Algorithmica

, Volume 79, Issue 2, pp 428–443

# Efficiently Correcting Matrix Products

• Leszek Gąsieniec
• Christos Levcopoulos
• Andrzej Lingas
• Rasmus Pagh
• Takeshi Tokuyama
Open Access
Article

## Abstract

We study the problem of efficiently correcting an erroneous product of two $$n\times n$$ matrices over a ring. Among other things, we provide a randomized algorithm for correcting a matrix product with at most k erroneous entries running in $${\tilde{O}}(n^2+kn)$$ time and a deterministic $${\tilde{O}}(kn^2)$$-time algorithm for this problem (where the notation $${\tilde{O}}$$ suppresses polylogarithmic terms in n and k).

## Keywords

Matrix multiplication Matrix product verification Matrix product correction Randomized algorithms Time complexity

## 1 Introduction

Matrix multiplication is a basic operation used in many scientific and engineering applications. There are several potential reasons for erroneous results of computation, in particular erroneous matrix products. They include software bugs, computational errors by logic circuits and bit-flips in memory. Or, if the computation is performed by remote computers or by parallel processors, some errors might be introduced due to faulty communication.

In 1977, Freivalds presented a randomized algorithm for verifying if a matrix $$C'$$ is the matrix product of two $$n\times n$$ matrices A and B,  running in $$O(n^2)$$ time [10]. His algorithm has been up today one of the most popular examples showing the power of randomization.

In spite of extensive efforts of the algorithmic community to derandomize this algorithm without substantially increasing its time complexity, one has solely succeeded partially, either decreasing the number of random bits to a logarithmic one [5, 15, 19] or using exponentially large numbers and the unrealistic BSS computational model [16]. One can argue that the latter solutions in different ways hide additional O(n) factors. By the way, if one can use quantum devices then even an $$O(n^{5/3})$$-time verification of an $$n\times n$$ matrix product over an integral domain is possible [2].

Interestingly, the problem of verifying matrix products over the $$(\min ,+ )$$ semi-ring seems to be much harder comparing to that over an arbitrary ring. Namely, it admits a truly subcubic algorithm if and only if there is a truly subcubic algorithm for the all-pairs shortest path problem on weighted digraphs (APSP) [24].

Freivalds’ algorithm has also pioneered a new subarea of the so called certifying algorithms [18]. Their purpose is to provide besides the output a certificate or easy to verify proof that the output is correct. The computational cost of the verification should be substantially lower than that incurred by recomputing the output (perhaps using a different method) from scratch.

In 1977, when Freivalds published his algorithm, the asymptotically fastest known algorithm for arithmetic matrix multiplication was that due to Strassen running in $$O(n^{2.81})$$ time [22]. Since then the asymptotic running time of fast matrix multiplication algorithms has been gradually improved to $$O(n^{2.3728639})$$ at present [6, 11, 23] which is still substantially super-quadratic.

In this paper, we go one step further and consider a more complex problem of not only verifying a computational result but also correcting it if necessary. Similarly as Freivalds, as a subject of our study we choose matrix multiplication.

Our approach is very different from that in fault tolerant setting, where one enriches input in order to control the correctness of computation (e.g., by check sums in the so called ABFT method) [8, 26, 27]. Instead, we use here an approach resembling methods from Combinatorial Group Testing where one keeps testing larger groups of items in search for multiple targets, see, e.g. [7, 9].
Table 1

The characteristics and time performances of the algorithms for correcting an $$n\times n$$ matrix product with at most k erroneous entries presented in this paper

The issue of adapting some of our randomized algorithms to unknown k is discussed Sect. 7

First, we provide a simple deterministic algorithm for correcting an $$n\times n$$ matrix product $$C'$$ over a ring, with at most one erroneous entry, in $$O(n^2)$$ time. It can be regarded as a deterministic version of Freivalds’ algorithm (Sect. 3). Next, we extend the aforementioned algorithm to include the case when $$C'$$ contains at most k erroneous entries. The extension relies on distributing erroneous entries of $$C'$$ into distinct submatrices by deterministically shuffling the columns of $$C'$$ and correspondingly the columns of B. The resulting deterministic algorithm runs in $${\tilde{O}}(k^2n^2)$$ time, where the notation $${\tilde{O}}$$ suppresses polylogarithmic terms in n and k (Sect. 4). Then we show how to reduce the time bound to $${\tilde{O}}(kn^2)$$ by applying this shuffling approach first with respect to the columns and then with respect to the rows of $$C'$$. In the same section, we discuss also a slightly randomized version of the aforementioned algorithm running in $${\tilde{O}}(\sqrt{k} n^2)$$ expected time using $$O(\log ^2k + \log k\log \log n)$$ random bits. Next, in Sect. 5, we present a faster randomized algorithm for correcting $$C'$$ in $$O((n\sqrt{\log n} +\sqrt{k}\min \{k,n\})n\sqrt{\log n})$$ time almost surely (i.e., with probability at least $$1-n^{-\alpha }$$ for any constant $$\alpha \ge 1$$), where k is the non-necessarily known number of erroneous entries of $$C'$$. A slight modification of this algorithm runs in $$O((n\log k +\sqrt{k} \min \{k,n\})n)$$ expected time provided that the number of erroneous entries is known. This is our fastest algorithm for correcting $$C'$$ when k is very small. Importantly, all our algorithms in Sects. 35 are combinatorial (thus, they do not rely on the known fast algorithms for matrix multiplication or fast polynomial multiplication) and easy to implement. In Sect. 6, we present a more advanced algebraic approach based on the compressed matrix multiplication technique from [20]. In effect, we obtain a randomized algorithm for correcting $$C'$$ in $$O((n + k\log k\log \log k)n\log n)$$ time almost surely. Roughly, it asymptotically subsumes the randomized algorithms of Sect. 5 for k larger than $$n^{2/3}$$ and asymptotically matches them up to a polylogarithmic factor for the remaining k. We conclude with Sect. 7, where we discuss how some of our randomized algorithms can be also adjusted to the situation when the number of erroneous entries is unknown. For a summary of our results, see Table 1.

## 2 Preliminaries

Let $$(U,+,\times )$$ be a semi-ring. For two n-dimensional vectors $$a=(a_0,\ldots ,a_{n-1})$$ and $$b=(b_0,\ldots ,b_{n-1})$$ with coordinates in U their dot product $$\sum _{i=0}^{n-1}a_i\times b_i$$ over the semi-ring is denoted by $$a\odot b.$$

For an $$p\times q$$ matrix $$A=(a_{ij})$$ with entries in U,  its ith row $$(a_{i1},\ldots ,a_{in})$$ is denoted by $$A(i,*).$$ Similarly, the jth column $$(a_{1j},\ldots ,a_{nj})$$ of A is denoted by $$A(*,j).$$ Given another $$q\times r$$ matrix B with entries in U,  the matrix product $$A\times B$$ of A with B over the semi-ring is a matrix $$C=(c_{ij})$$, where $$c_{ij}=A(i,*)\odot B(*,j)$$ for $$1\le i,j\le n.$$

## 3 Correcting a Matrix Product with a Single Error

Given two matrices $$A,\ B$$ of size $$p\times q$$ and $$q\times r,$$ respectively, and their possibly erroneous $$p\times r$$ matrix product $$C'$$ over a ring, Freivalds’ algorithm picks uniformly at random a vector x in $$\{0,\ 1\}^r$$ and checks if $$A(Bx^T)=C'x^T$$, where $$x^T$$ stands for a transpose of x, i.e., the column vector corresponding to x [10]. For $$i=1,\ldots ,p,$$ if the ith row of $$C'$$ contains an erroneous entry, the ith coordinates of the vectors $$A(Bx^T)$$ and $$C'x^T$$ will differ with probability at least 1 / 2.

In the special case, when $$C'$$ contains a single error, we can simply deterministically set x to the vector $$(1,\ldots ,1)\in \{0,\ 1\}^r$$ in the aforementioned Freivalds’ test. The vectors $$A(Bx^T)$$, $$C'x^T$$ will differ in exactly one coordinate whose number equals the number of the row of $$C'$$ containing the single erroneous entry. (Note that the assumption that there is only one error is crucial here since otherwise two or more errors in a row of $$C'$$ potentially could cancel out their effect so that the dot product of the row with x,  which in this case is just the sum of entries in the row, would be correct.) Then, we can simply compute the ith row of the matrix product of A and B in order to correct $$C'$$.

The time complexity is thus linear with respect to the total number of entries in all three matrices, i.e., $$O(pq+qr+pr)$$. More precisely, it takes time $$O(p \cdot r)$$ to compute $$C'x^T$$, $$O(q \cdot r)$$ to compute $$Bx^T$$, and finally $$O(p \cdot q)$$ to compute the product of A with $$Bx^T$$.

## Theorem 3.1

Let $$A,\ B, \ C'$$ be three matrices of size $$p\times q, q\times r$$ and $$p\times r$$, respectively, over a ring. Suppose that $$C'$$ is different from the matrix product C of A and B exactly in a single entry. We can identify this entry and correct it in time linear with respect to the total number of entries, i.e., in $$O(pq+qr+pr)$$ time.

## 4 Correcting a Matrix Product with at Most k Errors

In this section, we shall repeatedly use a generalization of the deterministic version of Freivalds’ test applied to detecting single erroneous entries in the previous section.

Let $$A,\ B$$ be two $$n\times n$$ matrices, and let $$C'$$ be their possibly faulty product matrix with at most k erroneous entries, over some ring. Let $$C^*$$ and $$B^*$$ denote matrices resulting from the same permutation of columns in the matrices $$C'$$ and B.

Similarly as in the previous section, the generalized deterministic version of Freivalds’ test verifies rows of $$C^*,$$ but only for a selected set of consecutive columns of the matrix. Such a set of columns will be called a strip.

We shall check each strip of $$C^*$$ independently for erroneous entries that occur in a single column of the strip. To do this, when we determine the vector v to be used in the coordinate-wise comparison of $$A(B^*v^T)$$ with $$C^*v^T$$, we set the ith coordinate of v to 1 if and only if the ith column of the matrix $$C^*$$ belongs to the strip we want to test. Otherwise, we set the coordinate to 0. (See Fig. 1.)

In this way, for each row in a strip, we can detect whether or not the strip row contains a single error. The time complexity for testing a whole strip in this way is $$O(n^2)$$, independently from the number of columns of the strip. If necessary, we can also correct a single row of a strip by recomputing all its entries in time proportional to n times the number of columns in the strip.

Our algorithm in this section relies also on the following number theoretical lemma.

## Lemma 4.1

Let $$P=\{i_1,\ldots ,i_l\}$$ be a set of l different indices in $$\{1,\ldots ,n\}.$$ There exists a constant c and for each $$i_m\in P,$$ a prime $$p_m$$ among the first $$cl\log n / \log \log n$$ primes such that for $$i_q\in P{\setminus } \{i_m\}$$, $$i_m \bmod p_m \ne i_q \bmod p_m .$$

## Proof

It follows from the Chinese remainder theorem, the density of primes and the fact that each index in P has $$O(\log n)$$ bits that there is a constant b such that for each pair $$i_m,\ i_q$$ of distinct indices in P there are at most $$b\log n/\log \log n$$ primes p such that $$i_m \bmod p = i_q \bmod p.$$ Consequently, for each $$i_m\in P$$ there are at most $$b(l-1)\log n / \log \log n$$ primes p for which there exists $$i_q\in P{\setminus } \{i_m\}$$ such that $$i_q \bmod p = i_m \bmod p .$$ Thus, it is sufficient to set the constant c to b in order to obtain the lemma. $$\square$$

Given the generalized deterministic version of Freivalds’ test and Lemma 4.1, the idea of our algorithm for correcting $$C'$$ is simple, see Fig. 2.

For each prime p among the first $$ck\log n/\log \log n$$ primes, for $$j=1,\ldots ,n,$$ the jth column is moved into a (vertical) strip corresponding to $$j\bmod p.$$ Correspondingly, the columns of the matrix B are permuted.

Let $$B^*$$ and $$C^*$$ denote the resulting shuffled matrices.

Next, for each strip V of $$C^*$$, we set v to the vector in $$\{0,\ 1\}^n$$ whose jth coordinate is 1 if and only if the jth column belongs to V. We compute and compare coordinate-wise the vectors $$A(B^*v^T)$$ and $$C^*v^T.$$ Note that for $$i=1,\ldots ,n,$$ if there is a single erroneous entry in the ith row of V then the vectors $$A(B^*v^T)$$, $$C^*v^T$$ are different in this coordinate. Simply, the ith coordinate of $$C^*v^T$$ is just the sum of the entries in the ith row of V while that coordinate of $$A(B^*v^T)$$ is the sum of the entries in the ith row of the vertical strip of the product of A and $$B^*$$ corresponding to V.

It follows in particular that for each strip which contains only one erroneous column, we shall find all erroneous rows in the strip. Furthermore, we can correct all the erroneous entries in a detected erroneous row of the vertical strip V in $$O(n^2/p)$$ time by computing O(n / p) dot products of rows of A and columns of $$B^*.$$ Thus, in particular the correction of a single error in a row of V takes $$O(n^2/p)$$ time.

It follows from Lemma 4.1, that for each erroneous column in $$C',$$ there is such a prime p that the column is a single erroneous column in one of the aforementioned vertical strips of the shuffled matrix $$C^*.$$ Hence, all the k errors can be localized and corrected.

## Lemma 4.2

Let $$A,\ B, \ C'$$ be three $$n\times n$$ matrices over a ring. Suppose that $$C'$$ is different from the matrix product C of A and B in at most k entries. Algorithm 1 identifies these erroneous entries and corrects them in $${\tilde{O}}(k^2n^2)$$ time.

## Proof

The correctness of Algorithm 1 (see Fig. 2) follows from the above discussion and Lemma 4.1.

Algorithm 1 iterates over $$ck\log n/\log \log n$$ smallest primes. Since an upper bound on the ith prime number is $$O(i \log i)$$ for any $$i>1$$, it follows that the largest prime considered by the algorithm has size $$O(ck\log n\log k)$$, and hence all these primes can be listed in $$O(c^2k^2\log ^2 n\log k)$$ time.

For a given prime p,  the algorithm tests p vertical strips V for the containment of rows with single errors by computing the vectors $$A(B^*v^T)$$ and $$C^*v^T$$. It takes $$O(n^2p)$$ time.

By the upper bounds on the number of considered primes and their size, the total time taken by the tests for all considered primes is $$O(c^2k^2n^2\log ^2 n \log k/\log \log n).$$

The correction of an erroneous entry in a detected erroneous row in a vertical strip V takes $$O(n^2p)$$ time. Hence, the correction of the at most k erroneous entries in $$C^*$$ takes $$O(ck^2n^2\log n \log k)$$ time.

The tests and corrections dominate the running time of the algorithm. $$\square$$

In a practical implementation of the algorithm above, one can of course implement the shuffling of the columns without actually copying data from one column to another. For this purpose one could also define the strips in a different way, i.e., they do not need to consist of consecutive columns.

### 4.1 Reducing the Time Bound to $${\tilde{O}}(kn^2)$$

In order to decrease the power of k in the upper bound of the time complexity from 2 to 1, we make the following observation. Consider any column i of $$C'$$. The number of erroneous entries in column i that are in rows that have at least $$\sqrt{k}$$ erroneous entries is at most $$\sqrt{k}$$.

We start by applying Algorithm 1 but only using the smallest $$c \sqrt{k} \log n / \log \log n$$ primes. In this way all rows that have at most $$\sqrt{k}$$ erroneous entries will be found in total $${\tilde{O}}((\sqrt{k})^2 n^2)$$ time, and will be fixed in $$O(n^2)$$ time for each detected erroneous row. So the time complexity up to this stage is dominated by $${\tilde{O}}(kn^2)$$.

Now, we let $$C''$$ be the partially corrected matrix and we apply the same procedure but reversing the roles of columns and rows, i.e., we work with $$B^T A^T$$ and $$C''^T$$. Since for any row of $$C''^T$$, all its erroneous entries that were in columns of $$C''^T$$ with at most $$\sqrt{k}$$ errors were already corrected, now by the observation, the number of erroneous entries in any row of $$C''^T$$ is at most $$\sqrt{k}$$. Thus Algorithm 1 will now find all remaining erroneous rows in time $${\tilde{O}}(kn^2)$$ and we can correct them in additional time $$O(kn^2)$$. Hence we obtain the following theorem:

## Theorem 4.3

Let $$A,\ B, \ C'$$ be three $$n\times n$$ matrices over a ring. Suppose that $$C'$$ is different from the matrix product C of A and B in at most k entries. We can identify these erroneous entries and correct them in $${\tilde{O}}(kn^2)$$ time.

### 4.2 Few Random Bits Help

We can decrease the power of k in the upper bound of Theorem 4.3 from 1 to 0.5 by using $$O(\log ^2k +\log k \log \log n)$$ random bits as follows and assuming that the exact number k of erroneous entries in $$C'$$ is known. (The removal of this assumption will be discussed later.) The idea is that instead of testing systematically a sequence of primes, we start by producing four times as many primes and then choose randomly among them in order to produce the strips.

We call a faulty entry in $$C'$$ 1-detectable if it lies in a row or column of $$C'$$ with at most $$2 \sqrt{k}$$ erroneous entries. From this definition it follows that most faulty entries are 1-detectable. More specifically, we call an entry in $$C'$$ 1-row-detectable, respectively 1-column-detectable, if it lies in a row, respectively column, with at most $$2 \sqrt{k}$$ erroneous entries.

We will aim at detecting first a constant fraction of the 1-row-detectable (false) entries, and then a constant fraction of the 1-column-detectable entries. For this purpose we start by producing, in a preprocessing phase, the smallest $$4c \sqrt{k}\log n /\log \log n$$ primes (i.e., four times as many primes as we did in the deterministic algorithm of Theorem 4.3).

To detect sufficiently many 1-row-detectable entries we run one iteration of Algorithm 1, with the difference that we use a prime chosen randomly among the produced $$4c \sqrt{k}\log n /\log \log n$$ smallest primes. In this way, for each 1-row-detectable entry there is at least a probability 1/2 that it will be detected.

Then we repeat once more this procedure but reversing the role of columns and rows, i.e., by working with $$B^T A^T$$ and $$C'^T$$. In this way for each 1-column-detectable entry there is at least a probability 1/2 that it will be detected.

In this way, now each 1-detectable entry has been detected with probability at least 1/2. By correcting all these detected entries, we thus reduce the total number of remaining false entries by an expected constant fraction.

Thus we can set k to the remaining number of false entries and start over again with the resulting, partially corrected matrix $$C'$$. We repeat in this way until all erroneous entries are corrected.

The expected time bound for the tests and corrections incurred by the first selected primes dominate the overall expected time complexity. Note that the bound is solely $$O(c \sqrt{k} n^2 \log n \log k).$$

The number of random bits needed to select such a random prime is only $$O(\log k + \log \log n).$$ For a small k,  this is much less than the logarithmic in n number of random bits used in the best known $$O(n^2)$$-time verification algorithms for matrix multiplication obtained by a partial derandomization of Freivalds’ algorithm [5, 15, 19].

The overall number of random bits, if we proceed in this way and use fresh random bits for every new selection of a prime number has to be multiplied by the expected number of the $$O(\log k)$$ iterations of the algorithm. Thus, it becomes $$O(\log ^2 k + \log k \log \log n).$$

Hence, we obtain the following slightly randomized version of Theorem 4.3.

## Theorem 4.4

Let $$A,\ B, \ C'$$ be three $$n\times n$$ matrices over a ring. Suppose that $$C'$$ is different from the matrix product C of A and B in exactly k entries. There is a randomized algorithm that identifies these erroneous entries and corrects them in $${\tilde{O}}(\sqrt{k} n^2)$$ expected time using $$O(\log ^2k + \log k\log \log n)$$ random bits.

If the number k or erroneous entries is not known, then our slightly randomized method can be adapted in order to estimate the number of erroneous columns. Since similar issues arise in connection to another randomized approaches presented in the next chapters, we postpone this discussion to Sect. 7.

## 5 A Simple Randomized Approach

In this section, similarly as in the previous one, we shall apply the original and modified Freivalds’ tests. First, we apply repeatedly the original Freivalds’ test to the input $$n\times n$$ matrices AB,  and $$C'$$ and then to their transposes. These tests allow us to extract a submatrix $$C_1$$ which very likely contains all erroneous entries of $$C'.$$ Finally, we apply modified Freivalds’ tests to (vertical) strips of the submatrix $$C_1$$ of $$C'.$$

In contrast with the previous section, the tests are randomized. The modified test is just a restriction of Freivalds’ original randomized algorithm [10] to a strip of $$C_1$$ that detects each erroneous row of a strip with probability at least 1 / 2 even if a row contains more than one erroneous entry.

More precisely, the vector v used to test a strip of $$C_1$$ by comparing $$A_1(B_1v^T)$$ with $$C_1v^T$$, where $$A_1$$ and $$B_1$$ are appropriate submatrices of A and B,  is set as follows. Suppose that $$C_1$$ is an $$q\times r$$ matrix. For $$j=1,\ldots ,r,$$ the jth coordinate of v is set to 1 independently with probability 1 / 2 if and only if the jth column of $$C'$$ belongs to the strip we want to test, otherwise the coordinate is set to 0. In this way, for each row in the strip, the test detects whether or not the strip row contains an erroneous entry with probability at least 1 / 2, even if the row contains more than one erroneous entry. The test for a whole strip takes $$O(n^2)$$ time, independently of the number of columns of the strip.

Using the aforementioned tests, we shall prove the following theorem.

## Theorem 5.1

Let A, B and $$C'$$ be three $$n \times n$$ matrices over a ring. Suppose that $$C'$$ is different from the matrix product C of A and B in k entries. There is a randomized algorithm that transforms $$C'$$ into the product $$A \times B$$ in $$O((n\sqrt{\log n} +\sqrt{k}\min \{k,n\})n\sqrt{\log n})$$ time almost surely without assuming any prior knowledge of k.

## Proof

Let us assume for the moment that k is known in advance (this assumption will be removed later). Our algorithm (see Algorithm 2 in Fig. 3) will successively correct the erroneous entries of $$C'$$ until $$C'$$ will become equal to $$A \times B$$.

Our algorithm consists of two main stages. In the first stage, the standard Freivalds’ algorithm is applied iteratively to $$A, B, C'$$ and then to the transposes of these matrices in order to filter out all the rows and all the columns of $$C'$$ containing erroneous entries almost certainly. If the number of the aforementioned rows or columns is less than $$\log n$$ (e.g., when $$k<\log n$$) then all the entries in the rows or columns of the product $$A \times B$$ are computed and the algorithm halts. The computation of the aforementioned entries takes $$O(\min \{k,n\}n^2)$$ time in total. Otherwise, a submatrix $$C_1$$ of $$C'$$ consisting of all entries on the intersection of the aforementioned rows and columns is formed. It has at most $$\min \{k,n\}$$ rows and at most $$\min \{k,n\}$$ columns.

In the second stage, we consider a partition of the columns of $$C_1$$ into at most $$\lceil \sqrt{\frac{k}{\log n}} \rceil$$ strips of equal size, i.e., consecutive groups of at most $$\min \{k,n\}/\lceil \sqrt{\frac{k}{\log n}}\rceil$$ columns of $$C_1.$$ We treat each such strip separately and independently. For each strip, we apply our modification of Freivalds’ test $$O(\log n)$$ times. In this way, we can identify almost surely which rows of the tested strip contain at least one error. (Recall that for each iteration and for each strip row, the chance of detecting an error, if it exists, is at least 1 / 2.) Finally, for each erroneous strip row, we compute the correct values for each one of its $$O(\min \{k,n\}/ \sqrt{\frac{k}{\log n}})$$ entries (Fig. 3).

In each iteration of the test in Step 1 in the algorithm, each erroneous row in $$C'$$ will be detected with a probability at least 1 / 2. Hence, for a sufficiently large constant c (e.g., c=3) all erroneous rows of $$C'$$ will be detected almost surely within $$c \cdot \log n$$ iterations in Step 1. Analogously, all erroneous columns of $$C'$$ will be detected almost surely within $$c \cdot \log n$$ iterations in Step 3. It follows that all the erroneous entries of $$C'$$ will belong to the submatrix $$C_1$$ consisting of all entries on the intersection of the aforementioned rows and columns of $$C',$$ almost surely. Recall that $$C_1$$ has at most $$\min \{k,n\}$$ rows and at most $$\min \{k,n\}$$ columns.

Next, similarly, in Step 7 in the algorithm, each erroneous row in each of the $$\lceil \sqrt{\frac{k}{\log n}} \rceil$$ strips of $$C_1$$ will be detected almost surely. If we use the straightforward method in order to compute the correct values of an erroneous strip row, then it will take O(n) time per entry. Since each strip row of $$C_1$$ contains $$O(\min \{k,n\}/\sqrt{\frac{k}{\log n}})$$ entries, the time taken by a strip row becomes $$O(n\min \{k,n\}/\sqrt{\frac{k}{\log n}})$$. Since there are at most k erroneous strip rows, the total time for correcting all the erroneous strip rows in all strips of $$C_1$$ is $$O(\sqrt{k} \min \{k,n\}n\sqrt{\log n})$$.

The total time taken by the logarithmic number of applications of Freivalds’ tests to $$A, B, C'$$ in Step 1 and to the transposes of these matrices in Step 3 is $$O(n^2\log n).$$ To estimate the total time taken by the logarithmic number of applications of the restrictions of Freivalds’ tests to the $$O(\sqrt{\frac{k}{\log n}})$$ vertical strips of $$C_1$$ in and matrices $$A_1$$ and $$B_1$$ in Step 7, recall that $$A_1$$ has at most $$\min \{k,n\}$$ rows and n columns, $$B_1$$ has n rows and at most $$\min \{k,n\}$$ columns, while $$C_1$$ has at most $$\min \{k,n\}$$ rows and columns. Hence, in particular multiplications of $$C_1$$ by the restricted test vectors take $$O(\min \{k,n\}\min \{k,n\}/\sqrt{k/\log n} \times \sqrt{k/\log n}\log n)$$ time in total, which is $$O(\sqrt{k} \min \{k,n\}n\sqrt{\log n})$$ since $$k\ge \log n$$ in the second stage. Similarly, multiplications of $$B_1$$ by the restricted test vectors take $$O(n\min \{k,n\}/\sqrt{k/\log n} \times \sqrt{k/\log n}\log n)$$ time in total, which is again $$O(\sqrt{k} \min \{k,n\}n\sqrt{\log n}).$$ Note that the n-coordinate vectors resulting from multiplications of $$B_1$$ with the restricted test vectors are not any more restricted and potentially each of their coordinates may be non-zero. Therefore, the multiplications of $$A_1$$ with the aforementioned vectors take $$O(\min \{k,n\}n \times \sqrt{k/\log n}\log n)$$ time in total, which is $$O(\sqrt{k} \cdot \min \{k,n\}n\cdot \sqrt{\log n})$$. All this yields an upper time bound of $$O(n^2 \cdot \log n +\sqrt{k} \cdot \min \{k,n\}n\cdot \sqrt{\log n})$$ on the total time taken by the tests in both stages.

In the second stage of Algorithm 2, if we use, instead of the correct number k of erroneous entries, a guessed number $$k'$$ which is larger than k, then the time complexity becomes $$O(n^2 \cdot \log n +\sqrt{k'} \cdot \min \{k',n\}n\cdot \sqrt{\log n})$$. This would be asymptotically fine as long as $$k'$$ is within a constant factor of k. On the other hand, if we guess $$k'$$ which is much smaller than k, then the length of each erroneous strip row in $$C_1$$ may become too large. For this reason, first we have to find an appropriate size $$k'$$ for the strips to be used by our algorithm. For this purpose, we perform the first stage of Algorithm 2, i.e., the logarithmic number of original Freivalds’ tests on the input matrices and their transposes. Next, we set $$k'$$ to the maximum $$k_0$$ of the number of erroneous rows and the number of erroneous columns reported by the aforementioned tests, and a small constant, e.g., 4. Then, we multiply our guess by 4,  until we reach a good balance. More precisely, for each such guessed $$k'$$, without correcting any errors, we consider the partition of the submatrix $$C_1$$ into $$O(\sqrt{\frac{k'}{\log n}})$$ strips, and apply our modified test to each strip. As soon as we discover more than $$k'$$ erroneous strip rows in $$C_1,$$ we break the procedure without correcting any errors, and we start over with a four times larger guess $$k'$$.

The aforementioned method of guessing $$k'$$ may result in at most $$O(\log k)$$ wrong guesses until we achieve a good guess. Since we multiply our guess every time with 4,  we obtain a geometric progression of the estimated costs of subsequent trials. In this way, the upper bound on the asymptotic complexity of the whole algorithm but the time complexity of the first logarithmic number of original Freivalds’ test is dominated by that of the iteration for the final $$k'.$$ In this iteration, we test each strip $$c \cdot \log n$$ times in order to detect almost surely all erroneous strip rows. $$\square$$

Algorithm 2 in the proof of Theorem 5.1 can be modified in order to achieve an expected time bound of $$O((n\log k +(\sqrt{k} \min \{k,n\})n)$$ for correcting all errors, if k is known in advance.

In the first stage, we perform only a single test for the matrices AB and a single test for their transposes. Note that each erroneous entry of C occurs with probability at least $$\frac{1}{2}$$ in a detected erroneous row of C as well as with probability at least $$\frac{1}{2}$$ in a detected erroneous column of C. Hence, an erroneous entry occurs with probability at least $$\frac{1}{4}$$ in the resulting matrix $$C_1.$$ It follows that the expected number of erroneous entries in $$C_1$$ is at least one fourth of those in C.

Next, we modify the second stage of Algorithm 2 as follows. We set the number of vertical strips to $$\lceil \sqrt{k} \rceil .$$ Next, instead of applying the strip restriction of Freivalds’ algorithm $$c \cdot \log n$$ times for each strip, we apply it only once for each strip and correct all erroneous rows which we detect. By counting how many errors we have corrected, we compute how many errors remain. Then we recurse in the same way on the partially corrected matrix C using as a parameter this new number of errors which remain to be corrected.

During each iteration of the algorithm, each remaining error in C will be detected and corrected with probability at least $$\frac{1}{2} \times \frac{1}{4}$$. Thus, the expected number of remaining errors will decrease at least by the multiplicative factor $$\frac{7}{8}$$ after each iteration. It follows that the expected number of iterations is $$O(\log k).$$ Consequently, the total cost of the tests in the first stage becomes $$O(n^2\log k).$$ For the total time cost of tests and corrections in the second stage, we obtain a geometric progression on the expected time complexity of each iteration, and so the total expected time complexity is dominated by the time taken by the first iteration, which is $$O(\sqrt{k}\min \{k,n\} \cdot n )$$. Thus we obtain the following theorem.

## Theorem 5.2

Let $$A,\ B, \ C'$$ be three $$n\times n$$ matrices over a ring. Suppose that $$C'$$ is different from the matrix product C of A and B in exactly k entries. There is a randomized algorithm that identifies these erroneous entries and corrects them in $$O((n\log k +\sqrt{k} \min \{k,n\})n)$$ expected time.

## 6 A Fast Algebraic Approach

In this section we present a fast randomized algorithm that makes use of the compressed matrix multiplication technique presented in [20]. We choose to give a self-contained and slightly simplified description because we do not need the full power of the framework of [20].

For integer parameters st to be chosen later, the construction uses t pairs of hash functions $$g_\ell , h_\ell : \{1,\dots ,n\}\rightarrow \{1,\dots ,s\}$$, with $$\ell =1,\dots , t$$, chosen independently from a strongly universal family of hash functions [4]. We will make use of the following property:

## Lemma 6.1

[20]

For $$(i_1,j_1), (i_2,j_2) \in \{1,\dots ,n\}^2$$ where $$(i_1,j_1)\ne (i_2,j_2)$$ we have
\begin{aligned} \Pr \left[ g_\ell (i_1)+h_\ell (j_1) = g_\ell (i_2)+h_\ell (j_2)\right] \le 1/s . \end{aligned}
Our algorithm first computes the following t polynomials based on the matrices $$A=(a_{ik})$$, $$B=(b_{kj})$$, and $$C'=(c'_{ij})$$:
\begin{aligned} p_\ell (x) = \sum _{k=1}^n \left( \sum _{i=1}^n a_{ik} x^{g_\ell (i)}\right) \left( \sum _{j=1}^n b_{kj} x^{h_\ell (j)}\right) - \sum _{i=1}^n \sum _{j=1}^n c'_{ij} x^{g_\ell (i)+h_\ell (j)}, \end{aligned}
(6.1)
for $$\ell =1,\dots ,t$$. Multiplication of the polynomials corresponding to A and B is done efficiently (over any ring) using the algorithm of Cantor and Kaltofen [3], based on the original polynomial multiplication algorithm of Schönhage and Strassen [21].

Let $$p(x)_m$$ denote the coefficient of $$x^m$$ in a polynomial p(x). For each entry ij of $$C'$$ we assess the error term that must be added to $$c'_{ij}$$ as the majority element of the sequence $$p_\ell (x)_{g_\ell (i)+h_\ell (j)}$$, $$\ell =1,\dots ,t$$. We will choose s and t such that with high probability the correction term (in most cases zero) appears more than t / 2 times in the sequence. If there is no such element for some entry ij the algorithm fails.

### 6.1 Correctness

Suppose $$C=AB=(c_{ij})$$ is the true matrix product. Expanding the sum (6.1) and reordering the order of summation we get:
\begin{aligned} p_\ell (x)&= \sum _{i=1}^n \sum _{j=1}^n \sum _{k=1}^n a_{ik} x^{g_\ell (i)} b_{kj} x^{h_\ell (j)} - \sum _{i=1}^n \sum _{j=1}^n c'_{ij} x^{g_\ell (i)+h_\ell (j)}\\&= \sum _{i=1}^n \sum _{j=1}^n (c_{ij} - c'_{ij})\, x^{g_\ell (i)+h_\ell (j)} . \end{aligned}
This means that each coefficient of $$p_\ell (x)$$ is a sum of error terms:
\begin{aligned} p_\ell (x)_m&= \sum _{\begin{array}{c} i,j\\ g(i)+h(j)=m \end{array}} c_{ij} - c'_{ij} . \end{aligned}
Let $$K\subseteq \{1,\dots ,n\}^2$$ be the set of positions of errors. For $$i^*,j^*\in \{1,\dots ,n\}$$:
\begin{aligned} p_\ell (x)_{g(i^*)+h(j^*)} = c_{i^* j^*} - c'_{i^* j^*} + \sum _{\begin{array}{c} (i,j)\in K\backslash \{(i^*,j^*)\}\\ g(i)+h(j)=g(i^*)+h(j^*) \end{array}} c_{ij} - c'_{ij} . \end{aligned}
(6.2)
Lemma 6.1 states that $$g(i)+h(j)=g(i^*)+h(j^*)$$ holds with probability at most 1 / s. By a union bound the probability that the sum in (6.2) has at least one nonzero term is at most k / s. Choosing $$s\ge 3k$$ we get that $$p_\ell (x)_{g(i^*)+h(j^*)} = c_{i^*j^*} - c'_{i^*j^*}$$ with probability at least 2 / 3. By Chernoff bounds this implies that after t repetitions the probability that $$p_\ell (x)_{g(i)+h(j)} = c_{ij} - c'_{ij}$$ does not hold for at least t / 2 values of $$\ell$$ is exponentially small in t. Choosing $$t=O(\log n)$$ we can achieve an arbitrarily small polynomial error probability in n (even when summed over all entries i, j).

### 6.2 Time Analysis

Strongly universal hash functions can be selected in constant time and space [4], and evaluated in constant time. This means that they will not dominate the running time. Time $$O(n^2+ns)$$ is used to compute the polynomials $$\sum _{i=1}^n a_{ik} x^{g_\ell (i)}$$, $$\sum _{j=1}^n b_{kj} x^{h_\ell (j)}$$, and $$\sum _{i=1}^n \sum _{j=1}^n c'_{ij} x^{g_\ell (i)+h_\ell (j)}$$ in (6.1). This can be seen by noticing that each entry of A, B, and $$C'$$ occur in one polynomial, and that there are $$2n+1$$ polynomials of degree s. Another component of the running time is the tn multiplications of degree-s polynomials, that each require $$O(s \log s \log \log s)$$ operations [3]. Finally, time $$O(tn^2)$$ is needed to compute the correction term for each entry ij of $$C'$$ based on the sequence $$p_\ell (x)_{g_\ell (i)+h_\ell (j)}$$. With the choices $$s=O(k)$$, $$t=O(\log n)$$ the combined number of operations (algebraic and logical) is $$O(n^2\log n + kn\log n\log k\log \log k).$$

## Theorem 6.2

Let A, B and $$C'$$ be three $$n \times n$$ matrices over a ring. Suppose that $$C'$$ is different from the matrix product C of A and B in at most k entries. There is a randomized algorithm that transforms $$C'$$ into the product $$A \times B$$ in $$O((n + k\log k\log \log k)n\log n)$$ time, i.e., $${\tilde{O}}(n^2 + kn)$$ time, almost surely.

While the above assumes prior knowledge of k, we observe in Sect. 7 that this assumption can be removed with only a slight increase in running time. Observe that the algorithm of Theorem 6.2 needs $$O(t\log n)$$ bits of space, which is $$O(\log ^2 n)$$.

## 7 Final Remarks

The majority of our randomized algorithms, in particular that from Sect. 6, can be efficiently adapted to the case when the number k of errors is unknown, proceeding similarly as in the proof of Theorem 5.1. First, observe that using a parameter value $$k'$$ that is larger than k by a constant factor will yield the same guarantee on correctness and asymptotic running time. This means that we can try geometrically increasing values of $$k'$$, for example $$k'=4^l$$ for $$l=1,2,3,\ldots$$ until the algorithm returns a correct answer within the stated time bound (using a suitably large constant in place of the big-O notation). Correctness is efficiently checked using Freivalds’ technique. This technique increases the time bound by at most a factor $$\log n$$ compared to the case where k is known. Furthermore, if $$k\ge n \log n$$ the time will be dominated by the last iteration, and we get time bounds identical to the case of known k.

A similar approach can also be used for refining the slightly randomized method of Theorem 4.4 when the number of errors k is not known in advance. However, if there is no knowledge at all concerning the number of errors, it may be difficult to handle the case when no errors are detected: does this happen because there are no errors at all, or because there are too many errors and we chose a random prime from a too small range, thus failing to isolate 1-detectable false entries? For this reason, if there is no known useful upper bound on the remaining number of errors, and we do not detect any errors during a series of iterations, we may have to resort to some of the known algorithms which test whether there are any errors at all [5, 15, 19]. All such known algorithms running in time $$O(n^2)$$ may need a logarithmic number of random bits, so if k is very small then this may be asymptotically larger than the low number of random bits stated in Theorem 4.4.

Note that the problem of correcting a matrix product is very general. In the extreme case, when all entries of the matrix $$C'$$ may be mistrusted, it includes the problem of computing the matrix product C from scratch. Also, when the matrix C is known to be sparse, i.e., mostly filled with zeros, then we can set $$C'$$ to the all-zeros matrix, and apply our matrix correction algorithms in order to obtain output-sensitive algorithms for matrix multiplication (the number of non-zero entries in C equals the number of erroneous entries in $$C'$$). They will be slower than those known in the literature based on fast rectangular matrix multiplication [1, 12, 13, 17] (cf. [25]).

Finally, the general idea of using linear sketches to compute compact summaries of matrix products may be useful in general for correcting matrix products. For example, Iwen and Spencer [14] show that for complex-valued matrix products there is an efficiently computable linear sketch that allows recovery of the matrix product if the number of nonzeros in each column is bounded by roughly $$n^{0.3}$$. Using linearity one can subtract the linear sketch for $$C'$$ to get the linear sketch of $$AB-C'$$, which has k nonzero entries. If the number of nonzeros in each column of $$AB-C$$ is bounded by $$n^{0.3}$$, they can all be computed in time $$n^{2+o(1)}$$. However, it is not clear for which rings this method will work, so while this is an interesting direction for future research we do not pursue it further here.

## Notes

### Acknowledgments

We thank anonymous referees for helping us to improve preliminary versions of this paper. Christos Levcopoulos and Andrzej Lingas were supported in part by Swedish Research Council Grant 621-2011-6179. Takeshi Tokuyama was supported by JSPS Grant Scientific Research (B) 15H02665, JSPS Scientific Research for Innovative Area 24106007, and JST ERATO Kawarabayashi Big-Graph Project.

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## Authors and Affiliations

• Leszek Gąsieniec
• 1
• Christos Levcopoulos
• 2
• Andrzej Lingas
• 2
• Rasmus Pagh
• 3
• Takeshi Tokuyama
• 4
1. 1.Department of Computer ScienceUniversity of LiverpoolLiverpoolUK
2. 2.Department of Computer ScienceLund UniversityLundSweden
3. 3.Theoretical Computer Science SectionIT University of CopenhagenCopenhagenDenmark
4. 4.Graduate School of Information SciencesTohoku UniversitySendaiJapan