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;
255 * The coefficients a_1 b_1 and a_0 b_0 just avoid overlapping
256 * in the output array, so we can compute them immediately in
261 printf("a1,a0 = 0x");
262 for (i = 0; i < len; i++) {
263 if (i == toplen) printf(", 0x");
264 printf("%0*x", BIGNUM_INT_BITS/4, a[i]);
267 printf("b1,b0 = 0x");
268 for (i = 0; i < len; i++) {
269 if (i == toplen) printf(", 0x");
270 printf("%0*x", BIGNUM_INT_BITS/4, b[i]);
276 internal_mul(a, b, c, toplen);
279 for (i = 0; i < 2*toplen; i++) {
280 printf("%0*x", BIGNUM_INT_BITS/4, c[i]);
286 internal_mul(a + toplen, b + toplen, c + 2*toplen, botlen);
289 for (i = 0; i < 2*botlen; i++) {
290 printf("%0*x", BIGNUM_INT_BITS/4, c[2*toplen+i]);
296 * We must allocate scratch space for the central coefficient,
297 * and also for the two input values that we multiply when
298 * computing it. Since either or both may carry into the
299 * (botlen+1)th word, we must use a slightly longer length
302 scratch = snewn(4 * midlen, BignumInt);
304 /* Zero padding. midlen exceeds toplen by at most 2, so just
305 * zero the first two words of each input and the rest will be
307 scratch[0] = scratch[1] = scratch[midlen] = scratch[midlen+1] = 0;
309 for (j = 0; j < toplen; j++) {
310 scratch[midlen - toplen + j] = a[j]; /* a_1 */
311 scratch[2*midlen - toplen + j] = b[j]; /* b_1 */
314 /* compute a_1 + a_0 */
315 scratch[0] = internal_add(scratch+1, a+toplen, scratch+1, botlen);
317 printf("a1plusa0 = 0x");
318 for (i = 0; i < midlen; i++) {
319 printf("%0*x", BIGNUM_INT_BITS/4, scratch[i]);
323 /* compute b_1 + b_0 */
324 scratch[midlen] = internal_add(scratch+midlen+1, b+toplen,
325 scratch+midlen+1, botlen);
327 printf("b1plusb0 = 0x");
328 for (i = 0; i < midlen; i++) {
329 printf("%0*x", BIGNUM_INT_BITS/4, scratch[midlen+i]);
335 * Now we can do the third multiplication.
337 internal_mul(scratch, scratch + midlen, scratch + 2*midlen, midlen);
339 printf("a1plusa0timesb1plusb0 = 0x");
340 for (i = 0; i < 2*midlen; i++) {
341 printf("%0*x", BIGNUM_INT_BITS/4, scratch[2*midlen+i]);
347 * Now we can reuse the first half of 'scratch' to compute the
348 * sum of the outer two coefficients, to subtract from that
349 * product to obtain the middle one.
351 scratch[0] = scratch[1] = scratch[2] = scratch[3] = 0;
352 for (j = 0; j < 2*toplen; j++)
353 scratch[2*midlen - 2*toplen + j] = c[j];
354 scratch[1] = internal_add(scratch+2, c + 2*toplen,
355 scratch+2, 2*botlen);
357 printf("a1b1plusa0b0 = 0x");
358 for (i = 0; i < 2*midlen; i++) {
359 printf("%0*x", BIGNUM_INT_BITS/4, scratch[i]);
364 internal_sub(scratch + 2*midlen, scratch,
365 scratch + 2*midlen, 2*midlen);
367 printf("a1b0plusa0b1 = 0x");
368 for (i = 0; i < 2*midlen; i++) {
369 printf("%0*x", BIGNUM_INT_BITS/4, scratch[2*midlen+i]);
375 * And now all we need to do is to add that middle coefficient
376 * back into the output. We may have to propagate a carry
377 * further up the output, but we can be sure it won't
378 * propagate right the way off the top.
380 carry = internal_add(c + 2*len - botlen - 2*midlen,
382 c + 2*len - botlen - 2*midlen, 2*midlen);
383 j = 2*len - botlen - 2*midlen - 1;
387 c[j] = (BignumInt)carry;
388 carry >>= BIGNUM_INT_BITS;
392 for (i = 0; i < 2*len; i++) {
393 printf("%0*x", BIGNUM_INT_BITS/4, c[i]);
399 for (j = 0; j < 4 * midlen; j++)
406 * Multiply in the ordinary O(N^2) way.
409 for (j = 0; j < 2 * len; j++)
412 for (i = len - 1; i >= 0; i--) {
414 for (j = len - 1; j >= 0; j--) {
415 t += MUL_WORD(a[i], (BignumDblInt) b[j]);
416 t += (BignumDblInt) c[i + j + 1];
417 c[i + j + 1] = (BignumInt) t;
418 t = t >> BIGNUM_INT_BITS;
420 c[i] = (BignumInt) t;
426 * Variant form of internal_mul used for the initial step of
427 * Montgomery reduction. Only bothers outputting 'len' words
428 * (everything above that is thrown away).
430 static void internal_mul_low(const BignumInt *a, const BignumInt *b,
431 BignumInt *c, int len)
436 if (len > KARATSUBA_THRESHOLD) {
439 * Karatsuba-aware version of internal_mul_low. As before, we
440 * express each input value as a shifted combination of two
446 * Then the full product is, as before,
448 * ab = a_1 b_1 D^2 + (a_1 b_0 + a_0 b_1) D + a_0 b_0
450 * Provided we choose D on the large side (so that a_0 and b_0
451 * are _at least_ as long as a_1 and b_1), we don't need the
452 * topmost term at all, and we only need half of the middle
453 * term. So there's no point in doing the proper Karatsuba
454 * optimisation which computes the middle term using the top
455 * one, because we'd take as long computing the top one as
456 * just computing the middle one directly.
458 * So instead, we do a much more obvious thing: we call the
459 * fully optimised internal_mul to compute a_0 b_0, and we
460 * recursively call ourself to compute the _bottom halves_ of
461 * a_1 b_0 and a_0 b_1, each of which we add into the result
462 * in the obvious way.
464 * In other words, there's no actual Karatsuba _optimisation_
465 * in this function; the only benefit in doing it this way is
466 * that we call internal_mul proper for a large part of the
467 * work, and _that_ can optimise its operation.
470 int toplen = len/2, botlen = len - toplen; /* botlen is the bigger */
474 * Allocate scratch space for the various bits and pieces
475 * we're going to be adding together. We need botlen*2 words
476 * for a_0 b_0 (though we may end up throwing away its topmost
477 * word), and toplen words for each of a_1 b_0 and a_0 b_1.
478 * That adds up to exactly 2*len.
480 scratch = snewn(len*2, BignumInt);
483 internal_mul(a + toplen, b + toplen, scratch + 2*toplen, botlen);
486 internal_mul_low(a, b + len - toplen, scratch + toplen, toplen);
489 internal_mul_low(a + len - toplen, b, scratch, toplen);
491 /* Copy the bottom half of the big coefficient into place */
492 for (j = 0; j < botlen; j++)
493 c[toplen + j] = scratch[2*toplen + botlen + j];
495 /* Add the two small coefficients, throwing away the returned carry */
496 internal_add(scratch, scratch + toplen, scratch, toplen);
498 /* And add that to the large coefficient, leaving the result in c. */
499 internal_add(scratch, scratch + 2*toplen + botlen - toplen,
503 for (j = 0; j < len*2; j++)
509 for (j = 0; j < len; j++)
512 for (i = len - 1; i >= 0; i--) {
514 for (j = len - 1; j >= len - i - 1; j--) {
515 t += MUL_WORD(a[i], (BignumDblInt) b[j]);
516 t += (BignumDblInt) c[i + j + 1 - len];
517 c[i + j + 1 - len] = (BignumInt) t;
518 t = t >> BIGNUM_INT_BITS;
526 * Montgomery reduction. Expects x to be a big-endian array of 2*len
527 * BignumInts whose value satisfies 0 <= x < rn (where r = 2^(len *
528 * BIGNUM_INT_BITS) is the Montgomery base). Returns in the same array
529 * a value x' which is congruent to xr^{-1} mod n, and satisfies 0 <=
532 * 'n' and 'mninv' should be big-endian arrays of 'len' BignumInts
533 * each, containing respectively n and the multiplicative inverse of
536 * 'tmp' is an array of at least '3*len' BignumInts used as scratch
539 static void monty_reduce(BignumInt *x, const BignumInt *n,
540 const BignumInt *mninv, BignumInt *tmp, int len)
546 * Multiply x by (-n)^{-1} mod r. This gives us a value m such
547 * that mn is congruent to -x mod r. Hence, mn+x is an exact
548 * multiple of r, and is also (obviously) congruent to x mod n.
550 internal_mul_low(x + len, mninv, tmp, len);
553 * Compute t = (mn+x)/r in ordinary, non-modular, integer
554 * arithmetic. By construction this is exact, and is congruent mod
555 * n to x * r^{-1}, i.e. the answer we want.
557 * The following multiply leaves that answer in the _most_
558 * significant half of the 'x' array, so then we must shift it
561 internal_mul(tmp, n, tmp+len, len);
562 carry = internal_add(x, tmp+len, x, 2*len);
563 for (i = 0; i < len; i++)
564 x[len + i] = x[i], x[i] = 0;
567 * Reduce t mod n. This doesn't require a full-on division by n,
568 * but merely a test and single optional subtraction, since we can
569 * show that 0 <= t < 2n.
572 * + we computed m mod r, so 0 <= m < r.
573 * + so 0 <= mn < rn, obviously
574 * + hence we only need 0 <= x < rn to guarantee that 0 <= mn+x < 2rn
575 * + yielding 0 <= (mn+x)/r < 2n as required.
578 for (i = 0; i < len; i++)
579 if (x[len + i] != n[i])
582 if (carry || i >= len || x[len + i] > n[i])
583 internal_sub(x+len, n, x+len, len);
586 static void internal_add_shifted(BignumInt *number,
587 unsigned n, int shift)
589 int word = 1 + (shift / BIGNUM_INT_BITS);
590 int bshift = shift % BIGNUM_INT_BITS;
593 addend = (BignumDblInt)n << bshift;
596 addend += number[word];
597 number[word] = (BignumInt) addend & BIGNUM_INT_MASK;
598 addend >>= BIGNUM_INT_BITS;
605 * Input in first alen words of a and first mlen words of m.
606 * Output in first alen words of a
607 * (of which first alen-mlen words will be zero).
608 * The MSW of m MUST have its high bit set.
609 * Quotient is accumulated in the `quotient' array, which is a Bignum
610 * rather than the internal bigendian format. Quotient parts are shifted
611 * left by `qshift' before adding into quot.
613 static void internal_mod(BignumInt *a, int alen,
614 BignumInt *m, int mlen,
615 BignumInt *quot, int qshift)
627 for (i = 0; i <= alen - mlen; i++) {
629 unsigned int q, r, c, ai1;
643 /* Find q = h:a[i] / m0 */
648 * To illustrate it, suppose a BignumInt is 8 bits, and
649 * we are dividing (say) A1:23:45:67 by A1:B2:C3. Then
650 * our initial division will be 0xA123 / 0xA1, which
651 * will give a quotient of 0x100 and a divide overflow.
652 * However, the invariants in this division algorithm
653 * are not violated, since the full number A1:23:... is
654 * _less_ than the quotient prefix A1:B2:... and so the
655 * following correction loop would have sorted it out.
657 * In this situation we set q to be the largest
658 * quotient we _can_ stomach (0xFF, of course).
662 /* Macro doesn't want an array subscript expression passed
663 * into it (see definition), so use a temporary. */
664 BignumInt tmplo = a[i];
665 DIVMOD_WORD(q, r, h, tmplo, m0);
667 /* Refine our estimate of q by looking at
668 h:a[i]:a[i+1] / m0:m1 */
670 if (t > ((BignumDblInt) r << BIGNUM_INT_BITS) + ai1) {
673 r = (r + m0) & BIGNUM_INT_MASK; /* overflow? */
674 if (r >= (BignumDblInt) m0 &&
675 t > ((BignumDblInt) r << BIGNUM_INT_BITS) + ai1) q--;
679 /* Subtract q * m from a[i...] */
681 for (k = mlen - 1; k >= 0; k--) {
682 t = MUL_WORD(q, m[k]);
684 c = (unsigned)(t >> BIGNUM_INT_BITS);
685 if ((BignumInt) t > a[i + k])
687 a[i + k] -= (BignumInt) t;
690 /* Add back m in case of borrow */
693 for (k = mlen - 1; k >= 0; k--) {
696 a[i + k] = (BignumInt) t;
697 t = t >> BIGNUM_INT_BITS;
702 internal_add_shifted(quot, q, qshift + BIGNUM_INT_BITS * (alen - mlen - i));
707 * Compute (base ^ exp) % mod. Uses the Montgomery multiplication
710 Bignum modpow(Bignum base_in, Bignum exp, Bignum mod)
712 BignumInt *a, *b, *x, *n, *mninv, *tmp;
714 Bignum base, base2, r, rn, inv, result;
717 * The most significant word of mod needs to be non-zero. It
718 * should already be, but let's make sure.
720 assert(mod[mod[0]] != 0);
723 * Make sure the base is smaller than the modulus, by reducing
724 * it modulo the modulus if not.
726 base = bigmod(base_in, mod);
729 * mod had better be odd, or we can't do Montgomery multiplication
730 * using a power of two at all.
735 * Compute the inverse of n mod r, for monty_reduce. (In fact we
736 * want the inverse of _minus_ n mod r, but we'll sort that out
740 r = bn_power_2(BIGNUM_INT_BITS * len);
741 inv = modinv(mod, r);
744 * Multiply the base by r mod n, to get it into Montgomery
747 base2 = modmul(base, r, mod);
751 rn = bigmod(r, mod); /* r mod n, i.e. Montgomerified 1 */
753 freebn(r); /* won't need this any more */
756 * Set up internal arrays of the right lengths, in big-endian
757 * format, containing the base, the modulus, and the modulus's
760 n = snewn(len, BignumInt);
761 for (j = 0; j < len; j++)
762 n[len - 1 - j] = mod[j + 1];
764 mninv = snewn(len, BignumInt);
765 for (j = 0; j < len; j++)
766 mninv[len - 1 - j] = (j < inv[0] ? inv[j + 1] : 0);
767 freebn(inv); /* we don't need this copy of it any more */
768 /* Now negate mninv mod r, so it's the inverse of -n rather than +n. */
769 x = snewn(len, BignumInt);
770 for (j = 0; j < len; j++)
772 internal_sub(x, mninv, mninv, len);
774 /* x = snewn(len, BignumInt); */ /* already done above */
775 for (j = 0; j < len; j++)
776 x[len - 1 - j] = (j < base[0] ? base[j + 1] : 0);
777 freebn(base); /* we don't need this copy of it any more */
779 a = snewn(2*len, BignumInt);
780 b = snewn(2*len, BignumInt);
781 for (j = 0; j < len; j++)
782 a[2*len - 1 - j] = (j < rn[0] ? rn[j + 1] : 0);
785 tmp = snewn(3*len, BignumInt);
787 /* Skip leading zero bits of exp. */
789 j = BIGNUM_INT_BITS-1;
790 while (i < (int)exp[0] && (exp[exp[0] - i] & (1 << j)) == 0) {
794 j = BIGNUM_INT_BITS-1;
798 /* Main computation */
799 while (i < (int)exp[0]) {
801 internal_mul(a + len, a + len, b, len);
802 monty_reduce(b, n, mninv, tmp, len);
803 if ((exp[exp[0] - i] & (1 << j)) != 0) {
804 internal_mul(b + len, x, a, len);
805 monty_reduce(a, n, mninv, tmp, len);
815 j = BIGNUM_INT_BITS-1;
819 * Final monty_reduce to get back from the adjusted Montgomery
822 monty_reduce(a, n, mninv, tmp, len);
824 /* Copy result to buffer */
825 result = newbn(mod[0]);
826 for (i = 0; i < len; i++)
827 result[result[0] - i] = a[i + len];
828 while (result[0] > 1 && result[result[0]] == 0)
831 /* Free temporary arrays */
832 for (i = 0; i < 3 * len; i++)
835 for (i = 0; i < 2 * len; i++)
838 for (i = 0; i < 2 * len; i++)
841 for (i = 0; i < len; i++)
844 for (i = 0; i < len; i++)
847 for (i = 0; i < len; i++)
855 * Compute (p * q) % mod.
856 * The most significant word of mod MUST be non-zero.
857 * We assume that the result array is the same size as the mod array.
859 Bignum modmul(Bignum p, Bignum q, Bignum mod)
861 BignumInt *a, *n, *m, *o;
863 int pqlen, mlen, rlen, i, j;
866 /* Allocate m of size mlen, copy mod to m */
867 /* We use big endian internally */
869 m = snewn(mlen, BignumInt);
870 for (j = 0; j < mlen; j++)
871 m[j] = mod[mod[0] - j];
873 /* Shift m left to make msb bit set */
874 for (mshift = 0; mshift < BIGNUM_INT_BITS-1; mshift++)
875 if ((m[0] << mshift) & BIGNUM_TOP_BIT)
878 for (i = 0; i < mlen - 1; i++)
879 m[i] = (m[i] << mshift) | (m[i + 1] >> (BIGNUM_INT_BITS - mshift));
880 m[mlen - 1] = m[mlen - 1] << mshift;
883 pqlen = (p[0] > q[0] ? p[0] : q[0]);
885 /* Allocate n of size pqlen, copy p to n */
886 n = snewn(pqlen, BignumInt);
888 for (j = 0; j < i; j++)
890 for (j = 0; j < (int)p[0]; j++)
891 n[i + j] = p[p[0] - j];
893 /* Allocate o of size pqlen, copy q to o */
894 o = snewn(pqlen, BignumInt);
896 for (j = 0; j < i; j++)
898 for (j = 0; j < (int)q[0]; j++)
899 o[i + j] = q[q[0] - j];
901 /* Allocate a of size 2*pqlen for result */
902 a = snewn(2 * pqlen, BignumInt);
904 /* Main computation */
905 internal_mul(n, o, a, pqlen);
906 internal_mod(a, pqlen * 2, m, mlen, NULL, 0);
908 /* Fixup result in case the modulus was shifted */
910 for (i = 2 * pqlen - mlen - 1; i < 2 * pqlen - 1; i++)
911 a[i] = (a[i] << mshift) | (a[i + 1] >> (BIGNUM_INT_BITS - mshift));
912 a[2 * pqlen - 1] = a[2 * pqlen - 1] << mshift;
913 internal_mod(a, pqlen * 2, m, mlen, NULL, 0);
914 for (i = 2 * pqlen - 1; i >= 2 * pqlen - mlen; i--)
915 a[i] = (a[i] >> mshift) | (a[i - 1] << (BIGNUM_INT_BITS - mshift));
918 /* Copy result to buffer */
919 rlen = (mlen < pqlen * 2 ? mlen : pqlen * 2);
920 result = newbn(rlen);
921 for (i = 0; i < rlen; i++)
922 result[result[0] - i] = a[i + 2 * pqlen - rlen];
923 while (result[0] > 1 && result[result[0]] == 0)
926 /* Free temporary arrays */
927 for (i = 0; i < 2 * pqlen; i++)
930 for (i = 0; i < mlen; i++)
933 for (i = 0; i < pqlen; i++)
936 for (i = 0; i < pqlen; i++)
945 * The most significant word of mod MUST be non-zero.
946 * We assume that the result array is the same size as the mod array.
947 * We optionally write out a quotient if `quotient' is non-NULL.
948 * We can avoid writing out the result if `result' is NULL.
950 static void bigdivmod(Bignum p, Bignum mod, Bignum result, Bignum quotient)
954 int plen, mlen, i, j;
956 /* Allocate m of size mlen, copy mod to m */
957 /* We use big endian internally */
959 m = snewn(mlen, BignumInt);
960 for (j = 0; j < mlen; j++)
961 m[j] = mod[mod[0] - j];
963 /* Shift m left to make msb bit set */
964 for (mshift = 0; mshift < BIGNUM_INT_BITS-1; mshift++)
965 if ((m[0] << mshift) & BIGNUM_TOP_BIT)
968 for (i = 0; i < mlen - 1; i++)
969 m[i] = (m[i] << mshift) | (m[i + 1] >> (BIGNUM_INT_BITS - mshift));
970 m[mlen - 1] = m[mlen - 1] << mshift;
974 /* Ensure plen > mlen */
978 /* Allocate n of size plen, copy p to n */
979 n = snewn(plen, BignumInt);
980 for (j = 0; j < plen; j++)
982 for (j = 1; j <= (int)p[0]; j++)
985 /* Main computation */
986 internal_mod(n, plen, m, mlen, quotient, mshift);
988 /* Fixup result in case the modulus was shifted */
990 for (i = plen - mlen - 1; i < plen - 1; i++)
991 n[i] = (n[i] << mshift) | (n[i + 1] >> (BIGNUM_INT_BITS - mshift));
992 n[plen - 1] = n[plen - 1] << mshift;
993 internal_mod(n, plen, m, mlen, quotient, 0);
994 for (i = plen - 1; i >= plen - mlen; i--)
995 n[i] = (n[i] >> mshift) | (n[i - 1] << (BIGNUM_INT_BITS - mshift));
998 /* Copy result to buffer */
1000 for (i = 1; i <= (int)result[0]; i++) {
1002 result[i] = j >= 0 ? n[j] : 0;
1006 /* Free temporary arrays */
1007 for (i = 0; i < mlen; i++)
1010 for (i = 0; i < plen; i++)
1016 * Decrement a number.
1018 void decbn(Bignum bn)
1021 while (i < (int)bn[0] && bn[i] == 0)
1022 bn[i++] = BIGNUM_INT_MASK;
1026 Bignum bignum_from_bytes(const unsigned char *data, int nbytes)
1031 w = (nbytes + BIGNUM_INT_BYTES - 1) / BIGNUM_INT_BYTES; /* bytes->words */
1034 for (i = 1; i <= w; i++)
1036 for (i = nbytes; i--;) {
1037 unsigned char byte = *data++;
1038 result[1 + i / BIGNUM_INT_BYTES] |= byte << (8*i % BIGNUM_INT_BITS);
1041 while (result[0] > 1 && result[result[0]] == 0)
1047 * Read an SSH-1-format bignum from a data buffer. Return the number
1048 * of bytes consumed, or -1 if there wasn't enough data.
1050 int ssh1_read_bignum(const unsigned char *data, int len, Bignum * result)
1052 const unsigned char *p = data;
1060 for (i = 0; i < 2; i++)
1061 w = (w << 8) + *p++;
1062 b = (w + 7) / 8; /* bits -> bytes */
1067 if (!result) /* just return length */
1070 *result = bignum_from_bytes(p, b);
1072 return p + b - data;
1076 * Return the bit count of a bignum, for SSH-1 encoding.
1078 int bignum_bitcount(Bignum bn)
1080 int bitcount = bn[0] * BIGNUM_INT_BITS - 1;
1081 while (bitcount >= 0
1082 && (bn[bitcount / BIGNUM_INT_BITS + 1] >> (bitcount % BIGNUM_INT_BITS)) == 0) bitcount--;
1083 return bitcount + 1;
1087 * Return the byte length of a bignum when SSH-1 encoded.
1089 int ssh1_bignum_length(Bignum bn)
1091 return 2 + (bignum_bitcount(bn) + 7) / 8;
1095 * Return the byte length of a bignum when SSH-2 encoded.
1097 int ssh2_bignum_length(Bignum bn)
1099 return 4 + (bignum_bitcount(bn) + 8) / 8;
1103 * Return a byte from a bignum; 0 is least significant, etc.
1105 int bignum_byte(Bignum bn, int i)
1107 if (i >= (int)(BIGNUM_INT_BYTES * bn[0]))
1108 return 0; /* beyond the end */
1110 return (bn[i / BIGNUM_INT_BYTES + 1] >>
1111 ((i % BIGNUM_INT_BYTES)*8)) & 0xFF;
1115 * Return a bit from a bignum; 0 is least significant, etc.
1117 int bignum_bit(Bignum bn, int i)
1119 if (i >= (int)(BIGNUM_INT_BITS * bn[0]))
1120 return 0; /* beyond the end */
1122 return (bn[i / BIGNUM_INT_BITS + 1] >> (i % BIGNUM_INT_BITS)) & 1;
1126 * Set a bit in a bignum; 0 is least significant, etc.
1128 void bignum_set_bit(Bignum bn, int bitnum, int value)
1130 if (bitnum >= (int)(BIGNUM_INT_BITS * bn[0]))
1131 abort(); /* beyond the end */
1133 int v = bitnum / BIGNUM_INT_BITS + 1;
1134 int mask = 1 << (bitnum % BIGNUM_INT_BITS);
1143 * Write a SSH-1-format bignum into a buffer. It is assumed the
1144 * buffer is big enough. Returns the number of bytes used.
1146 int ssh1_write_bignum(void *data, Bignum bn)
1148 unsigned char *p = data;
1149 int len = ssh1_bignum_length(bn);
1151 int bitc = bignum_bitcount(bn);
1153 *p++ = (bitc >> 8) & 0xFF;
1154 *p++ = (bitc) & 0xFF;
1155 for (i = len - 2; i--;)
1156 *p++ = bignum_byte(bn, i);
1161 * Compare two bignums. Returns like strcmp.
1163 int bignum_cmp(Bignum a, Bignum b)
1165 int amax = a[0], bmax = b[0];
1166 int i = (amax > bmax ? amax : bmax);
1168 BignumInt aval = (i > amax ? 0 : a[i]);
1169 BignumInt bval = (i > bmax ? 0 : b[i]);
1180 * Right-shift one bignum to form another.
1182 Bignum bignum_rshift(Bignum a, int shift)
1185 int i, shiftw, shiftb, shiftbb, bits;
1188 bits = bignum_bitcount(a) - shift;
1189 ret = newbn((bits + BIGNUM_INT_BITS - 1) / BIGNUM_INT_BITS);
1192 shiftw = shift / BIGNUM_INT_BITS;
1193 shiftb = shift % BIGNUM_INT_BITS;
1194 shiftbb = BIGNUM_INT_BITS - shiftb;
1196 ai1 = a[shiftw + 1];
1197 for (i = 1; i <= (int)ret[0]; i++) {
1199 ai1 = (i + shiftw + 1 <= (int)a[0] ? a[i + shiftw + 1] : 0);
1200 ret[i] = ((ai >> shiftb) | (ai1 << shiftbb)) & BIGNUM_INT_MASK;
1208 * Non-modular multiplication and addition.
1210 Bignum bigmuladd(Bignum a, Bignum b, Bignum addend)
1212 int alen = a[0], blen = b[0];
1213 int mlen = (alen > blen ? alen : blen);
1214 int rlen, i, maxspot;
1215 BignumInt *workspace;
1218 /* mlen space for a, mlen space for b, 2*mlen for result */
1219 workspace = snewn(mlen * 4, BignumInt);
1220 for (i = 0; i < mlen; i++) {
1221 workspace[0 * mlen + i] = (mlen - i <= (int)a[0] ? a[mlen - i] : 0);
1222 workspace[1 * mlen + i] = (mlen - i <= (int)b[0] ? b[mlen - i] : 0);
1225 internal_mul(workspace + 0 * mlen, workspace + 1 * mlen,
1226 workspace + 2 * mlen, mlen);
1228 /* now just copy the result back */
1229 rlen = alen + blen + 1;
1230 if (addend && rlen <= (int)addend[0])
1231 rlen = addend[0] + 1;
1234 for (i = 1; i <= (int)ret[0]; i++) {
1235 ret[i] = (i <= 2 * mlen ? workspace[4 * mlen - i] : 0);
1241 /* now add in the addend, if any */
1243 BignumDblInt carry = 0;
1244 for (i = 1; i <= rlen; i++) {
1245 carry += (i <= (int)ret[0] ? ret[i] : 0);
1246 carry += (i <= (int)addend[0] ? addend[i] : 0);
1247 ret[i] = (BignumInt) carry & BIGNUM_INT_MASK;
1248 carry >>= BIGNUM_INT_BITS;
1249 if (ret[i] != 0 && i > maxspot)
1260 * Non-modular multiplication.
1262 Bignum bigmul(Bignum a, Bignum b)
1264 return bigmuladd(a, b, NULL);
1270 Bignum bigadd(Bignum a, Bignum b)
1272 int alen = a[0], blen = b[0];
1273 int rlen = (alen > blen ? alen : blen) + 1;
1282 for (i = 1; i <= rlen; i++) {
1283 carry += (i <= (int)a[0] ? a[i] : 0);
1284 carry += (i <= (int)b[0] ? b[i] : 0);
1285 ret[i] = (BignumInt) carry & BIGNUM_INT_MASK;
1286 carry >>= BIGNUM_INT_BITS;
1287 if (ret[i] != 0 && i > maxspot)
1296 * Subtraction. Returns a-b, or NULL if the result would come out
1297 * negative (recall that this entire bignum module only handles
1298 * positive numbers).
1300 Bignum bigsub(Bignum a, Bignum b)
1302 int alen = a[0], blen = b[0];
1303 int rlen = (alen > blen ? alen : blen);
1312 for (i = 1; i <= rlen; i++) {
1313 carry += (i <= (int)a[0] ? a[i] : 0);
1314 carry += (i <= (int)b[0] ? b[i] ^ BIGNUM_INT_MASK : BIGNUM_INT_MASK);
1315 ret[i] = (BignumInt) carry & BIGNUM_INT_MASK;
1316 carry >>= BIGNUM_INT_BITS;
1317 if (ret[i] != 0 && i > maxspot)
1331 * Create a bignum which is the bitmask covering another one. That
1332 * is, the smallest integer which is >= N and is also one less than
1335 Bignum bignum_bitmask(Bignum n)
1337 Bignum ret = copybn(n);
1342 while (n[i] == 0 && i > 0)
1345 return ret; /* input was zero */
1351 ret[i] = BIGNUM_INT_MASK;
1356 * Convert a (max 32-bit) long into a bignum.
1358 Bignum bignum_from_long(unsigned long nn)
1361 BignumDblInt n = nn;
1364 ret[1] = (BignumInt)(n & BIGNUM_INT_MASK);
1365 ret[2] = (BignumInt)((n >> BIGNUM_INT_BITS) & BIGNUM_INT_MASK);
1367 ret[0] = (ret[2] ? 2 : 1);
1372 * Add a long to a bignum.
1374 Bignum bignum_add_long(Bignum number, unsigned long addendx)
1376 Bignum ret = newbn(number[0] + 1);
1378 BignumDblInt carry = 0, addend = addendx;
1380 for (i = 1; i <= (int)ret[0]; i++) {
1381 carry += addend & BIGNUM_INT_MASK;
1382 carry += (i <= (int)number[0] ? number[i] : 0);
1383 addend >>= BIGNUM_INT_BITS;
1384 ret[i] = (BignumInt) carry & BIGNUM_INT_MASK;
1385 carry >>= BIGNUM_INT_BITS;
1394 * Compute the residue of a bignum, modulo a (max 16-bit) short.
1396 unsigned short bignum_mod_short(Bignum number, unsigned short modulus)
1398 BignumDblInt mod, r;
1403 for (i = number[0]; i > 0; i--)
1404 r = (r * (BIGNUM_TOP_BIT % mod) * 2 + number[i] % mod) % mod;
1405 return (unsigned short) r;
1409 void diagbn(char *prefix, Bignum md)
1411 int i, nibbles, morenibbles;
1412 static const char hex[] = "0123456789ABCDEF";
1414 debug(("%s0x", prefix ? prefix : ""));
1416 nibbles = (3 + bignum_bitcount(md)) / 4;
1419 morenibbles = 4 * md[0] - nibbles;
1420 for (i = 0; i < morenibbles; i++)
1422 for (i = nibbles; i--;)
1424 hex[(bignum_byte(md, i / 2) >> (4 * (i % 2))) & 0xF]));
1434 Bignum bigdiv(Bignum a, Bignum b)
1436 Bignum q = newbn(a[0]);
1437 bigdivmod(a, b, NULL, q);
1444 Bignum bigmod(Bignum a, Bignum b)
1446 Bignum r = newbn(b[0]);
1447 bigdivmod(a, b, r, NULL);
1452 * Greatest common divisor.
1454 Bignum biggcd(Bignum av, Bignum bv)
1456 Bignum a = copybn(av);
1457 Bignum b = copybn(bv);
1459 while (bignum_cmp(b, Zero) != 0) {
1460 Bignum t = newbn(b[0]);
1461 bigdivmod(a, b, t, NULL);
1462 while (t[0] > 1 && t[t[0]] == 0)
1474 * Modular inverse, using Euclid's extended algorithm.
1476 Bignum modinv(Bignum number, Bignum modulus)
1478 Bignum a = copybn(modulus);
1479 Bignum b = copybn(number);
1480 Bignum xp = copybn(Zero);
1481 Bignum x = copybn(One);
1484 while (bignum_cmp(b, One) != 0) {
1485 Bignum t = newbn(b[0]);
1486 Bignum q = newbn(a[0]);
1487 bigdivmod(a, b, t, q);
1488 while (t[0] > 1 && t[t[0]] == 0)
1495 x = bigmuladd(q, xp, t);
1505 /* now we know that sign * x == 1, and that x < modulus */
1507 /* set a new x to be modulus - x */
1508 Bignum newx = newbn(modulus[0]);
1509 BignumInt carry = 0;
1513 for (i = 1; i <= (int)newx[0]; i++) {
1514 BignumInt aword = (i <= (int)modulus[0] ? modulus[i] : 0);
1515 BignumInt bword = (i <= (int)x[0] ? x[i] : 0);
1516 newx[i] = aword - bword - carry;
1518 carry = carry ? (newx[i] >= bword) : (newx[i] > bword);
1532 * Render a bignum into decimal. Return a malloced string holding
1533 * the decimal representation.
1535 char *bignum_decimal(Bignum x)
1537 int ndigits, ndigit;
1541 BignumInt *workspace;
1544 * First, estimate the number of digits. Since log(10)/log(2)
1545 * is just greater than 93/28 (the joys of continued fraction
1546 * approximations...) we know that for every 93 bits, we need
1547 * at most 28 digits. This will tell us how much to malloc.
1549 * Formally: if x has i bits, that means x is strictly less
1550 * than 2^i. Since 2 is less than 10^(28/93), this is less than
1551 * 10^(28i/93). We need an integer power of ten, so we must
1552 * round up (rounding down might make it less than x again).
1553 * Therefore if we multiply the bit count by 28/93, rounding
1554 * up, we will have enough digits.
1556 * i=0 (i.e., x=0) is an irritating special case.
1558 i = bignum_bitcount(x);
1560 ndigits = 1; /* x = 0 */
1562 ndigits = (28 * i + 92) / 93; /* multiply by 28/93 and round up */
1563 ndigits++; /* allow for trailing \0 */
1564 ret = snewn(ndigits, char);
1567 * Now allocate some workspace to hold the binary form as we
1568 * repeatedly divide it by ten. Initialise this to the
1569 * big-endian form of the number.
1571 workspace = snewn(x[0], BignumInt);
1572 for (i = 0; i < (int)x[0]; i++)
1573 workspace[i] = x[x[0] - i];
1576 * Next, write the decimal number starting with the last digit.
1577 * We use ordinary short division, dividing 10 into the
1580 ndigit = ndigits - 1;
1585 for (i = 0; i < (int)x[0]; i++) {
1586 carry = (carry << BIGNUM_INT_BITS) + workspace[i];
1587 workspace[i] = (BignumInt) (carry / 10);
1592 ret[--ndigit] = (char) (carry + '0');
1596 * There's a chance we've fallen short of the start of the
1597 * string. Correct if so.
1600 memmove(ret, ret + ndigit, ndigits - ndigit);
1616 * gcc -g -O0 -DTESTBN -o testbn sshbn.c misc.c -I unix -I charset
1619 void modalfatalbox(char *p, ...)
1622 fprintf(stderr, "FATAL ERROR: ");
1624 vfprintf(stderr, p, ap);
1626 fputc('\n', stderr);
1630 #define fromxdigit(c) ( (c)>'9' ? ((c)&0xDF) - 'A' + 10 : (c) - '0' )
1632 int main(int argc, char **argv)
1636 int passes = 0, fails = 0;
1638 while ((buf = fgetline(stdin)) != NULL) {
1639 int maxlen = strlen(buf);
1640 unsigned char *data = snewn(maxlen, unsigned char);
1641 unsigned char *ptrs[4], *q;
1654 while (*bufp && !isxdigit((unsigned char)*bufp))
1661 while (*bufp && isxdigit((unsigned char)*bufp))
1665 if (ptrnum >= lenof(ptrs))
1669 for (i = -((end - start) & 1); i < end-start; i += 2) {
1670 unsigned char val = (i < 0 ? 0 : fromxdigit(start[i]));
1671 val = val * 16 + fromxdigit(start[i+1]);
1679 Bignum a = bignum_from_bytes(ptrs[0], ptrs[1]-ptrs[0]);
1680 Bignum b = bignum_from_bytes(ptrs[1], ptrs[2]-ptrs[1]);
1681 Bignum c = bignum_from_bytes(ptrs[2], ptrs[3]-ptrs[2]);
1682 Bignum p = bigmul(a, b);
1684 if (bignum_cmp(c, p) == 0) {
1687 char *as = bignum_decimal(a);
1688 char *bs = bignum_decimal(b);
1689 char *cs = bignum_decimal(c);
1690 char *ps = bignum_decimal(p);
1692 printf("%d: fail: %s * %s gave %s expected %s\n",
1693 line, as, bs, ps, cs);
1710 printf("passed %d failed %d total %d\n", passes, fails, passes+fails);