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CSE20 Lecture 5 4/12/11

CSE20 Lecture 5 4/12/11. CK Cheng UC San Diego. Residual Numbers (NT-1 and Shaum’s Chapter 11). Introduction Definition Operations Range of numbers. Introduction. Applications: communication, cryptography, and high performance signal processing

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CSE20 Lecture 5 4/12/11

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  1. CSE20 Lecture 54/12/11 CK Cheng UC San Diego

  2. Residual Numbers(NT-1 and Shaum’s Chapter 11) • Introduction • Definition • Operations • Range of numbers

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

  4. Definition Mod (Modular) operation. • Given integer 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. Conversion to residual system: Given moduli (m1, m2, …, mk), where all mi are mutually prime, transform integer x to (r1, r2, …, rk), where ri=x%mi

  5. Definitions • 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 • Residual number: Given (m1, m2,…,mk) where mis are mutually prime and a positive integer x <M=m1xm2x…xmk (0 ≤x <M ) represent x as ( x%m1, x%m2,…, x%mk )

  6. Examples (x%mi=ri) Given (m1, m2, m3) = (3, 5, 7), convert x: (r1, r2, r3). • 0: (0, 0, 0); 0%3=0, 0%5=0, 0%7= 0 • 2: (2, 2, 2); 2%3=2, 2%5=2, 2%7=2 • 21: (0, 1, 0); 21%3=0 , 21%5=1, 21%7=0 • -2: (1, 3, 5); -2%3=1, -2%5=3, -2%7=5 • -3: (0, 2, 4); -3%3=0, -3%5=2, -3%7=4 • -21: (0, 4, 0); -21%3=0, -21%5= 4, -21%7=0 Hint: 0<= ri < mi

  7. Examples • k = 3 ( m1, m2, m3) = ( 2, 3, 7 ) • M = m1 x m2 x m3= 2 x 3 x 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 )

  8. 3. Modular Operations Theorem: Given three integers x,y,d, 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 =(rx+ry)%d Therefore, (x+y)%d = (x%d+y%d)%d

  9. 3. Modular Operations Theorem: Given three integers x,y,d, 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

  10. 3. Modular Operations • What about division? • Could we state the following equality? ((x%d)/(y%d))%d= (x/y)%d Answer: No! We have the following problems. • y%d can be zero. • (x%d)/(y%d) or x/y can be fractional.

  11. Range of Numbers Theorem: Given (m1, m2, …, mk), where all mi are mutually prime, let M=m1xm2x…xmk. For 0<= x< M, the residual number (x1, x2, …, xk) is distinct. Proof: By contradiction, let 0<= y<x < M. Suppose (x1, x2, …, xk)=(y1, y2, …, yk) then x-y : (0,0, …,0). However, for all numbers in the range of the interval, only 0 : (0,0, …,0) because the mods mi are mutually prime.

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