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New puzzle! Or rather, new-ish, because this one has been lying around

Browse Source 
in the 'unfinished' directory for a while, and has now been finished
up thanks to James Harvey putting in some effort and galvanising me to
put in the rest. This is 'Pearl', an implementation of Nikoli's 'Masyu'.

The code in Loopy that generates a random loop along grid edges to use
as the puzzle solution has been abstracted out into loopgen.[ch] so
that Pearl can use it for its puzzle solutions too. I've also
introduced a new utility module called 'tdq' (for 'to-do queue').

[originally from svn r9379]
This commit is contained in:
Simon Tatham
2012-01-22 14:14:26 +00:00
parent b2d7429d53
commit b16eece9fc
12 changed files with 3364 additions and 1907 deletions
Download Patch File Download Diff File Expand all files Collapse all files

5
icons/Makefile
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@ -2,8 +2,8 @@
PUZZLES = blackbox bridges cube dominosa fifteen filling flip galaxies guess \
inertia keen lightup loopy magnets map mines net netslide pattern \
pegs range rect samegame signpost singles sixteen slant solo tents \
towers twiddle unequal untangle
pearl pegs range rect samegame signpost singles sixteen slant solo \
tents towers twiddle unequal untangle
BASE = $(patsubst %,%-base.png,$(PUZZLES))
WEB = $(patsubst %,%-web.png,$(PUZZLES))
@ -69,6 +69,7 @@ mines-ibase.png : override CROP=240x240 110x110+130+130
net-ibase.png : override CROP=193x193 113x113+0+80
netslide-ibase.png : override CROP=289x289 144x144+0+0
pattern-ibase.png : override CROP=384x384 223x223+0+0
pearl-ibase.png : override CROP=216x216 94x94+108+15
pegs-ibase.png : override CROP=263x263 147x147+116+0
range-ibase.png : override CROP=256x256 98x98+111+15
rect-ibase.png : override CROP=205x205 115x115+90+0

23
icons/pearl.sav Normal file
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@ -0,0 +1,23 @@
SAVEFILE:41:Simon Tatham's Portable Puzzle Collection
VERSION :1:1
GAME :5:Pearl
PARAMS :5:6x6dt
CPARAMS :5:6x6dt
SEED :15:901944054393278
DESC :17:BbBfWcWbWBaBeWgWa
AUXINFO :72:f8bbe71b9be753d5fa143df207d7797ba62a9b3996eb8b8889487e1a2bd659d91a5e73e1
NSTATES :2:14
STATEPOS:1:7
MOVE :55:F4,2,0;F1,1,0;F4,1,0;F1,0,0;F8,0,0;F2,0,1;F8,0,1;F2,0,2
MOVE :27:F1,0,3;F4,1,3;F1,1,3;F4,2,3
MOVE :27:F8,3,0;F2,3,1;F8,3,1;F2,3,2
MOVE :97:F2,4,2;F8,4,1;F2,4,1;F8,4,0;F1,4,0;F4,5,0;F8,5,0;F2,5,1;F8,5,1;F2,5,2;F8,5,2;F2,5,3;F4,5,3;F1,4,3
MOVE :13:F4,4,2;F1,3,2
MOVE :13:F4,3,0;F1,2,0
MOVE :69:F2,2,3;F8,2,2;F2,2,2;F8,2,1;F4,2,1;F1,1,1;F8,1,1;F2,1,2;F4,1,2;F1,0,2
MOVE :41:F8,0,3;F2,0,4;F8,0,4;F2,0,5;F1,0,5;F4,1,5
MOVE :27:F1,1,4;F4,2,4;F1,2,4;F4,3,4
MOVE :13:F8,1,4;F2,1,5
MOVE :55:F1,3,5;F4,4,5;F1,4,5;F4,5,5;F2,5,5;F8,5,4;F4,5,4;F1,4,4
MOVE :13:F2,3,5;F8,3,4
MOVE :13:F2,4,4;F8,4,3

536
loopgen.c Normal file
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/*
* loopgen.c: loop generation functions for grid.[ch].
*/
#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"
/* We're going to store lists of current candidate faces for colouring black
* or white.
* Each face gets a 'score', which tells us how adding that face right
* now would affect the curliness of the solution loop. We're trying to
* maximise that quantity so will bias our random selection of faces to
* colour those with high scores */
struct face_score {
int white_score;
int black_score;
unsigned long random;
/* No need to store a grid_face* here. The 'face_scores' array will
* be a list of 'face_score' objects, one for each face of the grid, so
* the position (index) within the 'face_scores' array will determine
* which face corresponds to a particular face_score.
* Having a single 'face_scores' array for all faces simplifies memory
* management, and probably improves performance, because we don't have to
* malloc/free each individual face_score, and we don't have to maintain
* a mapping from grid_face* pointers to face_score* pointers.
*/
};
static int generic_sort_cmpfn(void *v1, void *v2, size_t offset)
{
struct face_score *f1 = v1;
struct face_score *f2 = v2;
int r;
r = *(int *)((char *)f2 + offset) - *(int *)((char *)f1 + offset);
if (r) {
return r;
}
if (f1->random < f2->random)
return -1;
else if (f1->random > f2->random)
return 1;
/*
* It's _just_ possible that two faces might have been given
* the same random value. In that situation, fall back to
* comparing based on the positions within the face_scores list.
* This introduces a tiny directional bias, but not a significant one.
*/
return f1 - f2;
}
static int white_sort_cmpfn(void *v1, void *v2)
{
return generic_sort_cmpfn(v1, v2, offsetof(struct face_score,white_score));
}
static int black_sort_cmpfn(void *v1, void *v2)
{
return generic_sort_cmpfn(v1, v2, offsetof(struct face_score,black_score));
}
/* 'board' is an array of enum face_colour, indicating which faces are
* currently black/white/grey. 'colour' is FACE_WHITE or FACE_BLACK.
* Returns whether it's legal to colour the given face with this colour. */
static int can_colour_face(grid *g, char* board, int face_index,
enum face_colour colour)
{
int i, j;
grid_face *test_face = g->faces + face_index;
grid_face *starting_face, *current_face;
grid_dot *starting_dot;
int transitions;
int current_state, s; /* booleans: equal or not-equal to 'colour' */
int found_same_coloured_neighbour = FALSE;
assert(board[face_index] != colour);
/* Can only consider a face for colouring if it's adjacent to a face
* with the same colour. */
for (i = 0; i < test_face->order; i++) {
grid_edge *e = test_face->edges[i];
grid_face *f = (e->face1 == test_face) ? e->face2 : e->face1;
if (FACE_COLOUR(f) == colour) {
found_same_coloured_neighbour = TRUE;
break;
}
}
if (!found_same_coloured_neighbour)
return FALSE;
/* Need to avoid creating a loop of faces of this colour around some
* differently-coloured faces.
* Also need to avoid meeting a same-coloured face at a corner, with
* other-coloured faces in between. Here's a simple test that (I believe)
* takes care of both these conditions:
*
* Take the circular path formed by this face's edges, and inflate it
* slightly outwards. Imagine walking around this path and consider
* the faces that you visit in sequence. This will include all faces
* touching the given face, either along an edge or just at a corner.
* Count the number of 'colour'/not-'colour' transitions you encounter, as
* you walk along the complete loop. This will obviously turn out to be
* an even number.
* If 0, we're either in the middle of an "island" of this colour (should
* be impossible as we're not supposed to create black or white loops),
* or we're about to start a new island - also not allowed.
* If 4 or greater, there are too many separate coloured regions touching
* this face, and colouring it would create a loop or a corner-violation.
* The only allowed case is when the count is exactly 2. */
/* i points to a dot around the test face.
* j points to a face around the i^th dot.
* The current face will always be:
* test_face->dots[i]->faces[j]
* We assume dots go clockwise around the test face,
* and faces go clockwise around dots. */
/*
* The end condition is slightly fiddly. In sufficiently strange
* degenerate grids, our test face may be adjacent to the same
* other face multiple times (typically if it's the exterior
* face). Consider this, in particular:
*
* +--+
* | |
* +--+--+
* | | |
* +--+--+
*
* The bottom left face there is adjacent to the exterior face
* twice, so we can't just terminate our iteration when we reach
* the same _face_ we started at. Furthermore, we can't
* condition on having the same (i,j) pair either, because
* several (i,j) pairs identify the bottom left contiguity with
* the exterior face! We canonicalise the (i,j) pair by taking
* one step around before we set the termination tracking.
*/
i = j = 0;
current_face = test_face->dots[0]->faces[0];
if (current_face == test_face) {
j = 1;
current_face = test_face->dots[0]->faces[1];
}
transitions = 0;
current_state = (FACE_COLOUR(current_face) == colour);
starting_dot = NULL;
starting_face = NULL;
while (TRUE) {
/* Advance to next face.
* Need to loop here because it might take several goes to
* find it. */
while (TRUE) {
j++;
if (j == test_face->dots[i]->order)
j = 0;
if (test_face->dots[i]->faces[j] == test_face) {
/* Advance to next dot round test_face, then
* find current_face around new dot
* and advance to the next face clockwise */
i++;
if (i == test_face->order)
i = 0;
for (j = 0; j < test_face->dots[i]->order; j++) {
if (test_face->dots[i]->faces[j] == current_face)
break;
}
/* Must actually find current_face around new dot,
* or else something's wrong with the grid. */
assert(j != test_face->dots[i]->order);
/* Found, so advance to next face and try again */
} else {
break;
}
}
/* (i,j) are now advanced to next face */
current_face = test_face->dots[i]->faces[j];
s = (FACE_COLOUR(current_face) == colour);
if (!starting_dot) {
starting_dot = test_face->dots[i];
starting_face = current_face;
current_state = s;
} else {
if (s != current_state) {
++transitions;
current_state = s;
if (transitions > 2)
break;
}
if (test_face->dots[i] == starting_dot &&
current_face == starting_face)
break;
}
}
return (transitions == 2) ? TRUE : FALSE;
}
/* Count the number of neighbours of 'face', having colour 'colour' */
static int face_num_neighbours(grid *g, char *board, grid_face *face,
enum face_colour colour)
{
int colour_count = 0;
int i;
grid_face *f;
grid_edge *e;
for (i = 0; i < face->order; i++) {
e = face->edges[i];
f = (e->face1 == face) ? e->face2 : e->face1;
if (FACE_COLOUR(f) == colour)
++colour_count;
}
return colour_count;
}
/* The 'score' of a face reflects its current desirability for selection
* as the next face to colour white or black. We want to encourage moving
* into grey areas and increasing loopiness, so we give scores according to
* how many of the face's neighbours are currently coloured the same as the
* proposed colour. */
static int face_score(grid *g, char *board, grid_face *face,
enum face_colour colour)
{
/* Simple formula: score = 0 - num. same-coloured neighbours,
* so a higher score means fewer same-coloured neighbours. */
return -face_num_neighbours(g, board, face, colour);
}
/*
* Generate a new complete random closed loop for the given grid.
*
* The method is to generate a WHITE/BLACK colouring of all the faces,
* such that the WHITE faces will define the inside of the path, and the
* BLACK faces define the outside.
* To do this, we initially colour all faces GREY. The infinite space outside
* the grid is coloured BLACK, and we choose a random face to colour WHITE.
* Then we gradually grow the BLACK and the WHITE regions, eliminating GREY
* faces, until the grid is filled with BLACK/WHITE. As we grow the regions,
* we avoid creating loops of a single colour, to preserve the topological
* shape of the WHITE and BLACK regions.
* We also try to make the boundary as loopy and twisty as possible, to avoid
* generating paths that are uninteresting.
* The algorithm works by choosing a BLACK/WHITE colour, then choosing a GREY
* face that can be coloured with that colour (without violating the
* topological shape of that region). It's not obvious, but I think this
* algorithm is guaranteed to terminate without leaving any GREY faces behind.
* Indeed, if there are any GREY faces at all, both the WHITE and BLACK
* regions can be grown.
* This is checked using assert()ions, and I haven't seen any failures yet.
*
* Hand-wavy proof: imagine what can go wrong...
*
* Could the white faces get completely cut off by the black faces, and still
* leave some grey faces remaining?
* No, because then the black faces would form a loop around both the white
* faces and the grey faces, which is disallowed because we continually
* maintain the correct topological shape of the black region.
* Similarly, the black faces can never get cut off by the white faces. That
* means both the WHITE and BLACK regions always have some room to grow into
* the GREY regions.
* Could it be that we can't colour some GREY face, because there are too many
* WHITE/BLACK transitions as we walk round the face? (see the
* can_colour_face() function for details)
* No. Imagine otherwise, and we see WHITE/BLACK/WHITE/BLACK as we walk
* around the face. The two WHITE faces would be connected by a WHITE path,
* and the BLACK faces would be connected by a BLACK path. These paths would
* have to cross, which is impossible.
* Another thing that could go wrong: perhaps we can't find any GREY face to
* colour WHITE, because it would create a loop-violation or a corner-violation
* with the other WHITE faces?
* This is a little bit tricky to prove impossible. Imagine you have such a
* GREY face (that is, if you coloured it WHITE, you would create a WHITE loop
* or corner violation).
* That would cut all the non-white area into two blobs. One of those blobs
* must be free of BLACK faces (because the BLACK stuff is a connected blob).
* So we have a connected GREY area, completely surrounded by WHITE
* (including the GREY face we've tentatively coloured WHITE).
* A well-known result in graph theory says that you can always find a GREY
* face whose removal leaves the remaining GREY area connected. And it says
* there are at least two such faces, so we can always choose the one that
* isn't the "tentative" GREY face. Colouring that face WHITE leaves
* everything nice and connected, including that "tentative" GREY face which
* acts as a gateway to the rest of the non-WHITE grid.
*/
void generate_loop(grid *g, char *board, random_state *rs,
loopgen_bias_fn_t bias, void *biasctx)
{
int i, j;
int num_faces = g->num_faces;
struct face_score *face_scores; /* Array of face_score objects */
struct face_score *fs; /* Points somewhere in the above list */
struct grid_face *cur_face;
tree234 *lightable_faces_sorted;
tree234 *darkable_faces_sorted;
int *face_list;
int do_random_pass;
/* Make a board */
memset(board, FACE_GREY, num_faces);
/* Create and initialise the list of face_scores */
face_scores = snewn(num_faces, struct face_score);
for (i = 0; i < num_faces; i++) {
face_scores[i].random = random_bits(rs, 31);
face_scores[i].black_score = face_scores[i].white_score = 0;
}
/* Colour a random, finite face white. The infinite face is implicitly
* coloured black. Together, they will seed the random growth process
* for the black and white areas. */
i = random_upto(rs, num_faces);
board[i] = FACE_WHITE;
/* We need a way of favouring faces that will increase our loopiness.
* We do this by maintaining a list of all candidate faces sorted by
* their score and choose randomly from that with appropriate skew.
* In order to avoid consistently biasing towards particular faces, we
* need the sort order _within_ each group of scores to be completely
* random. But it would be abusing the hospitality of the tree234 data
* structure if our comparison function were nondeterministic :-). So with
* each face we associate a random number that does not change during a
* particular run of the generator, and use that as a secondary sort key.
* Yes, this means we will be biased towards particular random faces in
* any one run but that doesn't actually matter. */
lightable_faces_sorted = newtree234(white_sort_cmpfn);
darkable_faces_sorted = newtree234(black_sort_cmpfn);
/* Initialise the lists of lightable and darkable faces. This is
* slightly different from the code inside the while-loop, because we need
* to check every face of the board (the grid structure does not keep a
* list of the infinite face's neighbours). */
for (i = 0; i < num_faces; i++) {
grid_face *f = g->faces + i;
struct face_score *fs = face_scores + i;
if (board[i] != FACE_GREY) continue;
/* We need the full colourability check here, it's not enough simply
* to check neighbourhood. On some grids, a neighbour of the infinite
* face is not necessarily darkable. */
if (can_colour_face(g, board, i, FACE_BLACK)) {
fs->black_score = face_score(g, board, f, FACE_BLACK);
add234(darkable_faces_sorted, fs);
}
if (can_colour_face(g, board, i, FACE_WHITE)) {
fs->white_score = face_score(g, board, f, FACE_WHITE);
add234(lightable_faces_sorted, fs);
}
}
/* Colour faces one at a time until no more faces are colourable. */
while (TRUE)
{
enum face_colour colour;
tree234 *faces_to_pick;
int c_lightable = count234(lightable_faces_sorted);
int c_darkable = count234(darkable_faces_sorted);
if (c_lightable == 0 && c_darkable == 0) {
/* No more faces we can use at all. */
break;
}
assert(c_lightable != 0 && c_darkable != 0);
/* Choose a colour, and colour the best available face
* with that colour. */
colour = random_upto(rs, 2) ? FACE_WHITE : FACE_BLACK;
if (colour == FACE_WHITE)
faces_to_pick = lightable_faces_sorted;
else
faces_to_pick = darkable_faces_sorted;
if (bias) {
/*
* Go through all the candidate faces and pick the one the
* bias function likes best, breaking ties using the
* ordering in our tree234 (which is why we replace only
* if score > bestscore, not >=).
*/
int j, k;
struct face_score *best = NULL;
int score, bestscore = 0;
for (j = 0;
(fs = (struct face_score *)index234(faces_to_pick, j))!=NULL;
j++) {
assert(fs);
k = fs - face_scores;
assert(board[k] == FACE_GREY);
board[k] = colour;
score = bias(biasctx, board, k);
board[k] = FACE_GREY;
bias(biasctx, board, k); /* let bias know we put it back */
if (!best || score > bestscore) {
bestscore = score;
best = fs;
}
}
fs = best;
} else {
fs = (struct face_score *)index234(faces_to_pick, 0);
}
assert(fs);
i = fs - face_scores;
assert(board[i] == FACE_GREY);
board[i] = colour;
if (bias)
bias(biasctx, board, i); /* notify bias function of the change */
/* Remove this newly-coloured face from the lists. These lists should
* only contain grey faces. */
del234(lightable_faces_sorted, fs);
del234(darkable_faces_sorted, fs);
/* Remember which face we've just coloured */
cur_face = g->faces + i;
/* The face we've just coloured potentially affects the colourability
* and the scores of any neighbouring faces (touching at a corner or
* edge). So the search needs to be conducted around all faces
* touching the one we've just lit. Iterate over its corners, then
* over each corner's faces. For each such face, we remove it from
* the lists, recalculate any scores, then add it back to the lists
* (depending on whether it is lightable, darkable or both). */
for (i = 0; i < cur_face->order; i++) {
grid_dot *d = cur_face->dots[i];
for (j = 0; j < d->order; j++) {
grid_face *f = d->faces[j];
int fi; /* face index of f */
if (f == NULL)
continue;
if (f == cur_face)
continue;
/* If the face is already coloured, it won't be on our
* lightable/darkable lists anyway, so we can skip it without
* bothering with the removal step. */
if (FACE_COLOUR(f) != FACE_GREY) continue;
/* Find the face index and face_score* corresponding to f */
fi = f - g->faces;
fs = face_scores + fi;
/* Remove from lightable list if it's in there. We do this,
* even if it is still lightable, because the score might
* be different, and we need to remove-then-add to maintain
* correct sort order. */
del234(lightable_faces_sorted, fs);
if (can_colour_face(g, board, fi, FACE_WHITE)) {
fs->white_score = face_score(g, board, f, FACE_WHITE);
add234(lightable_faces_sorted, fs);
}
/* Do the same for darkable list. */
del234(darkable_faces_sorted, fs);
if (can_colour_face(g, board, fi, FACE_BLACK)) {
fs->black_score = face_score(g, board, f, FACE_BLACK);
add234(darkable_faces_sorted, fs);
}
}
}
}
/* Clean up */
freetree234(lightable_faces_sorted);
freetree234(darkable_faces_sorted);
sfree(face_scores);
/* The next step requires a shuffled list of all faces */
face_list = snewn(num_faces, int);
for (i = 0; i < num_faces; ++i) {
face_list[i] = i;
}
shuffle(face_list, num_faces, sizeof(int), rs);
/* The above loop-generation algorithm can often leave large clumps
* of faces of one colour. In extreme cases, the resulting path can be
* degenerate and not very satisfying to solve.
* This next step alleviates this problem:
* Go through the shuffled list, and flip the colour of any face we can
* legally flip, and which is adjacent to only one face of the opposite
* colour - this tends to grow 'tendrils' into any clumps.
* Repeat until we can find no more faces to flip. This will
* eventually terminate, because each flip increases the loop's
* perimeter, which cannot increase for ever.
* The resulting path will have maximal loopiness (in the sense that it
* cannot be improved "locally". Unfortunately, this allows a player to
* make some illicit deductions. To combat this (and make the path more
* interesting), we do one final pass making random flips. */
/* Set to TRUE for final pass */
do_random_pass = FALSE;
while (TRUE) {
/* Remember whether a flip occurred during this pass */
int flipped = FALSE;
for (i = 0; i < num_faces; ++i) {
int j = face_list[i];
enum face_colour opp =
(board[j] == FACE_WHITE) ? FACE_BLACK : FACE_WHITE;
if (can_colour_face(g, board, j, opp)) {
grid_face *face = g->faces +j;
if (do_random_pass) {
/* final random pass */
if (!random_upto(rs, 10))
board[j] = opp;
} else {
/* normal pass - flip when neighbour count is 1 */
if (face_num_neighbours(g, board, face, opp) == 1) {
board[j] = opp;
flipped = TRUE;
}
}
}
}
if (do_random_pass) break;
if (!flipped) do_random_pass = TRUE;
}
sfree(face_list);
}

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loopgen.h Normal file
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/*
* loopgen.h: interface file for loop generation functions for grid.[ch].
*/
#ifndef _LOOPGEN_H
#define _LOOPGEN_H
#include "puzzles.h"
#include "grid.h"
enum face_colour { FACE_WHITE, FACE_GREY, FACE_BLACK };
/* face should be of type grid_face* here. */
#define FACE_COLOUR(face) \
( (face) == NULL ? FACE_BLACK : \
board[(face) - g->faces] )
typedef int (*loopgen_bias_fn_t)(void *ctx, char *board, int face);
/* 'board' should be a char array whose length is the same as
* g->num_faces: this will be filled in with FACE_WHITE or FACE_BLACK
* after loop generation.
*
* If 'bias' is non-null, it should be a user-provided function which
* rates a half-finished board (i.e. may include some FACE_GREYs) for
* desirability; this will cause the loop generator to bias in favour
* of loops with a high return value from that function. The 'face'
* parameter to the bias function indicates which face of the grid has
* been modified since the last call; it is guaranteed that only one
* will have been (so that bias functions can work incrementally
* rather than re-scanning the whole grid on every call). */
extern void generate_loop(grid *g, char *board, random_state *rs,
loopgen_bias_fn_t bias, void *biasctx);
#endif

2
loopy.R
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@ -1,6 +1,6 @@
# -*- makefile -*-
LOOPY_EXTRA = tree234 dsf grid penrose
LOOPY_EXTRA = tree234 dsf grid penrose loopgen
loopy : [X] GTK COMMON loopy LOOPY_EXTRA loopy-icon|no-icon

497
loopy.c
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@ -82,6 +82,7 @@
#include "puzzles.h"
#include "tree234.h"
#include "grid.h"
#include "loopgen.h"
/* Debugging options */
@ -1277,507 +1278,20 @@ static int face_setall(solver_state *sstate, int face,
* Loop generation and clue removal
*/
/* We're going to store lists of current candidate faces for colouring black
* or white.
* Each face gets a 'score', which tells us how adding that face right
* now would affect the curliness of the solution loop. We're trying to
* maximise that quantity so will bias our random selection of faces to
* colour those with high scores */
struct face_score {
int white_score;
int black_score;
unsigned long random;
/* No need to store a grid_face* here. The 'face_scores' array will
* be a list of 'face_score' objects, one for each face of the grid, so
* the position (index) within the 'face_scores' array will determine
* which face corresponds to a particular face_score.
* Having a single 'face_scores' array for all faces simplifies memory
* management, and probably improves performance, because we don't have to
* malloc/free each individual face_score, and we don't have to maintain
* a mapping from grid_face* pointers to face_score* pointers.
*/
};
static int generic_sort_cmpfn(void *v1, void *v2, size_t offset)
{
struct face_score *f1 = v1;
struct face_score *f2 = v2;
int r;
r = *(int *)((char *)f2 + offset) - *(int *)((char *)f1 + offset);
if (r) {
return r;
}
if (f1->random < f2->random)
return -1;
else if (f1->random > f2->random)
return 1;
/*
* It's _just_ possible that two faces might have been given
* the same random value. In that situation, fall back to
* comparing based on the positions within the face_scores list.
* This introduces a tiny directional bias, but not a significant one.
*/
return f1 - f2;
}
static int white_sort_cmpfn(void *v1, void *v2)
{
return generic_sort_cmpfn(v1, v2, offsetof(struct face_score,white_score));
}
static int black_sort_cmpfn(void *v1, void *v2)
{
return generic_sort_cmpfn(v1, v2, offsetof(struct face_score,black_score));
}
enum face_colour { FACE_WHITE, FACE_GREY, FACE_BLACK };
/* face should be of type grid_face* here. */
#define FACE_COLOUR(face) \
( (face) == NULL ? FACE_BLACK : \
board[(face) - g->faces] )
/* 'board' is an array of these enums, indicating which faces are
* currently black/white/grey. 'colour' is FACE_WHITE or FACE_BLACK.
* Returns whether it's legal to colour the given face with this colour. */
static int can_colour_face(grid *g, char* board, int face_index,
enum face_colour colour)
{
int i, j;
grid_face *test_face = g->faces + face_index;
grid_face *starting_face, *current_face;
grid_dot *starting_dot;
int transitions;
int current_state, s; /* booleans: equal or not-equal to 'colour' */
int found_same_coloured_neighbour = FALSE;
assert(board[face_index] != colour);
/* Can only consider a face for colouring if it's adjacent to a face
* with the same colour. */
for (i = 0; i < test_face->order; i++) {
grid_edge *e = test_face->edges[i];
grid_face *f = (e->face1 == test_face) ? e->face2 : e->face1;
if (FACE_COLOUR(f) == colour) {
found_same_coloured_neighbour = TRUE;
break;
}
}
if (!found_same_coloured_neighbour)
return FALSE;
/* Need to avoid creating a loop of faces of this colour around some
* differently-coloured faces.
* Also need to avoid meeting a same-coloured face at a corner, with
* other-coloured faces in between. Here's a simple test that (I believe)
* takes care of both these conditions:
*
* Take the circular path formed by this face's edges, and inflate it
* slightly outwards. Imagine walking around this path and consider
* the faces that you visit in sequence. This will include all faces
* touching the given face, either along an edge or just at a corner.
* Count the number of 'colour'/not-'colour' transitions you encounter, as
* you walk along the complete loop. This will obviously turn out to be
* an even number.
* If 0, we're either in the middle of an "island" of this colour (should
* be impossible as we're not supposed to create black or white loops),
* or we're about to start a new island - also not allowed.
* If 4 or greater, there are too many separate coloured regions touching
* this face, and colouring it would create a loop or a corner-violation.
* The only allowed case is when the count is exactly 2. */
/* i points to a dot around the test face.
* j points to a face around the i^th dot.
* The current face will always be:
* test_face->dots[i]->faces[j]
* We assume dots go clockwise around the test face,
* and faces go clockwise around dots. */
/*
* The end condition is slightly fiddly. In sufficiently strange
* degenerate grids, our test face may be adjacent to the same
* other face multiple times (typically if it's the exterior
* face). Consider this, in particular:
*
* +--+
* | |
* +--+--+
* | | |
* +--+--+
*
* The bottom left face there is adjacent to the exterior face
* twice, so we can't just terminate our iteration when we reach
* the same _face_ we started at. Furthermore, we can't
* condition on having the same (i,j) pair either, because
* several (i,j) pairs identify the bottom left contiguity with
* the exterior face! We canonicalise the (i,j) pair by taking
* one step around before we set the termination tracking.
*/
i = j = 0;
current_face = test_face->dots[0]->faces[0];
if (current_face == test_face) {
j = 1;
current_face = test_face->dots[0]->faces[1];
}
transitions = 0;
current_state = (FACE_COLOUR(current_face) == colour);
starting_dot = NULL;
starting_face = NULL;
while (TRUE) {
/* Advance to next face.
* Need to loop here because it might take several goes to
* find it. */
while (TRUE) {
j++;
if (j == test_face->dots[i]->order)
j = 0;
if (test_face->dots[i]->faces[j] == test_face) {
/* Advance to next dot round test_face, then
* find current_face around new dot
* and advance to the next face clockwise */
i++;
if (i == test_face->order)
i = 0;
for (j = 0; j < test_face->dots[i]->order; j++) {
if (test_face->dots[i]->faces[j] == current_face)
break;
}
/* Must actually find current_face around new dot,
* or else something's wrong with the grid. */
assert(j != test_face->dots[i]->order);
/* Found, so advance to next face and try again */
} else {
break;
}
}
/* (i,j) are now advanced to next face */
current_face = test_face->dots[i]->faces[j];
s = (FACE_COLOUR(current_face) == colour);
if (!starting_dot) {
starting_dot = test_face->dots[i];
starting_face = current_face;
current_state = s;
} else {
if (s != current_state) {
++transitions;
current_state = s;
if (transitions > 2)
break;
}
if (test_face->dots[i] == starting_dot &&
current_face == starting_face)
break;
}
}
return (transitions == 2) ? TRUE : FALSE;
}
/* Count the number of neighbours of 'face', having colour 'colour' */
static int face_num_neighbours(grid *g, char *board, grid_face *face,
enum face_colour colour)
{
int colour_count = 0;
int i;
grid_face *f;
grid_edge *e;
for (i = 0; i < face->order; i++) {
e = face->edges[i];
f = (e->face1 == face) ? e->face2 : e->face1;
if (FACE_COLOUR(f) == colour)
++colour_count;
}
return colour_count;
}
/* The 'score' of a face reflects its current desirability for selection
* as the next face to colour white or black. We want to encourage moving
* into grey areas and increasing loopiness, so we give scores according to
* how many of the face's neighbours are currently coloured the same as the
* proposed colour. */
static int face_score(grid *g, char *board, grid_face *face,
enum face_colour colour)
{
/* Simple formula: score = 0 - num. same-coloured neighbours,
* so a higher score means fewer same-coloured neighbours. */
return -face_num_neighbours(g, board, face, colour);
}
/* Generate a new complete set of clues for the given game_state.
* The method is to generate a WHITE/BLACK colouring of all the faces,
* such that the WHITE faces will define the inside of the path, and the
* BLACK faces define the outside.
* To do this, we initially colour all faces GREY. The infinite space outside
* the grid is coloured BLACK, and we choose a random face to colour WHITE.
* Then we gradually grow the BLACK and the WHITE regions, eliminating GREY
* faces, until the grid is filled with BLACK/WHITE. As we grow the regions,
* we avoid creating loops of a single colour, to preserve the topological
* shape of the WHITE and BLACK regions.
* We also try to make the boundary as loopy and twisty as possible, to avoid
* generating paths that are uninteresting.
* The algorithm works by choosing a BLACK/WHITE colour, then choosing a GREY
* face that can be coloured with that colour (without violating the
* topological shape of that region). It's not obvious, but I think this
* algorithm is guaranteed to terminate without leaving any GREY faces behind.
* Indeed, if there are any GREY faces at all, both the WHITE and BLACK
* regions can be grown.
* This is checked using assert()ions, and I haven't seen any failures yet.
*
* Hand-wavy proof: imagine what can go wrong...
*
* Could the white faces get completely cut off by the black faces, and still
* leave some grey faces remaining?
* No, because then the black faces would form a loop around both the white
* faces and the grey faces, which is disallowed because we continually
* maintain the correct topological shape of the black region.
* Similarly, the black faces can never get cut off by the white faces. That
* means both the WHITE and BLACK regions always have some room to grow into
* the GREY regions.
* Could it be that we can't colour some GREY face, because there are too many
* WHITE/BLACK transitions as we walk round the face? (see the
* can_colour_face() function for details)
* No. Imagine otherwise, and we see WHITE/BLACK/WHITE/BLACK as we walk
* around the face. The two WHITE faces would be connected by a WHITE path,
* and the BLACK faces would be connected by a BLACK path. These paths would
* have to cross, which is impossible.
* Another thing that could go wrong: perhaps we can't find any GREY face to
* colour WHITE, because it would create a loop-violation or a corner-violation
* with the other WHITE faces?
* This is a little bit tricky to prove impossible. Imagine you have such a
* GREY face (that is, if you coloured it WHITE, you would create a WHITE loop
* or corner violation).
* That would cut all the non-white area into two blobs. One of those blobs
* must be free of BLACK faces (because the BLACK stuff is a connected blob).
* So we have a connected GREY area, completely surrounded by WHITE
* (including the GREY face we've tentatively coloured WHITE).
* A well-known result in graph theory says that you can always find a GREY
* face whose removal leaves the remaining GREY area connected. And it says
* there are at least two such faces, so we can always choose the one that
* isn't the "tentative" GREY face. Colouring that face WHITE leaves
* everything nice and connected, including that "tentative" GREY face which
* acts as a gateway to the rest of the non-WHITE grid.
*/
static void add_full_clues(game_state *state, random_state *rs)
{
signed char *clues = state->clues;
char *board;
grid *g = state->game_grid;
int i, j;
int num_faces = g->num_faces;
struct face_score *face_scores; /* Array of face_score objects */
struct face_score *fs; /* Points somewhere in the above list */
struct grid_face *cur_face;
tree234 *lightable_faces_sorted;
tree234 *darkable_faces_sorted;
int *face_list;
int do_random_pass;
char *board = snewn(g->num_faces, char);
int i;
board = snewn(num_faces, char);
/* Make a board */
memset(board, FACE_GREY, num_faces);
/* Create and initialise the list of face_scores */
face_scores = snewn(num_faces, struct face_score);
for (i = 0; i < num_faces; i++) {
face_scores[i].random = random_bits(rs, 31);
face_scores[i].black_score = face_scores[i].white_score = 0;
}
/* Colour a random, finite face white. The infinite face is implicitly
* coloured black. Together, they will seed the random growth process
* for the black and white areas. */
i = random_upto(rs, num_faces);
board[i] = FACE_WHITE;
/* We need a way of favouring faces that will increase our loopiness.
* We do this by maintaining a list of all candidate faces sorted by
* their score and choose randomly from that with appropriate skew.
* In order to avoid consistently biasing towards particular faces, we
* need the sort order _within_ each group of scores to be completely
* random. But it would be abusing the hospitality of the tree234 data
* structure if our comparison function were nondeterministic :-). So with
* each face we associate a random number that does not change during a
* particular run of the generator, and use that as a secondary sort key.
* Yes, this means we will be biased towards particular random faces in
* any one run but that doesn't actually matter. */
lightable_faces_sorted = newtree234(white_sort_cmpfn);
darkable_faces_sorted = newtree234(black_sort_cmpfn);
/* Initialise the lists of lightable and darkable faces. This is
* slightly different from the code inside the while-loop, because we need
* to check every face of the board (the grid structure does not keep a
* list of the infinite face's neighbours). */
for (i = 0; i < num_faces; i++) {
grid_face *f = g->faces + i;
struct face_score *fs = face_scores + i;
if (board[i] != FACE_GREY) continue;
/* We need the full colourability check here, it's not enough simply
* to check neighbourhood. On some grids, a neighbour of the infinite
* face is not necessarily darkable. */
if (can_colour_face(g, board, i, FACE_BLACK)) {
fs->black_score = face_score(g, board, f, FACE_BLACK);
add234(darkable_faces_sorted, fs);
}
if (can_colour_face(g, board, i, FACE_WHITE)) {
fs->white_score = face_score(g, board, f, FACE_WHITE);
add234(lightable_faces_sorted, fs);
}
}
/* Colour faces one at a time until no more faces are colourable. */
while (TRUE)
{
enum face_colour colour;
struct face_score *fs_white, *fs_black;
int c_lightable = count234(lightable_faces_sorted);
int c_darkable = count234(darkable_faces_sorted);
if (c_lightable == 0 && c_darkable == 0) {
/* No more faces we can use at all. */
break;
}
assert(c_lightable != 0 && c_darkable != 0);
fs_white = (struct face_score *)index234(lightable_faces_sorted, 0);
fs_black = (struct face_score *)index234(darkable_faces_sorted, 0);
/* Choose a colour, and colour the best available face
* with that colour. */
colour = random_upto(rs, 2) ? FACE_WHITE : FACE_BLACK;
if (colour == FACE_WHITE)
fs = fs_white;
else
fs = fs_black;
assert(fs);
i = fs - face_scores;
assert(board[i] == FACE_GREY);
board[i] = colour;
/* Remove this newly-coloured face from the lists. These lists should
* only contain grey faces. */
del234(lightable_faces_sorted, fs);
del234(darkable_faces_sorted, fs);
/* Remember which face we've just coloured */
cur_face = g->faces + i;
/* The face we've just coloured potentially affects the colourability
* and the scores of any neighbouring faces (touching at a corner or
* edge). So the search needs to be conducted around all faces
* touching the one we've just lit. Iterate over its corners, then
* over each corner's faces. For each such face, we remove it from
* the lists, recalculate any scores, then add it back to the lists
* (depending on whether it is lightable, darkable or both). */
for (i = 0; i < cur_face->order; i++) {
grid_dot *d = cur_face->dots[i];
for (j = 0; j < d->order; j++) {
grid_face *f = d->faces[j];
int fi; /* face index of f */
if (f == NULL)
continue;
if (f == cur_face)
continue;
/* If the face is already coloured, it won't be on our
* lightable/darkable lists anyway, so we can skip it without
* bothering with the removal step. */
if (FACE_COLOUR(f) != FACE_GREY) continue;
/* Find the face index and face_score* corresponding to f */
fi = f - g->faces;
fs = face_scores + fi;
/* Remove from lightable list if it's in there. We do this,
* even if it is still lightable, because the score might
* be different, and we need to remove-then-add to maintain
* correct sort order. */
del234(lightable_faces_sorted, fs);
if (can_colour_face(g, board, fi, FACE_WHITE)) {
fs->white_score = face_score(g, board, f, FACE_WHITE);
add234(lightable_faces_sorted, fs);
}
/* Do the same for darkable list. */
del234(darkable_faces_sorted, fs);
if (can_colour_face(g, board, fi, FACE_BLACK)) {
fs->black_score = face_score(g, board, f, FACE_BLACK);
add234(darkable_faces_sorted, fs);
}
}
}
}
/* Clean up */
freetree234(lightable_faces_sorted);
freetree234(darkable_faces_sorted);
sfree(face_scores);
/* The next step requires a shuffled list of all faces */
face_list = snewn(num_faces, int);
for (i = 0; i < num_faces; ++i) {
face_list[i] = i;
}
shuffle(face_list, num_faces, sizeof(int), rs);
/* The above loop-generation algorithm can often leave large clumps
* of faces of one colour. In extreme cases, the resulting path can be
* degenerate and not very satisfying to solve.
* This next step alleviates this problem:
* Go through the shuffled list, and flip the colour of any face we can
* legally flip, and which is adjacent to only one face of the opposite
* colour - this tends to grow 'tendrils' into any clumps.
* Repeat until we can find no more faces to flip. This will
* eventually terminate, because each flip increases the loop's
* perimeter, which cannot increase for ever.
* The resulting path will have maximal loopiness (in the sense that it
* cannot be improved "locally". Unfortunately, this allows a player to
* make some illicit deductions. To combat this (and make the path more
* interesting), we do one final pass making random flips. */
/* Set to TRUE for final pass */
do_random_pass = FALSE;
while (TRUE) {
/* Remember whether a flip occurred during this pass */
int flipped = FALSE;
for (i = 0; i < num_faces; ++i) {
int j = face_list[i];
enum face_colour opp =
(board[j] == FACE_WHITE) ? FACE_BLACK : FACE_WHITE;
if (can_colour_face(g, board, j, opp)) {
grid_face *face = g->faces +j;
if (do_random_pass) {
/* final random pass */
if (!random_upto(rs, 10))
board[j] = opp;
} else {
/* normal pass - flip when neighbour count is 1 */
if (face_num_neighbours(g, board, face, opp) == 1) {
board[j] = opp;
flipped = TRUE;
}
}
}
}
if (do_random_pass) break;
if (!flipped) do_random_pass = TRUE;
}
sfree(face_list);
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, num_faces);
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;
@ -1791,7 +1305,6 @@ static void add_full_clues(game_state *state, random_state *rs)
if (f2) clues[f2 - g->faces]++;
}
}
sfree(board);
}

6
unfinished/pearl.R → pearl.R
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@ -1,11 +1,13 @@
# -*- makefile -*-
PEARL_EXTRA = dsf
PEARL_EXTRA = dsf tree234 grid penrose loopgen tdq
pearl : [X] GTK COMMON pearl PEARL_EXTRA pearl-icon|no-icon
pearl : [G] WINDOWS COMMON pearl PEARL_EXTRA pearl.res?
pearlbench : [U] pearl[STANDALONE_SOLVER] PEARL_EXTRA STANDALONE m.lib
pearlbench : [C] pearl[STANDALONE_SOLVER] PEARL_EXTRA STANDALONE
ALL += pearl[COMBINED] PEARL_EXTRA
!begin gtk

2567
pearl.c Normal file
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File diff suppressed because it is too large Load Diff

65
puzzles.but
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@ -2908,6 +2908,71 @@ These parameters are available from the \q{Custom...} option on the
\dd Size of grid in squares.
\C{pearl} \i{Pearl}
\cfg{winhelp-topic}{games.pearl}
You have a grid of squares. Your job is to draw lines between the
centres of horizontally or vertically adjacent squares, so that the
lines form a single closed loop. In the resulting grid, some of the
squares that the loop passes through will contain corners, and some
will be straight horizontal or vertical lines. (And some squares can
be completely empty \dash the loop doesn't have to pass through every
square.)
Some of the squares contain black and white circles, which are clues
that the loop must satisfy.
A black circle in a square indicates that that square is a corner, but
neither of the squares adjacent to it in the loop is also a corner.
A while circle indicates that the square is a straight edge, but \e{at
least one} of the squares adjacent to it in the loop is a corner.
(In both cases, the clue only constrains the two squares adjacent
\e{in the loop}, that is, the squares that the loop passes into after
leaving the clue square. The squares that are only adjacent \e{in the
grid} are not constrained.)
Credit for this puzzle goes to \i{Nikoli}, who call it \q{Masyu}.
\k{nikoli-pearl}.
Thanks to James Harvey for assistance with the implementation.
\B{nikoli-pearl}
\W{http://www.nikoli.co.jp/en/puzzles/masyu/}\cw{http://www.nikoli.co.jp/en/puzzles/masyu/}
\H{pearl-controls} \I{controls, for Pearl}Pearl controls
Click with the left button on a grid edge to draw a segment of the
loop through that edge, or to remove a segment once it is drawn.
Drag with the left button through a series of squares to draw more
than one segment of the loop in one go. Alternatively, drag over an
existing part of the loop to undraw it, or to undraw part of it and
then go in a different direction.
Click with the right button on a grid edge to mark it with a cross,
indicating that you are sure the loop does not go through that edge.
(For instance, if you have decided which of the squares adjacent to a
white clue has to be a corner, but don't yet know which way the corner
turns, you might mark the one way it \e{can't} go with a cross.)
(All the actions described in \k{common-actions} are also available.)
\H{pearl-parameters} \I{parameters, for Pearl}Pearl parameters
These parameters are available from the \q{Custom...} option on the
\q{Type} menu.
\dt \e{Width}, \e{Height}
\dd Size of grid in squares.
\dt \e{Difficulty}
\dd Controls the difficulty of the generated puzzle.
\A{licence} \I{MIT licence}\ii{Licence}
This software is \i{copyright} 2004-2010 Simon Tatham.

37
puzzles.h
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@ -347,6 +347,43 @@ void edsf_merge(int *dsf, int v1, int v2, int inverse);
void dsf_merge(int *dsf, int v1, int v2);
void dsf_init(int *dsf, int len);
/*
* tdq.c
*/
/*
* Data structure implementing a 'to-do queue', a simple
* de-duplicating to-do list mechanism.
*
* Specification: a tdq is a queue which can hold integers from 0 to
* n-1, where n was some constant specified at tdq creation time. No
* integer may appear in the queue's current contents more than once;
* an attempt to add an already-present integer again will do nothing,
* so that that integer is removed from the queue at the position
* where it was _first_ inserted. The add and remove operations take
* constant time.
*
* The idea is that you might use this in applications like solvers:
* keep a tdq listing the indices of grid squares that you currently
* need to process in some way. Whenever you modify a square in a way
* that will require you to re-scan its neighbours, add them to the
* list with tdq_add; meanwhile you're constantly taking elements off
* the list when you need another square to process. In solvers where
* deductions are mostly localised, this should prevent repeated
* O(N^2) loops over the whole grid looking for something to do. (But
* if only _most_ of the deductions are localised, then you should
* respond to an empty to-do list by re-adding everything using
* tdq_fill, so _then_ you rescan the whole grid looking for newly
* enabled non-local deductions. Only if you've done that and emptied
* the list again finding nothing new to do are you actually done.)
*/
typedef struct tdq tdq;
tdq *tdq_new(int n);
void tdq_free(tdq *tdq);
void tdq_add(tdq *tdq, int k);
int tdq_remove(tdq *tdq); /* returns -1 if nothing available */
void tdq_fill(tdq *tdq); /* add everything to the tdq at once */
/*
* laydomino.c
*/

88
tdq.c Normal file
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@ -0,0 +1,88 @@
/*
* tdq.c: implement a 'to-do queue', a simple de-duplicating to-do
* list mechanism.
*/
#include <assert.h>
#include "puzzles.h"
/*
* Implementation: a tdq consists of a circular buffer of size n
* storing the integers currently in the queue, plus an array of n
* booleans indicating whether each integer is already there.
*
* Using a circular buffer of size n to store between 0 and n items
* inclusive has an obvious failure mode: if the input and output
* pointers are the same, how do you know whether that means the
* buffer is full or empty?
*
* In this application we have a simple way to tell: in the former
* case, the flags array is all 1s, and in the latter case it's all
* 0s. So we could spot that case and check, say, flags[0].
*
* However, it's even easier to simply determine whether the queue is
* non-empty by testing flags[buffer[op]] - that way we don't even
* _have_ to compare ip against op.
*/
struct tdq {
int n;
int *queue;
int ip, op; /* in pointer, out pointer */
char *flags;
};
tdq *tdq_new(int n)
{
int i;
tdq *tdq = snew(struct tdq);
tdq->queue = snewn(n, int);
tdq->flags = snewn(n, char);
for (i = 0; i < n; i++) {
tdq->queue[i] = 0;
tdq->flags[i] = 0;
}
tdq->n = n;
tdq->ip = tdq->op = 0;
return tdq;
}
void tdq_free(tdq *tdq)
{
sfree(tdq->queue);
sfree(tdq->flags);
sfree(tdq);
}
void tdq_add(tdq *tdq, int k)
{
assert((unsigned)k < (unsigned)tdq->n);
if (!tdq->flags[k]) {
tdq->queue[tdq->ip] = k;
tdq->flags[k] = 1;
if (++tdq->ip == tdq->n)
tdq->ip = 0;
}
}
int tdq_remove(tdq *tdq)
{
int ret = tdq->queue[tdq->op];
if (!tdq->flags[ret])
return -1;
tdq->flags[ret] = 0;
if (++tdq->op == tdq->n)
tdq->op = 0;
return ret;
}
void tdq_fill(tdq *tdq)
{
int i;
for (i = 0; i < tdq->n; i++)
tdq_add(tdq, i);
}

1410
unfinished/pearl.c
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