2 * Bignum routines for RSA and DH and stuff.
\r
15 * * Do not call the DIVMOD_WORD macro with expressions such as array
\r
16 * subscripts, as some implementations object to this (see below).
\r
17 * * Note that none of the division methods below will cope if the
\r
18 * quotient won't fit into BIGNUM_INT_BITS. Callers should be careful
\r
19 * to avoid this case.
\r
20 * If this condition occurs, in the case of the x86 DIV instruction,
\r
21 * an overflow exception will occur, which (according to a correspondent)
\r
22 * will manifest on Windows as something like
\r
23 * 0xC0000095: Integer overflow
\r
24 * The C variant won't give the right answer, either.
\r
27 #if defined __GNUC__ && defined __i386__
\r
28 typedef unsigned long BignumInt;
\r
29 typedef unsigned long long BignumDblInt;
\r
30 #define BIGNUM_INT_MASK 0xFFFFFFFFUL
\r
31 #define BIGNUM_TOP_BIT 0x80000000UL
\r
32 #define BIGNUM_INT_BITS 32
\r
33 #define MUL_WORD(w1, w2) ((BignumDblInt)w1 * w2)
\r
34 #define DIVMOD_WORD(q, r, hi, lo, w) \
\r
35 __asm__("div %2" : \
\r
36 "=d" (r), "=a" (q) : \
\r
37 "r" (w), "d" (hi), "a" (lo))
\r
38 #elif defined _MSC_VER && defined _M_IX86
\r
39 typedef unsigned __int32 BignumInt;
\r
40 typedef unsigned __int64 BignumDblInt;
\r
41 #define BIGNUM_INT_MASK 0xFFFFFFFFUL
\r
42 #define BIGNUM_TOP_BIT 0x80000000UL
\r
43 #define BIGNUM_INT_BITS 32
\r
44 #define MUL_WORD(w1, w2) ((BignumDblInt)w1 * w2)
\r
45 /* Note: MASM interprets array subscripts in the macro arguments as
\r
46 * assembler syntax, which gives the wrong answer. Don't supply them.
\r
47 * <http://msdn2.microsoft.com/en-us/library/bf1dw62z.aspx> */
\r
48 #define DIVMOD_WORD(q, r, hi, lo, w) do { \
\r
56 /* 64-bit architectures can do 32x32->64 chunks at a time */
\r
57 typedef unsigned int BignumInt;
\r
58 typedef unsigned long BignumDblInt;
\r
59 #define BIGNUM_INT_MASK 0xFFFFFFFFU
\r
60 #define BIGNUM_TOP_BIT 0x80000000U
\r
61 #define BIGNUM_INT_BITS 32
\r
62 #define MUL_WORD(w1, w2) ((BignumDblInt)w1 * w2)
\r
63 #define DIVMOD_WORD(q, r, hi, lo, w) do { \
\r
64 BignumDblInt n = (((BignumDblInt)hi) << BIGNUM_INT_BITS) | lo; \
\r
68 #elif defined _LLP64
\r
69 /* 64-bit architectures in which unsigned long is 32 bits, not 64 */
\r
70 typedef unsigned long BignumInt;
\r
71 typedef unsigned long long BignumDblInt;
\r
72 #define BIGNUM_INT_MASK 0xFFFFFFFFUL
\r
73 #define BIGNUM_TOP_BIT 0x80000000UL
\r
74 #define BIGNUM_INT_BITS 32
\r
75 #define MUL_WORD(w1, w2) ((BignumDblInt)w1 * w2)
\r
76 #define DIVMOD_WORD(q, r, hi, lo, w) do { \
\r
77 BignumDblInt n = (((BignumDblInt)hi) << BIGNUM_INT_BITS) | lo; \
\r
82 /* Fallback for all other cases */
\r
83 typedef unsigned short BignumInt;
\r
84 typedef unsigned long BignumDblInt;
\r
85 #define BIGNUM_INT_MASK 0xFFFFU
\r
86 #define BIGNUM_TOP_BIT 0x8000U
\r
87 #define BIGNUM_INT_BITS 16
\r
88 #define MUL_WORD(w1, w2) ((BignumDblInt)w1 * w2)
\r
89 #define DIVMOD_WORD(q, r, hi, lo, w) do { \
\r
90 BignumDblInt n = (((BignumDblInt)hi) << BIGNUM_INT_BITS) | lo; \
\r
96 #define BIGNUM_INT_BYTES (BIGNUM_INT_BITS / 8)
\r
98 #define BIGNUM_INTERNAL
\r
99 typedef BignumInt *Bignum;
\r
103 BignumInt bnZero[1] = { 0 };
\r
104 BignumInt bnOne[2] = { 1, 1 };
\r
107 * The Bignum format is an array of `BignumInt'. The first
\r
108 * element of the array counts the remaining elements. The
\r
109 * remaining elements express the actual number, base 2^BIGNUM_INT_BITS, _least_
\r
110 * significant digit first. (So it's trivial to extract the bit
\r
111 * with value 2^n for any n.)
\r
113 * All Bignums in this module are positive. Negative numbers must
\r
114 * be dealt with outside it.
\r
116 * INVARIANT: the most significant word of any Bignum must be
\r
120 Bignum Zero = bnZero, One = bnOne;
\r
122 static Bignum newbn(int length)
\r
126 assert(length >= 0 && length < INT_MAX / BIGNUM_INT_BITS);
\r
128 b = snewn(length + 1, BignumInt);
\r
130 abort(); /* FIXME */
\r
131 memset(b, 0, (length + 1) * sizeof(*b));
\r
136 void bn_restore_invariant(Bignum b)
\r
138 while (b[0] > 1 && b[b[0]] == 0)
\r
142 Bignum copybn(Bignum orig)
\r
144 Bignum b = snewn(orig[0] + 1, BignumInt);
\r
146 abort(); /* FIXME */
\r
147 memcpy(b, orig, (orig[0] + 1) * sizeof(*b));
\r
151 void freebn(Bignum b)
\r
154 * Burn the evidence, just in case.
\r
156 smemclr(b, sizeof(b[0]) * (b[0] + 1));
\r
160 Bignum bn_power_2(int n)
\r
166 ret = newbn(n / BIGNUM_INT_BITS + 1);
\r
167 bignum_set_bit(ret, n, 1);
\r
172 * Internal addition. Sets c = a - b, where 'a', 'b' and 'c' are all
\r
173 * big-endian arrays of 'len' BignumInts. Returns a BignumInt carried
\r
176 static BignumInt internal_add(const BignumInt *a, const BignumInt *b,
\r
177 BignumInt *c, int len)
\r
180 BignumDblInt carry = 0;
\r
182 for (i = len-1; i >= 0; i--) {
\r
183 carry += (BignumDblInt)a[i] + b[i];
\r
184 c[i] = (BignumInt)carry;
\r
185 carry >>= BIGNUM_INT_BITS;
\r
188 return (BignumInt)carry;
\r
192 * Internal subtraction. Sets c = a - b, where 'a', 'b' and 'c' are
\r
193 * all big-endian arrays of 'len' BignumInts. Any borrow from the top
\r
196 static void internal_sub(const BignumInt *a, const BignumInt *b,
\r
197 BignumInt *c, int len)
\r
200 BignumDblInt carry = 1;
\r
202 for (i = len-1; i >= 0; i--) {
\r
203 carry += (BignumDblInt)a[i] + (b[i] ^ BIGNUM_INT_MASK);
\r
204 c[i] = (BignumInt)carry;
\r
205 carry >>= BIGNUM_INT_BITS;
\r
210 * Compute c = a * b.
\r
211 * Input is in the first len words of a and b.
\r
212 * Result is returned in the first 2*len words of c.
\r
214 * 'scratch' must point to an array of BignumInt of size at least
\r
215 * mul_compute_scratch(len). (This covers the needs of internal_mul
\r
216 * and all its recursive calls to itself.)
\r
218 #define KARATSUBA_THRESHOLD 50
\r
219 static int mul_compute_scratch(int len)
\r
222 while (len > KARATSUBA_THRESHOLD) {
\r
223 int toplen = len/2, botlen = len - toplen; /* botlen is the bigger */
\r
224 int midlen = botlen + 1;
\r
230 static void internal_mul(const BignumInt *a, const BignumInt *b,
\r
231 BignumInt *c, int len, BignumInt *scratch)
\r
233 if (len > KARATSUBA_THRESHOLD) {
\r
237 * Karatsuba divide-and-conquer algorithm. Cut each input in
\r
238 * half, so that it's expressed as two big 'digits' in a giant
\r
244 * Then the product is of course
\r
246 * ab = a_1 b_1 D^2 + (a_1 b_0 + a_0 b_1) D + a_0 b_0
\r
248 * and we compute the three coefficients by recursively
\r
249 * calling ourself to do half-length multiplications.
\r
251 * The clever bit that makes this worth doing is that we only
\r
252 * need _one_ half-length multiplication for the central
\r
253 * coefficient rather than the two that it obviouly looks
\r
254 * like, because we can use a single multiplication to compute
\r
256 * (a_1 + a_0) (b_1 + b_0) = a_1 b_1 + a_1 b_0 + a_0 b_1 + a_0 b_0
\r
258 * and then we subtract the other two coefficients (a_1 b_1
\r
259 * and a_0 b_0) which we were computing anyway.
\r
261 * Hence we get to multiply two numbers of length N in about
\r
262 * three times as much work as it takes to multiply numbers of
\r
263 * length N/2, which is obviously better than the four times
\r
264 * as much work it would take if we just did a long
\r
265 * conventional multiply.
\r
268 int toplen = len/2, botlen = len - toplen; /* botlen is the bigger */
\r
269 int midlen = botlen + 1;
\r
270 BignumDblInt carry;
\r
276 * The coefficients a_1 b_1 and a_0 b_0 just avoid overlapping
\r
277 * in the output array, so we can compute them immediately in
\r
282 printf("a1,a0 = 0x");
\r
283 for (i = 0; i < len; i++) {
\r
284 if (i == toplen) printf(", 0x");
\r
285 printf("%0*x", BIGNUM_INT_BITS/4, a[i]);
\r
288 printf("b1,b0 = 0x");
\r
289 for (i = 0; i < len; i++) {
\r
290 if (i == toplen) printf(", 0x");
\r
291 printf("%0*x", BIGNUM_INT_BITS/4, b[i]);
\r
297 internal_mul(a, b, c, toplen, scratch);
\r
299 printf("a1b1 = 0x");
\r
300 for (i = 0; i < 2*toplen; i++) {
\r
301 printf("%0*x", BIGNUM_INT_BITS/4, c[i]);
\r
307 internal_mul(a + toplen, b + toplen, c + 2*toplen, botlen, scratch);
\r
309 printf("a0b0 = 0x");
\r
310 for (i = 0; i < 2*botlen; i++) {
\r
311 printf("%0*x", BIGNUM_INT_BITS/4, c[2*toplen+i]);
\r
316 /* Zero padding. midlen exceeds toplen by at most 2, so just
\r
317 * zero the first two words of each input and the rest will be
\r
319 scratch[0] = scratch[1] = scratch[midlen] = scratch[midlen+1] = 0;
\r
321 for (i = 0; i < toplen; i++) {
\r
322 scratch[midlen - toplen + i] = a[i]; /* a_1 */
\r
323 scratch[2*midlen - toplen + i] = b[i]; /* b_1 */
\r
326 /* compute a_1 + a_0 */
\r
327 scratch[0] = internal_add(scratch+1, a+toplen, scratch+1, botlen);
\r
329 printf("a1plusa0 = 0x");
\r
330 for (i = 0; i < midlen; i++) {
\r
331 printf("%0*x", BIGNUM_INT_BITS/4, scratch[i]);
\r
335 /* compute b_1 + b_0 */
\r
336 scratch[midlen] = internal_add(scratch+midlen+1, b+toplen,
\r
337 scratch+midlen+1, botlen);
\r
339 printf("b1plusb0 = 0x");
\r
340 for (i = 0; i < midlen; i++) {
\r
341 printf("%0*x", BIGNUM_INT_BITS/4, scratch[midlen+i]);
\r
347 * Now we can do the third multiplication.
\r
349 internal_mul(scratch, scratch + midlen, scratch + 2*midlen, midlen,
\r
350 scratch + 4*midlen);
\r
352 printf("a1plusa0timesb1plusb0 = 0x");
\r
353 for (i = 0; i < 2*midlen; i++) {
\r
354 printf("%0*x", BIGNUM_INT_BITS/4, scratch[2*midlen+i]);
\r
360 * Now we can reuse the first half of 'scratch' to compute the
\r
361 * sum of the outer two coefficients, to subtract from that
\r
362 * product to obtain the middle one.
\r
364 scratch[0] = scratch[1] = scratch[2] = scratch[3] = 0;
\r
365 for (i = 0; i < 2*toplen; i++)
\r
366 scratch[2*midlen - 2*toplen + i] = c[i];
\r
367 scratch[1] = internal_add(scratch+2, c + 2*toplen,
\r
368 scratch+2, 2*botlen);
\r
370 printf("a1b1plusa0b0 = 0x");
\r
371 for (i = 0; i < 2*midlen; i++) {
\r
372 printf("%0*x", BIGNUM_INT_BITS/4, scratch[i]);
\r
377 internal_sub(scratch + 2*midlen, scratch,
\r
378 scratch + 2*midlen, 2*midlen);
\r
380 printf("a1b0plusa0b1 = 0x");
\r
381 for (i = 0; i < 2*midlen; i++) {
\r
382 printf("%0*x", BIGNUM_INT_BITS/4, scratch[2*midlen+i]);
\r
388 * And now all we need to do is to add that middle coefficient
\r
389 * back into the output. We may have to propagate a carry
\r
390 * further up the output, but we can be sure it won't
\r
391 * propagate right the way off the top.
\r
393 carry = internal_add(c + 2*len - botlen - 2*midlen,
\r
394 scratch + 2*midlen,
\r
395 c + 2*len - botlen - 2*midlen, 2*midlen);
\r
396 i = 2*len - botlen - 2*midlen - 1;
\r
400 c[i] = (BignumInt)carry;
\r
401 carry >>= BIGNUM_INT_BITS;
\r
406 for (i = 0; i < 2*len; i++) {
\r
407 printf("%0*x", BIGNUM_INT_BITS/4, c[i]);
\r
416 const BignumInt *ap, *bp;
\r
417 BignumInt *cp, *cps;
\r
420 * Multiply in the ordinary O(N^2) way.
\r
423 for (i = 0; i < 2 * len; i++)
\r
426 for (cps = c + 2*len, ap = a + len; ap-- > a; cps--) {
\r
428 for (cp = cps, bp = b + len; cp--, bp-- > b ;) {
\r
429 t = (MUL_WORD(*ap, *bp) + carry) + *cp;
\r
430 *cp = (BignumInt) t;
\r
431 carry = (BignumInt)(t >> BIGNUM_INT_BITS);
\r
439 * Variant form of internal_mul used for the initial step of
\r
440 * Montgomery reduction. Only bothers outputting 'len' words
\r
441 * (everything above that is thrown away).
\r
443 static void internal_mul_low(const BignumInt *a, const BignumInt *b,
\r
444 BignumInt *c, int len, BignumInt *scratch)
\r
446 if (len > KARATSUBA_THRESHOLD) {
\r
450 * Karatsuba-aware version of internal_mul_low. As before, we
\r
451 * express each input value as a shifted combination of two
\r
457 * Then the full product is, as before,
\r
459 * ab = a_1 b_1 D^2 + (a_1 b_0 + a_0 b_1) D + a_0 b_0
\r
461 * Provided we choose D on the large side (so that a_0 and b_0
\r
462 * are _at least_ as long as a_1 and b_1), we don't need the
\r
463 * topmost term at all, and we only need half of the middle
\r
464 * term. So there's no point in doing the proper Karatsuba
\r
465 * optimisation which computes the middle term using the top
\r
466 * one, because we'd take as long computing the top one as
\r
467 * just computing the middle one directly.
\r
469 * So instead, we do a much more obvious thing: we call the
\r
470 * fully optimised internal_mul to compute a_0 b_0, and we
\r
471 * recursively call ourself to compute the _bottom halves_ of
\r
472 * a_1 b_0 and a_0 b_1, each of which we add into the result
\r
473 * in the obvious way.
\r
475 * In other words, there's no actual Karatsuba _optimisation_
\r
476 * in this function; the only benefit in doing it this way is
\r
477 * that we call internal_mul proper for a large part of the
\r
478 * work, and _that_ can optimise its operation.
\r
481 int toplen = len/2, botlen = len - toplen; /* botlen is the bigger */
\r
484 * Scratch space for the various bits and pieces we're going
\r
485 * to be adding together: we need botlen*2 words for a_0 b_0
\r
486 * (though we may end up throwing away its topmost word), and
\r
487 * toplen words for each of a_1 b_0 and a_0 b_1. That adds up
\r
488 * to exactly 2*len.
\r
492 internal_mul(a + toplen, b + toplen, scratch + 2*toplen, botlen,
\r
496 internal_mul_low(a, b + len - toplen, scratch + toplen, toplen,
\r
500 internal_mul_low(a + len - toplen, b, scratch, toplen,
\r
503 /* Copy the bottom half of the big coefficient into place */
\r
504 for (i = 0; i < botlen; i++)
\r
505 c[toplen + i] = scratch[2*toplen + botlen + i];
\r
507 /* Add the two small coefficients, throwing away the returned carry */
\r
508 internal_add(scratch, scratch + toplen, scratch, toplen);
\r
510 /* And add that to the large coefficient, leaving the result in c. */
\r
511 internal_add(scratch, scratch + 2*toplen + botlen - toplen,
\r
518 const BignumInt *ap, *bp;
\r
519 BignumInt *cp, *cps;
\r
522 * Multiply in the ordinary O(N^2) way.
\r
525 for (i = 0; i < len; i++)
\r
528 for (cps = c + len, ap = a + len; ap-- > a; cps--) {
\r
530 for (cp = cps, bp = b + len; bp--, cp-- > c ;) {
\r
531 t = (MUL_WORD(*ap, *bp) + carry) + *cp;
\r
532 *cp = (BignumInt) t;
\r
533 carry = (BignumInt)(t >> BIGNUM_INT_BITS);
\r
540 * Montgomery reduction. Expects x to be a big-endian array of 2*len
\r
541 * BignumInts whose value satisfies 0 <= x < rn (where r = 2^(len *
\r
542 * BIGNUM_INT_BITS) is the Montgomery base). Returns in the same array
\r
543 * a value x' which is congruent to xr^{-1} mod n, and satisfies 0 <=
\r
546 * 'n' and 'mninv' should be big-endian arrays of 'len' BignumInts
\r
547 * each, containing respectively n and the multiplicative inverse of
\r
550 * 'tmp' is an array of BignumInt used as scratch space, of length at
\r
551 * least 3*len + mul_compute_scratch(len).
\r
553 static void monty_reduce(BignumInt *x, const BignumInt *n,
\r
554 const BignumInt *mninv, BignumInt *tmp, int len)
\r
560 * Multiply x by (-n)^{-1} mod r. This gives us a value m such
\r
561 * that mn is congruent to -x mod r. Hence, mn+x is an exact
\r
562 * multiple of r, and is also (obviously) congruent to x mod n.
\r
564 internal_mul_low(x + len, mninv, tmp, len, tmp + 3*len);
\r
567 * Compute t = (mn+x)/r in ordinary, non-modular, integer
\r
568 * arithmetic. By construction this is exact, and is congruent mod
\r
569 * n to x * r^{-1}, i.e. the answer we want.
\r
571 * The following multiply leaves that answer in the _most_
\r
572 * significant half of the 'x' array, so then we must shift it
\r
575 internal_mul(tmp, n, tmp+len, len, tmp + 3*len);
\r
576 carry = internal_add(x, tmp+len, x, 2*len);
\r
577 for (i = 0; i < len; i++)
\r
578 x[len + i] = x[i], x[i] = 0;
\r
581 * Reduce t mod n. This doesn't require a full-on division by n,
\r
582 * but merely a test and single optional subtraction, since we can
\r
583 * show that 0 <= t < 2n.
\r
586 * + we computed m mod r, so 0 <= m < r.
\r
587 * + so 0 <= mn < rn, obviously
\r
588 * + hence we only need 0 <= x < rn to guarantee that 0 <= mn+x < 2rn
\r
589 * + yielding 0 <= (mn+x)/r < 2n as required.
\r
592 for (i = 0; i < len; i++)
\r
593 if (x[len + i] != n[i])
\r
596 if (carry || i >= len || x[len + i] > n[i])
\r
597 internal_sub(x+len, n, x+len, len);
\r
600 static void internal_add_shifted(BignumInt *number,
\r
601 unsigned n, int shift)
\r
603 int word = 1 + (shift / BIGNUM_INT_BITS);
\r
604 int bshift = shift % BIGNUM_INT_BITS;
\r
605 BignumDblInt addend;
\r
607 addend = (BignumDblInt)n << bshift;
\r
610 assert(word <= number[0]);
\r
611 addend += number[word];
\r
612 number[word] = (BignumInt) addend & BIGNUM_INT_MASK;
\r
613 addend >>= BIGNUM_INT_BITS;
\r
619 * Compute a = a % m.
\r
620 * Input in first alen words of a and first mlen words of m.
\r
621 * Output in first alen words of a
\r
622 * (of which first alen-mlen words will be zero).
\r
623 * The MSW of m MUST have its high bit set.
\r
624 * Quotient is accumulated in the `quotient' array, which is a Bignum
\r
625 * rather than the internal bigendian format. Quotient parts are shifted
\r
626 * left by `qshift' before adding into quot.
\r
628 static void internal_mod(BignumInt *a, int alen,
\r
629 BignumInt *m, int mlen,
\r
630 BignumInt *quot, int qshift)
\r
637 assert(m0 >> (BIGNUM_INT_BITS-1) == 1);
\r
643 for (i = 0; i <= alen - mlen; i++) {
\r
645 unsigned int q, r, c, ai1;
\r
659 /* Find q = h:a[i] / m0 */
\r
664 * To illustrate it, suppose a BignumInt is 8 bits, and
\r
665 * we are dividing (say) A1:23:45:67 by A1:B2:C3. Then
\r
666 * our initial division will be 0xA123 / 0xA1, which
\r
667 * will give a quotient of 0x100 and a divide overflow.
\r
668 * However, the invariants in this division algorithm
\r
669 * are not violated, since the full number A1:23:... is
\r
670 * _less_ than the quotient prefix A1:B2:... and so the
\r
671 * following correction loop would have sorted it out.
\r
673 * In this situation we set q to be the largest
\r
674 * quotient we _can_ stomach (0xFF, of course).
\r
676 q = BIGNUM_INT_MASK;
\r
678 /* Macro doesn't want an array subscript expression passed
\r
679 * into it (see definition), so use a temporary. */
\r
680 BignumInt tmplo = a[i];
\r
681 DIVMOD_WORD(q, r, h, tmplo, m0);
\r
683 /* Refine our estimate of q by looking at
\r
684 h:a[i]:a[i+1] / m0:m1 */
\r
685 t = MUL_WORD(m1, q);
\r
686 if (t > ((BignumDblInt) r << BIGNUM_INT_BITS) + ai1) {
\r
689 r = (r + m0) & BIGNUM_INT_MASK; /* overflow? */
\r
690 if (r >= (BignumDblInt) m0 &&
\r
691 t > ((BignumDblInt) r << BIGNUM_INT_BITS) + ai1) q--;
\r
695 /* Subtract q * m from a[i...] */
\r
697 for (k = mlen - 1; k >= 0; k--) {
\r
698 t = MUL_WORD(q, m[k]);
\r
700 c = (unsigned)(t >> BIGNUM_INT_BITS);
\r
701 if ((BignumInt) t > a[i + k])
\r
703 a[i + k] -= (BignumInt) t;
\r
706 /* Add back m in case of borrow */
\r
709 for (k = mlen - 1; k >= 0; k--) {
\r
712 a[i + k] = (BignumInt) t;
\r
713 t = t >> BIGNUM_INT_BITS;
\r
718 internal_add_shifted(quot, q, qshift + BIGNUM_INT_BITS * (alen - mlen - i));
\r
723 * Compute (base ^ exp) % mod, the pedestrian way.
\r
725 Bignum modpow_simple(Bignum base_in, Bignum exp, Bignum mod)
\r
727 BignumInt *a, *b, *n, *m, *scratch;
\r
729 int mlen, scratchlen, i, j;
\r
730 Bignum base, result;
\r
733 * The most significant word of mod needs to be non-zero. It
\r
734 * should already be, but let's make sure.
\r
736 assert(mod[mod[0]] != 0);
\r
739 * Make sure the base is smaller than the modulus, by reducing
\r
740 * it modulo the modulus if not.
\r
742 base = bigmod(base_in, mod);
\r
744 /* Allocate m of size mlen, copy mod to m */
\r
745 /* We use big endian internally */
\r
747 m = snewn(mlen, BignumInt);
\r
748 for (j = 0; j < mlen; j++)
\r
749 m[j] = mod[mod[0] - j];
\r
751 /* Shift m left to make msb bit set */
\r
752 for (mshift = 0; mshift < BIGNUM_INT_BITS-1; mshift++)
\r
753 if ((m[0] << mshift) & BIGNUM_TOP_BIT)
\r
756 for (i = 0; i < mlen - 1; i++)
\r
757 m[i] = (m[i] << mshift) | (m[i + 1] >> (BIGNUM_INT_BITS - mshift));
\r
758 m[mlen - 1] = m[mlen - 1] << mshift;
\r
761 /* Allocate n of size mlen, copy base to n */
\r
762 n = snewn(mlen, BignumInt);
\r
763 i = mlen - base[0];
\r
764 for (j = 0; j < i; j++)
\r
766 for (j = 0; j < (int)base[0]; j++)
\r
767 n[i + j] = base[base[0] - j];
\r
769 /* Allocate a and b of size 2*mlen. Set a = 1 */
\r
770 a = snewn(2 * mlen, BignumInt);
\r
771 b = snewn(2 * mlen, BignumInt);
\r
772 for (i = 0; i < 2 * mlen; i++)
\r
774 a[2 * mlen - 1] = 1;
\r
776 /* Scratch space for multiplies */
\r
777 scratchlen = mul_compute_scratch(mlen);
\r
778 scratch = snewn(scratchlen, BignumInt);
\r
780 /* Skip leading zero bits of exp. */
\r
782 j = BIGNUM_INT_BITS-1;
\r
783 while (i < (int)exp[0] && (exp[exp[0] - i] & (1 << j)) == 0) {
\r
787 j = BIGNUM_INT_BITS-1;
\r
791 /* Main computation */
\r
792 while (i < (int)exp[0]) {
\r
794 internal_mul(a + mlen, a + mlen, b, mlen, scratch);
\r
795 internal_mod(b, mlen * 2, m, mlen, NULL, 0);
\r
796 if ((exp[exp[0] - i] & (1 << j)) != 0) {
\r
797 internal_mul(b + mlen, n, a, mlen, scratch);
\r
798 internal_mod(a, mlen * 2, m, mlen, NULL, 0);
\r
808 j = BIGNUM_INT_BITS-1;
\r
811 /* Fixup result in case the modulus was shifted */
\r
813 for (i = mlen - 1; i < 2 * mlen - 1; i++)
\r
814 a[i] = (a[i] << mshift) | (a[i + 1] >> (BIGNUM_INT_BITS - mshift));
\r
815 a[2 * mlen - 1] = a[2 * mlen - 1] << mshift;
\r
816 internal_mod(a, mlen * 2, m, mlen, NULL, 0);
\r
817 for (i = 2 * mlen - 1; i >= mlen; i--)
\r
818 a[i] = (a[i] >> mshift) | (a[i - 1] << (BIGNUM_INT_BITS - mshift));
\r
821 /* Copy result to buffer */
\r
822 result = newbn(mod[0]);
\r
823 for (i = 0; i < mlen; i++)
\r
824 result[result[0] - i] = a[i + mlen];
\r
825 while (result[0] > 1 && result[result[0]] == 0)
\r
828 /* Free temporary arrays */
\r
829 smemclr(a, 2 * mlen * sizeof(*a));
\r
831 smemclr(scratch, scratchlen * sizeof(*scratch));
\r
833 smemclr(b, 2 * mlen * sizeof(*b));
\r
835 smemclr(m, mlen * sizeof(*m));
\r
837 smemclr(n, mlen * sizeof(*n));
\r
846 * Compute (base ^ exp) % mod. Uses the Montgomery multiplication
\r
847 * technique where possible, falling back to modpow_simple otherwise.
\r
849 Bignum modpow(Bignum base_in, Bignum exp, Bignum mod)
\r
851 BignumInt *a, *b, *x, *n, *mninv, *scratch;
\r
852 int len, scratchlen, i, j;
\r
853 Bignum base, base2, r, rn, inv, result;
\r
856 * The most significant word of mod needs to be non-zero. It
\r
857 * should already be, but let's make sure.
\r
859 assert(mod[mod[0]] != 0);
\r
862 * mod had better be odd, or we can't do Montgomery multiplication
\r
863 * using a power of two at all.
\r
866 return modpow_simple(base_in, exp, mod);
\r
869 * Make sure the base is smaller than the modulus, by reducing
\r
870 * it modulo the modulus if not.
\r
872 base = bigmod(base_in, mod);
\r
875 * Compute the inverse of n mod r, for monty_reduce. (In fact we
\r
876 * want the inverse of _minus_ n mod r, but we'll sort that out
\r
880 r = bn_power_2(BIGNUM_INT_BITS * len);
\r
881 inv = modinv(mod, r);
\r
882 assert(inv); /* cannot fail, since mod is odd and r is a power of 2 */
\r
885 * Multiply the base by r mod n, to get it into Montgomery
\r
888 base2 = modmul(base, r, mod);
\r
892 rn = bigmod(r, mod); /* r mod n, i.e. Montgomerified 1 */
\r
894 freebn(r); /* won't need this any more */
\r
897 * Set up internal arrays of the right lengths, in big-endian
\r
898 * format, containing the base, the modulus, and the modulus's
\r
901 n = snewn(len, BignumInt);
\r
902 for (j = 0; j < len; j++)
\r
903 n[len - 1 - j] = mod[j + 1];
\r
905 mninv = snewn(len, BignumInt);
\r
906 for (j = 0; j < len; j++)
\r
907 mninv[len - 1 - j] = (j < (int)inv[0] ? inv[j + 1] : 0);
\r
908 freebn(inv); /* we don't need this copy of it any more */
\r
909 /* Now negate mninv mod r, so it's the inverse of -n rather than +n. */
\r
910 x = snewn(len, BignumInt);
\r
911 for (j = 0; j < len; j++)
\r
913 internal_sub(x, mninv, mninv, len);
\r
915 /* x = snewn(len, BignumInt); */ /* already done above */
\r
916 for (j = 0; j < len; j++)
\r
917 x[len - 1 - j] = (j < (int)base[0] ? base[j + 1] : 0);
\r
918 freebn(base); /* we don't need this copy of it any more */
\r
920 a = snewn(2*len, BignumInt);
\r
921 b = snewn(2*len, BignumInt);
\r
922 for (j = 0; j < len; j++)
\r
923 a[2*len - 1 - j] = (j < (int)rn[0] ? rn[j + 1] : 0);
\r
926 /* Scratch space for multiplies */
\r
927 scratchlen = 3*len + mul_compute_scratch(len);
\r
928 scratch = snewn(scratchlen, BignumInt);
\r
930 /* Skip leading zero bits of exp. */
\r
932 j = BIGNUM_INT_BITS-1;
\r
933 while (i < (int)exp[0] && (exp[exp[0] - i] & (1 << j)) == 0) {
\r
937 j = BIGNUM_INT_BITS-1;
\r
941 /* Main computation */
\r
942 while (i < (int)exp[0]) {
\r
944 internal_mul(a + len, a + len, b, len, scratch);
\r
945 monty_reduce(b, n, mninv, scratch, len);
\r
946 if ((exp[exp[0] - i] & (1 << j)) != 0) {
\r
947 internal_mul(b + len, x, a, len, scratch);
\r
948 monty_reduce(a, n, mninv, scratch, len);
\r
958 j = BIGNUM_INT_BITS-1;
\r
962 * Final monty_reduce to get back from the adjusted Montgomery
\r
965 monty_reduce(a, n, mninv, scratch, len);
\r
967 /* Copy result to buffer */
\r
968 result = newbn(mod[0]);
\r
969 for (i = 0; i < len; i++)
\r
970 result[result[0] - i] = a[i + len];
\r
971 while (result[0] > 1 && result[result[0]] == 0)
\r
974 /* Free temporary arrays */
\r
975 smemclr(scratch, scratchlen * sizeof(*scratch));
\r
977 smemclr(a, 2 * len * sizeof(*a));
\r
979 smemclr(b, 2 * len * sizeof(*b));
\r
981 smemclr(mninv, len * sizeof(*mninv));
\r
983 smemclr(n, len * sizeof(*n));
\r
985 smemclr(x, len * sizeof(*x));
\r
992 * Compute (p * q) % mod.
\r
993 * The most significant word of mod MUST be non-zero.
\r
994 * We assume that the result array is the same size as the mod array.
\r
996 Bignum modmul(Bignum p, Bignum q, Bignum mod)
\r
998 BignumInt *a, *n, *m, *o, *scratch;
\r
999 int mshift, scratchlen;
\r
1000 int pqlen, mlen, rlen, i, j;
\r
1004 * The most significant word of mod needs to be non-zero. It
\r
1005 * should already be, but let's make sure.
\r
1007 assert(mod[mod[0]] != 0);
\r
1009 /* Allocate m of size mlen, copy mod to m */
\r
1010 /* We use big endian internally */
\r
1012 m = snewn(mlen, BignumInt);
\r
1013 for (j = 0; j < mlen; j++)
\r
1014 m[j] = mod[mod[0] - j];
\r
1016 /* Shift m left to make msb bit set */
\r
1017 for (mshift = 0; mshift < BIGNUM_INT_BITS-1; mshift++)
\r
1018 if ((m[0] << mshift) & BIGNUM_TOP_BIT)
\r
1021 for (i = 0; i < mlen - 1; i++)
\r
1022 m[i] = (m[i] << mshift) | (m[i + 1] >> (BIGNUM_INT_BITS - mshift));
\r
1023 m[mlen - 1] = m[mlen - 1] << mshift;
\r
1026 pqlen = (p[0] > q[0] ? p[0] : q[0]);
\r
1029 * Make sure that we're allowing enough space. The shifting below
\r
1030 * will underflow the vectors we allocate if pqlen is too small.
\r
1032 if (2*pqlen <= mlen)
\r
1033 pqlen = mlen/2 + 1;
\r
1035 /* Allocate n of size pqlen, copy p to n */
\r
1036 n = snewn(pqlen, BignumInt);
\r
1038 for (j = 0; j < i; j++)
\r
1040 for (j = 0; j < (int)p[0]; j++)
\r
1041 n[i + j] = p[p[0] - j];
\r
1043 /* Allocate o of size pqlen, copy q to o */
\r
1044 o = snewn(pqlen, BignumInt);
\r
1046 for (j = 0; j < i; j++)
\r
1048 for (j = 0; j < (int)q[0]; j++)
\r
1049 o[i + j] = q[q[0] - j];
\r
1051 /* Allocate a of size 2*pqlen for result */
\r
1052 a = snewn(2 * pqlen, BignumInt);
\r
1054 /* Scratch space for multiplies */
\r
1055 scratchlen = mul_compute_scratch(pqlen);
\r
1056 scratch = snewn(scratchlen, BignumInt);
\r
1058 /* Main computation */
\r
1059 internal_mul(n, o, a, pqlen, scratch);
\r
1060 internal_mod(a, pqlen * 2, m, mlen, NULL, 0);
\r
1062 /* Fixup result in case the modulus was shifted */
\r
1064 for (i = 2 * pqlen - mlen - 1; i < 2 * pqlen - 1; i++)
\r
1065 a[i] = (a[i] << mshift) | (a[i + 1] >> (BIGNUM_INT_BITS - mshift));
\r
1066 a[2 * pqlen - 1] = a[2 * pqlen - 1] << mshift;
\r
1067 internal_mod(a, pqlen * 2, m, mlen, NULL, 0);
\r
1068 for (i = 2 * pqlen - 1; i >= 2 * pqlen - mlen; i--)
\r
1069 a[i] = (a[i] >> mshift) | (a[i - 1] << (BIGNUM_INT_BITS - mshift));
\r
1072 /* Copy result to buffer */
\r
1073 rlen = (mlen < pqlen * 2 ? mlen : pqlen * 2);
\r
1074 result = newbn(rlen);
\r
1075 for (i = 0; i < rlen; i++)
\r
1076 result[result[0] - i] = a[i + 2 * pqlen - rlen];
\r
1077 while (result[0] > 1 && result[result[0]] == 0)
\r
1080 /* Free temporary arrays */
\r
1081 smemclr(scratch, scratchlen * sizeof(*scratch));
\r
1083 smemclr(a, 2 * pqlen * sizeof(*a));
\r
1085 smemclr(m, mlen * sizeof(*m));
\r
1087 smemclr(n, pqlen * sizeof(*n));
\r
1089 smemclr(o, pqlen * sizeof(*o));
\r
1096 * Compute p % mod.
\r
1097 * The most significant word of mod MUST be non-zero.
\r
1098 * We assume that the result array is the same size as the mod array.
\r
1099 * We optionally write out a quotient if `quotient' is non-NULL.
\r
1100 * We can avoid writing out the result if `result' is NULL.
\r
1102 static void bigdivmod(Bignum p, Bignum mod, Bignum result, Bignum quotient)
\r
1106 int plen, mlen, i, j;
\r
1109 * The most significant word of mod needs to be non-zero. It
\r
1110 * should already be, but let's make sure.
\r
1112 assert(mod[mod[0]] != 0);
\r
1114 /* Allocate m of size mlen, copy mod to m */
\r
1115 /* We use big endian internally */
\r
1117 m = snewn(mlen, BignumInt);
\r
1118 for (j = 0; j < mlen; j++)
\r
1119 m[j] = mod[mod[0] - j];
\r
1121 /* Shift m left to make msb bit set */
\r
1122 for (mshift = 0; mshift < BIGNUM_INT_BITS-1; mshift++)
\r
1123 if ((m[0] << mshift) & BIGNUM_TOP_BIT)
\r
1126 for (i = 0; i < mlen - 1; i++)
\r
1127 m[i] = (m[i] << mshift) | (m[i + 1] >> (BIGNUM_INT_BITS - mshift));
\r
1128 m[mlen - 1] = m[mlen - 1] << mshift;
\r
1132 /* Ensure plen > mlen */
\r
1136 /* Allocate n of size plen, copy p to n */
\r
1137 n = snewn(plen, BignumInt);
\r
1138 for (j = 0; j < plen; j++)
\r
1140 for (j = 1; j <= (int)p[0]; j++)
\r
1141 n[plen - j] = p[j];
\r
1143 /* Main computation */
\r
1144 internal_mod(n, plen, m, mlen, quotient, mshift);
\r
1146 /* Fixup result in case the modulus was shifted */
\r
1148 for (i = plen - mlen - 1; i < plen - 1; i++)
\r
1149 n[i] = (n[i] << mshift) | (n[i + 1] >> (BIGNUM_INT_BITS - mshift));
\r
1150 n[plen - 1] = n[plen - 1] << mshift;
\r
1151 internal_mod(n, plen, m, mlen, quotient, 0);
\r
1152 for (i = plen - 1; i >= plen - mlen; i--)
\r
1153 n[i] = (n[i] >> mshift) | (n[i - 1] << (BIGNUM_INT_BITS - mshift));
\r
1156 /* Copy result to buffer */
\r
1158 for (i = 1; i <= (int)result[0]; i++) {
\r
1160 result[i] = j >= 0 ? n[j] : 0;
\r
1164 /* Free temporary arrays */
\r
1165 smemclr(m, mlen * sizeof(*m));
\r
1167 smemclr(n, plen * sizeof(*n));
\r
1172 * Decrement a number.
\r
1174 void decbn(Bignum bn)
\r
1177 while (i < (int)bn[0] && bn[i] == 0)
\r
1178 bn[i++] = BIGNUM_INT_MASK;
\r
1182 Bignum bignum_from_bytes(const unsigned char *data, int nbytes)
\r
1187 assert(nbytes >= 0 && nbytes < INT_MAX/8);
\r
1189 w = (nbytes + BIGNUM_INT_BYTES - 1) / BIGNUM_INT_BYTES; /* bytes->words */
\r
1191 result = newbn(w);
\r
1192 for (i = 1; i <= w; i++)
\r
1194 for (i = nbytes; i--;) {
\r
1195 unsigned char byte = *data++;
\r
1196 result[1 + i / BIGNUM_INT_BYTES] |= byte << (8*i % BIGNUM_INT_BITS);
\r
1199 while (result[0] > 1 && result[result[0]] == 0)
\r
1205 * Read an SSH-1-format bignum from a data buffer. Return the number
\r
1206 * of bytes consumed, or -1 if there wasn't enough data.
\r
1208 int ssh1_read_bignum(const unsigned char *data, int len, Bignum * result)
\r
1210 const unsigned char *p = data;
\r
1218 for (i = 0; i < 2; i++)
\r
1219 w = (w << 8) + *p++;
\r
1220 b = (w + 7) / 8; /* bits -> bytes */
\r
1225 if (!result) /* just return length */
\r
1228 *result = bignum_from_bytes(p, b);
\r
1230 return p + b - data;
\r
1234 * Return the bit count of a bignum, for SSH-1 encoding.
\r
1236 int bignum_bitcount(Bignum bn)
\r
1238 int bitcount = bn[0] * BIGNUM_INT_BITS - 1;
\r
1239 while (bitcount >= 0
\r
1240 && (bn[bitcount / BIGNUM_INT_BITS + 1] >> (bitcount % BIGNUM_INT_BITS)) == 0) bitcount--;
\r
1241 return bitcount + 1;
\r
1245 * Return the byte length of a bignum when SSH-1 encoded.
\r
1247 int ssh1_bignum_length(Bignum bn)
\r
1249 return 2 + (bignum_bitcount(bn) + 7) / 8;
\r
1253 * Return the byte length of a bignum when SSH-2 encoded.
\r
1255 int ssh2_bignum_length(Bignum bn)
\r
1257 return 4 + (bignum_bitcount(bn) + 8) / 8;
\r
1261 * Return a byte from a bignum; 0 is least significant, etc.
\r
1263 int bignum_byte(Bignum bn, int i)
\r
1265 if (i < 0 || i >= (int)(BIGNUM_INT_BYTES * bn[0]))
\r
1266 return 0; /* beyond the end */
\r
1268 return (bn[i / BIGNUM_INT_BYTES + 1] >>
\r
1269 ((i % BIGNUM_INT_BYTES)*8)) & 0xFF;
\r
1273 * Return a bit from a bignum; 0 is least significant, etc.
\r
1275 int bignum_bit(Bignum bn, int i)
\r
1277 if (i < 0 || i >= (int)(BIGNUM_INT_BITS * bn[0]))
\r
1278 return 0; /* beyond the end */
\r
1280 return (bn[i / BIGNUM_INT_BITS + 1] >> (i % BIGNUM_INT_BITS)) & 1;
\r
1284 * Set a bit in a bignum; 0 is least significant, etc.
\r
1286 void bignum_set_bit(Bignum bn, int bitnum, int value)
\r
1288 if (bitnum < 0 || bitnum >= (int)(BIGNUM_INT_BITS * bn[0]))
\r
1289 abort(); /* beyond the end */
\r
1291 int v = bitnum / BIGNUM_INT_BITS + 1;
\r
1292 int mask = 1 << (bitnum % BIGNUM_INT_BITS);
\r
1301 * Write a SSH-1-format bignum into a buffer. It is assumed the
\r
1302 * buffer is big enough. Returns the number of bytes used.
\r
1304 int ssh1_write_bignum(void *data, Bignum bn)
\r
1306 unsigned char *p = data;
\r
1307 int len = ssh1_bignum_length(bn);
\r
1309 int bitc = bignum_bitcount(bn);
\r
1311 *p++ = (bitc >> 8) & 0xFF;
\r
1312 *p++ = (bitc) & 0xFF;
\r
1313 for (i = len - 2; i--;)
\r
1314 *p++ = bignum_byte(bn, i);
\r
1319 * Compare two bignums. Returns like strcmp.
\r
1321 int bignum_cmp(Bignum a, Bignum b)
\r
1323 int amax = a[0], bmax = b[0];
\r
1326 /* Annoyingly we have two representations of zero */
\r
1327 if (amax == 1 && a[amax] == 0)
\r
1329 if (bmax == 1 && b[bmax] == 0)
\r
1332 assert(amax == 0 || a[amax] != 0);
\r
1333 assert(bmax == 0 || b[bmax] != 0);
\r
1335 i = (amax > bmax ? amax : bmax);
\r
1337 BignumInt aval = (i > amax ? 0 : a[i]);
\r
1338 BignumInt bval = (i > bmax ? 0 : b[i]);
\r
1349 * Right-shift one bignum to form another.
\r
1351 Bignum bignum_rshift(Bignum a, int shift)
\r
1354 int i, shiftw, shiftb, shiftbb, bits;
\r
1355 BignumInt ai, ai1;
\r
1357 assert(shift >= 0);
\r
1359 bits = bignum_bitcount(a) - shift;
\r
1360 ret = newbn((bits + BIGNUM_INT_BITS - 1) / BIGNUM_INT_BITS);
\r
1363 shiftw = shift / BIGNUM_INT_BITS;
\r
1364 shiftb = shift % BIGNUM_INT_BITS;
\r
1365 shiftbb = BIGNUM_INT_BITS - shiftb;
\r
1367 ai1 = a[shiftw + 1];
\r
1368 for (i = 1; i <= (int)ret[0]; i++) {
\r
1370 ai1 = (i + shiftw + 1 <= (int)a[0] ? a[i + shiftw + 1] : 0);
\r
1371 ret[i] = ((ai >> shiftb) | (ai1 << shiftbb)) & BIGNUM_INT_MASK;
\r
1379 * Non-modular multiplication and addition.
\r
1381 Bignum bigmuladd(Bignum a, Bignum b, Bignum addend)
\r
1383 int alen = a[0], blen = b[0];
\r
1384 int mlen = (alen > blen ? alen : blen);
\r
1385 int rlen, i, maxspot;
\r
1387 BignumInt *workspace;
\r
1390 /* mlen space for a, mlen space for b, 2*mlen for result,
\r
1391 * plus scratch space for multiplication */
\r
1392 wslen = mlen * 4 + mul_compute_scratch(mlen);
\r
1393 workspace = snewn(wslen, BignumInt);
\r
1394 for (i = 0; i < mlen; i++) {
\r
1395 workspace[0 * mlen + i] = (mlen - i <= (int)a[0] ? a[mlen - i] : 0);
\r
1396 workspace[1 * mlen + i] = (mlen - i <= (int)b[0] ? b[mlen - i] : 0);
\r
1399 internal_mul(workspace + 0 * mlen, workspace + 1 * mlen,
\r
1400 workspace + 2 * mlen, mlen, workspace + 4 * mlen);
\r
1402 /* now just copy the result back */
\r
1403 rlen = alen + blen + 1;
\r
1404 if (addend && rlen <= (int)addend[0])
\r
1405 rlen = addend[0] + 1;
\r
1406 ret = newbn(rlen);
\r
1408 for (i = 1; i <= (int)ret[0]; i++) {
\r
1409 ret[i] = (i <= 2 * mlen ? workspace[4 * mlen - i] : 0);
\r
1415 /* now add in the addend, if any */
\r
1417 BignumDblInt carry = 0;
\r
1418 for (i = 1; i <= rlen; i++) {
\r
1419 carry += (i <= (int)ret[0] ? ret[i] : 0);
\r
1420 carry += (i <= (int)addend[0] ? addend[i] : 0);
\r
1421 ret[i] = (BignumInt) carry & BIGNUM_INT_MASK;
\r
1422 carry >>= BIGNUM_INT_BITS;
\r
1423 if (ret[i] != 0 && i > maxspot)
\r
1429 smemclr(workspace, wslen * sizeof(*workspace));
\r
1435 * Non-modular multiplication.
\r
1437 Bignum bigmul(Bignum a, Bignum b)
\r
1439 return bigmuladd(a, b, NULL);
\r
1443 * Simple addition.
\r
1445 Bignum bigadd(Bignum a, Bignum b)
\r
1447 int alen = a[0], blen = b[0];
\r
1448 int rlen = (alen > blen ? alen : blen) + 1;
\r
1451 BignumDblInt carry;
\r
1453 ret = newbn(rlen);
\r
1457 for (i = 1; i <= rlen; i++) {
\r
1458 carry += (i <= (int)a[0] ? a[i] : 0);
\r
1459 carry += (i <= (int)b[0] ? b[i] : 0);
\r
1460 ret[i] = (BignumInt) carry & BIGNUM_INT_MASK;
\r
1461 carry >>= BIGNUM_INT_BITS;
\r
1462 if (ret[i] != 0 && i > maxspot)
\r
1471 * Subtraction. Returns a-b, or NULL if the result would come out
\r
1472 * negative (recall that this entire bignum module only handles
\r
1473 * positive numbers).
\r
1475 Bignum bigsub(Bignum a, Bignum b)
\r
1477 int alen = a[0], blen = b[0];
\r
1478 int rlen = (alen > blen ? alen : blen);
\r
1481 BignumDblInt carry;
\r
1483 ret = newbn(rlen);
\r
1487 for (i = 1; i <= rlen; i++) {
\r
1488 carry += (i <= (int)a[0] ? a[i] : 0);
\r
1489 carry += (i <= (int)b[0] ? b[i] ^ BIGNUM_INT_MASK : BIGNUM_INT_MASK);
\r
1490 ret[i] = (BignumInt) carry & BIGNUM_INT_MASK;
\r
1491 carry >>= BIGNUM_INT_BITS;
\r
1492 if (ret[i] != 0 && i > maxspot)
\r
1506 * Create a bignum which is the bitmask covering another one. That
\r
1507 * is, the smallest integer which is >= N and is also one less than
\r
1510 Bignum bignum_bitmask(Bignum n)
\r
1512 Bignum ret = copybn(n);
\r
1517 while (n[i] == 0 && i > 0)
\r
1520 return ret; /* input was zero */
\r
1526 ret[i] = BIGNUM_INT_MASK;
\r
1531 * Convert a (max 32-bit) long into a bignum.
\r
1533 Bignum bignum_from_long(unsigned long nn)
\r
1536 BignumDblInt n = nn;
\r
1539 ret[1] = (BignumInt)(n & BIGNUM_INT_MASK);
\r
1540 ret[2] = (BignumInt)((n >> BIGNUM_INT_BITS) & BIGNUM_INT_MASK);
\r
1542 ret[0] = (ret[2] ? 2 : 1);
\r
1547 * Add a long to a bignum.
\r
1549 Bignum bignum_add_long(Bignum number, unsigned long addendx)
\r
1551 Bignum ret = newbn(number[0] + 1);
\r
1552 int i, maxspot = 0;
\r
1553 BignumDblInt carry = 0, addend = addendx;
\r
1555 for (i = 1; i <= (int)ret[0]; i++) {
\r
1556 carry += addend & BIGNUM_INT_MASK;
\r
1557 carry += (i <= (int)number[0] ? number[i] : 0);
\r
1558 addend >>= BIGNUM_INT_BITS;
\r
1559 ret[i] = (BignumInt) carry & BIGNUM_INT_MASK;
\r
1560 carry >>= BIGNUM_INT_BITS;
\r
1569 * Compute the residue of a bignum, modulo a (max 16-bit) short.
\r
1571 unsigned short bignum_mod_short(Bignum number, unsigned short modulus)
\r
1573 BignumDblInt mod, r;
\r
1578 for (i = number[0]; i > 0; i--)
\r
1579 r = (r * (BIGNUM_TOP_BIT % mod) * 2 + number[i] % mod) % mod;
\r
1580 return (unsigned short) r;
\r
1584 void diagbn(char *prefix, Bignum md)
\r
1586 int i, nibbles, morenibbles;
\r
1587 static const char hex[] = "0123456789ABCDEF";
\r
1589 debug(("%s0x", prefix ? prefix : ""));
\r
1591 nibbles = (3 + bignum_bitcount(md)) / 4;
\r
1594 morenibbles = 4 * md[0] - nibbles;
\r
1595 for (i = 0; i < morenibbles; i++)
\r
1597 for (i = nibbles; i--;)
\r
1599 hex[(bignum_byte(md, i / 2) >> (4 * (i % 2))) & 0xF]));
\r
1607 * Simple division.
\r
1609 Bignum bigdiv(Bignum a, Bignum b)
\r
1611 Bignum q = newbn(a[0]);
\r
1612 bigdivmod(a, b, NULL, q);
\r
1617 * Simple remainder.
\r
1619 Bignum bigmod(Bignum a, Bignum b)
\r
1621 Bignum r = newbn(b[0]);
\r
1622 bigdivmod(a, b, r, NULL);
\r
1627 * Greatest common divisor.
\r
1629 Bignum biggcd(Bignum av, Bignum bv)
\r
1631 Bignum a = copybn(av);
\r
1632 Bignum b = copybn(bv);
\r
1634 while (bignum_cmp(b, Zero) != 0) {
\r
1635 Bignum t = newbn(b[0]);
\r
1636 bigdivmod(a, b, t, NULL);
\r
1637 while (t[0] > 1 && t[t[0]] == 0)
\r
1649 * Modular inverse, using Euclid's extended algorithm.
\r
1651 Bignum modinv(Bignum number, Bignum modulus)
\r
1653 Bignum a = copybn(modulus);
\r
1654 Bignum b = copybn(number);
\r
1655 Bignum xp = copybn(Zero);
\r
1656 Bignum x = copybn(One);
\r
1659 assert(number[number[0]] != 0);
\r
1660 assert(modulus[modulus[0]] != 0);
\r
1662 while (bignum_cmp(b, One) != 0) {
\r
1665 if (bignum_cmp(b, Zero) == 0) {
\r
1667 * Found a common factor between the inputs, so we cannot
\r
1668 * return a modular inverse at all.
\r
1679 bigdivmod(a, b, t, q);
\r
1680 while (t[0] > 1 && t[t[0]] == 0)
\r
1687 x = bigmuladd(q, xp, t);
\r
1697 /* now we know that sign * x == 1, and that x < modulus */
\r
1699 /* set a new x to be modulus - x */
\r
1700 Bignum newx = newbn(modulus[0]);
\r
1701 BignumInt carry = 0;
\r
1705 for (i = 1; i <= (int)newx[0]; i++) {
\r
1706 BignumInt aword = (i <= (int)modulus[0] ? modulus[i] : 0);
\r
1707 BignumInt bword = (i <= (int)x[0] ? x[i] : 0);
\r
1708 newx[i] = aword - bword - carry;
\r
1710 carry = carry ? (newx[i] >= bword) : (newx[i] > bword);
\r
1714 newx[0] = maxspot;
\r
1724 * Render a bignum into decimal. Return a malloced string holding
\r
1725 * the decimal representation.
\r
1727 char *bignum_decimal(Bignum x)
\r
1729 int ndigits, ndigit;
\r
1731 BignumDblInt carry;
\r
1733 BignumInt *workspace;
\r
1736 * First, estimate the number of digits. Since log(10)/log(2)
\r
1737 * is just greater than 93/28 (the joys of continued fraction
\r
1738 * approximations...) we know that for every 93 bits, we need
\r
1739 * at most 28 digits. This will tell us how much to malloc.
\r
1741 * Formally: if x has i bits, that means x is strictly less
\r
1742 * than 2^i. Since 2 is less than 10^(28/93), this is less than
\r
1743 * 10^(28i/93). We need an integer power of ten, so we must
\r
1744 * round up (rounding down might make it less than x again).
\r
1745 * Therefore if we multiply the bit count by 28/93, rounding
\r
1746 * up, we will have enough digits.
\r
1748 * i=0 (i.e., x=0) is an irritating special case.
\r
1750 i = bignum_bitcount(x);
\r
1752 ndigits = 1; /* x = 0 */
\r
1754 ndigits = (28 * i + 92) / 93; /* multiply by 28/93 and round up */
\r
1755 ndigits++; /* allow for trailing \0 */
\r
1756 ret = snewn(ndigits, char);
\r
1759 * Now allocate some workspace to hold the binary form as we
\r
1760 * repeatedly divide it by ten. Initialise this to the
\r
1761 * big-endian form of the number.
\r
1763 workspace = snewn(x[0], BignumInt);
\r
1764 for (i = 0; i < (int)x[0]; i++)
\r
1765 workspace[i] = x[x[0] - i];
\r
1768 * Next, write the decimal number starting with the last digit.
\r
1769 * We use ordinary short division, dividing 10 into the
\r
1772 ndigit = ndigits - 1;
\r
1773 ret[ndigit] = '\0';
\r
1777 for (i = 0; i < (int)x[0]; i++) {
\r
1778 carry = (carry << BIGNUM_INT_BITS) + workspace[i];
\r
1779 workspace[i] = (BignumInt) (carry / 10);
\r
1784 ret[--ndigit] = (char) (carry + '0');
\r
1785 } while (!iszero);
\r
1788 * There's a chance we've fallen short of the start of the
\r
1789 * string. Correct if so.
\r
1792 memmove(ret, ret + ndigit, ndigits - ndigit);
\r
1797 smemclr(workspace, x[0] * sizeof(*workspace));
\r
1804 #include <stdio.h>
\r
1805 #include <stdlib.h>
\r
1806 #include <ctype.h>
\r
1809 * gcc -Wall -g -O0 -DTESTBN -o testbn sshbn.c misc.c conf.c tree234.c unix/uxmisc.c -I. -I unix -I charset
\r
1811 * Then feed to this program's standard input the output of
\r
1812 * testdata/bignum.py .
\r
1815 void modalfatalbox(char *p, ...)
\r
1818 fprintf(stderr, "FATAL ERROR: ");
\r
1820 vfprintf(stderr, p, ap);
\r
1822 fputc('\n', stderr);
\r
1826 #define fromxdigit(c) ( (c)>'9' ? ((c)&0xDF) - 'A' + 10 : (c) - '0' )
\r
1828 int main(int argc, char **argv)
\r
1832 int passes = 0, fails = 0;
\r
1834 while ((buf = fgetline(stdin)) != NULL) {
\r
1835 int maxlen = strlen(buf);
\r
1836 unsigned char *data = snewn(maxlen, unsigned char);
\r
1837 unsigned char *ptrs[5], *q;
\r
1846 while (*bufp && !isspace((unsigned char)*bufp))
\r
1852 char *start, *end;
\r
1855 while (*bufp && !isxdigit((unsigned char)*bufp))
\r
1862 while (*bufp && isxdigit((unsigned char)*bufp))
\r
1866 if (ptrnum >= lenof(ptrs))
\r
1868 ptrs[ptrnum++] = q;
\r
1870 for (i = -((end - start) & 1); i < end-start; i += 2) {
\r
1871 unsigned char val = (i < 0 ? 0 : fromxdigit(start[i]));
\r
1872 val = val * 16 + fromxdigit(start[i+1]);
\r
1879 if (!strcmp(buf, "mul")) {
\r
1880 Bignum a, b, c, p;
\r
1882 if (ptrnum != 3) {
\r
1883 printf("%d: mul with %d parameters, expected 3\n", line, ptrnum);
\r
1886 a = bignum_from_bytes(ptrs[0], ptrs[1]-ptrs[0]);
\r
1887 b = bignum_from_bytes(ptrs[1], ptrs[2]-ptrs[1]);
\r
1888 c = bignum_from_bytes(ptrs[2], ptrs[3]-ptrs[2]);
\r
1891 if (bignum_cmp(c, p) == 0) {
\r
1894 char *as = bignum_decimal(a);
\r
1895 char *bs = bignum_decimal(b);
\r
1896 char *cs = bignum_decimal(c);
\r
1897 char *ps = bignum_decimal(p);
\r
1899 printf("%d: fail: %s * %s gave %s expected %s\n",
\r
1900 line, as, bs, ps, cs);
\r
1912 } else if (!strcmp(buf, "modmul")) {
\r
1913 Bignum a, b, m, c, p;
\r
1915 if (ptrnum != 4) {
\r
1916 printf("%d: modmul with %d parameters, expected 4\n",
\r
1920 a = bignum_from_bytes(ptrs[0], ptrs[1]-ptrs[0]);
\r
1921 b = bignum_from_bytes(ptrs[1], ptrs[2]-ptrs[1]);
\r
1922 m = bignum_from_bytes(ptrs[2], ptrs[3]-ptrs[2]);
\r
1923 c = bignum_from_bytes(ptrs[3], ptrs[4]-ptrs[3]);
\r
1924 p = modmul(a, b, m);
\r
1926 if (bignum_cmp(c, p) == 0) {
\r
1929 char *as = bignum_decimal(a);
\r
1930 char *bs = bignum_decimal(b);
\r
1931 char *ms = bignum_decimal(m);
\r
1932 char *cs = bignum_decimal(c);
\r
1933 char *ps = bignum_decimal(p);
\r
1935 printf("%d: fail: %s * %s mod %s gave %s expected %s\n",
\r
1936 line, as, bs, ms, ps, cs);
\r
1950 } else if (!strcmp(buf, "pow")) {
\r
1951 Bignum base, expt, modulus, expected, answer;
\r
1953 if (ptrnum != 4) {
\r
1954 printf("%d: mul with %d parameters, expected 4\n", line, ptrnum);
\r
1958 base = bignum_from_bytes(ptrs[0], ptrs[1]-ptrs[0]);
\r
1959 expt = bignum_from_bytes(ptrs[1], ptrs[2]-ptrs[1]);
\r
1960 modulus = bignum_from_bytes(ptrs[2], ptrs[3]-ptrs[2]);
\r
1961 expected = bignum_from_bytes(ptrs[3], ptrs[4]-ptrs[3]);
\r
1962 answer = modpow(base, expt, modulus);
\r
1964 if (bignum_cmp(expected, answer) == 0) {
\r
1967 char *as = bignum_decimal(base);
\r
1968 char *bs = bignum_decimal(expt);
\r
1969 char *cs = bignum_decimal(modulus);
\r
1970 char *ds = bignum_decimal(answer);
\r
1971 char *ps = bignum_decimal(expected);
\r
1973 printf("%d: fail: %s ^ %s mod %s gave %s expected %s\n",
\r
1974 line, as, bs, cs, ds, ps);
\r
1989 printf("%d: unrecognised test keyword: '%s'\n", line, buf);
\r
1997 printf("passed %d failed %d total %d\n", passes, fails, passes+fails);
\r
1998 return fails != 0;
\r