Homework 3. Assigned on 02/15 Due time: midnight on 02/21 (1 WEEK only!) B.2 B.11 B.14 (hint: use multiplexors) CSCI 402: Computer Architectures

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1 Homework 3 Assigned on 02/15 Due time: midnight on 02/21 (1 WEEK only!) B.2 B.11 B.14 (hint: use multiplexors) 1 CSCI 402: Computer Architectures Arithmetic for Computers (2) Fengguang Song Department of Computer & Information Science IUPUI 1
2 Today s Contents Have learned +, , x 3.3, 3.4 Optimization of multipliers Booth s Algorithm How to implement division 3 Recall: Multiplier P = 0; for i = 0 to 31 { P += A or Zero; P>>1; } Product 0 = 1 Start 1. Test Product 0 Product 0 = 0 Multiplier 1a. Add Multiplicand to left half of Product & place the result in left half. 2. Shift the Product register right 1 bit. 32nd repetition? No: < 32 repetitions Done Yes: 32 repetitions 4 2
3 Faster Hardware for i = 0 to 3 { C[i] = A[i] + B[i]; } Unroll loop We may use 32 adders C[0] = A[0] + B[0]; C[1] = A[1] + B[1]; C[2] = A[2] + B[2]; C[3] = A[3] + B[3]; Every adder outputs: 32 sum bits and a carryout bit 1) LSB of the intermediate sum is a bit in the final product. 2) The other 31 bits and the carryout bit are passed along to the next adder. Q: Why it is faster? 5 A 2bits Multiplier (C = B x A) The AND gates will produce the partial products. For a 2bit by 2bit multiplier, we can just use two half adders to sum the partial products. In general, though, we ll need full adders. Here C 3 C 0 are the product, not carries! B 1 B 0 x A 1 A 0 A 0 B 1 A 0 B 0 + A 1 B 1 A 1 B 0 C 3 C 2 C 1 C 0 6 3
4 A 4bits Multiplier (C = B x A) Input 1: Previous sum Input 2: Multiplier i AND Multiplicand 7 Faster Multiplication Hardware Uses multiple adders in parallel =(log 2 32 steps) Cost/performance tradeoff n Several adders can perform in parallel e.g., a0+a1+a2+a3+a4+a5+a6+a7? 8 4
5 A 3 A 2 A 1 A 0 x B 3 B 2 B 1 B 0 B 0 (A 3 A 2 A 1 A 0 ) B 1 (A 3 A 2 A 1 A 0 ) B 2 (A 3 A 2 A 1 A 0 ) B 3 (A 3 A 2 A 1 A 0 ) B 3 (A 3 A 2 A 1 A 0 ) B 2 (A 3 A 2 A 1 A 0 ) B 1 (A 3 A 2 A 1 A 0 ) B 0 (A 3 A 2 A 1 A 0 ) P2 P1 9 Signed Multiplication? Determine sign of the operands, make them positive Use the same hardware of unsigned multiplication Fix up the sign of the output 10 5
6 Multiply MIPS Instruction 32 bits x 32 bits à up to 64 bits Hi and Lo special registers // instead of creating 64bit register HI: stores mostsignificant 32 bits LO: stores leastsignificant 32 bits mfhi mflo Signed multiply mult (signed int), multu (unsigned int) MIPS Instructions mult rs, rt 64bit product goes to Hi and Lo mfhi rd or mflo rd Move from HI/LO to rd Users could test if HI=0 to see if product overflows 32 bits mul rd, rs, rt (pseudoinstruction) Leastsignificant 32 bits of product > rd Note: mul will ignore overflow! Note: even if they are Rtype operations, they only take 2 operands. 11 How to Make Multiply Even Faster? (A better algorithm) 89 x 9999 =? (calculate it by hand) = 89 ( ) = = = Similarly, apply the idea to binary numbers è Booth s algorithm! 12 6
7 Motivation for Booth s Algorithm Traditional way: x 0110: x nothing (0 in multiplier) + add (1 in multiplier) + add (1 in multiplier) nothing (0 in multiplier) ALU can get same result in more than one way: 0110 = 6 = = Replace a string of 1s with an initial subtract (when we first see a one), and then later add for the bit after the last 1. E.g., x nothing (0 in multiplier) sub (first 1 in multiplier) nothing (middle of string of 1s) + add (prior step had last 1) Booth s Algorithm middle end of run of run beginning of run Current Bit Bit to the Right Explanation Example Op 1 0 Begins run of 1s sub 1 1 Middle of run of 1s none 0 1 End of run of 1s add 0 0 Middle of run of 0s none
8 Booth s algorithm in Details 16 Booths Example 1 (2 x 7) (Assuming 4bit) Operation Multiplicand Product Extra bit What to do next? > sub 1b. P = P m  + = shift P (sign ext) > nop, shift > nop, shift > add 4b. + = shift Final Result 4 iterations in total 14=2x7 17 8
9 quotient divisor dividend remainder 10 nbit operands yield nbit quotient and remainder Division Check for 0 divisor Long division approach (by hand) If divisor dividend bits 1 bit in quotient, then subtract Otherwise (i.e., > dividend bits) 0 bit in quotient, bring down next dividend bit For computers Not as smart as a human Do the subtract, and if remainder < 0, add divisor back Signed integer division? Divide using absolute values (same as x) Adjust sign of quotient and remainder as required 19 Register Setup Remainder register R (initially place dividend in R) (64 bits) Divisor register D (place divisor in left half of D) (64 bits) We place divisor in the left half so we can start subtracting from the most significant dividend bits. Quotient register Q (32 bits) D Divisor 000 R 0000 Dividend 20 9
10 Division Hardware Version 1 D Divisor 000 R 0000 Dividend Initially divisor in left half Place divisor in left half of Divisor register D divisor Place dividend in the right half of Remainder register R So, we can subtract entire D from entire R The first quotient bit = 0. Initially = dividend in right half 21 Divide Algorithm Version 1 Takes n+1 steps for nbit Quotient & Remainder. Start: Place Dividend in Remainder 1. Subtract the Divisor register from the Remainder register, and place the result in the Remainder register. Remainder ³ 0 Test Remainder Remainder < 0 2a. Shift the Quotient register to the left, setting the new rightmost bit to 1. 2b. Restore the original value by adding the Divisor register to the Remainder register. Shift the Quotient register to the left, setting the new least significant bit to Shift the Divisor register right1 bit. n+1 repetition? No: < n+1 repetitions Yes: n+1 repetitions Done 22 10
11 7/2è quotient 3 remainder 1 Quotient Example Divisor Dividend in remainder 1 Initial values Q: 0000 D: 0000 R: D = : R = R D Q: 0000 D: 0000 R: ß R D < 0 2b: +D, sl Q, 0 Q: 0000 D: 0000 R: : Shr D Q: 0000 D: R: D = : R = R D Q: 0000 D: R: b: +D, sl Q, 0 Q: 0000 D: R: : Shr D Q: 0000 D: R: D = : R = R D Q: 0000 D: R: b: +D, sl Q, 0 Q: 0000 D: R: : Shr D Q: 0000 D: R: D = : R = R D Q: 0000 D: R: a: sl Q, 1 Q: 0001 D: R: : Shr D Q: 0001 D: 0000 R: D = : R = R D Q: 0001 D: 0000 R: a: sl Q, 1 Q: 0011 D: 0000 R: : Shr D Q: 0011 D: R: bits è 5 iterations: first iteration will always generate Q = 0 bit, because we are subtracting 0D iteration 23 Improved on Divide Version 2 Half of the bits in the divisor are always zero 32 bits of 64bit ALU are wasted! Instead of shifting Divisor to right, we can shift Remainder to left (they are equivalent) 1st step quotient bit is always zero, we can save 1 iteration by shift first and then subtract at the beginning. Can eliminate Quotient register by combining with Remainder Start by shifting Remainder left The consequence of combining the two registers together and the new order of the operations in the loop is that the Remainder will be shifted left one extra time. Thus, the final correction step must shift back the remainder in the left half of the register 25 11
12 Optimized Divider Divisor 32 bits 32bit ALU HI LO Remainder (Quotient) 64 bits Shift Left Write Control One cycle per partialremainder subtraction Looks a lot like a multiplier! Indeed, same hardware can be used for both 26 Divide Algorithm Version 2 Remainder Divisor Start: Place Dividend in Remainder 0. Shift the Remainder register left 1 bit. 1. Subtract the Divisor register from the left half of the Remainder register, & place the result in the left half of the Remainder register. Remainder ³ 0 Test Remainder Remainder < 0 2a. Shift the Remainder register to the left, setting the new rightmost bit to 1. 2b. Restore the original value by adding the Divisor register to the left half of the Remainder register. Shift the Remainder register to the left, setting the new rightmost bit to 0. nth repetition? No: < n repetitions Done. Shift left half of Remainder right 1 bit. Yes: n repetitions (n = 4 here) 27 12
13 Steps Input: Iteration 1 Divisor (D reg.) 7/2 =? Dividend (R reg.) Operation to do shl R sub D 2 = 1110 Start: Place Dividend in Remainder 0. Shift the Remainder register left 1 bit. Iteration a or 2b? 1. Subtract the Divisor register from the left half of the Remainder register, & place the result in the left half of the Remainder register. Iteration sub D Remainder 0 Test Remainder Remainder < 0 Iteration 2 Iteration a or 2b? sub D 2a. Shift the Remainder register to the left, setting the new rightmost bit to 1. 2b. Restore the original value by adding the Divisor register to the left half of the Remainder register. Shift the Remainder register to the left, setting the new rightmost bit to 0. Iteration a or 2b? nth repetition? No: < n repetitions Iteration sub D Done. Shift left half of Remainder right 1 bit. Yes: n repetitions (n = 4 here) Iteration a or 2b? Final: shift remainder to right 29 MIPS Division Also use special HI/LO registers for result HI: store 32bit Remainder LO: store 32bit Quotient Instructions div rs, rt or divu rs, rt No overflow checking, no divideby0 checking Software must perform checks if required Use mfhi, mflo to access result 31 13
14 Optimized Divider Divisor 32 bits 32bit ALU HI LO Dividend (Remainder) (Quotient) 64 bits Shift Left Write Control 32 Optimized Multiply Hardware Multiplicand 32 bits 32bit ALU HI LO Product (Multiplier) 64 bits Shift Right Read LSB Control 33 14
15 More About Divide Version 2 Same Hardware as Multiply: just let ALU do add or subtract, and let 64bit register shift left or shift right Hi and Lo registers combined to act as 64bit register for multiply and divide Signed divide: Simplest is to make both operands positive, then negate Quotient and Remainder if necessary Complexity: What are the signs of Quotient and Remainder? Note1: Quotient negated if Dividend & Divisor have different signs Note2: Dividend and Remainder must have same sign e.g., 7 2 = 3, remainder =? e.g., 72 = 3, remainder =? e.g., 72 = 3, remainder =? Because Dividend = Quotient x Divisor + Remainder 34 15
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