1 CSE20 Lecture 6 4/19/11 CK Cheng UC San Diego. Residual Numbers (NT-1 and Shaum’s Chapter 11) Introduction Definition Operations Range of numbers Conversion.

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Presentation transcript:

1 CSE20 Lecture 6 4/19/11 CK Cheng UC San Diego

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

3 Number x Residual number (x 1, x 2, …, x k ) +, -, x operations for each x i under m i Chinese Remainder Theorem Mod Operation Moduli (m 1, m 2, …, m k ) Results

Chinese Remainder Theorem Given a residual number (r 1, r 2, …, r k ) with moduli (m 1, m 2, …, m k ), where all m i are mutually prime, set M= m 1 ×m 2 × …×m k, and M i =M/m i. Let S i be the solution that (M i ×S i )%m i = 1 Then we have the corresponding number x = (∑ i=1,k (M i S i r i ))%M. 4

Example Given (m 1,m 2,m 3 )=(2,3,7), M=2×3×7=42, we have M 1 =m 2 ×m 3 =3×7=21(M 1 S 1 )%m 1 =(21S 1 )%2=1 M 2 =m 1 ×m 3 =2×7=14(M 2 S 2 )%m 2 =(14S 2 )%3=1 M 3 =m 1 ×m 2 =2×3=6(M 3 S 3 )%m 3 =(6S 3 )%7=1 Thus, (S 1, S 2, S 3 ) = (1,2,6) For a residual number (0,2,1): x=(M 1 S 1 r 1 + M 2 S 2 r 2 + M 3 S 3 r 3 )%M =(21×1×0 + 14×2×2 + 6×6×1 )%42 = ( )%42 = 92%42 = 8 5

Example For a residual number (1,2,5): x=(M 1 S 1 r 1 + M 2 S 2 r 2 + M 3 S 3 r 3 )%M = (21×1×1 + 14×2×2 + 6×6×5)%42 = ( )%42 = 257%42 = 5 6

Proof of Chinese Remainder Theorem Let A = ∑ i=1,k (M i S i r i ), we show that 1. A%m v = r v and 2. x=A%M is unique. 1. A%m v = (∑ i=1,k (M i S i r i ) )% m v = (Σ(M i S i r i ) % m v )%m v = (M v S v r v )%m v = [(M v S v )%m v × r v %m v ]%m v = r v %m v = r v 2. Proof was shown in lecture 5. 7