/*
* Add x*y to bn, which is usually (but not always) < 65536.
* Do it in a simple linear manner.
*/
static int
bnAddMult(BigNum *bn, unsigned x, unsigned y)
{
unsigned long z = (unsigned long)x * y;
while (z > 65535) {
if (bnAddQ(bn, 65535) < 0)
return -1;
z -= 65535;
}
return bnAddQ(bn, (unsigned)z);
}
/* dest = bn + sm, where 0 <= sm < 2^16 */
PGPError
PGPBigNumAddQ(
PGPBigNumRef bn,
PGPUInt16 sm,
PGPBigNumRef dest)
{
PGPError err = kPGPError_NoErr;
pgpValidateBigNum( bn );
pgpValidateBigNum( dest );
if ( IsBNError( bnAddQ( &dest->bn, sm ) ) )
{
err = kPGPError_OutOfMemory;
}
return( err );
}
int
bnPrint10(FILE *f, char const *prefix, BigNum const *bn,
char const *suffix)
{
BigNum pbig, pbig1;
BigNum ten, zero, rem;
char buf[3000]; /* up to 9000 bits */
int bufi = sizeof(buf)-1;
int n;
buf[bufi] = '\0';
bnBegin (&pbig);
bnBegin (&pbig1);
bnBegin (&rem);
bnBegin (&ten);
bnAddQ (&ten, 10);
bnBegin (&zero);
bnCopy (&pbig, bn);
while (bnCmp (&pbig, &zero) != 0) {
bnDivMod (&pbig1, &rem, &pbig, &ten);
bnCopy (&pbig, &pbig1);
n = bnLSWord (&rem);
buf[--bufi] = n + '0';
}
bnEnd (&pbig);
bnEnd (&pbig1);
bnEnd (&rem);
if (prefix && fputs(prefix, f) < 0)
return EOF;
fputs (buf+bufi, f);
pgpClearMemory (buf, sizeof(buf));
return suffix ? fputs(suffix, f) : 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;
}
/*
* 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! */
}
/*
* 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;
}
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;
}
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;
}
