9/6/2011. Multiplication. Binary Multipliers The key trick of multiplication is memorizing a digit-to-digit table Everything else was just adding

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1 9/6/2 Multiplication Binary Multipliers The key trick of multiplication is memorizing a digit-to-digit table Everything else was just adding

2 9/6/2 Multiplication Multiplication can be performed as a sequence of repeated additions. A * B is interpreted as add A, B times. However, such a scheme is very inefficient with a time complexity of O(m) where m is the magnitude of B. A better approach to multiplication, add-and-shift, produces a time complexity of O(n) where n is the length of the B. Binary Multiplication that looks like an AND gate The Binary Multiplication Table X Binary multiplication is implemented using the same basic longhand algorithm that you learned in grade school. x A 2 B 2 A B A A 3 B B 3 A j B i is a partial product A 3 B A 3 B A 2 B A B A 2 B A B A B A B 2 AB + A 3 B 3 A 3 B 2 A 2 B 2 A B 2 A 2 B 3 A B 3 A B 3 Multiplying N-digit number by M-digit number gives (N+M)-digit result Easy part: forming partial products (just an AND gate since B I is either or ) Hard part: adding M, N-bit partial products 2

3 9/6/2 ombinational Multiplier: accumulation of partial products A3 A2 A A B3 B2 B B A2 B A2 B A B A B A3 B A2 B A B A B A3 B2 A2 B2 A B2 A B2 A3 B3 A2 B3 A B3 A B Unsigned Multiplication Paper and Pencil Example: Multiplicand 2 = 2 Multiplier 2 = 3 Product 2 = 56 Binary multiplication is easy multiplicand = multiplicand = multiplicand m-bit multiplicand n-bit multiplier = (m+n)-bit product Accomplished via shifting and addition 3

4 9/6/2 Multiplication of unsigned numbers Product of 2 n-bit numbers is at most a 2n-bit number. Unsigned multiplication can be viewed as addition of shifted versions of the multiplicand. Array Multiplier: arry Forward y3 y2 y y Multiplicand x3 x2 x x Multiplier xy3 xy2 xy xy xy3 xy2 xy xy Partial x2y3 x2y2 x2y x2y Products x3y3 x3y2 x3y x3y p7 p6 p5 p4 p3 p2 p p Note: arry is added to the next partial product. Adding the carry from the final stage needs an extra stage. These additions are faster but we need four stages. 4

5 9/6/2 Array Multiplier: arry Forward y3 y2 y y Multiplicand x3 x2 x x Multiplier xy3 xy2 xy xy xy3 xy2 xy xy Partial x2y3 x2y2 x2y x2y Products x3y3 x3y2 x3y x3y p7 p6 p5 p4 p3 p2 p p Note: arry is added to the next partial product. Adding the carry from the final stage needs an extra stage. These additions are faster but we need four stages. Two-input AND Full-adder Basic Building Blocks Bi xk yi x kth partial product carry bits from (k-)th product Full adder pi = xyi th partial product carry bits to (k+)th product (k+)th partial product 5

6 9/6/2 Multiplication of unsigned numbers Typical multiplication cell Bit of incoming partial product (PPi) j th multiplicand bit i th multiplier bit i th multiplier bit carry out FA carry in Bit of outgoing partial product (PP(i+)) A j B i M U L T I P L I A N D M U L T I P L I E R A3 A2 A A B A3 B A2 B A B A B B A3 B A2 B A B A B B2 A3 B2 A2 B2 A B2 A B2 B3 A3 B3 A2 B3 A B3 A B3 P7 P6 P5 P4 P3 P2 P P 6

7 9/6/2 ombinatorial array multiplier ombinatorial array multiplier Multiplicand (PP) PP PP2 PP3 p 7 p 6 p 5 p 4 m 3 m 2 m m p 3 q 3 p 2 q 2 p, q p q Product is: p 7,p 6,..p Multiplicand is shifted by displacing it through an array of adders. ombinatorial array multiplier (contd..) ombinatorial array multipliers are: Extremely inefficient. Have a high gate count for multiplying numbers of practical size such as 32-bit or 64-bit numbers. Perform only one function, namely, unsigned integer product. Improve gate efficiency by using a mixture of combinatorial array techniques and sequential techniques requiring less combinational logic. 7

8 9/6/2 Add-and-shift Algorithm In each iteration the least-significant bit of multiplier is checked; if one, then multiplicand is added to the accumulator and the contents of accumulator and multiplier is shifted right one position. if zero, just shift accumulator and multiplier to the right. equential ircuit Multiplier Register A (initially ) hift right ontrol Algorithm:. A, M multiplicand, Q multiplier 2. If LB of Q== then add M to A else add 3. hift [A][Q] right 4. Repeat steps 2 and 3 n- times. 5. [A][Q] has product. a n - a q q n - Add/Noadd control Multiplier Q n-bit Adder MUX ontrol sequencer m n - m Multiplicand M 8

9 9/6/2 equential multiplication (contd..) M A Q Initial configuration Add hift ontrol Algorithm:. A, M multiplicand, Q multiplier 2. If LB of Q== then add M to A else add 3. hift [A][Q] right 4. Repeat steps 2 and 3 n- times. 5. [A][Q] has product. First cycle Add hift econd cycle No add hift Third cycle Add hift Fourth cycle Product igned Multiplication 9

10 9/6/2 igned Multiplication onsidering 2 s-complement signed operands, what will happen to (- 3) (+) if following the same method of unsigned multiplication? (- 3) ( +) ign extension is shown in blue (- 43) ign extension of negative multiplicand. igned multiplication multiplicand multiplier * (-3) (-5) (-3) Note: 2s complement ign extension +(-6) -(-24) (5)

11 9/6/2 igned Multiplication (Pencil & Paper) ase : Positive Multiplier Multiplicand 2 = -4 Multiplier 2 = +5 ign-extension Product 2 = -2 ase 2: Negative Multiplier Multiplicand 2 = -4 Multiplier 2 = -3 ign-extension (2's complement of ) Product 2 = +2 onsider decimal multiplication:

12 9/6/2 Examine multiplier: 999 = + Multiply as follows: What is the Trick? 457 = ( -) = = Booth Algorithm: Basic Idea onsider a multiplier, (3) We can write, 3 = 32 2, or (32) = ( 2) = 2 () 3 Interpret multiplier (scan right to left), check bit-pairs: Product, M 3 = M 2 5 -M 2 M: multiplicand Multiplication by 2 k means a k-bit left shift 2

13 9/6/2 Booth Algorithm In general, in the Booth scheme, -times the shifted multiplicand is selected when moving from to, and +times the shifted multiplicand is selected when moving from to, as the multiplier is scanned from right to left. () Booth recoding of a multiplier. Booth Algorithm ( +3) X (- 6) (- 78) Booth multiplication with a negative multiplier. 3

14 9/6/2 Booth Algorithm Multiplier Biti Biti- Version of multiplicand selected by bit i + Booth multiplier recoding table. Booth Algorithm Best case a long string of s (skipping over s) Worst case s and s are alternating Worst-case multiplier Ordinary multiplier Good multiplier

15 9/6/2 Booth Algorithm: Example 7 3 = 2 multiplicand = 7 () multiplier = 3 bit-pair, add -7 in two s com. bit-pair, do nothing bit-pair, add 7 bit-pair, do nothing 2 Booth Algorithm: Example 2 7 (-3) = -2 multiplicand = 7 () multiplier = -3 bit-pair, add -7 in two s com. bit-pair, add 7 bit-pair, add -7 in two s com. bit-pair, do nothing - 2 5

16 9/6/2 Booth Algorithm: Example = -2 multiplicand = -7 in two s com. () multiplier = 3 bit-pair, add 7 bit-pair, do nothing bit-pair, add -7 bit-pair, do nothing - 2 Booth Algorithm: Example 4-7 (-3) = 2 multiplicand = -7 in two s com. () multiplier = -3 in two s com. bit-pair, add 7 bit-pair, add -7 in two s com. bit-pair, add 7 bit-pair, do nothing 2 6

17 9/6/2 Booth Advantage erial multiplication Four partial product additions Booth algorithm Two partial product additions Fast Multiplication 7

18 9/6/2 Bit-Pair Recoding of Multipliers Bit-pair recoding halves the maximum number of summands (versions of the multiplicand). ign extension Implied to right of LB + 2 (a) Example of bit-pair recoding derived from Booth recoding Bit-Pair Recoding of Multipliers Multiplier bit-pair Multiplier bit on the right i + i i Multiplicand selected at position i (b) Table of multiplicand selection decisions 8

19 9/6/2 Bit-Pair Recoding of Multipliers Multiplier bit-pair i + i i Multiplicand selected at position i Multiplier bit on the right (- 78) ( +3) (- 6) Figure : Multiplication requiring only n/2 summands. Integer Division 9

20 9/6/2 Division imilar to multiplication, one can develop a routine to perform division as a sequence of subtractions. However, such an algorithm is very inefficient and slow. Instead one can develop an algorithm which performs division as a sequence of ompare, hift and ubtract operations. One should note that division in a binary system is much simpler than the division in decimal system, since the quotient digits are either or. In other words; division requires almost the same hardware modules as multiplication does. Methods of Division There are several different algorithms for division: Restoring Method Non-Restoring Method Direct omparison 2

21 9/6/2 restoring method In the restoring method, the contents of the partial remainder is restored whenever it is detected that the divisor is larger than the partial remainder. non-restoring method In non-restoring method, if it is detected that the divisor is larger than the partial remainder, its contents are not restored. However, in the next iteration, instead of subtracting the divisor from the partial remainder, it is added to the partial remainder. direct comparison or Longhand division In direct comparison approach, first the divisor is compared against the partial remainder. If it is smaller than or equal to the partial remainder, it will be subtracted and quotient digit is set to. Otherwise, just the quotient is set to zero. 2 Manual Division Longhand division examples. 2

22 9/6/2 Unsigned Division (Paper & Pencil) 2 = 9 Quotient Divisor 2 2 = 27 Dividend - - Dividend = Quotient Divisor + Remainder 27 = = 8 Remainder Binary division is accomplished via shifting and subtraction ircuit Arrangement hift left a n a n- A a q n- q Dividend Q Quotient etting N+ bit adder Add/ubtract ontrol equencer m n- m Divisor M Figure : ircuit arrangement for binary division. 22

23 9/6/2 Restoring Division hift A and Q left one binary position ubtract M from A, and place the answer back in A If the sign of A is, set q to and add M back to A (restore A); otherwise, set q to Repeat these steps n times Pseudo code is:. A, Q Dividend, M Divisor 2. hift left AQ by one position 3. A A-M 4. If sign of A is q A A+M Else q 5. Repeat step 2 to 4 for n- times Examples M. A, Q dividend, M divisor 2. hift left AQ by one position 3. A A-M 4. If sign of A is q A A+M Else q 5. Repeat step 2 to 4 for n- times A Initially hift ubtract etq Restore hift ubtract etq Restore hift ubtract etq hift ubtract etq Restore Q First cycle econd cycle Third cycle Fourth cycle Remainder Quotient Figure : A restoring-division example. 23

24 9/6/2 Nonrestoring Division Avoid the need for restoring A after an unsuccessful subtraction. Any idea? tep : (Repeat n times) If the sign of A is, shift A and Q left one bit position and subtract M from A; otherwise, shift A and Q left and add M to A. Now, if the sign of A is, set q to ; otherwise, set q to. tep2: If the sign of A is, add M to A Examples Initially hift ubtract etq First cycle hift Add etq econd cycle Add Restore remainder hift Add etq Third cycle Remainder hift ubtract etq Fourth cycle A nonrestoring-division example. Quotient 24

At the ith stage: Input: ci is the carry-in Output: si is the sum ci+1 carry-out to (i+1)st state

Chapter 4 xi yi Carry in ci Sum s i Carry out c i+ At the ith stage: Input: ci is the carry-in Output: si is the sum ci+ carry-out to (i+)st state si = xi yi ci + xi yi ci + xi yi ci + xi yi ci = x i yi