Possible Solution

Content Packing: Filling buffers for a fully cache-optimized GEMM Possible Solution Notes Before you continue Packing blocks of A Possible Solution Packing blocks of B Possible Solution

\[\nu(I,J) := M_r \cdot K \cdot \left\lfloor \frac{I}{M_r} \right\rfloor + J \cdot M_r + (I \bmod M_r)\]

Notes

Actually we get a nice formula due to the fact that indices for rows and columns start at zero:

• The \(I\)-th row of \(\overline{A}_Z\) belongs to panel \(A_i\) with \(i = \left\lfloor \frac{I}{M_r} \right\rfloor\). Panels \(A_0, \dots, A_{i-1}\) have a total of \(i \cdot M_r \cdot K\) elements. These must be stored in the buffer in front of the elements from panel \(A_i\).

  So the offset in the buffer is at least \(i \cdot M_r \cdot K = \left\lfloor \frac{I}{M_r} \right\rfloor \cdot M_r \cdot K\)

• In the panel \(A_i\) all elements with column index smaller than \(J\) must be stored in front of the elements with column index \(J\). These are in total \(M_r \cdot J\) elements.

  So the offset must be increased by \(M_r \cdot J\).

• The element \(a_{I, J}\) has the row index \(I \bmod M_r\) in the panel \(A_i\). All elements in the \(J\)-th column of \(A_i\) with smaller row-index must be stored in front of \(a_{I,J}\). These are \(I \bmod M_r\) elements.

  So the offset must be increased by \(I \bmod M_r\).

Before you continue

Make sure you understand the formula! :-)

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© Andreas F. Borchert and Michael C. Lehn