#include "puzzles.h"
#include "tree234.h"
#include "grid.h"
+#include "loopgen.h"
/* Debugging options */
};
struct game_state {
- grid *game_grid;
+ grid *game_grid; /* ref-counted (internally) */
/* Put -1 in a face that doesn't get a clue */
signed char *clues;
SOLVERLIST(SOLVER_FN_DECL)
static int (*(solver_fns[]))(solver_state *) = { SOLVERLIST(SOLVER_FN) };
static int const solver_diffs[] = { SOLVERLIST(SOLVER_DIFF) };
-const int NUM_SOLVERS = sizeof(solver_diffs)/sizeof(*solver_diffs);
+static const int NUM_SOLVERS = sizeof(solver_diffs)/sizeof(*solver_diffs);
struct game_params {
int w, h;
int diff;
int type;
-
- /* Grid generation is expensive, so keep a (ref-counted) reference to the
- * grid for these parameters, and only generate when required. */
- grid *game_grid;
};
/* line_drawstate is the same as line_state, but with the extra ERROR
char *clue_satisfied;
};
-static char *validate_desc(game_params *params, char *desc);
+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);
/* ------- List of grid generators ------- */
#define GRIDLIST(A) \
- A(Squares,grid_new_square,3,3) \
- A(Triangular,grid_new_triangular,3,3) \
- A(Honeycomb,grid_new_honeycomb,3,3) \
- A(Snub-Square,grid_new_snubsquare,3,3) \
- A(Cairo,grid_new_cairo,3,4) \
- A(Great-Hexagonal,grid_new_greathexagonal,3,3) \
- A(Octagonal,grid_new_octagonal,3,3) \
- A(Kites,grid_new_kites,3,3) \
- A(Floret,grid_new_floret,1,2) \
- A(Dodecagonal,grid_new_dodecagonal,2,2) \
- A(Great-Dodecagonal,grid_new_greatdodecagonal,2,2)
-
-#define GRID_NAME(title,fn,amin,omin) #title,
-#define GRID_CONFIG(title,fn,amin,omin) ":" #title
-#define GRID_FN(title,fn,amin,omin) &fn,
-#define GRID_SIZES(title,fn,amin,omin) \
+ A(Squares,GRID_SQUARE,3,3) \
+ A(Triangular,GRID_TRIANGULAR,3,3) \
+ A(Honeycomb,GRID_HONEYCOMB,3,3) \
+ A(Snub-Square,GRID_SNUBSQUARE,3,3) \
+ A(Cairo,GRID_CAIRO,3,4) \
+ A(Great-Hexagonal,GRID_GREATHEXAGONAL,3,3) \
+ A(Octagonal,GRID_OCTAGONAL,3,3) \
+ A(Kites,GRID_KITE,3,3) \
+ A(Floret,GRID_FLORET,1,2) \
+ A(Dodecagonal,GRID_DODECAGONAL,2,2) \
+ A(Great-Dodecagonal,GRID_GREATDODECAGONAL,2,2) \
+ A(Penrose (kite/dart),GRID_PENROSE_P2,3,3) \
+ A(Penrose (rhombs),GRID_PENROSE_P3,3,3)
+
+#define GRID_NAME(title,type,amin,omin) #title,
+#define GRID_CONFIG(title,type,amin,omin) ":" #title
+#define GRID_TYPE(title,type,amin,omin) type,
+#define GRID_SIZES(title,type,amin,omin) \
{amin, omin, \
"Width and height for this grid type must both be at least " #amin, \
"At least one of width and height for this grid type must be at least " #omin,},
static char const *const gridnames[] = { GRIDLIST(GRID_NAME) };
#define GRID_CONFIGS GRIDLIST(GRID_CONFIG)
-static grid * (*(grid_fns[]))(int w, int h) = { GRIDLIST(GRID_FN) };
-#define NUM_GRID_TYPES (sizeof(grid_fns) / sizeof(grid_fns[0]))
+static grid_type grid_types[] = { GRIDLIST(GRID_TYPE) };
+#define NUM_GRID_TYPES (sizeof(grid_types) / sizeof(grid_types[0]))
static const struct {
int amin, omin;
char *aerr, *oerr;
/* Generates a (dynamically allocated) new grid, according to the
* type and size requested in params. Does nothing if the grid is already
- * generated. The allocated grid is owned by the params object, and will be
- * freed in free_params(). */
-static void params_generate_grid(game_params *params)
+ * generated. */
+static grid *loopy_generate_grid(const game_params *params,
+ const char *grid_desc)
{
- if (!params->game_grid) {
- params->game_grid = grid_fns[params->type](params->w, params->h);
- }
+ return grid_new(grid_types[params->type], params->w, params->h, grid_desc);
}
/* ----------------------------------------------------------------------
* General struct manipulation and other straightforward code
*/
-static game_state *dup_game(game_state *state)
+static game_state *dup_game(const game_state *state)
{
game_state *ret = snew(game_state);
}
}
-static solver_state *new_solver_state(game_state *state, int diff) {
+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;
ret->diff = DIFF_EASY;
ret->type = 0;
- ret->game_grid = NULL;
-
return ret;
}
-static game_params *dup_params(game_params *params)
+static game_params *dup_params(const game_params *params)
{
game_params *ret = snew(game_params);
*ret = *params; /* structure copy */
- if (ret->game_grid) {
- ret->game_grid->refcount++;
- }
return ret;
}
static const game_params presets[] = {
#ifdef SMALL_SCREEN
- { 7, 7, DIFF_EASY, 0, NULL },
- { 7, 7, DIFF_NORMAL, 0, NULL },
- { 7, 7, DIFF_HARD, 0, NULL },
- { 7, 7, DIFF_HARD, 1, NULL },
- { 7, 7, DIFF_HARD, 2, NULL },
- { 5, 5, DIFF_HARD, 3, NULL },
- { 7, 7, DIFF_HARD, 4, NULL },
- { 5, 4, DIFF_HARD, 5, NULL },
- { 5, 5, DIFF_HARD, 6, NULL },
- { 5, 5, DIFF_HARD, 7, NULL },
- { 3, 3, DIFF_HARD, 8, NULL },
- { 3, 3, DIFF_HARD, 9, NULL },
- { 3, 3, DIFF_HARD, 10, NULL },
+ { 7, 7, DIFF_EASY, 0 },
+ { 7, 7, DIFF_NORMAL, 0 },
+ { 7, 7, DIFF_HARD, 0 },
+ { 7, 7, DIFF_HARD, 1 },
+ { 7, 7, DIFF_HARD, 2 },
+ { 5, 5, DIFF_HARD, 3 },
+ { 7, 7, DIFF_HARD, 4 },
+ { 5, 4, DIFF_HARD, 5 },
+ { 5, 5, DIFF_HARD, 6 },
+ { 5, 5, DIFF_HARD, 7 },
+ { 3, 3, DIFF_HARD, 8 },
+ { 3, 3, DIFF_HARD, 9 },
+ { 3, 3, DIFF_HARD, 10 },
+ { 6, 6, DIFF_HARD, 11 },
+ { 6, 6, DIFF_HARD, 12 },
#else
- { 7, 7, DIFF_EASY, 0, NULL },
- { 10, 10, DIFF_EASY, 0, NULL },
- { 7, 7, DIFF_NORMAL, 0, NULL },
- { 10, 10, DIFF_NORMAL, 0, NULL },
- { 7, 7, DIFF_HARD, 0, NULL },
- { 10, 10, DIFF_HARD, 0, NULL },
- { 10, 10, DIFF_HARD, 1, NULL },
- { 12, 10, DIFF_HARD, 2, NULL },
- { 7, 7, DIFF_HARD, 3, NULL },
- { 9, 9, DIFF_HARD, 4, NULL },
- { 5, 4, DIFF_HARD, 5, NULL },
- { 7, 7, DIFF_HARD, 6, NULL },
- { 5, 5, DIFF_HARD, 7, NULL },
- { 5, 5, DIFF_HARD, 8, NULL },
- { 5, 4, DIFF_HARD, 9, NULL },
- { 5, 4, DIFF_HARD, 10, NULL },
+ { 7, 7, DIFF_EASY, 0 },
+ { 10, 10, DIFF_EASY, 0 },
+ { 7, 7, DIFF_NORMAL, 0 },
+ { 10, 10, DIFF_NORMAL, 0 },
+ { 7, 7, DIFF_HARD, 0 },
+ { 10, 10, DIFF_HARD, 0 },
+ { 10, 10, DIFF_HARD, 1 },
+ { 12, 10, DIFF_HARD, 2 },
+ { 7, 7, DIFF_HARD, 3 },
+ { 9, 9, DIFF_HARD, 4 },
+ { 5, 4, DIFF_HARD, 5 },
+ { 7, 7, DIFF_HARD, 6 },
+ { 5, 5, DIFF_HARD, 7 },
+ { 5, 5, DIFF_HARD, 8 },
+ { 5, 4, DIFF_HARD, 9 },
+ { 5, 4, DIFF_HARD, 10 },
+ { 10, 10, DIFF_HARD, 11 },
+ { 10, 10, DIFF_HARD, 12 }
#endif
};
static void free_params(game_params *params)
{
- if (params->game_grid) {
- grid_free(params->game_grid);
- }
sfree(params);
}
static void decode_params(game_params *params, char const *string)
{
- if (params->game_grid) {
- grid_free(params->game_grid);
- params->game_grid = NULL;
- }
params->h = params->w = atoi(string);
params->diff = DIFF_EASY;
while (*string && isdigit((unsigned char)*string)) string++;
}
}
-static char *encode_params(game_params *params, int full)
+static char *encode_params(const game_params *params, int full)
{
char str[80];
sprintf(str, "%dx%dt%d", params->w, params->h, params->type);
return dupstr(str);
}
-static config_item *game_configure(game_params *params)
+static config_item *game_configure(const game_params *params)
{
config_item *ret;
char buf[80];
return ret;
}
-static game_params *custom_params(config_item *cfg)
+static game_params *custom_params(const config_item *cfg)
{
game_params *ret = snew(game_params);
ret->type = cfg[2].ival;
ret->diff = cfg[3].ival;
- ret->game_grid = NULL;
return ret;
}
-static char *validate_params(game_params *params, int full)
+static char *validate_params(const game_params *params, int full)
{
if (params->type < 0 || params->type >= NUM_GRID_TYPES)
return "Illegal grid type";
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(game_params *params, char *desc)
+static char *validate_desc(const game_params *params, const char *desc)
{
int count = 0;
grid *g;
- params_generate_grid(params);
- g = params->game_grid;
+ char *grid_desc, *ret;
+
+ /* It's pretty inefficient to do this just for validation. All we need to
+ * know is the precise number of faces. */
+ grid_desc = extract_grid_desc(&desc);
+ ret = grid_validate_desc(grid_types[params->type], params->w, params->h, grid_desc);
+ if (ret) return ret;
+
+ g = loopy_generate_grid(params, grid_desc);
+ if (grid_desc) sfree(grid_desc);
for (; *desc; ++desc) {
if ((*desc >= '0' && *desc <= '9') || (*desc >= 'A' && *desc <= 'Z')) {
if (count > g->num_faces)
return "Description too long for board size";
+ grid_free(g);
+
return NULL;
}
return ret;
}
-static game_ui *new_ui(game_state *state)
+static game_ui *new_ui(const game_state *state)
{
return NULL;
}
{
}
-static char *encode_ui(game_ui *ui)
+static char *encode_ui(const game_ui *ui)
{
return NULL;
}
-static void decode_ui(game_ui *ui, char *encoding)
+static void decode_ui(game_ui *ui, const char *encoding)
{
}
-static void game_changed_state(game_ui *ui, game_state *oldstate,
- game_state *newstate)
+static void game_changed_state(game_ui *ui, const game_state *oldstate,
+ const game_state *newstate)
{
}
-static void game_compute_size(game_params *params, int tilesize,
+static void game_compute_size(const game_params *params, int tilesize,
int *x, int *y)
{
- grid *g;
int grid_width, grid_height, rendered_width, rendered_height;
+ int g_tilesize;
+
+ grid_compute_size(grid_types[params->type], params->w, params->h,
+ &g_tilesize, &grid_width, &grid_height);
- params_generate_grid(params);
- g = params->game_grid;
- grid_width = g->highest_x - g->lowest_x;
- grid_height = g->highest_y - g->lowest_y;
/* multiply first to minimise rounding error on integer division */
- rendered_width = grid_width * tilesize / g->tilesize;
- rendered_height = grid_height * tilesize / g->tilesize;
+ rendered_width = grid_width * tilesize / g_tilesize;
+ rendered_height = grid_height * tilesize / g_tilesize;
*x = rendered_width + 2 * BORDER(tilesize) + 1;
*y = rendered_height + 2 * BORDER(tilesize) + 1;
}
static void game_set_size(drawing *dr, game_drawstate *ds,
- game_params *params, int tilesize)
+ const game_params *params, int tilesize)
{
ds->tilesize = tilesize;
}
static float *game_colours(frontend *fe, int *ncolours)
{
- float *ret = snewn(4 * NCOLOURS, float);
+ float *ret = snewn(3 * NCOLOURS, float);
frontend_default_colour(fe, &ret[COL_BACKGROUND * 3]);
return ret;
}
-static game_drawstate *game_new_drawstate(drawing *dr, game_state *state)
+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;
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(game_state *state, game_ui *ui)
+static int game_timing_state(const game_state *state, game_ui *ui)
{
return TRUE;
}
-static float game_anim_length(game_state *oldstate, game_state *newstate,
- int dir, game_ui *ui)
+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(game_params *params)
+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(game_state *state)
+static char *game_text_format(const game_state *state)
{
int w, h, W, H;
int x, y, i;
* Loop generation and clue removal
*/
-/* We're going to store lists of current candidate faces for colouring black
- * or white.
- * Each face gets a 'score', which tells us how adding that face right
- * now would affect the curliness of the solution loop. We're trying to
- * maximise that quantity so will bias our random selection of faces to
- * colour those with high scores */
-struct face_score {
- int white_score;
- int black_score;
- unsigned long random;
- /* No need to store a grid_face* here. The 'face_scores' array will
- * be a list of 'face_score' objects, one for each face of the grid, so
- * the position (index) within the 'face_scores' array will determine
- * which face corresponds to a particular face_score.
- * Having a single 'face_scores' array for all faces simplifies memory
- * management, and probably improves performance, because we don't have to
- * malloc/free each individual face_score, and we don't have to maintain
- * a mapping from grid_face* pointers to face_score* pointers.
- */
-};
-
-static int generic_sort_cmpfn(void *v1, void *v2, size_t offset)
-{
- struct face_score *f1 = v1;
- struct face_score *f2 = v2;
- int r;
-
- r = *(int *)((char *)f2 + offset) - *(int *)((char *)f1 + offset);
- if (r) {
- return r;
- }
-
- if (f1->random < f2->random)
- return -1;
- else if (f1->random > f2->random)
- return 1;
-
- /*
- * It's _just_ possible that two faces might have been given
- * the same random value. In that situation, fall back to
- * comparing based on the positions within the face_scores list.
- * This introduces a tiny directional bias, but not a significant one.
- */
- return f1 - f2;
-}
-
-static int white_sort_cmpfn(void *v1, void *v2)
-{
- return generic_sort_cmpfn(v1, v2, offsetof(struct face_score,white_score));
-}
-
-static int black_sort_cmpfn(void *v1, void *v2)
-{
- return generic_sort_cmpfn(v1, v2, offsetof(struct face_score,black_score));
-}
-
-enum face_colour { FACE_WHITE, FACE_GREY, FACE_BLACK };
-
-/* face should be of type grid_face* here. */
-#define FACE_COLOUR(face) \
- ( (face) == NULL ? FACE_BLACK : \
- board[(face) - g->faces] )
-
-/* 'board' is an array of these enums, indicating which faces are
- * currently black/white/grey. 'colour' is FACE_WHITE or FACE_BLACK.
- * Returns whether it's legal to colour the given face with this colour. */
-static int can_colour_face(grid *g, char* board, int face_index,
- enum face_colour colour)
-{
- int i, j;
- grid_face *test_face = g->faces + face_index;
- grid_face *starting_face, *current_face;
- grid_dot *starting_dot;
- int transitions;
- int current_state, s; /* booleans: equal or not-equal to 'colour' */
- int found_same_coloured_neighbour = FALSE;
- assert(board[face_index] != colour);
-
- /* Can only consider a face for colouring if it's adjacent to a face
- * with the same colour. */
- for (i = 0; i < test_face->order; i++) {
- grid_edge *e = test_face->edges[i];
- grid_face *f = (e->face1 == test_face) ? e->face2 : e->face1;
- if (FACE_COLOUR(f) == colour) {
- found_same_coloured_neighbour = TRUE;
- break;
- }
- }
- if (!found_same_coloured_neighbour)
- return FALSE;
-
- /* Need to avoid creating a loop of faces of this colour around some
- * differently-coloured faces.
- * Also need to avoid meeting a same-coloured face at a corner, with
- * other-coloured faces in between. Here's a simple test that (I believe)
- * takes care of both these conditions:
- *
- * Take the circular path formed by this face's edges, and inflate it
- * slightly outwards. Imagine walking around this path and consider
- * the faces that you visit in sequence. This will include all faces
- * touching the given face, either along an edge or just at a corner.
- * Count the number of 'colour'/not-'colour' transitions you encounter, as
- * you walk along the complete loop. This will obviously turn out to be
- * an even number.
- * If 0, we're either in the middle of an "island" of this colour (should
- * be impossible as we're not supposed to create black or white loops),
- * or we're about to start a new island - also not allowed.
- * If 4 or greater, there are too many separate coloured regions touching
- * this face, and colouring it would create a loop or a corner-violation.
- * The only allowed case is when the count is exactly 2. */
-
- /* i points to a dot around the test face.
- * j points to a face around the i^th dot.
- * The current face will always be:
- * test_face->dots[i]->faces[j]
- * We assume dots go clockwise around the test face,
- * and faces go clockwise around dots. */
-
- /*
- * The end condition is slightly fiddly. In sufficiently strange
- * degenerate grids, our test face may be adjacent to the same
- * other face multiple times (typically if it's the exterior
- * face). Consider this, in particular:
- *
- * +--+
- * | |
- * +--+--+
- * | | |
- * +--+--+
- *
- * The bottom left face there is adjacent to the exterior face
- * twice, so we can't just terminate our iteration when we reach
- * the same _face_ we started at. Furthermore, we can't
- * condition on having the same (i,j) pair either, because
- * several (i,j) pairs identify the bottom left contiguity with
- * the exterior face! We canonicalise the (i,j) pair by taking
- * one step around before we set the termination tracking.
- */
-
- i = j = 0;
- current_face = test_face->dots[0]->faces[0];
- if (current_face == test_face) {
- j = 1;
- current_face = test_face->dots[0]->faces[1];
- }
- transitions = 0;
- current_state = (FACE_COLOUR(current_face) == colour);
- starting_dot = NULL;
- starting_face = NULL;
- while (TRUE) {
- /* Advance to next face.
- * Need to loop here because it might take several goes to
- * find it. */
- while (TRUE) {
- j++;
- if (j == test_face->dots[i]->order)
- j = 0;
-
- if (test_face->dots[i]->faces[j] == test_face) {
- /* Advance to next dot round test_face, then
- * find current_face around new dot
- * and advance to the next face clockwise */
- i++;
- if (i == test_face->order)
- i = 0;
- for (j = 0; j < test_face->dots[i]->order; j++) {
- if (test_face->dots[i]->faces[j] == current_face)
- break;
- }
- /* Must actually find current_face around new dot,
- * or else something's wrong with the grid. */
- assert(j != test_face->dots[i]->order);
- /* Found, so advance to next face and try again */
- } else {
- break;
- }
- }
- /* (i,j) are now advanced to next face */
- current_face = test_face->dots[i]->faces[j];
- s = (FACE_COLOUR(current_face) == colour);
- if (!starting_dot) {
- starting_dot = test_face->dots[i];
- starting_face = current_face;
- current_state = s;
- } else {
- if (s != current_state) {
- ++transitions;
- current_state = s;
- if (transitions > 2)
- break;
- }
- if (test_face->dots[i] == starting_dot &&
- current_face == starting_face)
- break;
- }
- }
-
- return (transitions == 2) ? TRUE : FALSE;
-}
-
-/* Count the number of neighbours of 'face', having colour 'colour' */
-static int face_num_neighbours(grid *g, char *board, grid_face *face,
- enum face_colour colour)
-{
- int colour_count = 0;
- int i;
- grid_face *f;
- grid_edge *e;
- for (i = 0; i < face->order; i++) {
- e = face->edges[i];
- f = (e->face1 == face) ? e->face2 : e->face1;
- if (FACE_COLOUR(f) == colour)
- ++colour_count;
- }
- return colour_count;
-}
-
-/* The 'score' of a face reflects its current desirability for selection
- * as the next face to colour white or black. We want to encourage moving
- * into grey areas and increasing loopiness, so we give scores according to
- * how many of the face's neighbours are currently coloured the same as the
- * proposed colour. */
-static int face_score(grid *g, char *board, grid_face *face,
- enum face_colour colour)
-{
- /* Simple formula: score = 0 - num. same-coloured neighbours,
- * so a higher score means fewer same-coloured neighbours. */
- return -face_num_neighbours(g, board, face, colour);
-}
-
-/* Generate a new complete set of clues for the given game_state.
- * The method is to generate a WHITE/BLACK colouring of all the faces,
- * such that the WHITE faces will define the inside of the path, and the
- * BLACK faces define the outside.
- * To do this, we initially colour all faces GREY. The infinite space outside
- * the grid is coloured BLACK, and we choose a random face to colour WHITE.
- * Then we gradually grow the BLACK and the WHITE regions, eliminating GREY
- * faces, until the grid is filled with BLACK/WHITE. As we grow the regions,
- * we avoid creating loops of a single colour, to preserve the topological
- * shape of the WHITE and BLACK regions.
- * We also try to make the boundary as loopy and twisty as possible, to avoid
- * generating paths that are uninteresting.
- * The algorithm works by choosing a BLACK/WHITE colour, then choosing a GREY
- * face that can be coloured with that colour (without violating the
- * topological shape of that region). It's not obvious, but I think this
- * algorithm is guaranteed to terminate without leaving any GREY faces behind.
- * Indeed, if there are any GREY faces at all, both the WHITE and BLACK
- * regions can be grown.
- * This is checked using assert()ions, and I haven't seen any failures yet.
- *
- * Hand-wavy proof: imagine what can go wrong...
- *
- * Could the white faces get completely cut off by the black faces, and still
- * leave some grey faces remaining?
- * No, because then the black faces would form a loop around both the white
- * faces and the grey faces, which is disallowed because we continually
- * maintain the correct topological shape of the black region.
- * Similarly, the black faces can never get cut off by the white faces. That
- * means both the WHITE and BLACK regions always have some room to grow into
- * the GREY regions.
- * Could it be that we can't colour some GREY face, because there are too many
- * WHITE/BLACK transitions as we walk round the face? (see the
- * can_colour_face() function for details)
- * No. Imagine otherwise, and we see WHITE/BLACK/WHITE/BLACK as we walk
- * around the face. The two WHITE faces would be connected by a WHITE path,
- * and the BLACK faces would be connected by a BLACK path. These paths would
- * have to cross, which is impossible.
- * Another thing that could go wrong: perhaps we can't find any GREY face to
- * colour WHITE, because it would create a loop-violation or a corner-violation
- * with the other WHITE faces?
- * This is a little bit tricky to prove impossible. Imagine you have such a
- * GREY face (that is, if you coloured it WHITE, you would create a WHITE loop
- * or corner violation).
- * That would cut all the non-white area into two blobs. One of those blobs
- * must be free of BLACK faces (because the BLACK stuff is a connected blob).
- * So we have a connected GREY area, completely surrounded by WHITE
- * (including the GREY face we've tentatively coloured WHITE).
- * A well-known result in graph theory says that you can always find a GREY
- * face whose removal leaves the remaining GREY area connected. And it says
- * there are at least two such faces, so we can always choose the one that
- * isn't the "tentative" GREY face. Colouring that face WHITE leaves
- * everything nice and connected, including that "tentative" GREY face which
- * acts as a gateway to the rest of the non-WHITE grid.
- */
static void add_full_clues(game_state *state, random_state *rs)
{
signed char *clues = state->clues;
- char *board;
grid *g = state->game_grid;
- int i, j;
- int num_faces = g->num_faces;
- struct face_score *face_scores; /* Array of face_score objects */
- struct face_score *fs; /* Points somewhere in the above list */
- struct grid_face *cur_face;
- tree234 *lightable_faces_sorted;
- tree234 *darkable_faces_sorted;
- int *face_list;
- int do_random_pass;
-
- board = snewn(num_faces, char);
-
- /* Make a board */
- memset(board, FACE_GREY, num_faces);
-
- /* Create and initialise the list of face_scores */
- face_scores = snewn(num_faces, struct face_score);
- for (i = 0; i < num_faces; i++) {
- face_scores[i].random = random_bits(rs, 31);
- face_scores[i].black_score = face_scores[i].white_score = 0;
- }
-
- /* Colour a random, finite face white. The infinite face is implicitly
- * coloured black. Together, they will seed the random growth process
- * for the black and white areas. */
- i = random_upto(rs, num_faces);
- board[i] = FACE_WHITE;
-
- /* We need a way of favouring faces that will increase our loopiness.
- * We do this by maintaining a list of all candidate faces sorted by
- * their score and choose randomly from that with appropriate skew.
- * In order to avoid consistently biasing towards particular faces, we
- * need the sort order _within_ each group of scores to be completely
- * random. But it would be abusing the hospitality of the tree234 data
- * structure if our comparison function were nondeterministic :-). So with
- * each face we associate a random number that does not change during a
- * particular run of the generator, and use that as a secondary sort key.
- * Yes, this means we will be biased towards particular random faces in
- * any one run but that doesn't actually matter. */
-
- lightable_faces_sorted = newtree234(white_sort_cmpfn);
- darkable_faces_sorted = newtree234(black_sort_cmpfn);
-
- /* Initialise the lists of lightable and darkable faces. This is
- * slightly different from the code inside the while-loop, because we need
- * to check every face of the board (the grid structure does not keep a
- * list of the infinite face's neighbours). */
- for (i = 0; i < num_faces; i++) {
- grid_face *f = g->faces + i;
- struct face_score *fs = face_scores + i;
- if (board[i] != FACE_GREY) continue;
- /* We need the full colourability check here, it's not enough simply
- * to check neighbourhood. On some grids, a neighbour of the infinite
- * face is not necessarily darkable. */
- if (can_colour_face(g, board, i, FACE_BLACK)) {
- fs->black_score = face_score(g, board, f, FACE_BLACK);
- add234(darkable_faces_sorted, fs);
- }
- if (can_colour_face(g, board, i, FACE_WHITE)) {
- fs->white_score = face_score(g, board, f, FACE_WHITE);
- add234(lightable_faces_sorted, fs);
- }
- }
-
- /* Colour faces one at a time until no more faces are colourable. */
- while (TRUE)
- {
- enum face_colour colour;
- struct face_score *fs_white, *fs_black;
- int c_lightable = count234(lightable_faces_sorted);
- int c_darkable = count234(darkable_faces_sorted);
- if (c_lightable == 0 && c_darkable == 0) {
- /* No more faces we can use at all. */
- break;
- }
- assert(c_lightable != 0 && c_darkable != 0);
-
- fs_white = (struct face_score *)index234(lightable_faces_sorted, 0);
- fs_black = (struct face_score *)index234(darkable_faces_sorted, 0);
-
- /* Choose a colour, and colour the best available face
- * with that colour. */
- colour = random_upto(rs, 2) ? FACE_WHITE : FACE_BLACK;
-
- if (colour == FACE_WHITE)
- fs = fs_white;
- else
- fs = fs_black;
- assert(fs);
- i = fs - face_scores;
- assert(board[i] == FACE_GREY);
- board[i] = colour;
-
- /* Remove this newly-coloured face from the lists. These lists should
- * only contain grey faces. */
- del234(lightable_faces_sorted, fs);
- del234(darkable_faces_sorted, fs);
-
- /* Remember which face we've just coloured */
- cur_face = g->faces + i;
-
- /* The face we've just coloured potentially affects the colourability
- * and the scores of any neighbouring faces (touching at a corner or
- * edge). So the search needs to be conducted around all faces
- * touching the one we've just lit. Iterate over its corners, then
- * over each corner's faces. For each such face, we remove it from
- * the lists, recalculate any scores, then add it back to the lists
- * (depending on whether it is lightable, darkable or both). */
- for (i = 0; i < cur_face->order; i++) {
- grid_dot *d = cur_face->dots[i];
- for (j = 0; j < d->order; j++) {
- grid_face *f = d->faces[j];
- int fi; /* face index of f */
-
- if (f == NULL)
- continue;
- if (f == cur_face)
- continue;
-
- /* If the face is already coloured, it won't be on our
- * lightable/darkable lists anyway, so we can skip it without
- * bothering with the removal step. */
- if (FACE_COLOUR(f) != FACE_GREY) continue;
-
- /* Find the face index and face_score* corresponding to f */
- fi = f - g->faces;
- fs = face_scores + fi;
-
- /* Remove from lightable list if it's in there. We do this,
- * even if it is still lightable, because the score might
- * be different, and we need to remove-then-add to maintain
- * correct sort order. */
- del234(lightable_faces_sorted, fs);
- if (can_colour_face(g, board, fi, FACE_WHITE)) {
- fs->white_score = face_score(g, board, f, FACE_WHITE);
- add234(lightable_faces_sorted, fs);
- }
- /* Do the same for darkable list. */
- del234(darkable_faces_sorted, fs);
- if (can_colour_face(g, board, fi, FACE_BLACK)) {
- fs->black_score = face_score(g, board, f, FACE_BLACK);
- add234(darkable_faces_sorted, fs);
- }
- }
- }
- }
-
- /* Clean up */
- freetree234(lightable_faces_sorted);
- freetree234(darkable_faces_sorted);
- sfree(face_scores);
-
- /* The next step requires a shuffled list of all faces */
- face_list = snewn(num_faces, int);
- for (i = 0; i < num_faces; ++i) {
- face_list[i] = i;
- }
- shuffle(face_list, num_faces, sizeof(int), rs);
-
- /* The above loop-generation algorithm can often leave large clumps
- * of faces of one colour. In extreme cases, the resulting path can be
- * degenerate and not very satisfying to solve.
- * This next step alleviates this problem:
- * Go through the shuffled list, and flip the colour of any face we can
- * legally flip, and which is adjacent to only one face of the opposite
- * colour - this tends to grow 'tendrils' into any clumps.
- * Repeat until we can find no more faces to flip. This will
- * eventually terminate, because each flip increases the loop's
- * perimeter, which cannot increase for ever.
- * The resulting path will have maximal loopiness (in the sense that it
- * cannot be improved "locally". Unfortunately, this allows a player to
- * make some illicit deductions. To combat this (and make the path more
- * interesting), we do one final pass making random flips. */
-
- /* Set to TRUE for final pass */
- do_random_pass = FALSE;
-
- while (TRUE) {
- /* Remember whether a flip occurred during this pass */
- int flipped = FALSE;
-
- for (i = 0; i < num_faces; ++i) {
- int j = face_list[i];
- enum face_colour opp =
- (board[j] == FACE_WHITE) ? FACE_BLACK : FACE_WHITE;
- if (can_colour_face(g, board, j, opp)) {
- grid_face *face = g->faces +j;
- if (do_random_pass) {
- /* final random pass */
- if (!random_upto(rs, 10))
- board[j] = opp;
- } else {
- /* normal pass - flip when neighbour count is 1 */
- if (face_num_neighbours(g, board, face, opp) == 1) {
- board[j] = opp;
- flipped = TRUE;
- }
- }
- }
- }
-
- if (do_random_pass) break;
- if (!flipped) do_random_pass = TRUE;
- }
+ char *board = snewn(g->num_faces, char);
+ int i;
- sfree(face_list);
+ generate_loop(g, board, rs, NULL, NULL);
/* Fill out all the clues by initialising to 0, then iterating over
* all edges and incrementing each clue as we find edges that border
* between BLACK/WHITE faces. While we're at it, we verify that the
* algorithm does work, and there aren't any GREY faces still there. */
- memset(clues, 0, num_faces);
+ memset(clues, 0, g->num_faces);
for (i = 0; i < g->num_edges; i++) {
grid_edge *e = g->edges + i;
grid_face *f1 = e->face1;
if (f2) clues[f2 - g->faces]++;
}
}
-
sfree(board);
}
}
-static char *new_game_desc(game_params *params, random_state *rs,
+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;
+ char *retval, *game_desc, *grid_desc;
grid *g;
game_state *state = snew(game_state);
game_state *state_new;
- params_generate_grid(params);
- state->game_grid = g = params->game_grid;
- g->refcount++;
+
+ grid_desc = grid_new_desc(grid_types[params->type], params->w, params->h, rs);
+ state->game_grid = g = loopy_generate_grid(params, grid_desc);
+
state->clues = snewn(g->num_faces, signed char);
state->lines = snewn(g->num_edges, char);
state->line_errors = snewn(g->num_edges, unsigned char);
goto newboard_please;
}
- retval = state_to_text(state);
+ game_desc = state_to_text(state);
free_game(state);
+ if (grid_desc) {
+ retval = snewn(strlen(grid_desc) + 1 + strlen(game_desc) + 1, char);
+ sprintf(retval, "%s%c%s", grid_desc, (int)GRID_DESC_SEP, game_desc);
+ sfree(grid_desc);
+ sfree(game_desc);
+ } else {
+ retval = game_desc;
+ }
+
assert(!validate_desc(params, retval));
return retval;
}
-static game_state *new_game(midend *me, game_params *params, char *desc)
+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 = desc;
+ const char *dp;
+ char *grid_desc;
grid *g;
int num_faces, num_edges;
- params_generate_grid(params);
- state->game_grid = g = params->game_grid;
- g->refcount++;
+ grid_desc = extract_grid_desc(&desc);
+ state->game_grid = g = loopy_generate_grid(params, grid_desc);
+ if (grid_desc) sfree(grid_desc);
+
+ dp = desc;
+
num_faces = g->num_faces;
num_edges = g->num_edges;
return sstate;
}
-static char *solve_game(game_state *state, game_state *currstate,
- char *aux, char **error)
+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;
* Drawing and mouse-handling
*/
-static char *interpret_move(game_state *state, game_ui *ui, game_drawstate *ds,
+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;
return ret;
}
-static game_state *execute_move(game_state *state, char *move)
+static game_state *execute_move(const game_state *state, const char *move)
{
int i;
game_state *newstate = dup_game(state);
*y += BORDER(ds->tilesize);
}
-static int solve_2x2_matrix(double mx[4], double vin[2], double vout[2])
-{
- double inv[4];
- double det;
- det = (mx[0]*mx[3] - mx[1]*mx[2]);
- if (det == 0)
- return FALSE;
-
- inv[0] = mx[3] / det;
- inv[1] = -mx[1] / det;
- inv[2] = -mx[2] / det;
- inv[3] = mx[0] / det;
-
- vout[0] = inv[0]*vin[0] + inv[1]*vin[1];
- vout[1] = inv[2]*vin[0] + inv[3]*vin[1];
-
- return TRUE;
-}
-
-static int solve_3x3_matrix(double mx[9], double vin[3], double vout[3])
-{
- double inv[9];
- double det;
-
- det = (mx[0]*mx[4]*mx[8] + mx[1]*mx[5]*mx[6] + mx[2]*mx[3]*mx[7] -
- mx[0]*mx[5]*mx[7] - mx[1]*mx[3]*mx[8] - mx[2]*mx[4]*mx[6]);
- if (det == 0)
- return FALSE;
-
- inv[0] = (mx[4]*mx[8] - mx[5]*mx[7]) / det;
- inv[1] = (mx[2]*mx[7] - mx[1]*mx[8]) / det;
- inv[2] = (mx[1]*mx[5] - mx[2]*mx[4]) / det;
- inv[3] = (mx[5]*mx[6] - mx[3]*mx[8]) / det;
- inv[4] = (mx[0]*mx[8] - mx[2]*mx[6]) / det;
- inv[5] = (mx[2]*mx[3] - mx[0]*mx[5]) / det;
- inv[6] = (mx[3]*mx[7] - mx[4]*mx[6]) / det;
- inv[7] = (mx[1]*mx[6] - mx[0]*mx[7]) / det;
- inv[8] = (mx[0]*mx[4] - mx[1]*mx[3]) / det;
-
- vout[0] = inv[0]*vin[0] + inv[1]*vin[1] + inv[2]*vin[2];
- vout[1] = inv[3]*vin[0] + inv[4]*vin[1] + inv[5]*vin[2];
- vout[2] = inv[6]*vin[0] + inv[7]*vin[1] + inv[8]*vin[2];
-
- return TRUE;
-}
-
/* Returns (into x,y) position of centre of face for rendering the text clue.
*/
static void face_text_pos(const game_drawstate *ds, const grid *g,
- const grid_face *f, int *xret, int *yret)
+ grid_face *f, int *xret, int *yret)
{
- double xbest, ybest, bestdist;
- int i, j, k, m;
- grid_dot *edgedot1[3], *edgedot2[3];
- grid_dot *dots[3];
- int nedges, ndots;
int faceindex = f - g->faces;
/*
}
/*
- * Otherwise, try to find the point in the polygon with the
- * maximum distance to any edge or corner.
- *
- * This point must be in contact with at least three edges and/or
- * vertices; so we iterate through all combinations of three of
- * those, and find candidate points in each set.
- *
- * We don't actually iterate literally over _edges_, in the sense
- * of grid_edge structures. Instead, we fill in edgedot1[] and
- * edgedot2[] with a pair of dots adjacent in the face's list of
- * vertices. This ensures that we get the edges in consistent
- * orientation, which we could not do from the grid structure
- * alone. (A moment's consideration of an order-3 vertex should
- * make it clear that if a notional arrow was written on each
- * edge, _at least one_ of the three faces bordering that vertex
- * would have to have the two arrows tip-to-tip or tail-to-tail
- * rather than tip-to-tail.)
+ * Otherwise, use the incentre computed by grid.c and convert it
+ * to screen coordinates.
*/
- nedges = ndots = 0;
- bestdist = 0;
- xbest = ybest = 0;
-
- for (i = 0; i+2 < 2*f->order; i++) {
- if (i < f->order) {
- edgedot1[nedges] = f->dots[i];
- edgedot2[nedges++] = f->dots[(i+1)%f->order];
- } else
- dots[ndots++] = f->dots[i - f->order];
-
- for (j = i+1; j+1 < 2*f->order; j++) {
- if (j < f->order) {
- edgedot1[nedges] = f->dots[j];
- edgedot2[nedges++] = f->dots[(j+1)%f->order];
- } else
- dots[ndots++] = f->dots[j - f->order];
-
- for (k = j+1; k < 2*f->order; k++) {
- double cx[2], cy[2]; /* candidate positions */
- int cn = 0; /* number of candidates */
-
- if (k < f->order) {
- edgedot1[nedges] = f->dots[k];
- edgedot2[nedges++] = f->dots[(k+1)%f->order];
- } else
- dots[ndots++] = f->dots[k - f->order];
-
- /*
- * Find a point, or pair of points, equidistant from
- * all the specified edges and/or vertices.
- */
- if (nedges == 3) {
- /*
- * Three edges. This is a linear matrix equation:
- * each row of the matrix represents the fact that
- * the point (x,y) we seek is at distance r from
- * that edge, and we solve three of those
- * simultaneously to obtain x,y,r. (We ignore r.)
- */
- double matrix[9], vector[3], vector2[3];
- int m;
-
- for (m = 0; m < 3; m++) {
- int x1 = edgedot1[m]->x, x2 = edgedot2[m]->x;
- int y1 = edgedot1[m]->y, y2 = edgedot2[m]->y;
- int dx = x2-x1, dy = y2-y1;
-
- /*
- * ((x,y) - (x1,y1)) . (dy,-dx) = r |(dx,dy)|
- *
- * => x dy - y dx - r |(dx,dy)| = (x1 dy - y1 dx)
- */
- matrix[3*m+0] = dy;
- matrix[3*m+1] = -dx;
- matrix[3*m+2] = -sqrt((double)dx*dx+(double)dy*dy);
- vector[m] = (double)x1*dy - (double)y1*dx;
- }
-
- if (solve_3x3_matrix(matrix, vector, vector2)) {
- cx[cn] = vector2[0];
- cy[cn] = vector2[1];
- cn++;
- }
- } else if (nedges == 2) {
- /*
- * Two edges and a dot. This will end up in a
- * quadratic equation.
- *
- * First, look at the two edges. Having our point
- * be some distance r from both of them gives rise
- * to a pair of linear equations in x,y,r of the
- * form
- *
- * (x-x1) dy - (y-y1) dx = r sqrt(dx^2+dy^2)
- *
- * We eliminate r between those equations to give
- * us a single linear equation in x,y describing
- * the locus of points equidistant from both lines
- * - i.e. the angle bisector.
- *
- * We then choose one of x,y to be a parameter t,
- * and derive linear formulae for x,y,r in terms
- * of t. This enables us to write down the
- * circular equation (x-xd)^2+(y-yd)^2=r^2 as a
- * quadratic in t; solving that and substituting
- * in for x,y gives us two candidate points.
- */
- double eqs[2][4]; /* a,b,c,d : ax+by+cr=d */
- double eq[3]; /* a,b,c: ax+by=c */
- double xt[2], yt[2], rt[2]; /* a,b: {x,y,r}=at+b */
- double q[3]; /* a,b,c: at^2+bt+c=0 */
- double disc;
-
- /* Find equations of the two input lines. */
- for (m = 0; m < 2; m++) {
- int x1 = edgedot1[m]->x, x2 = edgedot2[m]->x;
- int y1 = edgedot1[m]->y, y2 = edgedot2[m]->y;
- int dx = x2-x1, dy = y2-y1;
-
- eqs[m][0] = dy;
- eqs[m][1] = -dx;
- eqs[m][2] = -sqrt(dx*dx+dy*dy);
- eqs[m][3] = x1*dy - y1*dx;
- }
-
- /* Derive the angle bisector by eliminating r. */
- eq[0] = eqs[0][0]*eqs[1][2] - eqs[1][0]*eqs[0][2];
- eq[1] = eqs[0][1]*eqs[1][2] - eqs[1][1]*eqs[0][2];
- eq[2] = eqs[0][3]*eqs[1][2] - eqs[1][3]*eqs[0][2];
-
- /* Parametrise x and y in terms of some t. */
- if (abs(eq[0]) < abs(eq[1])) {
- /* Parameter is x. */
- xt[0] = 1; xt[1] = 0;
- yt[0] = -eq[0]/eq[1]; yt[1] = eq[2]/eq[1];
- } else {
- /* Parameter is y. */
- yt[0] = 1; yt[1] = 0;
- xt[0] = -eq[1]/eq[0]; xt[1] = eq[2]/eq[0];
- }
-
- /* Find a linear representation of r using eqs[0]. */
- rt[0] = -(eqs[0][0]*xt[0] + eqs[0][1]*yt[0])/eqs[0][2];
- rt[1] = (eqs[0][3] - eqs[0][0]*xt[1] -
- eqs[0][1]*yt[1])/eqs[0][2];
-
- /* Construct the quadratic equation. */
- q[0] = -rt[0]*rt[0];
- q[1] = -2*rt[0]*rt[1];
- q[2] = -rt[1]*rt[1];
- q[0] += xt[0]*xt[0];
- q[1] += 2*xt[0]*(xt[1]-dots[0]->x);
- q[2] += (xt[1]-dots[0]->x)*(xt[1]-dots[0]->x);
- q[0] += yt[0]*yt[0];
- q[1] += 2*yt[0]*(yt[1]-dots[0]->y);
- q[2] += (yt[1]-dots[0]->y)*(yt[1]-dots[0]->y);
-
- /* And solve it. */
- disc = q[1]*q[1] - 4*q[0]*q[2];
- if (disc >= 0) {
- double t;
-
- disc = sqrt(disc);
-
- t = (-q[1] + disc) / (2*q[0]);
- cx[cn] = xt[0]*t + xt[1];
- cy[cn] = yt[0]*t + yt[1];
- cn++;
-
- t = (-q[1] - disc) / (2*q[0]);
- cx[cn] = xt[0]*t + xt[1];
- cy[cn] = yt[0]*t + yt[1];
- cn++;
- }
- } else if (nedges == 1) {
- /*
- * Two dots and an edge. This one's another
- * quadratic equation.
- *
- * The point we want must lie on the perpendicular
- * bisector of the two dots; that much is obvious.
- * So we can construct a parametrisation of that
- * bisecting line, giving linear formulae for x,y
- * in terms of t. We can also express the distance
- * from the edge as such a linear formula.
- *
- * Then we set that equal to the radius of the
- * circle passing through the two points, which is
- * a Pythagoras exercise; that gives rise to a
- * quadratic in t, which we solve.
- */
- double xt[2], yt[2], rt[2]; /* a,b: {x,y,r}=at+b */
- double q[3]; /* a,b,c: at^2+bt+c=0 */
- double disc;
- double halfsep;
-
- /* Find parametric formulae for x,y. */
- {
- int x1 = dots[0]->x, x2 = dots[1]->x;
- int y1 = dots[0]->y, y2 = dots[1]->y;
- int dx = x2-x1, dy = y2-y1;
- double d = sqrt((double)dx*dx + (double)dy*dy);
-
- xt[1] = (x1+x2)/2.0;
- yt[1] = (y1+y2)/2.0;
- /* It's convenient if we have t at standard scale. */
- xt[0] = -dy/d;
- yt[0] = dx/d;
-
- /* Also note down half the separation between
- * the dots, for use in computing the circle radius. */
- halfsep = 0.5*d;
- }
-
- /* Find a parametric formula for r. */
- {
- int x1 = edgedot1[0]->x, x2 = edgedot2[0]->x;
- int y1 = edgedot1[0]->y, y2 = edgedot2[0]->y;
- int dx = x2-x1, dy = y2-y1;
- double d = sqrt((double)dx*dx + (double)dy*dy);
- rt[0] = (xt[0]*dy - yt[0]*dx) / d;
- rt[1] = ((xt[1]-x1)*dy - (yt[1]-y1)*dx) / d;
- }
-
- /* Construct the quadratic equation. */
- q[0] = rt[0]*rt[0];
- q[1] = 2*rt[0]*rt[1];
- q[2] = rt[1]*rt[1];
- q[0] -= 1;
- q[2] -= halfsep*halfsep;
-
- /* And solve it. */
- disc = q[1]*q[1] - 4*q[0]*q[2];
- if (disc >= 0) {
- double t;
-
- disc = sqrt(disc);
-
- t = (-q[1] + disc) / (2*q[0]);
- cx[cn] = xt[0]*t + xt[1];
- cy[cn] = yt[0]*t + yt[1];
- cn++;
-
- t = (-q[1] - disc) / (2*q[0]);
- cx[cn] = xt[0]*t + xt[1];
- cy[cn] = yt[0]*t + yt[1];
- cn++;
- }
- } else if (nedges == 0) {
- /*
- * Three dots. This is another linear matrix
- * equation, this time with each row of the matrix
- * representing the perpendicular bisector between
- * two of the points. Of course we only need two
- * such lines to find their intersection, so we
- * need only solve a 2x2 matrix equation.
- */
-
- double matrix[4], vector[2], vector2[2];
- int m;
-
- for (m = 0; m < 2; m++) {
- int x1 = dots[m]->x, x2 = dots[m+1]->x;
- int y1 = dots[m]->y, y2 = dots[m+1]->y;
- int dx = x2-x1, dy = y2-y1;
-
- /*
- * ((x,y) - (x1,y1)) . (dx,dy) = 1/2 |(dx,dy)|^2
- *
- * => 2x dx + 2y dy = dx^2+dy^2 + (2 x1 dx + 2 y1 dy)
- */
- matrix[2*m+0] = 2*dx;
- matrix[2*m+1] = 2*dy;
- vector[m] = ((double)dx*dx + (double)dy*dy +
- 2.0*x1*dx + 2.0*y1*dy);
- }
-
- if (solve_2x2_matrix(matrix, vector, vector2)) {
- cx[cn] = vector2[0];
- cy[cn] = vector2[1];
- cn++;
- }
- }
-
- /*
- * Now go through our candidate points and see if any
- * of them are better than what we've got so far.
- */
- for (m = 0; m < cn; m++) {
- double x = cx[m], y = cy[m];
-
- /*
- * First, disqualify the point if it's not inside
- * the polygon, which we work out by counting the
- * edges to the right of the point. (For
- * tiebreaking purposes when edges start or end on
- * our y-coordinate or go right through it, we
- * consider our point to be offset by a small
- * _positive_ epsilon in both the x- and
- * y-direction.)
- */
- int e, in = 0;
- for (e = 0; e < f->order; e++) {
- int xs = f->edges[e]->dot1->x;
- int xe = f->edges[e]->dot2->x;
- int ys = f->edges[e]->dot1->y;
- int ye = f->edges[e]->dot2->y;
- if ((y >= ys && y < ye) || (y >= ye && y < ys)) {
- /*
- * The line goes past our y-position. Now we need
- * to know if its x-coordinate when it does so is
- * to our right.
- *
- * The x-coordinate in question is mathematically
- * (y - ys) * (xe - xs) / (ye - ys), and we want
- * to know whether (x - xs) >= that. Of course we
- * avoid the division, so we can work in integers;
- * to do this we must multiply both sides of the
- * inequality by ye - ys, which means we must
- * first check that's not negative.
- */
- int num = xe - xs, denom = ye - ys;
- if (denom < 0) {
- num = -num;
- denom = -denom;
- }
- if ((x - xs) * denom >= (y - ys) * num)
- in ^= 1;
- }
- }
-
- if (in) {
- double mindist = HUGE_VAL;
- int e, d;
-
- /*
- * This point is inside the polygon, so now we check
- * its minimum distance to every edge and corner.
- * First the corners ...
- */
- for (d = 0; d < f->order; d++) {
- int xp = f->dots[d]->x;
- int yp = f->dots[d]->y;
- double dx = x - xp, dy = y - yp;
- double dist = dx*dx + dy*dy;
- if (mindist > dist)
- mindist = dist;
- }
-
- /*
- * ... and now also check the perpendicular distance
- * to every edge, if the perpendicular lies between
- * the edge's endpoints.
- */
- for (e = 0; e < f->order; e++) {
- int xs = f->edges[e]->dot1->x;
- int xe = f->edges[e]->dot2->x;
- int ys = f->edges[e]->dot1->y;
- int ye = f->edges[e]->dot2->y;
-
- /*
- * If s and e are our endpoints, and p our
- * candidate circle centre, the foot of a
- * perpendicular from p to the line se lies
- * between s and e if and only if (p-s).(e-s) lies
- * strictly between 0 and (e-s).(e-s).
- */
- int edx = xe - xs, edy = ye - ys;
- double pdx = x - xs, pdy = y - ys;
- double pde = pdx * edx + pdy * edy;
- long ede = (long)edx * edx + (long)edy * edy;
- if (0 < pde && pde < ede) {
- /*
- * Yes, the nearest point on this edge is
- * closer than either endpoint, so we must
- * take it into account by measuring the
- * perpendicular distance to the edge and
- * checking its square against mindist.
- */
-
- double pdre = pdx * edy - pdy * edx;
- double sqlen = pdre * pdre / ede;
-
- if (mindist > sqlen)
- mindist = sqlen;
- }
- }
-
- /*
- * Right. Now we know the biggest circle around this
- * point, so we can check it against bestdist.
- */
- if (bestdist < mindist) {
- bestdist = mindist;
- xbest = x;
- ybest = y;
- }
- }
- }
-
- if (k < f->order)
- nedges--;
- else
- ndots--;
- }
- if (j < f->order)
- nedges--;
- else
- ndots--;
- }
- if (i < f->order)
- nedges--;
- else
- ndots--;
- }
-
- assert(bestdist > 0);
-
- /* convert to screen coordinates. Round doubles to nearest. */
- grid_to_screen(ds, g, xbest+0.5, ybest+0.5,
+ grid_find_incentre(f);
+ grid_to_screen(ds, g, f->ix, f->iy,
&ds->textx[faceindex], &ds->texty[faceindex]);
*xret = ds->textx[faceindex];
}
static void game_redraw_clue(drawing *dr, game_drawstate *ds,
- game_state *state, int i)
+ const game_state *state, int i)
{
grid *g = state->game_grid;
grid_face *f = g->faces + i;
int x, y;
- char c[3];
+ char c[20];
- if (state->clues[i] < 10) {
- c[0] = CLUE2CHAR(state->clues[i]);
- c[1] = '\0';
- } else {
- sprintf(c, "%d", state->clues[i]);
- }
+ sprintf(c, "%d", state->clues[i]);
face_text_pos(ds, g, f, &x, &y);
draw_text(dr, x, y,
#define NPHASES lenof(loopy_line_redraw_phases)
static void game_redraw_line(drawing *dr, game_drawstate *ds,
- game_state *state, int i, int phase)
+ 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 xmin, ymin, xmax, ymax;
int line_colour;
if (state->line_errors[i])
grid_to_screen(ds, g, e->dot1->x, e->dot1->y, &x1, &y1);
grid_to_screen(ds, g, e->dot2->x, e->dot2->y, &x2, &y2);
- xmin = min(x1, x2);
- xmax = max(x1, x2);
- ymin = min(y1, y2);
- ymax = max(y1, y2);
-
if (line_colour == COL_FAINT) {
static int draw_faint_lines = -1;
if (draw_faint_lines < 0) {
}
static void game_redraw_dot(drawing *dr, game_drawstate *ds,
- game_state *state, int i)
+ const game_state *state, int i)
{
grid *g = state->game_grid;
grid_dot *d = g->dots + i;
}
static void game_redraw_in_rect(drawing *dr, game_drawstate *ds,
- game_state *state, int x, int y, int w, int h)
+ const game_state *state,
+ int x, int y, int w, int h)
{
grid *g = state->game_grid;
int i, phase;
draw_update(dr, x, y, w, h);
}
-static void game_redraw(drawing *dr, game_drawstate *ds, game_state *oldstate,
- game_state *state, int dir, game_ui *ui,
+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 */
* what needs doing, and the second actually does it.
*/
- if (!ds->started)
+ if (!ds->started) {
redraw_everything = TRUE;
- else {
-
- /* First, trundle through the faces. */
- for (i = 0; i < g->num_faces; i++) {
- grid_face *f = g->faces + i;
- int sides = f->order;
- int clue_mistake;
- int clue_satisfied;
- int n = state->clues[i];
- if (n < 0)
- continue;
-
- clue_mistake = (face_order(state, i, LINE_YES) > n ||
- face_order(state, i, LINE_NO ) > (sides-n));
- clue_satisfied = (face_order(state, i, LINE_YES) == n &&
- face_order(state, i, LINE_NO ) == (sides-n));
-
- if (clue_mistake != ds->clue_error[i] ||
- clue_satisfied != ds->clue_satisfied[i]) {
- ds->clue_error[i] = clue_mistake;
- ds->clue_satisfied[i] = clue_satisfied;
- if (nfaces == REDRAW_OBJECTS_LIMIT)
- redraw_everything = TRUE;
- else
- faces[nfaces++] = i;
- }
- }
+ /*
+ * But we must still go through the upcoming loops, so that we
+ * set up stuff in ds correctly for the initial redraw.
+ */
+ }
- /* Work out what the flash state needs to be. */
- if (flashtime > 0 &&
- (flashtime <= FLASH_TIME/3 ||
- flashtime >= FLASH_TIME*2/3)) {
- flash_changed = !ds->flashing;
- ds->flashing = TRUE;
- } else {
- flash_changed = ds->flashing;
- ds->flashing = FALSE;
- }
+ /* First, trundle through the faces. */
+ for (i = 0; i < g->num_faces; i++) {
+ grid_face *f = g->faces + i;
+ int sides = f->order;
+ int clue_mistake;
+ int clue_satisfied;
+ int n = state->clues[i];
+ if (n < 0)
+ continue;
- /* Now, trundle through the edges. */
- for (i = 0; i < g->num_edges; i++) {
- char new_ds =
- state->line_errors[i] ? DS_LINE_ERROR : state->lines[i];
- if (new_ds != ds->lines[i] ||
- (flash_changed && state->lines[i] == LINE_YES)) {
- ds->lines[i] = new_ds;
- if (nedges == REDRAW_OBJECTS_LIMIT)
- redraw_everything = TRUE;
- else
- edges[nedges++] = i;
- }
- }
+ clue_mistake = (face_order(state, i, LINE_YES) > n ||
+ face_order(state, i, LINE_NO ) > (sides-n));
+ clue_satisfied = (face_order(state, i, LINE_YES) == n &&
+ face_order(state, i, LINE_NO ) == (sides-n));
+
+ if (clue_mistake != ds->clue_error[i] ||
+ clue_satisfied != ds->clue_satisfied[i]) {
+ ds->clue_error[i] = clue_mistake;
+ ds->clue_satisfied[i] = clue_satisfied;
+ if (nfaces == REDRAW_OBJECTS_LIMIT)
+ redraw_everything = TRUE;
+ else
+ faces[nfaces++] = i;
+ }
+ }
+
+ /* Work out what the flash state needs to be. */
+ if (flashtime > 0 &&
+ (flashtime <= FLASH_TIME/3 ||
+ flashtime >= FLASH_TIME*2/3)) {
+ flash_changed = !ds->flashing;
+ ds->flashing = TRUE;
+ } else {
+ flash_changed = ds->flashing;
+ ds->flashing = FALSE;
+ }
+
+ /* Now, trundle through the edges. */
+ for (i = 0; i < g->num_edges; i++) {
+ char new_ds =
+ state->line_errors[i] ? DS_LINE_ERROR : state->lines[i];
+ if (new_ds != ds->lines[i] ||
+ (flash_changed && state->lines[i] == LINE_YES)) {
+ ds->lines[i] = new_ds;
+ if (nedges == REDRAW_OBJECTS_LIMIT)
+ redraw_everything = TRUE;
+ else
+ edges[nedges++] = i;
+ }
}
/* Pass one is now done. Now we do the actual drawing. */
ds->started = TRUE;
}
-static float game_flash_length(game_state *oldstate, game_state *newstate,
- int dir, game_ui *ui)
+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 0.0F;
}
-static int game_is_solved(game_state *state)
+static int game_status(const game_state *state)
{
- return state->solved;
+ return state->solved ? +1 : 0;
}
-static void game_print_size(game_params *params, float *x, float *y)
+static void game_print_size(const game_params *params, float *x, float *y)
{
int pw, ph;
*y = ph / 100.0F;
}
-static void game_print(drawing *dr, game_state *state, int tilesize)
+static void game_print(drawing *dr, const game_state *state, int tilesize)
{
int ink = print_mono_colour(dr, 0);
int i;
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_face *f = g->faces + i;
int clue = state->clues[i];
if (clue >= 0) {
- char c[2];
+ char c[20];
int x, y;
- c[0] = CLUE2CHAR(clue);
- c[1] = '\0';
+ sprintf(c, "%d", state->clues[i]);
face_text_pos(ds, g, f, &x, &y);
draw_text(dr, x, y,
FONT_VARIABLE, ds->tilesize / 2,
}
}
}
+
+ sfree(ds->textx);
+ sfree(ds->texty);
}
#ifdef COMBINED
game_redraw,
game_anim_length,
game_flash_length,
- game_is_solved,
+ game_status,
TRUE, FALSE, game_print_size, game_print,
FALSE /* wants_statusbar */,
FALSE, game_timing_state,
}
#endif
+
+/* vim: set shiftwidth=4 tabstop=8: */