| 1 | /* |
| 2 | * This program implements a breadth-first search which |
| 3 | * exhaustively solves the Countdown numbers game, and related |
| 4 | * games with slightly different rule sets such as `Flippo'. |
| 5 | * |
| 6 | * Currently it is simply a standalone command-line utility to |
| 7 | * which you provide a set of numbers and it tells you everything |
| 8 | * it can make together with how many different ways it can be |
| 9 | * made. I would like ultimately to turn it into the generator for |
| 10 | * a Puzzles puzzle, but I haven't even started on writing a |
| 11 | * Puzzles user interface yet. |
| 12 | */ |
| 13 | |
| 14 | /* |
| 15 | * TODO: |
| 16 | * |
| 17 | * - start thinking about difficulty ratings |
| 18 | * + anything involving associative operations will be flagged |
| 19 | * as many-paths because of the associative options (e.g. |
| 20 | * 2*3*4 can be (2*3)*4 or 2*(3*4), or indeed (2*4)*3). This |
| 21 | * is probably a _good_ thing, since those are unusually |
| 22 | * easy. |
| 23 | * + tree-structured calculations ((a*b)/(c+d)) have multiple |
| 24 | * paths because the independent branches of the tree can be |
| 25 | * evaluated in either order, whereas straight-line |
| 26 | * calculations with no branches will be considered easier. |
| 27 | * Can we do anything about this? It's certainly not clear to |
| 28 | * me that tree-structure calculations are _easier_, although |
| 29 | * I'm also not convinced they're harder. |
| 30 | * + I think for a realistic difficulty assessment we must also |
| 31 | * consider the `obviousness' of the arithmetic operations in |
| 32 | * some heuristic sense, and also (in Countdown) how many |
| 33 | * numbers ended up being used. |
| 34 | * - actually try some generations |
| 35 | * - at this point we're probably ready to start on the Puzzles |
| 36 | * integration. |
| 37 | */ |
| 38 | |
| 39 | #include <stdio.h> |
| 40 | #include <limits.h> |
| 41 | #include <assert.h> |
| 42 | |
| 43 | #include "puzzles.h" |
| 44 | #include "tree234.h" |
| 45 | |
| 46 | /* |
| 47 | * To search for numbers we can make, we employ a breadth-first |
| 48 | * search across the space of sets of input numbers. That is, for |
| 49 | * example, we start with the set (3,6,25,50,75,100); we apply |
| 50 | * moves which involve combining two numbers (e.g. adding the 50 |
| 51 | * and the 75 takes us to the set (3,6,25,100,125); and then we see |
| 52 | * if we ever end up with a set containing (say) 952. |
| 53 | * |
| 54 | * If the rules are changed so that all the numbers must be used, |
| 55 | * this is easy to adjust to: we simply see if we end up with a set |
| 56 | * containing _only_ (say) 952. |
| 57 | * |
| 58 | * Obviously, we can vary the rules about permitted arithmetic |
| 59 | * operations simply by altering the set of valid moves in the bfs. |
| 60 | * However, there's one common rule in this sort of puzzle which |
| 61 | * takes a little more thought, and that's _concatenation_. For |
| 62 | * example, if you are given (say) four 4s and required to make 10, |
| 63 | * you are permitted to combine two of the 4s into a 44 to begin |
| 64 | * with, making (44-4)/4 = 10. However, you are generally not |
| 65 | * allowed to concatenate two numbers that _weren't_ both in the |
| 66 | * original input set (you couldn't multiply two 4s to get 16 and |
| 67 | * then concatenate a 4 on to it to make 164), so concatenation is |
| 68 | * not an operation which is valid in all situations. |
| 69 | * |
| 70 | * We could enforce this restriction by storing a flag alongside |
| 71 | * each number indicating whether or not it's an original number; |
| 72 | * the rules being that concatenation of two numbers is only valid |
| 73 | * if they both have the original flag, and that its output _also_ |
| 74 | * has the original flag (so that you can concatenate three 4s into |
| 75 | * a 444), but that applying any other arithmetic operation clears |
| 76 | * the original flag on the output. However, we can get marginally |
| 77 | * simpler than that by observing that since concatenation has to |
| 78 | * happen to a number before any other operation, we can simply |
| 79 | * place all the concatenations at the start of the search. In |
| 80 | * other words, we have a global flag on an entire number _set_ |
| 81 | * which indicates whether we are still permitted to perform |
| 82 | * concatenations; if so, we can concatenate any of the numbers in |
| 83 | * that set. Performing any other operation clears the flag. |
| 84 | */ |
| 85 | |
| 86 | #define SETFLAG_CONCAT 1 /* we can do concatenation */ |
| 87 | |
| 88 | struct sets; |
| 89 | |
| 90 | struct set { |
| 91 | int *numbers; /* rationals stored as n,d pairs */ |
| 92 | short nnumbers; /* # of rationals, so half # of ints */ |
| 93 | short flags; /* SETFLAG_CONCAT only, at present */ |
| 94 | struct set *prev; /* index of ancestor set in set list */ |
| 95 | unsigned char pa, pb, po, pr; /* operation that got here from prev */ |
| 96 | int npaths; /* number of ways to reach this set */ |
| 97 | }; |
| 98 | |
| 99 | struct output { |
| 100 | int number; |
| 101 | struct set *set; |
| 102 | int index; /* which number in the set is it? */ |
| 103 | int npaths; /* number of ways to reach this */ |
| 104 | }; |
| 105 | |
| 106 | #define SETLISTLEN 1024 |
| 107 | #define NUMBERLISTLEN 32768 |
| 108 | #define OUTPUTLISTLEN 1024 |
| 109 | struct operation; |
| 110 | struct sets { |
| 111 | struct set **setlists; |
| 112 | int nsets, nsetlists, setlistsize; |
| 113 | tree234 *settree; |
| 114 | int **numberlists; |
| 115 | int nnumbers, nnumberlists, numberlistsize; |
| 116 | struct output **outputlists; |
| 117 | int noutputs, noutputlists, outputlistsize; |
| 118 | tree234 *outputtree; |
| 119 | const struct operation *const *ops; |
| 120 | }; |
| 121 | |
| 122 | #define OPFLAG_NEEDS_CONCAT 1 |
| 123 | #define OPFLAG_KEEPS_CONCAT 2 |
| 124 | |
| 125 | struct operation { |
| 126 | /* |
| 127 | * Most operations should be shown in the output working, but |
| 128 | * concatenation should not; we just take the result of the |
| 129 | * concatenation and assume that it's obvious how it was |
| 130 | * derived. |
| 131 | */ |
| 132 | int display; |
| 133 | |
| 134 | /* |
| 135 | * Text display of the operator. |
| 136 | */ |
| 137 | char *text; |
| 138 | |
| 139 | /* |
| 140 | * Flags dictating when the operator can be applied. |
| 141 | */ |
| 142 | int flags; |
| 143 | |
| 144 | /* |
| 145 | * Priority of the operator (for avoiding unnecessary |
| 146 | * parentheses when formatting it into a string). |
| 147 | */ |
| 148 | int priority; |
| 149 | |
| 150 | /* |
| 151 | * Associativity of the operator. Bit 0 means we need parens |
| 152 | * when the left operand of one of these operators is another |
| 153 | * instance of it, e.g. (2^3)^4. Bit 1 means we need parens |
| 154 | * when the right operand is another instance of the same |
| 155 | * operator, e.g. 2-(3-4). Thus: |
| 156 | * |
| 157 | * - this field is 0 for a fully associative operator, since |
| 158 | * we never need parens. |
| 159 | * - it's 1 for a right-associative operator. |
| 160 | * - it's 2 for a left-associative operator. |
| 161 | * - it's 3 for a _non_-associative operator (which always |
| 162 | * uses parens just to be sure). |
| 163 | */ |
| 164 | int assoc; |
| 165 | |
| 166 | /* |
| 167 | * Whether the operator is commutative. Saves time in the |
| 168 | * search if we don't have to try it both ways round. |
| 169 | */ |
| 170 | int commutes; |
| 171 | |
| 172 | /* |
| 173 | * Function which implements the operator. Returns TRUE on |
| 174 | * success, FALSE on failure. Takes two rationals and writes |
| 175 | * out a third. |
| 176 | */ |
| 177 | int (*perform)(int *a, int *b, int *output); |
| 178 | }; |
| 179 | |
| 180 | struct rules { |
| 181 | const struct operation *const *ops; |
| 182 | int use_all; |
| 183 | }; |
| 184 | |
| 185 | #define MUL(r, a, b) do { \ |
| 186 | (r) = (a) * (b); \ |
| 187 | if ((b) && (a) && (r) / (b) != (a)) return FALSE; \ |
| 188 | } while (0) |
| 189 | |
| 190 | #define ADD(r, a, b) do { \ |
| 191 | (r) = (a) + (b); \ |
| 192 | if ((a) > 0 && (b) > 0 && (r) < 0) return FALSE; \ |
| 193 | if ((a) < 0 && (b) < 0 && (r) > 0) return FALSE; \ |
| 194 | } while (0) |
| 195 | |
| 196 | #define OUT(output, n, d) do { \ |
| 197 | int g = gcd((n),(d)); \ |
| 198 | if ((d) < 0) g = -g; \ |
| 199 | (output)[0] = (n)/g; \ |
| 200 | (output)[1] = (d)/g; \ |
| 201 | assert((output)[1] > 0); \ |
| 202 | } while (0) |
| 203 | |
| 204 | static int gcd(int x, int y) |
| 205 | { |
| 206 | while (x != 0 && y != 0) { |
| 207 | int t = x; |
| 208 | x = y; |
| 209 | y = t % y; |
| 210 | } |
| 211 | |
| 212 | return abs(x + y); /* i.e. whichever one isn't zero */ |
| 213 | } |
| 214 | |
| 215 | static int perform_add(int *a, int *b, int *output) |
| 216 | { |
| 217 | int at, bt, tn, bn; |
| 218 | /* |
| 219 | * a0/a1 + b0/b1 = (a0*b1 + b0*a1) / (a1*b1) |
| 220 | */ |
| 221 | MUL(at, a[0], b[1]); |
| 222 | MUL(bt, b[0], a[1]); |
| 223 | ADD(tn, at, bt); |
| 224 | MUL(bn, a[1], b[1]); |
| 225 | OUT(output, tn, bn); |
| 226 | return TRUE; |
| 227 | } |
| 228 | |
| 229 | static int perform_sub(int *a, int *b, int *output) |
| 230 | { |
| 231 | int at, bt, tn, bn; |
| 232 | /* |
| 233 | * a0/a1 - b0/b1 = (a0*b1 - b0*a1) / (a1*b1) |
| 234 | */ |
| 235 | MUL(at, a[0], b[1]); |
| 236 | MUL(bt, b[0], a[1]); |
| 237 | ADD(tn, at, -bt); |
| 238 | MUL(bn, a[1], b[1]); |
| 239 | OUT(output, tn, bn); |
| 240 | return TRUE; |
| 241 | } |
| 242 | |
| 243 | static int perform_mul(int *a, int *b, int *output) |
| 244 | { |
| 245 | int tn, bn; |
| 246 | /* |
| 247 | * a0/a1 * b0/b1 = (a0*b0) / (a1*b1) |
| 248 | */ |
| 249 | MUL(tn, a[0], b[0]); |
| 250 | MUL(bn, a[1], b[1]); |
| 251 | OUT(output, tn, bn); |
| 252 | return TRUE; |
| 253 | } |
| 254 | |
| 255 | static int perform_div(int *a, int *b, int *output) |
| 256 | { |
| 257 | int tn, bn; |
| 258 | |
| 259 | /* |
| 260 | * Division by zero is outlawed. |
| 261 | */ |
| 262 | if (b[0] == 0) |
| 263 | return FALSE; |
| 264 | |
| 265 | /* |
| 266 | * a0/a1 / b0/b1 = (a0*b1) / (a1*b0) |
| 267 | */ |
| 268 | MUL(tn, a[0], b[1]); |
| 269 | MUL(bn, a[1], b[0]); |
| 270 | OUT(output, tn, bn); |
| 271 | return TRUE; |
| 272 | } |
| 273 | |
| 274 | static int perform_exact_div(int *a, int *b, int *output) |
| 275 | { |
| 276 | int tn, bn; |
| 277 | |
| 278 | /* |
| 279 | * Division by zero is outlawed. |
| 280 | */ |
| 281 | if (b[0] == 0) |
| 282 | return FALSE; |
| 283 | |
| 284 | /* |
| 285 | * a0/a1 / b0/b1 = (a0*b1) / (a1*b0) |
| 286 | */ |
| 287 | MUL(tn, a[0], b[1]); |
| 288 | MUL(bn, a[1], b[0]); |
| 289 | OUT(output, tn, bn); |
| 290 | |
| 291 | /* |
| 292 | * Exact division means we require the result to be an integer. |
| 293 | */ |
| 294 | return (output[1] == 1); |
| 295 | } |
| 296 | |
| 297 | static int perform_concat(int *a, int *b, int *output) |
| 298 | { |
| 299 | int t1, t2, p10; |
| 300 | |
| 301 | /* |
| 302 | * We can't concatenate anything which isn't an integer. |
| 303 | */ |
| 304 | if (a[1] != 1 || b[1] != 1) |
| 305 | return FALSE; |
| 306 | |
| 307 | /* |
| 308 | * For concatenation, we can safely assume leading zeroes |
| 309 | * aren't an issue. It isn't clear whether they `should' be |
| 310 | * allowed, but it turns out not to matter: concatenating a |
| 311 | * leading zero on to a number in order to harmlessly get rid |
| 312 | * of the zero is never necessary because unwanted zeroes can |
| 313 | * be disposed of by adding them to something instead. So we |
| 314 | * disallow them always. |
| 315 | * |
| 316 | * The only other possibility is that you might want to |
| 317 | * concatenate a leading zero on to something and then |
| 318 | * concatenate another non-zero digit on to _that_ (to make, |
| 319 | * for example, 106); but that's also unnecessary, because you |
| 320 | * can make 106 just as easily by concatenating the 0 on to the |
| 321 | * _end_ of the 1 first. |
| 322 | */ |
| 323 | if (a[0] == 0) |
| 324 | return FALSE; |
| 325 | |
| 326 | /* |
| 327 | * Find the smallest power of ten strictly greater than b. This |
| 328 | * is the power of ten by which we'll multiply a. |
| 329 | * |
| 330 | * Special case: we must multiply a by at least 10, even if b |
| 331 | * is zero. |
| 332 | */ |
| 333 | p10 = 10; |
| 334 | while (p10 <= (INT_MAX/10) && p10 <= b[0]) |
| 335 | p10 *= 10; |
| 336 | if (p10 > INT_MAX/10) |
| 337 | return FALSE; /* integer overflow */ |
| 338 | MUL(t1, p10, a[0]); |
| 339 | ADD(t2, t1, b[0]); |
| 340 | OUT(output, t2, 1); |
| 341 | return TRUE; |
| 342 | } |
| 343 | |
| 344 | const static struct operation op_add = { |
| 345 | TRUE, "+", 0, 10, 0, TRUE, perform_add |
| 346 | }; |
| 347 | const static struct operation op_sub = { |
| 348 | TRUE, "-", 0, 10, 2, FALSE, perform_sub |
| 349 | }; |
| 350 | const static struct operation op_mul = { |
| 351 | TRUE, "*", 0, 20, 0, TRUE, perform_mul |
| 352 | }; |
| 353 | const static struct operation op_div = { |
| 354 | TRUE, "/", 0, 20, 2, FALSE, perform_div |
| 355 | }; |
| 356 | const static struct operation op_xdiv = { |
| 357 | TRUE, "/", 0, 20, 2, FALSE, perform_exact_div |
| 358 | }; |
| 359 | const static struct operation op_concat = { |
| 360 | FALSE, "", OPFLAG_NEEDS_CONCAT | OPFLAG_KEEPS_CONCAT, |
| 361 | 1000, 0, FALSE, perform_concat |
| 362 | }; |
| 363 | |
| 364 | /* |
| 365 | * In Countdown, divisions resulting in fractions are disallowed. |
| 366 | * http://www.askoxford.com/wordgames/countdown/rules/ |
| 367 | */ |
| 368 | const static struct operation *const ops_countdown[] = { |
| 369 | &op_add, &op_mul, &op_sub, &op_xdiv, NULL |
| 370 | }; |
| 371 | const static struct rules rules_countdown = { |
| 372 | ops_countdown, FALSE |
| 373 | }; |
| 374 | |
| 375 | /* |
| 376 | * A slightly different rule set which handles the reasonably well |
| 377 | * known puzzle of making 24 using two 3s and two 8s. For this we |
| 378 | * need rational rather than integer division. |
| 379 | */ |
| 380 | const static struct operation *const ops_3388[] = { |
| 381 | &op_add, &op_mul, &op_sub, &op_div, NULL |
| 382 | }; |
| 383 | const static struct rules rules_3388 = { |
| 384 | ops_3388, TRUE |
| 385 | }; |
| 386 | |
| 387 | /* |
| 388 | * A still more permissive rule set usable for the four-4s problem |
| 389 | * and similar things. Permits concatenation. |
| 390 | */ |
| 391 | const static struct operation *const ops_four4s[] = { |
| 392 | &op_add, &op_mul, &op_sub, &op_div, &op_concat, NULL |
| 393 | }; |
| 394 | const static struct rules rules_four4s = { |
| 395 | ops_four4s, TRUE |
| 396 | }; |
| 397 | |
| 398 | #define ratcmp(a,op,b) ( (long long)(a)[0] * (b)[1] op \ |
| 399 | (long long)(b)[0] * (a)[1] ) |
| 400 | |
| 401 | static int addtoset(struct set *set, int newnumber[2]) |
| 402 | { |
| 403 | int i, j; |
| 404 | |
| 405 | /* Find where we want to insert the new number */ |
| 406 | for (i = 0; i < set->nnumbers && |
| 407 | ratcmp(set->numbers+2*i, <, newnumber); i++); |
| 408 | |
| 409 | /* Move everything else up */ |
| 410 | for (j = set->nnumbers; j > i; j--) { |
| 411 | set->numbers[2*j] = set->numbers[2*j-2]; |
| 412 | set->numbers[2*j+1] = set->numbers[2*j-1]; |
| 413 | } |
| 414 | |
| 415 | /* Insert the new number */ |
| 416 | set->numbers[2*i] = newnumber[0]; |
| 417 | set->numbers[2*i+1] = newnumber[1]; |
| 418 | |
| 419 | set->nnumbers++; |
| 420 | |
| 421 | return i; |
| 422 | } |
| 423 | |
| 424 | #define ensure(array, size, newlen, type) do { \ |
| 425 | if ((newlen) > (size)) { \ |
| 426 | (size) = (newlen) + 512; \ |
| 427 | (array) = sresize((array), (size), type); \ |
| 428 | } \ |
| 429 | } while (0) |
| 430 | |
| 431 | static int setcmp(void *av, void *bv) |
| 432 | { |
| 433 | struct set *a = (struct set *)av; |
| 434 | struct set *b = (struct set *)bv; |
| 435 | int i; |
| 436 | |
| 437 | if (a->nnumbers < b->nnumbers) |
| 438 | return -1; |
| 439 | else if (a->nnumbers > b->nnumbers) |
| 440 | return +1; |
| 441 | |
| 442 | if (a->flags < b->flags) |
| 443 | return -1; |
| 444 | else if (a->flags > b->flags) |
| 445 | return +1; |
| 446 | |
| 447 | for (i = 0; i < a->nnumbers; i++) { |
| 448 | if (ratcmp(a->numbers+2*i, <, b->numbers+2*i)) |
| 449 | return -1; |
| 450 | else if (ratcmp(a->numbers+2*i, >, b->numbers+2*i)) |
| 451 | return +1; |
| 452 | } |
| 453 | |
| 454 | return 0; |
| 455 | } |
| 456 | |
| 457 | static int outputcmp(void *av, void *bv) |
| 458 | { |
| 459 | struct output *a = (struct output *)av; |
| 460 | struct output *b = (struct output *)bv; |
| 461 | |
| 462 | if (a->number < b->number) |
| 463 | return -1; |
| 464 | else if (a->number > b->number) |
| 465 | return +1; |
| 466 | |
| 467 | return 0; |
| 468 | } |
| 469 | |
| 470 | static int outputfindcmp(void *av, void *bv) |
| 471 | { |
| 472 | int *a = (int *)av; |
| 473 | struct output *b = (struct output *)bv; |
| 474 | |
| 475 | if (*a < b->number) |
| 476 | return -1; |
| 477 | else if (*a > b->number) |
| 478 | return +1; |
| 479 | |
| 480 | return 0; |
| 481 | } |
| 482 | |
| 483 | static void addset(struct sets *s, struct set *set, struct set *prev) |
| 484 | { |
| 485 | struct set *s2; |
| 486 | int npaths = (prev ? prev->npaths : 1); |
| 487 | |
| 488 | assert(set == s->setlists[s->nsets / SETLISTLEN] + s->nsets % SETLISTLEN); |
| 489 | s2 = add234(s->settree, set); |
| 490 | if (s2 == set) { |
| 491 | /* |
| 492 | * New set added to the tree. |
| 493 | */ |
| 494 | set->prev = prev; |
| 495 | set->npaths = npaths; |
| 496 | s->nsets++; |
| 497 | s->nnumbers += 2 * set->nnumbers; |
| 498 | } else { |
| 499 | /* |
| 500 | * Rediscovered an existing set. Update its npaths only. |
| 501 | */ |
| 502 | s2->npaths += npaths; |
| 503 | } |
| 504 | } |
| 505 | |
| 506 | static struct set *newset(struct sets *s, int nnumbers, int flags) |
| 507 | { |
| 508 | struct set *sn; |
| 509 | |
| 510 | ensure(s->setlists, s->setlistsize, s->nsets/SETLISTLEN+1, struct set *); |
| 511 | while (s->nsetlists <= s->nsets / SETLISTLEN) |
| 512 | s->setlists[s->nsetlists++] = snewn(SETLISTLEN, struct set); |
| 513 | sn = s->setlists[s->nsets / SETLISTLEN] + s->nsets % SETLISTLEN; |
| 514 | |
| 515 | if (s->nnumbers + nnumbers * 2 > s->nnumberlists * NUMBERLISTLEN) |
| 516 | s->nnumbers = s->nnumberlists * NUMBERLISTLEN; |
| 517 | ensure(s->numberlists, s->numberlistsize, |
| 518 | s->nnumbers/NUMBERLISTLEN+1, int *); |
| 519 | while (s->nnumberlists <= s->nnumbers / NUMBERLISTLEN) |
| 520 | s->numberlists[s->nnumberlists++] = snewn(NUMBERLISTLEN, int); |
| 521 | sn->numbers = s->numberlists[s->nnumbers / NUMBERLISTLEN] + |
| 522 | s->nnumbers % NUMBERLISTLEN; |
| 523 | |
| 524 | /* |
| 525 | * Start the set off empty. |
| 526 | */ |
| 527 | sn->nnumbers = 0; |
| 528 | |
| 529 | sn->flags = flags; |
| 530 | |
| 531 | return sn; |
| 532 | } |
| 533 | |
| 534 | static int addoutput(struct sets *s, struct set *ss, int index, int *n) |
| 535 | { |
| 536 | struct output *o, *o2; |
| 537 | |
| 538 | /* |
| 539 | * Target numbers are always integers. |
| 540 | */ |
| 541 | if (ss->numbers[2*index+1] != 1) |
| 542 | return FALSE; |
| 543 | |
| 544 | ensure(s->outputlists, s->outputlistsize, s->noutputs/OUTPUTLISTLEN+1, |
| 545 | struct output *); |
| 546 | while (s->noutputlists <= s->noutputs / OUTPUTLISTLEN) |
| 547 | s->outputlists[s->noutputlists++] = snewn(OUTPUTLISTLEN, |
| 548 | struct output); |
| 549 | o = s->outputlists[s->noutputs / OUTPUTLISTLEN] + |
| 550 | s->noutputs % OUTPUTLISTLEN; |
| 551 | |
| 552 | o->number = ss->numbers[2*index]; |
| 553 | o->set = ss; |
| 554 | o->index = index; |
| 555 | o->npaths = ss->npaths; |
| 556 | o2 = add234(s->outputtree, o); |
| 557 | if (o2 != o) { |
| 558 | o2->npaths += o->npaths; |
| 559 | } else { |
| 560 | s->noutputs++; |
| 561 | } |
| 562 | *n = o->number; |
| 563 | return TRUE; |
| 564 | } |
| 565 | |
| 566 | static struct sets *do_search(int ninputs, int *inputs, |
| 567 | const struct rules *rules, int *target) |
| 568 | { |
| 569 | struct sets *s; |
| 570 | struct set *sn; |
| 571 | int qpos, i; |
| 572 | const struct operation *const *ops = rules->ops; |
| 573 | |
| 574 | s = snew(struct sets); |
| 575 | s->setlists = NULL; |
| 576 | s->nsets = s->nsetlists = s->setlistsize = 0; |
| 577 | s->numberlists = NULL; |
| 578 | s->nnumbers = s->nnumberlists = s->numberlistsize = 0; |
| 579 | s->outputlists = NULL; |
| 580 | s->noutputs = s->noutputlists = s->outputlistsize = 0; |
| 581 | s->settree = newtree234(setcmp); |
| 582 | s->outputtree = newtree234(outputcmp); |
| 583 | s->ops = ops; |
| 584 | |
| 585 | /* |
| 586 | * Start with the input set. |
| 587 | */ |
| 588 | sn = newset(s, ninputs, SETFLAG_CONCAT); |
| 589 | for (i = 0; i < ninputs; i++) { |
| 590 | int newnumber[2]; |
| 591 | newnumber[0] = inputs[i]; |
| 592 | newnumber[1] = 1; |
| 593 | addtoset(sn, newnumber); |
| 594 | } |
| 595 | addset(s, sn, NULL); |
| 596 | |
| 597 | /* |
| 598 | * Now perform the breadth-first search: keep looping over sets |
| 599 | * until we run out of steam. |
| 600 | */ |
| 601 | qpos = 0; |
| 602 | while (qpos < s->nsets) { |
| 603 | struct set *ss = s->setlists[qpos / SETLISTLEN] + qpos % SETLISTLEN; |
| 604 | struct set *sn; |
| 605 | int i, j, k, m; |
| 606 | |
| 607 | /* |
| 608 | * Record all the valid output numbers in this state. We |
| 609 | * can always do this if there's only one number in the |
| 610 | * state; otherwise, we can only do it if we aren't |
| 611 | * required to use all the numbers in coming to our answer. |
| 612 | */ |
| 613 | if (ss->nnumbers == 1 || !rules->use_all) { |
| 614 | for (i = 0; i < ss->nnumbers; i++) { |
| 615 | int n; |
| 616 | |
| 617 | if (addoutput(s, ss, i, &n) && target && n == *target) |
| 618 | return s; |
| 619 | } |
| 620 | } |
| 621 | |
| 622 | /* |
| 623 | * Try every possible operation from this state. |
| 624 | */ |
| 625 | for (k = 0; ops[k] && ops[k]->perform; k++) { |
| 626 | if ((ops[k]->flags & OPFLAG_NEEDS_CONCAT) && |
| 627 | !(ss->flags & SETFLAG_CONCAT)) |
| 628 | continue; /* can't use this operation here */ |
| 629 | for (i = 0; i < ss->nnumbers; i++) { |
| 630 | for (j = 0; j < ss->nnumbers; j++) { |
| 631 | int n[2]; |
| 632 | |
| 633 | if (i == j) |
| 634 | continue; /* can't combine a number with itself */ |
| 635 | if (i > j && ops[k]->commutes) |
| 636 | continue; /* no need to do this both ways round */ |
| 637 | if (!ops[k]->perform(ss->numbers+2*i, ss->numbers+2*j, n)) |
| 638 | continue; /* operation failed */ |
| 639 | |
| 640 | sn = newset(s, ss->nnumbers-1, ss->flags); |
| 641 | |
| 642 | if (!(ops[k]->flags & OPFLAG_KEEPS_CONCAT)) |
| 643 | sn->flags &= ~SETFLAG_CONCAT; |
| 644 | |
| 645 | for (m = 0; m < ss->nnumbers; m++) { |
| 646 | if (m == i || m == j) |
| 647 | continue; |
| 648 | sn->numbers[2*sn->nnumbers] = ss->numbers[2*m]; |
| 649 | sn->numbers[2*sn->nnumbers + 1] = ss->numbers[2*m + 1]; |
| 650 | sn->nnumbers++; |
| 651 | } |
| 652 | sn->pa = i; |
| 653 | sn->pb = j; |
| 654 | sn->po = k; |
| 655 | sn->pr = addtoset(sn, n); |
| 656 | addset(s, sn, ss); |
| 657 | } |
| 658 | } |
| 659 | } |
| 660 | |
| 661 | qpos++; |
| 662 | } |
| 663 | |
| 664 | return s; |
| 665 | } |
| 666 | |
| 667 | static void free_sets(struct sets *s) |
| 668 | { |
| 669 | int i; |
| 670 | |
| 671 | freetree234(s->settree); |
| 672 | freetree234(s->outputtree); |
| 673 | for (i = 0; i < s->nsetlists; i++) |
| 674 | sfree(s->setlists[i]); |
| 675 | sfree(s->setlists); |
| 676 | for (i = 0; i < s->nnumberlists; i++) |
| 677 | sfree(s->numberlists[i]); |
| 678 | sfree(s->numberlists); |
| 679 | for (i = 0; i < s->noutputlists; i++) |
| 680 | sfree(s->outputlists[i]); |
| 681 | sfree(s->outputlists); |
| 682 | sfree(s); |
| 683 | } |
| 684 | |
| 685 | /* |
| 686 | * Construct a text formula for producing a given output. |
| 687 | */ |
| 688 | void mkstring_recurse(char **str, int *len, |
| 689 | struct sets *s, struct set *ss, int index, |
| 690 | int priority, int assoc, int child) |
| 691 | { |
| 692 | if (ss->prev && index != ss->pr) { |
| 693 | int pi; |
| 694 | |
| 695 | /* |
| 696 | * This number was passed straight down from this set's |
| 697 | * predecessor. Find its index in the previous set and |
| 698 | * recurse to there. |
| 699 | */ |
| 700 | pi = index; |
| 701 | assert(pi != ss->pr); |
| 702 | if (pi > ss->pr) |
| 703 | pi--; |
| 704 | if (pi >= min(ss->pa, ss->pb)) { |
| 705 | pi++; |
| 706 | if (pi >= max(ss->pa, ss->pb)) |
| 707 | pi++; |
| 708 | } |
| 709 | mkstring_recurse(str, len, s, ss->prev, pi, priority, assoc, child); |
| 710 | } else if (ss->prev && index == ss->pr && |
| 711 | s->ops[ss->po]->display) { |
| 712 | /* |
| 713 | * This number was created by a displayed operator in the |
| 714 | * transition from this set to its predecessor. Hence we |
| 715 | * write an open paren, then recurse into the first |
| 716 | * operand, then write the operator, then the second |
| 717 | * operand, and finally close the paren. |
| 718 | */ |
| 719 | char *op; |
| 720 | int parens, thispri, thisassoc; |
| 721 | |
| 722 | /* |
| 723 | * Determine whether we need parentheses. |
| 724 | */ |
| 725 | thispri = s->ops[ss->po]->priority; |
| 726 | thisassoc = s->ops[ss->po]->assoc; |
| 727 | parens = (thispri < priority || |
| 728 | (thispri == priority && (assoc & child))); |
| 729 | |
| 730 | if (parens) { |
| 731 | if (str) |
| 732 | *(*str)++ = '('; |
| 733 | if (len) |
| 734 | (*len)++; |
| 735 | } |
| 736 | mkstring_recurse(str, len, s, ss->prev, ss->pa, thispri, thisassoc, 1); |
| 737 | for (op = s->ops[ss->po]->text; *op; op++) { |
| 738 | if (str) |
| 739 | *(*str)++ = *op; |
| 740 | if (len) |
| 741 | (*len)++; |
| 742 | } |
| 743 | mkstring_recurse(str, len, s, ss->prev, ss->pb, thispri, thisassoc, 2); |
| 744 | if (parens) { |
| 745 | if (str) |
| 746 | *(*str)++ = ')'; |
| 747 | if (len) |
| 748 | (*len)++; |
| 749 | } |
| 750 | } else { |
| 751 | /* |
| 752 | * This number is either an original, or something formed |
| 753 | * by a non-displayed operator (concatenation). Either way, |
| 754 | * we display it as is. |
| 755 | */ |
| 756 | char buf[80], *p; |
| 757 | int blen; |
| 758 | blen = sprintf(buf, "%d", ss->numbers[2*index]); |
| 759 | if (ss->numbers[2*index+1] != 1) |
| 760 | blen += sprintf(buf+blen, "/%d", ss->numbers[2*index+1]); |
| 761 | assert(blen < lenof(buf)); |
| 762 | for (p = buf; *p; p++) { |
| 763 | if (str) |
| 764 | *(*str)++ = *p; |
| 765 | if (len) |
| 766 | (*len)++; |
| 767 | } |
| 768 | } |
| 769 | } |
| 770 | char *mkstring(struct sets *s, struct output *o) |
| 771 | { |
| 772 | int len; |
| 773 | char *str, *p; |
| 774 | |
| 775 | len = 0; |
| 776 | mkstring_recurse(NULL, &len, s, o->set, o->index, 0, 0, 0); |
| 777 | str = snewn(len+1, char); |
| 778 | p = str; |
| 779 | mkstring_recurse(&p, NULL, s, o->set, o->index, 0, 0, 0); |
| 780 | assert(p - str <= len); |
| 781 | *p = '\0'; |
| 782 | return str; |
| 783 | } |
| 784 | |
| 785 | int main(int argc, char **argv) |
| 786 | { |
| 787 | int doing_opts = TRUE; |
| 788 | const struct rules *rules = NULL; |
| 789 | char *pname = argv[0]; |
| 790 | int got_target = FALSE, target = 0; |
| 791 | int numbers[10], nnumbers = 0; |
| 792 | int verbose = FALSE; |
| 793 | int pathcounts = FALSE; |
| 794 | |
| 795 | struct output *o; |
| 796 | struct sets *s; |
| 797 | int i, start, limit; |
| 798 | |
| 799 | while (--argc) { |
| 800 | char *p = *++argv; |
| 801 | int c; |
| 802 | |
| 803 | if (doing_opts && *p == '-') { |
| 804 | p++; |
| 805 | |
| 806 | if (!strcmp(p, "-")) { |
| 807 | doing_opts = FALSE; |
| 808 | continue; |
| 809 | } else while (*p) switch (c = *p++) { |
| 810 | case 'C': |
| 811 | rules = &rules_countdown; |
| 812 | break; |
| 813 | case 'B': |
| 814 | rules = &rules_3388; |
| 815 | break; |
| 816 | case 'D': |
| 817 | rules = &rules_four4s; |
| 818 | break; |
| 819 | case 'v': |
| 820 | verbose = TRUE; |
| 821 | break; |
| 822 | case 'p': |
| 823 | pathcounts = TRUE; |
| 824 | break; |
| 825 | case 't': |
| 826 | { |
| 827 | char *v; |
| 828 | if (*p) { |
| 829 | v = p; |
| 830 | p = NULL; |
| 831 | } else if (--argc) { |
| 832 | v = *++argv; |
| 833 | } else { |
| 834 | fprintf(stderr, "%s: option '-%c' expects an" |
| 835 | " argument\n", pname, c); |
| 836 | return 1; |
| 837 | } |
| 838 | switch (c) { |
| 839 | case 't': |
| 840 | got_target = TRUE; |
| 841 | target = atoi(v); |
| 842 | break; |
| 843 | } |
| 844 | } |
| 845 | break; |
| 846 | default: |
| 847 | fprintf(stderr, "%s: option '-%c' not" |
| 848 | " recognised\n", pname, c); |
| 849 | return 1; |
| 850 | } |
| 851 | } else { |
| 852 | if (nnumbers >= lenof(numbers)) { |
| 853 | fprintf(stderr, "%s: internal limit of %d numbers exceeded\n", |
| 854 | pname, lenof(numbers)); |
| 855 | return 1; |
| 856 | } else { |
| 857 | numbers[nnumbers++] = atoi(p); |
| 858 | } |
| 859 | } |
| 860 | } |
| 861 | |
| 862 | if (!rules) { |
| 863 | fprintf(stderr, "%s: no rule set specified; use -C,-B,-D\n", pname); |
| 864 | return 1; |
| 865 | } |
| 866 | |
| 867 | if (!nnumbers) { |
| 868 | fprintf(stderr, "%s: no input numbers specified\n", pname); |
| 869 | return 1; |
| 870 | } |
| 871 | |
| 872 | s = do_search(nnumbers, numbers, rules, (got_target ? &target : NULL)); |
| 873 | |
| 874 | if (got_target) { |
| 875 | o = findrelpos234(s->outputtree, &target, outputfindcmp, |
| 876 | REL234_LE, &start); |
| 877 | if (!o) |
| 878 | start = -1; |
| 879 | o = findrelpos234(s->outputtree, &target, outputfindcmp, |
| 880 | REL234_GE, &limit); |
| 881 | if (!o) |
| 882 | limit = -1; |
| 883 | assert(start != -1 || limit != -1); |
| 884 | if (start == -1) |
| 885 | start = limit; |
| 886 | else if (limit == -1) |
| 887 | limit = start; |
| 888 | limit++; |
| 889 | } else { |
| 890 | start = 0; |
| 891 | limit = count234(s->outputtree); |
| 892 | } |
| 893 | |
| 894 | for (i = start; i < limit; i++) { |
| 895 | o = index234(s->outputtree, i); |
| 896 | |
| 897 | printf("%d", o->number); |
| 898 | |
| 899 | if (pathcounts) |
| 900 | printf(" [%d]", o->npaths); |
| 901 | |
| 902 | if (got_target || verbose) { |
| 903 | char *p = mkstring(s, o); |
| 904 | printf(" = %s", p); |
| 905 | sfree(p); |
| 906 | } |
| 907 | |
| 908 | printf("\n"); |
| 909 | } |
| 910 | |
| 911 | free_sets(s); |
| 912 | |
| 913 | return 0; |
| 914 | } |