ref: 48f6bfa474f091c661a281e36b390a52f998f94d
dir: /unfinished/pearl.c/
/*
* pearl.c: Nikoli's `Masyu' puzzle. Currently this is a blank
* puzzle file with nothing but a test solver-generator.
*/
/*
* TODO:
*
* - The generation method appears to be fundamentally flawed. I
* think generating a random loop and then choosing a clue set
* is simply not a viable approach, because on a test run of
* 10,000 attempts, it generated _six_ viable puzzles. All the
* rest of the randomly generated loops failed to be soluble
* even given a maximal clue set. Also, the vast majority of the
* clues were white circles (straight clues); black circles
* (corners) seem very uncommon.
* + So what can we do? One possible approach would be to
* adjust the random loop generation so that it created loops
* which were in some heuristic sense more likely to be
* viable Masyu puzzles. Certainly a good start on that would
* be to arrange that black clues actually _came up_ slightly
* more often, but I have no idea whether that would be
* sufficient.
* + A second option would be to throw the entire mechanism out
* and instead write a different generator from scratch which
* evolves the solution along with the puzzle: place a few
* clues, nail down a bit of the loop, place another clue,
* nail down some more, etc. It's unclear whether this can
* sensibly be done, though.
*
* - Puzzle playing UI and everything else apart from the
* generator...
*/
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <assert.h>
#include <ctype.h>
#include <math.h>
#include "puzzles.h"
#define NOCLUE 0
#define CORNER 1
#define STRAIGHT 2
#define R 1
#define U 2
#define L 4
#define D 8
#define DX(d) ( ((d)==R) - ((d)==L) )
#define DY(d) ( ((d)==D) - ((d)==U) )
#define F(d) (((d << 2) | (d >> 2)) & 0xF)
#define C(d) (((d << 3) | (d >> 1)) & 0xF)
#define A(d) (((d << 1) | (d >> 3)) & 0xF)
#define LR (L | R)
#define RL (R | L)
#define UD (U | D)
#define DU (D | U)
#define LU (L | U)
#define UL (U | L)
#define LD (L | D)
#define DL (D | L)
#define RU (R | U)
#define UR (U | R)
#define RD (R | D)
#define DR (D | R)
#define BLANK 0
#define UNKNOWN 15
#define bLR (1 << LR)
#define bRL (1 << RL)
#define bUD (1 << UD)
#define bDU (1 << DU)
#define bLU (1 << LU)
#define bUL (1 << UL)
#define bLD (1 << LD)
#define bDL (1 << DL)
#define bRU (1 << RU)
#define bUR (1 << UR)
#define bRD (1 << RD)
#define bDR (1 << DR)
#define bBLANK (1 << BLANK)
enum {
COL_BACKGROUND,
NCOLOURS
};
struct game_params {
int FIXME;
};
struct game_state {
int FIXME;
};
static game_params *default_params(void)
{
game_params *ret = snew(game_params);
ret->FIXME = 0;
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)
{
}
static char *encode_params(game_params *params, int full)
{
return dupstr("FIXME");
}
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.
*/
int pearl_solve(int w, int h, char *clues, char *result)
{
int W = 2*w+1, H = 2*h+1;
short *workspace;
int *dsf, *dsfsize;
int x, y, b, d;
int ret = -1;
/*
* workspace[(2*y+1)*W+(2*x+1)] indicates the possible nature
* of the square (x,y), as a logical OR of bitfields.
*
* workspace[(2*y)*W+(2*x+1)], for x odd and y even, indicates
* whether the horizontal edge between (x,y) and (x+1,y) is
* connected (1), disconnected (2) or unknown (3).
*
* workspace[(2*y+1)*W+(2*x)], indicates the same about the
* vertical edge between (x,y) and (x,y+1).
*
* Initially, every square is considered capable of being in
* any of the seven possible states (two straights, four
* corners and empty), except those corresponding to clue
* squares which are more restricted.
*
* Initially, all edges are unknown, except the ones around the
* grid border which are known to be disconnected.
*/
workspace = snewn(W*H, short);
for (x = 0; x < W*H; x++)
workspace[x] = 0;
/* Square states */
for (y = 0; y < h; y++)
for (x = 0; x < w; x++)
switch (clues[y*w+x]) {
case CORNER:
workspace[(2*y+1)*W+(2*x+1)] = bLU|bLD|bRU|bRD;
break;
case STRAIGHT:
workspace[(2*y+1)*W+(2*x+1)] = bLR|bUD;
break;
default:
workspace[(2*y+1)*W+(2*x+1)] = bLR|bUD|bLU|bLD|bRU|bRD|bBLANK;
break;
}
/* Horizontal edges */
for (y = 0; y <= h; y++)
for (x = 0; x < w; x++)
workspace[(2*y)*W+(2*x+1)] = (y==0 || y==h ? 2 : 3);
/* Vertical edges */
for (y = 0; y < h; y++)
for (x = 0; x <= w; x++)
workspace[(2*y+1)*W+(2*x)] = (x==0 || x==w ? 2 : 3);
/*
* We maintain a dsf of connected squares, together with a
* count of the size of each equivalence class.
*/
dsf = snewn(w*h, int);
dsfsize = snewn(w*h, int);
/*
* Now repeatedly try to find something we can do.
*/
while (1) {
int done_something = FALSE;
#ifdef SOLVER_DIAGNOSTICS
for (y = 0; y < H; y++) {
for (x = 0; x < W; x++)
printf("%*x", (x&1) ? 5 : 2, workspace[y*W+x]);
printf("\n");
}
#endif
/*
* Go through the square state words, and discard any
* square state which is inconsistent with known facts
* about the edges around the square.
*/
for (y = 0; y < h; y++)
for (x = 0; x < w; x++) {
for (b = 0; b < 0xD; b++)
if (workspace[(2*y+1)*W+(2*x+1)] & (1<<b)) {
/*
* If any edge of this square is known to
* be connected when state b would require
* it disconnected, or vice versa, discard
* the state.
*/
for (d = 1; d <= 8; d += d) {
int ex = 2*x+1 + DX(d), ey = 2*y+1 + DY(d);
if (workspace[ey*W+ex] ==
((b & d) ? 2 : 1)) {
workspace[(2*y+1)*W+(2*x+1)] &= ~(1<<b);
#ifdef SOLVER_DIAGNOSTICS
printf("edge (%d,%d)-(%d,%d) rules out state"
" %d for square (%d,%d)\n",
ex/2, ey/2, (ex+1)/2, (ey+1)/2,
b, x, y);
#endif
done_something = TRUE;
break;
}
}
}
/*
* Consistency check: each square must have at
* least one state left!
*/
if (!workspace[(2*y+1)*W+(2*x+1)]) {
#ifdef SOLVER_DIAGNOSTICS
printf("edge check at (%d,%d): inconsistency\n", x, y);
#endif
ret = 0;
goto cleanup;
}
}
/*
* Now go through the states array again, and nail down any
* unknown edge if one of its neighbouring squares makes it
* known.
*/
for (y = 0; y < h; y++)
for (x = 0; x < w; x++) {
int edgeor = 0, edgeand = 15;
for (b = 0; b < 0xD; b++)
if (workspace[(2*y+1)*W+(2*x+1)] & (1<<b)) {
edgeor |= b;
edgeand &= b;
}
/*
* Now any bit clear in edgeor marks a disconnected
* edge, and any bit set in edgeand marks a
* connected edge.
*/
/* First check consistency: neither bit is both! */
if (edgeand & ~edgeor) {
#ifdef SOLVER_DIAGNOSTICS
printf("square check at (%d,%d): inconsistency\n", x, y);
#endif
ret = 0;
goto cleanup;
}
for (d = 1; d <= 8; d += d) {
int ex = 2*x+1 + DX(d), ey = 2*y+1 + DY(d);
if (!(edgeor & d) && workspace[ey*W+ex] == 3) {
workspace[ey*W+ex] = 2;
done_something = TRUE;
#ifdef SOLVER_DIAGNOSTICS
printf("possible states of square (%d,%d) force edge"
" (%d,%d)-(%d,%d) to be disconnected\n",
x, y, ex/2, ey/2, (ex+1)/2, (ey+1)/2);
#endif
} else if ((edgeand & d) && workspace[ey*W+ex] == 3) {
workspace[ey*W+ex] = 1;
done_something = TRUE;
#ifdef SOLVER_DIAGNOSTICS
printf("possible states of square (%d,%d) force edge"
" (%d,%d)-(%d,%d) to be connected\n",
x, y, ex/2, ey/2, (ex+1)/2, (ey+1)/2);
#endif
}
}
}
if (done_something)
continue;
/*
* Now for longer-range clue-based deductions (using the
* rules that a corner clue must connect to two straight
* squares, and a straight clue must connect to at least
* one corner square).
*/
for (y = 0; y < h; y++)
for (x = 0; x < w; x++)
switch (clues[y*w+x]) {
case CORNER:
for (d = 1; d <= 8; d += d) {
int ex = 2*x+1 + DX(d), ey = 2*y+1 + DY(d);
int fx = ex + DX(d), fy = ey + DY(d);
int type = d | F(d);
if (workspace[ey*W+ex] == 1) {
/*
* If a corner clue is connected on any
* edge, then we can immediately nail
* down the square beyond that edge as
* being a straight in the appropriate
* direction.
*/
if (workspace[fy*W+fx] != (1<<type)) {
workspace[fy*W+fx] = (1<<type);
done_something = TRUE;
#ifdef SOLVER_DIAGNOSTICS
printf("corner clue at (%d,%d) forces square "
"(%d,%d) into state %d\n", x, y,
fx/2, fy/2, type);
#endif
}
} else if (workspace[ey*W+ex] == 3) {
/*
* Conversely, if a corner clue is
* separated by an unknown edge from a
* square which _cannot_ be a straight
* in the appropriate direction, we can
* mark that edge as disconnected.
*/
if (!(workspace[fy*W+fx] & (1<<type))) {
workspace[ey*W+ex] = 2;
done_something = TRUE;
#ifdef SOLVER_DIAGNOSTICS
printf("corner clue at (%d,%d), plus square "
"(%d,%d) not being state %d, "
"disconnects edge (%d,%d)-(%d,%d)\n",
x, y, fx/2, fy/2, type,
ex/2, ey/2, (ex+1)/2, (ey+1)/2);
#endif
}
}
}
break;
case STRAIGHT:
/*
* If a straight clue is between two squares
* neither of which is capable of being a
* corner connected to it, then the straight
* clue cannot point in that direction.
*/
for (d = 1; d <= 2; d += d) {
int fx = 2*x+1 + 2*DX(d), fy = 2*y+1 + 2*DY(d);
int gx = 2*x+1 - 2*DX(d), gy = 2*y+1 - 2*DY(d);
int type = d | F(d);
if (!(workspace[(2*y+1)*W+(2*x+1)] & (1<<type)))
continue;
if (!(workspace[fy*W+fx] & ((1<<(F(d)|A(d))) |
(1<<(F(d)|C(d))))) &&
!(workspace[gy*W+gx] & ((1<<( d |A(d))) |
(1<<( d |C(d)))))) {
workspace[(2*y+1)*W+(2*x+1)] &= ~(1<<type);
done_something = TRUE;
#ifdef SOLVER_DIAGNOSTICS
printf("straight clue at (%d,%d) cannot corner at "
"(%d,%d) or (%d,%d) so is not state %d\n",
x, y, fx/2, fy/2, gx/2, gy/2, type);
#endif
}
}
/*
* If a straight clue with known direction is
* connected on one side to a known straight,
* then on the other side it must be a corner.
*/
for (d = 1; d <= 8; d += d) {
int fx = 2*x+1 + 2*DX(d), fy = 2*y+1 + 2*DY(d);
int gx = 2*x+1 - 2*DX(d), gy = 2*y+1 - 2*DY(d);
int type = d | F(d);
if (workspace[(2*y+1)*W+(2*x+1)] != (1<<type))
continue;
if (!(workspace[fy*W+fx] &~ (bLR|bUD)) &&
(workspace[gy*W+gx] &~ (bLU|bLD|bRU|bRD))) {
workspace[gy*W+gx] &= (bLU|bLD|bRU|bRD);
done_something = TRUE;
#ifdef SOLVER_DIAGNOSTICS
printf("straight clue at (%d,%d) connecting to "
"straight at (%d,%d) makes (%d,%d) a "
"corner\n", x, y, fx/2, fy/2, gx/2, gy/2);
#endif
}
}
break;
}
if (done_something)
continue;
/*
* Now detect shortcut loops.
*/
{
int nonblanks, loopclass;
dsf_init(dsf, w*h);
for (x = 0; x < w*h; x++)
dsfsize[x] = 1;
/*
* First go through the edge entries and update the dsf
* of which squares are connected to which others. We
* also track the number of squares in each equivalence
* class, and count the overall number of
* known-non-blank squares.
*
* In the process of doing this, we must notice if a
* loop has already been formed. If it has, we blank
* out any square which isn't part of that loop
* (failing a consistency check if any such square does
* not have BLANK as one of its remaining options) and
* exit the deduction loop with success.
*/
nonblanks = 0;
loopclass = -1;
for (y = 1; y < H-1; y++)
for (x = 1; x < W-1; x++)
if ((y ^ x) & 1) {
/*
* (x,y) are the workspace coordinates of
* an edge field. Compute the normal-space
* coordinates of the squares it connects.
*/
int ax = (x-1)/2, ay = (y-1)/2, ac = ay*w+ax;
int bx = x/2, by = y/2, bc = by*w+bx;
/*
* If the edge is connected, do the dsf
* thing.
*/
if (workspace[y*W+x] == 1) {
int ae, be;
ae = dsf_canonify(dsf, ac);
be = dsf_canonify(dsf, bc);
if (ae == be) {
/*
* We have a loop!
*/
if (loopclass != -1) {
/*
* In fact, we have two
* separate loops, which is
* doom.
*/
#ifdef SOLVER_DIAGNOSTICS
printf("two loops found in grid!\n");
#endif
ret = 0;
goto cleanup;
}
loopclass = ae;
} else {
/*
* Merge the two equivalence
* classes.
*/
int size = dsfsize[ae] + dsfsize[be];
dsf_merge(dsf, ac, bc);
ae = dsf_canonify(dsf, ac);
dsfsize[ae] = size;
}
}
} else if ((y & x) & 1) {
/*
* (x,y) are the workspace coordinates of a
* square field. If the square is
* definitely not blank, count it.
*/
if (!(workspace[y*W+x] & bBLANK))
nonblanks++;
}
/*
* If we discovered an existing loop above, we must now
* blank every square not part of it, and exit the main
* deduction loop.
*/
if (loopclass != -1) {
#ifdef SOLVER_DIAGNOSTICS
printf("loop found in grid!\n");
#endif
for (y = 0; y < h; y++)
for (x = 0; x < w; x++)
if (dsf_canonify(dsf, y*w+x) != loopclass) {
if (workspace[(y*2+1)*W+(x*2+1)] & bBLANK) {
workspace[(y*2+1)*W+(x*2+1)] = bBLANK;
} else {
/*
* This square is not part of the
* loop, but is known non-blank. We
* have goofed.
*/
#ifdef SOLVER_DIAGNOSTICS
printf("non-blank square (%d,%d) found outside"
" loop!\n", x, y);
#endif
ret = 0;
goto cleanup;
}
}
/*
* And we're done.
*/
ret = 1;
break;
}
/*
* Now go through the workspace again and mark any edge
* which would cause a shortcut loop (i.e. would
* connect together two squares in the same equivalence
* class, and that equivalence class does not contain
* _all_ the known-non-blank squares currently in the
* grid) as disconnected. Also, mark any _square state_
* which would cause a shortcut loop as disconnected.
*/
for (y = 1; y < H-1; y++)
for (x = 1; x < W-1; x++)
if ((y ^ x) & 1) {
/*
* (x,y) are the workspace coordinates of
* an edge field. Compute the normal-space
* coordinates of the squares it connects.
*/
int ax = (x-1)/2, ay = (y-1)/2, ac = ay*w+ax;
int bx = x/2, by = y/2, bc = by*w+bx;
/*
* If the edge is currently unknown, and
* sits between two squares in the same
* equivalence class, and the size of that
* class is less than nonblanks, then
* connecting this edge would be a shortcut
* loop and so we must not do so.
*/
if (workspace[y*W+x] == 3) {
int ae, be;
ae = dsf_canonify(dsf, ac);
be = dsf_canonify(dsf, bc);
if (ae == be) {
/*
* We have a loop. Is it a shortcut?
*/
if (dsfsize[ae] < nonblanks) {
/*
* Yes! Mark this edge disconnected.
*/
workspace[y*W+x] = 2;
done_something = TRUE;
#ifdef SOLVER_DIAGNOSTICS
printf("edge (%d,%d)-(%d,%d) would create"
" a shortcut loop, hence must be"
" disconnected\n", x/2, y/2,
(x+1)/2, (y+1)/2);
#endif
}
}
}
} else if ((y & x) & 1) {
/*
* (x,y) are the workspace coordinates of a
* square field. Go through its possible
* (non-blank) states and see if any gives
* rise to a shortcut loop.
*
* This is slightly fiddly, because we have
* to check whether this square is already
* part of the same equivalence class as
* the things it's joining.
*/
int ae = dsf_canonify(dsf, (y/2)*w+(x/2));
for (b = 2; b < 0xD; b++)
if (workspace[y*W+x] & (1<<b)) {
/*
* Find the equivalence classes of
* the two squares this one would
* connect if it were in this
* state.
*/
int e = -1;
for (d = 1; d <= 8; d += d) if (b & d) {
int xx = x/2 + DX(d), yy = y/2 + DY(d);
int ee = dsf_canonify(dsf, yy*w+xx);
if (e == -1)
ee = e;
else if (e != ee)
e = -2;
}
if (e >= 0) {
/*
* This square state would form
* a loop on equivalence class
* e. Measure the size of that
* loop, and see if it's a
* shortcut.
*/
int loopsize = dsfsize[e];
if (e != ae)
loopsize++;/* add the square itself */
if (loopsize < nonblanks) {
/*
* It is! Mark this square
* state invalid.
*/
workspace[y*W+x] &= ~(1<<b);
done_something = TRUE;
#ifdef SOLVER_DIAGNOSTICS
printf("square (%d,%d) would create a "
"shortcut loop in state %d, "
"hence cannot be\n",
x/2, y/2, b);
#endif
}
}
}
}
}
if (done_something)
continue;
/*
* If we reach here, there is nothing left we can do.
* Return 2 for ambiguous puzzle.
*/
ret = 2;
goto cleanup;
}
/*
* If we reach _here_, it's by `break' out of the main loop,
* which means we've successfully achieved a solution. This
* means that we expect every square to be nailed down to
* exactly one possibility. Transcribe those possibilities into
* the result array.
*/
for (y = 0; y < h; y++)
for (x = 0; x < w; x++) {
for (b = 0; b < 0xD; b++)
if (workspace[(2*y+1)*W+(2*x+1)] == (1<<b)) {
result[y*w+x] = b;
break;
}
assert(b < 0xD); /* we should have had a break by now */
}
cleanup:
sfree(dsfsize);
sfree(dsf);
sfree(workspace);
assert(ret >= 0);
return ret;
}
/* ----------------------------------------------------------------------
* Loop generator.
*/
void pearl_loopgen(int w, int h, char *grid, random_state *rs)
{
int *options, *mindist, *maxdist, *list;
int x, y, d, total, n, area, limit;
/*
* We're eventually going to have to return a w-by-h array
* containing line segment data. However, it's more convenient
* while actually generating the loop to consider the problem
* as a (w-1) by (h-1) array in which some squares are `inside'
* and some `outside'.
*
* I'm going to use the top left corner of my return array in
* the latter manner until the end of the function.
*/
/*
* To begin with, all squares are outside (0), except for one
* randomly selected one which is inside (1).
*/
memset(grid, 0, w*h);
x = random_upto(rs, w-1);
y = random_upto(rs, h-1);
grid[y*w+x] = 1;
/*
* I'm also going to need an array to store the possible
* options for the next extension of the grid.
*/
options = snewn(w*h, int);
for (x = 0; x < w*h; x++)
options[x] = 0;
/*
* And some arrays and a list for breadth-first searching.
*/
mindist = snewn(w*h, int);
maxdist = snewn(w*h, int);
list = snewn(w*h, int);
/*
* Now we repeatedly scan the grid for feasible squares into
* which we can extend our loop, pick one, and do it.
*/
area = 1;
while (1) {
#ifdef LOOPGEN_DIAGNOSTICS
for (y = 0; y < h; y++) {
for (x = 0; x < w; x++)
printf("%d", grid[y*w+x]);
printf("\n");
}
printf("\n");
#endif
/*
* Our primary aim in growing this loop is to make it
* reasonably _dense_ in the target rectangle. That is, we
* want the maximum over all squares of the minimum
* distance from that square to the loop to be small.
*
* Therefore, we start with a breadth-first search of the
* grid to find those minimum distances.
*/
{
int head = 0, tail = 0;
int i;
for (i = 0; i < w*h; i++) {
mindist[i] = -1;
if (grid[i]) {
mindist[i] = 0;
list[tail++] = i;
}
}
while (head < tail) {
i = list[head++];
y = i / w;
x = i % w;
for (d = 1; d <= 8; d += d) {
int xx = x + DX(d), yy = y + DY(d);
if (xx >= 0 && xx < w && yy >= 0 && yy < h &&
mindist[yy*w+xx] < 0) {
mindist[yy*w+xx] = mindist[i] + 1;
list[tail++] = yy*w+xx;
}
}
}
/*
* Having done the BFS, we now backtrack along its path
* to determine the most distant square that each
* square is on the shortest path to. This tells us
* which of the loop extension candidates (all of which
* are squares marked 1) is most desirable to extend
* into in terms of minimising the maximum distance
* from any empty square to the nearest loop square.
*/
for (head = tail; head-- > 0 ;) {
int max;
i = list[head];
y = i / w;
x = i % w;
max = mindist[i];
for (d = 1; d <= 8; d += d) {
int xx = x + DX(d), yy = y + DY(d);
if (xx >= 0 && xx < w && yy >= 0 && yy < h &&
mindist[yy*w+xx] > mindist[i] &&
maxdist[yy*w+xx] > max) {
max = maxdist[yy*w+xx];
}
}
maxdist[i] = max;
}
}
/*
* A square is a viable candidate for extension of our loop
* if and only if the following conditions are all met:
* - It is currently labelled 0.
* - At least one of its four orthogonal neighbours is
* labelled 1.
* - If you consider its eight orthogonal and diagonal
* neighbours to form a ring, that ring contains at most
* one contiguous run of 1s. (It must also contain at
* _least_ one, of course, but that's already guaranteed
* by the previous condition so there's no need to test
* it separately.)
*/
total = 0;
for (y = 0; y < h-1; y++)
for (x = 0; x < w-1; x++) {
int ring[8];
int rx, neighbours, runs, dist;
dist = maxdist[y*w+x];
options[y*w+x] = 0;
if (grid[y*w+x])
continue; /* it isn't labelled 0 */
neighbours = 0;
for (rx = 0, d = 1; d <= 8; rx += 2, d += d) {
int x2 = x + DX(d), y2 = y + DY(d);
int x3 = x2 + DX(A(d)), y3 = y2 + DY(A(d));
int g2 = (x2 >= 0 && x2 < w && y2 >= 0 && y2 < h ?
grid[y2*w+x2] : 0);
int g3 = (x3 >= 0 && x3 < w && y3 >= 0 && y3 < h ?
grid[y3*w+x3] : 0);
ring[rx] = g2;
ring[rx+1] = g3;
if (g2)
neighbours++;
}
if (!neighbours)
continue; /* it doesn't have a 1 neighbour */
runs = 0;
for (rx = 0; rx < 8; rx++)
if (ring[rx] && !ring[(rx+1) & 7])
runs++;
if (runs > 1)
continue; /* too many runs of 1s */
/*
* Now we know this square is a viable extension
* candidate. Mark it.
*
* FIXME: probabilistic prioritisation based on
* perimeter perturbation? (Wow, must keep that
* phrase.)
*/
options[y*w+x] = dist * (4-neighbours) * (4-neighbours);
total += options[y*w+x];
}
if (!total)
break; /* nowhere to go! */
/*
* Now pick a random one of the viable extension squares,
* and extend into it.
*/
n = random_upto(rs, total);
for (y = 0; y < h-1; y++)
for (x = 0; x < w-1; x++) {
assert(n >= 0);
if (options[y*w+x] > n)
goto found; /* two-level break */
n -= options[y*w+x];
}
assert(!"We shouldn't ever get here");
found:
grid[y*w+x] = 1;
area++;
/*
* We terminate the loop when around 7/12 of the grid area
* is full, but we also require that the loop has reached
* all four edges.
*/
limit = random_upto(rs, (w-1)*(h-1)) + 13*(w-1)*(h-1);
if (24 * area > limit) {
int l = FALSE, r = FALSE, u = FALSE, d = FALSE;
for (x = 0; x < w; x++) {
if (grid[0*w+x])
u = TRUE;
if (grid[(h-2)*w+x])
d = TRUE;
}
for (y = 0; y < h; y++) {
if (grid[y*w+0])
l = TRUE;
if (grid[y*w+(w-2)])
r = TRUE;
}
if (l && r && u && d)
break;
}
}
sfree(list);
sfree(maxdist);
sfree(mindist);
sfree(options);
#ifdef LOOPGEN_DIAGNOSTICS
printf("final loop:\n");
for (y = 0; y < h; y++) {
for (x = 0; x < w; x++)
printf("%d", grid[y*w+x]);
printf("\n");
}
printf("\n");
#endif
/*
* Now convert this array of 0s and 1s into an array of path
* components.
*/
for (y = h; y-- > 0 ;) {
for (x = w; x-- > 0 ;) {
/*
* Examine the four grid squares of which (x,y) are in
* the bottom right, to determine the output for this
* square.
*/
int ul = (x > 0 && y > 0 ? grid[(y-1)*w+(x-1)] : 0);
int ur = (y > 0 ? grid[(y-1)*w+x] : 0);
int dl = (x > 0 ? grid[y*w+(x-1)] : 0);
int dr = grid[y*w+x];
int type = 0;
if (ul != ur) type |= U;
if (dl != dr) type |= D;
if (ul != dl) type |= L;
if (ur != dr) type |= R;
assert((bLR|bUD|bLU|bLD|bRU|bRD|bBLANK) & (1 << type));
grid[y*w+x] = type;
}
}
#if defined LOOPGEN_DIAGNOSTICS && !defined GENERATION_DIAGNOSTICS
printf("as returned:\n");
for (y = 0; y < h; y++) {
for (x = 0; x < w; x++) {
int type = grid[y*w+x];
char s[5], *p = s;
if (type & L) *p++ = 'L';
if (type & R) *p++ = 'R';
if (type & U) *p++ = 'U';
if (type & D) *p++ = 'D';
*p = '\0';
printf("%3s", s);
}
printf("\n");
}
printf("\n");
#endif
}
static char *new_game_desc(game_params *params, random_state *rs,
char **aux, int interactive)
{
char *grid, *clues;
int *clueorder;
int w = 10, h = 10;
int x, y, d, ret, i;
#if 0
clues = snewn(7*7, char);
memcpy(clues,
"\0\1\0\0\2\0\0"
"\0\0\0\2\0\0\0"
"\0\0\0\2\0\0\1"
"\2\0\0\2\0\0\0"
"\2\0\0\0\0\0\1"
"\0\0\1\0\0\2\0"
"\0\0\2\0\0\0\0", 7*7);
grid = snewn(7*7, char);
printf("%d\n", pearl_solve(7, 7, clues, grid));
#elif 0
clues = snewn(10*10, char);
memcpy(clues,
"\0\0\2\0\2\0\0\0\0\0"
"\0\0\0\0\2\0\0\0\1\0"
"\0\0\1\0\1\0\2\0\0\0"
"\0\0\0\2\0\0\2\0\0\0"
"\1\0\0\0\0\2\0\0\0\2"
"\0\0\2\0\0\0\0\2\0\0"
"\0\0\1\0\0\0\2\0\0\0"
"\2\0\0\0\1\0\0\0\0\2"
"\0\0\0\0\0\0\2\2\0\0"
"\0\0\1\0\0\0\0\0\0\1", 10*10);
grid = snewn(10*10, char);
printf("%d\n", pearl_solve(10, 10, clues, grid));
#elif 0
clues = snewn(10*10, char);
memcpy(clues,
"\0\0\0\0\0\0\1\0\0\0"
"\0\1\0\1\2\0\0\0\0\2"
"\0\0\0\0\0\0\0\0\0\1"
"\2\0\0\1\2\2\1\0\0\0"
"\1\0\0\0\0\0\0\1\0\0"
"\0\0\2\0\0\0\0\0\0\2"
"\0\0\0\2\1\2\1\0\0\2"
"\2\0\0\0\0\0\0\0\0\0"
"\2\0\0\0\0\1\1\0\2\0"
"\0\0\0\2\0\0\0\0\0\0", 10*10);
grid = snewn(10*10, char);
printf("%d\n", pearl_solve(10, 10, clues, grid));
#endif
grid = snewn(w*h, char);
clues = snewn(w*h, char);
clueorder = snewn(w*h, int);
while (1) {
pearl_loopgen(w, h, grid, rs);
#ifdef GENERATION_DIAGNOSTICS
printf("grid array:\n");
for (y = 0; y < h; y++) {
for (x = 0; x < w; x++) {
int type = grid[y*w+x];
char s[5], *p = s;
if (type & L) *p++ = 'L';
if (type & R) *p++ = 'R';
if (type & U) *p++ = 'U';
if (type & D) *p++ = 'D';
*p = '\0';
printf("%2s ", s);
}
printf("\n");
}
printf("\n");
#endif
/*
* Set up the maximal clue array.
*/
for (y = 0; y < h; y++)
for (x = 0; x < w; x++) {
int type = grid[y*w+x];
clues[y*w+x] = NOCLUE;
if ((bLR|bUD) & (1 << type)) {
/*
* This is a straight; see if it's a viable
* candidate for a straight clue. It qualifies if
* at least one of the squares it connects to is a
* corner.
*/
for (d = 1; d <= 8; d += d) if (type & d) {
int xx = x + DX(d), yy = y + DY(d);
assert(xx >= 0 && xx < w && yy >= 0 && yy < h);
if ((bLU|bLD|bRU|bRD) & (1 << grid[yy*w+xx]))
break;
}
if (d <= 8) /* we found one */
clues[y*w+x] = STRAIGHT;
} else if ((bLU|bLD|bRU|bRD) & (1 << type)) {
/*
* This is a corner; see if it's a viable candidate
* for a corner clue. It qualifies if all the
* squares it connects to are straights.
*/
for (d = 1; d <= 8; d += d) if (type & d) {
int xx = x + DX(d), yy = y + DY(d);
assert(xx >= 0 && xx < w && yy >= 0 && yy < h);
if (!((bLR|bUD) & (1 << grid[yy*w+xx])))
break;
}
if (d > 8) /* we didn't find a counterexample */
clues[y*w+x] = CORNER;
}
}
#ifdef GENERATION_DIAGNOSTICS
printf("clue array:\n");
for (y = 0; y < h; y++) {
for (x = 0; x < w; x++) {
printf("%c", " *O"[(unsigned char)clues[y*w+x]]);
}
printf("\n");
}
printf("\n");
#endif
/*
* See if we can solve the puzzle just like this.
*/
ret = pearl_solve(w, h, clues, grid);
assert(ret > 0); /* shouldn't be inconsistent! */
if (ret != 1)
continue; /* go round and try again */
/*
* Now shuffle the grid points and gradually remove the
* clues to find a minimal set which still leaves the
* puzzle soluble.
*/
for (i = 0; i < w*h; i++)
clueorder[i] = i;
shuffle(clueorder, w*h, sizeof(*clueorder), rs);
for (i = 0; i < w*h; i++) {
int clue;
y = clueorder[i] / w;
x = clueorder[i] % w;
if (clues[y*w+x] == 0)
continue;
clue = clues[y*w+x];
clues[y*w+x] = 0; /* try removing this clue */
ret = pearl_solve(w, h, clues, grid);
assert(ret > 0);
if (ret != 1)
clues[y*w+x] = clue; /* oops, put it back again */
}
#ifdef FINISHED_PUZZLE
printf("clue array:\n");
for (y = 0; y < h; y++) {
for (x = 0; x < w; x++) {
printf("%c", " *O"[(unsigned char)clues[y*w+x]]);
}
printf("\n");
}
printf("\n");
#endif
break; /* got it */
}
sfree(grid);
sfree(clues);
sfree(clueorder);
return dupstr("FIXME");
}
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 int game_can_format_as_text_now(game_params *params)
{
return TRUE;
}
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 pearl
#endif
const struct game thegame = {
"Pearl", 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_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,
FALSE, FALSE, game_print_size, game_print,
FALSE, /* wants_statusbar */
FALSE, game_timing_state,
0, /* flags */
};