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puzzles/loopy.c
Jonas Kölker 9b1b7e0f3a Don't overallocate colour memory in Loopy.
2015-10-03 16:58:05 +01:00

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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:
*/
/*
* Possible future solver enhancements:
*
* - There's an interesting deductive technique which makes use
* of topology rather than just graph theory. Each _face_ in
* the grid is either inside or outside the loop; you can tell
* that two faces are on the same side of the loop if they're
* separated by a LINE_NO (or, more generally, by a path
* crossing no LINE_UNKNOWNs and an even number of LINE_YESes),
* and on the opposite side of the loop if they're separated by
* a LINE_YES (or an odd number of LINE_YESes and no
* LINE_UNKNOWNs). Oh, and any face separated from the outside
* of the grid by a LINE_YES or a LINE_NO is on the inside or
* outside respectively. So if you can track this for all
* faces, you figure out the state of the line between a pair
* once their relative insideness is known.
* + The way I envisage this working is simply to keep an edsf
* of all _faces_, which indicates whether they're on
* opposite sides of the loop from one another. We also
* include a special entry in the edsf for the infinite
* exterior "face".
* + So, the simple way to do this is to just go through the
* edges: every time we see an edge in a state other than
* LINE_UNKNOWN which separates two faces that aren't in the
* same edsf class, we can rectify that by merging the
* classes. Then, conversely, an edge in LINE_UNKNOWN state
* which separates two faces that _are_ in the same edsf
* class can immediately have its state determined.
* + But you can go one better, if you're prepared to loop
* over all _pairs_ of edges. Suppose we have edges A and B,
* which respectively separate faces A1,A2 and B1,B2.
* Suppose that A,B are in the same edge-edsf class and that
* A1,B1 (wlog) are in the same face-edsf class; then we can
* immediately place A2,B2 into the same face-edsf class (as
* each other, not as A1 and A2) one way round or the other.
* And conversely again, if A1,B1 are in the same face-edsf
* class and so are A2,B2, then we can put A,B into the same
* face-edsf class.
* * Of course, this deduction requires a quadratic-time
* loop over all pairs of edges in the grid, so it should
* be reserved until there's nothing easier left to be
* done.
*
* - The generalised grid support has made me (SGT) notice a
* possible extension to the loop-avoidance code. When you have
* a path of connected edges such that no other edges at all
* are incident on any vertex in the middle of the path - or,
* alternatively, such that any such edges are already known to
* be LINE_NO - then you know those edges are either all
* LINE_YES or all LINE_NO. Hence you can mentally merge the
* entire path into a single long curly edge for the purposes
* of loop avoidance, and look directly at whether or not the
* extreme endpoints of the path are connected by some other
* route. I find this coming up fairly often when I play on the
* octagonal grid setting, so it might be worth implementing in
* the solver.
*
* - (Just a speed optimisation.) Consider some todo list queue where every
* time we modify something we mark it for consideration by other bits of
* the solver, to save iteration over things that have already been done.
*/
#include <stdio.h>
#include <stdlib.h>
#include <stddef.h>
#include <string.h>
#include <assert.h>
#include <ctype.h>
#include <math.h>
#include "puzzles.h"
#include "tree234.h"
#include "grid.h"
#include "loopgen.h"
/* Debugging options */
/*
#define DEBUG_CACHES
#define SHOW_WORKING
#define DEBUG_DLINES
*/
/* ----------------------------------------------------------------------
* Struct, enum and function declarations
*/
enum {
COL_BACKGROUND,
COL_FOREGROUND,
COL_LINEUNKNOWN,
COL_HIGHLIGHT,
COL_MISTAKE,
COL_SATISFIED,
COL_FAINT,
NCOLOURS
};
struct game_state {
grid *game_grid; /* ref-counted (internally) */
/* Put -1 in a face that doesn't get a clue */
signed char *clues;
/* Array of line states, to store whether each line is
* YES, NO or UNKNOWN */
char *lines;
unsigned char *line_errors;
int solved;
int cheated;
/* Used in game_text_format(), so that it knows what type of
* grid it's trying to render as ASCII text. */
int grid_type;
};
enum solver_status {
SOLVER_SOLVED, /* This is the only solution the solver could find */
SOLVER_MISTAKE, /* This is definitely not a solution */
SOLVER_AMBIGUOUS, /* This _might_ be an ambiguous solution */
SOLVER_INCOMPLETE /* This may be a partial solution */
};
/* ------ Solver state ------ */
typedef struct solver_state {
game_state *state;
enum solver_status solver_status;
/* NB looplen is the number of dots that are joined together at a point, ie a
* looplen of 1 means there are no lines to a particular dot */
int *looplen;
/* Difficulty level of solver. Used by solver functions that want to
* vary their behaviour depending on the requested difficulty level. */
int diff;
/* caches */
char *dot_yes_count;
char *dot_no_count;
char *face_yes_count;
char *face_no_count;
char *dot_solved, *face_solved;
int *dotdsf;
/* Information for Normal level deductions:
* For each dline, store a bitmask for whether we know:
* (bit 0) at least one is YES
* (bit 1) at most one is YES */
char *dlines;
/* Hard level information */
int *linedsf;
} solver_state;
/*
* Difficulty levels. I do some macro ickery here to ensure that my
* enum and the various forms of my name list always match up.
*/
#define DIFFLIST(A) \
A(EASY,Easy,e) \
A(NORMAL,Normal,n) \
A(TRICKY,Tricky,t) \
A(HARD,Hard,h)
#define ENUM(upper,title,lower) DIFF_ ## upper,
#define TITLE(upper,title,lower) #title,
#define ENCODE(upper,title,lower) #lower
#define CONFIG(upper,title,lower) ":" #title
enum { DIFFLIST(ENUM) DIFF_MAX };
static char const *const diffnames[] = { DIFFLIST(TITLE) };
static char const diffchars[] = DIFFLIST(ENCODE);
#define DIFFCONFIG DIFFLIST(CONFIG)
/*
* Solver routines, sorted roughly in order of computational cost.
* The solver will run the faster deductions first, and slower deductions are
* only invoked when the faster deductions are unable to make progress.
* Each function is associated with a difficulty level, so that the generated
* puzzles are solvable by applying only the functions with the chosen
* difficulty level or lower.
*/
#define SOLVERLIST(A) \
A(trivial_deductions, DIFF_EASY) \
A(dline_deductions, DIFF_NORMAL) \
A(linedsf_deductions, DIFF_HARD) \
A(loop_deductions, DIFF_EASY)
#define SOLVER_FN_DECL(fn,diff) static int fn(solver_state *);
#define SOLVER_FN(fn,diff) &fn,
#define SOLVER_DIFF(fn,diff) diff,
SOLVERLIST(SOLVER_FN_DECL)
static int (*(solver_fns[]))(solver_state *) = { SOLVERLIST(SOLVER_FN) };
static int const solver_diffs[] = { SOLVERLIST(SOLVER_DIFF) };
static const int NUM_SOLVERS = sizeof(solver_diffs)/sizeof(*solver_diffs);
struct game_params {
int w, h;
int diff;
int type;
};
/* line_drawstate is the same as line_state, but with the extra ERROR
* possibility. The drawing code copies line_state to line_drawstate,
* except in the case that the line is an error. */
enum line_state { LINE_YES, LINE_UNKNOWN, LINE_NO };
enum line_drawstate { DS_LINE_YES, DS_LINE_UNKNOWN,
DS_LINE_NO, DS_LINE_ERROR };
#define OPP(line_state) \
(2 - line_state)
struct game_drawstate {
int started;
int tilesize;
int flashing;
int *textx, *texty;
char *lines;
char *clue_error;
char *clue_satisfied;
};
static char *validate_desc(const game_params *params, const char *desc);
static int dot_order(const game_state* state, int i, char line_type);
static int face_order(const game_state* state, int i, char line_type);
static solver_state *solve_game_rec(const solver_state *sstate);
#ifdef DEBUG_CACHES
static void check_caches(const solver_state* sstate);
#else
#define check_caches(s)
#endif
/* ------- List of grid generators ------- */
#define GRIDLIST(A) \
A(Squares,GRID_SQUARE,3,3) \
A(Triangular,GRID_TRIANGULAR,3,3) \
A(Honeycomb,GRID_HONEYCOMB,3,3) \
A(Snub-Square,GRID_SNUBSQUARE,3,3) \
A(Cairo,GRID_CAIRO,3,4) \
A(Great-Hexagonal,GRID_GREATHEXAGONAL,3,3) \
A(Octagonal,GRID_OCTAGONAL,3,3) \
A(Kites,GRID_KITE,3,3) \
A(Floret,GRID_FLORET,1,2) \
A(Dodecagonal,GRID_DODECAGONAL,2,2) \
A(Great-Dodecagonal,GRID_GREATDODECAGONAL,2,2) \
A(Penrose (kite/dart),GRID_PENROSE_P2,3,3) \
A(Penrose (rhombs),GRID_PENROSE_P3,3,3)
#define GRID_NAME(title,type,amin,omin) #title,
#define GRID_CONFIG(title,type,amin,omin) ":" #title
#define GRID_TYPE(title,type,amin,omin) type,
#define GRID_SIZES(title,type,amin,omin) \
{amin, omin, \
"Width and height for this grid type must both be at least " #amin, \
"At least one of width and height for this grid type must be at least " #omin,},
static char const *const gridnames[] = { GRIDLIST(GRID_NAME) };
#define GRID_CONFIGS GRIDLIST(GRID_CONFIG)
static grid_type grid_types[] = { GRIDLIST(GRID_TYPE) };
#define NUM_GRID_TYPES (sizeof(grid_types) / sizeof(grid_types[0]))
static const struct {
int amin, omin;
char *aerr, *oerr;
} grid_size_limits[] = { GRIDLIST(GRID_SIZES) };
/* Generates a (dynamically allocated) new grid, according to the
* type and size requested in params. Does nothing if the grid is already
* generated. */
static grid *loopy_generate_grid(const game_params *params,
const char *grid_desc)
{
return grid_new(grid_types[params->type], params->w, params->h, grid_desc);
}
/* ----------------------------------------------------------------------
* Preprocessor magic
*/
/* General constants */
#define PREFERRED_TILE_SIZE 32
#define BORDER(tilesize) ((tilesize) / 2)
#define FLASH_TIME 0.5F
#define BIT_SET(field, bit) ((field) & (1<<(bit)))
#define SET_BIT(field, bit) (BIT_SET(field, bit) ? FALSE : \
((field) |= (1<<(bit)), TRUE))
#define CLEAR_BIT(field, bit) (BIT_SET(field, bit) ? \
((field) &= ~(1<<(bit)), TRUE) : FALSE)
#define CLUE2CHAR(c) \
((c < 0) ? ' ' : c < 10 ? c + '0' : c - 10 + 'A')
/* ----------------------------------------------------------------------
* General struct manipulation and other straightforward code
*/
static game_state *dup_game(const game_state *state)
{
game_state *ret = snew(game_state);
ret->game_grid = state->game_grid;
ret->game_grid->refcount++;
ret->solved = state->solved;
ret->cheated = state->cheated;
ret->clues = snewn(state->game_grid->num_faces, signed char);
memcpy(ret->clues, state->clues, state->game_grid->num_faces);
ret->lines = snewn(state->game_grid->num_edges, char);
memcpy(ret->lines, state->lines, state->game_grid->num_edges);
ret->line_errors = snewn(state->game_grid->num_edges, unsigned char);
memcpy(ret->line_errors, state->line_errors, state->game_grid->num_edges);
ret->grid_type = state->grid_type;
return ret;
}
static void free_game(game_state *state)
{
if (state) {
grid_free(state->game_grid);
sfree(state->clues);
sfree(state->lines);
sfree(state->line_errors);
sfree(state);
}
}
static solver_state *new_solver_state(const game_state *state, int diff) {
int i;
int num_dots = state->game_grid->num_dots;
int num_faces = state->game_grid->num_faces;
int num_edges = state->game_grid->num_edges;
solver_state *ret = snew(solver_state);
ret->state = dup_game(state);
ret->solver_status = SOLVER_INCOMPLETE;
ret->diff = diff;
ret->dotdsf = snew_dsf(num_dots);
ret->looplen = snewn(num_dots, int);
for (i = 0; i < num_dots; i++) {
ret->looplen[i] = 1;
}
ret->dot_solved = snewn(num_dots, char);
ret->face_solved = snewn(num_faces, char);
memset(ret->dot_solved, FALSE, num_dots);
memset(ret->face_solved, FALSE, num_faces);
ret->dot_yes_count = snewn(num_dots, char);
memset(ret->dot_yes_count, 0, num_dots);
ret->dot_no_count = snewn(num_dots, char);
memset(ret->dot_no_count, 0, num_dots);
ret->face_yes_count = snewn(num_faces, char);
memset(ret->face_yes_count, 0, num_faces);
ret->face_no_count = snewn(num_faces, char);
memset(ret->face_no_count, 0, num_faces);
if (diff < DIFF_NORMAL) {
ret->dlines = NULL;
} else {
ret->dlines = snewn(2*num_edges, char);
memset(ret->dlines, 0, 2*num_edges);
}
if (diff < DIFF_HARD) {
ret->linedsf = NULL;
} else {
ret->linedsf = snew_dsf(state->game_grid->num_edges);
}
return ret;
}
static void free_solver_state(solver_state *sstate) {
if (sstate) {
free_game(sstate->state);
sfree(sstate->dotdsf);
sfree(sstate->looplen);
sfree(sstate->dot_solved);
sfree(sstate->face_solved);
sfree(sstate->dot_yes_count);
sfree(sstate->dot_no_count);
sfree(sstate->face_yes_count);
sfree(sstate->face_no_count);
/* OK, because sfree(NULL) is a no-op */
sfree(sstate->dlines);
sfree(sstate->linedsf);
sfree(sstate);
}
}
static solver_state *dup_solver_state(const solver_state *sstate) {
game_state *state = sstate->state;
int num_dots = state->game_grid->num_dots;
int num_faces = state->game_grid->num_faces;
int num_edges = state->game_grid->num_edges;
solver_state *ret = snew(solver_state);
ret->state = state = dup_game(sstate->state);
ret->solver_status = sstate->solver_status;
ret->diff = sstate->diff;
ret->dotdsf = snewn(num_dots, int);
ret->looplen = snewn(num_dots, int);
memcpy(ret->dotdsf, sstate->dotdsf,
num_dots * sizeof(int));
memcpy(ret->looplen, sstate->looplen,
num_dots * sizeof(int));
ret->dot_solved = snewn(num_dots, char);
ret->face_solved = snewn(num_faces, char);
memcpy(ret->dot_solved, sstate->dot_solved, num_dots);
memcpy(ret->face_solved, sstate->face_solved, num_faces);
ret->dot_yes_count = snewn(num_dots, char);
memcpy(ret->dot_yes_count, sstate->dot_yes_count, num_dots);
ret->dot_no_count = snewn(num_dots, char);
memcpy(ret->dot_no_count, sstate->dot_no_count, num_dots);
ret->face_yes_count = snewn(num_faces, char);
memcpy(ret->face_yes_count, sstate->face_yes_count, num_faces);
ret->face_no_count = snewn(num_faces, char);
memcpy(ret->face_no_count, sstate->face_no_count, num_faces);
if (sstate->dlines) {
ret->dlines = snewn(2*num_edges, char);
memcpy(ret->dlines, sstate->dlines,
2*num_edges);
} else {
ret->dlines = NULL;
}
if (sstate->linedsf) {
ret->linedsf = snewn(num_edges, int);
memcpy(ret->linedsf, sstate->linedsf,
num_edges * sizeof(int));
} else {
ret->linedsf = NULL;
}
return ret;
}
static game_params *default_params(void)
{
game_params *ret = snew(game_params);
#ifdef SLOW_SYSTEM
ret->h = 7;
ret->w = 7;
#else
ret->h = 10;
ret->w = 10;
#endif
ret->diff = DIFF_EASY;
ret->type = 0;
return ret;
}
static game_params *dup_params(const game_params *params)
{
game_params *ret = snew(game_params);
*ret = *params; /* structure copy */
return ret;
}
static const game_params presets[] = {
#ifdef SMALL_SCREEN
{ 7, 7, DIFF_EASY, 0 },
{ 7, 7, DIFF_NORMAL, 0 },
{ 7, 7, DIFF_HARD, 0 },
{ 7, 7, DIFF_HARD, 1 },
{ 7, 7, DIFF_HARD, 2 },
{ 5, 5, DIFF_HARD, 3 },
{ 7, 7, DIFF_HARD, 4 },
{ 5, 4, DIFF_HARD, 5 },
{ 5, 5, DIFF_HARD, 6 },
{ 5, 5, DIFF_HARD, 7 },
{ 3, 3, DIFF_HARD, 8 },
{ 3, 3, DIFF_HARD, 9 },
{ 3, 3, DIFF_HARD, 10 },
{ 6, 6, DIFF_HARD, 11 },
{ 6, 6, DIFF_HARD, 12 },
#else
{ 7, 7, DIFF_EASY, 0 },
{ 10, 10, DIFF_EASY, 0 },
{ 7, 7, DIFF_NORMAL, 0 },
{ 10, 10, DIFF_NORMAL, 0 },
{ 7, 7, DIFF_HARD, 0 },
{ 10, 10, DIFF_HARD, 0 },
{ 10, 10, DIFF_HARD, 1 },
{ 12, 10, DIFF_HARD, 2 },
{ 7, 7, DIFF_HARD, 3 },
{ 9, 9, DIFF_HARD, 4 },
{ 5, 4, DIFF_HARD, 5 },
{ 7, 7, DIFF_HARD, 6 },
{ 5, 5, DIFF_HARD, 7 },
{ 5, 5, DIFF_HARD, 8 },
{ 5, 4, DIFF_HARD, 9 },
{ 5, 4, DIFF_HARD, 10 },
{ 10, 10, DIFF_HARD, 11 },
{ 10, 10, DIFF_HARD, 12 }
#endif
};
static int game_fetch_preset(int i, char **name, game_params **params)
{
game_params *tmppar;
char buf[80];
if (i < 0 || i >= lenof(presets))
return FALSE;
tmppar = snew(game_params);
*tmppar = presets[i];
*params = tmppar;
sprintf(buf, "%dx%d %s - %s", tmppar->h, tmppar->w,
gridnames[tmppar->type], diffnames[tmppar->diff]);
*name = dupstr(buf);
return TRUE;
}
static void free_params(game_params *params)
{
sfree(params);
}
static void decode_params(game_params *params, char const *string)
{
params->h = params->w = atoi(string);
params->diff = DIFF_EASY;
while (*string && isdigit((unsigned char)*string)) string++;
if (*string == 'x') {
string++;
params->h = atoi(string);
while (*string && isdigit((unsigned char)*string)) string++;
}
if (*string == 't') {
string++;
params->type = atoi(string);
while (*string && isdigit((unsigned char)*string)) string++;
}
if (*string == 'd') {
int i;
string++;
for (i = 0; i < DIFF_MAX; i++)
if (*string == diffchars[i])
params->diff = i;
if (*string) string++;
}
}
static char *encode_params(const game_params *params, int full)
{
char str[80];
sprintf(str, "%dx%dt%d", params->w, params->h, params->type);
if (full)
sprintf(str + strlen(str), "d%c", diffchars[params->diff]);
return dupstr(str);
}
static config_item *game_configure(const game_params *params)
{
config_item *ret;
char buf[80];
ret = snewn(5, config_item);
ret[0].name = "Width";
ret[0].type = C_STRING;
sprintf(buf, "%d", params->w);
ret[0].sval = dupstr(buf);
ret[0].ival = 0;
ret[1].name = "Height";
ret[1].type = C_STRING;
sprintf(buf, "%d", params->h);
ret[1].sval = dupstr(buf);
ret[1].ival = 0;
ret[2].name = "Grid type";
ret[2].type = C_CHOICES;
ret[2].sval = GRID_CONFIGS;
ret[2].ival = params->type;
ret[3].name = "Difficulty";
ret[3].type = C_CHOICES;
ret[3].sval = DIFFCONFIG;
ret[3].ival = params->diff;
ret[4].name = NULL;
ret[4].type = C_END;
ret[4].sval = NULL;
ret[4].ival = 0;
return ret;
}
static game_params *custom_params(const config_item *cfg)
{
game_params *ret = snew(game_params);
ret->w = atoi(cfg[0].sval);
ret->h = atoi(cfg[1].sval);
ret->type = cfg[2].ival;
ret->diff = cfg[3].ival;
return ret;
}
static char *validate_params(const game_params *params, int full)
{
if (params->type < 0 || params->type >= NUM_GRID_TYPES)
return "Illegal grid type";
if (params->w < grid_size_limits[params->type].amin ||
params->h < grid_size_limits[params->type].amin)
return grid_size_limits[params->type].aerr;
if (params->w < grid_size_limits[params->type].omin &&
params->h < grid_size_limits[params->type].omin)
return grid_size_limits[params->type].oerr;
/*
* This shouldn't be able to happen at all, since decode_params
* and custom_params will never generate anything that isn't
* within range.
*/
assert(params->diff < DIFF_MAX);
return NULL;
}
/* Returns a newly allocated string describing the current puzzle */
static char *state_to_text(const game_state *state)
{
grid *g = state->game_grid;
char *retval;
int num_faces = g->num_faces;
char *description = snewn(num_faces + 1, char);
char *dp = description;
int empty_count = 0;
int i;
for (i = 0; i < num_faces; i++) {
if (state->clues[i] < 0) {
if (empty_count > 25) {
dp += sprintf(dp, "%c", (int)(empty_count + 'a' - 1));
empty_count = 0;
}
empty_count++;
} else {
if (empty_count) {
dp += sprintf(dp, "%c", (int)(empty_count + 'a' - 1));
empty_count = 0;
}
dp += sprintf(dp, "%c", (int)CLUE2CHAR(state->clues[i]));
}
}
if (empty_count)
dp += sprintf(dp, "%c", (int)(empty_count + 'a' - 1));
retval = dupstr(description);
sfree(description);
return retval;
}
#define GRID_DESC_SEP '_'
/* Splits up a (optional) grid_desc from the game desc. Returns the
* grid_desc (which needs freeing) and updates the desc pointer to
* start of real desc, or returns NULL if no desc. */
static char *extract_grid_desc(const char **desc)
{
char *sep = strchr(*desc, GRID_DESC_SEP), *gd;
int gd_len;
if (!sep) return NULL;
gd_len = sep - (*desc);
gd = snewn(gd_len+1, char);
memcpy(gd, *desc, gd_len);
gd[gd_len] = '\0';
*desc = sep+1;
return gd;
}
/* We require that the params pass the test in validate_params and that the
* description fills the entire game area */
static char *validate_desc(const game_params *params, const char *desc)
{
int count = 0;
grid *g;
char *grid_desc, *ret;
/* It's pretty inefficient to do this just for validation. All we need to
* know is the precise number of faces. */
grid_desc = extract_grid_desc(&desc);
ret = grid_validate_desc(grid_types[params->type], params->w, params->h, grid_desc);
if (ret) return ret;
g = loopy_generate_grid(params, grid_desc);
if (grid_desc) sfree(grid_desc);
for (; *desc; ++desc) {
if ((*desc >= '0' && *desc <= '9') || (*desc >= 'A' && *desc <= 'Z')) {
count++;
continue;
}
if (*desc >= 'a') {
count += *desc - 'a' + 1;
continue;
}
return "Unknown character in description";
}
if (count < g->num_faces)
return "Description too short for board size";
if (count > g->num_faces)
return "Description too long for board size";
grid_free(g);
return NULL;
}
/* Sums the lengths of the numbers in range [0,n) */
/* See equivalent function in solo.c for justification of this. */
static int len_0_to_n(int n)
{
int len = 1; /* Counting 0 as a bit of a special case */
int i;
for (i = 1; i < n; i *= 10) {
len += max(n - i, 0);
}
return len;
}
static char *encode_solve_move(const game_state *state)
{
int len;
char *ret, *p;
int i;
int num_edges = state->game_grid->num_edges;
/* This is going to return a string representing the moves needed to set
* every line in a grid to be the same as the ones in 'state'. The exact
* length of this string is predictable. */
len = 1; /* Count the 'S' prefix */
/* Numbers in all lines */
len += len_0_to_n(num_edges);
/* For each line we also have a letter */
len += num_edges;
ret = snewn(len + 1, char);
p = ret;
p += sprintf(p, "S");
for (i = 0; i < num_edges; i++) {
switch (state->lines[i]) {
case LINE_YES:
p += sprintf(p, "%dy", i);
break;
case LINE_NO:
p += sprintf(p, "%dn", i);
break;
}
}
/* No point in doing sums like that if they're going to be wrong */
assert(strlen(ret) <= (size_t)len);
return ret;
}
static game_ui *new_ui(const game_state *state)
{
return NULL;
}
static void free_ui(game_ui *ui)
{
}
static char *encode_ui(const game_ui *ui)
{
return NULL;
}
static void decode_ui(game_ui *ui, const char *encoding)
{
}
static void game_changed_state(game_ui *ui, const game_state *oldstate,
const game_state *newstate)
{
}
static void game_compute_size(const game_params *params, int tilesize,
int *x, int *y)
{
int grid_width, grid_height, rendered_width, rendered_height;
int g_tilesize;
grid_compute_size(grid_types[params->type], params->w, params->h,
&g_tilesize, &grid_width, &grid_height);
/* multiply first to minimise rounding error on integer division */
rendered_width = grid_width * tilesize / g_tilesize;
rendered_height = grid_height * tilesize / g_tilesize;
*x = rendered_width + 2 * BORDER(tilesize) + 1;
*y = rendered_height + 2 * BORDER(tilesize) + 1;
}
static void game_set_size(drawing *dr, game_drawstate *ds,
const game_params *params, int tilesize)
{
ds->tilesize = tilesize;
}
static float *game_colours(frontend *fe, int *ncolours)
{
float *ret = snewn(3 * NCOLOURS, float);
frontend_default_colour(fe, &ret[COL_BACKGROUND * 3]);
ret[COL_FOREGROUND * 3 + 0] = 0.0F;
ret[COL_FOREGROUND * 3 + 1] = 0.0F;
ret[COL_FOREGROUND * 3 + 2] = 0.0F;
/*
* We want COL_LINEUNKNOWN to be a yellow which is a bit darker
* than the background. (I previously set it to 0.8,0.8,0, but
* found that this went badly with the 0.8,0.8,0.8 favoured as a
* background by the Java frontend.)
*/
ret[COL_LINEUNKNOWN * 3 + 0] = ret[COL_BACKGROUND * 3 + 0] * 0.9F;
ret[COL_LINEUNKNOWN * 3 + 1] = ret[COL_BACKGROUND * 3 + 1] * 0.9F;
ret[COL_LINEUNKNOWN * 3 + 2] = 0.0F;
ret[COL_HIGHLIGHT * 3 + 0] = 1.0F;
ret[COL_HIGHLIGHT * 3 + 1] = 1.0F;
ret[COL_HIGHLIGHT * 3 + 2] = 1.0F;
ret[COL_MISTAKE * 3 + 0] = 1.0F;
ret[COL_MISTAKE * 3 + 1] = 0.0F;
ret[COL_MISTAKE * 3 + 2] = 0.0F;
ret[COL_SATISFIED * 3 + 0] = 0.0F;
ret[COL_SATISFIED * 3 + 1] = 0.0F;
ret[COL_SATISFIED * 3 + 2] = 0.0F;
/* We want the faint lines to be a bit darker than the background.
* Except if the background is pretty dark already; then it ought to be a
* bit lighter. Oy vey.
*/
ret[COL_FAINT * 3 + 0] = ret[COL_BACKGROUND * 3 + 0] * 0.9F;
ret[COL_FAINT * 3 + 1] = ret[COL_BACKGROUND * 3 + 1] * 0.9F;
ret[COL_FAINT * 3 + 2] = ret[COL_BACKGROUND * 3 + 2] * 0.9F;
*ncolours = NCOLOURS;
return ret;
}
static game_drawstate *game_new_drawstate(drawing *dr, const game_state *state)
{
struct game_drawstate *ds = snew(struct game_drawstate);
int num_faces = state->game_grid->num_faces;
int num_edges = state->game_grid->num_edges;
int i;
ds->tilesize = 0;
ds->started = 0;
ds->lines = snewn(num_edges, char);
ds->clue_error = snewn(num_faces, char);
ds->clue_satisfied = snewn(num_faces, char);
ds->textx = snewn(num_faces, int);
ds->texty = snewn(num_faces, int);
ds->flashing = 0;
memset(ds->lines, LINE_UNKNOWN, num_edges);
memset(ds->clue_error, 0, num_faces);
memset(ds->clue_satisfied, 0, num_faces);
for (i = 0; i < num_faces; i++)
ds->textx[i] = ds->texty[i] = -1;
return ds;
}
static void game_free_drawstate(drawing *dr, game_drawstate *ds)
{
sfree(ds->textx);
sfree(ds->texty);
sfree(ds->clue_error);
sfree(ds->clue_satisfied);
sfree(ds->lines);
sfree(ds);
}
static int game_timing_state(const game_state *state, game_ui *ui)
{
return TRUE;
}
static float game_anim_length(const game_state *oldstate,
const game_state *newstate, int dir, game_ui *ui)
{
return 0.0F;
}
static int game_can_format_as_text_now(const game_params *params)
{
if (params->type != 0)
return FALSE;
return TRUE;
}
static char *game_text_format(const game_state *state)
{
int w, h, W, H;
int x, y, i;
int cell_size;
char *ret;
grid *g = state->game_grid;
grid_face *f;
assert(state->grid_type == 0);
/* Work out the basic size unit */
f = g->faces; /* first face */
assert(f->order == 4);
/* The dots are ordered clockwise, so the two opposite
* corners are guaranteed to span the square */
cell_size = abs(f->dots[0]->x - f->dots[2]->x);
w = (g->highest_x - g->lowest_x) / cell_size;
h = (g->highest_y - g->lowest_y) / cell_size;
/* Create a blank "canvas" to "draw" on */
W = 2 * w + 2;
H = 2 * h + 1;
ret = snewn(W * H + 1, char);
for (y = 0; y < H; y++) {
for (x = 0; x < W-1; x++) {
ret[y*W + x] = ' ';
}
ret[y*W + W-1] = '\n';
}
ret[H*W] = '\0';
/* Fill in edge info */
for (i = 0; i < g->num_edges; i++) {
grid_edge *e = g->edges + i;
/* Cell coordinates, from (0,0) to (w-1,h-1) */
int x1 = (e->dot1->x - g->lowest_x) / cell_size;
int x2 = (e->dot2->x - g->lowest_x) / cell_size;
int y1 = (e->dot1->y - g->lowest_y) / cell_size;
int y2 = (e->dot2->y - g->lowest_y) / cell_size;
/* Midpoint, in canvas coordinates (canvas coordinates are just twice
* cell coordinates) */
x = x1 + x2;
y = y1 + y2;
switch (state->lines[i]) {
case LINE_YES:
ret[y*W + x] = (y1 == y2) ? '-' : '|';
break;
case LINE_NO:
ret[y*W + x] = 'x';
break;
case LINE_UNKNOWN:
break; /* already a space */
default:
assert(!"Illegal line state");
}
}
/* Fill in clues */
for (i = 0; i < g->num_faces; i++) {
int x1, x2, y1, y2;
f = g->faces + i;
assert(f->order == 4);
/* Cell coordinates, from (0,0) to (w-1,h-1) */
x1 = (f->dots[0]->x - g->lowest_x) / cell_size;
x2 = (f->dots[2]->x - g->lowest_x) / cell_size;
y1 = (f->dots[0]->y - g->lowest_y) / cell_size;
y2 = (f->dots[2]->y - g->lowest_y) / cell_size;
/* Midpoint, in canvas coordinates */
x = x1 + x2;
y = y1 + y2;
ret[y*W + x] = CLUE2CHAR(state->clues[i]);
}
return ret;
}
/* ----------------------------------------------------------------------
* Debug code
*/
#ifdef DEBUG_CACHES
static void check_caches(const solver_state* sstate)
{
int i;
const game_state *state = sstate->state;
const grid *g = state->game_grid;
for (i = 0; i < g->num_dots; i++) {
assert(dot_order(state, i, LINE_YES) == sstate->dot_yes_count[i]);
assert(dot_order(state, i, LINE_NO) == sstate->dot_no_count[i]);
}
for (i = 0; i < g->num_faces; i++) {
assert(face_order(state, i, LINE_YES) == sstate->face_yes_count[i]);
assert(face_order(state, i, LINE_NO) == sstate->face_no_count[i]);
}
}
#if 0
#define check_caches(s) \
do { \
fprintf(stderr, "check_caches at line %d\n", __LINE__); \
check_caches(s); \
} while (0)
#endif
#endif /* DEBUG_CACHES */
/* ----------------------------------------------------------------------
* Solver utility functions
*/
/* Sets the line (with index i) to the new state 'line_new', and updates
* the cached counts of any affected faces and dots.
* Returns TRUE if this actually changed the line's state. */
static int solver_set_line(solver_state *sstate, int i,
enum line_state line_new
#ifdef SHOW_WORKING
, const char *reason
#endif
)
{
game_state *state = sstate->state;
grid *g;
grid_edge *e;
assert(line_new != LINE_UNKNOWN);
check_caches(sstate);
if (state->lines[i] == line_new) {
return FALSE; /* nothing changed */
}
state->lines[i] = line_new;
#ifdef SHOW_WORKING
fprintf(stderr, "solver: set line [%d] to %s (%s)\n",
i, line_new == LINE_YES ? "YES" : "NO",
reason);
#endif
g = state->game_grid;
e = g->edges + i;
/* Update the cache for both dots and both faces affected by this. */
if (line_new == LINE_YES) {
sstate->dot_yes_count[e->dot1 - g->dots]++;
sstate->dot_yes_count[e->dot2 - g->dots]++;
if (e->face1) {
sstate->face_yes_count[e->face1 - g->faces]++;
}
if (e->face2) {
sstate->face_yes_count[e->face2 - g->faces]++;
}
} else {
sstate->dot_no_count[e->dot1 - g->dots]++;
sstate->dot_no_count[e->dot2 - g->dots]++;
if (e->face1) {
sstate->face_no_count[e->face1 - g->faces]++;
}
if (e->face2) {
sstate->face_no_count[e->face2 - g->faces]++;
}
}
check_caches(sstate);
return TRUE;
}
#ifdef SHOW_WORKING
#define solver_set_line(a, b, c) \
solver_set_line(a, b, c, __FUNCTION__)
#endif
/*
* Merge two dots due to the existence of an edge between them.
* Updates the dsf tracking equivalence classes, and keeps track of
* the length of path each dot is currently a part of.
* Returns TRUE if the dots were already linked, ie if they are part of a
* closed loop, and false otherwise.
*/
static int merge_dots(solver_state *sstate, int edge_index)
{
int i, j, len;
grid *g = sstate->state->game_grid;
grid_edge *e = g->edges + edge_index;
i = e->dot1 - g->dots;
j = e->dot2 - g->dots;
i = dsf_canonify(sstate->dotdsf, i);
j = dsf_canonify(sstate->dotdsf, j);
if (i == j) {
return TRUE;
} else {
len = sstate->looplen[i] + sstate->looplen[j];
dsf_merge(sstate->dotdsf, i, j);
i = dsf_canonify(sstate->dotdsf, i);
sstate->looplen[i] = len;
return FALSE;
}
}
/* Merge two lines because the solver has deduced that they must be either
* identical or opposite. Returns TRUE if this is new information, otherwise
* FALSE. */
static int merge_lines(solver_state *sstate, int i, int j, int inverse
#ifdef SHOW_WORKING
, const char *reason
#endif
)
{
int inv_tmp;
assert(i < sstate->state->game_grid->num_edges);
assert(j < sstate->state->game_grid->num_edges);
i = edsf_canonify(sstate->linedsf, i, &inv_tmp);
inverse ^= inv_tmp;
j = edsf_canonify(sstate->linedsf, j, &inv_tmp);
inverse ^= inv_tmp;
edsf_merge(sstate->linedsf, i, j, inverse);
#ifdef SHOW_WORKING
if (i != j) {
fprintf(stderr, "%s [%d] [%d] %s(%s)\n",
__FUNCTION__, i, j,
inverse ? "inverse " : "", reason);
}
#endif
return (i != j);
}
#ifdef SHOW_WORKING
#define merge_lines(a, b, c, d) \
merge_lines(a, b, c, d, __FUNCTION__)
#endif
/* Count the number of lines of a particular type currently going into the
* given dot. */
static int dot_order(const game_state* state, int dot, char line_type)
{
int n = 0;
grid *g = state->game_grid;
grid_dot *d = g->dots + dot;
int i;
for (i = 0; i < d->order; i++) {
grid_edge *e = d->edges[i];
if (state->lines[e - g->edges] == line_type)
++n;
}
return n;
}
/* Count the number of lines of a particular type currently surrounding the
* given face */
static int face_order(const game_state* state, int face, char line_type)
{
int n = 0;
grid *g = state->game_grid;
grid_face *f = g->faces + face;
int i;
for (i = 0; i < f->order; i++) {
grid_edge *e = f->edges[i];
if (state->lines[e - g->edges] == line_type)
++n;
}
return n;
}
/* Set all lines bordering a dot of type old_type to type new_type
* Return value tells caller whether this function actually did anything */
static int dot_setall(solver_state *sstate, int dot,
char old_type, char new_type)
{
int retval = FALSE, r;
game_state *state = sstate->state;
grid *g;
grid_dot *d;
int i;
if (old_type == new_type)
return FALSE;
g = state->game_grid;
d = g->dots + dot;
for (i = 0; i < d->order; i++) {
int line_index = d->edges[i] - g->edges;
if (state->lines[line_index] == old_type) {
r = solver_set_line(sstate, line_index, new_type);
assert(r == TRUE);
retval = TRUE;
}
}
return retval;
}
/* Set all lines bordering a face of type old_type to type new_type */
static int face_setall(solver_state *sstate, int face,
char old_type, char new_type)
{
int retval = FALSE, r;
game_state *state = sstate->state;
grid *g;
grid_face *f;
int i;
if (old_type == new_type)
return FALSE;
g = state->game_grid;
f = g->faces + face;
for (i = 0; i < f->order; i++) {
int line_index = f->edges[i] - g->edges;
if (state->lines[line_index] == old_type) {
r = solver_set_line(sstate, line_index, new_type);
assert(r == TRUE);
retval = TRUE;
}
}
return retval;
}
/* ----------------------------------------------------------------------
* Loop generation and clue removal
*/
static void add_full_clues(game_state *state, random_state *rs)
{
signed char *clues = state->clues;
grid *g = state->game_grid;
char *board = snewn(g->num_faces, char);
int i;
generate_loop(g, board, rs, NULL, NULL);
/* Fill out all the clues by initialising to 0, then iterating over
* all edges and incrementing each clue as we find edges that border
* between BLACK/WHITE faces. While we're at it, we verify that the
* algorithm does work, and there aren't any GREY faces still there. */
memset(clues, 0, g->num_faces);
for (i = 0; i < g->num_edges; i++) {
grid_edge *e = g->edges + i;
grid_face *f1 = e->face1;
grid_face *f2 = e->face2;
enum face_colour c1 = FACE_COLOUR(f1);
enum face_colour c2 = FACE_COLOUR(f2);
assert(c1 != FACE_GREY);
assert(c2 != FACE_GREY);
if (c1 != c2) {
if (f1) clues[f1 - g->faces]++;
if (f2) clues[f2 - g->faces]++;
}
}
sfree(board);
}
static int game_has_unique_soln(const game_state *state, int diff)
{
int ret;
solver_state *sstate_new;
solver_state *sstate = new_solver_state((game_state *)state, diff);
sstate_new = solve_game_rec(sstate);
assert(sstate_new->solver_status != SOLVER_MISTAKE);
ret = (sstate_new->solver_status == SOLVER_SOLVED);
free_solver_state(sstate_new);
free_solver_state(sstate);
return ret;
}
/* Remove clues one at a time at random. */
static game_state *remove_clues(game_state *state, random_state *rs,
int diff)
{
int *face_list;
int num_faces = state->game_grid->num_faces;
game_state *ret = dup_game(state), *saved_ret;
int n;
/* We need to remove some clues. We'll do this by forming a list of all
* available clues, shuffling it, then going along one at a
* time clearing each clue in turn for which doing so doesn't render the
* board unsolvable. */
face_list = snewn(num_faces, int);
for (n = 0; n < num_faces; ++n) {
face_list[n] = n;
}
shuffle(face_list, num_faces, sizeof(int), rs);
for (n = 0; n < num_faces; ++n) {
saved_ret = dup_game(ret);
ret->clues[face_list[n]] = -1;
if (game_has_unique_soln(ret, diff)) {
free_game(saved_ret);
} else {
free_game(ret);
ret = saved_ret;
}
}
sfree(face_list);
return ret;
}
static char *new_game_desc(const game_params *params, random_state *rs,
char **aux, int interactive)
{
/* solution and description both use run-length encoding in obvious ways */
char *retval, *game_desc, *grid_desc;
grid *g;
game_state *state = snew(game_state);
game_state *state_new;
grid_desc = grid_new_desc(grid_types[params->type], params->w, params->h, rs);
state->game_grid = g = loopy_generate_grid(params, grid_desc);
state->clues = snewn(g->num_faces, signed char);
state->lines = snewn(g->num_edges, char);
state->line_errors = snewn(g->num_edges, unsigned char);
state->grid_type = params->type;
newboard_please:
memset(state->lines, LINE_UNKNOWN, g->num_edges);
memset(state->line_errors, 0, g->num_edges);
state->solved = state->cheated = FALSE;
/* Get a new random solvable board with all its clues filled in. Yes, this
* can loop for ever if the params are suitably unfavourable, but
* preventing games smaller than 4x4 seems to stop this happening */
do {
add_full_clues(state, rs);
} while (!game_has_unique_soln(state, params->diff));
state_new = remove_clues(state, rs, params->diff);
free_game(state);
state = state_new;
if (params->diff > 0 && game_has_unique_soln(state, params->diff-1)) {
#ifdef SHOW_WORKING
fprintf(stderr, "Rejecting board, it is too easy\n");
#endif
goto newboard_please;
}
game_desc = state_to_text(state);
free_game(state);
if (grid_desc) {
retval = snewn(strlen(grid_desc) + 1 + strlen(game_desc) + 1, char);
sprintf(retval, "%s%c%s", grid_desc, (int)GRID_DESC_SEP, game_desc);
sfree(grid_desc);
sfree(game_desc);
} else {
retval = game_desc;
}
assert(!validate_desc(params, retval));
return retval;
}
static game_state *new_game(midend *me, const game_params *params,
const char *desc)
{
int i;
game_state *state = snew(game_state);
int empties_to_make = 0;
int n,n2;
const char *dp;
char *grid_desc;
grid *g;
int num_faces, num_edges;
grid_desc = extract_grid_desc(&desc);
state->game_grid = g = loopy_generate_grid(params, grid_desc);
if (grid_desc) sfree(grid_desc);
dp = desc;
num_faces = g->num_faces;
num_edges = g->num_edges;
state->clues = snewn(num_faces, signed char);
state->lines = snewn(num_edges, char);
state->line_errors = snewn(num_edges, unsigned char);
state->solved = state->cheated = FALSE;
state->grid_type = params->type;
for (i = 0; i < num_faces; i++) {
if (empties_to_make) {
empties_to_make--;
state->clues[i] = -1;
continue;
}
assert(*dp);
n = *dp - '0';
n2 = *dp - 'A' + 10;
if (n >= 0 && n < 10) {
state->clues[i] = n;
} else if (n2 >= 10 && n2 < 36) {
state->clues[i] = n2;
} else {
n = *dp - 'a' + 1;
assert(n > 0);
state->clues[i] = -1;
empties_to_make = n - 1;
}
++dp;
}
memset(state->lines, LINE_UNKNOWN, num_edges);
memset(state->line_errors, 0, num_edges);
return state;
}
/* Calculates the line_errors data, and checks if the current state is a
* solution */
static int check_completion(game_state *state)
{
grid *g = state->game_grid;
int *dsf;
int num_faces = g->num_faces;
int i;
int infinite_area, finite_area;
int loops_found = 0;
int found_edge_not_in_loop = FALSE;
memset(state->line_errors, 0, g->num_edges);
/* LL implementation of SGT's idea:
* A loop will partition the grid into an inside and an outside.
* If there is more than one loop, the grid will be partitioned into
* even more distinct regions. We can therefore track equivalence of
* faces, by saying that two faces are equivalent when there is a non-YES
* edge between them.
* We could keep track of the number of connected components, by counting
* the number of dsf-merges that aren't no-ops.
* But we're only interested in 3 separate cases:
* no loops, one loop, more than one loop.
*
* No loops: all faces are equivalent to the infinite face.
* One loop: only two equivalence classes - finite and infinite.
* >= 2 loops: there are 2 distinct finite regions.
*
* So we simply make two passes through all the edges.
* In the first pass, we dsf-merge the two faces bordering each non-YES
* edge.
* In the second pass, we look for YES-edges bordering:
* a) two non-equivalent faces.
* b) two non-equivalent faces, and one of them is part of a different
* finite area from the first finite area we've seen.
*
* An occurrence of a) means there is at least one loop.
* An occurrence of b) means there is more than one loop.
* Edges satisfying a) are marked as errors.
*
* While we're at it, we set a flag if we find a YES edge that is not
* part of a loop.
* This information will help decide, if there's a single loop, whether it
* is a candidate for being a solution (that is, all YES edges are part of
* this loop).
*
* If there is a candidate loop, we then go through all clues and check
* they are all satisfied. If so, we have found a solution and we can
* unmark all line_errors.
*/
/* Infinite face is at the end - its index is num_faces.
* This macro is just to make this obvious! */
#define INF_FACE num_faces
dsf = snewn(num_faces + 1, int);
dsf_init(dsf, num_faces + 1);
/* First pass */
for (i = 0; i < g->num_edges; i++) {
grid_edge *e = g->edges + i;
int f1 = e->face1 ? e->face1 - g->faces : INF_FACE;
int f2 = e->face2 ? e->face2 - g->faces : INF_FACE;
if (state->lines[i] != LINE_YES)
dsf_merge(dsf, f1, f2);
}
/* Second pass */
infinite_area = dsf_canonify(dsf, INF_FACE);
finite_area = -1;
for (i = 0; i < g->num_edges; i++) {
grid_edge *e = g->edges + i;
int f1 = e->face1 ? e->face1 - g->faces : INF_FACE;
int can1 = dsf_canonify(dsf, f1);
int f2 = e->face2 ? e->face2 - g->faces : INF_FACE;
int can2 = dsf_canonify(dsf, f2);
if (state->lines[i] != LINE_YES) continue;
if (can1 == can2) {
/* Faces are equivalent, so this edge not part of a loop */
found_edge_not_in_loop = TRUE;
continue;
}
state->line_errors[i] = TRUE;
if (loops_found == 0) loops_found = 1;
/* Don't bother with further checks if we've already found 2 loops */
if (loops_found == 2) continue;
if (finite_area == -1) {
/* Found our first finite area */
if (can1 != infinite_area)
finite_area = can1;
else
finite_area = can2;
}
/* Have we found a second area? */
if (finite_area != -1) {
if (can1 != infinite_area && can1 != finite_area) {
loops_found = 2;
continue;
}
if (can2 != infinite_area && can2 != finite_area) {
loops_found = 2;
}
}
}
/*
printf("loops_found = %d\n", loops_found);
printf("found_edge_not_in_loop = %s\n",
found_edge_not_in_loop ? "TRUE" : "FALSE");
*/
sfree(dsf); /* No longer need the dsf */
/* Have we found a candidate loop? */
if (loops_found == 1 && !found_edge_not_in_loop) {
/* Yes, so check all clues are satisfied */
int found_clue_violation = FALSE;
for (i = 0; i < num_faces; i++) {
int c = state->clues[i];
if (c >= 0) {
if (face_order(state, i, LINE_YES) != c) {
found_clue_violation = TRUE;
break;
}
}
}
if (!found_clue_violation) {
/* The loop is good */
memset(state->line_errors, 0, g->num_edges);
return TRUE; /* No need to bother checking for dot violations */
}
}
/* Check for dot violations */
for (i = 0; i < g->num_dots; i++) {
int yes = dot_order(state, i, LINE_YES);
int unknown = dot_order(state, i, LINE_UNKNOWN);
if ((yes == 1 && unknown == 0) || (yes >= 3)) {
/* violation, so mark all YES edges as errors */
grid_dot *d = g->dots + i;
int j;
for (j = 0; j < d->order; j++) {
int e = d->edges[j] - g->edges;
if (state->lines[e] == LINE_YES)
state->line_errors[e] = TRUE;
}
}
}
return FALSE;
}
/* ----------------------------------------------------------------------
* Solver logic
*
* Our solver modes operate as follows. Each mode also uses the modes above it.
*
* Easy Mode
* Just implement the rules of the game.
*
* Normal and Tricky Modes
* For each (adjacent) pair of lines through each dot we store a bit for
* whether at least one of them is on and whether at most one is on. (If we
* know both or neither is on that's already stored more directly.)
*
* Advanced Mode
* Use edsf data structure to make equivalence classes of lines that are
* known identical to or opposite to one another.
*/
/* DLines:
* For general grids, we consider "dlines" to be pairs of lines joined
* at a dot. The lines must be adjacent around the dot, so we can think of
* a dline as being a dot+face combination. Or, a dot+edge combination where
* the second edge is taken to be the next clockwise edge from the dot.
* Original loopy code didn't have this extra restriction of the lines being
* adjacent. From my tests with square grids, this extra restriction seems to
* take little, if anything, away from the quality of the puzzles.
* A dline can be uniquely identified by an edge/dot combination, given that
* a dline-pair always goes clockwise around its common dot. The edge/dot
* combination can be represented by an edge/bool combination - if bool is
* TRUE, use edge->dot1 else use edge->dot2. So the total number of dlines is
* exactly twice the number of edges in the grid - although the dlines
* spanning the infinite face are not all that useful to the solver.
* Note that, by convention, a dline goes clockwise around its common dot,
* which means the dline goes anti-clockwise around its common face.
*/
/* Helper functions for obtaining an index into an array of dlines, given
* various information. We assume the grid layout conventions about how
* the various lists are interleaved - see grid_make_consistent() for
* details. */
/* i points to the first edge of the dline pair, reading clockwise around
* the dot. */
static int dline_index_from_dot(grid *g, grid_dot *d, int i)
{
grid_edge *e = d->edges[i];
int ret;
#ifdef DEBUG_DLINES
grid_edge *e2;
int i2 = i+1;
if (i2 == d->order) i2 = 0;
e2 = d->edges[i2];
#endif
ret = 2 * (e - g->edges) + ((e->dot1 == d) ? 1 : 0);
#ifdef DEBUG_DLINES
printf("dline_index_from_dot: d=%d,i=%d, edges [%d,%d] - %d\n",
(int)(d - g->dots), i, (int)(e - g->edges),
(int)(e2 - g->edges), ret);
#endif
return ret;
}
/* i points to the second edge of the dline pair, reading clockwise around
* the face. That is, the edges of the dline, starting at edge{i}, read
* anti-clockwise around the face. By layout conventions, the common dot
* of the dline will be f->dots[i] */
static int dline_index_from_face(grid *g, grid_face *f, int i)
{
grid_edge *e = f->edges[i];
grid_dot *d = f->dots[i];
int ret;
#ifdef DEBUG_DLINES
grid_edge *e2;
int i2 = i - 1;
if (i2 < 0) i2 += f->order;
e2 = f->edges[i2];
#endif
ret = 2 * (e - g->edges) + ((e->dot1 == d) ? 1 : 0);
#ifdef DEBUG_DLINES
printf("dline_index_from_face: f=%d,i=%d, edges [%d,%d] - %d\n",
(int)(f - g->faces), i, (int)(e - g->edges),
(int)(e2 - g->edges), ret);
#endif
return ret;
}
static int is_atleastone(const char *dline_array, int index)
{
return BIT_SET(dline_array[index], 0);
}
static int set_atleastone(char *dline_array, int index)
{
return SET_BIT(dline_array[index], 0);
}
static int is_atmostone(const char *dline_array, int index)
{
return BIT_SET(dline_array[index], 1);
}
static int set_atmostone(char *dline_array, int index)
{
return SET_BIT(dline_array[index], 1);
}
static void array_setall(char *array, char from, char to, int len)
{
char *p = array, *p_old = p;
int len_remaining = len;
while ((p = memchr(p, from, len_remaining))) {
*p = to;
len_remaining -= p - p_old;
p_old = p;
}
}
/* Helper, called when doing dline dot deductions, in the case where we
* have 4 UNKNOWNs, and two of them (adjacent) have *exactly* one YES between
* them (because of dline atmostone/atleastone).
* On entry, edge points to the first of these two UNKNOWNs. This function
* will find the opposite UNKNOWNS (if they are adjacent to one another)
* and set their corresponding dline to atleastone. (Setting atmostone
* already happens in earlier dline deductions) */
static int dline_set_opp_atleastone(solver_state *sstate,
grid_dot *d, int edge)
{
game_state *state = sstate->state;
grid *g = state->game_grid;
int N = d->order;
int opp, opp2;
for (opp = 0; opp < N; opp++) {
int opp_dline_index;
if (opp == edge || opp == edge+1 || opp == edge-1)
continue;
if (opp == 0 && edge == N-1)
continue;
if (opp == N-1 && edge == 0)
continue;
opp2 = opp + 1;
if (opp2 == N) opp2 = 0;
/* Check if opp, opp2 point to LINE_UNKNOWNs */
if (state->lines[d->edges[opp] - g->edges] != LINE_UNKNOWN)
continue;
if (state->lines[d->edges[opp2] - g->edges] != LINE_UNKNOWN)
continue;
/* Found opposite UNKNOWNS and they're next to each other */
opp_dline_index = dline_index_from_dot(g, d, opp);
return set_atleastone(sstate->dlines, opp_dline_index);
}
return FALSE;
}
/* Set pairs of lines around this face which are known to be identical, to
* the given line_state */
static int face_setall_identical(solver_state *sstate, int face_index,
enum line_state line_new)
{
/* can[dir] contains the canonical line associated with the line in
* direction dir from the square in question. Similarly inv[dir] is
* whether or not the line in question is inverse to its canonical
* element. */
int retval = FALSE;
game_state *state = sstate->state;
grid *g = state->game_grid;
grid_face *f = g->faces + face_index;
int N = f->order;
int i, j;
int can1, can2, inv1, inv2;
for (i = 0; i < N; i++) {
int line1_index = f->edges[i] - g->edges;
if (state->lines[line1_index] != LINE_UNKNOWN)
continue;
for (j = i + 1; j < N; j++) {
int line2_index = f->edges[j] - g->edges;
if (state->lines[line2_index] != LINE_UNKNOWN)
continue;
/* Found two UNKNOWNS */
can1 = edsf_canonify(sstate->linedsf, line1_index, &inv1);
can2 = edsf_canonify(sstate->linedsf, line2_index, &inv2);
if (can1 == can2 && inv1 == inv2) {
solver_set_line(sstate, line1_index, line_new);
solver_set_line(sstate, line2_index, line_new);
}
}
}
return retval;
}
/* Given a dot or face, and a count of LINE_UNKNOWNs, find them and
* return the edge indices into e. */
static void find_unknowns(game_state *state,
grid_edge **edge_list, /* Edge list to search (from a face or a dot) */
int expected_count, /* Number of UNKNOWNs (comes from solver's cache) */
int *e /* Returned edge indices */)
{
int c = 0;
grid *g = state->game_grid;
while (c < expected_count) {
int line_index = *edge_list - g->edges;
if (state->lines[line_index] == LINE_UNKNOWN) {
e[c] = line_index;
c++;
}
++edge_list;
}
}
/* If we have a list of edges, and we know whether the number of YESs should
* be odd or even, and there are only a few UNKNOWNs, we can do some simple
* linedsf deductions. This can be used for both face and dot deductions.
* Returns the difficulty level of the next solver that should be used,
* or DIFF_MAX if no progress was made. */
static int parity_deductions(solver_state *sstate,
grid_edge **edge_list, /* Edge list (from a face or a dot) */
int total_parity, /* Expected number of YESs modulo 2 (either 0 or 1) */
int unknown_count)
{
game_state *state = sstate->state;
int diff = DIFF_MAX;
int *linedsf = sstate->linedsf;
if (unknown_count == 2) {
/* Lines are known alike/opposite, depending on inv. */
int e[2];
find_unknowns(state, edge_list, 2, e);
if (merge_lines(sstate, e[0], e[1], total_parity))
diff = min(diff, DIFF_HARD);
} else if (unknown_count == 3) {
int e[3];
int can[3]; /* canonical edges */
int inv[3]; /* whether can[x] is inverse to e[x] */
find_unknowns(state, edge_list, 3, e);
can[0] = edsf_canonify(linedsf, e[0], inv);
can[1] = edsf_canonify(linedsf, e[1], inv+1);
can[2] = edsf_canonify(linedsf, e[2], inv+2);
if (can[0] == can[1]) {
if (solver_set_line(sstate, e[2], (total_parity^inv[0]^inv[1]) ?
LINE_YES : LINE_NO))
diff = min(diff, DIFF_EASY);
}
if (can[0] == can[2]) {
if (solver_set_line(sstate, e[1], (total_parity^inv[0]^inv[2]) ?
LINE_YES : LINE_NO))
diff = min(diff, DIFF_EASY);
}
if (can[1] == can[2]) {
if (solver_set_line(sstate, e[0], (total_parity^inv[1]^inv[2]) ?
LINE_YES : LINE_NO))
diff = min(diff, DIFF_EASY);
}
} else if (unknown_count == 4) {
int e[4];
int can[4]; /* canonical edges */
int inv[4]; /* whether can[x] is inverse to e[x] */
find_unknowns(state, edge_list, 4, e);
can[0] = edsf_canonify(linedsf, e[0], inv);
can[1] = edsf_canonify(linedsf, e[1], inv+1);
can[2] = edsf_canonify(linedsf, e[2], inv+2);
can[3] = edsf_canonify(linedsf, e[3], inv+3);
if (can[0] == can[1]) {
if (merge_lines(sstate, e[2], e[3], total_parity^inv[0]^inv[1]))
diff = min(diff, DIFF_HARD);
} else if (can[0] == can[2]) {
if (merge_lines(sstate, e[1], e[3], total_parity^inv[0]^inv[2]))
diff = min(diff, DIFF_HARD);
} else if (can[0] == can[3]) {
if (merge_lines(sstate, e[1], e[2], total_parity^inv[0]^inv[3]))
diff = min(diff, DIFF_HARD);
} else if (can[1] == can[2]) {
if (merge_lines(sstate, e[0], e[3], total_parity^inv[1]^inv[2]))
diff = min(diff, DIFF_HARD);
} else if (can[1] == can[3]) {
if (merge_lines(sstate, e[0], e[2], total_parity^inv[1]^inv[3]))
diff = min(diff, DIFF_HARD);
} else if (can[2] == can[3]) {
if (merge_lines(sstate, e[0], e[1], total_parity^inv[2]^inv[3]))
diff = min(diff, DIFF_HARD);
}
}
return diff;
}
/*
* These are the main solver functions.
*
* Their return values are diff values corresponding to the lowest mode solver
* that would notice the work that they have done. For example if the normal
* mode solver adds actual lines or crosses, it will return DIFF_EASY as the
* easy mode solver might be able to make progress using that. It doesn't make
* sense for one of them to return a diff value higher than that of the
* function itself.
*
* Each function returns the lowest value it can, as early as possible, in
* order to try and pass as much work as possible back to the lower level
* solvers which progress more quickly.
*/
/* PROPOSED NEW DESIGN:
* We have a work queue consisting of 'events' notifying us that something has
* happened that a particular solver mode might be interested in. For example
* the hard mode solver might do something that helps the normal mode solver at
* dot [x,y] in which case it will enqueue an event recording this fact. Then
* we pull events off the work queue, and hand each in turn to the solver that
* is interested in them. If a solver reports that it failed we pass the same
* event on to progressively more advanced solvers and the loop detector. Once
* we've exhausted an event, or it has helped us progress, we drop it and
* continue to the next one. The events are sorted first in order of solver
* complexity (easy first) then order of insertion (oldest first).
* Once we run out of events we loop over each permitted solver in turn
* (easiest first) until either a deduction is made (and an event therefore
* emerges) or no further deductions can be made (in which case we've failed).
*
* QUESTIONS:
* * How do we 'loop over' a solver when both dots and squares are concerned.
* Answer: first all squares then all dots.
*/
static int trivial_deductions(solver_state *sstate)
{
int i, current_yes, current_no;
game_state *state = sstate->state;
grid *g = state->game_grid;
int diff = DIFF_MAX;
/* Per-face deductions */
for (i = 0; i < g->num_faces; i++) {
grid_face *f = g->faces + i;
if (sstate->face_solved[i])
continue;
current_yes = sstate->face_yes_count[i];
current_no = sstate->face_no_count[i];
if (current_yes + current_no == f->order) {
sstate->face_solved[i] = TRUE;
continue;
}
if (state->clues[i] < 0)
continue;
/*
* This code checks whether the numeric clue on a face is so
* large as to permit all its remaining LINE_UNKNOWNs to be
* filled in as LINE_YES, or alternatively so small as to
* permit them all to be filled in as LINE_NO.
*/
if (state->clues[i] < current_yes) {
sstate->solver_status = SOLVER_MISTAKE;
return DIFF_EASY;
}
if (state->clues[i] == current_yes) {
if (face_setall(sstate, i, LINE_UNKNOWN, LINE_NO))
diff = min(diff, DIFF_EASY);
sstate->face_solved[i] = TRUE;
continue;
}
if (f->order - state->clues[i] < current_no) {
sstate->solver_status = SOLVER_MISTAKE;
return DIFF_EASY;
}
if (f->order - state->clues[i] == current_no) {
if (face_setall(sstate, i, LINE_UNKNOWN, LINE_YES))
diff = min(diff, DIFF_EASY);
sstate->face_solved[i] = TRUE;
continue;
}
if (f->order - state->clues[i] == current_no + 1 &&
f->order - current_yes - current_no > 2) {
/*
* One small refinement to the above: we also look for any
* adjacent pair of LINE_UNKNOWNs around the face with
* some LINE_YES incident on it from elsewhere. If we find
* one, then we know that pair of LINE_UNKNOWNs can't
* _both_ be LINE_YES, and hence that pushes us one line
* closer to being able to determine all the rest.
*/
int j, k, e1, e2, e, d;
for (j = 0; j < f->order; j++) {
e1 = f->edges[j] - g->edges;
e2 = f->edges[j+1 < f->order ? j+1 : 0] - g->edges;
if (g->edges[e1].dot1 == g->edges[e2].dot1 ||
g->edges[e1].dot1 == g->edges[e2].dot2) {
d = g->edges[e1].dot1 - g->dots;
} else {
assert(g->edges[e1].dot2 == g->edges[e2].dot1 ||
g->edges[e1].dot2 == g->edges[e2].dot2);
d = g->edges[e1].dot2 - g->dots;
}
if (state->lines[e1] == LINE_UNKNOWN &&
state->lines[e2] == LINE_UNKNOWN) {
for (k = 0; k < g->dots[d].order; k++) {
int e = g->dots[d].edges[k] - g->edges;
if (state->lines[e] == LINE_YES)
goto found; /* multi-level break */
}
}
}
continue;
found:
/*
* If we get here, we've found such a pair of edges, and
* they're e1 and e2.
*/
for (j = 0; j < f->order; j++) {
e = f->edges[j] - g->edges;
if (state->lines[e] == LINE_UNKNOWN && e != e1 && e != e2) {
int r = solver_set_line(sstate, e, LINE_YES);
assert(r);
diff = min(diff, DIFF_EASY);
}
}
}
}
check_caches(sstate);
/* Per-dot deductions */
for (i = 0; i < g->num_dots; i++) {
grid_dot *d = g->dots + i;
int yes, no, unknown;
if (sstate->dot_solved[i])
continue;
yes = sstate->dot_yes_count[i];
no = sstate->dot_no_count[i];
unknown = d->order - yes - no;
if (yes == 0) {
if (unknown == 0) {
sstate->dot_solved[i] = TRUE;
} else if (unknown == 1) {
dot_setall(sstate, i, LINE_UNKNOWN, LINE_NO);
diff = min(diff, DIFF_EASY);
sstate->dot_solved[i] = TRUE;
}
} else if (yes == 1) {
if (unknown == 0) {
sstate->solver_status = SOLVER_MISTAKE;
return DIFF_EASY;
} else if (unknown == 1) {
dot_setall(sstate, i, LINE_UNKNOWN, LINE_YES);
diff = min(diff, DIFF_EASY);
}
} else if (yes == 2) {
if (unknown > 0) {
dot_setall(sstate, i, LINE_UNKNOWN, LINE_NO);
diff = min(diff, DIFF_EASY);
}
sstate->dot_solved[i] = TRUE;
} else {
sstate->solver_status = SOLVER_MISTAKE;
return DIFF_EASY;
}
}
check_caches(sstate);
return diff;
}
static int dline_deductions(solver_state *sstate)
{
game_state *state = sstate->state;
grid *g = state->game_grid;
char *dlines = sstate->dlines;
int i;
int diff = DIFF_MAX;
/* ------ Face deductions ------ */
/* Given a set of dline atmostone/atleastone constraints, need to figure
* out if we can deduce any further info. For more general faces than
* squares, this turns out to be a tricky problem.
* The approach taken here is to define (per face) NxN matrices:
* "maxs" and "mins".
* The entries maxs(j,k) and mins(j,k) define the upper and lower limits
* for the possible number of edges that are YES between positions j and k
* going clockwise around the face. Can think of j and k as marking dots
* around the face (recall the labelling scheme: edge0 joins dot0 to dot1,
* edge1 joins dot1 to dot2 etc).
* Trivially, mins(j,j) = maxs(j,j) = 0, and we don't even bother storing
* these. mins(j,j+1) and maxs(j,j+1) are determined by whether edge{j}
* is YES, NO or UNKNOWN. mins(j,j+2) and maxs(j,j+2) are related to
* the dline atmostone/atleastone status for edges j and j+1.
*
* Then we calculate the remaining entries recursively. We definitely
* know that
* mins(j,k) >= { mins(j,u) + mins(u,k) } for any u between j and k.
* This is because any valid placement of YESs between j and k must give
* a valid placement between j and u, and also between u and k.
* I believe it's sufficient to use just the two values of u:
* j+1 and j+2. Seems to work well in practice - the bounds we compute
* are rigorous, even if they might not be best-possible.
*
* Once we have maxs and mins calculated, we can make inferences about
* each dline{j,j+1} by looking at the possible complementary edge-counts
* mins(j+2,j) and maxs(j+2,j) and comparing these with the face clue.
* As well as dlines, we can make similar inferences about single edges.
* For example, consider a pentagon with clue 3, and we know at most one
* of (edge0, edge1) is YES, and at most one of (edge2, edge3) is YES.
* We could then deduce edge4 is YES, because maxs(0,4) would be 2, so
* that final edge would have to be YES to make the count up to 3.
*/
/* Much quicker to allocate arrays on the stack than the heap, so
* define the largest possible face size, and base our array allocations
* on that. We check this with an assertion, in case someone decides to
* make a grid which has larger faces than this. Note, this algorithm
* could get quite expensive if there are many large faces. */
#define MAX_FACE_SIZE 12
for (i = 0; i < g->num_faces; i++) {
int maxs[MAX_FACE_SIZE][MAX_FACE_SIZE];
int mins[MAX_FACE_SIZE][MAX_FACE_SIZE];
grid_face *f = g->faces + i;
int N = f->order;
int j,m;
int clue = state->clues[i];
assert(N <= MAX_FACE_SIZE);
if (sstate->face_solved[i])
continue;
if (clue < 0) continue;
/* Calculate the (j,j+1) entries */
for (j = 0; j < N; j++) {
int edge_index = f->edges[j] - g->edges;
int dline_index;
enum line_state line1 = state->lines[edge_index];
enum line_state line2;
int tmp;
int k = j + 1;
if (k >= N) k = 0;
maxs[j][k] = (line1 == LINE_NO) ? 0 : 1;
mins[j][k] = (line1 == LINE_YES) ? 1 : 0;
/* Calculate the (j,j+2) entries */
dline_index = dline_index_from_face(g, f, k);
edge_index = f->edges[k] - g->edges;
line2 = state->lines[edge_index];
k++;
if (k >= N) k = 0;
/* max */
tmp = 2;
if (line1 == LINE_NO) tmp--;
if (line2 == LINE_NO) tmp--;
if (tmp == 2 && is_atmostone(dlines, dline_index))
tmp = 1;
maxs[j][k] = tmp;
/* min */
tmp = 0;
if (line1 == LINE_YES) tmp++;
if (line2 == LINE_YES) tmp++;
if (tmp == 0 && is_atleastone(dlines, dline_index))
tmp = 1;
mins[j][k] = tmp;
}
/* Calculate the (j,j+m) entries for m between 3 and N-1 */
for (m = 3; m < N; m++) {
for (j = 0; j < N; j++) {
int k = j + m;
int u = j + 1;
int v = j + 2;
int tmp;
if (k >= N) k -= N;
if (u >= N) u -= N;
if (v >= N) v -= N;
maxs[j][k] = maxs[j][u] + maxs[u][k];
mins[j][k] = mins[j][u] + mins[u][k];
tmp = maxs[j][v] + maxs[v][k];
maxs[j][k] = min(maxs[j][k], tmp);
tmp = mins[j][v] + mins[v][k];
mins[j][k] = max(mins[j][k], tmp);
}
}
/* See if we can make any deductions */
for (j = 0; j < N; j++) {
int k;
grid_edge *e = f->edges[j];
int line_index = e - g->edges;
int dline_index;
if (state->lines[line_index] != LINE_UNKNOWN)
continue;
k = j + 1;
if (k >= N) k = 0;
/* minimum YESs in the complement of this edge */
if (mins[k][j] > clue) {
sstate->solver_status = SOLVER_MISTAKE;
return DIFF_EASY;
}
if (mins[k][j] == clue) {
/* setting this edge to YES would make at least
* (clue+1) edges - contradiction */
solver_set_line(sstate, line_index, LINE_NO);
diff = min(diff, DIFF_EASY);
}
if (maxs[k][j] < clue - 1) {
sstate->solver_status = SOLVER_MISTAKE;
return DIFF_EASY;
}
if (maxs[k][j] == clue - 1) {
/* Only way to satisfy the clue is to set edge{j} as YES */
solver_set_line(sstate, line_index, LINE_YES);
diff = min(diff, DIFF_EASY);
}
/* More advanced deduction that allows propagation along diagonal
* chains of faces connected by dots, for example, 3-2-...-2-3
* in square grids. */
if (sstate->diff >= DIFF_TRICKY) {
/* Now see if we can make dline deduction for edges{j,j+1} */
e = f->edges[k];
if (state->lines[e - g->edges] != LINE_UNKNOWN)
/* Only worth doing this for an UNKNOWN,UNKNOWN pair.
* Dlines where one of the edges is known, are handled in the
* dot-deductions */
continue;
dline_index = dline_index_from_face(g, f, k);
k++;
if (k >= N) k = 0;
/* minimum YESs in the complement of this dline */
if (mins[k][j] > clue - 2) {
/* Adding 2 YESs would break the clue */
if (set_atmostone(dlines, dline_index))
diff = min(diff, DIFF_NORMAL);
}
/* maximum YESs in the complement of this dline */
if (maxs[k][j] < clue) {
/* Adding 2 NOs would mean not enough YESs */
if (set_atleastone(dlines, dline_index))
diff = min(diff, DIFF_NORMAL);
}
}
}
}
if (diff < DIFF_NORMAL)
return diff;
/* ------ Dot deductions ------ */
for (i = 0; i < g->num_dots; i++) {
grid_dot *d = g->dots + i;
int N = d->order;
int yes, no, unknown;
int j;
if (sstate->dot_solved[i])
continue;
yes = sstate->dot_yes_count[i];
no = sstate->dot_no_count[i];
unknown = N - yes - no;
for (j = 0; j < N; j++) {
int k;
int dline_index;
int line1_index, line2_index;
enum line_state line1, line2;
k = j + 1;
if (k >= N) k = 0;
dline_index = dline_index_from_dot(g, d, j);
line1_index = d->edges[j] - g->edges;
line2_index = d->edges[k] - g->edges;
line1 = state->lines[line1_index];
line2 = state->lines[line2_index];
/* Infer dline state from line state */
if (line1 == LINE_NO || line2 == LINE_NO) {
if (set_atmostone(dlines, dline_index))
diff = min(diff, DIFF_NORMAL);
}
if (line1 == LINE_YES || line2 == LINE_YES) {
if (set_atleastone(dlines, dline_index))
diff = min(diff, DIFF_NORMAL);
}
/* Infer line state from dline state */
if (is_atmostone(dlines, dline_index)) {
if (line1 == LINE_YES && line2 == LINE_UNKNOWN) {
solver_set_line(sstate, line2_index, LINE_NO);
diff = min(diff, DIFF_EASY);
}
if (line2 == LINE_YES && line1 == LINE_UNKNOWN) {
solver_set_line(sstate, line1_index, LINE_NO);
diff = min(diff, DIFF_EASY);
}
}
if (is_atleastone(dlines, dline_index)) {
if (line1 == LINE_NO && line2 == LINE_UNKNOWN) {
solver_set_line(sstate, line2_index, LINE_YES);
diff = min(diff, DIFF_EASY);
}
if (line2 == LINE_NO && line1 == LINE_UNKNOWN) {
solver_set_line(sstate, line1_index, LINE_YES);
diff = min(diff, DIFF_EASY);
}
}
/* Deductions that depend on the numbers of lines.
* Only bother if both lines are UNKNOWN, otherwise the
* easy-mode solver (or deductions above) would have taken
* care of it. */
if (line1 != LINE_UNKNOWN || line2 != LINE_UNKNOWN)
continue;
if (yes == 0 && unknown == 2) {
/* Both these unknowns must be identical. If we know
* atmostone or atleastone, we can make progress. */
if (is_atmostone(dlines, dline_index)) {
solver_set_line(sstate, line1_index, LINE_NO);
solver_set_line(sstate, line2_index, LINE_NO);
diff = min(diff, DIFF_EASY);
}
if (is_atleastone(dlines, dline_index)) {
solver_set_line(sstate, line1_index, LINE_YES);
solver_set_line(sstate, line2_index, LINE_YES);
diff = min(diff, DIFF_EASY);
}
}
if (yes == 1) {
if (set_atmostone(dlines, dline_index))
diff = min(diff, DIFF_NORMAL);
if (unknown == 2) {
if (set_atleastone(dlines, dline_index))
diff = min(diff, DIFF_NORMAL);
}
}
/* More advanced deduction that allows propagation along diagonal
* chains of faces connected by dots, for example: 3-2-...-2-3
* in square grids. */
if (sstate->diff >= DIFF_TRICKY) {
/* If we have atleastone set for this dline, infer
* atmostone for each "opposite" dline (that is, each
* dline without edges in common with this one).
* Again, this test is only worth doing if both these
* lines are UNKNOWN. For if one of these lines were YES,
* the (yes == 1) test above would kick in instead. */
if (is_atleastone(dlines, dline_index)) {
int opp;
for (opp = 0; opp < N; opp++) {
int opp_dline_index;
if (opp == j || opp == j+1 || opp == j-1)
continue;
if (j == 0 && opp == N-1)
continue;
if (j == N-1 && opp == 0)
continue;
opp_dline_index = dline_index_from_dot(g, d, opp);
if (set_atmostone(dlines, opp_dline_index))
diff = min(diff, DIFF_NORMAL);
}
if (yes == 0 && is_atmostone(dlines, dline_index)) {
/* This dline has *exactly* one YES and there are no
* other YESs. This allows more deductions. */
if (unknown == 3) {
/* Third unknown must be YES */
for (opp = 0; opp < N; opp++) {
int opp_index;
if (opp == j || opp == k)
continue;
opp_index = d->edges[opp] - g->edges;
if (state->lines[opp_index] == LINE_UNKNOWN) {
solver_set_line(sstate, opp_index,
LINE_YES);
diff = min(diff, DIFF_EASY);
}
}
} else if (unknown == 4) {
/* Exactly one of opposite UNKNOWNS is YES. We've
* already set atmostone, so set atleastone as
* well.
*/
if (dline_set_opp_atleastone(sstate, d, j))
diff = min(diff, DIFF_NORMAL);
}
}
}
}
}
}
return diff;
}
static int linedsf_deductions(solver_state *sstate)
{
game_state *state = sstate->state;
grid *g = state->game_grid;
char *dlines = sstate->dlines;
int i;
int diff = DIFF_MAX;
int diff_tmp;
/* ------ Face deductions ------ */
/* A fully-general linedsf deduction seems overly complicated
* (I suspect the problem is NP-complete, though in practice it might just
* be doable because faces are limited in size).
* For simplicity, we only consider *pairs* of LINE_UNKNOWNS that are
* known to be identical. If setting them both to YES (or NO) would break
* the clue, set them to NO (or YES). */
for (i = 0; i < g->num_faces; i++) {
int N, yes, no, unknown;
int clue;
if (sstate->face_solved[i])
continue;
clue = state->clues[i];
if (clue < 0)
continue;
N = g->faces[i].order;
yes = sstate->face_yes_count[i];
if (yes + 1 == clue) {
if (face_setall_identical(sstate, i, LINE_NO))
diff = min(diff, DIFF_EASY);
}
no = sstate->face_no_count[i];
if (no + 1 == N - clue) {
if (face_setall_identical(sstate, i, LINE_YES))
diff = min(diff, DIFF_EASY);
}
/* Reload YES count, it might have changed */
yes = sstate->face_yes_count[i];
unknown = N - no - yes;
/* Deductions with small number of LINE_UNKNOWNs, based on overall
* parity of lines. */
diff_tmp = parity_deductions(sstate, g->faces[i].edges,
(clue - yes) % 2, unknown);
diff = min(diff, diff_tmp);
}
/* ------ Dot deductions ------ */
for (i = 0; i < g->num_dots; i++) {
grid_dot *d = g->dots + i;
int N = d->order;
int j;
int yes, no, unknown;
/* Go through dlines, and do any dline<->linedsf deductions wherever
* we find two UNKNOWNS. */
for (j = 0; j < N; j++) {
int dline_index = dline_index_from_dot(g, d, j);
int line1_index;
int line2_index;
int can1, can2, inv1, inv2;
int j2;
line1_index = d->edges[j] - g->edges;
if (state->lines[line1_index] != LINE_UNKNOWN)
continue;
j2 = j + 1;
if (j2 == N) j2 = 0;
line2_index = d->edges[j2] - g->edges;
if (state->lines[line2_index] != LINE_UNKNOWN)
continue;
/* Infer dline flags from linedsf */
can1 = edsf_canonify(sstate->linedsf, line1_index, &inv1);
can2 = edsf_canonify(sstate->linedsf, line2_index, &inv2);
if (can1 == can2 && inv1 != inv2) {
/* These are opposites, so set dline atmostone/atleastone */
if (set_atmostone(dlines, dline_index))
diff = min(diff, DIFF_NORMAL);
if (set_atleastone(dlines, dline_index))
diff = min(diff, DIFF_NORMAL);
continue;
}
/* Infer linedsf from dline flags */
if (is_atmostone(dlines, dline_index)
&& is_atleastone(dlines, dline_index)) {
if (merge_lines(sstate, line1_index, line2_index, 1))
diff = min(diff, DIFF_HARD);
}
}
/* Deductions with small number of LINE_UNKNOWNs, based on overall
* parity of lines. */
yes = sstate->dot_yes_count[i];
no = sstate->dot_no_count[i];
unknown = N - yes - no;
diff_tmp = parity_deductions(sstate, d->edges,
yes % 2, unknown);
diff = min(diff, diff_tmp);
}
/* ------ Edge dsf deductions ------ */
/* If the state of a line is known, deduce the state of its canonical line
* too, and vice versa. */
for (i = 0; i < g->num_edges; i++) {
int can, inv;
enum line_state s;
can = edsf_canonify(sstate->linedsf, i, &inv);
if (can == i)
continue;
s = sstate->state->lines[can];
if (s != LINE_UNKNOWN) {
if (solver_set_line(sstate, i, inv ? OPP(s) : s))
diff = min(diff, DIFF_EASY);
} else {
s = sstate->state->lines[i];
if (s != LINE_UNKNOWN) {
if (solver_set_line(sstate, can, inv ? OPP(s) : s))
diff = min(diff, DIFF_EASY);
}
}
}
return diff;
}
static int loop_deductions(solver_state *sstate)
{
int edgecount = 0, clues = 0, satclues = 0, sm1clues = 0;
game_state *state = sstate->state;
grid *g = state->game_grid;
int shortest_chainlen = g->num_dots;
int loop_found = FALSE;
int dots_connected;
int progress = FALSE;
int i;
/*
* Go through the grid and update for all the new edges.
* Since merge_dots() is idempotent, the simplest way to
* do this is just to update for _all_ the edges.
* Also, while we're here, we count the edges.
*/
for (i = 0; i < g->num_edges; i++) {
if (state->lines[i] == LINE_YES) {
loop_found |= merge_dots(sstate, i);
edgecount++;
}
}
/*
* Count the clues, count the satisfied clues, and count the
* satisfied-minus-one clues.
*/
for (i = 0; i < g->num_faces; i++) {
int c = state->clues[i];
if (c >= 0) {
int o = sstate->face_yes_count[i];
if (o == c)
satclues++;
else if (o == c-1)
sm1clues++;
clues++;
}
}
for (i = 0; i < g->num_dots; ++i) {
dots_connected =
sstate->looplen[dsf_canonify(sstate->dotdsf, i)];
if (dots_connected > 1)
shortest_chainlen = min(shortest_chainlen, dots_connected);
}
assert(sstate->solver_status == SOLVER_INCOMPLETE);
if (satclues == clues && shortest_chainlen == edgecount) {
sstate->solver_status = SOLVER_SOLVED;
/* This discovery clearly counts as progress, even if we haven't
* just added any lines or anything */
progress = TRUE;
goto finished_loop_deductionsing;
}
/*
* Now go through looking for LINE_UNKNOWN edges which
* connect two dots that are already in the same
* equivalence class. If we find one, test to see if the
* loop it would create is a solution.
*/
for (i = 0; i < g->num_edges; i++) {
grid_edge *e = g->edges + i;
int d1 = e->dot1 - g->dots;
int d2 = e->dot2 - g->dots;
int eqclass, val;
if (state->lines[i] != LINE_UNKNOWN)
continue;
eqclass = dsf_canonify(sstate->dotdsf, d1);
if (eqclass != dsf_canonify(sstate->dotdsf, d2))
continue;
val = LINE_NO; /* loop is bad until proven otherwise */
/*
* This edge would form a loop. Next
* question: how long would the loop be?
* Would it equal the total number of edges
* (plus the one we'd be adding if we added
* it)?
*/
if (sstate->looplen[eqclass] == edgecount + 1) {
int sm1_nearby;
/*
* This edge would form a loop which
* took in all the edges in the entire
* grid. So now we need to work out
* whether it would be a valid solution
* to the puzzle, which means we have to
* check if it satisfies all the clues.
* This means that every clue must be
* either satisfied or satisfied-minus-
* 1, and also that the number of
* satisfied-minus-1 clues must be at
* most two and they must lie on either
* side of this edge.
*/
sm1_nearby = 0;
if (e->face1) {
int f = e->face1 - g->faces;
int c = state->clues[f];
if (c >= 0 && sstate->face_yes_count[f] == c - 1)
sm1_nearby++;
}
if (e->face2) {
int f = e->face2 - g->faces;
int c = state->clues[f];
if (c >= 0 && sstate->face_yes_count[f] == c - 1)
sm1_nearby++;
}
if (sm1clues == sm1_nearby &&
sm1clues + satclues == clues) {
val = LINE_YES; /* loop is good! */
}
}
/*
* Right. Now we know that adding this edge
* would form a loop, and we know whether
* that loop would be a viable solution or
* not.
*
* If adding this edge produces a solution,
* then we know we've found _a_ solution but
* we don't know that it's _the_ solution -
* if it were provably the solution then
* we'd have deduced this edge some time ago
* without the need to do loop detection. So
* in this state we return SOLVER_AMBIGUOUS,
* which has the effect that hitting Solve
* on a user-provided puzzle will fill in a
* solution but using the solver to
* construct new puzzles won't consider this
* a reasonable deduction for the user to
* make.
*/
progress = solver_set_line(sstate, i, val);
assert(progress == TRUE);
if (val == LINE_YES) {
sstate->solver_status = SOLVER_AMBIGUOUS;
goto finished_loop_deductionsing;
}
}
finished_loop_deductionsing:
return progress ? DIFF_EASY : DIFF_MAX;
}
/* This will return a dynamically allocated solver_state containing the (more)
* solved grid */
static solver_state *solve_game_rec(const solver_state *sstate_start)
{
solver_state *sstate;
/* Index of the solver we should call next. */
int i = 0;
/* As a speed-optimisation, we avoid re-running solvers that we know
* won't make any progress. This happens when a high-difficulty
* solver makes a deduction that can only help other high-difficulty
* solvers.
* For example: if a new 'dline' flag is set by dline_deductions, the
* trivial_deductions solver cannot do anything with this information.
* If we've already run the trivial_deductions solver (because it's
* earlier in the list), there's no point running it again.
*
* Therefore: if a solver is earlier in the list than "threshold_index",
* we don't bother running it if it's difficulty level is less than
* "threshold_diff".
*/
int threshold_diff = 0;
int threshold_index = 0;
sstate = dup_solver_state(sstate_start);
check_caches(sstate);
while (i < NUM_SOLVERS) {
if (sstate->solver_status == SOLVER_MISTAKE)
return sstate;
if (sstate->solver_status == SOLVER_SOLVED ||
sstate->solver_status == SOLVER_AMBIGUOUS) {
/* solver finished */
break;
}
if ((solver_diffs[i] >= threshold_diff || i >= threshold_index)
&& solver_diffs[i] <= sstate->diff) {
/* current_solver is eligible, so use it */
int next_diff = solver_fns[i](sstate);
if (next_diff != DIFF_MAX) {
/* solver made progress, so use new thresholds and
* start again at top of list. */
threshold_diff = next_diff;
threshold_index = i;
i = 0;
continue;
}
}
/* current_solver is ineligible, or failed to make progress, so
* go to the next solver in the list */
i++;
}
if (sstate->solver_status == SOLVER_SOLVED ||
sstate->solver_status == SOLVER_AMBIGUOUS) {
/* s/LINE_UNKNOWN/LINE_NO/g */
array_setall(sstate->state->lines, LINE_UNKNOWN, LINE_NO,
sstate->state->game_grid->num_edges);
return sstate;
}
return sstate;
}
static char *solve_game(const game_state *state, const game_state *currstate,
const char *aux, char **error)
{
char *soln = NULL;
solver_state *sstate, *new_sstate;
sstate = new_solver_state(state, DIFF_MAX);
new_sstate = solve_game_rec(sstate);
if (new_sstate->solver_status == SOLVER_SOLVED) {
soln = encode_solve_move(new_sstate->state);
} else if (new_sstate->solver_status == SOLVER_AMBIGUOUS) {
soln = encode_solve_move(new_sstate->state);
/**error = "Solver found ambiguous solutions"; */
} else {
soln = encode_solve_move(new_sstate->state);
/**error = "Solver failed"; */
}
free_solver_state(new_sstate);
free_solver_state(sstate);
return soln;
}
/* ----------------------------------------------------------------------
* Drawing and mouse-handling
*/
static char *interpret_move(const game_state *state, game_ui *ui,
const game_drawstate *ds,
int x, int y, int button)
{
grid *g = state->game_grid;
grid_edge *e;
int i;
char *ret, buf[80];
char button_char = ' ';
enum line_state old_state;
button &= ~MOD_MASK;
/* Convert mouse-click (x,y) to grid coordinates */
x -= BORDER(ds->tilesize);
y -= BORDER(ds->tilesize);
x = x * g->tilesize / ds->tilesize;
y = y * g->tilesize / ds->tilesize;
x += g->lowest_x;
y += g->lowest_y;
e = grid_nearest_edge(g, x, y);
if (e == NULL)
return NULL;
i = e - g->edges;
/* I think it's only possible to play this game with mouse clicks, sorry */
/* Maybe will add mouse drag support some time */
old_state = state->lines[i];
switch (button) {
case LEFT_BUTTON:
switch (old_state) {
case LINE_UNKNOWN:
button_char = 'y';
break;
case LINE_YES:
#ifdef STYLUS_BASED
button_char = 'n';
break;
#endif
case LINE_NO:
button_char = 'u';
break;
}
break;
case MIDDLE_BUTTON:
button_char = 'u';
break;
case RIGHT_BUTTON:
switch (old_state) {
case LINE_UNKNOWN:
button_char = 'n';
break;
case LINE_NO:
#ifdef STYLUS_BASED
button_char = 'y';
break;
#endif
case LINE_YES:
button_char = 'u';
break;
}
break;
default:
return NULL;
}
sprintf(buf, "%d%c", i, (int)button_char);
ret = dupstr(buf);
return ret;
}
static game_state *execute_move(const game_state *state, const char *move)
{
int i;
game_state *newstate = dup_game(state);
if (move[0] == 'S') {
move++;
newstate->cheated = TRUE;
}
while (*move) {
i = atoi(move);
if (i < 0 || i >= newstate->game_grid->num_edges)
goto fail;
move += strspn(move, "1234567890");
switch (*(move++)) {
case 'y':
newstate->lines[i] = LINE_YES;
break;
case 'n':
newstate->lines[i] = LINE_NO;
break;
case 'u':
newstate->lines[i] = LINE_UNKNOWN;
break;
default:
goto fail;
}
}
/*
* Check for completion.
*/
if (check_completion(newstate))
newstate->solved = TRUE;
return newstate;
fail:
free_game(newstate);
return NULL;
}
/* ----------------------------------------------------------------------
* Drawing routines.
*/
/* Convert from grid coordinates to screen coordinates */
static void grid_to_screen(const game_drawstate *ds, const grid *g,
int grid_x, int grid_y, int *x, int *y)
{
*x = grid_x - g->lowest_x;
*y = grid_y - g->lowest_y;
*x = *x * ds->tilesize / g->tilesize;
*y = *y * ds->tilesize / g->tilesize;
*x += BORDER(ds->tilesize);
*y += BORDER(ds->tilesize);
}
/* Returns (into x,y) position of centre of face for rendering the text clue.
*/
static void face_text_pos(const game_drawstate *ds, const grid *g,
grid_face *f, int *xret, int *yret)
{
int faceindex = f - g->faces;
/*
* Return the cached position for this face, if we've already
* worked it out.
*/
if (ds->textx[faceindex] >= 0) {
*xret = ds->textx[faceindex];
*yret = ds->texty[faceindex];
return;
}
/*
* Otherwise, use the incentre computed by grid.c and convert it
* to screen coordinates.
*/
grid_find_incentre(f);
grid_to_screen(ds, g, f->ix, f->iy,
&ds->textx[faceindex], &ds->texty[faceindex]);
*xret = ds->textx[faceindex];
*yret = ds->texty[faceindex];
}
static void face_text_bbox(game_drawstate *ds, grid *g, grid_face *f,
int *x, int *y, int *w, int *h)
{
int xx, yy;
face_text_pos(ds, g, f, &xx, &yy);
/* There seems to be a certain amount of trial-and-error involved
* in working out the correct bounding-box for the text. */
*x = xx - ds->tilesize/4 - 1;
*y = yy - ds->tilesize/4 - 3;
*w = ds->tilesize/2 + 2;
*h = ds->tilesize/2 + 5;
}
static void game_redraw_clue(drawing *dr, game_drawstate *ds,
const game_state *state, int i)
{
grid *g = state->game_grid;
grid_face *f = g->faces + i;
int x, y;
char c[20];
sprintf(c, "%d", state->clues[i]);
face_text_pos(ds, g, f, &x, &y);
draw_text(dr, x, y,
FONT_VARIABLE, ds->tilesize/2,
ALIGN_VCENTRE | ALIGN_HCENTRE,
ds->clue_error[i] ? COL_MISTAKE :
ds->clue_satisfied[i] ? COL_SATISFIED : COL_FOREGROUND, c);
}
static void edge_bbox(game_drawstate *ds, grid *g, grid_edge *e,
int *x, int *y, int *w, int *h)
{
int x1 = e->dot1->x;
int y1 = e->dot1->y;
int x2 = e->dot2->x;
int y2 = e->dot2->y;
int xmin, xmax, ymin, ymax;
grid_to_screen(ds, g, x1, y1, &x1, &y1);
grid_to_screen(ds, g, x2, y2, &x2, &y2);
/* Allow extra margin for dots, and thickness of lines */
xmin = min(x1, x2) - 2;
xmax = max(x1, x2) + 2;
ymin = min(y1, y2) - 2;
ymax = max(y1, y2) + 2;
*x = xmin;
*y = ymin;
*w = xmax - xmin + 1;
*h = ymax - ymin + 1;
}
static void dot_bbox(game_drawstate *ds, grid *g, grid_dot *d,
int *x, int *y, int *w, int *h)
{
int x1, y1;
grid_to_screen(ds, g, d->x, d->y, &x1, &y1);
*x = x1 - 2;
*y = y1 - 2;
*w = 5;
*h = 5;
}
static const int loopy_line_redraw_phases[] = {
COL_FAINT, COL_LINEUNKNOWN, COL_FOREGROUND, COL_HIGHLIGHT, COL_MISTAKE
};
#define NPHASES lenof(loopy_line_redraw_phases)
static void game_redraw_line(drawing *dr, game_drawstate *ds,
const game_state *state, int i, int phase)
{
grid *g = state->game_grid;
grid_edge *e = g->edges + i;
int x1, x2, y1, y2;
int line_colour;
if (state->line_errors[i])
line_colour = COL_MISTAKE;
else if (state->lines[i] == LINE_UNKNOWN)
line_colour = COL_LINEUNKNOWN;
else if (state->lines[i] == LINE_NO)
line_colour = COL_FAINT;
else if (ds->flashing)
line_colour = COL_HIGHLIGHT;
else
line_colour = COL_FOREGROUND;
if (line_colour != loopy_line_redraw_phases[phase])
return;
/* Convert from grid to screen coordinates */
grid_to_screen(ds, g, e->dot1->x, e->dot1->y, &x1, &y1);
grid_to_screen(ds, g, e->dot2->x, e->dot2->y, &x2, &y2);
if (line_colour == COL_FAINT) {
static int draw_faint_lines = -1;
if (draw_faint_lines < 0) {
char *env = getenv("LOOPY_FAINT_LINES");
draw_faint_lines = (!env || (env[0] == 'y' ||
env[0] == 'Y'));
}
if (draw_faint_lines)
draw_line(dr, x1, y1, x2, y2, line_colour);
} else {
draw_thick_line(dr, 3.0,
x1 + 0.5, y1 + 0.5,
x2 + 0.5, y2 + 0.5,
line_colour);
}
}
static void game_redraw_dot(drawing *dr, game_drawstate *ds,
const game_state *state, int i)
{
grid *g = state->game_grid;
grid_dot *d = g->dots + i;
int x, y;
grid_to_screen(ds, g, d->x, d->y, &x, &y);
draw_circle(dr, x, y, 2, COL_FOREGROUND, COL_FOREGROUND);
}
static int boxes_intersect(int x0, int y0, int w0, int h0,
int x1, int y1, int w1, int h1)
{
/*
* Two intervals intersect iff neither is wholly on one side of
* the other. Two boxes intersect iff their horizontal and
* vertical intervals both intersect.
*/
return (x0 < x1+w1 && x1 < x0+w0 && y0 < y1+h1 && y1 < y0+h0);
}
static void game_redraw_in_rect(drawing *dr, game_drawstate *ds,
const game_state *state,
int x, int y, int w, int h)
{
grid *g = state->game_grid;
int i, phase;
int bx, by, bw, bh;
clip(dr, x, y, w, h);
draw_rect(dr, x, y, w, h, COL_BACKGROUND);
for (i = 0; i < g->num_faces; i++) {
if (state->clues[i] >= 0) {
face_text_bbox(ds, g, &g->faces[i], &bx, &by, &bw, &bh);
if (boxes_intersect(x, y, w, h, bx, by, bw, bh))
game_redraw_clue(dr, ds, state, i);
}
}
for (phase = 0; phase < NPHASES; phase++) {
for (i = 0; i < g->num_edges; i++) {
edge_bbox(ds, g, &g->edges[i], &bx, &by, &bw, &bh);
if (boxes_intersect(x, y, w, h, bx, by, bw, bh))
game_redraw_line(dr, ds, state, i, phase);
}
}
for (i = 0; i < g->num_dots; i++) {
dot_bbox(ds, g, &g->dots[i], &bx, &by, &bw, &bh);
if (boxes_intersect(x, y, w, h, bx, by, bw, bh))
game_redraw_dot(dr, ds, state, i);
}
unclip(dr);
draw_update(dr, x, y, w, h);
}
static void game_redraw(drawing *dr, game_drawstate *ds,
const game_state *oldstate, const game_state *state,
int dir, const game_ui *ui,
float animtime, float flashtime)
{
#define REDRAW_OBJECTS_LIMIT 16 /* Somewhat arbitrary tradeoff */
grid *g = state->game_grid;
int border = BORDER(ds->tilesize);
int i;
int flash_changed;
int redraw_everything = FALSE;
int edges[REDRAW_OBJECTS_LIMIT], nedges = 0;
int faces[REDRAW_OBJECTS_LIMIT], nfaces = 0;
/* Redrawing is somewhat involved.
*
* An update can theoretically affect an arbitrary number of edges
* (consider, for example, completing or breaking a cycle which doesn't
* satisfy all the clues -- we'll switch many edges between error and
* normal states). On the other hand, redrawing the whole grid takes a
* while, making the game feel sluggish, and many updates are actually
* quite well localized.
*
* This redraw algorithm attempts to cope with both situations gracefully
* and correctly. For localized changes, we set a clip rectangle, fill
* it with background, and then redraw (a plausible but conservative
* guess at) the objects which intersect the rectangle; if several
* objects need redrawing, we'll do them individually. However, if lots
* of objects are affected, we'll just redraw everything.
*
* The reason for all of this is that it's just not safe to do the redraw
* piecemeal. If you try to draw an antialiased diagonal line over
* itself, you get a slightly thicker antialiased diagonal line, which
* looks rather ugly after a while.
*
* So, we take two passes over the grid. The first attempts to work out
* what needs doing, and the second actually does it.
*/
if (!ds->started) {
redraw_everything = TRUE;
/*
* But we must still go through the upcoming loops, so that we
* set up stuff in ds correctly for the initial redraw.
*/
}
/* First, trundle through the faces. */
for (i = 0; i < g->num_faces; i++) {
grid_face *f = g->faces + i;
int sides = f->order;
int clue_mistake;
int clue_satisfied;
int n = state->clues[i];
if (n < 0)
continue;
clue_mistake = (face_order(state, i, LINE_YES) > n ||
face_order(state, i, LINE_NO ) > (sides-n));
clue_satisfied = (face_order(state, i, LINE_YES) == n &&
face_order(state, i, LINE_NO ) == (sides-n));
if (clue_mistake != ds->clue_error[i] ||
clue_satisfied != ds->clue_satisfied[i]) {
ds->clue_error[i] = clue_mistake;
ds->clue_satisfied[i] = clue_satisfied;
if (nfaces == REDRAW_OBJECTS_LIMIT)
redraw_everything = TRUE;
else
faces[nfaces++] = i;
}
}
/* Work out what the flash state needs to be. */
if (flashtime > 0 &&
(flashtime <= FLASH_TIME/3 ||
flashtime >= FLASH_TIME*2/3)) {
flash_changed = !ds->flashing;
ds->flashing = TRUE;
} else {
flash_changed = ds->flashing;
ds->flashing = FALSE;
}
/* Now, trundle through the edges. */
for (i = 0; i < g->num_edges; i++) {
char new_ds =
state->line_errors[i] ? DS_LINE_ERROR : state->lines[i];
if (new_ds != ds->lines[i] ||
(flash_changed && state->lines[i] == LINE_YES)) {
ds->lines[i] = new_ds;
if (nedges == REDRAW_OBJECTS_LIMIT)
redraw_everything = TRUE;
else
edges[nedges++] = i;
}
}
/* Pass one is now done. Now we do the actual drawing. */
if (redraw_everything) {
int grid_width = g->highest_x - g->lowest_x;
int grid_height = g->highest_y - g->lowest_y;
int w = grid_width * ds->tilesize / g->tilesize;
int h = grid_height * ds->tilesize / g->tilesize;
game_redraw_in_rect(dr, ds, state,
0, 0, w + 2*border + 1, h + 2*border + 1);
} else {
/* Right. Now we roll up our sleeves. */
for (i = 0; i < nfaces; i++) {
grid_face *f = g->faces + faces[i];
int x, y, w, h;
face_text_bbox(ds, g, f, &x, &y, &w, &h);
game_redraw_in_rect(dr, ds, state, x, y, w, h);
}
for (i = 0; i < nedges; i++) {
grid_edge *e = g->edges + edges[i];
int x, y, w, h;
edge_bbox(ds, g, e, &x, &y, &w, &h);
game_redraw_in_rect(dr, ds, state, x, y, w, h);
}
}
ds->started = TRUE;
}
static float game_flash_length(const game_state *oldstate,
const game_state *newstate, int dir, game_ui *ui)
{
if (!oldstate->solved && newstate->solved &&
!oldstate->cheated && !newstate->cheated) {
return FLASH_TIME;
}
return 0.0F;
}
static int game_status(const game_state *state)
{
return state->solved ? +1 : 0;
}
static void game_print_size(const game_params *params, float *x, float *y)
{
int pw, ph;
/*
* I'll use 7mm "squares" by default.
*/
game_compute_size(params, 700, &pw, &ph);
*x = pw / 100.0F;
*y = ph / 100.0F;
}
static void game_print(drawing *dr, const game_state *state, int tilesize)
{
int ink = print_mono_colour(dr, 0);
int i;
game_drawstate ads, *ds = &ads;
grid *g = state->game_grid;
ds->tilesize = tilesize;
ds->textx = snewn(g->num_faces, int);
ds->texty = snewn(g->num_faces, int);
for (i = 0; i < g->num_faces; i++)
ds->textx[i] = ds->texty[i] = -1;
for (i = 0; i < g->num_dots; i++) {
int x, y;
grid_to_screen(ds, g, g->dots[i].x, g->dots[i].y, &x, &y);
draw_circle(dr, x, y, ds->tilesize / 15, ink, ink);
}
/*
* Clues.
*/
for (i = 0; i < g->num_faces; i++) {
grid_face *f = g->faces + i;
int clue = state->clues[i];
if (clue >= 0) {
char c[20];
int x, y;
sprintf(c, "%d", state->clues[i]);
face_text_pos(ds, g, f, &x, &y);
draw_text(dr, x, y,
FONT_VARIABLE, ds->tilesize / 2,
ALIGN_VCENTRE | ALIGN_HCENTRE, ink, c);
}
}
/*
* Lines.
*/
for (i = 0; i < g->num_edges; i++) {
int thickness = (state->lines[i] == LINE_YES) ? 30 : 150;
grid_edge *e = g->edges + i;
int x1, y1, x2, y2;
grid_to_screen(ds, g, e->dot1->x, e->dot1->y, &x1, &y1);
grid_to_screen(ds, g, e->dot2->x, e->dot2->y, &x2, &y2);
if (state->lines[i] == LINE_YES)
{
/* (dx, dy) points from (x1, y1) to (x2, y2).
* The line is then "fattened" in a perpendicular
* direction to create a thin rectangle. */
double d = sqrt(SQ((double)x1 - x2) + SQ((double)y1 - y2));
double dx = (x2 - x1) / d;
double dy = (y2 - y1) / d;
int points[8];
dx = (dx * ds->tilesize) / thickness;
dy = (dy * ds->tilesize) / thickness;
points[0] = x1 + (int)dy;
points[1] = y1 - (int)dx;
points[2] = x1 - (int)dy;
points[3] = y1 + (int)dx;
points[4] = x2 - (int)dy;
points[5] = y2 + (int)dx;
points[6] = x2 + (int)dy;
points[7] = y2 - (int)dx;
draw_polygon(dr, points, 4, ink, ink);
}
else
{
/* Draw a dotted line */
int divisions = 6;
int j;
for (j = 1; j < divisions; j++) {
/* Weighted average */
int x = (x1 * (divisions -j) + x2 * j) / divisions;
int y = (y1 * (divisions -j) + y2 * j) / divisions;
draw_circle(dr, x, y, ds->tilesize / thickness, ink, ink);
}
}
}
sfree(ds->textx);
sfree(ds->texty);
}
#ifdef COMBINED
#define thegame loopy
#endif
const struct game thegame = {
"Loopy", "games.loopy", "loopy",
default_params,
game_fetch_preset,
decode_params,
encode_params,
free_params,
dup_params,
TRUE, game_configure, custom_params,
validate_params,
new_game_desc,
validate_desc,
new_game,
dup_game,
free_game,
1, solve_game,
TRUE, game_can_format_as_text_now, game_text_format,
new_ui,
free_ui,
encode_ui,
decode_ui,
game_changed_state,
interpret_move,
execute_move,
PREFERRED_TILE_SIZE, game_compute_size, game_set_size,
game_colours,
game_new_drawstate,
game_free_drawstate,
game_redraw,
game_anim_length,
game_flash_length,
game_status,
TRUE, FALSE, game_print_size, game_print,
FALSE /* wants_statusbar */,
FALSE, game_timing_state,
0, /* mouse_priorities */
};
#ifdef STANDALONE_SOLVER
/*
* Half-hearted standalone solver. It can't output the solution to
* anything but a square puzzle, and it can't log the deductions
* it makes either. But it can solve square puzzles, and more
* importantly it can use its solver to grade the difficulty of
* any puzzle you give it.
*/
#include <stdarg.h>
int main(int argc, char **argv)
{
game_params *p;
game_state *s;
char *id = NULL, *desc, *err;
int grade = FALSE;
int ret, diff;
#if 0 /* verbose solver not supported here (yet) */
int really_verbose = FALSE;
#endif
while (--argc > 0) {
char *p = *++argv;
#if 0 /* verbose solver not supported here (yet) */
if (!strcmp(p, "-v")) {
really_verbose = TRUE;
} else
#endif
if (!strcmp(p, "-g")) {
grade = TRUE;
} else if (*p == '-') {
fprintf(stderr, "%s: unrecognised option `%s'\n", argv[0], p);
return 1;
} else {
id = p;
}
}
if (!id) {
fprintf(stderr, "usage: %s [-g | -v] <game_id>\n", argv[0]);
return 1;
}
desc = strchr(id, ':');
if (!desc) {
fprintf(stderr, "%s: game id expects a colon in it\n", argv[0]);
return 1;
}
*desc++ = '\0';
p = default_params();
decode_params(p, id);
err = validate_desc(p, desc);
if (err) {
fprintf(stderr, "%s: %s\n", argv[0], err);
return 1;
}
s = new_game(NULL, p, desc);
/*
* When solving an Easy puzzle, we don't want to bother the
* user with Hard-level deductions. For this reason, we grade
* the puzzle internally before doing anything else.
*/
ret = -1; /* placate optimiser */
for (diff = 0; diff < DIFF_MAX; diff++) {
solver_state *sstate_new;
solver_state *sstate = new_solver_state((game_state *)s, diff);
sstate_new = solve_game_rec(sstate);
if (sstate_new->solver_status == SOLVER_MISTAKE)
ret = 0;
else if (sstate_new->solver_status == SOLVER_SOLVED)
ret = 1;
else
ret = 2;
free_solver_state(sstate_new);
free_solver_state(sstate);
if (ret < 2)
break;
}
if (diff == DIFF_MAX) {
if (grade)
printf("Difficulty rating: harder than Hard, or ambiguous\n");
else
printf("Unable to find a unique solution\n");
} else {
if (grade) {
if (ret == 0)
printf("Difficulty rating: impossible (no solution exists)\n");
else if (ret == 1)
printf("Difficulty rating: %s\n", diffnames[diff]);
} else {
solver_state *sstate_new;
solver_state *sstate = new_solver_state((game_state *)s, diff);
/* If we supported a verbose solver, we'd set verbosity here */
sstate_new = solve_game_rec(sstate);
if (sstate_new->solver_status == SOLVER_MISTAKE)
printf("Puzzle is inconsistent\n");
else {
assert(sstate_new->solver_status == SOLVER_SOLVED);
if (s->grid_type == 0) {
fputs(game_text_format(sstate_new->state), stdout);
} else {
printf("Unable to output non-square grids\n");
}
}
free_solver_state(sstate_new);
free_solver_state(sstate);
}
}
return 0;
}
#endif
/* vim: set shiftwidth=4 tabstop=8: */
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