matrix multiplication as column major - c++

Question

I am trying to multiply as column major and I can't seem to find the right formula!
I want to have the matrices as 1D.
Let's say I have these matrices:
A=
1 3
2 4
and B=
5 2 1
6 3 7
The above matrices are assumed that are stored already in column major order.
I am trying:
int main(int argc, const char* argv[]) {
int rows=2;
int cols=3;
int A[rows*rows];
int B[rows*cols];
int res[rows*cols];
A[0]=1;
A[1]=3;
A[2]=2;
A[3]=4;
B[0]=5;
B[1]=2;
B[2]=1;
B[3]=6;
B[4]=3;
B[5]=7;
/*A[0]=1;
A[1]=2;
A[2]=3;
A[3]=4;
B[0]=5;
B[1]=6;
B[2]=2;
B[3]=3;
B[4]=1;
B[5]=7;
*/
//multiplication as column major
for (int i=0;i<rows;i++){
for (int j=0;j<cols;j++){
res[i+j*rows]=0;
for (int k=0;k<rows;k++){
res[i+j*rows]+=A[i+k*rows]*B[k+j*cols];
}
}
}
for (int i=0;i<rows*cols;i++){
printf("\n\nB[%d]=%d\t",i,res[i]);
}
return 0;
}
I am not getting the correct results.
Also,I can't understand (in the case where the matrices are stored in column major already) ,how to index the matrices A and B.
A[0]=1;
A[1]=3;
...
or
A[0]=1;
A[1]=2;
...
I don't want to transpose the matrices and then use row major.
I want to handle the data as column major.
Because the indices ,if stored as column major,will be different (hence,will matter in order to do the multiplication).

Answer 1

There are two things that lead to your confusion here.
First, the data in your contiguous one-dimensional vector is not in column-major order as you say, but in row-major order, as is the usual layout of two-dimensional contiguous arrays in C. The linear one-dimensional indices of row i and column j in a matrix with M rows and N columns (MxN) are:
A[i*N + j] // row major
A[i + M*j] // column major
The "major" refers to the dimension of the outer loop when traversing the array sequentially with two nested loops:
n = 0;
for (i = 0; i < M; i++) {
for (j = 0; j < N; j++) {
printf("%8d", A[n++]);
}
printf("\n");
}
Second, you use the two dimensions rows and columns which are the dimensions of the resulting matrix, which is confusing, because the number of columns in A is rows.
In fact, there are three different dimensions involved in matrix multiplication when you multiply an MxL matrix A with an LxN matrix B to get an MxN matrix C. In your case, M and L happen to be both 2:
L (k) | N (j)
|
| 5 2 1
L (k) |
| 6 3 7
|
-----------------+-------------
|
1 3 | 23 11 22
M (i) |
2 4 | 34 16 30
|
The letters in parentheses are the variables the code below uses to iterate over the respective dimension.
Now you can multiply your matrices in row-major format:
#define M 2
#define N 3
#define L 2
int A[M * L] = {1, 3, 2, 4};
int B[L * N] = {5, 2, 1, 6, 3, 7};
int res[M * N];
int i, j, k;
for (i = 0; i < M; i++) {
for (j = 0; j < N; j++) {
res[j + i * N] = 0;
for (k = 0; k < L; k++) {
res[j + i * N] += A[k + i * L] * B[j + k * N];
}
}
}
for (i = 0; i < M * N; i++) printf("[%d] = %d\n", i, res[i]);
or in column-major format:
#define M 2
#define N 3
#define L 2
int A[M * L] = {1, 2, 3, 4};
int B[L * N] = {5, 6, 2, 3, 1, 7};
int res[M * N];
int i, j, k;
for (i = 0; i < M; i++) {
for (j = 0; j < N; j++) {
res[j * M + i] = 0;
for (k = 0; k < L; k++) {
res[j * M + i] += A[k * M + i] * B[j * L + k];
}
}
}
for (i = 0; i < M * N; i++) printf("[%d] = %d\n", i, res[i]);
Both input and output are in the respective matrix representation and differ in the two cases, of course.

Answer 2

What do you think about
res[i+j*rows]+=A[i+k*rows]*B[k+j*cols];
what it will do?
It will access array res, A and B out of bound when i and k becomes 1 and j becomes 2.
res[1+2*2]+=A[1+1*2]*B[1+2*3] = res[5]+=A[4]*B[7];
This will invoke undefined behavior and you may get either expected or unexpected result.
I think you need this:
res[i*rows+j] += A[i*rows + k] * B[j + k*cols];