2 * map.c: Game involving four-colouring a map.
9 * - better four-colouring algorithm?
22 * In standalone solver mode, `verbose' is a variable which can be
23 * set by command-line option; in debugging mode it's simply always
26 #if defined STANDALONE_SOLVER
27 #define SOLVER_DIAGNOSTICS
29 #elif defined SOLVER_DIAGNOSTICS
34 * I don't seriously anticipate wanting to change the number of
35 * colours used in this game, but it doesn't cost much to use a
36 * #define just in case :-)
39 #define THREE (FOUR-1)
44 * Ghastly run-time configuration option, just for Gareth (again).
46 static int flash_type = -1;
47 static float flash_length;
50 * Difficulty levels. I do some macro ickery here to ensure that my
51 * enum and the various forms of my name list always match up.
57 A(RECURSE,Unreasonable,u)
58 #define ENUM(upper,title,lower) DIFF_ ## upper,
59 #define TITLE(upper,title,lower) #title,
60 #define ENCODE(upper,title,lower) #lower
61 #define CONFIG(upper,title,lower) ":" #title
62 enum { DIFFLIST(ENUM) DIFFCOUNT };
63 static char const *const map_diffnames[] = { DIFFLIST(TITLE) };
64 static char const map_diffchars[] = DIFFLIST(ENCODE);
65 #define DIFFCONFIG DIFFLIST(CONFIG)
67 enum { TE, BE, LE, RE }; /* top/bottom/left/right edges */
72 COL_0, COL_1, COL_2, COL_3,
73 COL_ERROR, COL_ERRTEXT,
88 int *edgex, *edgey; /* position of a point on each edge */
89 int *regionx, *regiony; /* position of a point in each region */
95 int *colouring, *pencil;
96 int completed, cheated;
99 static game_params *default_params(void)
101 game_params *ret = snew(game_params);
103 #ifdef PORTRAIT_SCREEN
111 ret->diff = DIFF_NORMAL;
116 static const struct game_params map_presets[] = {
117 #ifdef PORTRAIT_SCREEN
118 {16, 18, 30, DIFF_EASY},
119 {16, 18, 30, DIFF_NORMAL},
120 {16, 18, 30, DIFF_HARD},
121 {16, 18, 30, DIFF_RECURSE},
122 {25, 30, 75, DIFF_NORMAL},
123 {25, 30, 75, DIFF_HARD},
125 {20, 15, 30, DIFF_EASY},
126 {20, 15, 30, DIFF_NORMAL},
127 {20, 15, 30, DIFF_HARD},
128 {20, 15, 30, DIFF_RECURSE},
129 {30, 25, 75, DIFF_NORMAL},
130 {30, 25, 75, DIFF_HARD},
134 static int game_fetch_preset(int i, char **name, game_params **params)
139 if (i < 0 || i >= lenof(map_presets))
142 ret = snew(game_params);
143 *ret = map_presets[i];
145 sprintf(str, "%dx%d, %d regions, %s", ret->w, ret->h, ret->n,
146 map_diffnames[ret->diff]);
153 static void free_params(game_params *params)
158 static game_params *dup_params(game_params *params)
160 game_params *ret = snew(game_params);
161 *ret = *params; /* structure copy */
165 static void decode_params(game_params *params, char const *string)
167 char const *p = string;
170 while (*p && isdigit((unsigned char)*p)) p++;
174 while (*p && isdigit((unsigned char)*p)) p++;
176 params->h = params->w;
181 while (*p && (*p == '.' || isdigit((unsigned char)*p))) p++;
183 params->n = params->w * params->h / 8;
188 for (i = 0; i < DIFFCOUNT; i++)
189 if (*p == map_diffchars[i])
195 static char *encode_params(game_params *params, int full)
199 sprintf(ret, "%dx%dn%d", params->w, params->h, params->n);
201 sprintf(ret + strlen(ret), "d%c", map_diffchars[params->diff]);
206 static config_item *game_configure(game_params *params)
211 ret = snewn(5, config_item);
213 ret[0].name = "Width";
214 ret[0].type = C_STRING;
215 sprintf(buf, "%d", params->w);
216 ret[0].sval = dupstr(buf);
219 ret[1].name = "Height";
220 ret[1].type = C_STRING;
221 sprintf(buf, "%d", params->h);
222 ret[1].sval = dupstr(buf);
225 ret[2].name = "Regions";
226 ret[2].type = C_STRING;
227 sprintf(buf, "%d", params->n);
228 ret[2].sval = dupstr(buf);
231 ret[3].name = "Difficulty";
232 ret[3].type = C_CHOICES;
233 ret[3].sval = DIFFCONFIG;
234 ret[3].ival = params->diff;
244 static game_params *custom_params(config_item *cfg)
246 game_params *ret = snew(game_params);
248 ret->w = atoi(cfg[0].sval);
249 ret->h = atoi(cfg[1].sval);
250 ret->n = atoi(cfg[2].sval);
251 ret->diff = cfg[3].ival;
256 static char *validate_params(game_params *params, int full)
258 if (params->w < 2 || params->h < 2)
259 return "Width and height must be at least two";
261 return "Must have at least five regions";
262 if (params->n > params->w * params->h)
263 return "Too many regions to fit in grid";
267 /* ----------------------------------------------------------------------
268 * Cumulative frequency table functions.
272 * Initialise a cumulative frequency table. (Hardly worth writing
273 * this function; all it does is to initialise everything in the
276 static void cf_init(int *table, int n)
280 for (i = 0; i < n; i++)
285 * Increment the count of symbol `sym' by `count'.
287 static void cf_add(int *table, int n, int sym, int count)
304 * Cumulative frequency lookup: return the total count of symbols
305 * with value less than `sym'.
307 static int cf_clookup(int *table, int n, int sym)
309 int bit, index, limit, count;
314 assert(0 < sym && sym <= n);
316 count = table[0]; /* start with the whole table size */
326 * Find the least number with its lowest set bit in this
327 * position which is greater than or equal to sym.
329 index = ((sym + bit - 1) &~ (bit * 2 - 1)) + bit;
332 count -= table[index];
343 * Single frequency lookup: return the count of symbol `sym'.
345 static int cf_slookup(int *table, int n, int sym)
349 assert(0 <= sym && sym < n);
353 for (bit = 1; sym+bit < n && !(sym & bit); bit <<= 1)
354 count -= table[sym+bit];
360 * Return the largest symbol index such that the cumulative
361 * frequency up to that symbol is less than _or equal to_ count.
363 static int cf_whichsym(int *table, int n, int count) {
366 assert(count >= 0 && count < table[0]);
377 if (count >= top - table[sym+bit])
380 top -= table[sym+bit];
389 /* ----------------------------------------------------------------------
392 * FIXME: this isn't entirely optimal at present, because it
393 * inherently prioritises growing the largest region since there
394 * are more squares adjacent to it. This acts as a destabilising
395 * influence leading to a few large regions and mostly small ones.
396 * It might be better to do it some other way.
399 #define WEIGHT_INCREASED 2 /* for increased perimeter */
400 #define WEIGHT_DECREASED 4 /* for decreased perimeter */
401 #define WEIGHT_UNCHANGED 3 /* for unchanged perimeter */
404 * Look at a square and decide which colours can be extended into
407 * If called with index < 0, it adds together one of
408 * WEIGHT_INCREASED, WEIGHT_DECREASED or WEIGHT_UNCHANGED for each
409 * colour that has a valid extension (according to the effect that
410 * it would have on the perimeter of the region being extended) and
411 * returns the overall total.
413 * If called with index >= 0, it returns one of the possible
414 * colours depending on the value of index, in such a way that the
415 * number of possible inputs which would give rise to a given
416 * return value correspond to the weight of that value.
418 static int extend_options(int w, int h, int n, int *map,
419 int x, int y, int index)
425 if (map[y*w+x] >= 0) {
427 return 0; /* can't do this square at all */
431 * Fetch the eight neighbours of this square, in order around
434 for (dy = -1; dy <= +1; dy++)
435 for (dx = -1; dx <= +1; dx++) {
436 int index = (dy < 0 ? 6-dx : dy > 0 ? 2+dx : 2*(1+dx));
437 if (x+dx >= 0 && x+dx < w && y+dy >= 0 && y+dy < h)
438 col[index] = map[(y+dy)*w+(x+dx)];
444 * Iterate over each colour that might be feasible.
446 * FIXME: this routine currently has O(n) running time. We
447 * could turn it into O(FOUR) by only bothering to iterate over
448 * the colours mentioned in the four neighbouring squares.
451 for (c = 0; c < n; c++) {
452 int count, neighbours, runs;
455 * One of the even indices of col (representing the
456 * orthogonal neighbours of this square) must be equal to
457 * c, or else this square is not adjacent to region c and
458 * obviously cannot become an extension of it at this time.
461 for (i = 0; i < 8; i += 2)
468 * Now we know this square is adjacent to region c. The
469 * next question is, would extending it cause the region to
470 * become non-simply-connected? If so, we mustn't do it.
472 * We determine this by looking around col to see if we can
473 * find more than one separate run of colour c.
476 for (i = 0; i < 8; i++)
477 if (col[i] == c && col[(i+1) & 7] != c)
485 * This square is a possibility. Determine its effect on
486 * the region's perimeter (computed from the number of
487 * orthogonal neighbours - 1 means a perimeter increase, 3
488 * a decrease, 2 no change; 4 is impossible because the
489 * region would already not be simply connected) and we're
492 assert(neighbours > 0 && neighbours < 4);
493 count = (neighbours == 1 ? WEIGHT_INCREASED :
494 neighbours == 2 ? WEIGHT_UNCHANGED : WEIGHT_DECREASED);
497 if (index >= 0 && index < count)
508 static void genmap(int w, int h, int n, int *map, random_state *rs)
515 tmp = snewn(wh, int);
518 * Clear the map, and set up `tmp' as a list of grid indices.
520 for (i = 0; i < wh; i++) {
526 * Place the region seeds by selecting n members from `tmp'.
529 for (i = 0; i < n; i++) {
530 int j = random_upto(rs, k);
536 * Re-initialise `tmp' as a cumulative frequency table. This
537 * will store the number of possible region colours we can
538 * extend into each square.
543 * Go through the grid and set up the initial cumulative
546 for (y = 0; y < h; y++)
547 for (x = 0; x < w; x++)
548 cf_add(tmp, wh, y*w+x,
549 extend_options(w, h, n, map, x, y, -1));
552 * Now repeatedly choose a square we can extend a region into,
556 int k = random_upto(rs, tmp[0]);
561 sq = cf_whichsym(tmp, wh, k);
562 k -= cf_clookup(tmp, wh, sq);
565 colour = extend_options(w, h, n, map, x, y, k);
570 * Re-scan the nine cells around the one we've just
573 for (yy = max(y-1, 0); yy < min(y+2, h); yy++)
574 for (xx = max(x-1, 0); xx < min(x+2, w); xx++) {
575 cf_add(tmp, wh, yy*w+xx,
576 -cf_slookup(tmp, wh, yy*w+xx) +
577 extend_options(w, h, n, map, xx, yy, -1));
582 * Finally, go through and normalise the region labels into
583 * order, meaning that indistinguishable maps are actually
586 for (i = 0; i < n; i++)
589 for (i = 0; i < wh; i++) {
593 map[i] = tmp[map[i]];
599 /* ----------------------------------------------------------------------
600 * Functions to handle graphs.
604 * Having got a map in a square grid, convert it into a graph
607 static int gengraph(int w, int h, int n, int *map, int *graph)
612 * Start by setting the graph up as an adjacency matrix. We'll
613 * turn it into a list later.
615 for (i = 0; i < n*n; i++)
619 * Iterate over the map looking for all adjacencies.
621 for (y = 0; y < h; y++)
622 for (x = 0; x < w; x++) {
625 if (x+1 < w && (vx = map[y*w+(x+1)]) != v)
626 graph[v*n+vx] = graph[vx*n+v] = 1;
627 if (y+1 < h && (vy = map[(y+1)*w+x]) != v)
628 graph[v*n+vy] = graph[vy*n+v] = 1;
632 * Turn the matrix into a list.
634 for (i = j = 0; i < n*n; i++)
641 static int graph_edge_index(int *graph, int n, int ngraph, int i, int j)
648 while (top - bot > 1) {
649 mid = (top + bot) / 2;
652 else if (graph[mid] < v)
660 #define graph_adjacent(graph, n, ngraph, i, j) \
661 (graph_edge_index((graph), (n), (ngraph), (i), (j)) >= 0)
663 static int graph_vertex_start(int *graph, int n, int ngraph, int i)
670 while (top - bot > 1) {
671 mid = (top + bot) / 2;
680 /* ----------------------------------------------------------------------
681 * Generate a four-colouring of a graph.
683 * FIXME: it would be nice if we could convert this recursion into
684 * pseudo-recursion using some sort of explicit stack array, for
685 * the sake of the Palm port and its limited stack.
688 static int fourcolour_recurse(int *graph, int n, int ngraph,
689 int *colouring, int *scratch, random_state *rs)
691 int nfree, nvert, start, i, j, k, c, ci;
695 * Find the smallest number of free colours in any uncoloured
696 * vertex, and count the number of such vertices.
699 nfree = FIVE; /* start off bigger than FOUR! */
701 for (i = 0; i < n; i++)
702 if (colouring[i] < 0 && scratch[i*FIVE+FOUR] <= nfree) {
703 if (nfree > scratch[i*FIVE+FOUR]) {
704 nfree = scratch[i*FIVE+FOUR];
711 * If there aren't any uncoloured vertices at all, we're done.
714 return TRUE; /* we've got a colouring! */
717 * Pick a random vertex in that set.
719 j = random_upto(rs, nvert);
720 for (i = 0; i < n; i++)
721 if (colouring[i] < 0 && scratch[i*FIVE+FOUR] == nfree)
725 start = graph_vertex_start(graph, n, ngraph, i);
728 * Loop over the possible colours for i, and recurse for each
732 for (c = 0; c < FOUR; c++)
733 if (scratch[i*FIVE+c] == 0)
735 shuffle(cs, ci, sizeof(*cs), rs);
741 * Fill in this colour.
746 * Update the scratch space to reflect a new neighbour
747 * of this colour for each neighbour of vertex i.
749 for (j = start; j < ngraph && graph[j] < n*(i+1); j++) {
751 if (scratch[k*FIVE+c] == 0)
752 scratch[k*FIVE+FOUR]--;
759 if (fourcolour_recurse(graph, n, ngraph, colouring, scratch, rs))
760 return TRUE; /* got one! */
763 * If that didn't work, clean up and try again with a
766 for (j = start; j < ngraph && graph[j] < n*(i+1); j++) {
769 if (scratch[k*FIVE+c] == 0)
770 scratch[k*FIVE+FOUR]++;
776 * If we reach here, we were unable to find a colouring at all.
777 * (This doesn't necessarily mean the Four Colour Theorem is
778 * violated; it might just mean we've gone down a dead end and
779 * need to back up and look somewhere else. It's only an FCT
780 * violation if we get all the way back up to the top level and
786 static void fourcolour(int *graph, int n, int ngraph, int *colouring,
793 * For each vertex and each colour, we store the number of
794 * neighbours that have that colour. Also, we store the number
795 * of free colours for the vertex.
797 scratch = snewn(n * FIVE, int);
798 for (i = 0; i < n * FIVE; i++)
799 scratch[i] = (i % FIVE == FOUR ? FOUR : 0);
802 * Clear the colouring to start with.
804 for (i = 0; i < n; i++)
807 i = fourcolour_recurse(graph, n, ngraph, colouring, scratch, rs);
808 assert(i); /* by the Four Colour Theorem :-) */
813 /* ----------------------------------------------------------------------
814 * Non-recursive solver.
817 struct solver_scratch {
818 unsigned char *possible; /* bitmap of colours for each region */
826 #ifdef SOLVER_DIAGNOSTICS
833 static struct solver_scratch *new_scratch(int *graph, int n, int ngraph)
835 struct solver_scratch *sc;
837 sc = snew(struct solver_scratch);
841 sc->possible = snewn(n, unsigned char);
843 sc->bfsqueue = snewn(n, int);
844 sc->bfscolour = snewn(n, int);
845 #ifdef SOLVER_DIAGNOSTICS
846 sc->bfsprev = snewn(n, int);
852 static void free_scratch(struct solver_scratch *sc)
856 sfree(sc->bfscolour);
857 #ifdef SOLVER_DIAGNOSTICS
864 * Count the bits in a word. Only needs to cope with FOUR bits.
866 static int bitcount(int word)
868 assert(FOUR <= 4); /* or this needs changing */
869 word = ((word & 0xA) >> 1) + (word & 0x5);
870 word = ((word & 0xC) >> 2) + (word & 0x3);
874 #ifdef SOLVER_DIAGNOSTICS
875 static const char colnames[FOUR] = { 'R', 'Y', 'G', 'B' };
878 static int place_colour(struct solver_scratch *sc,
879 int *colouring, int index, int colour
880 #ifdef SOLVER_DIAGNOSTICS
885 int *graph = sc->graph, n = sc->n, ngraph = sc->ngraph;
888 if (!(sc->possible[index] & (1 << colour))) {
889 #ifdef SOLVER_DIAGNOSTICS
891 printf("%*scannot place %c in region %d\n", 2*sc->depth, "",
892 colnames[colour], index);
894 return FALSE; /* can't do it */
897 sc->possible[index] = 1 << colour;
898 colouring[index] = colour;
900 #ifdef SOLVER_DIAGNOSTICS
902 printf("%*s%s %c in region %d\n", 2*sc->depth, "",
903 verb, colnames[colour], index);
907 * Rule out this colour from all the region's neighbours.
909 for (j = graph_vertex_start(graph, n, ngraph, index);
910 j < ngraph && graph[j] < n*(index+1); j++) {
911 k = graph[j] - index*n;
912 #ifdef SOLVER_DIAGNOSTICS
913 if (verbose && (sc->possible[k] & (1 << colour)))
914 printf("%*s ruling out %c in region %d\n", 2*sc->depth, "",
915 colnames[colour], k);
917 sc->possible[k] &= ~(1 << colour);
923 #ifdef SOLVER_DIAGNOSTICS
924 static char *colourset(char *buf, int set)
930 for (i = 0; i < FOUR; i++)
931 if (set & (1 << i)) {
932 p += sprintf(p, "%s%c", sep, colnames[i]);
941 * Returns 0 for impossible, 1 for success, 2 for failure to
942 * converge (i.e. puzzle is either ambiguous or just too
945 static int map_solver(struct solver_scratch *sc,
946 int *graph, int n, int ngraph, int *colouring,
951 if (sc->depth == 0) {
953 * Initialise scratch space.
955 for (i = 0; i < n; i++)
956 sc->possible[i] = (1 << FOUR) - 1;
961 for (i = 0; i < n; i++)
962 if (colouring[i] >= 0) {
963 if (!place_colour(sc, colouring, i, colouring[i]
964 #ifdef SOLVER_DIAGNOSTICS
968 #ifdef SOLVER_DIAGNOSTICS
970 printf("%*sinitial clue set is inconsistent\n",
973 return 0; /* the clues aren't even consistent! */
979 * Now repeatedly loop until we find nothing further to do.
982 int done_something = FALSE;
984 if (difficulty < DIFF_EASY)
985 break; /* can't do anything at all! */
988 * Simplest possible deduction: find a region with only one
991 for (i = 0; i < n; i++) if (colouring[i] < 0) {
992 int p = sc->possible[i];
995 #ifdef SOLVER_DIAGNOSTICS
997 printf("%*sregion %d has no possible colours left\n",
1000 return 0; /* puzzle is inconsistent */
1003 if ((p & (p-1)) == 0) { /* p is a power of two */
1005 for (c = 0; c < FOUR; c++)
1009 ret = place_colour(sc, colouring, i, c
1010 #ifdef SOLVER_DIAGNOSTICS
1015 * place_colour() can only fail if colour c was not
1016 * even a _possibility_ for region i, and we're
1017 * pretty sure it was because we checked before
1018 * calling place_colour(). So we can safely assert
1019 * here rather than having to return a nice
1020 * friendly error code.
1023 done_something = TRUE;
1030 if (difficulty < DIFF_NORMAL)
1031 break; /* can't do anything harder */
1034 * Failing that, go up one level. Look for pairs of regions
1035 * which (a) both have the same pair of possible colours,
1036 * (b) are adjacent to one another, (c) are adjacent to the
1037 * same region, and (d) that region still thinks it has one
1038 * or both of those possible colours.
1040 * Simplest way to do this is by going through the graph
1041 * edge by edge, so that we start with property (b) and
1042 * then look for (a) and finally (c) and (d).
1044 for (i = 0; i < ngraph; i++) {
1045 int j1 = graph[i] / n, j2 = graph[i] % n;
1047 #ifdef SOLVER_DIAGNOSTICS
1048 int started = FALSE;
1052 continue; /* done it already, other way round */
1054 if (colouring[j1] >= 0 || colouring[j2] >= 0)
1055 continue; /* they're not undecided */
1057 if (sc->possible[j1] != sc->possible[j2])
1058 continue; /* they don't have the same possibles */
1060 v = sc->possible[j1];
1062 * See if v contains exactly two set bits.
1064 v2 = v & -v; /* find lowest set bit */
1065 v2 = v & ~v2; /* clear it */
1066 if (v2 == 0 || (v2 & (v2-1)) != 0) /* not power of 2 */
1070 * We've found regions j1 and j2 satisfying properties
1071 * (a) and (b): they have two possible colours between
1072 * them, and since they're adjacent to one another they
1073 * must use _both_ those colours between them.
1074 * Therefore, if they are both adjacent to any other
1075 * region then that region cannot be either colour.
1077 * Go through the neighbours of j1 and see if any are
1080 for (j = graph_vertex_start(graph, n, ngraph, j1);
1081 j < ngraph && graph[j] < n*(j1+1); j++) {
1082 k = graph[j] - j1*n;
1083 if (graph_adjacent(graph, n, ngraph, k, j2) &&
1084 (sc->possible[k] & v)) {
1085 #ifdef SOLVER_DIAGNOSTICS
1089 printf("%*sadjacent regions %d,%d share colours"
1090 " %s\n", 2*sc->depth, "", j1, j2,
1093 printf("%*s ruling out %s in region %d\n",2*sc->depth,
1094 "", colourset(buf, sc->possible[k] & v), k);
1097 sc->possible[k] &= ~v;
1098 done_something = TRUE;
1106 if (difficulty < DIFF_HARD)
1107 break; /* can't do anything harder */
1110 * Right; now we get creative. Now we're going to look for
1111 * `forcing chains'. A forcing chain is a path through the
1112 * graph with the following properties:
1114 * (a) Each vertex on the path has precisely two possible
1117 * (b) Each pair of vertices which are adjacent on the
1118 * path share at least one possible colour in common.
1120 * (c) Each vertex in the middle of the path shares _both_
1121 * of its colours with at least one of its neighbours
1122 * (not the same one with both neighbours).
1124 * These together imply that at least one of the possible
1125 * colour choices at one end of the path forces _all_ the
1126 * rest of the colours along the path. In order to make
1127 * real use of this, we need further properties:
1129 * (c) Ruling out some colour C from the vertex at one end
1130 * of the path forces the vertex at the other end to
1133 * (d) The two end vertices are mutually adjacent to some
1136 * (e) That third vertex currently has C as a possibility.
1138 * If we can find all of that lot, we can deduce that at
1139 * least one of the two ends of the forcing chain has
1140 * colour C, and that therefore the mutually adjacent third
1143 * To find forcing chains, we're going to start a bfs at
1144 * each suitable vertex of the graph, once for each of its
1145 * two possible colours.
1147 for (i = 0; i < n; i++) {
1150 if (colouring[i] >= 0 || bitcount(sc->possible[i]) != 2)
1153 for (c = 0; c < FOUR; c++)
1154 if (sc->possible[i] & (1 << c)) {
1155 int j, k, gi, origc, currc, head, tail;
1157 * Try a bfs from this vertex, ruling out
1160 * Within this loop, we work in colour bitmaps
1161 * rather than actual colours, because
1162 * converting back and forth is a needless
1163 * computational expense.
1168 for (j = 0; j < n; j++) {
1169 sc->bfscolour[j] = -1;
1170 #ifdef SOLVER_DIAGNOSTICS
1171 sc->bfsprev[j] = -1;
1175 sc->bfsqueue[tail++] = i;
1176 sc->bfscolour[i] = sc->possible[i] &~ origc;
1178 while (head < tail) {
1179 j = sc->bfsqueue[head++];
1180 currc = sc->bfscolour[j];
1183 * Try neighbours of j.
1185 for (gi = graph_vertex_start(graph, n, ngraph, j);
1186 gi < ngraph && graph[gi] < n*(j+1); gi++) {
1187 k = graph[gi] - j*n;
1190 * To continue with the bfs in vertex
1191 * k, we need k to be
1192 * (a) not already visited
1193 * (b) have two possible colours
1194 * (c) those colours include currc.
1197 if (sc->bfscolour[k] < 0 &&
1199 bitcount(sc->possible[k]) == 2 &&
1200 (sc->possible[k] & currc)) {
1201 sc->bfsqueue[tail++] = k;
1203 sc->possible[k] &~ currc;
1204 #ifdef SOLVER_DIAGNOSTICS
1210 * One other possibility is that k
1211 * might be the region in which we can
1212 * make a real deduction: if it's
1213 * adjacent to i, contains currc as a
1214 * possibility, and currc is equal to
1215 * the original colour we ruled out.
1217 if (currc == origc &&
1218 graph_adjacent(graph, n, ngraph, k, i) &&
1219 (sc->possible[k] & currc)) {
1220 #ifdef SOLVER_DIAGNOSTICS
1222 char buf[80], *sep = "";
1225 printf("%*sforcing chain, colour %s, ",
1227 colourset(buf, origc));
1228 for (r = j; r != -1; r = sc->bfsprev[r]) {
1229 printf("%s%d", sep, r);
1232 printf("\n%*s ruling out %s in region"
1233 " %d\n", 2*sc->depth, "",
1234 colourset(buf, origc), k);
1237 sc->possible[k] &= ~origc;
1238 done_something = TRUE;
1247 if (!done_something)
1252 * See if we've got a complete solution, and return if so.
1254 for (i = 0; i < n; i++)
1255 if (colouring[i] < 0)
1258 #ifdef SOLVER_DIAGNOSTICS
1260 printf("%*sone solution found\n", 2*sc->depth, "");
1262 return 1; /* success! */
1266 * If recursion is not permissible, we now give up.
1268 if (difficulty < DIFF_RECURSE) {
1269 #ifdef SOLVER_DIAGNOSTICS
1271 printf("%*sunable to proceed further without recursion\n",
1274 return 2; /* unable to complete */
1278 * Now we've got to do something recursive. So first hunt for a
1279 * currently-most-constrained region.
1283 struct solver_scratch *rsc;
1284 int *subcolouring, *origcolouring;
1286 int we_already_got_one;
1291 for (i = 0; i < n; i++) if (colouring[i] < 0) {
1292 int p = sc->possible[i];
1293 enum { compile_time_assertion = 1 / (FOUR <= 4) };
1296 /* Count the set bits. */
1297 c = (p & 5) + ((p >> 1) & 5);
1298 c = (c & 3) + ((c >> 2) & 3);
1299 assert(c > 1); /* or colouring[i] would be >= 0 */
1307 assert(best >= 0); /* or we'd be solved already */
1309 #ifdef SOLVER_DIAGNOSTICS
1311 printf("%*srecursing on region %d\n", 2*sc->depth, "", best);
1315 * Now iterate over the possible colours for this region.
1317 rsc = new_scratch(graph, n, ngraph);
1318 rsc->depth = sc->depth + 1;
1319 origcolouring = snewn(n, int);
1320 memcpy(origcolouring, colouring, n * sizeof(int));
1321 subcolouring = snewn(n, int);
1322 we_already_got_one = FALSE;
1325 for (i = 0; i < FOUR; i++) {
1326 if (!(sc->possible[best] & (1 << i)))
1329 memcpy(rsc->possible, sc->possible, n);
1330 memcpy(subcolouring, origcolouring, n * sizeof(int));
1332 place_colour(rsc, subcolouring, best, i
1333 #ifdef SOLVER_DIAGNOSTICS
1338 subret = map_solver(rsc, graph, n, ngraph,
1339 subcolouring, difficulty);
1341 #ifdef SOLVER_DIAGNOSTICS
1343 printf("%*sretracting %c in region %d; found %s\n",
1344 2*sc->depth, "", colnames[i], best,
1345 subret == 0 ? "no solutions" :
1346 subret == 1 ? "one solution" : "multiple solutions");
1351 * If this possibility turned up more than one valid
1352 * solution, or if it turned up one and we already had
1353 * one, we're definitely ambiguous.
1355 if (subret == 2 || (subret == 1 && we_already_got_one)) {
1361 * If this possibility turned up one valid solution and
1362 * it's the first we've seen, copy it into the output.
1365 memcpy(colouring, subcolouring, n * sizeof(int));
1366 we_already_got_one = TRUE;
1371 * Otherwise, this guess led to a contradiction, so we
1376 sfree(subcolouring);
1379 #ifdef SOLVER_DIAGNOSTICS
1380 if (verbose && sc->depth == 0) {
1381 printf("%*s%s found\n",
1383 ret == 0 ? "no solutions" :
1384 ret == 1 ? "one solution" : "multiple solutions");
1391 /* ----------------------------------------------------------------------
1392 * Game generation main function.
1395 static char *new_game_desc(game_params *params, random_state *rs,
1396 char **aux, int interactive)
1398 struct solver_scratch *sc = NULL;
1399 int *map, *graph, ngraph, *colouring, *colouring2, *regions;
1400 int i, j, w, h, n, solveret, cfreq[FOUR];
1403 #ifdef GENERATION_DIAGNOSTICS
1407 int retlen, retsize;
1416 map = snewn(wh, int);
1417 graph = snewn(n*n, int);
1418 colouring = snewn(n, int);
1419 colouring2 = snewn(n, int);
1420 regions = snewn(n, int);
1423 * This is the minimum difficulty below which we'll completely
1424 * reject a map design. Normally we set this to one below the
1425 * requested difficulty, ensuring that we have the right
1426 * result. However, for particularly dense maps or maps with
1427 * particularly few regions it might not be possible to get the
1428 * desired difficulty, so we will eventually drop this down to
1429 * -1 to indicate that any old map will do.
1431 mindiff = params->diff;
1439 genmap(w, h, n, map, rs);
1441 #ifdef GENERATION_DIAGNOSTICS
1442 for (y = 0; y < h; y++) {
1443 for (x = 0; x < w; x++) {
1448 putchar('a' + v-36);
1450 putchar('A' + v-10);
1459 * Convert the map into a graph.
1461 ngraph = gengraph(w, h, n, map, graph);
1463 #ifdef GENERATION_DIAGNOSTICS
1464 for (i = 0; i < ngraph; i++)
1465 printf("%d-%d\n", graph[i]/n, graph[i]%n);
1471 fourcolour(graph, n, ngraph, colouring, rs);
1473 #ifdef GENERATION_DIAGNOSTICS
1474 for (i = 0; i < n; i++)
1475 printf("%d: %d\n", i, colouring[i]);
1477 for (y = 0; y < h; y++) {
1478 for (x = 0; x < w; x++) {
1479 int v = colouring[map[y*w+x]];
1481 putchar('a' + v-36);
1483 putchar('A' + v-10);
1492 * Encode the solution as an aux string.
1494 if (*aux) /* in case we've come round again */
1496 retlen = retsize = 0;
1498 for (i = 0; i < n; i++) {
1501 if (colouring[i] < 0)
1504 len = sprintf(buf, "%s%d:%d", i ? ";" : "S;", colouring[i], i);
1505 if (retlen + len >= retsize) {
1506 retsize = retlen + len + 256;
1507 ret = sresize(ret, retsize, char);
1509 strcpy(ret + retlen, buf);
1515 * Remove the region colours one by one, keeping
1516 * solubility. Also ensure that there always remains at
1517 * least one region of every colour, so that the user can
1518 * drag from somewhere.
1520 for (i = 0; i < FOUR; i++)
1522 for (i = 0; i < n; i++) {
1524 cfreq[colouring[i]]++;
1526 for (i = 0; i < FOUR; i++)
1530 shuffle(regions, n, sizeof(*regions), rs);
1532 if (sc) free_scratch(sc);
1533 sc = new_scratch(graph, n, ngraph);
1535 for (i = 0; i < n; i++) {
1538 if (cfreq[colouring[j]] == 1)
1539 continue; /* can't remove last region of colour */
1541 memcpy(colouring2, colouring, n*sizeof(int));
1543 solveret = map_solver(sc, graph, n, ngraph, colouring2,
1545 assert(solveret >= 0); /* mustn't be impossible! */
1546 if (solveret == 1) {
1547 cfreq[colouring[j]]--;
1552 #ifdef GENERATION_DIAGNOSTICS
1553 for (i = 0; i < n; i++)
1554 if (colouring[i] >= 0) {
1558 putchar('a' + i-36);
1560 putchar('A' + i-10);
1563 printf(": %d\n", colouring[i]);
1568 * Finally, check that the puzzle is _at least_ as hard as
1569 * required, and indeed that it isn't already solved.
1570 * (Calling map_solver with negative difficulty ensures the
1571 * latter - if a solver which _does nothing_ can solve it,
1574 memcpy(colouring2, colouring, n*sizeof(int));
1575 if (map_solver(sc, graph, n, ngraph, colouring2,
1576 mindiff - 1) == 1) {
1578 * Drop minimum difficulty if necessary.
1580 if (mindiff > 0 && (n < 9 || n > 2*wh/3)) {
1582 mindiff = 0; /* give up and go for Easy */
1591 * Encode as a game ID. We do this by:
1593 * - first going along the horizontal edges row by row, and
1594 * then the vertical edges column by column
1595 * - encoding the lengths of runs of edges and runs of
1597 * - the decoder will reconstitute the region boundaries from
1598 * this and automatically number them the same way we did
1599 * - then we encode the initial region colours in a Slant-like
1600 * fashion (digits 0-3 interspersed with letters giving
1601 * lengths of runs of empty spaces).
1603 retlen = retsize = 0;
1610 * Start with a notional non-edge, so that there'll be an
1611 * explicit `a' to distinguish the case where we start with
1617 for (i = 0; i < w*(h-1) + (w-1)*h; i++) {
1618 int x, y, dx, dy, v;
1621 /* Horizontal edge. */
1627 /* Vertical edge. */
1628 x = (i - w*(h-1)) / h;
1629 y = (i - w*(h-1)) % h;
1634 if (retlen + 10 >= retsize) {
1635 retsize = retlen + 256;
1636 ret = sresize(ret, retsize, char);
1639 v = (map[y*w+x] != map[(y+dy)*w+(x+dx)]);
1642 ret[retlen++] = 'a'-1 + run;
1647 * 'z' is a special case in this encoding. Rather
1648 * than meaning a run of 26 and a state switch, it
1649 * means a run of 25 and _no_ state switch, because
1650 * otherwise there'd be no way to encode runs of
1654 ret[retlen++] = 'z';
1661 ret[retlen++] = 'a'-1 + run;
1662 ret[retlen++] = ',';
1665 for (i = 0; i < n; i++) {
1666 if (retlen + 10 >= retsize) {
1667 retsize = retlen + 256;
1668 ret = sresize(ret, retsize, char);
1671 if (colouring[i] < 0) {
1673 * In _this_ encoding, 'z' is a run of 26, since
1674 * there's no implicit state switch after each run.
1675 * Confusingly different, but more compact.
1678 ret[retlen++] = 'z';
1684 ret[retlen++] = 'a'-1 + run;
1685 ret[retlen++] = '0' + colouring[i];
1690 ret[retlen++] = 'a'-1 + run;
1693 assert(retlen < retsize);
1706 static char *parse_edge_list(game_params *params, char **desc, int *map)
1708 int w = params->w, h = params->h, wh = w*h, n = params->n;
1709 int i, k, pos, state;
1712 dsf_init(map+wh, wh);
1718 * Parse the game description to get the list of edges, and
1719 * build up a disjoint set forest as we go (by identifying
1720 * pairs of squares whenever the edge list shows a non-edge).
1722 while (*p && *p != ',') {
1723 if (*p < 'a' || *p > 'z')
1724 return "Unexpected character in edge list";
1735 } else if (pos < w*(h-1)) {
1736 /* Horizontal edge. */
1741 } else if (pos < 2*wh-w-h) {
1742 /* Vertical edge. */
1743 x = (pos - w*(h-1)) / h;
1744 y = (pos - w*(h-1)) % h;
1748 return "Too much data in edge list";
1750 dsf_merge(map+wh, y*w+x, (y+dy)*w+(x+dx));
1758 assert(pos <= 2*wh-w-h);
1760 return "Too little data in edge list";
1763 * Now go through again and allocate region numbers.
1766 for (i = 0; i < wh; i++)
1768 for (i = 0; i < wh; i++) {
1769 k = dsf_canonify(map+wh, i);
1775 return "Edge list defines the wrong number of regions";
1782 static char *validate_desc(game_params *params, char *desc)
1784 int w = params->w, h = params->h, wh = w*h, n = params->n;
1789 map = snewn(2*wh, int);
1790 ret = parse_edge_list(params, &desc, map);
1796 return "Expected comma before clue list";
1797 desc++; /* eat comma */
1801 if (*desc >= '0' && *desc < '0'+FOUR)
1803 else if (*desc >= 'a' && *desc <= 'z')
1804 area += *desc - 'a' + 1;
1806 return "Unexpected character in clue list";
1810 return "Too little data in clue list";
1812 return "Too much data in clue list";
1817 static game_state *new_game(midend *me, game_params *params, char *desc)
1819 int w = params->w, h = params->h, wh = w*h, n = params->n;
1822 game_state *state = snew(game_state);
1825 state->colouring = snewn(n, int);
1826 for (i = 0; i < n; i++)
1827 state->colouring[i] = -1;
1828 state->pencil = snewn(n, int);
1829 for (i = 0; i < n; i++)
1830 state->pencil[i] = 0;
1832 state->completed = state->cheated = FALSE;
1834 state->map = snew(struct map);
1835 state->map->refcount = 1;
1836 state->map->map = snewn(wh*4, int);
1837 state->map->graph = snewn(n*n, int);
1839 state->map->immutable = snewn(n, int);
1840 for (i = 0; i < n; i++)
1841 state->map->immutable[i] = FALSE;
1847 ret = parse_edge_list(params, &p, state->map->map);
1852 * Set up the other three quadrants in `map'.
1854 for (i = wh; i < 4*wh; i++)
1855 state->map->map[i] = state->map->map[i % wh];
1861 * Now process the clue list.
1865 if (*p >= '0' && *p < '0'+FOUR) {
1866 state->colouring[pos] = *p - '0';
1867 state->map->immutable[pos] = TRUE;
1870 assert(*p >= 'a' && *p <= 'z');
1871 pos += *p - 'a' + 1;
1877 state->map->ngraph = gengraph(w, h, n, state->map->map, state->map->graph);
1880 * Attempt to smooth out some of the more jagged region
1881 * outlines by the judicious use of diagonally divided squares.
1884 random_state *rs = random_new(desc, strlen(desc));
1885 int *squares = snewn(wh, int);
1888 for (i = 0; i < wh; i++)
1890 shuffle(squares, wh, sizeof(*squares), rs);
1893 done_something = FALSE;
1894 for (i = 0; i < wh; i++) {
1895 int y = squares[i] / w, x = squares[i] % w;
1896 int c = state->map->map[y*w+x];
1899 if (x == 0 || x == w-1 || y == 0 || y == h-1)
1902 if (state->map->map[TE * wh + y*w+x] !=
1903 state->map->map[BE * wh + y*w+x])
1906 tc = state->map->map[BE * wh + (y-1)*w+x];
1907 bc = state->map->map[TE * wh + (y+1)*w+x];
1908 lc = state->map->map[RE * wh + y*w+(x-1)];
1909 rc = state->map->map[LE * wh + y*w+(x+1)];
1912 * If this square is adjacent on two sides to one
1913 * region and on the other two sides to the other
1914 * region, and is itself one of the two regions, we can
1915 * adjust it so that it's a diagonal.
1917 if (tc != bc && (tc == c || bc == c)) {
1918 if ((lc == tc && rc == bc) ||
1919 (lc == bc && rc == tc)) {
1920 state->map->map[TE * wh + y*w+x] = tc;
1921 state->map->map[BE * wh + y*w+x] = bc;
1922 state->map->map[LE * wh + y*w+x] = lc;
1923 state->map->map[RE * wh + y*w+x] = rc;
1924 done_something = TRUE;
1928 } while (done_something);
1934 * Analyse the map to find a canonical line segment
1935 * corresponding to each edge, and a canonical point
1936 * corresponding to each region. The former are where we'll
1937 * eventually put error markers; the latter are where we'll put
1938 * per-region flags such as numbers (when in diagnostic mode).
1941 int *bestx, *besty, *an, pass;
1942 float *ax, *ay, *best;
1944 ax = snewn(state->map->ngraph + n, float);
1945 ay = snewn(state->map->ngraph + n, float);
1946 an = snewn(state->map->ngraph + n, int);
1947 bestx = snewn(state->map->ngraph + n, int);
1948 besty = snewn(state->map->ngraph + n, int);
1949 best = snewn(state->map->ngraph + n, float);
1951 for (i = 0; i < state->map->ngraph + n; i++) {
1952 bestx[i] = besty[i] = -1;
1953 best[i] = 2*(w+h)+1;
1954 ax[i] = ay[i] = 0.0F;
1959 * We make two passes over the map, finding all the line
1960 * segments separating regions and all the suitable points
1961 * within regions. In the first pass, we compute the
1962 * _average_ x and y coordinate of all the points in a
1963 * given class; in the second pass, for each such average
1964 * point, we find the candidate closest to it and call that
1967 * Line segments are considered to have coordinates in
1968 * their centre. Thus, at least one coordinate for any line
1969 * segment is always something-and-a-half; so we store our
1970 * coordinates as twice their normal value.
1972 for (pass = 0; pass < 2; pass++) {
1975 for (y = 0; y < h; y++)
1976 for (x = 0; x < w; x++) {
1977 int ex[4], ey[4], ea[4], eb[4], en = 0;
1980 * Look for an edge to the right of this
1981 * square, an edge below it, and an edge in the
1982 * middle of it. Also look to see if the point
1983 * at the bottom right of this square is on an
1984 * edge (and isn't a place where more than two
1989 ea[en] = state->map->map[RE * wh + y*w+x];
1990 eb[en] = state->map->map[LE * wh + y*w+(x+1)];
1997 ea[en] = state->map->map[BE * wh + y*w+x];
1998 eb[en] = state->map->map[TE * wh + (y+1)*w+x];
2004 ea[en] = state->map->map[TE * wh + y*w+x];
2005 eb[en] = state->map->map[BE * wh + y*w+x];
2010 if (x+1 < w && y+1 < h) {
2011 /* bottom right corner */
2012 int oct[8], othercol, nchanges;
2013 oct[0] = state->map->map[RE * wh + y*w+x];
2014 oct[1] = state->map->map[LE * wh + y*w+(x+1)];
2015 oct[2] = state->map->map[BE * wh + y*w+(x+1)];
2016 oct[3] = state->map->map[TE * wh + (y+1)*w+(x+1)];
2017 oct[4] = state->map->map[LE * wh + (y+1)*w+(x+1)];
2018 oct[5] = state->map->map[RE * wh + (y+1)*w+x];
2019 oct[6] = state->map->map[TE * wh + (y+1)*w+x];
2020 oct[7] = state->map->map[BE * wh + y*w+x];
2024 for (i = 0; i < 8; i++) {
2025 if (oct[i] != oct[0]) {
2028 else if (othercol != oct[i])
2029 break; /* three colours at this point */
2031 if (oct[i] != oct[(i+1) & 7])
2036 * Now if there are exactly two regions at
2037 * this point (not one, and not three or
2038 * more), and only two changes around the
2039 * loop, then this is a valid place to put
2042 if (i == 8 && othercol >= 0 && nchanges == 2) {
2051 * If there's exactly _one_ region at this
2052 * point, on the other hand, it's a valid
2053 * place to put a region centre.
2056 ea[en] = eb[en] = oct[0];
2064 * Now process the points we've found, one by
2067 for (i = 0; i < en; i++) {
2068 int emin = min(ea[i], eb[i]);
2069 int emax = max(ea[i], eb[i]);
2075 graph_edge_index(state->map->graph, n,
2076 state->map->ngraph, emin,
2080 gindex = state->map->ngraph + emin;
2083 assert(gindex >= 0);
2087 * In pass 0, accumulate the values
2088 * we'll use to compute the average
2091 ax[gindex] += ex[i];
2092 ay[gindex] += ey[i];
2096 * In pass 1, work out whether this
2097 * point is closer to the average than
2098 * the last one we've seen.
2102 assert(an[gindex] > 0);
2103 dx = ex[i] - ax[gindex];
2104 dy = ey[i] - ay[gindex];
2105 d = sqrt(dx*dx + dy*dy);
2106 if (d < best[gindex]) {
2108 bestx[gindex] = ex[i];
2109 besty[gindex] = ey[i];
2116 for (i = 0; i < state->map->ngraph + n; i++)
2124 state->map->edgex = snewn(state->map->ngraph, int);
2125 state->map->edgey = snewn(state->map->ngraph, int);
2126 memcpy(state->map->edgex, bestx, state->map->ngraph * sizeof(int));
2127 memcpy(state->map->edgey, besty, state->map->ngraph * sizeof(int));
2129 state->map->regionx = snewn(n, int);
2130 state->map->regiony = snewn(n, int);
2131 memcpy(state->map->regionx, bestx + state->map->ngraph, n*sizeof(int));
2132 memcpy(state->map->regiony, besty + state->map->ngraph, n*sizeof(int));
2134 for (i = 0; i < state->map->ngraph; i++)
2135 if (state->map->edgex[i] < 0) {
2136 /* Find the other representation of this edge. */
2137 int e = state->map->graph[i];
2138 int iprime = graph_edge_index(state->map->graph, n,
2139 state->map->ngraph, e%n, e/n);
2140 assert(state->map->edgex[iprime] >= 0);
2141 state->map->edgex[i] = state->map->edgex[iprime];
2142 state->map->edgey[i] = state->map->edgey[iprime];
2156 static game_state *dup_game(game_state *state)
2158 game_state *ret = snew(game_state);
2161 ret->colouring = snewn(state->p.n, int);
2162 memcpy(ret->colouring, state->colouring, state->p.n * sizeof(int));
2163 ret->pencil = snewn(state->p.n, int);
2164 memcpy(ret->pencil, state->pencil, state->p.n * sizeof(int));
2165 ret->map = state->map;
2166 ret->map->refcount++;
2167 ret->completed = state->completed;
2168 ret->cheated = state->cheated;
2173 static void free_game(game_state *state)
2175 if (--state->map->refcount <= 0) {
2176 sfree(state->map->map);
2177 sfree(state->map->graph);
2178 sfree(state->map->immutable);
2179 sfree(state->map->edgex);
2180 sfree(state->map->edgey);
2181 sfree(state->map->regionx);
2182 sfree(state->map->regiony);
2185 sfree(state->pencil);
2186 sfree(state->colouring);
2190 static char *solve_game(game_state *state, game_state *currstate,
2191 char *aux, char **error)
2198 struct solver_scratch *sc;
2202 int retlen, retsize;
2204 colouring = snewn(state->map->n, int);
2205 memcpy(colouring, state->colouring, state->map->n * sizeof(int));
2207 sc = new_scratch(state->map->graph, state->map->n, state->map->ngraph);
2208 sret = map_solver(sc, state->map->graph, state->map->n,
2209 state->map->ngraph, colouring, DIFFCOUNT-1);
2215 *error = "Puzzle is inconsistent";
2217 *error = "Unable to find a unique solution for this puzzle";
2222 ret = snewn(retsize, char);
2226 for (i = 0; i < state->map->n; i++) {
2229 assert(colouring[i] >= 0);
2230 if (colouring[i] == currstate->colouring[i])
2232 assert(!state->map->immutable[i]);
2234 len = sprintf(buf, ";%d:%d", colouring[i], i);
2235 if (retlen + len >= retsize) {
2236 retsize = retlen + len + 256;
2237 ret = sresize(ret, retsize, char);
2239 strcpy(ret + retlen, buf);
2250 static int game_can_format_as_text_now(game_params *params)
2255 static char *game_text_format(game_state *state)
2264 * - -2 means no drag currently active.
2265 * - >=0 means we're dragging a solid colour.
2266 * - -1 means we're dragging a blank space, and drag_pencil
2267 * might or might not add some pencil-mark stipples to that.
2275 static game_ui *new_ui(game_state *state)
2277 game_ui *ui = snew(game_ui);
2278 ui->dragx = ui->dragy = -1;
2279 ui->drag_colour = -2;
2280 ui->show_numbers = FALSE;
2284 static void free_ui(game_ui *ui)
2289 static char *encode_ui(game_ui *ui)
2294 static void decode_ui(game_ui *ui, char *encoding)
2298 static void game_changed_state(game_ui *ui, game_state *oldstate,
2299 game_state *newstate)
2303 struct game_drawstate {
2305 unsigned long *drawn, *todraw;
2307 int dragx, dragy, drag_visible;
2311 /* Flags in `drawn'. */
2312 #define ERR_BASE 0x00800000L
2313 #define ERR_MASK 0xFF800000L
2314 #define PENCIL_T_BASE 0x00080000L
2315 #define PENCIL_T_MASK 0x00780000L
2316 #define PENCIL_B_BASE 0x00008000L
2317 #define PENCIL_B_MASK 0x00078000L
2318 #define PENCIL_MASK 0x007F8000L
2319 #define SHOW_NUMBERS 0x00004000L
2321 #define TILESIZE (ds->tilesize)
2322 #define BORDER (TILESIZE)
2323 #define COORD(x) ( (x) * TILESIZE + BORDER )
2324 #define FROMCOORD(x) ( ((x) - BORDER + TILESIZE) / TILESIZE - 1 )
2326 static int region_from_coords(game_state *state, game_drawstate *ds,
2329 int w = state->p.w, h = state->p.h, wh = w*h /*, n = state->p.n */;
2330 int tx = FROMCOORD(x), ty = FROMCOORD(y);
2331 int dx = x - COORD(tx), dy = y - COORD(ty);
2334 if (tx < 0 || tx >= w || ty < 0 || ty >= h)
2335 return -1; /* border */
2337 quadrant = 2 * (dx > dy) + (TILESIZE - dx > dy);
2338 quadrant = (quadrant == 0 ? BE :
2339 quadrant == 1 ? LE :
2340 quadrant == 2 ? RE : TE);
2342 return state->map->map[quadrant * wh + ty*w+tx];
2345 static char *interpret_move(game_state *state, game_ui *ui, game_drawstate *ds,
2346 int x, int y, int button)
2348 char *bufp, buf[256];
2351 * Enable or disable numeric labels on regions.
2353 if (button == 'l' || button == 'L') {
2354 ui->show_numbers = !ui->show_numbers;
2358 if (button == LEFT_BUTTON || button == RIGHT_BUTTON) {
2359 int r = region_from_coords(state, ds, x, y);
2362 ui->drag_colour = state->colouring[r];
2363 ui->drag_pencil = state->pencil[r];
2364 if (ui->drag_colour >= 0)
2365 ui->drag_pencil = 0; /* should be already, but double-check */
2367 ui->drag_colour = -1;
2368 ui->drag_pencil = 0;
2375 if ((button == LEFT_DRAG || button == RIGHT_DRAG) &&
2376 ui->drag_colour > -2) {
2382 if ((button == LEFT_RELEASE || button == RIGHT_RELEASE) &&
2383 ui->drag_colour > -2) {
2384 int r = region_from_coords(state, ds, x, y);
2385 int c = ui->drag_colour;
2386 int p = ui->drag_pencil;
2390 * Cancel the drag, whatever happens.
2392 ui->drag_colour = -2;
2393 ui->dragx = ui->dragy = -1;
2396 return ""; /* drag into border; do nothing else */
2398 if (state->map->immutable[r])
2399 return ""; /* can't change this region */
2401 if (state->colouring[r] == c && state->pencil[r] == p)
2402 return ""; /* don't _need_ to change this region */
2404 if (button == RIGHT_RELEASE) {
2405 if (state->colouring[r] >= 0) {
2406 /* Can't pencil on a coloured region */
2408 } else if (c >= 0) {
2409 /* Right-dragging from colour to blank toggles one pencil */
2410 p = state->pencil[r] ^ (1 << c);
2413 /* Otherwise, right-dragging from blank to blank is equivalent
2414 * to left-dragging. */
2418 oldp = state->pencil[r];
2419 if (c != state->colouring[r]) {
2420 bufp += sprintf(bufp, ";%c:%d", (int)(c < 0 ? 'C' : '0' + c), r);
2426 for (i = 0; i < FOUR; i++)
2427 if ((oldp ^ p) & (1 << i))
2428 bufp += sprintf(bufp, ";p%c:%d", (int)('0' + i), r);
2431 return dupstr(buf+1); /* ignore first semicolon */
2437 static game_state *execute_move(game_state *state, char *move)
2440 game_state *ret = dup_game(state);
2451 if ((c == 'C' || (c >= '0' && c < '0'+FOUR)) &&
2452 sscanf(move+1, ":%d%n", &k, &adv) == 1 &&
2453 k >= 0 && k < state->p.n) {
2456 if (ret->colouring[k] >= 0) {
2463 ret->pencil[k] ^= 1 << (c - '0');
2465 ret->colouring[k] = (c == 'C' ? -1 : c - '0');
2468 } else if (*move == 'S') {
2470 ret->cheated = TRUE;
2476 if (*move && *move != ';') {
2485 * Check for completion.
2487 if (!ret->completed) {
2490 for (i = 0; i < n; i++)
2491 if (ret->colouring[i] < 0) {
2497 for (i = 0; i < ret->map->ngraph; i++) {
2498 int j = ret->map->graph[i] / n;
2499 int k = ret->map->graph[i] % n;
2500 if (ret->colouring[j] == ret->colouring[k]) {
2508 ret->completed = TRUE;
2514 /* ----------------------------------------------------------------------
2518 static void game_compute_size(game_params *params, int tilesize,
2521 /* Ick: fake up `ds->tilesize' for macro expansion purposes */
2522 struct { int tilesize; } ads, *ds = &ads;
2523 ads.tilesize = tilesize;
2525 *x = params->w * TILESIZE + 2 * BORDER + 1;
2526 *y = params->h * TILESIZE + 2 * BORDER + 1;
2529 static void game_set_size(drawing *dr, game_drawstate *ds,
2530 game_params *params, int tilesize)
2532 ds->tilesize = tilesize;
2534 assert(!ds->bl); /* set_size is never called twice */
2535 ds->bl = blitter_new(dr, TILESIZE+3, TILESIZE+3);
2538 const float map_colours[FOUR][3] = {
2539 #ifdef VIVID_COLOURS
2540 /* Use more vivid colours (e.g. on the Pocket PC) */
2541 {0.75F, 0.25F, 0.25F},
2544 {0.85F, 0.85F, 0.1F},
2549 {0.55F, 0.45F, 0.35F},
2552 const int map_hatching[FOUR] = {
2553 HATCH_VERT, HATCH_SLASH, HATCH_HORIZ, HATCH_BACKSLASH
2556 static float *game_colours(frontend *fe, int *ncolours)
2558 float *ret = snewn(3 * NCOLOURS, float);
2560 frontend_default_colour(fe, &ret[COL_BACKGROUND * 3]);
2562 ret[COL_GRID * 3 + 0] = 0.0F;
2563 ret[COL_GRID * 3 + 1] = 0.0F;
2564 ret[COL_GRID * 3 + 2] = 0.0F;
2566 memcpy(ret + COL_0 * 3, map_colours[0], 3 * sizeof(float));
2567 memcpy(ret + COL_1 * 3, map_colours[1], 3 * sizeof(float));
2568 memcpy(ret + COL_2 * 3, map_colours[2], 3 * sizeof(float));
2569 memcpy(ret + COL_3 * 3, map_colours[3], 3 * sizeof(float));
2571 ret[COL_ERROR * 3 + 0] = 1.0F;
2572 ret[COL_ERROR * 3 + 1] = 0.0F;
2573 ret[COL_ERROR * 3 + 2] = 0.0F;
2575 ret[COL_ERRTEXT * 3 + 0] = 1.0F;
2576 ret[COL_ERRTEXT * 3 + 1] = 1.0F;
2577 ret[COL_ERRTEXT * 3 + 2] = 1.0F;
2579 *ncolours = NCOLOURS;
2583 static game_drawstate *game_new_drawstate(drawing *dr, game_state *state)
2585 struct game_drawstate *ds = snew(struct game_drawstate);
2589 ds->drawn = snewn(state->p.w * state->p.h, unsigned long);
2590 for (i = 0; i < state->p.w * state->p.h; i++)
2591 ds->drawn[i] = 0xFFFFL;
2592 ds->todraw = snewn(state->p.w * state->p.h, unsigned long);
2593 ds->started = FALSE;
2595 ds->drag_visible = FALSE;
2596 ds->dragx = ds->dragy = -1;
2601 static void game_free_drawstate(drawing *dr, game_drawstate *ds)
2606 blitter_free(dr, ds->bl);
2610 static void draw_error(drawing *dr, game_drawstate *ds, int x, int y)
2618 coords[0] = x - TILESIZE*2/5;
2621 coords[3] = y - TILESIZE*2/5;
2622 coords[4] = x + TILESIZE*2/5;
2625 coords[7] = y + TILESIZE*2/5;
2626 draw_polygon(dr, coords, 4, COL_ERROR, COL_GRID);
2629 * Draw an exclamation mark in the diamond. This turns out to
2630 * look unpleasantly off-centre if done via draw_text, so I do
2631 * it by hand on the basis that exclamation marks aren't that
2632 * difficult to draw...
2635 yext = TILESIZE*2/5 - (xext*2+2);
2636 draw_rect(dr, x-xext, y-yext, xext*2+1, yext*2+1 - (xext*3),
2638 draw_rect(dr, x-xext, y+yext-xext*2+1, xext*2+1, xext*2, COL_ERRTEXT);
2641 static void draw_square(drawing *dr, game_drawstate *ds,
2642 game_params *params, struct map *map,
2643 int x, int y, unsigned long v)
2645 int w = params->w, h = params->h, wh = w*h;
2646 int tv, bv, xo, yo, i, j, oldj;
2647 unsigned long errs, pencil, show_numbers;
2649 errs = v & ERR_MASK;
2651 pencil = v & PENCIL_MASK;
2653 show_numbers = v & SHOW_NUMBERS;
2658 clip(dr, COORD(x), COORD(y), TILESIZE, TILESIZE);
2661 * Draw the region colour.
2663 draw_rect(dr, COORD(x), COORD(y), TILESIZE, TILESIZE,
2664 (tv == FOUR ? COL_BACKGROUND : COL_0 + tv));
2666 * Draw the second region colour, if this is a diagonally
2669 if (map->map[TE * wh + y*w+x] != map->map[BE * wh + y*w+x]) {
2671 coords[0] = COORD(x)-1;
2672 coords[1] = COORD(y+1)+1;
2673 if (map->map[LE * wh + y*w+x] == map->map[TE * wh + y*w+x])
2674 coords[2] = COORD(x+1)+1;
2676 coords[2] = COORD(x)-1;
2677 coords[3] = COORD(y)-1;
2678 coords[4] = COORD(x+1)+1;
2679 coords[5] = COORD(y+1)+1;
2680 draw_polygon(dr, coords, 3,
2681 (bv == FOUR ? COL_BACKGROUND : COL_0 + bv), COL_GRID);
2685 * Draw `pencil marks'. Currently we arrange these in a square
2686 * formation, which means we may be in trouble if the value of
2687 * FOUR changes later...
2690 for (yo = 0; yo < 4; yo++)
2691 for (xo = 0; xo < 4; xo++) {
2692 int te = map->map[TE * wh + y*w+x];
2695 e = (yo < xo && yo < 3-xo ? TE :
2696 yo > xo && yo > 3-xo ? BE :
2698 ee = map->map[e * wh + y*w+x];
2700 if (xo != (yo * 2 + 1) % 5)
2704 if (!(pencil & ((ee == te ? PENCIL_T_BASE : PENCIL_B_BASE) << c)))
2708 (map->map[TE * wh + y*w+x] != map->map[LE * wh + y*w+x]))
2709 continue; /* avoid TL-BR diagonal line */
2711 (map->map[TE * wh + y*w+x] != map->map[RE * wh + y*w+x]))
2712 continue; /* avoid BL-TR diagonal line */
2714 draw_circle(dr, COORD(x) + (xo+1)*TILESIZE/5,
2715 COORD(y) + (yo+1)*TILESIZE/5,
2716 TILESIZE/7, COL_0 + c, COL_0 + c);
2720 * Draw the grid lines, if required.
2722 if (x <= 0 || map->map[RE*wh+y*w+(x-1)] != map->map[LE*wh+y*w+x])
2723 draw_rect(dr, COORD(x), COORD(y), 1, TILESIZE, COL_GRID);
2724 if (y <= 0 || map->map[BE*wh+(y-1)*w+x] != map->map[TE*wh+y*w+x])
2725 draw_rect(dr, COORD(x), COORD(y), TILESIZE, 1, COL_GRID);
2726 if (x <= 0 || y <= 0 ||
2727 map->map[RE*wh+(y-1)*w+(x-1)] != map->map[TE*wh+y*w+x] ||
2728 map->map[BE*wh+(y-1)*w+(x-1)] != map->map[LE*wh+y*w+x])
2729 draw_rect(dr, COORD(x), COORD(y), 1, 1, COL_GRID);
2732 * Draw error markers.
2734 for (yo = 0; yo < 3; yo++)
2735 for (xo = 0; xo < 3; xo++)
2736 if (errs & (ERR_BASE << (yo*3+xo)))
2738 (COORD(x)*2+TILESIZE*xo)/2,
2739 (COORD(y)*2+TILESIZE*yo)/2);
2742 * Draw region numbers, if desired.
2746 for (i = 0; i < 2; i++) {
2747 j = map->map[(i?BE:TE)*wh+y*w+x];
2752 xo = map->regionx[j] - 2*x;
2753 yo = map->regiony[j] - 2*y;
2754 if (xo >= 0 && xo <= 2 && yo >= 0 && yo <= 2) {
2756 sprintf(buf, "%d", j);
2757 draw_text(dr, (COORD(x)*2+TILESIZE*xo)/2,
2758 (COORD(y)*2+TILESIZE*yo)/2,
2759 FONT_VARIABLE, 3*TILESIZE/5,
2760 ALIGN_HCENTRE|ALIGN_VCENTRE,
2768 draw_update(dr, COORD(x), COORD(y), TILESIZE, TILESIZE);
2771 static void game_redraw(drawing *dr, game_drawstate *ds, game_state *oldstate,
2772 game_state *state, int dir, game_ui *ui,
2773 float animtime, float flashtime)
2775 int w = state->p.w, h = state->p.h, wh = w*h, n = state->p.n;
2779 if (ds->drag_visible) {
2780 blitter_load(dr, ds->bl, ds->dragx, ds->dragy);
2781 draw_update(dr, ds->dragx, ds->dragy, TILESIZE + 3, TILESIZE + 3);
2782 ds->drag_visible = FALSE;
2786 * The initial contents of the window are not guaranteed and
2787 * can vary with front ends. To be on the safe side, all games
2788 * should start by drawing a big background-colour rectangle
2789 * covering the whole window.
2794 game_compute_size(&state->p, TILESIZE, &ww, &wh);
2795 draw_rect(dr, 0, 0, ww, wh, COL_BACKGROUND);
2796 draw_rect(dr, COORD(0), COORD(0), w*TILESIZE+1, h*TILESIZE+1,
2799 draw_update(dr, 0, 0, ww, wh);
2804 if (flash_type == 1)
2805 flash = (int)(flashtime * FOUR / flash_length);
2807 flash = 1 + (int)(flashtime * THREE / flash_length);
2812 * Set up the `todraw' array.
2814 for (y = 0; y < h; y++)
2815 for (x = 0; x < w; x++) {
2816 int tv = state->colouring[state->map->map[TE * wh + y*w+x]];
2817 int bv = state->colouring[state->map->map[BE * wh + y*w+x]];
2826 if (flash_type == 1) {
2831 } else if (flash_type == 2) {
2836 tv = (tv + flash) % FOUR;
2838 bv = (bv + flash) % FOUR;
2847 for (i = 0; i < FOUR; i++) {
2848 if (state->colouring[state->map->map[TE * wh + y*w+x]] < 0 &&
2849 (state->pencil[state->map->map[TE * wh + y*w+x]] & (1<<i)))
2850 v |= PENCIL_T_BASE << i;
2851 if (state->colouring[state->map->map[BE * wh + y*w+x]] < 0 &&
2852 (state->pencil[state->map->map[BE * wh + y*w+x]] & (1<<i)))
2853 v |= PENCIL_B_BASE << i;
2856 if (ui->show_numbers)
2859 ds->todraw[y*w+x] = v;
2863 * Add error markers to the `todraw' array.
2865 for (i = 0; i < state->map->ngraph; i++) {
2866 int v1 = state->map->graph[i] / n;
2867 int v2 = state->map->graph[i] % n;
2870 if (state->colouring[v1] < 0 || state->colouring[v2] < 0)
2872 if (state->colouring[v1] != state->colouring[v2])
2875 x = state->map->edgex[i];
2876 y = state->map->edgey[i];
2881 ds->todraw[y*w+x] |= ERR_BASE << (yo*3+xo);
2884 ds->todraw[y*w+(x-1)] |= ERR_BASE << (yo*3+2);
2888 ds->todraw[(y-1)*w+x] |= ERR_BASE << (2*3+xo);
2890 if (xo == 0 && yo == 0) {
2891 assert(x > 0 && y > 0);
2892 ds->todraw[(y-1)*w+(x-1)] |= ERR_BASE << (2*3+2);
2897 * Now actually draw everything.
2899 for (y = 0; y < h; y++)
2900 for (x = 0; x < w; x++) {
2901 unsigned long v = ds->todraw[y*w+x];
2902 if (ds->drawn[y*w+x] != v) {
2903 draw_square(dr, ds, &state->p, state->map, x, y, v);
2904 ds->drawn[y*w+x] = v;
2909 * Draw the dragged colour blob if any.
2911 if (ui->drag_colour > -2) {
2912 ds->dragx = ui->dragx - TILESIZE/2 - 2;
2913 ds->dragy = ui->dragy - TILESIZE/2 - 2;
2914 blitter_save(dr, ds->bl, ds->dragx, ds->dragy);
2915 draw_circle(dr, ui->dragx, ui->dragy, TILESIZE/2,
2916 (ui->drag_colour < 0 ? COL_BACKGROUND :
2917 COL_0 + ui->drag_colour), COL_GRID);
2918 for (i = 0; i < FOUR; i++)
2919 if (ui->drag_pencil & (1 << i))
2920 draw_circle(dr, ui->dragx + ((i*4+2)%10-3) * TILESIZE/10,
2921 ui->dragy + (i*2-3) * TILESIZE/10,
2922 TILESIZE/8, COL_0 + i, COL_0 + i);
2923 draw_update(dr, ds->dragx, ds->dragy, TILESIZE + 3, TILESIZE + 3);
2924 ds->drag_visible = TRUE;
2928 static float game_anim_length(game_state *oldstate, game_state *newstate,
2929 int dir, game_ui *ui)
2934 static float game_flash_length(game_state *oldstate, game_state *newstate,
2935 int dir, game_ui *ui)
2937 if (!oldstate->completed && newstate->completed &&
2938 !oldstate->cheated && !newstate->cheated) {
2939 if (flash_type < 0) {
2940 char *env = getenv("MAP_ALTERNATIVE_FLASH");
2942 flash_type = atoi(env);
2945 flash_length = (flash_type == 1 ? 0.50 : 0.30);
2947 return flash_length;
2952 static int game_timing_state(game_state *state, game_ui *ui)
2957 static void game_print_size(game_params *params, float *x, float *y)
2962 * I'll use 4mm squares by default, I think. Simplest way to
2963 * compute this size is to compute the pixel puzzle size at a
2964 * given tile size and then scale.
2966 game_compute_size(params, 400, &pw, &ph);
2971 static void game_print(drawing *dr, game_state *state, int tilesize)
2973 int w = state->p.w, h = state->p.h, wh = w*h, n = state->p.n;
2974 int ink, c[FOUR], i;
2976 int *coords, ncoords, coordsize;
2978 /* Ick: fake up `ds->tilesize' for macro expansion purposes */
2979 struct { int tilesize; } ads, *ds = &ads;
2980 /* We can't call game_set_size() here because we don't want a blitter */
2981 ads.tilesize = tilesize;
2983 ink = print_mono_colour(dr, 0);
2984 for (i = 0; i < FOUR; i++)
2985 c[i] = print_rgb_hatched_colour(dr, map_colours[i][0],
2986 map_colours[i][1], map_colours[i][2],
2992 print_line_width(dr, TILESIZE / 16);
2995 * Draw a single filled polygon around each region.
2997 for (r = 0; r < n; r++) {
2998 int octants[8], lastdir, d1, d2, ox, oy;
3001 * Start by finding a point on the region boundary. Any
3002 * point will do. To do this, we'll search for a square
3003 * containing the region and then decide which corner of it
3007 for (y = 0; y < h; y++) {
3008 for (x = 0; x < w; x++) {
3009 if (state->map->map[wh*0+y*w+x] == r ||
3010 state->map->map[wh*1+y*w+x] == r ||
3011 state->map->map[wh*2+y*w+x] == r ||
3012 state->map->map[wh*3+y*w+x] == r)
3018 assert(y < h && x < w); /* we must have found one somewhere */
3020 * This is the first square in lexicographic order which
3021 * contains part of this region. Therefore, one of the top
3022 * two corners of the square must be what we're after. The
3023 * only case in which it isn't the top left one is if the
3024 * square is diagonally divided and the region is in the
3025 * bottom right half.
3027 if (state->map->map[wh*TE+y*w+x] != r &&
3028 state->map->map[wh*LE+y*w+x] != r)
3029 x++; /* could just as well have done y++ */
3032 * Now we have a point on the region boundary. Trace around
3033 * the region until we come back to this point,
3034 * accumulating coordinates for a polygon draw operation as
3044 * There are eight possible directions we could head in
3045 * from here. We identify them by octant numbers, and
3046 * we also use octant numbers to identify the spaces
3059 octants[0] = x<w && y>0 ? state->map->map[wh*LE+(y-1)*w+x] : -1;
3060 octants[1] = x<w && y>0 ? state->map->map[wh*BE+(y-1)*w+x] : -1;
3061 octants[2] = x<w && y<h ? state->map->map[wh*TE+y*w+x] : -1;
3062 octants[3] = x<w && y<h ? state->map->map[wh*LE+y*w+x] : -1;
3063 octants[4] = x>0 && y<h ? state->map->map[wh*RE+y*w+(x-1)] : -1;
3064 octants[5] = x>0 && y<h ? state->map->map[wh*TE+y*w+(x-1)] : -1;
3065 octants[6] = x>0 && y>0 ? state->map->map[wh*BE+(y-1)*w+(x-1)] :-1;
3066 octants[7] = x>0 && y>0 ? state->map->map[wh*RE+(y-1)*w+(x-1)] :-1;
3069 for (i = 0; i < 8; i++)
3070 if ((octants[i] == r) ^ (octants[(i+1)%8] == r)) {
3078 assert(d1 != -1 && d2 != -1);
3083 * Now we're heading in direction d1. Save the current
3086 if (ncoords + 2 > coordsize) {
3088 coords = sresize(coords, coordsize, int);
3090 coords[ncoords++] = COORD(x);
3091 coords[ncoords++] = COORD(y);
3094 * Compute the new coordinates.
3096 x += (d1 % 4 == 3 ? 0 : d1 < 4 ? +1 : -1);
3097 y += (d1 % 4 == 1 ? 0 : d1 > 1 && d1 < 5 ? +1 : -1);
3098 assert(x >= 0 && x <= w && y >= 0 && y <= h);
3101 } while (x != ox || y != oy);
3103 draw_polygon(dr, coords, ncoords/2,
3104 state->colouring[r] >= 0 ?
3105 c[state->colouring[r]] : -1, ink);
3114 const struct game thegame = {
3115 "Map", "games.map", "map",
3122 TRUE, game_configure, custom_params,
3130 FALSE, game_can_format_as_text_now, game_text_format,
3138 20, game_compute_size, game_set_size,
3141 game_free_drawstate,
3145 TRUE, TRUE, game_print_size, game_print,
3146 FALSE, /* wants_statusbar */
3147 FALSE, game_timing_state,
3151 #ifdef STANDALONE_SOLVER
3153 int main(int argc, char **argv)
3157 char *id = NULL, *desc, *err;
3159 int ret, diff, really_verbose = FALSE;
3160 struct solver_scratch *sc;
3163 while (--argc > 0) {
3165 if (!strcmp(p, "-v")) {
3166 really_verbose = TRUE;
3167 } else if (!strcmp(p, "-g")) {
3169 } else if (*p == '-') {
3170 fprintf(stderr, "%s: unrecognised option `%s'\n", argv[0], p);
3178 fprintf(stderr, "usage: %s [-g | -v] <game_id>\n", argv[0]);
3182 desc = strchr(id, ':');
3184 fprintf(stderr, "%s: game id expects a colon in it\n", argv[0]);
3189 p = default_params();
3190 decode_params(p, id);
3191 err = validate_desc(p, desc);
3193 fprintf(stderr, "%s: %s\n", argv[0], err);
3196 s = new_game(NULL, p, desc);
3198 sc = new_scratch(s->map->graph, s->map->n, s->map->ngraph);
3201 * When solving an Easy puzzle, we don't want to bother the
3202 * user with Hard-level deductions. For this reason, we grade
3203 * the puzzle internally before doing anything else.
3205 ret = -1; /* placate optimiser */
3206 for (diff = 0; diff < DIFFCOUNT; diff++) {
3207 for (i = 0; i < s->map->n; i++)
3208 if (!s->map->immutable[i])
3209 s->colouring[i] = -1;
3210 ret = map_solver(sc, s->map->graph, s->map->n, s->map->ngraph,
3211 s->colouring, diff);
3216 if (diff == DIFFCOUNT) {
3218 printf("Difficulty rating: harder than Hard, or ambiguous\n");
3220 printf("Unable to find a unique solution\n");
3224 printf("Difficulty rating: impossible (no solution exists)\n");
3226 printf("Difficulty rating: %s\n", map_diffnames[diff]);
3228 verbose = really_verbose;
3229 for (i = 0; i < s->map->n; i++)
3230 if (!s->map->immutable[i])
3231 s->colouring[i] = -1;
3232 ret = map_solver(sc, s->map->graph, s->map->n, s->map->ngraph,
3233 s->colouring, diff);
3235 printf("Puzzle is inconsistent\n");
3239 for (i = 0; i < s->map->n; i++) {
3240 printf("%5d <- %c%c", i, colnames[s->colouring[i]],
3241 (col < 6 && i+1 < s->map->n ? ' ' : '\n'));