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 ${\vec {r}}$.
2. Compute ${\vec {P}}=A\times (B{\vec {r}})-C{\vec {r}}$.
3. Output "Yes" if ${\vec {P}}=(0,0,\ldots ,0)^{T}$; "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:

$AB={\begin{bmatrix}2&3\\3&4\end{bmatrix}}{\begin{bmatrix}1&0\\1&2\end{bmatrix}}{\stackrel {?}{=}}{\begin{bmatrix}6&5\\8&7\end{bmatrix}}=C.$

A random two-element vector with entries equal to 0 or 1 is selected — say ${\vec {r}}={\begin{bmatrix}1\\1\end{bmatrix}}$ — and used to compute:

${\begin{aligned}A\times (B{\vec {r}})-C{\vec {r}}&={\begin{bmatrix}2&3\\3&4\end{bmatrix}}\left({\begin{bmatrix}1&0\\1&2\end{bmatrix}}{\begin{bmatrix}1\\1\end{bmatrix}}\right)-{\begin{bmatrix}6&5\\8&7\end{bmatrix}}{\begin{bmatrix}1\\1\end{bmatrix}}\\&={\begin{bmatrix}2&3\\3&4\end{bmatrix}}{\begin{bmatrix}1\\3\end{bmatrix}}-{\begin{bmatrix}11\\15\end{bmatrix}}\\&={\begin{bmatrix}11\\15\end{bmatrix}}-{\begin{bmatrix}11\\15\end{bmatrix}}\\&={\begin{bmatrix}0\\0\end{bmatrix}}.\end{aligned}}$

This yields the zero vector, suggesting the possibility that AB = C. However, if in a second trial the vector ${\vec {r}}={\begin{bmatrix}1\\0\end{bmatrix}}$ is selected, the result becomes:

$A\times (B{\vec {r}})-C{\vec {r}}={\begin{bmatrix}2&3\\3&4\end{bmatrix}}\left({\begin{bmatrix}1&0\\1&2\end{bmatrix}}{\begin{bmatrix}1\\0\end{bmatrix}}\right)-{\begin{bmatrix}6&5\\8&7\end{bmatrix}}{\begin{bmatrix}1\\0\end{bmatrix}}={\begin{bmatrix}-1\\-1\end{bmatrix}}.$

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 (${\vec {r}}={\begin{bmatrix}0\\0\end{bmatrix}}$ and ${\vec {r}}={\begin{bmatrix}1\\1\end{bmatrix}}$), 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 ${\vec {P}}=A\times (B{\vec {r}})-C{\vec {r}}=(A\times B){\vec {r}}-C{\vec {r}}=(A\times B-C){\vec {r}}={\vec {0}}$, regardless of what our vector ${\vec {r}}$ was.

A × B ≠ C

Let $D=A\times B-C=(d_{ij})$, so ${\vec {P}}=D\times {\vec {r}}=(p_{1},p_{2},\dots ,p_{n})^{T}$. Since A × B ≠ C, we have A × B − C ≠ 0, so some element of D is nonzero.

Suppose that the element $d_{ij}\neq 0$. 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 $\Pr[p_{i}=0]=\Pr[p_{i}=0|y=0]\cdot \Pr[y=0]+\Pr[p_{i}=0|y\neq 0]\cdot \Pr[y\neq 0]$.

Also, note that:

$\Pr[p_{i}=0|y=0]=\Pr[r_{j}=0]={\frac {1}{2}}$

$\Pr[p_{i}=0|y\neq 0]\leq \Pr[r_{j}=1]={\frac {1}{2}}$

Plugging these in the above equation, we have:

$\Pr[p_{i}=0]\leq {\frac {1}{2}}\cdot \Pr[y=0]+{\frac {1}{2}}\cdot \Pr[y\neq 0]$

$\Pr[p_{i}=0]\leq {\frac {1}{2}}\cdot \Pr[y=0]+{\frac {1}{2}}\cdot (1-\Pr[y=0])$

$\Pr[p_{i}=0]\leq {\frac {1}{2}}$

Therefore,

$\Pr[{\vec {P}}=0]\leq \Pr[p_{i}=0]\leq {\frac {1}{2}}.$

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.