1. CS152 Computer Architecture and Engineering Lecture 6 Verilog (finish) Multiply, Divide, Shift February 11, 2004 John Kubiatowicz (www.cs.berkeley.edu/~kubitron) lecture slides: http://www-inst.eecs.berkeley.edu/~cs152/

2. CS152 / Kubiatowicz Lec6.2 2/11/03©UCB Spring 2004 Review from last time °Design Process Design Entry: Schematics, HDL, Compilers High Level Analysis: Simulation, Testing, Assertions Technology Mapping: Turn design into physical implementation Low Level Analysis: Check out Timing, Setup/Hold, etc °Verilog – Three programming styles Structural: Like a Netlist -Instantiation of modules + wires between them Dataflow: Higher Level -Expressions instead of gates Behavioral: Hardware programming -Full flow-control mechanisms -Registers, variables -File I/O, consol display, etc

3. CS152 / Kubiatowicz Lec6.3 2/11/03©UCB Spring 2004 °Blocking Assignments: Assignments happen more like programming language (sequential code) Both Right and left sides evaluated completely Wait until assignment before going on -Can cause unexpected results when connecting output to logic in other always blocks. -Also a bit strange with delays on left hand side (LHS) °Example: reg E, C; always @(posedge clk) begin E = ~A; C = ~E; end A E C Verilog subtlety: Blocking Assignments

4. CS152 / Kubiatowicz Lec6.4 2/11/03©UCB Spring 2004 °Non-blocking Assignments: All right-hand sides evaluated immediately Then assignments occur If no delays, often want output ports to be assigned with non-blocking assignments °Example: reg E, C; always @(posedge clk) begin E <= ~A; C <= ~E; end AC E Verilog subtlety: Non-Blocking Assignments

5. CS152 / Kubiatowicz Lec6.5 2/11/03©UCB Spring 2004 Sequential Logic (Revisited: better scheduling) module FF (CLK,Q,D); input D, CLK; output Q; reg Q; always @ (posedge CLK) Q <= D; endmodule // FF Good: Doesn’t output until “after edge” Must be careful mixing zero-time blocking assignments and edge-triggering: Probably won’t do what you expect when connecting it to other things! module FF (CLK,Q,D); input D, CLK; output Q; reg Q; always @ (posedge CLK) Q = #5 D; endmodule // FF Good: Outputs 5 units “after edge” module FF (CLK,Q,D); input D, CLK; output Q; reg Q; always @ (posedge CLK) #5 Q = D; endmodule // FF Probably Not what you Expect: Hold time of 5 units glitches < 5 units ignored

6. CS152 / Kubiatowicz Lec6.6 2/11/03©UCB Spring 2004 A final word on Verilog °Verilog does not turn hardware design into writing programs! Since Verilog looks similar to programming languages, some think that they can design hardware by writing programs. - NOT SO. Verilog is a hardware description language. -The best way to use it is to first figure out the circuit you want, then figure out how to describe it in Verilog. The behavioral construct hides a lot of the circuit details but you as the designer must still manage: - the structure -data-communication -Parallelism -timing of your design. -Not doing so leads to very inefficient designs! °Read the document on non-blocking assignment in Verilog that I put up on the handouts page. Lots of very interesting things!

7. CS152 / Kubiatowicz Lec6.7 2/11/03©UCB Spring 2004 How Program: FPGA Generic Design Flow °Design Entry: Create your design files using: - schematic editor or -hardware description language (Verilog, VHDL) °Design “implementation” on FPGA: Partition, place, and route (“PPR”) to create bit-stream file Divide into CLB-sized pieces, place into blocks, route to blocks °Design verification: Use Simulator to check function, Other software determines max clock frequency. Load onto FPGA device (cable connects PC to board) -check operation at full speed in real environment.

8. CS152 / Kubiatowicz Lec6.8 2/11/03©UCB Spring 2004 Idealized FPGA Logic Block °4-input Look Up Table (4-LUT) implements combinational logic functions °Register optionally stores output of LUT Latch determines whether read reg or LUT

9. CS152 / Kubiatowicz Lec6.9 2/11/03©UCB Spring 2004 4-LUT Implementation °n-bit LUT is actually implemented as a 2 n x 1 memory: inputs choose one of 2 n memory locations. memory locations (latches) are normally loaded with values from user’s configuration bit stream. Inputs to mux control are the CLB (Configurable Logic Block) inputs. °Result is a general purpose “logic gate”. n-LUT can implement any function of n inputs!

10. CS152 / Kubiatowicz Lec6.10 2/11/03©UCB Spring 2004 LUT as general logic gate °An n-lut as a direct implementation of a function truth-table °Each latch location holds value of function corresponding to one input combination Example: 4-lut Example: 2-lut Implements any function of 2 inputs. How many functions of n inputs?

11. CS152 / Kubiatowicz Lec6.11 2/11/03©UCB Spring 2004 RAM16X1S O D WE WCLK A0 A1 A2 A3 RAM32X1S O D WE WCLK A0 A1 A2 A3 A4 RAM16X2S O1 D0 WE WCLK A0 A1 A2 A3 D1 O0 = = LUT or LUT RAM16X1D SPO D WE WCLK A0 A1 A2 A3 DPRA0DPO DPRA1 DPRA2 DPRA3 or Additional application: Distributed RAM °CLB LUT configurable as Distributed RAM A LUT equals 16x1 RAM Implements Single and Dual-Ports Cascade LUTs to increase RAM size °Synchronous write °Synchronous/Asynchronous read Accompanying flip-flops used for synchronous read

12. CS152 / Kubiatowicz Lec6.12 2/11/03©UCB Spring 2004 Block RAM Spartan-IIE True Dual-Port Block RAM Port A Port B Block RAM (Extra RAM not using LUTs) °Most efficient memory implementation Dedicated blocks of memory °Ideal for most memory requirements Virtex-E XCV2000 has 160? blocks -4096 bits per blocks (4K x 1, 2K x 4, 512 x 8, 256 x 16) Use multiple blocks for larger memories °Builds both single and true dual-port RAMs °CORE Generator provides custom-sized block RAMs Quickly generates optimized RAM implementation

13. CS152 / Kubiatowicz Lec6.13 2/11/03©UCB Spring 2004 DQ CE DQ DQ DQ LUT IN CE CLK DEPTH[3:0] OUT LUT = Additional Application: Shift Register °Each LUT can be configured as shift register Serial in, serial out °Saves resources: can use less than 16 FFs °Faster: no routing °Note: CAD tools determine with CLB used as LUT, RAM, or shift register, rather than up to designer

14. CS152 / Kubiatowicz Lec6.14 2/11/03©UCB Spring 2004 Example Partition, Placement, and Route °Example Schematic Circuit: collection of gates and flip- flops °Idealized FPGA structure: Circuit combinational logic must be “covered” by 4-input 1-output “gates”. Flip-flops from circuit must map to FPGA flip-flops. (Best to preserve “closeness” to CL to minimize wiring.) Placement in general attempts to minimize wiring.

15. CS152 / Kubiatowicz Lec6.15 2/11/03©UCB Spring 2004 Xilinx Vittex-E Routing Hierarchy DIRECT CONNECTION INTERNAL BUSSES Single-length lines Buffered Hex lines (1/6 blocks) Direct connections Long lines and Global lines Internal 3-state Bus °24 single-length lines Route GRM signals to adjacent GRMs in 4 directions °96 buffered hex lines Route GRM (general routing matrix) signals to another GRMs six blocks away in each of the 4 directions °12 buffered Long lines Routing across top and bottom, left and right Note: CAD tools do PPR, not designers

16. CS152 / Kubiatowicz Lec6.16 2/11/03©UCB Spring 2004 Virtex-E Configurable Logic Block (CLB) 2 “logic slices” / CLB, two 4-LUTs / slice => Four 4-LUTs / CLB

17. CS152 / Kubiatowicz Lec6.17 2/11/03©UCB Spring 2004 Virtex-E CLB Slice Structure °Each slice contains two sets of the following: Four-input LUT -Any 4-input logic function -Or 16-bit x 1 sync RAM -Or 16-bit shift register Carry & Control -Fast arithmetic logic -Multiplexer logic -Multiplier logic Storage element -Latch or flip-flop -Set and reset -True or inverted inputs -Sync. or async. control

18. CS152 / Kubiatowicz Lec6.18 2/11/03©UCB Spring 2004 Details of Virtex-E Slice Very fast ripple carry: (24-bit @ 100 MHz) Multiplexors to help combine CLBs into larger multiplexor

19. CS152 / Kubiatowicz Lec6.19 2/11/03©UCB Spring 2004 CLB MUXF6 Slice LUT MUXF5 Slice LUT MUXF5 Virtex-E Dedicated Expansion Multiplexers °Since 4-LUT has 4 inputs, max is 2:1 Mux (2 inputs, 1 control line) °MUXF5 combines 2 LUTs to create 4x1 multiplexer Or any 5-input function (5-LUT) Or selected functions up to 9 inputs °MUXF6 combines 2 slices to form 8x1 multiplexer Or any 6-input function (6-LUT) Or selected functions up to 19 inputs °Dedicated muxes are faster and more space efficient

20. CS152 / Kubiatowicz Lec6.20 2/11/03©UCB Spring 2004 Administrivia °Change in Sections (sorry!): Section 1: 3107 Etcheverry  47 Evans Section 2: 3107 Etcheverry  105 Latimer °Starts tomorrow! °Hopefully well on the way with Lab 2! Some problem with schematics (which is fixed?) °Start Reading Chapter 5 No class on Monday (Holiday) °Next Time: Quick review of single-cycle processor Also: First homework quiz next Wednesday

21. CS152 / Kubiatowicz Lec6.21 2/11/03©UCB Spring 2004 MIPS arithmetic instructions °InstructionExampleMeaningComments °add add $1,$2,$3$1 = $2 + $33 operands; exception possible °subtractsub $1,$2,$3$1 = $2 – $33 operands; exception possible °add immediateaddi $1,$2,100$1 = $2 + 100+ constant; exception possible °add unsignedaddu $1,$2,$3$1 = $2 + $33 operands; no exceptions °subtract unsignedsubu $1,$2,$3$1 = $2 – $33 operands; no exceptions °add imm. unsign.addiu $1,$2,100$1 = $2 + 100+ constant; no exceptions °multiply mult $2,$3Hi, Lo = $2 x $364-bit signed product °multiply unsignedmultu$2,$3Hi, Lo = $2 x $3 64-bit unsigned product °divide div $2,$3Lo = $2 ÷ $3,Lo = quotient, Hi = remainder ° Hi = $2 mod $3 °divide unsigned divu $2,$3Lo = $2 ÷ $3,Unsigned quotient & remainder ° Hi = $2 mod $3 °Move from Himfhi $1$1 = HiUsed to get copy of Hi °Move from Lomflo $1$1 = LoUsed to get copy of Lo

22. CS152 / Kubiatowicz Lec6.22 2/11/03©UCB Spring 2004 MULTIPLY (unsigned) °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) °4 versions of multiply hardware & algorithm: successive refinement

23. CS152 / Kubiatowicz Lec6.23 2/11/03©UCB Spring 2004 Unsigned Combinational Multiplier °Stage i accumulates A * 2 i if B i == 1 °Q: How much hardware for 32 bit multiplier? Critical path? B0B0 A0A0 A1A1 A2A2 A3A3 A0A0 A1A1 A2A2 A3A3 A0A0 A1A1 A2A2 A3A3 A0A0 A1A1 A2A2 A3A3 B1B1 B2B2 B3B3 P0P0 P1P1 P2P2 P3P3 P4P4 P5P5 P6P6 P7P7 0000

24. CS152 / Kubiatowicz Lec6.24 2/11/03©UCB Spring 2004 How does it work? °At each stage shift A left ( x 2) °Use next bit of B to determine whether to add in shifted multiplicand °Accumulate 2n bit partial product at each stage B0B0 A0A0 A1A1 A2A2 A3A3 A0A0 A1A1 A2A2 A3A3 A0A0 A1A1 A2A2 A3A3 A0A0 A1A1 A2A2 A3A3 B1B1 B2B2 B3B3 P0P0 P1P1 P2P2 P3P3 P4P4 P5P5 P6P6 P7P7 0000 00 0

25. CS152 / Kubiatowicz Lec6.25 2/11/03©UCB Spring 2004 Carry Save addition of 4 integers °Adding:A 2 A 1 A 0 +B 2 B 1 B 0 +C 2 C 1 C 0 +D 2 D 1 D 0 ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ S 4 S 3 S 2 S 1 S 0 °Add Columns first, then rows! °Full Adder = 3  2 element °Can be used to reduce critical path of multiply °Example: 53 bit multiply (for floating point): At least 53 levels with naïve technique Only 9 with Carry save addition!

26. CS152 / Kubiatowicz Lec6.26 2/11/03©UCB Spring 2004 Unisigned shift-add multiplier (version 1) °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 Multiplier = datapath + control

27. CS152 / Kubiatowicz Lec6.27 2/11/03©UCB Spring 2004 Multiply Algorithm Version 1 ProductMultiplierMultiplicand 0000 0000 00110000 0010 1:0000 001000110000 0010 2:0000 001000110000 0100 3:0000 001000010000 0100 1:0000 011000010000 0100 2:0000 011000010000 1000 3:0000 011000000000 1000 0000 011000000000 1000 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 = 0 Multiplier0 = 1 1a. Add multiplicand to product & place the result in Product register 32nd repetition? Start

28. CS152 / Kubiatowicz Lec6.28 2/11/03©UCB Spring 2004 Observations on Multiply Version 1 °1 clock per cycle =>  100 clocks per multiply Ratio of multiply to add 5:1 to 100:1 °1/2 bits in multiplicand always 0 => 64-bit adder is wasted °0’s inserted in right of multiplicand as shifted => least significant bits of product never changed once formed °Instead of shifting multiplicand to left, shift product to right?

29. CS152 / Kubiatowicz Lec6.29 2/11/03©UCB Spring 2004 MULTIPLY HARDWARE Version 2 °32-bit Multiplicand reg, 32 -bit ALU, 64-bit Product reg, 32-bit Multiplier reg Product Multiplier Multiplicand 32-bit ALU Shift Right Write Control 32 bits 64 bits Shift Right

30. CS152 / Kubiatowicz Lec6.30 2/11/03©UCB Spring 2004 How to think of this? B0B0 A0A0 A1A1 A2A2 A3A3 A0A0 A1A1 A2A2 A3A3 A0A0 A1A1 A2A2 A3A3 A0A0 A1A1 A2A2 A3A3 B1B1 B2B2 B3B3 P0P0 P1P1 P2P2 P3P3 P4P4 P5P5 P6P6 P7P7 0000 Remember original combinational multiplier:

31. CS152 / Kubiatowicz Lec6.31 2/11/03©UCB Spring 2004 Simply warp to let product move right... °Multiplicand stay’s still and product moves right B0B0 B1B1 B2B2 B3B3 P0P0 P1P1 P2P2 P3P3 P4P4 P5P5 P6P6 P7P7 0000 A0A0 A1A1 A2A2 A3A3 A0A0 A1A1 A2A2 A3A3 A0A0 A1A1 A2A2 A3A3 A0A0 A1A1 A2A2 A3A3

32. CS152 / Kubiatowicz Lec6.32 2/11/03©UCB Spring 2004 Multiply Algorithm Version 2 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 = 0 Multiplier0 = 1 1a. Add multiplicand to the left half of product & place the result in the left half of Product register 32nd repetition? Start 0000 0000 0011 0010 1: 0010 0000 0011 0010 2: 0001 0000 0011 0010 3: 0001 0000 0001 0010 1: 0011 0000 0001 0010 2: 0001 1000 0001 0010 3: 0001 1000 0000 0010 1: 0001 1000 0000 0010 2: 0000 1100 0000 0010 3: 0000 1100 0000 0010 1: 0000 1100 0000 0010 2: 0000 0110 0000 0010 3: 0000 0110 0000 0010 0000 0110 0000 0010 Product Multiplier Multiplicand

33. CS152 / Kubiatowicz Lec6.33 2/11/03©UCB Spring 2004 Still more wasted space! 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 = 0 Multiplier0 = 1 1a. Add multiplicand to the left half of product & place the result in the left half of Product register 32nd repetition? Start 0000 0000 0011 0010 1: 0010 0000 0011 0010 2: 0001 0000 0011 0010 3: 0001 0000 0001 0010 1: 0011 0000 0001 0010 2: 0001 1000 0001 0010 3: 0001 1000 0000 0010 1: 0001 1000 0000 0010 2: 0000 1100 0000 0010 3: 0000 1100 0000 0010 1: 0000 1100 0000 0010 2: 0000 0110 0000 0010 3: 0000 0110 0000 0010 0000 0110 0000 0010 Product Multiplier Multiplicand

34. CS152 / Kubiatowicz Lec6.34 2/11/03©UCB Spring 2004 Observations on Multiply Version 2 °Product register wastes space that exactly matches size of multiplier => combine Multiplier register and Product register

35. CS152 / Kubiatowicz Lec6.35 2/11/03©UCB Spring 2004 MULTIPLY HARDWARE Version 3 °32-bit Multiplicand reg, 32 -bit ALU, 64-bit Product reg, (0-bit Multiplier reg) Product (Multiplier) Multiplicand 32-bit ALU Write Control 32 bits 64 bits Shift Right

36. CS152 / Kubiatowicz Lec6.36 2/11/03©UCB Spring 2004 Multiply Algorithm Version 3 Done Yes: 32 repetitions 2. Shift the Product register right 1 bit. No: < 32 repetitions 1. Test Product0 Product0 = 0 Product0 = 1 1a. Add multiplicand to the left half of product & place the result in the left half of Product register 32nd repetition? Start 0000 0011 0010 1: 0010 0011 0010 2: 0001 0001 0010 1: 0011 0000 0010 2: 0001 1000 0010 1: 0001 1000 0010 2: 0000 1100 0010 1: 0000 1100 0010 2: 0000 0110 0010 0000 0110 0010 Product Multiplicand

37. CS152 / Kubiatowicz Lec6.37 2/11/03©UCB Spring 2004 Observations on Multiply Version 3 °2 steps per bit because Multiplier & Product combined °MIPS registers Hi and Lo are left and right half of Product °Gives us MIPS instruction MultU °How can you make it faster? °What about signed multiplication? easiest solution is to make both positive & remember whether to complement product when done (leave out the sign bit, run for 31 steps) apply definition of 2’s complement -need to sign-extend partial products and subtract at the end Booth’s Algorithm is elegant way to multiply signed numbers using same hardware as before and save cycles -can handle multiple bits at a time

38. CS152 / Kubiatowicz Lec6.38 2/11/03©UCB Spring 2004 Motivation for Booth’s Algorithm °Example 2 x 6 = 0010 x 0110: 0010 x 0110 + 0000shift (0 in multiplier) + 0010 add (1 in multiplier) + 0010 add (1 in multiplier) + 0000 shift (0 in multiplier) 00001100 °ALU with add or subtract gets same result in more than one way: 6= – 2 + 8 0110 = – 00010 + 01000 = 11110 + 01000 °For example ° 0010 x 0110 0000 shift (0 in multiplier) – 0010 sub (first 1 in multpl.). 0000 shift (mid string of 1s). +0010 add (prior step had last 1) 00001100

39. CS152 / Kubiatowicz Lec6.39 2/11/03©UCB Spring 2004 Booth’s Algorithm Current BitBit to the RightExplanationExampleOp 10Begins run of 1s0001111000sub 11Middle of run of 1s0001111000none 01End of run of 1s0001111000add 00Middle of run of 0s0001111000none Originally for Speed (when shift was faster than add) °Replace a string of 1s in multiplier with an initial subtract when we first see a one and then later add for the bit after the last one 0 1 1 1 1 0 beginning of runend of run middle of run –1 + 10000 01111

40. CS152 / Kubiatowicz Lec6.40 2/11/03©UCB Spring 2004 Booths Example (2 x 7) 1a. P = P - m1110 +1110 1110 0111 0shift P (sign ext) 1b. 00101111 0011 111 -> nop, shift 2.00101111 1001 111 -> nop, shift 3.00101111 1100 101 -> add 4a.0010 +0010 0001 1100 1shift 4b.00100000 1110 0done OperationMultiplicandProductnext? 0. initial value00100000 0111 010 -> sub

41. CS152 / Kubiatowicz Lec6.41 2/11/03©UCB Spring 2004 Booths Example (2 x -3) 1a. P = P - m1110 +1110 1110 1101 0shift P (sign ext) 1b. 00101111 0110 101 -> add + 0010 2a.0001 0110 1shift P 2b.00100000 1011 010 -> sub +1110 3a.00101110 1011 0shift 3b.0010 1111 0101 111 -> nop 4a1111 0101 1 shift 4b.00101111 1010 1 done OperationMultiplicandProductnext? 0. initial value00100000 1101 010 -> sub

42. CS152 / Kubiatowicz Lec6.42 2/11/03©UCB Spring 2004 Radix-4 Modified Booth’s Algorithm Current Bit to theExplanationExampleRecode BitsRight 0 00Middle of zeros00 00 00 00 000 0 1 0Single one00 00 00 01 00 1 1 00Begins run of 1s00 01 11 10 00-2 1 10Begins run of 1s00 01 11 11 00 -1 0 01Ends run of 1s00 00 11 11 00 1 0 11Ends run of 1s00 01 11 11 00 2 1 01Isolated 000 11 10 11 00 -1 1 11Middle of run00 11 11 11 000 °Same insight as one-bit Booth’s, simply adjust for alignment of 2 bits. °Allows multiplication 2-bits at a time. –1 + 10000 01111

43. CS152 / Kubiatowicz Lec6.43 2/11/03©UCB Spring 2004 Divide: Paper & Pencil 1001 Quotient Divisor 1000 1001010 Dividend –1000 10 101 1010 –1000 10 Remainder (or Modulo result) See how big a number can be subtracted, creating quotient bit on each step Binary => 1 * divisor or 0 * divisor Dividend = Quotient x Divisor + Remainder => | Dividend | = | Quotient | + | Divisor | 3 versions of divide, successive refinement

44. CS152 / Kubiatowicz Lec6.44 2/11/03©UCB Spring 2004 DIVIDE HARDWARE Version 1 °64-bit Divisor reg, 64-bit ALU, 64-bit Remainder reg, 32-bit Quotient reg Remainder Quotient Divisor 64-bit ALU Shift Right Shift Left Write Control 32 bits 64 bits

45. CS152 / Kubiatowicz Lec6.45 2/11/03©UCB Spring 2004 2b. Restore the original value by adding the Divisor register to the Remainder register, & place the sum in the Remainder register. Also shift the Quotient register to the left, setting the new least significant bit to 0. Divide Algorithm Version 1 °Takes n+1 steps for n-bit Quotient & Rem. Remainder QuotientDivisor 0000 0111 00000010 0000 Test Remainder Remainder < 0 Remainder  0 1. Subtract the Divisor register from the Remainder register, and place the result in the Remainder register. 2a. Shift the Quotient register to the left setting the new rightmost bit to 1. 3. Shift the Divisor register right1 bit. Done Yes: n+1 repetitions (n = 4 here) Start: Place Dividend in Remainder n+1 repetition? No: < n+1 repetitions

46. CS152 / Kubiatowicz Lec6.46 2/11/03©UCB Spring 2004 Divide Algorithm I example (7 / 2) Remainder QuotientDivisor 0000 0111000000010 0000 1:1110 0111000000010 0000 2:0000 0111000000010 0000 3:0000 0111000000001 0000 1:1111 0111000000001 0000 2:0000 0111000000001 0000 3:0000 0111000000000 1000 1:1111 1111000000000 1000 2:0000 0111000000000 1000 3:0000 0111000000000 0100 1:0000 0011000000000 0100 2:0000 0011000010000 0100 3:0000 0011000010000 0010 1:0000 0001000010000 0010 2:0000 0001000110000 0010 3:0000 0001000110000 0001 Answer: Quotient = 3 Remainder = 1

47. CS152 / Kubiatowicz Lec6.47 2/11/03©UCB Spring 2004 Divide: Paper & Pencil 01010 Quotient Divisor 0001 00001010 Dividend 00001 –0001 0000 0001 –0001 0 00 Remainder (or Modulo result) No way to get a 1 in leading digit! (this is an overflow, i.e quotient would have n+1 bits)  switch order to shift first and then subtract, can save 1 iteration

48. CS152 / Kubiatowicz Lec6.48 2/11/03©UCB Spring 2004 Observations on Divide Version 1 °1/2 bits in divisor always 0 => 1/2 of 64-bit adder is wasted => 1/2 of divisor is wasted °Instead of shifting divisor to right, shift remainder to left?

49. CS152 / Kubiatowicz Lec6.49 2/11/03©UCB Spring 2004 Divide Algorithm I example: wasted space Remainder QuotientDivisor 0000 0111000000010 0000 1:1110 0111000000010 0000 2:0000 0111000000010 0000 3:0000 0111000000001 0000 1:1111 0111000000001 0000 2:0000 0111000000001 0000 3:0000 0111000000000 1000 1:1111 1111000000000 1000 2:0000 0111000000000 1000 3:0000 0111000000000 0100 1:0000 0011000000000 0100 2:0000 0011000010000 0100 3:0000 0011000010000 0010 1:0000 0001000010000 0010 2:0000 0001000110000 0010 3:0000 0001000110000 0010

50. CS152 / Kubiatowicz Lec6.50 2/11/03©UCB Spring 2004 DIVIDE HARDWARE Version 2 °32-bit Divisor reg, 32-bit ALU, 64-bit Remainder reg, 32-bit Quotient reg Remainder Quotient Divisor 32-bit ALU Shift Left Write Control 32 bits 64 bits Shift Left

51. CS152 / Kubiatowicz Lec6.51 2/11/03©UCB Spring 2004 Divide Algorithm Version 2 Remainder Quotient Divisor 0000 0111 0000 0010 3b. Restore the original value by adding the Divisor register to the left half of the Remainderregister, &place the sum in the left half of the Remainder register. Also shift the Quotient register to the left, setting the new least significant bit to 0. Test Remainder Remainder < 0 Remainder  0 2. Subtract the Divisor register from the left half of the Remainder register, & place the result in the left half of the Remainder register. 3a. Shift the Quotient register to the left setting the new rightmost bit to 1. 1. Shift the Remainder register left 1 bit. Done Yes: n repetitions (n = 4 here) nth repetition? No: < n repetitions Start: Place Dividend in Remainder

52. CS152 / Kubiatowicz Lec6.52 2/11/03©UCB Spring 2004 Observations on Divide Version 2 °Eliminate Quotient register by combining with Remainder as shifted left Start by shifting the Remainder left as before. Thereafter loop contains only two steps because the shifting of the Remainder register shifts both the remainder in the left half and the quotient in the right half The consequence of combining the two registers together and the new order of the operations in the loop is that the remainder will shifted left one time too many. Thus the final correction step must shift back only the remainder in the left half of the register

53. CS152 / Kubiatowicz Lec6.53 2/11/03©UCB Spring 2004 DIVIDE HARDWARE Version 3 °32-bit Divisor reg, 32 -bit ALU, 64-bit Remainder reg, (0-bit Quotient reg) Remainder (Quotient) Divisor 32-bit ALU Write Control 32 bits 64 bits Shift Left “HI”“LO”

54. CS152 / Kubiatowicz Lec6.54 2/11/03©UCB Spring 2004 Divide Algorithm Version 3 Remainder Divisor 0000 01110010 3b. Restore the original value by adding the Divisor register to the left half of the Remainderregister, &place the sum in the left half of the Remainder register. Also shift the Remainder register to the left, setting the new least significant bit to 0. Test Remainder Remainder < 0 Remainder  0 2. Subtract the Divisor register from the left half of the Remainder register, & place the result in the left half of the Remainder register. 3a. Shift the Remainder register to the left setting the new rightmost bit to 1. 1. Shift the Remainder register left 1 bit. Done. Shift left half of Remainder right 1 bit. Yes: n repetitions (n = 4 here) nth repetition? No: < n repetitions Start: Place Dividend in Remainder

55. CS152 / Kubiatowicz Lec6.55 2/11/03©UCB Spring 2004 Observations on Divide Version 3 °Same Hardware as Multiply: just need ALU to add or subtract, and 64-bit register to shift left or shift right °Hi and Lo registers in MIPS combine to act as 64-bit register for multiply and divide °Signed Divides: Simplest is to remember signs, make positive, and complement quotient and remainder if necessary Note: Dividend and Remainder must have same sign Note: Quotient negated if Divisor sign & Dividend sign disagree e.g., –7 ÷ 2 = –3, remainder = –1 What about? –7 ÷ 2 = –4, remainder = +1 °Possible for quotient to be too large: if divide 64-bit integer by 1, quotient is 64 bits (“called saturation”)

56. CS152 / Kubiatowicz Lec6.56 2/11/03©UCB Spring 2004 MIPS logical instructions °InstructionExampleMeaningComment °and and $1,$2,$3$1 = $2 & $33 reg. operands; Logical AND °or or $1,$2,$3$1 = $2 | $33 reg. operands; Logical OR °xor xor $1,$2,$3$1 = $2  $33 reg. operands; Logical XOR °nor nor $1,$2,$3$1 = ~($2 |$3)3 reg. operands; Logical NOR °and immediate andi $1,$2,10$1 = $2 & 10Logical AND reg, constant °or immediate ori $1,$2,10$1 = $2 | 10Logical OR reg, constant °xor immediate xori $1, $2,10 $1 = ~$2 &~10Logical XOR reg, constant °shift left logical sll $1,$2,10$1 = $2 << 10Shift left by constant °shift right logical srl $1,$2,10$1 = $2 >> 10Shift right by constant °shift right arithm. sra $1,$2,10$1 = $2 >> 10Shift right (sign extend) °shift left logical sllv $1,$2,$3$1 = $2 << $3 Shift left by variable °shift right logical srlv $1,$2, $3 $1 = $2 >> $3 Shift right by variable °shift right arithm. srav $1,$2, $3 $1 = $2 >> $3 Shift right arith. by variable

57. CS152 / Kubiatowicz Lec6.57 2/11/03©UCB Spring 2004 Shifters Two kinds: logical-- value shifted in is always "0" arithmetic-- on right shifts, sign extend msblsb"0" msblsb"0" Note: these are single bit shifts. A given instruction might request 0 to 32 bits to be shifted!

58. CS152 / Kubiatowicz Lec6.58 2/11/03©UCB Spring 2004 Combinational Shifter from MUXes °What comes in the MSBs? °How many levels for 32-bit shifter? °What if we use 4-1 Muxes ? 1 0 sel A B D Basic Building Block 8-bit right shifter 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 S 2 S 1 S 0 A0A0 A1A1 A2A2 A3A3 A4A4 A5A5 A6A6 A7A7 R0R0 R1R1 R2R2 R3R3 R4R4 R5R5 R6R6 R7R7

59. CS152 / Kubiatowicz Lec6.59 2/11/03©UCB Spring 2004 General Shift Right Scheme using 16 bit example If added Right-to-left connections could support Rotate (not in MIPS but found in ISAs)

60. CS152 / Kubiatowicz Lec6.60 2/11/03©UCB Spring 2004 Funnel Shifter XY R °Shift A by i bits (sa= shift right amount) °Logical: Y = 0, X=A, sa=i °Arithmetic? Y = _, X=_, sa=_ °Rotate? Y = _, X=_, sa=_ °Left shifts? Y = _, X=_, sa=_ Instead Extract 32 bits of 64. Shift Right 32 Y X R

61. CS152 / Kubiatowicz Lec6.61 2/11/03©UCB Spring 2004 Barrel Shifter Technology-dependent solutions: transistor per switch

62. CS152 / Kubiatowicz Lec6.62 2/11/03©UCB Spring 2004 Summary °Multiply: successive refinement to see final design 32-bit Adder, 64-bit shift register, 32-bit Multiplicand Register Booth’s algorithm to handle signed multiplies There are algorithms that calculate many bits of multiply per cycle (see exercises 4.36 to 4.39 in COD) °Booth Encoding: introducing multiple representations Numbers can be encoded in different ways  SRT Divide (for instance) Handles multiplication of negative numbers transparently °Divide can use same hardware as multiply: Hi & Lo registers in MIPS °Shifter: success refinement 1/bit at a time shift register to barrel shifter

63. CS152 / Kubiatowicz Lec6.63 2/11/03©UCB Spring 2004 To Get More Information °Chapter 4 of your text book: David Patterson & John Hennessy, “Computer Organization & Design,” Morgan Kaufmann Publishers, 1994. °David Winkel & Franklin Prosser, “The Art of Digital Design: An Introduction to Top-Down Design,” Prentice-Hall, Inc., 1980. °Kai Hwang, “Computer Arithmetic: Principles, archtiecture, and design”, Wiley 1979