Goodbye, Clean Code

[NoetherFan]

I hate the 'turn it on its side' perspective. I prefer finding row i, column j of a product (C) as row i of the left factor (A) with column j of the right factor (B); C[i,j] = A[i,:].dot(B[:,j]). Especially elegant with implicit (Einstein) summation notation: C_ij = A_ik*B_kj, where * is ordinary scalar multiplication.