Esempio n. 1
0
int main(int argc, char *argv[]) {
  uint32 n = 1000;
  FILE *ofp = stdout;

  if (argc >= 2) {
    if (sscanf(argv[1], "%d", &n) != 1) {
       n = 1000;
    }

    if (argc >= 3) {
	ofp = fopen(argv[2], "w");
    }

    if (ofp == NULL) // Reverting back to our standard out
       ofp = stdout;
  }

  printf("\033[94mGenerating the first: %d primes\033[00m\n", n);

  uint64 *primeList = primeGen(n);

  int i=0;
  fprintf(ofp, "%-20s %30s\n", "Index:", "Value");

  for (i = 0; i < n; ++i) {
     fprintf(ofp, "%-20d %30ld\n", i+1, primeList[i]);
  }

  if (primeList != NULL) free(primeList);

  fclose(ofp);
  return 0;
}
Esempio n. 2
0
/*
 * This program will sum N prime numbers.
 * Eg: 2+3+5+7 = 17 if 4 is the input.
 * This can be to a given number of primes (if given as an argument)
 * or will sum a default number of primes.
 */
int main(int argc, const char * argv[]) {

    int primeSum = 0;
    int numPrimesRequested = 1000;
    if (argc > 1) {
        // I'll just assume/hope an integer was given
        sscanf(argv[1], "%d", &numPrimesRequested);
    }

    // Note: The first run gives the number of primes generated
    int limit = primeGen(&numPrimesRequested);
    for ( int idx=0; (idx < limit); idx++)
    {
        primeSum += primeGen(NULL);
    }
    printf("%d\n", primeSum);
    return 0;
}
Esempio n. 3
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;
}
Esempio n. 4
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;
}