| 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 <string.h> |
| 41 | #include <limits.h> |
| 42 | #include <assert.h> |
| 43 | #include <math.h> |
| 44 | |
| 45 | #include "puzzles.h" |
| 46 | #include "tree234.h" |
| 47 | |
| 48 | /* |
| 49 | * To search for numbers we can make, we employ a breadth-first |
| 50 | * search across the space of sets of input numbers. That is, for |
| 51 | * example, we start with the set (3,6,25,50,75,100); we apply |
| 52 | * moves which involve combining two numbers (e.g. adding the 50 |
| 53 | * and the 75 takes us to the set (3,6,25,100,125); and then we see |
| 54 | * if we ever end up with a set containing (say) 952. |
| 55 | * |
| 56 | * If the rules are changed so that all the numbers must be used, |
| 57 | * this is easy to adjust to: we simply see if we end up with a set |
| 58 | * containing _only_ (say) 952. |
| 59 | * |
| 60 | * Obviously, we can vary the rules about permitted arithmetic |
| 61 | * operations simply by altering the set of valid moves in the bfs. |
| 62 | * However, there's one common rule in this sort of puzzle which |
| 63 | * takes a little more thought, and that's _concatenation_. For |
| 64 | * example, if you are given (say) four 4s and required to make 10, |
| 65 | * you are permitted to combine two of the 4s into a 44 to begin |
| 66 | * with, making (44-4)/4 = 10. However, you are generally not |
| 67 | * allowed to concatenate two numbers that _weren't_ both in the |
| 68 | * original input set (you couldn't multiply two 4s to get 16 and |
| 69 | * then concatenate a 4 on to it to make 164), so concatenation is |
| 70 | * not an operation which is valid in all situations. |
| 71 | * |
| 72 | * We could enforce this restriction by storing a flag alongside |
| 73 | * each number indicating whether or not it's an original number; |
| 74 | * the rules being that concatenation of two numbers is only valid |
| 75 | * if they both have the original flag, and that its output _also_ |
| 76 | * has the original flag (so that you can concatenate three 4s into |
| 77 | * a 444), but that applying any other arithmetic operation clears |
| 78 | * the original flag on the output. However, we can get marginally |
| 79 | * simpler than that by observing that since concatenation has to |
| 80 | * happen to a number before any other operation, we can simply |
| 81 | * place all the concatenations at the start of the search. In |
| 82 | * other words, we have a global flag on an entire number _set_ |
| 83 | * which indicates whether we are still permitted to perform |
| 84 | * concatenations; if so, we can concatenate any of the numbers in |
| 85 | * that set. Performing any other operation clears the flag. |
| 86 | */ |
| 87 | |
| 88 | #define SETFLAG_CONCAT 1 /* we can do concatenation */ |
| 89 | |
| 90 | struct sets; |
| 91 | |
| 92 | struct ancestor { |
| 93 | struct set *prev; /* index of ancestor set in set list */ |
| 94 | unsigned char pa, pb, po, pr; /* operation that got here from prev */ |
| 95 | }; |
| 96 | |
| 97 | struct set { |
| 98 | int *numbers; /* rationals stored as n,d pairs */ |
| 99 | short nnumbers; /* # of rationals, so half # of ints */ |
| 100 | short flags; /* SETFLAG_CONCAT only, at present */ |
| 101 | int npaths; /* number of ways to reach this set */ |
| 102 | struct ancestor a; /* primary ancestor */ |
| 103 | struct ancestor *as; /* further ancestors, if we care */ |
| 104 | int nas, assize; |
| 105 | }; |
| 106 | |
| 107 | struct output { |
| 108 | int number; |
| 109 | struct set *set; |
| 110 | int index; /* which number in the set is it? */ |
| 111 | int npaths; /* number of ways to reach this */ |
| 112 | }; |
| 113 | |
| 114 | #define SETLISTLEN 1024 |
| 115 | #define NUMBERLISTLEN 32768 |
| 116 | #define OUTPUTLISTLEN 1024 |
| 117 | struct operation; |
| 118 | struct sets { |
| 119 | struct set **setlists; |
| 120 | int nsets, nsetlists, setlistsize; |
| 121 | tree234 *settree; |
| 122 | int **numberlists; |
| 123 | int nnumbers, nnumberlists, numberlistsize; |
| 124 | struct output **outputlists; |
| 125 | int noutputs, noutputlists, outputlistsize; |
| 126 | tree234 *outputtree; |
| 127 | const struct operation *const *ops; |
| 128 | }; |
| 129 | |
| 130 | #define OPFLAG_NEEDS_CONCAT 1 |
| 131 | #define OPFLAG_KEEPS_CONCAT 2 |
| 132 | #define OPFLAG_UNARY 4 |
| 133 | #define OPFLAG_UNARYPREFIX 8 |
| 134 | |
| 135 | struct operation { |
| 136 | /* |
| 137 | * Most operations should be shown in the output working, but |
| 138 | * concatenation should not; we just take the result of the |
| 139 | * concatenation and assume that it's obvious how it was |
| 140 | * derived. |
| 141 | */ |
| 142 | int display; |
| 143 | |
| 144 | /* |
| 145 | * Text display of the operator, in expressions and for |
| 146 | * debugging respectively. |
| 147 | */ |
| 148 | char *text, *dbgtext; |
| 149 | |
| 150 | /* |
| 151 | * Flags dictating when the operator can be applied. |
| 152 | */ |
| 153 | int flags; |
| 154 | |
| 155 | /* |
| 156 | * Priority of the operator (for avoiding unnecessary |
| 157 | * parentheses when formatting it into a string). |
| 158 | */ |
| 159 | int priority; |
| 160 | |
| 161 | /* |
| 162 | * Associativity of the operator. Bit 0 means we need parens |
| 163 | * when the left operand of one of these operators is another |
| 164 | * instance of it, e.g. (2^3)^4. Bit 1 means we need parens |
| 165 | * when the right operand is another instance of the same |
| 166 | * operator, e.g. 2-(3-4). Thus: |
| 167 | * |
| 168 | * - this field is 0 for a fully associative operator, since |
| 169 | * we never need parens. |
| 170 | * - it's 1 for a right-associative operator. |
| 171 | * - it's 2 for a left-associative operator. |
| 172 | * - it's 3 for a _non_-associative operator (which always |
| 173 | * uses parens just to be sure). |
| 174 | */ |
| 175 | int assoc; |
| 176 | |
| 177 | /* |
| 178 | * Whether the operator is commutative. Saves time in the |
| 179 | * search if we don't have to try it both ways round. |
| 180 | */ |
| 181 | int commutes; |
| 182 | |
| 183 | /* |
| 184 | * Function which implements the operator. Returns TRUE on |
| 185 | * success, FALSE on failure. Takes two rationals and writes |
| 186 | * out a third. |
| 187 | */ |
| 188 | int (*perform)(int *a, int *b, int *output); |
| 189 | }; |
| 190 | |
| 191 | struct rules { |
| 192 | const struct operation *const *ops; |
| 193 | int use_all; |
| 194 | }; |
| 195 | |
| 196 | #define MUL(r, a, b) do { \ |
| 197 | (r) = (a) * (b); \ |
| 198 | if ((b) && (a) && (r) / (b) != (a)) return FALSE; \ |
| 199 | } while (0) |
| 200 | |
| 201 | #define ADD(r, a, b) do { \ |
| 202 | (r) = (a) + (b); \ |
| 203 | if ((a) > 0 && (b) > 0 && (r) < 0) return FALSE; \ |
| 204 | if ((a) < 0 && (b) < 0 && (r) > 0) return FALSE; \ |
| 205 | } while (0) |
| 206 | |
| 207 | #define OUT(output, n, d) do { \ |
| 208 | int g = gcd((n),(d)); \ |
| 209 | if (g < 0) g = -g; \ |
| 210 | if ((d) < 0) g = -g; \ |
| 211 | if (g == -1 && (n) < -INT_MAX) return FALSE; \ |
| 212 | if (g == -1 && (d) < -INT_MAX) return FALSE; \ |
| 213 | (output)[0] = (n)/g; \ |
| 214 | (output)[1] = (d)/g; \ |
| 215 | assert((output)[1] > 0); \ |
| 216 | } while (0) |
| 217 | |
| 218 | static int gcd(int x, int y) |
| 219 | { |
| 220 | while (x != 0 && y != 0) { |
| 221 | int t = x; |
| 222 | x = y; |
| 223 | y = t % y; |
| 224 | } |
| 225 | |
| 226 | return abs(x + y); /* i.e. whichever one isn't zero */ |
| 227 | } |
| 228 | |
| 229 | static int perform_add(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_sub(int *a, int *b, int *output) |
| 244 | { |
| 245 | int at, bt, tn, bn; |
| 246 | /* |
| 247 | * a0/a1 - b0/b1 = (a0*b1 - b0*a1) / (a1*b1) |
| 248 | */ |
| 249 | MUL(at, a[0], b[1]); |
| 250 | MUL(bt, b[0], a[1]); |
| 251 | ADD(tn, at, -bt); |
| 252 | MUL(bn, a[1], b[1]); |
| 253 | OUT(output, tn, bn); |
| 254 | return TRUE; |
| 255 | } |
| 256 | |
| 257 | static int perform_mul(int *a, int *b, int *output) |
| 258 | { |
| 259 | int tn, bn; |
| 260 | /* |
| 261 | * a0/a1 * b0/b1 = (a0*b0) / (a1*b1) |
| 262 | */ |
| 263 | MUL(tn, a[0], b[0]); |
| 264 | MUL(bn, a[1], b[1]); |
| 265 | OUT(output, tn, bn); |
| 266 | return TRUE; |
| 267 | } |
| 268 | |
| 269 | static int perform_div(int *a, int *b, int *output) |
| 270 | { |
| 271 | int tn, bn; |
| 272 | |
| 273 | /* |
| 274 | * Division by zero is outlawed. |
| 275 | */ |
| 276 | if (b[0] == 0) |
| 277 | return FALSE; |
| 278 | |
| 279 | /* |
| 280 | * a0/a1 / b0/b1 = (a0*b1) / (a1*b0) |
| 281 | */ |
| 282 | MUL(tn, a[0], b[1]); |
| 283 | MUL(bn, a[1], b[0]); |
| 284 | OUT(output, tn, bn); |
| 285 | return TRUE; |
| 286 | } |
| 287 | |
| 288 | static int perform_exact_div(int *a, int *b, int *output) |
| 289 | { |
| 290 | int tn, bn; |
| 291 | |
| 292 | /* |
| 293 | * Division by zero is outlawed. |
| 294 | */ |
| 295 | if (b[0] == 0) |
| 296 | return FALSE; |
| 297 | |
| 298 | /* |
| 299 | * a0/a1 / b0/b1 = (a0*b1) / (a1*b0) |
| 300 | */ |
| 301 | MUL(tn, a[0], b[1]); |
| 302 | MUL(bn, a[1], b[0]); |
| 303 | OUT(output, tn, bn); |
| 304 | |
| 305 | /* |
| 306 | * Exact division means we require the result to be an integer. |
| 307 | */ |
| 308 | return (output[1] == 1); |
| 309 | } |
| 310 | |
| 311 | static int max_p10(int n, int *p10_r) |
| 312 | { |
| 313 | /* |
| 314 | * Find the smallest power of ten strictly greater than n. |
| 315 | * |
| 316 | * Special case: we must return at least 10, even if n is |
| 317 | * zero. (This is because this function is used for finding |
| 318 | * the power of ten by which to multiply a number being |
| 319 | * concatenated to the front of n, and concatenating 1 to 0 |
| 320 | * should yield 10 and not 1.) |
| 321 | */ |
| 322 | int p10 = 10; |
| 323 | while (p10 <= (INT_MAX/10) && p10 <= n) |
| 324 | p10 *= 10; |
| 325 | if (p10 > INT_MAX/10) |
| 326 | return FALSE; /* integer overflow */ |
| 327 | *p10_r = p10; |
| 328 | return TRUE; |
| 329 | } |
| 330 | |
| 331 | static int perform_concat(int *a, int *b, int *output) |
| 332 | { |
| 333 | int t1, t2, p10; |
| 334 | |
| 335 | /* |
| 336 | * We can't concatenate anything which isn't a non-negative |
| 337 | * integer. |
| 338 | */ |
| 339 | if (a[1] != 1 || b[1] != 1 || a[0] < 0 || b[0] < 0) |
| 340 | return FALSE; |
| 341 | |
| 342 | /* |
| 343 | * For concatenation, we can safely assume leading zeroes |
| 344 | * aren't an issue. It isn't clear whether they `should' be |
| 345 | * allowed, but it turns out not to matter: concatenating a |
| 346 | * leading zero on to a number in order to harmlessly get rid |
| 347 | * of the zero is never necessary because unwanted zeroes can |
| 348 | * be disposed of by adding them to something instead. So we |
| 349 | * disallow them always. |
| 350 | * |
| 351 | * The only other possibility is that you might want to |
| 352 | * concatenate a leading zero on to something and then |
| 353 | * concatenate another non-zero digit on to _that_ (to make, |
| 354 | * for example, 106); but that's also unnecessary, because you |
| 355 | * can make 106 just as easily by concatenating the 0 on to the |
| 356 | * _end_ of the 1 first. |
| 357 | */ |
| 358 | if (a[0] == 0) |
| 359 | return FALSE; |
| 360 | |
| 361 | if (!max_p10(b[0], &p10)) return FALSE; |
| 362 | |
| 363 | MUL(t1, p10, a[0]); |
| 364 | ADD(t2, t1, b[0]); |
| 365 | OUT(output, t2, 1); |
| 366 | return TRUE; |
| 367 | } |
| 368 | |
| 369 | #define IPOW(ret, x, y) do { \ |
| 370 | int ipow_limit = (y); \ |
| 371 | if ((x) == 1 || (x) == 0) ipow_limit = 1; \ |
| 372 | else if ((x) == -1) ipow_limit &= 1; \ |
| 373 | (ret) = 1; \ |
| 374 | while (ipow_limit-- > 0) { \ |
| 375 | int tmp; \ |
| 376 | MUL(tmp, ret, x); \ |
| 377 | ret = tmp; \ |
| 378 | } \ |
| 379 | } while (0) |
| 380 | |
| 381 | static int perform_exp(int *a, int *b, int *output) |
| 382 | { |
| 383 | int an, ad, xn, xd; |
| 384 | |
| 385 | /* |
| 386 | * Exponentiation is permitted if the result is rational. This |
| 387 | * means that: |
| 388 | * |
| 389 | * - first we see whether we can take the (denominator-of-b)th |
| 390 | * root of a and get a rational; if not, we give up. |
| 391 | * |
| 392 | * - then we do take that root of a |
| 393 | * |
| 394 | * - then we multiply by itself (numerator-of-b) times. |
| 395 | */ |
| 396 | if (b[1] > 1) { |
| 397 | an = (int)(0.5 + pow(a[0], 1.0/b[1])); |
| 398 | ad = (int)(0.5 + pow(a[1], 1.0/b[1])); |
| 399 | IPOW(xn, an, b[1]); |
| 400 | IPOW(xd, ad, b[1]); |
| 401 | if (xn != a[0] || xd != a[1]) |
| 402 | return FALSE; |
| 403 | } else { |
| 404 | an = a[0]; |
| 405 | ad = a[1]; |
| 406 | } |
| 407 | if (b[0] >= 0) { |
| 408 | IPOW(xn, an, b[0]); |
| 409 | IPOW(xd, ad, b[0]); |
| 410 | } else { |
| 411 | IPOW(xd, an, -b[0]); |
| 412 | IPOW(xn, ad, -b[0]); |
| 413 | } |
| 414 | if (xd == 0) |
| 415 | return FALSE; |
| 416 | |
| 417 | OUT(output, xn, xd); |
| 418 | return TRUE; |
| 419 | } |
| 420 | |
| 421 | static int perform_factorial(int *a, int *b, int *output) |
| 422 | { |
| 423 | int ret, t, i; |
| 424 | |
| 425 | /* |
| 426 | * Factorials of non-negative integers are permitted. |
| 427 | */ |
| 428 | if (a[1] != 1 || a[0] < 0) |
| 429 | return FALSE; |
| 430 | |
| 431 | /* |
| 432 | * However, a special case: we don't take a factorial of |
| 433 | * anything which would thereby remain the same. |
| 434 | */ |
| 435 | if (a[0] == 1 || a[0] == 2) |
| 436 | return FALSE; |
| 437 | |
| 438 | ret = 1; |
| 439 | for (i = 1; i <= a[0]; i++) { |
| 440 | MUL(t, ret, i); |
| 441 | ret = t; |
| 442 | } |
| 443 | |
| 444 | OUT(output, ret, 1); |
| 445 | return TRUE; |
| 446 | } |
| 447 | |
| 448 | static int perform_decimal(int *a, int *b, int *output) |
| 449 | { |
| 450 | int p10; |
| 451 | |
| 452 | /* |
| 453 | * Add a decimal digit to the front of a number; |
| 454 | * fail if it's not an integer. |
| 455 | * So, 1 --> 0.1, 15 --> 0.15, |
| 456 | * or, rather, 1 --> 1/10, 15 --> 15/100, |
| 457 | * x --> x / (smallest power of 10 > than x) |
| 458 | * |
| 459 | */ |
| 460 | if (a[1] != 1) return FALSE; |
| 461 | |
| 462 | if (!max_p10(a[0], &p10)) return FALSE; |
| 463 | |
| 464 | OUT(output, a[0], p10); |
| 465 | return TRUE; |
| 466 | } |
| 467 | |
| 468 | static int perform_recur(int *a, int *b, int *output) |
| 469 | { |
| 470 | int p10, tn, bn; |
| 471 | |
| 472 | /* |
| 473 | * This converts a number like .4 to .44444..., or .45 to .45454... |
| 474 | * The input number must be -1 < a < 1. |
| 475 | * |
| 476 | * Calculate the smallest power of 10 that divides the denominator exactly, |
| 477 | * returning if no such power of 10 exists. Then multiply the numerator |
| 478 | * up accordingly, and the new denominator becomes that power of 10 - 1. |
| 479 | */ |
| 480 | if (abs(a[0]) >= abs(a[1])) return FALSE; /* -1 < a < 1 */ |
| 481 | |
| 482 | p10 = 10; |
| 483 | while (p10 <= (INT_MAX/10)) { |
| 484 | if ((a[1] <= p10) && (p10 % a[1]) == 0) goto found; |
| 485 | p10 *= 10; |
| 486 | } |
| 487 | return FALSE; |
| 488 | found: |
| 489 | tn = a[0] * (p10 / a[1]); |
| 490 | bn = p10 - 1; |
| 491 | |
| 492 | OUT(output, tn, bn); |
| 493 | return TRUE; |
| 494 | } |
| 495 | |
| 496 | static int perform_root(int *a, int *b, int *output) |
| 497 | { |
| 498 | /* |
| 499 | * A root B is: 1 iff a == 0 |
| 500 | * B ^ (1/A) otherwise |
| 501 | */ |
| 502 | int ainv[2], res; |
| 503 | |
| 504 | if (a[0] == 0) { |
| 505 | OUT(output, 1, 1); |
| 506 | return TRUE; |
| 507 | } |
| 508 | |
| 509 | OUT(ainv, a[1], a[0]); |
| 510 | res = perform_exp(b, ainv, output); |
| 511 | return res; |
| 512 | } |
| 513 | |
| 514 | const static struct operation op_add = { |
| 515 | TRUE, "+", "+", 0, 10, 0, TRUE, perform_add |
| 516 | }; |
| 517 | const static struct operation op_sub = { |
| 518 | TRUE, "-", "-", 0, 10, 2, FALSE, perform_sub |
| 519 | }; |
| 520 | const static struct operation op_mul = { |
| 521 | TRUE, "*", "*", 0, 20, 0, TRUE, perform_mul |
| 522 | }; |
| 523 | const static struct operation op_div = { |
| 524 | TRUE, "/", "/", 0, 20, 2, FALSE, perform_div |
| 525 | }; |
| 526 | const static struct operation op_xdiv = { |
| 527 | TRUE, "/", "/", 0, 20, 2, FALSE, perform_exact_div |
| 528 | }; |
| 529 | const static struct operation op_concat = { |
| 530 | FALSE, "", "concat", OPFLAG_NEEDS_CONCAT | OPFLAG_KEEPS_CONCAT, |
| 531 | 1000, 0, FALSE, perform_concat |
| 532 | }; |
| 533 | const static struct operation op_exp = { |
| 534 | TRUE, "^", "^", 0, 30, 1, FALSE, perform_exp |
| 535 | }; |
| 536 | const static struct operation op_factorial = { |
| 537 | TRUE, "!", "!", OPFLAG_UNARY, 40, 0, FALSE, perform_factorial |
| 538 | }; |
| 539 | const static struct operation op_decimal = { |
| 540 | TRUE, ".", ".", OPFLAG_UNARY | OPFLAG_UNARYPREFIX | OPFLAG_NEEDS_CONCAT | OPFLAG_KEEPS_CONCAT, 50, 0, FALSE, perform_decimal |
| 541 | }; |
| 542 | const static struct operation op_recur = { |
| 543 | TRUE, "...", "recur", OPFLAG_UNARY | OPFLAG_NEEDS_CONCAT, 45, 2, FALSE, perform_recur |
| 544 | }; |
| 545 | const static struct operation op_root = { |
| 546 | TRUE, "v~", "root", 0, 30, 1, FALSE, perform_root |
| 547 | }; |
| 548 | |
| 549 | /* |
| 550 | * In Countdown, divisions resulting in fractions are disallowed. |
| 551 | * http://www.askoxford.com/wordgames/countdown/rules/ |
| 552 | */ |
| 553 | const static struct operation *const ops_countdown[] = { |
| 554 | &op_add, &op_mul, &op_sub, &op_xdiv, NULL |
| 555 | }; |
| 556 | const static struct rules rules_countdown = { |
| 557 | ops_countdown, FALSE |
| 558 | }; |
| 559 | |
| 560 | /* |
| 561 | * A slightly different rule set which handles the reasonably well |
| 562 | * known puzzle of making 24 using two 3s and two 8s. For this we |
| 563 | * need rational rather than integer division. |
| 564 | */ |
| 565 | const static struct operation *const ops_3388[] = { |
| 566 | &op_add, &op_mul, &op_sub, &op_div, NULL |
| 567 | }; |
| 568 | const static struct rules rules_3388 = { |
| 569 | ops_3388, TRUE |
| 570 | }; |
| 571 | |
| 572 | /* |
| 573 | * A still more permissive rule set usable for the four-4s problem |
| 574 | * and similar things. Permits concatenation. |
| 575 | */ |
| 576 | const static struct operation *const ops_four4s[] = { |
| 577 | &op_add, &op_mul, &op_sub, &op_div, &op_concat, NULL |
| 578 | }; |
| 579 | const static struct rules rules_four4s = { |
| 580 | ops_four4s, TRUE |
| 581 | }; |
| 582 | |
| 583 | /* |
| 584 | * The most permissive ruleset I can think of. Permits |
| 585 | * exponentiation, and also silly unary operators like factorials. |
| 586 | */ |
| 587 | const static struct operation *const ops_anythinggoes[] = { |
| 588 | &op_add, &op_mul, &op_sub, &op_div, &op_concat, &op_exp, &op_factorial, |
| 589 | &op_decimal, &op_recur, &op_root, NULL |
| 590 | }; |
| 591 | const static struct rules rules_anythinggoes = { |
| 592 | ops_anythinggoes, TRUE |
| 593 | }; |
| 594 | |
| 595 | #define ratcmp(a,op,b) ( (long long)(a)[0] * (b)[1] op \ |
| 596 | (long long)(b)[0] * (a)[1] ) |
| 597 | |
| 598 | static int addtoset(struct set *set, int newnumber[2]) |
| 599 | { |
| 600 | int i, j; |
| 601 | |
| 602 | /* Find where we want to insert the new number */ |
| 603 | for (i = 0; i < set->nnumbers && |
| 604 | ratcmp(set->numbers+2*i, <, newnumber); i++); |
| 605 | |
| 606 | /* Move everything else up */ |
| 607 | for (j = set->nnumbers; j > i; j--) { |
| 608 | set->numbers[2*j] = set->numbers[2*j-2]; |
| 609 | set->numbers[2*j+1] = set->numbers[2*j-1]; |
| 610 | } |
| 611 | |
| 612 | /* Insert the new number */ |
| 613 | set->numbers[2*i] = newnumber[0]; |
| 614 | set->numbers[2*i+1] = newnumber[1]; |
| 615 | |
| 616 | set->nnumbers++; |
| 617 | |
| 618 | return i; |
| 619 | } |
| 620 | |
| 621 | #define ensure(array, size, newlen, type) do { \ |
| 622 | if ((newlen) > (size)) { \ |
| 623 | (size) = (newlen) + 512; \ |
| 624 | (array) = sresize((array), (size), type); \ |
| 625 | } \ |
| 626 | } while (0) |
| 627 | |
| 628 | static int setcmp(void *av, void *bv) |
| 629 | { |
| 630 | struct set *a = (struct set *)av; |
| 631 | struct set *b = (struct set *)bv; |
| 632 | int i; |
| 633 | |
| 634 | if (a->nnumbers < b->nnumbers) |
| 635 | return -1; |
| 636 | else if (a->nnumbers > b->nnumbers) |
| 637 | return +1; |
| 638 | |
| 639 | if (a->flags < b->flags) |
| 640 | return -1; |
| 641 | else if (a->flags > b->flags) |
| 642 | return +1; |
| 643 | |
| 644 | for (i = 0; i < a->nnumbers; i++) { |
| 645 | if (ratcmp(a->numbers+2*i, <, b->numbers+2*i)) |
| 646 | return -1; |
| 647 | else if (ratcmp(a->numbers+2*i, >, b->numbers+2*i)) |
| 648 | return +1; |
| 649 | } |
| 650 | |
| 651 | return 0; |
| 652 | } |
| 653 | |
| 654 | static int outputcmp(void *av, void *bv) |
| 655 | { |
| 656 | struct output *a = (struct output *)av; |
| 657 | struct output *b = (struct output *)bv; |
| 658 | |
| 659 | if (a->number < b->number) |
| 660 | return -1; |
| 661 | else if (a->number > b->number) |
| 662 | return +1; |
| 663 | |
| 664 | return 0; |
| 665 | } |
| 666 | |
| 667 | static int outputfindcmp(void *av, void *bv) |
| 668 | { |
| 669 | int *a = (int *)av; |
| 670 | struct output *b = (struct output *)bv; |
| 671 | |
| 672 | if (*a < b->number) |
| 673 | return -1; |
| 674 | else if (*a > b->number) |
| 675 | return +1; |
| 676 | |
| 677 | return 0; |
| 678 | } |
| 679 | |
| 680 | static void addset(struct sets *s, struct set *set, int multiple, |
| 681 | struct set *prev, int pa, int po, int pb, int pr) |
| 682 | { |
| 683 | struct set *s2; |
| 684 | int npaths = (prev ? prev->npaths : 1); |
| 685 | |
| 686 | assert(set == s->setlists[s->nsets / SETLISTLEN] + s->nsets % SETLISTLEN); |
| 687 | s2 = add234(s->settree, set); |
| 688 | if (s2 == set) { |
| 689 | /* |
| 690 | * New set added to the tree. |
| 691 | */ |
| 692 | set->a.prev = prev; |
| 693 | set->a.pa = pa; |
| 694 | set->a.po = po; |
| 695 | set->a.pb = pb; |
| 696 | set->a.pr = pr; |
| 697 | set->npaths = npaths; |
| 698 | s->nsets++; |
| 699 | s->nnumbers += 2 * set->nnumbers; |
| 700 | set->as = NULL; |
| 701 | set->nas = set->assize = 0; |
| 702 | } else { |
| 703 | /* |
| 704 | * Rediscovered an existing set. Update its npaths. |
| 705 | */ |
| 706 | s2->npaths += npaths; |
| 707 | /* |
| 708 | * And optionally enter it as an additional ancestor. |
| 709 | */ |
| 710 | if (multiple) { |
| 711 | if (s2->nas >= s2->assize) { |
| 712 | s2->assize = s2->nas * 3 / 2 + 4; |
| 713 | s2->as = sresize(s2->as, s2->assize, struct ancestor); |
| 714 | } |
| 715 | s2->as[s2->nas].prev = prev; |
| 716 | s2->as[s2->nas].pa = pa; |
| 717 | s2->as[s2->nas].po = po; |
| 718 | s2->as[s2->nas].pb = pb; |
| 719 | s2->as[s2->nas].pr = pr; |
| 720 | s2->nas++; |
| 721 | } |
| 722 | } |
| 723 | } |
| 724 | |
| 725 | static struct set *newset(struct sets *s, int nnumbers, int flags) |
| 726 | { |
| 727 | struct set *sn; |
| 728 | |
| 729 | ensure(s->setlists, s->setlistsize, s->nsets/SETLISTLEN+1, struct set *); |
| 730 | while (s->nsetlists <= s->nsets / SETLISTLEN) |
| 731 | s->setlists[s->nsetlists++] = snewn(SETLISTLEN, struct set); |
| 732 | sn = s->setlists[s->nsets / SETLISTLEN] + s->nsets % SETLISTLEN; |
| 733 | |
| 734 | if (s->nnumbers + nnumbers * 2 > s->nnumberlists * NUMBERLISTLEN) |
| 735 | s->nnumbers = s->nnumberlists * NUMBERLISTLEN; |
| 736 | ensure(s->numberlists, s->numberlistsize, |
| 737 | s->nnumbers/NUMBERLISTLEN+1, int *); |
| 738 | while (s->nnumberlists <= s->nnumbers / NUMBERLISTLEN) |
| 739 | s->numberlists[s->nnumberlists++] = snewn(NUMBERLISTLEN, int); |
| 740 | sn->numbers = s->numberlists[s->nnumbers / NUMBERLISTLEN] + |
| 741 | s->nnumbers % NUMBERLISTLEN; |
| 742 | |
| 743 | /* |
| 744 | * Start the set off empty. |
| 745 | */ |
| 746 | sn->nnumbers = 0; |
| 747 | |
| 748 | sn->flags = flags; |
| 749 | |
| 750 | return sn; |
| 751 | } |
| 752 | |
| 753 | static int addoutput(struct sets *s, struct set *ss, int index, int *n) |
| 754 | { |
| 755 | struct output *o, *o2; |
| 756 | |
| 757 | /* |
| 758 | * Target numbers are always integers. |
| 759 | */ |
| 760 | if (ss->numbers[2*index+1] != 1) |
| 761 | return FALSE; |
| 762 | |
| 763 | ensure(s->outputlists, s->outputlistsize, s->noutputs/OUTPUTLISTLEN+1, |
| 764 | struct output *); |
| 765 | while (s->noutputlists <= s->noutputs / OUTPUTLISTLEN) |
| 766 | s->outputlists[s->noutputlists++] = snewn(OUTPUTLISTLEN, |
| 767 | struct output); |
| 768 | o = s->outputlists[s->noutputs / OUTPUTLISTLEN] + |
| 769 | s->noutputs % OUTPUTLISTLEN; |
| 770 | |
| 771 | o->number = ss->numbers[2*index]; |
| 772 | o->set = ss; |
| 773 | o->index = index; |
| 774 | o->npaths = ss->npaths; |
| 775 | o2 = add234(s->outputtree, o); |
| 776 | if (o2 != o) { |
| 777 | o2->npaths += o->npaths; |
| 778 | } else { |
| 779 | s->noutputs++; |
| 780 | } |
| 781 | *n = o->number; |
| 782 | return TRUE; |
| 783 | } |
| 784 | |
| 785 | static struct sets *do_search(int ninputs, int *inputs, |
| 786 | const struct rules *rules, int *target, |
| 787 | int debug, int multiple) |
| 788 | { |
| 789 | struct sets *s; |
| 790 | struct set *sn; |
| 791 | int qpos, i; |
| 792 | const struct operation *const *ops = rules->ops; |
| 793 | |
| 794 | s = snew(struct sets); |
| 795 | s->setlists = NULL; |
| 796 | s->nsets = s->nsetlists = s->setlistsize = 0; |
| 797 | s->numberlists = NULL; |
| 798 | s->nnumbers = s->nnumberlists = s->numberlistsize = 0; |
| 799 | s->outputlists = NULL; |
| 800 | s->noutputs = s->noutputlists = s->outputlistsize = 0; |
| 801 | s->settree = newtree234(setcmp); |
| 802 | s->outputtree = newtree234(outputcmp); |
| 803 | s->ops = ops; |
| 804 | |
| 805 | /* |
| 806 | * Start with the input set. |
| 807 | */ |
| 808 | sn = newset(s, ninputs, SETFLAG_CONCAT); |
| 809 | for (i = 0; i < ninputs; i++) { |
| 810 | int newnumber[2]; |
| 811 | newnumber[0] = inputs[i]; |
| 812 | newnumber[1] = 1; |
| 813 | addtoset(sn, newnumber); |
| 814 | } |
| 815 | addset(s, sn, multiple, NULL, 0, 0, 0, 0); |
| 816 | |
| 817 | /* |
| 818 | * Now perform the breadth-first search: keep looping over sets |
| 819 | * until we run out of steam. |
| 820 | */ |
| 821 | qpos = 0; |
| 822 | while (qpos < s->nsets) { |
| 823 | struct set *ss = s->setlists[qpos / SETLISTLEN] + qpos % SETLISTLEN; |
| 824 | struct set *sn; |
| 825 | int i, j, k, m; |
| 826 | |
| 827 | if (debug) { |
| 828 | int i; |
| 829 | printf("processing set:"); |
| 830 | for (i = 0; i < ss->nnumbers; i++) { |
| 831 | printf(" %d", ss->numbers[2*i]); |
| 832 | if (ss->numbers[2*i+1] != 1) |
| 833 | printf("/%d", ss->numbers[2*i+1]); |
| 834 | } |
| 835 | printf("\n"); |
| 836 | } |
| 837 | |
| 838 | /* |
| 839 | * Record all the valid output numbers in this state. We |
| 840 | * can always do this if there's only one number in the |
| 841 | * state; otherwise, we can only do it if we aren't |
| 842 | * required to use all the numbers in coming to our answer. |
| 843 | */ |
| 844 | if (ss->nnumbers == 1 || !rules->use_all) { |
| 845 | for (i = 0; i < ss->nnumbers; i++) { |
| 846 | int n; |
| 847 | |
| 848 | if (addoutput(s, ss, i, &n) && target && n == *target) |
| 849 | return s; |
| 850 | } |
| 851 | } |
| 852 | |
| 853 | /* |
| 854 | * Try every possible operation from this state. |
| 855 | */ |
| 856 | for (k = 0; ops[k] && ops[k]->perform; k++) { |
| 857 | if ((ops[k]->flags & OPFLAG_NEEDS_CONCAT) && |
| 858 | !(ss->flags & SETFLAG_CONCAT)) |
| 859 | continue; /* can't use this operation here */ |
| 860 | for (i = 0; i < ss->nnumbers; i++) { |
| 861 | int jlimit = (ops[k]->flags & OPFLAG_UNARY ? 1 : ss->nnumbers); |
| 862 | for (j = 0; j < jlimit; j++) { |
| 863 | int n[2]; |
| 864 | int pa, po, pb, pr; |
| 865 | |
| 866 | if (!(ops[k]->flags & OPFLAG_UNARY)) { |
| 867 | if (i == j) |
| 868 | continue; /* can't combine a number with itself */ |
| 869 | if (i > j && ops[k]->commutes) |
| 870 | continue; /* no need to do this both ways round */ |
| 871 | } |
| 872 | if (!ops[k]->perform(ss->numbers+2*i, ss->numbers+2*j, n)) |
| 873 | continue; /* operation failed */ |
| 874 | |
| 875 | sn = newset(s, ss->nnumbers-1, ss->flags); |
| 876 | |
| 877 | if (!(ops[k]->flags & OPFLAG_KEEPS_CONCAT)) |
| 878 | sn->flags &= ~SETFLAG_CONCAT; |
| 879 | |
| 880 | for (m = 0; m < ss->nnumbers; m++) { |
| 881 | if (m == i || (!(ops[k]->flags & OPFLAG_UNARY) && |
| 882 | m == j)) |
| 883 | continue; |
| 884 | sn->numbers[2*sn->nnumbers] = ss->numbers[2*m]; |
| 885 | sn->numbers[2*sn->nnumbers + 1] = ss->numbers[2*m + 1]; |
| 886 | sn->nnumbers++; |
| 887 | } |
| 888 | pa = i; |
| 889 | if (ops[k]->flags & OPFLAG_UNARY) |
| 890 | pb = sn->nnumbers+10; |
| 891 | else |
| 892 | pb = j; |
| 893 | po = k; |
| 894 | pr = addtoset(sn, n); |
| 895 | addset(s, sn, multiple, ss, pa, po, pb, pr); |
| 896 | if (debug) { |
| 897 | int i; |
| 898 | if (ops[k]->flags & OPFLAG_UNARYPREFIX) |
| 899 | printf(" %s %d ->", ops[po]->dbgtext, pa); |
| 900 | else if (ops[k]->flags & OPFLAG_UNARY) |
| 901 | printf(" %d %s ->", pa, ops[po]->dbgtext); |
| 902 | else |
| 903 | printf(" %d %s %d ->", pa, ops[po]->dbgtext, pb); |
| 904 | for (i = 0; i < sn->nnumbers; i++) { |
| 905 | printf(" %d", sn->numbers[2*i]); |
| 906 | if (sn->numbers[2*i+1] != 1) |
| 907 | printf("/%d", sn->numbers[2*i+1]); |
| 908 | } |
| 909 | printf("\n"); |
| 910 | } |
| 911 | } |
| 912 | } |
| 913 | } |
| 914 | |
| 915 | qpos++; |
| 916 | } |
| 917 | |
| 918 | return s; |
| 919 | } |
| 920 | |
| 921 | static void free_sets(struct sets *s) |
| 922 | { |
| 923 | int i; |
| 924 | |
| 925 | freetree234(s->settree); |
| 926 | freetree234(s->outputtree); |
| 927 | for (i = 0; i < s->nsetlists; i++) |
| 928 | sfree(s->setlists[i]); |
| 929 | sfree(s->setlists); |
| 930 | for (i = 0; i < s->nnumberlists; i++) |
| 931 | sfree(s->numberlists[i]); |
| 932 | sfree(s->numberlists); |
| 933 | for (i = 0; i < s->noutputlists; i++) |
| 934 | sfree(s->outputlists[i]); |
| 935 | sfree(s->outputlists); |
| 936 | sfree(s); |
| 937 | } |
| 938 | |
| 939 | /* |
| 940 | * Print a text formula for producing a given output. |
| 941 | */ |
| 942 | void print_recurse(struct sets *s, struct set *ss, int pathindex, int index, |
| 943 | int priority, int assoc, int child); |
| 944 | void print_recurse_inner(struct sets *s, struct set *ss, |
| 945 | struct ancestor *a, int pathindex, int index, |
| 946 | int priority, int assoc, int child) |
| 947 | { |
| 948 | if (a->prev && index != a->pr) { |
| 949 | int pi; |
| 950 | |
| 951 | /* |
| 952 | * This number was passed straight down from this set's |
| 953 | * predecessor. Find its index in the previous set and |
| 954 | * recurse to there. |
| 955 | */ |
| 956 | pi = index; |
| 957 | assert(pi != a->pr); |
| 958 | if (pi > a->pr) |
| 959 | pi--; |
| 960 | if (pi >= min(a->pa, a->pb)) { |
| 961 | pi++; |
| 962 | if (pi >= max(a->pa, a->pb)) |
| 963 | pi++; |
| 964 | } |
| 965 | print_recurse(s, a->prev, pathindex, pi, priority, assoc, child); |
| 966 | } else if (a->prev && index == a->pr && |
| 967 | s->ops[a->po]->display) { |
| 968 | /* |
| 969 | * This number was created by a displayed operator in the |
| 970 | * transition from this set to its predecessor. Hence we |
| 971 | * write an open paren, then recurse into the first |
| 972 | * operand, then write the operator, then the second |
| 973 | * operand, and finally close the paren. |
| 974 | */ |
| 975 | char *op; |
| 976 | int parens, thispri, thisassoc; |
| 977 | |
| 978 | /* |
| 979 | * Determine whether we need parentheses. |
| 980 | */ |
| 981 | thispri = s->ops[a->po]->priority; |
| 982 | thisassoc = s->ops[a->po]->assoc; |
| 983 | parens = (thispri < priority || |
| 984 | (thispri == priority && (assoc & child))); |
| 985 | |
| 986 | if (parens) |
| 987 | putchar('('); |
| 988 | |
| 989 | if (s->ops[a->po]->flags & OPFLAG_UNARYPREFIX) |
| 990 | for (op = s->ops[a->po]->text; *op; op++) |
| 991 | putchar(*op); |
| 992 | |
| 993 | print_recurse(s, a->prev, pathindex, a->pa, thispri, thisassoc, 1); |
| 994 | |
| 995 | if (!(s->ops[a->po]->flags & OPFLAG_UNARYPREFIX)) |
| 996 | for (op = s->ops[a->po]->text; *op; op++) |
| 997 | putchar(*op); |
| 998 | |
| 999 | if (!(s->ops[a->po]->flags & OPFLAG_UNARY)) |
| 1000 | print_recurse(s, a->prev, pathindex, a->pb, thispri, thisassoc, 2); |
| 1001 | |
| 1002 | if (parens) |
| 1003 | putchar(')'); |
| 1004 | } else { |
| 1005 | /* |
| 1006 | * This number is either an original, or something formed |
| 1007 | * by a non-displayed operator (concatenation). Either way, |
| 1008 | * we display it as is. |
| 1009 | */ |
| 1010 | printf("%d", ss->numbers[2*index]); |
| 1011 | if (ss->numbers[2*index+1] != 1) |
| 1012 | printf("/%d", ss->numbers[2*index+1]); |
| 1013 | } |
| 1014 | } |
| 1015 | void print_recurse(struct sets *s, struct set *ss, int pathindex, int index, |
| 1016 | int priority, int assoc, int child) |
| 1017 | { |
| 1018 | if (!ss->a.prev || pathindex < ss->a.prev->npaths) { |
| 1019 | print_recurse_inner(s, ss, &ss->a, pathindex, |
| 1020 | index, priority, assoc, child); |
| 1021 | } else { |
| 1022 | int i; |
| 1023 | pathindex -= ss->a.prev->npaths; |
| 1024 | for (i = 0; i < ss->nas; i++) { |
| 1025 | if (pathindex < ss->as[i].prev->npaths) { |
| 1026 | print_recurse_inner(s, ss, &ss->as[i], pathindex, |
| 1027 | index, priority, assoc, child); |
| 1028 | break; |
| 1029 | } |
| 1030 | pathindex -= ss->as[i].prev->npaths; |
| 1031 | } |
| 1032 | } |
| 1033 | } |
| 1034 | void print(int pathindex, struct sets *s, struct output *o) |
| 1035 | { |
| 1036 | print_recurse(s, o->set, pathindex, o->index, 0, 0, 0); |
| 1037 | } |
| 1038 | |
| 1039 | /* |
| 1040 | * gcc -g -O0 -o numgame numgame.c -I.. ../{malloc,tree234,nullfe}.c -lm |
| 1041 | */ |
| 1042 | int main(int argc, char **argv) |
| 1043 | { |
| 1044 | int doing_opts = TRUE; |
| 1045 | const struct rules *rules = NULL; |
| 1046 | char *pname = argv[0]; |
| 1047 | int got_target = FALSE, target = 0; |
| 1048 | int numbers[10], nnumbers = 0; |
| 1049 | int verbose = FALSE; |
| 1050 | int pathcounts = FALSE; |
| 1051 | int multiple = FALSE; |
| 1052 | int debug_bfs = FALSE; |
| 1053 | |
| 1054 | struct output *o; |
| 1055 | struct sets *s; |
| 1056 | int i, start, limit; |
| 1057 | |
| 1058 | while (--argc) { |
| 1059 | char *p = *++argv; |
| 1060 | int c; |
| 1061 | |
| 1062 | if (doing_opts && *p == '-') { |
| 1063 | p++; |
| 1064 | |
| 1065 | if (!strcmp(p, "-")) { |
| 1066 | doing_opts = FALSE; |
| 1067 | continue; |
| 1068 | } else if (*p == '-') { |
| 1069 | p++; |
| 1070 | if (!strcmp(p, "debug-bfs")) { |
| 1071 | debug_bfs = TRUE; |
| 1072 | } else { |
| 1073 | fprintf(stderr, "%s: option '--%s' not recognised\n", |
| 1074 | pname, p); |
| 1075 | } |
| 1076 | } else while (*p) switch (c = *p++) { |
| 1077 | case 'C': |
| 1078 | rules = &rules_countdown; |
| 1079 | break; |
| 1080 | case 'B': |
| 1081 | rules = &rules_3388; |
| 1082 | break; |
| 1083 | case 'D': |
| 1084 | rules = &rules_four4s; |
| 1085 | break; |
| 1086 | case 'A': |
| 1087 | rules = &rules_anythinggoes; |
| 1088 | break; |
| 1089 | case 'v': |
| 1090 | verbose = TRUE; |
| 1091 | break; |
| 1092 | case 'p': |
| 1093 | pathcounts = TRUE; |
| 1094 | break; |
| 1095 | case 'm': |
| 1096 | multiple = TRUE; |
| 1097 | break; |
| 1098 | case 't': |
| 1099 | { |
| 1100 | char *v; |
| 1101 | if (*p) { |
| 1102 | v = p; |
| 1103 | p = NULL; |
| 1104 | } else if (--argc) { |
| 1105 | v = *++argv; |
| 1106 | } else { |
| 1107 | fprintf(stderr, "%s: option '-%c' expects an" |
| 1108 | " argument\n", pname, c); |
| 1109 | return 1; |
| 1110 | } |
| 1111 | switch (c) { |
| 1112 | case 't': |
| 1113 | got_target = TRUE; |
| 1114 | target = atoi(v); |
| 1115 | break; |
| 1116 | } |
| 1117 | } |
| 1118 | break; |
| 1119 | default: |
| 1120 | fprintf(stderr, "%s: option '-%c' not" |
| 1121 | " recognised\n", pname, c); |
| 1122 | return 1; |
| 1123 | } |
| 1124 | } else { |
| 1125 | if (nnumbers >= lenof(numbers)) { |
| 1126 | fprintf(stderr, "%s: internal limit of %d numbers exceeded\n", |
| 1127 | pname, lenof(numbers)); |
| 1128 | return 1; |
| 1129 | } else { |
| 1130 | numbers[nnumbers++] = atoi(p); |
| 1131 | } |
| 1132 | } |
| 1133 | } |
| 1134 | |
| 1135 | if (!rules) { |
| 1136 | fprintf(stderr, "%s: no rule set specified; use -C,-B,-D,-A\n", pname); |
| 1137 | return 1; |
| 1138 | } |
| 1139 | |
| 1140 | if (!nnumbers) { |
| 1141 | fprintf(stderr, "%s: no input numbers specified\n", pname); |
| 1142 | return 1; |
| 1143 | } |
| 1144 | |
| 1145 | s = do_search(nnumbers, numbers, rules, (got_target ? &target : NULL), |
| 1146 | debug_bfs, multiple); |
| 1147 | |
| 1148 | if (got_target) { |
| 1149 | o = findrelpos234(s->outputtree, &target, outputfindcmp, |
| 1150 | REL234_LE, &start); |
| 1151 | if (!o) |
| 1152 | start = -1; |
| 1153 | o = findrelpos234(s->outputtree, &target, outputfindcmp, |
| 1154 | REL234_GE, &limit); |
| 1155 | if (!o) |
| 1156 | limit = -1; |
| 1157 | assert(start != -1 || limit != -1); |
| 1158 | if (start == -1) |
| 1159 | start = limit; |
| 1160 | else if (limit == -1) |
| 1161 | limit = start; |
| 1162 | limit++; |
| 1163 | } else { |
| 1164 | start = 0; |
| 1165 | limit = count234(s->outputtree); |
| 1166 | } |
| 1167 | |
| 1168 | for (i = start; i < limit; i++) { |
| 1169 | char buf[256]; |
| 1170 | |
| 1171 | o = index234(s->outputtree, i); |
| 1172 | |
| 1173 | sprintf(buf, "%d", o->number); |
| 1174 | |
| 1175 | if (pathcounts) |
| 1176 | sprintf(buf + strlen(buf), " [%d]", o->npaths); |
| 1177 | |
| 1178 | if (got_target || verbose) { |
| 1179 | int j, npaths; |
| 1180 | |
| 1181 | if (multiple) |
| 1182 | npaths = o->npaths; |
| 1183 | else |
| 1184 | npaths = 1; |
| 1185 | |
| 1186 | for (j = 0; j < npaths; j++) { |
| 1187 | printf("%s = ", buf); |
| 1188 | print(j, s, o); |
| 1189 | putchar('\n'); |
| 1190 | } |
| 1191 | } else { |
| 1192 | printf("%s\n", buf); |
| 1193 | } |
| 1194 | } |
| 1195 | |
| 1196 | free_sets(s); |
| 1197 | |
| 1198 | return 0; |
| 1199 | } |