2 * Bignum routines for RSA and DH and stuff.
14 * * Do not call the DIVMOD_WORD macro with expressions such as array
15 * subscripts, as some implementations object to this (see below).
16 * * Note that none of the division methods below will cope if the
17 * quotient won't fit into BIGNUM_INT_BITS. Callers should be careful
19 * If this condition occurs, in the case of the x86 DIV instruction,
20 * an overflow exception will occur, which (according to a correspondent)
21 * will manifest on Windows as something like
22 * 0xC0000095: Integer overflow
23 * The C variant won't give the right answer, either.
26 #if defined __GNUC__ && defined __i386__
27 typedef unsigned long BignumInt;
28 typedef unsigned long long BignumDblInt;
29 #define BIGNUM_INT_MASK 0xFFFFFFFFUL
30 #define BIGNUM_TOP_BIT 0x80000000UL
31 #define BIGNUM_INT_BITS 32
32 #define MUL_WORD(w1, w2) ((BignumDblInt)w1 * w2)
33 #define DIVMOD_WORD(q, r, hi, lo, w) \
35 "=d" (r), "=a" (q) : \
36 "r" (w), "d" (hi), "a" (lo))
37 #elif defined _MSC_VER && defined _M_IX86
38 typedef unsigned __int32 BignumInt;
39 typedef unsigned __int64 BignumDblInt;
40 #define BIGNUM_INT_MASK 0xFFFFFFFFUL
41 #define BIGNUM_TOP_BIT 0x80000000UL
42 #define BIGNUM_INT_BITS 32
43 #define MUL_WORD(w1, w2) ((BignumDblInt)w1 * w2)
44 /* Note: MASM interprets array subscripts in the macro arguments as
45 * assembler syntax, which gives the wrong answer. Don't supply them.
46 * <http://msdn2.microsoft.com/en-us/library/bf1dw62z.aspx> */
47 #define DIVMOD_WORD(q, r, hi, lo, w) do { \
55 /* 64-bit architectures can do 32x32->64 chunks at a time */
56 typedef unsigned int BignumInt;
57 typedef unsigned long BignumDblInt;
58 #define BIGNUM_INT_MASK 0xFFFFFFFFU
59 #define BIGNUM_TOP_BIT 0x80000000U
60 #define BIGNUM_INT_BITS 32
61 #define MUL_WORD(w1, w2) ((BignumDblInt)w1 * w2)
62 #define DIVMOD_WORD(q, r, hi, lo, w) do { \
63 BignumDblInt n = (((BignumDblInt)hi) << BIGNUM_INT_BITS) | lo; \
68 /* 64-bit architectures in which unsigned long is 32 bits, not 64 */
69 typedef unsigned long BignumInt;
70 typedef unsigned long long BignumDblInt;
71 #define BIGNUM_INT_MASK 0xFFFFFFFFUL
72 #define BIGNUM_TOP_BIT 0x80000000UL
73 #define BIGNUM_INT_BITS 32
74 #define MUL_WORD(w1, w2) ((BignumDblInt)w1 * w2)
75 #define DIVMOD_WORD(q, r, hi, lo, w) do { \
76 BignumDblInt n = (((BignumDblInt)hi) << BIGNUM_INT_BITS) | lo; \
81 /* Fallback for all other cases */
82 typedef unsigned short BignumInt;
83 typedef unsigned long BignumDblInt;
84 #define BIGNUM_INT_MASK 0xFFFFU
85 #define BIGNUM_TOP_BIT 0x8000U
86 #define BIGNUM_INT_BITS 16
87 #define MUL_WORD(w1, w2) ((BignumDblInt)w1 * w2)
88 #define DIVMOD_WORD(q, r, hi, lo, w) do { \
89 BignumDblInt n = (((BignumDblInt)hi) << BIGNUM_INT_BITS) | lo; \
95 #define BIGNUM_INT_BYTES (BIGNUM_INT_BITS / 8)
97 #define BIGNUM_INTERNAL
98 typedef BignumInt *Bignum;
102 BignumInt bnZero[1] = { 0 };
103 BignumInt bnOne[2] = { 1, 1 };
106 * The Bignum format is an array of `BignumInt'. The first
107 * element of the array counts the remaining elements. The
108 * remaining elements express the actual number, base 2^BIGNUM_INT_BITS, _least_
109 * significant digit first. (So it's trivial to extract the bit
110 * with value 2^n for any n.)
112 * All Bignums in this module are positive. Negative numbers must
113 * be dealt with outside it.
115 * INVARIANT: the most significant word of any Bignum must be
119 Bignum Zero = bnZero, One = bnOne;
121 static Bignum newbn(int length)
123 Bignum b = snewn(length + 1, BignumInt);
126 memset(b, 0, (length + 1) * sizeof(*b));
131 void bn_restore_invariant(Bignum b)
133 while (b[0] > 1 && b[b[0]] == 0)
137 Bignum copybn(Bignum orig)
139 Bignum b = snewn(orig[0] + 1, BignumInt);
142 memcpy(b, orig, (orig[0] + 1) * sizeof(*b));
146 void freebn(Bignum b)
149 * Burn the evidence, just in case.
151 memset(b, 0, sizeof(b[0]) * (b[0] + 1));
155 Bignum bn_power_2(int n)
157 Bignum ret = newbn(n / BIGNUM_INT_BITS + 1);
158 bignum_set_bit(ret, n, 1);
163 * Internal addition. Sets c = a - b, where 'a', 'b' and 'c' are all
164 * big-endian arrays of 'len' BignumInts. Returns a BignumInt carried
167 static BignumInt internal_add(const BignumInt *a, const BignumInt *b,
168 BignumInt *c, int len)
171 BignumDblInt carry = 0;
173 for (i = len-1; i >= 0; i--) {
174 carry += (BignumDblInt)a[i] + b[i];
175 c[i] = (BignumInt)carry;
176 carry >>= BIGNUM_INT_BITS;
179 return (BignumInt)carry;
183 * Internal subtraction. Sets c = a - b, where 'a', 'b' and 'c' are
184 * all big-endian arrays of 'len' BignumInts. Any borrow from the top
187 static void internal_sub(const BignumInt *a, const BignumInt *b,
188 BignumInt *c, int len)
191 BignumDblInt carry = 1;
193 for (i = len-1; i >= 0; i--) {
194 carry += (BignumDblInt)a[i] + (b[i] ^ BIGNUM_INT_MASK);
195 c[i] = (BignumInt)carry;
196 carry >>= BIGNUM_INT_BITS;
202 * Input is in the first len words of a and b.
203 * Result is returned in the first 2*len words of c.
205 #define KARATSUBA_THRESHOLD 50
206 static void internal_mul(const BignumInt *a, const BignumInt *b,
207 BignumInt *c, int len)
212 if (len > KARATSUBA_THRESHOLD) {
215 * Karatsuba divide-and-conquer algorithm. Cut each input in
216 * half, so that it's expressed as two big 'digits' in a giant
222 * Then the product is of course
224 * ab = a_1 b_1 D^2 + (a_1 b_0 + a_0 b_1) D + a_0 b_0
226 * and we compute the three coefficients by recursively
227 * calling ourself to do half-length multiplications.
229 * The clever bit that makes this worth doing is that we only
230 * need _one_ half-length multiplication for the central
231 * coefficient rather than the two that it obviouly looks
232 * like, because we can use a single multiplication to compute
234 * (a_1 + a_0) (b_1 + b_0) = a_1 b_1 + a_1 b_0 + a_0 b_1 + a_0 b_0
236 * and then we subtract the other two coefficients (a_1 b_1
237 * and a_0 b_0) which we were computing anyway.
239 * Hence we get to multiply two numbers of length N in about
240 * three times as much work as it takes to multiply numbers of
241 * length N/2, which is obviously better than the four times
242 * as much work it would take if we just did a long
243 * conventional multiply.
246 int toplen = len/2, botlen = len - toplen; /* botlen is the bigger */
247 int midlen = botlen + 1;
252 * The coefficients a_1 b_1 and a_0 b_0 just avoid overlapping
253 * in the output array, so we can compute them immediately in
258 internal_mul(a, b, c, toplen);
261 internal_mul(a + toplen, b + toplen, c + 2*toplen, botlen);
264 * We must allocate scratch space for the central coefficient,
265 * and also for the two input values that we multiply when
266 * computing it. Since either or both may carry into the
267 * (botlen+1)th word, we must use a slightly longer length
270 scratch = snewn(4 * midlen, BignumInt);
272 /* Zero padding. midlen exceeds toplen by at most 2, so just
273 * zero the first two words of each input and the rest will be
275 scratch[0] = scratch[1] = scratch[midlen] = scratch[midlen+1] = 0;
277 for (j = 0; j < toplen; j++) {
278 scratch[midlen - toplen + j] = a[j]; /* a_1 */
279 scratch[2*midlen - toplen + j] = b[j]; /* b_1 */
282 /* compute a_1 + a_0 */
283 scratch[0] = internal_add(scratch+1, a+toplen, scratch+1, botlen);
284 /* compute b_1 + b_0 */
285 scratch[midlen] = internal_add(scratch+midlen+1, b+toplen,
286 scratch+midlen+1, botlen);
289 * Now we can do the third multiplication.
291 internal_mul(scratch, scratch + midlen, scratch + 2*midlen, midlen);
294 * Now we can reuse the first half of 'scratch' to compute the
295 * sum of the outer two coefficients, to subtract from that
296 * product to obtain the middle one.
298 scratch[0] = scratch[1] = scratch[2] = scratch[3] = 0;
299 for (j = 0; j < 2*toplen; j++)
300 scratch[2*midlen - 2*toplen + j] = c[j];
301 scratch[1] = internal_add(scratch+2, c + 2*toplen,
302 scratch+2, 2*botlen);
304 internal_sub(scratch + 2*midlen, scratch,
305 scratch + 2*midlen, 2*midlen);
308 * And now all we need to do is to add that middle coefficient
309 * back into the output. We may have to propagate a carry
310 * further up the output, but we can be sure it won't
311 * propagate right the way off the top.
313 carry = internal_add(c + 2*len - botlen - 2*midlen,
315 c + 2*len - botlen - 2*midlen, 2*midlen);
316 j = 2*len - botlen - 2*midlen - 1;
320 c[j] = (BignumInt)carry;
321 carry >>= BIGNUM_INT_BITS;
325 for (j = 0; j < 4 * midlen; j++)
332 * Multiply in the ordinary O(N^2) way.
335 for (j = 0; j < 2 * len; j++)
338 for (i = len - 1; i >= 0; i--) {
340 for (j = len - 1; j >= 0; j--) {
341 t += MUL_WORD(a[i], (BignumDblInt) b[j]);
342 t += (BignumDblInt) c[i + j + 1];
343 c[i + j + 1] = (BignumInt) t;
344 t = t >> BIGNUM_INT_BITS;
346 c[i] = (BignumInt) t;
352 * Variant form of internal_mul used for the initial step of
353 * Montgomery reduction. Only bothers outputting 'len' words
354 * (everything above that is thrown away).
356 static void internal_mul_low(const BignumInt *a, const BignumInt *b,
357 BignumInt *c, int len)
362 if (len > KARATSUBA_THRESHOLD) {
365 * Karatsuba-aware version of internal_mul_low. As before, we
366 * express each input value as a shifted combination of two
372 * Then the full product is, as before,
374 * ab = a_1 b_1 D^2 + (a_1 b_0 + a_0 b_1) D + a_0 b_0
376 * Provided we choose D on the large side (so that a_0 and b_0
377 * are _at least_ as long as a_1 and b_1), we don't need the
378 * topmost term at all, and we only need half of the middle
379 * term. So there's no point in doing the proper Karatsuba
380 * optimisation which computes the middle term using the top
381 * one, because we'd take as long computing the top one as
382 * just computing the middle one directly.
384 * So instead, we do a much more obvious thing: we call the
385 * fully optimised internal_mul to compute a_0 b_0, and we
386 * recursively call ourself to compute the _bottom halves_ of
387 * a_1 b_0 and a_0 b_1, each of which we add into the result
388 * in the obvious way.
390 * In other words, there's no actual Karatsuba _optimisation_
391 * in this function; the only benefit in doing it this way is
392 * that we call internal_mul proper for a large part of the
393 * work, and _that_ can optimise its operation.
396 int toplen = len/2, botlen = len - toplen; /* botlen is the bigger */
400 * Allocate scratch space for the various bits and pieces
401 * we're going to be adding together. We need botlen*2 words
402 * for a_0 b_0 (though we may end up throwing away its topmost
403 * word), and toplen words for each of a_1 b_0 and a_0 b_1.
404 * That adds up to exactly 2*len.
406 scratch = snewn(len*2, BignumInt);
409 internal_mul(a + toplen, b + toplen, scratch + 2*toplen, botlen);
412 internal_mul_low(a, b + len - toplen, scratch + toplen, toplen);
415 internal_mul_low(a + len - toplen, b, scratch, toplen);
417 /* Copy the bottom half of the big coefficient into place */
418 for (j = 0; j < botlen; j++)
419 c[toplen + j] = scratch[2*toplen + botlen + j];
421 /* Add the two small coefficients, throwing away the returned carry */
422 internal_add(scratch, scratch + toplen, scratch, toplen);
424 /* And add that to the large coefficient, leaving the result in c. */
425 internal_add(scratch, scratch + 2*toplen + botlen - toplen,
429 for (j = 0; j < len*2; j++)
435 for (j = 0; j < len; j++)
438 for (i = len - 1; i >= 0; i--) {
440 for (j = len - 1; j >= len - i - 1; j--) {
441 t += MUL_WORD(a[i], (BignumDblInt) b[j]);
442 t += (BignumDblInt) c[i + j + 1 - len];
443 c[i + j + 1 - len] = (BignumInt) t;
444 t = t >> BIGNUM_INT_BITS;
452 * Montgomery reduction. Expects x to be a big-endian array of 2*len
453 * BignumInts whose value satisfies 0 <= x < rn (where r = 2^(len *
454 * BIGNUM_INT_BITS) is the Montgomery base). Returns in the same array
455 * a value x' which is congruent to xr^{-1} mod n, and satisfies 0 <=
458 * 'n' and 'mninv' should be big-endian arrays of 'len' BignumInts
459 * each, containing respectively n and the multiplicative inverse of
462 * 'tmp' is an array of at least '3*len' BignumInts used as scratch
465 static void monty_reduce(BignumInt *x, const BignumInt *n,
466 const BignumInt *mninv, BignumInt *tmp, int len)
472 * Multiply x by (-n)^{-1} mod r. This gives us a value m such
473 * that mn is congruent to -x mod r. Hence, mn+x is an exact
474 * multiple of r, and is also (obviously) congruent to x mod n.
476 internal_mul_low(x + len, mninv, tmp, len);
479 * Compute t = (mn+x)/r in ordinary, non-modular, integer
480 * arithmetic. By construction this is exact, and is congruent mod
481 * n to x * r^{-1}, i.e. the answer we want.
483 * The following multiply leaves that answer in the _most_
484 * significant half of the 'x' array, so then we must shift it
487 internal_mul(tmp, n, tmp+len, len);
488 carry = internal_add(x, tmp+len, x, 2*len);
489 for (i = 0; i < len; i++)
490 x[len + i] = x[i], x[i] = 0;
493 * Reduce t mod n. This doesn't require a full-on division by n,
494 * but merely a test and single optional subtraction, since we can
495 * show that 0 <= t < 2n.
498 * + we computed m mod r, so 0 <= m < r.
499 * + so 0 <= mn < rn, obviously
500 * + hence we only need 0 <= x < rn to guarantee that 0 <= mn+x < 2rn
501 * + yielding 0 <= (mn+x)/r < 2n as required.
504 for (i = 0; i < len; i++)
505 if (x[len + i] != n[i])
508 if (carry || i >= len || x[len + i] > n[i])
509 internal_sub(x+len, n, x+len, len);
512 static void internal_add_shifted(BignumInt *number,
513 unsigned n, int shift)
515 int word = 1 + (shift / BIGNUM_INT_BITS);
516 int bshift = shift % BIGNUM_INT_BITS;
519 addend = (BignumDblInt)n << bshift;
522 addend += number[word];
523 number[word] = (BignumInt) addend & BIGNUM_INT_MASK;
524 addend >>= BIGNUM_INT_BITS;
531 * Input in first alen words of a and first mlen words of m.
532 * Output in first alen words of a
533 * (of which first alen-mlen words will be zero).
534 * The MSW of m MUST have its high bit set.
535 * Quotient is accumulated in the `quotient' array, which is a Bignum
536 * rather than the internal bigendian format. Quotient parts are shifted
537 * left by `qshift' before adding into quot.
539 static void internal_mod(BignumInt *a, int alen,
540 BignumInt *m, int mlen,
541 BignumInt *quot, int qshift)
553 for (i = 0; i <= alen - mlen; i++) {
555 unsigned int q, r, c, ai1;
569 /* Find q = h:a[i] / m0 */
574 * To illustrate it, suppose a BignumInt is 8 bits, and
575 * we are dividing (say) A1:23:45:67 by A1:B2:C3. Then
576 * our initial division will be 0xA123 / 0xA1, which
577 * will give a quotient of 0x100 and a divide overflow.
578 * However, the invariants in this division algorithm
579 * are not violated, since the full number A1:23:... is
580 * _less_ than the quotient prefix A1:B2:... and so the
581 * following correction loop would have sorted it out.
583 * In this situation we set q to be the largest
584 * quotient we _can_ stomach (0xFF, of course).
588 /* Macro doesn't want an array subscript expression passed
589 * into it (see definition), so use a temporary. */
590 BignumInt tmplo = a[i];
591 DIVMOD_WORD(q, r, h, tmplo, m0);
593 /* Refine our estimate of q by looking at
594 h:a[i]:a[i+1] / m0:m1 */
596 if (t > ((BignumDblInt) r << BIGNUM_INT_BITS) + ai1) {
599 r = (r + m0) & BIGNUM_INT_MASK; /* overflow? */
600 if (r >= (BignumDblInt) m0 &&
601 t > ((BignumDblInt) r << BIGNUM_INT_BITS) + ai1) q--;
605 /* Subtract q * m from a[i...] */
607 for (k = mlen - 1; k >= 0; k--) {
608 t = MUL_WORD(q, m[k]);
610 c = (unsigned)(t >> BIGNUM_INT_BITS);
611 if ((BignumInt) t > a[i + k])
613 a[i + k] -= (BignumInt) t;
616 /* Add back m in case of borrow */
619 for (k = mlen - 1; k >= 0; k--) {
622 a[i + k] = (BignumInt) t;
623 t = t >> BIGNUM_INT_BITS;
628 internal_add_shifted(quot, q, qshift + BIGNUM_INT_BITS * (alen - mlen - i));
633 * Compute (base ^ exp) % mod. Uses the Montgomery multiplication
636 Bignum modpow(Bignum base_in, Bignum exp, Bignum mod)
638 BignumInt *a, *b, *x, *n, *mninv, *tmp;
640 Bignum base, base2, r, rn, inv, result;
643 * The most significant word of mod needs to be non-zero. It
644 * should already be, but let's make sure.
646 assert(mod[mod[0]] != 0);
649 * Make sure the base is smaller than the modulus, by reducing
650 * it modulo the modulus if not.
652 base = bigmod(base_in, mod);
655 * mod had better be odd, or we can't do Montgomery multiplication
656 * using a power of two at all.
661 * Compute the inverse of n mod r, for monty_reduce. (In fact we
662 * want the inverse of _minus_ n mod r, but we'll sort that out
666 r = bn_power_2(BIGNUM_INT_BITS * len);
667 inv = modinv(mod, r);
670 * Multiply the base by r mod n, to get it into Montgomery
673 base2 = modmul(base, r, mod);
677 rn = bigmod(r, mod); /* r mod n, i.e. Montgomerified 1 */
679 freebn(r); /* won't need this any more */
682 * Set up internal arrays of the right lengths, in big-endian
683 * format, containing the base, the modulus, and the modulus's
686 n = snewn(len, BignumInt);
687 for (j = 0; j < len; j++)
688 n[len - 1 - j] = mod[j + 1];
690 mninv = snewn(len, BignumInt);
691 for (j = 0; j < len; j++)
692 mninv[len - 1 - j] = (j < inv[0] ? inv[j + 1] : 0);
693 freebn(inv); /* we don't need this copy of it any more */
694 /* Now negate mninv mod r, so it's the inverse of -n rather than +n. */
695 x = snewn(len, BignumInt);
696 for (j = 0; j < len; j++)
698 internal_sub(x, mninv, mninv, len);
700 /* x = snewn(len, BignumInt); */ /* already done above */
701 for (j = 0; j < len; j++)
702 x[len - 1 - j] = (j < base[0] ? base[j + 1] : 0);
703 freebn(base); /* we don't need this copy of it any more */
705 a = snewn(2*len, BignumInt);
706 b = snewn(2*len, BignumInt);
707 for (j = 0; j < len; j++)
708 a[2*len - 1 - j] = (j < rn[0] ? rn[j + 1] : 0);
711 tmp = snewn(3*len, BignumInt);
713 /* Skip leading zero bits of exp. */
715 j = BIGNUM_INT_BITS-1;
716 while (i < (int)exp[0] && (exp[exp[0] - i] & (1 << j)) == 0) {
720 j = BIGNUM_INT_BITS-1;
724 /* Main computation */
725 while (i < (int)exp[0]) {
727 internal_mul(a + len, a + len, b, len);
728 monty_reduce(b, n, mninv, tmp, len);
729 if ((exp[exp[0] - i] & (1 << j)) != 0) {
730 internal_mul(b + len, x, a, len);
731 monty_reduce(a, n, mninv, tmp, len);
741 j = BIGNUM_INT_BITS-1;
745 * Final monty_reduce to get back from the adjusted Montgomery
748 monty_reduce(a, n, mninv, tmp, len);
750 /* Copy result to buffer */
751 result = newbn(mod[0]);
752 for (i = 0; i < len; i++)
753 result[result[0] - i] = a[i + len];
754 while (result[0] > 1 && result[result[0]] == 0)
757 /* Free temporary arrays */
758 for (i = 0; i < 3 * len; i++)
761 for (i = 0; i < 2 * len; i++)
764 for (i = 0; i < 2 * len; i++)
767 for (i = 0; i < len; i++)
770 for (i = 0; i < len; i++)
773 for (i = 0; i < len; i++)
781 * Compute (p * q) % mod.
782 * The most significant word of mod MUST be non-zero.
783 * We assume that the result array is the same size as the mod array.
785 Bignum modmul(Bignum p, Bignum q, Bignum mod)
787 BignumInt *a, *n, *m, *o;
789 int pqlen, mlen, rlen, i, j;
792 /* Allocate m of size mlen, copy mod to m */
793 /* We use big endian internally */
795 m = snewn(mlen, BignumInt);
796 for (j = 0; j < mlen; j++)
797 m[j] = mod[mod[0] - j];
799 /* Shift m left to make msb bit set */
800 for (mshift = 0; mshift < BIGNUM_INT_BITS-1; mshift++)
801 if ((m[0] << mshift) & BIGNUM_TOP_BIT)
804 for (i = 0; i < mlen - 1; i++)
805 m[i] = (m[i] << mshift) | (m[i + 1] >> (BIGNUM_INT_BITS - mshift));
806 m[mlen - 1] = m[mlen - 1] << mshift;
809 pqlen = (p[0] > q[0] ? p[0] : q[0]);
811 /* Allocate n of size pqlen, copy p to n */
812 n = snewn(pqlen, BignumInt);
814 for (j = 0; j < i; j++)
816 for (j = 0; j < (int)p[0]; j++)
817 n[i + j] = p[p[0] - j];
819 /* Allocate o of size pqlen, copy q to o */
820 o = snewn(pqlen, BignumInt);
822 for (j = 0; j < i; j++)
824 for (j = 0; j < (int)q[0]; j++)
825 o[i + j] = q[q[0] - j];
827 /* Allocate a of size 2*pqlen for result */
828 a = snewn(2 * pqlen, BignumInt);
830 /* Main computation */
831 internal_mul(n, o, a, pqlen);
832 internal_mod(a, pqlen * 2, m, mlen, NULL, 0);
834 /* Fixup result in case the modulus was shifted */
836 for (i = 2 * pqlen - mlen - 1; i < 2 * pqlen - 1; i++)
837 a[i] = (a[i] << mshift) | (a[i + 1] >> (BIGNUM_INT_BITS - mshift));
838 a[2 * pqlen - 1] = a[2 * pqlen - 1] << mshift;
839 internal_mod(a, pqlen * 2, m, mlen, NULL, 0);
840 for (i = 2 * pqlen - 1; i >= 2 * pqlen - mlen; i--)
841 a[i] = (a[i] >> mshift) | (a[i - 1] << (BIGNUM_INT_BITS - mshift));
844 /* Copy result to buffer */
845 rlen = (mlen < pqlen * 2 ? mlen : pqlen * 2);
846 result = newbn(rlen);
847 for (i = 0; i < rlen; i++)
848 result[result[0] - i] = a[i + 2 * pqlen - rlen];
849 while (result[0] > 1 && result[result[0]] == 0)
852 /* Free temporary arrays */
853 for (i = 0; i < 2 * pqlen; i++)
856 for (i = 0; i < mlen; i++)
859 for (i = 0; i < pqlen; i++)
862 for (i = 0; i < pqlen; i++)
871 * The most significant word of mod MUST be non-zero.
872 * We assume that the result array is the same size as the mod array.
873 * We optionally write out a quotient if `quotient' is non-NULL.
874 * We can avoid writing out the result if `result' is NULL.
876 static void bigdivmod(Bignum p, Bignum mod, Bignum result, Bignum quotient)
880 int plen, mlen, i, j;
882 /* Allocate m of size mlen, copy mod to m */
883 /* We use big endian internally */
885 m = snewn(mlen, BignumInt);
886 for (j = 0; j < mlen; j++)
887 m[j] = mod[mod[0] - j];
889 /* Shift m left to make msb bit set */
890 for (mshift = 0; mshift < BIGNUM_INT_BITS-1; mshift++)
891 if ((m[0] << mshift) & BIGNUM_TOP_BIT)
894 for (i = 0; i < mlen - 1; i++)
895 m[i] = (m[i] << mshift) | (m[i + 1] >> (BIGNUM_INT_BITS - mshift));
896 m[mlen - 1] = m[mlen - 1] << mshift;
900 /* Ensure plen > mlen */
904 /* Allocate n of size plen, copy p to n */
905 n = snewn(plen, BignumInt);
906 for (j = 0; j < plen; j++)
908 for (j = 1; j <= (int)p[0]; j++)
911 /* Main computation */
912 internal_mod(n, plen, m, mlen, quotient, mshift);
914 /* Fixup result in case the modulus was shifted */
916 for (i = plen - mlen - 1; i < plen - 1; i++)
917 n[i] = (n[i] << mshift) | (n[i + 1] >> (BIGNUM_INT_BITS - mshift));
918 n[plen - 1] = n[plen - 1] << mshift;
919 internal_mod(n, plen, m, mlen, quotient, 0);
920 for (i = plen - 1; i >= plen - mlen; i--)
921 n[i] = (n[i] >> mshift) | (n[i - 1] << (BIGNUM_INT_BITS - mshift));
924 /* Copy result to buffer */
926 for (i = 1; i <= (int)result[0]; i++) {
928 result[i] = j >= 0 ? n[j] : 0;
932 /* Free temporary arrays */
933 for (i = 0; i < mlen; i++)
936 for (i = 0; i < plen; i++)
942 * Decrement a number.
944 void decbn(Bignum bn)
947 while (i < (int)bn[0] && bn[i] == 0)
948 bn[i++] = BIGNUM_INT_MASK;
952 Bignum bignum_from_bytes(const unsigned char *data, int nbytes)
957 w = (nbytes + BIGNUM_INT_BYTES - 1) / BIGNUM_INT_BYTES; /* bytes->words */
960 for (i = 1; i <= w; i++)
962 for (i = nbytes; i--;) {
963 unsigned char byte = *data++;
964 result[1 + i / BIGNUM_INT_BYTES] |= byte << (8*i % BIGNUM_INT_BITS);
967 while (result[0] > 1 && result[result[0]] == 0)
973 * Read an SSH-1-format bignum from a data buffer. Return the number
974 * of bytes consumed, or -1 if there wasn't enough data.
976 int ssh1_read_bignum(const unsigned char *data, int len, Bignum * result)
978 const unsigned char *p = data;
986 for (i = 0; i < 2; i++)
988 b = (w + 7) / 8; /* bits -> bytes */
993 if (!result) /* just return length */
996 *result = bignum_from_bytes(p, b);
1002 * Return the bit count of a bignum, for SSH-1 encoding.
1004 int bignum_bitcount(Bignum bn)
1006 int bitcount = bn[0] * BIGNUM_INT_BITS - 1;
1007 while (bitcount >= 0
1008 && (bn[bitcount / BIGNUM_INT_BITS + 1] >> (bitcount % BIGNUM_INT_BITS)) == 0) bitcount--;
1009 return bitcount + 1;
1013 * Return the byte length of a bignum when SSH-1 encoded.
1015 int ssh1_bignum_length(Bignum bn)
1017 return 2 + (bignum_bitcount(bn) + 7) / 8;
1021 * Return the byte length of a bignum when SSH-2 encoded.
1023 int ssh2_bignum_length(Bignum bn)
1025 return 4 + (bignum_bitcount(bn) + 8) / 8;
1029 * Return a byte from a bignum; 0 is least significant, etc.
1031 int bignum_byte(Bignum bn, int i)
1033 if (i >= (int)(BIGNUM_INT_BYTES * bn[0]))
1034 return 0; /* beyond the end */
1036 return (bn[i / BIGNUM_INT_BYTES + 1] >>
1037 ((i % BIGNUM_INT_BYTES)*8)) & 0xFF;
1041 * Return a bit from a bignum; 0 is least significant, etc.
1043 int bignum_bit(Bignum bn, int i)
1045 if (i >= (int)(BIGNUM_INT_BITS * bn[0]))
1046 return 0; /* beyond the end */
1048 return (bn[i / BIGNUM_INT_BITS + 1] >> (i % BIGNUM_INT_BITS)) & 1;
1052 * Set a bit in a bignum; 0 is least significant, etc.
1054 void bignum_set_bit(Bignum bn, int bitnum, int value)
1056 if (bitnum >= (int)(BIGNUM_INT_BITS * bn[0]))
1057 abort(); /* beyond the end */
1059 int v = bitnum / BIGNUM_INT_BITS + 1;
1060 int mask = 1 << (bitnum % BIGNUM_INT_BITS);
1069 * Write a SSH-1-format bignum into a buffer. It is assumed the
1070 * buffer is big enough. Returns the number of bytes used.
1072 int ssh1_write_bignum(void *data, Bignum bn)
1074 unsigned char *p = data;
1075 int len = ssh1_bignum_length(bn);
1077 int bitc = bignum_bitcount(bn);
1079 *p++ = (bitc >> 8) & 0xFF;
1080 *p++ = (bitc) & 0xFF;
1081 for (i = len - 2; i--;)
1082 *p++ = bignum_byte(bn, i);
1087 * Compare two bignums. Returns like strcmp.
1089 int bignum_cmp(Bignum a, Bignum b)
1091 int amax = a[0], bmax = b[0];
1092 int i = (amax > bmax ? amax : bmax);
1094 BignumInt aval = (i > amax ? 0 : a[i]);
1095 BignumInt bval = (i > bmax ? 0 : b[i]);
1106 * Right-shift one bignum to form another.
1108 Bignum bignum_rshift(Bignum a, int shift)
1111 int i, shiftw, shiftb, shiftbb, bits;
1114 bits = bignum_bitcount(a) - shift;
1115 ret = newbn((bits + BIGNUM_INT_BITS - 1) / BIGNUM_INT_BITS);
1118 shiftw = shift / BIGNUM_INT_BITS;
1119 shiftb = shift % BIGNUM_INT_BITS;
1120 shiftbb = BIGNUM_INT_BITS - shiftb;
1122 ai1 = a[shiftw + 1];
1123 for (i = 1; i <= (int)ret[0]; i++) {
1125 ai1 = (i + shiftw + 1 <= (int)a[0] ? a[i + shiftw + 1] : 0);
1126 ret[i] = ((ai >> shiftb) | (ai1 << shiftbb)) & BIGNUM_INT_MASK;
1134 * Non-modular multiplication and addition.
1136 Bignum bigmuladd(Bignum a, Bignum b, Bignum addend)
1138 int alen = a[0], blen = b[0];
1139 int mlen = (alen > blen ? alen : blen);
1140 int rlen, i, maxspot;
1141 BignumInt *workspace;
1144 /* mlen space for a, mlen space for b, 2*mlen for result */
1145 workspace = snewn(mlen * 4, BignumInt);
1146 for (i = 0; i < mlen; i++) {
1147 workspace[0 * mlen + i] = (mlen - i <= (int)a[0] ? a[mlen - i] : 0);
1148 workspace[1 * mlen + i] = (mlen - i <= (int)b[0] ? b[mlen - i] : 0);
1151 internal_mul(workspace + 0 * mlen, workspace + 1 * mlen,
1152 workspace + 2 * mlen, mlen);
1154 /* now just copy the result back */
1155 rlen = alen + blen + 1;
1156 if (addend && rlen <= (int)addend[0])
1157 rlen = addend[0] + 1;
1160 for (i = 1; i <= (int)ret[0]; i++) {
1161 ret[i] = (i <= 2 * mlen ? workspace[4 * mlen - i] : 0);
1167 /* now add in the addend, if any */
1169 BignumDblInt carry = 0;
1170 for (i = 1; i <= rlen; i++) {
1171 carry += (i <= (int)ret[0] ? ret[i] : 0);
1172 carry += (i <= (int)addend[0] ? addend[i] : 0);
1173 ret[i] = (BignumInt) carry & BIGNUM_INT_MASK;
1174 carry >>= BIGNUM_INT_BITS;
1175 if (ret[i] != 0 && i > maxspot)
1186 * Non-modular multiplication.
1188 Bignum bigmul(Bignum a, Bignum b)
1190 return bigmuladd(a, b, NULL);
1196 Bignum bigadd(Bignum a, Bignum b)
1198 int alen = a[0], blen = b[0];
1199 int rlen = (alen > blen ? alen : blen) + 1;
1208 for (i = 1; i <= rlen; i++) {
1209 carry += (i <= (int)a[0] ? a[i] : 0);
1210 carry += (i <= (int)b[0] ? b[i] : 0);
1211 ret[i] = (BignumInt) carry & BIGNUM_INT_MASK;
1212 carry >>= BIGNUM_INT_BITS;
1213 if (ret[i] != 0 && i > maxspot)
1222 * Subtraction. Returns a-b, or NULL if the result would come out
1223 * negative (recall that this entire bignum module only handles
1224 * positive numbers).
1226 Bignum bigsub(Bignum a, Bignum b)
1228 int alen = a[0], blen = b[0];
1229 int rlen = (alen > blen ? alen : blen);
1238 for (i = 1; i <= rlen; i++) {
1239 carry += (i <= (int)a[0] ? a[i] : 0);
1240 carry += (i <= (int)b[0] ? b[i] ^ BIGNUM_INT_MASK : BIGNUM_INT_MASK);
1241 ret[i] = (BignumInt) carry & BIGNUM_INT_MASK;
1242 carry >>= BIGNUM_INT_BITS;
1243 if (ret[i] != 0 && i > maxspot)
1257 * Create a bignum which is the bitmask covering another one. That
1258 * is, the smallest integer which is >= N and is also one less than
1261 Bignum bignum_bitmask(Bignum n)
1263 Bignum ret = copybn(n);
1268 while (n[i] == 0 && i > 0)
1271 return ret; /* input was zero */
1277 ret[i] = BIGNUM_INT_MASK;
1282 * Convert a (max 32-bit) long into a bignum.
1284 Bignum bignum_from_long(unsigned long nn)
1287 BignumDblInt n = nn;
1290 ret[1] = (BignumInt)(n & BIGNUM_INT_MASK);
1291 ret[2] = (BignumInt)((n >> BIGNUM_INT_BITS) & BIGNUM_INT_MASK);
1293 ret[0] = (ret[2] ? 2 : 1);
1298 * Add a long to a bignum.
1300 Bignum bignum_add_long(Bignum number, unsigned long addendx)
1302 Bignum ret = newbn(number[0] + 1);
1304 BignumDblInt carry = 0, addend = addendx;
1306 for (i = 1; i <= (int)ret[0]; i++) {
1307 carry += addend & BIGNUM_INT_MASK;
1308 carry += (i <= (int)number[0] ? number[i] : 0);
1309 addend >>= BIGNUM_INT_BITS;
1310 ret[i] = (BignumInt) carry & BIGNUM_INT_MASK;
1311 carry >>= BIGNUM_INT_BITS;
1320 * Compute the residue of a bignum, modulo a (max 16-bit) short.
1322 unsigned short bignum_mod_short(Bignum number, unsigned short modulus)
1324 BignumDblInt mod, r;
1329 for (i = number[0]; i > 0; i--)
1330 r = (r * (BIGNUM_TOP_BIT % mod) * 2 + number[i] % mod) % mod;
1331 return (unsigned short) r;
1335 void diagbn(char *prefix, Bignum md)
1337 int i, nibbles, morenibbles;
1338 static const char hex[] = "0123456789ABCDEF";
1340 debug(("%s0x", prefix ? prefix : ""));
1342 nibbles = (3 + bignum_bitcount(md)) / 4;
1345 morenibbles = 4 * md[0] - nibbles;
1346 for (i = 0; i < morenibbles; i++)
1348 for (i = nibbles; i--;)
1350 hex[(bignum_byte(md, i / 2) >> (4 * (i % 2))) & 0xF]));
1360 Bignum bigdiv(Bignum a, Bignum b)
1362 Bignum q = newbn(a[0]);
1363 bigdivmod(a, b, NULL, q);
1370 Bignum bigmod(Bignum a, Bignum b)
1372 Bignum r = newbn(b[0]);
1373 bigdivmod(a, b, r, NULL);
1378 * Greatest common divisor.
1380 Bignum biggcd(Bignum av, Bignum bv)
1382 Bignum a = copybn(av);
1383 Bignum b = copybn(bv);
1385 while (bignum_cmp(b, Zero) != 0) {
1386 Bignum t = newbn(b[0]);
1387 bigdivmod(a, b, t, NULL);
1388 while (t[0] > 1 && t[t[0]] == 0)
1400 * Modular inverse, using Euclid's extended algorithm.
1402 Bignum modinv(Bignum number, Bignum modulus)
1404 Bignum a = copybn(modulus);
1405 Bignum b = copybn(number);
1406 Bignum xp = copybn(Zero);
1407 Bignum x = copybn(One);
1410 while (bignum_cmp(b, One) != 0) {
1411 Bignum t = newbn(b[0]);
1412 Bignum q = newbn(a[0]);
1413 bigdivmod(a, b, t, q);
1414 while (t[0] > 1 && t[t[0]] == 0)
1421 x = bigmuladd(q, xp, t);
1431 /* now we know that sign * x == 1, and that x < modulus */
1433 /* set a new x to be modulus - x */
1434 Bignum newx = newbn(modulus[0]);
1435 BignumInt carry = 0;
1439 for (i = 1; i <= (int)newx[0]; i++) {
1440 BignumInt aword = (i <= (int)modulus[0] ? modulus[i] : 0);
1441 BignumInt bword = (i <= (int)x[0] ? x[i] : 0);
1442 newx[i] = aword - bword - carry;
1444 carry = carry ? (newx[i] >= bword) : (newx[i] > bword);
1458 * Render a bignum into decimal. Return a malloced string holding
1459 * the decimal representation.
1461 char *bignum_decimal(Bignum x)
1463 int ndigits, ndigit;
1467 BignumInt *workspace;
1470 * First, estimate the number of digits. Since log(10)/log(2)
1471 * is just greater than 93/28 (the joys of continued fraction
1472 * approximations...) we know that for every 93 bits, we need
1473 * at most 28 digits. This will tell us how much to malloc.
1475 * Formally: if x has i bits, that means x is strictly less
1476 * than 2^i. Since 2 is less than 10^(28/93), this is less than
1477 * 10^(28i/93). We need an integer power of ten, so we must
1478 * round up (rounding down might make it less than x again).
1479 * Therefore if we multiply the bit count by 28/93, rounding
1480 * up, we will have enough digits.
1482 * i=0 (i.e., x=0) is an irritating special case.
1484 i = bignum_bitcount(x);
1486 ndigits = 1; /* x = 0 */
1488 ndigits = (28 * i + 92) / 93; /* multiply by 28/93 and round up */
1489 ndigits++; /* allow for trailing \0 */
1490 ret = snewn(ndigits, char);
1493 * Now allocate some workspace to hold the binary form as we
1494 * repeatedly divide it by ten. Initialise this to the
1495 * big-endian form of the number.
1497 workspace = snewn(x[0], BignumInt);
1498 for (i = 0; i < (int)x[0]; i++)
1499 workspace[i] = x[x[0] - i];
1502 * Next, write the decimal number starting with the last digit.
1503 * We use ordinary short division, dividing 10 into the
1506 ndigit = ndigits - 1;
1511 for (i = 0; i < (int)x[0]; i++) {
1512 carry = (carry << BIGNUM_INT_BITS) + workspace[i];
1513 workspace[i] = (BignumInt) (carry / 10);
1518 ret[--ndigit] = (char) (carry + '0');
1522 * There's a chance we've fallen short of the start of the
1523 * string. Correct if so.
1526 memmove(ret, ret + ndigit, ndigits - ndigit);