Esempio n. 1
0
int rsaencrypt(unsigned char *data, int length, struct RSAKey *key)
{
    Bignum b1, b2;
    int i;
    unsigned char *p;

    if (key->bytes < length + 4)
	return 0;		       /* RSA key too short! */

    memmove(data + key->bytes - length, data, length);
    data[0] = 0;
    data[1] = 2;

    for (i = 2; i < key->bytes - length - 1; i++) {
	do {
	    data[i] = random_byte();
	} while (data[i] == 0);
    }
    data[key->bytes - length - 1] = 0;

    b1 = bignum_from_bytes(data, key->bytes);

    b2 = modpow(b1, key->exponent, key->modulus);

    p = data;
    for (i = key->bytes; i--;) {
	*p++ = bignum_byte(b2, i);
    }

    freebn(b1);
    freebn(b2);

    return 1;
}
Esempio n. 2
0
int ec_edgenerate(struct ec_key *key, int bits, progfn_t pfn, void *pfnparam)
{
  struct ec_point *publicKey;

  if (!ec_ed_alg_and_curve_by_bits(bits, &key->publicKey.curve, &key->signalg))
    return 0;

  {
    /* EdDSA secret keys are just 32 bytes of hash preimage; the
     * 64-byte SHA-512 hash of that key will be used when signing,
     * but the form of the key stored on disk is the preimage
     * only. */
    Bignum privMax = bn_power_2(bits);
    if (!privMax)
      return 0;
    key->privateKey = bignum_random_in_range(Zero, privMax);
    freebn(privMax);
    if (!key->privateKey)
      return 0;
  }

  publicKey = ec_public(key->privateKey, key->publicKey.curve);
  if (!publicKey) {
    freebn(key->privateKey);
    key->privateKey = NULL;
    return 0;
  }

  key->publicKey.x = publicKey->x;
  key->publicKey.y = publicKey->y;
  key->publicKey.z = NULL;
  sfree(publicKey);

  return 1;
}
Esempio n. 3
0
File: sshdss.c Progetto: rdebath/sgt
static void dss_freekey(void *key) {
    struct dss_key *dss = (struct dss_key *)key;
    freebn(dss->p);
    freebn(dss->q);
    freebn(dss->g);
    freebn(dss->y);
    sfree(dss);
}
Esempio n. 4
0
void freersakey(struct RSAKey *key)
{
    if (key->modulus)
	freebn(key->modulus);
    if (key->exponent)
	freebn(key->exponent);
    if (key->private_exponent)
	freebn(key->private_exponent);
    if (key->comment)
	sfree(key->comment);
}
Esempio n. 5
0
File: sshrsa.c Progetto: rdebath/sgt
void rsaencrypt(unsigned char *data, int length, struct RSAKey *key) {
    Bignum b1, b2;
    int w, i;
    unsigned char *p;

    debug(key->exponent);

    memmove(data+key->bytes-length, data, length);
    data[0] = 0;
    data[1] = 2;

    for (i = 2; i < key->bytes-length-1; i++) {
	do {
	    data[i] = random_byte();
	} while (data[i] == 0);
    }
    data[key->bytes-length-1] = 0;

    w = (key->bytes+1)/2;

    b1 = newbn(w);
    b2 = newbn(w);

    p = data;
    for (i=1; i<=w; i++)
	b1[i] = 0;
    for (i=0; i<key->bytes; i++) {
	unsigned char byte = *p++;
	if ((key->bytes-i) & 1)
	    b1[w-i/2] |= byte;
	else
	    b1[w-i/2] |= byte<<8;
    }

    debug(b1);

    modpow(b1, key->exponent, key->modulus, b2);

    debug(b2);

    p = data;
    for (i=0; i<key->bytes; i++) {
	unsigned char b;
	if (i & 1)
	    b = b2[w-i/2] & 0xFF;
	else
	    b = b2[w-i/2] >> 8;
	*p++ = b;
    }

    freebn(b1);
    freebn(b2);
}
Esempio n. 6
0
static void *dss_createkey(unsigned char *pub_blob, int pub_len,  unsigned char *priv_blob, int priv_len)
{
    dss_key *dss;
    char *pb = (char *) priv_blob;
    char *hash;
    int hashlen;
    SHA_State s;
    unsigned char digest[20];
    Bignum ytest;

    dss = dss_newkey((char *) pub_blob, pub_len);
    if (!dss)
        return NULL;
    dss->x = getmp(&pb, &priv_len);
    if (!dss->x) {
        dss_freekey(dss);
        return NULL;
    }

    /*
     * Check the obsolete hash in the old DSS key format.
     */
    hashlen = -1;
    getstring(&pb, &priv_len, &hash, &hashlen);
    if (hashlen == 20) 
	{
	SHA_Init(&s);
	sha_mpint(&s, dss->p);
	sha_mpint(&s, dss->q);
	sha_mpint(&s, dss->g);
	SHA_Final(&s, digest);

	if (0 != memcmp(hash, digest, 20)) 
	{
	    dss_freekey(dss);
	    return NULL;
	}
    }

    /*
     * Now ensure g^x mod p really is y.
     */
    ytest = modpow(dss->g, dss->x, dss->p);
    if (0 != bignum_cmp(ytest, dss->y)) 
	{
		dss_freekey(dss);
        freebn(ytest);
		return NULL;
    }
    freebn(ytest);

    return dss;
}
Esempio n. 7
0
int RSAKey::Generate( int bits )
{
	Bignum pm1, qm1, phi_n;

	/*
	 * We don't generate e; we just use a standard one always.
	 */
	this->exponent = bignum_from_long(RSA_EXPONENT);

	/*
	 * Generate p and q: primes with combined length `bits', not
	 * congruent to 1 modulo e. (Strictly speaking, we wanted (p-1)
	 * and e to be coprime, and (q-1) and e to be coprime, but in
	 * general that's slightly more fiddly to arrange. By choosing
	 * a prime e, we can simplify the criterion.)
	 */
	this->p = primegen(bits / 2, RSA_EXPONENT, 1, NULL, 1);
	this->q = primegen(bits - bits / 2, RSA_EXPONENT, 1, NULL, 2);

	/*
	 * Ensure p > q, by swapping them if not.
	 */
	if (bignum_cmp(this->p, this->q) < 0)
		swap( p, q );

	/*
	 * Now we have p, q and e. All we need to do now is work out
	 * the other helpful quantities: n=pq, d=e^-1 mod (p-1)(q-1),
	 * and (q^-1 mod p).
	 */
	this->modulus = bigmul(this->p, this->q);
	pm1 = copybn(this->p);
	decbn(pm1);
	qm1 = copybn(this->q);
	decbn(qm1);
	phi_n = bigmul(pm1, qm1);
	freebn(pm1);
	freebn(qm1);
	this->private_exponent = modinv(this->exponent, phi_n);
	this->iqmp = modinv(this->q, this->p);

	/*
	 * Clean up temporary numbers.
	 */
	freebn(phi_n);

	return 1;
}
Esempio n. 8
0
int ec_generate(struct ec_key *key, int bits, progfn_t pfn, void *pfnparam)
{
  struct ec_point *publicKey;

  if (!ec_nist_alg_and_curve_by_bits(
          bits, &key->publicKey.curve, &key->signalg))
    return 0;

  key->privateKey = bignum_random_in_range(One, key->publicKey.curve->w.n);
  if (!key->privateKey)
    return 0;

  publicKey = ec_public(key->privateKey, key->publicKey.curve);
  if (!publicKey) {
    freebn(key->privateKey);
    key->privateKey = NULL;
    return 0;
  }

  key->publicKey.x = publicKey->x;
  key->publicKey.y = publicKey->y;
  key->publicKey.z = NULL;
  sfree(publicKey);

  return 1;
}
Esempio n. 9
0
int ec_generate(struct ec_key *key, int bits, progfn_t pfn,
                void *pfnparam)
{
    struct ec_point *publicKey;

    if (bits == 256) {
        key->publicKey.curve = ec_p256();
    } else if (bits == 384) {
        key->publicKey.curve = ec_p384();
    } else if (bits == 521) {
        key->publicKey.curve = ec_p521();
    } else {
        return 0;
    }

    key->privateKey = bignum_random_in_range(One, key->publicKey.curve->w.n);
    if (!key->privateKey) return 0;

    publicKey = ec_public(key->privateKey, key->publicKey.curve);
    if (!publicKey) {
        freebn(key->privateKey);
        key->privateKey = NULL;
        return 0;
    }

    key->publicKey.x = publicKey->x;
    key->publicKey.y = publicKey->y;
    key->publicKey.z = NULL;
    sfree(publicKey);

    return 1;
}
Esempio n. 10
0
static int rsa2_verifysig(void *key, char *sig, int siglen,
			  char *data, int datalen)
{
    struct RSAKey *rsa = (struct RSAKey *) key;
    Bignum in, out;
    char *p;
    int slen;
    int bytes, i, j, ret;
    unsigned char hash[20];

    getstring(&sig, &siglen, &p, &slen);
    if (!p || slen != 7 || memcmp(p, "ssh-rsa", 7)) {
	return 0;
    }
    in = getmp(&sig, &siglen);
    out = modpow(in, rsa->exponent, rsa->modulus);
    freebn(in);

    ret = 1;

    bytes = (bignum_bitcount(rsa->modulus)+7) / 8;
    /* Top (partial) byte should be zero. */
    if (bignum_byte(out, bytes - 1) != 0)
	ret = 0;
    /* First whole byte should be 1. */
    if (bignum_byte(out, bytes - 2) != 1)
	ret = 0;
    /* Most of the rest should be FF. */
    for (i = bytes - 3; i >= 20 + ASN1_LEN; i--) {
	if (bignum_byte(out, i) != 0xFF)
	    ret = 0;
    }
    /* Then we expect to see the asn1_weird_stuff. */
    for (i = 20 + ASN1_LEN - 1, j = 0; i >= 20; i--, j++) {
	if (bignum_byte(out, i) != asn1_weird_stuff[j])
	    ret = 0;
    }
    /* Finally, we expect to see the SHA-1 hash of the signed data. */
    SHA_Simple(data, datalen, hash);
    for (i = 19, j = 0; i >= 0; i--, j++) {
	if (bignum_byte(out, i) != hash[j])
	    ret = 0;
    }
    freebn(out);

    return ret;
}
Esempio n. 11
0
File: sshrsa.c Progetto: rdebath/sgt
static void modmult(Bignum r1, Bignum r2, Bignum modulus, Bignum result) {
    Bignum temp = newbn(modulus[0]+1);
    Bignum tmp2 = newbn(modulus[0]+1);
    int i;
    int bit, bits, digit, smallbit;

    enter((">modmult\n"));
    debug(r1);
    debug(r2);
    debug(modulus);

    for (i=1; i<=result[0]; i++)
	result[i] = 0;		       /* result := 0 */
    for (i=1; i<=temp[0]; i++)
	temp[i] = (i > r2[0] ? 0 : r2[i]);   /* temp := r2 */

    bits = 1+msb(r1);

    for (bit = 0; bit < bits; bit++) {
	digit = 1 + bit / 16;
	smallbit = bit % 16;

	debug(temp);
	if (digit <= r1[0] && (r1[digit] & (1<<smallbit))) {
	    dmsg(("bit %d\n", bit));
	    add(temp, result, tmp2);
	    if (ge(tmp2, modulus))
		sub(tmp2, modulus, result);
	    else
		add(tmp2, Zero, result);
	    debug(result);
	}

	add(temp, temp, tmp2);
	if (ge(tmp2, modulus))
	    sub(tmp2, modulus, temp);
	else
	    add(tmp2, Zero, temp);
    }

    freebn(temp);
    freebn(tmp2);

    debug(result);
    leave(("<modmult\n"));
}
Esempio n. 12
0
static void dss_freekey(void *key)
{
    dss_key *dss = (dss_key *) key;

    if (dss->p)
        freebn(dss->p);
    if (dss->q)
        freebn(dss->q);
    if (dss->g)
        freebn(dss->g);
    if (dss->y)
        freebn(dss->y);
    if (dss->x)
        freebn(dss->x);

    sfree(dss);
}
Esempio n. 13
0
static unsigned char *rsa2_sign(void *key, char *data, int datalen,
				int *siglen)
{
    struct RSAKey *rsa = (struct RSAKey *) key;
    unsigned char *bytes;
    int nbytes;
    unsigned char hash[20];
    Bignum in, out;
    int i, j;

    SHA_Simple(data, datalen, hash);

    nbytes = (bignum_bitcount(rsa->modulus) - 1) / 8;
    assert(1 <= nbytes - 20 - ASN1_LEN);
    bytes = snewn(nbytes, unsigned char);

    bytes[0] = 1;
    for (i = 1; i < nbytes - 20 - ASN1_LEN; i++)
	bytes[i] = 0xFF;
    for (i = nbytes - 20 - ASN1_LEN, j = 0; i < nbytes - 20; i++, j++)
	bytes[i] = asn1_weird_stuff[j];
    for (i = nbytes - 20, j = 0; i < nbytes; i++, j++)
	bytes[i] = hash[j];

    in = bignum_from_bytes(bytes, nbytes);
    sfree(bytes);

    out = rsa_privkey_op(in, rsa);
    freebn(in);

    nbytes = (bignum_bitcount(out) + 7) / 8;
    bytes = snewn(4 + 7 + 4 + nbytes, unsigned char);
    PUT_32BIT(bytes, 7);
    memcpy(bytes + 4, "ssh-rsa", 7);
    PUT_32BIT(bytes + 4 + 7, nbytes);
    for (i = 0; i < nbytes; i++)
	bytes[4 + 7 + 4 + i] = bignum_byte(out, nbytes - 1 - i);
    freebn(out);

    *siglen = 4 + 7 + 4 + nbytes;
    return bytes;
}
Esempio n. 14
0
bool RSAKey::Verify( const CString &data, const CString &sig ) const
{
	Bignum in, out;
	int bytes, i, j;
	unsigned char hash[20];

	in = bignum_from_bytes( (const unsigned char *) sig.data(), sig.size() );

	/* Base (in) must be smaller than the modulus. */
	if( bignum_cmp(in, this->modulus) >= 0 )
	{
		freebn(in);
		return false;
	}
	out = modpow(in, this->exponent, this->modulus);
	freebn(in);

	bool ret = true;

	bytes = (bignum_bitcount(this->modulus)+7) / 8;
	/* Top (partial) byte should be zero. */
	if (bignum_byte(out, bytes - 1) != 0)
		ret = 0;
	/* First whole byte should be 1. */
	if (bignum_byte(out, bytes - 2) != 1)
		ret = 0;
	/* Most of the rest should be FF. */
	for (i = bytes - 3; i >= 20; i--) {
		if (bignum_byte(out, i) != 0xFF)
			ret = 0;
	}
	/* Finally, we expect to see the SHA-1 hash of the signed data. */
	SHA_Simple( data.data(), data.size(), hash );
	for (i = 19, j = 0; i >= 0; i--, j++) {
		if (bignum_byte(out, i) != hash[j])
			ret = false;
	}
	freebn(out);

	return ret;
}
Esempio n. 15
0
static unsigned char *dss_sign(void *key, char *data, int datalen, int *siglen)
{
    struct dss_key *dss = (struct dss_key *) key;
    Bignum k, gkp, hash, kinv, hxr, r, s;
    unsigned char digest[20];
    unsigned char *bytes;
    int nbytes, i;

    SHA_Simple(data, datalen, digest);

    k = dss_gen_k("DSA deterministic k generator", dss->q, dss->x,
                  digest, sizeof(digest));
    kinv = modinv(k, dss->q);	       /* k^-1 mod q */
    assert(kinv);

    /*
     * Now we have k, so just go ahead and compute the signature.
     */
    gkp = modpow(dss->g, k, dss->p);   /* g^k mod p */
    r = bigmod(gkp, dss->q);	       /* r = (g^k mod p) mod q */
    freebn(gkp);

    hash = bignum_from_bytes(digest, 20);
    hxr = bigmuladd(dss->x, r, hash);  /* hash + x*r */
    s = modmul(kinv, hxr, dss->q);     /* s = k^-1 * (hash + x*r) mod q */
    freebn(hxr);
    freebn(kinv);
    freebn(k);
    freebn(hash);

    /*
     * Signature blob is
     * 
     *   string  "ssh-dss"
     *   string  two 20-byte numbers r and s, end to end
     * 
     * i.e. 4+7 + 4+40 bytes.
     */
    nbytes = 4 + 7 + 4 + 40;
    bytes = snewn(nbytes, unsigned char);
    PUT_32BIT(bytes, 7);
    memcpy(bytes + 4, "ssh-dss", 7);
    PUT_32BIT(bytes + 4 + 7, 40);
    for (i = 0; i < 20; i++) {
	bytes[4 + 7 + 4 + i] = bignum_byte(r, 19 - i);
	bytes[4 + 7 + 4 + 20 + i] = bignum_byte(s, 19 - i);
    }
    freebn(r);
    freebn(s);

    *siglen = nbytes;
    return bytes;
}
Esempio n. 16
0
/*
 * DH stage 2-epsilon: given a number f, validate it to ensure it's in
 * range. (RFC 4253 section 8: "Values of 'e' or 'f' that are not in
 * the range [1, p-1] MUST NOT be sent or accepted by either side."
 * Also, we rule out 1 and p-1 too, since that's easy to do and since
 * they lead to obviously weak keys that even a passive eavesdropper
 * can figure out.)
 */
const char *dh_validate_f(void *handle, Bignum f)
{
    struct dh_ctx *ctx = (struct dh_ctx *)handle;
    if (bignum_cmp(f, One) <= 0) {
        return "f value received is too small";
    } else {
        Bignum pm1 = bigsub(ctx->p, One);
        int cmp = bignum_cmp(f, pm1);
        freebn(pm1);
        if (cmp >= 0)
            return "f value received is too large";
    }
    return NULL;
}
Esempio n. 17
0
/*
 * DH stage 1: invent a number x between 1 and q, and compute e =
 * g^x mod p. Return e.
 * 
 * If `nbits' is greater than zero, it is used as an upper limit
 * for the number of bits in x. This is safe provided that (a) you
 * use twice as many bits in x as the number of bits you expect to
 * use in your session key, and (b) the DH group is a safe prime
 * (which SSH demands that it must be).
 * 
 * P. C. van Oorschot, M. J. Wiener
 * "On Diffie-Hellman Key Agreement with Short Exponents".
 * Advances in Cryptology: Proceedings of Eurocrypt '96
 * Springer-Verlag, May 1996.
 */
Bignum dh_create_e(void *handle, int nbits)
{
    struct dh_ctx *ctx = (struct dh_ctx *)handle;
    int i;

    int nbytes;
    unsigned char *buf;

    nbytes = ssh1_bignum_length(ctx->qmask);
    buf = snewn(nbytes, unsigned char);

    do {
	/*
	 * Create a potential x, by ANDing a string of random bytes
	 * with qmask.
	 */
	if (ctx->x)
	    freebn(ctx->x);
	if (nbits == 0 || nbits > bignum_bitcount(ctx->qmask)) {
	    ssh1_write_bignum(buf, ctx->qmask);
	    for (i = 2; i < nbytes; i++)
		buf[i] &= random_byte();
	    ssh1_read_bignum(buf, nbytes, &ctx->x);   /* can't fail */
	} else {
	    int b, nb;
	    ctx->x = bn_power_2(nbits);
	    b = nb = 0;
	    for (i = 0; i < nbits; i++) {
		if (nb == 0) {
		    nb = 8;
		    b = random_byte();
		}
		bignum_set_bit(ctx->x, i, b & 1);
		b >>= 1;
		nb--;
	    }
	}
    } while (bignum_cmp(ctx->x, One) <= 0 || bignum_cmp(ctx->x, ctx->q) >= 0);

    sfree(buf);

    /*
     * Done. Now compute e = g^x mod p.
     */
    ctx->e = modpow(ctx->g, ctx->x, ctx->p);

    return ctx->e;
}
Esempio n. 18
0
/*
 * Clean up and free a context.
 */
void dh_cleanup(void *handle)
{
    struct dh_ctx *ctx = (struct dh_ctx *)handle;
    freebn(ctx->x);
    freebn(ctx->e);
    freebn(ctx->p);
    freebn(ctx->g);
    freebn(ctx->q);
    freebn(ctx->qmask);
    sfree(ctx);
}
Esempio n. 19
0
/*
 * Verify that the public data in an RSA key matches the private
 * data. We also check the private data itself: we ensure that p >
 * q and that iqmp really is the inverse of q mod p.
 */
bool RSAKey::Check() const
{
	Bignum n, ed, pm1, qm1;
	int cmp;

	/* n must equal pq. */
	n = bigmul(this->p, this->q);
	cmp = bignum_cmp(n, this->modulus);
	freebn(n);
	if (cmp != 0)
		return 0;

	/* e * d must be congruent to 1, modulo (p-1) and modulo (q-1). */
	pm1 = copybn(this->p);
	decbn(pm1);
	ed = modmul(this->exponent, this->private_exponent, pm1);
	cmp = bignum_cmp(ed, One);
	delete [] ed;
	if (cmp != 0)
		return 0;

	qm1 = copybn(this->q);
	decbn(qm1);
	ed = modmul(this->exponent, this->private_exponent, qm1);
	cmp = bignum_cmp(ed, One);
	delete [] ed;
	if (cmp != 0)
		return 0;

	/*
	 * Ensure p > q.
	 */
	if (bignum_cmp(this->p, this->q) <= 0)
		return 0;

	/*
	 * Ensure iqmp * q is congruent to 1, modulo p.
	 */
	n = modmul(this->iqmp, this->q, this->p);
	cmp = bignum_cmp(n, One);
	delete [] n;
	if (cmp != 0)
		return 0;

	return 1;
}
Esempio n. 20
0
/*
 * Verify that the public data in an RSA key matches the private
 * data. We also check the private data itself: we ensure that p >
 * q and that iqmp really is the inverse of q mod p.
 */
int rsa_verify(struct RSAKey *key)
{
    Bignum n, ed, pm1, qm1;
    int cmp;

    /* n must equal pq. */
    n = bigmul(key->p, key->q);
    cmp = bignum_cmp(n, key->modulus);
    freebn(n);
    if (cmp != 0)
	return 0;

    /* e * d must be congruent to 1, modulo (p-1) and modulo (q-1). */
    pm1 = copybn(key->p);
    decbn(pm1);
    ed = modmul(key->exponent, key->private_exponent, pm1);
    cmp = bignum_cmp(ed, One);
    sfree(ed);
    if (cmp != 0)
	return 0;

    qm1 = copybn(key->q);
    decbn(qm1);
    ed = modmul(key->exponent, key->private_exponent, qm1);
    cmp = bignum_cmp(ed, One);
    sfree(ed);
    if (cmp != 0)
	return 0;

    /*
     * Ensure p > q.
     */
    if (bignum_cmp(key->p, key->q) <= 0)
	return 0;

    /*
     * Ensure iqmp * q is congruent to 1, modulo p.
     */
    n = modmul(key->iqmp, key->q, key->p);
    cmp = bignum_cmp(n, One);
    sfree(n);
    if (cmp != 0)
	return 0;

    return 1;
}
Esempio n. 21
0
RSAKey::~RSAKey()
{
	if(p)
		freebn(p);
	if(q)
		freebn(q);
	if(modulus)
		freebn(modulus);
	if(exponent)
		freebn(exponent);
	if(private_exponent)
		freebn(private_exponent);
	if(iqmp)
		freebn(iqmp);
}
Esempio n. 22
0
int rsa_generate(struct RSAKey *key, struct RSAAux *aux, int bits,
                 progfn_t pfn, void *pfnparam) {
    Bignum pm1, qm1, phi_n;

    /*
     * Set up the phase limits for the progress report. We do this
     * by passing minus the phase number.
     *
     * For prime generation: our initial filter finds things
     * coprime to everything below 2^16. Computing the product of
     * (p-1)/p for all prime p below 2^16 gives about 20.33; so
     * among B-bit integers, one in every 20.33 will get through
     * the initial filter to be a candidate prime.
     *
     * Meanwhile, we are searching for primes in the region of 2^B;
     * since pi(x) ~ x/log(x), when x is in the region of 2^B, the
     * prime density will be d/dx pi(x) ~ 1/log(B), i.e. about
     * 1/0.6931B. So the chance of any given candidate being prime
     * is 20.33/0.6931B, which is roughly 29.34 divided by B.
     *
     * So now we have this probability P, we're looking at an
     * exponential distribution with parameter P: we will manage in
     * one attempt with probability P, in two with probability
     * P(1-P), in three with probability P(1-P)^2, etc. The
     * probability that we have still not managed to find a prime
     * after N attempts is (1-P)^N.
     * 
     * We therefore inform the progress indicator of the number B
     * (29.34/B), so that it knows how much to increment by each
     * time. We do this in 16-bit fixed point, so 29.34 becomes
     * 0x1D.57C4.
     */
    pfn(pfnparam, -1, -0x1D57C4/(bits/2));
    pfn(pfnparam, -2, -0x1D57C4/(bits-bits/2));
    pfn(pfnparam, -3, 5);

    /*
     * We don't generate e; we just use a standard one always.
     */
    key->exponent = bignum_from_short(RSA_EXPONENT);

    /*
     * Generate p and q: primes with combined length `bits', not
     * congruent to 1 modulo e. (Strictly speaking, we wanted (p-1)
     * and e to be coprime, and (q-1) and e to be coprime, but in
     * general that's slightly more fiddly to arrange. By choosing
     * a prime e, we can simplify the criterion.)
     */
    aux->p = primegen(bits/2, RSA_EXPONENT, 1, 1, pfn, pfnparam);
    aux->q = primegen(bits - bits/2, RSA_EXPONENT, 1, 2, pfn, pfnparam);

    /*
     * Ensure p > q, by swapping them if not.
     */
    if (bignum_cmp(aux->p, aux->q) < 0) {
        Bignum t = aux->p;
        aux->p = aux->q;
        aux->q = t;
    }

    /*
     * Now we have p, q and e. All we need to do now is work out
     * the other helpful quantities: n=pq, d=e^-1 mod (p-1)(q-1),
     * and (q^-1 mod p).
     */
    pfn(pfnparam, 3, 1);
    key->modulus = bigmul(aux->p, aux->q);
    pfn(pfnparam, 3, 2);
    pm1 = copybn(aux->p);
    decbn(pm1);
    qm1 = copybn(aux->q);
    decbn(qm1);
    phi_n = bigmul(pm1, qm1);
    pfn(pfnparam, 3, 3);
    freebn(pm1);
    freebn(qm1);
    key->private_exponent = modinv(key->exponent, phi_n);
    pfn(pfnparam, 3, 4);
    aux->iqmp = modinv(aux->q, aux->p);
    pfn(pfnparam, 3, 5);

    /*
     * Clean up temporary numbers.
     */
    freebn(phi_n);

    return 1;
}
Esempio n. 23
0
/*
 * This function is a wrapper on modpow(). It has the same effect
 * as modpow(), but employs RSA blinding to protect against timing
 * attacks.
 */
static Bignum rsa_privkey_op(Bignum input, struct RSAKey *key)
{
    Bignum random, random_encrypted, random_inverse;
    Bignum input_blinded, ret_blinded;
    Bignum ret;

    SHA512_State ss;
    unsigned char digest512[64];
    int digestused = lenof(digest512);
    int hashseq = 0;

    /*
     * Start by inventing a random number chosen uniformly from the
     * range 2..modulus-1. (We do this by preparing a random number
     * of the right length and retrying if it's greater than the
     * modulus, to prevent any potential Bleichenbacher-like
     * attacks making use of the uneven distribution within the
     * range that would arise from just reducing our number mod n.
     * There are timing implications to the potential retries, of
     * course, but all they tell you is the modulus, which you
     * already knew.)
     * 
     * To preserve determinism and avoid Pageant needing to share
     * the random number pool, we actually generate this `random'
     * number by hashing stuff with the private key.
     */
    while (1) {
	int bits, byte, bitsleft, v;
	random = copybn(key->modulus);
	/*
	 * Find the topmost set bit. (This function will return its
	 * index plus one.) Then we'll set all bits from that one
	 * downwards randomly.
	 */
	bits = bignum_bitcount(random);
	byte = 0;
	bitsleft = 0;
	while (bits--) {
	    if (bitsleft <= 0) {
		bitsleft = 8;
		/*
		 * Conceptually the following few lines are equivalent to
		 *    byte = random_byte();
		 */
		if (digestused >= lenof(digest512)) {
		    unsigned char seqbuf[4];
		    PUT_32BIT(seqbuf, hashseq);
		    pSHA512_Init(&ss);
		    SHA512_Bytes(&ss, "RSA deterministic blinding", 26);
		    SHA512_Bytes(&ss, seqbuf, sizeof(seqbuf));
		    sha512_mpint(&ss, key->private_exponent);
		    pSHA512_Final(&ss, digest512);
		    hashseq++;

		    /*
		     * Now hash that digest plus the signature
		     * input.
		     */
		    pSHA512_Init(&ss);
		    SHA512_Bytes(&ss, digest512, sizeof(digest512));
		    sha512_mpint(&ss, input);
		    pSHA512_Final(&ss, digest512);

		    digestused = 0;
		}
		byte = digest512[digestused++];
	    }
	    v = byte & 1;
	    byte >>= 1;
	    bitsleft--;
	    bignum_set_bit(random, bits, v);
	}

	/*
	 * Now check that this number is strictly greater than
	 * zero, and strictly less than modulus.
	 */
	if (bignum_cmp(random, Zero) <= 0 ||
	    bignum_cmp(random, key->modulus) >= 0) {
	    freebn(random);
	    continue;
	} else {
	    break;
	}
    }

    /*
     * RSA blinding relies on the fact that (xy)^d mod n is equal
     * to (x^d mod n) * (y^d mod n) mod n. We invent a random pair
     * y and y^d; then we multiply x by y, raise to the power d mod
     * n as usual, and divide by y^d to recover x^d. Thus an
     * attacker can't correlate the timing of the modpow with the
     * input, because they don't know anything about the number
     * that was input to the actual modpow.
     * 
     * The clever bit is that we don't have to do a huge modpow to
     * get y and y^d; we will use the number we just invented as
     * _y^d_, and use the _public_ exponent to compute (y^d)^e = y
     * from it, which is much faster to do.
     */
    random_encrypted = modpow(random, key->exponent, key->modulus);
    random_inverse = modinv(random, key->modulus);
    input_blinded = modmul(input, random_encrypted, key->modulus);
    ret_blinded = modpow(input_blinded, key->private_exponent, key->modulus);
    ret = modmul(ret_blinded, random_inverse, key->modulus);

    freebn(ret_blinded);
    freebn(input_blinded);
    freebn(random_inverse);
    freebn(random_encrypted);
    freebn(random);

    return ret;
}
Esempio n. 24
0
Bignum *dss_gen_k(const char *id_string, Bignum modulus, Bignum private_key,
                  unsigned char *digest, int digest_len)
{
    /*
     * The basic DSS signing algorithm is:
     * 
     *  - invent a random k between 1 and q-1 (exclusive).
     *  - Compute r = (g^k mod p) mod q.
     *  - Compute s = k^-1 * (hash + x*r) mod q.
     * 
     * This has the dangerous properties that:
     * 
     *  - if an attacker in possession of the public key _and_ the
     *    signature (for example, the host you just authenticated
     *    to) can guess your k, he can reverse the computation of s
     *    and work out x = r^-1 * (s*k - hash) mod q. That is, he
     *    can deduce the private half of your key, and masquerade
     *    as you for as long as the key is still valid.
     * 
     *  - since r is a function purely of k and the public key, if
     *    the attacker only has a _range of possibilities_ for k
     *    it's easy for him to work through them all and check each
     *    one against r; he'll never be unsure of whether he's got
     *    the right one.
     * 
     *  - if you ever sign two different hashes with the same k, it
     *    will be immediately obvious because the two signatures
     *    will have the same r, and moreover an attacker in
     *    possession of both signatures (and the public key of
     *    course) can compute k = (hash1-hash2) * (s1-s2)^-1 mod q,
     *    and from there deduce x as before.
     * 
     *  - the Bleichenbacher attack on DSA makes use of methods of
     *    generating k which are significantly non-uniformly
     *    distributed; in particular, generating a 160-bit random
     *    number and reducing it mod q is right out.
     * 
     * For this reason we must be pretty careful about how we
     * generate our k. Since this code runs on Windows, with no
     * particularly good system entropy sources, we can't trust our
     * RNG itself to produce properly unpredictable data. Hence, we
     * use a totally different scheme instead.
     * 
     * What we do is to take a SHA-512 (_big_) hash of the private
     * key x, and then feed this into another SHA-512 hash that
     * also includes the message hash being signed. That is:
     * 
     *   proto_k = SHA512 ( SHA512(x) || SHA160(message) )
     * 
     * This number is 512 bits long, so reducing it mod q won't be
     * noticeably non-uniform. So
     * 
     *   k = proto_k mod q
     * 
     * This has the interesting property that it's _deterministic_:
     * signing the same hash twice with the same key yields the
     * same signature.
     * 
     * Despite this determinism, it's still not predictable to an
     * attacker, because in order to repeat the SHA-512
     * construction that created it, the attacker would have to
     * know the private key value x - and by assumption he doesn't,
     * because if he knew that he wouldn't be attacking k!
     *
     * (This trick doesn't, _per se_, protect against reuse of k.
     * Reuse of k is left to chance; all it does is prevent
     * _excessively high_ chances of reuse of k due to entropy
     * problems.)
     * 
     * Thanks to Colin Plumb for the general idea of using x to
     * ensure k is hard to guess, and to the Cambridge University
     * Computer Security Group for helping to argue out all the
     * fine details.
     */
    SHA512_State ss;
    unsigned char digest512[64];
    Bignum proto_k, k;

    /*
     * Hash some identifying text plus x.
     */
    SHA512_Init(&ss);
    SHA512_Bytes(&ss, id_string, strlen(id_string) + 1);
    sha512_mpint(&ss, private_key);
    SHA512_Final(&ss, digest512);

    /*
     * Now hash that digest plus the message hash.
     */
    SHA512_Init(&ss);
    SHA512_Bytes(&ss, digest512, sizeof(digest512));
    SHA512_Bytes(&ss, digest, digest_len);

    while (1) {
        SHA512_State ss2 = ss;         /* structure copy */
        SHA512_Final(&ss2, digest512);

        smemclr(&ss2, sizeof(ss2));

        /*
         * Now convert the result into a bignum, and reduce it mod q.
         */
        proto_k = bignum_from_bytes(digest512, 64);
        k = bigmod(proto_k, modulus);
        freebn(proto_k);

        if (bignum_cmp(k, One) != 0 && bignum_cmp(k, Zero) != 0) {
            smemclr(&ss, sizeof(ss));
            smemclr(digest512, sizeof(digest512));
            return k;
        }

        /* Very unlikely we get here, but if so, k was unsuitable. */
        freebn(k);
        /* Perturb the hash to think of a different k. */
        SHA512_Bytes(&ss, "x", 1);
        /* Go round and try again. */
    }
}
Esempio n. 25
0
int dsa_generate(struct dss_key *key, int bits, progfn_t pfn,
		 void *pfnparam)
{
    Bignum qm1, power, g, h, tmp;
    unsigned pfirst, qfirst;
    int progress;

    /*
     * Set up the phase limits for the progress report. We do this
     * by passing minus the phase number.
     *
     * For prime generation: our initial filter finds things
     * coprime to everything below 2^16. Computing the product of
     * (p-1)/p for all prime p below 2^16 gives about 20.33; so
     * among B-bit integers, one in every 20.33 will get through
     * the initial filter to be a candidate prime.
     *
     * Meanwhile, we are searching for primes in the region of 2^B;
     * since pi(x) ~ x/log(x), when x is in the region of 2^B, the
     * prime density will be d/dx pi(x) ~ 1/log(B), i.e. about
     * 1/0.6931B. So the chance of any given candidate being prime
     * is 20.33/0.6931B, which is roughly 29.34 divided by B.
     *
     * So now we have this probability P, we're looking at an
     * exponential distribution with parameter P: we will manage in
     * one attempt with probability P, in two with probability
     * P(1-P), in three with probability P(1-P)^2, etc. The
     * probability that we have still not managed to find a prime
     * after N attempts is (1-P)^N.
     * 
     * We therefore inform the progress indicator of the number B
     * (29.34/B), so that it knows how much to increment by each
     * time. We do this in 16-bit fixed point, so 29.34 becomes
     * 0x1D.57C4.
     */
    pfn(pfnparam, PROGFN_PHASE_EXTENT, 1, 0x2800);
    pfn(pfnparam, PROGFN_EXP_PHASE, 1, -0x1D57C4 / 160);
    pfn(pfnparam, PROGFN_PHASE_EXTENT, 2, 0x40 * bits);
    pfn(pfnparam, PROGFN_EXP_PHASE, 2, -0x1D57C4 / bits);

    /*
     * In phase three we are finding an order-q element of the
     * multiplicative group of p, by finding an element whose order
     * is _divisible_ by q and raising it to the power of (p-1)/q.
     * _Most_ elements will have order divisible by q, since for a
     * start phi(p) of them will be primitive roots. So
     * realistically we don't need to set this much below 1 (64K).
     * Still, we'll set it to 1/2 (32K) to be on the safe side.
     */
    pfn(pfnparam, PROGFN_PHASE_EXTENT, 3, 0x2000);
    pfn(pfnparam, PROGFN_EXP_PHASE, 3, -32768);

    /*
     * In phase four we are finding an element x between 1 and q-1
     * (exclusive), by inventing 160 random bits and hoping they
     * come out to a plausible number; so assuming q is uniformly
     * distributed between 2^159 and 2^160, the chance of any given
     * attempt succeeding is somewhere between 0.5 and 1. Lacking
     * the energy to arrange to be able to specify this probability
     * _after_ generating q, we'll just set it to 0.75.
     */
    pfn(pfnparam, PROGFN_PHASE_EXTENT, 4, 0x2000);
    pfn(pfnparam, PROGFN_EXP_PHASE, 4, -49152);

    pfn(pfnparam, PROGFN_READY, 0, 0);

    invent_firstbits(&pfirst, &qfirst);
    /*
     * Generate q: a prime of length 160.
     */
    key->q = primegen(160, 2, 2, NULL, 1, pfn, pfnparam, qfirst);
    /*
     * Now generate p: a prime of length `bits', such that p-1 is
     * divisible by q.
     */
    key->p = primegen(bits-160, 2, 2, key->q, 2, pfn, pfnparam, pfirst);

    /*
     * Next we need g. Raise 2 to the power (p-1)/q modulo p, and
     * if that comes out to one then try 3, then 4 and so on. As
     * soon as we hit a non-unit (and non-zero!) one, that'll do
     * for g.
     */
    power = bigdiv(key->p, key->q);    /* this is floor(p/q) == (p-1)/q */
    h = bignum_from_long(1);
    progress = 0;
    while (1) {
	pfn(pfnparam, PROGFN_PROGRESS, 3, ++progress);
	g = modpow(h, power, key->p);
	if (bignum_cmp(g, One) > 0)
	    break;		       /* got one */
	tmp = h;
	h = bignum_add_long(h, 1);
	freebn(tmp);
    }
    key->g = g;
    freebn(h);

    /*
     * Now we're nearly done. All we need now is our private key x,
     * which should be a number between 1 and q-1 exclusive, and
     * our public key y = g^x mod p.
     */
    qm1 = copybn(key->q);
    decbn(qm1);
    progress = 0;
    while (1) {
	int i, v, byte, bitsleft;
	Bignum x;

	pfn(pfnparam, PROGFN_PROGRESS, 4, ++progress);
	x = bn_power_2(159);
	byte = 0;
	bitsleft = 0;

	for (i = 0; i < 160; i++) {
	    if (bitsleft <= 0)
		bitsleft = 8, byte = random_byte();
	    v = byte & 1;
	    byte >>= 1;
	    bitsleft--;
	    bignum_set_bit(x, i, v);
	}

	if (bignum_cmp(x, One) <= 0 || bignum_cmp(x, qm1) >= 0) {
	    freebn(x);
	    continue;
	} else {
	    key->x = x;
	    break;
	}
    }
    freebn(qm1);

    key->y = modpow(key->g, key->x, key->p);

    return 1;
}
Esempio n. 26
0
/*
 * Generate a prime. We arrange to select a prime with the property
 * (prime % modulus) != residue (to speed up use in RSA).
 */
Bignum primegen(int bits, int modulus, int residue,
                int phase, progfn_t pfn, void *pfnparam) {
    int i, k, v, byte, bitsleft, check, checks;
    unsigned long delta, moduli[NPRIMES+1], residues[NPRIMES+1];
    Bignum p, pm1, q, wqp, wqp2;
    int progress = 0;

    byte = 0; bitsleft = 0;

    STARTOVER:

    pfn(pfnparam, phase, ++progress);

    /*
     * Generate a k-bit random number with top and bottom bits set.
     */
    p = newbn((bits+15)/16);
    for (i = 0; i < bits; i++) {
        if (i == 0 || i == bits-1)
            v = 1;
        else {
            if (bitsleft <= 0)
                bitsleft = 8; byte = random_byte();
            v = byte & 1;
            byte >>= 1;
            bitsleft--;
        }
        bignum_set_bit(p, i, v);
    }

    /*
     * Ensure this random number is coprime to the first few
     * primes, by repeatedly adding 2 to it until it is.
     */
    for (i = 0; i < NPRIMES; i++) {
        moduli[i] = primes[i];
        residues[i] = bignum_mod_short(p, primes[i]);
    }
    moduli[NPRIMES] = modulus;
    residues[NPRIMES] = (bignum_mod_short(p, (unsigned short)modulus)
                         + modulus - residue);
    delta = 0;
    while (1) {
        for (i = 0; i < (sizeof(moduli) / sizeof(*moduli)); i++)
            if (!((residues[i] + delta) % moduli[i]))
                break;
        if (i < (sizeof(moduli) / sizeof(*moduli))) {/* we broke */
            delta += 2;
            if (delta < 2) {
                freebn(p);
                goto STARTOVER;
            }
            continue;
        }
        break;
    }
    q = p;
    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++);   /* 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 = newbn((bits+15)/16);
            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);
            }
            if (bignum_cmp(w, p) >= 0 || bignum_cmp(w, Zero) == 0) {
                freebn(w);
                continue;
            }
            break;
        }

        pfn(pfnparam, 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(p);
        freebn(pm1);
        freebn(q);
        goto STARTOVER;
    }

    /*
     * We have a prime!
     */
    freebn(q);
    freebn(pm1);
    return p;
}
Esempio n. 27
0
static int dss_verifysig(void *key, char *sig, int siglen,
			 char *data, int datalen)
{
    struct dss_key *dss = (struct dss_key *) key;
    char *p;
    int slen;
    char hash[20];
    Bignum r, s, w, gu1p, yu2p, gu1yu2p, u1, u2, sha, v;
    int ret;

    if (!dss->p)
	return 0;

#ifdef DEBUG_DSS
    {
	int i;
	printf("sig:");
	for (i = 0; i < siglen; i++)
	    printf("  %02x", (unsigned char) (sig[i]));
	printf("\n");
    }
#endif
    /*
     * Commercial SSH (2.0.13) and OpenSSH disagree over the format
     * of a DSA signature. OpenSSH is in line with the IETF drafts:
     * it uses a string "ssh-dss", followed by a 40-byte string
     * containing two 160-bit integers end-to-end. Commercial SSH
     * can't be bothered with the header bit, and considers a DSA
     * signature blob to be _just_ the 40-byte string containing
     * the two 160-bit integers. We tell them apart by measuring
     * the length: length 40 means the commercial-SSH bug, anything
     * else is assumed to be IETF-compliant.
     */
    if (siglen != 40) {		       /* bug not present; read admin fields */
	getstring(&sig, &siglen, &p, &slen);
	if (!p || slen != 7 || memcmp(p, "ssh-dss", 7)) {
	    return 0;
	}
	sig += 4, siglen -= 4;	       /* skip yet another length field */
    }
    r = get160(&sig, &siglen);
    s = get160(&sig, &siglen);
    if (!r || !s)
	return 0;

    /*
     * Step 1. w <- s^-1 mod q.
     */
    w = modinv(s, dss->q);

    /*
     * Step 2. u1 <- SHA(message) * w mod q.
     */
    SHA_Simple(data, datalen, (unsigned char *)hash);
    p = hash;
    slen = 20;
    sha = get160(&p, &slen);
    u1 = modmul(sha, w, dss->q);

    /*
     * Step 3. u2 <- r * w mod q.
     */
    u2 = modmul(r, w, dss->q);

    /*
     * Step 4. v <- (g^u1 * y^u2 mod p) mod q.
     */
    gu1p = modpow(dss->g, u1, dss->p);
    yu2p = modpow(dss->y, u2, dss->p);
    gu1yu2p = modmul(gu1p, yu2p, dss->p);
    v = modmul(gu1yu2p, One, dss->q);

    /*
     * Step 5. v should now be equal to r.
     */

    ret = !bignum_cmp(v, r);

    freebn(w);
    freebn(sha);
    freebn(gu1p);
    freebn(yu2p);
    freebn(gu1yu2p);
    freebn(v);
    freebn(r);
    freebn(s);

    return ret;
}
Esempio n. 28
0
static unsigned char *dss_sign(void *key, char *data, int datalen, int *siglen)
{
    /*
     * The basic DSS signing algorithm is:
     * 
     *  - invent a random k between 1 and q-1 (exclusive).
     *  - Compute r = (g^k mod p) mod q.
     *  - Compute s = k^-1 * (hash + x*r) mod q.
     * 
     * This has the dangerous properties that:
     * 
     *  - if an attacker in possession of the public key _and_ the
     *    signature (for example, the host you just authenticated
     *    to) can guess your k, he can reverse the computation of s
     *    and work out x = r^-1 * (s*k - hash) mod q. That is, he
     *    can deduce the private half of your key, and masquerade
     *    as you for as long as the key is still valid.
     * 
     *  - since r is a function purely of k and the public key, if
     *    the attacker only has a _range of possibilities_ for k
     *    it's easy for him to work through them all and check each
     *    one against r; he'll never be unsure of whether he's got
     *    the right one.
     * 
     *  - if you ever sign two different hashes with the same k, it
     *    will be immediately obvious because the two signatures
     *    will have the same r, and moreover an attacker in
     *    possession of both signatures (and the public key of
     *    course) can compute k = (hash1-hash2) * (s1-s2)^-1 mod q,
     *    and from there deduce x as before.
     * 
     *  - the Bleichenbacher attack on DSA makes use of methods of
     *    generating k which are significantly non-uniformly
     *    distributed; in particular, generating a 160-bit random
     *    number and reducing it mod q is right out.
     * 
     * For this reason we must be pretty careful about how we
     * generate our k. Since this code runs on Windows, with no
     * particularly good system entropy sources, we can't trust our
     * RNG itself to produce properly unpredictable data. Hence, we
     * use a totally different scheme instead.
     * 
     * What we do is to take a SHA-512 (_big_) hash of the private
     * key x, and then feed this into another SHA-512 hash that
     * also includes the message hash being signed. That is:
     * 
     *   proto_k = SHA512 ( SHA512(x) || SHA160(message) )
     * 
     * This number is 512 bits long, so reducing it mod q won't be
     * noticeably non-uniform. So
     * 
     *   k = proto_k mod q
     * 
     * This has the interesting property that it's _deterministic_:
     * signing the same hash twice with the same key yields the
     * same signature.
     * 
     * Despite this determinism, it's still not predictable to an
     * attacker, because in order to repeat the SHA-512
     * construction that created it, the attacker would have to
     * know the private key value x - and by assumption he doesn't,
     * because if he knew that he wouldn't be attacking k!
     *
     * (This trick doesn't, _per se_, protect against reuse of k.
     * Reuse of k is left to chance; all it does is prevent
     * _excessively high_ chances of reuse of k due to entropy
     * problems.)
     * 
     * Thanks to Colin Plumb for the general idea of using x to
     * ensure k is hard to guess, and to the Cambridge University
     * Computer Security Group for helping to argue out all the
     * fine details.
     */
    struct dss_key *dss = (struct dss_key *) key;
    SHA512_State ss;
    unsigned char digest[20], digest512[64];
    Bignum proto_k, k, gkp, hash, kinv, hxr, r, s;
    unsigned char *bytes;
    int nbytes, i;

    SHA_Simple(data, datalen, digest);

    /*
     * Hash some identifying text plus x.
     */
    SHA512_Init(&ss);
    SHA512_Bytes(&ss, "DSA deterministic k generator", 30);
    sha512_mpint(&ss, dss->x);
    SHA512_Final(&ss, digest512);

    /*
     * Now hash that digest plus the message hash.
     */
    SHA512_Init(&ss);
    SHA512_Bytes(&ss, digest512, sizeof(digest512));
    SHA512_Bytes(&ss, digest, sizeof(digest));
    SHA512_Final(&ss, digest512);

    memset(&ss, 0, sizeof(ss));

    /*
     * Now convert the result into a bignum, and reduce it mod q.
     */
    proto_k = bignum_from_bytes(digest512, 64);
    k = bigmod(proto_k, dss->q);
    freebn(proto_k);

    memset(digest512, 0, sizeof(digest512));

    /*
     * Now we have k, so just go ahead and compute the signature.
     */
    gkp = modpow(dss->g, k, dss->p);   /* g^k mod p */
    r = bigmod(gkp, dss->q);	       /* r = (g^k mod p) mod q */
    freebn(gkp);

    hash = bignum_from_bytes(digest, 20);
    kinv = modinv(k, dss->q);	       /* k^-1 mod q */
    hxr = bigmuladd(dss->x, r, hash);  /* hash + x*r */
    s = modmul(kinv, hxr, dss->q);     /* s = k^-1 * (hash + x*r) mod q */
    freebn(hxr);
    freebn(kinv);
    freebn(hash);

    /*
     * Signature blob is
     * 
     *   string  "ssh-dss"
     *   string  two 20-byte numbers r and s, end to end
     * 
     * i.e. 4+7 + 4+40 bytes.
     */
    nbytes = 4 + 7 + 4 + 40;
    bytes = snewn(nbytes, unsigned char);
    PUT_32BIT(bytes, 7);
    memcpy(bytes + 4, "ssh-dss", 7);
    PUT_32BIT(bytes + 4 + 7, 40);
    for (i = 0; i < 20; i++) {
	bytes[4 + 7 + 4 + i] = bignum_byte(r, 19 - i);
	bytes[4 + 7 + 4 + 20 + i] = bignum_byte(s, 19 - i);
    }
    freebn(r);
    freebn(s);

    *siglen = nbytes;
    return bytes;
}
Esempio n. 29
0
/*
 * Return a malloc'ed chunk of memory containing the public blob of
 * an RSA key, as given in the agent protocol (modulus bits,
 * exponent, modulus).
 */
int rsakey_pubblob(const Filename *filename, void **blob, int *bloblen,
		   char **commentptr, const char **errorstr)
{
    FILE *fp;
    char buf[64];
    struct RSAKey key;
    int ret;
    const char *error = NULL;

    /* Default return if we fail. */
    *blob = NULL;
    *bloblen = 0;
    ret = 0;

    fp = f_open(filename, "rb", FALSE);
    if (!fp) {
	error = "can't open file";
	goto end;
    }

    /*
     * Read the first line of the file and see if it's a v1 private
     * key file.
     */
    if (fgets(buf, sizeof(buf), fp) && !strcmp(buf, rsa_signature)) {
	memset(&key, 0, sizeof(key));
	if (loadrsakey_main(fp, &key, TRUE, commentptr, NULL, &error)) {
	    *blob = rsa_public_blob(&key, bloblen);
	    freersakey(&key);
	    ret = 1;
	}
	fp = NULL; /* loadrsakey_main unconditionally closes fp */
    } else {
        /*
         * Try interpreting the file as an SSH-1 public key.
         */
        char *line, *p, *bitsp, *expp, *modp, *commentp;

        rewind(fp);
        line = chomp(fgetline(fp));
        p = line;

        bitsp = p;
        p += strspn(p, "0123456789");
        if (*p != ' ')
            goto not_public_either;
        *p++ = '\0';

        expp = p;
        p += strspn(p, "0123456789");
        if (*p != ' ')
            goto not_public_either;
        *p++ = '\0';

        modp = p;
        p += strspn(p, "0123456789");
        if (*p) {
            if (*p != ' ')
                goto not_public_either;
            *p++ = '\0';
            commentp = p;
        } else {
            commentp = NULL;
        }

	memset(&key, 0, sizeof(key));
        key.exponent = bignum_from_decimal(expp);
        key.modulus = bignum_from_decimal(modp);
        if (atoi(bitsp) != bignum_bitcount(key.modulus)) {
            freebn(key.exponent);
            freebn(key.modulus);
            sfree(line);
            error = "key bit count does not match in SSH-1 public key file";
            goto end;
        }
        if (commentptr)
            *commentptr = commentp ? dupstr(commentp) : NULL;
        *blob = rsa_public_blob(&key, bloblen);
        freersakey(&key);
        return 1;

      not_public_either:
        sfree(line);
	error = "not an SSH-1 RSA file";
    }

  end:
    if (fp)
	fclose(fp);
    if ((ret != 1) && errorstr)
	*errorstr = error;
    return ret;
}