Possible Solution

Content

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

Notes

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

Before you continue

Make sure you understand the formula! :-)