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puzzles/loopy.c
Simon Tatham f38b711c73 Completely re-engineered version of Loopy, courtesy of Lambros 
Lambrou. Now capable of handling triangular and hexagonal grids as
well as square ones, and then a number of semiregular plane tilings
and duals of semiregular ones. In fact, most of the solver code
supports an _arbitrary_ planar graph (well, provided both the graph
and its dual have no self-edges), so it could easily be extended
further with only a little more effort.

[originally from svn r8162]
2008-09-06 15:19:47 +00:00

3304 lines
104 KiB
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/*
* 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:
*/
/*
*
* - There's an interesting deductive technique which makes use of topology
* rather than just graph theory. Each _square_ in the grid is either inside
* or outside the loop; you can tell that two squares are on the same side
* of the loop if they're separated by an x (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 (or an odd
* number of LINE_YESes and no LINE_UNKNOWNs). Oh, and any square 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 squares, you
* figure out the state of the line between a pair once their relative
* insideness is known.
*
* - (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 <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <assert.h>
#include <ctype.h>
#include <math.h>
#include "puzzles.h"
#include "tree234.h"
#include "grid.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,
NCOLOURS
};
struct game_state {
grid *game_grid;
/* 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;
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 normal {
/* 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;
} normal_mode_state;
typedef struct hard {
int *linedsf;
} hard_mode_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;
/* caches */
char *dot_yes_count;
char *dot_no_count;
char *face_yes_count;
char *face_no_count;
char *dot_solved, *face_solved;
int *dotdsf;
normal_mode_state *normal;
hard_mode_state *hard;
} 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,easy_mode_deductions) \
A(NORMAL,Normal,n,normal_mode_deductions) \
A(HARD,Hard,h,hard_mode_deductions)
#define ENUM(upper,title,lower,fn) DIFF_ ## upper,
#define TITLE(upper,title,lower,fn) #title,
#define ENCODE(upper,title,lower,fn) #lower
#define CONFIG(upper,title,lower,fn) ":" #title
#define SOLVER_FN_DECL(upper,title,lower,fn) static int fn(solver_state *);
#define SOLVER_FN(upper,title,lower,fn) &fn,
enum { DIFFLIST(ENUM) DIFF_MAX };
static char const *const diffnames[] = { DIFFLIST(TITLE) };
static char const diffchars[] = DIFFLIST(ENCODE);
#define DIFFCONFIG DIFFLIST(CONFIG)
DIFFLIST(SOLVER_FN_DECL);
static int (*(solver_fns[]))(solver_state *) = { DIFFLIST(SOLVER_FN) };
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;
};
enum line_state { LINE_YES, LINE_UNKNOWN, LINE_NO };
#define OPP(line_state) \
(2 - line_state)
struct game_drawstate {
int started;
int tilesize;
int flashing;
char *lines;
char *clue_error;
char *clue_satisfied;
};
static char *validate_desc(game_params *params, 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,
int diff);
#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_new_square) \
A(Triangular,grid_new_triangular) \
A(Honeycomb,grid_new_honeycomb) \
A(Snub-Square,grid_new_snubsquare) \
A(Cairo,grid_new_cairo) \
A(Great-Hexagonal,grid_new_greathexagonal) \
A(Octagonal,grid_new_octagonal) \
A(Kites,grid_new_kites)
#define GRID_NAME(title,fn) #title,
#define GRID_CONFIG(title,fn) ":" #title
#define GRID_FN(title,fn) &fn,
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) };
static const int NUM_GRID_TYPES = sizeof(grid_fns) / sizeof(grid_fns[0]);
/* 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)
{
if (!params->game_grid) {
params->game_grid = grid_fns[params->type](params->w, params->h);
}
}
/* ----------------------------------------------------------------------
* 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 + '0')
/* ----------------------------------------------------------------------
* General struct manipulation and other straightforward code
*/
static game_state *dup_game(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->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);
}
}
static solver_state *new_solver_state(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->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->normal = NULL;
} else {
ret->normal = snew(normal_mode_state);
ret->normal->dlines = snewn(2*num_edges, char);
memset(ret->normal->dlines, 0, 2*num_edges);
}
if (diff < DIFF_HARD) {
ret->hard = NULL;
} else {
ret->hard = snew(hard_mode_state);
ret->hard->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);
if (sstate->normal) {
sfree(sstate->normal->dlines);
sfree(sstate->normal);
}
if (sstate->hard) {
sfree(sstate->hard->linedsf);
sfree(sstate->hard);
}
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->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->normal) {
ret->normal = snew(normal_mode_state);
ret->normal->dlines = snewn(2*num_edges, char);
memcpy(ret->normal->dlines, sstate->normal->dlines,
2*num_edges);
} else {
ret->normal = NULL;
}
if (sstate->hard) {
ret->hard = snew(hard_mode_state);
ret->hard->linedsf = snewn(num_edges, int);
memcpy(ret->hard->linedsf, sstate->hard->linedsf,
num_edges * sizeof(int));
} else {
ret->hard = 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;
ret->game_grid = NULL;
return ret;
}
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[] = {
{ 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 },
};
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)
{
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++;
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(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(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(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;
ret->game_grid = NULL;
return ret;
}
static char *validate_params(game_params *params, int full)
{
if (params->w < 3 || params->h < 3)
return "Width and height must both be at least 3";
if (params->type < 0 || params->type >= NUM_GRID_TYPES)
return "Illegal grid type";
/*
* 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;
}
/* 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;
for (; *desc; ++desc) {
if (*desc >= '0' && *desc <= '9') {
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";
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(game_state *state)
{
return NULL;
}
static void free_ui(game_ui *ui)
{
}
static char *encode_ui(game_ui *ui)
{
return NULL;
}
static void decode_ui(game_ui *ui, char *encoding)
{
}
static void game_changed_state(game_ui *ui, game_state *oldstate,
game_state *newstate)
{
}
static void game_compute_size(game_params *params, int tilesize,
int *x, int *y)
{
grid *g;
params_generate_grid(params);
g = params->game_grid;
int grid_width = g->highest_x - g->lowest_x;
int grid_height = g->highest_y - g->lowest_y;
/* multiply first to minimise rounding error on integer division */
int rendered_width = grid_width * tilesize / g->tilesize;
int 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)
{
ds->tilesize = tilesize;
}
static float *game_colours(frontend *fe, int *ncolours)
{
float *ret = snewn(4 * 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;
ret[COL_LINEUNKNOWN * 3 + 0] = 0.8F;
ret[COL_LINEUNKNOWN * 3 + 1] = 0.8F;
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;
*ncolours = NCOLOURS;
return ret;
}
static game_drawstate *game_new_drawstate(drawing *dr, 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;
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->flashing = 0;
memset(ds->lines, LINE_UNKNOWN, num_edges);
memset(ds->clue_error, 0, num_faces);
memset(ds->clue_satisfied, 0, num_faces);
return ds;
}
static void game_free_drawstate(drawing *dr, game_drawstate *ds)
{
sfree(ds->clue_error);
sfree(ds->clue_satisfied);
sfree(ds->lines);
sfree(ds);
}
static int game_timing_state(game_state *state, game_ui *ui)
{
return TRUE;
}
static float game_anim_length(game_state *oldstate, game_state *newstate,
int dir, game_ui *ui)
{
return 0.0F;
}
static int game_can_format_as_text_now(game_params *params)
{
if (params->type != 0)
return FALSE;
return TRUE;
}
static char *game_text_format(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++) {
f = g->faces + i;
assert(f->order == 4);
/* Cell coordinates, from (0,0) to (w-1,h-1) */
int x1 = (f->dots[0]->x - g->lowest_x) / cell_size;
int x2 = (f->dots[2]->x - g->lowest_x) / cell_size;
int y1 = (f->dots[0]->y - g->lowest_y) / cell_size;
int 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->hard->linedsf, i, &inv_tmp);
inverse ^= inv_tmp;
j = edsf_canonify(sstate->hard->linedsf, j, &inv_tmp);
inverse ^= inv_tmp;
edsf_merge(sstate->hard->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
*/
/* We're going to store a list of current candidate faces for lighting.
* Each face gets a 'score', which tells us how adding that face right
* now would affect the length of the solution loop. We're trying to
* maximise that quantity so will bias our random selection of faces to
* light towards those with high scores */
struct face {
int score;
unsigned long random;
grid_face *f;
};
static int get_face_cmpfn(void *v1, void *v2)
{
struct face *f1 = v1;
struct face *f2 = v2;
/* These grid_face pointers always point into the same list of
* 'grid_face's, so it's valid to subtract them. */
return f1->f - f2->f;
}
static int face_sort_cmpfn(void *v1, void *v2)
{
struct face *f1 = v1;
struct face *f2 = v2;
int r;
r = f2->score - f1->score;
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 grid's face-list.
* This introduces a tiny directional bias, but not a significant one.
*/
return get_face_cmpfn(f1, f2);
}
enum { FACE_LIT, FACE_UNLIT };
/* face should be of type grid_face* here. */
#define FACE_LIT_STATE(face) \
( (face) == NULL ? FACE_UNLIT : \
board[(face) - g->faces] )
/* 'board' is an array of these enums, indicating which faces are
* currently lit. Returns whether it's legal to light up the
* given face. */
static int can_light_face(grid *g, char* board, int face_index)
{
int i, j;
grid_face *test_face = g->faces + face_index;
grid_face *starting_face, *current_face;
int transitions;
int current_state, s;
int found_lit_neighbour = FALSE;
assert(board[face_index] == FACE_UNLIT);
/* Can only consider a face for lighting if it's adjacent to an
* already lit face. */
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_LIT_STATE(f) == FACE_LIT) {
found_lit_neighbour = TRUE;
break;
}
}
if (!found_lit_neighbour)
return FALSE;
/* Need to avoid creating a loop of lit faces around some unlit faces.
* Also need to avoid meeting another lit face at a corner, with
* unlit 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 LIT/UNLIT 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 a completely unlit zone, or this face is a hole
* in a completely lit zone. If the former, we would create a brand new
* island by lighting this face. And the latter ought to be impossible -
* it would mean there's already a lit loop, so something went wrong
* earlier.
* If 4 or greater, there are too many separate lit regions touching this
* face, and lighting it up 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. */
i = j = 0;
starting_face = test_face->dots[0]->faces[0];
if (starting_face == test_face) {
j = 1;
starting_face = test_face->dots[0]->faces[1];
}
current_face = starting_face;
transitions = 0;
current_state = FACE_LIT_STATE(current_face);
do {
/* 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_LIT_STATE(current_face);
if (s != current_state) {
++transitions;
current_state = s;
if (transitions > 2)
return FALSE; /* no point in continuing */
}
} while (current_face != starting_face);
return (transitions == 2) ? TRUE : FALSE;
}
/* The 'score' of a face reflects its current desirability for selection
* as the next face to light. We want to encourage moving into uncharted
* areas so we give scores according to how many of the face's neighbours
* are currently unlit. */
static int face_score(grid *g, char *board, grid_face *face)
{
/* Simple formula: score = neighbours unlit - neighbours lit */
int lit_count = 0, unlit_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_LIT_STATE(f) == FACE_LIT)
++lit_count;
else
++unlit_count;
}
return unlit_count - lit_count;
}
/* Generate a new complete set of clues for the given game_state. */
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, c;
int num_faces = g->num_faces;
int first_time = TRUE;
struct face *face, *tmpface;
struct face face_pos;
/* These will contain exactly the same information, sorted into different
* orders */
tree234 *lightable_faces_sorted, *lightable_faces_gettable;
#define IS_LIGHTING_CANDIDATE(i) \
(board[i] == FACE_UNLIT && \
can_light_face(g, board, i))
board = snewn(num_faces, char);
/* Make a board */
memset(board, FACE_UNLIT, num_faces);
/* 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(face_sort_cmpfn);
lightable_faces_gettable = newtree234(get_face_cmpfn);
#define ADD_FACE(f) \
do { \
struct face *x = add234(lightable_faces_sorted, f); \
assert(x == f); \
x = add234(lightable_faces_gettable, f); \
assert(x == f); \
} while (0)
#define REMOVE_FACE(f) \
do { \
struct face *x = del234(lightable_faces_sorted, f); \
assert(x); \
x = del234(lightable_faces_gettable, f); \
assert(x); \
} while (0)
/* Light faces one at a time until the board is interesting enough */
while (TRUE)
{
if (first_time) {
first_time = FALSE;
/* lightable_faces_xxx are empty, so start the process by
* lighting up the middle face. These tree234s should
* remain empty, consistent with what would happen if
* first_time were FALSE. */
board[g->middle_face - g->faces] = FACE_LIT;
face = snew(struct face);
face->f = g->middle_face;
/* No need to initialise any more of 'face' here, no other fields
* are used in this case. */
} else {
/* We have count234(lightable_faces_gettable) possibilities, and in
* lightable_faces_sorted they are sorted with the most desirable
* first. */
c = count234(lightable_faces_sorted);
if (c == 0)
break;
assert(c == count234(lightable_faces_gettable));
/* Check that the best face available is any good */
face = (struct face *)index234(lightable_faces_sorted, 0);
assert(face);
/*
* The situation for a general grid is slightly different from
* a square grid. Decreasing the perimeter should be allowed
* sometimes (think about creating a hexagon of lit triangles,
* for example). For if it were _never_ done, then the user would
* be able to illicitly deduce certain things. So we do it
* sometimes but not always.
*/
if (face->score <= 0 && random_upto(rs, 2) == 0) {
break;
}
assert(face->f); /* not the infinite face */
assert(FACE_LIT_STATE(face->f) == FACE_UNLIT);
/* Update data structures */
/* Light up the face and remove it from the lists */
board[face->f - g->faces] = FACE_LIT;
REMOVE_FACE(face);
}
/* The face we've just lit up potentially affects the lightability
* 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 (i = 0; i < face->f->order; i++) {
grid_dot *d = face->f->dots[i];
for (j = 0; j < d->order; j++) {
grid_face *f2 = d->faces[j];
if (f2 == NULL)
continue;
if (f2 == face->f)
continue;
face_pos.f = f2;
tmpface = find234(lightable_faces_gettable, &face_pos, NULL);
if (tmpface) {
assert(tmpface->f == face_pos.f);
assert(FACE_LIT_STATE(tmpface->f) == FACE_UNLIT);
REMOVE_FACE(tmpface);
} else {
tmpface = snew(struct face);
tmpface->f = face_pos.f;
tmpface->random = random_bits(rs, 31);
}
tmpface->score = face_score(g, board, tmpface->f);
if (IS_LIGHTING_CANDIDATE(tmpface->f - g->faces)) {
ADD_FACE(tmpface);
} else {
sfree(tmpface);
}
}
}
sfree(face);
}
/* Clean up */
while ((face = delpos234(lightable_faces_gettable, 0)) != NULL)
sfree(face);
freetree234(lightable_faces_gettable);
freetree234(lightable_faces_sorted);
/* 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 LIT/UNLIT faces */
memset(clues, 0, 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;
if (FACE_LIT_STATE(f1) != FACE_LIT_STATE(f2)) {
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, diff);
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(game_params *params, random_state *rs,
char **aux, int interactive)
{
/* solution and description both use run-length encoding in obvious ways */
char *retval;
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++;
state->clues = snewn(g->num_faces, signed char);
state->lines = snewn(g->num_edges, char);
state->grid_type = params->type;
newboard_please:
memset(state->lines, LINE_UNKNOWN, 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;
}
retval = state_to_text(state);
free_game(state);
assert(!validate_desc(params, retval));
return retval;
}
static game_state *new_game(midend *me, game_params *params, char *desc)
{
int i;
game_state *state = snew(game_state);
int empties_to_make = 0;
int n;
const char *dp = desc;
grid *g;
params_generate_grid(params);
state->game_grid = g = params->game_grid;
g->refcount++;
int num_faces = g->num_faces;
int num_edges = g->num_edges;
state->clues = snewn(num_faces, signed char);
state->lines = snewn(num_edges, char);
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';
if (n >= 0 && n < 10) {
state->clues[i] = n;
} 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);
return state;
}
enum { LOOP_NONE=0, LOOP_SOLN, LOOP_NOT_SOLN };
/* ----------------------------------------------------------------------
* 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 Mode
* 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->normal->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->hard->linedsf, line1_index, &inv1);
can2 = edsf_canonify(sstate->hard->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->hard->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 easy_mode_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;
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;
}
}
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 normal_mode_deductions(solver_state *sstate)
{
game_state *state = sstate->state;
grid *g = state->game_grid;
char *dlines = sstate->normal->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 8
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);
}
/* 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);
}
}
/* 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 hard_mode_deductions(solver_state *sstate)
{
game_state *state = sstate->state;
grid *g = state->game_grid;
char *dlines = sstate->normal->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->hard->linedsf, line1_index, &inv1);
can2 = edsf_canonify(sstate->hard->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->hard->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,
int diff)
{
solver_state *sstate, *sstate_saved;
int solver_progress;
game_state *state;
/* Indicates which solver we should call next. This is a sensible starting
* point */
int current_solver = DIFF_EASY, next_solver;
sstate = dup_solver_state(sstate_start);
/* Cache the values of some variables for readability */
state = sstate->state;
sstate_saved = NULL;
solver_progress = FALSE;
check_caches(sstate);
do {
if (sstate->solver_status == SOLVER_MISTAKE)
return sstate;
next_solver = solver_fns[current_solver](sstate);
if (next_solver == DIFF_MAX) {
if (current_solver < diff && current_solver + 1 < DIFF_MAX) {
/* Try next beefier solver */
next_solver = current_solver + 1;
} else {
next_solver = loop_deductions(sstate);
}
}
if (sstate->solver_status == SOLVER_SOLVED ||
sstate->solver_status == SOLVER_AMBIGUOUS) {
/* fprintf(stderr, "Solver completed\n"); */
break;
}
/* Once we've looped over all permitted solvers then the loop
* deductions without making any progress, we'll exit this while loop */
current_solver = next_solver;
} while (current_solver < DIFF_MAX);
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(game_state *state, game_state *currstate,
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, DIFF_MAX);
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(game_state *state, game_ui *ui, 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:
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:
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(game_state *state, char *move)
{
int i;
game_state *newstate = dup_game(state);
grid *g = state->game_grid;
if (move[0] == 'S') {
move++;
newstate->cheated = TRUE;
}
while (*move) {
i = atoi(move);
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.
*/
for (i = 0; i < g->num_edges; i++) {
if (newstate->lines[i] == LINE_YES)
break;
}
if (i < g->num_edges) {
int looplen, count;
grid_edge *start_edge = g->edges + i;
grid_edge *e = start_edge;
grid_dot *d = e->dot1;
/*
* We've found an edge i. Follow it round
* to see if it's part of a loop.
*/
looplen = 0;
while (1) {
int j;
int order = dot_order(newstate, d - g->dots, LINE_YES);
if (order != 2)
goto completion_check_done;
/* Find other edge around this dot */
for (j = 0; j < d->order; j++) {
grid_edge *e2 = d->edges[j];
if (e2 != e && newstate->lines[e2 - g->edges] == LINE_YES)
break;
}
assert(j != d->order); /* dot_order guarantees success */
e = d->edges[j];
d = (e->dot1 == d) ? e->dot2 : e->dot1;
looplen++;
if (e == start_edge)
break;
}
/*
* We've traced our way round a loop, and we know how many
* line segments were involved. Count _all_ the line
* segments in the grid, to see if the loop includes them
* all.
*/
count = 0;
for (i = 0; i < g->num_edges; i++) {
if (newstate->lines[i] == LINE_YES)
count++;
}
assert(count >= looplen);
if (count != looplen)
goto completion_check_done;
/*
* The grid contains one closed loop and nothing else.
* Check that all the clues are satisfied.
*/
for (i = 0; i < g->num_faces; i++) {
int c = newstate->clues[i];
if (c >= 0) {
if (face_order(newstate, i, LINE_YES) != c) {
goto completion_check_done;
}
}
}
/*
* Completed!
*/
newstate->solved = TRUE;
}
completion_check_done:
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,
const grid_face *f, int *x, int *y)
{
int i;
/* Simplest solution is the centroid. Might not work in some cases. */
/* Another algorithm to look into:
* Find the midpoints of the sides, find the bounding-box,
* then take the centre of that. */
/* Best solution probably involves incentres (inscribed circles) */
int sx = 0, sy = 0; /* sums */
for (i = 0; i < f->order; i++) {
grid_dot *d = f->dots[i];
sx += d->x;
sy += d->y;
}
sx /= f->order;
sy /= f->order;
/* convert to screen coordinates */
grid_to_screen(ds, g, sx, sy, x, y);
}
static void game_redraw(drawing *dr, game_drawstate *ds, game_state *oldstate,
game_state *state, int dir, game_ui *ui,
float animtime, float flashtime)
{
grid *g = state->game_grid;
int border = BORDER(ds->tilesize);
int i, n;
char c[2];
int line_colour, flash_changed;
int clue_mistake;
int clue_satisfied;
if (!ds->started) {
/*
* The initial contents of the window are not guaranteed and
* can vary with front ends. To be on the safe side, all games
* should start by drawing a big background-colour rectangle
* covering the whole window.
*/
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;
draw_rect(dr, 0, 0, w + 2 * border, h + 2 * border, COL_BACKGROUND);
/* Draw clues */
for (i = 0; i < g->num_faces; i++) {
c[0] = CLUE2CHAR(state->clues[i]);
c[1] = '\0';
int x, y;
grid_face *f = g->faces + i;
face_text_pos(ds, g, f, &x, &y);
draw_text(dr, x, y, FONT_VARIABLE, ds->tilesize/2,
ALIGN_VCENTRE | ALIGN_HCENTRE, COL_FOREGROUND, c);
}
draw_update(dr, 0, 0, w + 2 * border, h + 2 * border);
}
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;
}
/* Some platforms may perform anti-aliasing, which may prevent clean
* repainting of lines when the colour is changed.
* If a line needs to be over-drawn in a different colour, erase a
* bounding-box around the line, then flag all nearby objects for redraw.
*/
if (ds->started) {
const char redraw_flag = 1<<7;
for (i = 0; i < g->num_edges; i++) {
/* If we're changing state, AND
* the previous state was a coloured line */
if ((state->lines[i] != (ds->lines[i] & ~redraw_flag)) &&
((ds->lines[i] & ~redraw_flag) != LINE_NO)) {
grid_edge *e = g->edges + i;
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;
int j;
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;
/* For testing, I find it helpful to change COL_BACKGROUND
* to COL_SATISFIED here. */
draw_rect(dr, xmin, ymin, xmax - xmin + 1, ymax - ymin + 1,
COL_BACKGROUND);
draw_update(dr, xmin, ymin, xmax - xmin + 1, ymax - ymin + 1);
/* Mark nearby lines for redraw */
for (j = 0; j < e->dot1->order; j++)
ds->lines[e->dot1->edges[j] - g->edges] |= redraw_flag;
for (j = 0; j < e->dot2->order; j++)
ds->lines[e->dot2->edges[j] - g->edges] |= redraw_flag;
/* Mark nearby clues for redraw. Use a value that is
* neither TRUE nor FALSE for this. */
if (e->face1)
ds->clue_error[e->face1 - g->faces] = 2;
if (e->face2)
ds->clue_error[e->face2 - g->faces] = 2;
}
}
}
/* Redraw clue colours if necessary */
for (i = 0; i < g->num_faces; i++) {
grid_face *f = g->faces + i;
int sides = f->order;
int j;
n = state->clues[i];
if (n < 0)
continue;
c[0] = CLUE2CHAR(n);
c[1] = '\0';
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]) {
int x, y;
face_text_pos(ds, g, f, &x, &y);
/* There seems to be a certain amount of trial-and-error
* involved in working out the correct bounding-box for
* the text. */
draw_rect(dr, x - ds->tilesize/4 - 1, y - ds->tilesize/4 - 3,
ds->tilesize/2 + 2, ds->tilesize/2 + 5,
COL_BACKGROUND);
draw_text(dr, x, y,
FONT_VARIABLE, ds->tilesize/2,
ALIGN_VCENTRE | ALIGN_HCENTRE,
clue_mistake ? COL_MISTAKE :
clue_satisfied ? COL_SATISFIED : COL_FOREGROUND, c);
draw_update(dr, x - ds->tilesize/4 - 1, y - ds->tilesize/4 - 3,
ds->tilesize/2 + 2, ds->tilesize/2 + 5);
ds->clue_error[i] = clue_mistake;
ds->clue_satisfied[i] = clue_satisfied;
/* Sometimes, the bounding-box encroaches into the surrounding
* lines (particularly if the window is resized fairly small).
* So redraw them. */
for (j = 0; j < f->order; j++)
ds->lines[f->edges[j] - g->edges] = -1;
}
}
/* I've also had a request to colour lines red if they make a non-solution
* loop, or if more than two lines go into any point. I think that would
* be good some time. */
/* Lines */
for (i = 0; i < g->num_edges; i++) {
grid_edge *e = g->edges + i;
int x1, x2, y1, y2;
int xmin, ymin, xmax, ymax;
int need_draw = (state->lines[i] != ds->lines[i]) ? TRUE : FALSE;
if (flash_changed && (state->lines[i] == LINE_YES))
need_draw = TRUE;
if (!ds->started)
need_draw = TRUE; /* draw everything at the start */
ds->lines[i] = state->lines[i];
if (!need_draw)
continue;
if (state->lines[i] == LINE_UNKNOWN)
line_colour = COL_LINEUNKNOWN;
else if (state->lines[i] == LINE_NO)
line_colour = COL_BACKGROUND;
else if (ds->flashing)
line_colour = COL_HIGHLIGHT;
else
line_colour = COL_FOREGROUND;
/* 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);
xmin = min(x1, x2);
xmax = max(x1, x2);
ymin = min(y1, y2);
ymax = max(y1, y2);
if (line_colour != COL_BACKGROUND) {
/* (dx, dy) points roughly from (x1, y1) to (x2, y2).
* The line is then "fattened" in a (roughly) perpendicular
* direction to create a thin rectangle. */
int dx = (x1 > x2) ? -1 : ((x1 < x2) ? 1 : 0);
int dy = (y1 > y2) ? -1 : ((y1 < y2) ? 1 : 0);
int points[] = {
x1 + dy, y1 - dx,
x1 - dy, y1 + dx,
x2 - dy, y2 + dx,
x2 + dy, y2 - dx
};
draw_polygon(dr, points, 4, line_colour, line_colour);
}
if (ds->started) {
/* Draw dots at ends of the line */
draw_circle(dr, x1, y1, 2, COL_FOREGROUND, COL_FOREGROUND);
draw_circle(dr, x2, y2, 2, COL_FOREGROUND, COL_FOREGROUND);
}
draw_update(dr, xmin-2, ymin-2, xmax - xmin + 4, ymax - ymin + 4);
}
/* Draw dots */
if (!ds->started) {
for (i = 0; i < g->num_dots; i++) {
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);
}
}
ds->started = TRUE;
}
static float game_flash_length(game_state *oldstate, game_state *newstate,
int dir, game_ui *ui)
{
if (!oldstate->solved && newstate->solved &&
!oldstate->cheated && !newstate->cheated) {
return FLASH_TIME;
}
return 0.0F;
}
static void game_print_size(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, game_state *state, int tilesize)
{
int ink = print_mono_colour(dr, 0);
int i;
game_drawstate ads, *ds = &ads;
grid *g = state->game_grid;
game_set_size(dr, ds, NULL, tilesize);
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[2];
int x, y;
c[0] = CLUE2CHAR(clue);
c[1] = '\0';
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;
dx = (dx * ds->tilesize) / thickness;
dy = (dy * ds->tilesize) / thickness;
int points[] = {
x1 + dy, y1 - dx,
x1 - dy, y1 + dx,
x2 - dy, y2 + dx,
x2 + dy, y2 - 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);
}
}
}
}
#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,
TRUE, FALSE, game_print_size, game_print,
FALSE /* wants_statusbar */,
FALSE, game_timing_state,
0, /* mouse_priorities */
};
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