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Ravi is an armchair futurist and an aspiring mad scientist. His mission is to create simplicity out of complexity and order out of chaos.

Friday, May 18, 2012

Freivalds' Algorithm


Freivalds' algorithm is used to verify correctness of matrix multiplication. It is a randomized Monte Carlo algorithm and has a small probability of error. By repeatedly running this algorithm, one can reduce the probability of error below any threshold.


One naive way to verify correctness of matrix multiplication is to actually redo the multiplication and compare the output with the given input. However, current matrix multiplication algorithms are quite slow - the fastest one being $O(n^{2.376})$. Freivalds' algorithm takes $\Theta(n^2)$ time to verify this multiplication and errs with probability $\leq 1/2$, only when the multiplication is incorrect. It therefore runs faster than matrix multiplication itself and is usually correct.

Freivalds' algorithm

  1. Input: $n\times{n}$ matrices $A$, $B$ and $C$. We would like to verify if $A\times{B}=C$.
  2. Choose a $n\times{1}$ column vector $\overrightarrow{r}\in\{0,1\}^n$, uniformly and at random. In other words, each component of the vector is either 0 or 1 and chosen randomly and uniformly.
  3. Compute $A\cdot(B\cdot\overrightarrow{r})$ and $C\cdot\overrightarrow{r}$. This takes $\Theta(n^2)$ time.
  4. Now comes the output
    1. If $A \cdot (B\cdot\overrightarrow{r}) \neq {C\cdot\overrightarrow{r}}$, output FALSE. In other words, we say that the given matrix multiplication is incorrect. This has a probability of correctness = 1.
    2. If $A \cdot (B\cdot\overrightarrow{r}) = {C\cdot\overrightarrow{r}}$, output TRUE. In other words, we say that the given matrix multiplication is correct. This has a probability of correctness $\geq{1/2}$. (See references for details.)


  • If Freivalds' algorithm says that:
    • $A\times{B}\neq{C}$, then this statement is always correct.
    • $A\times{B}={C}$, because we are only doing a small number of checks ($n$ - one for each row of the result), it is possible that $A\times{B}\neq{C}$, but the algorithm cannot determine it in the amount of time it has. (See references for details.)
  • Stated another way:
    • when $A\times{B}={C}$, Freivalds' algorithm is always correct.
    • When $A\times{B}\neq{C}$, the algorithm errs with probability $\leq{1/2}$.
  • By re-running this algorithm $k$ times and choosing a random $\overrightarrow{r}$ each time, we can reduce the probability of error to $\leq{1/2^{k}}$.
  • In other words, in $\Theta({kn^2})$ time, Freivalds' algorithm guarantees that the error in verifying multiplication of two $n\times{n}$ matrix is $\leq{1/2^{k}}$.


  1. Wikipedia - Freivalds' algorithm
  2. Motwani, R. and Raghavan, P., Randomized Algorithms, 2007.

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