Skip to content
GitLab
Projects
Groups
Snippets
/
Help
Help
Support
Community forum
Keyboard shortcuts
?
Submit feedback
Contribute to GitLab
Sign in / Register
Toggle navigation
Menu
Open sidebar
Marcus Hoffmann
nettle
Commits
6f5fc6a3
Commit
6f5fc6a3
authored
May 20, 2010
by
Niels Möller
Browse files
Added comment describing Pcklington's theorem.
Rev: nettle/bignum-random-prime.c:1.3
parent
a5b0a3c0
Changes
1
Hide whitespace changes
Inline
Side-by-side
bignum-random-prime.c
View file @
6f5fc6a3
...
...
@@ -180,6 +180,30 @@ miller_rabin_pocklington(mpz_t n, mpz_t nm1, mpz_t nm1dq, mpz_t a)
6.42 Handbook of applied cryptography), but with ratio = 1/2 (like
the variant in fips186-3). FIXME: Force primes to start with two
one bits? */
/* The algorithm is based on the following special case of
Pocklington's theorem:
Assume that n = 1 + r q, where q is a prime, q > sqrt(n) - 1. If we
can find an a such that
a^{n-1} = 1 (mod n)
gcd(a^r - 1, n) = 1
then n is prime.
Proof: Assume that n is composite, with smallest prime factor p <=
sqrt(n). Since q is prime, and q > sqrt(n) - 1 >= p - 1, q and p-1
are coprime, so that we can define u = q^{-1} (mod (p-1)). The
assumption a^{n-1} = 1 (mod n) implies that also a^{n-1} = 1 (mod
p). Since p is prime, we have a^{(p-1)} = 1 (mod p). Now, r =
(n-1)/q = (n-1) u (mod (p-1)), and it follows that a^r = a^{(n-1)
u} = 1 (mod p). Then p is a common factor of a^r - 1 and n. This
contradicts gcd(a^r - 1, n) = 1, and concludes the proof.
If n is specified as k bits, we need q of size ceil(k/2) + 1 bits
(or more) to make the theorem apply.
*/
void
nettle_random_prime
(
mpz_t
p
,
unsigned
bits
,
void
*
ctx
,
nettle_random_func
random
)
...
...
@@ -241,8 +265,9 @@ nettle_random_prime(mpz_t p, unsigned bits,
mpz_init
(
a
);
mpz_init
(
i
);
/* Bit size ceil(k/2) + 1, slightly larger than used in Alg.
4.62. */
/* Bit size ceil(k/2) + 1, slightly larger than used in Alg. 4.62
in Handbook of Applied Cryptography (which seems to be
incorrect for odd k). */
nettle_random_prime
(
q
,
(
bits
+
3
)
/
2
,
ctx
,
random
);
/* i = floor (2^{bits-2} / q) */
...
...
Write
Preview
Supports
Markdown
0%
Try again
or
attach a new file
.
Cancel
You are about to add
0
people
to the discussion. Proceed with caution.
Finish editing this message first!
Cancel
Please
register
or
sign in
to comment