Addition How fast can you add AB Addition
Addition How fast can you add A+B
Addition How fast can you add A+B 101011100100111
Addition How fast can you add A+B 101011100100111 0
Addition How fast can you add A+B 101011100100111 00
Addition How fast can you add A+B 101011100100111 100
Addition How fast can you add A+B 101011100100111 111100 n-bit numbers time = O(n)
Multiplication How fast can you multiply A*B 1010111001 *1 0 1 1
Multiplication How fast can you multiply A*B 1010111001 *1 0 1 1 1010111001
Multiplication How fast can you multiply A*B 1010111001 *1 0 1 1 1010111001 n-bit numbers time = O(n 2)
Karatsuba-Offman a=2 n/2 a 1 + a 0 b=2 n/2 b 1 + b 0 n/2 ab=(2 a 1+a 0 n/2 )(2 b 1+b 0) 2 n a 1 b 1 + 2 n/2 (a 1 b 0 + a 0 b 1) + a 0 b 0 =
Karatsuba-Offman a=2 n/2 a 1 + a 0 b=2 n/2 b 1 + b 0 Multiply(a, b, n) if n=1 return a*b else R 1 Multiply(a 1, b 1, n/2) R 2 Multiply(a 0, b 1, n/2) R 3 Multiply(a 1, b 0, n/2) R 4 Multiply(a 0, b 0, n/2) return 2 n R 1+ 2 n/2 (R 2+R 3) + R 4
Karatsuba-Offman Multiply(a, b, n) if n=1 return a*b else R 1 Multiply(a 1, b 1, n/2) R 2 Multiply(a 0, b 1, n/2) R 3 Multiply(a 1, b 0, n/2) R 4 Multiply(a 0, b 0, n/2) return 2 n R 1+ 2 n/2 (R 2+R 3) + R 4 Recurrence?
Karatsuba-Offman Multiply(a, b, n) if n=1 return a*b else R 1 Multiply(a 1, b 1, n/2) R 2 Multiply(a 0, b 1, n/2) R 3 Multiply(a 1, b 0, n/2) R 4 Multiply(a 0, b 0, n/2) return 2 n R 1+ 2 n/2 (R 2+R 3) + R 4 Recurrence? T(n) = 4 T(n/2) + O(n)
Karatsuba-Offman T(n) = 4 T(n/2) + O(n) T(n)=O(n 2)
Karatsuba-Offman ab=(2 n/2 a 1+a 0)(2 n/2 b 1+b 0) = 2 n a 1 b 1 + 2 n/2 (a 1 b 0 + a 0 b 1) + a 0 b 0 Can compute in less than 4 multiplications?
Karatsuba-Offman ab=(2 n/2 a 1+a 0)(2 n/2 b 1+b 0) = 2 n a 1 b 1 + 2 n/2 (a 1 b 0 + a 0 b 1) + a 0 b 0 Can compute using 3 multiplications: (a 0+a 1)(b 0+b 1) = a 0 b 0 + (a 1 b 0 + a 0 b 1) + a 1 b 1
Karatsuba-Offman Multiply(a, b, n) if n=1 return a*b else R 1 Multiply(a 1, b 1, n/2) R 2 Multiply(a 0, b 0, n/2) R 3 Multiply(a 1+a 0, b 1+b 0, n/2+1) R 4 R 3 – R 2 – R 1 return 2 n R 1+ 2 n/2 R 3 + R 2 Recurrence?
Karatsuba-Offman Multiply(a, b, n) if n=1 return a*b else R 1 Multiply(a 1, b 1, n/2) R 2 Multiply(a 0, b 0, n/2) R 3 Multiply(a 1+a 0, b 1+b 0, n/2+1) R 4 R 3 – R 2 – R 1 return 2 n R 1+ 2 n/2 R 3 + R 2 Recurrence? T(n) = 3 T(n/2) + O(n)
Karatsuba-Offman T(n) = 3 T(n/2) + O(n) T(n)=O(n. C) C=log 2 3 1. 58
Integer Division r=a mod b a, b q, r a = q*b + r 0 r<b Can be done in O(n 2) time.
d divides a DEFINITION: d divides a (denoted d | a) if there exists b such that b*d = a 3|6 3|0 0|3 0|0
d divides a DEFINITION: d divides a (denoted d | a) if there exists b such that b*d = a 3|6 3|0 0|3 0|0 yes, b=2 yes, b=0 no yes, b=?
d divides a 3|6 yes, b=2 3|0 yes, b=0 0|3 no 0|0 yes, b=? d|a a|c d|c Proof: a = b*d, c=b’*a c=(b*b’)*d
Divisibility poset 0 8 4 6 10 9 2 3 5 1 7
GCD (a, b) “largest” d such that d|a, d|
GCD (a, b) “largest” d such that d|a, d|b ( c; c|a, c|b) : c|d GCD(3, 6) GCD(0, 8) GCD(0, 0)
GCD (a, b) “largest” d such that d|a, d|b ( c; c|a, c|b) : c|d GCD(3, 6) = 3 GCD(0, 8) = 8 GCD(0, 0) = 0
GCD How quickly can we compute GCD (a, b) ?
GCD How quickly can we compute GCD (a, b) ? Euclid GCD(a, b) = GCD(b, a mod b)
GCD wlog a>b GCD(a, b) if b=0 then return a else return GCD(b, a mod b) Running time?
GCD wlog a>b GCD(a, b) if b=0 then return a else return GCD(b, a mod b) Running time? (a, b) (b, a mod b) (a mod b, ? ) (a mod b) < a/2
GCD (a, b) (b, a mod b) (a mod b, ? ) (a mod b) < a/2 2(log 2 a)=O(n) 2 O(n ) iterations each mod time O(n 3) time total
Modular exponentiation (a, b, m) ab mod m
Modular exponentiation (a, b, m) ab mod m a 2 mod m a 4 mod m 8 a mod m 16 a mod m. . . b = 10101 b a mod m
Modular exponentiation (a, b, m) ab mod m mod-ex(a, b, m) if b=0 then RETURN 1 else if b mod 2 = 0 then RETURN mod-ex(a, b/2, m)2 mod m else RETURN a*mod-ex(a, (b-1)/2, m)2 mod m
Algorithms so far a, b, m n-bit integers addition a+b O(n) time 1. 58 multiplication a*b O(n ) time division a/b, a mod b O(n 2) time gcd(a, b) O(n 3) time ab mod m O(n 3) time
GROUP (G, ) is a group if : G G G (a b) c = a (b c) exists G ( a G) a = a -1 a a-1=
Modular arithmetic G = {0, . . . , m-1} = Zm modulo m a b = a+b mod m (G, ) is a group if : G G G (a b) c = a (b c) exists G ( a G) a = a a a-1 a a-1=
Modular arithmetic G = {0, . . . , m-1} = Zm modulo m a b = a+b mod m (G, ) is a group if IS A GROUP : G G G (a b) c = a (b c) exists G ( a G) a = a a a-1 a a-1=
Modular arithmetic G = {0, . . . , m-1} = Zm modulo m a b = a*b mod m (G, ) is a group if : G G G (a b) c = a (b c) exists G ( a G) a = a a a-1 a a-1=
Modular arithmetic G = {0, . . . , m-1} = Zm modulo m a b = a*b mod m b; ab=1 [mod m] GCD(a, m)=1 (G, ) is a group if : G G G (a b) c = a (b c) exists G ( a G) a = a a a-1 a a-1=
Modular arithmetic modulo m G = Z*m ={a | GCD(a, m)=1 } a b = a*b mod m (G, ) is a group if IS A GROUP : G G G (a b) c = a (b c) exists G ( a G) a = a a a-1 a a-1=
![Fermat’s little Theorem p a prime ap-1 = 1 [mod p] {ak | k](http://slidetodoc.com/presentation_image_h/1b4db7e59c7fa0d69f8a0143c48f2a9d/image-43.jpg "Fermat’s little Theorem p a prime ap-1 = 1 [mod p] {ak | k")
Fermat’s little Theorem p a prime ap-1 = 1 [mod p] {ak | k Z} is a subgroup of Z*p
![Fermat’s little Theorem a (m)=1 [mod m] (m) = | Z*m | m=p 1](http://slidetodoc.com/presentation_image_h/1b4db7e59c7fa0d69f8a0143c48f2a9d/image-44.jpg "Fermat’s little Theorem a (m)=1 [mod m] (m) = | Z*m | m=p 1")
Fermat’s little Theorem a (m)=1 [mod m] (m) = | Z*m | m=p 1 a 1 p 2 a 2. . . pkak (m) = (1 -1/p 1). . . (1 -1/pk) m
Fermat’s little Theorem m=p 1 a 1 p 2 a 2. . . pkak (m) = (1 -1/p 1). . . (1 -1/pk) m E. g. if m=pq (m)= p, q primes
Fermat’s little Theorem m=p 1 a 1 p 2 a 2. . . pkak (m) = (1 -1/p 1). . . (1 -1/pk) m E. g. if m=pq p, q primes (m)=(p-1)(q-1)
![Fermat’s little Theorem a(p-1)(q-1) =1 [mod pq] E. g. if m=pq p, q primes](http://slidetodoc.com/presentation_image_h/1b4db7e59c7fa0d69f8a0143c48f2a9d/image-47.jpg "Fermat’s little Theorem a(p-1)(q-1) =1 [mod pq] E. g. if m=pq p, q primes")
Fermat’s little Theorem a(p-1)(q-1) =1 [mod pq] E. g. if m=pq p, q primes (m)=(p-1)(q-1)
RSA 1) choose primes p, q 2) let n pq 3) choose e 4) compute d=e-1 [mod (p-1)(q-1)] 5) announce n, e
RSA 1) choose primes p=13, q=17 2) let n pq 3) choose e 4) compute d=e-1 [mod (p-1)(q-1)] 5) announce n, e
RSA 1) choose primes p=13, q=17 2) let n pq=221 3) choose e 4) compute d=e-1 [mod (p-1)(q-1)] 5) announce n, e
RSA 1) choose primes p=13, q=17 2) let n pq=221 3) choose e=5 4) compute d=e-1 [mod (p-1)(q-1)] 5) announce n, e
RSA 1) choose primes p=13, q=17 2) let n pq=221 3) choose e=5 4) compute 77=d=e-1 [mod (p-1)(q-1)] 5) announce n, e
RSA 1) choose primes p=13, q=17 d 2) let n pq=221 3) choose e=5 4) compute 77=d=e-1 [mod (p-1)(q-1)] 5) announce n, e n=221 e=5 = 77
RSA d = 77 n=221 e=5 ENCODE: x xe mod n DECODE: x xd mod n
RSA m=42 d = 77 n=221 e=5 ENCODE: x xe mod n DECODE: x xd mod n
RSA m=42 425 (mod 221) = 9 d = 77 9 n=221 e=5 ENCODE: x xe mod n DECODE: x xd mod n
RSA m=42 425 (mod 221) = 9 d = 77 9 977 (mod 221) = 42 m=42 n=221 e=5 ENCODE: x xe mod n DECODE: x xd mod n
Primality testing
- Slides: 58
