CSE 246: Computer Arithmetic Algorithms and Hardware Design

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CSE 246 Computer Arithmetic Algorithms and
Hardware Design
Fall 2006 Lecture 2 Redundant and residue number
systems

• Instructor
• Prof. Chung-Kuan Cheng

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Topics

• Redundant Number Systems
• Residue Number Systems
• Mixed Radix Number Systems (as they pertain to
  conversions of Residue Number Systems)

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Redundant Number Systems

• Examine number systems that can be characterized
  by the 3-tuple (r, ?, ?)
• r radix, ?, ? set of digits
• The redundancy of such a system is said to be
• non-redundant if ? - ? 1 r
• redundant if ? - ? 1 gt r
• minimally redundant if ? - ? r
• maximally redundant if ? - ? 2r - 1
• over redundant if ? - ? gt 2r - 1

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Redundant Number Systems

• Ex (2,0,2)
• Redundant - furthermore, minimally redundant
• Addition example
• 1 0 2 (6)
• 1 1 1 (7)
• 2 1 3 3 not in radix
• 1 1
• 2 2 1 42 22 1 13 67

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Redundant Number Systems

• More addition examples
• 110 121 101121 111222
• 102 101 112210 112222
• 212 222 111 20
• 11 20
• 11 20
• 21 11
• 222211 22
• 11220
• 11
• 2
• 1121220

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Redundant Number Systems

• What has redundancy bought us?
• From the examples, it seems carry propagation has
  been somewhat reduced
• Intuitively - redundancy acts as a buffer against
  carry propagation
• However, carry propagation was not eliminated in
  (2,0,2)
• Need to formalize the degree to which a
  particular degree of redundancy eliminates carry
  propagation

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Redundant Number Systems

• To formalize
• xi1 xi where xi yi rti1 wi
• yi1 yi
• ti1 wi
• ti2 wi1
• ti1 wi1 we
  must bound ti1 wi1 for all i to eliminate
  carry propagation

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Redundant Number Systems

• As xi and yi are in ?, ? and xi yi r ti1
  wi1
• 2? ? rti1 wi ? 2?
• suppose there exist ? and ? such that
• ? ? ti1 ? ?
• we need
• ? - ? ? wi1 ? ? - ?
• to avoid propagation. Since this must be true
  for all i, whether the subscript is i or i1 is
  not important.
• From 2 3 we have
• r? ? - ? ? rti1 wi1 ? r? ? - ?
• From 1 4 we have
• 2? ? r? ? - ?

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Redundant Number Systems

• 2? ? r? ? - ?
• gives the following bounds for ? ? to insure no
  propagation
• ? ? ?/(r-1)
• ? ? ?/(r-1)
• For (2,0,2)
• ? ? ?/(r-1) -gt ? ? 2/(2-1)
• ? ? ?/(r-1) -gt ? ? 0/(2-1)
• which, from 3 means all wi must be in the range
  0,0 and there exist inputs requiring carry
  propagation

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Redundant Number Systems

• How then to ensure that no carry propagation is
  needed?
• Change (r, ?, ?)
• Consider (3,0,5)
• ? ? ?/(r-1) -gt ? ? 5/(3-1)
• ? ? ?/(r-1) -gt ? ? 0/(3-1) yields
• ? 3 (first integer after 5/2)
• ? 0
• From 3, this gives a range on wi of 0,2
• Thus, when presented with multiple possibilities
  for representing the sum of two digits, always
  choose terms in 0,2 and there will be no carry
  propagation

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Redundant Number Systems

• Ex for (3,0,5)
• 243
• 435 53 8 could use 15, but choose from 0,2
• 22 yielding 22 and no propagation
• 21
• 20
• 2232

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Redundant Number Systems

• Advantages
• Constant time addition/subtraction!
• Disadvantages
• More space
• Comparison
• no unique forms/representations
• conversion to canonical form as expensive as
  summation
• Uses
• excellent for intermediate results of addition
• Multiplication
• DSP loops that have no comparisons

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Residue Number Systems

• Define a Residue Number System as follows
• For any given integer x, x(x1x2...xk)RNS(P1P2
  ...Pk)
• where xi x mod Pi and ? i,j Pi is relatively
  prime to Pj
• EX 84 (040)RNS(753) 1
  (111)RNS(753)
• 2 (222)RNS(753)
• 3 (330)RNS(753)
• Residue numbers are not positional.
• Residue numbers have unique representations mod
• i.e. for (753) there are unique representations
  of 0-104.

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Residue Number Systems

• Need a conversion system to/from binary
• What benefits of doing operations in RNS?

Binary
RNS ops ,-,
RNS
Binary
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Residue Number Systems

• Addition (subtraction is similar)
• xy ((x1y1)p1 (x2y2)p2 ...
  (xkyk)pk)RNS(P1P2...Pk)
• where xi (x)pi and yi (y)pi
• Multiplication
• xy ((x1y1)p1 (x2y2)p2 ...
  (xkyk)pk)RNS(P1P2...Pk)
• where xi (x)pi and yi (y)pi
• Division ?
• Hard. What does a fraction look like in RNS?

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Residue Number Systems

• Advantages
• Parallel processing of ,,- on smaller numbers
• Adding more primes without increasing range
  allows for use of fields to assist in fault
  tolerance (Ex go from (753) to (7532) where
  last field is used for parity)
• Disadvantages
• No division
• Comparison non-trivial
• Conversion costs
• How to do conversion?

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Residue Number Systems

• Conversion RNS -gt Binary
• Given (x1x2...xk)RNS(P1P2..Pk)
• Binary number x
  mod
• where ?i inv ?
• One may consider, for the purposes of
  computation, all ?i as
• being a pre-defined part of each particular RNS
  system

See Chinese Remainder Theorem
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Residue Number Systems

• Conversion -gt Bin example
• 84 (040)RNS(753)
• For (753) (?1 53)7 1 ? ?1 1 (?2 73)5
  1 ? ?2 1 (?3 75)3 1 ? ?3 2
• x (0?1537/7 4?2735/5
  0?3753/3)105
• x (0 4?221 0)105
• x (4121)105 84105

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Residue Number Systems

• Conversion -gt Bin example
• 1 (111)RNS(753) (?1 53)7 1 ? ?1 1
  (?2 73)5 1 ? ?2 1 (?3 75)3 1 ? ?3 2
• x (1?1537/7 1?2735/5
  1?3753/3)105
• x (1?115 1?221 1?335)105
• x (111511211235)105(106)105 1105

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Residue Number Systems
Conversion -gt Chinese remainder theorem
Binary
RNS ops ,-,
How to choose primes?
RNS

• Minimize
• total length of primes (determining range)
• max length of any prime (determining HW
  cost/delay per unit)
• of primes (determining of parallel units)
• Popular choices 2r-1 (2,3 particularly popular)
• Thm 2a-1 2b-1 are relatively prime iff a b
  are relatively prime

Binary
What about Bin -gt RNS?
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Residue Number Systems

• Binary to RNS conversion
• Uses a lookup table
• Observation for n digit binary number y
  (yn-1,yn-2,...,y0)pi (2n-1)piyn-1
  (2n-2)piyn-2...(20)piy0pi
• Use a table to store (2i)pi
• Example for (753)
• (1011)(x1x2x3)RNS(753)
• x1 (121)7 4
• x2 (321)5 1
• x3 (221)3 2

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Residue Number Systems

• Comparison
• Could convert back and forth to/from binary.
• Another approach convert to a mixed radix
  system, as numbers in a mixed radix system are
  comparable.

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Mixed Radix Number Systems

• We shall describe a Mixed Radix System as
  follows
• x (Zk-1Zk-2...Z0)MRS(Pk-1Pk-2...P1)
• x Zk-1Pk-1Pk-2...P1 Zk-2Pk-2Pk-1...P0 ...
  Z1P1Z0

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Mixed Radix Number Systems

• Conversion from RNS to MRS
• Given x (xk-1xk-2...x0)RNS(Pk-1Pk-2...P0)
• We want x(Zk-1Zk-2...Z0)MRS(Pk-2Pk-3...P0)
• Observation The MRS digit Z0 is in units of 1,
  so
• fewer primes needed in MRS than in RNS

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Mixed Radix Number Systems

• Question what is the relationship between x0 and
  Z0?
• Can it be found by simple inspection?
• Yes.
• x0 Z0.
• Why?
• x0 is the residue left from x mod P0 - all other
  terms are multiples of P0 -gt x0 Z0

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Mixed Radix Number Systems

• This yields
• x-x0 (xk-1xk-2...x1-)RNS(Pk-1Pk-2...P1
  -) (Zk-1Zk-2...Z10)MRS(Pk-2Pk-3..
  .P0)where xi (xi-x0)pi Note
  that this is the only change in MRS

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Mixed Radix Number Systems

• Which leads to
• (x-x0)/P0 (xk-1xk-2...x1-)RNS(Pk-1Pk-2.
  ..P1-) (Zk-1Zk-2...Z1) MRS
  (Pk-2Pk-1...P1)Z1 x1 and so forth.
  (Deduction, division, repeat)
• However, it was earlier noted that division in
  RNS is
• hard - yet here we are doing division.
• The trick? In this case, we know that we will
  always
• get integer results.

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Back to RNS

• Division in RNS
• xi xi (P0-1)pi
• where
• (Pj-1)pi is the multiplicative inverse of Pj
  with respect to Pi
• Ex (3-1)7 5 --gt (3(3-1)7)7 1 (3-1)5 2
  --gt (3(3-1)5)5 1

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Example RNS -gt MRS

• Y(132)RNS(753) (Z2Z1Z0)MRS(53)
• That is, Y Z253Z13Z0
• by 1 (see slide 25)
• Z0 x0 2
• from 2 (see slide 26) we have
• y-x0 y-2 (x2x10)RNS(753)
  (Z2Z10)MRS(53)
• x2 (x2-x0)p2 (1-2)7 6
• x1 (3-2)5 1
• y-2 (610)RNS(753)(Z2Z10)MRS(53)

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Example continued.

• 3 (See slide 27) then gives
• (y-x0)/P0 (y-2)/3 (x2x1-)RNS(753)
  (Z2Z1)MRS(5)
• then
• Q How does one derive x2 ?
• A Its hard. One has to try values one by one
  up to the modulus
• remembering that (3-1)7 5, (3-1)5 2
• x2 (56)7 2, x1 (12)5 2
• (y-2)/3 (22-)RNS(753) (Z2Z1)MRS(5)

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Ex. Cont.

• Apply 1 again,
• Z1x12
• gives by 2
• (y-x0)/P0 - x1 (x20-)RNS(753)
  (Z20)MRS(5)
• x2 (x2-2)7 (2-2)7 0
• (00-)RNS(753)(Z00)MRS(5)
• By 3
• x2
  (x2(5-1)7)7 0 Z2
• Yields final result
• (022)MRS(53)

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Ex. Concluded.

• Check correctness
• (022)MRS(53) 053232 8
• 87 1, 85 3, 83 2 -gt (132)RNS(753)
• Correct!
• Closing remark / Moral of the examples
  -Inversion is key