MULTIPLICATION ALGORITHM

With the important capability of decision making in our repertoire we move on to the discussion of an algorithm, which will help us uncover an important set of instructions in our processor used for bit manipulations.

Multiplication is a common process that we use, and we were trained to do in early schooling. Remember multiplying by a digit and then putting a cross and then multiplying with the next digit and putting two crosses and so on and summing the intermediate results in the end. Very familiar process but we never saw the process as an algorithm, and we need to see it as an algorithm to convey it to the processor.

To highlight the important thing in the algorithm we revise it on two 4bit binary numbers. The numbers are 1101 i.e. 13 and 0101 i.e. 5. The answer should be 65 or in binary 01000001. Observe that the answer is twice as long as the multiplier and the multiplicand. The multiplication is shown in the following figure.

1101 = 13

0101 = 5
-----
1101

0000x 1101xx 0000xxx
--------
01000001 = 65

We take the first digit of the multiplier and multiply it with the multiplicand. As the digit is one the answer is the multiplicand itself. So we place the multiplicand below the bar. Before multiplying with the next digit a cross is placed at the right most place on the next line and the result is placed shifted one digit left. However since the digit is zero, the result is zero. Next digit is one, multiplying with which, the answer is 1101. We put two crosses on the next line at the right most positions and place the result there shifted two places to the left. The fourth digit is zero, so the answer 0000 is placed with three crosses to its right.

Observe the beauty of binary base, as no real multiplication is needed at the digit level. If the digit is 0 the answer is 0 and if the digit is 1 the answer is the multiplicand itself. Also observe that for every next digit in the multiplier the answer is written shifted one more place to the left. No shifting for the first digit, once for the second, twice for the third and thrice for the fourth one. Adding all the intermediate answers the result is 01000001=65 as desired. Crosses are treated as zero in this addition.

Before formulating the algorithm for this problem, we need some more instructions that can shift a number so that we use this instruction for our multiplicand shifting and also some way to check the bits of the multiplier one by one.