ref: c746f6d0e3996092de94ecbf72d039632594f7e0
parent: 4ff90bd47f4bd3b11c4fb8b1ea25afb7b95f76da
author: Simon Tatham <anakin@pobox.com>
date: Sat Aug 25 10:10:49 EDT 2007
Commit my work so far on a generator for Nikoli's `Block Puzzle'. It works, but it's slow, and the puzzles are currently at a relatively low level of difficulty. Also this is a generator only: no UI yet (because I'm waiting to see if I can make the generator practical before bothering to write the rest). [originally from svn r7700]
--- /dev/null
+++ b/unfinished/separate.R
@@ -1,0 +1,21 @@
+# -*- makefile -*-
+
+SEPARATE = separate divvy dsf
+
+separate : [X] GTK COMMON SEPARATE separate-icon|no-icon
+
+separate : [G] WINDOWS COMMON SEPARATE separate.res|noicon.res
+
+ALL += separate
+
+!begin gtk
+GAMES += separate
+!end
+
+!begin >list.c
+ A(separate) \
+!end
+
+!begin >wingames.lst
+separate.exe:Separate
+!end
--- /dev/null
+++ b/unfinished/separate.c
@@ -1,0 +1,845 @@
+/*
+ * 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(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(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(game_params *params)
+{+ return NULL;
+}
+
+static game_params *custom_params(config_item *cfg)
+{+ return NULL;
+}
+
+static char *validate_params(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(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 char *validate_desc(game_params *params, char *desc)
+{+ return NULL;
+}
+
+static game_state *new_game(midend *me, game_params *params, char *desc)
+{+ game_state *state = snew(game_state);
+
+ state->FIXME = 0;
+
+ return state;
+}
+
+static game_state *dup_game(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(game_state *state, game_state *currstate,
+ char *aux, char **error)
+{+ return NULL;
+}
+
+static char *game_text_format(game_state *state)
+{+ return NULL;
+}
+
+static game_ui *new_ui(game_state *state)
+{+ return NULL;
+}
+
+static void free_ui(game_ui *ui)
+{+}
+
+static char *encode_ui(game_ui *ui)
+{+ return NULL;
+}
+
+static void decode_ui(game_ui *ui, char *encoding)
+{+}
+
+static void game_changed_state(game_ui *ui, game_state *oldstate,
+ game_state *newstate)
+{+}
+
+struct game_drawstate {+ int tilesize;
+ int FIXME;
+};
+
+static char *interpret_move(game_state *state, game_ui *ui, game_drawstate *ds,
+ int x, int y, int button)
+{+ return NULL;
+}
+
+static game_state *execute_move(game_state *state, char *move)
+{+ return NULL;
+}
+
+/* ----------------------------------------------------------------------
+ * Drawing routines.
+ */
+
+static void game_compute_size(game_params *params, int tilesize,
+ int *x, int *y)
+{+ *x = *y = 10 * tilesize; /* FIXME */
+}
+
+static void game_set_size(drawing *dr, game_drawstate *ds,
+ game_params *params, int tilesize)
+{+ ds->tilesize = tilesize;
+}
+
+static float *game_colours(frontend *fe, int *ncolours)
+{+ float *ret = snewn(3 * NCOLOURS, float);
+
+ frontend_default_colour(fe, &ret[COL_BACKGROUND * 3]);
+
+ *ncolours = NCOLOURS;
+ return ret;
+}
+
+static game_drawstate *game_new_drawstate(drawing *dr, 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, game_state *oldstate,
+ game_state *state, int dir, 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(game_state *oldstate, game_state *newstate,
+ int dir, game_ui *ui)
+{+ return 0.0F;
+}
+
+static float game_flash_length(game_state *oldstate, game_state *newstate,
+ int dir, game_ui *ui)
+{+ return 0.0F;
+}
+
+static int game_timing_state(game_state *state, game_ui *ui)
+{+ return TRUE;
+}
+
+static void game_print_size(game_params *params, float *x, float *y)
+{+}
+
+static void game_print(drawing *dr, game_state *state, int tilesize)
+{+}
+
+#ifdef COMBINED
+#define thegame nullgame
+#endif
+
+const struct game thegame = {+ "Null Game", NULL, NULL,
+ default_params,
+ game_fetch_preset,
+ 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_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,
+ FALSE, FALSE, game_print_size, game_print,
+ FALSE, /* wants_statusbar */
+ FALSE, game_timing_state,
+ 0, /* flags */
+};