Radix –Two Division Most complex of the four arithmetic operations (addition, subtraction, multiplication and division). Requires the most in terms of computation power compared to the other operations

Radix –Two Division The division can be represented as: X = Q.D + R with R < D Where X -> dividend D -> divisor R -> remainder Q -> quotient Assume that X, Q, R and D >0 X may occupy a double length register (such an accumulator holding partial results: X is a 32 bit register while R, D and Q are 16-bit registers.

Radix –Two Division As such, we need to insure that Q < 2n-1 to prevent overflow. If Q < 2n-1 and R < D then: X < (2n-1+1)D If Q < 2n-1 and R = 0 then: X = QD < 2n-1D If fractional division is performed, then to prevent overflow condition  X < D. Division is obtained by a series of subtractions and shifts.

Sequential Division Algorithm At each step i, the value (2 x remainder) is compared to the divisor D. If this value is the larger of the two, then the quotient bit qi is set to 1, otherwise set to 0. ri = 2ri-1 - qiD , i = 1, 2, …m, Equ(1) where m is the number of fractional bits. where ri is the new remainder. where ri-1 is the previous remainder. where ro is set to be the dividend X How the above equation will perform the division?

Sequential Division Algorithm Let rm be the remainder in the last step then: rm = 2rm-1 – qm.D Substituting recursively using Equ(1) we get: rm = 2(2rm-2 – qm-1.D) - qm.D = ….. rm = 2mr0 – (qm + 2qm-1 + … + 2m-1q1).D Substituting r0 by the dividend or X and dividing both sides by 2m we get: 2-mrm = X – (q12-1 + q22-2 + … + qm2-m).D 2-mrm = X – Q.D

Division: Example Let X = 0.10000 (0.5), D = 0.110 (3/4) X < D is satisfied -> no overflow Remainder = r32-3 (1/32) r0 = X 2 r0 Add -D 0.100 01.000 11.010 000 00 Set q1 = 1 r1 = 2r0 – D 2 r1 r2 = 2r1 2 r2 00.010 00.100 00 0 Set q2 = 0 Set q3 = 1 r3 = 2r2 – D Quotient = 0. q1 q2 q3 Quotient = 0.101 (5/8) Note for signed division, get the absolute values, perform division and apply sign at the end