X-Git-Url: https://git.distorted.org.uk/~mdw/sgt/puzzles/blobdiff_plain/a2f35d71b745ec2a03de58976c6434437c5f303e..b21345d458b5ea10f110a59b7522fe602cc1b9ea:/loopy.c diff --git a/loopy.c b/loopy.c index 7d03ed2..8617c72 100644 --- a/loopy.c +++ b/loopy.c @@ -107,7 +107,7 @@ enum { }; struct game_state { - grid *game_grid; + grid *game_grid; /* ref-counted (internally) */ /* Put -1 in a face that doesn't get a clue */ signed char *clues; @@ -207,10 +207,6 @@ struct game_params { int w, h; int diff; int type; - - /* Grid generation is expensive, so keep a (ref-counted) reference to the - * grid for these parameters, and only generate when required. */ - grid *game_grid; }; /* line_drawstate is the same as line_state, but with the extra ERROR @@ -247,29 +243,31 @@ static void check_caches(const solver_state* sstate); /* ------- List of grid generators ------- */ #define GRIDLIST(A) \ - A(Squares,grid_new_square,3,3) \ - A(Triangular,grid_new_triangular,3,3) \ - A(Honeycomb,grid_new_honeycomb,3,3) \ - A(Snub-Square,grid_new_snubsquare,3,3) \ - A(Cairo,grid_new_cairo,3,4) \ - A(Great-Hexagonal,grid_new_greathexagonal,3,3) \ - A(Octagonal,grid_new_octagonal,3,3) \ - A(Kites,grid_new_kites,3,3) \ - A(Floret,grid_new_floret,1,2) \ - A(Dodecagonal,grid_new_dodecagonal,2,2) \ - A(Great-Dodecagonal,grid_new_greatdodecagonal,2,2) - -#define GRID_NAME(title,fn,amin,omin) #title, -#define GRID_CONFIG(title,fn,amin,omin) ":" #title -#define GRID_FN(title,fn,amin,omin) &fn, -#define GRID_SIZES(title,fn,amin,omin) \ + A(Squares,GRID_SQUARE,3,3) \ + A(Triangular,GRID_TRIANGULAR,3,3) \ + A(Honeycomb,GRID_HONEYCOMB,3,3) \ + A(Snub-Square,GRID_SNUBSQUARE,3,3) \ + A(Cairo,GRID_CAIRO,3,4) \ + A(Great-Hexagonal,GRID_GREATHEXAGONAL,3,3) \ + A(Octagonal,GRID_OCTAGONAL,3,3) \ + A(Kites,GRID_KITE,3,3) \ + A(Floret,GRID_FLORET,1,2) \ + A(Dodecagonal,GRID_DODECAGONAL,2,2) \ + A(Great-Dodecagonal,GRID_GREATDODECAGONAL,2,2) \ + A(Penrose (kite/dart),GRID_PENROSE_P2,3,3) \ + A(Penrose (rhombs),GRID_PENROSE_P3,3,3) + +#define GRID_NAME(title,type,amin,omin) #title, +#define GRID_CONFIG(title,type,amin,omin) ":" #title +#define GRID_TYPE(title,type,amin,omin) type, +#define GRID_SIZES(title,type,amin,omin) \ {amin, omin, \ "Width and height for this grid type must both be at least " #amin, \ "At least one of width and height for this grid type must be at least " #omin,}, static char const *const gridnames[] = { GRIDLIST(GRID_NAME) }; #define GRID_CONFIGS GRIDLIST(GRID_CONFIG) -static grid * (*(grid_fns[]))(int w, int h) = { GRIDLIST(GRID_FN) }; -#define NUM_GRID_TYPES (sizeof(grid_fns) / sizeof(grid_fns[0])) +static grid_type grid_types[] = { GRIDLIST(GRID_TYPE) }; +#define NUM_GRID_TYPES (sizeof(grid_types) / sizeof(grid_types[0])) static const struct { int amin, omin; char *aerr, *oerr; @@ -277,13 +275,10 @@ static const struct { /* Generates a (dynamically allocated) new grid, according to the * type and size requested in params. Does nothing if the grid is already - * generated. The allocated grid is owned by the params object, and will be - * freed in free_params(). */ -static void params_generate_grid(game_params *params) + * generated. */ +static grid *loopy_generate_grid(game_params *params, char *grid_desc) { - if (!params->game_grid) { - params->game_grid = grid_fns[params->type](params->w, params->h); - } + return grid_new(grid_types[params->type], params->w, params->h, grid_desc); } /* ---------------------------------------------------------------------- @@ -480,8 +475,6 @@ static game_params *default_params(void) ret->diff = DIFF_EASY; ret->type = 0; - ret->game_grid = NULL; - return ret; } @@ -490,44 +483,45 @@ static game_params *dup_params(game_params *params) game_params *ret = snew(game_params); *ret = *params; /* structure copy */ - if (ret->game_grid) { - ret->game_grid->refcount++; - } return ret; } static const game_params presets[] = { #ifdef SMALL_SCREEN - { 7, 7, DIFF_EASY, 0, NULL }, - { 7, 7, DIFF_NORMAL, 0, NULL }, - { 7, 7, DIFF_HARD, 0, NULL }, - { 7, 7, DIFF_HARD, 1, NULL }, - { 7, 7, DIFF_HARD, 2, NULL }, - { 5, 5, DIFF_HARD, 3, NULL }, - { 7, 7, DIFF_HARD, 4, NULL }, - { 5, 4, DIFF_HARD, 5, NULL }, - { 5, 5, DIFF_HARD, 6, NULL }, - { 5, 5, DIFF_HARD, 7, NULL }, - { 3, 3, DIFF_HARD, 8, NULL }, - { 3, 3, DIFF_HARD, 9, NULL }, - { 3, 3, DIFF_HARD, 10, NULL }, + { 7, 7, DIFF_EASY, 0 }, + { 7, 7, DIFF_NORMAL, 0 }, + { 7, 7, DIFF_HARD, 0 }, + { 7, 7, DIFF_HARD, 1 }, + { 7, 7, DIFF_HARD, 2 }, + { 5, 5, DIFF_HARD, 3 }, + { 7, 7, DIFF_HARD, 4 }, + { 5, 4, DIFF_HARD, 5 }, + { 5, 5, DIFF_HARD, 6 }, + { 5, 5, DIFF_HARD, 7 }, + { 3, 3, DIFF_HARD, 8 }, + { 3, 3, DIFF_HARD, 9 }, + { 3, 3, DIFF_HARD, 10 }, + { 6, 6, DIFF_HARD, 11 }, + { 6, 6, DIFF_HARD, 12 }, #else - { 7, 7, DIFF_EASY, 0, NULL }, - { 10, 10, DIFF_EASY, 0, NULL }, - { 7, 7, DIFF_NORMAL, 0, NULL }, - { 10, 10, DIFF_NORMAL, 0, NULL }, - { 7, 7, DIFF_HARD, 0, NULL }, - { 10, 10, DIFF_HARD, 0, NULL }, - { 10, 10, DIFF_HARD, 1, NULL }, - { 12, 10, DIFF_HARD, 2, NULL }, - { 7, 7, DIFF_HARD, 3, NULL }, - { 9, 9, DIFF_HARD, 4, NULL }, - { 5, 4, DIFF_HARD, 5, NULL }, - { 7, 7, DIFF_HARD, 6, NULL }, - { 5, 5, DIFF_HARD, 7, NULL }, - { 5, 5, DIFF_HARD, 8, NULL }, - { 5, 4, DIFF_HARD, 9, NULL }, - { 5, 4, DIFF_HARD, 10, NULL }, + { 7, 7, DIFF_EASY, 0 }, + { 10, 10, DIFF_EASY, 0 }, + { 7, 7, DIFF_NORMAL, 0 }, + { 10, 10, DIFF_NORMAL, 0 }, + { 7, 7, DIFF_HARD, 0 }, + { 10, 10, DIFF_HARD, 0 }, + { 10, 10, DIFF_HARD, 1 }, + { 12, 10, DIFF_HARD, 2 }, + { 7, 7, DIFF_HARD, 3 }, + { 9, 9, DIFF_HARD, 4 }, + { 5, 4, DIFF_HARD, 5 }, + { 7, 7, DIFF_HARD, 6 }, + { 5, 5, DIFF_HARD, 7 }, + { 5, 5, DIFF_HARD, 8 }, + { 5, 4, DIFF_HARD, 9 }, + { 5, 4, DIFF_HARD, 10 }, + { 10, 10, DIFF_HARD, 11 }, + { 10, 10, DIFF_HARD, 12 } #endif }; @@ -551,18 +545,11 @@ static int game_fetch_preset(int i, char **name, game_params **params) static void free_params(game_params *params) { - if (params->game_grid) { - grid_free(params->game_grid); - } sfree(params); } static void decode_params(game_params *params, char const *string) { - if (params->game_grid) { - grid_free(params->game_grid); - params->game_grid = NULL; - } params->h = params->w = atoi(string); params->diff = DIFF_EASY; while (*string && isdigit((unsigned char)*string)) string++; @@ -641,7 +628,6 @@ static game_params *custom_params(config_item *cfg) ret->type = cfg[2].ival; ret->diff = cfg[3].ival; - ret->game_grid = NULL; return ret; } @@ -702,14 +688,44 @@ static char *state_to_text(const game_state *state) return retval; } +#define GRID_DESC_SEP '_' + +/* Splits up a (optional) grid_desc from the game desc. Returns the + * grid_desc (which needs freeing) and updates the desc pointer to + * start of real desc, or returns NULL if no desc. */ +static char *extract_grid_desc(char **desc) +{ + char *sep = strchr(*desc, GRID_DESC_SEP), *gd; + int gd_len; + + if (!sep) return NULL; + + gd_len = sep - (*desc); + gd = snewn(gd_len+1, char); + memcpy(gd, *desc, gd_len); + gd[gd_len] = '\0'; + + *desc = sep+1; + + return gd; +} + /* We require that the params pass the test in validate_params and that the * description fills the entire game area */ static char *validate_desc(game_params *params, char *desc) { int count = 0; grid *g; - params_generate_grid(params); - g = params->game_grid; + char *grid_desc, *ret; + + /* It's pretty inefficient to do this just for validation. All we need to + * know is the precise number of faces. */ + grid_desc = extract_grid_desc(&desc); + ret = grid_validate_desc(grid_types[params->type], params->w, params->h, grid_desc); + if (ret) return ret; + + g = loopy_generate_grid(params, grid_desc); + if (grid_desc) sfree(grid_desc); for (; *desc; ++desc) { if ((*desc >= '0' && *desc <= '9') || (*desc >= 'A' && *desc <= 'Z')) { @@ -728,6 +744,8 @@ static char *validate_desc(game_params *params, char *desc) if (count > g->num_faces) return "Description too long for board size"; + grid_free(g); + return NULL; } @@ -809,16 +827,15 @@ static void game_changed_state(game_ui *ui, game_state *oldstate, static void game_compute_size(game_params *params, int tilesize, int *x, int *y) { - grid *g; int grid_width, grid_height, rendered_width, rendered_height; + int g_tilesize; + + grid_compute_size(grid_types[params->type], params->w, params->h, + &g_tilesize, &grid_width, &grid_height); - params_generate_grid(params); - g = params->game_grid; - grid_width = g->highest_x - g->lowest_x; - grid_height = g->highest_y - g->lowest_y; /* multiply first to minimise rounding error on integer division */ - rendered_width = grid_width * tilesize / g->tilesize; - rendered_height = grid_height * tilesize / g->tilesize; + rendered_width = grid_width * tilesize / g_tilesize; + rendered_height = grid_height * tilesize / g_tilesize; *x = rendered_width + 2 * BORDER(tilesize) + 1; *y = rendered_height + 2 * BORDER(tilesize) + 1; } @@ -900,6 +917,8 @@ static game_drawstate *game_new_drawstate(drawing *dr, game_state *state) static void game_free_drawstate(drawing *dr, game_drawstate *ds) { + sfree(ds->textx); + sfree(ds->texty); sfree(ds->clue_error); sfree(ds->clue_satisfied); sfree(ds->lines); @@ -1836,13 +1855,14 @@ static char *new_game_desc(game_params *params, random_state *rs, char **aux, int interactive) { /* solution and description both use run-length encoding in obvious ways */ - char *retval; + char *retval, *game_desc, *grid_desc; grid *g; game_state *state = snew(game_state); game_state *state_new; - params_generate_grid(params); - state->game_grid = g = params->game_grid; - g->refcount++; + + grid_desc = grid_new_desc(grid_types[params->type], params->w, params->h, rs); + state->game_grid = g = loopy_generate_grid(params, grid_desc); + state->clues = snewn(g->num_faces, signed char); state->lines = snewn(g->num_edges, char); state->line_errors = snewn(g->num_edges, unsigned char); @@ -1875,10 +1895,19 @@ static char *new_game_desc(game_params *params, random_state *rs, goto newboard_please; } - retval = state_to_text(state); + game_desc = state_to_text(state); free_game(state); + if (grid_desc) { + retval = snewn(strlen(grid_desc) + 1 + strlen(game_desc) + 1, char); + sprintf(retval, "%s%c%s", grid_desc, GRID_DESC_SEP, game_desc); + sfree(grid_desc); + sfree(game_desc); + } else { + retval = game_desc; + } + assert(!validate_desc(params, retval)); return retval; @@ -1890,13 +1919,17 @@ static game_state *new_game(midend *me, game_params *params, char *desc) game_state *state = snew(game_state); int empties_to_make = 0; int n,n2; - const char *dp = desc; + const char *dp; + char *grid_desc; grid *g; int num_faces, num_edges; - params_generate_grid(params); - state->game_grid = g = params->game_grid; - g->refcount++; + grid_desc = extract_grid_desc(&desc); + state->game_grid = g = loopy_generate_grid(params, grid_desc); + if (grid_desc) sfree(grid_desc); + + dp = desc; + num_faces = g->num_faces; num_edges = g->num_edges; @@ -3401,62 +3434,11 @@ static void grid_to_screen(const game_drawstate *ds, const grid *g, *y += BORDER(ds->tilesize); } -static int solve_2x2_matrix(double mx[4], double vin[2], double vout[2]) -{ - double inv[4]; - double det; - det = (mx[0]*mx[3] - mx[1]*mx[2]); - if (det == 0) - return FALSE; - - inv[0] = mx[3] / det; - inv[1] = -mx[1] / det; - inv[2] = -mx[2] / det; - inv[3] = mx[0] / det; - - vout[0] = inv[0]*vin[0] + inv[1]*vin[1]; - vout[1] = inv[2]*vin[0] + inv[3]*vin[1]; - - return TRUE; -} - -static int solve_3x3_matrix(double mx[9], double vin[3], double vout[3]) -{ - double inv[9]; - double det; - - det = (mx[0]*mx[4]*mx[8] + mx[1]*mx[5]*mx[6] + mx[2]*mx[3]*mx[7] - - mx[0]*mx[5]*mx[7] - mx[1]*mx[3]*mx[8] - mx[2]*mx[4]*mx[6]); - if (det == 0) - return FALSE; - - inv[0] = (mx[4]*mx[8] - mx[5]*mx[7]) / det; - inv[1] = (mx[2]*mx[7] - mx[1]*mx[8]) / det; - inv[2] = (mx[1]*mx[5] - mx[2]*mx[4]) / det; - inv[3] = (mx[5]*mx[6] - mx[3]*mx[8]) / det; - inv[4] = (mx[0]*mx[8] - mx[2]*mx[6]) / det; - inv[5] = (mx[2]*mx[3] - mx[0]*mx[5]) / det; - inv[6] = (mx[3]*mx[7] - mx[4]*mx[6]) / det; - inv[7] = (mx[1]*mx[6] - mx[0]*mx[7]) / det; - inv[8] = (mx[0]*mx[4] - mx[1]*mx[3]) / det; - - vout[0] = inv[0]*vin[0] + inv[1]*vin[1] + inv[2]*vin[2]; - vout[1] = inv[3]*vin[0] + inv[4]*vin[1] + inv[5]*vin[2]; - vout[2] = inv[6]*vin[0] + inv[7]*vin[1] + inv[8]*vin[2]; - - return TRUE; -} - /* Returns (into x,y) position of centre of face for rendering the text clue. */ static void face_text_pos(const game_drawstate *ds, const grid *g, - const grid_face *f, int *xret, int *yret) + grid_face *f, int *xret, int *yret) { - double xbest, ybest, bestdist; - int i, j, k, m; - grid_dot *edgedot1[3], *edgedot2[3]; - grid_dot *dots[3]; - int nedges, ndots; int faceindex = f - g->faces; /* @@ -3470,425 +3452,11 @@ static void face_text_pos(const game_drawstate *ds, const grid *g, } /* - * Otherwise, try to find the point in the polygon with the - * maximum distance to any edge or corner. - * - * This point must be in contact with at least three edges and/or - * vertices; so we iterate through all combinations of three of - * those, and find candidate points in each set. - * - * We don't actually iterate literally over _edges_, in the sense - * of grid_edge structures. Instead, we fill in edgedot1[] and - * edgedot2[] with a pair of dots adjacent in the face's list of - * vertices. This ensures that we get the edges in consistent - * orientation, which we could not do from the grid structure - * alone. (A moment's consideration of an order-3 vertex should - * make it clear that if a notional arrow was written on each - * edge, _at least one_ of the three faces bordering that vertex - * would have to have the two arrows tip-to-tip or tail-to-tail - * rather than tip-to-tail.) + * Otherwise, use the incentre computed by grid.c and convert it + * to screen coordinates. */ - nedges = ndots = 0; - bestdist = 0; - xbest = ybest = 0; - - for (i = 0; i+2 < 2*f->order; i++) { - if (i < f->order) { - edgedot1[nedges] = f->dots[i]; - edgedot2[nedges++] = f->dots[(i+1)%f->order]; - } else - dots[ndots++] = f->dots[i - f->order]; - - for (j = i+1; j+1 < 2*f->order; j++) { - if (j < f->order) { - edgedot1[nedges] = f->dots[j]; - edgedot2[nedges++] = f->dots[(j+1)%f->order]; - } else - dots[ndots++] = f->dots[j - f->order]; - - for (k = j+1; k < 2*f->order; k++) { - double cx[2], cy[2]; /* candidate positions */ - int cn = 0; /* number of candidates */ - - if (k < f->order) { - edgedot1[nedges] = f->dots[k]; - edgedot2[nedges++] = f->dots[(k+1)%f->order]; - } else - dots[ndots++] = f->dots[k - f->order]; - - /* - * Find a point, or pair of points, equidistant from - * all the specified edges and/or vertices. - */ - if (nedges == 3) { - /* - * Three edges. This is a linear matrix equation: - * each row of the matrix represents the fact that - * the point (x,y) we seek is at distance r from - * that edge, and we solve three of those - * simultaneously to obtain x,y,r. (We ignore r.) - */ - double matrix[9], vector[3], vector2[3]; - int m; - - for (m = 0; m < 3; m++) { - int x1 = edgedot1[m]->x, x2 = edgedot2[m]->x; - int y1 = edgedot1[m]->y, y2 = edgedot2[m]->y; - int dx = x2-x1, dy = y2-y1; - - /* - * ((x,y) - (x1,y1)) . (dy,-dx) = r |(dx,dy)| - * - * => x dy - y dx - r |(dx,dy)| = (x1 dy - y1 dx) - */ - matrix[3*m+0] = dy; - matrix[3*m+1] = -dx; - matrix[3*m+2] = -sqrt((double)dx*dx+(double)dy*dy); - vector[m] = (double)x1*dy - (double)y1*dx; - } - - if (solve_3x3_matrix(matrix, vector, vector2)) { - cx[cn] = vector2[0]; - cy[cn] = vector2[1]; - cn++; - } - } else if (nedges == 2) { - /* - * Two edges and a dot. This will end up in a - * quadratic equation. - * - * First, look at the two edges. Having our point - * be some distance r from both of them gives rise - * to a pair of linear equations in x,y,r of the - * form - * - * (x-x1) dy - (y-y1) dx = r sqrt(dx^2+dy^2) - * - * We eliminate r between those equations to give - * us a single linear equation in x,y describing - * the locus of points equidistant from both lines - * - i.e. the angle bisector. - * - * We then choose one of x,y to be a parameter t, - * and derive linear formulae for x,y,r in terms - * of t. This enables us to write down the - * circular equation (x-xd)^2+(y-yd)^2=r^2 as a - * quadratic in t; solving that and substituting - * in for x,y gives us two candidate points. - */ - double eqs[2][4]; /* a,b,c,d : ax+by+cr=d */ - double eq[3]; /* a,b,c: ax+by=c */ - double xt[2], yt[2], rt[2]; /* a,b: {x,y,r}=at+b */ - double q[3]; /* a,b,c: at^2+bt+c=0 */ - double disc; - - /* Find equations of the two input lines. */ - for (m = 0; m < 2; m++) { - int x1 = edgedot1[m]->x, x2 = edgedot2[m]->x; - int y1 = edgedot1[m]->y, y2 = edgedot2[m]->y; - int dx = x2-x1, dy = y2-y1; - - eqs[m][0] = dy; - eqs[m][1] = -dx; - eqs[m][2] = -sqrt(dx*dx+dy*dy); - eqs[m][3] = x1*dy - y1*dx; - } - - /* Derive the angle bisector by eliminating r. */ - eq[0] = eqs[0][0]*eqs[1][2] - eqs[1][0]*eqs[0][2]; - eq[1] = eqs[0][1]*eqs[1][2] - eqs[1][1]*eqs[0][2]; - eq[2] = eqs[0][3]*eqs[1][2] - eqs[1][3]*eqs[0][2]; - - /* Parametrise x and y in terms of some t. */ - if (abs(eq[0]) < abs(eq[1])) { - /* Parameter is x. */ - xt[0] = 1; xt[1] = 0; - yt[0] = -eq[0]/eq[1]; yt[1] = eq[2]/eq[1]; - } else { - /* Parameter is y. */ - yt[0] = 1; yt[1] = 0; - xt[0] = -eq[1]/eq[0]; xt[1] = eq[2]/eq[0]; - } - - /* Find a linear representation of r using eqs[0]. */ - rt[0] = -(eqs[0][0]*xt[0] + eqs[0][1]*yt[0])/eqs[0][2]; - rt[1] = (eqs[0][3] - eqs[0][0]*xt[1] - - eqs[0][1]*yt[1])/eqs[0][2]; - - /* Construct the quadratic equation. */ - q[0] = -rt[0]*rt[0]; - q[1] = -2*rt[0]*rt[1]; - q[2] = -rt[1]*rt[1]; - q[0] += xt[0]*xt[0]; - q[1] += 2*xt[0]*(xt[1]-dots[0]->x); - q[2] += (xt[1]-dots[0]->x)*(xt[1]-dots[0]->x); - q[0] += yt[0]*yt[0]; - q[1] += 2*yt[0]*(yt[1]-dots[0]->y); - q[2] += (yt[1]-dots[0]->y)*(yt[1]-dots[0]->y); - - /* And solve it. */ - disc = q[1]*q[1] - 4*q[0]*q[2]; - if (disc >= 0) { - double t; - - disc = sqrt(disc); - - t = (-q[1] + disc) / (2*q[0]); - cx[cn] = xt[0]*t + xt[1]; - cy[cn] = yt[0]*t + yt[1]; - cn++; - - t = (-q[1] - disc) / (2*q[0]); - cx[cn] = xt[0]*t + xt[1]; - cy[cn] = yt[0]*t + yt[1]; - cn++; - } - } else if (nedges == 1) { - /* - * Two dots and an edge. This one's another - * quadratic equation. - * - * The point we want must lie on the perpendicular - * bisector of the two dots; that much is obvious. - * So we can construct a parametrisation of that - * bisecting line, giving linear formulae for x,y - * in terms of t. We can also express the distance - * from the edge as such a linear formula. - * - * Then we set that equal to the radius of the - * circle passing through the two points, which is - * a Pythagoras exercise; that gives rise to a - * quadratic in t, which we solve. - */ - double xt[2], yt[2], rt[2]; /* a,b: {x,y,r}=at+b */ - double q[3]; /* a,b,c: at^2+bt+c=0 */ - double disc; - double halfsep; - - /* Find parametric formulae for x,y. */ - { - int x1 = dots[0]->x, x2 = dots[1]->x; - int y1 = dots[0]->y, y2 = dots[1]->y; - int dx = x2-x1, dy = y2-y1; - double d = sqrt((double)dx*dx + (double)dy*dy); - - xt[1] = (x1+x2)/2.0; - yt[1] = (y1+y2)/2.0; - /* It's convenient if we have t at standard scale. */ - xt[0] = -dy/d; - yt[0] = dx/d; - - /* Also note down half the separation between - * the dots, for use in computing the circle radius. */ - halfsep = 0.5*d; - } - - /* Find a parametric formula for r. */ - { - int x1 = edgedot1[0]->x, x2 = edgedot2[0]->x; - int y1 = edgedot1[0]->y, y2 = edgedot2[0]->y; - int dx = x2-x1, dy = y2-y1; - double d = sqrt((double)dx*dx + (double)dy*dy); - rt[0] = (xt[0]*dy - yt[0]*dx) / d; - rt[1] = ((xt[1]-x1)*dy - (yt[1]-y1)*dx) / d; - } - - /* Construct the quadratic equation. */ - q[0] = rt[0]*rt[0]; - q[1] = 2*rt[0]*rt[1]; - q[2] = rt[1]*rt[1]; - q[0] -= 1; - q[2] -= halfsep*halfsep; - - /* And solve it. */ - disc = q[1]*q[1] - 4*q[0]*q[2]; - if (disc >= 0) { - double t; - - disc = sqrt(disc); - - t = (-q[1] + disc) / (2*q[0]); - cx[cn] = xt[0]*t + xt[1]; - cy[cn] = yt[0]*t + yt[1]; - cn++; - - t = (-q[1] - disc) / (2*q[0]); - cx[cn] = xt[0]*t + xt[1]; - cy[cn] = yt[0]*t + yt[1]; - cn++; - } - } else if (nedges == 0) { - /* - * Three dots. This is another linear matrix - * equation, this time with each row of the matrix - * representing the perpendicular bisector between - * two of the points. Of course we only need two - * such lines to find their intersection, so we - * need only solve a 2x2 matrix equation. - */ - - double matrix[4], vector[2], vector2[2]; - int m; - - for (m = 0; m < 2; m++) { - int x1 = dots[m]->x, x2 = dots[m+1]->x; - int y1 = dots[m]->y, y2 = dots[m+1]->y; - int dx = x2-x1, dy = y2-y1; - - /* - * ((x,y) - (x1,y1)) . (dx,dy) = 1/2 |(dx,dy)|^2 - * - * => 2x dx + 2y dy = dx^2+dy^2 + (2 x1 dx + 2 y1 dy) - */ - matrix[2*m+0] = 2*dx; - matrix[2*m+1] = 2*dy; - vector[m] = ((double)dx*dx + (double)dy*dy + - 2.0*x1*dx + 2.0*y1*dy); - } - - if (solve_2x2_matrix(matrix, vector, vector2)) { - cx[cn] = vector2[0]; - cy[cn] = vector2[1]; - cn++; - } - } - - /* - * Now go through our candidate points and see if any - * of them are better than what we've got so far. - */ - for (m = 0; m < cn; m++) { - double x = cx[m], y = cy[m]; - - /* - * First, disqualify the point if it's not inside - * the polygon, which we work out by counting the - * edges to the right of the point. (For - * tiebreaking purposes when edges start or end on - * our y-coordinate or go right through it, we - * consider our point to be offset by a small - * _positive_ epsilon in both the x- and - * y-direction.) - */ - int e, in = 0; - for (e = 0; e < f->order; e++) { - int xs = f->edges[e]->dot1->x; - int xe = f->edges[e]->dot2->x; - int ys = f->edges[e]->dot1->y; - int ye = f->edges[e]->dot2->y; - if ((y >= ys && y < ye) || (y >= ye && y < ys)) { - /* - * The line goes past our y-position. Now we need - * to know if its x-coordinate when it does so is - * to our right. - * - * The x-coordinate in question is mathematically - * (y - ys) * (xe - xs) / (ye - ys), and we want - * to know whether (x - xs) >= that. Of course we - * avoid the division, so we can work in integers; - * to do this we must multiply both sides of the - * inequality by ye - ys, which means we must - * first check that's not negative. - */ - int num = xe - xs, denom = ye - ys; - if (denom < 0) { - num = -num; - denom = -denom; - } - if ((x - xs) * denom >= (y - ys) * num) - in ^= 1; - } - } - - if (in) { - double mindist = HUGE_VAL; - int e, d; - - /* - * This point is inside the polygon, so now we check - * its minimum distance to every edge and corner. - * First the corners ... - */ - for (d = 0; d < f->order; d++) { - int xp = f->dots[d]->x; - int yp = f->dots[d]->y; - double dx = x - xp, dy = y - yp; - double dist = dx*dx + dy*dy; - if (mindist > dist) - mindist = dist; - } - - /* - * ... and now also check the perpendicular distance - * to every edge, if the perpendicular lies between - * the edge's endpoints. - */ - for (e = 0; e < f->order; e++) { - int xs = f->edges[e]->dot1->x; - int xe = f->edges[e]->dot2->x; - int ys = f->edges[e]->dot1->y; - int ye = f->edges[e]->dot2->y; - - /* - * If s and e are our endpoints, and p our - * candidate circle centre, the foot of a - * perpendicular from p to the line se lies - * between s and e if and only if (p-s).(e-s) lies - * strictly between 0 and (e-s).(e-s). - */ - int edx = xe - xs, edy = ye - ys; - double pdx = x - xs, pdy = y - ys; - double pde = pdx * edx + pdy * edy; - long ede = (long)edx * edx + (long)edy * edy; - if (0 < pde && pde < ede) { - /* - * Yes, the nearest point on this edge is - * closer than either endpoint, so we must - * take it into account by measuring the - * perpendicular distance to the edge and - * checking its square against mindist. - */ - - double pdre = pdx * edy - pdy * edx; - double sqlen = pdre * pdre / ede; - - if (mindist > sqlen) - mindist = sqlen; - } - } - - /* - * Right. Now we know the biggest circle around this - * point, so we can check it against bestdist. - */ - if (bestdist < mindist) { - bestdist = mindist; - xbest = x; - ybest = y; - } - } - } - - if (k < f->order) - nedges--; - else - ndots--; - } - if (j < f->order) - nedges--; - else - ndots--; - } - if (i < f->order) - nedges--; - else - ndots--; - } - - assert(bestdist > 0); - - /* convert to screen coordinates. Round doubles to nearest. */ - grid_to_screen(ds, g, xbest+0.5, ybest+0.5, + grid_find_incentre(f); + grid_to_screen(ds, g, f->ix, f->iy, &ds->textx[faceindex], &ds->texty[faceindex]); *xret = ds->textx[faceindex]; @@ -3980,7 +3548,6 @@ static void game_redraw_line(drawing *dr, game_drawstate *ds, grid *g = state->game_grid; grid_edge *e = g->edges + i; int x1, x2, y1, y2; - int xmin, ymin, xmax, ymax; int line_colour; if (state->line_errors[i]) @@ -4000,11 +3567,6 @@ static void game_redraw_line(drawing *dr, game_drawstate *ds, grid_to_screen(ds, g, e->dot1->x, e->dot1->y, &x1, &y1); grid_to_screen(ds, g, e->dot2->x, e->dot2->y, &x2, &y2); - xmin = min(x1, x2); - xmax = max(x1, x2); - ymin = min(y1, y2); - ymax = max(y1, y2); - if (line_colour == COL_FAINT) { static int draw_faint_lines = -1; if (draw_faint_lines < 0) { @@ -4482,3 +4044,5 @@ int main(int argc, char **argv) } #endif + +/* vim: set shiftwidth=4 tabstop=8: */