| 1 | /* |
| 2 | * Bignum routines for RSA and DH and stuff. |
| 3 | */ |
| 4 | |
| 5 | #include <stdio.h> |
| 6 | #include <assert.h> |
| 7 | #include <stdlib.h> |
| 8 | #include <string.h> |
| 9 | |
| 10 | #include "misc.h" |
| 11 | |
| 12 | /* |
| 13 | * Usage notes: |
| 14 | * * Do not call the DIVMOD_WORD macro with expressions such as array |
| 15 | * subscripts, as some implementations object to this (see below). |
| 16 | * * Note that none of the division methods below will cope if the |
| 17 | * quotient won't fit into BIGNUM_INT_BITS. Callers should be careful |
| 18 | * to avoid this case. |
| 19 | * If this condition occurs, in the case of the x86 DIV instruction, |
| 20 | * an overflow exception will occur, which (according to a correspondent) |
| 21 | * will manifest on Windows as something like |
| 22 | * 0xC0000095: Integer overflow |
| 23 | * The C variant won't give the right answer, either. |
| 24 | */ |
| 25 | |
| 26 | #if defined __GNUC__ && defined __i386__ |
| 27 | typedef unsigned long BignumInt; |
| 28 | typedef unsigned long long BignumDblInt; |
| 29 | #define BIGNUM_INT_MASK 0xFFFFFFFFUL |
| 30 | #define BIGNUM_TOP_BIT 0x80000000UL |
| 31 | #define BIGNUM_INT_BITS 32 |
| 32 | #define MUL_WORD(w1, w2) ((BignumDblInt)w1 * w2) |
| 33 | #define DIVMOD_WORD(q, r, hi, lo, w) \ |
| 34 | __asm__("div %2" : \ |
| 35 | "=d" (r), "=a" (q) : \ |
| 36 | "r" (w), "d" (hi), "a" (lo)) |
| 37 | #elif defined _MSC_VER && defined _M_IX86 |
| 38 | typedef unsigned __int32 BignumInt; |
| 39 | typedef unsigned __int64 BignumDblInt; |
| 40 | #define BIGNUM_INT_MASK 0xFFFFFFFFUL |
| 41 | #define BIGNUM_TOP_BIT 0x80000000UL |
| 42 | #define BIGNUM_INT_BITS 32 |
| 43 | #define MUL_WORD(w1, w2) ((BignumDblInt)w1 * w2) |
| 44 | /* Note: MASM interprets array subscripts in the macro arguments as |
| 45 | * assembler syntax, which gives the wrong answer. Don't supply them. |
| 46 | * <http://msdn2.microsoft.com/en-us/library/bf1dw62z.aspx> */ |
| 47 | #define DIVMOD_WORD(q, r, hi, lo, w) do { \ |
| 48 | __asm mov edx, hi \ |
| 49 | __asm mov eax, lo \ |
| 50 | __asm div w \ |
| 51 | __asm mov r, edx \ |
| 52 | __asm mov q, eax \ |
| 53 | } while(0) |
| 54 | #elif defined _LP64 |
| 55 | /* 64-bit architectures can do 32x32->64 chunks at a time */ |
| 56 | typedef unsigned int BignumInt; |
| 57 | typedef unsigned long BignumDblInt; |
| 58 | #define BIGNUM_INT_MASK 0xFFFFFFFFU |
| 59 | #define BIGNUM_TOP_BIT 0x80000000U |
| 60 | #define BIGNUM_INT_BITS 32 |
| 61 | #define MUL_WORD(w1, w2) ((BignumDblInt)w1 * w2) |
| 62 | #define DIVMOD_WORD(q, r, hi, lo, w) do { \ |
| 63 | BignumDblInt n = (((BignumDblInt)hi) << BIGNUM_INT_BITS) | lo; \ |
| 64 | q = n / w; \ |
| 65 | r = n % w; \ |
| 66 | } while (0) |
| 67 | #elif defined _LLP64 |
| 68 | /* 64-bit architectures in which unsigned long is 32 bits, not 64 */ |
| 69 | typedef unsigned long BignumInt; |
| 70 | typedef unsigned long long BignumDblInt; |
| 71 | #define BIGNUM_INT_MASK 0xFFFFFFFFUL |
| 72 | #define BIGNUM_TOP_BIT 0x80000000UL |
| 73 | #define BIGNUM_INT_BITS 32 |
| 74 | #define MUL_WORD(w1, w2) ((BignumDblInt)w1 * w2) |
| 75 | #define DIVMOD_WORD(q, r, hi, lo, w) do { \ |
| 76 | BignumDblInt n = (((BignumDblInt)hi) << BIGNUM_INT_BITS) | lo; \ |
| 77 | q = n / w; \ |
| 78 | r = n % w; \ |
| 79 | } while (0) |
| 80 | #else |
| 81 | /* Fallback for all other cases */ |
| 82 | typedef unsigned short BignumInt; |
| 83 | typedef unsigned long BignumDblInt; |
| 84 | #define BIGNUM_INT_MASK 0xFFFFU |
| 85 | #define BIGNUM_TOP_BIT 0x8000U |
| 86 | #define BIGNUM_INT_BITS 16 |
| 87 | #define MUL_WORD(w1, w2) ((BignumDblInt)w1 * w2) |
| 88 | #define DIVMOD_WORD(q, r, hi, lo, w) do { \ |
| 89 | BignumDblInt n = (((BignumDblInt)hi) << BIGNUM_INT_BITS) | lo; \ |
| 90 | q = n / w; \ |
| 91 | r = n % w; \ |
| 92 | } while (0) |
| 93 | #endif |
| 94 | |
| 95 | #define BIGNUM_INT_BYTES (BIGNUM_INT_BITS / 8) |
| 96 | |
| 97 | #define BIGNUM_INTERNAL |
| 98 | typedef BignumInt *Bignum; |
| 99 | |
| 100 | #include "ssh.h" |
| 101 | |
| 102 | BignumInt bnZero[1] = { 0 }; |
| 103 | BignumInt bnOne[2] = { 1, 1 }; |
| 104 | |
| 105 | /* |
| 106 | * The Bignum format is an array of `BignumInt'. The first |
| 107 | * element of the array counts the remaining elements. The |
| 108 | * remaining elements express the actual number, base 2^BIGNUM_INT_BITS, _least_ |
| 109 | * significant digit first. (So it's trivial to extract the bit |
| 110 | * with value 2^n for any n.) |
| 111 | * |
| 112 | * All Bignums in this module are positive. Negative numbers must |
| 113 | * be dealt with outside it. |
| 114 | * |
| 115 | * INVARIANT: the most significant word of any Bignum must be |
| 116 | * nonzero. |
| 117 | */ |
| 118 | |
| 119 | Bignum Zero = bnZero, One = bnOne; |
| 120 | |
| 121 | static Bignum newbn(int length) |
| 122 | { |
| 123 | Bignum b = snewn(length + 1, BignumInt); |
| 124 | if (!b) |
| 125 | abort(); /* FIXME */ |
| 126 | memset(b, 0, (length + 1) * sizeof(*b)); |
| 127 | b[0] = length; |
| 128 | return b; |
| 129 | } |
| 130 | |
| 131 | void bn_restore_invariant(Bignum b) |
| 132 | { |
| 133 | while (b[0] > 1 && b[b[0]] == 0) |
| 134 | b[0]--; |
| 135 | } |
| 136 | |
| 137 | Bignum copybn(Bignum orig) |
| 138 | { |
| 139 | Bignum b = snewn(orig[0] + 1, BignumInt); |
| 140 | if (!b) |
| 141 | abort(); /* FIXME */ |
| 142 | memcpy(b, orig, (orig[0] + 1) * sizeof(*b)); |
| 143 | return b; |
| 144 | } |
| 145 | |
| 146 | void freebn(Bignum b) |
| 147 | { |
| 148 | /* |
| 149 | * Burn the evidence, just in case. |
| 150 | */ |
| 151 | memset(b, 0, sizeof(b[0]) * (b[0] + 1)); |
| 152 | sfree(b); |
| 153 | } |
| 154 | |
| 155 | Bignum bn_power_2(int n) |
| 156 | { |
| 157 | Bignum ret = newbn(n / BIGNUM_INT_BITS + 1); |
| 158 | bignum_set_bit(ret, n, 1); |
| 159 | return ret; |
| 160 | } |
| 161 | |
| 162 | /* |
| 163 | * Internal addition. Sets c = a - b, where 'a', 'b' and 'c' are all |
| 164 | * big-endian arrays of 'len' BignumInts. Returns a BignumInt carried |
| 165 | * off the top. |
| 166 | */ |
| 167 | static BignumInt internal_add(const BignumInt *a, const BignumInt *b, |
| 168 | BignumInt *c, int len) |
| 169 | { |
| 170 | int i; |
| 171 | BignumDblInt carry = 0; |
| 172 | |
| 173 | for (i = len-1; i >= 0; i--) { |
| 174 | carry += (BignumDblInt)a[i] + b[i]; |
| 175 | c[i] = (BignumInt)carry; |
| 176 | carry >>= BIGNUM_INT_BITS; |
| 177 | } |
| 178 | |
| 179 | return (BignumInt)carry; |
| 180 | } |
| 181 | |
| 182 | /* |
| 183 | * Internal subtraction. Sets c = a - b, where 'a', 'b' and 'c' are |
| 184 | * all big-endian arrays of 'len' BignumInts. Any borrow from the top |
| 185 | * is ignored. |
| 186 | */ |
| 187 | static void internal_sub(const BignumInt *a, const BignumInt *b, |
| 188 | BignumInt *c, int len) |
| 189 | { |
| 190 | int i; |
| 191 | BignumDblInt carry = 1; |
| 192 | |
| 193 | for (i = len-1; i >= 0; i--) { |
| 194 | carry += (BignumDblInt)a[i] + (b[i] ^ BIGNUM_INT_MASK); |
| 195 | c[i] = (BignumInt)carry; |
| 196 | carry >>= BIGNUM_INT_BITS; |
| 197 | } |
| 198 | } |
| 199 | |
| 200 | /* |
| 201 | * Compute c = a * b. |
| 202 | * Input is in the first len words of a and b. |
| 203 | * Result is returned in the first 2*len words of c. |
| 204 | */ |
| 205 | #define KARATSUBA_THRESHOLD 50 |
| 206 | static void internal_mul(const BignumInt *a, const BignumInt *b, |
| 207 | BignumInt *c, int len) |
| 208 | { |
| 209 | int i, j; |
| 210 | BignumDblInt t; |
| 211 | |
| 212 | if (len > KARATSUBA_THRESHOLD) { |
| 213 | |
| 214 | /* |
| 215 | * Karatsuba divide-and-conquer algorithm. Cut each input in |
| 216 | * half, so that it's expressed as two big 'digits' in a giant |
| 217 | * base D: |
| 218 | * |
| 219 | * a = a_1 D + a_0 |
| 220 | * b = b_1 D + b_0 |
| 221 | * |
| 222 | * Then the product is of course |
| 223 | * |
| 224 | * ab = a_1 b_1 D^2 + (a_1 b_0 + a_0 b_1) D + a_0 b_0 |
| 225 | * |
| 226 | * and we compute the three coefficients by recursively |
| 227 | * calling ourself to do half-length multiplications. |
| 228 | * |
| 229 | * The clever bit that makes this worth doing is that we only |
| 230 | * need _one_ half-length multiplication for the central |
| 231 | * coefficient rather than the two that it obviouly looks |
| 232 | * like, because we can use a single multiplication to compute |
| 233 | * |
| 234 | * (a_1 + a_0) (b_1 + b_0) = a_1 b_1 + a_1 b_0 + a_0 b_1 + a_0 b_0 |
| 235 | * |
| 236 | * and then we subtract the other two coefficients (a_1 b_1 |
| 237 | * and a_0 b_0) which we were computing anyway. |
| 238 | * |
| 239 | * Hence we get to multiply two numbers of length N in about |
| 240 | * three times as much work as it takes to multiply numbers of |
| 241 | * length N/2, which is obviously better than the four times |
| 242 | * as much work it would take if we just did a long |
| 243 | * conventional multiply. |
| 244 | */ |
| 245 | |
| 246 | int toplen = len/2, botlen = len - toplen; /* botlen is the bigger */ |
| 247 | int midlen = botlen + 1; |
| 248 | BignumInt *scratch; |
| 249 | BignumDblInt carry; |
| 250 | |
| 251 | /* |
| 252 | * The coefficients a_1 b_1 and a_0 b_0 just avoid overlapping |
| 253 | * in the output array, so we can compute them immediately in |
| 254 | * place. |
| 255 | */ |
| 256 | |
| 257 | /* a_1 b_1 */ |
| 258 | internal_mul(a, b, c, toplen); |
| 259 | |
| 260 | /* a_0 b_0 */ |
| 261 | internal_mul(a + toplen, b + toplen, c + 2*toplen, botlen); |
| 262 | |
| 263 | /* |
| 264 | * We must allocate scratch space for the central coefficient, |
| 265 | * and also for the two input values that we multiply when |
| 266 | * computing it. Since either or both may carry into the |
| 267 | * (botlen+1)th word, we must use a slightly longer length |
| 268 | * 'midlen'. |
| 269 | */ |
| 270 | scratch = snewn(4 * midlen, BignumInt); |
| 271 | |
| 272 | /* Zero padding. midlen exceeds toplen by at most 2, so just |
| 273 | * zero the first two words of each input and the rest will be |
| 274 | * copied over. */ |
| 275 | scratch[0] = scratch[1] = scratch[midlen] = scratch[midlen+1] = 0; |
| 276 | |
| 277 | for (j = 0; j < toplen; j++) { |
| 278 | scratch[midlen - toplen + j] = a[j]; /* a_1 */ |
| 279 | scratch[2*midlen - toplen + j] = b[j]; /* b_1 */ |
| 280 | } |
| 281 | |
| 282 | /* compute a_1 + a_0 */ |
| 283 | scratch[0] = internal_add(scratch+1, a+toplen, scratch+1, botlen); |
| 284 | /* compute b_1 + b_0 */ |
| 285 | scratch[midlen] = internal_add(scratch+midlen+1, b+toplen, |
| 286 | scratch+midlen+1, botlen); |
| 287 | |
| 288 | /* |
| 289 | * Now we can do the third multiplication. |
| 290 | */ |
| 291 | internal_mul(scratch, scratch + midlen, scratch + 2*midlen, midlen); |
| 292 | |
| 293 | /* |
| 294 | * Now we can reuse the first half of 'scratch' to compute the |
| 295 | * sum of the outer two coefficients, to subtract from that |
| 296 | * product to obtain the middle one. |
| 297 | */ |
| 298 | scratch[0] = scratch[1] = scratch[2] = scratch[3] = 0; |
| 299 | for (j = 0; j < 2*toplen; j++) |
| 300 | scratch[2*midlen - 2*toplen + j] = c[j]; |
| 301 | scratch[1] = internal_add(scratch+2, c + 2*toplen, |
| 302 | scratch+2, 2*botlen); |
| 303 | |
| 304 | internal_sub(scratch + 2*midlen, scratch, |
| 305 | scratch + 2*midlen, 2*midlen); |
| 306 | |
| 307 | /* |
| 308 | * And now all we need to do is to add that middle coefficient |
| 309 | * back into the output. We may have to propagate a carry |
| 310 | * further up the output, but we can be sure it won't |
| 311 | * propagate right the way off the top. |
| 312 | */ |
| 313 | carry = internal_add(c + 2*len - botlen - 2*midlen, |
| 314 | scratch + 2*midlen, |
| 315 | c + 2*len - botlen - 2*midlen, 2*midlen); |
| 316 | j = 2*len - botlen - 2*midlen - 1; |
| 317 | while (carry) { |
| 318 | assert(j >= 0); |
| 319 | carry += c[j]; |
| 320 | c[j] = (BignumInt)carry; |
| 321 | carry >>= BIGNUM_INT_BITS; |
| 322 | } |
| 323 | |
| 324 | /* Free scratch. */ |
| 325 | for (j = 0; j < 4 * midlen; j++) |
| 326 | scratch[j] = 0; |
| 327 | sfree(scratch); |
| 328 | |
| 329 | } else { |
| 330 | |
| 331 | /* |
| 332 | * Multiply in the ordinary O(N^2) way. |
| 333 | */ |
| 334 | |
| 335 | for (j = 0; j < 2 * len; j++) |
| 336 | c[j] = 0; |
| 337 | |
| 338 | for (i = len - 1; i >= 0; i--) { |
| 339 | t = 0; |
| 340 | for (j = len - 1; j >= 0; j--) { |
| 341 | t += MUL_WORD(a[i], (BignumDblInt) b[j]); |
| 342 | t += (BignumDblInt) c[i + j + 1]; |
| 343 | c[i + j + 1] = (BignumInt) t; |
| 344 | t = t >> BIGNUM_INT_BITS; |
| 345 | } |
| 346 | c[i] = (BignumInt) t; |
| 347 | } |
| 348 | } |
| 349 | } |
| 350 | |
| 351 | /* |
| 352 | * Variant form of internal_mul used for the initial step of |
| 353 | * Montgomery reduction. Only bothers outputting 'len' words |
| 354 | * (everything above that is thrown away). |
| 355 | */ |
| 356 | static void internal_mul_low(const BignumInt *a, const BignumInt *b, |
| 357 | BignumInt *c, int len) |
| 358 | { |
| 359 | int i, j; |
| 360 | BignumDblInt t; |
| 361 | |
| 362 | if (len > KARATSUBA_THRESHOLD) { |
| 363 | |
| 364 | /* |
| 365 | * Karatsuba-aware version of internal_mul_low. As before, we |
| 366 | * express each input value as a shifted combination of two |
| 367 | * halves: |
| 368 | * |
| 369 | * a = a_1 D + a_0 |
| 370 | * b = b_1 D + b_0 |
| 371 | * |
| 372 | * Then the full product is, as before, |
| 373 | * |
| 374 | * ab = a_1 b_1 D^2 + (a_1 b_0 + a_0 b_1) D + a_0 b_0 |
| 375 | * |
| 376 | * Provided we choose D on the large side (so that a_0 and b_0 |
| 377 | * are _at least_ as long as a_1 and b_1), we don't need the |
| 378 | * topmost term at all, and we only need half of the middle |
| 379 | * term. So there's no point in doing the proper Karatsuba |
| 380 | * optimisation which computes the middle term using the top |
| 381 | * one, because we'd take as long computing the top one as |
| 382 | * just computing the middle one directly. |
| 383 | * |
| 384 | * So instead, we do a much more obvious thing: we call the |
| 385 | * fully optimised internal_mul to compute a_0 b_0, and we |
| 386 | * recursively call ourself to compute the _bottom halves_ of |
| 387 | * a_1 b_0 and a_0 b_1, each of which we add into the result |
| 388 | * in the obvious way. |
| 389 | * |
| 390 | * In other words, there's no actual Karatsuba _optimisation_ |
| 391 | * in this function; the only benefit in doing it this way is |
| 392 | * that we call internal_mul proper for a large part of the |
| 393 | * work, and _that_ can optimise its operation. |
| 394 | */ |
| 395 | |
| 396 | int toplen = len/2, botlen = len - toplen; /* botlen is the bigger */ |
| 397 | BignumInt *scratch; |
| 398 | |
| 399 | /* |
| 400 | * Allocate scratch space for the various bits and pieces |
| 401 | * we're going to be adding together. We need botlen*2 words |
| 402 | * for a_0 b_0 (though we may end up throwing away its topmost |
| 403 | * word), and toplen words for each of a_1 b_0 and a_0 b_1. |
| 404 | * That adds up to exactly 2*len. |
| 405 | */ |
| 406 | scratch = snewn(len*2, BignumInt); |
| 407 | |
| 408 | /* a_0 b_0 */ |
| 409 | internal_mul(a + toplen, b + toplen, scratch + 2*toplen, botlen); |
| 410 | |
| 411 | /* a_1 b_0 */ |
| 412 | internal_mul_low(a, b + len - toplen, scratch + toplen, toplen); |
| 413 | |
| 414 | /* a_0 b_1 */ |
| 415 | internal_mul_low(a + len - toplen, b, scratch, toplen); |
| 416 | |
| 417 | /* Copy the bottom half of the big coefficient into place */ |
| 418 | for (j = 0; j < botlen; j++) |
| 419 | c[toplen + j] = scratch[2*toplen + botlen + j]; |
| 420 | |
| 421 | /* Add the two small coefficients, throwing away the returned carry */ |
| 422 | internal_add(scratch, scratch + toplen, scratch, toplen); |
| 423 | |
| 424 | /* And add that to the large coefficient, leaving the result in c. */ |
| 425 | internal_add(scratch, scratch + 2*toplen + botlen - toplen, |
| 426 | c, toplen); |
| 427 | |
| 428 | /* Free scratch. */ |
| 429 | for (j = 0; j < len*2; j++) |
| 430 | scratch[j] = 0; |
| 431 | sfree(scratch); |
| 432 | |
| 433 | } else { |
| 434 | |
| 435 | for (j = 0; j < len; j++) |
| 436 | c[j] = 0; |
| 437 | |
| 438 | for (i = len - 1; i >= 0; i--) { |
| 439 | t = 0; |
| 440 | for (j = len - 1; j >= len - i - 1; j--) { |
| 441 | t += MUL_WORD(a[i], (BignumDblInt) b[j]); |
| 442 | t += (BignumDblInt) c[i + j + 1 - len]; |
| 443 | c[i + j + 1 - len] = (BignumInt) t; |
| 444 | t = t >> BIGNUM_INT_BITS; |
| 445 | } |
| 446 | } |
| 447 | |
| 448 | } |
| 449 | } |
| 450 | |
| 451 | /* |
| 452 | * Montgomery reduction. Expects x to be a big-endian array of 2*len |
| 453 | * BignumInts whose value satisfies 0 <= x < rn (where r = 2^(len * |
| 454 | * BIGNUM_INT_BITS) is the Montgomery base). Returns in the same array |
| 455 | * a value x' which is congruent to xr^{-1} mod n, and satisfies 0 <= |
| 456 | * x' < n. |
| 457 | * |
| 458 | * 'n' and 'mninv' should be big-endian arrays of 'len' BignumInts |
| 459 | * each, containing respectively n and the multiplicative inverse of |
| 460 | * -n mod r. |
| 461 | * |
| 462 | * 'tmp' is an array of at least '3*len' BignumInts used as scratch |
| 463 | * space. |
| 464 | */ |
| 465 | static void monty_reduce(BignumInt *x, const BignumInt *n, |
| 466 | const BignumInt *mninv, BignumInt *tmp, int len) |
| 467 | { |
| 468 | int i; |
| 469 | BignumInt carry; |
| 470 | |
| 471 | /* |
| 472 | * Multiply x by (-n)^{-1} mod r. This gives us a value m such |
| 473 | * that mn is congruent to -x mod r. Hence, mn+x is an exact |
| 474 | * multiple of r, and is also (obviously) congruent to x mod n. |
| 475 | */ |
| 476 | internal_mul_low(x + len, mninv, tmp, len); |
| 477 | |
| 478 | /* |
| 479 | * Compute t = (mn+x)/r in ordinary, non-modular, integer |
| 480 | * arithmetic. By construction this is exact, and is congruent mod |
| 481 | * n to x * r^{-1}, i.e. the answer we want. |
| 482 | * |
| 483 | * The following multiply leaves that answer in the _most_ |
| 484 | * significant half of the 'x' array, so then we must shift it |
| 485 | * down. |
| 486 | */ |
| 487 | internal_mul(tmp, n, tmp+len, len); |
| 488 | carry = internal_add(x, tmp+len, x, 2*len); |
| 489 | for (i = 0; i < len; i++) |
| 490 | x[len + i] = x[i], x[i] = 0; |
| 491 | |
| 492 | /* |
| 493 | * Reduce t mod n. This doesn't require a full-on division by n, |
| 494 | * but merely a test and single optional subtraction, since we can |
| 495 | * show that 0 <= t < 2n. |
| 496 | * |
| 497 | * Proof: |
| 498 | * + we computed m mod r, so 0 <= m < r. |
| 499 | * + so 0 <= mn < rn, obviously |
| 500 | * + hence we only need 0 <= x < rn to guarantee that 0 <= mn+x < 2rn |
| 501 | * + yielding 0 <= (mn+x)/r < 2n as required. |
| 502 | */ |
| 503 | if (!carry) { |
| 504 | for (i = 0; i < len; i++) |
| 505 | if (x[len + i] != n[i]) |
| 506 | break; |
| 507 | } |
| 508 | if (carry || i >= len || x[len + i] > n[i]) |
| 509 | internal_sub(x+len, n, x+len, len); |
| 510 | } |
| 511 | |
| 512 | static void internal_add_shifted(BignumInt *number, |
| 513 | unsigned n, int shift) |
| 514 | { |
| 515 | int word = 1 + (shift / BIGNUM_INT_BITS); |
| 516 | int bshift = shift % BIGNUM_INT_BITS; |
| 517 | BignumDblInt addend; |
| 518 | |
| 519 | addend = (BignumDblInt)n << bshift; |
| 520 | |
| 521 | while (addend) { |
| 522 | addend += number[word]; |
| 523 | number[word] = (BignumInt) addend & BIGNUM_INT_MASK; |
| 524 | addend >>= BIGNUM_INT_BITS; |
| 525 | word++; |
| 526 | } |
| 527 | } |
| 528 | |
| 529 | /* |
| 530 | * Compute a = a % m. |
| 531 | * Input in first alen words of a and first mlen words of m. |
| 532 | * Output in first alen words of a |
| 533 | * (of which first alen-mlen words will be zero). |
| 534 | * The MSW of m MUST have its high bit set. |
| 535 | * Quotient is accumulated in the `quotient' array, which is a Bignum |
| 536 | * rather than the internal bigendian format. Quotient parts are shifted |
| 537 | * left by `qshift' before adding into quot. |
| 538 | */ |
| 539 | static void internal_mod(BignumInt *a, int alen, |
| 540 | BignumInt *m, int mlen, |
| 541 | BignumInt *quot, int qshift) |
| 542 | { |
| 543 | BignumInt m0, m1; |
| 544 | unsigned int h; |
| 545 | int i, k; |
| 546 | |
| 547 | m0 = m[0]; |
| 548 | if (mlen > 1) |
| 549 | m1 = m[1]; |
| 550 | else |
| 551 | m1 = 0; |
| 552 | |
| 553 | for (i = 0; i <= alen - mlen; i++) { |
| 554 | BignumDblInt t; |
| 555 | unsigned int q, r, c, ai1; |
| 556 | |
| 557 | if (i == 0) { |
| 558 | h = 0; |
| 559 | } else { |
| 560 | h = a[i - 1]; |
| 561 | a[i - 1] = 0; |
| 562 | } |
| 563 | |
| 564 | if (i == alen - 1) |
| 565 | ai1 = 0; |
| 566 | else |
| 567 | ai1 = a[i + 1]; |
| 568 | |
| 569 | /* Find q = h:a[i] / m0 */ |
| 570 | if (h >= m0) { |
| 571 | /* |
| 572 | * Special case. |
| 573 | * |
| 574 | * To illustrate it, suppose a BignumInt is 8 bits, and |
| 575 | * we are dividing (say) A1:23:45:67 by A1:B2:C3. Then |
| 576 | * our initial division will be 0xA123 / 0xA1, which |
| 577 | * will give a quotient of 0x100 and a divide overflow. |
| 578 | * However, the invariants in this division algorithm |
| 579 | * are not violated, since the full number A1:23:... is |
| 580 | * _less_ than the quotient prefix A1:B2:... and so the |
| 581 | * following correction loop would have sorted it out. |
| 582 | * |
| 583 | * In this situation we set q to be the largest |
| 584 | * quotient we _can_ stomach (0xFF, of course). |
| 585 | */ |
| 586 | q = BIGNUM_INT_MASK; |
| 587 | } else { |
| 588 | /* Macro doesn't want an array subscript expression passed |
| 589 | * into it (see definition), so use a temporary. */ |
| 590 | BignumInt tmplo = a[i]; |
| 591 | DIVMOD_WORD(q, r, h, tmplo, m0); |
| 592 | |
| 593 | /* Refine our estimate of q by looking at |
| 594 | h:a[i]:a[i+1] / m0:m1 */ |
| 595 | t = MUL_WORD(m1, q); |
| 596 | if (t > ((BignumDblInt) r << BIGNUM_INT_BITS) + ai1) { |
| 597 | q--; |
| 598 | t -= m1; |
| 599 | r = (r + m0) & BIGNUM_INT_MASK; /* overflow? */ |
| 600 | if (r >= (BignumDblInt) m0 && |
| 601 | t > ((BignumDblInt) r << BIGNUM_INT_BITS) + ai1) q--; |
| 602 | } |
| 603 | } |
| 604 | |
| 605 | /* Subtract q * m from a[i...] */ |
| 606 | c = 0; |
| 607 | for (k = mlen - 1; k >= 0; k--) { |
| 608 | t = MUL_WORD(q, m[k]); |
| 609 | t += c; |
| 610 | c = (unsigned)(t >> BIGNUM_INT_BITS); |
| 611 | if ((BignumInt) t > a[i + k]) |
| 612 | c++; |
| 613 | a[i + k] -= (BignumInt) t; |
| 614 | } |
| 615 | |
| 616 | /* Add back m in case of borrow */ |
| 617 | if (c != h) { |
| 618 | t = 0; |
| 619 | for (k = mlen - 1; k >= 0; k--) { |
| 620 | t += m[k]; |
| 621 | t += a[i + k]; |
| 622 | a[i + k] = (BignumInt) t; |
| 623 | t = t >> BIGNUM_INT_BITS; |
| 624 | } |
| 625 | q--; |
| 626 | } |
| 627 | if (quot) |
| 628 | internal_add_shifted(quot, q, qshift + BIGNUM_INT_BITS * (alen - mlen - i)); |
| 629 | } |
| 630 | } |
| 631 | |
| 632 | /* |
| 633 | * Compute (base ^ exp) % mod. Uses the Montgomery multiplication |
| 634 | * technique. |
| 635 | */ |
| 636 | Bignum modpow(Bignum base_in, Bignum exp, Bignum mod) |
| 637 | { |
| 638 | BignumInt *a, *b, *x, *n, *mninv, *tmp; |
| 639 | int len, i, j; |
| 640 | Bignum base, base2, r, rn, inv, result; |
| 641 | |
| 642 | /* |
| 643 | * The most significant word of mod needs to be non-zero. It |
| 644 | * should already be, but let's make sure. |
| 645 | */ |
| 646 | assert(mod[mod[0]] != 0); |
| 647 | |
| 648 | /* |
| 649 | * Make sure the base is smaller than the modulus, by reducing |
| 650 | * it modulo the modulus if not. |
| 651 | */ |
| 652 | base = bigmod(base_in, mod); |
| 653 | |
| 654 | /* |
| 655 | * mod had better be odd, or we can't do Montgomery multiplication |
| 656 | * using a power of two at all. |
| 657 | */ |
| 658 | assert(mod[1] & 1); |
| 659 | |
| 660 | /* |
| 661 | * Compute the inverse of n mod r, for monty_reduce. (In fact we |
| 662 | * want the inverse of _minus_ n mod r, but we'll sort that out |
| 663 | * below.) |
| 664 | */ |
| 665 | len = mod[0]; |
| 666 | r = bn_power_2(BIGNUM_INT_BITS * len); |
| 667 | inv = modinv(mod, r); |
| 668 | |
| 669 | /* |
| 670 | * Multiply the base by r mod n, to get it into Montgomery |
| 671 | * representation. |
| 672 | */ |
| 673 | base2 = modmul(base, r, mod); |
| 674 | freebn(base); |
| 675 | base = base2; |
| 676 | |
| 677 | rn = bigmod(r, mod); /* r mod n, i.e. Montgomerified 1 */ |
| 678 | |
| 679 | freebn(r); /* won't need this any more */ |
| 680 | |
| 681 | /* |
| 682 | * Set up internal arrays of the right lengths, in big-endian |
| 683 | * format, containing the base, the modulus, and the modulus's |
| 684 | * inverse. |
| 685 | */ |
| 686 | n = snewn(len, BignumInt); |
| 687 | for (j = 0; j < len; j++) |
| 688 | n[len - 1 - j] = mod[j + 1]; |
| 689 | |
| 690 | mninv = snewn(len, BignumInt); |
| 691 | for (j = 0; j < len; j++) |
| 692 | mninv[len - 1 - j] = (j < inv[0] ? inv[j + 1] : 0); |
| 693 | freebn(inv); /* we don't need this copy of it any more */ |
| 694 | /* Now negate mninv mod r, so it's the inverse of -n rather than +n. */ |
| 695 | x = snewn(len, BignumInt); |
| 696 | for (j = 0; j < len; j++) |
| 697 | x[j] = 0; |
| 698 | internal_sub(x, mninv, mninv, len); |
| 699 | |
| 700 | /* x = snewn(len, BignumInt); */ /* already done above */ |
| 701 | for (j = 0; j < len; j++) |
| 702 | x[len - 1 - j] = (j < base[0] ? base[j + 1] : 0); |
| 703 | freebn(base); /* we don't need this copy of it any more */ |
| 704 | |
| 705 | a = snewn(2*len, BignumInt); |
| 706 | b = snewn(2*len, BignumInt); |
| 707 | for (j = 0; j < len; j++) |
| 708 | a[2*len - 1 - j] = (j < rn[0] ? rn[j + 1] : 0); |
| 709 | freebn(rn); |
| 710 | |
| 711 | tmp = snewn(3*len, BignumInt); |
| 712 | |
| 713 | /* Skip leading zero bits of exp. */ |
| 714 | i = 0; |
| 715 | j = BIGNUM_INT_BITS-1; |
| 716 | while (i < (int)exp[0] && (exp[exp[0] - i] & (1 << j)) == 0) { |
| 717 | j--; |
| 718 | if (j < 0) { |
| 719 | i++; |
| 720 | j = BIGNUM_INT_BITS-1; |
| 721 | } |
| 722 | } |
| 723 | |
| 724 | /* Main computation */ |
| 725 | while (i < (int)exp[0]) { |
| 726 | while (j >= 0) { |
| 727 | internal_mul(a + len, a + len, b, len); |
| 728 | monty_reduce(b, n, mninv, tmp, len); |
| 729 | if ((exp[exp[0] - i] & (1 << j)) != 0) { |
| 730 | internal_mul(b + len, x, a, len); |
| 731 | monty_reduce(a, n, mninv, tmp, len); |
| 732 | } else { |
| 733 | BignumInt *t; |
| 734 | t = a; |
| 735 | a = b; |
| 736 | b = t; |
| 737 | } |
| 738 | j--; |
| 739 | } |
| 740 | i++; |
| 741 | j = BIGNUM_INT_BITS-1; |
| 742 | } |
| 743 | |
| 744 | /* |
| 745 | * Final monty_reduce to get back from the adjusted Montgomery |
| 746 | * representation. |
| 747 | */ |
| 748 | monty_reduce(a, n, mninv, tmp, len); |
| 749 | |
| 750 | /* Copy result to buffer */ |
| 751 | result = newbn(mod[0]); |
| 752 | for (i = 0; i < len; i++) |
| 753 | result[result[0] - i] = a[i + len]; |
| 754 | while (result[0] > 1 && result[result[0]] == 0) |
| 755 | result[0]--; |
| 756 | |
| 757 | /* Free temporary arrays */ |
| 758 | for (i = 0; i < 3 * len; i++) |
| 759 | tmp[i] = 0; |
| 760 | sfree(tmp); |
| 761 | for (i = 0; i < 2 * len; i++) |
| 762 | a[i] = 0; |
| 763 | sfree(a); |
| 764 | for (i = 0; i < 2 * len; i++) |
| 765 | b[i] = 0; |
| 766 | sfree(b); |
| 767 | for (i = 0; i < len; i++) |
| 768 | mninv[i] = 0; |
| 769 | sfree(mninv); |
| 770 | for (i = 0; i < len; i++) |
| 771 | n[i] = 0; |
| 772 | sfree(n); |
| 773 | for (i = 0; i < len; i++) |
| 774 | x[i] = 0; |
| 775 | sfree(x); |
| 776 | |
| 777 | return result; |
| 778 | } |
| 779 | |
| 780 | /* |
| 781 | * Compute (p * q) % mod. |
| 782 | * The most significant word of mod MUST be non-zero. |
| 783 | * We assume that the result array is the same size as the mod array. |
| 784 | */ |
| 785 | Bignum modmul(Bignum p, Bignum q, Bignum mod) |
| 786 | { |
| 787 | BignumInt *a, *n, *m, *o; |
| 788 | int mshift; |
| 789 | int pqlen, mlen, rlen, i, j; |
| 790 | Bignum result; |
| 791 | |
| 792 | /* Allocate m of size mlen, copy mod to m */ |
| 793 | /* We use big endian internally */ |
| 794 | mlen = mod[0]; |
| 795 | m = snewn(mlen, BignumInt); |
| 796 | for (j = 0; j < mlen; j++) |
| 797 | m[j] = mod[mod[0] - j]; |
| 798 | |
| 799 | /* Shift m left to make msb bit set */ |
| 800 | for (mshift = 0; mshift < BIGNUM_INT_BITS-1; mshift++) |
| 801 | if ((m[0] << mshift) & BIGNUM_TOP_BIT) |
| 802 | break; |
| 803 | if (mshift) { |
| 804 | for (i = 0; i < mlen - 1; i++) |
| 805 | m[i] = (m[i] << mshift) | (m[i + 1] >> (BIGNUM_INT_BITS - mshift)); |
| 806 | m[mlen - 1] = m[mlen - 1] << mshift; |
| 807 | } |
| 808 | |
| 809 | pqlen = (p[0] > q[0] ? p[0] : q[0]); |
| 810 | |
| 811 | /* Allocate n of size pqlen, copy p to n */ |
| 812 | n = snewn(pqlen, BignumInt); |
| 813 | i = pqlen - p[0]; |
| 814 | for (j = 0; j < i; j++) |
| 815 | n[j] = 0; |
| 816 | for (j = 0; j < (int)p[0]; j++) |
| 817 | n[i + j] = p[p[0] - j]; |
| 818 | |
| 819 | /* Allocate o of size pqlen, copy q to o */ |
| 820 | o = snewn(pqlen, BignumInt); |
| 821 | i = pqlen - q[0]; |
| 822 | for (j = 0; j < i; j++) |
| 823 | o[j] = 0; |
| 824 | for (j = 0; j < (int)q[0]; j++) |
| 825 | o[i + j] = q[q[0] - j]; |
| 826 | |
| 827 | /* Allocate a of size 2*pqlen for result */ |
| 828 | a = snewn(2 * pqlen, BignumInt); |
| 829 | |
| 830 | /* Main computation */ |
| 831 | internal_mul(n, o, a, pqlen); |
| 832 | internal_mod(a, pqlen * 2, m, mlen, NULL, 0); |
| 833 | |
| 834 | /* Fixup result in case the modulus was shifted */ |
| 835 | if (mshift) { |
| 836 | for (i = 2 * pqlen - mlen - 1; i < 2 * pqlen - 1; i++) |
| 837 | a[i] = (a[i] << mshift) | (a[i + 1] >> (BIGNUM_INT_BITS - mshift)); |
| 838 | a[2 * pqlen - 1] = a[2 * pqlen - 1] << mshift; |
| 839 | internal_mod(a, pqlen * 2, m, mlen, NULL, 0); |
| 840 | for (i = 2 * pqlen - 1; i >= 2 * pqlen - mlen; i--) |
| 841 | a[i] = (a[i] >> mshift) | (a[i - 1] << (BIGNUM_INT_BITS - mshift)); |
| 842 | } |
| 843 | |
| 844 | /* Copy result to buffer */ |
| 845 | rlen = (mlen < pqlen * 2 ? mlen : pqlen * 2); |
| 846 | result = newbn(rlen); |
| 847 | for (i = 0; i < rlen; i++) |
| 848 | result[result[0] - i] = a[i + 2 * pqlen - rlen]; |
| 849 | while (result[0] > 1 && result[result[0]] == 0) |
| 850 | result[0]--; |
| 851 | |
| 852 | /* Free temporary arrays */ |
| 853 | for (i = 0; i < 2 * pqlen; i++) |
| 854 | a[i] = 0; |
| 855 | sfree(a); |
| 856 | for (i = 0; i < mlen; i++) |
| 857 | m[i] = 0; |
| 858 | sfree(m); |
| 859 | for (i = 0; i < pqlen; i++) |
| 860 | n[i] = 0; |
| 861 | sfree(n); |
| 862 | for (i = 0; i < pqlen; i++) |
| 863 | o[i] = 0; |
| 864 | sfree(o); |
| 865 | |
| 866 | return result; |
| 867 | } |
| 868 | |
| 869 | /* |
| 870 | * Compute p % mod. |
| 871 | * The most significant word of mod MUST be non-zero. |
| 872 | * We assume that the result array is the same size as the mod array. |
| 873 | * We optionally write out a quotient if `quotient' is non-NULL. |
| 874 | * We can avoid writing out the result if `result' is NULL. |
| 875 | */ |
| 876 | static void bigdivmod(Bignum p, Bignum mod, Bignum result, Bignum quotient) |
| 877 | { |
| 878 | BignumInt *n, *m; |
| 879 | int mshift; |
| 880 | int plen, mlen, i, j; |
| 881 | |
| 882 | /* Allocate m of size mlen, copy mod to m */ |
| 883 | /* We use big endian internally */ |
| 884 | mlen = mod[0]; |
| 885 | m = snewn(mlen, BignumInt); |
| 886 | for (j = 0; j < mlen; j++) |
| 887 | m[j] = mod[mod[0] - j]; |
| 888 | |
| 889 | /* Shift m left to make msb bit set */ |
| 890 | for (mshift = 0; mshift < BIGNUM_INT_BITS-1; mshift++) |
| 891 | if ((m[0] << mshift) & BIGNUM_TOP_BIT) |
| 892 | break; |
| 893 | if (mshift) { |
| 894 | for (i = 0; i < mlen - 1; i++) |
| 895 | m[i] = (m[i] << mshift) | (m[i + 1] >> (BIGNUM_INT_BITS - mshift)); |
| 896 | m[mlen - 1] = m[mlen - 1] << mshift; |
| 897 | } |
| 898 | |
| 899 | plen = p[0]; |
| 900 | /* Ensure plen > mlen */ |
| 901 | if (plen <= mlen) |
| 902 | plen = mlen + 1; |
| 903 | |
| 904 | /* Allocate n of size plen, copy p to n */ |
| 905 | n = snewn(plen, BignumInt); |
| 906 | for (j = 0; j < plen; j++) |
| 907 | n[j] = 0; |
| 908 | for (j = 1; j <= (int)p[0]; j++) |
| 909 | n[plen - j] = p[j]; |
| 910 | |
| 911 | /* Main computation */ |
| 912 | internal_mod(n, plen, m, mlen, quotient, mshift); |
| 913 | |
| 914 | /* Fixup result in case the modulus was shifted */ |
| 915 | if (mshift) { |
| 916 | for (i = plen - mlen - 1; i < plen - 1; i++) |
| 917 | n[i] = (n[i] << mshift) | (n[i + 1] >> (BIGNUM_INT_BITS - mshift)); |
| 918 | n[plen - 1] = n[plen - 1] << mshift; |
| 919 | internal_mod(n, plen, m, mlen, quotient, 0); |
| 920 | for (i = plen - 1; i >= plen - mlen; i--) |
| 921 | n[i] = (n[i] >> mshift) | (n[i - 1] << (BIGNUM_INT_BITS - mshift)); |
| 922 | } |
| 923 | |
| 924 | /* Copy result to buffer */ |
| 925 | if (result) { |
| 926 | for (i = 1; i <= (int)result[0]; i++) { |
| 927 | int j = plen - i; |
| 928 | result[i] = j >= 0 ? n[j] : 0; |
| 929 | } |
| 930 | } |
| 931 | |
| 932 | /* Free temporary arrays */ |
| 933 | for (i = 0; i < mlen; i++) |
| 934 | m[i] = 0; |
| 935 | sfree(m); |
| 936 | for (i = 0; i < plen; i++) |
| 937 | n[i] = 0; |
| 938 | sfree(n); |
| 939 | } |
| 940 | |
| 941 | /* |
| 942 | * Decrement a number. |
| 943 | */ |
| 944 | void decbn(Bignum bn) |
| 945 | { |
| 946 | int i = 1; |
| 947 | while (i < (int)bn[0] && bn[i] == 0) |
| 948 | bn[i++] = BIGNUM_INT_MASK; |
| 949 | bn[i]--; |
| 950 | } |
| 951 | |
| 952 | Bignum bignum_from_bytes(const unsigned char *data, int nbytes) |
| 953 | { |
| 954 | Bignum result; |
| 955 | int w, i; |
| 956 | |
| 957 | w = (nbytes + BIGNUM_INT_BYTES - 1) / BIGNUM_INT_BYTES; /* bytes->words */ |
| 958 | |
| 959 | result = newbn(w); |
| 960 | for (i = 1; i <= w; i++) |
| 961 | result[i] = 0; |
| 962 | for (i = nbytes; i--;) { |
| 963 | unsigned char byte = *data++; |
| 964 | result[1 + i / BIGNUM_INT_BYTES] |= byte << (8*i % BIGNUM_INT_BITS); |
| 965 | } |
| 966 | |
| 967 | while (result[0] > 1 && result[result[0]] == 0) |
| 968 | result[0]--; |
| 969 | return result; |
| 970 | } |
| 971 | |
| 972 | /* |
| 973 | * Read an SSH-1-format bignum from a data buffer. Return the number |
| 974 | * of bytes consumed, or -1 if there wasn't enough data. |
| 975 | */ |
| 976 | int ssh1_read_bignum(const unsigned char *data, int len, Bignum * result) |
| 977 | { |
| 978 | const unsigned char *p = data; |
| 979 | int i; |
| 980 | int w, b; |
| 981 | |
| 982 | if (len < 2) |
| 983 | return -1; |
| 984 | |
| 985 | w = 0; |
| 986 | for (i = 0; i < 2; i++) |
| 987 | w = (w << 8) + *p++; |
| 988 | b = (w + 7) / 8; /* bits -> bytes */ |
| 989 | |
| 990 | if (len < b+2) |
| 991 | return -1; |
| 992 | |
| 993 | if (!result) /* just return length */ |
| 994 | return b + 2; |
| 995 | |
| 996 | *result = bignum_from_bytes(p, b); |
| 997 | |
| 998 | return p + b - data; |
| 999 | } |
| 1000 | |
| 1001 | /* |
| 1002 | * Return the bit count of a bignum, for SSH-1 encoding. |
| 1003 | */ |
| 1004 | int bignum_bitcount(Bignum bn) |
| 1005 | { |
| 1006 | int bitcount = bn[0] * BIGNUM_INT_BITS - 1; |
| 1007 | while (bitcount >= 0 |
| 1008 | && (bn[bitcount / BIGNUM_INT_BITS + 1] >> (bitcount % BIGNUM_INT_BITS)) == 0) bitcount--; |
| 1009 | return bitcount + 1; |
| 1010 | } |
| 1011 | |
| 1012 | /* |
| 1013 | * Return the byte length of a bignum when SSH-1 encoded. |
| 1014 | */ |
| 1015 | int ssh1_bignum_length(Bignum bn) |
| 1016 | { |
| 1017 | return 2 + (bignum_bitcount(bn) + 7) / 8; |
| 1018 | } |
| 1019 | |
| 1020 | /* |
| 1021 | * Return the byte length of a bignum when SSH-2 encoded. |
| 1022 | */ |
| 1023 | int ssh2_bignum_length(Bignum bn) |
| 1024 | { |
| 1025 | return 4 + (bignum_bitcount(bn) + 8) / 8; |
| 1026 | } |
| 1027 | |
| 1028 | /* |
| 1029 | * Return a byte from a bignum; 0 is least significant, etc. |
| 1030 | */ |
| 1031 | int bignum_byte(Bignum bn, int i) |
| 1032 | { |
| 1033 | if (i >= (int)(BIGNUM_INT_BYTES * bn[0])) |
| 1034 | return 0; /* beyond the end */ |
| 1035 | else |
| 1036 | return (bn[i / BIGNUM_INT_BYTES + 1] >> |
| 1037 | ((i % BIGNUM_INT_BYTES)*8)) & 0xFF; |
| 1038 | } |
| 1039 | |
| 1040 | /* |
| 1041 | * Return a bit from a bignum; 0 is least significant, etc. |
| 1042 | */ |
| 1043 | int bignum_bit(Bignum bn, int i) |
| 1044 | { |
| 1045 | if (i >= (int)(BIGNUM_INT_BITS * bn[0])) |
| 1046 | return 0; /* beyond the end */ |
| 1047 | else |
| 1048 | return (bn[i / BIGNUM_INT_BITS + 1] >> (i % BIGNUM_INT_BITS)) & 1; |
| 1049 | } |
| 1050 | |
| 1051 | /* |
| 1052 | * Set a bit in a bignum; 0 is least significant, etc. |
| 1053 | */ |
| 1054 | void bignum_set_bit(Bignum bn, int bitnum, int value) |
| 1055 | { |
| 1056 | if (bitnum >= (int)(BIGNUM_INT_BITS * bn[0])) |
| 1057 | abort(); /* beyond the end */ |
| 1058 | else { |
| 1059 | int v = bitnum / BIGNUM_INT_BITS + 1; |
| 1060 | int mask = 1 << (bitnum % BIGNUM_INT_BITS); |
| 1061 | if (value) |
| 1062 | bn[v] |= mask; |
| 1063 | else |
| 1064 | bn[v] &= ~mask; |
| 1065 | } |
| 1066 | } |
| 1067 | |
| 1068 | /* |
| 1069 | * Write a SSH-1-format bignum into a buffer. It is assumed the |
| 1070 | * buffer is big enough. Returns the number of bytes used. |
| 1071 | */ |
| 1072 | int ssh1_write_bignum(void *data, Bignum bn) |
| 1073 | { |
| 1074 | unsigned char *p = data; |
| 1075 | int len = ssh1_bignum_length(bn); |
| 1076 | int i; |
| 1077 | int bitc = bignum_bitcount(bn); |
| 1078 | |
| 1079 | *p++ = (bitc >> 8) & 0xFF; |
| 1080 | *p++ = (bitc) & 0xFF; |
| 1081 | for (i = len - 2; i--;) |
| 1082 | *p++ = bignum_byte(bn, i); |
| 1083 | return len; |
| 1084 | } |
| 1085 | |
| 1086 | /* |
| 1087 | * Compare two bignums. Returns like strcmp. |
| 1088 | */ |
| 1089 | int bignum_cmp(Bignum a, Bignum b) |
| 1090 | { |
| 1091 | int amax = a[0], bmax = b[0]; |
| 1092 | int i = (amax > bmax ? amax : bmax); |
| 1093 | while (i) { |
| 1094 | BignumInt aval = (i > amax ? 0 : a[i]); |
| 1095 | BignumInt bval = (i > bmax ? 0 : b[i]); |
| 1096 | if (aval < bval) |
| 1097 | return -1; |
| 1098 | if (aval > bval) |
| 1099 | return +1; |
| 1100 | i--; |
| 1101 | } |
| 1102 | return 0; |
| 1103 | } |
| 1104 | |
| 1105 | /* |
| 1106 | * Right-shift one bignum to form another. |
| 1107 | */ |
| 1108 | Bignum bignum_rshift(Bignum a, int shift) |
| 1109 | { |
| 1110 | Bignum ret; |
| 1111 | int i, shiftw, shiftb, shiftbb, bits; |
| 1112 | BignumInt ai, ai1; |
| 1113 | |
| 1114 | bits = bignum_bitcount(a) - shift; |
| 1115 | ret = newbn((bits + BIGNUM_INT_BITS - 1) / BIGNUM_INT_BITS); |
| 1116 | |
| 1117 | if (ret) { |
| 1118 | shiftw = shift / BIGNUM_INT_BITS; |
| 1119 | shiftb = shift % BIGNUM_INT_BITS; |
| 1120 | shiftbb = BIGNUM_INT_BITS - shiftb; |
| 1121 | |
| 1122 | ai1 = a[shiftw + 1]; |
| 1123 | for (i = 1; i <= (int)ret[0]; i++) { |
| 1124 | ai = ai1; |
| 1125 | ai1 = (i + shiftw + 1 <= (int)a[0] ? a[i + shiftw + 1] : 0); |
| 1126 | ret[i] = ((ai >> shiftb) | (ai1 << shiftbb)) & BIGNUM_INT_MASK; |
| 1127 | } |
| 1128 | } |
| 1129 | |
| 1130 | return ret; |
| 1131 | } |
| 1132 | |
| 1133 | /* |
| 1134 | * Non-modular multiplication and addition. |
| 1135 | */ |
| 1136 | Bignum bigmuladd(Bignum a, Bignum b, Bignum addend) |
| 1137 | { |
| 1138 | int alen = a[0], blen = b[0]; |
| 1139 | int mlen = (alen > blen ? alen : blen); |
| 1140 | int rlen, i, maxspot; |
| 1141 | BignumInt *workspace; |
| 1142 | Bignum ret; |
| 1143 | |
| 1144 | /* mlen space for a, mlen space for b, 2*mlen for result */ |
| 1145 | workspace = snewn(mlen * 4, BignumInt); |
| 1146 | for (i = 0; i < mlen; i++) { |
| 1147 | workspace[0 * mlen + i] = (mlen - i <= (int)a[0] ? a[mlen - i] : 0); |
| 1148 | workspace[1 * mlen + i] = (mlen - i <= (int)b[0] ? b[mlen - i] : 0); |
| 1149 | } |
| 1150 | |
| 1151 | internal_mul(workspace + 0 * mlen, workspace + 1 * mlen, |
| 1152 | workspace + 2 * mlen, mlen); |
| 1153 | |
| 1154 | /* now just copy the result back */ |
| 1155 | rlen = alen + blen + 1; |
| 1156 | if (addend && rlen <= (int)addend[0]) |
| 1157 | rlen = addend[0] + 1; |
| 1158 | ret = newbn(rlen); |
| 1159 | maxspot = 0; |
| 1160 | for (i = 1; i <= (int)ret[0]; i++) { |
| 1161 | ret[i] = (i <= 2 * mlen ? workspace[4 * mlen - i] : 0); |
| 1162 | if (ret[i] != 0) |
| 1163 | maxspot = i; |
| 1164 | } |
| 1165 | ret[0] = maxspot; |
| 1166 | |
| 1167 | /* now add in the addend, if any */ |
| 1168 | if (addend) { |
| 1169 | BignumDblInt carry = 0; |
| 1170 | for (i = 1; i <= rlen; i++) { |
| 1171 | carry += (i <= (int)ret[0] ? ret[i] : 0); |
| 1172 | carry += (i <= (int)addend[0] ? addend[i] : 0); |
| 1173 | ret[i] = (BignumInt) carry & BIGNUM_INT_MASK; |
| 1174 | carry >>= BIGNUM_INT_BITS; |
| 1175 | if (ret[i] != 0 && i > maxspot) |
| 1176 | maxspot = i; |
| 1177 | } |
| 1178 | } |
| 1179 | ret[0] = maxspot; |
| 1180 | |
| 1181 | sfree(workspace); |
| 1182 | return ret; |
| 1183 | } |
| 1184 | |
| 1185 | /* |
| 1186 | * Non-modular multiplication. |
| 1187 | */ |
| 1188 | Bignum bigmul(Bignum a, Bignum b) |
| 1189 | { |
| 1190 | return bigmuladd(a, b, NULL); |
| 1191 | } |
| 1192 | |
| 1193 | /* |
| 1194 | * Create a bignum which is the bitmask covering another one. That |
| 1195 | * is, the smallest integer which is >= N and is also one less than |
| 1196 | * a power of two. |
| 1197 | */ |
| 1198 | Bignum bignum_bitmask(Bignum n) |
| 1199 | { |
| 1200 | Bignum ret = copybn(n); |
| 1201 | int i; |
| 1202 | BignumInt j; |
| 1203 | |
| 1204 | i = ret[0]; |
| 1205 | while (n[i] == 0 && i > 0) |
| 1206 | i--; |
| 1207 | if (i <= 0) |
| 1208 | return ret; /* input was zero */ |
| 1209 | j = 1; |
| 1210 | while (j < n[i]) |
| 1211 | j = 2 * j + 1; |
| 1212 | ret[i] = j; |
| 1213 | while (--i > 0) |
| 1214 | ret[i] = BIGNUM_INT_MASK; |
| 1215 | return ret; |
| 1216 | } |
| 1217 | |
| 1218 | /* |
| 1219 | * Convert a (max 32-bit) long into a bignum. |
| 1220 | */ |
| 1221 | Bignum bignum_from_long(unsigned long nn) |
| 1222 | { |
| 1223 | Bignum ret; |
| 1224 | BignumDblInt n = nn; |
| 1225 | |
| 1226 | ret = newbn(3); |
| 1227 | ret[1] = (BignumInt)(n & BIGNUM_INT_MASK); |
| 1228 | ret[2] = (BignumInt)((n >> BIGNUM_INT_BITS) & BIGNUM_INT_MASK); |
| 1229 | ret[3] = 0; |
| 1230 | ret[0] = (ret[2] ? 2 : 1); |
| 1231 | return ret; |
| 1232 | } |
| 1233 | |
| 1234 | /* |
| 1235 | * Add a long to a bignum. |
| 1236 | */ |
| 1237 | Bignum bignum_add_long(Bignum number, unsigned long addendx) |
| 1238 | { |
| 1239 | Bignum ret = newbn(number[0] + 1); |
| 1240 | int i, maxspot = 0; |
| 1241 | BignumDblInt carry = 0, addend = addendx; |
| 1242 | |
| 1243 | for (i = 1; i <= (int)ret[0]; i++) { |
| 1244 | carry += addend & BIGNUM_INT_MASK; |
| 1245 | carry += (i <= (int)number[0] ? number[i] : 0); |
| 1246 | addend >>= BIGNUM_INT_BITS; |
| 1247 | ret[i] = (BignumInt) carry & BIGNUM_INT_MASK; |
| 1248 | carry >>= BIGNUM_INT_BITS; |
| 1249 | if (ret[i] != 0) |
| 1250 | maxspot = i; |
| 1251 | } |
| 1252 | ret[0] = maxspot; |
| 1253 | return ret; |
| 1254 | } |
| 1255 | |
| 1256 | /* |
| 1257 | * Compute the residue of a bignum, modulo a (max 16-bit) short. |
| 1258 | */ |
| 1259 | unsigned short bignum_mod_short(Bignum number, unsigned short modulus) |
| 1260 | { |
| 1261 | BignumDblInt mod, r; |
| 1262 | int i; |
| 1263 | |
| 1264 | r = 0; |
| 1265 | mod = modulus; |
| 1266 | for (i = number[0]; i > 0; i--) |
| 1267 | r = (r * (BIGNUM_TOP_BIT % mod) * 2 + number[i] % mod) % mod; |
| 1268 | return (unsigned short) r; |
| 1269 | } |
| 1270 | |
| 1271 | #ifdef DEBUG |
| 1272 | void diagbn(char *prefix, Bignum md) |
| 1273 | { |
| 1274 | int i, nibbles, morenibbles; |
| 1275 | static const char hex[] = "0123456789ABCDEF"; |
| 1276 | |
| 1277 | debug(("%s0x", prefix ? prefix : "")); |
| 1278 | |
| 1279 | nibbles = (3 + bignum_bitcount(md)) / 4; |
| 1280 | if (nibbles < 1) |
| 1281 | nibbles = 1; |
| 1282 | morenibbles = 4 * md[0] - nibbles; |
| 1283 | for (i = 0; i < morenibbles; i++) |
| 1284 | debug(("-")); |
| 1285 | for (i = nibbles; i--;) |
| 1286 | debug(("%c", |
| 1287 | hex[(bignum_byte(md, i / 2) >> (4 * (i % 2))) & 0xF])); |
| 1288 | |
| 1289 | if (prefix) |
| 1290 | debug(("\n")); |
| 1291 | } |
| 1292 | #endif |
| 1293 | |
| 1294 | /* |
| 1295 | * Simple division. |
| 1296 | */ |
| 1297 | Bignum bigdiv(Bignum a, Bignum b) |
| 1298 | { |
| 1299 | Bignum q = newbn(a[0]); |
| 1300 | bigdivmod(a, b, NULL, q); |
| 1301 | return q; |
| 1302 | } |
| 1303 | |
| 1304 | /* |
| 1305 | * Simple remainder. |
| 1306 | */ |
| 1307 | Bignum bigmod(Bignum a, Bignum b) |
| 1308 | { |
| 1309 | Bignum r = newbn(b[0]); |
| 1310 | bigdivmod(a, b, r, NULL); |
| 1311 | return r; |
| 1312 | } |
| 1313 | |
| 1314 | /* |
| 1315 | * Greatest common divisor. |
| 1316 | */ |
| 1317 | Bignum biggcd(Bignum av, Bignum bv) |
| 1318 | { |
| 1319 | Bignum a = copybn(av); |
| 1320 | Bignum b = copybn(bv); |
| 1321 | |
| 1322 | while (bignum_cmp(b, Zero) != 0) { |
| 1323 | Bignum t = newbn(b[0]); |
| 1324 | bigdivmod(a, b, t, NULL); |
| 1325 | while (t[0] > 1 && t[t[0]] == 0) |
| 1326 | t[0]--; |
| 1327 | freebn(a); |
| 1328 | a = b; |
| 1329 | b = t; |
| 1330 | } |
| 1331 | |
| 1332 | freebn(b); |
| 1333 | return a; |
| 1334 | } |
| 1335 | |
| 1336 | /* |
| 1337 | * Modular inverse, using Euclid's extended algorithm. |
| 1338 | */ |
| 1339 | Bignum modinv(Bignum number, Bignum modulus) |
| 1340 | { |
| 1341 | Bignum a = copybn(modulus); |
| 1342 | Bignum b = copybn(number); |
| 1343 | Bignum xp = copybn(Zero); |
| 1344 | Bignum x = copybn(One); |
| 1345 | int sign = +1; |
| 1346 | |
| 1347 | while (bignum_cmp(b, One) != 0) { |
| 1348 | Bignum t = newbn(b[0]); |
| 1349 | Bignum q = newbn(a[0]); |
| 1350 | bigdivmod(a, b, t, q); |
| 1351 | while (t[0] > 1 && t[t[0]] == 0) |
| 1352 | t[0]--; |
| 1353 | freebn(a); |
| 1354 | a = b; |
| 1355 | b = t; |
| 1356 | t = xp; |
| 1357 | xp = x; |
| 1358 | x = bigmuladd(q, xp, t); |
| 1359 | sign = -sign; |
| 1360 | freebn(t); |
| 1361 | freebn(q); |
| 1362 | } |
| 1363 | |
| 1364 | freebn(b); |
| 1365 | freebn(a); |
| 1366 | freebn(xp); |
| 1367 | |
| 1368 | /* now we know that sign * x == 1, and that x < modulus */ |
| 1369 | if (sign < 0) { |
| 1370 | /* set a new x to be modulus - x */ |
| 1371 | Bignum newx = newbn(modulus[0]); |
| 1372 | BignumInt carry = 0; |
| 1373 | int maxspot = 1; |
| 1374 | int i; |
| 1375 | |
| 1376 | for (i = 1; i <= (int)newx[0]; i++) { |
| 1377 | BignumInt aword = (i <= (int)modulus[0] ? modulus[i] : 0); |
| 1378 | BignumInt bword = (i <= (int)x[0] ? x[i] : 0); |
| 1379 | newx[i] = aword - bword - carry; |
| 1380 | bword = ~bword; |
| 1381 | carry = carry ? (newx[i] >= bword) : (newx[i] > bword); |
| 1382 | if (newx[i] != 0) |
| 1383 | maxspot = i; |
| 1384 | } |
| 1385 | newx[0] = maxspot; |
| 1386 | freebn(x); |
| 1387 | x = newx; |
| 1388 | } |
| 1389 | |
| 1390 | /* and return. */ |
| 1391 | return x; |
| 1392 | } |
| 1393 | |
| 1394 | /* |
| 1395 | * Render a bignum into decimal. Return a malloced string holding |
| 1396 | * the decimal representation. |
| 1397 | */ |
| 1398 | char *bignum_decimal(Bignum x) |
| 1399 | { |
| 1400 | int ndigits, ndigit; |
| 1401 | int i, iszero; |
| 1402 | BignumDblInt carry; |
| 1403 | char *ret; |
| 1404 | BignumInt *workspace; |
| 1405 | |
| 1406 | /* |
| 1407 | * First, estimate the number of digits. Since log(10)/log(2) |
| 1408 | * is just greater than 93/28 (the joys of continued fraction |
| 1409 | * approximations...) we know that for every 93 bits, we need |
| 1410 | * at most 28 digits. This will tell us how much to malloc. |
| 1411 | * |
| 1412 | * Formally: if x has i bits, that means x is strictly less |
| 1413 | * than 2^i. Since 2 is less than 10^(28/93), this is less than |
| 1414 | * 10^(28i/93). We need an integer power of ten, so we must |
| 1415 | * round up (rounding down might make it less than x again). |
| 1416 | * Therefore if we multiply the bit count by 28/93, rounding |
| 1417 | * up, we will have enough digits. |
| 1418 | * |
| 1419 | * i=0 (i.e., x=0) is an irritating special case. |
| 1420 | */ |
| 1421 | i = bignum_bitcount(x); |
| 1422 | if (!i) |
| 1423 | ndigits = 1; /* x = 0 */ |
| 1424 | else |
| 1425 | ndigits = (28 * i + 92) / 93; /* multiply by 28/93 and round up */ |
| 1426 | ndigits++; /* allow for trailing \0 */ |
| 1427 | ret = snewn(ndigits, char); |
| 1428 | |
| 1429 | /* |
| 1430 | * Now allocate some workspace to hold the binary form as we |
| 1431 | * repeatedly divide it by ten. Initialise this to the |
| 1432 | * big-endian form of the number. |
| 1433 | */ |
| 1434 | workspace = snewn(x[0], BignumInt); |
| 1435 | for (i = 0; i < (int)x[0]; i++) |
| 1436 | workspace[i] = x[x[0] - i]; |
| 1437 | |
| 1438 | /* |
| 1439 | * Next, write the decimal number starting with the last digit. |
| 1440 | * We use ordinary short division, dividing 10 into the |
| 1441 | * workspace. |
| 1442 | */ |
| 1443 | ndigit = ndigits - 1; |
| 1444 | ret[ndigit] = '\0'; |
| 1445 | do { |
| 1446 | iszero = 1; |
| 1447 | carry = 0; |
| 1448 | for (i = 0; i < (int)x[0]; i++) { |
| 1449 | carry = (carry << BIGNUM_INT_BITS) + workspace[i]; |
| 1450 | workspace[i] = (BignumInt) (carry / 10); |
| 1451 | if (workspace[i]) |
| 1452 | iszero = 0; |
| 1453 | carry %= 10; |
| 1454 | } |
| 1455 | ret[--ndigit] = (char) (carry + '0'); |
| 1456 | } while (!iszero); |
| 1457 | |
| 1458 | /* |
| 1459 | * There's a chance we've fallen short of the start of the |
| 1460 | * string. Correct if so. |
| 1461 | */ |
| 1462 | if (ndigit > 0) |
| 1463 | memmove(ret, ret + ndigit, ndigits - ndigit); |
| 1464 | |
| 1465 | /* |
| 1466 | * Done. |
| 1467 | */ |
| 1468 | sfree(workspace); |
| 1469 | return ret; |
| 1470 | } |