Multiplication of signed-operands

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Multiplication of signed-operands Recall we discussed multiplication of unsigned numbers: Combinatorial array multiplier. Sequential multiplier. Need an approach that works uniformly with unsigned and signed (positive and negative 2 s complement) n-bit operands. Booth s algorithm treats positive and negative 2 s complement operands uniformly. 1

2 Booth s algorithm Booth s algorithm applies uniformly to both unsigned and 2 s complement signed integers. Basis of other fast product algorithms. Fundamental observation: Division of an integer into the sum of block-1 s integers. Suppose we have a 16-bit binary number: 0110011011110110 This number can be represented as the sum of 4 block-1 integers: 0110000000000000 0000011000000000 0000000011110000 +0000000000000110 0110011011110110

3 Booth s algorithm (contd..) Suppose Q is a block-1 integer:q = 0000000001111000 = 120 Then: X.Q = X.120 Now: 120 = 128-8, so that X.Q = X.120 = X.(128-8) = X.128 - X.8 And: Q = 0000000001111000 128 = 0000000010000000 8 = 0000000000001000 If we label the LSB as 0, then the first 1 in the block of 1 s is at position 3 and the last one in the block of 1 s is at position 6. As a result: X.Q = X.120 = X.128 - X.8 = X.2 7 - X.2 3

4 Booth s algorithm (contd..) Representing Block-1 integers Q is an n-bit block-1 unsigned integer: -Bit position 0 is LSB. -First 1 is in bit position j -Last 1 is in bit position k Then: Q = 2 k+1-2 j Q.X = X.(2 k+1-2 j ) = X. 2 k+1 - X. 2 j

5 Booth s algorithm (contd..) Let Q be the block-1 integer: Q = 01111111111111110 To form the product X.Q using normal multiplication would involve 14 add/shifts (one for each 1-valued bit in multiplier Q). Since: Q = 2 15 2 1 X.Q = X.(2 15 2 1 ) Product X.Q can be computed as follows: 1. Set the Partial Product (PP) to 0. 2. Subtract X.2 1 from PP. 3. Add X.2 15 to PP. Note that X.2 j is equivalent to shifting X left j times.

6 Booth s algorithm (contd..) If Q is not a block-1 integer, Q can be decomposed so that it can be represented as a sum of block-1 integers. Suppose: Q = 0110011011110110 Q can be decomposed as: 0110000000000000 = 2 15-2 13 0000011000000000= 2 11-2 9 0000000011110000= 2 7-2 4 +0000000000000110 = 2 3-2 1 0110011011110110 Thus, Q.X = X.(2 15-2 13 + 2 11-2 9 +2 7-2 4 + 2 3-2 1 )

7 Booth s algorithm (contd..) Inputs: n-bit multiplier Q n-bit multiplicand x 2n-bit current Partial Product (PP) initially set to 0. (Upper half of PP is bits n-1 through n) Q has an added 0 bit attached to the LSB (Q has n+1 bits). Algorithm: For every bit in Q: 1. Examine the bit and its neighbor to the immediate right. If the bit pair is: 00 do nothing. 01 Add the multiplicand to the upper half of PP. 10 Sub the multiplicand from the upper half of PP. 11 Do nothing. 2. Shift the PP right by one bit, extending the sign bit.

8 Booth s algorithm (contd..) Example #1 01001100 Multiplicand X 01110111 Multiplier Q 00000000 00000000 Double-length PP register Step 0 bit pair 10 action: SUB 1 0 1 1 0 1 0 0 0 0 0 0 0 0 0 0 Shift Right 1 1 0 1 1 0 1 0 0 0 0 0 0 0 0 0 Step 1 bit pair 11 action: --- 1 1 0 1 1 0 1 0 0 0 0 0 0 0 0 0 Shift Right 1 1 1 0 1 1 0 1 0 0 0 0 0 0 0 0 Step 2 bit pair 11 action: --- 1 1 1 0 1 1 0 1 0 0 0 0 0 0 0 0 Shift Right 1 1 1 1 0 1 1 0 1 0 0 0 0 0 0 0 Step 3 bit pair 01 action: ADD 0 1 0 0 0 0 1 0 1 0 0 0 0 0 0 0 Shift Right 0 0 1 0 0 0 0 1 0 1 0 0 0 0 0 0 Step 4 bit pair 10 action: SUB 1 1 0 1 0 1 0 1 0 1 0 0 0 0 0 0 Shift Right 1 1 1 0 1 0 1 0 1 0 1 0 0 0 0 0 Step 5 bit pair 11 action: --- 1 1 1 0 1 0 1 0 1 0 1 0 0 0 0 0 Shift Right 1 1 1 1 0 1 0 1 0 1 0 1 0 0 0 0 Step 6 bit pair 11 action: --- 1 1 1 1 0 1 0 1 0 1 0 1 0 0 0 0 Shift Right 1 1 1 1 1 0 1 0 1 0 1 0 1 0 0 0 Step 7 bit pair 01 action: ADD 0 1 0 0 0 1 1 0 1 0 1 0 1 0 0 0 Shift Right 0 0 1 0 0 0 1 1 0 1 0 1 0 1 0 0 = Product: 9044

9 Booth s algorithm (contd..) Example #2 Multiplicand in Binary: 1 0 1 1 0 1 0 0 = -119 Multiplier in Binary: 0 1 1 1 0 1 1 1 = 76 Step 0 bit pair 10 action: SUB 0 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 Shift Right 0 0 1 0 0 1 1 0 0 0 0 0 0 0 0 0 Step 1 bit pair 11 action: --- 0 0 1 0 0 1 1 0 0 0 0 0 0 0 0 0 Shift Right 0 0 0 1 0 0 1 1 0 0 0 0 0 0 0 0 Multiplier: Step 2 bit pair 11 Positive action: --- 0 0 0 1 0 0 1 1 0 0 0 0 0 0 0 0 Shift Right 0 0 0 0 1 0 0 1 1 0 0 0 0 0 0 0 Multiplicand: Step 3 bit pair 01 action: ADD 1 0 1 1 1 1 0 1 1 0 0 0 0 0 0 0 2 s comp. -ve Shift Right 1 1 0 1 1 1 1 0 1 1 0 0 0 0 0 0 Step 4 bit pair 10 action: SUB 0 0 1 0 1 0 1 0 1 1 0 0 0 0 0 0 Shift Right 0 0 0 1 0 1 0 1 0 1 1 0 0 0 0 0 Step 5 bit pair 11 action: --- 0 0 0 1 0 1 0 1 0 1 1 0 0 0 0 0 Shift Right 0 0 0 0 1 0 1 0 1 0 1 1 0 0 0 0 Step 6 bit pair 11 action: --- 0 0 0 0 1 0 1 0 1 0 1 1 0 0 0 0 Shift Right 0 0 0 0 0 1 0 1 0 1 0 1 1 0 0 0 Step 7 bit pair 01 action: ADD 1 0 1 1 1 0 0 1 0 1 0 1 1 0 0 0 Shift Right 1 1 0 1 1 1 0 0 1 0 1 0 1 1 0 0 = Product: -9044

10 Booth s algorithm (contd..) Example #3 Multiplicand in Binary: 1 0 1 1 0 1 0 0 = -119 Multiplier in Binary: 1 0 0 0 1 0 0 1= -76 Step 0 bit pair 10 action: SUB 0 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 Shift Right 0 0 1 0 0 1 1 0 0 0 0 0 0 0 0 0 Step 1 bit pair 01 action: ADD 1 1 0 1 1 0 1 0 0 0 0 0 0 0 0 0 Shift Right 1 1 1 0 1 1 0 1 0 0 0 0 0 0 0 0 Multiplier: Step 2 bit pair 00 action: --- 1 1 1 0 1 1 0 1 0 0 0 0 0 0 0 0 Shift Right 1 1 1 1 0 1 1 0 1 0 0 0 0 0 0 0 Step 3 bit pair 10 action: SUB 0 1 0 0 0 0 1 0 1 0 0 0 0 0 0 0 Shift Right 0 0 1 0 0 0 0 1 0 1 0 0 0 0 0 0 Step 4 bit pair 01 action: ADD 1 1 0 1 0 1 0 1 0 1 0 0 0 0 0 0 Shift Right 1 1 1 0 1 0 1 0 1 0 1 0 0 0 0 0 Step 5 bit pair 00 action: --- 1 1 1 0 1 0 1 0 1 0 1 0 0 0 0 0 Shift Right 1 1 1 1 0 1 0 1 0 1 0 1 0 0 0 0 Step 6 bit pair 00 action: --- 1 1 1 1 0 1 0 1 0 1 0 1 0 0 0 0 Shift Right 1 1 1 1 1 0 1 0 1 0 1 0 1 0 0 0 Step 7 bit pair 10 action: SUB 0 1 0 0 0 1 1 0 1 0 1 0 1 0 0 0 Shift Right 0 0 1 0 0 0 1 1 0 1 0 1 0 1 0 0 = Product: 9044 2 s comp. -ve Multiplicand: 2 s comp. -ve

11 Booth s algorithm and signed integers Booth s algorithm works correctly for any combination of 2 s complement numbers. If Q is a 2 s complement negative number, the Q is of the form: Q = 1 q n-2 q n-3 q n-4... q 1 q 0 And: V(Q) = -2 n-1 + q n-2 2 n-2 + q n-3 2 n-3 +... + q 2 2 2 + q 1 2 1 + q 0 2 0 If Q has no zeros, that is: Q = 111...11 = -1 Booth technique will: - See the first pair as 10 (subtract the multiplicand) - See all other pairs as 11 (do nothing) Thus, Booth technique will compute the result as required: 0 - X = -X = -1.X

Booth s algorithm and signed integers (contd..) Booth s technique is correct for an arbitrary ve 2 s complement number Q is a 2 s complement number: Q = 1 q n-2 q n-3 q n-4... q 1 q 0 V(Q) = -2 n-1 + q n-2 2 n-2 + q n-3 2 n-3 +... + q 2 2 2 + q 1 2 1 + q 0 2 0 If we read Q from MSB (left hand side), Q will have certain number of 1s followed by a 0. If the first 0 appears in the m th bit of Q. V(Q) = -2 n-1 + q n-2 2 n-2 + q n-3 2 n-3 +... + q 2 2 2 + q 1 2 1 + q 0 2 0 = ( -2 n-1 + 2 n-2 + 2 n-3 +... 2 m+1 ) + (q m-1 2 m-1 +... + q 2 2 2 + q 1 2 1 + q 0 2 0 ) = -2 m+1 + (q m-1 2 m-1 +... + q 2 2 2 + q 1 2 1 + q 0 2 0 ) Where ( -2 n-1 + 2 n-2 + 2 n-3 +... 2 m+1 ) = -2 m+1 Bit m is 0, so m-bit number starting from 0 is a positive number. Booth s algorithm will work correctly. Transitioning from bit m to m+1, algo sees the bit pair 10, causing it to subtract the multiplicand 2 m+1 from the PP. All other transition see pairs 11 which have no effect. 12

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