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puzzles/unfinished/separate.c
Simon Tatham b3243d7504 Return error messages as 'const char *', not 'char *'. 
They're never dynamically allocated, and are almost always string
literals, so const is more appropriate.
2017-10-01 16:34:41 +01:00

860 lines
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/*
* separate.c: Implementation of `Block Puzzle', a Japanese-only
* Nikoli puzzle seen at
* http://www.nikoli.co.jp/ja/puzzles/block_puzzle/
*
* It's difficult to be absolutely sure of the rules since online
* Japanese translators are so bad, but looking at the sample
* puzzle it seems fairly clear that the rules of this one are
* very simple. You have an mxn grid in which every square
* contains a letter, there are k distinct letters with k dividing
* mn, and every letter occurs the same number of times; your aim
* is to find a partition of the grid into disjoint k-ominoes such
* that each k-omino contains exactly one of each letter.
*
* (It may be that Nikoli always have m,n,k equal to one another.
* However, I don't see that that's critical to the puzzle; k|mn
* is the only really important constraint, and even that could
* probably be dispensed with if some squares were marked as
* unused.)
*/
/*
* Current status: only the solver/generator is yet written, and
* although working in principle it's _very_ slow. It generates
* 5x5n5 or 6x6n4 readily enough, 6x6n6 with a bit of effort, and
* 7x7n7 only with a serious strain. I haven't dared try it higher
* than that yet.
*
* One idea to speed it up is to implement more of the solver.
* Ideas I've so far had include:
*
* - Generalise the deduction currently expressed as `an
* undersized chain with only one direction to extend must take
* it'. More generally, the deduction should say `if all the
* possible k-ominoes containing a given chain also contain
* square x, then mark square x as part of that k-omino'.
* + For example, consider this case:
*
* a ? b This represents the top left of a board; the letters
* ? ? ? a,b,c do not represent the letters used in the puzzle,
* c ? ? but indicate that those three squares are known to be
* of different ominoes. Now if k >= 4, we can immediately
* deduce that the square midway between b and c belongs to the
* same omino as a, because there is no way we can make a 4-or-
* more-omino containing a which does not also contain that square.
* (Most easily seen by imagining cutting that square out of the
* grid; then, clearly, the omino containing a has only two
* squares to expand into, and needs at least three.)
*
* The key difficulty with this mode of reasoning is
* identifying such squares. I can't immediately think of a
* simple algorithm for finding them on a wholesale basis.
*
* - Bfs out from a chain looking for the letters it lacks. For
* example, in this situation (top three rows of a 7x7n7 grid):
*
* +-----------+-+
* |E-A-F-B-C D|D|
* +------- ||
* |E-C-G-D G|G E|
* +-+--- |
* |E|E G A B F A|
*
* In this situation we can be sure that the top left chain
* E-A-F-B-C does extend rightwards to the D, because there is
* no other D within reach of that chain. Note also that the
* bfs can skip squares which are known to belong to other
* ominoes than this one.
*
* (This deduction, I fear, should only be used in an
* emergency, because it relies on _all_ squares within range
* of the bfs having particular values and so using it during
* incremental generation rather nails down a lot of the grid.)
*
* It's conceivable that another thing we could do would be to
* increase the flexibility in the grid generator: instead of
* nailing down the _value_ of any square depended on, merely nail
* down its equivalence to other squares. Unfortunately this turns
* the letter-selection phase of generation into a general graph
* colouring problem (we must draw a graph with equivalence
* classes of squares as the vertices, and an edge between any two
* vertices representing equivalence classes which contain squares
* that share an omino, and then k-colour the result) and hence
* requires recursion, which bodes ill for something we're doing
* that many times per generation.
*
* I suppose a simple thing I could try would be tuning the retry
* count, just in case it's set too high or too low for efficient
* generation.
*/
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <assert.h>
#include <ctype.h>
#include <math.h>
#include "puzzles.h"
enum {
COL_BACKGROUND,
NCOLOURS
};
struct game_params {
int w, h, k;
};
struct game_state {
int FIXME;
};
static game_params *default_params(void)
{
game_params *ret = snew(game_params);
ret->w = ret->h = ret->k = 5; /* FIXME: a bit bigger? */
return ret;
}
static int game_fetch_preset(int i, char **name, game_params **params)
{
return FALSE;
}
static void free_params(game_params *params)
{
sfree(params);
}
static game_params *dup_params(const game_params *params)
{
game_params *ret = snew(game_params);
*ret = *params; /* structure copy */
return ret;
}
static void decode_params(game_params *params, char const *string)
{
params->w = params->h = params->k = atoi(string);
while (*string && isdigit((unsigned char)*string)) string++;
if (*string == 'x') {
string++;
params->h = atoi(string);
while (*string && isdigit((unsigned char)*string)) string++;
}
if (*string == 'n') {
string++;
params->k = atoi(string);
while (*string && isdigit((unsigned char)*string)) string++;
}
}
static char *encode_params(const game_params *params, int full)
{
char buf[256];
sprintf(buf, "%dx%dn%d", params->w, params->h, params->k);
return dupstr(buf);
}
static config_item *game_configure(const game_params *params)
{
return NULL;
}
static game_params *custom_params(const config_item *cfg)
{
return NULL;
}
static const char *validate_params(const game_params *params, int full)
{
return NULL;
}
/* ----------------------------------------------------------------------
* Solver and generator.
*/
struct solver_scratch {
int w, h, k;
/*
* Tracks connectedness between squares.
*/
int *dsf;
/*
* size[dsf_canonify(dsf, yx)] tracks the size of the
* connected component containing yx.
*/
int *size;
/*
* contents[dsf_canonify(dsf, yx)*k+i] tracks whether or not
* the connected component containing yx includes letter i. If
* the value is -1, it doesn't; otherwise its value is the
* index in the main grid of the square which contributes that
* letter to the component.
*/
int *contents;
/*
* disconnect[dsf_canonify(dsf, yx1)*w*h + dsf_canonify(dsf, yx2)]
* tracks whether or not the connected components containing
* yx1 and yx2 are known to be distinct.
*/
unsigned char *disconnect;
/*
* Temporary space used only inside particular solver loops.
*/
int *tmp;
};
struct solver_scratch *solver_scratch_new(int w, int h, int k)
{
int wh = w*h;
struct solver_scratch *sc = snew(struct solver_scratch);
sc->w = w;
sc->h = h;
sc->k = k;
sc->dsf = snew_dsf(wh);
sc->size = snewn(wh, int);
sc->contents = snewn(wh * k, int);
sc->disconnect = snewn(wh*wh, unsigned char);
sc->tmp = snewn(wh, int);
return sc;
}
void solver_scratch_free(struct solver_scratch *sc)
{
sfree(sc->dsf);
sfree(sc->size);
sfree(sc->contents);
sfree(sc->disconnect);
sfree(sc->tmp);
sfree(sc);
}
void solver_connect(struct solver_scratch *sc, int yx1, int yx2)
{
int w = sc->w, h = sc->h, k = sc->k;
int wh = w*h;
int i, yxnew;
yx1 = dsf_canonify(sc->dsf, yx1);
yx2 = dsf_canonify(sc->dsf, yx2);
assert(yx1 != yx2);
/*
* To connect two components together into a bigger one, we
* start by merging them in the dsf itself.
*/
dsf_merge(sc->dsf, yx1, yx2);
yxnew = dsf_canonify(sc->dsf, yx2);
/*
* The size of the new component is the sum of the sizes of the
* old ones.
*/
sc->size[yxnew] = sc->size[yx1] + sc->size[yx2];
/*
* The contents bitmap of the new component is the union of the
* contents of the old ones.
*
* Given two numbers at most one of which is not -1, we can
* find the other one by adding the two and adding 1; this
* will yield -1 if both were -1 to begin with, otherwise the
* other.
*
* (A neater approach would be to take their bitwise AND, but
* this is unfortunately not well-defined standard C when done
* to signed integers.)
*/
for (i = 0; i < k; i++) {
assert(sc->contents[yx1*k+i] < 0 || sc->contents[yx2*k+i] < 0);
sc->contents[yxnew*k+i] = (sc->contents[yx1*k+i] +
sc->contents[yx2*k+i] + 1);
}
/*
* We must combine the rows _and_ the columns in the disconnect
* matrix.
*/
for (i = 0; i < wh; i++)
sc->disconnect[yxnew*wh+i] = (sc->disconnect[yx1*wh+i] ||
sc->disconnect[yx2*wh+i]);
for (i = 0; i < wh; i++)
sc->disconnect[i*wh+yxnew] = (sc->disconnect[i*wh+yx1] ||
sc->disconnect[i*wh+yx2]);
}
void solver_disconnect(struct solver_scratch *sc, int yx1, int yx2)
{
int w = sc->w, h = sc->h;
int wh = w*h;
yx1 = dsf_canonify(sc->dsf, yx1);
yx2 = dsf_canonify(sc->dsf, yx2);
assert(yx1 != yx2);
assert(!sc->disconnect[yx1*wh+yx2]);
assert(!sc->disconnect[yx2*wh+yx1]);
/*
* Mark the components as disconnected from each other in the
* disconnect matrix.
*/
sc->disconnect[yx1*wh+yx2] = sc->disconnect[yx2*wh+yx1] = 1;
}
void solver_init(struct solver_scratch *sc)
{
int w = sc->w, h = sc->h;
int wh = w*h;
int i;
/*
* Set up most of the scratch space. We don't set up the
* contents array, however, because this will change if we
* adjust the letter arrangement and re-run the solver.
*/
dsf_init(sc->dsf, wh);
for (i = 0; i < wh; i++) sc->size[i] = 1;
memset(sc->disconnect, 0, wh*wh);
}
int solver_attempt(struct solver_scratch *sc, const unsigned char *grid,
unsigned char *gen_lock)
{
int w = sc->w, h = sc->h, k = sc->k;
int wh = w*h;
int i, x, y;
int done_something_overall = FALSE;
/*
* Set up the contents array from the grid.
*/
for (i = 0; i < wh*k; i++)
sc->contents[i] = -1;
for (i = 0; i < wh; i++)
sc->contents[dsf_canonify(sc->dsf, i)*k+grid[i]] = i;
while (1) {
int done_something = FALSE;
/*
* Go over the grid looking for reasons to add to the
* disconnect matrix. We're after pairs of squares which:
*
* - are adjacent in the grid
* - belong to distinct dsf components
* - their components are not already marked as
* disconnected
* - their components share a letter in common.
*/
for (y = 0; y < h; y++) {
for (x = 0; x < w; x++) {
int dir;
for (dir = 0; dir < 2; dir++) {
int x2 = x + dir, y2 = y + 1 - dir;
int yx = y*w+x, yx2 = y2*w+x2;
if (x2 >= w || y2 >= h)
continue; /* one square is outside the grid */
yx = dsf_canonify(sc->dsf, yx);
yx2 = dsf_canonify(sc->dsf, yx2);
if (yx == yx2)
continue; /* same dsf component */
if (sc->disconnect[yx*wh+yx2])
continue; /* already known disconnected */
for (i = 0; i < k; i++)
if (sc->contents[yx*k+i] >= 0 &&
sc->contents[yx2*k+i] >= 0)
break;
if (i == k)
continue; /* no letter in common */
/*
* We've found one. Mark yx and yx2 as
* disconnected from each other.
*/
#ifdef SOLVER_DIAGNOSTICS
printf("Disconnecting %d and %d (%c)\n", yx, yx2, 'A'+i);
#endif
solver_disconnect(sc, yx, yx2);
done_something = done_something_overall = TRUE;
/*
* We have just made a deduction which hinges
* on two particular grid squares being the
* same. If we are feeding back to a generator
* loop, we must therefore mark those squares
* as fixed in the generator, so that future
* rearrangement of the grid will not break
* the information on which we have already
* based deductions.
*/
if (gen_lock) {
gen_lock[sc->contents[yx*k+i]] = 1;
gen_lock[sc->contents[yx2*k+i]] = 1;
}
}
}
}
/*
* Now go over the grid looking for dsf components which
* are below maximum size and only have one way to extend,
* and extending them.
*/
for (i = 0; i < wh; i++)
sc->tmp[i] = -1;
for (y = 0; y < h; y++) {
for (x = 0; x < w; x++) {
int yx = dsf_canonify(sc->dsf, y*w+x);
int dir;
if (sc->size[yx] == k)
continue;
for (dir = 0; dir < 4; dir++) {
int x2 = x + (dir==0 ? -1 : dir==2 ? 1 : 0);
int y2 = y + (dir==1 ? -1 : dir==3 ? 1 : 0);
int yx2, yx2c;
if (y2 < 0 || y2 >= h || x2 < 0 || x2 >= w)
continue;
yx2 = y2*w+x2;
yx2c = dsf_canonify(sc->dsf, yx2);
if (yx2c != yx && !sc->disconnect[yx2c*wh+yx]) {
/*
* Component yx can be extended into square
* yx2.
*/
if (sc->tmp[yx] == -1)
sc->tmp[yx] = yx2;
else if (sc->tmp[yx] != yx2)
sc->tmp[yx] = -2; /* multiple choices found */
}
}
}
}
for (i = 0; i < wh; i++) {
if (sc->tmp[i] >= 0) {
/*
* Make sure we haven't connected the two already
* during this loop (which could happen if for
* _both_ components this was the only way to
* extend them).
*/
if (dsf_canonify(sc->dsf, i) ==
dsf_canonify(sc->dsf, sc->tmp[i]))
continue;
#ifdef SOLVER_DIAGNOSTICS
printf("Connecting %d and %d\n", i, sc->tmp[i]);
#endif
solver_connect(sc, i, sc->tmp[i]);
done_something = done_something_overall = TRUE;
break;
}
}
if (!done_something)
break;
}
/*
* Return 0 if we haven't made any progress; 1 if we've done
* something but not solved it completely; 2 if we've solved
* it completely.
*/
for (i = 0; i < wh; i++)
if (sc->size[dsf_canonify(sc->dsf, i)] != k)
break;
if (i == wh)
return 2;
if (done_something_overall)
return 1;
return 0;
}
unsigned char *generate(int w, int h, int k, random_state *rs)
{
int wh = w*h;
int n = wh/k;
struct solver_scratch *sc;
unsigned char *grid;
unsigned char *shuffled;
int i, j, m, retries;
int *permutation;
unsigned char *gen_lock;
extern int *divvy_rectangle(int w, int h, int k, random_state *rs);
sc = solver_scratch_new(w, h, k);
grid = snewn(wh, unsigned char);
shuffled = snewn(k, unsigned char);
permutation = snewn(wh, int);
gen_lock = snewn(wh, unsigned char);
do {
int *dsf = divvy_rectangle(w, h, k, rs);
/*
* Go through the dsf and find the indices of all the
* squares involved in each omino, in a manner conducive
* to per-omino indexing. We set permutation[i*k+j] to be
* the index of the jth square (ordered arbitrarily) in
* omino i.
*/
for (i = j = 0; i < wh; i++)
if (dsf_canonify(dsf, i) == i) {
sc->tmp[i] = j;
/*
* During this loop and the following one, we use
* the last element of each row of permutation[]
* as a counter of the number of indices so far
* placed in it. When we place the final index of
* an omino, that counter is overwritten, but that
* doesn't matter because we'll never use it
* again. Of course this depends critically on
* divvy_rectangle() having returned correct
* results, or else chaos would ensue.
*/
permutation[j*k+k-1] = 0;
j++;
}
for (i = 0; i < wh; i++) {
j = sc->tmp[dsf_canonify(dsf, i)];
m = permutation[j*k+k-1]++;
permutation[j*k+m] = i;
}
/*
* Track which squares' letters we have already depended
* on for deductions. This is gradually updated by
* solver_attempt().
*/
memset(gen_lock, 0, wh);
/*
* Now repeatedly fill the grid with letters, and attempt
* to solve it. If the solver makes progress but does not
* fail completely, then gen_lock will have been updated
* and we try again. On a complete failure, though, we
* have no option but to give up and abandon this set of
* ominoes.
*/
solver_init(sc);
retries = k*k;
while (1) {
/*
* Fill the grid with letters. We can safely use
* sc->tmp to hold the set of letters required at each
* stage, since it's at least size k and is currently
* unused.
*/
for (i = 0; i < n; i++) {
/*
* First, determine the set of letters already
* placed in this omino by gen_lock.
*/
for (j = 0; j < k; j++)
sc->tmp[j] = j;
for (j = 0; j < k; j++) {
int index = permutation[i*k+j];
int letter = grid[index];
if (gen_lock[index])
sc->tmp[letter] = -1;
}
/*
* Now collect together all the remaining letters
* and randomly shuffle them.
*/
for (j = m = 0; j < k; j++)
if (sc->tmp[j] >= 0)
sc->tmp[m++] = sc->tmp[j];
shuffle(sc->tmp, m, sizeof(*sc->tmp), rs);
/*
* Finally, write the shuffled letters into the
* grid.
*/
for (j = 0; j < k; j++) {
int index = permutation[i*k+j];
if (!gen_lock[index])
grid[index] = sc->tmp[--m];
}
assert(m == 0);
}
/*
* Now we have a candidate grid. Attempt to progress
* the solution.
*/
m = solver_attempt(sc, grid, gen_lock);
if (m == 2 || /* success */
(m == 0 && retries-- <= 0)) /* failure */
break;
if (m == 1)
retries = k*k; /* reset this counter, and continue */
}
sfree(dsf);
} while (m == 0);
sfree(gen_lock);
sfree(permutation);
sfree(shuffled);
solver_scratch_free(sc);
return grid;
}
/* ----------------------------------------------------------------------
* End of solver/generator code.
*/
static char *new_game_desc(const game_params *params, random_state *rs,
char **aux, int interactive)
{
int w = params->w, h = params->h, wh = w*h, k = params->k;
unsigned char *grid;
char *desc;
int i;
grid = generate(w, h, k, rs);
desc = snewn(wh+1, char);
for (i = 0; i < wh; i++)
desc[i] = 'A' + grid[i];
desc[wh] = '\0';
sfree(grid);
return desc;
}
static const char *validate_desc(const game_params *params, const char *desc)
{
return NULL;
}
static game_state *new_game(midend *me, const game_params *params,
const char *desc)
{
game_state *state = snew(game_state);
state->FIXME = 0;
return state;
}
static game_state *dup_game(const game_state *state)
{
game_state *ret = snew(game_state);
ret->FIXME = state->FIXME;
return ret;
}
static void free_game(game_state *state)
{
sfree(state);
}
static char *solve_game(const game_state *state, const game_state *currstate,
const char *aux, const char **error)
{
return NULL;
}
static int game_can_format_as_text_now(const game_params *params)
{
return TRUE;
}
static char *game_text_format(const game_state *state)
{
return NULL;
}
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)
{
}
struct game_drawstate {
int tilesize;
int FIXME;
};
static char *interpret_move(const game_state *state, game_ui *ui,
const game_drawstate *ds,
int x, int y, int button)
{
return NULL;
}
static game_state *execute_move(const game_state *state, const char *move)
{
return NULL;
}
/* ----------------------------------------------------------------------
* Drawing routines.
*/
static void game_compute_size(const game_params *params, int tilesize,
int *x, int *y)
{
*x = *y = 10 * tilesize; /* FIXME */
}
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]);
*ncolours = NCOLOURS;
return ret;
}
static game_drawstate *game_new_drawstate(drawing *dr, const game_state *state)
{
struct game_drawstate *ds = snew(struct game_drawstate);
ds->tilesize = 0;
ds->FIXME = 0;
return ds;
}
static void game_free_drawstate(drawing *dr, game_drawstate *ds)
{
sfree(ds);
}
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)
{
/*
* The initial contents of the window are not guaranteed and
* can vary with front ends. To be on the safe side, all games
* should start by drawing a big background-colour rectangle
* covering the whole window.
*/
draw_rect(dr, 0, 0, 10*ds->tilesize, 10*ds->tilesize, COL_BACKGROUND);
}
static float game_anim_length(const game_state *oldstate,
const game_state *newstate, int dir, game_ui *ui)
{
return 0.0F;
}
static float game_flash_length(const game_state *oldstate,
const game_state *newstate, int dir, game_ui *ui)
{
return 0.0F;
}
static int game_status(const game_state *state)
{
return 0;
}
static int game_timing_state(const game_state *state, game_ui *ui)
{
return TRUE;
}
static void game_print_size(const game_params *params, float *x, float *y)
{
}
static void game_print(drawing *dr, const game_state *state, int tilesize)
{
}
#ifdef COMBINED
#define thegame separate
#endif
const struct game thegame = {
"Separate", NULL, NULL,
default_params,
game_fetch_preset, NULL,
decode_params,
encode_params,
free_params,
dup_params,
FALSE, game_configure, custom_params,
validate_params,
new_game_desc,
validate_desc,
new_game,
dup_game,
free_game,
FALSE, solve_game,
FALSE, game_can_format_as_text_now, game_text_format,
new_ui,
free_ui,
encode_ui,
decode_ui,
game_changed_state,
interpret_move,
execute_move,
20 /* FIXME */, game_compute_size, game_set_size,
game_colours,
game_new_drawstate,
game_free_drawstate,
game_redraw,
game_anim_length,
game_flash_length,
game_status,
FALSE, FALSE, game_print_size, game_print,
FALSE, /* wants_statusbar */
FALSE, game_timing_state,
0, /* flags */
};
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