2 * solo.c: the number-placing puzzle most popularly known as `Sudoku'.
6 * - reports from users are that `Trivial'-mode puzzles are still
7 * rather hard compared to newspapers' easy ones, so some better
8 * low-end difficulty grading would be nice
9 * + it's possible that really easy puzzles always have
10 * _several_ things you can do, so don't make you hunt too
11 * hard for the one deduction you can currently make
12 * + it's also possible that easy puzzles require fewer
13 * cross-eliminations: perhaps there's a higher incidence of
14 * things you can deduce by looking only at (say) rows,
15 * rather than things you have to check both rows and columns
17 * + but really, what I need to do is find some really easy
18 * puzzles and _play_ them, to see what's actually easy about
20 * + while I'm revamping this area, filling in the _last_
21 * number in a nearly-full row or column should certainly be
22 * permitted even at the lowest difficulty level.
23 * + also Owen noticed that `Basic' grids requiring numeric
24 * elimination are actually very hard, so I wonder if a
25 * difficulty gradation between that and positional-
26 * elimination-only might be in order
27 * + but it's not good to have _too_ many difficulty levels, or
28 * it'll take too long to randomly generate a given level.
30 * - it might still be nice to do some prioritisation on the
31 * removal of numbers from the grid
32 * + one possibility is to try to minimise the maximum number
33 * of filled squares in any block, which in particular ought
34 * to enforce never leaving a completely filled block in the
35 * puzzle as presented.
37 * - alternative interface modes
38 * + sudoku.com's Windows program has a palette of possible
39 * entries; you select a palette entry first and then click
40 * on the square you want it to go in, thus enabling
41 * mouse-only play. Useful for PDAs! I don't think it's
42 * actually incompatible with the current highlight-then-type
43 * approach: you _either_ highlight a palette entry and then
44 * click, _or_ you highlight a square and then type. At most
45 * one thing is ever highlighted at a time, so there's no way
47 * + then again, I don't actually like sudoku.com's interface;
48 * it's too much like a paint package whereas I prefer to
49 * think of Solo as a text editor.
50 * + another PDA-friendly possibility is a drag interface:
51 * _drag_ numbers from the palette into the grid squares.
52 * Thought experiments suggest I'd prefer that to the
53 * sudoku.com approach, but I haven't actually tried it.
57 * Solo puzzles need to be square overall (since each row and each
58 * column must contain one of every digit), but they need not be
59 * subdivided the same way internally. I am going to adopt a
60 * convention whereby I _always_ refer to `r' as the number of rows
61 * of _big_ divisions, and `c' as the number of columns of _big_
62 * divisions. Thus, a 2c by 3r puzzle looks something like this:
66 * ------+------ (Of course, you can't subdivide it the other way
67 * 1 4 5 | 6 3 2 or you'll get clashes; observe that the 4 in the
68 * 3 2 6 | 4 1 5 top left would conflict with the 4 in the second
69 * ------+------ box down on the left-hand side.)
73 * The need for a strong naming convention should now be clear:
74 * each small box is two rows of digits by three columns, while the
75 * overall puzzle has three rows of small boxes by two columns. So
76 * I will (hopefully) consistently use `r' to denote the number of
77 * rows _of small boxes_ (here 3), which is also the number of
78 * columns of digits in each small box; and `c' vice versa (here
81 * I'm also going to choose arbitrarily to list c first wherever
82 * possible: the above is a 2x3 puzzle, not a 3x2 one.
92 #ifdef STANDALONE_SOLVER
94 int solver_show_working
, solver_recurse_depth
;
100 * To save space, I store digits internally as unsigned char. This
101 * imposes a hard limit of 255 on the order of the puzzle. Since
102 * even a 5x5 takes unacceptably long to generate, I don't see this
103 * as a serious limitation unless something _really_ impressive
104 * happens in computing technology; but here's a typedef anyway for
105 * general good practice.
107 typedef unsigned char digit
;
108 #define ORDER_MAX 255
110 #define PREFERRED_TILE_SIZE 32
111 #define TILE_SIZE (ds->tilesize)
112 #define BORDER (TILE_SIZE / 2)
114 #define FLASH_TIME 0.4F
116 enum { SYMM_NONE
, SYMM_ROT2
, SYMM_ROT4
, SYMM_REF2
, SYMM_REF2D
, SYMM_REF4
,
117 SYMM_REF4D
, SYMM_REF8
};
119 enum { DIFF_BLOCK
, DIFF_SIMPLE
, DIFF_INTERSECT
, DIFF_SET
, DIFF_EXTREME
,
120 DIFF_RECURSIVE
, DIFF_AMBIGUOUS
, DIFF_IMPOSSIBLE
};
134 int c
, r
, symm
, diff
;
140 unsigned char *pencil
; /* c*r*c*r elements */
141 unsigned char *immutable
; /* marks which digits are clues */
142 int completed
, cheated
;
145 static game_params
*default_params(void)
147 game_params
*ret
= snew(game_params
);
150 ret
->symm
= SYMM_ROT2
; /* a plausible default */
151 ret
->diff
= DIFF_BLOCK
; /* so is this */
156 static void free_params(game_params
*params
)
161 static game_params
*dup_params(game_params
*params
)
163 game_params
*ret
= snew(game_params
);
164 *ret
= *params
; /* structure copy */
168 static int game_fetch_preset(int i
, char **name
, game_params
**params
)
174 { "2x2 Trivial", { 2, 2, SYMM_ROT2
, DIFF_BLOCK
} },
175 { "2x3 Basic", { 2, 3, SYMM_ROT2
, DIFF_SIMPLE
} },
176 { "3x3 Trivial", { 3, 3, SYMM_ROT2
, DIFF_BLOCK
} },
177 { "3x3 Basic", { 3, 3, SYMM_ROT2
, DIFF_SIMPLE
} },
178 { "3x3 Intermediate", { 3, 3, SYMM_ROT2
, DIFF_INTERSECT
} },
179 { "3x3 Advanced", { 3, 3, SYMM_ROT2
, DIFF_SET
} },
180 { "3x3 Extreme", { 3, 3, SYMM_ROT2
, DIFF_EXTREME
} },
181 { "3x3 Unreasonable", { 3, 3, SYMM_ROT2
, DIFF_RECURSIVE
} },
183 { "3x4 Basic", { 3, 4, SYMM_ROT2
, DIFF_SIMPLE
} },
184 { "4x4 Basic", { 4, 4, SYMM_ROT2
, DIFF_SIMPLE
} },
188 if (i
< 0 || i
>= lenof(presets
))
191 *name
= dupstr(presets
[i
].title
);
192 *params
= dup_params(&presets
[i
].params
);
197 static void decode_params(game_params
*ret
, char const *string
)
199 ret
->c
= ret
->r
= atoi(string
);
200 while (*string
&& isdigit((unsigned char)*string
)) string
++;
201 if (*string
== 'x') {
203 ret
->r
= atoi(string
);
204 while (*string
&& isdigit((unsigned char)*string
)) string
++;
207 if (*string
== 'r' || *string
== 'm' || *string
== 'a') {
210 if (*string
== 'd') {
217 while (*string
&& isdigit((unsigned char)*string
)) string
++;
218 if (sc
== 'm' && sn
== 8)
219 ret
->symm
= SYMM_REF8
;
220 if (sc
== 'm' && sn
== 4)
221 ret
->symm
= sd ? SYMM_REF4D
: SYMM_REF4
;
222 if (sc
== 'm' && sn
== 2)
223 ret
->symm
= sd ? SYMM_REF2D
: SYMM_REF2
;
224 if (sc
== 'r' && sn
== 4)
225 ret
->symm
= SYMM_ROT4
;
226 if (sc
== 'r' && sn
== 2)
227 ret
->symm
= SYMM_ROT2
;
229 ret
->symm
= SYMM_NONE
;
230 } else if (*string
== 'd') {
232 if (*string
== 't') /* trivial */
233 string
++, ret
->diff
= DIFF_BLOCK
;
234 else if (*string
== 'b') /* basic */
235 string
++, ret
->diff
= DIFF_SIMPLE
;
236 else if (*string
== 'i') /* intermediate */
237 string
++, ret
->diff
= DIFF_INTERSECT
;
238 else if (*string
== 'a') /* advanced */
239 string
++, ret
->diff
= DIFF_SET
;
240 else if (*string
== 'e') /* extreme */
241 string
++, ret
->diff
= DIFF_EXTREME
;
242 else if (*string
== 'u') /* unreasonable */
243 string
++, ret
->diff
= DIFF_RECURSIVE
;
245 string
++; /* eat unknown character */
249 static char *encode_params(game_params
*params
, int full
)
253 sprintf(str
, "%dx%d", params
->c
, params
->r
);
255 switch (params
->symm
) {
256 case SYMM_REF8
: strcat(str
, "m8"); break;
257 case SYMM_REF4
: strcat(str
, "m4"); break;
258 case SYMM_REF4D
: strcat(str
, "md4"); break;
259 case SYMM_REF2
: strcat(str
, "m2"); break;
260 case SYMM_REF2D
: strcat(str
, "md2"); break;
261 case SYMM_ROT4
: strcat(str
, "r4"); break;
262 /* case SYMM_ROT2: strcat(str, "r2"); break; [default] */
263 case SYMM_NONE
: strcat(str
, "a"); break;
265 switch (params
->diff
) {
266 /* case DIFF_BLOCK: strcat(str, "dt"); break; [default] */
267 case DIFF_SIMPLE
: strcat(str
, "db"); break;
268 case DIFF_INTERSECT
: strcat(str
, "di"); break;
269 case DIFF_SET
: strcat(str
, "da"); break;
270 case DIFF_EXTREME
: strcat(str
, "de"); break;
271 case DIFF_RECURSIVE
: strcat(str
, "du"); break;
277 static config_item
*game_configure(game_params
*params
)
282 ret
= snewn(5, config_item
);
284 ret
[0].name
= "Columns of sub-blocks";
285 ret
[0].type
= C_STRING
;
286 sprintf(buf
, "%d", params
->c
);
287 ret
[0].sval
= dupstr(buf
);
290 ret
[1].name
= "Rows of sub-blocks";
291 ret
[1].type
= C_STRING
;
292 sprintf(buf
, "%d", params
->r
);
293 ret
[1].sval
= dupstr(buf
);
296 ret
[2].name
= "Symmetry";
297 ret
[2].type
= C_CHOICES
;
298 ret
[2].sval
= ":None:2-way rotation:4-way rotation:2-way mirror:"
299 "2-way diagonal mirror:4-way mirror:4-way diagonal mirror:"
301 ret
[2].ival
= params
->symm
;
303 ret
[3].name
= "Difficulty";
304 ret
[3].type
= C_CHOICES
;
305 ret
[3].sval
= ":Trivial:Basic:Intermediate:Advanced:Extreme:Unreasonable";
306 ret
[3].ival
= params
->diff
;
316 static game_params
*custom_params(config_item
*cfg
)
318 game_params
*ret
= snew(game_params
);
320 ret
->c
= atoi(cfg
[0].sval
);
321 ret
->r
= atoi(cfg
[1].sval
);
322 ret
->symm
= cfg
[2].ival
;
323 ret
->diff
= cfg
[3].ival
;
328 static char *validate_params(game_params
*params
, int full
)
330 if (params
->c
< 2 || params
->r
< 2)
331 return "Both dimensions must be at least 2";
332 if (params
->c
> ORDER_MAX
|| params
->r
> ORDER_MAX
)
333 return "Dimensions greater than "STR(ORDER_MAX
)" are not supported";
334 if ((params
->c
* params
->r
) > 36)
335 return "Unable to support more than 36 distinct symbols in a puzzle";
339 /* ----------------------------------------------------------------------
342 * This solver is used for two purposes:
343 * + to check solubility of a grid as we gradually remove numbers
345 * + to solve an externally generated puzzle when the user selects
348 * It supports a variety of specific modes of reasoning. By
349 * enabling or disabling subsets of these modes we can arrange a
350 * range of difficulty levels.
354 * Modes of reasoning currently supported:
356 * - Positional elimination: a number must go in a particular
357 * square because all the other empty squares in a given
358 * row/col/blk are ruled out.
360 * - Numeric elimination: a square must have a particular number
361 * in because all the other numbers that could go in it are
364 * - Intersectional analysis: given two domains which overlap
365 * (hence one must be a block, and the other can be a row or
366 * col), if the possible locations for a particular number in
367 * one of the domains can be narrowed down to the overlap, then
368 * that number can be ruled out everywhere but the overlap in
369 * the other domain too.
371 * - Set elimination: if there is a subset of the empty squares
372 * within a domain such that the union of the possible numbers
373 * in that subset has the same size as the subset itself, then
374 * those numbers can be ruled out everywhere else in the domain.
375 * (For example, if there are five empty squares and the
376 * possible numbers in each are 12, 23, 13, 134 and 1345, then
377 * the first three empty squares form such a subset: the numbers
378 * 1, 2 and 3 _must_ be in those three squares in some
379 * permutation, and hence we can deduce none of them can be in
380 * the fourth or fifth squares.)
381 * + You can also see this the other way round, concentrating
382 * on numbers rather than squares: if there is a subset of
383 * the unplaced numbers within a domain such that the union
384 * of all their possible positions has the same size as the
385 * subset itself, then all other numbers can be ruled out for
386 * those positions. However, it turns out that this is
387 * exactly equivalent to the first formulation at all times:
388 * there is a 1-1 correspondence between suitable subsets of
389 * the unplaced numbers and suitable subsets of the unfilled
390 * places, found by taking the _complement_ of the union of
391 * the numbers' possible positions (or the spaces' possible
394 * - Mutual neighbour elimination: find two squares A,B and a
395 * number N in the possible set of A, such that putting N in A
396 * would rule out enough possibilities from the mutual
397 * neighbours of A and B that there would be no possibilities
398 * left for B. Thereby rule out N in A.
399 * + The simplest case of this is if B has two possibilities
400 * (wlog {1,2}), and there are two mutual neighbours of A and
401 * B which have possibilities {1,3} and {2,3}. Thus, if A
402 * were to be 3, then those neighbours would contain 1 and 2,
403 * and hence there would be nothing left which could go in B.
404 * + There can be more complex cases of it too: if A and B are
405 * in the same column of large blocks, then they can have
406 * more than two mutual neighbours, some of which can also be
407 * neighbours of one another. Suppose, for example, that B
408 * has possibilities {1,2,3}; there's one square P in the
409 * same column as B and the same block as A, with
410 * possibilities {1,4}; and there are _two_ squares Q,R in
411 * the same column as A and the same block as B with
412 * possibilities {2,3,4}. Then if A contained 4, P would
413 * contain 1, and Q and R would have to contain 2 and 3 in
414 * _some_ order; therefore, once again, B would have no
415 * remaining possibilities.
417 * - Recursion. If all else fails, we pick one of the currently
418 * most constrained empty squares and take a random guess at its
419 * contents, then continue solving on that basis and see if we
424 * Within this solver, I'm going to transform all y-coordinates by
425 * inverting the significance of the block number and the position
426 * within the block. That is, we will start with the top row of
427 * each block in order, then the second row of each block in order,
430 * This transformation has the enormous advantage that it means
431 * every row, column _and_ block is described by an arithmetic
432 * progression of coordinates within the cubic array, so that I can
433 * use the same very simple function to do blockwise, row-wise and
434 * column-wise elimination.
436 #define YTRANS(y) (((y)%c)*r+(y)/c)
437 #define YUNTRANS(y) (((y)%r)*c+(y)/r)
439 struct solver_usage
{
442 * We set up a cubic array, indexed by x, y and digit; each
443 * element of this array is TRUE or FALSE according to whether
444 * or not that digit _could_ in principle go in that position.
446 * The way to index this array is cube[(x*cr+y)*cr+n-1].
447 * y-coordinates in here are transformed.
451 * This is the grid in which we write down our final
452 * deductions. y-coordinates in here are _not_ transformed.
456 * Now we keep track, at a slightly higher level, of what we
457 * have yet to work out, to prevent doing the same deduction
460 /* row[y*cr+n-1] TRUE if digit n has been placed in row y */
462 /* col[x*cr+n-1] TRUE if digit n has been placed in row x */
464 /* blk[(y*c+x)*cr+n-1] TRUE if digit n has been placed in block (x,y) */
467 #define cubepos(x,y,n) (((x)*usage->cr+(y))*usage->cr+(n)-1)
468 #define cube(x,y,n) (usage->cube[cubepos(x,y,n)])
471 * Function called when we are certain that a particular square has
472 * a particular number in it. The y-coordinate passed in here is
475 static void solver_place(struct solver_usage
*usage
, int x
, int y
, int n
)
477 int c
= usage
->c
, r
= usage
->r
, cr
= usage
->cr
;
483 * Rule out all other numbers in this square.
485 for (i
= 1; i
<= cr
; i
++)
490 * Rule out this number in all other positions in the row.
492 for (i
= 0; i
< cr
; i
++)
497 * Rule out this number in all other positions in the column.
499 for (i
= 0; i
< cr
; i
++)
504 * Rule out this number in all other positions in the block.
508 for (i
= 0; i
< r
; i
++)
509 for (j
= 0; j
< c
; j
++)
510 if (bx
+i
!= x
|| by
+j
*r
!= y
)
511 cube(bx
+i
,by
+j
*r
,n
) = FALSE
;
514 * Enter the number in the result grid.
516 usage
->grid
[YUNTRANS(y
)*cr
+x
] = n
;
519 * Cross out this number from the list of numbers left to place
520 * in its row, its column and its block.
522 usage
->row
[y
*cr
+n
-1] = usage
->col
[x
*cr
+n
-1] =
523 usage
->blk
[((y
%r
)*c
+(x
/r
))*cr
+n
-1] = TRUE
;
526 static int solver_elim(struct solver_usage
*usage
, int start
, int step
527 #ifdef STANDALONE_SOLVER
532 int c
= usage
->c
, r
= usage
->r
, cr
= c
*r
;
536 * Count the number of set bits within this section of the
541 for (i
= 0; i
< cr
; i
++)
542 if (usage
->cube
[start
+i
*step
]) {
556 if (!usage
->grid
[YUNTRANS(y
)*cr
+x
]) {
557 #ifdef STANDALONE_SOLVER
558 if (solver_show_working
) {
560 printf("%*s", solver_recurse_depth
*4, "");
564 printf(":\n%*s placing %d at (%d,%d)\n",
565 solver_recurse_depth
*4, "", n
, 1+x
, 1+YUNTRANS(y
));
568 solver_place(usage
, x
, y
, n
);
572 #ifdef STANDALONE_SOLVER
573 if (solver_show_working
) {
575 printf("%*s", solver_recurse_depth
*4, "");
579 printf(":\n%*s no possibilities available\n",
580 solver_recurse_depth
*4, "");
589 static int solver_intersect(struct solver_usage
*usage
,
590 int start1
, int step1
, int start2
, int step2
591 #ifdef STANDALONE_SOLVER
596 int c
= usage
->c
, r
= usage
->r
, cr
= c
*r
;
600 * Loop over the first domain and see if there's any set bit
601 * not also in the second.
603 for (i
= 0; i
< cr
; i
++) {
604 int p
= start1
+i
*step1
;
605 if (usage
->cube
[p
] &&
606 !(p
>= start2
&& p
< start2
+cr
*step2
&&
607 (p
- start2
) % step2
== 0))
608 return 0; /* there is, so we can't deduce */
612 * We have determined that all set bits in the first domain are
613 * within its overlap with the second. So loop over the second
614 * domain and remove all set bits that aren't also in that
615 * overlap; return +1 iff we actually _did_ anything.
618 for (i
= 0; i
< cr
; i
++) {
619 int p
= start2
+i
*step2
;
620 if (usage
->cube
[p
] &&
621 !(p
>= start1
&& p
< start1
+cr
*step1
&& (p
- start1
) % step1
== 0))
623 #ifdef STANDALONE_SOLVER
624 if (solver_show_working
) {
629 printf("%*s", solver_recurse_depth
*4, "");
641 printf("%*s ruling out %d at (%d,%d)\n",
642 solver_recurse_depth
*4, "", pn
, 1+px
, 1+YUNTRANS(py
));
645 ret
= +1; /* we did something */
653 struct solver_scratch
{
654 unsigned char *grid
, *rowidx
, *colidx
, *set
;
655 int *neighbours
, *bfsqueue
;
656 #ifdef STANDALONE_SOLVER
661 static int solver_set(struct solver_usage
*usage
,
662 struct solver_scratch
*scratch
,
663 int start
, int step1
, int step2
664 #ifdef STANDALONE_SOLVER
669 int c
= usage
->c
, r
= usage
->r
, cr
= c
*r
;
671 unsigned char *grid
= scratch
->grid
;
672 unsigned char *rowidx
= scratch
->rowidx
;
673 unsigned char *colidx
= scratch
->colidx
;
674 unsigned char *set
= scratch
->set
;
677 * We are passed a cr-by-cr matrix of booleans. Our first job
678 * is to winnow it by finding any definite placements - i.e.
679 * any row with a solitary 1 - and discarding that row and the
680 * column containing the 1.
682 memset(rowidx
, TRUE
, cr
);
683 memset(colidx
, TRUE
, cr
);
684 for (i
= 0; i
< cr
; i
++) {
685 int count
= 0, first
= -1;
686 for (j
= 0; j
< cr
; j
++)
687 if (usage
->cube
[start
+i
*step1
+j
*step2
])
691 * If count == 0, then there's a row with no 1s at all and
692 * the puzzle is internally inconsistent. However, we ought
693 * to have caught this already during the simpler reasoning
694 * methods, so we can safely fail an assertion if we reach
699 rowidx
[i
] = colidx
[first
] = FALSE
;
703 * Convert each of rowidx/colidx from a list of 0s and 1s to a
704 * list of the indices of the 1s.
706 for (i
= j
= 0; i
< cr
; i
++)
710 for (i
= j
= 0; i
< cr
; i
++)
716 * And create the smaller matrix.
718 for (i
= 0; i
< n
; i
++)
719 for (j
= 0; j
< n
; j
++)
720 grid
[i
*cr
+j
] = usage
->cube
[start
+rowidx
[i
]*step1
+colidx
[j
]*step2
];
723 * Having done that, we now have a matrix in which every row
724 * has at least two 1s in. Now we search to see if we can find
725 * a rectangle of zeroes (in the set-theoretic sense of
726 * `rectangle', i.e. a subset of rows crossed with a subset of
727 * columns) whose width and height add up to n.
734 * We have a candidate set. If its size is <=1 or >=n-1
735 * then we move on immediately.
737 if (count
> 1 && count
< n
-1) {
739 * The number of rows we need is n-count. See if we can
740 * find that many rows which each have a zero in all
741 * the positions listed in `set'.
744 for (i
= 0; i
< n
; i
++) {
746 for (j
= 0; j
< n
; j
++)
747 if (set
[j
] && grid
[i
*cr
+j
]) {
756 * We expect never to be able to get _more_ than
757 * n-count suitable rows: this would imply that (for
758 * example) there are four numbers which between them
759 * have at most three possible positions, and hence it
760 * indicates a faulty deduction before this point or
763 if (rows
> n
- count
) {
764 #ifdef STANDALONE_SOLVER
765 if (solver_show_working
) {
767 printf("%*s", solver_recurse_depth
*4,
772 printf(":\n%*s contradiction reached\n",
773 solver_recurse_depth
*4, "");
779 if (rows
>= n
- count
) {
780 int progress
= FALSE
;
783 * We've got one! Now, for each row which _doesn't_
784 * satisfy the criterion, eliminate all its set
785 * bits in the positions _not_ listed in `set'.
786 * Return +1 (meaning progress has been made) if we
787 * successfully eliminated anything at all.
789 * This involves referring back through
790 * rowidx/colidx in order to work out which actual
791 * positions in the cube to meddle with.
793 for (i
= 0; i
< n
; i
++) {
795 for (j
= 0; j
< n
; j
++)
796 if (set
[j
] && grid
[i
*cr
+j
]) {
801 for (j
= 0; j
< n
; j
++)
802 if (!set
[j
] && grid
[i
*cr
+j
]) {
803 int fpos
= (start
+rowidx
[i
]*step1
+
805 #ifdef STANDALONE_SOLVER
806 if (solver_show_working
) {
811 printf("%*s", solver_recurse_depth
*4,
824 printf("%*s ruling out %d at (%d,%d)\n",
825 solver_recurse_depth
*4, "",
826 pn
, 1+px
, 1+YUNTRANS(py
));
830 usage
->cube
[fpos
] = FALSE
;
842 * Binary increment: change the rightmost 0 to a 1, and
843 * change all 1s to the right of it to 0s.
846 while (i
> 0 && set
[i
-1])
847 set
[--i
] = 0, count
--;
849 set
[--i
] = 1, count
++;
858 * Try to find a number in the possible set of (x1,y1) which can be
859 * ruled out because it would leave no possibilities for (x2,y2).
861 static int solver_mne(struct solver_usage
*usage
,
862 struct solver_scratch
*scratch
,
863 int x1
, int y1
, int x2
, int y2
)
865 int c
= usage
->c
, r
= usage
->r
, cr
= c
*r
;
867 unsigned char *set
= scratch
->set
;
868 unsigned char *numbers
= scratch
->rowidx
;
869 unsigned char *numbersleft
= scratch
->colidx
;
873 nb
[0] = scratch
->neighbours
;
874 nb
[1] = scratch
->neighbours
+ cr
;
877 * First, work out the mutual neighbour squares of the two. We
878 * can assert that they're not actually in the same block,
879 * which leaves two possibilities: they're in different block
880 * rows _and_ different block columns (thus their mutual
881 * neighbours are precisely the other two corners of the
882 * rectangle), or they're in the same row (WLOG) and different
883 * columns, in which case their mutual neighbours are the
884 * column of each block aligned with the other square.
886 * We divide the mutual neighbours into two separate subsets
887 * nb[0] and nb[1]; squares in the same subset are not only
888 * adjacent to both our key squares, but are also always
889 * adjacent to one another.
891 if (x1
/ r
!= x2
/ r
&& y1
% r
!= y2
% r
) {
892 /* Corners of the rectangle. */
894 nb
[0][0] = cubepos(x2
, y1
, 1);
895 nb
[1][0] = cubepos(x1
, y2
, 1);
896 } else if (x1
/ r
!= x2
/ r
) {
897 /* Same row of blocks; different blocks within that row. */
898 int x1b
= x1
- (x1
% r
);
899 int x2b
= x2
- (x2
% r
);
902 for (i
= 0; i
< r
; i
++) {
903 nb
[0][i
] = cubepos(x2b
+i
, y1
, 1);
904 nb
[1][i
] = cubepos(x1b
+i
, y2
, 1);
907 /* Same column of blocks; different blocks within that column. */
911 assert(y1
% r
!= y2
% r
);
914 for (i
= 0; i
< c
; i
++) {
915 nb
[0][i
] = cubepos(x2
, y1b
+i
*r
, 1);
916 nb
[1][i
] = cubepos(x1
, y2b
+i
*r
, 1);
921 * Right. Now loop over each possible number.
923 for (n
= 1; n
<= cr
; n
++) {
924 if (!cube(x1
, y1
, n
))
926 for (j
= 0; j
< cr
; j
++)
927 numbersleft
[j
] = cube(x2
, y2
, j
+1);
930 * Go over every possible subset of each neighbour list,
931 * and see if its union of possible numbers minus n has the
932 * same size as the subset. If so, add the numbers in that
933 * subset to the set of things which would be ruled out
934 * from (x2,y2) if n were placed at (x1,y1).
940 * Binary increment: change the rightmost 0 to a 1, and
941 * change all 1s to the right of it to 0s.
944 while (i
> 0 && set
[i
-1])
945 set
[--i
] = 0, count
--;
947 set
[--i
] = 1, count
++;
952 * Examine this subset of each neighbour set.
954 for (nbi
= 0; nbi
< 2; nbi
++) {
957 memset(numbers
, 0, cr
);
959 for (i
= 0; i
< nnb
; i
++)
961 for (j
= 0; j
< cr
; j
++)
962 if (j
!= n
-1 && usage
->cube
[nbs
[i
] + j
])
965 for (i
= j
= 0; j
< cr
; j
++)
970 * Got one. This subset of nbs, in the absence
971 * of n, would definitely contain all the
972 * numbers listed in `numbers'. Rule them out
975 for (j
= 0; j
< cr
; j
++)
983 * If we've got nothing left in `numbersleft', we have a
984 * successful mutual neighbour elimination.
986 for (j
= 0; j
< cr
; j
++)
991 #ifdef STANDALONE_SOLVER
992 if (solver_show_working
) {
993 printf("%*smutual neighbour elimination, (%d,%d) vs (%d,%d):\n",
994 solver_recurse_depth
*4, "",
995 1+x1
, 1+YUNTRANS(y1
), 1+x2
, 1+YUNTRANS(y2
));
996 printf("%*s ruling out %d at (%d,%d)\n",
997 solver_recurse_depth
*4, "",
998 n
, 1+x1
, 1+YUNTRANS(y1
));
1001 cube(x1
, y1
, n
) = FALSE
;
1006 return 0; /* nothing found */
1010 * Look for forcing chains. A forcing chain is a path of
1011 * pairwise-exclusive squares (i.e. each pair of adjacent squares
1012 * in the path are in the same row, column or block) with the
1013 * following properties:
1015 * (a) Each square on the path has precisely two possible numbers.
1017 * (b) Each pair of squares which are adjacent on the path share
1018 * at least one possible number in common.
1020 * (c) Each square in the middle of the path shares _both_ of its
1021 * numbers with at least one of its neighbours (not the same
1022 * one with both neighbours).
1024 * These together imply that at least one of the possible number
1025 * choices at one end of the path forces _all_ the rest of the
1026 * numbers along the path. In order to make real use of this, we
1027 * need further properties:
1029 * (c) Ruling out some number N from the square at one end
1030 * of the path forces the square at the other end to
1033 * (d) The two end squares are both in line with some third
1036 * (e) That third square currently has N as a possibility.
1038 * If we can find all of that lot, we can deduce that at least one
1039 * of the two ends of the forcing chain has number N, and that
1040 * therefore the mutually adjacent third square does not.
1042 * To find forcing chains, we're going to start a bfs at each
1043 * suitable square, once for each of its two possible numbers.
1045 static int solver_forcing(struct solver_usage
*usage
,
1046 struct solver_scratch
*scratch
)
1048 int c
= usage
->c
, r
= usage
->r
, cr
= c
*r
;
1049 int *bfsqueue
= scratch
->bfsqueue
;
1050 #ifdef STANDALONE_SOLVER
1051 int *bfsprev
= scratch
->bfsprev
;
1053 unsigned char *number
= scratch
->grid
;
1054 int *neighbours
= scratch
->neighbours
;
1057 for (y
= 0; y
< cr
; y
++)
1058 for (x
= 0; x
< cr
; x
++) {
1062 * If this square doesn't have exactly two candidate
1063 * numbers, don't try it.
1065 * In this loop we also sum the candidate numbers,
1066 * which is a nasty hack to allow us to quickly find
1067 * `the other one' (since we will shortly know there
1070 for (count
= t
= 0, n
= 1; n
<= cr
; n
++)
1077 * Now attempt a bfs for each candidate.
1079 for (n
= 1; n
<= cr
; n
++)
1080 if (cube(x
, y
, n
)) {
1081 int orign
, currn
, head
, tail
;
1088 memset(number
, cr
+1, cr
*cr
);
1090 bfsqueue
[tail
++] = y
*cr
+x
;
1091 #ifdef STANDALONE_SOLVER
1092 bfsprev
[y
*cr
+x
] = -1;
1094 number
[y
*cr
+x
] = t
- n
;
1096 while (head
< tail
) {
1097 int xx
, yy
, nneighbours
, xt
, yt
, xblk
, i
;
1099 xx
= bfsqueue
[head
++];
1103 currn
= number
[yy
*cr
+xx
];
1106 * Find neighbours of yy,xx.
1109 for (yt
= 0; yt
< cr
; yt
++)
1110 neighbours
[nneighbours
++] = yt
*cr
+xx
;
1111 for (xt
= 0; xt
< cr
; xt
++)
1112 neighbours
[nneighbours
++] = yy
*cr
+xt
;
1113 xblk
= xx
- (xx
% r
);
1114 for (yt
= yy
% r
; yt
< cr
; yt
+= r
)
1115 for (xt
= xblk
; xt
< xblk
+r
; xt
++)
1116 neighbours
[nneighbours
++] = yt
*cr
+xt
;
1119 * Try visiting each of those neighbours.
1121 for (i
= 0; i
< nneighbours
; i
++) {
1124 xt
= neighbours
[i
] % cr
;
1125 yt
= neighbours
[i
] / cr
;
1128 * We need this square to not be
1129 * already visited, and to include
1130 * currn as a possible number.
1132 if (number
[yt
*cr
+xt
] <= cr
)
1134 if (!cube(xt
, yt
, currn
))
1138 * Don't visit _this_ square a second
1141 if (xt
== xx
&& yt
== yy
)
1145 * To continue with the bfs, we need
1146 * this square to have exactly two
1149 for (cc
= tt
= 0, nn
= 1; nn
<= cr
; nn
++)
1150 if (cube(xt
, yt
, nn
))
1153 bfsqueue
[tail
++] = yt
*cr
+xt
;
1154 #ifdef STANDALONE_SOLVER
1155 bfsprev
[yt
*cr
+xt
] = yy
*cr
+xx
;
1157 number
[yt
*cr
+xt
] = tt
- currn
;
1161 * One other possibility is that this
1162 * might be the square in which we can
1163 * make a real deduction: if it's
1164 * adjacent to x,y, and currn is equal
1165 * to the original number we ruled out.
1167 if (currn
== orign
&&
1168 (xt
== x
|| yt
== y
||
1169 (xt
/ r
== x
/ r
&& yt
% r
== y
% r
))) {
1170 #ifdef STANDALONE_SOLVER
1171 if (solver_show_working
) {
1174 printf("%*sforcing chain, %d at ends of ",
1175 solver_recurse_depth
*4, "", orign
);
1179 printf("%s(%d,%d)", sep
, 1+xl
,
1181 xl
= bfsprev
[yl
*cr
+xl
];
1188 printf("\n%*s ruling out %d at (%d,%d)\n",
1189 solver_recurse_depth
*4, "",
1190 orign
, 1+xt
, 1+YUNTRANS(yt
));
1193 cube(xt
, yt
, orign
) = FALSE
;
1204 static struct solver_scratch
*solver_new_scratch(struct solver_usage
*usage
)
1206 struct solver_scratch
*scratch
= snew(struct solver_scratch
);
1208 scratch
->grid
= snewn(cr
*cr
, unsigned char);
1209 scratch
->rowidx
= snewn(cr
, unsigned char);
1210 scratch
->colidx
= snewn(cr
, unsigned char);
1211 scratch
->set
= snewn(cr
, unsigned char);
1212 scratch
->neighbours
= snewn(3*cr
, int);
1213 scratch
->bfsqueue
= snewn(cr
*cr
, int);
1214 #ifdef STANDALONE_SOLVER
1215 scratch
->bfsprev
= snewn(cr
*cr
, int);
1220 static void solver_free_scratch(struct solver_scratch
*scratch
)
1222 #ifdef STANDALONE_SOLVER
1223 sfree(scratch
->bfsprev
);
1225 sfree(scratch
->bfsqueue
);
1226 sfree(scratch
->neighbours
);
1227 sfree(scratch
->set
);
1228 sfree(scratch
->colidx
);
1229 sfree(scratch
->rowidx
);
1230 sfree(scratch
->grid
);
1234 static int solver(int c
, int r
, digit
*grid
, int maxdiff
)
1236 struct solver_usage
*usage
;
1237 struct solver_scratch
*scratch
;
1239 int x
, y
, x2
, y2
, n
, ret
;
1240 int diff
= DIFF_BLOCK
;
1243 * Set up a usage structure as a clean slate (everything
1246 usage
= snew(struct solver_usage
);
1250 usage
->cube
= snewn(cr
*cr
*cr
, unsigned char);
1251 usage
->grid
= grid
; /* write straight back to the input */
1252 memset(usage
->cube
, TRUE
, cr
*cr
*cr
);
1254 usage
->row
= snewn(cr
* cr
, unsigned char);
1255 usage
->col
= snewn(cr
* cr
, unsigned char);
1256 usage
->blk
= snewn(cr
* cr
, unsigned char);
1257 memset(usage
->row
, FALSE
, cr
* cr
);
1258 memset(usage
->col
, FALSE
, cr
* cr
);
1259 memset(usage
->blk
, FALSE
, cr
* cr
);
1261 scratch
= solver_new_scratch(usage
);
1264 * Place all the clue numbers we are given.
1266 for (x
= 0; x
< cr
; x
++)
1267 for (y
= 0; y
< cr
; y
++)
1269 solver_place(usage
, x
, YTRANS(y
), grid
[y
*cr
+x
]);
1272 * Now loop over the grid repeatedly trying all permitted modes
1273 * of reasoning. The loop terminates if we complete an
1274 * iteration without making any progress; we then return
1275 * failure or success depending on whether the grid is full or
1280 * I'd like to write `continue;' inside each of the
1281 * following loops, so that the solver returns here after
1282 * making some progress. However, I can't specify that I
1283 * want to continue an outer loop rather than the innermost
1284 * one, so I'm apologetically resorting to a goto.
1289 * Blockwise positional elimination.
1291 for (x
= 0; x
< cr
; x
+= r
)
1292 for (y
= 0; y
< r
; y
++)
1293 for (n
= 1; n
<= cr
; n
++)
1294 if (!usage
->blk
[(y
*c
+(x
/r
))*cr
+n
-1]) {
1295 ret
= solver_elim(usage
, cubepos(x
,y
,n
), r
*cr
1296 #ifdef STANDALONE_SOLVER
1297 , "positional elimination,"
1298 " %d in block (%d,%d)", n
, 1+x
/r
, 1+y
1302 diff
= DIFF_IMPOSSIBLE
;
1304 } else if (ret
> 0) {
1305 diff
= max(diff
, DIFF_BLOCK
);
1310 if (maxdiff
<= DIFF_BLOCK
)
1314 * Row-wise positional elimination.
1316 for (y
= 0; y
< cr
; y
++)
1317 for (n
= 1; n
<= cr
; n
++)
1318 if (!usage
->row
[y
*cr
+n
-1]) {
1319 ret
= solver_elim(usage
, cubepos(0,y
,n
), cr
*cr
1320 #ifdef STANDALONE_SOLVER
1321 , "positional elimination,"
1322 " %d in row %d", n
, 1+YUNTRANS(y
)
1326 diff
= DIFF_IMPOSSIBLE
;
1328 } else if (ret
> 0) {
1329 diff
= max(diff
, DIFF_SIMPLE
);
1334 * Column-wise positional elimination.
1336 for (x
= 0; x
< cr
; x
++)
1337 for (n
= 1; n
<= cr
; n
++)
1338 if (!usage
->col
[x
*cr
+n
-1]) {
1339 ret
= solver_elim(usage
, cubepos(x
,0,n
), cr
1340 #ifdef STANDALONE_SOLVER
1341 , "positional elimination,"
1342 " %d in column %d", n
, 1+x
1346 diff
= DIFF_IMPOSSIBLE
;
1348 } else if (ret
> 0) {
1349 diff
= max(diff
, DIFF_SIMPLE
);
1355 * Numeric elimination.
1357 for (x
= 0; x
< cr
; x
++)
1358 for (y
= 0; y
< cr
; y
++)
1359 if (!usage
->grid
[YUNTRANS(y
)*cr
+x
]) {
1360 ret
= solver_elim(usage
, cubepos(x
,y
,1), 1
1361 #ifdef STANDALONE_SOLVER
1362 , "numeric elimination at (%d,%d)", 1+x
,
1367 diff
= DIFF_IMPOSSIBLE
;
1369 } else if (ret
> 0) {
1370 diff
= max(diff
, DIFF_SIMPLE
);
1375 if (maxdiff
<= DIFF_SIMPLE
)
1379 * Intersectional analysis, rows vs blocks.
1381 for (y
= 0; y
< cr
; y
++)
1382 for (x
= 0; x
< cr
; x
+= r
)
1383 for (n
= 1; n
<= cr
; n
++)
1385 * solver_intersect() never returns -1.
1387 if (!usage
->row
[y
*cr
+n
-1] &&
1388 !usage
->blk
[((y
%r
)*c
+(x
/r
))*cr
+n
-1] &&
1389 (solver_intersect(usage
, cubepos(0,y
,n
), cr
*cr
,
1390 cubepos(x
,y
%r
,n
), r
*cr
1391 #ifdef STANDALONE_SOLVER
1392 , "intersectional analysis,"
1393 " %d in row %d vs block (%d,%d)",
1394 n
, 1+YUNTRANS(y
), 1+x
/r
, 1+y
%r
1397 solver_intersect(usage
, cubepos(x
,y
%r
,n
), r
*cr
,
1398 cubepos(0,y
,n
), cr
*cr
1399 #ifdef STANDALONE_SOLVER
1400 , "intersectional analysis,"
1401 " %d in block (%d,%d) vs row %d",
1402 n
, 1+x
/r
, 1+y
%r
, 1+YUNTRANS(y
)
1405 diff
= max(diff
, DIFF_INTERSECT
);
1410 * Intersectional analysis, columns vs blocks.
1412 for (x
= 0; x
< cr
; x
++)
1413 for (y
= 0; y
< r
; y
++)
1414 for (n
= 1; n
<= cr
; n
++)
1415 if (!usage
->col
[x
*cr
+n
-1] &&
1416 !usage
->blk
[(y
*c
+(x
/r
))*cr
+n
-1] &&
1417 (solver_intersect(usage
, cubepos(x
,0,n
), cr
,
1418 cubepos((x
/r
)*r
,y
,n
), r
*cr
1419 #ifdef STANDALONE_SOLVER
1420 , "intersectional analysis,"
1421 " %d in column %d vs block (%d,%d)",
1425 solver_intersect(usage
, cubepos((x
/r
)*r
,y
,n
), r
*cr
,
1427 #ifdef STANDALONE_SOLVER
1428 , "intersectional analysis,"
1429 " %d in block (%d,%d) vs column %d",
1433 diff
= max(diff
, DIFF_INTERSECT
);
1437 if (maxdiff
<= DIFF_INTERSECT
)
1441 * Blockwise set elimination.
1443 for (x
= 0; x
< cr
; x
+= r
)
1444 for (y
= 0; y
< r
; y
++) {
1445 ret
= solver_set(usage
, scratch
, cubepos(x
,y
,1), r
*cr
, 1
1446 #ifdef STANDALONE_SOLVER
1447 , "set elimination, block (%d,%d)", 1+x
/r
, 1+y
1451 diff
= DIFF_IMPOSSIBLE
;
1453 } else if (ret
> 0) {
1454 diff
= max(diff
, DIFF_SET
);
1460 * Row-wise set elimination.
1462 for (y
= 0; y
< cr
; y
++) {
1463 ret
= solver_set(usage
, scratch
, cubepos(0,y
,1), cr
*cr
, 1
1464 #ifdef STANDALONE_SOLVER
1465 , "set elimination, row %d", 1+YUNTRANS(y
)
1469 diff
= DIFF_IMPOSSIBLE
;
1471 } else if (ret
> 0) {
1472 diff
= max(diff
, DIFF_SET
);
1478 * Column-wise set elimination.
1480 for (x
= 0; x
< cr
; x
++) {
1481 ret
= solver_set(usage
, scratch
, cubepos(x
,0,1), cr
, 1
1482 #ifdef STANDALONE_SOLVER
1483 , "set elimination, column %d", 1+x
1487 diff
= DIFF_IMPOSSIBLE
;
1489 } else if (ret
> 0) {
1490 diff
= max(diff
, DIFF_SET
);
1496 * Row-vs-column set elimination on a single number.
1498 for (n
= 1; n
<= cr
; n
++) {
1499 ret
= solver_set(usage
, scratch
, cubepos(0,0,n
), cr
*cr
, cr
1500 #ifdef STANDALONE_SOLVER
1501 , "positional set elimination, number %d", n
1505 diff
= DIFF_IMPOSSIBLE
;
1507 } else if (ret
> 0) {
1508 diff
= max(diff
, DIFF_EXTREME
);
1514 * Mutual neighbour elimination.
1516 for (y
= 0; y
+1 < cr
; y
++) {
1517 for (x
= 0; x
+1 < cr
; x
++) {
1518 for (y2
= y
+1; y2
< cr
; y2
++) {
1519 for (x2
= x
+1; x2
< cr
; x2
++) {
1521 * Can't do mutual neighbour elimination
1522 * between elements of the same actual
1525 if (x
/r
== x2
/r
&& y
%r
== y2
%r
)
1529 * Otherwise, try (x,y) vs (x2,y2) in both
1530 * directions, and likewise (x2,y) vs
1533 if (!usage
->grid
[YUNTRANS(y
)*cr
+x
] &&
1534 !usage
->grid
[YUNTRANS(y2
)*cr
+x2
] &&
1535 (solver_mne(usage
, scratch
, x
, y
, x2
, y2
) ||
1536 solver_mne(usage
, scratch
, x2
, y2
, x
, y
))) {
1537 diff
= max(diff
, DIFF_EXTREME
);
1540 if (!usage
->grid
[YUNTRANS(y
)*cr
+x2
] &&
1541 !usage
->grid
[YUNTRANS(y2
)*cr
+x
] &&
1542 (solver_mne(usage
, scratch
, x2
, y
, x
, y2
) ||
1543 solver_mne(usage
, scratch
, x
, y2
, x2
, y
))) {
1544 diff
= max(diff
, DIFF_EXTREME
);
1555 if (solver_forcing(usage
, scratch
)) {
1556 diff
= max(diff
, DIFF_EXTREME
);
1561 * If we reach here, we have made no deductions in this
1562 * iteration, so the algorithm terminates.
1568 * Last chance: if we haven't fully solved the puzzle yet, try
1569 * recursing based on guesses for a particular square. We pick
1570 * one of the most constrained empty squares we can find, which
1571 * has the effect of pruning the search tree as much as
1574 if (maxdiff
>= DIFF_RECURSIVE
) {
1575 int best
, bestcount
;
1580 for (y
= 0; y
< cr
; y
++)
1581 for (x
= 0; x
< cr
; x
++)
1582 if (!grid
[y
*cr
+x
]) {
1586 * An unfilled square. Count the number of
1587 * possible digits in it.
1590 for (n
= 1; n
<= cr
; n
++)
1591 if (cube(x
,YTRANS(y
),n
))
1595 * We should have found any impossibilities
1596 * already, so this can safely be an assert.
1600 if (count
< bestcount
) {
1608 digit
*list
, *ingrid
, *outgrid
;
1610 diff
= DIFF_IMPOSSIBLE
; /* no solution found yet */
1613 * Attempt recursion.
1618 list
= snewn(cr
, digit
);
1619 ingrid
= snewn(cr
* cr
, digit
);
1620 outgrid
= snewn(cr
* cr
, digit
);
1621 memcpy(ingrid
, grid
, cr
* cr
);
1623 /* Make a list of the possible digits. */
1624 for (j
= 0, n
= 1; n
<= cr
; n
++)
1625 if (cube(x
,YTRANS(y
),n
))
1628 #ifdef STANDALONE_SOLVER
1629 if (solver_show_working
) {
1631 printf("%*srecursing on (%d,%d) [",
1632 solver_recurse_depth
*4, "", x
, y
);
1633 for (i
= 0; i
< j
; i
++) {
1634 printf("%s%d", sep
, list
[i
]);
1642 * And step along the list, recursing back into the
1643 * main solver at every stage.
1645 for (i
= 0; i
< j
; i
++) {
1648 memcpy(outgrid
, ingrid
, cr
* cr
);
1649 outgrid
[y
*cr
+x
] = list
[i
];
1651 #ifdef STANDALONE_SOLVER
1652 if (solver_show_working
)
1653 printf("%*sguessing %d at (%d,%d)\n",
1654 solver_recurse_depth
*4, "", list
[i
], x
, y
);
1655 solver_recurse_depth
++;
1658 ret
= solver(c
, r
, outgrid
, maxdiff
);
1660 #ifdef STANDALONE_SOLVER
1661 solver_recurse_depth
--;
1662 if (solver_show_working
) {
1663 printf("%*sretracting %d at (%d,%d)\n",
1664 solver_recurse_depth
*4, "", list
[i
], x
, y
);
1669 * If we have our first solution, copy it into the
1670 * grid we will return.
1672 if (diff
== DIFF_IMPOSSIBLE
&& ret
!= DIFF_IMPOSSIBLE
)
1673 memcpy(grid
, outgrid
, cr
*cr
);
1675 if (ret
== DIFF_AMBIGUOUS
)
1676 diff
= DIFF_AMBIGUOUS
;
1677 else if (ret
== DIFF_IMPOSSIBLE
)
1678 /* do not change our return value */;
1680 /* the recursion turned up exactly one solution */
1681 if (diff
== DIFF_IMPOSSIBLE
)
1682 diff
= DIFF_RECURSIVE
;
1684 diff
= DIFF_AMBIGUOUS
;
1688 * As soon as we've found more than one solution,
1689 * give up immediately.
1691 if (diff
== DIFF_AMBIGUOUS
)
1702 * We're forbidden to use recursion, so we just see whether
1703 * our grid is fully solved, and return DIFF_IMPOSSIBLE
1706 for (y
= 0; y
< cr
; y
++)
1707 for (x
= 0; x
< cr
; x
++)
1709 diff
= DIFF_IMPOSSIBLE
;
1714 #ifdef STANDALONE_SOLVER
1715 if (solver_show_working
)
1716 printf("%*s%s found\n",
1717 solver_recurse_depth
*4, "",
1718 diff
== DIFF_IMPOSSIBLE ?
"no solution" :
1719 diff
== DIFF_AMBIGUOUS ?
"multiple solutions" :
1729 solver_free_scratch(scratch
);
1734 /* ----------------------------------------------------------------------
1735 * End of solver code.
1738 /* ----------------------------------------------------------------------
1739 * Solo filled-grid generator.
1741 * This grid generator works by essentially trying to solve a grid
1742 * starting from no clues, and not worrying that there's more than
1743 * one possible solution. Unfortunately, it isn't computationally
1744 * feasible to do this by calling the above solver with an empty
1745 * grid, because that one needs to allocate a lot of scratch space
1746 * at every recursion level. Instead, I have a much simpler
1747 * algorithm which I shamelessly copied from a Python solver
1748 * written by Andrew Wilkinson (which is GPLed, but I've reused
1749 * only ideas and no code). It mostly just does the obvious
1750 * recursive thing: pick an empty square, put one of the possible
1751 * digits in it, recurse until all squares are filled, backtrack
1752 * and change some choices if necessary.
1754 * The clever bit is that every time it chooses which square to
1755 * fill in next, it does so by counting the number of _possible_
1756 * numbers that can go in each square, and it prioritises so that
1757 * it picks a square with the _lowest_ number of possibilities. The
1758 * idea is that filling in lots of the obvious bits (particularly
1759 * any squares with only one possibility) will cut down on the list
1760 * of possibilities for other squares and hence reduce the enormous
1761 * search space as much as possible as early as possible.
1765 * Internal data structure used in gridgen to keep track of
1768 struct gridgen_coord
{ int x
, y
, r
; };
1769 struct gridgen_usage
{
1770 int c
, r
, cr
; /* cr == c*r */
1771 /* grid is a copy of the input grid, modified as we go along */
1773 /* row[y*cr+n-1] TRUE if digit n has been placed in row y */
1775 /* col[x*cr+n-1] TRUE if digit n has been placed in row x */
1777 /* blk[(y*c+x)*cr+n-1] TRUE if digit n has been placed in block (x,y) */
1779 /* This lists all the empty spaces remaining in the grid. */
1780 struct gridgen_coord
*spaces
;
1782 /* If we need randomisation in the solve, this is our random state. */
1787 * The real recursive step in the generating function.
1789 static int gridgen_real(struct gridgen_usage
*usage
, digit
*grid
)
1791 int c
= usage
->c
, r
= usage
->r
, cr
= usage
->cr
;
1792 int i
, j
, n
, sx
, sy
, bestm
, bestr
, ret
;
1796 * Firstly, check for completion! If there are no spaces left
1797 * in the grid, we have a solution.
1799 if (usage
->nspaces
== 0) {
1800 memcpy(grid
, usage
->grid
, cr
* cr
);
1805 * Otherwise, there must be at least one space. Find the most
1806 * constrained space, using the `r' field as a tie-breaker.
1808 bestm
= cr
+1; /* so that any space will beat it */
1811 for (j
= 0; j
< usage
->nspaces
; j
++) {
1812 int x
= usage
->spaces
[j
].x
, y
= usage
->spaces
[j
].y
;
1816 * Find the number of digits that could go in this space.
1819 for (n
= 0; n
< cr
; n
++)
1820 if (!usage
->row
[y
*cr
+n
] && !usage
->col
[x
*cr
+n
] &&
1821 !usage
->blk
[((y
/c
)*c
+(x
/r
))*cr
+n
])
1824 if (m
< bestm
|| (m
== bestm
&& usage
->spaces
[j
].r
< bestr
)) {
1826 bestr
= usage
->spaces
[j
].r
;
1834 * Swap that square into the final place in the spaces array,
1835 * so that decrementing nspaces will remove it from the list.
1837 if (i
!= usage
->nspaces
-1) {
1838 struct gridgen_coord t
;
1839 t
= usage
->spaces
[usage
->nspaces
-1];
1840 usage
->spaces
[usage
->nspaces
-1] = usage
->spaces
[i
];
1841 usage
->spaces
[i
] = t
;
1845 * Now we've decided which square to start our recursion at,
1846 * simply go through all possible values, shuffling them
1847 * randomly first if necessary.
1849 digits
= snewn(bestm
, int);
1851 for (n
= 0; n
< cr
; n
++)
1852 if (!usage
->row
[sy
*cr
+n
] && !usage
->col
[sx
*cr
+n
] &&
1853 !usage
->blk
[((sy
/c
)*c
+(sx
/r
))*cr
+n
]) {
1858 shuffle(digits
, j
, sizeof(*digits
), usage
->rs
);
1860 /* And finally, go through the digit list and actually recurse. */
1862 for (i
= 0; i
< j
; i
++) {
1865 /* Update the usage structure to reflect the placing of this digit. */
1866 usage
->row
[sy
*cr
+n
-1] = usage
->col
[sx
*cr
+n
-1] =
1867 usage
->blk
[((sy
/c
)*c
+(sx
/r
))*cr
+n
-1] = TRUE
;
1868 usage
->grid
[sy
*cr
+sx
] = n
;
1871 /* Call the solver recursively. Stop when we find a solution. */
1872 if (gridgen_real(usage
, grid
))
1875 /* Revert the usage structure. */
1876 usage
->row
[sy
*cr
+n
-1] = usage
->col
[sx
*cr
+n
-1] =
1877 usage
->blk
[((sy
/c
)*c
+(sx
/r
))*cr
+n
-1] = FALSE
;
1878 usage
->grid
[sy
*cr
+sx
] = 0;
1890 * Entry point to generator. You give it dimensions and a starting
1891 * grid, which is simply an array of cr*cr digits.
1893 static void gridgen(int c
, int r
, digit
*grid
, random_state
*rs
)
1895 struct gridgen_usage
*usage
;
1899 * Clear the grid to start with.
1901 memset(grid
, 0, cr
*cr
);
1904 * Create a gridgen_usage structure.
1906 usage
= snew(struct gridgen_usage
);
1912 usage
->grid
= snewn(cr
* cr
, digit
);
1913 memcpy(usage
->grid
, grid
, cr
* cr
);
1915 usage
->row
= snewn(cr
* cr
, unsigned char);
1916 usage
->col
= snewn(cr
* cr
, unsigned char);
1917 usage
->blk
= snewn(cr
* cr
, unsigned char);
1918 memset(usage
->row
, FALSE
, cr
* cr
);
1919 memset(usage
->col
, FALSE
, cr
* cr
);
1920 memset(usage
->blk
, FALSE
, cr
* cr
);
1922 usage
->spaces
= snewn(cr
* cr
, struct gridgen_coord
);
1928 * Initialise the list of grid spaces.
1930 for (y
= 0; y
< cr
; y
++) {
1931 for (x
= 0; x
< cr
; x
++) {
1932 usage
->spaces
[usage
->nspaces
].x
= x
;
1933 usage
->spaces
[usage
->nspaces
].y
= y
;
1934 usage
->spaces
[usage
->nspaces
].r
= random_bits(rs
, 31);
1940 * Run the real generator function.
1942 gridgen_real(usage
, grid
);
1945 * Clean up the usage structure now we have our answer.
1947 sfree(usage
->spaces
);
1955 /* ----------------------------------------------------------------------
1956 * End of grid generator code.
1960 * Check whether a grid contains a valid complete puzzle.
1962 static int check_valid(int c
, int r
, digit
*grid
)
1965 unsigned char *used
;
1968 used
= snewn(cr
, unsigned char);
1971 * Check that each row contains precisely one of everything.
1973 for (y
= 0; y
< cr
; y
++) {
1974 memset(used
, FALSE
, cr
);
1975 for (x
= 0; x
< cr
; x
++)
1976 if (grid
[y
*cr
+x
] > 0 && grid
[y
*cr
+x
] <= cr
)
1977 used
[grid
[y
*cr
+x
]-1] = TRUE
;
1978 for (n
= 0; n
< cr
; n
++)
1986 * Check that each column contains precisely one of everything.
1988 for (x
= 0; x
< cr
; x
++) {
1989 memset(used
, FALSE
, cr
);
1990 for (y
= 0; y
< cr
; y
++)
1991 if (grid
[y
*cr
+x
] > 0 && grid
[y
*cr
+x
] <= cr
)
1992 used
[grid
[y
*cr
+x
]-1] = TRUE
;
1993 for (n
= 0; n
< cr
; n
++)
2001 * Check that each block contains precisely one of everything.
2003 for (x
= 0; x
< cr
; x
+= r
) {
2004 for (y
= 0; y
< cr
; y
+= c
) {
2006 memset(used
, FALSE
, cr
);
2007 for (xx
= x
; xx
< x
+r
; xx
++)
2008 for (yy
= 0; yy
< y
+c
; yy
++)
2009 if (grid
[yy
*cr
+xx
] > 0 && grid
[yy
*cr
+xx
] <= cr
)
2010 used
[grid
[yy
*cr
+xx
]-1] = TRUE
;
2011 for (n
= 0; n
< cr
; n
++)
2023 static int symmetries(game_params
*params
, int x
, int y
, int *output
, int s
)
2025 int c
= params
->c
, r
= params
->r
, cr
= c
*r
;
2028 #define ADD(x,y) (*output++ = (x), *output++ = (y), i++)
2034 break; /* just x,y is all we need */
2036 ADD(cr
- 1 - x
, cr
- 1 - y
);
2041 ADD(cr
- 1 - x
, cr
- 1 - y
);
2052 ADD(cr
- 1 - x
, cr
- 1 - y
);
2056 ADD(cr
- 1 - x
, cr
- 1 - y
);
2057 ADD(cr
- 1 - y
, cr
- 1 - x
);
2062 ADD(cr
- 1 - x
, cr
- 1 - y
);
2066 ADD(cr
- 1 - y
, cr
- 1 - x
);
2075 static char *encode_solve_move(int cr
, digit
*grid
)
2078 char *ret
, *p
, *sep
;
2081 * It's surprisingly easy to work out _exactly_ how long this
2082 * string needs to be. To decimal-encode all the numbers from 1
2085 * - every number has a units digit; total is n.
2086 * - all numbers above 9 have a tens digit; total is max(n-9,0).
2087 * - all numbers above 99 have a hundreds digit; total is max(n-99,0).
2091 for (i
= 1; i
<= cr
; i
*= 10)
2092 len
+= max(cr
- i
+ 1, 0);
2093 len
+= cr
; /* don't forget the commas */
2094 len
*= cr
; /* there are cr rows of these */
2097 * Now len is one bigger than the total size of the
2098 * comma-separated numbers (because we counted an
2099 * additional leading comma). We need to have a leading S
2100 * and a trailing NUL, so we're off by one in total.
2104 ret
= snewn(len
, char);
2108 for (i
= 0; i
< cr
*cr
; i
++) {
2109 p
+= sprintf(p
, "%s%d", sep
, grid
[i
]);
2113 assert(p
- ret
== len
);
2118 static char *new_game_desc(game_params
*params
, random_state
*rs
,
2119 char **aux
, int interactive
)
2121 int c
= params
->c
, r
= params
->r
, cr
= c
*r
;
2123 digit
*grid
, *grid2
;
2124 struct xy
{ int x
, y
; } *locs
;
2127 int coords
[16], ncoords
;
2132 * Adjust the maximum difficulty level to be consistent with
2133 * the puzzle size: all 2x2 puzzles appear to be Trivial
2134 * (DIFF_BLOCK) so we cannot hold out for even a Basic
2135 * (DIFF_SIMPLE) one.
2137 maxdiff
= params
->diff
;
2138 if (c
== 2 && r
== 2)
2139 maxdiff
= DIFF_BLOCK
;
2141 grid
= snewn(area
, digit
);
2142 locs
= snewn(area
, struct xy
);
2143 grid2
= snewn(area
, digit
);
2146 * Loop until we get a grid of the required difficulty. This is
2147 * nasty, but it seems to be unpleasantly hard to generate
2148 * difficult grids otherwise.
2152 * Generate a random solved state.
2154 gridgen(c
, r
, grid
, rs
);
2155 assert(check_valid(c
, r
, grid
));
2158 * Save the solved grid in aux.
2162 * We might already have written *aux the last time we
2163 * went round this loop, in which case we should free
2164 * the old aux before overwriting it with the new one.
2170 *aux
= encode_solve_move(cr
, grid
);
2174 * Now we have a solved grid, start removing things from it
2175 * while preserving solubility.
2179 * Find the set of equivalence classes of squares permitted
2180 * by the selected symmetry. We do this by enumerating all
2181 * the grid squares which have no symmetric companion
2182 * sorting lower than themselves.
2185 for (y
= 0; y
< cr
; y
++)
2186 for (x
= 0; x
< cr
; x
++) {
2190 ncoords
= symmetries(params
, x
, y
, coords
, params
->symm
);
2191 for (j
= 0; j
< ncoords
; j
++)
2192 if (coords
[2*j
+1]*cr
+coords
[2*j
] < i
)
2202 * Now shuffle that list.
2204 shuffle(locs
, nlocs
, sizeof(*locs
), rs
);
2207 * Now loop over the shuffled list and, for each element,
2208 * see whether removing that element (and its reflections)
2209 * from the grid will still leave the grid soluble.
2211 for (i
= 0; i
< nlocs
; i
++) {
2217 memcpy(grid2
, grid
, area
);
2218 ncoords
= symmetries(params
, x
, y
, coords
, params
->symm
);
2219 for (j
= 0; j
< ncoords
; j
++)
2220 grid2
[coords
[2*j
+1]*cr
+coords
[2*j
]] = 0;
2222 ret
= solver(c
, r
, grid2
, maxdiff
);
2223 if (ret
<= maxdiff
) {
2224 for (j
= 0; j
< ncoords
; j
++)
2225 grid
[coords
[2*j
+1]*cr
+coords
[2*j
]] = 0;
2229 memcpy(grid2
, grid
, area
);
2230 } while (solver(c
, r
, grid2
, maxdiff
) < maxdiff
);
2236 * Now we have the grid as it will be presented to the user.
2237 * Encode it in a game desc.
2243 desc
= snewn(5 * area
, char);
2246 for (i
= 0; i
<= area
; i
++) {
2247 int n
= (i
< area ? grid
[i
] : -1);
2254 int c
= 'a' - 1 + run
;
2258 run
-= c
- ('a' - 1);
2262 * If there's a number in the very top left or
2263 * bottom right, there's no point putting an
2264 * unnecessary _ before or after it.
2266 if (p
> desc
&& n
> 0)
2270 p
+= sprintf(p
, "%d", n
);
2274 assert(p
- desc
< 5 * area
);
2276 desc
= sresize(desc
, p
- desc
, char);
2284 static char *validate_desc(game_params
*params
, char *desc
)
2286 int area
= params
->r
* params
->r
* params
->c
* params
->c
;
2291 if (n
>= 'a' && n
<= 'z') {
2292 squares
+= n
- 'a' + 1;
2293 } else if (n
== '_') {
2295 } else if (n
> '0' && n
<= '9') {
2296 int val
= atoi(desc
-1);
2297 if (val
< 1 || val
> params
->c
* params
->r
)
2298 return "Out-of-range number in game description";
2300 while (*desc
>= '0' && *desc
<= '9')
2303 return "Invalid character in game description";
2307 return "Not enough data to fill grid";
2310 return "Too much data to fit in grid";
2315 static game_state
*new_game(midend
*me
, game_params
*params
, char *desc
)
2317 game_state
*state
= snew(game_state
);
2318 int c
= params
->c
, r
= params
->r
, cr
= c
*r
, area
= cr
* cr
;
2321 state
->c
= params
->c
;
2322 state
->r
= params
->r
;
2324 state
->grid
= snewn(area
, digit
);
2325 state
->pencil
= snewn(area
* cr
, unsigned char);
2326 memset(state
->pencil
, 0, area
* cr
);
2327 state
->immutable
= snewn(area
, unsigned char);
2328 memset(state
->immutable
, FALSE
, area
);
2330 state
->completed
= state
->cheated
= FALSE
;
2335 if (n
>= 'a' && n
<= 'z') {
2336 int run
= n
- 'a' + 1;
2337 assert(i
+ run
<= area
);
2339 state
->grid
[i
++] = 0;
2340 } else if (n
== '_') {
2342 } else if (n
> '0' && n
<= '9') {
2344 state
->immutable
[i
] = TRUE
;
2345 state
->grid
[i
++] = atoi(desc
-1);
2346 while (*desc
>= '0' && *desc
<= '9')
2349 assert(!"We can't get here");
2357 static game_state
*dup_game(game_state
*state
)
2359 game_state
*ret
= snew(game_state
);
2360 int c
= state
->c
, r
= state
->r
, cr
= c
*r
, area
= cr
* cr
;
2365 ret
->grid
= snewn(area
, digit
);
2366 memcpy(ret
->grid
, state
->grid
, area
);
2368 ret
->pencil
= snewn(area
* cr
, unsigned char);
2369 memcpy(ret
->pencil
, state
->pencil
, area
* cr
);
2371 ret
->immutable
= snewn(area
, unsigned char);
2372 memcpy(ret
->immutable
, state
->immutable
, area
);
2374 ret
->completed
= state
->completed
;
2375 ret
->cheated
= state
->cheated
;
2380 static void free_game(game_state
*state
)
2382 sfree(state
->immutable
);
2383 sfree(state
->pencil
);
2388 static char *solve_game(game_state
*state
, game_state
*currstate
,
2389 char *ai
, char **error
)
2391 int c
= state
->c
, r
= state
->r
, cr
= c
*r
;
2397 * If we already have the solution in ai, save ourselves some
2403 grid
= snewn(cr
*cr
, digit
);
2404 memcpy(grid
, state
->grid
, cr
*cr
);
2405 solve_ret
= solver(c
, r
, grid
, DIFF_RECURSIVE
);
2409 if (solve_ret
== DIFF_IMPOSSIBLE
)
2410 *error
= "No solution exists for this puzzle";
2411 else if (solve_ret
== DIFF_AMBIGUOUS
)
2412 *error
= "Multiple solutions exist for this puzzle";
2419 ret
= encode_solve_move(cr
, grid
);
2426 static char *grid_text_format(int c
, int r
, digit
*grid
)
2434 * There are cr lines of digits, plus r-1 lines of block
2435 * separators. Each line contains cr digits, cr-1 separating
2436 * spaces, and c-1 two-character block separators. Thus, the
2437 * total length of a line is 2*cr+2*c-3 (not counting the
2438 * newline), and there are cr+r-1 of them.
2440 maxlen
= (cr
+r
-1) * (2*cr
+2*c
-2);
2441 ret
= snewn(maxlen
+1, char);
2444 for (y
= 0; y
< cr
; y
++) {
2445 for (x
= 0; x
< cr
; x
++) {
2446 int ch
= grid
[y
* cr
+ x
];
2456 if ((x
+1) % r
== 0) {
2463 if (y
+1 < cr
&& (y
+1) % c
== 0) {
2464 for (x
= 0; x
< cr
; x
++) {
2468 if ((x
+1) % r
== 0) {
2478 assert(p
- ret
== maxlen
);
2483 static char *game_text_format(game_state
*state
)
2485 return grid_text_format(state
->c
, state
->r
, state
->grid
);
2490 * These are the coordinates of the currently highlighted
2491 * square on the grid, or -1,-1 if there isn't one. When there
2492 * is, pressing a valid number or letter key or Space will
2493 * enter that number or letter in the grid.
2497 * This indicates whether the current highlight is a
2498 * pencil-mark one or a real one.
2503 static game_ui
*new_ui(game_state
*state
)
2505 game_ui
*ui
= snew(game_ui
);
2507 ui
->hx
= ui
->hy
= -1;
2513 static void free_ui(game_ui
*ui
)
2518 static char *encode_ui(game_ui
*ui
)
2523 static void decode_ui(game_ui
*ui
, char *encoding
)
2527 static void game_changed_state(game_ui
*ui
, game_state
*oldstate
,
2528 game_state
*newstate
)
2530 int c
= newstate
->c
, r
= newstate
->r
, cr
= c
*r
;
2532 * We prevent pencil-mode highlighting of a filled square. So
2533 * if the user has just filled in a square which we had a
2534 * pencil-mode highlight in (by Undo, or by Redo, or by Solve),
2535 * then we cancel the highlight.
2537 if (ui
->hx
>= 0 && ui
->hy
>= 0 && ui
->hpencil
&&
2538 newstate
->grid
[ui
->hy
* cr
+ ui
->hx
] != 0) {
2539 ui
->hx
= ui
->hy
= -1;
2543 struct game_drawstate
{
2548 unsigned char *pencil
;
2550 /* This is scratch space used within a single call to game_redraw. */
2554 static char *interpret_move(game_state
*state
, game_ui
*ui
, game_drawstate
*ds
,
2555 int x
, int y
, int button
)
2557 int c
= state
->c
, r
= state
->r
, cr
= c
*r
;
2561 button
&= ~MOD_MASK
;
2563 tx
= (x
+ TILE_SIZE
- BORDER
) / TILE_SIZE
- 1;
2564 ty
= (y
+ TILE_SIZE
- BORDER
) / TILE_SIZE
- 1;
2566 if (tx
>= 0 && tx
< cr
&& ty
>= 0 && ty
< cr
) {
2567 if (button
== LEFT_BUTTON
) {
2568 if (state
->immutable
[ty
*cr
+tx
]) {
2569 ui
->hx
= ui
->hy
= -1;
2570 } else if (tx
== ui
->hx
&& ty
== ui
->hy
&& ui
->hpencil
== 0) {
2571 ui
->hx
= ui
->hy
= -1;
2577 return ""; /* UI activity occurred */
2579 if (button
== RIGHT_BUTTON
) {
2581 * Pencil-mode highlighting for non filled squares.
2583 if (state
->grid
[ty
*cr
+tx
] == 0) {
2584 if (tx
== ui
->hx
&& ty
== ui
->hy
&& ui
->hpencil
) {
2585 ui
->hx
= ui
->hy
= -1;
2592 ui
->hx
= ui
->hy
= -1;
2594 return ""; /* UI activity occurred */
2598 if (ui
->hx
!= -1 && ui
->hy
!= -1 &&
2599 ((button
>= '1' && button
<= '9' && button
- '0' <= cr
) ||
2600 (button
>= 'a' && button
<= 'z' && button
- 'a' + 10 <= cr
) ||
2601 (button
>= 'A' && button
<= 'Z' && button
- 'A' + 10 <= cr
) ||
2602 button
== ' ' || button
== '\010' || button
== '\177')) {
2603 int n
= button
- '0';
2604 if (button
>= 'A' && button
<= 'Z')
2605 n
= button
- 'A' + 10;
2606 if (button
>= 'a' && button
<= 'z')
2607 n
= button
- 'a' + 10;
2608 if (button
== ' ' || button
== '\010' || button
== '\177')
2612 * Can't overwrite this square. In principle this shouldn't
2613 * happen anyway because we should never have even been
2614 * able to highlight the square, but it never hurts to be
2617 if (state
->immutable
[ui
->hy
*cr
+ui
->hx
])
2621 * Can't make pencil marks in a filled square. In principle
2622 * this shouldn't happen anyway because we should never
2623 * have even been able to pencil-highlight the square, but
2624 * it never hurts to be careful.
2626 if (ui
->hpencil
&& state
->grid
[ui
->hy
*cr
+ui
->hx
])
2629 sprintf(buf
, "%c%d,%d,%d",
2630 (char)(ui
->hpencil
&& n
> 0 ?
'P' : 'R'), ui
->hx
, ui
->hy
, n
);
2632 ui
->hx
= ui
->hy
= -1;
2640 static game_state
*execute_move(game_state
*from
, char *move
)
2642 int c
= from
->c
, r
= from
->r
, cr
= c
*r
;
2646 if (move
[0] == 'S') {
2649 ret
= dup_game(from
);
2650 ret
->completed
= ret
->cheated
= TRUE
;
2653 for (n
= 0; n
< cr
*cr
; n
++) {
2654 ret
->grid
[n
] = atoi(p
);
2656 if (!*p
|| ret
->grid
[n
] < 1 || ret
->grid
[n
] > cr
) {
2661 while (*p
&& isdigit((unsigned char)*p
)) p
++;
2666 } else if ((move
[0] == 'P' || move
[0] == 'R') &&
2667 sscanf(move
+1, "%d,%d,%d", &x
, &y
, &n
) == 3 &&
2668 x
>= 0 && x
< cr
&& y
>= 0 && y
< cr
&& n
>= 0 && n
<= cr
) {
2670 ret
= dup_game(from
);
2671 if (move
[0] == 'P' && n
> 0) {
2672 int index
= (y
*cr
+x
) * cr
+ (n
-1);
2673 ret
->pencil
[index
] = !ret
->pencil
[index
];
2675 ret
->grid
[y
*cr
+x
] = n
;
2676 memset(ret
->pencil
+ (y
*cr
+x
)*cr
, 0, cr
);
2679 * We've made a real change to the grid. Check to see
2680 * if the game has been completed.
2682 if (!ret
->completed
&& check_valid(c
, r
, ret
->grid
)) {
2683 ret
->completed
= TRUE
;
2688 return NULL
; /* couldn't parse move string */
2691 /* ----------------------------------------------------------------------
2695 #define SIZE(cr) ((cr) * TILE_SIZE + 2*BORDER + 1)
2696 #define GETTILESIZE(cr, w) ( (double)(w-1) / (double)(cr+1) )
2698 static void game_compute_size(game_params
*params
, int tilesize
,
2701 /* Ick: fake up `ds->tilesize' for macro expansion purposes */
2702 struct { int tilesize
; } ads
, *ds
= &ads
;
2703 ads
.tilesize
= tilesize
;
2705 *x
= SIZE(params
->c
* params
->r
);
2706 *y
= SIZE(params
->c
* params
->r
);
2709 static void game_set_size(drawing
*dr
, game_drawstate
*ds
,
2710 game_params
*params
, int tilesize
)
2712 ds
->tilesize
= tilesize
;
2715 static float *game_colours(frontend
*fe
, int *ncolours
)
2717 float *ret
= snewn(3 * NCOLOURS
, float);
2719 frontend_default_colour(fe
, &ret
[COL_BACKGROUND
* 3]);
2721 ret
[COL_GRID
* 3 + 0] = 0.0F
;
2722 ret
[COL_GRID
* 3 + 1] = 0.0F
;
2723 ret
[COL_GRID
* 3 + 2] = 0.0F
;
2725 ret
[COL_CLUE
* 3 + 0] = 0.0F
;
2726 ret
[COL_CLUE
* 3 + 1] = 0.0F
;
2727 ret
[COL_CLUE
* 3 + 2] = 0.0F
;
2729 ret
[COL_USER
* 3 + 0] = 0.0F
;
2730 ret
[COL_USER
* 3 + 1] = 0.6F
* ret
[COL_BACKGROUND
* 3 + 1];
2731 ret
[COL_USER
* 3 + 2] = 0.0F
;
2733 ret
[COL_HIGHLIGHT
* 3 + 0] = 0.85F
* ret
[COL_BACKGROUND
* 3 + 0];
2734 ret
[COL_HIGHLIGHT
* 3 + 1] = 0.85F
* ret
[COL_BACKGROUND
* 3 + 1];
2735 ret
[COL_HIGHLIGHT
* 3 + 2] = 0.85F
* ret
[COL_BACKGROUND
* 3 + 2];
2737 ret
[COL_ERROR
* 3 + 0] = 1.0F
;
2738 ret
[COL_ERROR
* 3 + 1] = 0.0F
;
2739 ret
[COL_ERROR
* 3 + 2] = 0.0F
;
2741 ret
[COL_PENCIL
* 3 + 0] = 0.5F
* ret
[COL_BACKGROUND
* 3 + 0];
2742 ret
[COL_PENCIL
* 3 + 1] = 0.5F
* ret
[COL_BACKGROUND
* 3 + 1];
2743 ret
[COL_PENCIL
* 3 + 2] = ret
[COL_BACKGROUND
* 3 + 2];
2745 *ncolours
= NCOLOURS
;
2749 static game_drawstate
*game_new_drawstate(drawing
*dr
, game_state
*state
)
2751 struct game_drawstate
*ds
= snew(struct game_drawstate
);
2752 int c
= state
->c
, r
= state
->r
, cr
= c
*r
;
2754 ds
->started
= FALSE
;
2758 ds
->grid
= snewn(cr
*cr
, digit
);
2759 memset(ds
->grid
, 0, cr
*cr
);
2760 ds
->pencil
= snewn(cr
*cr
*cr
, digit
);
2761 memset(ds
->pencil
, 0, cr
*cr
*cr
);
2762 ds
->hl
= snewn(cr
*cr
, unsigned char);
2763 memset(ds
->hl
, 0, cr
*cr
);
2764 ds
->entered_items
= snewn(cr
*cr
, int);
2765 ds
->tilesize
= 0; /* not decided yet */
2769 static void game_free_drawstate(drawing
*dr
, game_drawstate
*ds
)
2774 sfree(ds
->entered_items
);
2778 static void draw_number(drawing
*dr
, game_drawstate
*ds
, game_state
*state
,
2779 int x
, int y
, int hl
)
2781 int c
= state
->c
, r
= state
->r
, cr
= c
*r
;
2786 if (ds
->grid
[y
*cr
+x
] == state
->grid
[y
*cr
+x
] &&
2787 ds
->hl
[y
*cr
+x
] == hl
&&
2788 !memcmp(ds
->pencil
+(y
*cr
+x
)*cr
, state
->pencil
+(y
*cr
+x
)*cr
, cr
))
2789 return; /* no change required */
2791 tx
= BORDER
+ x
* TILE_SIZE
+ 2;
2792 ty
= BORDER
+ y
* TILE_SIZE
+ 2;
2808 clip(dr
, cx
, cy
, cw
, ch
);
2810 /* background needs erasing */
2811 draw_rect(dr
, cx
, cy
, cw
, ch
, (hl
& 15) == 1 ? COL_HIGHLIGHT
: COL_BACKGROUND
);
2813 /* pencil-mode highlight */
2814 if ((hl
& 15) == 2) {
2818 coords
[2] = cx
+cw
/2;
2821 coords
[5] = cy
+ch
/2;
2822 draw_polygon(dr
, coords
, 3, COL_HIGHLIGHT
, COL_HIGHLIGHT
);
2825 /* new number needs drawing? */
2826 if (state
->grid
[y
*cr
+x
]) {
2828 str
[0] = state
->grid
[y
*cr
+x
] + '0';
2830 str
[0] += 'a' - ('9'+1);
2831 draw_text(dr
, tx
+ TILE_SIZE
/2, ty
+ TILE_SIZE
/2,
2832 FONT_VARIABLE
, TILE_SIZE
/2, ALIGN_VCENTRE
| ALIGN_HCENTRE
,
2833 state
->immutable
[y
*cr
+x
] ? COL_CLUE
: (hl
& 16) ? COL_ERROR
: COL_USER
, str
);
2836 int pw
, ph
, pmax
, fontsize
;
2838 /* count the pencil marks required */
2839 for (i
= npencil
= 0; i
< cr
; i
++)
2840 if (state
->pencil
[(y
*cr
+x
)*cr
+i
])
2844 * It's not sensible to arrange pencil marks in the same
2845 * layout as the squares within a block, because this leads
2846 * to the font being too small. Instead, we arrange pencil
2847 * marks in the nearest thing we can to a square layout,
2848 * and we adjust the square layout depending on the number
2849 * of pencil marks in the square.
2851 for (pw
= 1; pw
* pw
< npencil
; pw
++);
2852 if (pw
< 3) pw
= 3; /* otherwise it just looks _silly_ */
2853 ph
= (npencil
+ pw
- 1) / pw
;
2854 if (ph
< 2) ph
= 2; /* likewise */
2856 fontsize
= TILE_SIZE
/(pmax
*(11-pmax
)/8);
2858 for (i
= j
= 0; i
< cr
; i
++)
2859 if (state
->pencil
[(y
*cr
+x
)*cr
+i
]) {
2860 int dx
= j
% pw
, dy
= j
/ pw
;
2865 str
[0] += 'a' - ('9'+1);
2866 draw_text(dr
, tx
+ (4*dx
+3) * TILE_SIZE
/ (4*pw
+2),
2867 ty
+ (4*dy
+3) * TILE_SIZE
/ (4*ph
+2),
2868 FONT_VARIABLE
, fontsize
,
2869 ALIGN_VCENTRE
| ALIGN_HCENTRE
, COL_PENCIL
, str
);
2876 draw_update(dr
, cx
, cy
, cw
, ch
);
2878 ds
->grid
[y
*cr
+x
] = state
->grid
[y
*cr
+x
];
2879 memcpy(ds
->pencil
+(y
*cr
+x
)*cr
, state
->pencil
+(y
*cr
+x
)*cr
, cr
);
2880 ds
->hl
[y
*cr
+x
] = hl
;
2883 static void game_redraw(drawing
*dr
, game_drawstate
*ds
, game_state
*oldstate
,
2884 game_state
*state
, int dir
, game_ui
*ui
,
2885 float animtime
, float flashtime
)
2887 int c
= state
->c
, r
= state
->r
, cr
= c
*r
;
2892 * The initial contents of the window are not guaranteed
2893 * and can vary with front ends. To be on the safe side,
2894 * all games should start by drawing a big
2895 * background-colour rectangle covering the whole window.
2897 draw_rect(dr
, 0, 0, SIZE(cr
), SIZE(cr
), COL_BACKGROUND
);
2902 for (x
= 0; x
<= cr
; x
++) {
2903 int thick
= (x
% r ?
0 : 1);
2904 draw_rect(dr
, BORDER
+ x
*TILE_SIZE
- thick
, BORDER
-1,
2905 1+2*thick
, cr
*TILE_SIZE
+3, COL_GRID
);
2907 for (y
= 0; y
<= cr
; y
++) {
2908 int thick
= (y
% c ?
0 : 1);
2909 draw_rect(dr
, BORDER
-1, BORDER
+ y
*TILE_SIZE
- thick
,
2910 cr
*TILE_SIZE
+3, 1+2*thick
, COL_GRID
);
2915 * This array is used to keep track of rows, columns and boxes
2916 * which contain a number more than once.
2918 for (x
= 0; x
< cr
* cr
; x
++)
2919 ds
->entered_items
[x
] = 0;
2920 for (x
= 0; x
< cr
; x
++)
2921 for (y
= 0; y
< cr
; y
++) {
2922 digit d
= state
->grid
[y
*cr
+x
];
2924 int box
= (x
/r
)+(y
/c
)*c
;
2925 ds
->entered_items
[x
*cr
+d
-1] |= ((ds
->entered_items
[x
*cr
+d
-1] & 1) << 1) | 1;
2926 ds
->entered_items
[y
*cr
+d
-1] |= ((ds
->entered_items
[y
*cr
+d
-1] & 4) << 1) | 4;
2927 ds
->entered_items
[box
*cr
+d
-1] |= ((ds
->entered_items
[box
*cr
+d
-1] & 16) << 1) | 16;
2932 * Draw any numbers which need redrawing.
2934 for (x
= 0; x
< cr
; x
++) {
2935 for (y
= 0; y
< cr
; y
++) {
2937 digit d
= state
->grid
[y
*cr
+x
];
2939 if (flashtime
> 0 &&
2940 (flashtime
<= FLASH_TIME
/3 ||
2941 flashtime
>= FLASH_TIME
*2/3))
2944 /* Highlight active input areas. */
2945 if (x
== ui
->hx
&& y
== ui
->hy
)
2946 highlight
= ui
->hpencil ?
2 : 1;
2948 /* Mark obvious errors (ie, numbers which occur more than once
2949 * in a single row, column, or box). */
2950 if (d
&& ((ds
->entered_items
[x
*cr
+d
-1] & 2) ||
2951 (ds
->entered_items
[y
*cr
+d
-1] & 8) ||
2952 (ds
->entered_items
[((x
/r
)+(y
/c
)*c
)*cr
+d
-1] & 32)))
2955 draw_number(dr
, ds
, state
, x
, y
, highlight
);
2960 * Update the _entire_ grid if necessary.
2963 draw_update(dr
, 0, 0, SIZE(cr
), SIZE(cr
));
2968 static float game_anim_length(game_state
*oldstate
, game_state
*newstate
,
2969 int dir
, game_ui
*ui
)
2974 static float game_flash_length(game_state
*oldstate
, game_state
*newstate
,
2975 int dir
, game_ui
*ui
)
2977 if (!oldstate
->completed
&& newstate
->completed
&&
2978 !oldstate
->cheated
&& !newstate
->cheated
)
2983 static int game_timing_state(game_state
*state
, game_ui
*ui
)
2988 static void game_print_size(game_params
*params
, float *x
, float *y
)
2993 * I'll use 9mm squares by default. They should be quite big
2994 * for this game, because players will want to jot down no end
2995 * of pencil marks in the squares.
2997 game_compute_size(params
, 900, &pw
, &ph
);
3002 static void game_print(drawing
*dr
, game_state
*state
, int tilesize
)
3004 int c
= state
->c
, r
= state
->r
, cr
= c
*r
;
3005 int ink
= print_mono_colour(dr
, 0);
3008 /* Ick: fake up `ds->tilesize' for macro expansion purposes */
3009 game_drawstate ads
, *ds
= &ads
;
3010 game_set_size(dr
, ds
, NULL
, tilesize
);
3015 print_line_width(dr
, 3 * TILE_SIZE
/ 40);
3016 draw_rect_outline(dr
, BORDER
, BORDER
, cr
*TILE_SIZE
, cr
*TILE_SIZE
, ink
);
3021 for (x
= 1; x
< cr
; x
++) {
3022 print_line_width(dr
, (x
% r ?
1 : 3) * TILE_SIZE
/ 40);
3023 draw_line(dr
, BORDER
+x
*TILE_SIZE
, BORDER
,
3024 BORDER
+x
*TILE_SIZE
, BORDER
+cr
*TILE_SIZE
, ink
);
3026 for (y
= 1; y
< cr
; y
++) {
3027 print_line_width(dr
, (y
% c ?
1 : 3) * TILE_SIZE
/ 40);
3028 draw_line(dr
, BORDER
, BORDER
+y
*TILE_SIZE
,
3029 BORDER
+cr
*TILE_SIZE
, BORDER
+y
*TILE_SIZE
, ink
);
3035 for (y
= 0; y
< cr
; y
++)
3036 for (x
= 0; x
< cr
; x
++)
3037 if (state
->grid
[y
*cr
+x
]) {
3040 str
[0] = state
->grid
[y
*cr
+x
] + '0';
3042 str
[0] += 'a' - ('9'+1);
3043 draw_text(dr
, BORDER
+ x
*TILE_SIZE
+ TILE_SIZE
/2,
3044 BORDER
+ y
*TILE_SIZE
+ TILE_SIZE
/2,
3045 FONT_VARIABLE
, TILE_SIZE
/2,
3046 ALIGN_VCENTRE
| ALIGN_HCENTRE
, ink
, str
);
3051 #define thegame solo
3054 const struct game thegame
= {
3055 "Solo", "games.solo", "solo",
3062 TRUE
, game_configure
, custom_params
,
3070 TRUE
, game_text_format
,
3078 PREFERRED_TILE_SIZE
, game_compute_size
, game_set_size
,
3081 game_free_drawstate
,
3085 TRUE
, FALSE
, game_print_size
, game_print
,
3086 FALSE
, /* wants_statusbar */
3087 FALSE
, game_timing_state
,
3091 #ifdef STANDALONE_SOLVER
3093 int main(int argc
, char **argv
)
3097 char *id
= NULL
, *desc
, *err
;
3101 while (--argc
> 0) {
3103 if (!strcmp(p
, "-v")) {
3104 solver_show_working
= TRUE
;
3105 } else if (!strcmp(p
, "-g")) {
3107 } else if (*p
== '-') {
3108 fprintf(stderr
, "%s: unrecognised option `%s'\n", argv
[0], p
);
3116 fprintf(stderr
, "usage: %s [-g | -v] <game_id>\n", argv
[0]);
3120 desc
= strchr(id
, ':');
3122 fprintf(stderr
, "%s: game id expects a colon in it\n", argv
[0]);
3127 p
= default_params();
3128 decode_params(p
, id
);
3129 err
= validate_desc(p
, desc
);
3131 fprintf(stderr
, "%s: %s\n", argv
[0], err
);
3134 s
= new_game(NULL
, p
, desc
);
3136 ret
= solver(p
->c
, p
->r
, s
->grid
, DIFF_RECURSIVE
);
3138 printf("Difficulty rating: %s\n",
3139 ret
==DIFF_BLOCK ?
"Trivial (blockwise positional elimination only)":
3140 ret
==DIFF_SIMPLE ?
"Basic (row/column/number elimination required)":
3141 ret
==DIFF_INTERSECT ?
"Intermediate (intersectional analysis required)":
3142 ret
==DIFF_SET ?
"Advanced (set elimination required)":
3143 ret
==DIFF_EXTREME ?
"Extreme (complex non-recursive techniques required)":
3144 ret
==DIFF_RECURSIVE ?
"Unreasonable (guesswork and backtracking required)":
3145 ret
==DIFF_AMBIGUOUS ?
"Ambiguous (multiple solutions exist)":
3146 ret
==DIFF_IMPOSSIBLE ?
"Impossible (no solution exists)":
3147 "INTERNAL ERROR: unrecognised difficulty code");
3149 printf("%s\n", grid_text_format(p
->c
, p
->r
, s
->grid
));