int
rsaPublicEncrypt(BigNum *bn, PGPByte const *in, unsigned len,
RSApub const *pub, PGPRandomContext const *rc)
{
unsigned bytes = (bnBits(&pub->n)+7)/8;
pgpPKCSPack(bn, in, len, PKCS_PAD_ENCRYPTED, bytes, rc);
bndPrintf("RSA encrypting.\n");
bndPut("plaintext = ", bn);
return bnExpMod(bn, bn, &pub->e, &pub->n);
}
/*
* Decrypt a message with a public key.
* These destroy (actually, replace with a decrypted version) the
* input bignum bn.
*
* It recongizes the PGP 2.2 format, but not in all its generality;
* only the one case (framing byte = 1, length = 16) which was ever
* generated. It fakes the DER prefix in that case.
*
* Performs an RSA signature check. Returns a prefix of the unwrapped
* data in the given buf. Returns the length of the untruncated
* data, which may exceed "len". Returns <0 on error.
*/
int
rsaPublicDecrypt(PGPByte *buf, unsigned len, BigNum *bn,
RSApub const *pub)
{
unsigned bytes;
bndPrintf("RSA signature checking.\n");
if (bnExpMod(bn, bn, &pub->e, &pub->n) < 0)
return kPGPError_OutOfMemory;
bndPut("decrypted = ", bn);
bytes = (bnBits(&pub->n)+7)/8;
return pgpPKCSUnpack(buf, len, bn, PKCS_PAD_SIGNED, bytes);
}
/*
* Performs an RSA decryption. Returns a prefix of the unwrapped
* data in the given buf. Returns the length of the untruncated
* data, which may exceed "len". Returns <0 on error.
*/
int
rsaPrivateDecrypt(PGPByte *buf, unsigned len, BigNum *bn,
RSAsec const *sec)
{
unsigned bytes;
bndPrintf("RSA decrypting\n");
if (bnExpModCRA(bn, &sec->d, &sec->p, &sec->q, &sec->u) < 0)
return kPGPError_OutOfMemory;
bndPut("decrypted = ", bn);
bytes = (bnBits(&sec->n)+7)/8;
return pgpPKCSUnpack(buf, len, bn, PKCS_PAD_ENCRYPTED, bytes);
}
int
rsaPrivateEncrypt(BigNum *bn, PGPByte const *in, unsigned len,
RSAsec const *sec)
{
unsigned bytes = (bnBits(&sec->n)+7)/8;
pgpPKCSPack(bn, in, len, PKCS_PAD_SIGNED, bytes,
(PGPRandomContext const *)NULL);
bndPrintf("RSA signing.\n");
bndPut("plaintext = ", bn);
return bnExpModCRA(bn, &sec->d, &sec->p, &sec->q, &sec->u);
}
/*
* 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;
}
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;
}
static int
genDH(struct BigNum *bn, unsigned bits, unsigned char *seed, unsigned len,
FILE *f)
{
#if CLOCK_AVAIL
timetype start, stop;
unsigned long s;
#endif
int i;
unsigned char s1[1024], s2[1024];
unsigned p1, p2;
struct BigNum step;
struct Progress progress;
if (f)
fprintf(f, "Generating a %u-bit D-H prime with \"%.*s\"\n",
bits, (int)len, (char *)seed);
progress.f = f;
progress.column = 0;
progress.wrap = 78;
/* Find p - choose a starting place */
if (genRandBn(bn, bits, 0xC0, 3, seed, len) < 0)
return -1;
#if BNDEBUG /* DEBUG - check that sieve works properly */
bnBegin(&step);
bnSetQ(&step, 2);
sieveBuild(s1, 1024, bn, 2, 0);
sieveBuildBig(s2, 1024, bn, &step, 0);
p1 = p2 = 0;
if (s1[0] != s2[0])
printf("Difference: s1[0] = %x s2[0] = %x\n", s1[0], s2[0]);
do {
p1 = sieveSearch(s1, 1024, p1);
p2 = sieveSearch(s2, 1024, p2);
if (p1 != p2)
printf("Difference: p1 = %u p2 = %u\n", p1, p2);
} while (p1 && p2);
bnEnd(&step);
#endif
/* And search for a prime */
#if CLOCK_AVAIL
gettime(&start);
#endif
i = germainPrimeGen(bn, 1, f ? genProgress : 0, (void *)&progress);
if (i < 0)
return -1;
#if CLOCK_AVAIL
gettime(&stop);
#endif
if (f) {
putc('\n', f);
fprintf(f, "%d modular exponentiations performed.\n", i);
}
#if CLOCK_AVAIL
subtime(stop, start);
s = sec(stop);
bndPrintf("%u-bit time = %lu.%03u sec.", bits, s, msec(stop));
if (s > 60) {
putchar(' ');
putchar('(');
if (s > 3600)
printf("%u:%02u", (unsigned)(s/3600),
(unsigned)(s/60%60));
else
printf("%u", (unsigned)(s/60));
printf(":%02u)", (unsigned)(s%60));
}
putchar('\n');
#endif
bndPut("p = ", bn);
return 0;
}
