diff options
Diffstat (limited to 'tools/plink/sshbn.c')
| -rw-r--r-- | tools/plink/sshbn.c | 884 |
1 files changed, 855 insertions, 29 deletions
diff --git a/tools/plink/sshbn.c b/tools/plink/sshbn.c index e9ff0cde4..51cecdf2b 100644 --- a/tools/plink/sshbn.c +++ b/tools/plink/sshbn.c @@ -51,7 +51,34 @@ typedef unsigned __int64 BignumDblInt; __asm mov r, edx \
__asm mov q, eax \
} while(0)
+#elif defined _LP64
+/* 64-bit architectures can do 32x32->64 chunks at a time */
+typedef unsigned int BignumInt;
+typedef unsigned long BignumDblInt;
+#define BIGNUM_INT_MASK 0xFFFFFFFFU
+#define BIGNUM_TOP_BIT 0x80000000U
+#define BIGNUM_INT_BITS 32
+#define MUL_WORD(w1, w2) ((BignumDblInt)w1 * w2)
+#define DIVMOD_WORD(q, r, hi, lo, w) do { \
+ BignumDblInt n = (((BignumDblInt)hi) << BIGNUM_INT_BITS) | lo; \
+ q = n / w; \
+ r = n % w; \
+} while (0)
+#elif defined _LLP64
+/* 64-bit architectures in which unsigned long is 32 bits, not 64 */
+typedef unsigned long BignumInt;
+typedef unsigned long long BignumDblInt;
+#define BIGNUM_INT_MASK 0xFFFFFFFFUL
+#define BIGNUM_TOP_BIT 0x80000000UL
+#define BIGNUM_INT_BITS 32
+#define MUL_WORD(w1, w2) ((BignumDblInt)w1 * w2)
+#define DIVMOD_WORD(q, r, hi, lo, w) do { \
+ BignumDblInt n = (((BignumDblInt)hi) << BIGNUM_INT_BITS) | lo; \
+ q = n / w; \
+ r = n % w; \
+} while (0)
#else
+/* Fallback for all other cases */
typedef unsigned short BignumInt;
typedef unsigned long BignumDblInt;
#define BIGNUM_INT_MASK 0xFFFFU
@@ -133,29 +160,432 @@ Bignum bn_power_2(int n) }
/*
+ * Internal addition. Sets c = a - b, where 'a', 'b' and 'c' are all
+ * big-endian arrays of 'len' BignumInts. Returns a BignumInt carried
+ * off the top.
+ */
+static BignumInt internal_add(const BignumInt *a, const BignumInt *b,
+ BignumInt *c, int len)
+{
+ int i;
+ BignumDblInt carry = 0;
+
+ for (i = len-1; i >= 0; i--) {
+ carry += (BignumDblInt)a[i] + b[i];
+ c[i] = (BignumInt)carry;
+ carry >>= BIGNUM_INT_BITS;
+ }
+
+ return (BignumInt)carry;
+}
+
+/*
+ * Internal subtraction. Sets c = a - b, where 'a', 'b' and 'c' are
+ * all big-endian arrays of 'len' BignumInts. Any borrow from the top
+ * is ignored.
+ */
+static void internal_sub(const BignumInt *a, const BignumInt *b,
+ BignumInt *c, int len)
+{
+ int i;
+ BignumDblInt carry = 1;
+
+ for (i = len-1; i >= 0; i--) {
+ carry += (BignumDblInt)a[i] + (b[i] ^ BIGNUM_INT_MASK);
+ c[i] = (BignumInt)carry;
+ carry >>= BIGNUM_INT_BITS;
+ }
+}
+
+/*
* Compute c = a * b.
* Input is in the first len words of a and b.
* Result is returned in the first 2*len words of c.
+ *
+ * 'scratch' must point to an array of BignumInt of size at least
+ * mul_compute_scratch(len). (This covers the needs of internal_mul
+ * and all its recursive calls to itself.)
*/
-static void internal_mul(BignumInt *a, BignumInt *b,
- BignumInt *c, int len)
+#define KARATSUBA_THRESHOLD 50
+static int mul_compute_scratch(int len)
{
- int i, j;
- BignumDblInt t;
-
- for (j = 0; j < 2 * len; j++)
- c[j] = 0;
-
- for (i = len - 1; i >= 0; i--) {
- t = 0;
- for (j = len - 1; j >= 0; j--) {
- t += MUL_WORD(a[i], (BignumDblInt) b[j]);
- t += (BignumDblInt) c[i + j + 1];
- c[i + j + 1] = (BignumInt) t;
- t = t >> BIGNUM_INT_BITS;
- }
- c[i] = (BignumInt) t;
+ int ret = 0;
+ while (len > KARATSUBA_THRESHOLD) {
+ int toplen = len/2, botlen = len - toplen; /* botlen is the bigger */
+ int midlen = botlen + 1;
+ ret += 4*midlen;
+ len = midlen;
}
+ return ret;
+}
+static void internal_mul(const BignumInt *a, const BignumInt *b,
+ BignumInt *c, int len, BignumInt *scratch)
+{
+ if (len > KARATSUBA_THRESHOLD) {
+ int i;
+
+ /*
+ * Karatsuba divide-and-conquer algorithm. Cut each input in
+ * half, so that it's expressed as two big 'digits' in a giant
+ * base D:
+ *
+ * a = a_1 D + a_0
+ * b = b_1 D + b_0
+ *
+ * Then the product is of course
+ *
+ * ab = a_1 b_1 D^2 + (a_1 b_0 + a_0 b_1) D + a_0 b_0
+ *
+ * and we compute the three coefficients by recursively
+ * calling ourself to do half-length multiplications.
+ *
+ * The clever bit that makes this worth doing is that we only
+ * need _one_ half-length multiplication for the central
+ * coefficient rather than the two that it obviouly looks
+ * like, because we can use a single multiplication to compute
+ *
+ * (a_1 + a_0) (b_1 + b_0) = a_1 b_1 + a_1 b_0 + a_0 b_1 + a_0 b_0
+ *
+ * and then we subtract the other two coefficients (a_1 b_1
+ * and a_0 b_0) which we were computing anyway.
+ *
+ * Hence we get to multiply two numbers of length N in about
+ * three times as much work as it takes to multiply numbers of
+ * length N/2, which is obviously better than the four times
+ * as much work it would take if we just did a long
+ * conventional multiply.
+ */
+
+ int toplen = len/2, botlen = len - toplen; /* botlen is the bigger */
+ int midlen = botlen + 1;
+ BignumDblInt carry;
+#ifdef KARA_DEBUG
+ int i;
+#endif
+
+ /*
+ * The coefficients a_1 b_1 and a_0 b_0 just avoid overlapping
+ * in the output array, so we can compute them immediately in
+ * place.
+ */
+
+#ifdef KARA_DEBUG
+ printf("a1,a0 = 0x");
+ for (i = 0; i < len; i++) {
+ if (i == toplen) printf(", 0x");
+ printf("%0*x", BIGNUM_INT_BITS/4, a[i]);
+ }
+ printf("\n");
+ printf("b1,b0 = 0x");
+ for (i = 0; i < len; i++) {
+ if (i == toplen) printf(", 0x");
+ printf("%0*x", BIGNUM_INT_BITS/4, b[i]);
+ }
+ printf("\n");
+#endif
+
+ /* a_1 b_1 */
+ internal_mul(a, b, c, toplen, scratch);
+#ifdef KARA_DEBUG
+ printf("a1b1 = 0x");
+ for (i = 0; i < 2*toplen; i++) {
+ printf("%0*x", BIGNUM_INT_BITS/4, c[i]);
+ }
+ printf("\n");
+#endif
+
+ /* a_0 b_0 */
+ internal_mul(a + toplen, b + toplen, c + 2*toplen, botlen, scratch);
+#ifdef KARA_DEBUG
+ printf("a0b0 = 0x");
+ for (i = 0; i < 2*botlen; i++) {
+ printf("%0*x", BIGNUM_INT_BITS/4, c[2*toplen+i]);
+ }
+ printf("\n");
+#endif
+
+ /* Zero padding. midlen exceeds toplen by at most 2, so just
+ * zero the first two words of each input and the rest will be
+ * copied over. */
+ scratch[0] = scratch[1] = scratch[midlen] = scratch[midlen+1] = 0;
+
+ for (i = 0; i < toplen; i++) {
+ scratch[midlen - toplen + i] = a[i]; /* a_1 */
+ scratch[2*midlen - toplen + i] = b[i]; /* b_1 */
+ }
+
+ /* compute a_1 + a_0 */
+ scratch[0] = internal_add(scratch+1, a+toplen, scratch+1, botlen);
+#ifdef KARA_DEBUG
+ printf("a1plusa0 = 0x");
+ for (i = 0; i < midlen; i++) {
+ printf("%0*x", BIGNUM_INT_BITS/4, scratch[i]);
+ }
+ printf("\n");
+#endif
+ /* compute b_1 + b_0 */
+ scratch[midlen] = internal_add(scratch+midlen+1, b+toplen,
+ scratch+midlen+1, botlen);
+#ifdef KARA_DEBUG
+ printf("b1plusb0 = 0x");
+ for (i = 0; i < midlen; i++) {
+ printf("%0*x", BIGNUM_INT_BITS/4, scratch[midlen+i]);
+ }
+ printf("\n");
+#endif
+
+ /*
+ * Now we can do the third multiplication.
+ */
+ internal_mul(scratch, scratch + midlen, scratch + 2*midlen, midlen,
+ scratch + 4*midlen);
+#ifdef KARA_DEBUG
+ printf("a1plusa0timesb1plusb0 = 0x");
+ for (i = 0; i < 2*midlen; i++) {
+ printf("%0*x", BIGNUM_INT_BITS/4, scratch[2*midlen+i]);
+ }
+ printf("\n");
+#endif
+
+ /*
+ * Now we can reuse the first half of 'scratch' to compute the
+ * sum of the outer two coefficients, to subtract from that
+ * product to obtain the middle one.
+ */
+ scratch[0] = scratch[1] = scratch[2] = scratch[3] = 0;
+ for (i = 0; i < 2*toplen; i++)
+ scratch[2*midlen - 2*toplen + i] = c[i];
+ scratch[1] = internal_add(scratch+2, c + 2*toplen,
+ scratch+2, 2*botlen);
+#ifdef KARA_DEBUG
+ printf("a1b1plusa0b0 = 0x");
+ for (i = 0; i < 2*midlen; i++) {
+ printf("%0*x", BIGNUM_INT_BITS/4, scratch[i]);
+ }
+ printf("\n");
+#endif
+
+ internal_sub(scratch + 2*midlen, scratch,
+ scratch + 2*midlen, 2*midlen);
+#ifdef KARA_DEBUG
+ printf("a1b0plusa0b1 = 0x");
+ for (i = 0; i < 2*midlen; i++) {
+ printf("%0*x", BIGNUM_INT_BITS/4, scratch[2*midlen+i]);
+ }
+ printf("\n");
+#endif
+
+ /*
+ * And now all we need to do is to add that middle coefficient
+ * back into the output. We may have to propagate a carry
+ * further up the output, but we can be sure it won't
+ * propagate right the way off the top.
+ */
+ carry = internal_add(c + 2*len - botlen - 2*midlen,
+ scratch + 2*midlen,
+ c + 2*len - botlen - 2*midlen, 2*midlen);
+ i = 2*len - botlen - 2*midlen - 1;
+ while (carry) {
+ assert(i >= 0);
+ carry += c[i];
+ c[i] = (BignumInt)carry;
+ carry >>= BIGNUM_INT_BITS;
+ i--;
+ }
+#ifdef KARA_DEBUG
+ printf("ab = 0x");
+ for (i = 0; i < 2*len; i++) {
+ printf("%0*x", BIGNUM_INT_BITS/4, c[i]);
+ }
+ printf("\n");
+#endif
+
+ } else {
+ int i;
+ BignumInt carry;
+ BignumDblInt t;
+ const BignumInt *ap, *bp;
+ BignumInt *cp, *cps;
+
+ /*
+ * Multiply in the ordinary O(N^2) way.
+ */
+
+ for (i = 0; i < 2 * len; i++)
+ c[i] = 0;
+
+ for (cps = c + 2*len, ap = a + len; ap-- > a; cps--) {
+ carry = 0;
+ for (cp = cps, bp = b + len; cp--, bp-- > b ;) {
+ t = (MUL_WORD(*ap, *bp) + carry) + *cp;
+ *cp = (BignumInt) t;
+ carry = (BignumInt)(t >> BIGNUM_INT_BITS);
+ }
+ *cp = carry;
+ }
+ }
+}
+
+/*
+ * Variant form of internal_mul used for the initial step of
+ * Montgomery reduction. Only bothers outputting 'len' words
+ * (everything above that is thrown away).
+ */
+static void internal_mul_low(const BignumInt *a, const BignumInt *b,
+ BignumInt *c, int len, BignumInt *scratch)
+{
+ if (len > KARATSUBA_THRESHOLD) {
+ int i;
+
+ /*
+ * Karatsuba-aware version of internal_mul_low. As before, we
+ * express each input value as a shifted combination of two
+ * halves:
+ *
+ * a = a_1 D + a_0
+ * b = b_1 D + b_0
+ *
+ * Then the full product is, as before,
+ *
+ * ab = a_1 b_1 D^2 + (a_1 b_0 + a_0 b_1) D + a_0 b_0
+ *
+ * Provided we choose D on the large side (so that a_0 and b_0
+ * are _at least_ as long as a_1 and b_1), we don't need the
+ * topmost term at all, and we only need half of the middle
+ * term. So there's no point in doing the proper Karatsuba
+ * optimisation which computes the middle term using the top
+ * one, because we'd take as long computing the top one as
+ * just computing the middle one directly.
+ *
+ * So instead, we do a much more obvious thing: we call the
+ * fully optimised internal_mul to compute a_0 b_0, and we
+ * recursively call ourself to compute the _bottom halves_ of
+ * a_1 b_0 and a_0 b_1, each of which we add into the result
+ * in the obvious way.
+ *
+ * In other words, there's no actual Karatsuba _optimisation_
+ * in this function; the only benefit in doing it this way is
+ * that we call internal_mul proper for a large part of the
+ * work, and _that_ can optimise its operation.
+ */
+
+ int toplen = len/2, botlen = len - toplen; /* botlen is the bigger */
+
+ /*
+ * Scratch space for the various bits and pieces we're going
+ * to be adding together: we need botlen*2 words for a_0 b_0
+ * (though we may end up throwing away its topmost word), and
+ * toplen words for each of a_1 b_0 and a_0 b_1. That adds up
+ * to exactly 2*len.
+ */
+
+ /* a_0 b_0 */
+ internal_mul(a + toplen, b + toplen, scratch + 2*toplen, botlen,
+ scratch + 2*len);
+
+ /* a_1 b_0 */
+ internal_mul_low(a, b + len - toplen, scratch + toplen, toplen,
+ scratch + 2*len);
+
+ /* a_0 b_1 */
+ internal_mul_low(a + len - toplen, b, scratch, toplen,
+ scratch + 2*len);
+
+ /* Copy the bottom half of the big coefficient into place */
+ for (i = 0; i < botlen; i++)
+ c[toplen + i] = scratch[2*toplen + botlen + i];
+
+ /* Add the two small coefficients, throwing away the returned carry */
+ internal_add(scratch, scratch + toplen, scratch, toplen);
+
+ /* And add that to the large coefficient, leaving the result in c. */
+ internal_add(scratch, scratch + 2*toplen + botlen - toplen,
+ c, toplen);
+
+ } else {
+ int i;
+ BignumInt carry;
+ BignumDblInt t;
+ const BignumInt *ap, *bp;
+ BignumInt *cp, *cps;
+
+ /*
+ * Multiply in the ordinary O(N^2) way.
+ */
+
+ for (i = 0; i < len; i++)
+ c[i] = 0;
+
+ for (cps = c + len, ap = a + len; ap-- > a; cps--) {
+ carry = 0;
+ for (cp = cps, bp = b + len; bp--, cp-- > c ;) {
+ t = (MUL_WORD(*ap, *bp) + carry) + *cp;
+ *cp = (BignumInt) t;
+ carry = (BignumInt)(t >> BIGNUM_INT_BITS);
+ }
+ }
+ }
+}
+
+/*
+ * Montgomery reduction. Expects x to be a big-endian array of 2*len
+ * BignumInts whose value satisfies 0 <= x < rn (where r = 2^(len *
+ * BIGNUM_INT_BITS) is the Montgomery base). Returns in the same array
+ * a value x' which is congruent to xr^{-1} mod n, and satisfies 0 <=
+ * x' < n.
+ *
+ * 'n' and 'mninv' should be big-endian arrays of 'len' BignumInts
+ * each, containing respectively n and the multiplicative inverse of
+ * -n mod r.
+ *
+ * 'tmp' is an array of BignumInt used as scratch space, of length at
+ * least 3*len + mul_compute_scratch(len).
+ */
+static void monty_reduce(BignumInt *x, const BignumInt *n,
+ const BignumInt *mninv, BignumInt *tmp, int len)
+{
+ int i;
+ BignumInt carry;
+
+ /*
+ * Multiply x by (-n)^{-1} mod r. This gives us a value m such
+ * that mn is congruent to -x mod r. Hence, mn+x is an exact
+ * multiple of r, and is also (obviously) congruent to x mod n.
+ */
+ internal_mul_low(x + len, mninv, tmp, len, tmp + 3*len);
+
+ /*
+ * Compute t = (mn+x)/r in ordinary, non-modular, integer
+ * arithmetic. By construction this is exact, and is congruent mod
+ * n to x * r^{-1}, i.e. the answer we want.
+ *
+ * The following multiply leaves that answer in the _most_
+ * significant half of the 'x' array, so then we must shift it
+ * down.
+ */
+ internal_mul(tmp, n, tmp+len, len, tmp + 3*len);
+ carry = internal_add(x, tmp+len, x, 2*len);
+ for (i = 0; i < len; i++)
+ x[len + i] = x[i], x[i] = 0;
+
+ /*
+ * Reduce t mod n. This doesn't require a full-on division by n,
+ * but merely a test and single optional subtraction, since we can
+ * show that 0 <= t < 2n.
+ *
+ * Proof:
+ * + we computed m mod r, so 0 <= m < r.
+ * + so 0 <= mn < rn, obviously
+ * + hence we only need 0 <= x < rn to guarantee that 0 <= mn+x < 2rn
+ * + yielding 0 <= (mn+x)/r < 2n as required.
+ */
+ if (!carry) {
+ for (i = 0; i < len; i++)
+ if (x[len + i] != n[i])
+ break;
+ }
+ if (carry || i >= len || x[len + i] > n[i])
+ internal_sub(x+len, n, x+len, len);
}
static void internal_add_shifted(BignumInt *number,
@@ -279,13 +709,13 @@ static void internal_mod(BignumInt *a, int alen, }
/*
- * Compute (base ^ exp) % mod.
+ * Compute (base ^ exp) % mod, the pedestrian way.
*/
-Bignum modpow(Bignum base_in, Bignum exp, Bignum mod)
+Bignum modpow_simple(Bignum base_in, Bignum exp, Bignum mod)
{
- BignumInt *a, *b, *n, *m;
+ BignumInt *a, *b, *n, *m, *scratch;
int mshift;
- int mlen, i, j;
+ int mlen, scratchlen, i, j;
Bignum base, result;
/*
@@ -332,6 +762,10 @@ Bignum modpow(Bignum base_in, Bignum exp, Bignum mod) a[i] = 0;
a[2 * mlen - 1] = 1;
+ /* Scratch space for multiplies */
+ scratchlen = mul_compute_scratch(mlen);
+ scratch = snewn(scratchlen, BignumInt);
+
/* Skip leading zero bits of exp. */
i = 0;
j = BIGNUM_INT_BITS-1;
@@ -346,10 +780,10 @@ Bignum modpow(Bignum base_in, Bignum exp, Bignum mod) /* Main computation */
while (i < (int)exp[0]) {
while (j >= 0) {
- internal_mul(a + mlen, a + mlen, b, mlen);
+ internal_mul(a + mlen, a + mlen, b, mlen, scratch);
internal_mod(b, mlen * 2, m, mlen, NULL, 0);
if ((exp[exp[0] - i] & (1 << j)) != 0) {
- internal_mul(b + mlen, n, a, mlen);
+ internal_mul(b + mlen, n, a, mlen, scratch);
internal_mod(a, mlen * 2, m, mlen, NULL, 0);
} else {
BignumInt *t;
@@ -384,6 +818,9 @@ Bignum modpow(Bignum base_in, Bignum exp, Bignum mod) for (i = 0; i < 2 * mlen; i++)
a[i] = 0;
sfree(a);
+ for (i = 0; i < scratchlen; i++)
+ scratch[i] = 0;
+ sfree(scratch);
for (i = 0; i < 2 * mlen; i++)
b[i] = 0;
sfree(b);
@@ -400,14 +837,165 @@ Bignum modpow(Bignum base_in, Bignum exp, Bignum mod) }
/*
+ * Compute (base ^ exp) % mod. Uses the Montgomery multiplication
+ * technique where possible, falling back to modpow_simple otherwise.
+ */
+Bignum modpow(Bignum base_in, Bignum exp, Bignum mod)
+{
+ BignumInt *a, *b, *x, *n, *mninv, *scratch;
+ int len, scratchlen, i, j;
+ Bignum base, base2, r, rn, inv, result;
+
+ /*
+ * The most significant word of mod needs to be non-zero. It
+ * should already be, but let's make sure.
+ */
+ assert(mod[mod[0]] != 0);
+
+ /*
+ * mod had better be odd, or we can't do Montgomery multiplication
+ * using a power of two at all.
+ */
+ if (!(mod[1] & 1))
+ return modpow_simple(base_in, exp, mod);
+
+ /*
+ * Make sure the base is smaller than the modulus, by reducing
+ * it modulo the modulus if not.
+ */
+ base = bigmod(base_in, mod);
+
+ /*
+ * Compute the inverse of n mod r, for monty_reduce. (In fact we
+ * want the inverse of _minus_ n mod r, but we'll sort that out
+ * below.)
+ */
+ len = mod[0];
+ r = bn_power_2(BIGNUM_INT_BITS * len);
+ inv = modinv(mod, r);
+
+ /*
+ * Multiply the base by r mod n, to get it into Montgomery
+ * representation.
+ */
+ base2 = modmul(base, r, mod);
+ freebn(base);
+ base = base2;
+
+ rn = bigmod(r, mod); /* r mod n, i.e. Montgomerified 1 */
+
+ freebn(r); /* won't need this any more */
+
+ /*
+ * Set up internal arrays of the right lengths, in big-endian
+ * format, containing the base, the modulus, and the modulus's
+ * inverse.
+ */
+ n = snewn(len, BignumInt);
+ for (j = 0; j < len; j++)
+ n[len - 1 - j] = mod[j + 1];
+
+ mninv = snewn(len, BignumInt);
+ for (j = 0; j < len; j++)
+ mninv[len - 1 - j] = (j < (int)inv[0] ? inv[j + 1] : 0);
+ freebn(inv); /* we don't need this copy of it any more */
+ /* Now negate mninv mod r, so it's the inverse of -n rather than +n. */
+ x = snewn(len, BignumInt);
+ for (j = 0; j < len; j++)
+ x[j] = 0;
+ internal_sub(x, mninv, mninv, len);
+
+ /* x = snewn(len, BignumInt); */ /* already done above */
+ for (j = 0; j < len; j++)
+ x[len - 1 - j] = (j < (int)base[0] ? base[j + 1] : 0);
+ freebn(base); /* we don't need this copy of it any more */
+
+ a = snewn(2*len, BignumInt);
+ b = snewn(2*len, BignumInt);
+ for (j = 0; j < len; j++)
+ a[2*len - 1 - j] = (j < (int)rn[0] ? rn[j + 1] : 0);
+ freebn(rn);
+
+ /* Scratch space for multiplies */
+ scratchlen = 3*len + mul_compute_scratch(len);
+ scratch = snewn(scratchlen, BignumInt);
+
+ /* Skip leading zero bits of exp. */
+ i = 0;
+ j = BIGNUM_INT_BITS-1;
+ while (i < (int)exp[0] && (exp[exp[0] - i] & (1 << j)) == 0) {
+ j--;
+ if (j < 0) {
+ i++;
+ j = BIGNUM_INT_BITS-1;
+ }
+ }
+
+ /* Main computation */
+ while (i < (int)exp[0]) {
+ while (j >= 0) {
+ internal_mul(a + len, a + len, b, len, scratch);
+ monty_reduce(b, n, mninv, scratch, len);
+ if ((exp[exp[0] - i] & (1 << j)) != 0) {
+ internal_mul(b + len, x, a, len, scratch);
+ monty_reduce(a, n, mninv, scratch, len);
+ } else {
+ BignumInt *t;
+ t = a;
+ a = b;
+ b = t;
+ }
+ j--;
+ }
+ i++;
+ j = BIGNUM_INT_BITS-1;
+ }
+
+ /*
+ * Final monty_reduce to get back from the adjusted Montgomery
+ * representation.
+ */
+ monty_reduce(a, n, mninv, scratch, len);
+
+ /* Copy result to buffer */
+ result = newbn(mod[0]);
+ for (i = 0; i < len; i++)
+ result[result[0] - i] = a[i + len];
+ while (result[0] > 1 && result[result[0]] == 0)
+ result[0]--;
+
+ /* Free temporary arrays */
+ for (i = 0; i < scratchlen; i++)
+ scratch[i] = 0;
+ sfree(scratch);
+ for (i = 0; i < 2 * len; i++)
+ a[i] = 0;
+ sfree(a);
+ for (i = 0; i < 2 * len; i++)
+ b[i] = 0;
+ sfree(b);
+ for (i = 0; i < len; i++)
+ mninv[i] = 0;
+ sfree(mninv);
+ for (i = 0; i < len; i++)
+ n[i] = 0;
+ sfree(n);
+ for (i = 0; i < len; i++)
+ x[i] = 0;
+ sfree(x);
+
+ return result;
+}
+
+/*
* Compute (p * q) % mod.
* The most significant word of mod MUST be non-zero.
* We assume that the result array is the same size as the mod array.
*/
Bignum modmul(Bignum p, Bignum q, Bignum mod)
{
- BignumInt *a, *n, *m, *o;
- int mshift;
+ BignumInt *a, *n, *m, *o, *scratch;
+ int mshift, scratchlen;
int pqlen, mlen, rlen, i, j;
Bignum result;
@@ -449,8 +1037,12 @@ Bignum modmul(Bignum p, Bignum q, Bignum mod) /* Allocate a of size 2*pqlen for result */
a = snewn(2 * pqlen, BignumInt);
+ /* Scratch space for multiplies */
+ scratchlen = mul_compute_scratch(pqlen);
+ scratch = snewn(scratchlen, BignumInt);
+
/* Main computation */
- internal_mul(n, o, a, pqlen);
+ internal_mul(n, o, a, pqlen, scratch);
internal_mod(a, pqlen * 2, m, mlen, NULL, 0);
/* Fixup result in case the modulus was shifted */
@@ -472,6 +1064,9 @@ Bignum modmul(Bignum p, Bignum q, Bignum mod) result[0]--;
/* Free temporary arrays */
+ for (i = 0; i < scratchlen; i++)
+ scratch[i] = 0;
+ sfree(scratch);
for (i = 0; i < 2 * pqlen; i++)
a[i] = 0;
sfree(a);
@@ -760,18 +1355,21 @@ Bignum bigmuladd(Bignum a, Bignum b, Bignum addend) int alen = a[0], blen = b[0];
int mlen = (alen > blen ? alen : blen);
int rlen, i, maxspot;
+ int wslen;
BignumInt *workspace;
Bignum ret;
- /* mlen space for a, mlen space for b, 2*mlen for result */
- workspace = snewn(mlen * 4, BignumInt);
+ /* mlen space for a, mlen space for b, 2*mlen for result,
+ * plus scratch space for multiplication */
+ wslen = mlen * 4 + mul_compute_scratch(mlen);
+ workspace = snewn(wslen, BignumInt);
for (i = 0; i < mlen; i++) {
workspace[0 * mlen + i] = (mlen - i <= (int)a[0] ? a[mlen - i] : 0);
workspace[1 * mlen + i] = (mlen - i <= (int)b[0] ? b[mlen - i] : 0);
}
internal_mul(workspace + 0 * mlen, workspace + 1 * mlen,
- workspace + 2 * mlen, mlen);
+ workspace + 2 * mlen, mlen, workspace + 4 * mlen);
/* now just copy the result back */
rlen = alen + blen + 1;
@@ -800,6 +1398,8 @@ Bignum bigmuladd(Bignum a, Bignum b, Bignum addend) }
ret[0] = maxspot;
+ for (i = 0; i < wslen; i++)
+ workspace[i] = 0;
sfree(workspace);
return ret;
}
@@ -813,6 +1413,69 @@ Bignum bigmul(Bignum a, Bignum b) }
/*
+ * Simple addition.
+ */
+Bignum bigadd(Bignum a, Bignum b)
+{
+ int alen = a[0], blen = b[0];
+ int rlen = (alen > blen ? alen : blen) + 1;
+ int i, maxspot;
+ Bignum ret;
+ BignumDblInt carry;
+
+ ret = newbn(rlen);
+
+ carry = 0;
+ maxspot = 0;
+ for (i = 1; i <= rlen; i++) {
+ carry += (i <= (int)a[0] ? a[i] : 0);
+ carry += (i <= (int)b[0] ? b[i] : 0);
+ ret[i] = (BignumInt) carry & BIGNUM_INT_MASK;
+ carry >>= BIGNUM_INT_BITS;
+ if (ret[i] != 0 && i > maxspot)
+ maxspot = i;
+ }
+ ret[0] = maxspot;
+
+ return ret;
+}
+
+/*
+ * Subtraction. Returns a-b, or NULL if the result would come out
+ * negative (recall that this entire bignum module only handles
+ * positive numbers).
+ */
+Bignum bigsub(Bignum a, Bignum b)
+{
+ int alen = a[0], blen = b[0];
+ int rlen = (alen > blen ? alen : blen);
+ int i, maxspot;
+ Bignum ret;
+ BignumDblInt carry;
+
+ ret = newbn(rlen);
+
+ carry = 1;
+ maxspot = 0;
+ for (i = 1; i <= rlen; i++) {
+ carry += (i <= (int)a[0] ? a[i] : 0);
+ carry += (i <= (int)b[0] ? b[i] ^ BIGNUM_INT_MASK : BIGNUM_INT_MASK);
+ ret[i] = (BignumInt) carry & BIGNUM_INT_MASK;
+ carry >>= BIGNUM_INT_BITS;
+ if (ret[i] != 0 && i > maxspot)
+ maxspot = i;
+ }
+ ret[0] = maxspot;
+
+ if (!carry) {
+ freebn(ret);
+ return NULL;
+ }
+
+ return ret;
+}
+
+/*
* Create a bignum which is the bitmask covering another one. That
* is, the smallest integer which is >= N and is also one less than
* a power of two.
@@ -1090,3 +1753,166 @@ char *bignum_decimal(Bignum x) sfree(workspace);
return ret;
}
+
+#ifdef TESTBN
+
+#include <stdio.h>
+#include <stdlib.h>
+#include <ctype.h>
+
+/*
+ * gcc -g -O0 -DTESTBN -o testbn sshbn.c misc.c -I unix -I charset
+ *
+ * Then feed to this program's standard input the output of
+ * testdata/bignum.py .
+ */
+
+void modalfatalbox(char *p, ...)
+{
+ va_list ap;
+ fprintf(stderr, "FATAL ERROR: ");
+ va_start(ap, p);
+ vfprintf(stderr, p, ap);
+ va_end(ap);
+ fputc('\n', stderr);
+ exit(1);
+}
+
+#define fromxdigit(c) ( (c)>'9' ? ((c)&0xDF) - 'A' + 10 : (c) - '0' )
+
+int main(int argc, char **argv)
+{
+ char *buf;
+ int line = 0;
+ int passes = 0, fails = 0;
+
+ while ((buf = fgetline(stdin)) != NULL) {
+ int maxlen = strlen(buf);
+ unsigned char *data = snewn(maxlen, unsigned char);
+ unsigned char *ptrs[5], *q;
+ int ptrnum;
+ char *bufp = buf;
+
+ line++;
+
+ q = data;
+ ptrnum = 0;
+
+ while (*bufp && !isspace((unsigned char)*bufp))
+ bufp++;
+ if (bufp)
+ *bufp++ = '\0';
+
+ while (*bufp) {
+ char *start, *end;
+ int i;
+
+ while (*bufp && !isxdigit((unsigned char)*bufp))
+ bufp++;
+ start = bufp;
+
+ if (!*bufp)
+ break;
+
+ while (*bufp && isxdigit((unsigned char)*bufp))
+ bufp++;
+ end = bufp;
+
+ if (ptrnum >= lenof(ptrs))
+ break;
+ ptrs[ptrnum++] = q;
+
+ for (i = -((end - start) & 1); i < end-start; i += 2) {
+ unsigned char val = (i < 0 ? 0 : fromxdigit(start[i]));
+ val = val * 16 + fromxdigit(start[i+1]);
+ *q++ = val;
+ }
+
+ ptrs[ptrnum] = q;
+ }
+
+ if (!strcmp(buf, "mul")) {
+ Bignum a, b, c, p;
+
+ if (ptrnum != 3) {
+ printf("%d: mul with %d parameters, expected 3\n", line);
+ exit(1);
+ }
+ a = bignum_from_bytes(ptrs[0], ptrs[1]-ptrs[0]);
+ b = bignum_from_bytes(ptrs[1], ptrs[2]-ptrs[1]);
+ c = bignum_from_bytes(ptrs[2], ptrs[3]-ptrs[2]);
+ p = bigmul(a, b);
+
+ if (bignum_cmp(c, p) == 0) {
+ passes++;
+ } else {
+ char *as = bignum_decimal(a);
+ char *bs = bignum_decimal(b);
+ char *cs = bignum_decimal(c);
+ char *ps = bignum_decimal(p);
+
+ printf("%d: fail: %s * %s gave %s expected %s\n",
+ line, as, bs, ps, cs);
+ fails++;
+
+ sfree(as);
+ sfree(bs);
+ sfree(cs);
+ sfree(ps);
+ }
+ freebn(a);
+ freebn(b);
+ freebn(c);
+ freebn(p);
+ } else if (!strcmp(buf, "pow")) {
+ Bignum base, expt, modulus, expected, answer;
+
+ if (ptrnum != 4) {
+ printf("%d: mul with %d parameters, expected 3\n", line);
+ exit(1);
+ }
+
+ base = bignum_from_bytes(ptrs[0], ptrs[1]-ptrs[0]);
+ expt = bignum_from_bytes(ptrs[1], ptrs[2]-ptrs[1]);
+ modulus = bignum_from_bytes(ptrs[2], ptrs[3]-ptrs[2]);
+ expected = bignum_from_bytes(ptrs[3], ptrs[4]-ptrs[3]);
+ answer = modpow(base, expt, modulus);
+
+ if (bignum_cmp(expected, answer) == 0) {
+ passes++;
+ } else {
+ char *as = bignum_decimal(base);
+ char *bs = bignum_decimal(expt);
+ char *cs = bignum_decimal(modulus);
+ char *ds = bignum_decimal(answer);
+ char *ps = bignum_decimal(expected);
+
+ printf("%d: fail: %s ^ %s mod %s gave %s expected %s\n",
+ line, as, bs, cs, ds, ps);
+ fails++;
+
+ sfree(as);
+ sfree(bs);
+ sfree(cs);
+ sfree(ds);
+ sfree(ps);
+ }
+ freebn(base);
+ freebn(expt);
+ freebn(modulus);
+ freebn(expected);
+ freebn(answer);
+ } else {
+ printf("%d: unrecognised test keyword: '%s'\n", line, buf);
+ exit(1);
+ }
+
+ sfree(buf);
+ sfree(data);
+ }
+
+ printf("passed %d failed %d total %d\n", passes, fails, passes+fails);
+ return fails != 0;
+}
+
+#endif
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