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I have three matrices A,B and C: enter image description here

And the matrix-matrix product for general matrices:

void cblas_dgemm(const enum CBLAS_ORDER __Order, const enum CBLAS_TRANSPOSE __TransA, const enum CBLAS_TRANSPOSE __TransB, const int __M, const int __N, const int __K, const double __alpha, const double *__A, const int __lda, const double *__B, const int __ldb, const double __beta, double *__C, const int __ldc);

For using the cblas_dgemm-command I need to know the leading dimension. For me it is clear that in the case of the total matrix A (or its transpose form) we have: M=5, N=4, lda=4.

In the case of submatrix C I think i have to overgive &A[5] and set M=3, N=2, ldc=4

But I have no idea how this could work in the case of red submatrix B with M=4, N=2. Can someone explain this to me. Thanks a lot.

Suslik
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1 Answers1

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This article pretty much nails it: https://petewarden.com/2015/10/25/an-engineers-guide-to-gemm/

The reason for the apparent complexity of BLAS routines is that they allow a lot of flexibility and are optimised to perform extremely well. Both goals are achieved, if the routines can be applied on matrices consisting of submatrices of interest etc. Oftentimes they can do more than you need. xGEMM class is an outstanding example. You may perform A * B but also A * B + c*C ...

In you above cases:

A: M=5, N=5, LDA = 5
B: M=4, N=1, LDB = 10
C: M=3, N=2, LDB = 5

And you are correct in that your first entry in C is &C[6]

The leading dimension would in other words be usually the length on the column of the outer matrix if you are column major and the length of the row if you are computing row-major.

In case B, it's a little trickier, as you have to jump over two columns of 5 each, i.e. 10, when going from one column of the submatrix to the next.

All BLAS wants to now is (column/row major):

  1. Where to start (&A[0], &B[0], &C[6]/&A[0], &B[0], &C[5])
  2. How long is the first dimension, M (5, 4, 3 / 4, 2, 2)
  3. How many of these columns/rows there are, N (4, 2, 2 / 5, 4, 3)
  4. How far to jump through the outer matrix to get to next column/row ldx (5, 10, 5)
Kaveh Vahedipour
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  • Thank you, also for the article! Is it not `A: M=5 N=4, LDA=4` and `B:M=4, N=2, LDB=10`. But how do you find the number `ldb=10` – Suslik Jan 03 '19 at 17:01
  • I thought the leading dimension is like the steps the pointer has to do to get from the first element of a row to the first element of the next row. – Suslik Jan 03 '19 at 17:02
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    Exactly. Only, when row-major. All above I did is column-major. Most libraries out there do column-major naturally. That's why I always use column-major data even in C/C++. – Kaveh Vahedipour Jan 03 '19 at 17:07
  • Ahha now I see it! Then in such a case I will use `CblasColMajor`. Thanks a lot and good evening – Suslik Jan 03 '19 at 17:10
  • I dealt with this stuff in my PhD. I was working on multidimensional MRI data. (x,y,z,time,phases, ...) The code is here.: github.com/kvahed/codeare Specifically the BLAS/Lapack code is here: https://github.com/kvahed/codeare/tree/master/src/matrix/linalg – Kaveh Vahedipour Jan 03 '19 at 17:14
  • Wow and congratulation for your PhD! thank you for your detailed help. – Suslik Jan 04 '19 at 09:14
  • Hey guys, I'm a little confused, hope you can clear something up for me. Why is N != 4 for A? And N != 2 for B? -Thanks – mriera Jul 22 '20 at 09:24