matrix multiplication as column major

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).

You have uninitialized fields in the `a` array, fields you use in your calculation. Since they are uninitialized, using them leads to *undefined behavior*. — Some programmer dude, Mar 03 '14 at 11:07

score 8 · Accepted Answer · answered Mar 03 '14 at 14:15

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.

@M Oehm:Hmm..I just noticed that in the column major example ,I am not getting the same results as for row. — George, Mar 03 '14 at 20:01
@George: The resulting matrix should be the same, only the one-dimensional representation should be different. Note that also the input matrices are different in their element order. — M Oehm, Mar 03 '14 at 20:12
@M Oehm:I just tried to do the multiplication by hand and that's why the results are different.But I am not taking into account that at the column multiplication we take every other element in order to do the multiplication ,so we result to the same initial matrix — George, Mar 04 '14 at 07:59
@M Oehm:Check here if you want . :) http://stackoverflow.com/questions/22182221/multiplication-of-3-matrices-in-a-sum/22182474?noredirect=1#22182474 — George, Mar 04 '14 at 21:41

haccks · Answer 2 · 2014-03-03T11:27:28.493

0

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];

edited Mar 03 '14 at 11:27

answered Mar 03 '14 at 11:17

haccks

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:But this solution is for row major order.I want to multiple as column major – George Mar 03 '14 at 11:28
Then also store your matrices in column-major order: `int A[] = {1, 2, 3, 4}` and `int B[] = {5, 6, 2, 3, 1, 7}`. – M Oehm Mar 03 '14 at 11:39
@M Oehm:Matrices A and B are already stored as column major data and i want to perform a multiplication on them.Because the indices are different if they are already stored as column – George Mar 03 '14 at 11:58

matrix multiplication as column major

2 Answers2