings.