Booth Algo 4 Multiplication
Example
Find 3 × (−4), with m = 3 and r = −4, and x = 4 and y = 4: * m = 0011, -m = 1101, r = 1100 * A = 0011 0000 0 * S = 1101 0000 0 * P = 0000 1100 0 * Perform the loop four times : 1. P = 0000 1100 0. The last two bits are 00. * P = 0000 0110 0. Arithmetic right shift. 2. P = 0000 0110 0. The last two bits are 00. * P = 0000 0011 0. Arithmetic right shift. 3. P = 0000 0011 0. The last two bits are 10. * P = 1101 0011 0. P = P + S. * P = 1110 1001 1. Arithmetic right shift. 4. P = 1110 1001 1. The last two bits are 11. * P = 1111 0100 1. Arithmetic right shift. * The product is 1111 0100, which is −12.
The above mentioned technique is inadequate when the multiplicand is the largest negative number that can be represented (e.g. if the multiplicand has 4 bits then this value is −8). One possible correction to this problem is to add one more bit to the left of A, S and P. Below, we demonstrate the improved technique by multiplying −8 by 2 using 4 bits for the multiplicand and the multiplier: * A = 1 1000 0000 0 * S = 0 1000 0000 0 * P = 0 0000 0010 0 * Perform the loop four times : 1. P = 0 0000 0010 0. The last two bits are 00. * P = 0 0000 0001 0. Right shift. 2. P = 0 0000 0001 0. The last two bits are 10. * P = 0 1000 0001 0. P = P + S. * P = 0 0100 0000 1. Right shift. 3. P = 0 0100 0000 1. The last two bits are 01. * P = 1 1100 0000 1. P = P + A. * P = 1 1110 0000 0. Right shift. 4. P = 1 1110 0000 0. The last two bits are 00. * P = 1 1111 0000 0. Right shift. * The product is 11110000 (after discarding the first and the last bit) which is
Find 3 × (−4), with m = 3 and r = −4, and x = 4 and y = 4: * m = 0011, -m = 1101, r = 1100 * A = 0011 0000 0 * S = 1101 0000 0 * P = 0000 1100 0 * Perform the loop four times : 1. P = 0000 1100 0. The last two bits are 00. * P = 0000 0110 0. Arithmetic right shift. 2. P = 0000 0110 0. The last two bits are 00. * P = 0000 0011 0. Arithmetic right shift. 3. P = 0000 0011 0. The last two bits are 10. * P = 1101 0011 0. P = P + S. * P = 1110 1001 1. Arithmetic right shift. 4. P = 1110 1001 1. The last two bits are 11. * P = 1111 0100 1. Arithmetic right shift. * The product is 1111 0100, which is −12.
The above mentioned technique is inadequate when the multiplicand is the largest negative number that can be represented (e.g. if the multiplicand has 4 bits then this value is −8). One possible correction to this problem is to add one more bit to the left of A, S and P. Below, we demonstrate the improved technique by multiplying −8 by 2 using 4 bits for the multiplicand and the multiplier: * A = 1 1000 0000 0 * S = 0 1000 0000 0 * P = 0 0000 0010 0 * Perform the loop four times : 1. P = 0 0000 0010 0. The last two bits are 00. * P = 0 0000 0001 0. Right shift. 2. P = 0 0000 0001 0. The last two bits are 10. * P = 0 1000 0001 0. P = P + S. * P = 0 0100 0000 1. Right shift. 3. P = 0 0100 0000 1. The last two bits are 01. * P = 1 1100 0000 1. P = P + A. * P = 1 1110 0000 0. Right shift. 4. P = 1 1110 0000 0. The last two bits are 00. * P = 1 1111 0000 0. Right shift. * The product is 11110000 (after discarding the first and the last bit) which is