/* * loopy.c: * * An implementation of the Nikoli game 'Loop the loop'. * (c) Mike Pinna, 2005, 2006 * Substantially rewritten to allowing for more general types of grid. * (c) Lambros Lambrou 2008 * * vim: set shiftwidth=4 :set textwidth=80: */ /* * Possible future solver enhancements: * * - There's an interesting deductive technique which makes use * of topology rather than just graph theory. Each _face_ in * the grid is either inside or outside the loop; you can tell * that two faces are on the same side of the loop if they're * separated by a LINE_NO (or, more generally, by a path * crossing no LINE_UNKNOWNs and an even number of LINE_YESes), * and on the opposite side of the loop if they're separated by * a LINE_YES (or an odd number of LINE_YESes and no * LINE_UNKNOWNs). Oh, and any face separated from the outside * of the grid by a LINE_YES or a LINE_NO is on the inside or * outside respectively. So if you can track this for all * faces, you figure out the state of the line between a pair * once their relative insideness is known. * + The way I envisage this working is simply to keep an edsf * of all _faces_, which indicates whether they're on * opposite sides of the loop from one another. We also * include a special entry in the edsf for the infinite * exterior "face". * + So, the simple way to do this is to just go through the * edges: every time we see an edge in a state other than * LINE_UNKNOWN which separates two faces that aren't in the * same edsf class, we can rectify that by merging the * classes. Then, conversely, an edge in LINE_UNKNOWN state * which separates two faces that _are_ in the same edsf * class can immediately have its state determined. * + But you can go one better, if you're prepared to loop * over all _pairs_ of edges. Suppose we have edges A and B, * which respectively separate faces A1,A2 and B1,B2. * Suppose that A,B are in the same edge-edsf class and that * A1,B1 (wlog) are in the same face-edsf class; then we can * immediately place A2,B2 into the same face-edsf class (as * each other, not as A1 and A2) one way round or the other. * And conversely again, if A1,B1 are in the same face-edsf * class and so are A2,B2, then we can put A,B into the same * face-edsf class. * * Of course, this deduction requires a quadratic-time * loop over all pairs of edges in the grid, so it should * be reserved until there's nothing easier left to be * done. * * - The generalised grid support has made me (SGT) notice a * possible extension to the loop-avoidance code. When you have * a path of connected edges such that no other edges at all * are incident on any vertex in the middle of the path - or, * alternatively, such that any such edges are already known to * be LINE_NO - then you know those edges are either all * LINE_YES or all LINE_NO. Hence you can mentally merge the * entire path into a single long curly edge for the purposes * of loop avoidance, and look directly at whether or not the * extreme endpoints of the path are connected by some other * route. I find this coming up fairly often when I play on the * octagonal grid setting, so it might be worth implementing in * the solver. * * - (Just a speed optimisation.) Consider some todo list queue where every * time we modify something we mark it for consideration by other bits of * the solver, to save iteration over things that have already been done. */ #include #include #include #include #include "rbassert.h" #include #include #include "puzzles.h" #include "tree234.h" #include "grid.h" #include "loopgen.h" /* Debugging options */ /* #define DEBUG_CACHES #define SHOW_WORKING #define DEBUG_DLINES */ /* ---------------------------------------------------------------------- * Struct, enum and function declarations */ enum { COL_BACKGROUND, COL_FOREGROUND, COL_LINEUNKNOWN, COL_HIGHLIGHT, COL_MISTAKE, COL_SATISFIED, COL_FAINT, NCOLOURS }; struct game_state { grid *game_grid; /* ref-counted (internally) */ /* Put -1 in a face that doesn't get a clue */ signed char *clues; /* Array of line states, to store whether each line is * YES, NO or UNKNOWN */ char *lines; unsigned char *line_errors; int exactly_one_loop; int solved; int cheated; /* Used in game_text_format(), so that it knows what type of * grid it's trying to render as ASCII text. */ int grid_type; }; enum solver_status { SOLVER_SOLVED, /* This is the only solution the solver could find */ SOLVER_MISTAKE, /* This is definitely not a solution */ SOLVER_AMBIGUOUS, /* This _might_ be an ambiguous solution */ SOLVER_INCOMPLETE /* This may be a partial solution */ }; /* ------ Solver state ------ */ typedef struct solver_state { game_state *state; enum solver_status solver_status; /* NB looplen is the number of dots that are joined together at a point, ie a * looplen of 1 means there are no lines to a particular dot */ int *looplen; /* Difficulty level of solver. Used by solver functions that want to * vary their behaviour depending on the requested difficulty level. */ int diff; /* caches */ char *dot_yes_count; char *dot_no_count; char *face_yes_count; char *face_no_count; char *dot_solved, *face_solved; int *dotdsf; /* Information for Normal level deductions: * For each dline, store a bitmask for whether we know: * (bit 0) at least one is YES * (bit 1) at most one is YES */ char *dlines; /* Hard level information */ int *linedsf; } solver_state; /* * Difficulty levels. I do some macro ickery here to ensure that my * enum and the various forms of my name list always match up. */ #define DIFFLIST(A) \ A(EASY,Easy,e) \ A(NORMAL,Normal,n) \ A(TRICKY,Tricky,t) \ A(HARD,Hard,h) #define ENUM(upper,title,lower) DIFF_ ## upper, #define TITLE(upper,title,lower) #title, #define ENCODE(upper,title,lower) #lower #define CONFIG(upper,title,lower) ":" #title enum { DIFFLIST(ENUM) DIFF_MAX }; static char const *const diffnames[] = { DIFFLIST(TITLE) }; static char const diffchars[] = DIFFLIST(ENCODE); #define DIFFCONFIG DIFFLIST(CONFIG) /* * Solver routines, sorted roughly in order of computational cost. * The solver will run the faster deductions first, and slower deductions are * only invoked when the faster deductions are unable to make progress. * Each function is associated with a difficulty level, so that the generated * puzzles are solvable by applying only the functions with the chosen * difficulty level or lower. */ #define SOLVERLIST(A) \ A(trivial_deductions, DIFF_EASY) \ A(dline_deductions, DIFF_NORMAL) \ A(linedsf_deductions, DIFF_HARD) \ A(loop_deductions, DIFF_EASY) #define SOLVER_FN_DECL(fn,diff) static int fn(solver_state *); #define SOLVER_FN(fn,diff) &fn, #define SOLVER_DIFF(fn,diff) diff, SOLVERLIST(SOLVER_FN_DECL) static int (*(solver_fns[]))(solver_state *) = { SOLVERLIST(SOLVER_FN) }; static int const solver_diffs[] = { SOLVERLIST(SOLVER_DIFF) }; static const int NUM_SOLVERS = sizeof(solver_diffs)/sizeof(*solver_diffs); struct game_params { int w, h; int diff; int type; }; /* line_drawstate is the same as line_state, but with the extra ERROR * possibility. The drawing code copies line_state to line_drawstate, * except in the case that the line is an error. */ enum line_state { LINE_YES, LINE_UNKNOWN, LINE_NO }; enum line_drawstate { DS_LINE_YES, DS_LINE_UNKNOWN, DS_LINE_NO, DS_LINE_ERROR }; #define OPP(line_state) \ (2 - line_state) struct game_drawstate { int started; int tilesize; int flashing; int *textx, *texty; char *lines; char *clue_error; char *clue_satisfied; }; static char *validate_desc(const game_params *params, const char *desc); static int dot_order(const game_state* state, int i, char line_type); static int face_order(const game_state* state, int i, char line_type); static solver_state *solve_game_rec(const solver_state *sstate); #ifdef DEBUG_CACHES static void check_caches(const solver_state* sstate); #else #define check_caches(s) #endif /* ------- List of grid generators ------- */ #define GRIDLIST(A) \ 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_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; } grid_size_limits[] = { GRIDLIST(GRID_SIZES) }; /* Generates a (dynamically allocated) new grid, according to the * type and size requested in params. Does nothing if the grid is already * generated. */ static grid *loopy_generate_grid(const game_params *params, const char *grid_desc) { return grid_new(grid_types[params->type], params->w, params->h, grid_desc); } /* ---------------------------------------------------------------------- * Preprocessor magic */ /* General constants */ #define PREFERRED_TILE_SIZE 32 #define BORDER(tilesize) ((tilesize) / 2) #define FLASH_TIME 0.5F #define BIT_SET(field, bit) ((field) & (1<<(bit))) #define SET_BIT(field, bit) (BIT_SET(field, bit) ? FALSE : \ ((field) |= (1<<(bit)), TRUE)) #define CLEAR_BIT(field, bit) (BIT_SET(field, bit) ? \ ((field) &= ~(1<<(bit)), TRUE) : FALSE) #define CLUE2CHAR(c) \ ((c < 0) ? ' ' : c < 10 ? c + '0' : c - 10 + 'A') /* ---------------------------------------------------------------------- * General struct manipulation and other straightforward code */ static game_state *dup_game(const game_state *state) { game_state *ret = snew(game_state); ret->game_grid = state->game_grid; ret->game_grid->refcount++; ret->solved = state->solved; ret->cheated = state->cheated; ret->clues = snewn(state->game_grid->num_faces, signed char); memcpy(ret->clues, state->clues, state->game_grid->num_faces); ret->lines = snewn(state->game_grid->num_edges, char); memcpy(ret->lines, state->lines, state->game_grid->num_edges); ret->line_errors = snewn(state->game_grid->num_edges, unsigned char); memcpy(ret->line_errors, state->line_errors, state->game_grid->num_edges); ret->exactly_one_loop = state->exactly_one_loop; ret->grid_type = state->grid_type; return ret; } static void free_game(game_state *state) { if (state) { grid_free(state->game_grid); sfree(state->clues); sfree(state->lines); sfree(state->line_errors); sfree(state); } } static solver_state *new_solver_state(const game_state *state, int diff) { int i; int num_dots = state->game_grid->num_dots; int num_faces = state->game_grid->num_faces; int num_edges = state->game_grid->num_edges; solver_state *ret = snew(solver_state); ret->state = dup_game(state); ret->solver_status = SOLVER_INCOMPLETE; ret->diff = diff; ret->dotdsf = snew_dsf(num_dots); ret->looplen = snewn(num_dots, int); for (i = 0; i < num_dots; i++) { ret->looplen[i] = 1; } ret->dot_solved = snewn(num_dots, char); ret->face_solved = snewn(num_faces, char); memset(ret->dot_solved, FALSE, num_dots); memset(ret->face_solved, FALSE, num_faces); ret->dot_yes_count = snewn(num_dots, char); memset(ret->dot_yes_count, 0, num_dots); ret->dot_no_count = snewn(num_dots, char); memset(ret->dot_no_count, 0, num_dots); ret->face_yes_count = snewn(num_faces, char); memset(ret->face_yes_count, 0, num_faces); ret->face_no_count = snewn(num_faces, char); memset(ret->face_no_count, 0, num_faces); if (diff < DIFF_NORMAL) { ret->dlines = NULL; } else { ret->dlines = snewn(2*num_edges, char); memset(ret->dlines, 0, 2*num_edges); } if (diff < DIFF_HARD) { ret->linedsf = NULL; } else { ret->linedsf = snew_dsf(state->game_grid->num_edges); } return ret; } static void free_solver_state(solver_state *sstate) { if (sstate) { free_game(sstate->state); sfree(sstate->dotdsf); sfree(sstate->looplen); sfree(sstate->dot_solved); sfree(sstate->face_solved); sfree(sstate->dot_yes_count); sfree(sstate->dot_no_count); sfree(sstate->face_yes_count); sfree(sstate->face_no_count); /* OK, because sfree(NULL) is a no-op */ sfree(sstate->dlines); sfree(sstate->linedsf); sfree(sstate); } } static solver_state *dup_solver_state(const solver_state *sstate) { game_state *state = sstate->state; int num_dots = state->game_grid->num_dots; int num_faces = state->game_grid->num_faces; int num_edges = state->game_grid->num_edges; solver_state *ret = snew(solver_state); ret->state = state = dup_game(sstate->state); ret->solver_status = sstate->solver_status; ret->diff = sstate->diff; ret->dotdsf = snewn(num_dots, int); ret->looplen = snewn(num_dots, int); memcpy(ret->dotdsf, sstate->dotdsf, num_dots * sizeof(int)); memcpy(ret->looplen, sstate->looplen, num_dots * sizeof(int)); ret->dot_solved = snewn(num_dots, char); ret->face_solved = snewn(num_faces, char); memcpy(ret->dot_solved, sstate->dot_solved, num_dots); memcpy(ret->face_solved, sstate->face_solved, num_faces); ret->dot_yes_count = snewn(num_dots, char); memcpy(ret->dot_yes_count, sstate->dot_yes_count, num_dots); ret->dot_no_count = snewn(num_dots, char); memcpy(ret->dot_no_count, sstate->dot_no_count, num_dots); ret->face_yes_count = snewn(num_faces, char); memcpy(ret->face_yes_count, sstate->face_yes_count, num_faces); ret->face_no_count = snewn(num_faces, char); memcpy(ret->face_no_count, sstate->face_no_count, num_faces); if (sstate->dlines) { ret->dlines = snewn(2*num_edges, char); memcpy(ret->dlines, sstate->dlines, 2*num_edges); } else { ret->dlines = NULL; } if (sstate->linedsf) { ret->linedsf = snewn(num_edges, int); memcpy(ret->linedsf, sstate->linedsf, num_edges * sizeof(int)); } else { ret->linedsf = NULL; } return ret; } static game_params *default_params(void) { game_params *ret = snew(game_params); #ifdef SLOW_SYSTEM ret->h = 7; ret->w = 7; #else ret->h = 10; ret->w = 10; #endif ret->diff = DIFF_EASY; ret->type = 0; return ret; } static game_params *dup_params(const game_params *params) { game_params *ret = snew(game_params); *ret = *params; /* structure copy */ return ret; } static const game_params presets[] = { #ifdef SMALL_SCREEN { 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 }, { 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 }; static int game_fetch_preset(int i, char **name, game_params **params) { game_params *tmppar; char buf[80]; if (i < 0 || i >= lenof(presets)) return FALSE; tmppar = snew(game_params); *tmppar = presets[i]; *params = tmppar; sprintf(buf, "%dx%d %s - %s", tmppar->h, tmppar->w, gridnames[tmppar->type], diffnames[tmppar->diff]); *name = dupstr(buf); return TRUE; } static void free_params(game_params *params) { sfree(params); } static void decode_params(game_params *params, char const *string) { params->h = params->w = atoi(string); params->diff = DIFF_EASY; while (*string && isdigit((unsigned char)*string)) string++; if (*string == 'x') { string++; params->h = atoi(string); while (*string && isdigit((unsigned char)*string)) string++; } if (*string == 't') { string++; params->type = atoi(string); while (*string && isdigit((unsigned char)*string)) string++; } if (*string == 'd') { int i; string++; for (i = 0; i < DIFF_MAX; i++) if (*string == diffchars[i]) params->diff = i; if (*string) string++; } } static char *encode_params(const game_params *params, int full) { char str[80]; sprintf(str, "%dx%dt%d", params->w, params->h, params->type); if (full) sprintf(str + strlen(str), "d%c", diffchars[params->diff]); return dupstr(str); } static config_item *game_configure(const game_params *params) { config_item *ret; char buf[80]; ret = snewn(5, config_item); ret[0].name = "Width"; ret[0].type = C_STRING; sprintf(buf, "%d", params->w); ret[0].sval = dupstr(buf); ret[0].ival = 0; ret[1].name = "Height"; ret[1].type = C_STRING; sprintf(buf, "%d", params->h); ret[1].sval = dupstr(buf); ret[1].ival = 0; ret[2].name = "Grid type"; ret[2].type = C_CHOICES; ret[2].sval = GRID_CONFIGS; ret[2].ival = params->type; ret[3].name = "Difficulty"; ret[3].type = C_CHOICES; ret[3].sval = DIFFCONFIG; ret[3].ival = params->diff; ret[4].name = NULL; ret[4].type = C_END; ret[4].sval = NULL; ret[4].ival = 0; return ret; } static game_params *custom_params(const config_item *cfg) { game_params *ret = snew(game_params); ret->w = atoi(cfg[0].sval); ret->h = atoi(cfg[1].sval); ret->type = cfg[2].ival; ret->diff = cfg[3].ival; return ret; } static char *validate_params(const game_params *params, int full) { if (params->type < 0 || params->type >= NUM_GRID_TYPES) return "Illegal grid type"; if (params->w < grid_size_limits[params->type].amin || params->h < grid_size_limits[params->type].amin) return grid_size_limits[params->type].aerr; if (params->w < grid_size_limits[params->type].omin && params->h < grid_size_limits[params->type].omin) return grid_size_limits[params->type].oerr; /* * This shouldn't be able to happen at all, since decode_params * and custom_params will never generate anything that isn't * within range. */ assert(params->diff < DIFF_MAX); return NULL; } /* Returns a newly allocated string describing the current puzzle */ static char *state_to_text(const game_state *state) { grid *g = state->game_grid; char *retval; int num_faces = g->num_faces; char *description = snewn(num_faces + 1, char); char *dp = description; int empty_count = 0; int i; for (i = 0; i < num_faces; i++) { if (state->clues[i] < 0) { if (empty_count > 25) { dp += sprintf(dp, "%c", (int)(empty_count + 'a' - 1)); empty_count = 0; } empty_count++; } else { if (empty_count) { dp += sprintf(dp, "%c", (int)(empty_count + 'a' - 1)); empty_count = 0; } dp += sprintf(dp, "%c", (int)CLUE2CHAR(state->clues[i])); } } if (empty_count) dp += sprintf(dp, "%c", (int)(empty_count + 'a' - 1)); retval = dupstr(description); sfree(description); 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(const 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(const game_params *params, const char *desc) { int count = 0; grid *g; 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')) { count++; continue; } if (*desc >= 'a') { count += *desc - 'a' + 1; continue; } return "Unknown character in description"; } if (count < g->num_faces) return "Description too short for board size"; if (count > g->num_faces) return "Description too long for board size"; grid_free(g); return NULL; } /* Sums the lengths of the numbers in range [0,n) */ /* See equivalent function in solo.c for justification of this. */ static int len_0_to_n(int n) { int len = 1; /* Counting 0 as a bit of a special case */ int i; for (i = 1; i < n; i *= 10) { len += max(n - i, 0); } return len; } static char *encode_solve_move(const game_state *state) { int len; char *ret, *p; int i; int num_edges = state->game_grid->num_edges; /* This is going to return a string representing the moves needed to set * every line in a grid to be the same as the ones in 'state'. The exact * length of this string is predictable. */ len = 1; /* Count the 'S' prefix */ /* Numbers in all lines */ len += len_0_to_n(num_edges); /* For each line we also have a letter */ len += num_edges; ret = snewn(len + 1, char); p = ret; p += sprintf(p, "S"); for (i = 0; i < num_edges; i++) { switch (state->lines[i]) { case LINE_YES: p += sprintf(p, "%dy", i); break; case LINE_NO: p += sprintf(p, "%dn", i); break; } } /* No point in doing sums like that if they're going to be wrong */ assert(strlen(ret) <= (size_t)len); return ret; } static game_ui *new_ui(const game_state *state) { return NULL; } static void free_ui(game_ui *ui) { } static char *encode_ui(const game_ui *ui) { return NULL; } static void decode_ui(game_ui *ui, const char *encoding) { } static void game_changed_state(game_ui *ui, const game_state *oldstate, const game_state *newstate) { } static void game_compute_size(const game_params *params, int tilesize, int *x, int *y) { 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); /* multiply first to minimise rounding error on integer division */ 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; } static void game_set_size(drawing *dr, game_drawstate *ds, const game_params *params, int tilesize) { ds->tilesize = tilesize; } static float *game_colours(frontend *fe, int *ncolours) { float *ret = snewn(3 * NCOLOURS, float); frontend_default_colour(fe, &ret[COL_BACKGROUND * 3]); ret[COL_FOREGROUND * 3 + 0] = 0.0F; ret[COL_FOREGROUND * 3 + 1] = 0.0F; ret[COL_FOREGROUND * 3 + 2] = 0.0F; /* * We want COL_LINEUNKNOWN to be a yellow which is a bit darker * than the background. (I previously set it to 0.8,0.8,0, but * found that this went badly with the 0.8,0.8,0.8 favoured as a * background by the Java frontend.) */ ret[COL_LINEUNKNOWN * 3 + 0] = ret[COL_BACKGROUND * 3 + 0] * 0.9F; ret[COL_LINEUNKNOWN * 3 + 1] = ret[COL_BACKGROUND * 3 + 1] * 0.9F; ret[COL_LINEUNKNOWN * 3 + 2] = 0.0F; ret[COL_HIGHLIGHT * 3 + 0] = 1.0F; ret[COL_HIGHLIGHT * 3 + 1] = 1.0F; ret[COL_HIGHLIGHT * 3 + 2] = 1.0F; ret[COL_MISTAKE * 3 + 0] = 1.0F; ret[COL_MISTAKE * 3 + 1] = 0.0F; ret[COL_MISTAKE * 3 + 2] = 0.0F; ret[COL_SATISFIED * 3 + 0] = 0.0F; ret[COL_SATISFIED * 3 + 1] = 0.0F; ret[COL_SATISFIED * 3 + 2] = 0.0F; /* We want the faint lines to be a bit darker than the background. * Except if the background is pretty dark already; then it ought to be a * bit lighter. Oy vey. */ ret[COL_FAINT * 3 + 0] = ret[COL_BACKGROUND * 3 + 0] * 0.9F; ret[COL_FAINT * 3 + 1] = ret[COL_BACKGROUND * 3 + 1] * 0.9F; ret[COL_FAINT * 3 + 2] = ret[COL_BACKGROUND * 3 + 2] * 0.9F; *ncolours = NCOLOURS; return ret; } static game_drawstate *game_new_drawstate(drawing *dr, const game_state *state) { struct game_drawstate *ds = snew(struct game_drawstate); int num_faces = state->game_grid->num_faces; int num_edges = state->game_grid->num_edges; int i; ds->tilesize = 0; ds->started = 0; ds->lines = snewn(num_edges, char); ds->clue_error = snewn(num_faces, char); ds->clue_satisfied = snewn(num_faces, char); ds->textx = snewn(num_faces, int); ds->texty = snewn(num_faces, int); ds->flashing = 0; memset(ds->lines, LINE_UNKNOWN, num_edges); memset(ds->clue_error, 0, num_faces); memset(ds->clue_satisfied, 0, num_faces); for (i = 0; i < num_faces; i++) ds->textx[i] = ds->texty[i] = -1; return ds; } 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); sfree(ds); } static int game_timing_state(const game_state *state, game_ui *ui) { return TRUE; } static float game_anim_length(const game_state *oldstate, const game_state *newstate, int dir, game_ui *ui) { return 0.0F; } static int game_can_format_as_text_now(const game_params *params) { if (params->type != 0) return FALSE; return TRUE; } static char *game_text_format(const game_state *state) { int w, h, W, H; int x, y, i; int cell_size; char *ret; grid *g = state->game_grid; grid_face *f; assert(state->grid_type == 0); /* Work out the basic size unit */ f = g->faces; /* first face */ assert(f->order == 4); /* The dots are ordered clockwise, so the two opposite * corners are guaranteed to span the square */ cell_size = abs(f->dots[0]->x - f->dots[2]->x); w = (g->highest_x - g->lowest_x) / cell_size; h = (g->highest_y - g->lowest_y) / cell_size; /* Create a blank "canvas" to "draw" on */ W = 2 * w + 2; H = 2 * h + 1; ret = snewn(W * H + 1, char); for (y = 0; y < H; y++) { for (x = 0; x < W-1; x++) { ret[y*W + x] = ' '; } ret[y*W + W-1] = '\n'; } ret[H*W] = '\0'; /* Fill in edge info */ for (i = 0; i < g->num_edges; i++) { grid_edge *e = g->edges + i; /* Cell coordinates, from (0,0) to (w-1,h-1) */ int x1 = (e->dot1->x - g->lowest_x) / cell_size; int x2 = (e->dot2->x - g->lowest_x) / cell_size; int y1 = (e->dot1->y - g->lowest_y) / cell_size; int y2 = (e->dot2->y - g->lowest_y) / cell_size; /* Midpoint, in canvas coordinates (canvas coordinates are just twice * cell coordinates) */ x = x1 + x2; y = y1 + y2; switch (state->lines[i]) { case LINE_YES: ret[y*W + x] = (y1 == y2) ? '-' : '|'; break; case LINE_NO: ret[y*W + x] = 'x'; break; case LINE_UNKNOWN: break; /* already a space */ default: assert(!"Illegal line state"); } } /* Fill in clues */ for (i = 0; i < g->num_faces; i++) { int x1, x2, y1, y2; f = g->faces + i; assert(f->order == 4); /* Cell coordinates, from (0,0) to (w-1,h-1) */ x1 = (f->dots[0]->x - g->lowest_x) / cell_size; x2 = (f->dots[2]->x - g->lowest_x) / cell_size; y1 = (f->dots[0]->y - g->lowest_y) / cell_size; y2 = (f->dots[2]->y - g->lowest_y) / cell_size; /* Midpoint, in canvas coordinates */ x = x1 + x2; y = y1 + y2; ret[y*W + x] = CLUE2CHAR(state->clues[i]); } return ret; } /* ---------------------------------------------------------------------- * Debug code */ #ifdef DEBUG_CACHES static void check_caches(const solver_state* sstate) { int i; const game_state *state = sstate->state; const grid *g = state->game_grid; for (i = 0; i < g->num_dots; i++) { assert(dot_order(state, i, LINE_YES) == sstate->dot_yes_count[i]); assert(dot_order(state, i, LINE_NO) == sstate->dot_no_count[i]); } for (i = 0; i < g->num_faces; i++) { assert(face_order(state, i, LINE_YES) == sstate->face_yes_count[i]); assert(face_order(state, i, LINE_NO) == sstate->face_no_count[i]); } } #if 0 #define check_caches(s) \ do { \ fprintf(stderr, "check_caches at line %d\n", __LINE__); \ check_caches(s); \ } while (0) #endif #endif /* DEBUG_CACHES */ /* ---------------------------------------------------------------------- * Solver utility functions */ /* Sets the line (with index i) to the new state 'line_new', and updates * the cached counts of any affected faces and dots. * Returns TRUE if this actually changed the line's state. */ static int solver_set_line(solver_state *sstate, int i, enum line_state line_new #ifdef SHOW_WORKING , const char *reason #endif ) { game_state *state = sstate->state; grid *g; grid_edge *e; assert(line_new != LINE_UNKNOWN); check_caches(sstate); if (state->lines[i] == line_new) { return FALSE; /* nothing changed */ } state->lines[i] = line_new; #ifdef SHOW_WORKING fprintf(stderr, "solver: set line [%d] to %s (%s)\n", i, line_new == LINE_YES ? "YES" : "NO", reason); #endif g = state->game_grid; e = g->edges + i; /* Update the cache for both dots and both faces affected by this. */ if (line_new == LINE_YES) { sstate->dot_yes_count[e->dot1 - g->dots]++; sstate->dot_yes_count[e->dot2 - g->dots]++; if (e->face1) { sstate->face_yes_count[e->face1 - g->faces]++; } if (e->face2) { sstate->face_yes_count[e->face2 - g->faces]++; } } else { sstate->dot_no_count[e->dot1 - g->dots]++; sstate->dot_no_count[e->dot2 - g->dots]++; if (e->face1) { sstate->face_no_count[e->face1 - g->faces]++; } if (e->face2) { sstate->face_no_count[e->face2 - g->faces]++; } } check_caches(sstate); return TRUE; } #ifdef SHOW_WORKING #define solver_set_line(a, b, c) \ solver_set_line(a, b, c, __FUNCTION__) #endif /* * Merge two dots due to the existence of an edge between them. * Updates the dsf tracking equivalence classes, and keeps track of * the length of path each dot is currently a part of. * Returns TRUE if the dots were already linked, ie if they are part of a * closed loop, and false otherwise. */ static int merge_dots(solver_state *sstate, int edge_index) { int i, j, len; grid *g = sstate->state->game_grid; grid_edge *e = g->edges + edge_index; i = e->dot1 - g->dots; j = e->dot2 - g->dots; i = dsf_canonify(sstate->dotdsf, i); j = dsf_canonify(sstate->dotdsf, j); if (i == j) { return TRUE; } else { len = sstate->looplen[i] + sstate->looplen[j]; dsf_merge(sstate->dotdsf, i, j); i = dsf_canonify(sstate->dotdsf, i); sstate->looplen[i] = len; return FALSE; } } /* Merge two lines because the solver has deduced that they must be either * identical or opposite. Returns TRUE if this is new information, otherwise * FALSE. */ static int merge_lines(solver_state *sstate, int i, int j, int inverse #ifdef SHOW_WORKING , const char *reason #endif ) { int inv_tmp; assert(i < sstate->state->game_grid->num_edges); assert(j < sstate->state->game_grid->num_edges); i = edsf_canonify(sstate->linedsf, i, &inv_tmp); inverse ^= inv_tmp; j = edsf_canonify(sstate->linedsf, j, &inv_tmp); inverse ^= inv_tmp; edsf_merge(sstate->linedsf, i, j, inverse); #ifdef SHOW_WORKING if (i != j) { fprintf(stderr, "%s [%d] [%d] %s(%s)\n", __FUNCTION__, i, j, inverse ? "inverse " : "", reason); } #endif return (i != j); } #ifdef SHOW_WORKING #define merge_lines(a, b, c, d) \ merge_lines(a, b, c, d, __FUNCTION__) #endif /* Count the number of lines of a particular type currently going into the * given dot. */ static int dot_order(const game_state* state, int dot, char line_type) { int n = 0; grid *g = state->game_grid; grid_dot *d = g->dots + dot; int i; for (i = 0; i < d->order; i++) { grid_edge *e = d->edges[i]; if (state->lines[e - g->edges] == line_type) ++n; } return n; } /* Count the number of lines of a particular type currently surrounding the * given face */ static int face_order(const game_state* state, int face, char line_type) { int n = 0; grid *g = state->game_grid; grid_face *f = g->faces + face; int i; for (i = 0; i < f->order; i++) { grid_edge *e = f->edges[i]; if (state->lines[e - g->edges] == line_type) ++n; } return n; } /* Set all lines bordering a dot of type old_type to type new_type * Return value tells caller whether this function actually did anything */ static int dot_setall(solver_state *sstate, int dot, char old_type, char new_type) { int retval = FALSE, r; game_state *state = sstate->state; grid *g; grid_dot *d; int i; if (old_type == new_type) return FALSE; g = state->game_grid; d = g->dots + dot; for (i = 0; i < d->order; i++) { int line_index = d->edges[i] - g->edges; if (state->lines[line_index] == old_type) { r = solver_set_line(sstate, line_index, new_type); assert(r == TRUE); retval = TRUE; } } return retval; } /* Set all lines bordering a face of type old_type to type new_type */ static int face_setall(solver_state *sstate, int face, char old_type, char new_type) { int retval = FALSE, r; game_state *state = sstate->state; grid *g; grid_face *f; int i; if (old_type == new_type) return FALSE; g = state->game_grid; f = g->faces + face; for (i = 0; i < f->order; i++) { int line_index = f->edges[i] - g->edges; if (state->lines[line_index] == old_type) { r = solver_set_line(sstate, line_index, new_type); assert(r == TRUE); retval = TRUE; } } return retval; } /* ---------------------------------------------------------------------- * Loop generation and clue removal */ static void add_full_clues(game_state *state, random_state *rs) { signed char *clues = state->clues; grid *g = state->game_grid; char *board = snewn(g->num_faces, char); int i; 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, g->num_faces); for (i = 0; i < g->num_edges; i++) { grid_edge *e = g->edges + i; grid_face *f1 = e->face1; grid_face *f2 = e->face2; enum face_colour c1 = FACE_COLOUR(f1); enum face_colour c2 = FACE_COLOUR(f2); assert(c1 != FACE_GREY); assert(c2 != FACE_GREY); if (c1 != c2) { if (f1) clues[f1 - g->faces]++; if (f2) clues[f2 - g->faces]++; } } sfree(board); } static int game_has_unique_soln(const game_state *state, int diff) { int ret; solver_state *sstate_new; solver_state *sstate = new_solver_state((game_state *)state, diff); sstate_new = solve_game_rec(sstate); assert(sstate_new->solver_status != SOLVER_MISTAKE); ret = (sstate_new->solver_status == SOLVER_SOLVED); free_solver_state(sstate_new); free_solver_state(sstate); return ret; } /* Remove clues one at a time at random. */ static game_state *remove_clues(game_state *state, random_state *rs, int diff) { int *face_list; int num_faces = state->game_grid->num_faces; game_state *ret = dup_game(state), *saved_ret; int n; /* We need to remove some clues. We'll do this by forming a list of all * available clues, shuffling it, then going along one at a * time clearing each clue in turn for which doing so doesn't render the * board unsolvable. */ face_list = snewn(num_faces, int); for (n = 0; n < num_faces; ++n) { face_list[n] = n; } shuffle(face_list, num_faces, sizeof(int), rs); for (n = 0; n < num_faces; ++n) { saved_ret = dup_game(ret); ret->clues[face_list[n]] = -1; if (game_has_unique_soln(ret, diff)) { free_game(saved_ret); } else { free_game(ret); ret = saved_ret; } } sfree(face_list); return ret; } static char *new_game_desc(const game_params *params, random_state *rs, char **aux, int interactive) { /* solution and description both use run-length encoding in obvious ways */ char *retval, *game_desc, *grid_desc; grid *g; game_state *state = snew(game_state); game_state *state_new; 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); state->exactly_one_loop = FALSE; state->grid_type = params->type; newboard_please: memset(state->lines, LINE_UNKNOWN, g->num_edges); memset(state->line_errors, 0, g->num_edges); state->solved = state->cheated = FALSE; /* Get a new random solvable board with all its clues filled in. Yes, this * can loop for ever if the params are suitably unfavourable, but * preventing games smaller than 4x4 seems to stop this happening */ do { add_full_clues(state, rs); } while (!game_has_unique_soln(state, params->diff)); state_new = remove_clues(state, rs, params->diff); free_game(state); state = state_new; if (params->diff > 0 && game_has_unique_soln(state, params->diff-1)) { #ifdef SHOW_WORKING fprintf(stderr, "Rejecting board, it is too easy\n"); #endif goto newboard_please; } 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; } static game_state *new_game(midend *me, const game_params *params, const char *desc) { int i; game_state *state = snew(game_state); int empties_to_make = 0; int n,n2; const char *dp; char *grid_desc; grid *g; int num_faces, num_edges; 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; state->clues = snewn(num_faces, signed char); state->lines = snewn(num_edges, char); state->line_errors = snewn(num_edges, unsigned char); state->exactly_one_loop = FALSE; state->solved = state->cheated = FALSE; state->grid_type = params->type; for (i = 0; i < num_faces; i++) { if (empties_to_make) { empties_to_make--; state->clues[i] = -1; continue; } assert(*dp); n = *dp - '0'; n2 = *dp - 'A' + 10; if (n >= 0 && n < 10) { state->clues[i] = n; } else if (n2 >= 10 && n2 < 36) { state->clues[i] = n2; } else { n = *dp - 'a' + 1; assert(n > 0); state->clues[i] = -1; empties_to_make = n - 1; } ++dp; } memset(state->lines, LINE_UNKNOWN, num_edges); memset(state->line_errors, 0, num_edges); return state; } /* Calculates the line_errors data, and checks if the current state is a * solution */ static int check_completion(game_state *state) { grid *g = state->game_grid; int i, ret; int *dsf, *component_state; int nsilly, nloop, npath, largest_comp, largest_size, total_pathsize; enum { COMP_NONE, COMP_LOOP, COMP_PATH, COMP_SILLY, COMP_EMPTY }; memset(state->line_errors, 0, g->num_edges); /* * Find loops in the grid, and determine whether the puzzle is * solved. * * Loopy is a bit more complicated than most puzzles that care * about loop detection. In most of them, loops are simply * _forbidden_; so the obviously right way to do * error-highlighting during play is to light up a graph edge red * iff it is part of a loop, which is exactly what the centralised * findloop.c makes easy. * * But Loopy is unusual in that you're _supposed_ to be making a * loop - and yet _some_ loops are not the right loop. So we need * to be more discriminating, by identifying loops one by one and * then thinking about which ones to highlight, and so findloop.c * isn't quite the right tool for the job in this case. * * Worse still, consider situations in which the grid contains a * loop and also some non-loop edges: there are some cases like * this in which the user's intuitive expectation would be to * highlight the loop (if you're only about half way through the * puzzle and have accidentally made a little loop in some corner * of the grid), and others in which they'd be more likely to * expect you to highlight the non-loop edges (if you've just * closed off a whole loop that you thought was the entire * solution, but forgot some disconnected edges in a corner * somewhere). So while it's easy enough to check whether the * solution is _right_, highlighting the wrong parts is a tricky * problem for this puzzle! * * I'd quite like, in some situations, to identify the largest * loop among the player's YES edges, and then light up everything * other than that. But finding the longest cycle in a graph is an * NP-complete problem (because, in particular, it must return a * Hamilton cycle if one exists). * * However, I think we can make the problem tractable by * exercising the Puzzles principle that it isn't absolutely * necessary to highlight _all_ errors: the key point is that by * the time the user has filled in the whole grid, they should * either have seen a completion flash, or have _some_ error * highlight showing them why the solution isn't right. So in * principle it would be *just about* good enough to highlight * just one error in the whole grid, if there was really no better * way. But we'd like to highlight as many errors as possible. * * In this case, I think the simple approach is to make use of the * fact that no vertex may have degree > 2, and that's really * simple to detect. So the plan goes like this: * * - Form the dsf of connected components of the graph vertices. * * - Highlight an error at any vertex with degree > 2. (It so * happens that we do this by lighting up all the edges * incident to that vertex, but that's an output detail.) * * - Any component that contains such a vertex is now excluded * from further consideration, because it already has a * highlight. * * - The remaining components have no vertex with degree > 2, and * hence they all consist of either a simple loop, or a simple * path with two endpoints. * * - For these purposes, group together all the paths and imagine * them to be a single component (because in most normal * situations the player will gradually build up the solution * _not_ all in one connected segment, but as lots of separate * little path pieces that gradually connect to each other). * * - After doing that, if there is exactly one (sensible) * component - be it a collection of paths or a loop - then * highlight no further edge errors. (The former case is normal * during play, and the latter is a potentially solved puzzle.) * * - Otherwise, find the largest of the sensible components, * leave that one unhighlighted, and light the rest up in red. */ dsf = snew_dsf(g->num_dots); /* Build the dsf. */ for (i = 0; i < g->num_edges; i++) { if (state->lines[i] == LINE_YES) { grid_edge *e = g->edges + i; int d1 = e->dot1 - g->dots, d2 = e->dot2 - g->dots; dsf_merge(dsf, d1, d2); } } /* Initialise a state variable for each connected component. */ component_state = snewn(g->num_dots, int); for (i = 0; i < g->num_dots; i++) { if (dsf_canonify(dsf, i) == i) component_state[i] = COMP_LOOP; else component_state[i] = COMP_NONE; } /* Check for dots with degree > 3. Here we also spot dots of * degree 1 in which the user has marked all the non-edges as * LINE_NO, because those are also clear vertex-level errors, so * we give them the same treatment of excluding their connected * component from the subsequent loop analysis. */ for (i = 0; i < g->num_dots; i++) { int comp = dsf_canonify(dsf, i); int yes = dot_order(state, i, LINE_YES); int unknown = dot_order(state, i, LINE_UNKNOWN); if ((yes == 1 && unknown == 0) || (yes >= 3)) { /* violation, so mark all YES edges as errors */ grid_dot *d = g->dots + i; int j; for (j = 0; j < d->order; j++) { int e = d->edges[j] - g->edges; if (state->lines[e] == LINE_YES) state->line_errors[e] = TRUE; } /* And mark this component as not worthy of further * consideration. */ component_state[comp] = COMP_SILLY; } else if (yes == 0) { /* A completely isolated dot must also be excluded it from * the subsequent loop highlighting pass, but we tag it * with a different enum value to avoid it counting * towards the components that inhibit returning a win * status. */ component_state[comp] = COMP_EMPTY; } else if (yes == 1) { /* A dot with degree 1 that didn't fall into the 'clearly * erroneous' case above indicates that this connected * component will be a path rather than a loop - unless * something worse elsewhere in the component has * classified it as silly. */ if (component_state[comp] != COMP_SILLY) component_state[comp] = COMP_PATH; } } /* Count up the components. Also, find the largest sensible * component. (Tie-breaking condition is derived from the order of * vertices in the grid data structure, which is fairly arbitrary * but at least stays stable throughout the game.) */ nsilly = nloop = npath = 0; total_pathsize = 0; largest_comp = largest_size = -1; for (i = 0; i < g->num_dots; i++) { if (component_state[i] == COMP_SILLY) { nsilly++; } else if (component_state[i] == COMP_PATH) { total_pathsize += dsf_size(dsf, i); npath = 1; } else if (component_state[i] == COMP_LOOP) { int this_size; nloop++; if ((this_size = dsf_size(dsf, i)) > largest_size) { largest_comp = i; largest_size = this_size; } } } if (largest_size < total_pathsize) { largest_comp = -1; /* means the paths */ largest_size = total_pathsize; } if (nloop > 0 && nloop + npath > 1) { /* * If there are at least two sensible components including at * least one loop, highlight all edges in every sensible * component that is not the largest one. */ for (i = 0; i < g->num_edges; i++) { if (state->lines[i] == LINE_YES) { grid_edge *e = g->edges + i; int d1 = e->dot1 - g->dots; /* either endpoint is good enough */ int comp = dsf_canonify(dsf, d1); if ((component_state[comp] == COMP_PATH && -1 != largest_comp) || (component_state[comp] == COMP_LOOP && comp != largest_comp)) state->line_errors[i] = TRUE; } } } if (nloop == 1 && npath == 0 && nsilly == 0) { /* * If there is exactly one component and it is a loop, then * the puzzle is potentially complete, so check the clues. */ ret = TRUE; for (i = 0; i < g->num_faces; i++) { int c = state->clues[i]; if (c >= 0 && face_order(state, i, LINE_YES) != c) { ret = FALSE; break; } } /* * Also, whether or not the puzzle is actually complete, set * the flag that says this game_state has exactly one loop and * nothing else, which will be used to vary the semantics of * clue highlighting at display time. */ state->exactly_one_loop = TRUE; } else { ret = FALSE; state->exactly_one_loop = FALSE; } sfree(component_state); sfree(dsf); return ret; } /* ---------------------------------------------------------------------- * Solver logic * * Our solver modes operate as follows. Each mode also uses the modes above it. * * Easy Mode * Just implement the rules of the game. * * Normal and Tricky Modes * For each (adjacent) pair of lines through each dot we store a bit for * whether at least one of them is on and whether at most one is on. (If we * know both or neither is on that's already stored more directly.) * * Advanced Mode * Use edsf data structure to make equivalence classes of lines that are * known identical to or opposite to one another. */ /* DLines: * For general grids, we consider "dlines" to be pairs of lines joined * at a dot. The lines must be adjacent around the dot, so we can think of * a dline as being a dot+face combination. Or, a dot+edge combination where * the second edge is taken to be the next clockwise edge from the dot. * Original loopy code didn't have this extra restriction of the lines being * adjacent. From my tests with square grids, this extra restriction seems to * take little, if anything, away from the quality of the puzzles. * A dline can be uniquely identified by an edge/dot combination, given that * a dline-pair always goes clockwise around its common dot. The edge/dot * combination can be represented by an edge/bool combination - if bool is * TRUE, use edge->dot1 else use edge->dot2. So the total number of dlines is * exactly twice the number of edges in the grid - although the dlines * spanning the infinite face are not all that useful to the solver. * Note that, by convention, a dline goes clockwise around its common dot, * which means the dline goes anti-clockwise around its common face. */ /* Helper functions for obtaining an index into an array of dlines, given * various information. We assume the grid layout conventions about how * the various lists are interleaved - see grid_make_consistent() for * details. */ /* i points to the first edge of the dline pair, reading clockwise around * the dot. */ static int dline_index_from_dot(grid *g, grid_dot *d, int i) { grid_edge *e = d->edges[i]; int ret; #ifdef DEBUG_DLINES grid_edge *e2; int i2 = i+1; if (i2 == d->order) i2 = 0; e2 = d->edges[i2]; #endif ret = 2 * (e - g->edges) + ((e->dot1 == d) ? 1 : 0); #ifdef DEBUG_DLINES printf("dline_index_from_dot: d=%d,i=%d, edges [%d,%d] - %d\n", (int)(d - g->dots), i, (int)(e - g->edges), (int)(e2 - g->edges), ret); #endif return ret; } /* i points to the second edge of the dline pair, reading clockwise around * the face. That is, the edges of the dline, starting at edge{i}, read * anti-clockwise around the face. By layout conventions, the common dot * of the dline will be f->dots[i] */ static int dline_index_from_face(grid *g, grid_face *f, int i) { grid_edge *e = f->edges[i]; grid_dot *d = f->dots[i]; int ret; #ifdef DEBUG_DLINES grid_edge *e2; int i2 = i - 1; if (i2 < 0) i2 += f->order; e2 = f->edges[i2]; #endif ret = 2 * (e - g->edges) + ((e->dot1 == d) ? 1 : 0); #ifdef DEBUG_DLINES printf("dline_index_from_face: f=%d,i=%d, edges [%d,%d] - %d\n", (int)(f - g->faces), i, (int)(e - g->edges), (int)(e2 - g->edges), ret); #endif return ret; } static int is_atleastone(const char *dline_array, int index) { return BIT_SET(dline_array[index], 0); } static int set_atleastone(char *dline_array, int index) { return SET_BIT(dline_array[index], 0); } static int is_atmostone(const char *dline_array, int index) { return BIT_SET(dline_array[index], 1); } static int set_atmostone(char *dline_array, int index) { return SET_BIT(dline_array[index], 1); } static void array_setall(char *array, char from, char to, int len) { char *p = array, *p_old = p; int len_remaining = len; while ((p = memchr(p, from, len_remaining))) { *p = to; len_remaining -= p - p_old; p_old = p; } } /* Helper, called when doing dline dot deductions, in the case where we * have 4 UNKNOWNs, and two of them (adjacent) have *exactly* one YES between * them (because of dline atmostone/atleastone). * On entry, edge points to the first of these two UNKNOWNs. This function * will find the opposite UNKNOWNS (if they are adjacent to one another) * and set their corresponding dline to atleastone. (Setting atmostone * already happens in earlier dline deductions) */ static int dline_set_opp_atleastone(solver_state *sstate, grid_dot *d, int edge) { game_state *state = sstate->state; grid *g = state->game_grid; int N = d->order; int opp, opp2; for (opp = 0; opp < N; opp++) { int opp_dline_index; if (opp == edge || opp == edge+1 || opp == edge-1) continue; if (opp == 0 && edge == N-1) continue; if (opp == N-1 && edge == 0) continue; opp2 = opp + 1; if (opp2 == N) opp2 = 0; /* Check if opp, opp2 point to LINE_UNKNOWNs */ if (state->lines[d->edges[opp] - g->edges] != LINE_UNKNOWN) continue; if (state->lines[d->edges[opp2] - g->edges] != LINE_UNKNOWN) continue; /* Found opposite UNKNOWNS and they're next to each other */ opp_dline_index = dline_index_from_dot(g, d, opp); return set_atleastone(sstate->dlines, opp_dline_index); } return FALSE; } /* Set pairs of lines around this face which are known to be identical, to * the given line_state */ static int face_setall_identical(solver_state *sstate, int face_index, enum line_state line_new) { /* can[dir] contains the canonical line associated with the line in * direction dir from the square in question. Similarly inv[dir] is * whether or not the line in question is inverse to its canonical * element. */ int retval = FALSE; game_state *state = sstate->state; grid *g = state->game_grid; grid_face *f = g->faces + face_index; int N = f->order; int i, j; int can1, can2, inv1, inv2; for (i = 0; i < N; i++) { int line1_index = f->edges[i] - g->edges; if (state->lines[line1_index] != LINE_UNKNOWN) continue; for (j = i + 1; j < N; j++) { int line2_index = f->edges[j] - g->edges; if (state->lines[line2_index] != LINE_UNKNOWN) continue; /* Found two UNKNOWNS */ can1 = edsf_canonify(sstate->linedsf, line1_index, &inv1); can2 = edsf_canonify(sstate->linedsf, line2_index, &inv2); if (can1 == can2 && inv1 == inv2) { solver_set_line(sstate, line1_index, line_new); solver_set_line(sstate, line2_index, line_new); } } } return retval; } /* Given a dot or face, and a count of LINE_UNKNOWNs, find them and * return the edge indices into e. */ static void find_unknowns(game_state *state, grid_edge **edge_list, /* Edge list to search (from a face or a dot) */ int expected_count, /* Number of UNKNOWNs (comes from solver's cache) */ int *e /* Returned edge indices */) { int c = 0; grid *g = state->game_grid; while (c < expected_count) { int line_index = *edge_list - g->edges; if (state->lines[line_index] == LINE_UNKNOWN) { e[c] = line_index; c++; } ++edge_list; } } /* If we have a list of edges, and we know whether the number of YESs should * be odd or even, and there are only a few UNKNOWNs, we can do some simple * linedsf deductions. This can be used for both face and dot deductions. * Returns the difficulty level of the next solver that should be used, * or DIFF_MAX if no progress was made. */ static int parity_deductions(solver_state *sstate, grid_edge **edge_list, /* Edge list (from a face or a dot) */ int total_parity, /* Expected number of YESs modulo 2 (either 0 or 1) */ int unknown_count) { game_state *state = sstate->state; int diff = DIFF_MAX; int *linedsf = sstate->linedsf; if (unknown_count == 2) { /* Lines are known alike/opposite, depending on inv. */ int e[2]; find_unknowns(state, edge_list, 2, e); if (merge_lines(sstate, e[0], e[1], total_parity)) diff = min(diff, DIFF_HARD); } else if (unknown_count == 3) { int e[3]; int can[3]; /* canonical edges */ int inv[3]; /* whether can[x] is inverse to e[x] */ find_unknowns(state, edge_list, 3, e); can[0] = edsf_canonify(linedsf, e[0], inv); can[1] = edsf_canonify(linedsf, e[1], inv+1); can[2] = edsf_canonify(linedsf, e[2], inv+2); if (can[0] == can[1]) { if (solver_set_line(sstate, e[2], (total_parity^inv[0]^inv[1]) ? LINE_YES : LINE_NO)) diff = min(diff, DIFF_EASY); } if (can[0] == can[2]) { if (solver_set_line(sstate, e[1], (total_parity^inv[0]^inv[2]) ? LINE_YES : LINE_NO)) diff = min(diff, DIFF_EASY); } if (can[1] == can[2]) { if (solver_set_line(sstate, e[0], (total_parity^inv[1]^inv[2]) ? LINE_YES : LINE_NO)) diff = min(diff, DIFF_EASY); } } else if (unknown_count == 4) { int e[4]; int can[4]; /* canonical edges */ int inv[4]; /* whether can[x] is inverse to e[x] */ find_unknowns(state, edge_list, 4, e); can[0] = edsf_canonify(linedsf, e[0], inv); can[1] = edsf_canonify(linedsf, e[1], inv+1); can[2] = edsf_canonify(linedsf, e[2], inv+2); can[3] = edsf_canonify(linedsf, e[3], inv+3); if (can[0] == can[1]) { if (merge_lines(sstate, e[2], e[3], total_parity^inv[0]^inv[1])) diff = min(diff, DIFF_HARD); } else if (can[0] == can[2]) { if (merge_lines(sstate, e[1], e[3], total_parity^inv[0]^inv[2])) diff = min(diff, DIFF_HARD); } else if (can[0] == can[3]) { if (merge_lines(sstate, e[1], e[2], total_parity^inv[0]^inv[3])) diff = min(diff, DIFF_HARD); } else if (can[1] == can[2]) { if (merge_lines(sstate, e[0], e[3], total_parity^inv[1]^inv[2])) diff = min(diff, DIFF_HARD); } else if (can[1] == can[3]) { if (merge_lines(sstate, e[0], e[2], total_parity^inv[1]^inv[3])) diff = min(diff, DIFF_HARD); } else if (can[2] == can[3]) { if (merge_lines(sstate, e[0], e[1], total_parity^inv[2]^inv[3])) diff = min(diff, DIFF_HARD); } } return diff; } /* * These are the main solver functions. * * Their return values are diff values corresponding to the lowest mode solver * that would notice the work that they have done. For example if the normal * mode solver adds actual lines or crosses, it will return DIFF_EASY as the * easy mode solver might be able to make progress using that. It doesn't make * sense for one of them to return a diff value higher than that of the * function itself. * * Each function returns the lowest value it can, as early as possible, in * order to try and pass as much work as possible back to the lower level * solvers which progress more quickly. */ /* PROPOSED NEW DESIGN: * We have a work queue consisting of 'events' notifying us that something has * happened that a particular solver mode might be interested in. For example * the hard mode solver might do something that helps the normal mode solver at * dot [x,y] in which case it will enqueue an event recording this fact. Then * we pull events off the work queue, and hand each in turn to the solver that * is interested in them. If a solver reports that it failed we pass the same * event on to progressively more advanced solvers and the loop detector. Once * we've exhausted an event, or it has helped us progress, we drop it and * continue to the next one. The events are sorted first in order of solver * complexity (easy first) then order of insertion (oldest first). * Once we run out of events we loop over each permitted solver in turn * (easiest first) until either a deduction is made (and an event therefore * emerges) or no further deductions can be made (in which case we've failed). * * QUESTIONS: * * How do we 'loop over' a solver when both dots and squares are concerned. * Answer: first all squares then all dots. */ static int trivial_deductions(solver_state *sstate) { int i, current_yes, current_no; game_state *state = sstate->state; grid *g = state->game_grid; int diff = DIFF_MAX; /* Per-face deductions */ for (i = 0; i < g->num_faces; i++) { grid_face *f = g->faces + i; if (sstate->face_solved[i]) continue; current_yes = sstate->face_yes_count[i]; current_no = sstate->face_no_count[i]; if (current_yes + current_no == f->order) { sstate->face_solved[i] = TRUE; continue; } if (state->clues[i] < 0) continue; /* * This code checks whether the numeric clue on a face is so * large as to permit all its remaining LINE_UNKNOWNs to be * filled in as LINE_YES, or alternatively so small as to * permit them all to be filled in as LINE_NO. */ if (state->clues[i] < current_yes) { sstate->solver_status = SOLVER_MISTAKE; return DIFF_EASY; } if (state->clues[i] == current_yes) { if (face_setall(sstate, i, LINE_UNKNOWN, LINE_NO)) diff = min(diff, DIFF_EASY); sstate->face_solved[i] = TRUE; continue; } if (f->order - state->clues[i] < current_no) { sstate->solver_status = SOLVER_MISTAKE; return DIFF_EASY; } if (f->order - state->clues[i] == current_no) { if (face_setall(sstate, i, LINE_UNKNOWN, LINE_YES)) diff = min(diff, DIFF_EASY); sstate->face_solved[i] = TRUE; continue; } if (f->order - state->clues[i] == current_no + 1 && f->order - current_yes - current_no > 2) { /* * One small refinement to the above: we also look for any * adjacent pair of LINE_UNKNOWNs around the face with * some LINE_YES incident on it from elsewhere. If we find * one, then we know that pair of LINE_UNKNOWNs can't * _both_ be LINE_YES, and hence that pushes us one line * closer to being able to determine all the rest. */ int j, k, e1, e2, e, d; for (j = 0; j < f->order; j++) { e1 = f->edges[j] - g->edges; e2 = f->edges[j+1 < f->order ? j+1 : 0] - g->edges; if (g->edges[e1].dot1 == g->edges[e2].dot1 || g->edges[e1].dot1 == g->edges[e2].dot2) { d = g->edges[e1].dot1 - g->dots; } else { assert(g->edges[e1].dot2 == g->edges[e2].dot1 || g->edges[e1].dot2 == g->edges[e2].dot2); d = g->edges[e1].dot2 - g->dots; } if (state->lines[e1] == LINE_UNKNOWN && state->lines[e2] == LINE_UNKNOWN) { for (k = 0; k < g->dots[d].order; k++) { int e = g->dots[d].edges[k] - g->edges; if (state->lines[e] == LINE_YES) goto found; /* multi-level break */ } } } continue; found: /* * If we get here, we've found such a pair of edges, and * they're e1 and e2. */ for (j = 0; j < f->order; j++) { e = f->edges[j] - g->edges; if (state->lines[e] == LINE_UNKNOWN && e != e1 && e != e2) { int r = solver_set_line(sstate, e, LINE_YES); assert(r); diff = min(diff, DIFF_EASY); } } } } check_caches(sstate); /* Per-dot deductions */ for (i = 0; i < g->num_dots; i++) { grid_dot *d = g->dots + i; int yes, no, unknown; if (sstate->dot_solved[i]) continue; yes = sstate->dot_yes_count[i]; no = sstate->dot_no_count[i]; unknown = d->order - yes - no; if (yes == 0) { if (unknown == 0) { sstate->dot_solved[i] = TRUE; } else if (unknown == 1) { dot_setall(sstate, i, LINE_UNKNOWN, LINE_NO); diff = min(diff, DIFF_EASY); sstate->dot_solved[i] = TRUE; } } else if (yes == 1) { if (unknown == 0) { sstate->solver_status = SOLVER_MISTAKE; return DIFF_EASY; } else if (unknown == 1) { dot_setall(sstate, i, LINE_UNKNOWN, LINE_YES); diff = min(diff, DIFF_EASY); } } else if (yes == 2) { if (unknown > 0) { dot_setall(sstate, i, LINE_UNKNOWN, LINE_NO); diff = min(diff, DIFF_EASY); } sstate->dot_solved[i] = TRUE; } else { sstate->solver_status = SOLVER_MISTAKE; return DIFF_EASY; } } check_caches(sstate); return diff; } static int dline_deductions(solver_state *sstate) { game_state *state = sstate->state; grid *g = state->game_grid; char *dlines = sstate->dlines; int i; int diff = DIFF_MAX; /* ------ Face deductions ------ */ /* Given a set of dline atmostone/atleastone constraints, need to figure * out if we can deduce any further info. For more general faces than * squares, this turns out to be a tricky problem. * The approach taken here is to define (per face) NxN matrices: * "maxs" and "mins". * The entries maxs(j,k) and mins(j,k) define the upper and lower limits * for the possible number of edges that are YES between positions j and k * going clockwise around the face. Can think of j and k as marking dots * around the face (recall the labelling scheme: edge0 joins dot0 to dot1, * edge1 joins dot1 to dot2 etc). * Trivially, mins(j,j) = maxs(j,j) = 0, and we don't even bother storing * these. mins(j,j+1) and maxs(j,j+1) are determined by whether edge{j} * is YES, NO or UNKNOWN. mins(j,j+2) and maxs(j,j+2) are related to * the dline atmostone/atleastone status for edges j and j+1. * * Then we calculate the remaining entries recursively. We definitely * know that * mins(j,k) >= { mins(j,u) + mins(u,k) } for any u between j and k. * This is because any valid placement of YESs between j and k must give * a valid placement between j and u, and also between u and k. * I believe it's sufficient to use just the two values of u: * j+1 and j+2. Seems to work well in practice - the bounds we compute * are rigorous, even if they might not be best-possible. * * Once we have maxs and mins calculated, we can make inferences about * each dline{j,j+1} by looking at the possible complementary edge-counts * mins(j+2,j) and maxs(j+2,j) and comparing these with the face clue. * As well as dlines, we can make similar inferences about single edges. * For example, consider a pentagon with clue 3, and we know at most one * of (edge0, edge1) is YES, and at most one of (edge2, edge3) is YES. * We could then deduce edge4 is YES, because maxs(0,4) would be 2, so * that final edge would have to be YES to make the count up to 3. */ /* Much quicker to allocate arrays on the stack than the heap, so * define the largest possible face size, and base our array allocations * on that. We check this with an assertion, in case someone decides to * make a grid which has larger faces than this. Note, this algorithm * could get quite expensive if there are many large faces. */ #define MAX_FACE_SIZE 12 for (i = 0; i < g->num_faces; i++) { int maxs[MAX_FACE_SIZE][MAX_FACE_SIZE]; int mins[MAX_FACE_SIZE][MAX_FACE_SIZE]; grid_face *f = g->faces + i; int N = f->order; int j,m; int clue = state->clues[i]; assert(N <= MAX_FACE_SIZE); if (sstate->face_solved[i]) continue; if (clue < 0) continue; /* Calculate the (j,j+1) entries */ for (j = 0; j < N; j++) { int edge_index = f->edges[j] - g->edges; int dline_index; enum line_state line1 = state->lines[edge_index]; enum line_state line2; int tmp; int k = j + 1; if (k >= N) k = 0; maxs[j][k] = (line1 == LINE_NO) ? 0 : 1; mins[j][k] = (line1 == LINE_YES) ? 1 : 0; /* Calculate the (j,j+2) entries */ dline_index = dline_index_from_face(g, f, k); edge_index = f->edges[k] - g->edges; line2 = state->lines[edge_index]; k++; if (k >= N) k = 0; /* max */ tmp = 2; if (line1 == LINE_NO) tmp--; if (line2 == LINE_NO) tmp--; if (tmp == 2 && is_atmostone(dlines, dline_index)) tmp = 1; maxs[j][k] = tmp; /* min */ tmp = 0; if (line1 == LINE_YES) tmp++; if (line2 == LINE_YES) tmp++; if (tmp == 0 && is_atleastone(dlines, dline_index)) tmp = 1; mins[j][k] = tmp; } /* Calculate the (j,j+m) entries for m between 3 and N-1 */ for (m = 3; m < N; m++) { for (j = 0; j < N; j++) { int k = j + m; int u = j + 1; int v = j + 2; int tmp; if (k >= N) k -= N; if (u >= N) u -= N; if (v >= N) v -= N; maxs[j][k] = maxs[j][u] + maxs[u][k]; mins[j][k] = mins[j][u] + mins[u][k]; tmp = maxs[j][v] + maxs[v][k]; maxs[j][k] = min(maxs[j][k], tmp); tmp = mins[j][v] + mins[v][k]; mins[j][k] = max(mins[j][k], tmp); } } /* See if we can make any deductions */ for (j = 0; j < N; j++) { int k; grid_edge *e = f->edges[j]; int line_index = e - g->edges; int dline_index; if (state->lines[line_index] != LINE_UNKNOWN) continue; k = j + 1; if (k >= N) k = 0; /* minimum YESs in the complement of this edge */ if (mins[k][j] > clue) { sstate->solver_status = SOLVER_MISTAKE; return DIFF_EASY; } if (mins[k][j] == clue) { /* setting this edge to YES would make at least * (clue+1) edges - contradiction */ solver_set_line(sstate, line_index, LINE_NO); diff = min(diff, DIFF_EASY); } if (maxs[k][j] < clue - 1) { sstate->solver_status = SOLVER_MISTAKE; return DIFF_EASY; } if (maxs[k][j] == clue - 1) { /* Only way to satisfy the clue is to set edge{j} as YES */ solver_set_line(sstate, line_index, LINE_YES); diff = min(diff, DIFF_EASY); } /* More advanced deduction that allows propagation along diagonal * chains of faces connected by dots, for example, 3-2-...-2-3 * in square grids. */ if (sstate->diff >= DIFF_TRICKY) { /* Now see if we can make dline deduction for edges{j,j+1} */ e = f->edges[k]; if (state->lines[e - g->edges] != LINE_UNKNOWN) /* Only worth doing this for an UNKNOWN,UNKNOWN pair. * Dlines where one of the edges is known, are handled in the * dot-deductions */ continue; dline_index = dline_index_from_face(g, f, k); k++; if (k >= N) k = 0; /* minimum YESs in the complement of this dline */ if (mins[k][j] > clue - 2) { /* Adding 2 YESs would break the clue */ if (set_atmostone(dlines, dline_index)) diff = min(diff, DIFF_NORMAL); } /* maximum YESs in the complement of this dline */ if (maxs[k][j] < clue) { /* Adding 2 NOs would mean not enough YESs */ if (set_atleastone(dlines, dline_index)) diff = min(diff, DIFF_NORMAL); } } } } if (diff < DIFF_NORMAL) return diff; /* ------ Dot deductions ------ */ for (i = 0; i < g->num_dots; i++) { grid_dot *d = g->dots + i; int N = d->order; int yes, no, unknown; int j; if (sstate->dot_solved[i]) continue; yes = sstate->dot_yes_count[i]; no = sstate->dot_no_count[i]; unknown = N - yes - no; for (j = 0; j < N; j++) { int k; int dline_index; int line1_index, line2_index; enum line_state line1, line2; k = j + 1; if (k >= N) k = 0; dline_index = dline_index_from_dot(g, d, j); line1_index = d->edges[j] - g->edges; line2_index = d->edges[k] - g->edges; line1 = state->lines[line1_index]; line2 = state->lines[line2_index]; /* Infer dline state from line state */ if (line1 == LINE_NO || line2 == LINE_NO) { if (set_atmostone(dlines, dline_index)) diff = min(diff, DIFF_NORMAL); } if (line1 == LINE_YES || line2 == LINE_YES) { if (set_atleastone(dlines, dline_index)) diff = min(diff, DIFF_NORMAL); } /* Infer line state from dline state */ if (is_atmostone(dlines, dline_index)) { if (line1 == LINE_YES && line2 == LINE_UNKNOWN) { solver_set_line(sstate, line2_index, LINE_NO); diff = min(diff, DIFF_EASY); } if (line2 == LINE_YES && line1 == LINE_UNKNOWN) { solver_set_line(sstate, line1_index, LINE_NO); diff = min(diff, DIFF_EASY); } } if (is_atleastone(dlines, dline_index)) { if (line1 == LINE_NO && line2 == LINE_UNKNOWN) { solver_set_line(sstate, line2_index, LINE_YES); diff = min(diff, DIFF_EASY); } if (line2 == LINE_NO && line1 == LINE_UNKNOWN) { solver_set_line(sstate, line1_index, LINE_YES); diff = min(diff, DIFF_EASY); } } /* Deductions that depend on the numbers of lines. * Only bother if both lines are UNKNOWN, otherwise the * easy-mode solver (or deductions above) would have taken * care of it. */ if (line1 != LINE_UNKNOWN || line2 != LINE_UNKNOWN) continue; if (yes == 0 && unknown == 2) { /* Both these unknowns must be identical. If we know * atmostone or atleastone, we can make progress. */ if (is_atmostone(dlines, dline_index)) { solver_set_line(sstate, line1_index, LINE_NO); solver_set_line(sstate, line2_index, LINE_NO); diff = min(diff, DIFF_EASY); } if (is_atleastone(dlines, dline_index)) { solver_set_line(sstate, line1_index, LINE_YES); solver_set_line(sstate, line2_index, LINE_YES); diff = min(diff, DIFF_EASY); } } if (yes == 1) { if (set_atmostone(dlines, dline_index)) diff = min(diff, DIFF_NORMAL); if (unknown == 2) { if (set_atleastone(dlines, dline_index)) diff = min(diff, DIFF_NORMAL); } } /* More advanced deduction that allows propagation along diagonal * chains of faces connected by dots, for example: 3-2-...-2-3 * in square grids. */ if (sstate->diff >= DIFF_TRICKY) { /* If we have atleastone set for this dline, infer * atmostone for each "opposite" dline (that is, each * dline without edges in common with this one). * Again, this test is only worth doing if both these * lines are UNKNOWN. For if one of these lines were YES, * the (yes == 1) test above would kick in instead. */ if (is_atleastone(dlines, dline_index)) { int opp; for (opp = 0; opp < N; opp++) { int opp_dline_index; if (opp == j || opp == j+1 || opp == j-1) continue; if (j == 0 && opp == N-1) continue; if (j == N-1 && opp == 0) continue; opp_dline_index = dline_index_from_dot(g, d, opp); if (set_atmostone(dlines, opp_dline_index)) diff = min(diff, DIFF_NORMAL); } if (yes == 0 && is_atmostone(dlines, dline_index)) { /* This dline has *exactly* one YES and there are no * other YESs. This allows more deductions. */ if (unknown == 3) { /* Third unknown must be YES */ for (opp = 0; opp < N; opp++) { int opp_index; if (opp == j || opp == k) continue; opp_index = d->edges[opp] - g->edges; if (state->lines[opp_index] == LINE_UNKNOWN) { solver_set_line(sstate, opp_index, LINE_YES); diff = min(diff, DIFF_EASY); } } } else if (unknown == 4) { /* Exactly one of opposite UNKNOWNS is YES. We've * already set atmostone, so set atleastone as * well. */ if (dline_set_opp_atleastone(sstate, d, j)) diff = min(diff, DIFF_NORMAL); } } } } } } return diff; } static int linedsf_deductions(solver_state *sstate) { game_state *state = sstate->state; grid *g = state->game_grid; char *dlines = sstate->dlines; int i; int diff = DIFF_MAX; int diff_tmp; /* ------ Face deductions ------ */ /* A fully-general linedsf deduction seems overly complicated * (I suspect the problem is NP-complete, though in practice it might just * be doable because faces are limited in size). * For simplicity, we only consider *pairs* of LINE_UNKNOWNS that are * known to be identical. If setting them both to YES (or NO) would break * the clue, set them to NO (or YES). */ for (i = 0; i < g->num_faces; i++) { int N, yes, no, unknown; int clue; if (sstate->face_solved[i]) continue; clue = state->clues[i]; if (clue < 0) continue; N = g->faces[i].order; yes = sstate->face_yes_count[i]; if (yes + 1 == clue) { if (face_setall_identical(sstate, i, LINE_NO)) diff = min(diff, DIFF_EASY); } no = sstate->face_no_count[i]; if (no + 1 == N - clue) { if (face_setall_identical(sstate, i, LINE_YES)) diff = min(diff, DIFF_EASY); } /* Reload YES count, it might have changed */ yes = sstate->face_yes_count[i]; unknown = N - no - yes; /* Deductions with small number of LINE_UNKNOWNs, based on overall * parity of lines. */ diff_tmp = parity_deductions(sstate, g->faces[i].edges, (clue - yes) % 2, unknown); diff = min(diff, diff_tmp); } /* ------ Dot deductions ------ */ for (i = 0; i < g->num_dots; i++) { grid_dot *d = g->dots + i; int N = d->order; int j; int yes, no, unknown; /* Go through dlines, and do any dline<->linedsf deductions wherever * we find two UNKNOWNS. */ for (j = 0; j < N; j++) { int dline_index = dline_index_from_dot(g, d, j); int line1_index; int line2_index; int can1, can2, inv1, inv2; int j2; line1_index = d->edges[j] - g->edges; if (state->lines[line1_index] != LINE_UNKNOWN) continue; j2 = j + 1; if (j2 == N) j2 = 0; line2_index = d->edges[j2] - g->edges; if (state->lines[line2_index] != LINE_UNKNOWN) continue; /* Infer dline flags from linedsf */ can1 = edsf_canonify(sstate->linedsf, line1_index, &inv1); can2 = edsf_canonify(sstate->linedsf, line2_index, &inv2); if (can1 == can2 && inv1 != inv2) { /* These are opposites, so set dline atmostone/atleastone */ if (set_atmostone(dlines, dline_index)) diff = min(diff, DIFF_NORMAL); if (set_atleastone(dlines, dline_index)) diff = min(diff, DIFF_NORMAL); continue; } /* Infer linedsf from dline flags */ if (is_atmostone(dlines, dline_index) && is_atleastone(dlines, dline_index)) { if (merge_lines(sstate, line1_index, line2_index, 1)) diff = min(diff, DIFF_HARD); } } /* Deductions with small number of LINE_UNKNOWNs, based on overall * parity of lines. */ yes = sstate->dot_yes_count[i]; no = sstate->dot_no_count[i]; unknown = N - yes - no; diff_tmp = parity_deductions(sstate, d->edges, yes % 2, unknown); diff = min(diff, diff_tmp); } /* ------ Edge dsf deductions ------ */ /* If the state of a line is known, deduce the state of its canonical line * too, and vice versa. */ for (i = 0; i < g->num_edges; i++) { int can, inv; enum line_state s; can = edsf_canonify(sstate->linedsf, i, &inv); if (can == i) continue; s = sstate->state->lines[can]; if (s != LINE_UNKNOWN) { if (solver_set_line(sstate, i, inv ? OPP(s) : s)) diff = min(diff, DIFF_EASY); } else { s = sstate->state->lines[i]; if (s != LINE_UNKNOWN) { if (solver_set_line(sstate, can, inv ? OPP(s) : s)) diff = min(diff, DIFF_EASY); } } } return diff; } static int loop_deductions(solver_state *sstate) { int edgecount = 0, clues = 0, satclues = 0, sm1clues = 0; game_state *state = sstate->state; grid *g = state->game_grid; int shortest_chainlen = g->num_dots; int loop_found = FALSE; int dots_connected; int progress = FALSE; int i; /* * Go through the grid and update for all the new edges. * Since merge_dots() is idempotent, the simplest way to * do this is just to update for _all_ the edges. * Also, while we're here, we count the edges. */ for (i = 0; i < g->num_edges; i++) { if (state->lines[i] == LINE_YES) { loop_found |= merge_dots(sstate, i); edgecount++; } } /* * Count the clues, count the satisfied clues, and count the * satisfied-minus-one clues. */ for (i = 0; i < g->num_faces; i++) { int c = state->clues[i]; if (c >= 0) { int o = sstate->face_yes_count[i]; if (o == c) satclues++; else if (o == c-1) sm1clues++; clues++; } } for (i = 0; i < g->num_dots; ++i) { dots_connected = sstate->looplen[dsf_canonify(sstate->dotdsf, i)]; if (dots_connected > 1) shortest_chainlen = min(shortest_chainlen, dots_connected); } assert(sstate->solver_status == SOLVER_INCOMPLETE); if (satclues == clues && shortest_chainlen == edgecount) { sstate->solver_status = SOLVER_SOLVED; /* This discovery clearly counts as progress, even if we haven't * just added any lines or anything */ progress = TRUE; goto finished_loop_deductionsing; } /* * Now go through looking for LINE_UNKNOWN edges which * connect two dots that are already in the same * equivalence class. If we find one, test to see if the * loop it would create is a solution. */ for (i = 0; i < g->num_edges; i++) { grid_edge *e = g->edges + i; int d1 = e->dot1 - g->dots; int d2 = e->dot2 - g->dots; int eqclass, val; if (state->lines[i] != LINE_UNKNOWN) continue; eqclass = dsf_canonify(sstate->dotdsf, d1); if (eqclass != dsf_canonify(sstate->dotdsf, d2)) continue; val = LINE_NO; /* loop is bad until proven otherwise */ /* * This edge would form a loop. Next * question: how long would the loop be? * Would it equal the total number of edges * (plus the one we'd be adding if we added * it)? */ if (sstate->looplen[eqclass] == edgecount + 1) { int sm1_nearby; /* * This edge would form a loop which * took in all the edges in the entire * grid. So now we need to work out * whether it would be a valid solution * to the puzzle, which means we have to * check if it satisfies all the clues. * This means that every clue must be * either satisfied or satisfied-minus- * 1, and also that the number of * satisfied-minus-1 clues must be at * most two and they must lie on either * side of this edge. */ sm1_nearby = 0; if (e->face1) { int f = e->face1 - g->faces; int c = state->clues[f]; if (c >= 0 && sstate->face_yes_count[f] == c - 1) sm1_nearby++; } if (e->face2) { int f = e->face2 - g->faces; int c = state->clues[f]; if (c >= 0 && sstate->face_yes_count[f] == c - 1) sm1_nearby++; } if (sm1clues == sm1_nearby && sm1clues + satclues == clues) { val = LINE_YES; /* loop is good! */ } } /* * Right. Now we know that adding this edge * would form a loop, and we know whether * that loop would be a viable solution or * not. * * If adding this edge produces a solution, * then we know we've found _a_ solution but * we don't know that it's _the_ solution - * if it were provably the solution then * we'd have deduced this edge some time ago * without the need to do loop detection. So * in this state we return SOLVER_AMBIGUOUS, * which has the effect that hitting Solve * on a user-provided puzzle will fill in a * solution but using the solver to * construct new puzzles won't consider this * a reasonable deduction for the user to * make. */ progress = solver_set_line(sstate, i, val); assert(progress == TRUE); if (val == LINE_YES) { sstate->solver_status = SOLVER_AMBIGUOUS; goto finished_loop_deductionsing; } } finished_loop_deductionsing: return progress ? DIFF_EASY : DIFF_MAX; } /* This will return a dynamically allocated solver_state containing the (more) * solved grid */ static solver_state *solve_game_rec(const solver_state *sstate_start) { solver_state *sstate; /* Index of the solver we should call next. */ int i = 0; /* As a speed-optimisation, we avoid re-running solvers that we know * won't make any progress. This happens when a high-difficulty * solver makes a deduction that can only help other high-difficulty * solvers. * For example: if a new 'dline' flag is set by dline_deductions, the * trivial_deductions solver cannot do anything with this information. * If we've already run the trivial_deductions solver (because it's * earlier in the list), there's no point running it again. * * Therefore: if a solver is earlier in the list than "threshold_index", * we don't bother running it if it's difficulty level is less than * "threshold_diff". */ int threshold_diff = 0; int threshold_index = 0; sstate = dup_solver_state(sstate_start); check_caches(sstate); while (i < NUM_SOLVERS) { if (sstate->solver_status == SOLVER_MISTAKE) return sstate; if (sstate->solver_status == SOLVER_SOLVED || sstate->solver_status == SOLVER_AMBIGUOUS) { /* solver finished */ break; } if ((solver_diffs[i] >= threshold_diff || i >= threshold_index) && solver_diffs[i] <= sstate->diff) { /* current_solver is eligible, so use it */ int next_diff = solver_fns[i](sstate); if (next_diff != DIFF_MAX) { /* solver made progress, so use new thresholds and * start again at top of list. */ threshold_diff = next_diff; threshold_index = i; i = 0; continue; } } /* current_solver is ineligible, or failed to make progress, so * go to the next solver in the list */ i++; } if (sstate->solver_status == SOLVER_SOLVED || sstate->solver_status == SOLVER_AMBIGUOUS) { /* s/LINE_UNKNOWN/LINE_NO/g */ array_setall(sstate->state->lines, LINE_UNKNOWN, LINE_NO, sstate->state->game_grid->num_edges); return sstate; } return sstate; } static char *solve_game(const game_state *state, const game_state *currstate, const char *aux, char **error) { char *soln = NULL; solver_state *sstate, *new_sstate; sstate = new_solver_state(state, DIFF_MAX); new_sstate = solve_game_rec(sstate); if (new_sstate->solver_status == SOLVER_SOLVED) { soln = encode_solve_move(new_sstate->state); } else if (new_sstate->solver_status == SOLVER_AMBIGUOUS) { soln = encode_solve_move(new_sstate->state); /**error = "Solver found ambiguous solutions"; */ } else { soln = encode_solve_move(new_sstate->state); /**error = "Solver failed"; */ } free_solver_state(new_sstate); free_solver_state(sstate); return soln; } /* ---------------------------------------------------------------------- * Drawing and mouse-handling */ static char *interpret_move(const game_state *state, game_ui *ui, const game_drawstate *ds, int x, int y, int button) { grid *g = state->game_grid; grid_edge *e; int i; char *ret, buf[80]; char button_char = ' '; enum line_state old_state; button &= ~MOD_MASK; /* Convert mouse-click (x,y) to grid coordinates */ x -= BORDER(ds->tilesize); y -= BORDER(ds->tilesize); x = x * g->tilesize / ds->tilesize; y = y * g->tilesize / ds->tilesize; x += g->lowest_x; y += g->lowest_y; e = grid_nearest_edge(g, x, y); if (e == NULL) return NULL; i = e - g->edges; /* I think it's only possible to play this game with mouse clicks, sorry */ /* Maybe will add mouse drag support some time */ old_state = state->lines[i]; switch (button) { case LEFT_BUTTON: switch (old_state) { case LINE_UNKNOWN: button_char = 'y'; break; case LINE_YES: #ifdef STYLUS_BASED button_char = 'n'; break; #endif case LINE_NO: button_char = 'u'; break; } break; case MIDDLE_BUTTON: button_char = 'u'; break; case RIGHT_BUTTON: switch (old_state) { case LINE_UNKNOWN: button_char = 'n'; break; case LINE_NO: #ifdef STYLUS_BASED button_char = 'y'; break; #endif case LINE_YES: button_char = 'u'; break; } break; default: return NULL; } sprintf(buf, "%d%c", i, (int)button_char); ret = dupstr(buf); return ret; } static game_state *execute_move(const game_state *state, const char *move) { int i; game_state *newstate = dup_game(state); if (move[0] == 'S') { move++; newstate->cheated = TRUE; } while (*move) { i = atoi(move); if (i < 0 || i >= newstate->game_grid->num_edges) goto fail; move += strspn(move, "1234567890"); switch (*(move++)) { case 'y': newstate->lines[i] = LINE_YES; break; case 'n': newstate->lines[i] = LINE_NO; break; case 'u': newstate->lines[i] = LINE_UNKNOWN; break; default: goto fail; } } /* * Check for completion. */ if (check_completion(newstate)) newstate->solved = TRUE; return newstate; fail: free_game(newstate); return NULL; } /* ---------------------------------------------------------------------- * Drawing routines. */ /* Convert from grid coordinates to screen coordinates */ static void grid_to_screen(const game_drawstate *ds, const grid *g, int grid_x, int grid_y, int *x, int *y) { *x = grid_x - g->lowest_x; *y = grid_y - g->lowest_y; *x = *x * ds->tilesize / g->tilesize; *y = *y * ds->tilesize / g->tilesize; *x += BORDER(ds->tilesize); *y += BORDER(ds->tilesize); } /* 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, grid_face *f, int *xret, int *yret) { int faceindex = f - g->faces; /* * Return the cached position for this face, if we've already * worked it out. */ if (ds->textx[faceindex] >= 0) { *xret = ds->textx[faceindex]; *yret = ds->texty[faceindex]; return; } /* * Otherwise, use the incentre computed by grid.c and convert it * to screen coordinates. */ grid_find_incentre(f); grid_to_screen(ds, g, f->ix, f->iy, &ds->textx[faceindex], &ds->texty[faceindex]); *xret = ds->textx[faceindex]; *yret = ds->texty[faceindex]; } static void face_text_bbox(game_drawstate *ds, grid *g, grid_face *f, int *x, int *y, int *w, int *h) { int xx, yy; face_text_pos(ds, g, f, &xx, &yy); /* There seems to be a certain amount of trial-and-error involved * in working out the correct bounding-box for the text. */ *x = xx - ds->tilesize/4 - 1; *y = yy - ds->tilesize/4 - 3; *w = ds->tilesize/2 + 2; *h = ds->tilesize/2 + 5; } static void game_redraw_clue(drawing *dr, game_drawstate *ds, const game_state *state, int i) { grid *g = state->game_grid; grid_face *f = g->faces + i; int x, y; char c[20]; sprintf(c, "%d", state->clues[i]); face_text_pos(ds, g, f, &x, &y); draw_text(dr, x, y, FONT_VARIABLE, ds->tilesize/2, ALIGN_VCENTRE | ALIGN_HCENTRE, ds->clue_error[i] ? COL_MISTAKE : ds->clue_satisfied[i] ? COL_SATISFIED : COL_FOREGROUND, c); } static void edge_bbox(game_drawstate *ds, grid *g, grid_edge *e, int *x, int *y, int *w, int *h) { int x1 = e->dot1->x; int y1 = e->dot1->y; int x2 = e->dot2->x; int y2 = e->dot2->y; int xmin, xmax, ymin, ymax; grid_to_screen(ds, g, x1, y1, &x1, &y1); grid_to_screen(ds, g, x2, y2, &x2, &y2); /* Allow extra margin for dots, and thickness of lines */ xmin = min(x1, x2) - 2; xmax = max(x1, x2) + 2; ymin = min(y1, y2) - 2; ymax = max(y1, y2) + 2; *x = xmin; *y = ymin; *w = xmax - xmin + 1; *h = ymax - ymin + 1; } static void dot_bbox(game_drawstate *ds, grid *g, grid_dot *d, int *x, int *y, int *w, int *h) { int x1, y1; grid_to_screen(ds, g, d->x, d->y, &x1, &y1); *x = x1 - 2; *y = y1 - 2; *w = 5; *h = 5; } static const int loopy_line_redraw_phases[] = { COL_FAINT, COL_LINEUNKNOWN, COL_FOREGROUND, COL_HIGHLIGHT, COL_MISTAKE }; #define NPHASES lenof(loopy_line_redraw_phases) static void game_redraw_line(drawing *dr, game_drawstate *ds, const game_state *state, int i, int phase) { grid *g = state->game_grid; grid_edge *e = g->edges + i; int x1, x2, y1, y2; int line_colour; if (state->line_errors[i]) line_colour = COL_MISTAKE; else if (state->lines[i] == LINE_UNKNOWN) line_colour = COL_LINEUNKNOWN; else if (state->lines[i] == LINE_NO) line_colour = COL_FAINT; else if (ds->flashing) line_colour = COL_HIGHLIGHT; else line_colour = COL_FOREGROUND; if (line_colour != loopy_line_redraw_phases[phase]) return; /* Convert from grid to screen coordinates */ 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); if (line_colour == COL_FAINT) { static int draw_faint_lines = -1; if (draw_faint_lines < 0) { char *env = getenv("LOOPY_FAINT_LINES"); draw_faint_lines = (!env || (env[0] == 'y' || env[0] == 'Y')); } if (draw_faint_lines) draw_line(dr, x1, y1, x2, y2, line_colour); } else { draw_thick_line(dr, 3.0, x1 + 0.5, y1 + 0.5, x2 + 0.5, y2 + 0.5, line_colour); } } static void game_redraw_dot(drawing *dr, game_drawstate *ds, const game_state *state, int i) { grid *g = state->game_grid; grid_dot *d = g->dots + i; int x, y; grid_to_screen(ds, g, d->x, d->y, &x, &y); draw_circle(dr, x, y, 2, COL_FOREGROUND, COL_FOREGROUND); } static int boxes_intersect(int x0, int y0, int w0, int h0, int x1, int y1, int w1, int h1) { /* * Two intervals intersect iff neither is wholly on one side of * the other. Two boxes intersect iff their horizontal and * vertical intervals both intersect. */ return (x0 < x1+w1 && x1 < x0+w0 && y0 < y1+h1 && y1 < y0+h0); } static void game_redraw_in_rect(drawing *dr, game_drawstate *ds, const game_state *state, int x, int y, int w, int h) { grid *g = state->game_grid; int i, phase; int bx, by, bw, bh; clip(dr, x, y, w, h); draw_rect(dr, x, y, w, h, COL_BACKGROUND); for (i = 0; i < g->num_faces; i++) { if (state->clues[i] >= 0) { face_text_bbox(ds, g, &g->faces[i], &bx, &by, &bw, &bh); if (boxes_intersect(x, y, w, h, bx, by, bw, bh)) game_redraw_clue(dr, ds, state, i); } } for (phase = 0; phase < NPHASES; phase++) { for (i = 0; i < g->num_edges; i++) { edge_bbox(ds, g, &g->edges[i], &bx, &by, &bw, &bh); if (boxes_intersect(x, y, w, h, bx, by, bw, bh)) game_redraw_line(dr, ds, state, i, phase); } } for (i = 0; i < g->num_dots; i++) { dot_bbox(ds, g, &g->dots[i], &bx, &by, &bw, &bh); if (boxes_intersect(x, y, w, h, bx, by, bw, bh)) game_redraw_dot(dr, ds, state, i); } unclip(dr); draw_update(dr, x, y, w, h); } static void game_redraw(drawing *dr, game_drawstate *ds, const game_state *oldstate, const game_state *state, int dir, const game_ui *ui, float animtime, float flashtime) { #define REDRAW_OBJECTS_LIMIT 16 /* Somewhat arbitrary tradeoff */ grid *g = state->game_grid; int border = BORDER(ds->tilesize); int i; int flash_changed; int redraw_everything = FALSE; int edges[REDRAW_OBJECTS_LIMIT], nedges = 0; int faces[REDRAW_OBJECTS_LIMIT], nfaces = 0; /* Redrawing is somewhat involved. * * An update can theoretically affect an arbitrary number of edges * (consider, for example, completing or breaking a cycle which doesn't * satisfy all the clues -- we'll switch many edges between error and * normal states). On the other hand, redrawing the whole grid takes a * while, making the game feel sluggish, and many updates are actually * quite well localized. * * This redraw algorithm attempts to cope with both situations gracefully * and correctly. For localized changes, we set a clip rectangle, fill * it with background, and then redraw (a plausible but conservative * guess at) the objects which intersect the rectangle; if several * objects need redrawing, we'll do them individually. However, if lots * of objects are affected, we'll just redraw everything. * * The reason for all of this is that it's just not safe to do the redraw * piecemeal. If you try to draw an antialiased diagonal line over * itself, you get a slightly thicker antialiased diagonal line, which * looks rather ugly after a while. * * So, we take two passes over the grid. The first attempts to work out * what needs doing, and the second actually does it. */ if (!ds->started) { redraw_everything = TRUE; /* * But we must still go through the upcoming loops, so that we * set up stuff in ds correctly for the initial redraw. */ } /* First, trundle through the faces. */ for (i = 0; i < g->num_faces; i++) { grid_face *f = g->faces + i; int sides = f->order; int yes_order, no_order; int clue_mistake; int clue_satisfied; int n = state->clues[i]; if (n < 0) continue; yes_order = face_order(state, i, LINE_YES); if (state->exactly_one_loop) { /* * Special case: if the set of LINE_YES edges in the grid * consists of exactly one loop and nothing else, then we * switch to treating LINE_UNKNOWN the same as LINE_NO for * purposes of clue checking. * * This is because some people like to play Loopy without * using the right-click, i.e. never setting anything to * LINE_NO. Without this special case, if a person playing * in that style fills in what they think is a correct * solution loop but in fact it has an underfilled clue, * then we will display no victory flash and also no error * highlight explaining why not. With this special case, * we light up underfilled clues at the instant the loop * is closed. (Of course, *overfilled* clues are fine * either way.) * * (It might still be considered unfortunate that we can't * warn this style of player any earlier, if they make a * mistake very near the beginning which doesn't show up * until they close the last edge of the loop. One other * thing we _could_ do here is to treat any LINE_UNKNOWN * as LINE_NO if either of its endpoints has yes-degree 2, * reflecting the fact that setting that line to YES would * be an obvious error. But I don't think even that could * catch _all_ clue errors in a timely manner; I think * there are some that won't be displayed until the loop * is filled in, even so, and there's no way to avoid that * with complete reliability except to switch to being a * player who sets things to LINE_NO.) */ no_order = sides - yes_order; } else { no_order = face_order(state, i, LINE_NO); } clue_mistake = (yes_order > n || no_order > (sides-n)); clue_satisfied = (yes_order == n && no_order == (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. */ if (redraw_everything) { int grid_width = g->highest_x - g->lowest_x; int grid_height = g->highest_y - g->lowest_y; int w = grid_width * ds->tilesize / g->tilesize; int h = grid_height * ds->tilesize / g->tilesize; game_redraw_in_rect(dr, ds, state, 0, 0, w + 2*border + 1, h + 2*border + 1); } else { /* Right. Now we roll up our sleeves. */ for (i = 0; i < nfaces; i++) { grid_face *f = g->faces + faces[i]; int x, y, w, h; face_text_bbox(ds, g, f, &x, &y, &w, &h); game_redraw_in_rect(dr, ds, state, x, y, w, h); } for (i = 0; i < nedges; i++) { grid_edge *e = g->edges + edges[i]; int x, y, w, h; edge_bbox(ds, g, e, &x, &y, &w, &h); game_redraw_in_rect(dr, ds, state, x, y, w, h); } } ds->started = TRUE; } static float game_flash_length(const game_state *oldstate, const game_state *newstate, int dir, game_ui *ui) { if (!oldstate->solved && newstate->solved && !oldstate->cheated && !newstate->cheated) { return FLASH_TIME; } return 0.0F; } static int game_status(const game_state *state) { return state->solved ? +1 : 0; } static void game_print_size(const game_params *params, float *x, float *y) { int pw, ph; /* * I'll use 7mm "squares" by default. */ game_compute_size(params, 700, &pw, &ph); *x = pw / 100.0F; *y = ph / 100.0F; } static void game_print(drawing *dr, const game_state *state, int tilesize) { int ink = print_mono_colour(dr, 0); int i; game_drawstate ads, *ds = &ads; 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; grid_to_screen(ds, g, g->dots[i].x, g->dots[i].y, &x, &y); draw_circle(dr, x, y, ds->tilesize / 15, ink, ink); } /* * Clues. */ for (i = 0; i < g->num_faces; i++) { grid_face *f = g->faces + i; int clue = state->clues[i]; if (clue >= 0) { char c[20]; int x, y; sprintf(c, "%d", state->clues[i]); face_text_pos(ds, g, f, &x, &y); draw_text(dr, x, y, FONT_VARIABLE, ds->tilesize / 2, ALIGN_VCENTRE | ALIGN_HCENTRE, ink, c); } } /* * Lines. */ for (i = 0; i < g->num_edges; i++) { int thickness = (state->lines[i] == LINE_YES) ? 30 : 150; grid_edge *e = g->edges + i; int x1, y1, x2, y2; 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); if (state->lines[i] == LINE_YES) { /* (dx, dy) points from (x1, y1) to (x2, y2). * The line is then "fattened" in a perpendicular * direction to create a thin rectangle. */ double d = sqrt(SQ((double)x1 - x2) + SQ((double)y1 - y2)); double dx = (x2 - x1) / d; double dy = (y2 - y1) / d; int points[8]; dx = (dx * ds->tilesize) / thickness; dy = (dy * ds->tilesize) / thickness; points[0] = x1 + (int)dy; points[1] = y1 - (int)dx; points[2] = x1 - (int)dy; points[3] = y1 + (int)dx; points[4] = x2 - (int)dy; points[5] = y2 + (int)dx; points[6] = x2 + (int)dy; points[7] = y2 - (int)dx; draw_polygon(dr, points, 4, ink, ink); } else { /* Draw a dotted line */ int divisions = 6; int j; for (j = 1; j < divisions; j++) { /* Weighted average */ int x = (x1 * (divisions -j) + x2 * j) / divisions; int y = (y1 * (divisions -j) + y2 * j) / divisions; draw_circle(dr, x, y, ds->tilesize / thickness, ink, ink); } } } sfree(ds->textx); sfree(ds->texty); } #ifdef COMBINED #define thegame loopy #endif const struct game thegame = { "Loopy", "games.loopy", "loopy", default_params, game_fetch_preset, decode_params, encode_params, free_params, dup_params, TRUE, game_configure, custom_params, validate_params, new_game_desc, validate_desc, new_game, dup_game, free_game, 1, solve_game, TRUE, game_can_format_as_text_now, game_text_format, new_ui, free_ui, encode_ui, decode_ui, game_changed_state, interpret_move, execute_move, PREFERRED_TILE_SIZE, game_compute_size, game_set_size, game_colours, game_new_drawstate, game_free_drawstate, game_redraw, game_anim_length, game_flash_length, game_status, TRUE, FALSE, game_print_size, game_print, FALSE /* wants_statusbar */, FALSE, game_timing_state, 0, /* mouse_priorities */ }; #ifdef STANDALONE_SOLVER /* * Half-hearted standalone solver. It can't output the solution to * anything but a square puzzle, and it can't log the deductions * it makes either. But it can solve square puzzles, and more * importantly it can use its solver to grade the difficulty of * any puzzle you give it. */ #include int main(int argc, char **argv) { game_params *p; game_state *s; char *id = NULL, *desc, *err; int grade = FALSE; int ret, diff; #if 0 /* verbose solver not supported here (yet) */ int really_verbose = FALSE; #endif while (--argc > 0) { char *p = *++argv; #if 0 /* verbose solver not supported here (yet) */ if (!strcmp(p, "-v")) { really_verbose = TRUE; } else #endif if (!strcmp(p, "-g")) { grade = TRUE; } else if (*p == '-') { fprintf(stderr, "%s: unrecognised option `%s'\n", argv[0], p); return 1; } else { id = p; } } if (!id) { fprintf(stderr, "usage: %s [-g | -v] \n", argv[0]); return 1; } desc = strchr(id, ':'); if (!desc) { fprintf(stderr, "%s: game id expects a colon in it\n", argv[0]); return 1; } *desc++ = '\0'; p = default_params(); decode_params(p, id); err = validate_desc(p, desc); if (err) { fprintf(stderr, "%s: %s\n", argv[0], err); return 1; } s = new_game(NULL, p, desc); /* * When solving an Easy puzzle, we don't want to bother the * user with Hard-level deductions. For this reason, we grade * the puzzle internally before doing anything else. */ ret = -1; /* placate optimiser */ for (diff = 0; diff < DIFF_MAX; diff++) { solver_state *sstate_new; solver_state *sstate = new_solver_state((game_state *)s, diff); sstate_new = solve_game_rec(sstate); if (sstate_new->solver_status == SOLVER_MISTAKE) ret = 0; else if (sstate_new->solver_status == SOLVER_SOLVED) ret = 1; else ret = 2; free_solver_state(sstate_new); free_solver_state(sstate); if (ret < 2) break; } if (diff == DIFF_MAX) { if (grade) printf("Difficulty rating: harder than Hard, or ambiguous\n"); else printf("Unable to find a unique solution\n"); } else { if (grade) { if (ret == 0) printf("Difficulty rating: impossible (no solution exists)\n"); else if (ret == 1) printf("Difficulty rating: %s\n", diffnames[diff]); } else { solver_state *sstate_new; solver_state *sstate = new_solver_state((game_state *)s, diff); /* If we supported a verbose solver, we'd set verbosity here */ sstate_new = solve_game_rec(sstate); if (sstate_new->solver_status == SOLVER_MISTAKE) printf("Puzzle is inconsistent\n"); else { assert(sstate_new->solver_status == SOLVER_SOLVED); if (s->grid_type == 0) { fputs(game_text_format(sstate_new->state), stdout); } else { printf("Unable to output non-square grids\n"); } } free_solver_state(sstate_new); free_solver_state(sstate); } } return 0; } #endif /* vim: set shiftwidth=4 tabstop=8: */