CSE20 Lecture 5: Number Systems 5. Residual Numbers

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CSE20 Lecture 5: Number Systems 5. Residual Numbers CK Cheng UC San Diego

5. Residual Numbers (NT-1 and Shaum’s Chapter 11) Introduction (parallel processing on huge numbers) Definition (conversion) Operations (+, -, x) Inverse Conversion (recover the solution)

5. Residual Numbers: iClicker The statement about residual number system. A counting system in Sun Tsu’s Arithmetic Manual in the 4th century AD. The system can be used for error correction. The system is efficient for numbers with wide bit width. All of the above Two of the above.

5.1 Introduction Applications: communication, cryptography, and high performance signal processing Goal: Simplify arithmetic operations (+, -, x) when bit width n is huge, e.g. n= 1000. Note no division is involved. Usage in CSE: CSE20, CSE107, CSE127

5.1 Introduction: Flow Results Number x Residual number Residual Number System Mod Operation Moduli (m1, m2, …, mk) Number x Residual number (x1, x2, …, xk) +, -, x operations for each xi under mi Results Chinese Remainder Theorem

5.2 Definition Mod (Modular) operation: Given integers x and d (d> 0), find q and r such that x = q*d+r, 0<= r <d, where q: quotient, d: divisor, and r: remainder. We define x%d= r. Examples: 20=2*7+6 => 20%7=6 11=2*5+1 => 11%5=1 -8=(-2)*5+2 => (-8)%5=2

5.2 Definitions Residual number: Conversion Given (m1, m2,…,mk) where mi are mutually prime and a positive integer x <M=m1m2…mk (0 ≤x <M ) represent x as ( x%m1, x%m2,…, x%mk ) Mutually Prime: Two integers a & b are mutually (or relatively) prime if their greatest common divisor is 1. e.g. 3 & 8, 4 & 9, but not 6 & 9

Examples (X%mi=ri) Given moduli (m1, m2, m3) = (3, 5, 7), convert X: (x1, x2, x3), where xi=X%mi. X=0; x1=x%3=0, x2=x%5=0, x3=x%7= 0. =>(x1,x2,x3)=(0,0,0) X=21: x1=x%3=0, x2=x%5=1, x3=x%7=0. =>(x1,x2,x3)=(0,1,0) X=18: r1=x%3= , r2=x%5= , r3=x%7= . =>(x1,x2,x3)=( , , ) X=-9 : x1=x%3= , x2=x%5= , x3=x%7= =>(x1,x2,x3)=( , , )

5.2 Examples: iClicker Can we compare the value of the residual numbers by comparing the residuals? iClicker (A: Yes, B: No) Given moduli (m1, m2, m3) = (2, 5, 7) and X=17, derive the corresponding residual number (x1, x2, x3), where ri=X%mi. A. (1, 3, 2) B. (1, 2, 3) C. (0, 1, 2) D. None of the above

Examples A residual number system with (m1,m2,m3)=(2,3,7) M = m1 * m2 * m3 = 2 * 3 * 7 = 42 Given X=30, ( X%m1, X%m2, X%m3 ) = ( 30%2, 30%3, 30%7 ) = ( 0, 0, 2 ) Given Y=4, ( Y%m1, Y%m2, Y%m3 ) = ( 4%2, 4%3, 4%7 ) = ( 0, 1, 4 ) Given X+Y=34, ((X+Y)%m1,(X+Y)%m2,(X+Y)%m3 ) = ( 34%2, 34%3, 34%7 ) = ( 0, 1, 6 )

5.3 Modular Operations: Addition Theorem 1: Given three integers x,y,d (d>0), we have (x+y)%d=(x%d+y%d)%d. Proof: Let x = qxd + rx and y = qyd + ry We have (x+y)%d = (qxd + rx + qyd + ry)%d =(rx+ry)%d Therefore, (x+y)%d = (x%d+y%d)%d

5.3 Modular Operations: Multiplication Theorem 2: Given three integers x,y,d (d>0), we have (x*y)%d=(x%d * y%d)%d. Proof: Let x = qxd + rx, y = qyd + ry We have (x*y)%d = (qxd + rx )*(qyd + ry)%d =(qxqyd2+ryqxd+rxqyd+rx*ry)%d = (rx*ry)%d Therefore, (x*y)%d = (x%d * y%d)%d

5.3 Modular Operations What about division? Could we state the following equality? ((x%d)/(y%d))%d= (x/y)%d E.g. try the case that x=6, y=3 & d=3

5.3 Modular Operations What about division? Could we state the following equality? ((x%d)/(y%d))%d= (x/y)%d Answer: Not always! We have the following problems. y%d can be zero. (x%d)/(y%d) or x/y can be fractional.

5.3 Modular Operations: Examples Given x=24 and y=19, derive x+y in the residual number system (m1,m2,m3)=(3,5,8) . I. x+y=24+19=43, r1=43%3=1, r2=43%5=3, r3=43%8=3. =>(r1,r2,r3)=(1,3,3) II. Apply theorem 1. x=> (x1,x2,x3)= y=> (y1,y2,y3)= x+y=> ((x1+y1)%m1,(x2+y2)%m2,(x3+y3)%m3) =

5.3 Modular Operations: Examples Given x=9 and y=12, derive x*y in the residual number system (m1,m2,m3)=(3,5,8) . I. x*y=9*12=108, r1=108%3=0, r2=108%5=3, r3=108%8=4. =>(r1,r2,r3)=(0,3,4) II. Apply theorem 2. x=> (x1,x2,x3)= y=> (y1,y2,y3)= x*y=> ((x1*y1)%m1,(x2*y2)%m2,(x3*y3)%m3) =

5.3 Range of Numbers Theorem 3: In a residual number system with (m1, m2, …, mk), where all mi are mutually prime, for any 0<=y<x<M=m1m2…mk, we have the inequality (x1, x2, …, xk) != (y1, y2, …, yk), i.e. there exists an 0<i<=k such that xi!= yi. Proof: By contradiction. Suppose that (x1, x2, …, xk)=(y1, y2, …, yk), then x-y : (0,0, …,0) (theorem 1). However, for all numbers in the range of the interval, only 0 : (0,0, …,0) because the mods mi are mutually prime.

5.3 Range of Numbers Example: Let k=3 & (m1, m2, m3)=(3,4,5) M=3x4x5=60 Given X=45 and Y=40 (x1,x2,x3)=(45%3, 45%4, 45%5)=(0,1,0) and (y1,y2,y3)=(40%3, 40%4, 40%5)=(1,0,0) 3. Suppose Y=105 then (y1,y2,y3)=(105%3, 105%4, 105%5)=(0,1,0)