Question

A matrix representation stores matrices such that the offset address of the element in row i and column j can be calculated by concatenating the binary representations of i and j where row and column indicies start from 0. The size of the bit fields for row and column number should be big enough to accommodate the highest possible row and column numbers. For example: A 3 x 3 matrix, A, will have row and column field widths of 2 bits each. The element in row 2 and column 1 will have an offset address 1001. A matrix, M, has 57 rows and 10 columns. Find the offset address of the element in row 16 and column 3.

Answer 1

ANSWER 9:

The offset address of an element in row 16 and column 3 = 0100000011 (10 bits address)

Explanation:

The size of bit fields for row and column numbers should be big enough to accommodate the highest possible row and column numbers.  

For 57 × 10 matrix M will have 6 bits for rows (26=64 so 6 bits can accommodate 57 values) and 4 bits for columns (24 =16 so 4 bits can accommodate 10 values). For offset address of an element in row 16 and column 3:

6 bits of row address will be the binary value of 16 = 010000

4 bits of column address will be the binary value of 3 = 0011  

Now, for offset address of an element in row 16 and column 3 will be the concatenation of row and column address = 010000 concat 0011 = 0100000011 (10 bits address)