| blob | 964a90a1bb459720c126c8073cb9d9fd61b5ba61 |
1 /*\r
2 * Bignum routines for RSA and DH and stuff.\r
3 */\r
4 \r
5 #include <stdio.h>\r
6 #include <assert.h>\r
7 #include <stdlib.h>\r
8 #include <string.h>\r
9 #include <limits.h>\r
10 #include <ctype.h>\r
11 \r
12 #include "misc.h"\r
13 \r
14 #include "sshbn.h"\r
15 \r
16 #define BIGNUM_INTERNAL\r
17 typedef BignumInt *Bignum;\r
18 \r
19 #include "ssh.h"\r
20 \r
21 BignumInt bnZero[1] = { 0 };\r
22 BignumInt bnOne[2] = { 1, 1 };\r
23 BignumInt bnTen[2] = { 1, 10 };\r
24 \r
25 /*\r
26 * The Bignum format is an array of `BignumInt'. The first\r
27 * element of the array counts the remaining elements. The\r
28 * remaining elements express the actual number, base 2^BIGNUM_INT_BITS, _least_\r
29 * significant digit first. (So it's trivial to extract the bit\r
30 * with value 2^n for any n.)\r
31 *\r
32 * All Bignums in this module are positive. Negative numbers must\r
33 * be dealt with outside it.\r
34 *\r
35 * INVARIANT: the most significant word of any Bignum must be\r
36 * nonzero.\r
37 */\r
38 \r
39 Bignum Zero = bnZero, One = bnOne, Ten = bnTen;\r
40 \r
41 static Bignum newbn(int length)\r
42 {\r
43 Bignum b;\r
44 \r
45 assert(length >= 0 && length < INT_MAX / BIGNUM_INT_BITS);\r
46 \r
47 b = snewn(length + 1, BignumInt);\r
48 memset(b, 0, (length + 1) * sizeof(*b));\r
49 b[0] = length;\r
50 return b;\r
51 }\r
52 \r
53 void bn_restore_invariant(Bignum b)\r
54 {\r
55 while (b[0] > 1 && b[b[0]] == 0)\r
56 b[0]--;\r
57 }\r
58 \r
59 Bignum copybn(Bignum orig)\r
60 {\r
61 Bignum b = snewn(orig[0] + 1, BignumInt);\r
62 if (!b)\r
63 abort(); /* FIXME */\r
64 memcpy(b, orig, (orig[0] + 1) * sizeof(*b));\r
65 return b;\r
66 }\r
67 \r
68 void freebn(Bignum b)\r
69 {\r
70 /*\r
71 * Burn the evidence, just in case.\r
72 */\r
73 smemclr(b, sizeof(b[0]) * (b[0] + 1));\r
74 sfree(b);\r
75 }\r
76 \r
77 Bignum bn_power_2(int n)\r
78 {\r
79 Bignum ret;\r
80 \r
81 assert(n >= 0);\r
82 \r
83 ret = newbn(n / BIGNUM_INT_BITS + 1);\r
84 bignum_set_bit(ret, n, 1);\r
85 return ret;\r
86 }\r
87 \r
88 /*\r
89 * Internal addition. Sets c = a - b, where 'a', 'b' and 'c' are all\r
90 * big-endian arrays of 'len' BignumInts. Returns the carry off the\r
91 * top.\r
92 */\r
93 static BignumCarry internal_add(const BignumInt *a, const BignumInt *b,\r
94 BignumInt *c, int len)\r
95 {\r
96 int i;\r
97 BignumCarry carry = 0;\r
98 \r
99 for (i = len-1; i >= 0; i--)\r
100 BignumADC(c[i], carry, a[i], b[i], carry);\r
101 \r
102 return (BignumInt)carry;\r
103 }\r
104 \r
105 /*\r
106 * Internal subtraction. Sets c = a - b, where 'a', 'b' and 'c' are\r
107 * all big-endian arrays of 'len' BignumInts. Any borrow from the top\r
108 * is ignored.\r
109 */\r
110 static void internal_sub(const BignumInt *a, const BignumInt *b,\r
111 BignumInt *c, int len)\r
112 {\r
113 int i;\r
114 BignumCarry carry = 1;\r
115 \r
116 for (i = len-1; i >= 0; i--)\r
117 BignumADC(c[i], carry, a[i], ~b[i], carry);\r
118 }\r
119 \r
120 /*\r
121 * Compute c = a * b.\r
122 * Input is in the first len words of a and b.\r
123 * Result is returned in the first 2*len words of c.\r
124 *\r
125 * 'scratch' must point to an array of BignumInt of size at least\r
126 * mul_compute_scratch(len). (This covers the needs of internal_mul\r
127 * and all its recursive calls to itself.)\r
128 */\r
129 #define KARATSUBA_THRESHOLD 50\r
130 static int mul_compute_scratch(int len)\r
131 {\r
132 int ret = 0;\r
133 while (len > KARATSUBA_THRESHOLD) {\r
134 int toplen = len/2, botlen = len - toplen; /* botlen is the bigger */\r
135 int midlen = botlen + 1;\r
136 ret += 4*midlen;\r
137 len = midlen;\r
138 }\r
139 return ret;\r
140 }\r
141 static void internal_mul(const BignumInt *a, const BignumInt *b,\r
142 BignumInt *c, int len, BignumInt *scratch)\r
143 {\r
144 if (len > KARATSUBA_THRESHOLD) {\r
145 int i;\r
146 \r
147 /*\r
148 * Karatsuba divide-and-conquer algorithm. Cut each input in\r
149 * half, so that it's expressed as two big 'digits' in a giant\r
150 * base D:\r
151 *\r
152 * a = a_1 D + a_0\r
153 * b = b_1 D + b_0\r
154 *\r
155 * Then the product is of course\r
156 *\r
157 * ab = a_1 b_1 D^2 + (a_1 b_0 + a_0 b_1) D + a_0 b_0\r
158 *\r
159 * and we compute the three coefficients by recursively\r
160 * calling ourself to do half-length multiplications.\r
161 *\r
162 * The clever bit that makes this worth doing is that we only\r
163 * need _one_ half-length multiplication for the central\r
164 * coefficient rather than the two that it obviouly looks\r
165 * like, because we can use a single multiplication to compute\r
166 *\r
167 * (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
168 *\r
169 * and then we subtract the other two coefficients (a_1 b_1\r
170 * and a_0 b_0) which we were computing anyway.\r
171 *\r
172 * Hence we get to multiply two numbers of length N in about\r
173 * three times as much work as it takes to multiply numbers of\r
174 * length N/2, which is obviously better than the four times\r
175 * as much work it would take if we just did a long\r
176 * conventional multiply.\r
177 */\r
178 \r
179 int toplen = len/2, botlen = len - toplen; /* botlen is the bigger */\r
180 int midlen = botlen + 1;\r
181 BignumCarry carry;\r
182 #ifdef KARA_DEBUG\r
183 int i;\r
184 #endif\r
185 \r
186 /*\r
187 * The coefficients a_1 b_1 and a_0 b_0 just avoid overlapping\r
188 * in the output array, so we can compute them immediately in\r
189 * place.\r
190 */\r
191 \r
192 #ifdef KARA_DEBUG\r
193 printf("a1,a0 = 0x");\r
194 for (i = 0; i < len; i++) {\r
195 if (i == toplen) printf(", 0x");\r
196 printf("%0*x", BIGNUM_INT_BITS/4, a[i]);\r
197 }\r
198 printf("\n");\r
199 printf("b1,b0 = 0x");\r
200 for (i = 0; i < len; i++) {\r
201 if (i == toplen) printf(", 0x");\r
202 printf("%0*x", BIGNUM_INT_BITS/4, b[i]);\r
203 }\r
204 printf("\n");\r
205 #endif\r
206 \r
207 /* a_1 b_1 */\r
208 internal_mul(a, b, c, toplen, scratch);\r
209 #ifdef KARA_DEBUG\r
210 printf("a1b1 = 0x");\r
211 for (i = 0; i < 2*toplen; i++) {\r
212 printf("%0*x", BIGNUM_INT_BITS/4, c[i]);\r
213 }\r
214 printf("\n");\r
215 #endif\r
216 \r
217 /* a_0 b_0 */\r
218 internal_mul(a + toplen, b + toplen, c + 2*toplen, botlen, scratch);\r
219 #ifdef KARA_DEBUG\r
220 printf("a0b0 = 0x");\r
221 for (i = 0; i < 2*botlen; i++) {\r
222 printf("%0*x", BIGNUM_INT_BITS/4, c[2*toplen+i]);\r
223 }\r
224 printf("\n");\r
225 #endif\r
226 \r
227 /* Zero padding. midlen exceeds toplen by at most 2, so just\r
228 * zero the first two words of each input and the rest will be\r
229 * copied over. */\r
230 scratch[0] = scratch[1] = scratch[midlen] = scratch[midlen+1] = 0;\r
231 \r
232 for (i = 0; i < toplen; i++) {\r
233 scratch[midlen - toplen + i] = a[i]; /* a_1 */\r
234 scratch[2*midlen - toplen + i] = b[i]; /* b_1 */\r
235 }\r
236 \r
237 /* compute a_1 + a_0 */\r
238 scratch[0] = internal_add(scratch+1, a+toplen, scratch+1, botlen);\r
239 #ifdef KARA_DEBUG\r
240 printf("a1plusa0 = 0x");\r
241 for (i = 0; i < midlen; i++) {\r
242 printf("%0*x", BIGNUM_INT_BITS/4, scratch[i]);\r
243 }\r
244 printf("\n");\r
245 #endif\r
246 /* compute b_1 + b_0 */\r
247 scratch[midlen] = internal_add(scratch+midlen+1, b+toplen,\r
248 scratch+midlen+1, botlen);\r
249 #ifdef KARA_DEBUG\r
250 printf("b1plusb0 = 0x");\r
251 for (i = 0; i < midlen; i++) {\r
252 printf("%0*x", BIGNUM_INT_BITS/4, scratch[midlen+i]);\r
253 }\r
254 printf("\n");\r
255 #endif\r
256 \r
257 /*\r
258 * Now we can do the third multiplication.\r
259 */\r
260 internal_mul(scratch, scratch + midlen, scratch + 2*midlen, midlen,\r
261 scratch + 4*midlen);\r
262 #ifdef KARA_DEBUG\r
263 printf("a1plusa0timesb1plusb0 = 0x");\r
264 for (i = 0; i < 2*midlen; i++) {\r
265 printf("%0*x", BIGNUM_INT_BITS/4, scratch[2*midlen+i]);\r
266 }\r
267 printf("\n");\r
268 #endif\r
269 \r
270 /*\r
271 * Now we can reuse the first half of 'scratch' to compute the\r
272 * sum of the outer two coefficients, to subtract from that\r
273 * product to obtain the middle one.\r
274 */\r
275 scratch[0] = scratch[1] = scratch[2] = scratch[3] = 0;\r
276 for (i = 0; i < 2*toplen; i++)\r
277 scratch[2*midlen - 2*toplen + i] = c[i];\r
278 scratch[1] = internal_add(scratch+2, c + 2*toplen,\r
279 scratch+2, 2*botlen);\r
280 #ifdef KARA_DEBUG\r
281 printf("a1b1plusa0b0 = 0x");\r
282 for (i = 0; i < 2*midlen; i++) {\r
283 printf("%0*x", BIGNUM_INT_BITS/4, scratch[i]);\r
284 }\r
285 printf("\n");\r
286 #endif\r
287 \r
288 internal_sub(scratch + 2*midlen, scratch,\r
289 scratch + 2*midlen, 2*midlen);\r
290 #ifdef KARA_DEBUG\r
291 printf("a1b0plusa0b1 = 0x");\r
292 for (i = 0; i < 2*midlen; i++) {\r
293 printf("%0*x", BIGNUM_INT_BITS/4, scratch[2*midlen+i]);\r
294 }\r
295 printf("\n");\r
296 #endif\r
297 \r
298 /*\r
299 * And now all we need to do is to add that middle coefficient\r
300 * back into the output. We may have to propagate a carry\r
301 * further up the output, but we can be sure it won't\r
302 * propagate right the way off the top.\r
303 */\r
304 carry = internal_add(c + 2*len - botlen - 2*midlen,\r
305 scratch + 2*midlen,\r
306 c + 2*len - botlen - 2*midlen, 2*midlen);\r
307 i = 2*len - botlen - 2*midlen - 1;\r
308 while (carry) {\r
309 assert(i >= 0);\r
310 BignumADC(c[i], carry, c[i], 0, carry);\r
311 i--;\r
312 }\r
313 #ifdef KARA_DEBUG\r
314 printf("ab = 0x");\r
315 for (i = 0; i < 2*len; i++) {\r
316 printf("%0*x", BIGNUM_INT_BITS/4, c[i]);\r
317 }\r
318 printf("\n");\r
319 #endif\r
320 \r
321 } else {\r
322 int i;\r
323 BignumInt carry;\r
324 const BignumInt *ap, *bp;\r
325 BignumInt *cp, *cps;\r
326 \r
327 /*\r
328 * Multiply in the ordinary O(N^2) way.\r
329 */\r
330 \r
331 for (i = 0; i < 2 * len; i++)\r
332 c[i] = 0;\r
333 \r
334 for (cps = c + 2*len, ap = a + len; ap-- > a; cps--) {\r
335 carry = 0;\r
336 for (cp = cps, bp = b + len; cp--, bp-- > b ;)\r
337 BignumMULADD2(carry, *cp, *ap, *bp, *cp, carry);\r
338 *cp = carry;\r
339 }\r
340 }\r
341 }\r
342 \r
343 /*\r
344 * Variant form of internal_mul used for the initial step of\r
345 * Montgomery reduction. Only bothers outputting 'len' words\r
346 * (everything above that is thrown away).\r
347 */\r
348 static void internal_mul_low(const BignumInt *a, const BignumInt *b,\r
349 BignumInt *c, int len, BignumInt *scratch)\r
350 {\r
351 if (len > KARATSUBA_THRESHOLD) {\r
352 int i;\r
353 \r
354 /*\r
355 * Karatsuba-aware version of internal_mul_low. As before, we\r
356 * express each input value as a shifted combination of two\r
357 * halves:\r
358 *\r
359 * a = a_1 D + a_0\r
360 * b = b_1 D + b_0\r
361 *\r
362 * Then the full product is, as before,\r
363 *\r
364 * ab = a_1 b_1 D^2 + (a_1 b_0 + a_0 b_1) D + a_0 b_0\r
365 *\r
366 * Provided we choose D on the large side (so that a_0 and b_0\r
367 * are _at least_ as long as a_1 and b_1), we don't need the\r
368 * topmost term at all, and we only need half of the middle\r
369 * term. So there's no point in doing the proper Karatsuba\r
370 * optimisation which computes the middle term using the top\r
371 * one, because we'd take as long computing the top one as\r
372 * just computing the middle one directly.\r
373 *\r
374 * So instead, we do a much more obvious thing: we call the\r
375 * fully optimised internal_mul to compute a_0 b_0, and we\r
376 * recursively call ourself to compute the _bottom halves_ of\r
377 * a_1 b_0 and a_0 b_1, each of which we add into the result\r
378 * in the obvious way.\r
379 *\r
380 * In other words, there's no actual Karatsuba _optimisation_\r
381 * in this function; the only benefit in doing it this way is\r
382 * that we call internal_mul proper for a large part of the\r
383 * work, and _that_ can optimise its operation.\r
384 */\r
385 \r
386 int toplen = len/2, botlen = len - toplen; /* botlen is the bigger */\r
387 \r
388 /*\r
389 * Scratch space for the various bits and pieces we're going\r
390 * to be adding together: we need botlen*2 words for a_0 b_0\r
391 * (though we may end up throwing away its topmost word), and\r
392 * toplen words for each of a_1 b_0 and a_0 b_1. That adds up\r
393 * to exactly 2*len.\r
394 */\r
395 \r
396 /* a_0 b_0 */\r
397 internal_mul(a + toplen, b + toplen, scratch + 2*toplen, botlen,\r
398 scratch + 2*len);\r
399 \r
400 /* a_1 b_0 */\r
401 internal_mul_low(a, b + len - toplen, scratch + toplen, toplen,\r
402 scratch + 2*len);\r
403 \r
404 /* a_0 b_1 */\r
405 internal_mul_low(a + len - toplen, b, scratch, toplen,\r
406 scratch + 2*len);\r
407 \r
408 /* Copy the bottom half of the big coefficient into place */\r
409 for (i = 0; i < botlen; i++)\r
410 c[toplen + i] = scratch[2*toplen + botlen + i];\r
411 \r
412 /* Add the two small coefficients, throwing away the returned carry */\r
413 internal_add(scratch, scratch + toplen, scratch, toplen);\r
414 \r
415 /* And add that to the large coefficient, leaving the result in c. */\r
416 internal_add(scratch, scratch + 2*toplen + botlen - toplen,\r
417 c, toplen);\r
418 \r
419 } else {\r
420 int i;\r
421 BignumInt carry;\r
422 const BignumInt *ap, *bp;\r
423 BignumInt *cp, *cps;\r
424 \r
425 /*\r
426 * Multiply in the ordinary O(N^2) way.\r
427 */\r
428 \r
429 for (i = 0; i < len; i++)\r
430 c[i] = 0;\r
431 \r
432 for (cps = c + len, ap = a + len; ap-- > a; cps--) {\r
433 carry = 0;\r
434 for (cp = cps, bp = b + len; bp--, cp-- > c ;)\r
435 BignumMULADD2(carry, *cp, *ap, *bp, *cp, carry);\r
436 }\r
437 }\r
438 }\r
439 \r
440 /*\r
441 * Montgomery reduction. Expects x to be a big-endian array of 2*len\r
442 * BignumInts whose value satisfies 0 <= x < rn (where r = 2^(len *\r
443 * BIGNUM_INT_BITS) is the Montgomery base). Returns in the same array\r
444 * a value x' which is congruent to xr^{-1} mod n, and satisfies 0 <=\r
445 * x' < n.\r
446 *\r
447 * 'n' and 'mninv' should be big-endian arrays of 'len' BignumInts\r
448 * each, containing respectively n and the multiplicative inverse of\r
449 * -n mod r.\r
450 *\r
451 * 'tmp' is an array of BignumInt used as scratch space, of length at\r
452 * least 3*len + mul_compute_scratch(len).\r
453 */\r
454 static void monty_reduce(BignumInt *x, const BignumInt *n,\r
455 const BignumInt *mninv, BignumInt *tmp, int len)\r
456 {\r
457 int i;\r
458 BignumInt carry;\r
459 \r
460 /*\r
461 * Multiply x by (-n)^{-1} mod r. This gives us a value m such\r
462 * that mn is congruent to -x mod r. Hence, mn+x is an exact\r
463 * multiple of r, and is also (obviously) congruent to x mod n.\r
464 */\r
465 internal_mul_low(x + len, mninv, tmp, len, tmp + 3*len);\r
466 \r
467 /*\r
468 * Compute t = (mn+x)/r in ordinary, non-modular, integer\r
469 * arithmetic. By construction this is exact, and is congruent mod\r
470 * n to x * r^{-1}, i.e. the answer we want.\r
471 *\r
472 * The following multiply leaves that answer in the _most_\r
473 * significant half of the 'x' array, so then we must shift it\r
474 * down.\r
475 */\r
476 internal_mul(tmp, n, tmp+len, len, tmp + 3*len);\r
477 carry = internal_add(x, tmp+len, x, 2*len);\r
478 for (i = 0; i < len; i++)\r
479 x[len + i] = x[i], x[i] = 0;\r
480 \r
481 /*\r
482 * Reduce t mod n. This doesn't require a full-on division by n,\r
483 * but merely a test and single optional subtraction, since we can\r
484 * show that 0 <= t < 2n.\r
485 *\r
486 * Proof:\r
487 * + we computed m mod r, so 0 <= m < r.\r
488 * + so 0 <= mn < rn, obviously\r
489 * + hence we only need 0 <= x < rn to guarantee that 0 <= mn+x < 2rn\r
490 * + yielding 0 <= (mn+x)/r < 2n as required.\r
491 */\r
492 if (!carry) {\r
493 for (i = 0; i < len; i++)\r
494 if (x[len + i] != n[i])\r
495 break;\r
496 }\r
497 if (carry || i >= len || x[len + i] > n[i])\r
498 internal_sub(x+len, n, x+len, len);\r
499 }\r
500 \r
501 static void internal_add_shifted(BignumInt *number,\r
502 BignumInt n, int shift)\r
503 {\r
504 int word = 1 + (shift / BIGNUM_INT_BITS);\r
505 int bshift = shift % BIGNUM_INT_BITS;\r
506 BignumInt addendh, addendl;\r
507 BignumCarry carry;\r
508 \r
509 addendl = n << bshift;\r
510 addendh = (bshift == 0 ? 0 : n >> (BIGNUM_INT_BITS - bshift));\r
511 \r
512 assert(word <= number[0]);\r
513 BignumADC(number[word], carry, number[word], addendl, 0);\r
514 word++;\r
515 if (!addendh && !carry)\r
516 return;\r
517 assert(word <= number[0]);\r
518 BignumADC(number[word], carry, number[word], addendh, carry);\r
519 word++;\r
520 while (carry) {\r
521 assert(word <= number[0]);\r
522 BignumADC(number[word], carry, number[word], 0, carry);\r
523 word++;\r
524 }\r
525 }\r
526 \r
527 static int bn_clz(BignumInt x)\r
528 {\r
529 /*\r
530 * Count the leading zero bits in x. Equivalently, how far left\r
531 * would we need to shift x to make its top bit set?\r
532 *\r
533 * Precondition: x != 0.\r
534 */\r
535 \r
536 /* FIXME: would be nice to put in some compiler intrinsics under\r
537 * ifdef here */\r
538 int i, ret = 0;\r
539 for (i = BIGNUM_INT_BITS / 2; i != 0; i >>= 1) {\r
540 if ((x >> (BIGNUM_INT_BITS-i)) == 0) {\r
541 x <<= i;\r
542 ret += i;\r
543 }\r
544 }\r
545 return ret;\r
546 }\r
547 \r
548 static BignumInt reciprocal_word(BignumInt d)\r
549 {\r
550 BignumInt dshort, recip, prodh, prodl;\r
551 int corrections;\r
552 \r
553 /*\r
554 * Input: a BignumInt value d, with its top bit set.\r
555 */\r
556 assert(d >> (BIGNUM_INT_BITS-1) == 1);\r
557 \r
558 /*\r
559 * Output: a value, shifted to fill a BignumInt, which is strictly\r
560 * less than 1/(d+1), i.e. is an *under*-estimate (but by as\r
561 * little as possible within the constraints) of the reciprocal of\r
562 * any number whose first BIGNUM_INT_BITS bits match d.\r
563 *\r
564 * Ideally we'd like to _totally_ fill BignumInt, i.e. always\r
565 * return a value with the top bit set. Unfortunately we can't\r
566 * quite guarantee that for all inputs and also return a fixed\r
567 * exponent. So instead we take our reciprocal to be\r
568 * 2^(BIGNUM_INT_BITS*2-1) / d, so that it has the top bit clear\r
569 * only in the exceptional case where d takes exactly the maximum\r
570 * value BIGNUM_INT_MASK; in that case, the top bit is clear and\r
571 * the next bit down is set.\r
572 */\r
573 \r
574 /*\r
575 * Start by computing a half-length version of the answer, by\r
576 * straightforward division within a BignumInt.\r
577 */\r
578 dshort = (d >> (BIGNUM_INT_BITS/2)) + 1;\r
579 recip = (BIGNUM_TOP_BIT + dshort - 1) / dshort;\r
580 recip <<= BIGNUM_INT_BITS - BIGNUM_INT_BITS/2;\r
581 \r
582 /*\r
583 * Newton-Raphson iteration to improve that starting reciprocal\r
584 * estimate: take f(x) = d - 1/x, and then the N-R formula gives\r
585 * x_new = x - f(x)/f'(x) = x - (d-1/x)/(1/x^2) = x(2-d*x). Or,\r
586 * taking our fixed-point representation into account, take f(x)\r
587 * to be d - K/x (where K = 2^(BIGNUM_INT_BITS*2-1) as discussed\r
588 * above) and then we get (2K - d*x) * x/K.\r
589 *\r
590 * Newton-Raphson doubles the number of correct bits at every\r
591 * iteration, and the initial division above already gave us half\r
592 * the output word, so it's only worth doing one iteration.\r
593 */\r
594 BignumMULADD(prodh, prodl, recip, d, recip);\r
595 prodl = ~prodl;\r
596 prodh = ~prodh;\r
597 {\r
598 BignumCarry c;\r
599 BignumADC(prodl, c, prodl, 1, 0);\r
600 prodh += c;\r
601 }\r
602 BignumMUL(prodh, prodl, prodh, recip);\r
603 recip = (prodh << 1) | (prodl >> (BIGNUM_INT_BITS-1));\r
604 \r
605 /*\r
606 * Now make sure we have the best possible reciprocal estimate,\r
607 * before we return it. We might have been off by a handful either\r
608 * way - not enough to bother with any better-thought-out kind of\r
609 * correction loop.\r
610 */\r
611 BignumMULADD(prodh, prodl, recip, d, recip);\r
612 corrections = 0;\r
613 if (prodh >= BIGNUM_TOP_BIT) {\r
614 do {\r
615 BignumCarry c = 1;\r
616 BignumADC(prodl, c, prodl, ~d, c); prodh += BIGNUM_INT_MASK + c;\r
617 recip--;\r
618 corrections++;\r
619 } while (prodh >= ((BignumInt)1 << (BIGNUM_INT_BITS-1)));\r
620 } else {\r
621 while (1) {\r
622 BignumInt newprodh, newprodl;\r
623 BignumCarry c = 0;\r
624 BignumADC(newprodl, c, prodl, d, c); newprodh = prodh + c;\r
625 if (newprodh >= BIGNUM_TOP_BIT)\r
626 break;\r
627 prodh = newprodh;\r
628 prodl = newprodl;\r
629 recip++;\r
630 corrections++;\r
631 }\r
632 }\r
633 \r
634 return recip;\r
635 }\r
636 \r
637 /*\r
638 * Compute a = a % m.\r
639 * Input in first alen words of a and first mlen words of m.\r
640 * Output in first alen words of a\r
641 * (of which first alen-mlen words will be zero).\r
642 * Quotient is accumulated in the `quotient' array, which is a Bignum\r
643 * rather than the internal bigendian format.\r
644 *\r
645 * 'recip' must be the result of calling reciprocal_word() on the top\r
646 * BIGNUM_INT_BITS of the modulus (denoted m0 in comments below), with\r
647 * the topmost set bit normalised to the MSB of the input to\r
648 * reciprocal_word. 'rshift' is how far left the top nonzero word of\r
649 * the modulus had to be shifted to set that top bit.\r
650 */\r
651 static void internal_mod(BignumInt *a, int alen,\r
652 BignumInt *m, int mlen,\r
653 BignumInt *quot, BignumInt recip, int rshift)\r
654 {\r
655 int i, k;\r
656 \r
657 #ifdef DIVISION_DEBUG\r
658 {\r
659 int d;\r
660 printf("start division, m=0x");\r
661 for (d = 0; d < mlen; d++)\r
662 printf("%0*llx", BIGNUM_INT_BITS/4, (unsigned long long)m[d]);\r
663 printf(", recip=%#0*llx, rshift=%d\n",\r
664 BIGNUM_INT_BITS/4, (unsigned long long)recip, rshift);\r
665 }\r
666 #endif\r
667 \r
668 /*\r
669 * Repeatedly use that reciprocal estimate to get a decent number\r
670 * of quotient bits, and subtract off the resulting multiple of m.\r
671 *\r
672 * Normally we expect to terminate this loop by means of finding\r
673 * out q=0 part way through, but one way in which we might not get\r
674 * that far in the first place is if the input a is actually zero,\r
675 * in which case we'll discard zero words from the front of a\r
676 * until we reach the termination condition in the for statement\r
677 * here.\r
678 */\r
679 for (i = 0; i <= alen - mlen ;) {\r
680 BignumInt product;\r
681 BignumInt aword, q;\r
682 int shift, full_bitoffset, bitoffset, wordoffset;\r
683 \r
684 #ifdef DIVISION_DEBUG\r
685 {\r
686 int d;\r
687 printf("main loop, a=0x");\r
688 for (d = 0; d < alen; d++)\r
689 printf("%0*llx", BIGNUM_INT_BITS/4, (unsigned long long)a[d]);\r
690 printf("\n");\r
691 }\r
692 #endif\r
693 \r
694 if (a[i] == 0) {\r
695 #ifdef DIVISION_DEBUG\r
696 printf("zero word at i=%d\n", i);\r
697 #endif\r
698 i++;\r
699 continue;\r
700 }\r
701 \r
702 aword = a[i];\r
703 shift = bn_clz(aword);\r
704 aword <<= shift;\r
705 if (shift > 0 && i+1 < alen)\r
706 aword |= a[i+1] >> (BIGNUM_INT_BITS - shift);\r
707 \r
708 {\r
709 BignumInt unused;\r
710 BignumMUL(q, unused, recip, aword);\r
711 (void)unused;\r
712 }\r
713 \r
714 #ifdef DIVISION_DEBUG\r
715 printf("i=%d, aword=%#0*llx, shift=%d, q=%#0*llx\n",\r
716 i, BIGNUM_INT_BITS/4, (unsigned long long)aword,\r
717 shift, BIGNUM_INT_BITS/4, (unsigned long long)q);\r
718 #endif\r
719 \r
720 /*\r
721 * Work out the right bit and word offsets to use when\r
722 * subtracting q*m from a.\r
723 *\r
724 * aword was taken from a[i], which means its LSB was at bit\r
725 * position (alen-1-i) * BIGNUM_INT_BITS. But then we shifted\r
726 * it left by 'shift', so now the low bit of aword corresponds\r
727 * to bit position (alen-1-i) * BIGNUM_INT_BITS - shift, i.e.\r
728 * aword is approximately equal to a / 2^(that).\r
729 *\r
730 * m0 comes from the top word of mod, so its LSB is at bit\r
731 * position (mlen-1) * BIGNUM_INT_BITS - rshift, i.e. it can\r
732 * be considered to be m / 2^(that power). 'recip' is the\r
733 * reciprocal of m0, times 2^(BIGNUM_INT_BITS*2-1), i.e. it's\r
734 * about 2^((mlen+1) * BIGNUM_INT_BITS - rshift - 1) / m.\r
735 *\r
736 * Hence, recip * aword is approximately equal to the product\r
737 * of those, which simplifies to\r
738 *\r
739 * a/m * 2^((mlen+2+i-alen)*BIGNUM_INT_BITS + shift - rshift - 1)\r
740 *\r
741 * But we've also shifted recip*aword down by BIGNUM_INT_BITS\r
742 * to form q, so we have\r
743 *\r
744 * q ~= a/m * 2^((mlen+1+i-alen)*BIGNUM_INT_BITS + shift - rshift - 1)\r
745 *\r
746 * and hence, when we now compute q*m, it will be about\r
747 * a*2^(all that lot), i.e. the negation of that expression is\r
748 * how far left we have to shift the product q*m to make it\r
749 * approximately equal to a.\r
750 */\r
751 full_bitoffset = -((mlen+1+i-alen)*BIGNUM_INT_BITS + shift-rshift-1);\r
752 #ifdef DIVISION_DEBUG\r
753 printf("full_bitoffset=%d\n", full_bitoffset);\r
754 #endif\r
755 \r
756 if (full_bitoffset < 0) {\r
757 /*\r
758 * If we find ourselves needing to shift q*m _right_, that\r
759 * means we've reached the bottom of the quotient. Clip q\r
760 * so that its right shift becomes zero, and if that means\r
761 * q becomes _actually_ zero, this loop is done.\r
762 */\r
763 if (full_bitoffset <= -BIGNUM_INT_BITS)\r
764 break;\r
765 q >>= -full_bitoffset;\r
766 full_bitoffset = 0;\r
767 if (!q)\r
768 break;\r
769 #ifdef DIVISION_DEBUG\r
770 printf("now full_bitoffset=%d, q=%#0*llx\n",\r
771 full_bitoffset, BIGNUM_INT_BITS/4, (unsigned long long)q);\r
772 #endif\r
773 }\r
774 \r
775 wordoffset = full_bitoffset / BIGNUM_INT_BITS;\r
776 bitoffset = full_bitoffset % BIGNUM_INT_BITS;\r
777 #ifdef DIVISION_DEBUG\r
778 printf("wordoffset=%d, bitoffset=%d\n", wordoffset, bitoffset);\r
779 #endif\r
780 \r
781 /* wordoffset as computed above is the offset between the LSWs\r
782 * of m and a. But in fact m and a are stored MSW-first, so we\r
783 * need to adjust it to be the offset between the actual array\r
784 * indices, and flip the sign too. */\r
785 wordoffset = alen - mlen - wordoffset;\r
786 \r
787 if (bitoffset == 0) {\r
788 BignumCarry c = 1;\r
789 BignumInt prev_hi_word = 0;\r
790 for (k = mlen - 1; wordoffset+k >= i; k--) {\r
791 BignumInt mword = k<0 ? 0 : m[k];\r
792 BignumMULADD(prev_hi_word, product, q, mword, prev_hi_word);\r
793 #ifdef DIVISION_DEBUG\r
794 printf(" aligned sub: product word for m[%d] = %#0*llx\n",\r
795 k, BIGNUM_INT_BITS/4,\r
796 (unsigned long long)product);\r
797 #endif\r
798 #ifdef DIVISION_DEBUG\r
799 printf(" aligned sub: subtrahend for a[%d] = %#0*llx\n",\r
800 wordoffset+k, BIGNUM_INT_BITS/4,\r
801 (unsigned long long)product);\r
802 #endif\r
803 BignumADC(a[wordoffset+k], c, a[wordoffset+k], ~product, c);\r
804 }\r
805 } else {\r
806 BignumInt add_word = 0;\r
807 BignumInt c = 1;\r
808 BignumInt prev_hi_word = 0;\r
809 for (k = mlen - 1; wordoffset+k >= i; k--) {\r
810 BignumInt mword = k<0 ? 0 : m[k];\r
811 BignumMULADD(prev_hi_word, product, q, mword, prev_hi_word);\r
812 #ifdef DIVISION_DEBUG\r
813 printf(" unaligned sub: product word for m[%d] = %#0*llx\n",\r
814 k, BIGNUM_INT_BITS/4,\r
815 (unsigned long long)product);\r
816 #endif\r
817 \r
818 add_word |= product << bitoffset;\r
819 \r
820 #ifdef DIVISION_DEBUG\r
821 printf(" unaligned sub: subtrahend for a[%d] = %#0*llx\n",\r
822 wordoffset+k,\r
823 BIGNUM_INT_BITS/4, (unsigned long long)add_word);\r
824 #endif\r
825 BignumADC(a[wordoffset+k], c, a[wordoffset+k], ~add_word, c);\r
826 \r
827 add_word = product >> (BIGNUM_INT_BITS - bitoffset);\r
828 }\r
829 }\r
830 \r
831 if (quot) {\r
832 #ifdef DIVISION_DEBUG\r
833 printf("adding quotient word %#0*llx << %d\n",\r
834 BIGNUM_INT_BITS/4, (unsigned long long)q, full_bitoffset);\r
835 #endif\r
836 internal_add_shifted(quot, q, full_bitoffset);\r
837 #ifdef DIVISION_DEBUG\r
838 {\r
839 int d;\r
840 printf("now quot=0x");\r
841 for (d = quot[0]; d > 0; d--)\r
842 printf("%0*llx", BIGNUM_INT_BITS/4,\r
843 (unsigned long long)quot[d]);\r
844 printf("\n");\r
845 }\r
846 #endif\r
847 }\r
848 }\r
849 \r
850 #ifdef DIVISION_DEBUG\r
851 {\r
852 int d;\r
853 printf("end main loop, a=0x");\r
854 for (d = 0; d < alen; d++)\r
855 printf("%0*llx", BIGNUM_INT_BITS/4, (unsigned long long)a[d]);\r
856 if (quot) {\r
857 printf(", quot=0x");\r
858 for (d = quot[0]; d > 0; d--)\r
859 printf("%0*llx", BIGNUM_INT_BITS/4,\r
860 (unsigned long long)quot[d]);\r
861 }\r
862 printf("\n");\r
863 }\r
864 #endif\r
865 \r
866 /*\r
867 * The above loop should terminate with the remaining value in a\r
868 * being strictly less than 2*m (if a >= 2*m then we should always\r
869 * have managed to get a nonzero q word), but we can't guarantee\r
870 * that it will be strictly less than m: consider a case where the\r
871 * remainder is 1, and another where the remainder is m-1. By the\r
872 * time a contains a value that's _about m_, you clearly can't\r
873 * distinguish those cases by looking at only the top word of a -\r
874 * you have to go all the way down to the bottom before you find\r
875 * out whether it's just less or just more than m.\r
876 *\r
877 * Hence, we now do a final fixup in which we subtract one last\r
878 * copy of m, or don't, accordingly. We should never have to\r
879 * subtract more than one copy of m here.\r
880 */\r
881 for (i = 0; i < alen; i++) {\r
882 /* Compare a with m, word by word, from the MSW down. As soon\r
883 * as we encounter a difference, we know whether we need the\r
884 * fixup. */\r
885 int mindex = mlen-alen+i;\r
886 BignumInt mword = mindex < 0 ? 0 : m[mindex];\r
887 if (a[i] < mword) {\r
888 #ifdef DIVISION_DEBUG\r
889 printf("final fixup not needed, a < m\n");\r
890 #endif\r
891 return;\r
892 } else if (a[i] > mword) {\r
893 #ifdef DIVISION_DEBUG\r
894 printf("final fixup is needed, a > m\n");\r
895 #endif\r
896 break;\r
897 }\r
898 /* If neither of those cases happened, the words are the same,\r
899 * so keep going and look at the next one. */\r
900 }\r
901 #ifdef DIVISION_DEBUG\r
902 if (i == mlen) /* if we printed neither of the above diagnostics */\r
903 printf("final fixup is needed, a == m\n");\r
904 #endif\r
905 \r
906 /*\r
907 * If we got here without returning, then a >= m, so we must\r
908 * subtract m, and increment the quotient.\r
909 */\r
910 {\r
911 BignumCarry c = 1;\r
912 for (i = alen - 1; i >= 0; i--) {\r
913 int mindex = mlen-alen+i;\r
914 BignumInt mword = mindex < 0 ? 0 : m[mindex];\r
915 BignumADC(a[i], c, a[i], ~mword, c);\r
916 }\r
917 }\r
918 if (quot)\r
919 internal_add_shifted(quot, 1, 0);\r
920 \r
921 #ifdef DIVISION_DEBUG\r
922 {\r
923 int d;\r
924 printf("after final fixup, a=0x");\r
925 for (d = 0; d < alen; d++)\r
926 printf("%0*llx", BIGNUM_INT_BITS/4, (unsigned long long)a[d]);\r
927 if (quot) {\r
928 printf(", quot=0x");\r
929 for (d = quot[0]; d > 0; d--)\r
930 printf("%0*llx", BIGNUM_INT_BITS/4,\r
931 (unsigned long long)quot[d]);\r
932 }\r
933 printf("\n");\r
934 }\r
935 #endif\r
936 }\r
937 \r
938 /*\r
939 * Compute (base ^ exp) % mod, the pedestrian way.\r
940 */\r
941 Bignum modpow_simple(Bignum base_in, Bignum exp, Bignum mod)\r
942 {\r
943 BignumInt *a, *b, *n, *m, *scratch;\r
944 BignumInt recip;\r
945 int rshift;\r
946 int mlen, scratchlen, i, j;\r
947 Bignum base, result;\r
948 \r
949 /*\r
950 * The most significant word of mod needs to be non-zero. It\r
951 * should already be, but let's make sure.\r
952 */\r
953 assert(mod[mod[0]] != 0);\r
954 \r
955 /*\r
956 * Make sure the base is smaller than the modulus, by reducing\r
957 * it modulo the modulus if not.\r
958 */\r
959 base = bigmod(base_in, mod);\r
960 \r
961 /* Allocate m of size mlen, copy mod to m */\r
962 /* We use big endian internally */\r
963 mlen = mod[0];\r
964 m = snewn(mlen, BignumInt);\r
965 for (j = 0; j < mlen; j++)\r
966 m[j] = mod[mod[0] - j];\r
967 \r
968 /* Allocate n of size mlen, copy base to n */\r
969 n = snewn(mlen, BignumInt);\r
970 i = mlen - base[0];\r
971 for (j = 0; j < i; j++)\r
972 n[j] = 0;\r
973 for (j = 0; j < (int)base[0]; j++)\r
974 n[i + j] = base[base[0] - j];\r
975 \r
976 /* Allocate a and b of size 2*mlen. Set a = 1 */\r
977 a = snewn(2 * mlen, BignumInt);\r
978 b = snewn(2 * mlen, BignumInt);\r
979 for (i = 0; i < 2 * mlen; i++)\r
980 a[i] = 0;\r
981 a[2 * mlen - 1] = 1;\r
982 \r
983 /* Scratch space for multiplies */\r
984 scratchlen = mul_compute_scratch(mlen);\r
985 scratch = snewn(scratchlen, BignumInt);\r
986 \r
987 /* Skip leading zero bits of exp. */\r
988 i = 0;\r
989 j = BIGNUM_INT_BITS-1;\r
990 while (i < (int)exp[0] && (exp[exp[0] - i] & ((BignumInt)1 << j)) == 0) {\r
991 j--;\r
992 if (j < 0) {\r
993 i++;\r
994 j = BIGNUM_INT_BITS-1;\r
995 }\r
996 }\r
997 \r
998 /* Compute reciprocal of the top full word of the modulus */\r
999 {\r
1000 BignumInt m0 = m[0];\r
1001 rshift = bn_clz(m0);\r
1002 if (rshift) {\r
1003 m0 <<= rshift;\r
1004 if (mlen > 1)\r
1005 m0 |= m[1] >> (BIGNUM_INT_BITS - rshift);\r
1006 }\r
1007 recip = reciprocal_word(m0);\r
1008 }\r
1009 \r
1010 /* Main computation */\r
1011 while (i < (int)exp[0]) {\r
1012 while (j >= 0) {\r
1013 internal_mul(a + mlen, a + mlen, b, mlen, scratch);\r
1014 internal_mod(b, mlen * 2, m, mlen, NULL, recip, rshift);\r
1015 if ((exp[exp[0] - i] & ((BignumInt)1 << j)) != 0) {\r
1016 internal_mul(b + mlen, n, a, mlen, scratch);\r
1017 internal_mod(a, mlen * 2, m, mlen, NULL, recip, rshift);\r
1018 } else {\r
1019 BignumInt *t;\r
1020 t = a;\r
1021 a = b;\r
1022 b = t;\r
1023 }\r
1024 j--;\r
1025 }\r
1026 i++;\r
1027 j = BIGNUM_INT_BITS-1;\r
1028 }\r
1029 \r
1030 /* Copy result to buffer */\r
1031 result = newbn(mod[0]);\r
1032 for (i = 0; i < mlen; i++)\r
1033 result[result[0] - i] = a[i + mlen];\r
1034 while (result[0] > 1 && result[result[0]] == 0)\r
1035 result[0]--;\r
1036 \r
1037 /* Free temporary arrays */\r
1038 smemclr(a, 2 * mlen * sizeof(*a));\r
1039 sfree(a);\r
1040 smemclr(scratch, scratchlen * sizeof(*scratch));\r
1041 sfree(scratch);\r
1042 smemclr(b, 2 * mlen * sizeof(*b));\r
1043 sfree(b);\r
1044 smemclr(m, mlen * sizeof(*m));\r
1045 sfree(m);\r
1046 smemclr(n, mlen * sizeof(*n));\r
1047 sfree(n);\r
1048 \r
1049 freebn(base);\r
1050 \r
1051 return result;\r
1052 }\r
1053 \r
1054 /*\r
1055 * Compute (base ^ exp) % mod. Uses the Montgomery multiplication\r
1056 * technique where possible, falling back to modpow_simple otherwise.\r
1057 */\r
1058 Bignum modpow(Bignum base_in, Bignum exp, Bignum mod)\r
1059 {\r
1060 BignumInt *a, *b, *x, *n, *mninv, *scratch;\r
1061 int len, scratchlen, i, j;\r
1062 Bignum base, base2, r, rn, inv, result;\r
1063 \r
1064 /*\r
1065 * The most significant word of mod needs to be non-zero. It\r
1066 * should already be, but let's make sure.\r
1067 */\r
1068 assert(mod[mod[0]] != 0);\r
1069 \r
1070 /*\r
1071 * mod had better be odd, or we can't do Montgomery multiplication\r
1072 * using a power of two at all.\r
1073 */\r
1074 if (!(mod[1] & 1))\r
1075 return modpow_simple(base_in, exp, mod);\r
1076 \r
1077 /*\r
1078 * Make sure the base is smaller than the modulus, by reducing\r
1079 * it modulo the modulus if not.\r
1080 */\r
1081 base = bigmod(base_in, mod);\r
1082 \r
1083 /*\r
1084 * Compute the inverse of n mod r, for monty_reduce. (In fact we\r
1085 * want the inverse of _minus_ n mod r, but we'll sort that out\r
1086 * below.)\r
1087 */\r
1088 len = mod[0];\r
1089 r = bn_power_2(BIGNUM_INT_BITS * len);\r
1090 inv = modinv(mod, r);\r
1091 assert(inv); /* cannot fail, since mod is odd and r is a power of 2 */\r
1092 \r
1093 /*\r
1094 * Multiply the base by r mod n, to get it into Montgomery\r
1095 * representation.\r
1096 */\r
1097 base2 = modmul(base, r, mod);\r
1098 freebn(base);\r
1099 base = base2;\r
1100 \r
1101 rn = bigmod(r, mod); /* r mod n, i.e. Montgomerified 1 */\r
1102 \r
1103 freebn(r); /* won't need this any more */\r
1104 \r
1105 /*\r
1106 * Set up internal arrays of the right lengths, in big-endian\r
1107 * format, containing the base, the modulus, and the modulus's\r
1108 * inverse.\r
1109 */\r
1110 n = snewn(len, BignumInt);\r
1111 for (j = 0; j < len; j++)\r
1112 n[len - 1 - j] = mod[j + 1];\r
1113 \r
1114 mninv = snewn(len, BignumInt);\r
1115 for (j = 0; j < len; j++)\r
1116 mninv[len - 1 - j] = (j < (int)inv[0] ? inv[j + 1] : 0);\r
1117 freebn(inv); /* we don't need this copy of it any more */\r
1118 /* Now negate mninv mod r, so it's the inverse of -n rather than +n. */\r
1119 x = snewn(len, BignumInt);\r
1120 for (j = 0; j < len; j++)\r
1121 x[j] = 0;\r
1122 internal_sub(x, mninv, mninv, len);\r
1123 \r
1124 /* x = snewn(len, BignumInt); */ /* already done above */\r
1125 for (j = 0; j < len; j++)\r
1126 x[len - 1 - j] = (j < (int)base[0] ? base[j + 1] : 0);\r
1127 freebn(base); /* we don't need this copy of it any more */\r
1128 \r
1129 a = snewn(2*len, BignumInt);\r
1130 b = snewn(2*len, BignumInt);\r
1131 for (j = 0; j < len; j++)\r
1132 a[2*len - 1 - j] = (j < (int)rn[0] ? rn[j + 1] : 0);\r
1133 freebn(rn);\r
1134 \r
1135 /* Scratch space for multiplies */\r
1136 scratchlen = 3*len + mul_compute_scratch(len);\r
1137 scratch = snewn(scratchlen, BignumInt);\r
1138 \r
1139 /* Skip leading zero bits of exp. */\r
1140 i = 0;\r
1141 j = BIGNUM_INT_BITS-1;\r
1142 while (i < (int)exp[0] && (exp[exp[0] - i] & ((BignumInt)1 << j)) == 0) {\r
1143 j--;\r
1144 if (j < 0) {\r
1145 i++;\r
1146 j = BIGNUM_INT_BITS-1;\r
1147 }\r
1148 }\r
1149 \r
1150 /* Main computation */\r
1151 while (i < (int)exp[0]) {\r
1152 while (j >= 0) {\r
1153 internal_mul(a + len, a + len, b, len, scratch);\r
1154 monty_reduce(b, n, mninv, scratch, len);\r
1155 if ((exp[exp[0] - i] & ((BignumInt)1 << j)) != 0) {\r
1156 internal_mul(b + len, x, a, len, scratch);\r
1157 monty_reduce(a, n, mninv, scratch, len);\r
1158 } else {\r
1159 BignumInt *t;\r
1160 t = a;\r
1161 a = b;\r
1162 b = t;\r
1163 }\r
1164 j--;\r
1165 }\r
1166 i++;\r
1167 j = BIGNUM_INT_BITS-1;\r
1168 }\r
1169 \r
1170 /*\r
1171 * Final monty_reduce to get back from the adjusted Montgomery\r
1172 * representation.\r
1173 */\r
1174 monty_reduce(a, n, mninv, scratch, len);\r
1175 \r
1176 /* Copy result to buffer */\r
1177 result = newbn(mod[0]);\r
1178 for (i = 0; i < len; i++)\r
1179 result[result[0] - i] = a[i + len];\r
1180 while (result[0] > 1 && result[result[0]] == 0)\r
1181 result[0]--;\r
1182 \r
1183 /* Free temporary arrays */\r
1184 smemclr(scratch, scratchlen * sizeof(*scratch));\r
1185 sfree(scratch);\r
1186 smemclr(a, 2 * len * sizeof(*a));\r
1187 sfree(a);\r
1188 smemclr(b, 2 * len * sizeof(*b));\r
1189 sfree(b);\r
1190 smemclr(mninv, len * sizeof(*mninv));\r
1191 sfree(mninv);\r
1192 smemclr(n, len * sizeof(*n));\r
1193 sfree(n);\r
1194 smemclr(x, len * sizeof(*x));\r
1195 sfree(x);\r
1196 \r
1197 return result;\r
1198 }\r
1199 \r
1200 /*\r
1201 * Compute (p * q) % mod.\r
1202 * The most significant word of mod MUST be non-zero.\r
1203 * We assume that the result array is the same size as the mod array.\r
1204 */\r
1205 Bignum modmul(Bignum p, Bignum q, Bignum mod)\r
1206 {\r
1207 BignumInt *a, *n, *m, *o, *scratch;\r
1208 BignumInt recip;\r
1209 int rshift, scratchlen;\r
1210 int pqlen, mlen, rlen, i, j;\r
1211 Bignum result;\r
1212 \r
1213 /*\r
1214 * The most significant word of mod needs to be non-zero. It\r
1215 * should already be, but let's make sure.\r
1216 */\r
1217 assert(mod[mod[0]] != 0);\r
1218 \r
1219 /* Allocate m of size mlen, copy mod to m */\r
1220 /* We use big endian internally */\r
1221 mlen = mod[0];\r
1222 m = snewn(mlen, BignumInt);\r
1223 for (j = 0; j < mlen; j++)\r
1224 m[j] = mod[mod[0] - j];\r
1225 \r
1226 pqlen = (p[0] > q[0] ? p[0] : q[0]);\r
1227 \r
1228 /*\r
1229 * Make sure that we're allowing enough space. The shifting below\r
1230 * will underflow the vectors we allocate if pqlen is too small.\r
1231 */\r
1232 if (2*pqlen <= mlen)\r
1233 pqlen = mlen/2 + 1;\r
1234 \r
1235 /* Allocate n of size pqlen, copy p to n */\r
1236 n = snewn(pqlen, BignumInt);\r
1237 i = pqlen - p[0];\r
1238 for (j = 0; j < i; j++)\r
1239 n[j] = 0;\r
1240 for (j = 0; j < (int)p[0]; j++)\r
1241 n[i + j] = p[p[0] - j];\r
1242 \r
1243 /* Allocate o of size pqlen, copy q to o */\r
1244 o = snewn(pqlen, BignumInt);\r
1245 i = pqlen - q[0];\r
1246 for (j = 0; j < i; j++)\r
1247 o[j] = 0;\r
1248 for (j = 0; j < (int)q[0]; j++)\r
1249 o[i + j] = q[q[0] - j];\r
1250 \r
1251 /* Allocate a of size 2*pqlen for result */\r
1252 a = snewn(2 * pqlen, BignumInt);\r
1253 \r
1254 /* Scratch space for multiplies */\r
1255 scratchlen = mul_compute_scratch(pqlen);\r
1256 scratch = snewn(scratchlen, BignumInt);\r
1257 \r
1258 /* Compute reciprocal of the top full word of the modulus */\r
1259 {\r
1260 BignumInt m0 = m[0];\r
1261 rshift = bn_clz(m0);\r
1262 if (rshift) {\r
1263 m0 <<= rshift;\r
1264 if (mlen > 1)\r
1265 m0 |= m[1] >> (BIGNUM_INT_BITS - rshift);\r
1266 }\r
1267 recip = reciprocal_word(m0);\r
1268 }\r
1269 \r
1270 /* Main computation */\r
1271 internal_mul(n, o, a, pqlen, scratch);\r
1272 internal_mod(a, pqlen * 2, m, mlen, NULL, recip, rshift);\r
1273 \r
1274 /* Copy result to buffer */\r
1275 rlen = (mlen < pqlen * 2 ? mlen : pqlen * 2);\r
1276 result = newbn(rlen);\r
1277 for (i = 0; i < rlen; i++)\r
1278 result[result[0] - i] = a[i + 2 * pqlen - rlen];\r
1279 while (result[0] > 1 && result[result[0]] == 0)\r
1280 result[0]--;\r
1281 \r
1282 /* Free temporary arrays */\r
1283 smemclr(scratch, scratchlen * sizeof(*scratch));\r
1284 sfree(scratch);\r
1285 smemclr(a, 2 * pqlen * sizeof(*a));\r
1286 sfree(a);\r
1287 smemclr(m, mlen * sizeof(*m));\r
1288 sfree(m);\r
1289 smemclr(n, pqlen * sizeof(*n));\r
1290 sfree(n);\r
1291 smemclr(o, pqlen * sizeof(*o));\r
1292 sfree(o);\r
1293 \r
1294 return result;\r
1295 }\r
1296 \r
1297 Bignum modsub(const Bignum a, const Bignum b, const Bignum n)\r
1298 {\r
1299 Bignum a1, b1, ret;\r
1300 \r
1301 if (bignum_cmp(a, n) >= 0) a1 = bigmod(a, n);\r
1302 else a1 = a;\r
1303 if (bignum_cmp(b, n) >= 0) b1 = bigmod(b, n);\r
1304 else b1 = b;\r
1305 \r
1306 if (bignum_cmp(a1, b1) >= 0) /* a >= b */\r
1307 {\r
1308 ret = bigsub(a1, b1);\r
1309 }\r
1310 else\r
1311 {\r
1312 /* Handle going round the corner of the modulus without having\r
1313 * negative support in Bignum */\r
1314 Bignum tmp = bigsub(n, b1);\r
1315 assert(tmp);\r
1316 ret = bigadd(tmp, a1);\r
1317 freebn(tmp);\r
1318 }\r
1319 \r
1320 if (a != a1) freebn(a1);\r
1321 if (b != b1) freebn(b1);\r
1322 \r
1323 return ret;\r
1324 }\r
1325 \r
1326 /*\r
1327 * Compute p % mod.\r
1328 * The most significant word of mod MUST be non-zero.\r
1329 * We assume that the result array is the same size as the mod array.\r
1330 * We optionally write out a quotient if `quotient' is non-NULL.\r
1331 * We can avoid writing out the result if `result' is NULL.\r
1332 */\r
1333 static void bigdivmod(Bignum p, Bignum mod, Bignum result, Bignum quotient)\r
1334 {\r
1335 BignumInt *n, *m;\r
1336 BignumInt recip;\r
1337 int rshift;\r
1338 int plen, mlen, i, j;\r
1339 \r
1340 /*\r
1341 * The most significant word of mod needs to be non-zero. It\r
1342 * should already be, but let's make sure.\r
1343 */\r
1344 assert(mod[mod[0]] != 0);\r
1345 \r
1346 /* Allocate m of size mlen, copy mod to m */\r
1347 /* We use big endian internally */\r
1348 mlen = mod[0];\r
1349 m = snewn(mlen, BignumInt);\r
1350 for (j = 0; j < mlen; j++)\r
1351 m[j] = mod[mod[0] - j];\r
1352 \r
1353 plen = p[0];\r
1354 /* Ensure plen > mlen */\r
1355 if (plen <= mlen)\r
1356 plen = mlen + 1;\r
1357 \r
1358 /* Allocate n of size plen, copy p to n */\r
1359 n = snewn(plen, BignumInt);\r
1360 for (j = 0; j < plen; j++)\r
1361 n[j] = 0;\r
1362 for (j = 1; j <= (int)p[0]; j++)\r
1363 n[plen - j] = p[j];\r
1364 \r
1365 /* Compute reciprocal of the top full word of the modulus */\r
1366 {\r
1367 BignumInt m0 = m[0];\r
1368 rshift = bn_clz(m0);\r
1369 if (rshift) {\r
1370 m0 <<= rshift;\r
1371 if (mlen > 1)\r
1372 m0 |= m[1] >> (BIGNUM_INT_BITS - rshift);\r
1373 }\r
1374 recip = reciprocal_word(m0);\r
1375 }\r
1376 \r
1377 /* Main computation */\r
1378 internal_mod(n, plen, m, mlen, quotient, recip, rshift);\r
1379 \r
1380 /* Copy result to buffer */\r
1381 if (result) {\r
1382 for (i = 1; i <= (int)result[0]; i++) {\r
1383 int j = plen - i;\r
1384 result[i] = j >= 0 ? n[j] : 0;\r
1385 }\r
1386 }\r
1387 \r
1388 /* Free temporary arrays */\r
1389 smemclr(m, mlen * sizeof(*m));\r
1390 sfree(m);\r
1391 smemclr(n, plen * sizeof(*n));\r
1392 sfree(n);\r
1393 }\r
1394 \r
1395 /*\r
1396 * Decrement a number.\r
1397 */\r
1398 void decbn(Bignum bn)\r
1399 {\r
1400 int i = 1;\r
1401 while (i < (int)bn[0] && bn[i] == 0)\r
1402 bn[i++] = BIGNUM_INT_MASK;\r
1403 bn[i]--;\r
1404 }\r
1405 \r
1406 Bignum bignum_from_bytes(const unsigned char *data, int nbytes)\r
1407 {\r
1408 Bignum result;\r
1409 int w, i;\r
1410 \r
1411 assert(nbytes >= 0 && nbytes < INT_MAX/8);\r
1412 \r
1413 w = (nbytes + BIGNUM_INT_BYTES - 1) / BIGNUM_INT_BYTES; /* bytes->words */\r
1414 \r
1415 result = newbn(w);\r
1416 for (i = 1; i <= w; i++)\r
1417 result[i] = 0;\r
1418 for (i = nbytes; i--;) {\r
1419 unsigned char byte = *data++;\r
1420 result[1 + i / BIGNUM_INT_BYTES] |=\r
1421 (BignumInt)byte << (8*i % BIGNUM_INT_BITS);\r
1422 }\r
1423 \r
1424 bn_restore_invariant(result);\r
1425 return result;\r
1426 }\r
1427 \r
1428 Bignum bignum_from_bytes_le(const unsigned char *data, int nbytes)\r
1429 {\r
1430 Bignum result;\r
1431 int w, i;\r
1432 \r
1433 assert(nbytes >= 0 && nbytes < INT_MAX/8);\r
1434 \r
1435 w = (nbytes + BIGNUM_INT_BYTES - 1) / BIGNUM_INT_BYTES; /* bytes->words */\r
1436 \r
1437 result = newbn(w);\r
1438 for (i = 1; i <= w; i++)\r
1439 result[i] = 0;\r
1440 for (i = 0; i < nbytes; ++i) {\r
1441 unsigned char byte = *data++;\r
1442 result[1 + i / BIGNUM_INT_BYTES] |=\r
1443 (BignumInt)byte << (8*i % BIGNUM_INT_BITS);\r
1444 }\r
1445 \r
1446 bn_restore_invariant(result);\r
1447 return result;\r
1448 }\r
1449 \r
1450 Bignum bignum_from_decimal(const char *decimal)\r
1451 {\r
1452 Bignum result = copybn(Zero);\r
1453 \r
1454 while (*decimal) {\r
1455 Bignum tmp, tmp2;\r
1456 \r
1457 if (!isdigit((unsigned char)*decimal)) {\r
1458 freebn(result);\r
1459 return 0;\r
1460 }\r
1461 \r
1462 tmp = bigmul(result, Ten);\r
1463 tmp2 = bignum_from_long(*decimal - '0');\r
1464 freebn(result);\r
1465 result = bigadd(tmp, tmp2);\r
1466 freebn(tmp);\r
1467 freebn(tmp2);\r
1468 \r
1469 decimal++;\r
1470 }\r
1471 \r
1472 return result;\r
1473 }\r
1474 \r
1475 Bignum bignum_random_in_range(const Bignum lower, const Bignum upper)\r
1476 {\r
1477 Bignum ret = NULL;\r
1478 unsigned char *bytes;\r
1479 int upper_len = bignum_bitcount(upper);\r
1480 int upper_bytes = upper_len / 8;\r
1481 int upper_bits = upper_len % 8;\r
1482 if (upper_bits) ++upper_bytes;\r
1483 \r
1484 bytes = snewn(upper_bytes, unsigned char);\r
1485 do {\r
1486 int i;\r
1487 \r
1488 if (ret) freebn(ret);\r
1489 \r
1490 for (i = 0; i < upper_bytes; ++i)\r
1491 {\r
1492 bytes[i] = (unsigned char)random_byte();\r
1493 }\r
1494 /* Mask the top to reduce failure rate to 50/50 */\r
1495 if (upper_bits)\r
1496 {\r
1497 bytes[i - 1] &= 0xFF >> (8 - upper_bits);\r
1498 }\r
1499 \r
1500 ret = bignum_from_bytes(bytes, upper_bytes);\r
1501 } while (bignum_cmp(ret, lower) < 0 || bignum_cmp(ret, upper) > 0);\r
1502 smemclr(bytes, upper_bytes);\r
1503 sfree(bytes);\r
1504 \r
1505 return ret;\r
1506 }\r
1507 \r
1508 /*\r
1509 * Read an SSH-1-format bignum from a data buffer. Return the number\r
1510 * of bytes consumed, or -1 if there wasn't enough data.\r
1511 */\r
1512 int ssh1_read_bignum(const unsigned char *data, int len, Bignum * result)\r
1513 {\r
1514 const unsigned char *p = data;\r
1515 int i;\r
1516 int w, b;\r
1517 \r
1518 if (len < 2)\r
1519 return -1;\r
1520 \r
1521 w = 0;\r
1522 for (i = 0; i < 2; i++)\r
1523 w = (w << 8) + *p++;\r
1524 b = (w + 7) / 8; /* bits -> bytes */\r
1525 \r
1526 if (len < b+2)\r
1527 return -1;\r
1528 \r
1529 if (!result) /* just return length */\r
1530 return b + 2;\r
1531 \r
1532 *result = bignum_from_bytes(p, b);\r
1533 \r
1534 return p + b - data;\r
1535 }\r
1536 \r
1537 /*\r
1538 * Return the bit count of a bignum, for SSH-1 encoding.\r
1539 */\r
1540 int bignum_bitcount(Bignum bn)\r
1541 {\r
1542 int bitcount = bn[0] * BIGNUM_INT_BITS - 1;\r
1543 while (bitcount >= 0\r
1544 && (bn[bitcount / BIGNUM_INT_BITS + 1] >> (bitcount % BIGNUM_INT_BITS)) == 0) bitcount--;\r
1545 return bitcount + 1;\r
1546 }\r
1547 \r
1548 /*\r
1549 * Return the byte length of a bignum when SSH-1 encoded.\r
1550 */\r
1551 int ssh1_bignum_length(Bignum bn)\r
1552 {\r
1553 return 2 + (bignum_bitcount(bn) + 7) / 8;\r
1554 }\r
1555 \r
1556 /*\r
1557 * Return the byte length of a bignum when SSH-2 encoded.\r
1558 */\r
1559 int ssh2_bignum_length(Bignum bn)\r
1560 {\r
1561 return 4 + (bignum_bitcount(bn) + 8) / 8;\r
1562 }\r
1563 \r
1564 /*\r
1565 * Return a byte from a bignum; 0 is least significant, etc.\r
1566 */\r
1567 int bignum_byte(Bignum bn, int i)\r
1568 {\r
1569 if (i < 0 || i >= (int)(BIGNUM_INT_BYTES * bn[0]))\r
1570 return 0; /* beyond the end */\r
1571 else\r
1572 return (bn[i / BIGNUM_INT_BYTES + 1] >>\r
1573 ((i % BIGNUM_INT_BYTES)*8)) & 0xFF;\r
1574 }\r
1575 \r
1576 /*\r
1577 * Return a bit from a bignum; 0 is least significant, etc.\r
1578 */\r
1579 int bignum_bit(Bignum bn, int i)\r
1580 {\r
1581 if (i < 0 || i >= (int)(BIGNUM_INT_BITS * bn[0]))\r
1582 return 0; /* beyond the end */\r
1583 else\r
1584 return (bn[i / BIGNUM_INT_BITS + 1] >> (i % BIGNUM_INT_BITS)) & 1;\r
1585 }\r
1586 \r
1587 /*\r
1588 * Set a bit in a bignum; 0 is least significant, etc.\r
1589 */\r
1590 void bignum_set_bit(Bignum bn, int bitnum, int value)\r
1591 {\r
1592 if (bitnum < 0 || bitnum >= (int)(BIGNUM_INT_BITS * bn[0])) {\r
1593 if (value) abort(); /* beyond the end */\r
1594 } else {\r
1595 int v = bitnum / BIGNUM_INT_BITS + 1;\r
1596 BignumInt mask = (BignumInt)1 << (bitnum % BIGNUM_INT_BITS);\r
1597 if (value)\r
1598 bn[v] |= mask;\r
1599 else\r
1600 bn[v] &= ~mask;\r
1601 }\r
1602 }\r
1603 \r
1604 /*\r
1605 * Write a SSH-1-format bignum into a buffer. It is assumed the\r
1606 * buffer is big enough. Returns the number of bytes used.\r
1607 */\r
1608 int ssh1_write_bignum(void *data, Bignum bn)\r
1609 {\r
1610 unsigned char *p = data;\r
1611 int len = ssh1_bignum_length(bn);\r
1612 int i;\r
1613 int bitc = bignum_bitcount(bn);\r
1614 \r
1615 *p++ = (bitc >> 8) & 0xFF;\r
1616 *p++ = (bitc) & 0xFF;\r
1617 for (i = len - 2; i--;)\r
1618 *p++ = bignum_byte(bn, i);\r
1619 return len;\r
1620 }\r
1621 \r
1622 /*\r
1623 * Compare two bignums. Returns like strcmp.\r
1624 */\r
1625 int bignum_cmp(Bignum a, Bignum b)\r
1626 {\r
1627 int amax = a[0], bmax = b[0];\r
1628 int i;\r
1629 \r
1630 /* Annoyingly we have two representations of zero */\r
1631 if (amax == 1 && a[amax] == 0)\r
1632 amax = 0;\r
1633 if (bmax == 1 && b[bmax] == 0)\r
1634 bmax = 0;\r
1635 \r
1636 assert(amax == 0 || a[amax] != 0);\r
1637 assert(bmax == 0 || b[bmax] != 0);\r
1638 \r
1639 i = (amax > bmax ? amax : bmax);\r
1640 while (i) {\r
1641 BignumInt aval = (i > amax ? 0 : a[i]);\r
1642 BignumInt bval = (i > bmax ? 0 : b[i]);\r
1643 if (aval < bval)\r
1644 return -1;\r
1645 if (aval > bval)\r
1646 return +1;\r
1647 i--;\r
1648 }\r
1649 return 0;\r
1650 }\r
1651 \r
1652 /*\r
1653 * Right-shift one bignum to form another.\r
1654 */\r
1655 Bignum bignum_rshift(Bignum a, int shift)\r
1656 {\r
1657 Bignum ret;\r
1658 int i, shiftw, shiftb, shiftbb, bits;\r
1659 BignumInt ai, ai1;\r
1660 \r
1661 assert(shift >= 0);\r
1662 \r
1663 bits = bignum_bitcount(a) - shift;\r
1664 ret = newbn((bits + BIGNUM_INT_BITS - 1) / BIGNUM_INT_BITS);\r
1665 \r
1666 if (ret) {\r
1667 shiftw = shift / BIGNUM_INT_BITS;\r
1668 shiftb = shift % BIGNUM_INT_BITS;\r
1669 shiftbb = BIGNUM_INT_BITS - shiftb;\r
1670 \r
1671 ai1 = a[shiftw + 1];\r
1672 for (i = 1; i <= (int)ret[0]; i++) {\r
1673 ai = ai1;\r
1674 ai1 = (i + shiftw + 1 <= (int)a[0] ? a[i + shiftw + 1] : 0);\r
1675 ret[i] = ((ai >> shiftb) | (ai1 << shiftbb)) & BIGNUM_INT_MASK;\r
1676 }\r
1677 }\r
1678 \r
1679 return ret;\r
1680 }\r
1681 \r
1682 /*\r
1683 * Left-shift one bignum to form another.\r
1684 */\r
1685 Bignum bignum_lshift(Bignum a, int shift)\r
1686 {\r
1687 Bignum ret;\r
1688 int bits, shiftWords, shiftBits;\r
1689 \r
1690 assert(shift >= 0);\r
1691 \r
1692 bits = bignum_bitcount(a) + shift;\r
1693 ret = newbn((bits + BIGNUM_INT_BITS - 1) / BIGNUM_INT_BITS);\r
1694 \r
1695 shiftWords = shift / BIGNUM_INT_BITS;\r
1696 shiftBits = shift % BIGNUM_INT_BITS;\r
1697 \r
1698 if (shiftBits == 0)\r
1699 {\r
1700 memcpy(&ret[1 + shiftWords], &a[1], sizeof(BignumInt) * a[0]);\r
1701 }\r
1702 else\r
1703 {\r
1704 int i;\r
1705 BignumInt carry = 0;\r
1706 \r
1707 /* Remember that Bignum[0] is length, so add 1 */\r
1708 for (i = shiftWords + 1; i < ((int)a[0]) + shiftWords + 1; ++i)\r
1709 {\r
1710 BignumInt from = a[i - shiftWords];\r
1711 ret[i] = (from << shiftBits) | carry;\r
1712 carry = from >> (BIGNUM_INT_BITS - shiftBits);\r
1713 }\r
1714 if (carry) ret[i] = carry;\r
1715 }\r
1716 \r
1717 return ret;\r
1718 }\r
1719 \r
1720 /*\r
1721 * Non-modular multiplication and addition.\r
1722 */\r
1723 Bignum bigmuladd(Bignum a, Bignum b, Bignum addend)\r
1724 {\r
1725 int alen = a[0], blen = b[0];\r
1726 int mlen = (alen > blen ? alen : blen);\r
1727 int rlen, i, maxspot;\r
1728 int wslen;\r
1729 BignumInt *workspace;\r
1730 Bignum ret;\r
1731 \r
1732 /* mlen space for a, mlen space for b, 2*mlen for result,\r
1733 * plus scratch space for multiplication */\r
1734 wslen = mlen * 4 + mul_compute_scratch(mlen);\r
1735 workspace = snewn(wslen, BignumInt);\r
1736 for (i = 0; i < mlen; i++) {\r
1737 workspace[0 * mlen + i] = (mlen - i <= (int)a[0] ? a[mlen - i] : 0);\r
1738 workspace[1 * mlen + i] = (mlen - i <= (int)b[0] ? b[mlen - i] : 0);\r
1739 }\r
1740 \r
1741 internal_mul(workspace + 0 * mlen, workspace + 1 * mlen,\r
1742 workspace + 2 * mlen, mlen, workspace + 4 * mlen);\r
1743 \r
1744 /* now just copy the result back */\r
1745 rlen = alen + blen + 1;\r
1746 if (addend && rlen <= (int)addend[0])\r
1747 rlen = addend[0] + 1;\r
1748 ret = newbn(rlen);\r
1749 maxspot = 0;\r
1750 for (i = 1; i <= (int)ret[0]; i++) {\r
1751 ret[i] = (i <= 2 * mlen ? workspace[4 * mlen - i] : 0);\r
1752 if (ret[i] != 0)\r
1753 maxspot = i;\r
1754 }\r
1755 ret[0] = maxspot;\r
1756 \r
1757 /* now add in the addend, if any */\r
1758 if (addend) {\r
1759 BignumCarry carry = 0;\r
1760 for (i = 1; i <= rlen; i++) {\r
1761 BignumInt retword = (i <= (int)ret[0] ? ret[i] : 0);\r
1762 BignumInt addword = (i <= (int)addend[0] ? addend[i] : 0);\r
1763 BignumADC(ret[i], carry, retword, addword, carry);\r
1764 if (ret[i] != 0 && i > maxspot)\r
1765 maxspot = i;\r
1766 }\r
1767 }\r
1768 ret[0] = maxspot;\r
1769 \r
1770 smemclr(workspace, wslen * sizeof(*workspace));\r
1771 sfree(workspace);\r
1772 return ret;\r
1773 }\r
1774 \r
1775 /*\r
1776 * Non-modular multiplication.\r
1777 */\r
1778 Bignum bigmul(Bignum a, Bignum b)\r
1779 {\r
1780 return bigmuladd(a, b, NULL);\r
1781 }\r
1782 \r
1783 /*\r
1784 * Simple addition.\r
1785 */\r
1786 Bignum bigadd(Bignum a, Bignum b)\r
1787 {\r
1788 int alen = a[0], blen = b[0];\r
1789 int rlen = (alen > blen ? alen : blen) + 1;\r
1790 int i, maxspot;\r
1791 Bignum ret;\r
1792 BignumCarry carry;\r
1793 \r
1794 ret = newbn(rlen);\r
1795 \r
1796 carry = 0;\r
1797 maxspot = 0;\r
1798 for (i = 1; i <= rlen; i++) {\r
1799 BignumInt aword = (i <= (int)a[0] ? a[i] : 0);\r
1800 BignumInt bword = (i <= (int)b[0] ? b[i] : 0);\r
1801 BignumADC(ret[i], carry, aword, bword, carry);\r
1802 if (ret[i] != 0 && i > maxspot)\r
1803 maxspot = i;\r
1804 }\r
1805 ret[0] = maxspot;\r
1806 \r
1807 return ret;\r
1808 }\r
1809 \r
1810 /*\r
1811 * Subtraction. Returns a-b, or NULL if the result would come out\r
1812 * negative (recall that this entire bignum module only handles\r
1813 * positive numbers).\r
1814 */\r
1815 Bignum bigsub(Bignum a, Bignum b)\r
1816 {\r
1817 int alen = a[0], blen = b[0];\r
1818 int rlen = (alen > blen ? alen : blen);\r
1819 int i, maxspot;\r
1820 Bignum ret;\r
1821 BignumCarry carry;\r
1822 \r
1823 ret = newbn(rlen);\r
1824 \r
1825 carry = 1;\r
1826 maxspot = 0;\r
1827 for (i = 1; i <= rlen; i++) {\r
1828 BignumInt aword = (i <= (int)a[0] ? a[i] : 0);\r
1829 BignumInt bword = (i <= (int)b[0] ? b[i] : 0);\r
1830 BignumADC(ret[i], carry, aword, ~bword, carry);\r
1831 if (ret[i] != 0 && i > maxspot)\r
1832 maxspot = i;\r
1833 }\r
1834 ret[0] = maxspot;\r
1835 \r
1836 if (!carry) {\r
1837 freebn(ret);\r
1838 return NULL;\r
1839 }\r
1840 \r
1841 return ret;\r
1842 }\r
1843 \r
1844 /*\r
1845 * Create a bignum which is the bitmask covering another one. That\r
1846 * is, the smallest integer which is >= N and is also one less than\r
1847 * a power of two.\r
1848 */\r
1849 Bignum bignum_bitmask(Bignum n)\r
1850 {\r
1851 Bignum ret = copybn(n);\r
1852 int i;\r
1853 BignumInt j;\r
1854 \r
1855 i = ret[0];\r
1856 while (n[i] == 0 && i > 0)\r
1857 i--;\r
1858 if (i <= 0)\r
1859 return ret; /* input was zero */\r
1860 j = 1;\r
1861 while (j < n[i])\r
1862 j = 2 * j + 1;\r
1863 ret[i] = j;\r
1864 while (--i > 0)\r
1865 ret[i] = BIGNUM_INT_MASK;\r
1866 return ret;\r
1867 }\r
1868 \r
1869 /*\r
1870 * Convert an unsigned long into a bignum.\r
1871 */\r
1872 Bignum bignum_from_long(unsigned long n)\r
1873 {\r
1874 const int maxwords =\r
1875 (sizeof(unsigned long) + sizeof(BignumInt) - 1) / sizeof(BignumInt);\r
1876 Bignum ret;\r
1877 int i;\r
1878 \r
1879 ret = newbn(maxwords);\r
1880 ret[0] = 0;\r
1881 for (i = 0; i < maxwords; i++) {\r
1882 ret[i+1] = n >> (i * BIGNUM_INT_BITS);\r
1883 if (ret[i+1] != 0)\r
1884 ret[0] = i+1;\r
1885 }\r
1886 \r
1887 return ret;\r
1888 }\r
1889 \r
1890 /*\r
1891 * Add a long to a bignum.\r
1892 */\r
1893 Bignum bignum_add_long(Bignum number, unsigned long n)\r
1894 {\r
1895 const int maxwords =\r
1896 (sizeof(unsigned long) + sizeof(BignumInt) - 1) / sizeof(BignumInt);\r
1897 Bignum ret;\r
1898 int words, i;\r
1899 BignumCarry carry;\r
1900 \r
1901 words = number[0];\r
1902 if (words < maxwords)\r
1903 words = maxwords;\r
1904 words++;\r
1905 ret = newbn(words);\r
1906 \r
1907 carry = 0;\r
1908 ret[0] = 0;\r
1909 for (i = 0; i < words; i++) {\r
1910 BignumInt nword = (i < maxwords ? n >> (i * BIGNUM_INT_BITS) : 0);\r
1911 BignumInt numword = (i < number[0] ? number[i+1] : 0);\r
1912 BignumADC(ret[i+1], carry, numword, nword, carry);\r
1913 if (ret[i+1] != 0)\r
1914 ret[0] = i+1;\r
1915 }\r
1916 return ret;\r
1917 }\r
1918 \r
1919 /*\r
1920 * Compute the residue of a bignum, modulo a (max 16-bit) short.\r
1921 */\r
1922 unsigned short bignum_mod_short(Bignum number, unsigned short modulus)\r
1923 {\r
1924 unsigned long mod = modulus, r = 0;\r
1925 /* Precompute (BIGNUM_INT_MASK+1) % mod */\r
1926 unsigned long base_r = (BIGNUM_INT_MASK - modulus + 1) % mod;\r
1927 int i;\r
1928 \r
1929 for (i = number[0]; i > 0; i--) {\r
1930 /*\r
1931 * Conceptually, ((r << BIGNUM_INT_BITS) + number[i]) % mod\r
1932 */\r
1933 r = ((r * base_r) + (number[i] % mod)) % mod;\r
1934 }\r
1935 return (unsigned short) r;\r
1936 }\r
1937 \r
1938 #ifdef DEBUG\r
1939 void diagbn(char *prefix, Bignum md)\r
1940 {\r
1941 int i, nibbles, morenibbles;\r
1942 static const char hex[] = "0123456789ABCDEF";\r
1943 \r
1944 debug(("%s0x", prefix ? prefix : ""));\r
1945 \r
1946 nibbles = (3 + bignum_bitcount(md)) / 4;\r
1947 if (nibbles < 1)\r
1948 nibbles = 1;\r
1949 morenibbles = 4 * md[0] - nibbles;\r
1950 for (i = 0; i < morenibbles; i++)\r
1951 debug(("-"));\r
1952 for (i = nibbles; i--;)\r
1953 debug(("%c",\r
1954 hex[(bignum_byte(md, i / 2) >> (4 * (i % 2))) & 0xF]));\r
1955 \r
1956 if (prefix)\r
1957 debug(("\n"));\r
1958 }\r
1959 #endif\r
1960 \r
1961 /*\r
1962 * Simple division.\r
1963 */\r
1964 Bignum bigdiv(Bignum a, Bignum b)\r
1965 {\r
1966 Bignum q = newbn(a[0]);\r
1967 bigdivmod(a, b, NULL, q);\r
1968 while (q[0] > 1 && q[q[0]] == 0)\r
1969 q[0]--;\r
1970 return q;\r
1971 }\r
1972 \r
1973 /*\r
1974 * Simple remainder.\r
1975 */\r
1976 Bignum bigmod(Bignum a, Bignum b)\r
1977 {\r
1978 Bignum r = newbn(b[0]);\r
1979 bigdivmod(a, b, r, NULL);\r
1980 while (r[0] > 1 && r[r[0]] == 0)\r
1981 r[0]--;\r
1982 return r;\r
1983 }\r
1984 \r
1985 /*\r
1986 * Greatest common divisor.\r
1987 */\r
1988 Bignum biggcd(Bignum av, Bignum bv)\r
1989 {\r
1990 Bignum a = copybn(av);\r
1991 Bignum b = copybn(bv);\r
1992 \r
1993 while (bignum_cmp(b, Zero) != 0) {\r
1994 Bignum t = newbn(b[0]);\r
1995 bigdivmod(a, b, t, NULL);\r
1996 while (t[0] > 1 && t[t[0]] == 0)\r
1997 t[0]--;\r
1998 freebn(a);\r
1999 a = b;\r
2000 b = t;\r
2001 }\r
2002 \r
2003 freebn(b);\r
2004 return a;\r
2005 }\r
2006 \r
2007 /*\r
2008 * Modular inverse, using Euclid's extended algorithm.\r
2009 */\r
2010 Bignum modinv(Bignum number, Bignum modulus)\r
2011 {\r
2012 Bignum a = copybn(modulus);\r
2013 Bignum b = copybn(number);\r
2014 Bignum xp = copybn(Zero);\r
2015 Bignum x = copybn(One);\r
2016 int sign = +1;\r
2017 \r
2018 assert(number[number[0]] != 0);\r
2019 assert(modulus[modulus[0]] != 0);\r
2020 \r
2021 while (bignum_cmp(b, One) != 0) {\r
2022 Bignum t, q;\r
2023 \r
2024 if (bignum_cmp(b, Zero) == 0) {\r
2025 /*\r
2026 * Found a common factor between the inputs, so we cannot\r
2027 * return a modular inverse at all.\r
2028 */\r
2029 freebn(b);\r
2030 freebn(a);\r
2031 freebn(xp);\r
2032 freebn(x);\r
2033 return NULL;\r
2034 }\r
2035 \r
2036 t = newbn(b[0]);\r
2037 q = newbn(a[0]);\r
2038 bigdivmod(a, b, t, q);\r
2039 while (t[0] > 1 && t[t[0]] == 0)\r
2040 t[0]--;\r
2041 while (q[0] > 1 && q[q[0]] == 0)\r
2042 q[0]--;\r
2043 freebn(a);\r
2044 a = b;\r
2045 b = t;\r
2046 t = xp;\r
2047 xp = x;\r
2048 x = bigmuladd(q, xp, t);\r
2049 sign = -sign;\r
2050 freebn(t);\r
2051 freebn(q);\r
2052 }\r
2053 \r
2054 freebn(b);\r
2055 freebn(a);\r
2056 freebn(xp);\r
2057 \r
2058 /* now we know that sign * x == 1, and that x < modulus */\r
2059 if (sign < 0) {\r
2060 /* set a new x to be modulus - x */\r
2061 Bignum newx = newbn(modulus[0]);\r
2062 BignumInt carry = 0;\r
2063 int maxspot = 1;\r
2064 int i;\r
2065 \r
2066 for (i = 1; i <= (int)newx[0]; i++) {\r
2067 BignumInt aword = (i <= (int)modulus[0] ? modulus[i] : 0);\r
2068 BignumInt bword = (i <= (int)x[0] ? x[i] : 0);\r
2069 newx[i] = aword - bword - carry;\r
2070 bword = ~bword;\r
2071 carry = carry ? (newx[i] >= bword) : (newx[i] > bword);\r
2072 if (newx[i] != 0)\r
2073 maxspot = i;\r
2074 }\r
2075 newx[0] = maxspot;\r
2076 freebn(x);\r
2077 x = newx;\r
2078 }\r
2079 \r
2080 /* and return. */\r
2081 return x;\r
2082 }\r
2083 \r
2084 /*\r
2085 * Render a bignum into decimal. Return a malloced string holding\r
2086 * the decimal representation.\r
2087 */\r
2088 char *bignum_decimal(Bignum x)\r
2089 {\r
2090 int ndigits, ndigit;\r
2091 int i, iszero;\r
2092 BignumInt carry;\r
2093 char *ret;\r
2094 BignumInt *workspace;\r
2095 \r
2096 /*\r
2097 * First, estimate the number of digits. Since log(10)/log(2)\r
2098 * is just greater than 93/28 (the joys of continued fraction\r
2099 * approximations...) we know that for every 93 bits, we need\r
2100 * at most 28 digits. This will tell us how much to malloc.\r
2101 *\r
2102 * Formally: if x has i bits, that means x is strictly less\r
2103 * than 2^i. Since 2 is less than 10^(28/93), this is less than\r
2104 * 10^(28i/93). We need an integer power of ten, so we must\r
2105 * round up (rounding down might make it less than x again).\r
2106 * Therefore if we multiply the bit count by 28/93, rounding\r
2107 * up, we will have enough digits.\r
2108 *\r
2109 * i=0 (i.e., x=0) is an irritating special case.\r
2110 */\r
2111 i = bignum_bitcount(x);\r
2112 if (!i)\r
2113 ndigits = 1; /* x = 0 */\r
2114 else\r
2115 ndigits = (28 * i + 92) / 93; /* multiply by 28/93 and round up */\r
2116 ndigits++; /* allow for trailing \0 */\r
2117 ret = snewn(ndigits, char);\r
2118 \r
2119 /*\r
2120 * Now allocate some workspace to hold the binary form as we\r
2121 * repeatedly divide it by ten. Initialise this to the\r
2122 * big-endian form of the number.\r
2123 */\r
2124 workspace = snewn(x[0], BignumInt);\r
2125 for (i = 0; i < (int)x[0]; i++)\r
2126 workspace[i] = x[x[0] - i];\r
2127 \r
2128 /*\r
2129 * Next, write the decimal number starting with the last digit.\r
2130 * We use ordinary short division, dividing 10 into the\r
2131 * workspace.\r
2132 */\r
2133 ndigit = ndigits - 1;\r
2134 ret[ndigit] = '\0';\r
2135 do {\r
2136 iszero = 1;\r
2137 carry = 0;\r
2138 for (i = 0; i < (int)x[0]; i++) {\r
2139 /*\r
2140 * Conceptually, we want to compute\r
2141 *\r
2142 * (carry << BIGNUM_INT_BITS) + workspace[i]\r
2143 * -----------------------------------------\r
2144 * 10\r
2145 *\r
2146 * but we don't have an integer type longer than BignumInt\r
2147 * to work with. So we have to do it in pieces.\r
2148 */\r
2149 \r
2150 BignumInt q, r;\r
2151 q = workspace[i] / 10;\r
2152 r = workspace[i] % 10;\r
2153 \r
2154 /* I want (BIGNUM_INT_MASK+1)/10 but can't say so directly! */\r
2155 q += carry * ((BIGNUM_INT_MASK-9) / 10 + 1);\r
2156 r += carry * ((BIGNUM_INT_MASK-9) % 10);\r
2157 \r
2158 q += r / 10;\r
2159 r %= 10;\r
2160 \r
2161 workspace[i] = q;\r
2162 carry = r;\r
2163 \r
2164 if (workspace[i])\r
2165 iszero = 0;\r
2166 }\r
2167 ret[--ndigit] = (char) (carry + '0');\r
2168 } while (!iszero);\r
2169 \r
2170 /*\r
2171 * There's a chance we've fallen short of the start of the\r
2172 * string. Correct if so.\r
2173 */\r
2174 if (ndigit > 0)\r
2175 memmove(ret, ret + ndigit, ndigits - ndigit);\r
2176 \r
2177 /*\r
2178 * Done.\r
2179 */\r
2180 smemclr(workspace, x[0] * sizeof(*workspace));\r
2181 sfree(workspace);\r
2182 return ret;\r
2183 }\r