Multiplication algorithm, hardware and flowchart
- 1. Computer Organization And Architecture
- 3. Multiplication (often denoted by x) is the mathematical operation of scaling one number by another. It is a basic arithmetic operation. Example: 3 x 4= 3+3+3+3=12 5 x 3 ½ =5+5+5+(half of 5)= 17.5 The basic idea of multiplication is repeated addition.
- 4. *Paper and pencil example (unsigned): Multiplicand 1000 Multiplier 1001 1000 0000 0000 1000 Product 01001000 *m bits x n bits = m+n bit product *Binary makes it easy: *0 => place 0 ( 0 x multiplicand) *1 => place a copy ( 1 x multiplicand) Place a copy Place 0
- 5. *64-bit Multiplicand reg, 64-bit ALU, 64-bit Product reg, 32-bit multiplier reg Product Multiplier Multiplicand 64-bit ALU Shift Left Shift Right Write Control 32 bits 64 bits 64 bits
- 6. *Product Multiplier Multiplicand 0000 0000 0011 0000 0010 *0000 0010 0001 0000 0100 *0000 0110 0000 0000 1000 *0000 0110 3. Shift the Multiplier register right 1 bit. Done Yes: 32 repetitions 2. Shift the Multiplicand register left 1 bit. No: < 32 repetitions 1. Test Multiplier0 Multiplier0 = 0Multiplier0 = 1 1a. Add multiplicand to product & place the result in Product register 32nd repetition? Start
- 8. *Half of the shifted multiplicand register contains 0. Why not shift the product right? *32-bit Multiplicand reg, 32 -bit ALU, 64-bit Product reg, 32-bit Multiplier reg Product Multiplier Multiplican d 32-bit ALU Shift Right Write Control 32 bits 32 bits 64 bits Shift Right
- 9. 3. Shift the Multiplier register right 1 bit. Done Yes: 32 repetitions 2. Shift the Product register right 1 bit. No: < 32 repetitions 1. Test Multiplier0 Multiplier0 = 0Multiplier0 = 1 1a. Add multiplicand to the left half of product & place the result in the left half of Product register 32nd repetition? Start ° Product Multiplier Multiplicand 0000 0000 0011 0010
- 10. *Product Multiplier Multiplicand 0000 0000 0011 0010 *0010 0000 *0001 0000 0001 0010 *0011 0000 0001 0010 *0001 1000 0000 0010 *0000 1100 0000 0010 *0000 0110 0000 0010 3. Shift the Multiplier register right 1 bit. Done Yes: 32 repetitions 2. Shift the Product register right 1 bit. No: < 32 repetitions 1. Test Multiplier0 Multiplier0 = 0Multiplier0 = 1 1a. Add multiplicand to the left half of product & place the result in the left half of Product register 32nd repetition? Start
- 12. *Combine Multiplier register and Product register *32-bit Multiplicand reg, 32 -bit ALU, 64-bit Product reg, (0-bit Multiplier reg) Product (Multiplier) Multiplican d 32-bit ALU Write Control 32 bits 64 bits Shift Right
- 13. Multiplicand Product 0010 0000 0011 Done Yes: 32 repetitions 2. Shift the Product register right 1 bit. No: < 32 repetitions 1. Test Product0 Product0 = 0Product0 = 1 1a. Add multiplicand to the left half of product & place the result in the left half of Product register 32nd repetition? Start Missing Multiplier
- 14. Mcand: 0010 P: 0000 0011 1 P=P+Mcand Mcand: 0010 P: 0010 0011 Shr P Mcand: 0010 P: 0001 0001 2 P=P+Mcand Mcand: 0010 P: 0011 0001 Shr P Mcand: 0010 P: 0001 1000 3 0=>nop Mcand: 0010 P: 0001 1000 Shr P Mcand: 0010 P: 0000 1100 4 0=>nop Mcand: 0010 P: 0000 1100 Shr P Mcand: 0010 P: 0000 0110
- 16. Booth algorithm gives a procedure for multiplying binary integers in signed – 2’s complement representation. I will illustrate the booth algorithm with the following example: Example, 2 x (- 4) 0010 * 1100 Step 1: Making the Booth table 1. From the two numbers, pick the number with the smallest difference between a series of consecutive numbers, and make it a multiplier. i.e., 0010 -- From 0 to 0 no change, 0 to 1 one change, 1 to 0 another change ,so there are two changes on this one 1100 -- From 1 to 1 no change, 1 to 0 one change, 0 to 0 no change, so there is only one change on this one. Therefore, multiplication of 2 x (– 4), where 2 ten (0010 two) is the multiplicand and (– 4) ten (1100two) is the multiplier.
- 17. 2. Let X = 1100 (multiplier) Let Y = 0010 (multiplicand) Take the 2’s complement of Y and call it –Y –Y = 1110 3. Load the X value in the table. 4. Load 0 for X-1 value it should be the previous first least significant bit of X 5. Load 0 in U and V rows which will have the product of X and Y at the end of operation. 6. Make four rows for each cycle; this is because we are multiplying four bits numbers.
- 18. Step 2: Booth Algorithm Look at the first least significant bits of the multiplier “X”, and the previous least significant bits of the multiplier “X - 1”. a. 0 0 Shift only 1 1 Shift only. 0 1 Add Y to U, and shift 1 0 Subtract Y from U, and shift or add (-Y) to U and shift b. Take U & V together and shift arithmetic right shift which preserves the sign bit of 2’s complement number. Thus a positive number remains positive, and a negative number remains negative. c. Shift X circular right shift because this will prevent us from using two registers for the X value.
- 20. We have finished four cycles, so the answer is shown, in the last rows of U and V which is: 11111000 Now to easily check our calculations, we take the original question, 2 x -4. This becomes -8. -8 is the two's compliment of 8. 8 in eight bit binary is 00001000. Taking the two's compliment by the method previously described, we get the result 11111000, which is exactly the same as our Booths algorithm answer.
- 22. Questions...