RESOLFT

Freivalds' algorithm (named after Rusins Freivalds) is a probabilistic randomized algorithm used to verify matrix multiplication. Given three n × n matrices A, B, and C, a general problem is to verify whether A × B = C. A naïve algorithm would compute the product A × B explicitly and compare term by term whether this product equals C. However, the best known matrix multiplication algorithm runs in O(n^2.3727) time. Freivalds' algorithm utilizes randomization in order to reduce this time bound to O(n²) with high probability. In O(kn²) time the algorithm can verify a matrix product with probability of failure less than $2^{-k}$.

The algorithm

Input

Three n × n matrices A, B, C.

Output

Yes, if A × B = C; No, otherwise.

Procedure

1. Generate an n × 1 random 0/1 vector $\overrightarrow{r}$.
2. Compute $\overrightarrow{P}=A\times (B\overrightarrow{r})-C\overrightarrow{r}$.
3. Output "Yes" if $\overrightarrow{P}=(0,0,\dots ,0{)}^T$; "No," otherwise.

Error

If A × B = C, then the algorithm always returns "Yes". If A × B ≠ C, then the probability that the algorithm returns "Yes" is less than or equal to one half. This is called one-sided error.

By iterating the algorithm k times and returning "Yes" only if all iterations yield "Yes", a runtime of O(kn²) and error probability of ≤ 1/2^k is achieved.

Example

Suppose one wished to determine whether:

$AB=\left[\begin{array}{cc}2& 3\\ 3& 4\end{array}\right]\left[\begin{array}{cc}1& 0\\ 1& 2\end{array}\right]\stackrel{?}{=}\left[\begin{array}{cc}6& 5\\ 8& 7\end{array}\right]=C.$

A random two-element vector with entries equal to 0 or 1 is selected — say $\overrightarrow{r}=\left[\begin{array}{c}1\\ 1\end{array}\right]$ — and used to compute:

$\begin{array}{rl}A\times (B\overrightarrow{r})-C\overrightarrow{r}& =\left[\begin{array}{cc}2& 3\\ 3& 4\end{array}\right]\left(\left[\begin{array}{cc}1& 0\\ 1& 2\end{array}\right]\left[\begin{array}{c}1\\ 1\end{array}\right]\right)-\left[\begin{array}{cc}6& 5\\ 8& 7\end{array}\right]\left[\begin{array}{c}1\\ 1\end{array}\right]\\ & =\left[\begin{array}{cc}2& 3\\ 3& 4\end{array}\right]\left[\begin{array}{c}1\\ 3\end{array}\right]-\left[\begin{array}{c}11\\ 15\end{array}\right]\\ & =\left[\begin{array}{c}11\\ 15\end{array}\right]-\left[\begin{array}{c}11\\ 15\end{array}\right]\\ & =\left[\begin{array}{c}0\\ 0\end{array}\right].\end{array}$

This yields the zero vector, suggesting the possibility that AB = C. However, if in a second trial the vector $\overrightarrow{r}=\left[\begin{array}{c}1\\ 0\end{array}\right]$ is selected, the result becomes:

$A\times (B\overrightarrow{r})-C\overrightarrow{r}=\left[\begin{array}{cc}2& 3\\ 3& 4\end{array}\right]\left(\left[\begin{array}{cc}1& 0\\ 1& 2\end{array}\right]\left[\begin{array}{c}1\\ 0\end{array}\right]\right)-\left[\begin{array}{cc}6& 5\\ 8& 7\end{array}\right]\left[\begin{array}{c}1\\ 0\end{array}\right]=\left[\begin{array}{c}-1\\ -1\end{array}\right].$

The result is nonzero, proving that in fact AB ≠ C.

There are four two-element 0/1 vectors, and half of them give the zero vector in this case ($\overrightarrow{r}=\left[\begin{array}{c}0\\ 0\end{array}\right]$ and $\overrightarrow{r}=\left[\begin{array}{c}1\\ 1\end{array}\right]$), so the chance of randomly selecting these in two trials (and falsely concluding that AB=C) is 1/2² or 1/4. In the general case, the proportion of r yielding the zero vector may be less than 1/2, and a larger number of trials (such as 20) would be used, rendering the probability of error very small.

Error analysis

Let p equal the probability of error. We claim that if A × B = C, then p = 0, and if A × B ≠ C, then p ≤ 1/2.

A × B = C

If A × B = C, then A × B − C = 0, and so $\overrightarrow{P}=A\times (B\overrightarrow{r})-C\overrightarrow{r}=(A\times B)\overrightarrow{r}-C\overrightarrow{r}=(A\times B-C)\overrightarrow{r}=\overrightarrow{0}$, regardless of what our vector $\overrightarrow{r}$ was.

A × B ≠ C

Let $D=A\times B-C=(d_{ij})$, so $\overrightarrow{P}=D\times \overrightarrow{r}=(p_1,p_2,\dots ,p_n{)}^T$. Since A × B ≠ C, we have A × B − C ≠ 0, so some element of D is nonzero.

Suppose that the element $d_{ij}\ne 0$. By the definition of matrix multiplication, we have $p_i=\sum _{k=1}^n{d}_{ik}{r}_k=d_{i1}{r}_1+\cdots +d_{ij}{r}_j+\cdots +d_{in}{r}_n=d_{ij}{r}_j+y$.

Using Bayes' Theorem, we have $\mathrm{Pr}[p_i=0]=\mathrm{Pr}[p_i=0|y=0]\cdot \mathrm{Pr}[y=0]+\mathrm{Pr}[p_i=0|y\ne 0]\cdot \mathrm{Pr}[y\ne 0]$.

Also, note that:

$\mathrm{Pr}[p_i=0|y=0]=\mathrm{Pr}[r_j=0]=\frac{1}{2}$

$\mathrm{Pr}[p_i=0|y\ne 0]\le \mathrm{Pr}[r_j=1]=\frac{1}{2}$

Plugging these in the above equation, we have:

$\mathrm{Pr}[p_i=0]\le \frac{1}{2}\cdot \mathrm{Pr}[y=0]+\frac{1}{2}\cdot \mathrm{Pr}[y\ne 0]$

$\mathrm{Pr}[p_i=0]\le \frac{1}{2}\cdot \mathrm{Pr}[y=0]+\frac{1}{2}\cdot (1-\mathrm{Pr}[y=0])$

$\mathrm{Pr}[p_i=0]\le \frac{1}{2}$

Therefore,

$\mathrm{Pr}[\overrightarrow{P}=0]\le \mathrm{Pr}[p_i=0]\le \frac{1}{2}.$

This completes the proof.

Ramifications

Simple algorithmic analysis shows that the running time of this algorithm is O(n²), beating the classical deterministic algorithm's bound of O(n³). The error analysis also shows that if we run our algorithm k times, we can achieve an error bound of less than $\frac{1}{2^k}$, an exponentially small quantity. The algorithm is also fast in practice due to wide availability of fast implementations for matrix-vector products. Therefore, utilization of randomized algorithms can speed up a very slow deterministic algorithm. In fact, the best known deterministic matrix multiplication verification algorithm known at the current time is a variant of the Coppersmith-Winograd algorithm with an asymptotic running time of O(n^2.3727).

Freivalds' algorithm frequently arises in introductions to probabilistic algorithms due to its simplicity and how it illustrates the superiority of probabilistic algorithms in practice for some problems.

See also

• Schwartz–Zippel lemma

References

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• Freivalds, R. (1977), “Probabilistic Machines Can Use Less Running Time”, IFIP Congress 1977, pp. 839–842.