X-Git-Url: https://git.distorted.org.uk/~mdw/sgt/puzzles/blobdiff_plain/a2f35d71b745ec2a03de58976c6434437c5f303e..HEAD:/loopy.c diff --git a/loopy.c b/loopy.c index 7d03ed2..0ee1098 100644 --- a/loopy.c +++ b/loopy.c @@ -82,6 +82,7 @@ #include "puzzles.h" #include "tree234.h" #include "grid.h" +#include "loopgen.h" /* Debugging options */ @@ -107,7 +108,7 @@ enum { }; struct game_state { - grid *game_grid; + grid *game_grid; /* ref-counted (internally) */ /* Put -1 in a face that doesn't get a clue */ signed char *clues; @@ -201,16 +202,12 @@ static char const diffchars[] = DIFFLIST(ENCODE); SOLVERLIST(SOLVER_FN_DECL) static int (*(solver_fns[]))(solver_state *) = { SOLVERLIST(SOLVER_FN) }; static int const solver_diffs[] = { SOLVERLIST(SOLVER_DIFF) }; -const int NUM_SOLVERS = sizeof(solver_diffs)/sizeof(*solver_diffs); +static const int NUM_SOLVERS = sizeof(solver_diffs)/sizeof(*solver_diffs); struct game_params { int w, h; int diff; int type; - - /* Grid generation is expensive, so keep a (ref-counted) reference to the - * grid for these parameters, and only generate when required. */ - grid *game_grid; }; /* line_drawstate is the same as line_state, but with the extra ERROR @@ -247,29 +244,31 @@ static void check_caches(const solver_state* sstate); /* ------- List of grid generators ------- */ #define GRIDLIST(A) \ - A(Squares,grid_new_square,3,3) \ - A(Triangular,grid_new_triangular,3,3) \ - A(Honeycomb,grid_new_honeycomb,3,3) \ - A(Snub-Square,grid_new_snubsquare,3,3) \ - A(Cairo,grid_new_cairo,3,4) \ - A(Great-Hexagonal,grid_new_greathexagonal,3,3) \ - A(Octagonal,grid_new_octagonal,3,3) \ - A(Kites,grid_new_kites,3,3) \ - A(Floret,grid_new_floret,1,2) \ - A(Dodecagonal,grid_new_dodecagonal,2,2) \ - A(Great-Dodecagonal,grid_new_greatdodecagonal,2,2) - -#define GRID_NAME(title,fn,amin,omin) #title, -#define GRID_CONFIG(title,fn,amin,omin) ":" #title -#define GRID_FN(title,fn,amin,omin) &fn, -#define GRID_SIZES(title,fn,amin,omin) \ + A(Squares,GRID_SQUARE,3,3) \ + A(Triangular,GRID_TRIANGULAR,3,3) \ + A(Honeycomb,GRID_HONEYCOMB,3,3) \ + A(Snub-Square,GRID_SNUBSQUARE,3,3) \ + A(Cairo,GRID_CAIRO,3,4) \ + A(Great-Hexagonal,GRID_GREATHEXAGONAL,3,3) \ + A(Octagonal,GRID_OCTAGONAL,3,3) \ + A(Kites,GRID_KITE,3,3) \ + A(Floret,GRID_FLORET,1,2) \ + A(Dodecagonal,GRID_DODECAGONAL,2,2) \ + A(Great-Dodecagonal,GRID_GREATDODECAGONAL,2,2) \ + A(Penrose (kite/dart),GRID_PENROSE_P2,3,3) \ + A(Penrose (rhombs),GRID_PENROSE_P3,3,3) + +#define GRID_NAME(title,type,amin,omin) #title, +#define GRID_CONFIG(title,type,amin,omin) ":" #title +#define GRID_TYPE(title,type,amin,omin) type, +#define GRID_SIZES(title,type,amin,omin) \ {amin, omin, \ "Width and height for this grid type must both be at least " #amin, \ "At least one of width and height for this grid type must be at least " #omin,}, static char const *const gridnames[] = { GRIDLIST(GRID_NAME) }; #define GRID_CONFIGS GRIDLIST(GRID_CONFIG) -static grid * (*(grid_fns[]))(int w, int h) = { GRIDLIST(GRID_FN) }; -#define NUM_GRID_TYPES (sizeof(grid_fns) / sizeof(grid_fns[0])) +static grid_type grid_types[] = { GRIDLIST(GRID_TYPE) }; +#define NUM_GRID_TYPES (sizeof(grid_types) / sizeof(grid_types[0])) static const struct { int amin, omin; char *aerr, *oerr; @@ -277,13 +276,10 @@ static const struct { /* Generates a (dynamically allocated) new grid, according to the * type and size requested in params. Does nothing if the grid is already - * generated. The allocated grid is owned by the params object, and will be - * freed in free_params(). */ -static void params_generate_grid(game_params *params) + * generated. */ +static grid *loopy_generate_grid(game_params *params, char *grid_desc) { - if (!params->game_grid) { - params->game_grid = grid_fns[params->type](params->w, params->h); - } + return grid_new(grid_types[params->type], params->w, params->h, grid_desc); } /* ---------------------------------------------------------------------- @@ -480,8 +476,6 @@ static game_params *default_params(void) ret->diff = DIFF_EASY; ret->type = 0; - ret->game_grid = NULL; - return ret; } @@ -490,44 +484,45 @@ static game_params *dup_params(game_params *params) game_params *ret = snew(game_params); *ret = *params; /* structure copy */ - if (ret->game_grid) { - ret->game_grid->refcount++; - } return ret; } static const game_params presets[] = { #ifdef SMALL_SCREEN - { 7, 7, DIFF_EASY, 0, NULL }, - { 7, 7, DIFF_NORMAL, 0, NULL }, - { 7, 7, DIFF_HARD, 0, NULL }, - { 7, 7, DIFF_HARD, 1, NULL }, - { 7, 7, DIFF_HARD, 2, NULL }, - { 5, 5, DIFF_HARD, 3, NULL }, - { 7, 7, DIFF_HARD, 4, NULL }, - { 5, 4, DIFF_HARD, 5, NULL }, - { 5, 5, DIFF_HARD, 6, NULL }, - { 5, 5, DIFF_HARD, 7, NULL }, - { 3, 3, DIFF_HARD, 8, NULL }, - { 3, 3, DIFF_HARD, 9, NULL }, - { 3, 3, DIFF_HARD, 10, NULL }, + { 7, 7, DIFF_EASY, 0 }, + { 7, 7, DIFF_NORMAL, 0 }, + { 7, 7, DIFF_HARD, 0 }, + { 7, 7, DIFF_HARD, 1 }, + { 7, 7, DIFF_HARD, 2 }, + { 5, 5, DIFF_HARD, 3 }, + { 7, 7, DIFF_HARD, 4 }, + { 5, 4, DIFF_HARD, 5 }, + { 5, 5, DIFF_HARD, 6 }, + { 5, 5, DIFF_HARD, 7 }, + { 3, 3, DIFF_HARD, 8 }, + { 3, 3, DIFF_HARD, 9 }, + { 3, 3, DIFF_HARD, 10 }, + { 6, 6, DIFF_HARD, 11 }, + { 6, 6, DIFF_HARD, 12 }, #else - { 7, 7, DIFF_EASY, 0, NULL }, - { 10, 10, DIFF_EASY, 0, NULL }, - { 7, 7, DIFF_NORMAL, 0, NULL }, - { 10, 10, DIFF_NORMAL, 0, NULL }, - { 7, 7, DIFF_HARD, 0, NULL }, - { 10, 10, DIFF_HARD, 0, NULL }, - { 10, 10, DIFF_HARD, 1, NULL }, - { 12, 10, DIFF_HARD, 2, NULL }, - { 7, 7, DIFF_HARD, 3, NULL }, - { 9, 9, DIFF_HARD, 4, NULL }, - { 5, 4, DIFF_HARD, 5, NULL }, - { 7, 7, DIFF_HARD, 6, NULL }, - { 5, 5, DIFF_HARD, 7, NULL }, - { 5, 5, DIFF_HARD, 8, NULL }, - { 5, 4, DIFF_HARD, 9, NULL }, - { 5, 4, DIFF_HARD, 10, NULL }, + { 7, 7, DIFF_EASY, 0 }, + { 10, 10, DIFF_EASY, 0 }, + { 7, 7, DIFF_NORMAL, 0 }, + { 10, 10, DIFF_NORMAL, 0 }, + { 7, 7, DIFF_HARD, 0 }, + { 10, 10, DIFF_HARD, 0 }, + { 10, 10, DIFF_HARD, 1 }, + { 12, 10, DIFF_HARD, 2 }, + { 7, 7, DIFF_HARD, 3 }, + { 9, 9, DIFF_HARD, 4 }, + { 5, 4, DIFF_HARD, 5 }, + { 7, 7, DIFF_HARD, 6 }, + { 5, 5, DIFF_HARD, 7 }, + { 5, 5, DIFF_HARD, 8 }, + { 5, 4, DIFF_HARD, 9 }, + { 5, 4, DIFF_HARD, 10 }, + { 10, 10, DIFF_HARD, 11 }, + { 10, 10, DIFF_HARD, 12 } #endif }; @@ -551,18 +546,11 @@ static int game_fetch_preset(int i, char **name, game_params **params) static void free_params(game_params *params) { - if (params->game_grid) { - grid_free(params->game_grid); - } sfree(params); } static void decode_params(game_params *params, char const *string) { - if (params->game_grid) { - grid_free(params->game_grid); - params->game_grid = NULL; - } params->h = params->w = atoi(string); params->diff = DIFF_EASY; while (*string && isdigit((unsigned char)*string)) string++; @@ -641,7 +629,6 @@ static game_params *custom_params(config_item *cfg) ret->type = cfg[2].ival; ret->diff = cfg[3].ival; - ret->game_grid = NULL; return ret; } @@ -702,14 +689,44 @@ static char *state_to_text(const game_state *state) return retval; } +#define GRID_DESC_SEP '_' + +/* Splits up a (optional) grid_desc from the game desc. Returns the + * grid_desc (which needs freeing) and updates the desc pointer to + * start of real desc, or returns NULL if no desc. */ +static char *extract_grid_desc(char **desc) +{ + char *sep = strchr(*desc, GRID_DESC_SEP), *gd; + int gd_len; + + if (!sep) return NULL; + + gd_len = sep - (*desc); + gd = snewn(gd_len+1, char); + memcpy(gd, *desc, gd_len); + gd[gd_len] = '\0'; + + *desc = sep+1; + + return gd; +} + /* We require that the params pass the test in validate_params and that the * description fills the entire game area */ static char *validate_desc(game_params *params, char *desc) { int count = 0; grid *g; - params_generate_grid(params); - g = params->game_grid; + char *grid_desc, *ret; + + /* It's pretty inefficient to do this just for validation. All we need to + * know is the precise number of faces. */ + grid_desc = extract_grid_desc(&desc); + ret = grid_validate_desc(grid_types[params->type], params->w, params->h, grid_desc); + if (ret) return ret; + + g = loopy_generate_grid(params, grid_desc); + if (grid_desc) sfree(grid_desc); for (; *desc; ++desc) { if ((*desc >= '0' && *desc <= '9') || (*desc >= 'A' && *desc <= 'Z')) { @@ -728,6 +745,8 @@ static char *validate_desc(game_params *params, char *desc) if (count > g->num_faces) return "Description too long for board size"; + grid_free(g); + return NULL; } @@ -809,16 +828,15 @@ static void game_changed_state(game_ui *ui, game_state *oldstate, static void game_compute_size(game_params *params, int tilesize, int *x, int *y) { - grid *g; int grid_width, grid_height, rendered_width, rendered_height; + int g_tilesize; + + grid_compute_size(grid_types[params->type], params->w, params->h, + &g_tilesize, &grid_width, &grid_height); - params_generate_grid(params); - g = params->game_grid; - grid_width = g->highest_x - g->lowest_x; - grid_height = g->highest_y - g->lowest_y; /* multiply first to minimise rounding error on integer division */ - rendered_width = grid_width * tilesize / g->tilesize; - rendered_height = grid_height * tilesize / g->tilesize; + rendered_width = grid_width * tilesize / g_tilesize; + rendered_height = grid_height * tilesize / g_tilesize; *x = rendered_width + 2 * BORDER(tilesize) + 1; *y = rendered_height + 2 * BORDER(tilesize) + 1; } @@ -900,6 +918,8 @@ static game_drawstate *game_new_drawstate(drawing *dr, game_state *state) static void game_free_drawstate(drawing *dr, game_drawstate *ds) { + sfree(ds->textx); + sfree(ds->texty); sfree(ds->clue_error); sfree(ds->clue_satisfied); sfree(ds->lines); @@ -1258,507 +1278,20 @@ static int face_setall(solver_state *sstate, int face, * Loop generation and clue removal */ -/* We're going to store lists of current candidate faces for colouring black - * or white. - * Each face gets a 'score', which tells us how adding that face right - * now would affect the curliness of the solution loop. We're trying to - * maximise that quantity so will bias our random selection of faces to - * colour those with high scores */ -struct face_score { - int white_score; - int black_score; - unsigned long random; - /* No need to store a grid_face* here. The 'face_scores' array will - * be a list of 'face_score' objects, one for each face of the grid, so - * the position (index) within the 'face_scores' array will determine - * which face corresponds to a particular face_score. - * Having a single 'face_scores' array for all faces simplifies memory - * management, and probably improves performance, because we don't have to - * malloc/free each individual face_score, and we don't have to maintain - * a mapping from grid_face* pointers to face_score* pointers. - */ -}; - -static int generic_sort_cmpfn(void *v1, void *v2, size_t offset) -{ - struct face_score *f1 = v1; - struct face_score *f2 = v2; - int r; - - r = *(int *)((char *)f2 + offset) - *(int *)((char *)f1 + offset); - if (r) { - return r; - } - - if (f1->random < f2->random) - return -1; - else if (f1->random > f2->random) - return 1; - - /* - * It's _just_ possible that two faces might have been given - * the same random value. In that situation, fall back to - * comparing based on the positions within the face_scores list. - * This introduces a tiny directional bias, but not a significant one. - */ - return f1 - f2; -} - -static int white_sort_cmpfn(void *v1, void *v2) -{ - return generic_sort_cmpfn(v1, v2, offsetof(struct face_score,white_score)); -} - -static int black_sort_cmpfn(void *v1, void *v2) -{ - return generic_sort_cmpfn(v1, v2, offsetof(struct face_score,black_score)); -} - -enum face_colour { FACE_WHITE, FACE_GREY, FACE_BLACK }; - -/* face should be of type grid_face* here. */ -#define FACE_COLOUR(face) \ - ( (face) == NULL ? FACE_BLACK : \ - board[(face) - g->faces] ) - -/* 'board' is an array of these enums, indicating which faces are - * currently black/white/grey. 'colour' is FACE_WHITE or FACE_BLACK. - * Returns whether it's legal to colour the given face with this colour. */ -static int can_colour_face(grid *g, char* board, int face_index, - enum face_colour colour) -{ - int i, j; - grid_face *test_face = g->faces + face_index; - grid_face *starting_face, *current_face; - grid_dot *starting_dot; - int transitions; - int current_state, s; /* booleans: equal or not-equal to 'colour' */ - int found_same_coloured_neighbour = FALSE; - assert(board[face_index] != colour); - - /* Can only consider a face for colouring if it's adjacent to a face - * with the same colour. */ - for (i = 0; i < test_face->order; i++) { - grid_edge *e = test_face->edges[i]; - grid_face *f = (e->face1 == test_face) ? e->face2 : e->face1; - if (FACE_COLOUR(f) == colour) { - found_same_coloured_neighbour = TRUE; - break; - } - } - if (!found_same_coloured_neighbour) - return FALSE; - - /* Need to avoid creating a loop of faces of this colour around some - * differently-coloured faces. - * Also need to avoid meeting a same-coloured face at a corner, with - * other-coloured faces in between. Here's a simple test that (I believe) - * takes care of both these conditions: - * - * Take the circular path formed by this face's edges, and inflate it - * slightly outwards. Imagine walking around this path and consider - * the faces that you visit in sequence. This will include all faces - * touching the given face, either along an edge or just at a corner. - * Count the number of 'colour'/not-'colour' transitions you encounter, as - * you walk along the complete loop. This will obviously turn out to be - * an even number. - * If 0, we're either in the middle of an "island" of this colour (should - * be impossible as we're not supposed to create black or white loops), - * or we're about to start a new island - also not allowed. - * If 4 or greater, there are too many separate coloured regions touching - * this face, and colouring it would create a loop or a corner-violation. - * The only allowed case is when the count is exactly 2. */ - - /* i points to a dot around the test face. - * j points to a face around the i^th dot. - * The current face will always be: - * test_face->dots[i]->faces[j] - * We assume dots go clockwise around the test face, - * and faces go clockwise around dots. */ - - /* - * The end condition is slightly fiddly. In sufficiently strange - * degenerate grids, our test face may be adjacent to the same - * other face multiple times (typically if it's the exterior - * face). Consider this, in particular: - * - * +--+ - * | | - * +--+--+ - * | | | - * +--+--+ - * - * The bottom left face there is adjacent to the exterior face - * twice, so we can't just terminate our iteration when we reach - * the same _face_ we started at. Furthermore, we can't - * condition on having the same (i,j) pair either, because - * several (i,j) pairs identify the bottom left contiguity with - * the exterior face! We canonicalise the (i,j) pair by taking - * one step around before we set the termination tracking. - */ - - i = j = 0; - current_face = test_face->dots[0]->faces[0]; - if (current_face == test_face) { - j = 1; - current_face = test_face->dots[0]->faces[1]; - } - transitions = 0; - current_state = (FACE_COLOUR(current_face) == colour); - starting_dot = NULL; - starting_face = NULL; - while (TRUE) { - /* Advance to next face. - * Need to loop here because it might take several goes to - * find it. */ - while (TRUE) { - j++; - if (j == test_face->dots[i]->order) - j = 0; - - if (test_face->dots[i]->faces[j] == test_face) { - /* Advance to next dot round test_face, then - * find current_face around new dot - * and advance to the next face clockwise */ - i++; - if (i == test_face->order) - i = 0; - for (j = 0; j < test_face->dots[i]->order; j++) { - if (test_face->dots[i]->faces[j] == current_face) - break; - } - /* Must actually find current_face around new dot, - * or else something's wrong with the grid. */ - assert(j != test_face->dots[i]->order); - /* Found, so advance to next face and try again */ - } else { - break; - } - } - /* (i,j) are now advanced to next face */ - current_face = test_face->dots[i]->faces[j]; - s = (FACE_COLOUR(current_face) == colour); - if (!starting_dot) { - starting_dot = test_face->dots[i]; - starting_face = current_face; - current_state = s; - } else { - if (s != current_state) { - ++transitions; - current_state = s; - if (transitions > 2) - break; - } - if (test_face->dots[i] == starting_dot && - current_face == starting_face) - break; - } - } - - return (transitions == 2) ? TRUE : FALSE; -} - -/* Count the number of neighbours of 'face', having colour 'colour' */ -static int face_num_neighbours(grid *g, char *board, grid_face *face, - enum face_colour colour) -{ - int colour_count = 0; - int i; - grid_face *f; - grid_edge *e; - for (i = 0; i < face->order; i++) { - e = face->edges[i]; - f = (e->face1 == face) ? e->face2 : e->face1; - if (FACE_COLOUR(f) == colour) - ++colour_count; - } - return colour_count; -} - -/* The 'score' of a face reflects its current desirability for selection - * as the next face to colour white or black. We want to encourage moving - * into grey areas and increasing loopiness, so we give scores according to - * how many of the face's neighbours are currently coloured the same as the - * proposed colour. */ -static int face_score(grid *g, char *board, grid_face *face, - enum face_colour colour) -{ - /* Simple formula: score = 0 - num. same-coloured neighbours, - * so a higher score means fewer same-coloured neighbours. */ - return -face_num_neighbours(g, board, face, colour); -} - -/* Generate a new complete set of clues for the given game_state. - * The method is to generate a WHITE/BLACK colouring of all the faces, - * such that the WHITE faces will define the inside of the path, and the - * BLACK faces define the outside. - * To do this, we initially colour all faces GREY. The infinite space outside - * the grid is coloured BLACK, and we choose a random face to colour WHITE. - * Then we gradually grow the BLACK and the WHITE regions, eliminating GREY - * faces, until the grid is filled with BLACK/WHITE. As we grow the regions, - * we avoid creating loops of a single colour, to preserve the topological - * shape of the WHITE and BLACK regions. - * We also try to make the boundary as loopy and twisty as possible, to avoid - * generating paths that are uninteresting. - * The algorithm works by choosing a BLACK/WHITE colour, then choosing a GREY - * face that can be coloured with that colour (without violating the - * topological shape of that region). It's not obvious, but I think this - * algorithm is guaranteed to terminate without leaving any GREY faces behind. - * Indeed, if there are any GREY faces at all, both the WHITE and BLACK - * regions can be grown. - * This is checked using assert()ions, and I haven't seen any failures yet. - * - * Hand-wavy proof: imagine what can go wrong... - * - * Could the white faces get completely cut off by the black faces, and still - * leave some grey faces remaining? - * No, because then the black faces would form a loop around both the white - * faces and the grey faces, which is disallowed because we continually - * maintain the correct topological shape of the black region. - * Similarly, the black faces can never get cut off by the white faces. That - * means both the WHITE and BLACK regions always have some room to grow into - * the GREY regions. - * Could it be that we can't colour some GREY face, because there are too many - * WHITE/BLACK transitions as we walk round the face? (see the - * can_colour_face() function for details) - * No. Imagine otherwise, and we see WHITE/BLACK/WHITE/BLACK as we walk - * around the face. The two WHITE faces would be connected by a WHITE path, - * and the BLACK faces would be connected by a BLACK path. These paths would - * have to cross, which is impossible. - * Another thing that could go wrong: perhaps we can't find any GREY face to - * colour WHITE, because it would create a loop-violation or a corner-violation - * with the other WHITE faces? - * This is a little bit tricky to prove impossible. Imagine you have such a - * GREY face (that is, if you coloured it WHITE, you would create a WHITE loop - * or corner violation). - * That would cut all the non-white area into two blobs. One of those blobs - * must be free of BLACK faces (because the BLACK stuff is a connected blob). - * So we have a connected GREY area, completely surrounded by WHITE - * (including the GREY face we've tentatively coloured WHITE). - * A well-known result in graph theory says that you can always find a GREY - * face whose removal leaves the remaining GREY area connected. And it says - * there are at least two such faces, so we can always choose the one that - * isn't the "tentative" GREY face. Colouring that face WHITE leaves - * everything nice and connected, including that "tentative" GREY face which - * acts as a gateway to the rest of the non-WHITE grid. - */ static void add_full_clues(game_state *state, random_state *rs) { signed char *clues = state->clues; - char *board; grid *g = state->game_grid; - int i, j; - int num_faces = g->num_faces; - struct face_score *face_scores; /* Array of face_score objects */ - struct face_score *fs; /* Points somewhere in the above list */ - struct grid_face *cur_face; - tree234 *lightable_faces_sorted; - tree234 *darkable_faces_sorted; - int *face_list; - int do_random_pass; - - board = snewn(num_faces, char); - - /* Make a board */ - memset(board, FACE_GREY, num_faces); - - /* Create and initialise the list of face_scores */ - face_scores = snewn(num_faces, struct face_score); - for (i = 0; i < num_faces; i++) { - face_scores[i].random = random_bits(rs, 31); - face_scores[i].black_score = face_scores[i].white_score = 0; - } - - /* Colour a random, finite face white. The infinite face is implicitly - * coloured black. Together, they will seed the random growth process - * for the black and white areas. */ - i = random_upto(rs, num_faces); - board[i] = FACE_WHITE; - - /* We need a way of favouring faces that will increase our loopiness. - * We do this by maintaining a list of all candidate faces sorted by - * their score and choose randomly from that with appropriate skew. - * In order to avoid consistently biasing towards particular faces, we - * need the sort order _within_ each group of scores to be completely - * random. But it would be abusing the hospitality of the tree234 data - * structure if our comparison function were nondeterministic :-). So with - * each face we associate a random number that does not change during a - * particular run of the generator, and use that as a secondary sort key. - * Yes, this means we will be biased towards particular random faces in - * any one run but that doesn't actually matter. */ - - lightable_faces_sorted = newtree234(white_sort_cmpfn); - darkable_faces_sorted = newtree234(black_sort_cmpfn); - - /* Initialise the lists of lightable and darkable faces. This is - * slightly different from the code inside the while-loop, because we need - * to check every face of the board (the grid structure does not keep a - * list of the infinite face's neighbours). */ - for (i = 0; i < num_faces; i++) { - grid_face *f = g->faces + i; - struct face_score *fs = face_scores + i; - if (board[i] != FACE_GREY) continue; - /* We need the full colourability check here, it's not enough simply - * to check neighbourhood. On some grids, a neighbour of the infinite - * face is not necessarily darkable. */ - if (can_colour_face(g, board, i, FACE_BLACK)) { - fs->black_score = face_score(g, board, f, FACE_BLACK); - add234(darkable_faces_sorted, fs); - } - if (can_colour_face(g, board, i, FACE_WHITE)) { - fs->white_score = face_score(g, board, f, FACE_WHITE); - add234(lightable_faces_sorted, fs); - } - } - - /* Colour faces one at a time until no more faces are colourable. */ - while (TRUE) - { - enum face_colour colour; - struct face_score *fs_white, *fs_black; - int c_lightable = count234(lightable_faces_sorted); - int c_darkable = count234(darkable_faces_sorted); - if (c_lightable == 0 && c_darkable == 0) { - /* No more faces we can use at all. */ - break; - } - assert(c_lightable != 0 && c_darkable != 0); - - fs_white = (struct face_score *)index234(lightable_faces_sorted, 0); - fs_black = (struct face_score *)index234(darkable_faces_sorted, 0); - - /* Choose a colour, and colour the best available face - * with that colour. */ - colour = random_upto(rs, 2) ? FACE_WHITE : FACE_BLACK; - - if (colour == FACE_WHITE) - fs = fs_white; - else - fs = fs_black; - assert(fs); - i = fs - face_scores; - assert(board[i] == FACE_GREY); - board[i] = colour; - - /* Remove this newly-coloured face from the lists. These lists should - * only contain grey faces. */ - del234(lightable_faces_sorted, fs); - del234(darkable_faces_sorted, fs); - - /* Remember which face we've just coloured */ - cur_face = g->faces + i; - - /* The face we've just coloured potentially affects the colourability - * and the scores of any neighbouring faces (touching at a corner or - * edge). So the search needs to be conducted around all faces - * touching the one we've just lit. Iterate over its corners, then - * over each corner's faces. For each such face, we remove it from - * the lists, recalculate any scores, then add it back to the lists - * (depending on whether it is lightable, darkable or both). */ - for (i = 0; i < cur_face->order; i++) { - grid_dot *d = cur_face->dots[i]; - for (j = 0; j < d->order; j++) { - grid_face *f = d->faces[j]; - int fi; /* face index of f */ - - if (f == NULL) - continue; - if (f == cur_face) - continue; - - /* If the face is already coloured, it won't be on our - * lightable/darkable lists anyway, so we can skip it without - * bothering with the removal step. */ - if (FACE_COLOUR(f) != FACE_GREY) continue; - - /* Find the face index and face_score* corresponding to f */ - fi = f - g->faces; - fs = face_scores + fi; - - /* Remove from lightable list if it's in there. We do this, - * even if it is still lightable, because the score might - * be different, and we need to remove-then-add to maintain - * correct sort order. */ - del234(lightable_faces_sorted, fs); - if (can_colour_face(g, board, fi, FACE_WHITE)) { - fs->white_score = face_score(g, board, f, FACE_WHITE); - add234(lightable_faces_sorted, fs); - } - /* Do the same for darkable list. */ - del234(darkable_faces_sorted, fs); - if (can_colour_face(g, board, fi, FACE_BLACK)) { - fs->black_score = face_score(g, board, f, FACE_BLACK); - add234(darkable_faces_sorted, fs); - } - } - } - } - - /* Clean up */ - freetree234(lightable_faces_sorted); - freetree234(darkable_faces_sorted); - sfree(face_scores); - - /* The next step requires a shuffled list of all faces */ - face_list = snewn(num_faces, int); - for (i = 0; i < num_faces; ++i) { - face_list[i] = i; - } - shuffle(face_list, num_faces, sizeof(int), rs); - - /* The above loop-generation algorithm can often leave large clumps - * of faces of one colour. In extreme cases, the resulting path can be - * degenerate and not very satisfying to solve. - * This next step alleviates this problem: - * Go through the shuffled list, and flip the colour of any face we can - * legally flip, and which is adjacent to only one face of the opposite - * colour - this tends to grow 'tendrils' into any clumps. - * Repeat until we can find no more faces to flip. This will - * eventually terminate, because each flip increases the loop's - * perimeter, which cannot increase for ever. - * The resulting path will have maximal loopiness (in the sense that it - * cannot be improved "locally". Unfortunately, this allows a player to - * make some illicit deductions. To combat this (and make the path more - * interesting), we do one final pass making random flips. */ - - /* Set to TRUE for final pass */ - do_random_pass = FALSE; - - while (TRUE) { - /* Remember whether a flip occurred during this pass */ - int flipped = FALSE; - - for (i = 0; i < num_faces; ++i) { - int j = face_list[i]; - enum face_colour opp = - (board[j] == FACE_WHITE) ? FACE_BLACK : FACE_WHITE; - if (can_colour_face(g, board, j, opp)) { - grid_face *face = g->faces +j; - if (do_random_pass) { - /* final random pass */ - if (!random_upto(rs, 10)) - board[j] = opp; - } else { - /* normal pass - flip when neighbour count is 1 */ - if (face_num_neighbours(g, board, face, opp) == 1) { - board[j] = opp; - flipped = TRUE; - } - } - } - } + char *board = snewn(g->num_faces, char); + int i; - if (do_random_pass) break; - if (!flipped) do_random_pass = TRUE; - } - - sfree(face_list); + generate_loop(g, board, rs, NULL, NULL); /* Fill out all the clues by initialising to 0, then iterating over * all edges and incrementing each clue as we find edges that border * between BLACK/WHITE faces. While we're at it, we verify that the * algorithm does work, and there aren't any GREY faces still there. */ - memset(clues, 0, num_faces); + memset(clues, 0, g->num_faces); for (i = 0; i < g->num_edges; i++) { grid_edge *e = g->edges + i; grid_face *f1 = e->face1; @@ -1772,7 +1305,6 @@ static void add_full_clues(game_state *state, random_state *rs) if (f2) clues[f2 - g->faces]++; } } - sfree(board); } @@ -1836,13 +1368,14 @@ static char *new_game_desc(game_params *params, random_state *rs, char **aux, int interactive) { /* solution and description both use run-length encoding in obvious ways */ - char *retval; + char *retval, *game_desc, *grid_desc; grid *g; game_state *state = snew(game_state); game_state *state_new; - params_generate_grid(params); - state->game_grid = g = params->game_grid; - g->refcount++; + + grid_desc = grid_new_desc(grid_types[params->type], params->w, params->h, rs); + state->game_grid = g = loopy_generate_grid(params, grid_desc); + state->clues = snewn(g->num_faces, signed char); state->lines = snewn(g->num_edges, char); state->line_errors = snewn(g->num_edges, unsigned char); @@ -1875,10 +1408,19 @@ static char *new_game_desc(game_params *params, random_state *rs, goto newboard_please; } - retval = state_to_text(state); + game_desc = state_to_text(state); free_game(state); + if (grid_desc) { + retval = snewn(strlen(grid_desc) + 1 + strlen(game_desc) + 1, char); + sprintf(retval, "%s%c%s", grid_desc, (int)GRID_DESC_SEP, game_desc); + sfree(grid_desc); + sfree(game_desc); + } else { + retval = game_desc; + } + assert(!validate_desc(params, retval)); return retval; @@ -1890,13 +1432,17 @@ static game_state *new_game(midend *me, game_params *params, char *desc) game_state *state = snew(game_state); int empties_to_make = 0; int n,n2; - const char *dp = desc; + const char *dp; + char *grid_desc; grid *g; int num_faces, num_edges; - params_generate_grid(params); - state->game_grid = g = params->game_grid; - g->refcount++; + grid_desc = extract_grid_desc(&desc); + state->game_grid = g = loopy_generate_grid(params, grid_desc); + if (grid_desc) sfree(grid_desc); + + dp = desc; + num_faces = g->num_faces; num_edges = g->num_edges; @@ -3267,7 +2813,7 @@ static char *solve_game(game_state *state, game_state *currstate, * Drawing and mouse-handling */ -static char *interpret_move(game_state *state, game_ui *ui, game_drawstate *ds, +static char *interpret_move(game_state *state, game_ui *ui, const game_drawstate *ds, int x, int y, int button) { grid *g = state->game_grid; @@ -3401,62 +2947,11 @@ static void grid_to_screen(const game_drawstate *ds, const grid *g, *y += BORDER(ds->tilesize); } -static int solve_2x2_matrix(double mx[4], double vin[2], double vout[2]) -{ - double inv[4]; - double det; - det = (mx[0]*mx[3] - mx[1]*mx[2]); - if (det == 0) - return FALSE; - - inv[0] = mx[3] / det; - inv[1] = -mx[1] / det; - inv[2] = -mx[2] / det; - inv[3] = mx[0] / det; - - vout[0] = inv[0]*vin[0] + inv[1]*vin[1]; - vout[1] = inv[2]*vin[0] + inv[3]*vin[1]; - - return TRUE; -} - -static int solve_3x3_matrix(double mx[9], double vin[3], double vout[3]) -{ - double inv[9]; - double det; - - det = (mx[0]*mx[4]*mx[8] + mx[1]*mx[5]*mx[6] + mx[2]*mx[3]*mx[7] - - mx[0]*mx[5]*mx[7] - mx[1]*mx[3]*mx[8] - mx[2]*mx[4]*mx[6]); - if (det == 0) - return FALSE; - - inv[0] = (mx[4]*mx[8] - mx[5]*mx[7]) / det; - inv[1] = (mx[2]*mx[7] - mx[1]*mx[8]) / det; - inv[2] = (mx[1]*mx[5] - mx[2]*mx[4]) / det; - inv[3] = (mx[5]*mx[6] - mx[3]*mx[8]) / det; - inv[4] = (mx[0]*mx[8] - mx[2]*mx[6]) / det; - inv[5] = (mx[2]*mx[3] - mx[0]*mx[5]) / det; - inv[6] = (mx[3]*mx[7] - mx[4]*mx[6]) / det; - inv[7] = (mx[1]*mx[6] - mx[0]*mx[7]) / det; - inv[8] = (mx[0]*mx[4] - mx[1]*mx[3]) / det; - - vout[0] = inv[0]*vin[0] + inv[1]*vin[1] + inv[2]*vin[2]; - vout[1] = inv[3]*vin[0] + inv[4]*vin[1] + inv[5]*vin[2]; - vout[2] = inv[6]*vin[0] + inv[7]*vin[1] + inv[8]*vin[2]; - - return TRUE; -} - /* Returns (into x,y) position of centre of face for rendering the text clue. */ static void face_text_pos(const game_drawstate *ds, const grid *g, - const grid_face *f, int *xret, int *yret) + grid_face *f, int *xret, int *yret) { - double xbest, ybest, bestdist; - int i, j, k, m; - grid_dot *edgedot1[3], *edgedot2[3]; - grid_dot *dots[3]; - int nedges, ndots; int faceindex = f - g->faces; /* @@ -3470,425 +2965,11 @@ static void face_text_pos(const game_drawstate *ds, const grid *g, } /* - * Otherwise, try to find the point in the polygon with the - * maximum distance to any edge or corner. - * - * This point must be in contact with at least three edges and/or - * vertices; so we iterate through all combinations of three of - * those, and find candidate points in each set. - * - * We don't actually iterate literally over _edges_, in the sense - * of grid_edge structures. Instead, we fill in edgedot1[] and - * edgedot2[] with a pair of dots adjacent in the face's list of - * vertices. This ensures that we get the edges in consistent - * orientation, which we could not do from the grid structure - * alone. (A moment's consideration of an order-3 vertex should - * make it clear that if a notional arrow was written on each - * edge, _at least one_ of the three faces bordering that vertex - * would have to have the two arrows tip-to-tip or tail-to-tail - * rather than tip-to-tail.) + * Otherwise, use the incentre computed by grid.c and convert it + * to screen coordinates. */ - nedges = ndots = 0; - bestdist = 0; - xbest = ybest = 0; - - for (i = 0; i+2 < 2*f->order; i++) { - if (i < f->order) { - edgedot1[nedges] = f->dots[i]; - edgedot2[nedges++] = f->dots[(i+1)%f->order]; - } else - dots[ndots++] = f->dots[i - f->order]; - - for (j = i+1; j+1 < 2*f->order; j++) { - if (j < f->order) { - edgedot1[nedges] = f->dots[j]; - edgedot2[nedges++] = f->dots[(j+1)%f->order]; - } else - dots[ndots++] = f->dots[j - f->order]; - - for (k = j+1; k < 2*f->order; k++) { - double cx[2], cy[2]; /* candidate positions */ - int cn = 0; /* number of candidates */ - - if (k < f->order) { - edgedot1[nedges] = f->dots[k]; - edgedot2[nedges++] = f->dots[(k+1)%f->order]; - } else - dots[ndots++] = f->dots[k - f->order]; - - /* - * Find a point, or pair of points, equidistant from - * all the specified edges and/or vertices. - */ - if (nedges == 3) { - /* - * Three edges. This is a linear matrix equation: - * each row of the matrix represents the fact that - * the point (x,y) we seek is at distance r from - * that edge, and we solve three of those - * simultaneously to obtain x,y,r. (We ignore r.) - */ - double matrix[9], vector[3], vector2[3]; - int m; - - for (m = 0; m < 3; m++) { - int x1 = edgedot1[m]->x, x2 = edgedot2[m]->x; - int y1 = edgedot1[m]->y, y2 = edgedot2[m]->y; - int dx = x2-x1, dy = y2-y1; - - /* - * ((x,y) - (x1,y1)) . (dy,-dx) = r |(dx,dy)| - * - * => x dy - y dx - r |(dx,dy)| = (x1 dy - y1 dx) - */ - matrix[3*m+0] = dy; - matrix[3*m+1] = -dx; - matrix[3*m+2] = -sqrt((double)dx*dx+(double)dy*dy); - vector[m] = (double)x1*dy - (double)y1*dx; - } - - if (solve_3x3_matrix(matrix, vector, vector2)) { - cx[cn] = vector2[0]; - cy[cn] = vector2[1]; - cn++; - } - } else if (nedges == 2) { - /* - * Two edges and a dot. This will end up in a - * quadratic equation. - * - * First, look at the two edges. Having our point - * be some distance r from both of them gives rise - * to a pair of linear equations in x,y,r of the - * form - * - * (x-x1) dy - (y-y1) dx = r sqrt(dx^2+dy^2) - * - * We eliminate r between those equations to give - * us a single linear equation in x,y describing - * the locus of points equidistant from both lines - * - i.e. the angle bisector. - * - * We then choose one of x,y to be a parameter t, - * and derive linear formulae for x,y,r in terms - * of t. This enables us to write down the - * circular equation (x-xd)^2+(y-yd)^2=r^2 as a - * quadratic in t; solving that and substituting - * in for x,y gives us two candidate points. - */ - double eqs[2][4]; /* a,b,c,d : ax+by+cr=d */ - double eq[3]; /* a,b,c: ax+by=c */ - double xt[2], yt[2], rt[2]; /* a,b: {x,y,r}=at+b */ - double q[3]; /* a,b,c: at^2+bt+c=0 */ - double disc; - - /* Find equations of the two input lines. */ - for (m = 0; m < 2; m++) { - int x1 = edgedot1[m]->x, x2 = edgedot2[m]->x; - int y1 = edgedot1[m]->y, y2 = edgedot2[m]->y; - int dx = x2-x1, dy = y2-y1; - - eqs[m][0] = dy; - eqs[m][1] = -dx; - eqs[m][2] = -sqrt(dx*dx+dy*dy); - eqs[m][3] = x1*dy - y1*dx; - } - - /* Derive the angle bisector by eliminating r. */ - eq[0] = eqs[0][0]*eqs[1][2] - eqs[1][0]*eqs[0][2]; - eq[1] = eqs[0][1]*eqs[1][2] - eqs[1][1]*eqs[0][2]; - eq[2] = eqs[0][3]*eqs[1][2] - eqs[1][3]*eqs[0][2]; - - /* Parametrise x and y in terms of some t. */ - if (abs(eq[0]) < abs(eq[1])) { - /* Parameter is x. */ - xt[0] = 1; xt[1] = 0; - yt[0] = -eq[0]/eq[1]; yt[1] = eq[2]/eq[1]; - } else { - /* Parameter is y. */ - yt[0] = 1; yt[1] = 0; - xt[0] = -eq[1]/eq[0]; xt[1] = eq[2]/eq[0]; - } - - /* Find a linear representation of r using eqs[0]. */ - rt[0] = -(eqs[0][0]*xt[0] + eqs[0][1]*yt[0])/eqs[0][2]; - rt[1] = (eqs[0][3] - eqs[0][0]*xt[1] - - eqs[0][1]*yt[1])/eqs[0][2]; - - /* Construct the quadratic equation. */ - q[0] = -rt[0]*rt[0]; - q[1] = -2*rt[0]*rt[1]; - q[2] = -rt[1]*rt[1]; - q[0] += xt[0]*xt[0]; - q[1] += 2*xt[0]*(xt[1]-dots[0]->x); - q[2] += (xt[1]-dots[0]->x)*(xt[1]-dots[0]->x); - q[0] += yt[0]*yt[0]; - q[1] += 2*yt[0]*(yt[1]-dots[0]->y); - q[2] += (yt[1]-dots[0]->y)*(yt[1]-dots[0]->y); - - /* And solve it. */ - disc = q[1]*q[1] - 4*q[0]*q[2]; - if (disc >= 0) { - double t; - - disc = sqrt(disc); - - t = (-q[1] + disc) / (2*q[0]); - cx[cn] = xt[0]*t + xt[1]; - cy[cn] = yt[0]*t + yt[1]; - cn++; - - t = (-q[1] - disc) / (2*q[0]); - cx[cn] = xt[0]*t + xt[1]; - cy[cn] = yt[0]*t + yt[1]; - cn++; - } - } else if (nedges == 1) { - /* - * Two dots and an edge. This one's another - * quadratic equation. - * - * The point we want must lie on the perpendicular - * bisector of the two dots; that much is obvious. - * So we can construct a parametrisation of that - * bisecting line, giving linear formulae for x,y - * in terms of t. We can also express the distance - * from the edge as such a linear formula. - * - * Then we set that equal to the radius of the - * circle passing through the two points, which is - * a Pythagoras exercise; that gives rise to a - * quadratic in t, which we solve. - */ - double xt[2], yt[2], rt[2]; /* a,b: {x,y,r}=at+b */ - double q[3]; /* a,b,c: at^2+bt+c=0 */ - double disc; - double halfsep; - - /* Find parametric formulae for x,y. */ - { - int x1 = dots[0]->x, x2 = dots[1]->x; - int y1 = dots[0]->y, y2 = dots[1]->y; - int dx = x2-x1, dy = y2-y1; - double d = sqrt((double)dx*dx + (double)dy*dy); - - xt[1] = (x1+x2)/2.0; - yt[1] = (y1+y2)/2.0; - /* It's convenient if we have t at standard scale. */ - xt[0] = -dy/d; - yt[0] = dx/d; - - /* Also note down half the separation between - * the dots, for use in computing the circle radius. */ - halfsep = 0.5*d; - } - - /* Find a parametric formula for r. */ - { - int x1 = edgedot1[0]->x, x2 = edgedot2[0]->x; - int y1 = edgedot1[0]->y, y2 = edgedot2[0]->y; - int dx = x2-x1, dy = y2-y1; - double d = sqrt((double)dx*dx + (double)dy*dy); - rt[0] = (xt[0]*dy - yt[0]*dx) / d; - rt[1] = ((xt[1]-x1)*dy - (yt[1]-y1)*dx) / d; - } - - /* Construct the quadratic equation. */ - q[0] = rt[0]*rt[0]; - q[1] = 2*rt[0]*rt[1]; - q[2] = rt[1]*rt[1]; - q[0] -= 1; - q[2] -= halfsep*halfsep; - - /* And solve it. */ - disc = q[1]*q[1] - 4*q[0]*q[2]; - if (disc >= 0) { - double t; - - disc = sqrt(disc); - - t = (-q[1] + disc) / (2*q[0]); - cx[cn] = xt[0]*t + xt[1]; - cy[cn] = yt[0]*t + yt[1]; - cn++; - - t = (-q[1] - disc) / (2*q[0]); - cx[cn] = xt[0]*t + xt[1]; - cy[cn] = yt[0]*t + yt[1]; - cn++; - } - } else if (nedges == 0) { - /* - * Three dots. This is another linear matrix - * equation, this time with each row of the matrix - * representing the perpendicular bisector between - * two of the points. Of course we only need two - * such lines to find their intersection, so we - * need only solve a 2x2 matrix equation. - */ - - double matrix[4], vector[2], vector2[2]; - int m; - - for (m = 0; m < 2; m++) { - int x1 = dots[m]->x, x2 = dots[m+1]->x; - int y1 = dots[m]->y, y2 = dots[m+1]->y; - int dx = x2-x1, dy = y2-y1; - - /* - * ((x,y) - (x1,y1)) . (dx,dy) = 1/2 |(dx,dy)|^2 - * - * => 2x dx + 2y dy = dx^2+dy^2 + (2 x1 dx + 2 y1 dy) - */ - matrix[2*m+0] = 2*dx; - matrix[2*m+1] = 2*dy; - vector[m] = ((double)dx*dx + (double)dy*dy + - 2.0*x1*dx + 2.0*y1*dy); - } - - if (solve_2x2_matrix(matrix, vector, vector2)) { - cx[cn] = vector2[0]; - cy[cn] = vector2[1]; - cn++; - } - } - - /* - * Now go through our candidate points and see if any - * of them are better than what we've got so far. - */ - for (m = 0; m < cn; m++) { - double x = cx[m], y = cy[m]; - - /* - * First, disqualify the point if it's not inside - * the polygon, which we work out by counting the - * edges to the right of the point. (For - * tiebreaking purposes when edges start or end on - * our y-coordinate or go right through it, we - * consider our point to be offset by a small - * _positive_ epsilon in both the x- and - * y-direction.) - */ - int e, in = 0; - for (e = 0; e < f->order; e++) { - int xs = f->edges[e]->dot1->x; - int xe = f->edges[e]->dot2->x; - int ys = f->edges[e]->dot1->y; - int ye = f->edges[e]->dot2->y; - if ((y >= ys && y < ye) || (y >= ye && y < ys)) { - /* - * The line goes past our y-position. Now we need - * to know if its x-coordinate when it does so is - * to our right. - * - * The x-coordinate in question is mathematically - * (y - ys) * (xe - xs) / (ye - ys), and we want - * to know whether (x - xs) >= that. Of course we - * avoid the division, so we can work in integers; - * to do this we must multiply both sides of the - * inequality by ye - ys, which means we must - * first check that's not negative. - */ - int num = xe - xs, denom = ye - ys; - if (denom < 0) { - num = -num; - denom = -denom; - } - if ((x - xs) * denom >= (y - ys) * num) - in ^= 1; - } - } - - if (in) { - double mindist = HUGE_VAL; - int e, d; - - /* - * This point is inside the polygon, so now we check - * its minimum distance to every edge and corner. - * First the corners ... - */ - for (d = 0; d < f->order; d++) { - int xp = f->dots[d]->x; - int yp = f->dots[d]->y; - double dx = x - xp, dy = y - yp; - double dist = dx*dx + dy*dy; - if (mindist > dist) - mindist = dist; - } - - /* - * ... and now also check the perpendicular distance - * to every edge, if the perpendicular lies between - * the edge's endpoints. - */ - for (e = 0; e < f->order; e++) { - int xs = f->edges[e]->dot1->x; - int xe = f->edges[e]->dot2->x; - int ys = f->edges[e]->dot1->y; - int ye = f->edges[e]->dot2->y; - - /* - * If s and e are our endpoints, and p our - * candidate circle centre, the foot of a - * perpendicular from p to the line se lies - * between s and e if and only if (p-s).(e-s) lies - * strictly between 0 and (e-s).(e-s). - */ - int edx = xe - xs, edy = ye - ys; - double pdx = x - xs, pdy = y - ys; - double pde = pdx * edx + pdy * edy; - long ede = (long)edx * edx + (long)edy * edy; - if (0 < pde && pde < ede) { - /* - * Yes, the nearest point on this edge is - * closer than either endpoint, so we must - * take it into account by measuring the - * perpendicular distance to the edge and - * checking its square against mindist. - */ - - double pdre = pdx * edy - pdy * edx; - double sqlen = pdre * pdre / ede; - - if (mindist > sqlen) - mindist = sqlen; - } - } - - /* - * Right. Now we know the biggest circle around this - * point, so we can check it against bestdist. - */ - if (bestdist < mindist) { - bestdist = mindist; - xbest = x; - ybest = y; - } - } - } - - if (k < f->order) - nedges--; - else - ndots--; - } - if (j < f->order) - nedges--; - else - ndots--; - } - if (i < f->order) - nedges--; - else - ndots--; - } - - assert(bestdist > 0); - - /* convert to screen coordinates. Round doubles to nearest. */ - grid_to_screen(ds, g, xbest+0.5, ybest+0.5, + grid_find_incentre(f); + grid_to_screen(ds, g, f->ix, f->iy, &ds->textx[faceindex], &ds->texty[faceindex]); *xret = ds->textx[faceindex]; @@ -3980,7 +3061,6 @@ static void game_redraw_line(drawing *dr, game_drawstate *ds, grid *g = state->game_grid; grid_edge *e = g->edges + i; int x1, x2, y1, y2; - int xmin, ymin, xmax, ymax; int line_colour; if (state->line_errors[i]) @@ -4000,11 +3080,6 @@ static void game_redraw_line(drawing *dr, game_drawstate *ds, grid_to_screen(ds, g, e->dot1->x, e->dot1->y, &x1, &y1); grid_to_screen(ds, g, e->dot2->x, e->dot2->y, &x2, &y2); - xmin = min(x1, x2); - xmax = max(x1, x2); - ymin = min(y1, y2); - ymax = max(y1, y2); - if (line_colour == COL_FAINT) { static int draw_faint_lines = -1; if (draw_faint_lines < 0) { @@ -4118,60 +3193,63 @@ static void game_redraw(drawing *dr, game_drawstate *ds, game_state *oldstate, * what needs doing, and the second actually does it. */ - if (!ds->started) + if (!ds->started) { redraw_everything = TRUE; - else { - - /* First, trundle through the faces. */ - for (i = 0; i < g->num_faces; i++) { - grid_face *f = g->faces + i; - int sides = f->order; - int clue_mistake; - int clue_satisfied; - int n = state->clues[i]; - if (n < 0) - continue; - - clue_mistake = (face_order(state, i, LINE_YES) > n || - face_order(state, i, LINE_NO ) > (sides-n)); - clue_satisfied = (face_order(state, i, LINE_YES) == n && - face_order(state, i, LINE_NO ) == (sides-n)); - - if (clue_mistake != ds->clue_error[i] || - clue_satisfied != ds->clue_satisfied[i]) { - ds->clue_error[i] = clue_mistake; - ds->clue_satisfied[i] = clue_satisfied; - if (nfaces == REDRAW_OBJECTS_LIMIT) - redraw_everything = TRUE; - else - faces[nfaces++] = i; - } - } + /* + * But we must still go through the upcoming loops, so that we + * set up stuff in ds correctly for the initial redraw. + */ + } - /* Work out what the flash state needs to be. */ - if (flashtime > 0 && - (flashtime <= FLASH_TIME/3 || - flashtime >= FLASH_TIME*2/3)) { - flash_changed = !ds->flashing; - ds->flashing = TRUE; - } else { - flash_changed = ds->flashing; - ds->flashing = FALSE; - } + /* First, trundle through the faces. */ + for (i = 0; i < g->num_faces; i++) { + grid_face *f = g->faces + i; + int sides = f->order; + int clue_mistake; + int clue_satisfied; + int n = state->clues[i]; + if (n < 0) + continue; - /* Now, trundle through the edges. */ - for (i = 0; i < g->num_edges; i++) { - char new_ds = - state->line_errors[i] ? DS_LINE_ERROR : state->lines[i]; - if (new_ds != ds->lines[i] || - (flash_changed && state->lines[i] == LINE_YES)) { - ds->lines[i] = new_ds; - if (nedges == REDRAW_OBJECTS_LIMIT) - redraw_everything = TRUE; - else - edges[nedges++] = i; - } - } + clue_mistake = (face_order(state, i, LINE_YES) > n || + face_order(state, i, LINE_NO ) > (sides-n)); + clue_satisfied = (face_order(state, i, LINE_YES) == n && + face_order(state, i, LINE_NO ) == (sides-n)); + + if (clue_mistake != ds->clue_error[i] || + clue_satisfied != ds->clue_satisfied[i]) { + ds->clue_error[i] = clue_mistake; + ds->clue_satisfied[i] = clue_satisfied; + if (nfaces == REDRAW_OBJECTS_LIMIT) + redraw_everything = TRUE; + else + faces[nfaces++] = i; + } + } + + /* Work out what the flash state needs to be. */ + if (flashtime > 0 && + (flashtime <= FLASH_TIME/3 || + flashtime >= FLASH_TIME*2/3)) { + flash_changed = !ds->flashing; + ds->flashing = TRUE; + } else { + flash_changed = ds->flashing; + ds->flashing = FALSE; + } + + /* Now, trundle through the edges. */ + for (i = 0; i < g->num_edges; i++) { + char new_ds = + state->line_errors[i] ? DS_LINE_ERROR : state->lines[i]; + if (new_ds != ds->lines[i] || + (flash_changed && state->lines[i] == LINE_YES)) { + ds->lines[i] = new_ds; + if (nedges == REDRAW_OBJECTS_LIMIT) + redraw_everything = TRUE; + else + edges[nedges++] = i; + } } /* Pass one is now done. Now we do the actual drawing. */ @@ -4218,9 +3296,9 @@ static float game_flash_length(game_state *oldstate, game_state *newstate, return 0.0F; } -static int game_is_solved(game_state *state) +static int game_status(game_state *state) { - return state->solved; + return state->solved ? +1 : 0; } static void game_print_size(game_params *params, float *x, float *y) @@ -4243,6 +3321,10 @@ static void game_print(drawing *dr, game_state *state, int tilesize) grid *g = state->game_grid; ds->tilesize = tilesize; + ds->textx = snewn(g->num_faces, int); + ds->texty = snewn(g->num_faces, int); + for (i = 0; i < g->num_faces; i++) + ds->textx[i] = ds->texty[i] = -1; for (i = 0; i < g->num_dots; i++) { int x, y; @@ -4312,6 +3394,9 @@ static void game_print(drawing *dr, game_state *state, int tilesize) } } } + + sfree(ds->textx); + sfree(ds->texty); } #ifdef COMBINED @@ -4349,7 +3434,7 @@ const struct game thegame = { game_redraw, game_anim_length, game_flash_length, - game_is_solved, + game_status, TRUE, FALSE, game_print_size, game_print, FALSE /* wants_statusbar */, FALSE, game_timing_state, @@ -4482,3 +3567,5 @@ int main(int argc, char **argv) } #endif + +/* vim: set shiftwidth=4 tabstop=8: */