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@@ -63,7 +63,34 @@ typedef unsigned __int64 BignumDblInt;
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__asm mov q, eax \
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} while(0)
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#endif
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+#elif defined _LP64
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+/* 64-bit architectures can do 32x32->64 chunks at a time */
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+typedef unsigned int BignumInt;
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+typedef unsigned long BignumDblInt;
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+#define BIGNUM_INT_MASK 0xFFFFFFFFU
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+#define BIGNUM_TOP_BIT 0x80000000U
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+#define BIGNUM_INT_BITS 32
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+#define MUL_WORD(w1, w2) ((BignumDblInt)w1 * w2)
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+#define DIVMOD_WORD(q, r, hi, lo, w) do { \
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+ BignumDblInt n = (((BignumDblInt)hi) << BIGNUM_INT_BITS) | lo; \
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+ q = n / w; \
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+ r = n % w; \
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+} while (0)
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+#elif defined _LLP64
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+/* 64-bit architectures in which unsigned long is 32 bits, not 64 */
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+typedef unsigned long BignumInt;
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+typedef unsigned long long BignumDblInt;
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+#define BIGNUM_INT_MASK 0xFFFFFFFFUL
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+#define BIGNUM_TOP_BIT 0x80000000UL
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+#define BIGNUM_INT_BITS 32
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+#define MUL_WORD(w1, w2) ((BignumDblInt)w1 * w2)
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+#define DIVMOD_WORD(q, r, hi, lo, w) do { \
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+ BignumDblInt n = (((BignumDblInt)hi) << BIGNUM_INT_BITS) | lo; \
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+ q = n / w; \
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+ r = n % w; \
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+} while (0)
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#else
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+/* Fallback for all other cases */
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typedef unsigned short BignumInt;
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typedef unsigned long BignumDblInt;
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#define BIGNUM_INT_MASK 0xFFFFU
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@@ -144,32 +171,431 @@ Bignum bn_power_2(int n)
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return ret;
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}
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+/*
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+ * Internal addition. Sets c = a - b, where 'a', 'b' and 'c' are all
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+ * big-endian arrays of 'len' BignumInts. Returns a BignumInt carried
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+ * off the top.
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+ */
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+static BignumInt internal_add(const BignumInt *a, const BignumInt *b,
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+ BignumInt *c, int len)
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+{
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+ int i;
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+ BignumDblInt carry = 0;
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+
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+ for (i = len-1; i >= 0; i--) {
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+ carry += (BignumDblInt)a[i] + b[i];
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+ c[i] = (BignumInt)carry;
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+ carry >>= BIGNUM_INT_BITS;
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+ }
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+
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+ return (BignumInt)carry;
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+}
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+
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+/*
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+ * Internal subtraction. Sets c = a - b, where 'a', 'b' and 'c' are
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+ * all big-endian arrays of 'len' BignumInts. Any borrow from the top
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+ * is ignored.
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+ */
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+static void internal_sub(const BignumInt *a, const BignumInt *b,
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+ BignumInt *c, int len)
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+{
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+ int i;
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+ BignumDblInt carry = 1;
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+
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+ for (i = len-1; i >= 0; i--) {
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+ carry += (BignumDblInt)a[i] + (b[i] ^ BIGNUM_INT_MASK);
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+ c[i] = (BignumInt)carry;
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+ carry >>= BIGNUM_INT_BITS;
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+ }
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+}
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+
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/*
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* Compute c = a * b.
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* Input is in the first len words of a and b.
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* Result is returned in the first 2*len words of c.
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*/
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-static void internal_mul(BignumInt *a, BignumInt *b,
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+#define KARATSUBA_THRESHOLD 50
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+static void internal_mul(const BignumInt *a, const BignumInt *b,
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BignumInt *c, int len)
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{
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int i, j;
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BignumDblInt t;
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- for (j = 0; j < 2 * len; j++)
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- c[j] = 0;
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+ if (len > KARATSUBA_THRESHOLD) {
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+
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+ /*
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+ * Karatsuba divide-and-conquer algorithm. Cut each input in
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+ * half, so that it's expressed as two big 'digits' in a giant
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+ * base D:
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+ *
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+ * a = a_1 D + a_0
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+ * b = b_1 D + b_0
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+ *
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+ * Then the product is of course
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+ *
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+ * ab = a_1 b_1 D^2 + (a_1 b_0 + a_0 b_1) D + a_0 b_0
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+ *
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+ * and we compute the three coefficients by recursively
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+ * calling ourself to do half-length multiplications.
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+ *
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+ * The clever bit that makes this worth doing is that we only
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+ * need _one_ half-length multiplication for the central
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+ * coefficient rather than the two that it obviouly looks
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+ * like, because we can use a single multiplication to compute
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+ *
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+ * (a_1 + a_0) (b_1 + b_0) = a_1 b_1 + a_1 b_0 + a_0 b_1 + a_0 b_0
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+ *
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+ * and then we subtract the other two coefficients (a_1 b_1
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+ * and a_0 b_0) which we were computing anyway.
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+ *
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+ * Hence we get to multiply two numbers of length N in about
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+ * three times as much work as it takes to multiply numbers of
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+ * length N/2, which is obviously better than the four times
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+ * as much work it would take if we just did a long
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+ * conventional multiply.
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+ */
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+
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+ int toplen = len/2, botlen = len - toplen; /* botlen is the bigger */
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+ int midlen = botlen + 1;
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+ BignumInt *scratch;
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+ BignumDblInt carry;
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+#ifdef KARA_DEBUG
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+ int i;
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+#endif
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- for (i = len - 1; i >= 0; i--) {
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- t = 0;
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- for (j = len - 1; j >= 0; j--) {
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- t += MUL_WORD(a[i], (BignumDblInt) b[j]);
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- t += (BignumDblInt) c[i + j + 1];
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- c[i + j + 1] = (BignumInt) t;
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- t = t >> BIGNUM_INT_BITS;
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- }
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- c[i] = (BignumInt) t;
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+ /*
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+ * The coefficients a_1 b_1 and a_0 b_0 just avoid overlapping
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+ * in the output array, so we can compute them immediately in
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+ * place.
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+ */
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+
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+#ifdef KARA_DEBUG
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+ printf("a1,a0 = 0x");
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+ for (i = 0; i < len; i++) {
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+ if (i == toplen) printf(", 0x");
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+ printf("%0*x", BIGNUM_INT_BITS/4, a[i]);
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+ }
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+ printf("\n");
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+ printf("b1,b0 = 0x");
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+ for (i = 0; i < len; i++) {
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+ if (i == toplen) printf(", 0x");
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+ printf("%0*x", BIGNUM_INT_BITS/4, b[i]);
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+ }
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+ printf("\n");
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+#endif
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+
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+ /* a_1 b_1 */
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+ internal_mul(a, b, c, toplen);
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+#ifdef KARA_DEBUG
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+ printf("a1b1 = 0x");
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+ for (i = 0; i < 2*toplen; i++) {
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+ printf("%0*x", BIGNUM_INT_BITS/4, c[i]);
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+ }
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+ printf("\n");
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+#endif
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+
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+ /* a_0 b_0 */
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+ internal_mul(a + toplen, b + toplen, c + 2*toplen, botlen);
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+#ifdef KARA_DEBUG
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+ printf("a0b0 = 0x");
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+ for (i = 0; i < 2*botlen; i++) {
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+ printf("%0*x", BIGNUM_INT_BITS/4, c[2*toplen+i]);
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+ }
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+ printf("\n");
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+#endif
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+
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+ /*
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+ * We must allocate scratch space for the central coefficient,
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+ * and also for the two input values that we multiply when
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+ * computing it. Since either or both may carry into the
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+ * (botlen+1)th word, we must use a slightly longer length
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+ * 'midlen'.
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+ */
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+ scratch = snewn(4 * midlen, BignumInt);
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+
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+ /* Zero padding. midlen exceeds toplen by at most 2, so just
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+ * zero the first two words of each input and the rest will be
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+ * copied over. */
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+ scratch[0] = scratch[1] = scratch[midlen] = scratch[midlen+1] = 0;
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+
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+ for (j = 0; j < toplen; j++) {
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+ scratch[midlen - toplen + j] = a[j]; /* a_1 */
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+ scratch[2*midlen - toplen + j] = b[j]; /* b_1 */
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+ }
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+
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+ /* compute a_1 + a_0 */
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+ scratch[0] = internal_add(scratch+1, a+toplen, scratch+1, botlen);
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+#ifdef KARA_DEBUG
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+ printf("a1plusa0 = 0x");
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+ for (i = 0; i < midlen; i++) {
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+ printf("%0*x", BIGNUM_INT_BITS/4, scratch[i]);
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+ }
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+ printf("\n");
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+#endif
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+ /* compute b_1 + b_0 */
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+ scratch[midlen] = internal_add(scratch+midlen+1, b+toplen,
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+ scratch+midlen+1, botlen);
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+#ifdef KARA_DEBUG
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+ printf("b1plusb0 = 0x");
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+ for (i = 0; i < midlen; i++) {
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+ printf("%0*x", BIGNUM_INT_BITS/4, scratch[midlen+i]);
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+ }
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+ printf("\n");
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+#endif
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+
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+ /*
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+ * Now we can do the third multiplication.
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+ */
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+ internal_mul(scratch, scratch + midlen, scratch + 2*midlen, midlen);
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+#ifdef KARA_DEBUG
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+ printf("a1plusa0timesb1plusb0 = 0x");
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+ for (i = 0; i < 2*midlen; i++) {
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+ printf("%0*x", BIGNUM_INT_BITS/4, scratch[2*midlen+i]);
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+ }
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+ printf("\n");
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+#endif
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+
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+ /*
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+ * Now we can reuse the first half of 'scratch' to compute the
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+ * sum of the outer two coefficients, to subtract from that
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+ * product to obtain the middle one.
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+ */
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+ scratch[0] = scratch[1] = scratch[2] = scratch[3] = 0;
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+ for (j = 0; j < 2*toplen; j++)
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+ scratch[2*midlen - 2*toplen + j] = c[j];
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+ scratch[1] = internal_add(scratch+2, c + 2*toplen,
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+ scratch+2, 2*botlen);
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+#ifdef KARA_DEBUG
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+ printf("a1b1plusa0b0 = 0x");
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+ for (i = 0; i < 2*midlen; i++) {
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+ printf("%0*x", BIGNUM_INT_BITS/4, scratch[i]);
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+ }
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+ printf("\n");
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+#endif
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+
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+ internal_sub(scratch + 2*midlen, scratch,
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+ scratch + 2*midlen, 2*midlen);
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+#ifdef KARA_DEBUG
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+ printf("a1b0plusa0b1 = 0x");
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+ for (i = 0; i < 2*midlen; i++) {
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+ printf("%0*x", BIGNUM_INT_BITS/4, scratch[2*midlen+i]);
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+ }
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+ printf("\n");
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+#endif
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+
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+ /*
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+ * And now all we need to do is to add that middle coefficient
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+ * back into the output. We may have to propagate a carry
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+ * further up the output, but we can be sure it won't
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+ * propagate right the way off the top.
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+ */
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+ carry = internal_add(c + 2*len - botlen - 2*midlen,
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+ scratch + 2*midlen,
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+ c + 2*len - botlen - 2*midlen, 2*midlen);
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+ j = 2*len - botlen - 2*midlen - 1;
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+ while (carry) {
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+ assert(j >= 0);
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+ carry += c[j];
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+ c[j] = (BignumInt)carry;
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+ carry >>= BIGNUM_INT_BITS;
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+ j--;
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+ }
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+#ifdef KARA_DEBUG
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+ printf("ab = 0x");
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+ for (i = 0; i < 2*len; i++) {
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+ printf("%0*x", BIGNUM_INT_BITS/4, c[i]);
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+ }
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+ printf("\n");
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+#endif
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+
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+ /* Free scratch. */
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+ for (j = 0; j < 4 * midlen; j++)
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+ scratch[j] = 0;
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+ sfree(scratch);
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+
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+ } else {
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+
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+ /*
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+ * Multiply in the ordinary O(N^2) way.
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+ */
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+
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+ for (j = 0; j < 2 * len; j++)
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+ c[j] = 0;
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+
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+ for (i = len - 1; i >= 0; i--) {
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+ t = 0;
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+ for (j = len - 1; j >= 0; j--) {
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+ t += MUL_WORD(a[i], (BignumDblInt) b[j]);
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+ t += (BignumDblInt) c[i + j + 1];
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+ c[i + j + 1] = (BignumInt) t;
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+ t = t >> BIGNUM_INT_BITS;
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+ }
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+ c[i] = (BignumInt) t;
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+ }
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}
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}
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+/*
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+ * Variant form of internal_mul used for the initial step of
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+ * Montgomery reduction. Only bothers outputting 'len' words
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+ * (everything above that is thrown away).
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+ */
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+static void internal_mul_low(const BignumInt *a, const BignumInt *b,
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+ BignumInt *c, int len)
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+{
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+ int i, j;
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+ BignumDblInt t;
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+
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+ if (len > KARATSUBA_THRESHOLD) {
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+
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+ /*
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+ * Karatsuba-aware version of internal_mul_low. As before, we
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+ * express each input value as a shifted combination of two
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+ * halves:
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+ *
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+ * a = a_1 D + a_0
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+ * b = b_1 D + b_0
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+ *
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+ * Then the full product is, as before,
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+ *
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+ * ab = a_1 b_1 D^2 + (a_1 b_0 + a_0 b_1) D + a_0 b_0
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+ *
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+ * Provided we choose D on the large side (so that a_0 and b_0
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+ * are _at least_ as long as a_1 and b_1), we don't need the
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+ * topmost term at all, and we only need half of the middle
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+ * term. So there's no point in doing the proper Karatsuba
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+ * optimisation which computes the middle term using the top
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+ * one, because we'd take as long computing the top one as
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+ * just computing the middle one directly.
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+ *
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+ * So instead, we do a much more obvious thing: we call the
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+ * fully optimised internal_mul to compute a_0 b_0, and we
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+ * recursively call ourself to compute the _bottom halves_ of
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+ * a_1 b_0 and a_0 b_1, each of which we add into the result
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+ * in the obvious way.
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+ *
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+ * In other words, there's no actual Karatsuba _optimisation_
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+ * in this function; the only benefit in doing it this way is
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+ * that we call internal_mul proper for a large part of the
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+ * work, and _that_ can optimise its operation.
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+ */
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+
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+ int toplen = len/2, botlen = len - toplen; /* botlen is the bigger */
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+ BignumInt *scratch;
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+
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+ /*
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+ * Allocate scratch space for the various bits and pieces
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+ * we're going to be adding together. We need botlen*2 words
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+ * for a_0 b_0 (though we may end up throwing away its topmost
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+ * word), and toplen words for each of a_1 b_0 and a_0 b_1.
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+ * That adds up to exactly 2*len.
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+ */
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+ scratch = snewn(len*2, BignumInt);
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+
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+ /* a_0 b_0 */
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+ internal_mul(a + toplen, b + toplen, scratch + 2*toplen, botlen);
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+
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+ /* a_1 b_0 */
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+ internal_mul_low(a, b + len - toplen, scratch + toplen, toplen);
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+
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+ /* a_0 b_1 */
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|
|
+ internal_mul_low(a + len - toplen, b, scratch, toplen);
|
|
|
+
|
|
|
+ /* Copy the bottom half of the big coefficient into place */
|
|
|
+ for (j = 0; j < botlen; j++)
|
|
|
+ c[toplen + j] = scratch[2*toplen + botlen + j];
|
|
|
+
|
|
|
+ /* 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);
|
|
|
+
|
|
|
+ /* Free scratch. */
|
|
|
+ for (j = 0; j < len*2; j++)
|
|
|
+ scratch[j] = 0;
|
|
|
+ sfree(scratch);
|
|
|
+
|
|
|
+ } else {
|
|
|
+
|
|
|
+ for (j = 0; j < len; j++)
|
|
|
+ c[j] = 0;
|
|
|
+
|
|
|
+ for (i = len - 1; i >= 0; i--) {
|
|
|
+ t = 0;
|
|
|
+ for (j = len - 1; j >= len - i - 1; j--) {
|
|
|
+ t += MUL_WORD(a[i], (BignumDblInt) b[j]);
|
|
|
+ t += (BignumDblInt) c[i + j + 1 - len];
|
|
|
+ c[i + j + 1 - len] = (BignumInt) t;
|
|
|
+ t = 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 at least '3*len' BignumInts used as scratch
|
|
|
+ * space.
|
|
|
+ */
|
|
|
+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);
|
|
|
+
|
|
|
+ /*
|
|
|
+ * 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);
|
|
|
+ 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,
|
|
|
unsigned n, int shift)
|
|
|
{
|
|
|
@@ -291,9 +717,9 @@ 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;
|
|
|
int mshift;
|
|
|
@@ -411,6 +837,155 @@ Bignum modpow(Bignum base_in, Bignum exp, Bignum mod)
|
|
|
return result;
|
|
|
}
|
|
|
|
|
|
+/*
|
|
|
+ * 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, *tmp;
|
|
|
+ int len, 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 < 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 < 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 < rn[0] ? rn[j + 1] : 0);
|
|
|
+ freebn(rn);
|
|
|
+
|
|
|
+ tmp = snewn(3*len, 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);
|
|
|
+ monty_reduce(b, n, mninv, tmp, len);
|
|
|
+ if ((exp[exp[0] - i] & (1 << j)) != 0) {
|
|
|
+ internal_mul(b + len, x, a, len);
|
|
|
+ monty_reduce(a, n, mninv, tmp, 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, tmp, 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 < 3 * len; i++)
|
|
|
+ tmp[i] = 0;
|
|
|
+ sfree(tmp);
|
|
|
+ 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.
|
|
|
@@ -824,6 +1399,69 @@ Bignum bigmul(Bignum a, Bignum b)
|
|
|
return bigmuladd(a, b, NULL);
|
|
|
}
|
|
|
|
|
|
+/*
|
|
|
+ * 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
|
|
|
@@ -1102,3 +1740,166 @@ 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
|