1. Choosing RNS Moduli Assume we wish to represent 100,000 10 Values Standard Binary  lg 2 (100,000) 10  =  16.6096 10  =17 bits RNS(13|11|7|5|3|2), Dynamic Range  M=30,030 10 –Insufficient Dynamic Range –Maximum Digit Width = 4 bits, Total = 17 bits RNS(17|13|11|7|5|3|2), Dynamic Range  M=510,510 10 – Dynamic Range 5.1 Times Too Large – Maximum Digit Width = 5 bits, Total = 22 bits Adding More Prime Moduli is Inefficient

2. Choosing RNS Moduli Remove m i =5 From RNS(17|13|11|7|5|3|2) RNS(17|13|11|7|3|2), Dynamic Range  M=102,102 10 Still Have Relatively Prime Moduli – Maximum Digit Width = 5 bits, Total = 19 bits – 1 5-bit, 2 4-bit, 1 3-bit, 1 2-bit and 1 1-bit Modulo Units Required Maximum Delay 5-bit Carry-Propagate Can Combine (3,7) and (2,13) Moduli With no Speed Penalty RNS(26|21|17|11), Dynamic Range  M=102,102 10 – Maximum Digit Width = 5 bits, Total = 19 bits – 3 5-bit and 1 4-bit Modulo Units Required

3. Relatively Prime Values Powers of Smaller Primes are Relatively Prime Example gcd(3 2,2 2 )=1 But gcd(3 2,3)=3 – Can REPLACE a Modulus With its’ Power – Try Use Sequence of SMALLEST Valued Moduli RNS(2 2 |3), Dynamic Range  M=12 10 RNS(3 2 |2 3 |7|5), Dynamic Range  M=2,520 10 RNS(11|3 2 |2 3 |7|5), Dynamic Range  M=27,720 10 RNS(13|11|3 2 |2 3 |7|5), Dynamic Range  M=360,360 10 – Maximum Digit Width = 4 bits, Total = 21 bits – Dynamic Range 3.6 times that Needed

4. Relatively Prime Values RNS(13|11|3 2 |2 3 |7|5), Dynamic Range  M=360,360 10 – Maximum Digit Width = 4 bits, Total = 21 bits – Dynamic Range 3.6 times that Needed Reduce the Above by Factor of 3 Replace 3 2 with 3 and Combine 3 and 5 to Get 15 RNS(15|13|11 |2 3 |7), Dynamic Range  M=120,120 10 – Maximum Digit Width = 4 bits, Total = 18 bits – Dynamic Range 1.2 times that Needed Using This Strategy Can Generally Find the “Best” Moduli in Terms of Speed and Representation Efficiency

5. Moduli Choice for Simple Arithmetic Unit Design Simple Units Also Lead to Speed and Cost Benefits Modulo-ADD,SUBTRACT, MULTIPLY Units Simple to Design if m i =2 ai OR 2 ai -1 Power of 2 Moduli Lead to Simple Design – Standard a-bit Binary Adder – Example: Use 16 Instead of 13 – Exception in Case of Lookup Table Implementation Power of 2 a -1 Moduli Lead to Simple Design – Standard a-bit Binary Adder with End-around Carry – Referred to as “Low-cost” Moduli

6. RNS Low-Cost Moduli Theorem: A sufficient condition for 2 a -1 and 2 b -1 to be a relatively prime pair is that a and b are relatively prime. Any List of Relatively Prime Numbers: a k-2 > a k-1 >...>a 1 >a 0 Can be Used as a BASIS of k-modulus RNS: RNS(2 ak-2 |2 ak-2 -1|...|2 a1 -1|2 a0 -1) Widest Residues (Longest Carry-chain) is a k-2 -bit Values

7. Low-Cost Moduli Example Consider the Example From Earlier X=[0,100,000] Choosing the Moduli From Smallest to Largest: RNS(2 3 | 2 3 -1| 2 2 -1)Basis:3, 2M=168 10 RNS(2 4 | 2 4 -1| 2 3 -1)Basis:4, 3M=1680 10 RNS(2 5 | 2 5 -1 | 2 3 -1| 2 2 -1)Basis:5, 3, 2M=20,832 10 RNS(2 5 | 2 5 -1 | 2 4 -1| 2 3 -1)Basis:5, 4, 3M=104,160 10 Can’t Include 2 and 4 in Same Basis Set, gcd(2,4)=2

8. Low-Cost Moduli Example RNS(2 5 | 2 5 -1 | 2 4 -1| 2 3 -1)Basis:5, 4, 3M=104,160 10 = RNS(32 | 31 | 15| 7) Requires 5+5+4+3=17 bits Requires 2 5-bit, 1 4-bit and 1 3-bit Module 4 RNS Digits Efficiency = (100,001/104,160)=0.96004  100% Comparing With Unrestricted Moduli: RNS(2 5 | 2 5 -1 | 2 4 -1| 2 3 -1)17 bits M=104,160 10 5-bit Carry-ripple but Simpler Circuit, Fewer Digits RNS( 15|13|11 |2 3 |7 )18 bits M=120,120 10 4-bit Carry-ripple, 1 Extra Digit

9. Encoding and Decoding Advantages of Alternative Number Systems Must Not be Outweighed By Conversions to/from the System Encoding From Fixed Positional System to RNS Easily Accomplished Using a Table-Lookup and Modulo Addition Circuits

10. Encoding with Lookup Table Conversion of Signed-Magnitude or 2’s Complement Accomplished by Converting Magnitude and Taking RNS Complement Consider the Following Identity: Idea is to Compute a Table of All Terms and Store in a Table for all i, j Then Add

11. Example Lookup Table Use Default RNS=(8|7|5|3) For m i =8 We Can Use 3 LSbs of Value

12. Example Encoding

13. RNS to Mixed Radix Form CRT States That a Mixed-Radix Number System (MRS) is Associated with any RNS MRS is k-digit Weighted Positional Number System (m k-1 |m k-2 |...|m 2 |m 1 |m 0 ) MRS Weights are Products: (m k-2...m 2 m 1 m 0,...,m 2 m 1 m 0, m 1 m 0, m 0,1) MRS Digit Sets in Each of k Positions: [0,m k-1 -1],...,[0,m 2 -1],[0,m 1 -1],[m 0 -1] MRS Digits in Same Range as RNS Digits

14. RNS to MRS Example Example Position Weights MRS (8|7|5|3) (7)(5)(3)=105, (5)(3)=15, 3, 1 (0|3|1|0) MRS(8|7|5|3) =(0)(105)+(3)(15)+(1)(3)+(0)(1)=48 10 RNS to MRS Conversion Requires Finding the z i that Correspond to the y i in:

15. RNS to MRS Conversion From MRS Definition we Have: Easy to See that z 0 = y 0, Subtracting This Value From RNS and MRS Values Results in:

16. RNS to MRS Conversion (cont) Thus, if We Can Divide by m 0, We Have an Iterative Approach for Conversion Dividing y' (a Multiple of m 0 ) by m 0 is SCALING Easier Than Normal RNS Division Accomplished by Multiplying by Muliplicative Inverse of m 0 Next, Divide Both Representations by m 0 :

17. Multiplicative Inverses Multiplicative Inverse is a Value When Multiplied by Given Quantity Yields a Product of 1 Example Multiplicative Inverses of 3 Relative to m i =8, 7, 5: Thus, Multiplicative Inverses are 3, 5 and 2 Can Build a Lookup Table Circuit to Store Inverses

18. CRT LUT

19. Multiplicative Inverses Example Divide the Number Y' = (0|6|3|0) RNS by 3 Accomplish Through Multiplication by (3|5|2|-) RNS

20. RNS/MRS Conversion Example Convert Y=(0|6|3|0) RNS to MRS z 0 =y 0 =0 Divide by 3 Now, We Have z 1 =1, Subtract by 1 and Divide by 5

21. RNS/MRS Conversion Consider Conversion of (3|2|4|2) RNS from RNS(8|7|5|3) to Decimal Need to Determine Values of (1|0|0|0) RNS, (0|1|0|0) RNS, (0|0|1|0) RNS and (0|0|0|1) RNS

22. RNS/MRS Conversion From Definition of RNS, Positions with 0 are Multiples of RNS(8|7|5|3) and Position with 1 are m i =1

23. Chinese Remainder Theorem How Did We Find w 3 = (1|0|0|0) RNS = 105? Since Digits in 7, 5, 3 Places are 0, w 3 Must be a Multiple of (7)(5)(3)=105 Must Pick the Multiple of 105 Such That its Residue With Respect to 8 is 1 Accomplished by Multiplying 105 by its’ Multiplicative Inverse with Respect to 8 This Process is Formalized in Chinese Remainder Theorem

24. Chinese Remainder Theorem THEOREM: Chinese Remainder Theorem (CRT) The magnitude of an RNS number can be obtained from the CRT formula: where, by definition, M i = M/m i and  I = mi is the multiplicative inverse of M i with respect to m i.

25. Chinese Remainder Theorem Can Avoid Multiplications in Conversion Process by Storing mi > M in a Table Example Table Given on page 64 of Textbook