Пример #1
0
static int
bnSubMult(BigNum *bn, unsigned x, unsigned y)
{
	unsigned long z = (unsigned long)x * y;

	while (z > 65535) {
		if (bnSubQ(bn, 65535) < 0)
			return -1;
		z -= 65535;
	}
	return bnSubQ(bn, (unsigned)z);
}
Пример #2
0
/* dest = bn + sm, where 0 <= sm < 2^16 */
	PGPError
PGPBigNumSubtractQ(
	PGPBigNumRef	bn,
	PGPUInt16		sm,
	PGPBigNumRef	dest,
	PGPBoolean *	underflowPtr )
{
	PGPError	err	= kPGPError_NoErr;
	int			bnError	= 0;
	PGPBoolean	underflow	= FALSE;
	
	if ( IsntNull( underflowPtr ) )
		*underflowPtr	= FALSE;
		
	pgpValidateBigNum( bn );
	pgpValidateBigNum( dest );

	if ( bn != dest )
	{
		bnError	= bnCopy( &dest->bn, &bn->bn );
		if ( IsBNError( bnError ) )
			err	= kPGPError_OutOfMemory;
	}
	
	if ( IsntBNError( bnError ) )
	{
		bnError	= bnSubQ( &dest->bn, sm );	
		if ( IsBNError( bnError ) )
		{
			underflow	= TRUE;
		}
	}
	
	if ( IsntNull( underflowPtr ) )
		*underflowPtr	= underflow;
	
	return( err );
}
Пример #3
0
/*
 * This performs a modular exponentiation using the Chinese Remainder
 * Algorithm when the modulus is known to have two relatively prime
 * factors n = p * q, and u = p^-1 (mod q) has been precomputed.
 *
 * The chinese remainder algorithm lets a computation mod n be performed
 * mod p and mod q, and the results combined.  Since it takes
 * (considerably) more than twice as long to perform modular exponentiation
 * mod n as it does to perform it mod p and mod q, time is saved.
 *
 * If x is the desired result, let xp and xq be the values of x mod p
 * and mod q, respectively.  Obviously, x = xp + p * k for some k.
 * Taking this mod q, xq == xp + p*k (mod q), so p*k == xq-xp (mod q)
 * and k == p^-1 * (xq-xp) (mod q), so k = u * (xq-xp mod q) mod q.
 * After that, x = xp + p * k.
 *
 * Another savings comes from reducing the exponent d modulo phi(p)
 * and phi(q).  Here, we assume that p and q are prime, so phi(p) = p-1
 * and phi(q) = q-1.
 */
static int
bnExpModCRA(BigNum *x, BigNum const *d,
	BigNum const *p, BigNum const *q, BigNum const *u)
{
	BigNum xp, xq, k;
	int i;
	PGPMemoryMgrRef	mgr	= x->mgr;

bndPrintf("Performing Chinese Remainder Algorithm\n");
bndPut("x = ", x);
bndPut("p = ", p);
bndPut("q = ", q);
bndPut("d = ", d);
bndPut("u = ", u);

	bnBegin(&xp, mgr, TRUE);
	bnBegin(&xq, mgr, TRUE);
	bnBegin(&k, mgr, TRUE);

	/* Compute xp = (x mod p) ^ (d mod p-1) mod p */
	if (bnCopy(&xp, p) < 0)		/* First, use xp to hold p-1 */
		goto fail;
	(void)bnSubQ(&xp, 1);		/* p > 1, so subtracting is safe. */
	if (bnMod(&k, d, &xp) < 0)	/* k = d mod (p-1) */
		goto fail;
bndPut("d mod p-1 = ", &k);
	if (bnMod(&xp, x, p) < 0)	/* Now xp = (x mod p) */
		goto fail;
bndPut("x mod p = ", &xp);
	if (bnExpMod(&xp, &xp, &k, p) < 0)	/* xp = (x mod p)^k mod p */
		goto fail;
bndPut("xp = x^d mod p = ", &xp);

	/* Compute xq = (x mod q) ^ (d mod q-1) mod q */
	if (bnCopy(&xq, q) < 0)		/* First, use xq to hold q-1 */
		goto fail;
	(void)bnSubQ(&xq, 1);		/* q > 1, so subtracting is safe. */
	if (bnMod(&k, d, &xq) < 0)	/* k = d mod (q-1) */
		goto fail;
bndPut("d mod q-1 = ", &k);
	if (bnMod(&xq, x, q) < 0)	/* Now xq = (x mod q) */
		goto fail;
bndPut("x mod q = ", &xq);
	if (bnExpMod(&xq, &xq, &k, q) < 0)	/* xq = (x mod q)^k mod q */
		goto fail;
bndPut("xq = x^d mod q = ", &xq);

	/* xp < p and PGP has p < q, so this is a no-op, but just in case */
	if (bnMod(&k, &xp, q) < 0)
		goto fail;	
bndPut("xp mod q = ", &k);
	
	i = bnSub(&xq, &k);
bndPut("xq - xp = ", &xq);
bndPrintf("With sign %d\n", i);
	if (i < 0)
		goto fail;
	if (i) {
		/*
		 * Borrow out - xq-xp is negative, so bnSub returned
		 * xp-xq instead, the negative of the true answer.
		 * Add q back (which is subtracting from the negative)
		 * so the sign flips again.
		 */
		i = bnSub(&xq, q);
		if (i < 0)
			goto fail;
bndPut("xq - xp mod q = ", &xq);
bndPrintf("With sign %d\n", i);		/* Must be 1 */
	}

	/* Compute k = xq * u mod q */
	if (bnMul(&k, u, &xq) < 0)
		goto fail;
bndPut("(xq-xp) * u = ", &k);
	if (bnMod(&k, &k, q) < 0)
		goto fail;
bndPut("k = (xq-xp)*u % q = ", &k);

	/* Now x = k * p + xp is the final answer */
	if (bnMul(x, &k, p) < 0)
		goto fail;
bndPut("k * p = ", x);
	if (bnAdd(x, &xp) < 0)
		goto fail;
bndPut("k*p + xp = ", x);

#if BNDEBUG	/* @@@ DEBUG - do it the slow way for comparison */
	if (bnMul(&xq, p, q) < 0)
		goto fail;
bndPut("n = p*q = ", &xq);
	if (bnExpMod(&xp, x, d, &xq) < 0)
		goto fail;
bndPut("x^d mod n = ", &xp);
	if (bnCmp(x, &xp) != 0) {
bndPrintf("Nasty!!!\n");
		goto fail;
	}
	bnSetQ(&k, 17);
	bnExpMod(&xp, &xp, &k, &xq);
bndPut("x^17 mod n = ", &xp);
#endif

	bnEnd(&xp);
	bnEnd(&xq);
	bnEnd(&k);
	return 0;

fail:
	bnEnd(&xp);
	bnEnd(&xq);
	bnEnd(&k);
	return kPGPError_OutOfMemory;
}
Пример #4
0
/*
 * Modifies the bignum to return a nearby (slightly larger) number which
 * is a probable prime.  Returns >=0 on success or -1 on failure (out of
 * memory).  The return value is the number of unsuccessful modular
 * exponentiations performed.  This never gives up searching.
 *
 * All other arguments are optional.  They may be NULL.  They are:
 *
 * unsigned (*randfunc)(unsigned limit)
 * For better distributed numbers, supply a non-null pointer to a
 * function which returns a random x, 0 <= x < limit.  (It may make it
 * simpler to know that 0 < limit <= SHUFFLE, so you need at most a byte.)
 * The program generates a large window of sieve data and then does
 * pseudoprimality tests on the data.  If a randfunc function is supplied,
 * the candidates which survive sieving are shuffled with a window of
 * size SHUFFLE before testing to increase the uniformity of the prime
 * selection.  This isn't perfect, but it reduces the correlation between
 * the size of the prime-free gap before a prime and the probability
 * that that prime will be found by a sequential search.
 *
 * If randfunc is NULL, sequential search is used.  If you want sequential
 * search, note that the search begins with the given number; if you're
 * trying to generate consecutive primes, you must increment the previous
 * one by two before calling this again.
 *
 * int (*f)(void *arg, int c), void *arg
 * The function f argument, if non-NULL, is called with progress indicator
 * characters for printing.  A dot (.) is written every time a primality test
 * is failed, a star (*) every time one is passed, and a slash (/) in the
 * (very rare) case that the sieve was emptied without finding a prime
 * and is being refilled.  f is also passed the void *arg argument for
 * private context storage.  If f returns < 0, the test aborts and returns
 * that value immediately.  (bn is set to the last value tested, so you
 * can increment bn and continue.)
 *
 * The "exponent" argument, and following unsigned numbers, are exponents
 * for which an inverse is desired, modulo p.  For a d to exist such that
 * (x^e)^d == x (mod p), then d*e == 1 (mod p-1), so gcd(e,p-1) must be 1.
 * The prime returned is constrained to not be congruent to 1 modulo
 * any of the zero-terminated list of 16-bit numbers.  Note that this list
 * should contain all the small prime factors of e.  (You'll have to test
 * for large prime factors of e elsewhere, but the chances of needing to
 * generate another prime are low.)
 *
 * The list is terminated by a 0, and may be empty.
 */
int
bnPrimeGen(BigNum *bn, unsigned (*randfunc)(unsigned),
         int (*f)(void *arg, int c), void *arg, unsigned exponent, ...)
{
	int retval;
	int modexps = 0;
	unsigned short offsets[SHUFFLE];
	unsigned i, j;
	unsigned p, q, prev;
	BigNum a, e;
#ifdef MSDOS
	unsigned char *sieve;
#else
	unsigned char sieve[SIEVE];
#endif

#ifdef MSDOS
	sieve = bniMemAlloc(SIEVE);
	if (!sieve)
		return -1;
#endif
	PGPBoolean		isSecure	= TRUE;
	PGPMemoryMgrRef	mgr	= bn->mgr;

	bnBegin(&a, mgr, isSecure);
	bnBegin(&e, mgr, isSecure);

#if 0	/* Self-test (not used for production) */
{
	BigNum t;
	static unsigned char const prime1[] = {5};
	static unsigned char const prime2[] = {7};
	static unsigned char const prime3[] = {11};
	static unsigned char const prime4[] = {1, 1}; /* 257 */
	static unsigned char const prime5[] = {0xFF, 0xF1}; /* 65521 */
	static unsigned char const prime6[] = {1, 0, 1}; /* 65537 */
	static unsigned char const prime7[] = {1, 0, 3}; /* 65539 */
	/* A small prime: 1234567891 */
	static unsigned char const prime8[] = {0x49, 0x96, 0x02, 0xD3};
	/* A slightly larger prime: 12345678901234567891 */
	static unsigned char const prime9[] = {
		0xAB, 0x54, 0xA9, 0x8C, 0xEB, 0x1F, 0x0A, 0xD3 };
	/*
	 * No, 123456789012345678901234567891 isn't prime; it's just a
	 * lucky, easy-to-remember conicidence.  (You have to go to
	 * ...4567907 for a prime.)
	 */
	static struct {
		unsigned char const *prime;
		unsigned size;
	} const primelist[] = {
		{ prime1, sizeof(prime1) },
		{ prime2, sizeof(prime2) },
		{ prime3, sizeof(prime3) },
		{ prime4, sizeof(prime4) },
		{ prime5, sizeof(prime5) },
		{ prime6, sizeof(prime6) },
		{ prime7, sizeof(prime7) },
		{ prime8, sizeof(prime8) },
		{ prime9, sizeof(prime9) } };

	bnBegin(&t);

	for (i = 0; i < sizeof(primelist)/sizeof(primelist[0]); i++) {
			bnInsertBytes(&t, primelist[i].prime, 0,
				      primelist[i].size);
			bnCopy(&e, &t);
			(void)bnSubQ(&e, 1);
			bnTwoExpMod(&a, &e, &t);
			p = bnBits(&a);
			if (p != 1) {
				printf(
			"Bug: Fermat(2) %u-bit output (1 expected)\n", p);
				fputs("Prime = 0x", stdout);
				for (j = 0; j < primelist[i].size; j++)
					printf("%02X", primelist[i].prime[j]);
				putchar('\n');
			}
			bnSetQ(&a, 3);
			bnExpMod(&a, &a, &e, &t);
			p = bnBits(&a);
			if (p != 1) {
				printf(
			"Bug: Fermat(3) %u-bit output (1 expected)\n", p);
				fputs("Prime = 0x", stdout);
				for (j = 0; j < primelist[i].size; j++)
					printf("%02X", primelist[i].prime[j]);
				putchar('\n');
			}
		}

	bnEnd(&t);
}
#endif

	/* First, make sure that bn is odd. */
	if ((bnLSWord(bn) & 1) == 0)
		(void)bnAddQ(bn, 1);

retry:
	/* Then build a sieve starting at bn. */
	sieveBuild(sieve, SIEVE, bn, 2, 0);

	/* Do the extra exponent sieving */
	if (exponent) {
		va_list ap;
		unsigned t = exponent;

		va_start(ap, exponent);

		do {
			/* The exponent had better be odd! */
			pgpAssert(t & 1);

			i = bnModQ(bn, t);
			/* Find 1-i */
			if (i == 0)
				i = 1;
			else if (--i)
				i = t - i;

			/* Divide by 2, modulo the exponent */
			i = (i & 1) ? i/2 + t/2 + 1 : i/2;

			/* Remove all following multiples from the sieve. */
			sieveSingle(sieve, SIEVE, i, t);

			/* Get the next exponent value */
			t = va_arg(ap, unsigned);
		} while (t);

		va_end(ap);
	}

	/* Fill up the offsets array with the first SHUFFLE candidates */
	i = p = 0;
	/* Get first prime */
	if (sieve[0] & 1 || (p = sieveSearch(sieve, SIEVE, p)) != 0) {
		offsets[i++] = p;
		p = sieveSearch(sieve, SIEVE, p);
	}
	/*
	 * Okay, from this point onwards, p is always the next entry
	 * from the sieve, that has not been added to the shuffle table,
	 * and is 0 iff the sieve has been exhausted.
	 *
	 * If we want to shuffle, then fill the shuffle table until the
	 * sieve is exhausted or the table is full.
	 */
	if (randfunc && p) {
		do {
			offsets[i++] = p;
			p = sieveSearch(sieve, SIEVE, p);
		} while (p && i < SHUFFLE);
	}

	/* Choose a random candidate for experimentation */
	prev = 0;
	while (i) {
		/* Pick a random entry from the shuffle table */
		j = randfunc ? randfunc(i) : 0;
		q = offsets[j];	/* The entry to use */

		/* Replace the entry with some more data, if possible */
		if (p) {
			offsets[j] = p;
			p = sieveSearch(sieve, SIEVE, p);
		} else {
			offsets[j] = offsets[--i];
			offsets[i] = 0;
		}

		/* Adjust bn to have the right value */
		if ((q > prev ? bnAddMult(bn, q-prev, 2)
		              : bnSubMult(bn, prev-q, 2)) < 0)
			goto failed;
		prev = q;

		/* Now do the Fermat tests */
		retval = primeTest(bn, &e, &a, f, arg);
		if (retval <= 0)
			goto done;	/* Success or error */
		modexps += retval;
		if (f && (retval = f(arg, '.')) < 0)
			goto done;
	}

	/* Ran out of sieve space - increase bn and keep trying. */
	if (bnAddMult(bn, SIEVE*8-prev, 2) < 0)
		goto failed;
	if (f && (retval = f(arg, '/')) < 0)
		goto done;
	goto retry;

failed:
	retval = -1;
done:
	bnEnd(&e);
	bnEnd(&a);
	bniMemWipe(offsets, sizeof(offsets));
#ifdef MSDOS
	bniMemFree(sieve, SIEVE);
#else
	bniMemWipe(sieve, sizeof(sieve));
#endif

	return retval < 0 ? retval : modexps + CONFIRMTESTS;
}
Пример #5
0
/*
 * Helper function that does the slow primality test.
 * bn is the input bignum; a and e are temporary buffers that are
 * allocated by the caller to save overhead.
 *
 * Returns 0 if prime, >0 if not prime, and -1 on error (out of memory).
 * If not prime, returns the number of modular exponentiations performed.
 * Calls the given progress function with a '*' for each primality test
 * that is passed.
 *
 * The testing consists of strong pseudoprimality tests, to the bases given
 * in the confirm[] array above.  (Also called Miller-Rabin, although that's
 * not technically correct if we're using fixed bases.)  Some people worry
 * that this might not be enough.  Number theorists may wish to generate
 * primality proofs, but for random inputs, this returns non-primes with
 * a probability which is quite negligible, which is good enough.
 *
 * It has been proved (see Carl Pomerance, "On the Distribution of
 * Pseudoprimes", Math. Comp. v.37 (1981) pp. 587-593) that the number of
 * pseudoprimes (composite numbers that pass a Fermat test to the base 2)
 * less than x is bounded by:
 * exp(ln(x)^(5/14)) <= P_2(x)	### CHECK THIS FORMULA - it looks wrong! ###
 * P_2(x) <= x * exp(-1/2 * ln(x) * ln(ln(ln(x))) / ln(ln(x))).
 * Thus, the local density of Pseudoprimes near x is at most
 * exp(-1/2 * ln(x) * ln(ln(ln(x))) / ln(ln(x))), and at least
 * exp(ln(x)^(5/14) - ln(x)).  Here are some values of this function
 * for various k-bit numbers x = 2^k:
 * Bits	Density <=	Bit equivalent	Density >=	Bit equivalent
 *  128	3.577869e-07	 21.414396	4.202213e-37	 120.840190
 *  192	4.175629e-10	 31.157288	4.936250e-56	 183.724558
 *  256 5.804314e-13	 40.647940	4.977813e-75	 246.829095
 *  384 1.578039e-18	 59.136573	3.938861e-113	 373.400096
 *  512 5.858255e-24	 77.175803	2.563353e-151	 500.253110
 *  768 1.489276e-34	112.370944	7.872825e-228	 754.422724
 * 1024 6.633188e-45	146.757062	1.882404e-304	1008.953565
 *
 * As you can see, there's quite a bit of slop between these estimates.
 * In fact, the density of pseudoprimes is conjectured to be closer to the
 * square of that upper bound.  E.g. the density of pseudoprimes of size
 * 256 is around 3 * 10^-27.  The density of primes is very high, from
 * 0.005636 at 256 bits to 0.001409 at 1024 bits, i.e.  more than 10^-3.
 *
 * For those people used to cryptographic levels of security where the
 * 56 bits of DES key space is too small because it's exhaustible with
 * custom hardware searching engines, note that you are not generating
 * 50,000,000 primes per second on each of 56,000 custom hardware chips
 * for several hours.  The chances that another Dinosaur Killer asteroid
 * will land today is about 10^-11 or 2^-36, so it would be better to
 * spend your time worrying about *that*.  Well, okay, there should be
 * some derating for the chance that astronomers haven't seen it yet,
 * but I think you get the idea.  For a good feel about the probability
 * of various events, I have heard that a good book is by E'mile Borel,
 * "Les Probabilite's et la vie".  (The 's are accents, not apostrophes.)
 *
 * For more on the subject, try "Finding Four Million Large Random Primes",
 * by Ronald Rivest, in Advancess in Cryptology: Proceedings of Crypto
 * '90.  He used a small-divisor test, then a Fermat test to the base 2,
 * and then 8 iterations of a Miller-Rabin test.  About 718 million random
 * 256-bit integers were generated, 43,741,404 passed the small divisor
 * test, 4,058,000 passed the Fermat test, and all 4,058,000 passed all
 * 8 iterations of the Miller-Rabin test, proving their primality beyond
 * most reasonable doubts.
 *
 * If the probability of getting a pseudoprime is some small p, then the
 * probability of not getting it in t trials is (1-p)^t.  Remember that,
 * for small p, (1-p)^(1/p) ~ 1/e, the base of natural logarithms.
 * (This is more commonly expressed as e = lim_{x\to\infty} (1+1/x)^x.)
 * Thus, (1-p)^t ~ e^(-p*t) = exp(-p*t).  So the odds of being able to
 * do this many tests without seeing a pseudoprime if you assume that
 * p = 10^-6 (one in a million) is one in 57.86.  If you assume that
 * p = 2*10^-6, it's one in 3347.6.  So it's implausible that the density
 * of pseudoprimes is much more than one millionth the density of primes.
 *
 * He also gives a theoretical argument that the chance of finding a
 * 256-bit non-prime which satisfies one Fermat test to the base 2 is
 * less than 10^-22.  The small divisor test improves this number, and
 * if the numbers are 512 bits (as needed for a 1024-bit key) the odds
 * of failure shrink to about 10^-44.  Thus, he concludes, for practical
 * purposes *one* Fermat test to the base 2 is sufficient.
 */
static int
primeTest(BigNum const *bn, BigNum *e, BigNum *a,
	int (*f)(void *arg, int c), void *arg)
{
	unsigned i, j;
	unsigned k, l;
	int err;

#if BNDEBUG	/* Debugging */
	/*
	 * This is debugging code to test the sieving stage.
	 * If the sieving is wrong, it will let past numbers with
	 * small divisors.  The prime test here will still work, and
	 * weed them out, but you'll be doing a lot more slow tests,
	 * and presumably excluding from consideration some other numbers
	 * which might be prime.  This check just verifies that none
	 * of the candidates have any small divisors.  If this
	 * code is enabled and never triggers, you can feel quite
	 * confident that the sieving is doing its job.
	 */
	i = bnLSWord(bn);
	if (!(i % 2)) printf("bn div by 2!");
	i = bnModQ(bn, 51051);	/* 51051 = 3 * 7 * 11 * 13 * 17 */
	if (!(i % 3)) printf("bn div by 3!");
	if (!(i % 7)) printf("bn div by 7!");
	if (!(i % 11)) printf("bn div by 11!");
	if (!(i % 13)) printf("bn div by 13!");
	if (!(i % 17)) printf("bn div by 17!");
	i = bnModQ(bn, 63365);	/* 63365 = 5 * 19 * 23 * 29 */
	if (!(i % 5)) printf("bn div by 5!");
	if (!(i % 19)) printf("bn div by 19!");
	if (!(i % 23)) printf("bn div by 23!");
	if (!(i % 29)) printf("bn div by 29!");
	i = bnModQ(bn, 47027);	/* 47027 = 31 * 37 * 41 */
	if (!(i % 31)) printf("bn div by 31!");
	if (!(i % 37)) printf("bn div by 37!");
	if (!(i % 41)) printf("bn div by 41!");
#endif

	/*
	 * Now, check that bn is prime.  If it passes to the base 2,
	 * it's prime beyond all reasonable doubt, and everything else
	 * is just gravy, but it gives people warm fuzzies to do it.
	 *
	 * This starts with verifying Euler's criterion for a base of 2.
	 * This is the fastest pseudoprimality test that I know of,
	 * saving a modular squaring over a Fermat test, as well as
	 * being stronger.  7/8 of the time, it's as strong as a strong
	 * pseudoprimality test, too.  (The exception being when bn ==
	 * 1 mod 8 and 2 is a quartic residue, i.e. bn is of the form
	 * a^2 + (8*b)^2.)  The precise series of tricks used here is
	 * not documented anywhere, so here's an explanation.
	 * Euler's criterion states that if p is prime then a^((p-1)/2)
	 * is congruent to Jacobi(a,p), modulo p.  Jacobi(a,p) is
	 * a function which is +1 if a is a square modulo p, and -1 if
	 * it is not.  For a = 2, this is particularly simple.  It's
	 * +1 if p == +/-1 (mod 8), and -1 if m == +/-3 (mod 8).
	 * If p == 3 mod 4, then all a strong test does is compute
	 * 2^((p-1)/2). and see if it's +1 or -1.  (Euler's criterion
	 * says *which* it should be.)  If p == 5 (mod 8), then
	 * 2^((p-1)/2) is -1, so the initial step in a strong test,
	 * looking at 2^((p-1)/4), is wasted - you're not going to
	 * find a +/-1 before then if it *is* prime, and it shouldn't
	 * have either of those values if it isn't.  So don't bother.
	 *
	 * The remaining case is p == 1 (mod 8).  In this case, we
	 * expect 2^((p-1)/2) == 1 (mod p), so we expect that the
	 * square root of this, 2^((p-1)/4), will be +/-1 (mod p).
	 * Evaluating this saves us a modular squaring 1/4 of the time.
	 * If it's -1, a strong pseudoprimality test would call p
	 * prime as well.  Only if the result is +1, indicating that
	 * 2 is not only a quadratic residue, but a quartic one as well,
	 * does a strong pseudoprimality test verify more things than
	 * this test does.  Good enough.
	 *
	 * We could back that down another step, looking at 2^((p-1)/8)
	 * if there was a cheap way to determine if 2 were expected to
	 * be a quartic residue or not.  Dirichlet proved that 2 is
	 * a quartic residue iff p is of the form a^2 + (8*b^2).
	 * All primes == 1 (mod 4) can be expressed as a^2 + (2*b)^2,
	 * but I see no cheap way to evaluate this condition.
	 */
	if (bnCopy(e, bn) < 0)
		return -1;
	(void)bnSubQ(e, 1);
	l = bnLSWord(e);

	j = 1;	/* Where to start in prime array for strong prime tests */

	if (l & 7) {
		bnRShift(e, 1);
		if (bnTwoExpMod(a, e, bn) < 0)
			return -1;
		if ((l & 7) == 6) {
			/* bn == 7 mod 8, expect +1 */
			if (bnBits(a) != 1)
				return 1;	/* Not prime */
			k = 1;
		} else {
			/* bn == 3 or 5 mod 8, expect -1 == bn-1 */
			if (bnAddQ(a, 1) < 0)
				return -1;
			if (bnCmp(a, bn) != 0)
				return 1;	/* Not prime */
			k = 1;
			if (l & 4) {
				/* bn == 5 mod 8, make odd for strong tests */
				bnRShift(e, 1);
				k = 2;
			}
		}
	} else {
		/* bn == 1 mod 8, expect 2^((bn-1)/4) == +/-1 mod bn */
		bnRShift(e, 2);
		if (bnTwoExpMod(a, e, bn) < 0)
			return -1;
		if (bnBits(a) == 1) {
			j = 0;	/* Re-do strong prime test to base 2 */
		} else {
			if (bnAddQ(a, 1) < 0)
				return -1;
			if (bnCmp(a, bn) != 0)
				return 1;	/* Not prime */
		}
		k = 2 + bnMakeOdd(e);
	}
	/* It's prime!  Now go on to confirmation tests */

	/*
	 * Now, e = (bn-1)/2^k is odd.  k >= 1, and has a given value
	 * with probability 2^-k, so its expected value is 2.
	 * j = 1 in the usual case when the previous test was as good as
	 * a strong prime test, but 1/8 of the time, j = 0 because
	 * the strong prime test to the base 2 needs to be re-done.
	 */
	for (i = j; i < CONFIRMTESTS; i++) {
		if (f && (err = f(arg, '*')) < 0)
			return err;
		(void)bnSetQ(a, confirm[i]);
		if (bnExpMod(a, a, e, bn) < 0)
			return -1;
		if (bnBits(a) == 1)
			continue;	/* Passed this test */

		l = k;
		for (;;) {
			if (bnAddQ(a, 1) < 0)
				return -1;
			if (bnCmp(a, bn) == 0)	/* Was result bn-1? */
				break;	/* Prime */
			if (!--l)	/* Reached end, not -1? luck? */
				return i+2-j;	/* Failed, not prime */
			/* This portion is executed, on average, once. */
			(void)bnSubQ(a, 1);	/* Put a back where it was. */
			if (bnSquare(a, a) < 0 || bnMod(a, a, bn) < 0)
				return -1;
			if (bnBits(a) == 1)
				return i+2-j;	/* Failed, not prime */
		}
		/* It worked (to the base confirm[i]) */
	}
	
	/* Yes, we've decided that it's prime. */
	if (f && (err = f(arg, '*')) < 0)
		return err;
	return 0;	/* Prime! */
}
Пример #6
0
int rsa_GENKEYS(RSA_CTX *ctx,unsigned char *s1,unsigned char *s2){
int t,c;
unsigned long int MB=(ctx->bits)/2;
struct BigNum bn;
struct BigNum p,q;
int len=MB/8;
if(ctx->bits==0)
	return BADDATALEN;
bnBegin(&bn);
bnBegin(&p);
bnBegin(&q);
t=bnInsertBigBytes(&p,(void*)s1,0,len);
c=bnInsertBigBytes(&q,(void*)s2,0,len);
if(t<0 || c<0){
	bnEnd(&bn);
	bnEnd(&p);
	bnEnd(&q);
	return BADTHINGS;
}
t=primeGen(&p,NULL,NULL,NULL,1,3,5,0);
c=primeGen(&q,NULL,NULL,NULL,1,3,5,0);
if(t<0 || c<0){
	bnEnd(&bn);
	bnEnd(&p);
	bnEnd(&q);
	return BADTHINGS;
}

t=bnMul(&ctx->n,&p,&q);
if(t<0){
	bnEnd(&bn);
	bnEnd(&p);
	bnEnd(&q);
	return BADTHINGS;
}
t=bnSubQ(&p,1);
if(t<0){
	bnEnd(&bn);
	bnEnd(&p);
	bnEnd(&q);
        return BADTHINGS;
}
t=bnSubQ(&q,1);
if(t<0){
	bnEnd(&bn);
	bnEnd(&p);
	bnEnd(&q);
        return BADTHINGS;
}
t=bnMul(&bn,&p,&q);
if(t<0){
	bnEnd(&bn);
	bnEnd(&p);
	bnEnd(&q);
        return BADTHINGS;
}
t=bnInv(&ctx->d,&ctx->e,&bn);
if(t<0){
	bnEnd(&bn);
	bnEnd(&p);
	bnEnd(&q);
        return BADTHINGS;
}
bnEnd(&bn);
bnEnd(&p);
bnEnd(&q);
return OK;
}
Пример #7
0
/*
 * Modifies the given bnq to return a prime slightly larger, and then
 * modifies the given bnp to be a prime which is == 1 (mod bnq).
 * This is done by decreasing bnp until it is == 1 (mod 2*bnq), and
 * then searching forward in steps of 2*bnq.
 * Returns >=0 on success or -1 on failure (out of memory).  On
 * success, the return value is the number of modular exponentiations
 * performed (excluding the final confirmation).
 * This never gives up searching.
 *
 * int (*f)(void *arg, int c), void *arg
 * The function f argument, if non-NULL, is called with progress indicator
 * characters for printing.  A dot (.) is written every time a primality test
 * is failed, a star (*) every time one is passed, and a slash (/) in the
 * case that the sieve was emptied without finding a prime and is being
 * refilled.  f is also passed the void *arg argument for private
 * context storage.  If f returns < 0, the test aborts and returns
 * that value immediately.
 *
 * Apologies to structured programmers for all the GOTOs.
 */
int
dsaPrimeGen(BigNum *bnq, BigNum *bnp,
	int (*f)(void *arg, int c), void *arg)
{
	int retval;
	unsigned p, prev;
	BigNum a, e;
	int modexps = 0;
#ifdef MSDOS
	unsigned char *sieve;
#else
	unsigned char sieve[SIEVE];
#endif

#ifdef MSDOS
	sieve = bniMemAlloc(SIEVE);
	if (!sieve)
		return -1;
#endif

	bnBegin(&a);
	bnBegin(&e);

	/* Phase 1: Search forwards from bnq for a suitable prime. */

	/* First, make sure that bnq is odd. */
	(void)bnAddQ(bnq, ~bnLSWord(bnq) & 1);

	for (;;) {
		if (sieveBuild(sieve, SIEVE, bnq, 2, 1) < 0)
			goto failed;

		p = prev = 0;
		if (sieve[0] & 1 || (p = sieveSearch(sieve, SIEVE, p)) != 0) {
			do {
				/*
				 * Adjust bn to have the right value,
				 * incrementing in steps of < 65536.
				 * 32767 = 65535/2.
				 */
				pgpAssert(p >= prev);
				prev = p-prev;	/* Delta - add 2*prev to bn */
#if SIEVE*8*2 >= 65536
				while (prev > 32767) {
					if (bnAddQ(bnq, 2*32767) < 0)
						goto failed;
					prev -= 32767;
				}
#endif
				if (bnAddQ(bnq, 2*prev) < 0)
					goto failed;
				prev = p;

				retval = primeTest(bnq, &e, &a, f, arg);
				if (retval <= 0)
					goto phase2;	/* Success! */
				modexps += retval;
				if (f && (retval = f(arg, '.')) < 0)
					goto done;

				/* And try again */
				p = sieveSearch(sieve, SIEVE, p);
			} while (p);
		}

		/* Ran out of sieve space - increase bn and keep trying. */
#if SIEVE*8*2 >= 65536
		p = ((SIEVE-1)*8+7) - prev;	/* Number of steps (of 2) */
		while (p >= 32737) {
			if (bnAddQ(bnq, 2*32767) < 0)
				goto failed;
			p -= 32767;
		}
		if (bnAddQ(bnq, 2*(p+1)) < 0)
			goto failed;
#else
		if (bnAddQ(bnq, SIEVE*8*2 - prev) < 0)
			goto failed;
#endif
		if (f && (retval = f(arg, '/')) < 0)
			goto done;
	} /* for (;;) */

	/*
	 * Phase 2: find a suitable prime bnp == 1 (mod bnq).
	 */

	/*
	 * Since bnp will be, and bnq is, odd, bnp-1 must be a multiple
	 * of 2*bnq.  So start by subtracting the excess.
	 */

phase2:
	/* Double bnq until end of bnp search. */
	if (bnAdd(bnq, bnq) < 0)
		goto failed;

	bnMod(&a, bnp, bnq);
	if (bnBits(&a)) {	/* Will always be true, but... */
		(void)bnSubQ(&a, 1);
		if (bnSub(bnp, &a)) 	/* Also error on underflow */
			goto failed;
	}

	/* Okay, now we're ready. */

	for (;;) {
		if (sieveBuildBig(sieve, SIEVE, bnp, bnq, 0) < 0)
			goto failed;
		if (f && (retval = f(arg, '/')) < 0)
			goto done;

		p = prev = 0;
		if (sieve[0] & 1 || (p = sieveSearch(sieve, SIEVE, p)) != 0) {
			do {
				/*
				 * Adjust bn to have the right value,
				 * adding (p-prev) * 2*bnq.
				 */
				pgpAssert(p >= prev);
				/* Compute delta into a */
				if (bnMulQ(&a, bnq, p-prev) < 0)
					goto failed;
				if (bnAdd(bnp, &a) < 0)
					goto failed;
				prev = p;

				retval = primeTest(bnp, &e, &a, f, arg);
				if (retval <= 0)
					goto done;	/* Success! */
				modexps += retval;
				if (f && (retval = f(arg, '.')) < 0)
					goto done;

				/* And try again */
				p = sieveSearch(sieve, SIEVE, p);
			} while (p);
		}

		/* Ran out of sieve space - increase bn and keep trying. */
#if SIEVE*8 == 65536
		if (prev) {
			p = (unsigned)(SIEVE*8ul - prev);
		} else {
			/* Corner case that will never actually happen */
			if (bnAdd(bnp, bnq) < 0)
				goto failed;
			p = 65535;
		}
#else
		p = SIEVE*8 - prev;
#endif
		/* Add p * bnq to bnp */
		if (bnMulQ(&a, bnq, p) < 0)
			goto failed;
		if (bnAdd(bnp, &a) < 0)
			goto failed;
	} /* for (;;) */

failed:
	retval = -1;

done:
	/* Shift bnq back down by the extra bit again. */
	bnRShift(bnq, 1);	/* Harmless even if bnq is random */

	bnEnd(&e);
	bnEnd(&a);
#ifdef MSDOS
	bniMemFree(sieve, SIEVE);
#else
	bniMemWipe(sieve, sizeof(sieve));
#endif
	return retval < 0 ? retval : modexps + 2*CONFIRMTESTS;
}
Пример #8
0
ZrtpDH::ZrtpDH(const char* type) {

    uint8_t random[64];

    dhCtx* tmpCtx = new dhCtx;
    ctx = static_cast<void*>(tmpCtx);

    // Well - the algo type is only 4 char thus cast to int32 and compare
    if (*(int32_t*)type == *(int32_t*)dh2k) {
        pkType = DH2K;
    }
    else if (*(int32_t*)type == *(int32_t*)dh3k) {
        pkType = DH3K;
    }
    else if (*(int32_t*)type == *(int32_t*)ec25) {
        pkType = EC25;
    }
    else if (*(int32_t*)type == *(int32_t*)ec38) {
        pkType = EC38;
    }
    else if (*(int32_t*)type == *(int32_t*)e255) {
        pkType = E255;
    }
    else if (*(int32_t*)type == *(int32_t*)e414) {
        pkType = E414;
    }
    else {
        return;
    }

    randomZRTP(random, sizeof(random));

    if (!dhinit) {
        bnBegin(&two);
        bnSetQ(&two, 2);

        bnBegin(&bnP2048);
        bnInsertBigBytes(&bnP2048, P2048, 0, sizeof(P2048));
        bnBegin(&bnP3072);
        bnInsertBigBytes(&bnP3072, P3072, 0, sizeof(P3072));

        bnBegin(&bnP2048MinusOne);
        bnCopy(&bnP2048MinusOne, &bnP2048);
        bnSubQ(&bnP2048MinusOne, 1);

        bnBegin(&bnP3072MinusOne);
        bnCopy(&bnP3072MinusOne, &bnP3072);
        bnSubQ(&bnP3072MinusOne, 1);

        dhinit = 1;
    }

    bnBegin(&tmpCtx->privKey);
    INIT_EC_POINT(&tmpCtx->pubPoint);

    switch (pkType) {
    case DH2K:
    case DH3K:
        bnInsertBigBytes(&tmpCtx->privKey, random, 0, 256/8);
        break;

    case EC25:
        ecGetCurveNistECp(NIST256P, &tmpCtx->curve);
        ecGenerateRandomNumber(&tmpCtx->curve, &tmpCtx->privKey);
        break;

    case EC38:
        ecGetCurveNistECp(NIST384P, &tmpCtx->curve);
        ecGenerateRandomNumber(&tmpCtx->curve, &tmpCtx->privKey);
        break;

    case E255:
        ecGetCurvesCurve(Curve25519, &tmpCtx->curve);
        ecGenerateRandomNumber(&tmpCtx->curve, &tmpCtx->privKey);
        break;

    case E414:
        ecGetCurvesCurve(Curve3617, &tmpCtx->curve);
        ecGenerateRandomNumber(&tmpCtx->curve, &tmpCtx->privKey);
        break;
    }
}
Пример #9
0
int
genRsaKey(struct PubKey *pub, struct SecKey *sec,
	  unsigned bits, unsigned exp, FILE *file)
{
	int modexps = 0;
	struct BigNum t;	/* Temporary */
	int i;
	struct Progress progress;

	progress.f = file;
	progress.column = 0;
	progress.wrap = 78;

	if (bnSetQ(&pub->e, exp))
		return -1;

	/* Find p - choose a starting place */
	if (genRandBn(&sec->p, bits/2, 0xC0, 1) < 0)
		return -1;
	/* And search for a prime */
	i = primeGen(&sec->p, randRange, file ? genProgress : 0, &progress,
	             exp, 0);
	if (i < 0)
		goto error;
	modexps = i;
	assert(bnModQ(&sec->p, exp) != 1);
bndPut("p = ", &sec->p);

	do {
		/* Visual separator between the two progress indicators */
		if (file)
			genProgress(&progress, ' ');

		if (genRandBn(&sec->q, (bits+1)/2, 0xC0, 1) < 0)
			goto error;
		if (bnCopy(&pub->n, &sec->q) < 0)
			goto error;
		if (bnSub(&pub->n, &sec->p) < 0)
			goto error;
		/* Note that bnSub(a,b) returns abs(a-b) */
	} while (bnBits(&pub->n) < bits/2-5);

	if (file)
		fflush(file);	/* Ensure the separators are visible */

	i = primeGen(&sec->q, randRange, file ? genProgress : 0, &progress,
	             exp, 0);
	if (i < 0)
		goto error;
	modexps += i;
	assert(bnModQ(&sec->p, exp) != 1);
bndPut("q = ", &sec->q);

	/* Wash the random number pool. */
	randFlush();

	/* Ensure that q is larger */
	if (bnCmp(&sec->p, &sec->q) > 0)
		bnSwap(&sec->p, &sec->q);
bndPut("p = ", &sec->p);
bndPut("q = ", &sec->q);


	/*
	 * Now we dive into a large amount of fiddling to compute d,
	 * the decryption exponent, from the encryption exponent.
	 * We require that e*d == 1 (mod p-1) and e*d == 1 (mod q-1).
	 * This can alomost be done via the Chinese Remainder Algorithm,
	 * but it doesn't quite apply, because p-1 and q-1 are not
	 * realitvely prime.  Our task is to massage these into
	 * two numbers a and b such that a*b = lcm(p-1,q-1) and
	 * gcd(a,b) = 1.  The technique is not well documented,
	 * so I'll describe it here.
	 * First, let d = gcd(p-1,q-1), then let a' = (p-1)/d and
	 * b' = (q-1)/d.  By the definition of the gcd, gcd(a',b') at
	 * this point is 1, but a'*b' is a factor of d shy of the desired
	 * value.  We have to produce a = a' * d1 and b = b' * d2 such
	 * d1*d2 = d and gcd(a,b) is 1.  This will be the case iff
	 * gcd(a,d2) = gcd(b,d1) = 1.  Since GCD is associative and
	 * (gcd(x,y,z) = gcd(x,gcd(y,z)) = gcd(gcd(x,y),z), etc.),
	 * gcd(a',b') = 1 implies that gcd(a',b',d) = 1 which implies
	 * that gcd(a',gcd(b',d)) = gcd(gcd(a',d),b') = 1.  So you can
	 * extract gcd(b',d) from d and make it part of d2, and the
	 * same for d1.  And iterate?  A pessimal example is x = 2*6^k
	 * and y = 3*6^k.  gcd(x,y) = 6^k and we have to divvy it up
	 * somehow so that all the factors of 2 go to x and all the
	 * factors of 3 go to y, ending up with a = 2*2^k and b = 3*3^k.
	 *
	 * Aah, f**k it.  It's simpler to do one big inverse for now.
	 * Later I'll figure out how to get this to work properly.
	 */

	/* Decrement q temporarily */
	(void)bnSubQ(&sec->q, 1);
	/* And u = p-1, to be divided by gcd(p-1,q-1) */
	if (bnCopy(&sec->u, &sec->p) < 0)
		goto error;
	(void)bnSubQ(&sec->u, 1);
bndPut("p-1 = ", &sec->u);
bndPut("q-1 = ", &sec->q);
	/* Use t to store gcd(p-1,q-1) */
	bnBegin(&t);
	if (bnGcd(&t, &sec->q, &sec->u) < 0) {
		bnEnd(&t);
		goto error;
	}
bndPut("t = gcd(p-1,q-1) = ", &t);

	/* Let d = (p-1) / gcd(p-1,q-1) (n is scratch for the remainder) */
	i = bnDivMod(&sec->d, &pub->n, &sec->u, &t);
bndPut("(p-1)/t = ", &sec->d);
bndPut("(p-1)%t = ", &pub->n);
	bnEnd(&t);
	if (i < 0)
		goto error;
	assert(bnBits(&pub->n) == 0);
	/* Now we have q-1 and d = (p-1) / gcd(p-1,q-1) */
	/* Find the product, n = lcm(p-1,q-1) = c * d */
	if (bnMul(&pub->n, &sec->q, &sec->d) < 0)
		goto error;
bndPut("(p-1)*(q-1)/t = ", &pub->n);
	/* Find the inverse of the exponent mod n */
	i = bnInv(&sec->d, &pub->e, &pub->n);
bndPut("e = ", &pub->e);
bndPut("d = ", &sec->d);
	if (i < 0)
		goto error;
	assert(!i);	/* We should NOT get an error here */
	/*
	 * Now we have the comparatively simple task of computing
	 * u = p^-1 mod q.
	 */
#if BNDEBUG
	bnMul(&sec->u, &sec->d, &pub->e);
bndPut("d * e = ", &sec->u);
	bnMod(&pub->n, &sec->u, &sec->q);
bndPut("d * e = ", &sec->u);
bndPut("q-1 = ", &sec->q);
bndPut("d * e % (q-1)= ", &pub->n);
	bnNorm(&pub->n);
	bnSubQ(&sec->p, 1);
bndPut("d * e = ", &sec->u);
	bnMod(&sec->u, &sec->u, &sec->p);
bndPut("p-1 = ", &sec->p);
bndPut("d * e % (p-1)= ", &sec->u);
	bnNorm(&sec->u);
	bnAddQ(&sec->p, 1);
#endif

	/* But it *would* be nice to have q back first. */
	(void)bnAddQ(&sec->q, 1);

bndPut("p = ", &sec->p);
bndPut("q = ", &sec->q);

	/* Now compute u = p^-1 mod q */
	i = bnInv(&sec->u, &sec->p, &sec->q);
	if (i < 0)
		goto error;
bndPut("u = p^-1 % q = ", &sec->u);
	assert(!i);	/* p and q had better be relatively prime! */

#if BNDEBUG
	bnMul(&pub->n, &sec->u, &sec->p);
bndPut("u * p = ", &pub->n);
	bnMod(&pub->n, &pub->n, &sec->q);
bndPut("u * p % q = ", &pub->n);
	bnNorm(&pub->n);
#endif
	/* And finally,  n = p * q */
	if (bnMul(&pub->n, &sec->p, &sec->q) < 0)
		goto error;
bndPut("n = p * q = ", &pub->n);
	/* And that's it... success! */
	if (file)
		putc('\n', file);	/* Signal done */
	return modexps;

error:
	if (file)
		fputs("?\n", file);	/* Signal error */

	return -1;
}
Пример #10
0
/* Initialize BSafe privkey structure from a RSAsec. */
static int
rprivk_init(B_KEY_OBJ rprivk, RSAsec const *sec,
	PGPMemoryMgrRef	mgr)
{
	BigNum dmodp, dmodq, tmp;
	A_RSA_CRT_KEY kdata;
	PGPByte *buf;
	PGPSize bufsize;
	int err;

	/* Calculate d mod p-1 and d mod q-1 */
	bnBegin(&dmodp, mgr, TRUE);
	bnBegin(&dmodq, mgr, TRUE);
	bnBegin(&tmp, mgr, TRUE);
	bnCopy(&tmp, &sec->p);
	bnSubQ(&tmp, 1);
	bnMod(&dmodp, &sec->d, &tmp);
	bnCopy(&tmp, &sec->q);
	bnSubQ(&tmp, 1);
	bnMod(&dmodq, &sec->d, &tmp);

	bufsize = bnBytes(&sec->n) + bnBytes(&sec->q) + bnBytes(&sec->p)
			+ bnBytes(&dmodq) + bnBytes(&dmodp) + bnBytes(&sec->u);
	buf = PGPNewSecureData( mgr, bufsize, 0 );
	kdata.modulus.data = buf;
	kdata.modulus.len = bnBytes(&sec->n);
	kdata.prime[0].data = kdata.modulus.data + kdata.modulus.len;
	kdata.prime[0].len = bnBytes(&sec->q);
	kdata.prime[1].data = kdata.prime[0].data + kdata.prime[0].len;
	kdata.prime[1].len = bnBytes(&sec->p);
	kdata.primeExponent[0].data = kdata.prime[1].data
				+ kdata.prime[1].len;
	kdata.primeExponent[0].len = bnBytes(&dmodq);
	kdata.primeExponent[1].data = kdata.primeExponent[0].data
				+ kdata.primeExponent[0].len;
	kdata.primeExponent[1].len = bnBytes(&dmodp);
	kdata.coefficient.data = kdata.primeExponent[1].data
				+ kdata.primeExponent[1].len;
	kdata.coefficient.len = bnBytes(&sec->u);
	bnExtractBigBytes (&sec->n, kdata.modulus.data, 0,
						kdata.modulus.len);
	bnExtractBigBytes (&sec->q, kdata.prime[0].data, 0,
						kdata.prime[0].len);
	bnExtractBigBytes (&sec->p, kdata.prime[1].data, 0,
						kdata.prime[1].len);
	bnExtractBigBytes (&dmodq, kdata.primeExponent[0].data, 0,
						kdata.primeExponent[0].len);
	bnExtractBigBytes (&dmodp, kdata.primeExponent[1].data, 0,
						kdata.primeExponent[1].len);
	bnExtractBigBytes (&sec->u, kdata.coefficient.data, 0,
						kdata.coefficient.len);
	err = B_SetKeyInfo (rprivk, KI_RSA_CRT, (POINTER)&kdata);
	pgpAssert (err == 0);
	pgpClearMemory (buf, bufsize);
	PGPFreeData (buf);

	bnEnd(&dmodp);
	bnEnd(&dmodq);
	bnEnd(&tmp);
	return err;
}
Пример #11
0
static int
testDH(struct BigNum *bn)
{
    struct BigNum pub1, pub2, sec1, sec2;
    unsigned bits;
    int i = 0;
    char buf[4];

    bnBegin(&pub1);
    bnBegin(&pub2);
    bnBegin(&sec1);
    bnBegin(&sec2);

    /* Bits of secret - add a few to ensure an even distribution */
    bits = bnBits(bn)+4;
    /* Temporarily decrement bn for some operations */
    (void)bnSubQ(bn, 1);

    strcpy(buf, "foo");
    i = genRandBn(&sec1, bits, 0, 0, (unsigned char *)buf, 4);
    if (i < 0)
        goto done;
    /* Reduce sec1 to the correct range */
    i = bnMod(&sec1, &sec1, bn);
    if (i < 0)
        goto done;

    strcpy(buf, "bar");
    i = genRandBn(&sec2, bits, 0, 0, (unsigned char *)buf, 4);
    if (i < 0)
        goto done;
    /* Reduce sec2 to the correct range */
    i = bnMod(&sec2, &sec2, bn);
    if (i < 0)
        goto done;

    /* Re-increment bn */
    (void)bnAddQ(bn, 1);

    puts("Doing first half for party 1");
    i = bnTwoExpMod(&pub1, &sec1, bn);
    if (i < 0)
        goto done;
    puts("Doing first half for party 2");
    i = bnTwoExpMod(&pub2, &sec2, bn);
    if (i < 0)
        goto done;

    /* In a real protocol, pub1 and pub2 are now exchanged */

    puts("Doing second half for party 1");
    i = bnExpMod(&pub2, &pub2, &sec1, bn);
    if (i < 0)
        goto done;
    bndPut("shared = ", &pub2);
    puts("Doing second half for party 2");
    i = bnExpMod(&pub1, &pub1, &sec2, bn);
    if (i < 0)
        goto done;
    bndPut("shared = ", &pub1);

    if (bnCmp(&pub1, &pub2) != 0) {
        puts("Diffie-Hellman failed!");
        i = -1;
    } else {
        puts("Test successful.");
    }
done:
    bnEnd(&sec2);
    bnEnd(&sec1);
    bnEnd(&pub2);
    bnEnd(&pub1);

    return i;
}