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Residue Number systems

Residue Number systems. P.V. Ananda Mohan FNAE, Fellow IEEE pvam@vsnl.net IEEE CAS Chapter 8 th March 2008 Bangalore. Binary to RNS converter. Binary to RNS converter. Binary to RNS converter. Why RNS. Input Binary Number.

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Residue Number systems

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  1. Residue Number systems P.V. Ananda Mohan FNAE, Fellow IEEE pvam@vsnl.net IEEE CAS Chapter 8th March 2008 Bangalore

  2. Binary to RNS converter Binary to RNS converter Binary to RNS converter Why RNS Input Binary Number • Using several processors in parallel, some operations can be faster. rj r1 r2 Mod mj Processor Mod m1 Processor Mod m2 Processor Instruction Oj O1 O2 RNS to Binary Converter Result

  3. Points to be considered • Choice of moduli set • Computation time and area requirements for the following blocks: • Binary to RNS conversion • RNS to Binary conversion • Multiplication • Scaling • Base extension • Sign detection • Comparison

  4. Binary to RNS conversion • (a) Conventional method: division to get residue throwing away quotient • --Very time consuming. • Example (1000 0001 1010) mod 13? • 2074 mod 13 = 7.

  5. (b) Iterative reduction mod mi • (Capocelli and Giancarlo) • Start with LSBs. Store residues of powers of two in memory go on accumulating till end mod 13: • 1,2,4,8,3,6,12,11,9,5,10,7 • Example (1000 0001 1010) mod 13? • Last three bits you can skip. • 2+23 mod 13 = 2+8 = 10 • 10+24 mod 13 = 10+3=0 and so on • Hardware needed : a modulo adder, Memory containing residues of Powers of 2 mod 13.

  6. (c) Use periodic properties of moduli • For example consider modulus 18. • Residues of powers of two are (1,2,4,8,3,6), (12,11,9,5,10,7),(1,2,4,8..) etc • Note the periodic property • (1,2,4,8,3,6), (-1,-2,-4,-8,-3,-6), (1,2,4,8,3,6), (-1,-2,-4,-8,-3,-6)

  7. Consider mod 89 • Residues of successive powers of two are 1,2,4,8,16,32,64,39,78,67,45, 1,2,4,8,16,32,64,39,78,67,45, • Thus period (or order) is 11 • i.e. 211 mod 89=1

  8. Implementation: Group input bits based on period or half period. • If based on period, add all words with same period mod 211 and have one Binary to RNS converter of Capocelli and Giancarlo. • If based on half-period add all odd fields and add all even fields, Compute odd-even and use Capocelli and Giancarlo method

  9. Example • 2074 mod 13= (100000 011010) mod 13 • = (26-32) mod 13 = -6 mod 13 = 7. • 2074 mod 7 = (100 000 011 010) mod7 • = (4+0+3+2) mod 7=2 • Use for full period case, Adders with end around carry (EAC) and for half period case, two adders with EAC

  10. 100 • 000 • 011 • ----- • 111Sum • 0000 Carry • 010 • ------- • 101 Sum • 0100 Carry • ------ • 1001 • 1 • ------ • 010 • Delay is (2+3+2)DFA 1 0 0 0 0 0 0 0 1 1 0 0 1 0

  11. X Y n bit Adder Two’s complement of mi or (2n-mi) (X+Y) Sign bit (n +1) bit Adder Delay = nDFA+(n+1)DFA+DMUX Area = nAFA+(n+1)AFA+n D2:1MUX 2:1 MUX select (X+Y) mod mi Modulo adders and subtractors • (X+Y) mod mi = (X+Y) or (X+Y-mi) • (X-Y) mod mi = (X-Y) or (X-Y+mi) Cascade of Adders

  12. Faster Adder Implementations X Y Two’s complement of mi or (2n-mi) • Subtractor is same bur two’s compliment of input to be added. Sign bit n bit Adder (n +1) bit Adder (X+Y) 2:1 MUX select Delay = (n+2)DFA+DMUX Area = nAFA+2(n+1)AFA+n D2:1MUX (X+Y) mod mi

  13. Modulo Multipliers X Y Multiplier • Area Multiplier+divider • Delay Multiplier+divider • Divider can be restoring or non-restoring. • Word length of the processor 2n bits mi XY Divider Quotient Throw it. Reminder

  14. Brickell’s Algorithm based Modulo Multipliers • Maximum word length (n+1) bits for taking one bit at a time. • Higher radix feasible. • Area intensive • Other methods exist such as using Redundant Arithmetic, non-overlapping multibit recoding

  15. 13.15 mod 23 • We do not want to do in a straight forward manner . • Write b = 13 in binary form: • b3b2b1b0 =1101 • Do repeatedly starting from MSB: • Old= (2.Old + bi.A) mod 23

  16. EXAMPLE • b3b2b1b0 =1101; A =15, mi = 23 • P= (2.0 + 1.15) mod 23 = 15 • P=(2.15 + 1.15) mod 23 = 22 • P=(2.22 + 0.15) mod 23 = 21 • P=(2.21+ 1.15) mod 23 = 11 • Maximum value of P <3(23) i.e. 3mi • Modulo subtraction is by two comparisons: • Is P>N? or Is P>2n? • Answer is either P, P-mi, P-2mi; choosebased on sign of P-mi, P-2mi. • Example 45 mod 23, anwers are 45,45-23=22,45-46=-1; since P-2mi is negative and P-mi is positive, P-mi is the correct result. • Multiple precision arithmetic to be used in PC based implementations

  17. Architecture for Modmul Old A bi TC of 2mi LSB of Zero TC of mi 2Old Adder Adder (n+2) bit adder 3:1 Mux Latch Latch

  18. ModMUL • Computation time= n[(n+2)DFA+DMux] • Area = 3(n+2)AFA+A3:1MUX+nAAND

  19. Modmul for IDEA • IDEA (International Data Encryption Algorithm) uses (xy) mod (216+1) as a programmable S-Box (Substitution Box), where x and y are 16 bit words. • Ideal for DSPs • Get P=xy a 32 bit word. • Subtract MSB 16 bit word from LSB 16 bit word. If negative, add (216+1)

  20. RNS to Binary Conversion • CRT based • MRC based • CRT: RNS {m1,m2,m3} Residues {x1,x2,x3} • Define Mi=M/mi and M=m1m2m3 • Decoded Binary number X • = [M1{(1/M1) mod m1}x1+ {M2 (1/M2) mod m2}x2+ M3{(1/M3) mod m3}x3]mod M • e.g. {3,5,7} M=105, M1=35,M2=21,M3=15 • (1/35) mod 3 = 2, (1/21) mod 5=1, (1/15) mod 7=1. • X= [70x1+21x2+15x3] mod 105 • Consider (1,2,3), X = (70+42+45) mod 105 = 157 mod 105 = 52 • Generally, Mi are large, Mi{(1/Mi) mod mi} are stored,involves multiplication of these large numbers by xi in parallel and adding.

  21. Multiplier Multiplier Multiplier CRT Implementation [M2(1/M2) mod m2] [M3(1/M3) mod m3] X1 [M1(1/M1) mod m1] • Modulo M adder may involve n subtractions for a n moduli system • Delay = DMult+ DMODADD X2 X3 Mod M adder X

  22. MRC • Example RNS {7,8,9} • 7 8 9 • 2 3 • -3 -3 • 7 • x4 x1 • 7 • -7 • 6 • x1 • 6 • X = 6.72+7.9+3 = 498 m1 m2 m3 r1 r2 r3 - r3- r3 (r1-r3) mod m1 = p (r2-r3) mod m2 =q XAXB UA UB -UB (UA-UB) mod m1 =r XC UC • Note XA= (1/m3) mod m1 and • XB= (1/m3) mod m2,XC= (1/m2) mod m1 • UC, UB and r3 are known as MRC digits. • X = UCm2m3+UBm3+r3 is always less than M.

  23. MRC versus CRT • MRC is sequential but avoids reduction modulo a large number needed in CRT . • MRC needs storage of multiplicative inverses, Modulo subtraction and modulo multiplication, final addition of n numbers for a n moduli RNS, • Multiplicative inverses can be powers of two small numbers such as 6 or 9 for powers of two related moduli sets. • Moduli set with all MIs of value unity also suggested e.g {3,7,22}, Only modulo subtractions will do for evaluating MRC digits; But multipliers are cumbersome. • Generally need ROMs.

  24. ′ Modulo 17 adder Architecture for XY mod 17 x3 x2 x1 x0 y3 y2 y1 y0 y0x3 yox2 y0x1 yox0 y1x3 y1x2 y1x1 y1x0 (y1x3)′ added 1 y2x3 y2x2 y2x1 y2x0 (y2x3)′ (y2x2)′ added 3 y3x3 y3x2 y3x1 y3x0 (y3x3)′ (y3x2)′ (y3x1)′ added 7 Write MSBs bi as (1- bi′) • 1011 • 1101 • 1011 • 00001 • 101101 • 1011010 • Adding 4 words in a CSA • 1011 • 0001 • 1101 • 0111 • 10010 Added 1 • 1010 • 1111 • 00101 Added 1 • 0100 add 4 (correction • 0111 term in a modulo • 17 adder)

  25. Scaling • Division by a number • E.g. RNS given {3,5,7}. Divide 99 (0,4,1) by 11 (2,1,4). • If division is exact, multiply 99 by multiplicative inverse of 11. • (1/11) = (2,1,2) =86 (Note (1/11) mod 3 = 2 etc. • (99/11) = (0,4,1)x(2,1,2)= (0,1,4) =9

  26. Scaling by arbitrary number when division is not exact • Example 1 : 100/13 in RNS {3,5,7} • 100 = (1,0,2} • Direct method by multiplying with (1/13) will not work. • 100 = 1,0,2 • (1/13) = 1,2,6 • 100/13 = 1,0,5 = 40 wrong. • First you need to find residue of 100mod 13 = 9. • Subtract from 100 to get (100-9)=91 • 100 = 1,0,2 • 9 = 0,4,2 • 91 = 1,1,0 • (1/13) = 1,2,6 • 91/13 = 1,2,0 = 7.

  27. Scaling by one modulus • Divide 100/7 • 100 = 1,0,2 • Subtract residue 100mod 7 first =2 • 100 = 1, 0, 2 • 2 = 2, 2, 2 • 98 = 2, 3, 0 • x(1/7) = x1 x3 • = 2 4 • Now you need to do base extension to get RNS number again (2,4,0) • Scaling by another modulus aso feasible in the same way. • Note that MRC does this.

  28. Scaled Residue /Montgomery’s Modular Multiplication • Example: To evaluate (5.6) mod 13 = 4. • Prescaling by 16: 5 = (5.16) mod 13 = 2, (6.16) mod 13 = 5 • Montgomery step = [(5.16)(6.16)/16] mod 13 = (2.5/16) mod 13 = (10/3) mod 13 = (10.9) mod 13 = 12. • Result is obtained by post scaling: (12/16) mod 13 = (12/3) mod 13 = 4. • Prescaling is Binary to RNS conversion: Successive multiplication by 2 and modulo reduction , (5.2) mod 13= 10, (10.2) mod 13 = (7.2) mod 13= 1, (1.2) mod 13 = 2. • Post scaling is another Montgomery step.

  29. Montgomery step avoids modulo reduction. Only conditional addition. If LSB is 1 add modulus, ignore LSB. Example (2.5/16) mod 13. Four steps are needed. Each step a partial product is added and result scaled by two. 2 = 0010 (binary) Computation of (0010)x5/16: Formula: (old value+ bix5)/2 Old value =0. (0+0.5)/2= 0 (0+1x5)/2 = (5+13)/2 = 9 since LSB of current result in brackets is 1. (9+0.5)/2 = (9+13)/2 = 11 (11+0.5)/2 = (11+13)/2 = 12. Addition of two numbers using a (n+1)-bit CPA, n AND gates, n Flip-flops

  30. Higher Radix Montgomery’s Technique • Higher Radix possible. • 16 or 8 or 4 bits at a time can be considered. • Example considering 4 bits at a time: • Consider [(10001100)/16] mod 23 • Find (-1/23) mod 16=(-1/7) mod16 = 9 ((-1/mi) mod 2k) • Find 10001100 mod 16 = four LSBs= 12 (X mod 2k) • Find (12x9) mod 16 = 12 α= [(-X/mi) mod 2k] • Find 10001100+12(23) = 11010 0000 (X+ αmi) • Ignore last 4 bits to get 26. (X+ αmi)/2k • Need a multiplier mod 16 to get the multiple to be added. • Then addition of shifted versions of modulus (in this case of radix 16, four shifted versions) using a CASA tree followed by CPA.

  31. Popular Powers-of-two related moduli set • (2n-1, 2n, 2n+1) • Dynamic range <3n bits. • Example 16 bit DSP needs n = 6; RNS {63,64,65} • RNS to binary conversion using CRT can be done very fast. • .

  32. The beauty is these are powers of two related facilitating easy implementation. The various multiplicative inverses used above are as follows:

  33. Example {7,8,9} [(32+4)x1-8x2+(36-1)x3] mod 63 yields 6 MSBs Subtract x2 from both sides Divide by 2n to get 2n MSBs of the result as

  34. Realization • Andraros and Ahmad : Four 2n-bit words to be added using two levels of Adders of rotated bits. • Piestrak suggested using CSA two level with CPA using end around carry for adding four 2n-bit words • Delay - (4n+2) DFA, Area = (6n) AFA • Suggested Low delay version (2n+2) DFA+DMUX also, 2n A2:1MUXes needed. • Dhurkadas (NPOL, Cochin) suggested simplification to three 2n-bit inputs to be added • Delay – (4n+2) DFA, Area = (4n) AFA • Bhardwaj, Premkumar, Srikanthan [1998] suggested using n-bit adders e.g Carry select adders n-bit • Wang et al [2002] 2n-bit as well as n-bit adders three converters.

  35. {7,8,9} example (x1,x2,x3) x1, x2 3 bit, x3 4 bit x12x11x10, x22x21x20, x33x32x31x30 • [(32+4)x1-8x2+(36-1)x3] mod 63 : Dhurkadas Simplified as x10x12 x11 x10 x12 x11 x22′ x21′ x20′ y x31′ x30 ′ X3x x32 x31 x30 x32 x31 Y= (x33+x32)′ x10x12 x11 x10 x12 x11 x22′ x21′ x20′ 1 1 1 X3x x32 x31 x3x x32 x31 1 1 x33 ′ x32 ′ x31′ x30 ′ X3x= x30+x33 since either x30 or x33 exist

  36. Other three, Four and Five moduli sets • {2n,2n-1,2n-1-1} Hiasat and Abdel-Aty-Zohdy, Wang, Wang, Swamy and Ahmad: not better than popular moduli set, multipliers etc are simpler • {2n,2n-1,2n+1-1} Ananda Mohan better in area or time, multipliers are simpler • {2n,22n-1,22n+1} Ananda Mohan better than Cao et al four moduli set, one large modulus • {2n,2n-1,2n+1, 2n+1-1 } Vinod and Premkumar • {2n,2n-1,2n+1, 2n+1-1 } Bhardwaj, Srikanthan, Ananda Mohan and Premkumar Area and Time intensive • {2n,2n-1,2n+1, 22n+1} Cao et al better than other four moduli sets but one modulus bigger in size. • {2n-3,2n-1,2n+1,2n+3} Sheu et al uses ROM not attractive • {2n-1-1, 2n-1,2n,2n+1,2n+1-1} Cao et al 2007 Increases cardinality to 5, DR of 5n bits but RNS to Binary conversion is slower/area consuming

  37. M2 {2k,2k-1,2k-1-1}, M1{2k-1,2k,2k+1}, M4{2k,2k-1,2k+1-1}, M3{2k-1,2k,2k+1,2k+1-1} Comparison of various converters for three moduli sets

  38. Base Extension • Needed in scaling or division. • Uses MRC fist to divide followed by base extension. • CRT can be used but is cumbersome. • Example: {3,5,7} 52= (1,2,3) Scale by 7 • 3 5 7 • 1 2 3 • -3 -3 • 4 • x1 x3 • 2 2 First Base Extension step • -2 • 2 • X2 • +(1x5)mod 7 Base Extension step • 0

  39. RSA using RNS/ECC • Needs computation of PQ mod N • e.g 1023 mod 37 = (1016)(104)(102)(101) mod 37 • Successive squaring mod 37 and Multiplications mod 37 of selected results. • Needs (XY) mod N ass basic step where X,Y,N are 1024 bit numbers. • RNS can be used. • Montgomery technique has been used to find (X′Y′/M) mod N where M is the product of Moduli in RNS. • Needs two RNS dynamic ranges M and M′ which are mutually Prime and a redundant modulus • Determine q such that (X′Y′+qN) is a multiple of M. • Extend q to RNS with Dynamic range M′. • Find r = (X′Y′+qN)/M in second RNS • Do base extension to First RNS

  40. Sign Detection and Comparison • Is difficult • Needed to go to Binary number to detect sign • Comparison is also difficult Needed to go to Binary numbers or sequential techniques such as comparing Mixed Radix Digits.

  41. Applications • FIR Filters (ensure that RNS dynamic range is larger than that of the filter) • Digital Frequency Synthesis • Video Filters • 2-D filters • NTTs (Number Theoretic Transforms) • Cryptography

  42. Applications of RNS • [5] Freking, W.L., and Parhi, K.K., "Low-power FIR digital filters using residue arithmetic, " in Conf. Record 31st Asil. Conf. Signals, Syst. and Comput. (ACSSC 1997), vol. 1, Pacific Grove, CA USA [1997], 739-43. • [6] D'Amora, A. et al., "Reducing power dissipation in complex digital filters by using the quadratic residue number system, " in Conf. Record 34th Asil. Conf. Signals, Syst. Comput. (ACSSC 2000), vol. 2, Pacific Grove, CA USA [2000], 879-83. • [7] Cardarilli, G.C. et al., "Low-power implementation of polyphase filters in Quadratic Residue Number system," in Proc. IEEE Int. Symp. Circuits Syst. (ISCAS 2004), vol. 2, Vancouver, BC, Canada [2004], 725-728. • [8] Shanbag, N.R., and Siferd, R.E., A single-chip pipelined 2-D FIR filter using residue Arithmetic, IEEE JSSC -26[1991], 796-805. • [9] Tuukka Toivonen., and Janne Heikkilä., Video Filtering With Fermat Number Theoretic Transforms Using Residue Number System, IEEE CSVT-16[2006], 128-135. • [10] Schwemmlein, J., and Posch, K.C., Reinhard Posch. RNS-modulo reduction upon a restricted base value set and its applicability to RSA cryptography, Computer & Security [1998], 17, 637-650. • [11]Hanae Nozaki., Masahiko Motoyama., Atsushi Shimbo., and Shinichi Kawamura., Implementation of RSA algorithm based on RNS Montgomery multiplication, In C. Paar (ed). Cryptographic Hardware and Embedded Systems – CHES, Springer-Verlag, Berlin, Germany [2001], 364-376.

  43. [12] Jean-Claude Bajard., Laurent Stephane Didier., Peter Kornerup., An RNS Montgomery modular multiplication Algorithm, IEEE C-47 [1998], 766-776. [13] Jean-Claude Bajard., and Laurent Imbert., A Full RNS Implementation of RSA, IEEE C-53[2004],769-774. [14] Schinianakis, D.M., Kakarountas. A.P., and Stouraitis. T., A New Approach to Elliptic Curve Cryptography: an RNS Architecture, IEEE MELECON, May 16-19, Benalmádena (Málaga), Spain [2006], 1241-1245. [15] Lie-Liang Yang.,and Lajos Hanzo., A Residue Number System Based Parallel Communication Scheme Using Orthogonal Signaling: Part I—System Outline, IEEE VT-51[2002],1534-1546. [16] Chaves, R., and Sousa, L., “RDSP: A RISC DSP based on residue number system,” in Proc. Euro. Symp. Digital System Design: Architectures, Methods, and Tools, Antalya, Turkey [2003], 128-135. [17] Wei, W. et al., "RNS application for digital image processing," in 4th IEEE Int. Workshop Syst.-on-Chip for Real Time Applications, Banff, Alta., Canada [2004],77-80.

  44. Conclusion • Very mature today • Can be used in place of Custom DSP blocks • Research on newer moduli sets with high cardinality and Faster Reverse Conversion is of interest

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