When the matrix size of Strassen matrix multiplication is not the form of 2^k, is the time complexity better than the naïve algorithm?

Tag: greater than the use of str greater than or equal to log good other complexity time

It turns out to be N, to find a number m greater than or equal to N and 2^k form.
The matrix of the n*n is the matrix of the M*m, the original matrix is placed at the top left, and the value of the other positions is 0.
Naïve method: N^3
Now: m^2.8
That is, m/n needs to be less than e^ (3/2.8) =2.919 to be good, and n<=m<2*n, that is, using this method is better.

When the matrix size of Strassen matrix multiplication is not the form of 2^k, is the time complexity better than the naïve algorithm?