[PDF] - A Mec hanical Pro of of the Chinese Remainder Theorem Da PDF Document

SLIDE 1 A Mec hanical Pro

f
f

the Chinese Remainder Theorem Da vid M. Russino Adv anced Micro Devices, Inc. david.russinoff@a md. co m http://www.onr.c

m

/us er /ru ss /da vid

SLIDE 2 Informal Statemen t Theorem L et m 1 ; : : : ; m k 2 N b e p airwise r elatively prime mo duli and let a 1 ; : : : ; a k 2 N . Ther e exists x 2 N such that x

a

1 (mo d m 1 ) x

a

2 (mo d m 2 ) . . . x

a

k (mo d m k ): If x satises the same c

ngruenc

es, then x

x

(mo d m 1 m 2

m

k ): 1

SLIDE 3 A CL2 F

rmalization

(defun g-c-d (x y) (declare (xargs :measure (nfix (+ x y)))) (if (zp x) y (if (zp y) x (if (<= x y) (g-c-d x (- y x)) (g-c-d (- x y) y))))) (defun rel-prime (x y) (= (g-c-d x y) 1)) (defun congruent (x y m) (= (rem x m) (rem y m))) (defun congruent-all (x a m) (if (endp m) t (and (congruent x (car a) (car m)) (congruent-al l x (cdr a) (cdr m))))) (defthm chinese-remain de r-t he

r

em (implies (and (natp-all a) (rel-prime-modu li m) (= (len a) (len m))) (and (natp (crt-witness a m)) (congruent-all (crt-witness a m) a m)))) 2

SLIDE 4 Informal Pro

f

Lemma 1 If x; y 2 N ar e r elatively prime, then ther e exists s 2 Z such that sy

1

(mo d x). Lemma 2 If x; y ; z 2 N and x is r elatively prime to b

th

y and z , then x is r elatively prime to y z . Pro

f
f

CR T: Let M = m 1 m 2

m

k . F

r

i = 1; : : : ; k , let M i = M =m i and nd s i suc h that s i M i

1

(mo d m i ). Let x = a 1 s 1 M 1 + a 2 s 2 M 2 +

+

a k s k M k : Then x

a

i s i M i

a

i (mo d m i ): 3

SLIDE 5

SLIDE 6 Example Supp

se

w e ha v e 10000

N
50000

and N

6

(mo d 25) N

13

(mo d 36) N

28

(mo d 49) Then w e ma y solv e for N as follo ws: M = 25

36
49

= 44100 M 1 = 36

49

= 1764 M 2 = 25

49

= 1225 M 3 = 25

36

= 900 1764s 1

1

(mo d 25) , 14s 1

1

(mo d 25) , s 1

9

(mo d 25) 1225s 2

1

(mo d 36) , s 2

1

(mo d 36) 900s 3

1

(mo d 49) , 18s 3

1

(mo d 49) , s 3

30

(mo d 49) a 1 = 6, a 2 = 13, a 3 = 28 x = a 1 M 1 s 1 + a 2 M 2 s 2 + a 3 M 3 s 3 = 6

1764
9

+ 13

1225
1

+ 28

900
30

= 867281

29281

(mo d 44100) N = 29281 5

SLIDE 7 Pro

f
f

Lemma 1 Lemma 1 If x; y 2 N ar e r elatively prime, then ther e exists s 2 Z such that sy

1

(mo d x). This is a sp ecial case

f

the follo wing: F

r

al l x; y 2 N , ther e exist r ; s 2 Z such that r x + sy = g cd(x; y ). The pro

f

is b y induction

n

x + y : (1) If x = 0, then r = and s = 1. (2) If y = 0, then r = 1 and s = 0. (3) If < x

y

, then nd r and s suc h that r x + s (y

x)

= g cd(x; y

x)

= g cd(x; y ) and let r = r

s

and s = s . Then r x + sy = (r

s

)x + s y = r x + s (y

x)

= g cd(x; y ): (4) If < y < x, then nd r and s suc h that r (x

y

) + s y = g cd(x

y

; y ) = g cd(x; y ) and let r = r and s = s

r

. 6

SLIDE 8 F

rmal

Pro

f

(mutual-recurs io n (defun r (x y) (declare (xargs :measure (nfix (+ x y)))) (if (zp x) (if (zp y) 1 (if (<= x y) (- (r x (- y x)) (s x (- y x))) (r (- x y) y))))) (defun s (x y) (declare (xargs :measure (nfix (+ x y)))) (if (zp x) 1 (if (zp y) (if (<= x y) (s x (- y x)) (- (s (- x y) y) (r (- x y) y)))))) ) (defthm r-s-lemma (implies (and (natp x) (natp y)) (= (+ (* (r x y) x) (* (s x y) y)) (g-c-d x y)))) 7

SLIDE 9 Pro

f
f

Lemma 2 Lemma 2 If x; y ; z 2 N and x is r elatively prime to b

th

y and z , then x is r elatively prime to y z . This is a consequence

f

the follo wing basic prop erties

f

g cd and primes: (1) g cd(x; y ) divides b

th

x and y . (2) If d divides b

th

x and y , then d divides g cd(x; y ). (3) If x > 1, then some prime divides x. (4) If a prime p divides ab, then p divides either a

r

b. It w

uld

tak e some w

rk

to pro v e these in A CL2. F

rtunately

, there is a more direct route to CR T. 8

SLIDE 10 Alternate Approac h Lemma 3 L et x; y 1 ; y 2 ; : : : ; y k 2 N and p = y 1

y

k . If x is r elatively prime to e ach y i , then ther e exist c; d 2 Z such that cx + dp = 1. Pro

f:

Let p = y 1

y

k 1 . Assume that r x + sy k = 1 and, b y induction, that c x + d p = 1: Then (sd )p = (sy k )(d p ) = (1

r

x)(1

c

x) = 1

(r

+ c

r

c x)x: Th us, if c = r + c

r

c x and d = sd , then cx + dp = 1: 9

SLIDE 11 F

rmal

Pro

f

(defun c (x l) (if (endp l) (- (+ (r x (car l)) (c x (cdr l))) (* (r x (car l)) (c x (cdr l)) x)))) (defun d (x l) (if (endp l) 1 (* (s x (car l)) (d x (cdr l))))) (defthm c-d-lemma (implies (and (natp x) (natp-all l) (rel-prime-all x l)) (= (+ (* (c x l) x) (* (d x l) (prod l))) 1))) 10

SLIDE 12 Denition

f

crt-witness (defun

ne-mod

(x l) (* (d x l) (prod l) (d x l) (prod l))) (defthm rem-one-mod-1 (implies (and (natp x) (> x 1) (natp-all l) (rel-prime-all x l)) (= (rem (one-mod x l) x) 1))) (defthm rem-one-mod-0 (implies (and (natp x) (> x 1) (rel-prime-modu li l) (rel-prime-all x l) (member y l)) (= (rem (one-mod x l) y) 0))) (defun crt1 (a m l) (if (endp a) (+ (* (car a) (one-mod (car m) (remove (car m) l))) (crt1 (cdr a) (cdr m) l)))) (defun crt-witness (a m) (crt1 a m m)) 11

SLIDE 13 The Main Lemma W e pro v e the follo wing generalization

f

CR T: (defthm crt1-lemma (implies (and (natp-all a) (rel-prime-modu li l) (sublistp m l) (= (len a) (len m))) (congruent-all (crt1 a m l) a m))) The pro

f

is b y induction, as suggested b y the denition: (defun crt1 (a m l) (if (endp a) (+ (* (car a) (one-mod (car m) (remove (car m) l))) (crt1 (cdr a) (cdr m) l)))) In the inductiv e case, the conclusion

f

the lemma expands as follo ws: (and (congruent (+ (* (car a) (one-mod (car m) (remove (car m) l))) (crt1 (cdr a) (cdr m) l)) (car a) (car m)) (congruent-al l (+ (* (car a) (one-mod (car m) (remove (car m) l))) (crt1 (cdr a) (cdr m) l)) (cdr a) (cdr m))). 12

SLIDE 14 The Final Result CR T is deriv ed as an instance

f

crt1-lemma: (defthm crt1-lemma (implies (and (natp-all a) (rel-prime-modu li l) (sublistp m l) (= (len a) (len m))) (congruent-all (crt1 a m l) a m))) (defthm chinese-remain de r-t he

r

em (implies (and (natp-all a) (rel-prime-modu li m) (= (len a) (len m))) (and (natp (crt-witness a m)) (congruent-all (crt a m) a m)))) 13