applied cryptography - protocols, algorithms, and source code in c

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Unformatted text preview: ome primes can appear more than once. For example, p1 might be equal to p2 .) In other words, a number (less than the product of some primes) is uniquely identified by its residues mod those primes. For example, use 3 and 5 as primes, and 14 as the number. 14 mod 3 = 2, and 14 mod 5 = 4. There is only one number less than 3*5 = 15 which has those residues: 14. The two residues uniquely determine the number. So, for an arbitrary a < p and b < q (where p and q are prime), there exists a unique x, where x is less than pq, such that x a a (mod p), and x a b (mod q) To find this x, first use Euclid’s algorithm to find u, such that u*q a 1 (mod p) Then compute: x = (((a - b)*u) mod p)*q + b Here is the Chinese remainder theorem in C: /* r is the number of elements in arrays m and u; m is the array of (pairwise relatively prime) moduli u is the array of coefficients return value is n such than n == u[k]%m[k] (k=0..r-1) and n < m[0]*m[1]*...*m[r-1] */ /* totient() is left as an exercise to the reader. */ int chinese_remainder (size_t r, int *m, int *u) { size_t i; int modulus; int n; modulus = 1; for (i=0; i<r; ++i) modulus *= m[i]; n = 0; for (i=0; i<r; ++i) { n += u[i] * modexp(modulus / m[i], totient(m[i]), m[i]); n %= modulus; } return n; } The converse of the Chinese remainder theorem can also be used to find the solution to the problem: if p and q are primes, and p is less than q, then there exists a unique x less than pq, such that a a x (mod p), and b a x (mod q) If a e b mod p, then x = (((a - (b mod p))*u) mod p)*q + b If a < b mod p, then x = (((a + p - (b mod p))*u) mod p)*q + b Quadratic Residues If p is prime, and a is greater than 0 and less than p, then a is a quadratic residue mod p if x2 a a (mod p), for some x Not all values of a satisfy this property. For a to be a quadratic residue modulo n, it must be a quadratic residue modulo all the prime factors of n. For example, if p = 7, the quadratic residues are 1, 2, and 4: 12 = 1 a 1 (mod 7) 22 = 4 a 4 (m...
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