Fast multiplication of k x k boolean matrices, where 8 <= k <= 16
Given two 4x4 matrices a= 0010,0100,1111,0001, b=1100,0001,0100,0100, one could first calculate the transpose b' = 1000,1011,0000,0100.
Then the resulting matrix M(i,j)=a x b mod 2 == popcount(a[i]&b[j]) & 1; // or parity
From that one can notice that the complexity only grows in n^2, as long as the bitvector fits a computer word.
This can be speed up for 8x8 matrices at least, provided that some special permutation and bit selection operations are available. One can iterate exactly N times with NxN bits in a vector. (so 16x16 is pretty much the limit).
Each step consists of accumulating i.e. Result(n+1) = Result(n) XOR A(n) .& B(n), where Result(0) = 0, A(n) is A <<< n, and '<<<' == columnwise rotation of elements and where B(n) copies diagonal elements from the matrix B:
a b c a e i d h c g b f
B= d e f B(0) = a e i B(1) = d h c B(2) = g b f
g h i a e i d h c g b f
And after thinking it a bit further, a better option is to ^^^
(row wise rotate) matrix B and select A(n) == column copied diagonals from A:
a b c a a a b b b c c c
A= d e f A(0) = e e e , A(1) = f f f, A(2) = d d d
g h i i i i g g g h h h
EDIT To benefit later readers, I'd propose the full solution for W<=16 bit matrix multiplications in portable C.
#include <stdint.h>
void matrix_mul_gf2(uint16_t *a, uint16_t *b, uint16_t *c)
{
// these arrays can be read in two successive xmm registers or in a single ymm
uint16_t D[16]; // Temporary
uint16_t C[16]={0}; // result
uint16_t B[16];
uint16_t A[16];
int i,j;
uint16_t top_row;
// Preprocess B (while reading from input)
// -- "un-tilt" the diagonal to bit position 0x8000
for (i=0;i<W;i++) B[i]=(b[i]<<i) | (b[i]>>(W-i));
for (i=0;i<W;i++) A[i]=a[i]; // Just read in matrix 'a'
// Loop W times
// Can be parallelized 4x with MMX, 8x with XMM and 16x with YMM instructions
for (j=0;j<W;j++) {
for (i=0;i<W;i++) D[i]=((int16_t)B[i])>>15; // copy sign bit to rows
for (i=0;i<W;i++) B[i]<<=1; // Prepare B for next round
for (i=0;i<W;i++) C[i]^= A[i]&D[i]; // Add the partial product
top_row=A[0];
for (i=0;i<W-1;i++) A[i]=A[i+1];
A[W-1]=top_row;
}
for (i=0;i<W;i++) c[i]=C[i]; // return result
}
How about padding it out to the next "clever" (e.g. 8 or 16) size, with all '1' on the diagonal?
Depending on your application, storing both the matrix and its transpose together might help. You will save a lot of time that otherwise would be used to transpose during matrix multiplications, at the expense of some memory and some more operations.