#include <string.h>
#include "misc.h"
-
-/*
- * Usage notes:
- * * Do not call the DIVMOD_WORD macro with expressions such as array
- * subscripts, as some implementations object to this (see below).
- * * Note that none of the division methods below will cope if the
- * quotient won't fit into BIGNUM_INT_BITS. Callers should be careful
- * to avoid this case.
- * If this condition occurs, in the case of the x86 DIV instruction,
- * an overflow exception will occur, which (according to a correspondent)
- * will manifest on Windows as something like
- * 0xC0000095: Integer overflow
- * The C variant won't give the right answer, either.
- */
-
-#if defined __GNUC__ && defined __i386__
-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) \
- __asm__("div %2" : \
- "=d" (r), "=a" (q) : \
- "r" (w), "d" (hi), "a" (lo))
-#elif defined _MSC_VER && defined _M_IX86
-typedef unsigned __int32 BignumInt;
-typedef unsigned __int64 BignumDblInt;
-#define BIGNUM_INT_MASK 0xFFFFFFFFUL
-#define BIGNUM_TOP_BIT 0x80000000UL
-#define BIGNUM_INT_BITS 32
-#define MUL_WORD(w1, w2) ((BignumDblInt)w1 * w2)
-/* Note: MASM interprets array subscripts in the macro arguments as
- * assembler syntax, which gives the wrong answer. Don't supply them.
- * <http://msdn2.microsoft.com/en-us/library/bf1dw62z.aspx> */
-#define DIVMOD_WORD(q, r, hi, lo, w) do { \
- __asm mov edx, hi \
- __asm mov eax, lo \
- __asm div w \
- __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
-#define BIGNUM_TOP_BIT 0x8000U
-#define BIGNUM_INT_BITS 16
-#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)
-#endif
-
-#define BIGNUM_INT_BYTES (BIGNUM_INT_BITS / 8)
-
-#define BIGNUM_INTERNAL
-typedef BignumInt *Bignum;
-
+#include "bn-internal.h"
#include "ssh.h"
BignumInt bnZero[1] = { 0 };
}
/*
+ * Return a bignum which is the result of shifting another left by N bits.
+ * If N is negative then you get a right shift instead.
+ */
+Bignum biglsl(Bignum x, int n)
+{
+ Bignum d;
+ unsigned o, i;
+
+ if (!n || !x[0])
+ return copybn(x);
+ else if (n < 0)
+ return biglsr(x, -n);
+
+ o = n/BIGNUM_INT_BITS;
+ n %= BIGNUM_INT_BITS;
+ d = newbn(x[0] + o + !!n);
+
+ for (i = 1; i <= o; i++)
+ d[i] = 0;
+
+ if (!n) {
+ for (i = 1; i <= x[0]; i++)
+ d[o + i] = x[i];
+ } else {
+ d[o + 1] = x[1] << n;
+ for (i = 2; i <= x[0]; i--)
+ d[o + i] = (x[i] << n) | (x[i - 1] >> (BIGNUM_INT_BITS - n));
+ d[o + x[0] + 1] = x[x[0]] >> (BIGNUM_INT_BITS - n);
+ }
+
+ bn_restore_invariant(d);
+ return d;
+}
+
+/*
+ * Return a bignum which is the result of shifting another right by N bits
+ * (discarding the least significant N bits, and shifting zeroes in at the
+ * most significant end). If N is negative then you get a left shift
+ * instead.
+ */
+Bignum biglsr(Bignum x, int n)
+{
+ Bignum d;
+ unsigned o, i;
+
+ if (!n || !x[0])
+ return copybn(x);
+ else if (n < 0)
+ return biglsl(x, -n);
+
+ o = n/BIGNUM_INT_BITS;
+ n %= BIGNUM_INT_BITS;
+ d = newbn(x[0]);
+
+ if (!n) {
+ for (i = o + 1; i <= x[0]; i++)
+ d[i - o] = x[i];
+ } else {
+ d[1] = x[o + 1] >> n;
+ for (i = o + 2; i < x[0]; i++)
+ d[i - o] = x[
+ d[o + x[0] + 1] = x[x[0]] >> (BIGNUM_INT_BITS - n);
+ for (i = x[0]; i > 1; i--)
+ d[o + i] = (x[i] << n) | (x[i - 1] >> (BIGNUM_INT_BITS - n));
+ d[o + 1] = x[1] << n;
+ }
+
+ bn_restore_invariant(d);
+ return d;
+}
+
+/*
* 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.
}
/*
+ * Extract the largest power of 2 dividing x, storing it in p2, and returning
+ * the product of the remaining factors.
+ */
+static Bignum extract_p2(Bignum x, unsigned *p2)
+{
+ unsigned i, j, k, n;
+ Bignum y;
+
+ /* If x is zero then the following won't work. And if x is odd then
+ * there's nothing very useful to do.
+ */
+ if (!x[0] || (x[1] & 1)) {
+ *p2 = 0;
+ return copybn(x);
+ }
+
+ /* Find the power of two. */
+ for (i = 0; !x[i + 1]; i++);
+ for (j = 0; !((x[i + 1] >> j) & 1); j++);
+ *p2 = i*BIGNUM_INT_BITS + j;
+
+ /* Work out how big the copy should be. */
+ n = x[0] - i - 1;
+ if (x[x[0]] >> j) n++;
+
+ /* Copy and shift down. */
+ y = newbn(n);
+ for (k = 1; k <= n; k++) {
+ y[k] = x[k + i] >> j;
+ if (j && k < x[0]) y[k] |= x[k + i + 1] << (BIGNUM_INT_BITS - j);
+ }
+
+ /* Done. */
+ return y;
+}
+
+/*
+ * Kronecker symbol (a|n). The result is always in { -1, 0, +1 }, and is
+ * zero if and only if a and n have a nontrivial common factor. Most
+ * usefully, if n is prime, this is the Legendre symbol, taking the value +1
+ * if a is a quadratic residue mod n, and -1 otherwise; i.e., (a|p) ==
+ * a^{(p-1)/2} (mod p).
+ */
+int kronecker(Bignum a, Bignum n)
+{
+ unsigned s, nn;
+ int r = +1;
+ Bignum t;
+
+ /* Special case for n = 0. This is the same convention PARI uses,
+ * except that we can't represent negative numbers.
+ */
+ if (bignum_cmp(n, Zero) == 0) {
+ if (bignum_cmp(a, One) == 0) return +1;
+ else return 0;
+ }
+
+ /* Write n = 2^s t, with t odd. If s > 0 and a is even, then the answer
+ * is zero; otherwise throw in a factor of (-1)^s if a == 3 or 5 (mod 8).
+ *
+ * At this point, we have a copy of n, and must remember to free it when
+ * we're done. It's convenient to take a copy of a at the same time.
+ */
+ a = copybn(a);
+ n = extract_p2(n, &s);
+
+ if (s && (!a[0] || !(a[1] & 1))) { r = 0; goto done; }
+ else if ((s & 1) && ((a[1] & 7) == 3 || (a[1] & 7) == 5)) r = -r;
+
+ /* If n is (now) a unit then we're done. */
+ if (bignum_cmp(n, One) == 0) goto done;
+
+ /* Reduce a modulo n before we go any further. */
+ if (bignum_cmp(a, n) >= 0) { t = bigmod(a, n); freebn(a); a = t; }
+
+ /* Main loop. */
+ for (;;) {
+ if (bignum_cmp(a, Zero) == 0) { r = 0; goto done; }
+
+ /* Strip out and handle powers of two from a. */
+ t = extract_p2(a, &s); freebn(a); a = t;
+ nn = n[1] & 7;
+ if ((s & 1) && (nn == 3 || nn == 5)) r = -r;
+ if (bignum_cmp(a, One) == 0) break;
+
+ /* Swap, applying quadratic reciprocity. */
+ if ((nn & 3) == 3 && (a[1] & 3) == 3) r = -r;
+ t = bigmod(n, a); freebn(n); n = a; a = t;
+ }
+
+ /* Tidy up: we're done. */
+done:
+ freebn(a); freebn(n);
+ return r;
+}
+
+/*
+ * Modular square root. We must have p prime: extracting square roots modulo
+ * composites is equivalent to factoring (but we don't check: you'll just get
+ * the wrong answer). Returns NULL if x is not a quadratic residue mod p.
+ */
+Bignum modsqrt(Bignum x, Bignum p)
+{
+ Bignum xinv, b, c, r, t, z, X, mone;
+ unsigned i, j, s;
+
+ /* If x is not a quadratic residue then we will not go to space today. */
+ if (kronecker(x, p) != +1) return NULL;
+
+ /* We need a quadratic nonresidue from somewhere. Exactly half of all
+ * units mod p are quadratic residues, but no efficient deterministic
+ * algorithm for finding one is known. So pick at random: we don't
+ * expect this to take long.
+ */
+ z = newbn(p[0]);
+ do {
+ for (i = 1; i <= p[0]; i++) z[i] = rand();
+ z[0] = p[0]; bn_restore_invariant(z);
+ } while (kronecker(z, p) != -1);
+ b = bigmod(z, p); freebn(z);
+
+ /* We need to compute a few things before we really get started. */
+ xinv = modinv(x, p); /* x^{-1} mod p */
+ mone = bigsub(p, One); /* p - 1 == -1 (mod p) */
+ t = extract_p2(mone, &s); /* 2^s t = p - 1 */
+ c = modpow(b, t, p); /* b^t (mod p) */
+ z = bigadd(t, One); freebn(t); t = z; /* (t + 1) */
+ shift_right(t + 1, t[0], 1); if (!t[t[0]]) t[0]--;
+ r = modpow(x, t, p); /* x^{(t+1)/2} (mod p) */
+ freebn(b); freebn(mone); freebn(t);
+
+ /* OK, so how does this work anyway?
+ *
+ * We know that x^t is somewhere in the order-2^s subgroup of GF(p)^*;
+ * and g = c^{-1} is a generator for this subgroup (since we know that
+ * g^{2^{s-1}} = b^{(p-1)/2} = (b|p) = -1); so x^t = g^m for some m. In
+ * fact, we know that m is even because x is a square. Suppose we can
+ * determine m; then we know that x^t/g^m = 1, so x^{t+1}/c^m = x -- but
+ * both t + 1 and m are even, so x^{(t+1)/2}/g^{m/2} is a square root of
+ * x.
+ *
+ * Conveniently, finding the discrete log of an element X in a group of
+ * order 2^s is easy. Write X = g^m = g^{m_0+2k'}; then X^{2^{s-1}} =
+ * g^{m_0 2^{s-1}} c^{m' 2^s} = g^{m_0 2^{s-1}} is either -1 or +1,
+ * telling us that m_0 is 1 or 0 respectively. Then X/g^{m_0} =
+ * (g^2)^{m'} has order 2^{s-1} so we can continue inductively. What we
+ * end up with at the end is X/g^m.
+ *
+ * There are a few wrinkles. As we proceed through the induction, the
+ * generator for the subgroup will be c^{-2}, since we know that m is
+ * even. While we want the discrete log of X = x^t, we're actually going
+ * to keep track of r, which will eventually be x^{(t+1)/2}/g^{m/2} =
+ * x^{(t+1)/2} c^m, recovering X/g^m = r^2/x as we go. We don't actually
+ * form the discrete log explicitly, because the final result will
+ * actually be the square root we want.
+ */
+ for (i = 1; i < s; i++) {
+
+ /* Determine X. We could optimize this, only recomputing it when
+ * it's been invalidated, but that's fiddlier and this isn't
+ * performance critical.
+ */
+ z = modmul(r, r, p);
+ X = modmul(z, xinv, p);
+ freebn(z);
+
+ /* Determine X^{2^{s-1-i}}. */
+ for (j = i + 1; j < s; j++)
+ z = modmul(X, X, p), freebn(X), X = z;
+
+ /* Maybe accumulate a factor of c. */
+ if (bignum_cmp(X, One) != 0)
+ z = modmul(r, c, p), freebn(r), r = z;
+
+ /* Move on to the next smaller subgroup. */
+ z = modmul(c, c, p), freebn(c), c = z;
+ freebn(X);
+ }
+
+ /* Of course, there are two square roots of x. For predictability's sake
+ * we'll always return the one in [1..(p - 1)/2]. The other is, of
+ * course, p - r.
+ */
+ z = bigsub(p, r);
+ if (bignum_cmp(r, z) < 0)
+ freebn(z);
+ else {
+ freebn(r);
+ r = z;
+ }
+
+ /* We're done. */
+ freebn(xinv); freebn(c);
+ return r;
+}
+
+/*
* Render a bignum into decimal. Return a malloced string holding
* the decimal representation.
*/
freebn(modulus);
freebn(expected);
freebn(answer);
+ } else if (!strcmp(buf, "modsqrt")) {
+ Bignum x, p, expected, answer;
+
+ if (ptrnum != 3) {
+ printf("%d: modsqrt with %d parameters, expected 3\n", line, ptrnum);
+ exit(1);
+ }
+
+ x = bignum_from_bytes(ptrs[0], ptrs[1]-ptrs[0]);
+ p = bignum_from_bytes(ptrs[1], ptrs[2]-ptrs[1]);
+ expected = bignum_from_bytes(ptrs[2], ptrs[3]-ptrs[2]);
+ answer = modsqrt(x, p);
+ if (!answer)
+ answer = copybn(Zero);
+
+ if (bignum_cmp(expected, answer) == 0) {
+ passes++;
+ } else {
+ char *xs = bignum_decimal(x);
+ char *ps = bignum_decimal(p);
+ char *qs = bignum_decimal(answer);
+ char *ws = bignum_decimal(expected);
+
+ printf("%d: fail: sqrt(%s) mod %s gave %s expected %s\n",
+ line, xs, ps, qs, ws);
+ fails++;
+
+ sfree(xs);
+ sfree(ps);
+ sfree(qs);
+ sfree(ws);
+ }
+ freebn(p);
+ freebn(x);
+ freebn(expected);
+ freebn(answer);
} else {
printf("%d: unrecognised test keyword: '%s'\n", line, buf);
exit(1);
projects / u / mdw / putty / blobdiff