In Strassen's algorithm, why does padding the matrices with zeros not affect the asymptopic complexity? In Strassen's algorithm, why does padding the matrices with zeros, in order to multiply matrices that are not powers of 2, not affect the asymptopic complexity?
Hi, I was reading this question but I do not follow Yuval Filmus's answer completely.
He offers two ways of padding the matrix with zeroes, I am interested in his first suggestion, padding the entire matrix with zeroes at the start such that the new matrix has dimensions $N\times N$ where $N = 2^c$.
He says &quot;$N &lt; 2n$ so this doesn't affect the asymptotic complexity.&quot;
Could someone please elaborate as I do not follow, I understand calculating time complexity with the master theorem and $T(n) = aT(n/b) + f(n)$ so if someone could explain how this works in reference to that it would be greatly appreciated.
Suppose that if $N = 2^c$ then you can multiply two $N \times N$ matrices in time $O(N^{\log_2 7})$. For concreteness, let us say that two such matrices can be multiplied in time at most $CN^{\log_27}$.
Now suppose that we are given two $n\times n$ matrices. We pad them to $N \times N$ matrices, where $N = 2^{\lceil \log_2 n \rceil} &lt; 2n$. We can extract the product of the original matrices from the product of the new matrices. The two new matrices can be multiplied using at most this many operations:
$$
CN^{\log_2 7} &lt; C (2n)^{\log_2 7} = 7C n^{\log_2 7} = O(n^{\log_2 7}).
$$