2 * solo.c: the number-placing puzzle most popularly known as `Sudoku'.
6 * - Jigsaw Sudoku is currently an undocumented feature enabled
7 * by setting r (`Rows of sub-blocks' in the GUI configurer) to
8 * 1. The reason it's undocumented is because they're rather
9 * erratic to generate, because gridgen tends to hang up for
10 * ages. I think this is because some jigsaw block layouts
11 * simply do not admit very many valid filled grids (and
12 * perhaps some have none at all).
13 * + To fix this, I think probably the solution is a change in
14 * grid generation policy: gridgen needs to have less of an
15 * all-or-nothing attitude and instead make only a limited
16 * amount of effort to construct a filled grid before giving
17 * up and trying a new layout. (Come to think of it, this
18 * same change might also make 5x5 standard Sudoku more
19 * practical to generate, if correctly tuned.)
20 * + If I get this fixed, other work needed on jigsaw mode is:
21 * * introduce a GUI config checkbox. game_configure()
22 * ticks this box iff r==1; if it's ticked in a call to
23 * custom_params(), we replace (c, r) with (c*r, 1).
26 * - reports from users are that `Trivial'-mode puzzles are still
27 * rather hard compared to newspapers' easy ones, so some better
28 * low-end difficulty grading would be nice
29 * + it's possible that really easy puzzles always have
30 * _several_ things you can do, so don't make you hunt too
31 * hard for the one deduction you can currently make
32 * + it's also possible that easy puzzles require fewer
33 * cross-eliminations: perhaps there's a higher incidence of
34 * things you can deduce by looking only at (say) rows,
35 * rather than things you have to check both rows and columns
37 * + but really, what I need to do is find some really easy
38 * puzzles and _play_ them, to see what's actually easy about
40 * + while I'm revamping this area, filling in the _last_
41 * number in a nearly-full row or column should certainly be
42 * permitted even at the lowest difficulty level.
43 * + also Owen noticed that `Basic' grids requiring numeric
44 * elimination are actually very hard, so I wonder if a
45 * difficulty gradation between that and positional-
46 * elimination-only might be in order
47 * + but it's not good to have _too_ many difficulty levels, or
48 * it'll take too long to randomly generate a given level.
50 * - it might still be nice to do some prioritisation on the
51 * removal of numbers from the grid
52 * + one possibility is to try to minimise the maximum number
53 * of filled squares in any block, which in particular ought
54 * to enforce never leaving a completely filled block in the
55 * puzzle as presented.
57 * - alternative interface modes
58 * + sudoku.com's Windows program has a palette of possible
59 * entries; you select a palette entry first and then click
60 * on the square you want it to go in, thus enabling
61 * mouse-only play. Useful for PDAs! I don't think it's
62 * actually incompatible with the current highlight-then-type
63 * approach: you _either_ highlight a palette entry and then
64 * click, _or_ you highlight a square and then type. At most
65 * one thing is ever highlighted at a time, so there's no way
67 * + then again, I don't actually like sudoku.com's interface;
68 * it's too much like a paint package whereas I prefer to
69 * think of Solo as a text editor.
70 * + another PDA-friendly possibility is a drag interface:
71 * _drag_ numbers from the palette into the grid squares.
72 * Thought experiments suggest I'd prefer that to the
73 * sudoku.com approach, but I haven't actually tried it.
77 * Solo puzzles need to be square overall (since each row and each
78 * column must contain one of every digit), but they need not be
79 * subdivided the same way internally. I am going to adopt a
80 * convention whereby I _always_ refer to `r' as the number of rows
81 * of _big_ divisions, and `c' as the number of columns of _big_
82 * divisions. Thus, a 2c by 3r puzzle looks something like this:
86 * ------+------ (Of course, you can't subdivide it the other way
87 * 1 4 5 | 6 3 2 or you'll get clashes; observe that the 4 in the
88 * 3 2 6 | 4 1 5 top left would conflict with the 4 in the second
89 * ------+------ box down on the left-hand side.)
93 * The need for a strong naming convention should now be clear:
94 * each small box is two rows of digits by three columns, while the
95 * overall puzzle has three rows of small boxes by two columns. So
96 * I will (hopefully) consistently use `r' to denote the number of
97 * rows _of small boxes_ (here 3), which is also the number of
98 * columns of digits in each small box; and `c' vice versa (here
101 * I'm also going to choose arbitrarily to list c first wherever
102 * possible: the above is a 2x3 puzzle, not a 3x2 one.
112 #ifdef STANDALONE_SOLVER
114 int solver_show_working, solver_recurse_depth;
120 * To save space, I store digits internally as unsigned char. This
121 * imposes a hard limit of 255 on the order of the puzzle. Since
122 * even a 5x5 takes unacceptably long to generate, I don't see this
123 * as a serious limitation unless something _really_ impressive
124 * happens in computing technology; but here's a typedef anyway for
125 * general good practice.
127 typedef unsigned char digit;
128 #define ORDER_MAX 255
130 #define PREFERRED_TILE_SIZE 32
131 #define TILE_SIZE (ds->tilesize)
132 #define BORDER (TILE_SIZE / 2)
133 #define GRIDEXTRA (TILE_SIZE / 32)
135 #define FLASH_TIME 0.4F
137 enum { SYMM_NONE, SYMM_ROT2, SYMM_ROT4, SYMM_REF2, SYMM_REF2D, SYMM_REF4,
138 SYMM_REF4D, SYMM_REF8 };
140 enum { DIFF_BLOCK, DIFF_SIMPLE, DIFF_INTERSECT, DIFF_SET, DIFF_EXTREME,
141 DIFF_RECURSIVE, DIFF_AMBIGUOUS, DIFF_IMPOSSIBLE };
157 * For a square puzzle, `c' and `r' indicate the puzzle
158 * parameters as described above.
160 * A jigsaw-style puzzle is indicated by r==1, in which case c
161 * can be whatever it likes (there is no constraint on
162 * compositeness - a 7x7 jigsaw sudoku makes perfect sense).
164 int c, r, symm, diff;
165 int xtype; /* require all digits in X-diagonals */
168 struct block_structure {
172 * For text formatting, we do need c and r here.
177 * For any square index, whichblock[i] gives its block index.
179 * For 0 <= b,i < cr, blocks[b][i] gives the index of the ith
182 * whichblock and blocks are each dynamically allocated in
183 * their own right, but the subarrays in blocks are appended
184 * to the whichblock array, so shouldn't be freed
187 int *whichblock, **blocks;
189 #ifdef STANDALONE_SOLVER
191 * Textual descriptions of each block. For normal Sudoku these
192 * are of the form "(1,3)"; for jigsaw they are "starting at
193 * (5,7)". So the sensible usage in both cases is to say
194 * "elimination within block %s" with one of these strings.
196 * Only blocknames itself needs individually freeing; it's all
205 * For historical reasons, I use `cr' to denote the overall
206 * width/height of the puzzle. It was a natural notation when
207 * all puzzles were divided into blocks in a grid, but doesn't
208 * really make much sense given jigsaw puzzles. However, the
209 * obvious `n' is heavily used in the solver to describe the
210 * index of a number being placed, so `cr' will have to stay.
213 struct block_structure *blocks;
216 unsigned char *pencil; /* c*r*c*r elements */
217 unsigned char *immutable; /* marks which digits are clues */
218 int completed, cheated;
221 static game_params *default_params(void)
223 game_params *ret = snew(game_params);
227 ret->symm = SYMM_ROT2; /* a plausible default */
228 ret->diff = DIFF_BLOCK; /* so is this */
233 static void free_params(game_params *params)
238 static game_params *dup_params(game_params *params)
240 game_params *ret = snew(game_params);
241 *ret = *params; /* structure copy */
245 static int game_fetch_preset(int i, char **name, game_params **params)
251 { "2x2 Trivial", { 2, 2, SYMM_ROT2, DIFF_BLOCK, FALSE } },
252 { "2x3 Basic", { 2, 3, SYMM_ROT2, DIFF_SIMPLE, FALSE } },
253 { "3x3 Trivial", { 3, 3, SYMM_ROT2, DIFF_BLOCK, FALSE } },
254 { "3x3 Basic", { 3, 3, SYMM_ROT2, DIFF_SIMPLE, FALSE } },
255 { "3x3 Basic X", { 3, 3, SYMM_ROT2, DIFF_SIMPLE, TRUE } },
256 { "3x3 Intermediate", { 3, 3, SYMM_ROT2, DIFF_INTERSECT, FALSE } },
257 { "3x3 Advanced", { 3, 3, SYMM_ROT2, DIFF_SET, FALSE } },
258 { "3x3 Advanced X", { 3, 3, SYMM_ROT2, DIFF_SET, TRUE } },
259 { "3x3 Extreme", { 3, 3, SYMM_ROT2, DIFF_EXTREME, FALSE } },
260 { "3x3 Unreasonable", { 3, 3, SYMM_ROT2, DIFF_RECURSIVE, FALSE } },
262 { "3x4 Basic", { 3, 4, SYMM_ROT2, DIFF_SIMPLE, FALSE } },
263 { "4x4 Basic", { 4, 4, SYMM_ROT2, DIFF_SIMPLE, FALSE } },
267 if (i < 0 || i >= lenof(presets))
270 *name = dupstr(presets[i].title);
271 *params = dup_params(&presets[i].params);
276 static void decode_params(game_params *ret, char const *string)
280 ret->c = ret->r = atoi(string);
282 while (*string && isdigit((unsigned char)*string)) string++;
283 if (*string == 'x') {
285 ret->r = atoi(string);
287 while (*string && isdigit((unsigned char)*string)) string++;
290 if (*string == 'j') {
295 } else if (*string == 'x') {
298 } else if (*string == 'r' || *string == 'm' || *string == 'a') {
301 if (sc == 'm' && *string == 'd') {
308 while (*string && isdigit((unsigned char)*string)) string++;
309 if (sc == 'm' && sn == 8)
310 ret->symm = SYMM_REF8;
311 if (sc == 'm' && sn == 4)
312 ret->symm = sd ? SYMM_REF4D : SYMM_REF4;
313 if (sc == 'm' && sn == 2)
314 ret->symm = sd ? SYMM_REF2D : SYMM_REF2;
315 if (sc == 'r' && sn == 4)
316 ret->symm = SYMM_ROT4;
317 if (sc == 'r' && sn == 2)
318 ret->symm = SYMM_ROT2;
320 ret->symm = SYMM_NONE;
321 } else if (*string == 'd') {
323 if (*string == 't') /* trivial */
324 string++, ret->diff = DIFF_BLOCK;
325 else if (*string == 'b') /* basic */
326 string++, ret->diff = DIFF_SIMPLE;
327 else if (*string == 'i') /* intermediate */
328 string++, ret->diff = DIFF_INTERSECT;
329 else if (*string == 'a') /* advanced */
330 string++, ret->diff = DIFF_SET;
331 else if (*string == 'e') /* extreme */
332 string++, ret->diff = DIFF_EXTREME;
333 else if (*string == 'u') /* unreasonable */
334 string++, ret->diff = DIFF_RECURSIVE;
336 string++; /* eat unknown character */
340 static char *encode_params(game_params *params, int full)
345 sprintf(str, "%dx%d", params->c, params->r);
347 sprintf(str, "%dj", params->c);
352 switch (params->symm) {
353 case SYMM_REF8: strcat(str, "m8"); break;
354 case SYMM_REF4: strcat(str, "m4"); break;
355 case SYMM_REF4D: strcat(str, "md4"); break;
356 case SYMM_REF2: strcat(str, "m2"); break;
357 case SYMM_REF2D: strcat(str, "md2"); break;
358 case SYMM_ROT4: strcat(str, "r4"); break;
359 /* case SYMM_ROT2: strcat(str, "r2"); break; [default] */
360 case SYMM_NONE: strcat(str, "a"); break;
362 switch (params->diff) {
363 /* case DIFF_BLOCK: strcat(str, "dt"); break; [default] */
364 case DIFF_SIMPLE: strcat(str, "db"); break;
365 case DIFF_INTERSECT: strcat(str, "di"); break;
366 case DIFF_SET: strcat(str, "da"); break;
367 case DIFF_EXTREME: strcat(str, "de"); break;
368 case DIFF_RECURSIVE: strcat(str, "du"); break;
374 static config_item *game_configure(game_params *params)
379 ret = snewn(6, config_item);
381 ret[0].name = "Columns of sub-blocks";
382 ret[0].type = C_STRING;
383 sprintf(buf, "%d", params->c);
384 ret[0].sval = dupstr(buf);
387 ret[1].name = "Rows of sub-blocks";
388 ret[1].type = C_STRING;
389 sprintf(buf, "%d", params->r);
390 ret[1].sval = dupstr(buf);
393 ret[2].name = "\"X\" (require every number in each main diagonal)";
394 ret[2].type = C_BOOLEAN;
396 ret[2].ival = params->xtype;
398 ret[3].name = "Symmetry";
399 ret[3].type = C_CHOICES;
400 ret[3].sval = ":None:2-way rotation:4-way rotation:2-way mirror:"
401 "2-way diagonal mirror:4-way mirror:4-way diagonal mirror:"
403 ret[3].ival = params->symm;
405 ret[4].name = "Difficulty";
406 ret[4].type = C_CHOICES;
407 ret[4].sval = ":Trivial:Basic:Intermediate:Advanced:Extreme:Unreasonable";
408 ret[4].ival = params->diff;
418 static game_params *custom_params(config_item *cfg)
420 game_params *ret = snew(game_params);
422 ret->c = atoi(cfg[0].sval);
423 ret->r = atoi(cfg[1].sval);
424 ret->xtype = cfg[2].ival;
425 ret->symm = cfg[3].ival;
426 ret->diff = cfg[4].ival;
431 static char *validate_params(game_params *params, int full)
434 return "Both dimensions must be at least 2";
435 if (params->c > ORDER_MAX || params->r > ORDER_MAX)
436 return "Dimensions greater than "STR(ORDER_MAX)" are not supported";
437 if ((params->c * params->r) > 35)
438 return "Unable to support more than 35 distinct symbols in a puzzle";
442 /* ----------------------------------------------------------------------
445 * This solver is used for two purposes:
446 * + to check solubility of a grid as we gradually remove numbers
448 * + to solve an externally generated puzzle when the user selects
451 * It supports a variety of specific modes of reasoning. By
452 * enabling or disabling subsets of these modes we can arrange a
453 * range of difficulty levels.
457 * Modes of reasoning currently supported:
459 * - Positional elimination: a number must go in a particular
460 * square because all the other empty squares in a given
461 * row/col/blk are ruled out.
463 * - Numeric elimination: a square must have a particular number
464 * in because all the other numbers that could go in it are
467 * - Intersectional analysis: given two domains which overlap
468 * (hence one must be a block, and the other can be a row or
469 * col), if the possible locations for a particular number in
470 * one of the domains can be narrowed down to the overlap, then
471 * that number can be ruled out everywhere but the overlap in
472 * the other domain too.
474 * - Set elimination: if there is a subset of the empty squares
475 * within a domain such that the union of the possible numbers
476 * in that subset has the same size as the subset itself, then
477 * those numbers can be ruled out everywhere else in the domain.
478 * (For example, if there are five empty squares and the
479 * possible numbers in each are 12, 23, 13, 134 and 1345, then
480 * the first three empty squares form such a subset: the numbers
481 * 1, 2 and 3 _must_ be in those three squares in some
482 * permutation, and hence we can deduce none of them can be in
483 * the fourth or fifth squares.)
484 * + You can also see this the other way round, concentrating
485 * on numbers rather than squares: if there is a subset of
486 * the unplaced numbers within a domain such that the union
487 * of all their possible positions has the same size as the
488 * subset itself, then all other numbers can be ruled out for
489 * those positions. However, it turns out that this is
490 * exactly equivalent to the first formulation at all times:
491 * there is a 1-1 correspondence between suitable subsets of
492 * the unplaced numbers and suitable subsets of the unfilled
493 * places, found by taking the _complement_ of the union of
494 * the numbers' possible positions (or the spaces' possible
497 * - Forcing chains (see comment for solver_forcing().)
499 * - Recursion. If all else fails, we pick one of the currently
500 * most constrained empty squares and take a random guess at its
501 * contents, then continue solving on that basis and see if we
505 struct solver_usage {
507 struct block_structure *blocks;
509 * We set up a cubic array, indexed by x, y and digit; each
510 * element of this array is TRUE or FALSE according to whether
511 * or not that digit _could_ in principle go in that position.
513 * The way to index this array is cube[(y*cr+x)*cr+n-1]; there
514 * are macros below to help with this.
518 * This is the grid in which we write down our final
519 * deductions. y-coordinates in here are _not_ transformed.
523 * Now we keep track, at a slightly higher level, of what we
524 * have yet to work out, to prevent doing the same deduction
527 /* row[y*cr+n-1] TRUE if digit n has been placed in row y */
529 /* col[x*cr+n-1] TRUE if digit n has been placed in row x */
531 /* blk[i*cr+n-1] TRUE if digit n has been placed in block i */
533 /* diag[i*cr+n-1] TRUE if digit n has been placed in diagonal i */
534 unsigned char *diag; /* diag 0 is \, 1 is / */
536 #define cubepos2(xy,n) ((xy)*usage->cr+(n)-1)
537 #define cubepos(x,y,n) cubepos2((y)*usage->cr+(x),n)
538 #define cube(x,y,n) (usage->cube[cubepos(x,y,n)])
539 #define cube2(xy,n) (usage->cube[cubepos2(xy,n)])
541 #define ondiag0(xy) ((xy) % (cr+1) == 0)
542 #define ondiag1(xy) ((xy) % (cr-1) == 0 && (xy) > 0 && (xy) < cr*cr-1)
543 #define diag0(i) ((i) * (cr+1))
544 #define diag1(i) ((i+1) * (cr-1))
547 * Function called when we are certain that a particular square has
548 * a particular number in it. The y-coordinate passed in here is
551 static void solver_place(struct solver_usage *usage, int x, int y, int n)
554 int sqindex = y*cr+x;
560 * Rule out all other numbers in this square.
562 for (i = 1; i <= cr; i++)
567 * Rule out this number in all other positions in the row.
569 for (i = 0; i < cr; i++)
574 * Rule out this number in all other positions in the column.
576 for (i = 0; i < cr; i++)
581 * Rule out this number in all other positions in the block.
583 bi = usage->blocks->whichblock[sqindex];
584 for (i = 0; i < cr; i++) {
585 int bp = usage->blocks->blocks[bi][i];
591 * Enter the number in the result grid.
593 usage->grid[sqindex] = n;
596 * Cross out this number from the list of numbers left to place
597 * in its row, its column and its block.
599 usage->row[y*cr+n-1] = usage->col[x*cr+n-1] =
600 usage->blk[bi*cr+n-1] = TRUE;
603 if (ondiag0(sqindex)) {
604 for (i = 0; i < cr; i++)
605 if (diag0(i) != sqindex)
606 cube2(diag0(i),n) = FALSE;
607 usage->diag[n-1] = TRUE;
609 if (ondiag1(sqindex)) {
610 for (i = 0; i < cr; i++)
611 if (diag1(i) != sqindex)
612 cube2(diag1(i),n) = FALSE;
613 usage->diag[cr+n-1] = TRUE;
618 static int solver_elim(struct solver_usage *usage, int *indices
619 #ifdef STANDALONE_SOLVER
628 * Count the number of set bits within this section of the
633 for (i = 0; i < cr; i++)
634 if (usage->cube[indices[i]]) {
648 if (!usage->grid[y*cr+x]) {
649 #ifdef STANDALONE_SOLVER
650 if (solver_show_working) {
652 printf("%*s", solver_recurse_depth*4, "");
656 printf(":\n%*s placing %d at (%d,%d)\n",
657 solver_recurse_depth*4, "", n, 1+x, 1+y);
660 solver_place(usage, x, y, n);
664 #ifdef STANDALONE_SOLVER
665 if (solver_show_working) {
667 printf("%*s", solver_recurse_depth*4, "");
671 printf(":\n%*s no possibilities available\n",
672 solver_recurse_depth*4, "");
681 static int solver_intersect(struct solver_usage *usage,
682 int *indices1, int *indices2
683 #ifdef STANDALONE_SOLVER
692 * Loop over the first domain and see if there's any set bit
693 * not also in the second.
695 for (i = j = 0; i < cr; i++) {
697 while (j < cr && indices2[j] < p)
699 if (usage->cube[p]) {
700 if (j < cr && indices2[j] == p)
701 continue; /* both domains contain this index */
703 return 0; /* there is, so we can't deduce */
708 * We have determined that all set bits in the first domain are
709 * within its overlap with the second. So loop over the second
710 * domain and remove all set bits that aren't also in that
711 * overlap; return +1 iff we actually _did_ anything.
714 for (i = j = 0; i < cr; i++) {
716 while (j < cr && indices1[j] < p)
718 if (usage->cube[p] && (j >= cr || indices1[j] != p)) {
719 #ifdef STANDALONE_SOLVER
720 if (solver_show_working) {
725 printf("%*s", solver_recurse_depth*4, "");
737 printf("%*s ruling out %d at (%d,%d)\n",
738 solver_recurse_depth*4, "", pn, 1+px, 1+py);
741 ret = +1; /* we did something */
749 struct solver_scratch {
750 unsigned char *grid, *rowidx, *colidx, *set;
751 int *neighbours, *bfsqueue;
752 int *indexlist, *indexlist2;
753 #ifdef STANDALONE_SOLVER
758 static int solver_set(struct solver_usage *usage,
759 struct solver_scratch *scratch,
761 #ifdef STANDALONE_SOLVER
768 unsigned char *grid = scratch->grid;
769 unsigned char *rowidx = scratch->rowidx;
770 unsigned char *colidx = scratch->colidx;
771 unsigned char *set = scratch->set;
774 * We are passed a cr-by-cr matrix of booleans. Our first job
775 * is to winnow it by finding any definite placements - i.e.
776 * any row with a solitary 1 - and discarding that row and the
777 * column containing the 1.
779 memset(rowidx, TRUE, cr);
780 memset(colidx, TRUE, cr);
781 for (i = 0; i < cr; i++) {
782 int count = 0, first = -1;
783 for (j = 0; j < cr; j++)
784 if (usage->cube[indices[i*cr+j]])
788 * If count == 0, then there's a row with no 1s at all and
789 * the puzzle is internally inconsistent. However, we ought
790 * to have caught this already during the simpler reasoning
791 * methods, so we can safely fail an assertion if we reach
796 rowidx[i] = colidx[first] = FALSE;
800 * Convert each of rowidx/colidx from a list of 0s and 1s to a
801 * list of the indices of the 1s.
803 for (i = j = 0; i < cr; i++)
807 for (i = j = 0; i < cr; i++)
813 * And create the smaller matrix.
815 for (i = 0; i < n; i++)
816 for (j = 0; j < n; j++)
817 grid[i*cr+j] = usage->cube[indices[rowidx[i]*cr+colidx[j]]];
820 * Having done that, we now have a matrix in which every row
821 * has at least two 1s in. Now we search to see if we can find
822 * a rectangle of zeroes (in the set-theoretic sense of
823 * `rectangle', i.e. a subset of rows crossed with a subset of
824 * columns) whose width and height add up to n.
831 * We have a candidate set. If its size is <=1 or >=n-1
832 * then we move on immediately.
834 if (count > 1 && count < n-1) {
836 * The number of rows we need is n-count. See if we can
837 * find that many rows which each have a zero in all
838 * the positions listed in `set'.
841 for (i = 0; i < n; i++) {
843 for (j = 0; j < n; j++)
844 if (set[j] && grid[i*cr+j]) {
853 * We expect never to be able to get _more_ than
854 * n-count suitable rows: this would imply that (for
855 * example) there are four numbers which between them
856 * have at most three possible positions, and hence it
857 * indicates a faulty deduction before this point or
860 if (rows > n - count) {
861 #ifdef STANDALONE_SOLVER
862 if (solver_show_working) {
864 printf("%*s", solver_recurse_depth*4,
869 printf(":\n%*s contradiction reached\n",
870 solver_recurse_depth*4, "");
876 if (rows >= n - count) {
877 int progress = FALSE;
880 * We've got one! Now, for each row which _doesn't_
881 * satisfy the criterion, eliminate all its set
882 * bits in the positions _not_ listed in `set'.
883 * Return +1 (meaning progress has been made) if we
884 * successfully eliminated anything at all.
886 * This involves referring back through
887 * rowidx/colidx in order to work out which actual
888 * positions in the cube to meddle with.
890 for (i = 0; i < n; i++) {
892 for (j = 0; j < n; j++)
893 if (set[j] && grid[i*cr+j]) {
898 for (j = 0; j < n; j++)
899 if (!set[j] && grid[i*cr+j]) {
900 int fpos = indices[rowidx[i]*cr+colidx[j]];
901 #ifdef STANDALONE_SOLVER
902 if (solver_show_working) {
907 printf("%*s", solver_recurse_depth*4,
920 printf("%*s ruling out %d at (%d,%d)\n",
921 solver_recurse_depth*4, "",
926 usage->cube[fpos] = FALSE;
938 * Binary increment: change the rightmost 0 to a 1, and
939 * change all 1s to the right of it to 0s.
942 while (i > 0 && set[i-1])
943 set[--i] = 0, count--;
945 set[--i] = 1, count++;
954 * Look for forcing chains. A forcing chain is a path of
955 * pairwise-exclusive squares (i.e. each pair of adjacent squares
956 * in the path are in the same row, column or block) with the
957 * following properties:
959 * (a) Each square on the path has precisely two possible numbers.
961 * (b) Each pair of squares which are adjacent on the path share
962 * at least one possible number in common.
964 * (c) Each square in the middle of the path shares _both_ of its
965 * numbers with at least one of its neighbours (not the same
966 * one with both neighbours).
968 * These together imply that at least one of the possible number
969 * choices at one end of the path forces _all_ the rest of the
970 * numbers along the path. In order to make real use of this, we
971 * need further properties:
973 * (c) Ruling out some number N from the square at one end of the
974 * path forces the square at the other end to take the same
977 * (d) The two end squares are both in line with some third
980 * (e) That third square currently has N as a possibility.
982 * If we can find all of that lot, we can deduce that at least one
983 * of the two ends of the forcing chain has number N, and that
984 * therefore the mutually adjacent third square does not.
986 * To find forcing chains, we're going to start a bfs at each
987 * suitable square, once for each of its two possible numbers.
989 static int solver_forcing(struct solver_usage *usage,
990 struct solver_scratch *scratch)
993 int *bfsqueue = scratch->bfsqueue;
994 #ifdef STANDALONE_SOLVER
995 int *bfsprev = scratch->bfsprev;
997 unsigned char *number = scratch->grid;
998 int *neighbours = scratch->neighbours;
1001 for (y = 0; y < cr; y++)
1002 for (x = 0; x < cr; x++) {
1006 * If this square doesn't have exactly two candidate
1007 * numbers, don't try it.
1009 * In this loop we also sum the candidate numbers,
1010 * which is a nasty hack to allow us to quickly find
1011 * `the other one' (since we will shortly know there
1014 for (count = t = 0, n = 1; n <= cr; n++)
1021 * Now attempt a bfs for each candidate.
1023 for (n = 1; n <= cr; n++)
1024 if (cube(x, y, n)) {
1025 int orign, currn, head, tail;
1032 memset(number, cr+1, cr*cr);
1034 bfsqueue[tail++] = y*cr+x;
1035 #ifdef STANDALONE_SOLVER
1036 bfsprev[y*cr+x] = -1;
1038 number[y*cr+x] = t - n;
1040 while (head < tail) {
1041 int xx, yy, nneighbours, xt, yt, i;
1043 xx = bfsqueue[head++];
1047 currn = number[yy*cr+xx];
1050 * Find neighbours of yy,xx.
1053 for (yt = 0; yt < cr; yt++)
1054 neighbours[nneighbours++] = yt*cr+xx;
1055 for (xt = 0; xt < cr; xt++)
1056 neighbours[nneighbours++] = yy*cr+xt;
1057 xt = usage->blocks->whichblock[yy*cr+xx];
1058 for (yt = 0; yt < cr; yt++)
1059 neighbours[nneighbours++] = usage->blocks->blocks[xt][yt];
1061 int sqindex = yy*cr+xx;
1062 if (ondiag0(sqindex)) {
1063 for (i = 0; i < cr; i++)
1064 neighbours[nneighbours++] = diag0(i);
1066 if (ondiag1(sqindex)) {
1067 for (i = 0; i < cr; i++)
1068 neighbours[nneighbours++] = diag1(i);
1073 * Try visiting each of those neighbours.
1075 for (i = 0; i < nneighbours; i++) {
1078 xt = neighbours[i] % cr;
1079 yt = neighbours[i] / cr;
1082 * We need this square to not be
1083 * already visited, and to include
1084 * currn as a possible number.
1086 if (number[yt*cr+xt] <= cr)
1088 if (!cube(xt, yt, currn))
1092 * Don't visit _this_ square a second
1095 if (xt == xx && yt == yy)
1099 * To continue with the bfs, we need
1100 * this square to have exactly two
1103 for (cc = tt = 0, nn = 1; nn <= cr; nn++)
1104 if (cube(xt, yt, nn))
1107 bfsqueue[tail++] = yt*cr+xt;
1108 #ifdef STANDALONE_SOLVER
1109 bfsprev[yt*cr+xt] = yy*cr+xx;
1111 number[yt*cr+xt] = tt - currn;
1115 * One other possibility is that this
1116 * might be the square in which we can
1117 * make a real deduction: if it's
1118 * adjacent to x,y, and currn is equal
1119 * to the original number we ruled out.
1121 if (currn == orign &&
1122 (xt == x || yt == y ||
1123 (usage->blocks->whichblock[yt*cr+xt] == usage->blocks->whichblock[y*cr+x]) ||
1124 (usage->diag && ((ondiag0(yt*cr+xt) && ondiag0(y*cr+x)) ||
1125 (ondiag1(yt*cr+xt) && ondiag1(y*cr+x)))))) {
1126 #ifdef STANDALONE_SOLVER
1127 if (solver_show_working) {
1130 printf("%*sforcing chain, %d at ends of ",
1131 solver_recurse_depth*4, "", orign);
1135 printf("%s(%d,%d)", sep, 1+xl,
1137 xl = bfsprev[yl*cr+xl];
1144 printf("\n%*s ruling out %d at (%d,%d)\n",
1145 solver_recurse_depth*4, "",
1149 cube(xt, yt, orign) = FALSE;
1160 static struct solver_scratch *solver_new_scratch(struct solver_usage *usage)
1162 struct solver_scratch *scratch = snew(struct solver_scratch);
1164 scratch->grid = snewn(cr*cr, unsigned char);
1165 scratch->rowidx = snewn(cr, unsigned char);
1166 scratch->colidx = snewn(cr, unsigned char);
1167 scratch->set = snewn(cr, unsigned char);
1168 scratch->neighbours = snewn(5*cr, int);
1169 scratch->bfsqueue = snewn(cr*cr, int);
1170 #ifdef STANDALONE_SOLVER
1171 scratch->bfsprev = snewn(cr*cr, int);
1173 scratch->indexlist = snewn(cr*cr, int); /* used for set elimination */
1174 scratch->indexlist2 = snewn(cr, int); /* only used for intersect() */
1178 static void solver_free_scratch(struct solver_scratch *scratch)
1180 #ifdef STANDALONE_SOLVER
1181 sfree(scratch->bfsprev);
1183 sfree(scratch->bfsqueue);
1184 sfree(scratch->neighbours);
1185 sfree(scratch->set);
1186 sfree(scratch->colidx);
1187 sfree(scratch->rowidx);
1188 sfree(scratch->grid);
1189 sfree(scratch->indexlist);
1190 sfree(scratch->indexlist2);
1194 static int solver(int cr, struct block_structure *blocks, int xtype,
1195 digit *grid, int maxdiff)
1197 struct solver_usage *usage;
1198 struct solver_scratch *scratch;
1199 int x, y, b, i, n, ret;
1200 int diff = DIFF_BLOCK;
1203 * Set up a usage structure as a clean slate (everything
1206 usage = snew(struct solver_usage);
1208 usage->blocks = blocks;
1209 usage->cube = snewn(cr*cr*cr, unsigned char);
1210 usage->grid = grid; /* write straight back to the input */
1211 memset(usage->cube, TRUE, cr*cr*cr);
1213 usage->row = snewn(cr * cr, unsigned char);
1214 usage->col = snewn(cr * cr, unsigned char);
1215 usage->blk = snewn(cr * cr, unsigned char);
1216 memset(usage->row, FALSE, cr * cr);
1217 memset(usage->col, FALSE, cr * cr);
1218 memset(usage->blk, FALSE, cr * cr);
1221 usage->diag = snewn(cr * 2, unsigned char);
1222 memset(usage->diag, FALSE, cr * 2);
1226 scratch = solver_new_scratch(usage);
1229 * Place all the clue numbers we are given.
1231 for (x = 0; x < cr; x++)
1232 for (y = 0; y < cr; y++)
1234 solver_place(usage, x, y, grid[y*cr+x]);
1237 * Now loop over the grid repeatedly trying all permitted modes
1238 * of reasoning. The loop terminates if we complete an
1239 * iteration without making any progress; we then return
1240 * failure or success depending on whether the grid is full or
1245 * I'd like to write `continue;' inside each of the
1246 * following loops, so that the solver returns here after
1247 * making some progress. However, I can't specify that I
1248 * want to continue an outer loop rather than the innermost
1249 * one, so I'm apologetically resorting to a goto.
1254 * Blockwise positional elimination.
1256 for (b = 0; b < cr; b++)
1257 for (n = 1; n <= cr; n++)
1258 if (!usage->blk[b*cr+n-1]) {
1259 for (i = 0; i < cr; i++)
1260 scratch->indexlist[i] = cubepos2(usage->blocks->blocks[b][i],n);
1261 ret = solver_elim(usage, scratch->indexlist
1262 #ifdef STANDALONE_SOLVER
1263 , "positional elimination,"
1264 " %d in block %s", n,
1265 usage->blocks->blocknames[b]
1269 diff = DIFF_IMPOSSIBLE;
1271 } else if (ret > 0) {
1272 diff = max(diff, DIFF_BLOCK);
1277 if (maxdiff <= DIFF_BLOCK)
1281 * Row-wise positional elimination.
1283 for (y = 0; y < cr; y++)
1284 for (n = 1; n <= cr; n++)
1285 if (!usage->row[y*cr+n-1]) {
1286 for (x = 0; x < cr; x++)
1287 scratch->indexlist[x] = cubepos(x, y, n);
1288 ret = solver_elim(usage, scratch->indexlist
1289 #ifdef STANDALONE_SOLVER
1290 , "positional elimination,"
1291 " %d in row %d", n, 1+y
1295 diff = DIFF_IMPOSSIBLE;
1297 } else if (ret > 0) {
1298 diff = max(diff, DIFF_SIMPLE);
1303 * Column-wise positional elimination.
1305 for (x = 0; x < cr; x++)
1306 for (n = 1; n <= cr; n++)
1307 if (!usage->col[x*cr+n-1]) {
1308 for (y = 0; y < cr; y++)
1309 scratch->indexlist[y] = cubepos(x, y, n);
1310 ret = solver_elim(usage, scratch->indexlist
1311 #ifdef STANDALONE_SOLVER
1312 , "positional elimination,"
1313 " %d in column %d", n, 1+x
1317 diff = DIFF_IMPOSSIBLE;
1319 } else if (ret > 0) {
1320 diff = max(diff, DIFF_SIMPLE);
1326 * X-diagonal positional elimination.
1329 for (n = 1; n <= cr; n++)
1330 if (!usage->diag[n-1]) {
1331 for (i = 0; i < cr; i++)
1332 scratch->indexlist[i] = cubepos2(diag0(i), n);
1333 ret = solver_elim(usage, scratch->indexlist
1334 #ifdef STANDALONE_SOLVER
1335 , "positional elimination,"
1336 " %d in \\-diagonal", n
1340 diff = DIFF_IMPOSSIBLE;
1342 } else if (ret > 0) {
1343 diff = max(diff, DIFF_SIMPLE);
1347 for (n = 1; n <= cr; n++)
1348 if (!usage->diag[cr+n-1]) {
1349 for (i = 0; i < cr; i++)
1350 scratch->indexlist[i] = cubepos2(diag1(i), n);
1351 ret = solver_elim(usage, scratch->indexlist
1352 #ifdef STANDALONE_SOLVER
1353 , "positional elimination,"
1354 " %d in /-diagonal", n
1358 diff = DIFF_IMPOSSIBLE;
1360 } else if (ret > 0) {
1361 diff = max(diff, DIFF_SIMPLE);
1368 * Numeric elimination.
1370 for (x = 0; x < cr; x++)
1371 for (y = 0; y < cr; y++)
1372 if (!usage->grid[y*cr+x]) {
1373 for (n = 1; n <= cr; n++)
1374 scratch->indexlist[n-1] = cubepos(x, y, n);
1375 ret = solver_elim(usage, scratch->indexlist
1376 #ifdef STANDALONE_SOLVER
1377 , "numeric elimination at (%d,%d)",
1382 diff = DIFF_IMPOSSIBLE;
1384 } else if (ret > 0) {
1385 diff = max(diff, DIFF_SIMPLE);
1390 if (maxdiff <= DIFF_SIMPLE)
1394 * Intersectional analysis, rows vs blocks.
1396 for (y = 0; y < cr; y++)
1397 for (b = 0; b < cr; b++)
1398 for (n = 1; n <= cr; n++) {
1399 if (usage->row[y*cr+n-1] ||
1400 usage->blk[b*cr+n-1])
1402 for (i = 0; i < cr; i++) {
1403 scratch->indexlist[i] = cubepos(i, y, n);
1404 scratch->indexlist2[i] = cubepos2(usage->blocks->blocks[b][i], n);
1407 * solver_intersect() never returns -1.
1409 if (solver_intersect(usage, scratch->indexlist,
1411 #ifdef STANDALONE_SOLVER
1412 , "intersectional analysis,"
1413 " %d in row %d vs block %s",
1414 n, 1+y, usage->blocks->blocknames[b]
1417 solver_intersect(usage, scratch->indexlist2,
1419 #ifdef STANDALONE_SOLVER
1420 , "intersectional analysis,"
1421 " %d in block %s vs row %d",
1422 n, usage->blocks->blocknames[b], 1+y
1425 diff = max(diff, DIFF_INTERSECT);
1431 * Intersectional analysis, columns vs blocks.
1433 for (x = 0; x < cr; x++)
1434 for (b = 0; b < cr; b++)
1435 for (n = 1; n <= cr; n++) {
1436 if (usage->col[x*cr+n-1] ||
1437 usage->blk[b*cr+n-1])
1439 for (i = 0; i < cr; i++) {
1440 scratch->indexlist[i] = cubepos(x, i, n);
1441 scratch->indexlist2[i] = cubepos2(usage->blocks->blocks[b][i], n);
1443 if (solver_intersect(usage, scratch->indexlist,
1445 #ifdef STANDALONE_SOLVER
1446 , "intersectional analysis,"
1447 " %d in column %d vs block %s",
1448 n, 1+x, usage->blocks->blocknames[b]
1451 solver_intersect(usage, scratch->indexlist2,
1453 #ifdef STANDALONE_SOLVER
1454 , "intersectional analysis,"
1455 " %d in block %s vs column %d",
1456 n, usage->blocks->blocknames[b], 1+x
1459 diff = max(diff, DIFF_INTERSECT);
1466 * Intersectional analysis, \-diagonal vs blocks.
1468 for (b = 0; b < cr; b++)
1469 for (n = 1; n <= cr; n++) {
1470 if (usage->diag[n-1] ||
1471 usage->blk[b*cr+n-1])
1473 for (i = 0; i < cr; i++) {
1474 scratch->indexlist[i] = cubepos2(diag0(i), n);
1475 scratch->indexlist2[i] = cubepos2(usage->blocks->blocks[b][i], n);
1477 if (solver_intersect(usage, scratch->indexlist,
1479 #ifdef STANDALONE_SOLVER
1480 , "intersectional analysis,"
1481 " %d in \\-diagonal vs block %s",
1482 n, 1+x, usage->blocks->blocknames[b]
1485 solver_intersect(usage, scratch->indexlist2,
1487 #ifdef STANDALONE_SOLVER
1488 , "intersectional analysis,"
1489 " %d in block %s vs \\-diagonal",
1490 n, usage->blocks->blocknames[b], 1+x
1493 diff = max(diff, DIFF_INTERSECT);
1499 * Intersectional analysis, /-diagonal vs blocks.
1501 for (b = 0; b < cr; b++)
1502 for (n = 1; n <= cr; n++) {
1503 if (usage->diag[cr+n-1] ||
1504 usage->blk[b*cr+n-1])
1506 for (i = 0; i < cr; i++) {
1507 scratch->indexlist[i] = cubepos2(diag1(i), n);
1508 scratch->indexlist2[i] = cubepos2(usage->blocks->blocks[b][i], n);
1510 if (solver_intersect(usage, scratch->indexlist,
1512 #ifdef STANDALONE_SOLVER
1513 , "intersectional analysis,"
1514 " %d in /-diagonal vs block %s",
1515 n, 1+x, usage->blocks->blocknames[b]
1518 solver_intersect(usage, scratch->indexlist2,
1520 #ifdef STANDALONE_SOLVER
1521 , "intersectional analysis,"
1522 " %d in block %s vs /-diagonal",
1523 n, usage->blocks->blocknames[b], 1+x
1526 diff = max(diff, DIFF_INTERSECT);
1532 if (maxdiff <= DIFF_INTERSECT)
1536 * Blockwise set elimination.
1538 for (b = 0; b < cr; b++) {
1539 for (i = 0; i < cr; i++)
1540 for (n = 1; n <= cr; n++)
1541 scratch->indexlist[i*cr+n-1] = cubepos2(usage->blocks->blocks[b][i], n);
1542 ret = solver_set(usage, scratch, scratch->indexlist
1543 #ifdef STANDALONE_SOLVER
1544 , "set elimination, block %s",
1545 usage->blocks->blocknames[b]
1549 diff = DIFF_IMPOSSIBLE;
1551 } else if (ret > 0) {
1552 diff = max(diff, DIFF_SET);
1558 * Row-wise set elimination.
1560 for (y = 0; y < cr; y++) {
1561 for (x = 0; x < cr; x++)
1562 for (n = 1; n <= cr; n++)
1563 scratch->indexlist[x*cr+n-1] = cubepos(x, y, n);
1564 ret = solver_set(usage, scratch, scratch->indexlist
1565 #ifdef STANDALONE_SOLVER
1566 , "set elimination, row %d", 1+y
1570 diff = DIFF_IMPOSSIBLE;
1572 } else if (ret > 0) {
1573 diff = max(diff, DIFF_SET);
1579 * Column-wise set elimination.
1581 for (x = 0; x < cr; x++) {
1582 for (y = 0; y < cr; y++)
1583 for (n = 1; n <= cr; n++)
1584 scratch->indexlist[y*cr+n-1] = cubepos(x, y, n);
1585 ret = solver_set(usage, scratch, scratch->indexlist
1586 #ifdef STANDALONE_SOLVER
1587 , "set elimination, column %d", 1+x
1591 diff = DIFF_IMPOSSIBLE;
1593 } else if (ret > 0) {
1594 diff = max(diff, DIFF_SET);
1601 * \-diagonal set elimination.
1603 for (i = 0; i < cr; i++)
1604 for (n = 1; n <= cr; n++)
1605 scratch->indexlist[i*cr+n-1] = cubepos2(diag0(i), n);
1606 ret = solver_set(usage, scratch, scratch->indexlist
1607 #ifdef STANDALONE_SOLVER
1608 , "set elimination, \\-diagonal"
1612 diff = DIFF_IMPOSSIBLE;
1614 } else if (ret > 0) {
1615 diff = max(diff, DIFF_SET);
1620 * /-diagonal set elimination.
1622 for (i = 0; i < cr; i++)
1623 for (n = 1; n <= cr; n++)
1624 scratch->indexlist[i*cr+n-1] = cubepos2(diag1(i), n);
1625 ret = solver_set(usage, scratch, scratch->indexlist
1626 #ifdef STANDALONE_SOLVER
1627 , "set elimination, \\-diagonal"
1631 diff = DIFF_IMPOSSIBLE;
1633 } else if (ret > 0) {
1634 diff = max(diff, DIFF_SET);
1639 if (maxdiff <= DIFF_SET)
1643 * Row-vs-column set elimination on a single number.
1645 for (n = 1; n <= cr; n++) {
1646 for (y = 0; y < cr; y++)
1647 for (x = 0; x < cr; x++)
1648 scratch->indexlist[y*cr+x] = cubepos(x, y, n);
1649 ret = solver_set(usage, scratch, scratch->indexlist
1650 #ifdef STANDALONE_SOLVER
1651 , "positional set elimination, number %d", n
1655 diff = DIFF_IMPOSSIBLE;
1657 } else if (ret > 0) {
1658 diff = max(diff, DIFF_EXTREME);
1666 if (solver_forcing(usage, scratch)) {
1667 diff = max(diff, DIFF_EXTREME);
1672 * If we reach here, we have made no deductions in this
1673 * iteration, so the algorithm terminates.
1679 * Last chance: if we haven't fully solved the puzzle yet, try
1680 * recursing based on guesses for a particular square. We pick
1681 * one of the most constrained empty squares we can find, which
1682 * has the effect of pruning the search tree as much as
1685 if (maxdiff >= DIFF_RECURSIVE) {
1686 int best, bestcount;
1691 for (y = 0; y < cr; y++)
1692 for (x = 0; x < cr; x++)
1693 if (!grid[y*cr+x]) {
1697 * An unfilled square. Count the number of
1698 * possible digits in it.
1701 for (n = 1; n <= cr; n++)
1706 * We should have found any impossibilities
1707 * already, so this can safely be an assert.
1711 if (count < bestcount) {
1719 digit *list, *ingrid, *outgrid;
1721 diff = DIFF_IMPOSSIBLE; /* no solution found yet */
1724 * Attempt recursion.
1729 list = snewn(cr, digit);
1730 ingrid = snewn(cr * cr, digit);
1731 outgrid = snewn(cr * cr, digit);
1732 memcpy(ingrid, grid, cr * cr);
1734 /* Make a list of the possible digits. */
1735 for (j = 0, n = 1; n <= cr; n++)
1739 #ifdef STANDALONE_SOLVER
1740 if (solver_show_working) {
1742 printf("%*srecursing on (%d,%d) [",
1743 solver_recurse_depth*4, "", x + 1, y + 1);
1744 for (i = 0; i < j; i++) {
1745 printf("%s%d", sep, list[i]);
1753 * And step along the list, recursing back into the
1754 * main solver at every stage.
1756 for (i = 0; i < j; i++) {
1759 memcpy(outgrid, ingrid, cr * cr);
1760 outgrid[y*cr+x] = list[i];
1762 #ifdef STANDALONE_SOLVER
1763 if (solver_show_working)
1764 printf("%*sguessing %d at (%d,%d)\n",
1765 solver_recurse_depth*4, "", list[i], x + 1, y + 1);
1766 solver_recurse_depth++;
1769 ret = solver(cr, blocks, xtype, outgrid, maxdiff);
1771 #ifdef STANDALONE_SOLVER
1772 solver_recurse_depth--;
1773 if (solver_show_working) {
1774 printf("%*sretracting %d at (%d,%d)\n",
1775 solver_recurse_depth*4, "", list[i], x + 1, y + 1);
1780 * If we have our first solution, copy it into the
1781 * grid we will return.
1783 if (diff == DIFF_IMPOSSIBLE && ret != DIFF_IMPOSSIBLE)
1784 memcpy(grid, outgrid, cr*cr);
1786 if (ret == DIFF_AMBIGUOUS)
1787 diff = DIFF_AMBIGUOUS;
1788 else if (ret == DIFF_IMPOSSIBLE)
1789 /* do not change our return value */;
1791 /* the recursion turned up exactly one solution */
1792 if (diff == DIFF_IMPOSSIBLE)
1793 diff = DIFF_RECURSIVE;
1795 diff = DIFF_AMBIGUOUS;
1799 * As soon as we've found more than one solution,
1800 * give up immediately.
1802 if (diff == DIFF_AMBIGUOUS)
1813 * We're forbidden to use recursion, so we just see whether
1814 * our grid is fully solved, and return DIFF_IMPOSSIBLE
1817 for (y = 0; y < cr; y++)
1818 for (x = 0; x < cr; x++)
1820 diff = DIFF_IMPOSSIBLE;
1825 #ifdef STANDALONE_SOLVER
1826 if (solver_show_working)
1827 printf("%*s%s found\n",
1828 solver_recurse_depth*4, "",
1829 diff == DIFF_IMPOSSIBLE ? "no solution" :
1830 diff == DIFF_AMBIGUOUS ? "multiple solutions" :
1840 solver_free_scratch(scratch);
1845 /* ----------------------------------------------------------------------
1846 * End of solver code.
1849 /* ----------------------------------------------------------------------
1850 * Solo filled-grid generator.
1852 * This grid generator works by essentially trying to solve a grid
1853 * starting from no clues, and not worrying that there's more than
1854 * one possible solution. Unfortunately, it isn't computationally
1855 * feasible to do this by calling the above solver with an empty
1856 * grid, because that one needs to allocate a lot of scratch space
1857 * at every recursion level. Instead, I have a much simpler
1858 * algorithm which I shamelessly copied from a Python solver
1859 * written by Andrew Wilkinson (which is GPLed, but I've reused
1860 * only ideas and no code). It mostly just does the obvious
1861 * recursive thing: pick an empty square, put one of the possible
1862 * digits in it, recurse until all squares are filled, backtrack
1863 * and change some choices if necessary.
1865 * The clever bit is that every time it chooses which square to
1866 * fill in next, it does so by counting the number of _possible_
1867 * numbers that can go in each square, and it prioritises so that
1868 * it picks a square with the _lowest_ number of possibilities. The
1869 * idea is that filling in lots of the obvious bits (particularly
1870 * any squares with only one possibility) will cut down on the list
1871 * of possibilities for other squares and hence reduce the enormous
1872 * search space as much as possible as early as possible.
1876 * Internal data structure used in gridgen to keep track of
1879 struct gridgen_coord { int x, y, r; };
1880 struct gridgen_usage {
1882 struct block_structure *blocks;
1883 /* grid is a copy of the input grid, modified as we go along */
1885 /* row[y*cr+n-1] TRUE if digit n has been placed in row y */
1887 /* col[x*cr+n-1] TRUE if digit n has been placed in row x */
1889 /* blk[(y*c+x)*cr+n-1] TRUE if digit n has been placed in block (x,y) */
1891 /* diag[i*cr+n-1] TRUE if digit n has been placed in diagonal i */
1892 unsigned char *diag;
1893 /* This lists all the empty spaces remaining in the grid. */
1894 struct gridgen_coord *spaces;
1896 /* If we need randomisation in the solve, this is our random state. */
1900 static void gridgen_place(struct gridgen_usage *usage, int x, int y, digit n,
1904 usage->row[y*cr+n-1] = usage->col[x*cr+n-1] =
1905 usage->blk[usage->blocks->whichblock[y*cr+x]*cr+n-1] = placing;
1907 if (ondiag0(y*cr+x))
1908 usage->diag[n-1] = placing;
1909 if (ondiag1(y*cr+x))
1910 usage->diag[cr+n-1] = placing;
1912 usage->grid[y*cr+x] = placing ? n : 0;
1916 * The real recursive step in the generating function.
1918 * Return values: 1 means solution found, 0 means no solution
1919 * found on this branch.
1921 static int gridgen_real(struct gridgen_usage *usage, digit *grid, int *steps)
1924 int i, j, n, sx, sy, bestm, bestr, ret;
1928 * Firstly, check for completion! If there are no spaces left
1929 * in the grid, we have a solution.
1931 if (usage->nspaces == 0)
1935 * Next, abandon generation if we went over our steps limit.
1942 * Otherwise, there must be at least one space. Find the most
1943 * constrained space, using the `r' field as a tie-breaker.
1945 bestm = cr+1; /* so that any space will beat it */
1948 for (j = 0; j < usage->nspaces; j++) {
1949 int x = usage->spaces[j].x, y = usage->spaces[j].y;
1953 * Find the number of digits that could go in this space.
1956 for (n = 0; n < cr; n++)
1957 if (!usage->row[y*cr+n] && !usage->col[x*cr+n] &&
1958 !usage->blk[usage->blocks->whichblock[y*cr+x]*cr+n] &&
1959 (!usage->diag || ((!ondiag0(y*cr+x) || !usage->diag[n]) &&
1960 (!ondiag1(y*cr+x) || !usage->diag[cr+n]))))
1963 if (m < bestm || (m == bestm && usage->spaces[j].r < bestr)) {
1965 bestr = usage->spaces[j].r;
1973 * Swap that square into the final place in the spaces array,
1974 * so that decrementing nspaces will remove it from the list.
1976 if (i != usage->nspaces-1) {
1977 struct gridgen_coord t;
1978 t = usage->spaces[usage->nspaces-1];
1979 usage->spaces[usage->nspaces-1] = usage->spaces[i];
1980 usage->spaces[i] = t;
1984 * Now we've decided which square to start our recursion at,
1985 * simply go through all possible values, shuffling them
1986 * randomly first if necessary.
1988 digits = snewn(bestm, int);
1990 for (n = 0; n < cr; n++)
1991 if (!usage->row[sy*cr+n] && !usage->col[sx*cr+n] &&
1992 !usage->blk[usage->blocks->whichblock[sy*cr+sx]*cr+n] &&
1993 (!usage->diag || ((!ondiag0(sy*cr+sx) || !usage->diag[n]) &&
1994 (!ondiag1(sy*cr+sx) || !usage->diag[cr+n])))) {
1999 shuffle(digits, j, sizeof(*digits), usage->rs);
2001 /* And finally, go through the digit list and actually recurse. */
2003 for (i = 0; i < j; i++) {
2006 /* Update the usage structure to reflect the placing of this digit. */
2007 gridgen_place(usage, sx, sy, n, TRUE);
2010 /* Call the solver recursively. Stop when we find a solution. */
2011 if (gridgen_real(usage, grid, steps)) {
2016 /* Revert the usage structure. */
2017 gridgen_place(usage, sx, sy, n, FALSE);
2026 * Entry point to generator. You give it parameters and a starting
2027 * grid, which is simply an array of cr*cr digits.
2029 static int gridgen(int cr, struct block_structure *blocks, int xtype,
2030 digit *grid, random_state *rs, int maxsteps)
2032 struct gridgen_usage *usage;
2036 * Clear the grid to start with.
2038 memset(grid, 0, cr*cr);
2041 * Create a gridgen_usage structure.
2043 usage = snew(struct gridgen_usage);
2046 usage->blocks = blocks;
2050 usage->row = snewn(cr * cr, unsigned char);
2051 usage->col = snewn(cr * cr, unsigned char);
2052 usage->blk = snewn(cr * cr, unsigned char);
2053 memset(usage->row, FALSE, cr * cr);
2054 memset(usage->col, FALSE, cr * cr);
2055 memset(usage->blk, FALSE, cr * cr);
2058 usage->diag = snewn(2 * cr, unsigned char);
2059 memset(usage->diag, FALSE, 2 * cr);
2065 * Begin by filling in the whole top row with randomly chosen
2066 * numbers. This cannot introduce any bias or restriction on
2067 * the available grids, since we already know those numbers
2068 * are all distinct so all we're doing is choosing their
2071 for (x = 0; x < cr; x++)
2073 shuffle(grid, cr, sizeof(*grid), rs);
2074 for (x = 0; x < cr; x++)
2075 gridgen_place(usage, x, 0, grid[x], TRUE);
2077 usage->spaces = snewn(cr * cr, struct gridgen_coord);
2083 * Initialise the list of grid spaces, taking care to leave
2084 * out the row I've already filled in above.
2086 for (y = 1; y < cr; y++) {
2087 for (x = 0; x < cr; x++) {
2088 usage->spaces[usage->nspaces].x = x;
2089 usage->spaces[usage->nspaces].y = y;
2090 usage->spaces[usage->nspaces].r = random_bits(rs, 31);
2096 * Run the real generator function.
2098 ret = gridgen_real(usage, grid, &maxsteps);
2101 * Clean up the usage structure now we have our answer.
2103 sfree(usage->spaces);
2112 /* ----------------------------------------------------------------------
2113 * End of grid generator code.
2117 * Check whether a grid contains a valid complete puzzle.
2119 static int check_valid(int cr, struct block_structure *blocks, int xtype,
2122 unsigned char *used;
2125 used = snewn(cr, unsigned char);
2128 * Check that each row contains precisely one of everything.
2130 for (y = 0; y < cr; y++) {
2131 memset(used, FALSE, cr);
2132 for (x = 0; x < cr; x++)
2133 if (grid[y*cr+x] > 0 && grid[y*cr+x] <= cr)
2134 used[grid[y*cr+x]-1] = TRUE;
2135 for (n = 0; n < cr; n++)
2143 * Check that each column contains precisely one of everything.
2145 for (x = 0; x < cr; x++) {
2146 memset(used, FALSE, cr);
2147 for (y = 0; y < cr; y++)
2148 if (grid[y*cr+x] > 0 && grid[y*cr+x] <= cr)
2149 used[grid[y*cr+x]-1] = TRUE;
2150 for (n = 0; n < cr; n++)
2158 * Check that each block contains precisely one of everything.
2160 for (i = 0; i < cr; i++) {
2161 memset(used, FALSE, cr);
2162 for (j = 0; j < cr; j++)
2163 if (grid[blocks->blocks[i][j]] > 0 &&
2164 grid[blocks->blocks[i][j]] <= cr)
2165 used[grid[blocks->blocks[i][j]]-1] = TRUE;
2166 for (n = 0; n < cr; n++)
2174 * Check that each diagonal contains precisely one of everything.
2177 memset(used, FALSE, cr);
2178 for (i = 0; i < cr; i++)
2179 if (grid[diag0(i)] > 0 && grid[diag0(i)] <= cr)
2180 used[grid[diag0(i)]-1] = TRUE;
2181 for (n = 0; n < cr; n++)
2186 for (i = 0; i < cr; i++)
2187 if (grid[diag1(i)] > 0 && grid[diag1(i)] <= cr)
2188 used[grid[diag1(i)]-1] = TRUE;
2189 for (n = 0; n < cr; n++)
2200 static int symmetries(game_params *params, int x, int y, int *output, int s)
2202 int c = params->c, r = params->r, cr = c*r;
2205 #define ADD(x,y) (*output++ = (x), *output++ = (y), i++)
2211 break; /* just x,y is all we need */
2213 ADD(cr - 1 - x, cr - 1 - y);
2218 ADD(cr - 1 - x, cr - 1 - y);
2229 ADD(cr - 1 - x, cr - 1 - y);
2233 ADD(cr - 1 - x, cr - 1 - y);
2234 ADD(cr - 1 - y, cr - 1 - x);
2239 ADD(cr - 1 - x, cr - 1 - y);
2243 ADD(cr - 1 - y, cr - 1 - x);
2252 static char *encode_solve_move(int cr, digit *grid)
2255 char *ret, *p, *sep;
2258 * It's surprisingly easy to work out _exactly_ how long this
2259 * string needs to be. To decimal-encode all the numbers from 1
2262 * - every number has a units digit; total is n.
2263 * - all numbers above 9 have a tens digit; total is max(n-9,0).
2264 * - all numbers above 99 have a hundreds digit; total is max(n-99,0).
2268 for (i = 1; i <= cr; i *= 10)
2269 len += max(cr - i + 1, 0);
2270 len += cr; /* don't forget the commas */
2271 len *= cr; /* there are cr rows of these */
2274 * Now len is one bigger than the total size of the
2275 * comma-separated numbers (because we counted an
2276 * additional leading comma). We need to have a leading S
2277 * and a trailing NUL, so we're off by one in total.
2281 ret = snewn(len, char);
2285 for (i = 0; i < cr*cr; i++) {
2286 p += sprintf(p, "%s%d", sep, grid[i]);
2290 assert(p - ret == len);
2295 static char *new_game_desc(game_params *params, random_state *rs,
2296 char **aux, int interactive)
2298 int c = params->c, r = params->r, cr = c*r;
2300 struct block_structure *blocks;
2301 digit *grid, *grid2;
2302 struct xy { int x, y; } *locs;
2305 int coords[16], ncoords;
2310 * Adjust the maximum difficulty level to be consistent with
2311 * the puzzle size: all 2x2 puzzles appear to be Trivial
2312 * (DIFF_BLOCK) so we cannot hold out for even a Basic
2313 * (DIFF_SIMPLE) one.
2315 maxdiff = params->diff;
2316 if (c == 2 && r == 2)
2317 maxdiff = DIFF_BLOCK;
2319 grid = snewn(area, digit);
2320 locs = snewn(area, struct xy);
2321 grid2 = snewn(area, digit);
2323 blocks = snew(struct block_structure);
2324 blocks->c = params->c; blocks->r = params->r;
2325 blocks->whichblock = snewn(area*2, int);
2326 blocks->blocks = snewn(cr, int *);
2327 for (i = 0; i < cr; i++)
2328 blocks->blocks[i] = blocks->whichblock + area + i*cr;
2329 #ifdef STANDALONE_SOLVER
2330 assert(!"This should never happen, so we don't need to create blocknames");
2334 * Loop until we get a grid of the required difficulty. This is
2335 * nasty, but it seems to be unpleasantly hard to generate
2336 * difficult grids otherwise.
2340 * Generate a random solved state, starting by
2341 * constructing the block structure.
2343 if (r == 1) { /* jigsaw mode */
2344 int *dsf = divvy_rectangle(cr, cr, cr, rs);
2347 for (i = 0; i < area; i++)
2348 blocks->whichblock[i] = -1;
2349 for (i = 0; i < area; i++) {
2350 int j = dsf_canonify(dsf, i);
2351 if (blocks->whichblock[j] < 0)
2352 blocks->whichblock[j] = nb++;
2353 blocks->whichblock[i] = blocks->whichblock[j];
2358 } else { /* basic Sudoku mode */
2359 for (y = 0; y < cr; y++)
2360 for (x = 0; x < cr; x++)
2361 blocks->whichblock[y*cr+x] = (y/c) * c + (x/r);
2363 for (i = 0; i < cr; i++)
2364 blocks->blocks[i][cr-1] = 0;
2365 for (i = 0; i < area; i++) {
2366 int b = blocks->whichblock[i];
2367 j = blocks->blocks[b][cr-1]++;
2369 blocks->blocks[b][j] = i;
2372 if (!gridgen(cr, blocks, params->xtype, grid, rs, area*area))
2374 assert(check_valid(cr, blocks, params->xtype, grid));
2377 * Save the solved grid in aux.
2381 * We might already have written *aux the last time we
2382 * went round this loop, in which case we should free
2383 * the old aux before overwriting it with the new one.
2389 *aux = encode_solve_move(cr, grid);
2393 * Now we have a solved grid, start removing things from it
2394 * while preserving solubility.
2398 * Find the set of equivalence classes of squares permitted
2399 * by the selected symmetry. We do this by enumerating all
2400 * the grid squares which have no symmetric companion
2401 * sorting lower than themselves.
2404 for (y = 0; y < cr; y++)
2405 for (x = 0; x < cr; x++) {
2409 ncoords = symmetries(params, x, y, coords, params->symm);
2410 for (j = 0; j < ncoords; j++)
2411 if (coords[2*j+1]*cr+coords[2*j] < i)
2421 * Now shuffle that list.
2423 shuffle(locs, nlocs, sizeof(*locs), rs);
2426 * Now loop over the shuffled list and, for each element,
2427 * see whether removing that element (and its reflections)
2428 * from the grid will still leave the grid soluble.
2430 for (i = 0; i < nlocs; i++) {
2436 memcpy(grid2, grid, area);
2437 ncoords = symmetries(params, x, y, coords, params->symm);
2438 for (j = 0; j < ncoords; j++)
2439 grid2[coords[2*j+1]*cr+coords[2*j]] = 0;
2441 ret = solver(cr, blocks, params->xtype, grid2, maxdiff);
2442 if (ret <= maxdiff) {
2443 for (j = 0; j < ncoords; j++)
2444 grid[coords[2*j+1]*cr+coords[2*j]] = 0;
2448 memcpy(grid2, grid, area);
2450 if (solver(cr, blocks, params->xtype, grid2, maxdiff) == maxdiff)
2451 break; /* found one! */
2458 * Now we have the grid as it will be presented to the user.
2459 * Encode it in a game desc.
2465 desc = snewn(7 * area, char);
2468 for (i = 0; i <= area; i++) {
2469 int n = (i < area ? grid[i] : -1);
2476 int c = 'a' - 1 + run;
2480 run -= c - ('a' - 1);
2484 * If there's a number in the very top left or
2485 * bottom right, there's no point putting an
2486 * unnecessary _ before or after it.
2488 if (p > desc && n > 0)
2492 p += sprintf(p, "%d", n);
2503 * Encode the block structure. We do this by encoding
2504 * the pattern of dividing lines: first we iterate
2505 * over the cr*(cr-1) internal vertical grid lines in
2506 * ordinary reading order, then over the cr*(cr-1)
2507 * internal horizontal ones in transposed reading
2510 * We encode the number of non-lines between the
2511 * lines; _ means zero (two adjacent divisions), a
2512 * means 1, ..., y means 25, and z means 25 non-lines
2513 * _and no following line_ (so that za means 26, zb 27
2516 for (i = 0; i <= 2*cr*(cr-1); i++) {
2519 if (i == 2*cr*(cr-1)) {
2520 edge = TRUE; /* terminating virtual edge */
2522 if (i < cr*(cr-1)) {
2533 edge = (blocks->whichblock[p0] != blocks->whichblock[p1]);
2537 while (currrun > 25)
2538 *p++ = 'z', currrun -= 25;
2540 *p++ = 'a'-1 + currrun;
2549 assert(p - desc < 7 * area);
2551 desc = sresize(desc, p - desc, char);
2559 static char *validate_desc(game_params *params, char *desc)
2561 int cr = params->c * params->r, area = cr*cr;
2565 while (*desc && *desc != ',') {
2567 if (n >= 'a' && n <= 'z') {
2568 squares += n - 'a' + 1;
2569 } else if (n == '_') {
2571 } else if (n > '0' && n <= '9') {
2572 int val = atoi(desc-1);
2573 if (val < 1 || val > params->c * params->r)
2574 return "Out-of-range number in game description";
2576 while (*desc >= '0' && *desc <= '9')
2579 return "Invalid character in game description";
2583 return "Not enough data to fill grid";
2586 return "Too much data to fit in grid";
2588 if (params->r == 1) {
2592 * Now we expect a suffix giving the jigsaw block
2593 * structure. Parse it and validate that it divides the
2594 * grid into the right number of regions which are the
2598 return "Expected jigsaw block structure in game description";
2601 dsf = snew_dsf(area);
2609 else if (*desc >= 'a' && *desc <= 'z')
2610 c = *desc - 'a' + 1;
2613 return "Invalid character in game description";
2617 adv = (c != 25); /* 'z' is a special case */
2623 * Non-edge; merge the two dsf classes on either
2626 if (pos >= 2*cr*(cr-1)) {
2628 return "Too much data in block structure specification";
2629 } else if (pos < cr*(cr-1)) {
2635 int x = pos/(cr-1) - cr;
2640 dsf_merge(dsf, p0, p1);
2649 * When desc is exhausted, we expect to have gone exactly
2650 * one space _past_ the end of the grid, due to the dummy
2653 if (pos != 2*cr*(cr-1)+1) {
2655 return "Not enough data in block structure specification";
2659 * Now we've got our dsf. Verify that it matches
2663 int *canons, *counts;
2664 int i, j, c, ncanons = 0;
2666 canons = snewn(cr, int);
2667 counts = snewn(cr, int);
2669 for (i = 0; i < area; i++) {
2670 j = dsf_canonify(dsf, i);
2672 for (c = 0; c < ncanons; c++)
2673 if (canons[c] == j) {
2675 if (counts[c] > cr) {
2679 return "A jigsaw block is too big";
2685 if (ncanons >= cr) {
2689 return "Too many distinct jigsaw blocks";
2691 canons[ncanons] = j;
2692 counts[ncanons] = 1;
2698 * If we've managed to get through that loop without
2699 * tripping either of the error conditions, then we
2700 * must have partitioned the entire grid into at most
2701 * cr blocks of at most cr squares each; therefore we
2702 * must have _exactly_ cr blocks of _exactly_ cr
2703 * squares each. I'll verify that by assertion just in
2704 * case something has gone horribly wrong, but it
2705 * shouldn't have been able to happen by duff input,
2706 * only by a bug in the above code.
2708 assert(ncanons == cr);
2709 for (c = 0; c < ncanons; c++)
2710 assert(counts[c] == cr);
2719 return "Unexpected jigsaw block structure in game description";
2725 static game_state *new_game(midend *me, game_params *params, char *desc)
2727 game_state *state = snew(game_state);
2728 int c = params->c, r = params->r, cr = c*r, area = cr * cr;
2732 state->xtype = params->xtype;
2734 state->grid = snewn(area, digit);
2735 state->pencil = snewn(area * cr, unsigned char);
2736 memset(state->pencil, 0, area * cr);
2737 state->immutable = snewn(area, unsigned char);
2738 memset(state->immutable, FALSE, area);
2740 state->blocks = snew(struct block_structure);
2741 state->blocks->c = c; state->blocks->r = r;
2742 state->blocks->refcount = 1;
2743 state->blocks->whichblock = snewn(area*2, int);
2744 state->blocks->blocks = snewn(cr, int *);
2745 for (i = 0; i < cr; i++)
2746 state->blocks->blocks[i] = state->blocks->whichblock + area + i*cr;
2747 #ifdef STANDALONE_SOLVER
2748 state->blocks->blocknames = (char **)smalloc(cr*(sizeof(char *)+80));
2751 state->completed = state->cheated = FALSE;
2754 while (*desc && *desc != ',') {
2756 if (n >= 'a' && n <= 'z') {
2757 int run = n - 'a' + 1;
2758 assert(i + run <= area);
2760 state->grid[i++] = 0;
2761 } else if (n == '_') {
2763 } else if (n > '0' && n <= '9') {
2765 state->immutable[i] = TRUE;
2766 state->grid[i++] = atoi(desc-1);
2767 while (*desc >= '0' && *desc <= '9')
2770 assert(!"We can't get here");
2780 assert(*desc == ',');
2782 dsf = snew_dsf(area);
2790 else if (*desc >= 'a' && *desc <= 'z')
2791 c = *desc - 'a' + 1;
2793 assert(!"Shouldn't get here");
2796 adv = (c != 25); /* 'z' is a special case */
2802 * Non-edge; merge the two dsf classes on either
2805 assert(pos < 2*cr*(cr-1));
2806 if (pos < cr*(cr-1)) {
2812 int x = pos/(cr-1) - cr;
2817 dsf_merge(dsf, p0, p1);
2826 * When desc is exhausted, we expect to have gone exactly
2827 * one space _past_ the end of the grid, due to the dummy
2830 assert(pos == 2*cr*(cr-1)+1);
2833 * Now we've got our dsf. Translate it into a block
2837 for (i = 0; i < area; i++)
2838 state->blocks->whichblock[i] = -1;
2839 for (i = 0; i < area; i++) {
2840 int j = dsf_canonify(dsf, i);
2841 if (state->blocks->whichblock[j] < 0)
2842 state->blocks->whichblock[j] = nb++;
2843 state->blocks->whichblock[i] = state->blocks->whichblock[j];
2853 for (y = 0; y < cr; y++)
2854 for (x = 0; x < cr; x++)
2855 state->blocks->whichblock[y*cr+x] = (y/c) * c + (x/r);
2859 * Having sorted out whichblock[], set up the block index arrays.
2861 for (i = 0; i < cr; i++)
2862 state->blocks->blocks[i][cr-1] = 0;
2863 for (i = 0; i < area; i++) {
2864 int b = state->blocks->whichblock[i];
2865 int j = state->blocks->blocks[b][cr-1]++;
2867 state->blocks->blocks[b][j] = i;
2870 #ifdef STANDALONE_SOLVER
2872 * Set up the block names for solver diagnostic output.
2875 char *p = (char *)(state->blocks->blocknames + cr);
2878 for (i = 0; i < cr; i++)
2879 state->blocks->blocknames[i] = NULL;
2881 for (i = 0; i < area; i++) {
2882 int j = state->blocks->whichblock[i];
2883 if (!state->blocks->blocknames[j]) {
2884 state->blocks->blocknames[j] = p;
2885 p += 1 + sprintf(p, "starting at (%d,%d)",
2886 1 + i%cr, 1 + i/cr);
2891 for (by = 0; by < r; by++)
2892 for (bx = 0; bx < c; bx++) {
2893 state->blocks->blocknames[by*c+bx] = p;
2894 p += 1 + sprintf(p, "(%d,%d)", bx+1, by+1);
2897 assert(p - (char *)state->blocks->blocknames < cr*(sizeof(char *)+80));
2898 for (i = 0; i < cr; i++)
2899 assert(state->blocks->blocknames[i]);
2906 static game_state *dup_game(game_state *state)
2908 game_state *ret = snew(game_state);
2909 int cr = state->cr, area = cr * cr;
2911 ret->cr = state->cr;
2912 ret->xtype = state->xtype;
2914 ret->blocks = state->blocks;
2915 ret->blocks->refcount++;
2917 ret->grid = snewn(area, digit);
2918 memcpy(ret->grid, state->grid, area);
2920 ret->pencil = snewn(area * cr, unsigned char);
2921 memcpy(ret->pencil, state->pencil, area * cr);
2923 ret->immutable = snewn(area, unsigned char);
2924 memcpy(ret->immutable, state->immutable, area);
2926 ret->completed = state->completed;
2927 ret->cheated = state->cheated;
2932 static void free_game(game_state *state)
2934 if (--state->blocks->refcount == 0) {
2935 sfree(state->blocks->whichblock);
2936 sfree(state->blocks->blocks);
2937 #ifdef STANDALONE_SOLVER
2938 sfree(state->blocks->blocknames);
2940 sfree(state->blocks);
2942 sfree(state->immutable);
2943 sfree(state->pencil);
2948 static char *solve_game(game_state *state, game_state *currstate,
2949 char *ai, char **error)
2957 * If we already have the solution in ai, save ourselves some
2963 grid = snewn(cr*cr, digit);
2964 memcpy(grid, state->grid, cr*cr);
2965 solve_ret = solver(cr, state->blocks, state->xtype, grid, DIFF_RECURSIVE);
2969 if (solve_ret == DIFF_IMPOSSIBLE)
2970 *error = "No solution exists for this puzzle";
2971 else if (solve_ret == DIFF_AMBIGUOUS)
2972 *error = "Multiple solutions exist for this puzzle";
2979 ret = encode_solve_move(cr, grid);
2986 static char *grid_text_format(int cr, struct block_structure *blocks,
2987 int xtype, digit *grid)
2991 int totallen, linelen, nlines;
2995 * For non-jigsaw Sudoku, we format in the way we always have,
2996 * by having the digits unevenly spaced so that the dividing
3005 * For jigsaw puzzles, however, we must leave space between
3006 * _all_ pairs of digits for an optional dividing line, so we
3007 * have to move to the rather ugly
3017 * We deal with both cases using the same formatting code; we
3018 * simply invent a vmod value such that there's a vertical
3019 * dividing line before column i iff i is divisible by vmod
3020 * (so it's r in the first case and 1 in the second), and hmod
3021 * likewise for horizontal dividing lines.
3024 if (blocks->r != 1) {
3032 * Line length: we have cr digits, each with a space after it,
3033 * and (cr-1)/vmod dividing lines, each with a space after it.
3034 * The final space is replaced by a newline, but that doesn't
3035 * affect the length.
3037 linelen = 2*(cr + (cr-1)/vmod);
3040 * Number of lines: we have cr rows of digits, and (cr-1)/hmod
3043 nlines = cr + (cr-1)/hmod;
3046 * Allocate the space.
3048 totallen = linelen * nlines;
3049 ret = snewn(totallen+1, char); /* leave room for terminating NUL */
3055 for (y = 0; y < cr; y++) {
3059 for (x = 0; x < cr; x++) {
3063 digit d = grid[y*cr+x];
3067 * Empty space: we usually write a dot, but we'll
3068 * highlight spaces on the X-diagonals (in X mode)
3069 * by using underscores instead.
3071 if (xtype && (ondiag0(y*cr+x) || ondiag1(y*cr+x)))
3075 } else if (d <= 9) {
3092 * Optional dividing line.
3094 if (blocks->whichblock[y*cr+x] != blocks->whichblock[y*cr+x+1])
3101 if (y == cr-1 || (y+1) % hmod)
3107 for (x = 0; x < cr; x++) {
3112 * Division between two squares. This varies
3113 * complicatedly in length.
3115 dwid = 2; /* digit and its following space */
3117 dwid--; /* no following space at end of line */
3118 if (x > 0 && x % vmod == 0)
3119 dwid++; /* preceding space after a divider */
3121 if (blocks->whichblock[y*cr+x] != blocks->whichblock[(y+1)*cr+x])
3138 * Corner square. This is:
3139 * - a space if all four surrounding squares are in
3141 * - a vertical line if the two left ones are in one
3142 * block and the two right in another
3143 * - a horizontal line if the two top ones are in one
3144 * block and the two bottom in another
3145 * - a plus sign in all other cases. (If we had a
3146 * richer character set available we could break
3147 * this case up further by doing fun things with
3148 * line-drawing T-pieces.)
3150 tl = blocks->whichblock[y*cr+x];
3151 tr = blocks->whichblock[y*cr+x+1];
3152 bl = blocks->whichblock[(y+1)*cr+x];
3153 br = blocks->whichblock[(y+1)*cr+x+1];
3155 if (tl == tr && tr == bl && bl == br)
3157 else if (tl == bl && tr == br)
3159 else if (tl == tr && bl == br)
3168 assert(p - ret == totallen);
3173 static char *game_text_format(game_state *state)
3175 return grid_text_format(state->cr, state->blocks, state->xtype,
3181 * These are the coordinates of the currently highlighted
3182 * square on the grid, or -1,-1 if there isn't one. When there
3183 * is, pressing a valid number or letter key or Space will
3184 * enter that number or letter in the grid.
3188 * This indicates whether the current highlight is a
3189 * pencil-mark one or a real one.
3194 static game_ui *new_ui(game_state *state)
3196 game_ui *ui = snew(game_ui);
3198 ui->hx = ui->hy = -1;
3204 static void free_ui(game_ui *ui)
3209 static char *encode_ui(game_ui *ui)
3214 static void decode_ui(game_ui *ui, char *encoding)
3218 static void game_changed_state(game_ui *ui, game_state *oldstate,
3219 game_state *newstate)
3221 int cr = newstate->cr;
3223 * We prevent pencil-mode highlighting of a filled square. So
3224 * if the user has just filled in a square which we had a
3225 * pencil-mode highlight in (by Undo, or by Redo, or by Solve),
3226 * then we cancel the highlight.
3228 if (ui->hx >= 0 && ui->hy >= 0 && ui->hpencil &&
3229 newstate->grid[ui->hy * cr + ui->hx] != 0) {
3230 ui->hx = ui->hy = -1;
3234 struct game_drawstate {
3239 unsigned char *pencil;
3241 /* This is scratch space used within a single call to game_redraw. */
3245 static char *interpret_move(game_state *state, game_ui *ui, game_drawstate *ds,
3246 int x, int y, int button)
3252 button &= ~MOD_MASK;
3254 tx = (x + TILE_SIZE - BORDER) / TILE_SIZE - 1;
3255 ty = (y + TILE_SIZE - BORDER) / TILE_SIZE - 1;
3257 if (tx >= 0 && tx < cr && ty >= 0 && ty < cr) {
3258 if (button == LEFT_BUTTON) {
3259 if (state->immutable[ty*cr+tx]) {
3260 ui->hx = ui->hy = -1;
3261 } else if (tx == ui->hx && ty == ui->hy && ui->hpencil == 0) {
3262 ui->hx = ui->hy = -1;
3268 return ""; /* UI activity occurred */
3270 if (button == RIGHT_BUTTON) {
3272 * Pencil-mode highlighting for non filled squares.
3274 if (state->grid[ty*cr+tx] == 0) {
3275 if (tx == ui->hx && ty == ui->hy && ui->hpencil) {
3276 ui->hx = ui->hy = -1;
3283 ui->hx = ui->hy = -1;
3285 return ""; /* UI activity occurred */
3289 if (ui->hx != -1 && ui->hy != -1 &&
3290 ((button >= '1' && button <= '9' && button - '0' <= cr) ||
3291 (button >= 'a' && button <= 'z' && button - 'a' + 10 <= cr) ||
3292 (button >= 'A' && button <= 'Z' && button - 'A' + 10 <= cr) ||
3293 button == ' ' || button == '\010' || button == '\177')) {
3294 int n = button - '0';
3295 if (button >= 'A' && button <= 'Z')
3296 n = button - 'A' + 10;
3297 if (button >= 'a' && button <= 'z')
3298 n = button - 'a' + 10;
3299 if (button == ' ' || button == '\010' || button == '\177')
3303 * Can't overwrite this square. In principle this shouldn't
3304 * happen anyway because we should never have even been
3305 * able to highlight the square, but it never hurts to be
3308 if (state->immutable[ui->hy*cr+ui->hx])
3312 * Can't make pencil marks in a filled square. In principle
3313 * this shouldn't happen anyway because we should never
3314 * have even been able to pencil-highlight the square, but
3315 * it never hurts to be careful.
3317 if (ui->hpencil && state->grid[ui->hy*cr+ui->hx])
3320 sprintf(buf, "%c%d,%d,%d",
3321 (char)(ui->hpencil && n > 0 ? 'P' : 'R'), ui->hx, ui->hy, n);
3323 ui->hx = ui->hy = -1;
3331 static game_state *execute_move(game_state *from, char *move)
3337 if (move[0] == 'S') {
3340 ret = dup_game(from);
3341 ret->completed = ret->cheated = TRUE;
3344 for (n = 0; n < cr*cr; n++) {
3345 ret->grid[n] = atoi(p);
3347 if (!*p || ret->grid[n] < 1 || ret->grid[n] > cr) {
3352 while (*p && isdigit((unsigned char)*p)) p++;
3357 } else if ((move[0] == 'P' || move[0] == 'R') &&
3358 sscanf(move+1, "%d,%d,%d", &x, &y, &n) == 3 &&
3359 x >= 0 && x < cr && y >= 0 && y < cr && n >= 0 && n <= cr) {
3361 ret = dup_game(from);
3362 if (move[0] == 'P' && n > 0) {
3363 int index = (y*cr+x) * cr + (n-1);
3364 ret->pencil[index] = !ret->pencil[index];
3366 ret->grid[y*cr+x] = n;
3367 memset(ret->pencil + (y*cr+x)*cr, 0, cr);
3370 * We've made a real change to the grid. Check to see
3371 * if the game has been completed.
3373 if (!ret->completed && check_valid(cr, ret->blocks, ret->xtype,
3375 ret->completed = TRUE;
3380 return NULL; /* couldn't parse move string */
3383 /* ----------------------------------------------------------------------
3387 #define SIZE(cr) ((cr) * TILE_SIZE + 2*BORDER + 1)
3388 #define GETTILESIZE(cr, w) ( (double)(w-1) / (double)(cr+1) )
3390 static void game_compute_size(game_params *params, int tilesize,
3393 /* Ick: fake up `ds->tilesize' for macro expansion purposes */
3394 struct { int tilesize; } ads, *ds = &ads;
3395 ads.tilesize = tilesize;
3397 *x = SIZE(params->c * params->r);
3398 *y = SIZE(params->c * params->r);
3401 static void game_set_size(drawing *dr, game_drawstate *ds,
3402 game_params *params, int tilesize)
3404 ds->tilesize = tilesize;
3407 static float *game_colours(frontend *fe, int *ncolours)
3409 float *ret = snewn(3 * NCOLOURS, float);
3411 frontend_default_colour(fe, &ret[COL_BACKGROUND * 3]);
3413 ret[COL_XDIAGONALS * 3 + 0] = 0.9F * ret[COL_BACKGROUND * 3 + 0];
3414 ret[COL_XDIAGONALS * 3 + 1] = 0.9F * ret[COL_BACKGROUND * 3 + 1];
3415 ret[COL_XDIAGONALS * 3 + 2] = 0.9F * ret[COL_BACKGROUND * 3 + 2];
3417 ret[COL_GRID * 3 + 0] = 0.0F;
3418 ret[COL_GRID * 3 + 1] = 0.0F;
3419 ret[COL_GRID * 3 + 2] = 0.0F;
3421 ret[COL_CLUE * 3 + 0] = 0.0F;
3422 ret[COL_CLUE * 3 + 1] = 0.0F;
3423 ret[COL_CLUE * 3 + 2] = 0.0F;
3425 ret[COL_USER * 3 + 0] = 0.0F;
3426 ret[COL_USER * 3 + 1] = 0.6F * ret[COL_BACKGROUND * 3 + 1];
3427 ret[COL_USER * 3 + 2] = 0.0F;
3429 ret[COL_HIGHLIGHT * 3 + 0] = 0.78F * ret[COL_BACKGROUND * 3 + 0];
3430 ret[COL_HIGHLIGHT * 3 + 1] = 0.78F * ret[COL_BACKGROUND * 3 + 1];
3431 ret[COL_HIGHLIGHT * 3 + 2] = 0.78F * ret[COL_BACKGROUND * 3 + 2];
3433 ret[COL_ERROR * 3 + 0] = 1.0F;
3434 ret[COL_ERROR * 3 + 1] = 0.0F;
3435 ret[COL_ERROR * 3 + 2] = 0.0F;
3437 ret[COL_PENCIL * 3 + 0] = 0.5F * ret[COL_BACKGROUND * 3 + 0];
3438 ret[COL_PENCIL * 3 + 1] = 0.5F * ret[COL_BACKGROUND * 3 + 1];
3439 ret[COL_PENCIL * 3 + 2] = ret[COL_BACKGROUND * 3 + 2];
3441 *ncolours = NCOLOURS;
3445 static game_drawstate *game_new_drawstate(drawing *dr, game_state *state)
3447 struct game_drawstate *ds = snew(struct game_drawstate);
3450 ds->started = FALSE;
3452 ds->xtype = state->xtype;
3453 ds->grid = snewn(cr*cr, digit);
3454 memset(ds->grid, cr+2, cr*cr);
3455 ds->pencil = snewn(cr*cr*cr, digit);
3456 memset(ds->pencil, 0, cr*cr*cr);
3457 ds->hl = snewn(cr*cr, unsigned char);
3458 memset(ds->hl, 0, cr*cr);
3459 ds->entered_items = snewn(cr*cr, int);
3460 ds->tilesize = 0; /* not decided yet */
3464 static void game_free_drawstate(drawing *dr, game_drawstate *ds)
3469 sfree(ds->entered_items);
3473 static void draw_number(drawing *dr, game_drawstate *ds, game_state *state,
3474 int x, int y, int hl)
3481 if (ds->grid[y*cr+x] == state->grid[y*cr+x] &&
3482 ds->hl[y*cr+x] == hl &&
3483 !memcmp(ds->pencil+(y*cr+x)*cr, state->pencil+(y*cr+x)*cr, cr))
3484 return; /* no change required */
3486 tx = BORDER + x * TILE_SIZE + 1 + GRIDEXTRA;
3487 ty = BORDER + y * TILE_SIZE + 1 + GRIDEXTRA;
3491 cw = TILE_SIZE-1-2*GRIDEXTRA;
3492 ch = TILE_SIZE-1-2*GRIDEXTRA;
3494 if (x > 0 && state->blocks->whichblock[y*cr+x] == state->blocks->whichblock[y*cr+x-1])
3495 cx -= GRIDEXTRA, cw += GRIDEXTRA;
3496 if (x+1 < cr && state->blocks->whichblock[y*cr+x] == state->blocks->whichblock[y*cr+x+1])
3498 if (y > 0 && state->blocks->whichblock[y*cr+x] == state->blocks->whichblock[(y-1)*cr+x])
3499 cy -= GRIDEXTRA, ch += GRIDEXTRA;
3500 if (y+1 < cr && state->blocks->whichblock[y*cr+x] == state->blocks->whichblock[(y+1)*cr+x])
3503 clip(dr, cx, cy, cw, ch);
3505 /* background needs erasing */
3506 draw_rect(dr, cx, cy, cw, ch,
3507 ((hl & 15) == 1 ? COL_HIGHLIGHT :
3508 (ds->xtype && (ondiag0(y*cr+x) || ondiag1(y*cr+x))) ? COL_XDIAGONALS :
3512 * Draw the corners of thick lines in corner-adjacent squares,
3513 * which jut into this square by one pixel.
3515 if (x > 0 && y > 0 && state->blocks->whichblock[y*cr+x] != state->blocks->whichblock[(y-1)*cr+x-1])
3516 draw_rect(dr, tx-GRIDEXTRA, ty-GRIDEXTRA, GRIDEXTRA, GRIDEXTRA, COL_GRID);
3517 if (x+1 < cr && y > 0 && state->blocks->whichblock[y*cr+x] != state->blocks->whichblock[(y-1)*cr+x+1])
3518 draw_rect(dr, tx+TILE_SIZE-1-2*GRIDEXTRA, ty-GRIDEXTRA, GRIDEXTRA, GRIDEXTRA, COL_GRID);
3519 if (x > 0 && y+1 < cr && state->blocks->whichblock[y*cr+x] != state->blocks->whichblock[(y+1)*cr+x-1])
3520 draw_rect(dr, tx-GRIDEXTRA, ty+TILE_SIZE-1-2*GRIDEXTRA, GRIDEXTRA, GRIDEXTRA, COL_GRID);
3521 if (x+1 < cr && y+1 < cr && state->blocks->whichblock[y*cr+x] != state->blocks->whichblock[(y+1)*cr+x+1])
3522 draw_rect(dr, tx+TILE_SIZE-1-2*GRIDEXTRA, ty+TILE_SIZE-1-2*GRIDEXTRA, GRIDEXTRA, GRIDEXTRA, COL_GRID);
3524 /* pencil-mode highlight */
3525 if ((hl & 15) == 2) {
3529 coords[2] = cx+cw/2;
3532 coords[5] = cy+ch/2;
3533 draw_polygon(dr, coords, 3, COL_HIGHLIGHT, COL_HIGHLIGHT);
3536 /* new number needs drawing? */
3537 if (state->grid[y*cr+x]) {
3539 str[0] = state->grid[y*cr+x] + '0';
3541 str[0] += 'a' - ('9'+1);
3542 draw_text(dr, tx + TILE_SIZE/2, ty + TILE_SIZE/2,
3543 FONT_VARIABLE, TILE_SIZE/2, ALIGN_VCENTRE | ALIGN_HCENTRE,
3544 state->immutable[y*cr+x] ? COL_CLUE : (hl & 16) ? COL_ERROR : COL_USER, str);
3547 int pw, ph, pmax, fontsize;
3549 /* count the pencil marks required */
3550 for (i = npencil = 0; i < cr; i++)
3551 if (state->pencil[(y*cr+x)*cr+i])
3555 * It's not sensible to arrange pencil marks in the same
3556 * layout as the squares within a block, because this leads
3557 * to the font being too small. Instead, we arrange pencil
3558 * marks in the nearest thing we can to a square layout,
3559 * and we adjust the square layout depending on the number
3560 * of pencil marks in the square.
3562 for (pw = 1; pw * pw < npencil; pw++);
3563 if (pw < 3) pw = 3; /* otherwise it just looks _silly_ */
3564 ph = (npencil + pw - 1) / pw;
3565 if (ph < 2) ph = 2; /* likewise */
3567 fontsize = TILE_SIZE/(pmax*(11-pmax)/8);
3569 for (i = j = 0; i < cr; i++)
3570 if (state->pencil[(y*cr+x)*cr+i]) {
3571 int dx = j % pw, dy = j / pw;
3576 str[0] += 'a' - ('9'+1);
3577 draw_text(dr, tx + (4*dx+3) * TILE_SIZE / (4*pw+2),
3578 ty + (4*dy+3) * TILE_SIZE / (4*ph+2),
3579 FONT_VARIABLE, fontsize,
3580 ALIGN_VCENTRE | ALIGN_HCENTRE, COL_PENCIL, str);
3587 draw_update(dr, cx, cy, cw, ch);
3589 ds->grid[y*cr+x] = state->grid[y*cr+x];
3590 memcpy(ds->pencil+(y*cr+x)*cr, state->pencil+(y*cr+x)*cr, cr);
3591 ds->hl[y*cr+x] = hl;
3594 static void game_redraw(drawing *dr, game_drawstate *ds, game_state *oldstate,
3595 game_state *state, int dir, game_ui *ui,
3596 float animtime, float flashtime)
3603 * The initial contents of the window are not guaranteed
3604 * and can vary with front ends. To be on the safe side,
3605 * all games should start by drawing a big
3606 * background-colour rectangle covering the whole window.
3608 draw_rect(dr, 0, 0, SIZE(cr), SIZE(cr), COL_BACKGROUND);
3611 * Draw the grid. We draw it as a big thick rectangle of
3612 * COL_GRID initially; individual calls to draw_number()
3613 * will poke the right-shaped holes in it.
3615 draw_rect(dr, BORDER-GRIDEXTRA, BORDER-GRIDEXTRA,
3616 cr*TILE_SIZE+1+2*GRIDEXTRA, cr*TILE_SIZE+1+2*GRIDEXTRA,
3621 * This array is used to keep track of rows, columns and boxes
3622 * which contain a number more than once.
3624 for (x = 0; x < cr * cr; x++)
3625 ds->entered_items[x] = 0;
3626 for (x = 0; x < cr; x++)
3627 for (y = 0; y < cr; y++) {
3628 digit d = state->grid[y*cr+x];
3630 int box = state->blocks->whichblock[y*cr+x];
3631 ds->entered_items[x*cr+d-1] |= ((ds->entered_items[x*cr+d-1] & 1) << 1) | 1;
3632 ds->entered_items[y*cr+d-1] |= ((ds->entered_items[y*cr+d-1] & 4) << 1) | 4;
3633 ds->entered_items[box*cr+d-1] |= ((ds->entered_items[box*cr+d-1] & 16) << 1) | 16;
3635 if (ondiag0(y*cr+x))
3636 ds->entered_items[d-1] |= ((ds->entered_items[d-1] & 64) << 1) | 64;
3637 if (ondiag1(y*cr+x))
3638 ds->entered_items[cr+d-1] |= ((ds->entered_items[cr+d-1] & 64) << 1) | 64;
3644 * Draw any numbers which need redrawing.
3646 for (x = 0; x < cr; x++) {
3647 for (y = 0; y < cr; y++) {
3649 digit d = state->grid[y*cr+x];
3651 if (flashtime > 0 &&
3652 (flashtime <= FLASH_TIME/3 ||
3653 flashtime >= FLASH_TIME*2/3))
3656 /* Highlight active input areas. */
3657 if (x == ui->hx && y == ui->hy)
3658 highlight = ui->hpencil ? 2 : 1;
3660 /* Mark obvious errors (ie, numbers which occur more than once
3661 * in a single row, column, or box). */
3662 if (d && ((ds->entered_items[x*cr+d-1] & 2) ||
3663 (ds->entered_items[y*cr+d-1] & 8) ||
3664 (ds->entered_items[state->blocks->whichblock[y*cr+x]*cr+d-1] & 32) ||
3665 (ds->xtype && ((ondiag0(y*cr+x) && (ds->entered_items[d-1] & 128)) ||
3666 (ondiag1(y*cr+x) && (ds->entered_items[cr+d-1] & 128))))))
3669 draw_number(dr, ds, state, x, y, highlight);
3674 * Update the _entire_ grid if necessary.
3677 draw_update(dr, 0, 0, SIZE(cr), SIZE(cr));
3682 static float game_anim_length(game_state *oldstate, game_state *newstate,
3683 int dir, game_ui *ui)
3688 static float game_flash_length(game_state *oldstate, game_state *newstate,
3689 int dir, game_ui *ui)
3691 if (!oldstate->completed && newstate->completed &&
3692 !oldstate->cheated && !newstate->cheated)
3697 static int game_timing_state(game_state *state, game_ui *ui)
3702 static void game_print_size(game_params *params, float *x, float *y)
3707 * I'll use 9mm squares by default. They should be quite big
3708 * for this game, because players will want to jot down no end
3709 * of pencil marks in the squares.
3711 game_compute_size(params, 900, &pw, &ph);
3716 static void game_print(drawing *dr, game_state *state, int tilesize)
3719 int ink = print_mono_colour(dr, 0);
3722 /* Ick: fake up `ds->tilesize' for macro expansion purposes */
3723 game_drawstate ads, *ds = &ads;
3724 game_set_size(dr, ds, NULL, tilesize);
3729 print_line_width(dr, 3 * TILE_SIZE / 40);
3730 draw_rect_outline(dr, BORDER, BORDER, cr*TILE_SIZE, cr*TILE_SIZE, ink);
3733 * Highlight X-diagonal squares.
3737 int xhighlight = print_grey_colour(dr, 0.90F);
3739 for (i = 0; i < cr; i++)
3740 draw_rect(dr, BORDER + i*TILE_SIZE, BORDER + i*TILE_SIZE,
3741 TILE_SIZE, TILE_SIZE, xhighlight);
3742 for (i = 0; i < cr; i++)
3743 if (i*2 != cr-1) /* avoid redoing centre square, just for fun */
3744 draw_rect(dr, BORDER + i*TILE_SIZE,
3745 BORDER + (cr-1-i)*TILE_SIZE,
3746 TILE_SIZE, TILE_SIZE, xhighlight);
3752 for (x = 1; x < cr; x++) {
3753 print_line_width(dr, TILE_SIZE / 40);
3754 draw_line(dr, BORDER+x*TILE_SIZE, BORDER,
3755 BORDER+x*TILE_SIZE, BORDER+cr*TILE_SIZE, ink);
3757 for (y = 1; y < cr; y++) {
3758 print_line_width(dr, TILE_SIZE / 40);
3759 draw_line(dr, BORDER, BORDER+y*TILE_SIZE,
3760 BORDER+cr*TILE_SIZE, BORDER+y*TILE_SIZE, ink);
3764 * Thick lines between cells. In order to do this using the
3765 * line-drawing rather than rectangle-drawing API (so as to
3766 * get line thicknesses to scale correctly) and yet have
3767 * correctly mitred joins between lines, we must do this by
3768 * tracing the boundary of each sub-block and drawing it in
3769 * one go as a single polygon.
3774 int x, y, dx, dy, sx, sy, sdx, sdy;
3776 print_line_width(dr, 3 * TILE_SIZE / 40);
3779 * Maximum perimeter of a k-omino is 2k+2. (Proof: start
3780 * with k unconnected squares, with total perimeter 4k.
3781 * Now repeatedly join two disconnected components
3782 * together into a larger one; every time you do so you
3783 * remove at least two unit edges, and you require k-1 of
3784 * these operations to create a single connected piece, so
3785 * you must have at most 4k-2(k-1) = 2k+2 unit edges left
3788 coords = snewn(4*cr+4, int); /* 2k+2 points, 2 coords per point */
3791 * Iterate over all the blocks.
3793 for (bi = 0; bi < cr; bi++) {
3796 * For each block, find a starting square within it
3797 * which has a boundary at the left.
3799 for (i = 0; i < cr; i++) {
3800 int j = state->blocks->blocks[bi][i];
3801 if (j % cr == 0 || state->blocks->whichblock[j-1] != bi)
3804 assert(i < cr); /* every block must have _some_ leftmost square */
3805 x = state->blocks->blocks[bi][i] % cr;
3806 y = state->blocks->blocks[bi][i] / cr;
3811 * Now begin tracing round the perimeter. At all
3812 * times, (x,y) describes some square within the
3813 * block, and (x+dx,y+dy) is some adjacent square
3814 * outside it; so the edge between those two squares
3815 * is always an edge of the block.
3817 sx = x, sy = y, sdx = dx, sdy = dy; /* save starting position */
3820 int cx, cy, tx, ty, nin;
3823 * To begin with, record the point at one end of
3824 * the edge. To do this, we translate (x,y) down
3825 * and right by half a unit (so they're describing
3826 * a point in the _centre_ of the square) and then
3827 * translate back again in a manner rotated by dy
3831 cx = ((2*x+1) + dy + dx) / 2;
3832 cy = ((2*y+1) - dx + dy) / 2;
3833 coords[2*n+0] = BORDER + cx * TILE_SIZE;
3834 coords[2*n+1] = BORDER + cy * TILE_SIZE;
3838 * Now advance to the next edge, by looking at the
3839 * two squares beyond it. If they're both outside
3840 * the block, we turn right (by leaving x,y the
3841 * same and rotating dx,dy clockwise); if they're
3842 * both inside, we turn left (by rotating dx,dy
3843 * anticlockwise and contriving to leave x+dx,y+dy
3844 * unchanged); if one of each, we go straight on
3845 * (and may enforce by assertion that they're one
3846 * of each the _right_ way round).
3851 nin += (tx >= 0 && tx < cr && ty >= 0 && ty < cr &&
3852 state->blocks->whichblock[ty*cr+tx] == bi);
3855 nin += (tx >= 0 && tx < cr && ty >= 0 && ty < cr &&
3856 state->blocks->whichblock[ty*cr+tx] == bi);
3865 } else if (nin == 2) {
3889 * Now enforce by assertion that we ended up
3890 * somewhere sensible.
3892 assert(x >= 0 && x < cr && y >= 0 && y < cr &&
3893 state->blocks->whichblock[y*cr+x] == bi);
3894 assert(x+dx < 0 || x+dx >= cr || y+dy < 0 || y+dy >= cr ||
3895 state->blocks->whichblock[(y+dy)*cr+(x+dx)] != bi);
3897 } while (x != sx || y != sy || dx != sdx || dy != sdy);
3900 * That's our polygon; now draw it.
3902 draw_polygon(dr, coords, n, -1, ink);
3911 for (y = 0; y < cr; y++)
3912 for (x = 0; x < cr; x++)
3913 if (state->grid[y*cr+x]) {
3916 str[0] = state->grid[y*cr+x] + '0';
3918 str[0] += 'a' - ('9'+1);
3919 draw_text(dr, BORDER + x*TILE_SIZE + TILE_SIZE/2,
3920 BORDER + y*TILE_SIZE + TILE_SIZE/2,
3921 FONT_VARIABLE, TILE_SIZE/2,
3922 ALIGN_VCENTRE | ALIGN_HCENTRE, ink, str);
3927 #define thegame solo
3930 const struct game thegame = {
3931 "Solo", "games.solo", "solo",
3938 TRUE, game_configure, custom_params,
3946 TRUE, game_text_format,
3954 PREFERRED_TILE_SIZE, game_compute_size, game_set_size,
3957 game_free_drawstate,
3961 TRUE, FALSE, game_print_size, game_print,
3962 FALSE, /* wants_statusbar */
3963 FALSE, game_timing_state,
3964 REQUIRE_RBUTTON | REQUIRE_NUMPAD, /* flags */
3967 #ifdef STANDALONE_SOLVER
3969 int main(int argc, char **argv)
3973 char *id = NULL, *desc, *err;
3977 while (--argc > 0) {
3979 if (!strcmp(p, "-v")) {
3980 solver_show_working = TRUE;
3981 } else if (!strcmp(p, "-g")) {
3983 } else if (*p == '-') {
3984 fprintf(stderr, "%s: unrecognised option `%s'\n", argv[0], p);
3992 fprintf(stderr, "usage: %s [-g | -v] <game_id>\n", argv[0]);
3996 desc = strchr(id, ':');
3998 fprintf(stderr, "%s: game id expects a colon in it\n", argv[0]);
4003 p = default_params();
4004 decode_params(p, id);
4005 err = validate_desc(p, desc);
4007 fprintf(stderr, "%s: %s\n", argv[0], err);
4010 s = new_game(NULL, p, desc);
4012 ret = solver(s->cr, s->blocks, s->xtype, s->grid, DIFF_RECURSIVE);
4014 printf("Difficulty rating: %s\n",
4015 ret==DIFF_BLOCK ? "Trivial (blockwise positional elimination only)":
4016 ret==DIFF_SIMPLE ? "Basic (row/column/number elimination required)":
4017 ret==DIFF_INTERSECT ? "Intermediate (intersectional analysis required)":
4018 ret==DIFF_SET ? "Advanced (set elimination required)":
4019 ret==DIFF_EXTREME ? "Extreme (complex non-recursive techniques required)":
4020 ret==DIFF_RECURSIVE ? "Unreasonable (guesswork and backtracking required)":
4021 ret==DIFF_AMBIGUOUS ? "Ambiguous (multiple solutions exist)":
4022 ret==DIFF_IMPOSSIBLE ? "Impossible (no solution exists)":
4023 "INTERNAL ERROR: unrecognised difficulty code");
4025 printf("%s\n", grid_text_format(s->cr, s->blocks, s->xtype, s->grid));