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CSE 246: Computer Arithmetic Algorithms and Hardware Design

This lecture delves into Redundant Number Systems, Residue Number Systems, and Mixed Radix Number Systems, outlining their characteristics and advantages for arithmetic operations and hardware design. Understanding the levels of redundancy and formalizing the elimination of carry propagation are key topics covered.

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CSE 246: Computer Arithmetic Algorithms and Hardware Design

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  1. CSE 246: Computer Arithmetic Algorithms and Hardware Design Winter 2004 Lecture 2 Instructor: Prof. Chung-Kuan Cheng

  2. Topics: • Redundant Number Systems • Residue Number Systems • Mixed Radix Number Systems (as they pertain to conversions of Residue Number Systems)

  3. 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  -  = r • redundant if  -  >r • minimally redundant if  -  = r + 1 • maximally redundant if  -  =2r - 1 • over redundant if  -  > 2r - 1

  4. 4 2 1 0 0 0 0 0 1 0 0 2 0 1 0 0 1 1 0 1 2 0 2 0 0 2 1 0 2 2 1 0 0 0 1 2 2 3 4 4 5 6 4 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 = 4*2 + 2*2 +1 = 13 = 6+7

  5. 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

  6. 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

  7. Redundant Number Systems • To formalize: xi+1 xi where xi + yi = r*ti+1 + wi + yi+1 yi ti+1 wi ti+2 wi+1 ti+1 + wi+1 we must bound ti+1 + wi+1 for all i to eliminate carry propagation

  8. Redundant Number Systems As xi and yi are in [, ] and xi + yi = r * ti+1 + wi+1 • 2  r*ti+1 + wi  2 suppose there exist  and  such that •   ti+1   we need •  -   wi+1   -  to avoid propagation. Since this must be true for all i, whether the subscript is i or i+1 is not important. From 2 & 3 we have • r* +  -   r*ti+1 + wi+1  r* +  -  From 1 & 4 we have • 2  r* +  - 

  9. Redundant Number Systems • 2  r* +  -  gives the following bounds for  &  to insure no propagation   /(r-1)   /(r-1) For (2,0,2):   /(r-1) ->   2/(2-1)   /(r-1) ->   0/(2-1) which, from 3 means all wi must be in the range [0,0] and there exist inputs requiring carry propagation

  10. Redundant Number Systems • How then to ensure that no carry propagation is needed? • Change (r, , ) Consider (3,0,5)   /(r-1) ->   5/(3-1)   /(r-1) ->   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

  11. Redundant Number Systems Ex: for (3,0,5) 243 +435 5+3 = 8 could use 15, but choose from [0,2] 22 yielding 22 and no propagation 21 20 2232

  12. 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

  13. Residue Number Systems • Define a Residue Number System as follows For any given integer x, x=(x1|x2|...|xk)RNS(P1|P2|...|Pk) where xi = x mod Pi and  i,j Pi is relatively prime to Pj EX: 84 = (0|4|0)RNS(7|5|3) 1 = (1|1|1)RNS(7|5|3) 2 = (2|2|2)RNS(7|5|3) 3 = (3|3|0)RNS(7|5|3) Residue numbers are not positional. Residue numbers have unique representations mod i.e. for (7|5|3) there are unique representations of 0-104.

  14. Residue Number Systems • Need a conversion system to/from binary • What benefit to doing operations in RNS? Binary # RNS ops +,-,* RNS # Binary #

  15. Residue Number Systems • Addition (subtraction is similar) x+y = ((x1+y1)p1| (x2+y2)p2| ... | (xk+yk)pk)RNS(P1|P2|...|Pk) where xi = (x)pi and yi = (y)pi • Multiplication x*y = ((x1*y1)p1| (x2*y2)p2| ... | (xk*yk)pk)RNS(P1|P2|...|Pk) where xi = (x)pi and yi = (y)pi • Division ? • Hard. What does a fraction look like in RNS?

  16. 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 (7|5|3) to (7|5|3|2) where last field is used for parity) • Disadvantages: • No division • Comparison non-trivial • Conversion costs • How to do conversion?

  17. Residue Number Systems • Conversion: RNS -> Binary # Given (x1|x2|...|xk)RNS(P1|P2|..|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

  18. Residue Number Systems • Conversion -> Bin# example: 84 = (0|4|0)RNS(7|5|3) For (7|5|3): (1* 5*3/7)7 = 1  1 : 1 (2* 7*3/5)5 = 1  2 : 1 (3* 7*5/3)3 = 1  3 : 2 x = (0*1*5*3*7/7 + 4*2*7*3*5/5 + 0*3*7*5*3/3)105 x = (0 + 4*2*21 + 0)105 x = (4*1*21)105 = 84105

  19. Residue Number Systems • Conversion -> Bin # example: 1 = (1|1|1)RNS(7|5|3) (1* 5*3/7)7 = 1  1 : 1 (2* 7*3/5)5 = 1  2 : 1 (3* 7*5/3)3 = 1  3 : 2 x = (1*1*5*3*7/7 + 1*2*7*3*5/5 + 1*3*7*5*3/3)105 x = (1*1*15 + 1*2*21 + 1*3*35)105 x = (1*1*15+1*1*21+1*2*35)105=(106)105 = 1105

  20. Residue Number Systems • 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 Conversion -> Chinese remainder theorem Binary # RNS ops +,-,* How to choose primes? RNS # Binary # What about Bin -> RNS?

  21. 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)pi*yn-1+ (2n-2)pi*yn-2+...+(20)pi*y0]pi Use a table to store (2i)pi Example for (7|5|3) (1011)=(x1|x2|x3)RNS(7|5|3) x1 = (1+2+1)7 = 4 x2 = (3+2+1)5 = 1 x3 = (2+2+1)3 = 2

  22. 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.

  23. Mixed Radix Number Systems • We shall describe a Mixed Radix System as follows: x =(Zk-1|Zk-2|...Z0)MRS(Pk-1|Pk-2|...|P1) x = Zk-1Pk-1Pk-2...P1 + Zk-2Pk-2Pk-1...P0 + ... + Z1P1+Z0

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

  25. 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 -> x0 = Z0

  26. Mixed Radix Number Systems • This yields • x-x0 = (x’k-1|x’k-2|...|x’1|-)RNS(Pk-1|Pk-2|...|P1|-) = (Zk-1|Zk-2|...|Z1|0)MRS(Pk-2|Pk-3|...|P0)where x’i = (xi-x0)pi Note that this is the only change in MRS

  27. Mixed Radix Number Systems • Which leads to • (x-x0)/P0 = (x”k-1|x”k-2|...|x”1|-)RNS(Pk-1|Pk-2|...|P1|-) = (Zk-1|Zk-2|...|Z1) MRS (Pk-2|Pk-1|...|P1)Z1 = x”1and 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.

  28. Back to RNS • Division in RNS x”i =xi * (P0-1)pi where (Pj-1)pi is the multiplicative inverse of Pj with respect to Pi Ex: (3-1)7 = 5 --> (3*(3-1)7)7 = 1 (3-1)5 = 2 --> (3*(3-1)5)5 = 1

  29. Example: RNS -> MRS Y=(1|3|2)RNS(7|5|3) = (Z2|Z1|Z0)MRS(5|3) That is, Y = Z2*5*3+Z1*3+Z0 by 1 (see slide #25) Z0 = x0 = 2 from 2 (see slide #26) we have y-x0 = y-2 =(x’2|x’1|0)RNS(7|5|3) = (Z2|Z1|0)MRS(5|3) x’2 = (x2-x0)p2= (1-2)7 = 6 x’1 = (3-2)5 = 1 y-2 = (6|1|0)RNS(7|5|3)=(Z2|Z1|0)MRS(5|3)

  30. Example continued. 3 (See slide #27) then gives (y-x0)/P0= (y-2)/3 = (x”2|x”1|-)RNS(7|5|3) = (Z2|Z1)MRS(5) then Q: How does one derive x”2 ? A: “It’s hard.” One has to try values one by one up to the modulus remembering that (3-1)7 = 5, (3-1)5 = 2 x”2= (5*6)7 = 2, x”1 =(1*2)5 = 2 (y-2)/3 = (2|2|-)RNS(7|5|3) = (Z2|Z1)MRS(5)

  31. Ex. Cont. Apply 1 again, Z1=x”1=2 gives by 2 (y-x0)/P0 - x”1 = (x”’2|0|-)RNS(7|5|3) = (Z2|0)MRS(5) x”’2 =(x”2-2)7 = (2-2)7 = 0 (0|0|-)RNS(7|5|3)=(Z0|0)MRS(5) By 3 x””2 = (x”’2*(5-1)7)7 = 0 = Z2 Yields final result (0|2|2)MRS(5|3)

  32. Ex. Concluded. Check correctness (0|2|2)MRS(5|3) = 0*5*3+2*3+2 = 8 87 = 1, 85 = 3,83 = 2 -> (1|3|2)RNS(7|5|3) Correct! • Closing remark /Moral of the examples -“Inversion is key”

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