ref: 5550660f13add636dacb189ccebea9beb47230ec
parent: f01f82105e5feb2586a2ca257947d76e9b982d04
author: Simon Tatham <anakin@pobox.com>
date: Fri Jun 17 14:55:36 EDT 2005
Solver for Flip. [originally from svn r5970]
--- a/flip.c
+++ b/flip.c
@@ -3,15 +3,6 @@
* where each click toggles an overlapping set of lights.
*/
-/*
- * TODO:
- *
- * - `Solve' could mark the squares you must click to solve
- * + infrastructure change: this would mean the Solve operation
- * must receive the current game_state as well as the initial
- * one, which I've been wondering about for a while
- */
-
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
@@ -28,6 +19,7 @@
COL_RIGHT,
COL_GRID,
COL_DIAG,
+ COL_HINT,
NCOLOURS
};
@@ -65,7 +57,7 @@
struct game_state {int w, h;
- int moves, completed;
+ int moves, completed, cheated, hints_active;
unsigned char *grid; /* array of w*h */
struct matrix *matrix;
};
@@ -633,6 +625,8 @@
state->w = w;
state->h = h;
state->completed = FALSE;
+ state->cheated = FALSE;
+ state->hints_active = FALSE;
state->moves = 0;
state->matrix = snew(struct matrix);
state->matrix->refcount = 1;
@@ -651,6 +645,8 @@
ret->w = state->w;
ret->h = state->h;
ret->completed = state->completed;
+ ret->cheated = state->cheated;
+ ret->hints_active = state->hints_active;
ret->moves = state->moves;
ret->matrix = state->matrix;
state->matrix->refcount++;
@@ -670,10 +666,194 @@
sfree(state);
}
+static void rowxor(unsigned char *row1, unsigned char *row2, int len)
+{+ int i;
+ for (i = 0; i < len; i++)
+ row1[i] ^= row2[i];
+}
+
static game_state *solve_game(game_state *state, game_state *currstate,
game_aux_info *aux, char **error)
{- return NULL;
+ int w = state->w, h = state->h, wh = w * h;
+ unsigned char *equations, *solution, *shortest;
+ int *und, nund;
+ int rowsdone, colsdone;
+ int i, j, k, len, bestlen;
+ game_state *ret;
+
+ /*
+ * Set up a list of simultaneous equations. Each one is of
+ * length (wh+1) and has wh coefficients followed by a value.
+ */
+ equations = snewn((wh + 1) * wh, unsigned char);
+ for (i = 0; i < wh; i++) {+ for (j = 0; j < wh; j++)
+ equations[i * (wh+1) + j] = currstate->matrix->matrix[j*wh+i];
+ equations[i * (wh+1) + wh] = currstate->grid[i] & 1;
+ }
+
+ /*
+ * Perform Gaussian elimination over GF(2).
+ */
+ rowsdone = colsdone = 0;
+ nund = 0;
+ und = snewn(wh, int);
+ do {+ /*
+ * Find the leftmost column which has a 1 in it somewhere
+ * outside the first `rowsdone' rows.
+ */
+ j = -1;
+ for (i = colsdone; i < wh; i++) {+ for (j = rowsdone; j < wh; j++)
+ if (equations[j * (wh+1) + i])
+ break;
+ if (j < wh)
+ break; /* found one */
+ /*
+ * This is a column which will not have an equation
+ * controlling it. Mark it as undetermined.
+ */
+ und[nund++] = i;
+ }
+
+ /*
+ * If there wasn't one, then we've finished: all remaining
+ * equations are of the form 0 = constant. Check to see if
+ * any of them wants 0 to be equal to 1; this is the
+ * condition which indicates an insoluble problem
+ * (therefore _hopefully_ one typed in by a user!).
+ */
+ if (i == wh) {+ for (j = rowsdone; j < wh; j++)
+ if (equations[j * (wh+1) + wh]) {+ *error = "No solution exists for this position";
+ sfree(equations);
+ return NULL;
+ }
+ break;
+ }
+
+ /*
+ * We've found a 1. It's in column i, and the topmost 1 in
+ * that column is in row j. Do a row-XOR to move it up to
+ * the topmost row if it isn't already there.
+ */
+ assert(j != -1);
+ if (j > rowsdone)
+ rowxor(equations + rowsdone*(wh+1), equations + j*(wh+1), wh+1);
+
+ /*
+ * Do row-XORs to eliminate that 1 from all rows below the
+ * topmost row.
+ */
+ for (j = rowsdone + 1; j < wh; j++)
+ if (equations[j*(wh+1) + i])
+ rowxor(equations + j*(wh+1),
+ equations + rowsdone*(wh+1), wh+1);
+
+ /*
+ * Mark this row and column as done.
+ */
+ rowsdone++;
+ colsdone = i+1;
+
+ /*
+ * If we've done all the rows, terminate.
+ */
+ } while (rowsdone < wh);
+
+ /*
+ * If we reach here, we have the ability to produce a solution.
+ * So we go through _all_ possible solutions (each
+ * corresponding to a set of arbitrary choices of those
+ * components not directly determined by an equation), and pick
+ * one requiring the smallest number of flips.
+ */
+ solution = snewn(wh, unsigned char);
+ shortest = snewn(wh, unsigned char);
+ memset(solution, 0, wh);
+ bestlen = wh + 1;
+ while (1) {+ /*
+ * Find a solution based on the current values of the
+ * undetermined variables.
+ */
+ for (j = rowsdone; j-- ;) {+ int v;
+
+ /*
+ * Find the leftmost set bit in this equation.
+ */
+ for (i = 0; i < wh; i++)
+ if (equations[j * (wh+1) + i])
+ break;
+ assert(i < wh); /* there must have been one! */
+
+ /*
+ * Compute this variable using the rest.
+ */
+ v = equations[j * (wh+1) + wh];
+ for (k = i+1; k < wh; k++)
+ if (equations[j * (wh+1) + k])
+ v ^= solution[k];
+
+ solution[i] = v;
+ }
+
+ /*
+ * Compare this solution to the current best one, and
+ * replace the best one if this one is shorter.
+ */
+ len = 0;
+ for (i = 0; i < wh; i++)
+ if (solution[i])
+ len++;
+ if (len < bestlen) {+ bestlen = len;
+ memcpy(shortest, solution, wh);
+ }
+
+ /*
+ * Now increment the binary number given by the
+ * undetermined variables: turn all 1s into 0s until we see
+ * a 0, at which point we turn it into a 1.
+ */
+ for (i = 0; i < nund; i++) {+ solution[und[i]] = !solution[und[i]];
+ if (solution[und[i]])
+ break;
+ }
+
+ /*
+ * If we didn't find a 0 at any point, we have wrapped
+ * round and are back at the start, i.e. we have enumerated
+ * all solutions.
+ */
+ if (i == nund)
+ break;
+ }
+
+ /*
+ * We have a solution. Produce a game state with the solution
+ * marked in annotations.
+ */
+ ret = dup_game(currstate);
+ ret->hints_active = TRUE;
+ ret->cheated = TRUE;
+ for (i = 0; i < wh; i++) {+ ret->grid[i] &= ~2;
+ if (shortest[i])
+ ret->grid[i] |= 2;
+ }
+
+ sfree(shortest);
+ sfree(solution);
+ sfree(equations);
+
+ return ret;
}
static char *game_text_format(game_state *state)
@@ -725,8 +905,11 @@
if (ret->grid[j] & 1)
done = FALSE;
}
- if (done)
+ ret->grid[i] ^= 2; /* toggle hint */
+ if (done) {ret->completed = TRUE;
+ ret->hints_active = FALSE;
+ }
return ret;
}
@@ -782,6 +965,10 @@
ret[COL_DIAG * 3 + 1] = ret[COL_GRID * 3 + 1];
ret[COL_DIAG * 3 + 2] = ret[COL_GRID * 3 + 2];
+ ret[COL_HINT * 3 + 0] = 1.0F;
+ ret[COL_HINT * 3 + 1] = 0.0F;
+ ret[COL_HINT * 3 + 2] = 0.0F;
+
*ncolours = NCOLOURS;
return ret;
}
@@ -865,6 +1052,14 @@
}
}
+ /*
+ * Draw a hint blob if required.
+ */
+ if (tile & 2) {+ draw_rect(fe, bx + TILE_SIZE/20, by + TILE_SIZE / 20,
+ TILE_SIZE / 6, TILE_SIZE / 6, COL_HINT);
+ }
+
unclip(fe);
draw_update(fe, bx+1, by+1, TILE_SIZE-1, TILE_SIZE-1);
@@ -922,6 +1117,9 @@
v &= ~1;
}
+ if (!state->hints_active)
+ v &= ~2;
+
if (oldstate && state->grid[i] != oldstate->grid[i])
vv = 255; /* means `animated' */
else
@@ -936,7 +1134,10 @@
{char buf[256];
- sprintf(buf, "%sMoves: %d", state->completed ? "COMPLETED! " : "",
+ sprintf(buf, "%sMoves: %d",
+ (state->completed ?
+ (state->cheated ? "Auto-solved. " : "COMPLETED! ") :
+ (state->cheated ? "Auto-solver used. " : "")),
state->moves);
status_bar(fe, buf);
@@ -988,7 +1189,7 @@
new_game,
dup_game,
free_game,
- FALSE, solve_game,
+ TRUE, solve_game,
FALSE, game_text_format,
new_ui,
free_ui,
--- a/puzzles.but
+++ b/puzzles.but
@@ -1011,8 +1011,14 @@
\IM{Flip controls} keys, for Flip \IM{Flip controls} shortcuts (keyboard), for Flip-Left-click in a square to flip it and its associated squares. That's
-all!
+Left-click in a square to flip it and its associated squares.
+
+If you use the \q{Solve} function on this game, it will highlight+some of the squares with red blobs. If you click once in every
+square with a red blob, the game should be solved. (If you click in
+a square \e{without} a red blob, a red blob will appear in it to+indicate that you will need to reverse that operation to reach the
+solution.)
\H{flip-parameters} \I{parameters, for flip}Flip parameters