Chapter 8.3, Problem 3QE

Program Plan Intro

Address polynomial for converting references in terms of row and column positions into memory addresses:

• Finding the address of entry in i^th row and j^th column using the below formula

$\text{x}\text{+}\left(\text{c}\text{×}\left(\text{i}\text{-}\text{1}\right)\right)\text{+}\left(\text{j}\text{-}\text{1}\right)$

• From the given expression, “$\left(\text{c}\text{×}\left(\text{i}\text{-}\text{1}\right)\right)\text{+}\left(\text{j}\text{-}\text{1}\right)$” is called as address polynomial.
• From the above formula, “c” denotes the number of columns in given array.
• Then “x” denotes the address of the cell containing the entry in the first row and first column.
  • That is, user must move above “i – 1” rows, each of which contains “c” entries, to reach i^th row and then “j – 1” more entries to reach the j^th entry in this row.

Example:

The example for finding the address polynomial is shown below:

• Assume, the number of column in an array is “5” that is “c = 5” and “i = 2” and “j = 3”.
• Hence if the array were stored beginning at address “x”, then entry in the second row, third column would be at address

 $\begin{array}{c}\text{= x}\text{+}\left(\text{5}\text{×}\left(\text{2}\text{-}\text{1}\right)\right)\text{+}\left(\text{3-}\text{1}\right)\\ \text{=}\text{x}\text{+}\left(\text{5}\text{×}\text{1}\right)\text{+ 2}\\ \text{=}\text{x}\text{+}\text{5 + 2}\\ \text{= x + 7}\end{array}$

• Therefore, to reach the entry in the second row and third column, user must move above “7” entries from the beginning of the block.