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putty-source/sshprime.c

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/*
* Prime generation.
*/
#include <assert.h>
#include "ssh.h"
/*
* This prime generation algorithm is pretty much cribbed from
* OpenSSL. The algorithm is:
*
* - invent a B-bit random number and ensure the top and bottom
* bits are set (so it's definitely B-bit, and it's definitely
* odd)
*
* - see if it's coprime to all primes below 2^16; increment it by
* two until it is (this shouldn't take long in general)
*
* - perform the Miller-Rabin primality test enough times to
* ensure the probability of it being composite is 2^-80 or
* less
*
* - go back to square one if any M-R test fails.
*/
/*
* The Miller-Rabin primality test is an extension to the Fermat
* test. The Fermat test just checks that a^(p-1) == 1 mod p; this
* is vulnerable to Carmichael numbers. Miller-Rabin considers how
* that 1 is derived as well.
*
* Lemma: if a^2 == 1 (mod p), and p is prime, then either a == 1
* or a == -1 (mod p).
*
* Proof: p divides a^2-1, i.e. p divides (a+1)(a-1). Hence,
* since p is prime, either p divides (a+1) or p divides (a-1).
* But this is the same as saying that either a is congruent to
* -1 mod p or a is congruent to +1 mod p. []
*
* Comment: This fails when p is not prime. Consider p=mn, so
* that mn divides (a+1)(a-1). Now we could have m dividing (a+1)
* and n dividing (a-1), without the whole of mn dividing either.
* For example, consider a=10 and p=99. 99 = 9 * 11; 9 divides
* 10-1 and 11 divides 10+1, so a^2 is congruent to 1 mod p
* without a having to be congruent to either 1 or -1.
*
* So the Miller-Rabin test, as well as considering a^(p-1),
* considers a^((p-1)/2), a^((p-1)/4), and so on as far as it can
* go. In other words. we write p-1 as q * 2^k, with k as large as
* possible (i.e. q must be odd), and we consider the powers
*
* a^(q*2^0) a^(q*2^1) ... a^(q*2^(k-1)) a^(q*2^k)
* i.e. a^((n-1)/2^k) a^((n-1)/2^(k-1)) ... a^((n-1)/2) a^(n-1)
*
* If p is to be prime, the last of these must be 1. Therefore, by
* the above lemma, the one before it must be either 1 or -1. And
* _if_ it's 1, then the one before that must be either 1 or -1,
* and so on ... In other words, we expect to see a trailing chain
* of 1s preceded by a -1. (If we're unlucky, our trailing chain of
* 1s will be as long as the list so we'll never get to see what
* lies before it. This doesn't count as a test failure because it
* hasn't _proved_ that p is not prime.)
*
* For example, consider a=2 and p=1729. 1729 is a Carmichael
* number: although it's not prime, it satisfies a^(p-1) == 1 mod p
* for any a coprime to it. So the Fermat test wouldn't have a
* problem with it at all, unless we happened to stumble on an a
* which had a common factor.
*
* So. 1729 - 1 equals 27 * 2^6. So we look at
*
* 2^27 mod 1729 == 645
* 2^108 mod 1729 == 1065
* 2^216 mod 1729 == 1
* 2^432 mod 1729 == 1
* 2^864 mod 1729 == 1
* 2^1728 mod 1729 == 1
*
* We do have a trailing string of 1s, so the Fermat test would
* have been happy. But this trailing string of 1s is preceded by
* 1065; whereas if 1729 were prime, we'd expect to see it preceded
* by -1 (i.e. 1728.). Guards! Seize this impostor.
*
* (If we were unlucky, we might have tried a=16 instead of a=2;
* now 16^27 mod 1729 == 1, so we would have seen a long string of
* 1s and wouldn't have seen the thing _before_ the 1s. So, just
* like the Fermat test, for a given p there may well exist values
* of a which fail to show up its compositeness. So we try several,
* just like the Fermat test. The difference is that Miller-Rabin
* is not _in general_ fooled by Carmichael numbers.)
*
* Put simply, then, the Miller-Rabin test requires us to:
*
* 1. write p-1 as q * 2^k, with q odd
* 2. compute z = (a^q) mod p.
* 3. report success if z == 1 or z == -1.
* 4. square z at most k-1 times, and report success if it becomes
* -1 at any point.
* 5. report failure otherwise.
*
* (We expect z to become -1 after at most k-1 squarings, because
* if it became -1 after k squarings then a^(p-1) would fail to be
* 1. And we don't need to investigate what happens after we see a
* -1, because we _know_ that -1 squared is 1 modulo anything at
* all, so after we've seen a -1 we can be sure of seeing nothing
* but 1s.)
*/
static unsigned short primes[6542]; /* # primes < 65536 */
#define NPRIMES (lenof(primes))
static void init_primes_array(void)
{
if (primes[0])
return; /* already done */
bool A[65536];
for (size_t i = 2; i < lenof(A); i++)
A[i] = true;
for (size_t i = 2; i < lenof(A); i++) {
if (!A[i])
continue;
for (size_t j = 2*i; j < lenof(A); j += i)
A[j] = false;
}
size_t pos = 0;
for (size_t i = 2; i < lenof(A); i++)
if (A[i])
primes[pos++] = i;
assert(pos == NPRIMES);
}
/*
* Generate a prime. We can deal with various extra properties of
* the prime:
*
* - to speed up use in RSA, we can arrange to select a prime with
* the property (prime % modulus) != residue.
*
* - for use in DSA, we can arrange to select a prime which is one
* more than a multiple of a dirty great bignum. In this case
* `bits' gives the size of the factor by which we _multiply_
* that bignum, rather than the size of the whole number.
*
* - for the basically cosmetic purposes of generating keys of the
* length actually specified rather than off by one bit, we permit
* the caller to provide an unsigned integer 'firstbits' which will
* match the top few bits of the returned prime. (That is, there
* will exist some n such that (returnvalue >> n) == firstbits.) If
* 'firstbits' is not needed, specifying it to either 0 or 1 is
* an adequate no-op.
*/
Bignum primegen(int bits, int modulus, int residue, Bignum factor,
int phase, progfn_t pfn, void *pfnparam, unsigned firstbits)
{
int i, k, v, byte, bitsleft, check, checks, fbsize;
unsigned long delta;
unsigned long moduli[NPRIMES + 1];
unsigned long residues[NPRIMES + 1];
unsigned long multipliers[NPRIMES + 1];
Bignum p, pm1, q, wqp, wqp2;
int progress = 0;
init_primes_array();
byte = 0;
bitsleft = 0;
fbsize = 0;
while (firstbits >> fbsize) /* work out how to align this */
fbsize++;
STARTOVER:
pfn(pfnparam, PROGFN_PROGRESS, phase, ++progress);
/*
* Generate a k-bit random number with top and bottom bits set.
* Alternatively, if `factor' is nonzero, generate a k-bit
* random number with the top bit set and the bottom bit clear,
* multiply it by `factor', and add one.
*/
p = bn_power_2(bits - 1);
for (i = 0; i < bits; i++) {
if (i == 0 || i == bits - 1) {
v = (i != 0 || !factor) ? 1 : 0;
} else if (i >= bits - fbsize) {
v = (firstbits >> (i - (bits - fbsize))) & 1;
} else {
if (bitsleft <= 0)
bitsleft = 8, byte = random_byte();
v = byte & 1;
byte >>= 1;
bitsleft--;
}
bignum_set_bit(p, i, v);
}
if (factor) {
Bignum tmp = p;
p = bigmul(tmp, factor);
freebn(tmp);
assert(bignum_bit(p, 0) == 0);
bignum_set_bit(p, 0, 1);
}
/*
* Ensure this random number is coprime to the first few
* primes, by repeatedly adding either 2 or 2*factor to it
* until it is.
*/
for (i = 0; i < NPRIMES; i++) {
moduli[i] = primes[i];
residues[i] = bignum_mod_short(p, primes[i]);
if (factor)
multipliers[i] = bignum_mod_short(factor, primes[i]);
else
multipliers[i] = 1;
}
moduli[NPRIMES] = modulus;
residues[NPRIMES] = (bignum_mod_short(p, (unsigned short) modulus)
+ modulus - residue);
if (factor)
multipliers[NPRIMES] = bignum_mod_short(factor, modulus);
else
multipliers[NPRIMES] = 1;
delta = 0;
while (1) {
for (i = 0; i < (sizeof(moduli) / sizeof(*moduli)); i++)
if (!((residues[i] + delta * multipliers[i]) % moduli[i]))
break;
if (i < (sizeof(moduli) / sizeof(*moduli))) { /* we broke */
delta += 2;
if (delta > 65536) {
freebn(p);
goto STARTOVER;
}
continue;
}
break;
}
q = p;
if (factor) {
Bignum tmp;
tmp = bignum_from_long(delta);
p = bigmuladd(tmp, factor, q);
freebn(tmp);
} else {
p = bignum_add_long(q, delta);
}
freebn(q);
/*
* Now apply the Miller-Rabin primality test a few times. First
* work out how many checks are needed.
*/
checks = 27;
if (bits >= 150)
checks = 18;
if (bits >= 200)
checks = 15;
if (bits >= 250)
checks = 12;
if (bits >= 300)
checks = 9;
if (bits >= 350)
checks = 8;
if (bits >= 400)
checks = 7;
if (bits >= 450)
checks = 6;
if (bits >= 550)
checks = 5;
if (bits >= 650)
checks = 4;
if (bits >= 850)
checks = 3;
if (bits >= 1300)
checks = 2;
/*
* Next, write p-1 as q*2^k.
*/
for (k = 0; bignum_bit(p, k) == !k; k++)
continue; /* find first 1 bit in p-1 */
q = bignum_rshift(p, k);
/* And store p-1 itself, which we'll need. */
pm1 = copybn(p);
decbn(pm1);
/*
* Now, for each check ...
*/
for (check = 0; check < checks; check++) {
Bignum w;
/*
* Invent a random number between 1 and p-1 inclusive.
*/
while (1) {
w = bn_power_2(bits - 1);
for (i = 0; i < bits; i++) {
if (bitsleft <= 0)
bitsleft = 8, byte = random_byte();
v = byte & 1;
byte >>= 1;
bitsleft--;
bignum_set_bit(w, i, v);
}
bn_restore_invariant(w);
if (bignum_cmp(w, p) >= 0 || bignum_cmp(w, Zero) == 0) {
freebn(w);
continue;
}
break;
}
pfn(pfnparam, PROGFN_PROGRESS, phase, ++progress);
/*
* Compute w^q mod p.
*/
wqp = modpow(w, q, p);
freebn(w);
/*
* See if this is 1, or if it is -1, or if it becomes -1
* when squared at most k-1 times.
*/
if (bignum_cmp(wqp, One) == 0 || bignum_cmp(wqp, pm1) == 0) {
freebn(wqp);
continue;
}
for (i = 0; i < k - 1; i++) {
wqp2 = modmul(wqp, wqp, p);
freebn(wqp);
wqp = wqp2;
if (bignum_cmp(wqp, pm1) == 0)
break;
}
if (i < k - 1) {
freebn(wqp);
continue;
}
/*
* It didn't. Therefore, w is a witness for the
* compositeness of p.
*/
freebn(wqp);
freebn(p);
freebn(pm1);
freebn(q);
goto STARTOVER;
}
/*
* We have a prime!
*/
freebn(q);
freebn(pm1);
return p;
}
/*
* Invent a pair of values suitable for use as 'firstbits' in the
* above function, such that their product is at least 2.
*
* This is used for generating both RSA and DSA keys which have
* exactly the specified number of bits rather than one fewer - if you
* generate an a-bit and a b-bit number completely at random and
* multiply them together, you could end up with either an (ab-1)-bit
* number or an (ab)-bit number. The former happens log(2)*2-1 of the
* time (about 39%) and, though actually harmless, every time it
* occurs it has a non-zero probability of sparking a user email along
* the lines of 'Hey, I asked PuTTYgen for a 2048-bit key and I only
* got 2047 bits! Bug!'
*/
void invent_firstbits(unsigned *one, unsigned *two)
{
/*
* Our criterion is that any number in the range [one,one+1)
* multiplied by any number in the range [two,two+1) should have
* the highest bit set. It should be clear that we can trivially
* test this by multiplying the smallest values in each interval,
* i.e. the ones we actually invented.
*/
do {
*one = 0x100 | random_byte();
*two = 0x100 | random_byte();
} while (*one * *two < 0x20000);
}