Arithmetic III CPSC 321

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Arithmetic IIICPSC 321

• Andreas Klappenecker

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Any Questions?
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Todays Menu

• Addition
• Multiplication
• Floating Point Numbers

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Recall Full Adder
cin
s
a
b
cout
3 gates delay for first adder, 2(n-1) for
remaining adders
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Ripple Carry Adders

• Each gates causes a delay
• our example 3 gates for carry generation
• book has example with 2 gates
• Carry might ripple through all n adders
• O(n) gates causing delay
• intolerable delay if n is large
• Carry lookahead adders

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Faster Adders
Why are they called like that?
cin a b cout s
0 0 0 0 0
0 0 1 0 1
0 1 0 0 1
0 1 1 1 0
1 0 0 0 1
1 0 1 1 0
1 1 0 1 0
1 1 1 1 1
coutabcin(a xor b) abacinbcin
ab(ab)cin g p cin Generate g
ab Propagate p ab
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Fast Adders

• Iterate the idea, generate and propagate
• ci1 gi pici
• gi pi(gi-1 pi-1 ci-1)
• gi pigi-1 pipi-1ci-1
• gi pigi-1 pipi-1gi-2 pipi-1 p1g0
• 
  pipi-1 p1p0c0
• Two level AND-OR circuit
• Carry is known early!

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A Simple ALU for MIPS

• Need to support the set-on-less-than instruction
  (slt)
• remember slt is an arithmetic instruction
• produces 1 if rs lt rt and 0 otherwise
• use subtraction (a-b) lt 0 implies a lt b
• Need to support test for equality (beq t5, t6,
  t7)
• use subtraction (a-b) 0 implies a b

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ALU

• 000 and001 or010 add110 subtract111
  slt

• Note zero is a 1 when the result is zero!

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Multipliers
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Multiplication

• More complicated than addition
• accomplished via shifting and addition
• Let's look at 3 versions based on the grade
  school algorithm 0010
  (multiplicand) __ x_1011 (multiplier)
• 0010 x 1
• 00100 x 1
• 001000 x 0
• 0010000 x 1
• 00010110
• Shift and add if multiplier bit equals 1

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Multiplication

• 0010 (multiplicand)
• __ x_1011 (multiplier)
  
• 0010 x 1
• 00100 x 1
• 001000 x 0 0010000 x 1
  0010110

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Multiplication

• If each step took a clock cycle, this algorithm
  would use almost 100 clock cycles to multiply two
  32-bit numbers.
• Requires 64-bit wide adder
• Multiplicand register 64-bit wide

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Variations on a Theme

• Product register has to be 64-bit
• Nothing we can do about that!
• Can we take advantage of that fact?
• Yes! Add multiplicand to 32 MSBs
• product product gtgt 1
• Repeat last steps

0010 (multiplicand) __ x_1011 (multiplier)
0010 x 1
00100 x 1 001000 x 0
0010000 x 1 0010110

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Second Version
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Version 1 versus Version 2
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Critique

• Registers needed for
• multiplicand
• multiplier
• product
• Use lower 32 bits of product register
• place multiplier in lower 32 bits
• add multiplicand to higher 32 bits
• product product gtgt 1
• repeat

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Final Version
Multiplier (shifts right)
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Summary

• It was possible to improve upon the
• well-known grade school algorithm by
• reducing the adder from 64 to 32 bits
• keeping the multiplicand fixed
• shifting the product register
• omitting the multiplier register

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The Booth Multiplier
Lets kick it up a notch!
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Runs of 1s

• 011102 14 842 16 2
• Runs of 1s
• (current bit, bit to the right)
• 10 beginning of run
• 11 middle of a run
• 01 end of a run of 1s
• 00 middle of a run of 0s

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Runs of 1s

• 0111 1111 11002 2044
• How do you get this conversion quickly?
• 0111 11112 128 1 127
• 0111 1111 11112 2048 1
• 0111 1111 11002 2048 1 3 2048 4

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Example
0010 0110 0000 shift -0010 sub
0000 shift 0010 add 00001100

• 0010
• 0110
• 0000 shift
• 0010 add
• 0010 add
• 0000 shift
• 00001100

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Booth Multiplication

• Current and previous bit
• 00 middle of run of 0s, no action
• 01 end of a run of 1s, add multiplicand
• 10 beginning of a run of 1s, subtract mcnd
• 11 middle of string of 1s, no action

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Example 0010 x 0110
Iteration Mcand Step Product
0 0010 Initial values 0000 0110,0
1 0010 0010 00 no op arithgtgt 1 0000 0110,0 0000 0011,0
2 0010 0010 10 prod-Mcand arithgtgt 1 1110 0011,0 1111 0001,1
3 0010 0010 11 no op arithgtgt 1 1111 0001,1 1111 1000,1
4 0010 0010 01 prodMcand arithgtgt 1 0001 1000,1 0000 1100,0
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Negative numbers

• Booths multiplication works also with negative
  numbers
• 2 x -3 -6
• 00102 x 11012 1111 10102

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Negative Numbers
00102 x 11012 1111 10102 0) Mcnd 0010 Prod
0000 1101,0 1) Mcnd 0010 Prod 1110 1101,1 sub 1)
Mcnd 0010 Prod 1111 0110,1 gtgt 2) Mcnd 0010 Prod
0001 0110,1 add 2) Mcnd 0010 Prod 0000 1011,0
gtgt 3) Mcnd 0010 Prod 1110 1011,0 sub 3) Mcnd 0010
Prod 1111 0101,1 gtgt 4) Mcnd 0010 Prod 1111 0101,1
nop 4) Mcnd 0010 Prod 1111 1010,1 gtgt
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Summary

• Extends the final version of the grade school
  algorithm
• Simple change add, subtract, or do nothing if
  last and previous bit respectively satisfy 0,1
  1,0 or 0,0 1,1
• 0111 11002 128 4
• 1000 0002 0000 01002

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Floating Point Numbers
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Floating Point Numbers

• We often use calculations based on real numbers,
  such as
• e 2.71828
• Pi 3.14592
• We represent approximations to such numbers by
  floating point numbers
• 1.xxxxxxxxxx2 x 2yyyy

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Floating-Point Representation float

• We need to distribute the 32 bits among sign,
  exponent, and significand
• seeeeeeeexxxxxxxxxxxxxxxxxxxxxxx
• The general form of such a number is
• (-1)s x F x 2E
• s is the sign, F is derived from the
  significand field, and E is derived from the
  exponent field

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Floating Point Representation double

• 1 bit sign, 11 bits for exponent, 52 bits for
  significand
• seeeeeeeeeeexxxxxxxxxxxxxxxxxxxx
• xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
• Range of float 2.0 x 10-38 2.0 x 1038
• Range of double 2.0 x 10-308 2.0 x 10308

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IEEE 754 Floating-Point Standard

• Makes leading bit of normalized binary number
  implicit 1 significand
• If significand is s1 s2 s3 s4 s5 s6
• then the value is
• (-1)s x (1 s1/2 s2/4 s3/8 ) 2E
• Design goal of IEEE 754
  Integer comparisons should yield meaningful
  comparisons for floating point numbers

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IEEE 754 Standard

• Negative exponents are a difficulty for sorting
• Idea most positive most negative
• 1111 1111 0000 0000
• IEEE 754 uses a bias of 127 for single precision.
  
• Exponent -1 is represented by
• -1 127 126

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IEEE 754 Example

• Represent -0.75 in single precision format.
• -0.75 -3/4 -112 / 4 -0.112
• In scientific notation -0.11 x 20 -1.1 x 2-1
• the latter form is normalized sc. notation
• Value (-1)s x (1 significand) x 2(Expnt 127)

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Example (contd)

• -1.1 x 2-1
• (-1)1 x (1 .1000 0000 0000 0000 0000 000) x
  2(126 127)
• The single precision representation is
• 1 0111 1110 1000 0000 0000 0000 0000 000

BAM!
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Conclusion

• We learned how to multiply
• Three variations on the grade school algorithm
• Booth multiplication
• Floating point representation
• a la IEEE 754
• (Photos are courtesy of www.emerils.com,
• some graphs are due to Patterson and Hennessy)