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fee17c3704
people have generally seemed to think Loopy is one of the more difficult puzzles in the collection. There's a new level called Tricky, between Normal and Hard, which is equivalent to the old Normal. [originally from svn r8398]
3791 lines
124 KiB
C
3791 lines
124 KiB
C
/*
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* loopy.c:
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*
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* An implementation of the Nikoli game 'Loop the loop'.
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* (c) Mike Pinna, 2005, 2006
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* Substantially rewritten to allowing for more general types of grid.
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* (c) Lambros Lambrou 2008
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*
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* vim: set shiftwidth=4 :set textwidth=80:
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*/
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/*
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* Possible future solver enhancements:
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*
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* - There's an interesting deductive technique which makes use
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* of topology rather than just graph theory. Each _face_ in
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* the grid is either inside or outside the loop; you can tell
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* that two faces are on the same side of the loop if they're
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* separated by a LINE_NO (or, more generally, by a path
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* crossing no LINE_UNKNOWNs and an even number of LINE_YESes),
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* and on the opposite side of the loop if they're separated by
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* a LINE_YES (or an odd number of LINE_YESes and no
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* LINE_UNKNOWNs). Oh, and any face separated from the outside
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* of the grid by a LINE_YES or a LINE_NO is on the inside or
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* outside respectively. So if you can track this for all
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* faces, you figure out the state of the line between a pair
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* once their relative insideness is known.
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* + The way I envisage this working is simply to keep an edsf
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* of all _faces_, which indicates whether they're on
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* opposite sides of the loop from one another. We also
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* include a special entry in the edsf for the infinite
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* exterior "face".
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* + So, the simple way to do this is to just go through the
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* edges: every time we see an edge in a state other than
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* LINE_UNKNOWN which separates two faces that aren't in the
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* same edsf class, we can rectify that by merging the
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* classes. Then, conversely, an edge in LINE_UNKNOWN state
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* which separates two faces that _are_ in the same edsf
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* class can immediately have its state determined.
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* + But you can go one better, if you're prepared to loop
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* over all _pairs_ of edges. Suppose we have edges A and B,
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* which respectively separate faces A1,A2 and B1,B2.
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* Suppose that A,B are in the same edge-edsf class and that
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* A1,B1 (wlog) are in the same face-edsf class; then we can
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* immediately place A2,B2 into the same face-edsf class (as
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* each other, not as A1 and A2) one way round or the other.
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* And conversely again, if A1,B1 are in the same face-edsf
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* class and so are A2,B2, then we can put A,B into the same
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* face-edsf class.
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* * Of course, this deduction requires a quadratic-time
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* loop over all pairs of edges in the grid, so it should
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* be reserved until there's nothing easier left to be
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* done.
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*
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* - The generalised grid support has made me (SGT) notice a
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* possible extension to the loop-avoidance code. When you have
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* a path of connected edges such that no other edges at all
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* are incident on any vertex in the middle of the path - or,
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* alternatively, such that any such edges are already known to
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* be LINE_NO - then you know those edges are either all
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* LINE_YES or all LINE_NO. Hence you can mentally merge the
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* entire path into a single long curly edge for the purposes
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* of loop avoidance, and look directly at whether or not the
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* extreme endpoints of the path are connected by some other
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* route. I find this coming up fairly often when I play on the
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* octagonal grid setting, so it might be worth implementing in
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* the solver.
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*
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* - (Just a speed optimisation.) Consider some todo list queue where every
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* time we modify something we mark it for consideration by other bits of
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* the solver, to save iteration over things that have already been done.
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*/
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#include <stdio.h>
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#include <stdlib.h>
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#include <stddef.h>
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#include <string.h>
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#include <assert.h>
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#include <ctype.h>
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#include <math.h>
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#include "puzzles.h"
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#include "tree234.h"
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#include "grid.h"
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/* Debugging options */
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/*
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#define DEBUG_CACHES
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#define SHOW_WORKING
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#define DEBUG_DLINES
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*/
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/* ----------------------------------------------------------------------
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* Struct, enum and function declarations
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*/
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enum {
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COL_BACKGROUND,
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COL_FOREGROUND,
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COL_LINEUNKNOWN,
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COL_HIGHLIGHT,
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COL_MISTAKE,
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COL_SATISFIED,
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NCOLOURS
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};
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struct game_state {
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grid *game_grid;
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/* Put -1 in a face that doesn't get a clue */
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signed char *clues;
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/* Array of line states, to store whether each line is
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* YES, NO or UNKNOWN */
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char *lines;
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unsigned char *line_errors;
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int solved;
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int cheated;
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/* Used in game_text_format(), so that it knows what type of
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* grid it's trying to render as ASCII text. */
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int grid_type;
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};
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enum solver_status {
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SOLVER_SOLVED, /* This is the only solution the solver could find */
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SOLVER_MISTAKE, /* This is definitely not a solution */
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SOLVER_AMBIGUOUS, /* This _might_ be an ambiguous solution */
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SOLVER_INCOMPLETE /* This may be a partial solution */
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};
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/* ------ Solver state ------ */
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typedef struct solver_state {
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game_state *state;
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enum solver_status solver_status;
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/* NB looplen is the number of dots that are joined together at a point, ie a
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* looplen of 1 means there are no lines to a particular dot */
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int *looplen;
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/* Difficulty level of solver. Used by solver functions that want to
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* vary their behaviour depending on the requested difficulty level. */
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int diff;
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/* caches */
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char *dot_yes_count;
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char *dot_no_count;
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char *face_yes_count;
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char *face_no_count;
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char *dot_solved, *face_solved;
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int *dotdsf;
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/* Information for Normal level deductions:
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* For each dline, store a bitmask for whether we know:
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* (bit 0) at least one is YES
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* (bit 1) at most one is YES */
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char *dlines;
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/* Hard level information */
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int *linedsf;
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} solver_state;
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/*
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* Difficulty levels. I do some macro ickery here to ensure that my
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* enum and the various forms of my name list always match up.
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*/
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#define DIFFLIST(A) \
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A(EASY,Easy,e) \
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A(NORMAL,Normal,n) \
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A(TRICKY,Tricky,t) \
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A(HARD,Hard,h)
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#define ENUM(upper,title,lower) DIFF_ ## upper,
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#define TITLE(upper,title,lower) #title,
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#define ENCODE(upper,title,lower) #lower
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#define CONFIG(upper,title,lower) ":" #title
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enum { DIFFLIST(ENUM) DIFF_MAX };
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static char const *const diffnames[] = { DIFFLIST(TITLE) };
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static char const diffchars[] = DIFFLIST(ENCODE);
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#define DIFFCONFIG DIFFLIST(CONFIG)
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/*
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* Solver routines, sorted roughly in order of computational cost.
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* The solver will run the faster deductions first, and slower deductions are
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* only invoked when the faster deductions are unable to make progress.
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* Each function is associated with a difficulty level, so that the generated
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* puzzles are solvable by applying only the functions with the chosen
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* difficulty level or lower.
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*/
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#define SOLVERLIST(A) \
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A(trivial_deductions, DIFF_EASY) \
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A(dline_deductions, DIFF_NORMAL) \
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A(linedsf_deductions, DIFF_HARD) \
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A(loop_deductions, DIFF_EASY)
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#define SOLVER_FN_DECL(fn,diff) static int fn(solver_state *);
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#define SOLVER_FN(fn,diff) &fn,
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#define SOLVER_DIFF(fn,diff) diff,
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SOLVERLIST(SOLVER_FN_DECL)
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static int (*(solver_fns[]))(solver_state *) = { SOLVERLIST(SOLVER_FN) };
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static int const solver_diffs[] = { SOLVERLIST(SOLVER_DIFF) };
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const int NUM_SOLVERS = sizeof(solver_diffs)/sizeof(*solver_diffs);
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struct game_params {
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int w, h;
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int diff;
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int type;
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/* Grid generation is expensive, so keep a (ref-counted) reference to the
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* grid for these parameters, and only generate when required. */
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grid *game_grid;
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};
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/* line_drawstate is the same as line_state, but with the extra ERROR
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* possibility. The drawing code copies line_state to line_drawstate,
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* except in the case that the line is an error. */
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enum line_state { LINE_YES, LINE_UNKNOWN, LINE_NO };
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enum line_drawstate { DS_LINE_YES, DS_LINE_UNKNOWN,
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DS_LINE_NO, DS_LINE_ERROR };
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#define OPP(line_state) \
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(2 - line_state)
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struct game_drawstate {
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int started;
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int tilesize;
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int flashing;
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char *lines;
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char *clue_error;
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char *clue_satisfied;
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};
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static char *validate_desc(game_params *params, char *desc);
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static int dot_order(const game_state* state, int i, char line_type);
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static int face_order(const game_state* state, int i, char line_type);
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static solver_state *solve_game_rec(const solver_state *sstate);
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#ifdef DEBUG_CACHES
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static void check_caches(const solver_state* sstate);
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#else
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#define check_caches(s)
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#endif
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/* ------- List of grid generators ------- */
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#define GRIDLIST(A) \
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A(Squares,grid_new_square,3,3) \
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A(Triangular,grid_new_triangular,3,3) \
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A(Honeycomb,grid_new_honeycomb,3,3) \
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A(Snub-Square,grid_new_snubsquare,3,3) \
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A(Cairo,grid_new_cairo,3,4) \
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A(Great-Hexagonal,grid_new_greathexagonal,3,3) \
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A(Octagonal,grid_new_octagonal,3,3) \
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A(Kites,grid_new_kites,3,3)
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#define GRID_NAME(title,fn,amin,omin) #title,
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#define GRID_CONFIG(title,fn,amin,omin) ":" #title
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#define GRID_FN(title,fn,amin,omin) &fn,
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#define GRID_SIZES(title,fn,amin,omin) \
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{amin, omin, \
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"Width and height for this grid type must both be at least " #amin, \
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"At least one of width and height for this grid type must be at least " #omin,},
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static char const *const gridnames[] = { GRIDLIST(GRID_NAME) };
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#define GRID_CONFIGS GRIDLIST(GRID_CONFIG)
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static grid * (*(grid_fns[]))(int w, int h) = { GRIDLIST(GRID_FN) };
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#define NUM_GRID_TYPES (sizeof(grid_fns) / sizeof(grid_fns[0]))
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static const struct {
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int amin, omin;
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char *aerr, *oerr;
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} grid_size_limits[] = { GRIDLIST(GRID_SIZES) };
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/* Generates a (dynamically allocated) new grid, according to the
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* type and size requested in params. Does nothing if the grid is already
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* generated. The allocated grid is owned by the params object, and will be
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* freed in free_params(). */
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static void params_generate_grid(game_params *params)
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{
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if (!params->game_grid) {
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params->game_grid = grid_fns[params->type](params->w, params->h);
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}
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}
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/* ----------------------------------------------------------------------
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* Preprocessor magic
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*/
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/* General constants */
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#define PREFERRED_TILE_SIZE 32
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#define BORDER(tilesize) ((tilesize) / 2)
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#define FLASH_TIME 0.5F
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#define BIT_SET(field, bit) ((field) & (1<<(bit)))
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#define SET_BIT(field, bit) (BIT_SET(field, bit) ? FALSE : \
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((field) |= (1<<(bit)), TRUE))
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#define CLEAR_BIT(field, bit) (BIT_SET(field, bit) ? \
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((field) &= ~(1<<(bit)), TRUE) : FALSE)
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#define CLUE2CHAR(c) \
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((c < 0) ? ' ' : c + '0')
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/* ----------------------------------------------------------------------
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* General struct manipulation and other straightforward code
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*/
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static game_state *dup_game(game_state *state)
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{
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game_state *ret = snew(game_state);
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ret->game_grid = state->game_grid;
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ret->game_grid->refcount++;
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ret->solved = state->solved;
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ret->cheated = state->cheated;
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ret->clues = snewn(state->game_grid->num_faces, signed char);
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memcpy(ret->clues, state->clues, state->game_grid->num_faces);
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ret->lines = snewn(state->game_grid->num_edges, char);
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memcpy(ret->lines, state->lines, state->game_grid->num_edges);
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ret->line_errors = snewn(state->game_grid->num_edges, unsigned char);
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memcpy(ret->line_errors, state->line_errors, state->game_grid->num_edges);
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ret->grid_type = state->grid_type;
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return ret;
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}
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static void free_game(game_state *state)
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{
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if (state) {
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grid_free(state->game_grid);
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sfree(state->clues);
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sfree(state->lines);
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sfree(state->line_errors);
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sfree(state);
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}
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}
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static solver_state *new_solver_state(game_state *state, int diff) {
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int i;
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int num_dots = state->game_grid->num_dots;
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int num_faces = state->game_grid->num_faces;
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int num_edges = state->game_grid->num_edges;
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solver_state *ret = snew(solver_state);
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ret->state = dup_game(state);
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ret->solver_status = SOLVER_INCOMPLETE;
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ret->diff = diff;
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ret->dotdsf = snew_dsf(num_dots);
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ret->looplen = snewn(num_dots, int);
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for (i = 0; i < num_dots; i++) {
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ret->looplen[i] = 1;
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}
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ret->dot_solved = snewn(num_dots, char);
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ret->face_solved = snewn(num_faces, char);
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memset(ret->dot_solved, FALSE, num_dots);
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memset(ret->face_solved, FALSE, num_faces);
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ret->dot_yes_count = snewn(num_dots, char);
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memset(ret->dot_yes_count, 0, num_dots);
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ret->dot_no_count = snewn(num_dots, char);
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memset(ret->dot_no_count, 0, num_dots);
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ret->face_yes_count = snewn(num_faces, char);
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memset(ret->face_yes_count, 0, num_faces);
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ret->face_no_count = snewn(num_faces, char);
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memset(ret->face_no_count, 0, num_faces);
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if (diff < DIFF_NORMAL) {
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ret->dlines = NULL;
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} else {
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ret->dlines = snewn(2*num_edges, char);
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memset(ret->dlines, 0, 2*num_edges);
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}
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if (diff < DIFF_HARD) {
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ret->linedsf = NULL;
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} else {
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ret->linedsf = snew_dsf(state->game_grid->num_edges);
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}
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return ret;
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}
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static void free_solver_state(solver_state *sstate) {
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if (sstate) {
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free_game(sstate->state);
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sfree(sstate->dotdsf);
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sfree(sstate->looplen);
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sfree(sstate->dot_solved);
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sfree(sstate->face_solved);
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sfree(sstate->dot_yes_count);
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sfree(sstate->dot_no_count);
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sfree(sstate->face_yes_count);
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sfree(sstate->face_no_count);
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/* OK, because sfree(NULL) is a no-op */
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sfree(sstate->dlines);
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sfree(sstate->linedsf);
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sfree(sstate);
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}
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}
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static solver_state *dup_solver_state(const solver_state *sstate) {
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game_state *state = sstate->state;
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int num_dots = state->game_grid->num_dots;
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int num_faces = state->game_grid->num_faces;
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int num_edges = state->game_grid->num_edges;
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solver_state *ret = snew(solver_state);
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ret->state = state = dup_game(sstate->state);
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ret->solver_status = sstate->solver_status;
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ret->diff = sstate->diff;
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ret->dotdsf = snewn(num_dots, int);
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ret->looplen = snewn(num_dots, int);
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memcpy(ret->dotdsf, sstate->dotdsf,
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num_dots * sizeof(int));
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memcpy(ret->looplen, sstate->looplen,
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num_dots * sizeof(int));
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ret->dot_solved = snewn(num_dots, char);
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ret->face_solved = snewn(num_faces, char);
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memcpy(ret->dot_solved, sstate->dot_solved, num_dots);
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memcpy(ret->face_solved, sstate->face_solved, num_faces);
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ret->dot_yes_count = snewn(num_dots, char);
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memcpy(ret->dot_yes_count, sstate->dot_yes_count, num_dots);
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ret->dot_no_count = snewn(num_dots, char);
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memcpy(ret->dot_no_count, sstate->dot_no_count, num_dots);
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ret->face_yes_count = snewn(num_faces, char);
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memcpy(ret->face_yes_count, sstate->face_yes_count, num_faces);
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ret->face_no_count = snewn(num_faces, char);
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memcpy(ret->face_no_count, sstate->face_no_count, num_faces);
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if (sstate->dlines) {
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ret->dlines = snewn(2*num_edges, char);
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memcpy(ret->dlines, sstate->dlines,
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2*num_edges);
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} else {
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ret->dlines = NULL;
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}
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if (sstate->linedsf) {
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ret->linedsf = snewn(num_edges, int);
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memcpy(ret->linedsf, sstate->linedsf,
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num_edges * sizeof(int));
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} else {
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ret->linedsf = NULL;
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}
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return ret;
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}
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static game_params *default_params(void)
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{
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game_params *ret = snew(game_params);
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#ifdef SLOW_SYSTEM
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ret->h = 7;
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ret->w = 7;
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#else
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ret->h = 10;
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|
ret->w = 10;
|
|
#endif
|
|
ret->diff = DIFF_EASY;
|
|
ret->type = 0;
|
|
|
|
ret->game_grid = NULL;
|
|
|
|
return ret;
|
|
}
|
|
|
|
static game_params *dup_params(game_params *params)
|
|
{
|
|
game_params *ret = snew(game_params);
|
|
|
|
*ret = *params; /* structure copy */
|
|
if (ret->game_grid) {
|
|
ret->game_grid->refcount++;
|
|
}
|
|
return ret;
|
|
}
|
|
|
|
static const game_params presets[] = {
|
|
#ifdef SMALL_SCREEN
|
|
{ 7, 7, DIFF_EASY, 0, NULL },
|
|
{ 7, 7, DIFF_NORMAL, 0, NULL },
|
|
{ 7, 7, DIFF_HARD, 0, NULL },
|
|
{ 7, 7, DIFF_HARD, 1, NULL },
|
|
{ 7, 7, DIFF_HARD, 2, NULL },
|
|
{ 5, 5, DIFF_HARD, 3, NULL },
|
|
{ 7, 7, DIFF_HARD, 4, NULL },
|
|
{ 5, 4, DIFF_HARD, 5, NULL },
|
|
{ 5, 5, DIFF_HARD, 6, NULL },
|
|
{ 5, 5, DIFF_HARD, 7, NULL },
|
|
#else
|
|
{ 7, 7, DIFF_EASY, 0, NULL },
|
|
{ 10, 10, DIFF_EASY, 0, NULL },
|
|
{ 7, 7, DIFF_NORMAL, 0, NULL },
|
|
{ 10, 10, DIFF_NORMAL, 0, NULL },
|
|
{ 7, 7, DIFF_HARD, 0, NULL },
|
|
{ 10, 10, DIFF_HARD, 0, NULL },
|
|
{ 10, 10, DIFF_HARD, 1, NULL },
|
|
{ 12, 10, DIFF_HARD, 2, NULL },
|
|
{ 7, 7, DIFF_HARD, 3, NULL },
|
|
{ 9, 9, DIFF_HARD, 4, NULL },
|
|
{ 5, 4, DIFF_HARD, 5, NULL },
|
|
{ 7, 7, DIFF_HARD, 6, NULL },
|
|
{ 5, 5, DIFF_HARD, 7, NULL },
|
|
#endif
|
|
};
|
|
|
|
static int game_fetch_preset(int i, char **name, game_params **params)
|
|
{
|
|
game_params *tmppar;
|
|
char buf[80];
|
|
|
|
if (i < 0 || i >= lenof(presets))
|
|
return FALSE;
|
|
|
|
tmppar = snew(game_params);
|
|
*tmppar = presets[i];
|
|
*params = tmppar;
|
|
sprintf(buf, "%dx%d %s - %s", tmppar->h, tmppar->w,
|
|
gridnames[tmppar->type], diffnames[tmppar->diff]);
|
|
*name = dupstr(buf);
|
|
|
|
return TRUE;
|
|
}
|
|
|
|
static void free_params(game_params *params)
|
|
{
|
|
if (params->game_grid) {
|
|
grid_free(params->game_grid);
|
|
}
|
|
sfree(params);
|
|
}
|
|
|
|
static void decode_params(game_params *params, char const *string)
|
|
{
|
|
if (params->game_grid) {
|
|
grid_free(params->game_grid);
|
|
params->game_grid = NULL;
|
|
}
|
|
params->h = params->w = atoi(string);
|
|
params->diff = DIFF_EASY;
|
|
while (*string && isdigit((unsigned char)*string)) string++;
|
|
if (*string == 'x') {
|
|
string++;
|
|
params->h = atoi(string);
|
|
while (*string && isdigit((unsigned char)*string)) string++;
|
|
}
|
|
if (*string == 't') {
|
|
string++;
|
|
params->type = atoi(string);
|
|
while (*string && isdigit((unsigned char)*string)) string++;
|
|
}
|
|
if (*string == 'd') {
|
|
int i;
|
|
string++;
|
|
for (i = 0; i < DIFF_MAX; i++)
|
|
if (*string == diffchars[i])
|
|
params->diff = i;
|
|
if (*string) string++;
|
|
}
|
|
}
|
|
|
|
static char *encode_params(game_params *params, int full)
|
|
{
|
|
char str[80];
|
|
sprintf(str, "%dx%dt%d", params->w, params->h, params->type);
|
|
if (full)
|
|
sprintf(str + strlen(str), "d%c", diffchars[params->diff]);
|
|
return dupstr(str);
|
|
}
|
|
|
|
static config_item *game_configure(game_params *params)
|
|
{
|
|
config_item *ret;
|
|
char buf[80];
|
|
|
|
ret = snewn(5, config_item);
|
|
|
|
ret[0].name = "Width";
|
|
ret[0].type = C_STRING;
|
|
sprintf(buf, "%d", params->w);
|
|
ret[0].sval = dupstr(buf);
|
|
ret[0].ival = 0;
|
|
|
|
ret[1].name = "Height";
|
|
ret[1].type = C_STRING;
|
|
sprintf(buf, "%d", params->h);
|
|
ret[1].sval = dupstr(buf);
|
|
ret[1].ival = 0;
|
|
|
|
ret[2].name = "Grid type";
|
|
ret[2].type = C_CHOICES;
|
|
ret[2].sval = GRID_CONFIGS;
|
|
ret[2].ival = params->type;
|
|
|
|
ret[3].name = "Difficulty";
|
|
ret[3].type = C_CHOICES;
|
|
ret[3].sval = DIFFCONFIG;
|
|
ret[3].ival = params->diff;
|
|
|
|
ret[4].name = NULL;
|
|
ret[4].type = C_END;
|
|
ret[4].sval = NULL;
|
|
ret[4].ival = 0;
|
|
|
|
return ret;
|
|
}
|
|
|
|
static game_params *custom_params(config_item *cfg)
|
|
{
|
|
game_params *ret = snew(game_params);
|
|
|
|
ret->w = atoi(cfg[0].sval);
|
|
ret->h = atoi(cfg[1].sval);
|
|
ret->type = cfg[2].ival;
|
|
ret->diff = cfg[3].ival;
|
|
|
|
ret->game_grid = NULL;
|
|
return ret;
|
|
}
|
|
|
|
static char *validate_params(game_params *params, int full)
|
|
{
|
|
if (params->type < 0 || params->type >= NUM_GRID_TYPES)
|
|
return "Illegal grid type";
|
|
if (params->w < grid_size_limits[params->type].amin ||
|
|
params->h < grid_size_limits[params->type].amin)
|
|
return grid_size_limits[params->type].aerr;
|
|
if (params->w < grid_size_limits[params->type].omin &&
|
|
params->h < grid_size_limits[params->type].omin)
|
|
return grid_size_limits[params->type].oerr;
|
|
|
|
/*
|
|
* This shouldn't be able to happen at all, since decode_params
|
|
* and custom_params will never generate anything that isn't
|
|
* within range.
|
|
*/
|
|
assert(params->diff < DIFF_MAX);
|
|
|
|
return NULL;
|
|
}
|
|
|
|
/* Returns a newly allocated string describing the current puzzle */
|
|
static char *state_to_text(const game_state *state)
|
|
{
|
|
grid *g = state->game_grid;
|
|
char *retval;
|
|
int num_faces = g->num_faces;
|
|
char *description = snewn(num_faces + 1, char);
|
|
char *dp = description;
|
|
int empty_count = 0;
|
|
int i;
|
|
|
|
for (i = 0; i < num_faces; i++) {
|
|
if (state->clues[i] < 0) {
|
|
if (empty_count > 25) {
|
|
dp += sprintf(dp, "%c", (int)(empty_count + 'a' - 1));
|
|
empty_count = 0;
|
|
}
|
|
empty_count++;
|
|
} else {
|
|
if (empty_count) {
|
|
dp += sprintf(dp, "%c", (int)(empty_count + 'a' - 1));
|
|
empty_count = 0;
|
|
}
|
|
dp += sprintf(dp, "%c", (int)CLUE2CHAR(state->clues[i]));
|
|
}
|
|
}
|
|
|
|
if (empty_count)
|
|
dp += sprintf(dp, "%c", (int)(empty_count + 'a' - 1));
|
|
|
|
retval = dupstr(description);
|
|
sfree(description);
|
|
|
|
return retval;
|
|
}
|
|
|
|
/* We require that the params pass the test in validate_params and that the
|
|
* description fills the entire game area */
|
|
static char *validate_desc(game_params *params, char *desc)
|
|
{
|
|
int count = 0;
|
|
grid *g;
|
|
params_generate_grid(params);
|
|
g = params->game_grid;
|
|
|
|
for (; *desc; ++desc) {
|
|
if (*desc >= '0' && *desc <= '9') {
|
|
count++;
|
|
continue;
|
|
}
|
|
if (*desc >= 'a') {
|
|
count += *desc - 'a' + 1;
|
|
continue;
|
|
}
|
|
return "Unknown character in description";
|
|
}
|
|
|
|
if (count < g->num_faces)
|
|
return "Description too short for board size";
|
|
if (count > g->num_faces)
|
|
return "Description too long for board size";
|
|
|
|
return NULL;
|
|
}
|
|
|
|
/* Sums the lengths of the numbers in range [0,n) */
|
|
/* See equivalent function in solo.c for justification of this. */
|
|
static int len_0_to_n(int n)
|
|
{
|
|
int len = 1; /* Counting 0 as a bit of a special case */
|
|
int i;
|
|
|
|
for (i = 1; i < n; i *= 10) {
|
|
len += max(n - i, 0);
|
|
}
|
|
|
|
return len;
|
|
}
|
|
|
|
static char *encode_solve_move(const game_state *state)
|
|
{
|
|
int len;
|
|
char *ret, *p;
|
|
int i;
|
|
int num_edges = state->game_grid->num_edges;
|
|
|
|
/* This is going to return a string representing the moves needed to set
|
|
* every line in a grid to be the same as the ones in 'state'. The exact
|
|
* length of this string is predictable. */
|
|
|
|
len = 1; /* Count the 'S' prefix */
|
|
/* Numbers in all lines */
|
|
len += len_0_to_n(num_edges);
|
|
/* For each line we also have a letter */
|
|
len += num_edges;
|
|
|
|
ret = snewn(len + 1, char);
|
|
p = ret;
|
|
|
|
p += sprintf(p, "S");
|
|
|
|
for (i = 0; i < num_edges; i++) {
|
|
switch (state->lines[i]) {
|
|
case LINE_YES:
|
|
p += sprintf(p, "%dy", i);
|
|
break;
|
|
case LINE_NO:
|
|
p += sprintf(p, "%dn", i);
|
|
break;
|
|
}
|
|
}
|
|
|
|
/* No point in doing sums like that if they're going to be wrong */
|
|
assert(strlen(ret) <= (size_t)len);
|
|
return ret;
|
|
}
|
|
|
|
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)
|
|
{
|
|
}
|
|
|
|
static void game_compute_size(game_params *params, int tilesize,
|
|
int *x, int *y)
|
|
{
|
|
grid *g;
|
|
int grid_width, grid_height, rendered_width, rendered_height;
|
|
|
|
params_generate_grid(params);
|
|
g = params->game_grid;
|
|
grid_width = g->highest_x - g->lowest_x;
|
|
grid_height = g->highest_y - g->lowest_y;
|
|
/* multiply first to minimise rounding error on integer division */
|
|
rendered_width = grid_width * tilesize / g->tilesize;
|
|
rendered_height = grid_height * tilesize / g->tilesize;
|
|
*x = rendered_width + 2 * BORDER(tilesize) + 1;
|
|
*y = rendered_height + 2 * BORDER(tilesize) + 1;
|
|
}
|
|
|
|
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(4 * NCOLOURS, float);
|
|
|
|
frontend_default_colour(fe, &ret[COL_BACKGROUND * 3]);
|
|
|
|
ret[COL_FOREGROUND * 3 + 0] = 0.0F;
|
|
ret[COL_FOREGROUND * 3 + 1] = 0.0F;
|
|
ret[COL_FOREGROUND * 3 + 2] = 0.0F;
|
|
|
|
ret[COL_LINEUNKNOWN * 3 + 0] = 0.8F;
|
|
ret[COL_LINEUNKNOWN * 3 + 1] = 0.8F;
|
|
ret[COL_LINEUNKNOWN * 3 + 2] = 0.0F;
|
|
|
|
ret[COL_HIGHLIGHT * 3 + 0] = 1.0F;
|
|
ret[COL_HIGHLIGHT * 3 + 1] = 1.0F;
|
|
ret[COL_HIGHLIGHT * 3 + 2] = 1.0F;
|
|
|
|
ret[COL_MISTAKE * 3 + 0] = 1.0F;
|
|
ret[COL_MISTAKE * 3 + 1] = 0.0F;
|
|
ret[COL_MISTAKE * 3 + 2] = 0.0F;
|
|
|
|
ret[COL_SATISFIED * 3 + 0] = 0.0F;
|
|
ret[COL_SATISFIED * 3 + 1] = 0.0F;
|
|
ret[COL_SATISFIED * 3 + 2] = 0.0F;
|
|
|
|
*ncolours = NCOLOURS;
|
|
return ret;
|
|
}
|
|
|
|
static game_drawstate *game_new_drawstate(drawing *dr, game_state *state)
|
|
{
|
|
struct game_drawstate *ds = snew(struct game_drawstate);
|
|
int num_faces = state->game_grid->num_faces;
|
|
int num_edges = state->game_grid->num_edges;
|
|
|
|
ds->tilesize = 0;
|
|
ds->started = 0;
|
|
ds->lines = snewn(num_edges, char);
|
|
ds->clue_error = snewn(num_faces, char);
|
|
ds->clue_satisfied = snewn(num_faces, char);
|
|
ds->flashing = 0;
|
|
|
|
memset(ds->lines, LINE_UNKNOWN, num_edges);
|
|
memset(ds->clue_error, 0, num_faces);
|
|
memset(ds->clue_satisfied, 0, num_faces);
|
|
|
|
return ds;
|
|
}
|
|
|
|
static void game_free_drawstate(drawing *dr, game_drawstate *ds)
|
|
{
|
|
sfree(ds->clue_error);
|
|
sfree(ds->clue_satisfied);
|
|
sfree(ds->lines);
|
|
sfree(ds);
|
|
}
|
|
|
|
static int game_timing_state(game_state *state, game_ui *ui)
|
|
{
|
|
return TRUE;
|
|
}
|
|
|
|
static float game_anim_length(game_state *oldstate, game_state *newstate,
|
|
int dir, game_ui *ui)
|
|
{
|
|
return 0.0F;
|
|
}
|
|
|
|
static int game_can_format_as_text_now(game_params *params)
|
|
{
|
|
if (params->type != 0)
|
|
return FALSE;
|
|
return TRUE;
|
|
}
|
|
|
|
static char *game_text_format(game_state *state)
|
|
{
|
|
int w, h, W, H;
|
|
int x, y, i;
|
|
int cell_size;
|
|
char *ret;
|
|
grid *g = state->game_grid;
|
|
grid_face *f;
|
|
|
|
assert(state->grid_type == 0);
|
|
|
|
/* Work out the basic size unit */
|
|
f = g->faces; /* first face */
|
|
assert(f->order == 4);
|
|
/* The dots are ordered clockwise, so the two opposite
|
|
* corners are guaranteed to span the square */
|
|
cell_size = abs(f->dots[0]->x - f->dots[2]->x);
|
|
|
|
w = (g->highest_x - g->lowest_x) / cell_size;
|
|
h = (g->highest_y - g->lowest_y) / cell_size;
|
|
|
|
/* Create a blank "canvas" to "draw" on */
|
|
W = 2 * w + 2;
|
|
H = 2 * h + 1;
|
|
ret = snewn(W * H + 1, char);
|
|
for (y = 0; y < H; y++) {
|
|
for (x = 0; x < W-1; x++) {
|
|
ret[y*W + x] = ' ';
|
|
}
|
|
ret[y*W + W-1] = '\n';
|
|
}
|
|
ret[H*W] = '\0';
|
|
|
|
/* Fill in edge info */
|
|
for (i = 0; i < g->num_edges; i++) {
|
|
grid_edge *e = g->edges + i;
|
|
/* Cell coordinates, from (0,0) to (w-1,h-1) */
|
|
int x1 = (e->dot1->x - g->lowest_x) / cell_size;
|
|
int x2 = (e->dot2->x - g->lowest_x) / cell_size;
|
|
int y1 = (e->dot1->y - g->lowest_y) / cell_size;
|
|
int y2 = (e->dot2->y - g->lowest_y) / cell_size;
|
|
/* Midpoint, in canvas coordinates (canvas coordinates are just twice
|
|
* cell coordinates) */
|
|
x = x1 + x2;
|
|
y = y1 + y2;
|
|
switch (state->lines[i]) {
|
|
case LINE_YES:
|
|
ret[y*W + x] = (y1 == y2) ? '-' : '|';
|
|
break;
|
|
case LINE_NO:
|
|
ret[y*W + x] = 'x';
|
|
break;
|
|
case LINE_UNKNOWN:
|
|
break; /* already a space */
|
|
default:
|
|
assert(!"Illegal line state");
|
|
}
|
|
}
|
|
|
|
/* Fill in clues */
|
|
for (i = 0; i < g->num_faces; i++) {
|
|
int x1, x2, y1, y2;
|
|
|
|
f = g->faces + i;
|
|
assert(f->order == 4);
|
|
/* Cell coordinates, from (0,0) to (w-1,h-1) */
|
|
x1 = (f->dots[0]->x - g->lowest_x) / cell_size;
|
|
x2 = (f->dots[2]->x - g->lowest_x) / cell_size;
|
|
y1 = (f->dots[0]->y - g->lowest_y) / cell_size;
|
|
y2 = (f->dots[2]->y - g->lowest_y) / cell_size;
|
|
/* Midpoint, in canvas coordinates */
|
|
x = x1 + x2;
|
|
y = y1 + y2;
|
|
ret[y*W + x] = CLUE2CHAR(state->clues[i]);
|
|
}
|
|
return ret;
|
|
}
|
|
|
|
/* ----------------------------------------------------------------------
|
|
* Debug code
|
|
*/
|
|
|
|
#ifdef DEBUG_CACHES
|
|
static void check_caches(const solver_state* sstate)
|
|
{
|
|
int i;
|
|
const game_state *state = sstate->state;
|
|
const grid *g = state->game_grid;
|
|
|
|
for (i = 0; i < g->num_dots; i++) {
|
|
assert(dot_order(state, i, LINE_YES) == sstate->dot_yes_count[i]);
|
|
assert(dot_order(state, i, LINE_NO) == sstate->dot_no_count[i]);
|
|
}
|
|
|
|
for (i = 0; i < g->num_faces; i++) {
|
|
assert(face_order(state, i, LINE_YES) == sstate->face_yes_count[i]);
|
|
assert(face_order(state, i, LINE_NO) == sstate->face_no_count[i]);
|
|
}
|
|
}
|
|
|
|
#if 0
|
|
#define check_caches(s) \
|
|
do { \
|
|
fprintf(stderr, "check_caches at line %d\n", __LINE__); \
|
|
check_caches(s); \
|
|
} while (0)
|
|
#endif
|
|
#endif /* DEBUG_CACHES */
|
|
|
|
/* ----------------------------------------------------------------------
|
|
* Solver utility functions
|
|
*/
|
|
|
|
/* Sets the line (with index i) to the new state 'line_new', and updates
|
|
* the cached counts of any affected faces and dots.
|
|
* Returns TRUE if this actually changed the line's state. */
|
|
static int solver_set_line(solver_state *sstate, int i,
|
|
enum line_state line_new
|
|
#ifdef SHOW_WORKING
|
|
, const char *reason
|
|
#endif
|
|
)
|
|
{
|
|
game_state *state = sstate->state;
|
|
grid *g;
|
|
grid_edge *e;
|
|
|
|
assert(line_new != LINE_UNKNOWN);
|
|
|
|
check_caches(sstate);
|
|
|
|
if (state->lines[i] == line_new) {
|
|
return FALSE; /* nothing changed */
|
|
}
|
|
state->lines[i] = line_new;
|
|
|
|
#ifdef SHOW_WORKING
|
|
fprintf(stderr, "solver: set line [%d] to %s (%s)\n",
|
|
i, line_new == LINE_YES ? "YES" : "NO",
|
|
reason);
|
|
#endif
|
|
|
|
g = state->game_grid;
|
|
e = g->edges + i;
|
|
|
|
/* Update the cache for both dots and both faces affected by this. */
|
|
if (line_new == LINE_YES) {
|
|
sstate->dot_yes_count[e->dot1 - g->dots]++;
|
|
sstate->dot_yes_count[e->dot2 - g->dots]++;
|
|
if (e->face1) {
|
|
sstate->face_yes_count[e->face1 - g->faces]++;
|
|
}
|
|
if (e->face2) {
|
|
sstate->face_yes_count[e->face2 - g->faces]++;
|
|
}
|
|
} else {
|
|
sstate->dot_no_count[e->dot1 - g->dots]++;
|
|
sstate->dot_no_count[e->dot2 - g->dots]++;
|
|
if (e->face1) {
|
|
sstate->face_no_count[e->face1 - g->faces]++;
|
|
}
|
|
if (e->face2) {
|
|
sstate->face_no_count[e->face2 - g->faces]++;
|
|
}
|
|
}
|
|
|
|
check_caches(sstate);
|
|
return TRUE;
|
|
}
|
|
|
|
#ifdef SHOW_WORKING
|
|
#define solver_set_line(a, b, c) \
|
|
solver_set_line(a, b, c, __FUNCTION__)
|
|
#endif
|
|
|
|
/*
|
|
* Merge two dots due to the existence of an edge between them.
|
|
* Updates the dsf tracking equivalence classes, and keeps track of
|
|
* the length of path each dot is currently a part of.
|
|
* Returns TRUE if the dots were already linked, ie if they are part of a
|
|
* closed loop, and false otherwise.
|
|
*/
|
|
static int merge_dots(solver_state *sstate, int edge_index)
|
|
{
|
|
int i, j, len;
|
|
grid *g = sstate->state->game_grid;
|
|
grid_edge *e = g->edges + edge_index;
|
|
|
|
i = e->dot1 - g->dots;
|
|
j = e->dot2 - g->dots;
|
|
|
|
i = dsf_canonify(sstate->dotdsf, i);
|
|
j = dsf_canonify(sstate->dotdsf, j);
|
|
|
|
if (i == j) {
|
|
return TRUE;
|
|
} else {
|
|
len = sstate->looplen[i] + sstate->looplen[j];
|
|
dsf_merge(sstate->dotdsf, i, j);
|
|
i = dsf_canonify(sstate->dotdsf, i);
|
|
sstate->looplen[i] = len;
|
|
return FALSE;
|
|
}
|
|
}
|
|
|
|
/* Merge two lines because the solver has deduced that they must be either
|
|
* identical or opposite. Returns TRUE if this is new information, otherwise
|
|
* FALSE. */
|
|
static int merge_lines(solver_state *sstate, int i, int j, int inverse
|
|
#ifdef SHOW_WORKING
|
|
, const char *reason
|
|
#endif
|
|
)
|
|
{
|
|
int inv_tmp;
|
|
|
|
assert(i < sstate->state->game_grid->num_edges);
|
|
assert(j < sstate->state->game_grid->num_edges);
|
|
|
|
i = edsf_canonify(sstate->linedsf, i, &inv_tmp);
|
|
inverse ^= inv_tmp;
|
|
j = edsf_canonify(sstate->linedsf, j, &inv_tmp);
|
|
inverse ^= inv_tmp;
|
|
|
|
edsf_merge(sstate->linedsf, i, j, inverse);
|
|
|
|
#ifdef SHOW_WORKING
|
|
if (i != j) {
|
|
fprintf(stderr, "%s [%d] [%d] %s(%s)\n",
|
|
__FUNCTION__, i, j,
|
|
inverse ? "inverse " : "", reason);
|
|
}
|
|
#endif
|
|
return (i != j);
|
|
}
|
|
|
|
#ifdef SHOW_WORKING
|
|
#define merge_lines(a, b, c, d) \
|
|
merge_lines(a, b, c, d, __FUNCTION__)
|
|
#endif
|
|
|
|
/* Count the number of lines of a particular type currently going into the
|
|
* given dot. */
|
|
static int dot_order(const game_state* state, int dot, char line_type)
|
|
{
|
|
int n = 0;
|
|
grid *g = state->game_grid;
|
|
grid_dot *d = g->dots + dot;
|
|
int i;
|
|
|
|
for (i = 0; i < d->order; i++) {
|
|
grid_edge *e = d->edges[i];
|
|
if (state->lines[e - g->edges] == line_type)
|
|
++n;
|
|
}
|
|
return n;
|
|
}
|
|
|
|
/* Count the number of lines of a particular type currently surrounding the
|
|
* given face */
|
|
static int face_order(const game_state* state, int face, char line_type)
|
|
{
|
|
int n = 0;
|
|
grid *g = state->game_grid;
|
|
grid_face *f = g->faces + face;
|
|
int i;
|
|
|
|
for (i = 0; i < f->order; i++) {
|
|
grid_edge *e = f->edges[i];
|
|
if (state->lines[e - g->edges] == line_type)
|
|
++n;
|
|
}
|
|
return n;
|
|
}
|
|
|
|
/* Set all lines bordering a dot of type old_type to type new_type
|
|
* Return value tells caller whether this function actually did anything */
|
|
static int dot_setall(solver_state *sstate, int dot,
|
|
char old_type, char new_type)
|
|
{
|
|
int retval = FALSE, r;
|
|
game_state *state = sstate->state;
|
|
grid *g;
|
|
grid_dot *d;
|
|
int i;
|
|
|
|
if (old_type == new_type)
|
|
return FALSE;
|
|
|
|
g = state->game_grid;
|
|
d = g->dots + dot;
|
|
|
|
for (i = 0; i < d->order; i++) {
|
|
int line_index = d->edges[i] - g->edges;
|
|
if (state->lines[line_index] == old_type) {
|
|
r = solver_set_line(sstate, line_index, new_type);
|
|
assert(r == TRUE);
|
|
retval = TRUE;
|
|
}
|
|
}
|
|
return retval;
|
|
}
|
|
|
|
/* Set all lines bordering a face of type old_type to type new_type */
|
|
static int face_setall(solver_state *sstate, int face,
|
|
char old_type, char new_type)
|
|
{
|
|
int retval = FALSE, r;
|
|
game_state *state = sstate->state;
|
|
grid *g;
|
|
grid_face *f;
|
|
int i;
|
|
|
|
if (old_type == new_type)
|
|
return FALSE;
|
|
|
|
g = state->game_grid;
|
|
f = g->faces + face;
|
|
|
|
for (i = 0; i < f->order; i++) {
|
|
int line_index = f->edges[i] - g->edges;
|
|
if (state->lines[line_index] == old_type) {
|
|
r = solver_set_line(sstate, line_index, new_type);
|
|
assert(r == TRUE);
|
|
retval = TRUE;
|
|
}
|
|
}
|
|
return retval;
|
|
}
|
|
|
|
/* ----------------------------------------------------------------------
|
|
* 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;
|
|
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. */
|
|
i = j = 0;
|
|
starting_face = test_face->dots[0]->faces[0];
|
|
if (starting_face == test_face) {
|
|
j = 1;
|
|
starting_face = test_face->dots[0]->faces[1];
|
|
}
|
|
current_face = starting_face;
|
|
transitions = 0;
|
|
current_state = (FACE_COLOUR(current_face) == colour);
|
|
|
|
do {
|
|
/* 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 (s != current_state) {
|
|
++transitions;
|
|
current_state = s;
|
|
if (transitions > 2)
|
|
return FALSE; /* no point in continuing */
|
|
}
|
|
} while (current_face != starting_face);
|
|
|
|
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;
|
|
|
|
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);
|
|
}
|
|
|
|
/* 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) {
|
|
/* No more lightable faces. Because of how the algorithm
|
|
* works, there should be no more darkable faces either. */
|
|
assert(c_darkable == 0);
|
|
break;
|
|
}
|
|
|
|
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);
|
|
|
|
/* 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);
|
|
for (i = 0; i < g->num_edges; i++) {
|
|
grid_edge *e = g->edges + i;
|
|
grid_face *f1 = e->face1;
|
|
grid_face *f2 = e->face2;
|
|
enum face_colour c1 = FACE_COLOUR(f1);
|
|
enum face_colour c2 = FACE_COLOUR(f2);
|
|
assert(c1 != FACE_GREY);
|
|
assert(c2 != FACE_GREY);
|
|
if (c1 != c2) {
|
|
if (f1) clues[f1 - g->faces]++;
|
|
if (f2) clues[f2 - g->faces]++;
|
|
}
|
|
}
|
|
|
|
sfree(board);
|
|
}
|
|
|
|
|
|
static int game_has_unique_soln(const game_state *state, int diff)
|
|
{
|
|
int ret;
|
|
solver_state *sstate_new;
|
|
solver_state *sstate = new_solver_state((game_state *)state, diff);
|
|
|
|
sstate_new = solve_game_rec(sstate);
|
|
|
|
assert(sstate_new->solver_status != SOLVER_MISTAKE);
|
|
ret = (sstate_new->solver_status == SOLVER_SOLVED);
|
|
|
|
free_solver_state(sstate_new);
|
|
free_solver_state(sstate);
|
|
|
|
return ret;
|
|
}
|
|
|
|
|
|
/* Remove clues one at a time at random. */
|
|
static game_state *remove_clues(game_state *state, random_state *rs,
|
|
int diff)
|
|
{
|
|
int *face_list;
|
|
int num_faces = state->game_grid->num_faces;
|
|
game_state *ret = dup_game(state), *saved_ret;
|
|
int n;
|
|
|
|
/* We need to remove some clues. We'll do this by forming a list of all
|
|
* available clues, shuffling it, then going along one at a
|
|
* time clearing each clue in turn for which doing so doesn't render the
|
|
* board unsolvable. */
|
|
face_list = snewn(num_faces, int);
|
|
for (n = 0; n < num_faces; ++n) {
|
|
face_list[n] = n;
|
|
}
|
|
|
|
shuffle(face_list, num_faces, sizeof(int), rs);
|
|
|
|
for (n = 0; n < num_faces; ++n) {
|
|
saved_ret = dup_game(ret);
|
|
ret->clues[face_list[n]] = -1;
|
|
|
|
if (game_has_unique_soln(ret, diff)) {
|
|
free_game(saved_ret);
|
|
} else {
|
|
free_game(ret);
|
|
ret = saved_ret;
|
|
}
|
|
}
|
|
sfree(face_list);
|
|
|
|
return ret;
|
|
}
|
|
|
|
|
|
static char *new_game_desc(game_params *params, random_state *rs,
|
|
char **aux, int interactive)
|
|
{
|
|
/* solution and description both use run-length encoding in obvious ways */
|
|
char *retval;
|
|
grid *g;
|
|
game_state *state = snew(game_state);
|
|
game_state *state_new;
|
|
params_generate_grid(params);
|
|
state->game_grid = g = params->game_grid;
|
|
g->refcount++;
|
|
state->clues = snewn(g->num_faces, signed char);
|
|
state->lines = snewn(g->num_edges, char);
|
|
state->line_errors = snewn(g->num_edges, unsigned char);
|
|
|
|
state->grid_type = params->type;
|
|
|
|
newboard_please:
|
|
|
|
memset(state->lines, LINE_UNKNOWN, g->num_edges);
|
|
memset(state->line_errors, 0, g->num_edges);
|
|
|
|
state->solved = state->cheated = FALSE;
|
|
|
|
/* Get a new random solvable board with all its clues filled in. Yes, this
|
|
* can loop for ever if the params are suitably unfavourable, but
|
|
* preventing games smaller than 4x4 seems to stop this happening */
|
|
do {
|
|
add_full_clues(state, rs);
|
|
} while (!game_has_unique_soln(state, params->diff));
|
|
|
|
state_new = remove_clues(state, rs, params->diff);
|
|
free_game(state);
|
|
state = state_new;
|
|
|
|
|
|
if (params->diff > 0 && game_has_unique_soln(state, params->diff-1)) {
|
|
#ifdef SHOW_WORKING
|
|
fprintf(stderr, "Rejecting board, it is too easy\n");
|
|
#endif
|
|
goto newboard_please;
|
|
}
|
|
|
|
retval = state_to_text(state);
|
|
|
|
free_game(state);
|
|
|
|
assert(!validate_desc(params, retval));
|
|
|
|
return retval;
|
|
}
|
|
|
|
static game_state *new_game(midend *me, game_params *params, char *desc)
|
|
{
|
|
int i;
|
|
game_state *state = snew(game_state);
|
|
int empties_to_make = 0;
|
|
int n;
|
|
const char *dp = desc;
|
|
grid *g;
|
|
int num_faces, num_edges;
|
|
|
|
params_generate_grid(params);
|
|
state->game_grid = g = params->game_grid;
|
|
g->refcount++;
|
|
num_faces = g->num_faces;
|
|
num_edges = g->num_edges;
|
|
|
|
state->clues = snewn(num_faces, signed char);
|
|
state->lines = snewn(num_edges, char);
|
|
state->line_errors = snewn(num_edges, unsigned char);
|
|
|
|
state->solved = state->cheated = FALSE;
|
|
|
|
state->grid_type = params->type;
|
|
|
|
for (i = 0; i < num_faces; i++) {
|
|
if (empties_to_make) {
|
|
empties_to_make--;
|
|
state->clues[i] = -1;
|
|
continue;
|
|
}
|
|
|
|
assert(*dp);
|
|
n = *dp - '0';
|
|
if (n >= 0 && n < 10) {
|
|
state->clues[i] = n;
|
|
} else {
|
|
n = *dp - 'a' + 1;
|
|
assert(n > 0);
|
|
state->clues[i] = -1;
|
|
empties_to_make = n - 1;
|
|
}
|
|
++dp;
|
|
}
|
|
|
|
memset(state->lines, LINE_UNKNOWN, num_edges);
|
|
memset(state->line_errors, 0, num_edges);
|
|
return state;
|
|
}
|
|
|
|
/* Calculates the line_errors data, and checks if the current state is a
|
|
* solution */
|
|
static int check_completion(game_state *state)
|
|
{
|
|
grid *g = state->game_grid;
|
|
int *dsf;
|
|
int num_faces = g->num_faces;
|
|
int i;
|
|
int infinite_area, finite_area;
|
|
int loops_found = 0;
|
|
int found_edge_not_in_loop = FALSE;
|
|
|
|
memset(state->line_errors, 0, g->num_edges);
|
|
|
|
/* LL implementation of SGT's idea:
|
|
* A loop will partition the grid into an inside and an outside.
|
|
* If there is more than one loop, the grid will be partitioned into
|
|
* even more distinct regions. We can therefore track equivalence of
|
|
* faces, by saying that two faces are equivalent when there is a non-YES
|
|
* edge between them.
|
|
* We could keep track of the number of connected components, by counting
|
|
* the number of dsf-merges that aren't no-ops.
|
|
* But we're only interested in 3 separate cases:
|
|
* no loops, one loop, more than one loop.
|
|
*
|
|
* No loops: all faces are equivalent to the infinite face.
|
|
* One loop: only two equivalence classes - finite and infinite.
|
|
* >= 2 loops: there are 2 distinct finite regions.
|
|
*
|
|
* So we simply make two passes through all the edges.
|
|
* In the first pass, we dsf-merge the two faces bordering each non-YES
|
|
* edge.
|
|
* In the second pass, we look for YES-edges bordering:
|
|
* a) two non-equivalent faces.
|
|
* b) two non-equivalent faces, and one of them is part of a different
|
|
* finite area from the first finite area we've seen.
|
|
*
|
|
* An occurrence of a) means there is at least one loop.
|
|
* An occurrence of b) means there is more than one loop.
|
|
* Edges satisfying a) are marked as errors.
|
|
*
|
|
* While we're at it, we set a flag if we find a YES edge that is not
|
|
* part of a loop.
|
|
* This information will help decide, if there's a single loop, whether it
|
|
* is a candidate for being a solution (that is, all YES edges are part of
|
|
* this loop).
|
|
*
|
|
* If there is a candidate loop, we then go through all clues and check
|
|
* they are all satisfied. If so, we have found a solution and we can
|
|
* unmark all line_errors.
|
|
*/
|
|
|
|
/* Infinite face is at the end - its index is num_faces.
|
|
* This macro is just to make this obvious! */
|
|
#define INF_FACE num_faces
|
|
dsf = snewn(num_faces + 1, int);
|
|
dsf_init(dsf, num_faces + 1);
|
|
|
|
/* First pass */
|
|
for (i = 0; i < g->num_edges; i++) {
|
|
grid_edge *e = g->edges + i;
|
|
int f1 = e->face1 ? e->face1 - g->faces : INF_FACE;
|
|
int f2 = e->face2 ? e->face2 - g->faces : INF_FACE;
|
|
if (state->lines[i] != LINE_YES)
|
|
dsf_merge(dsf, f1, f2);
|
|
}
|
|
|
|
/* Second pass */
|
|
infinite_area = dsf_canonify(dsf, INF_FACE);
|
|
finite_area = -1;
|
|
for (i = 0; i < g->num_edges; i++) {
|
|
grid_edge *e = g->edges + i;
|
|
int f1 = e->face1 ? e->face1 - g->faces : INF_FACE;
|
|
int can1 = dsf_canonify(dsf, f1);
|
|
int f2 = e->face2 ? e->face2 - g->faces : INF_FACE;
|
|
int can2 = dsf_canonify(dsf, f2);
|
|
if (state->lines[i] != LINE_YES) continue;
|
|
|
|
if (can1 == can2) {
|
|
/* Faces are equivalent, so this edge not part of a loop */
|
|
found_edge_not_in_loop = TRUE;
|
|
continue;
|
|
}
|
|
state->line_errors[i] = TRUE;
|
|
if (loops_found == 0) loops_found = 1;
|
|
|
|
/* Don't bother with further checks if we've already found 2 loops */
|
|
if (loops_found == 2) continue;
|
|
|
|
if (finite_area == -1) {
|
|
/* Found our first finite area */
|
|
if (can1 != infinite_area)
|
|
finite_area = can1;
|
|
else
|
|
finite_area = can2;
|
|
}
|
|
|
|
/* Have we found a second area? */
|
|
if (finite_area != -1) {
|
|
if (can1 != infinite_area && can1 != finite_area) {
|
|
loops_found = 2;
|
|
continue;
|
|
}
|
|
if (can2 != infinite_area && can2 != finite_area) {
|
|
loops_found = 2;
|
|
}
|
|
}
|
|
}
|
|
|
|
/*
|
|
printf("loops_found = %d\n", loops_found);
|
|
printf("found_edge_not_in_loop = %s\n",
|
|
found_edge_not_in_loop ? "TRUE" : "FALSE");
|
|
*/
|
|
|
|
sfree(dsf); /* No longer need the dsf */
|
|
|
|
/* Have we found a candidate loop? */
|
|
if (loops_found == 1 && !found_edge_not_in_loop) {
|
|
/* Yes, so check all clues are satisfied */
|
|
int found_clue_violation = FALSE;
|
|
for (i = 0; i < num_faces; i++) {
|
|
int c = state->clues[i];
|
|
if (c >= 0) {
|
|
if (face_order(state, i, LINE_YES) != c) {
|
|
found_clue_violation = TRUE;
|
|
break;
|
|
}
|
|
}
|
|
}
|
|
|
|
if (!found_clue_violation) {
|
|
/* The loop is good */
|
|
memset(state->line_errors, 0, g->num_edges);
|
|
return TRUE; /* No need to bother checking for dot violations */
|
|
}
|
|
}
|
|
|
|
/* Check for dot violations */
|
|
for (i = 0; i < g->num_dots; i++) {
|
|
int yes = dot_order(state, i, LINE_YES);
|
|
int unknown = dot_order(state, i, LINE_UNKNOWN);
|
|
if ((yes == 1 && unknown == 0) || (yes >= 3)) {
|
|
/* violation, so mark all YES edges as errors */
|
|
grid_dot *d = g->dots + i;
|
|
int j;
|
|
for (j = 0; j < d->order; j++) {
|
|
int e = d->edges[j] - g->edges;
|
|
if (state->lines[e] == LINE_YES)
|
|
state->line_errors[e] = TRUE;
|
|
}
|
|
}
|
|
}
|
|
return FALSE;
|
|
}
|
|
|
|
/* ----------------------------------------------------------------------
|
|
* Solver logic
|
|
*
|
|
* Our solver modes operate as follows. Each mode also uses the modes above it.
|
|
*
|
|
* Easy Mode
|
|
* Just implement the rules of the game.
|
|
*
|
|
* Normal and Tricky Modes
|
|
* For each (adjacent) pair of lines through each dot we store a bit for
|
|
* whether at least one of them is on and whether at most one is on. (If we
|
|
* know both or neither is on that's already stored more directly.)
|
|
*
|
|
* Advanced Mode
|
|
* Use edsf data structure to make equivalence classes of lines that are
|
|
* known identical to or opposite to one another.
|
|
*/
|
|
|
|
|
|
/* DLines:
|
|
* For general grids, we consider "dlines" to be pairs of lines joined
|
|
* at a dot. The lines must be adjacent around the dot, so we can think of
|
|
* a dline as being a dot+face combination. Or, a dot+edge combination where
|
|
* the second edge is taken to be the next clockwise edge from the dot.
|
|
* Original loopy code didn't have this extra restriction of the lines being
|
|
* adjacent. From my tests with square grids, this extra restriction seems to
|
|
* take little, if anything, away from the quality of the puzzles.
|
|
* A dline can be uniquely identified by an edge/dot combination, given that
|
|
* a dline-pair always goes clockwise around its common dot. The edge/dot
|
|
* combination can be represented by an edge/bool combination - if bool is
|
|
* TRUE, use edge->dot1 else use edge->dot2. So the total number of dlines is
|
|
* exactly twice the number of edges in the grid - although the dlines
|
|
* spanning the infinite face are not all that useful to the solver.
|
|
* Note that, by convention, a dline goes clockwise around its common dot,
|
|
* which means the dline goes anti-clockwise around its common face.
|
|
*/
|
|
|
|
/* Helper functions for obtaining an index into an array of dlines, given
|
|
* various information. We assume the grid layout conventions about how
|
|
* the various lists are interleaved - see grid_make_consistent() for
|
|
* details. */
|
|
|
|
/* i points to the first edge of the dline pair, reading clockwise around
|
|
* the dot. */
|
|
static int dline_index_from_dot(grid *g, grid_dot *d, int i)
|
|
{
|
|
grid_edge *e = d->edges[i];
|
|
int ret;
|
|
#ifdef DEBUG_DLINES
|
|
grid_edge *e2;
|
|
int i2 = i+1;
|
|
if (i2 == d->order) i2 = 0;
|
|
e2 = d->edges[i2];
|
|
#endif
|
|
ret = 2 * (e - g->edges) + ((e->dot1 == d) ? 1 : 0);
|
|
#ifdef DEBUG_DLINES
|
|
printf("dline_index_from_dot: d=%d,i=%d, edges [%d,%d] - %d\n",
|
|
(int)(d - g->dots), i, (int)(e - g->edges),
|
|
(int)(e2 - g->edges), ret);
|
|
#endif
|
|
return ret;
|
|
}
|
|
/* i points to the second edge of the dline pair, reading clockwise around
|
|
* the face. That is, the edges of the dline, starting at edge{i}, read
|
|
* anti-clockwise around the face. By layout conventions, the common dot
|
|
* of the dline will be f->dots[i] */
|
|
static int dline_index_from_face(grid *g, grid_face *f, int i)
|
|
{
|
|
grid_edge *e = f->edges[i];
|
|
grid_dot *d = f->dots[i];
|
|
int ret;
|
|
#ifdef DEBUG_DLINES
|
|
grid_edge *e2;
|
|
int i2 = i - 1;
|
|
if (i2 < 0) i2 += f->order;
|
|
e2 = f->edges[i2];
|
|
#endif
|
|
ret = 2 * (e - g->edges) + ((e->dot1 == d) ? 1 : 0);
|
|
#ifdef DEBUG_DLINES
|
|
printf("dline_index_from_face: f=%d,i=%d, edges [%d,%d] - %d\n",
|
|
(int)(f - g->faces), i, (int)(e - g->edges),
|
|
(int)(e2 - g->edges), ret);
|
|
#endif
|
|
return ret;
|
|
}
|
|
static int is_atleastone(const char *dline_array, int index)
|
|
{
|
|
return BIT_SET(dline_array[index], 0);
|
|
}
|
|
static int set_atleastone(char *dline_array, int index)
|
|
{
|
|
return SET_BIT(dline_array[index], 0);
|
|
}
|
|
static int is_atmostone(const char *dline_array, int index)
|
|
{
|
|
return BIT_SET(dline_array[index], 1);
|
|
}
|
|
static int set_atmostone(char *dline_array, int index)
|
|
{
|
|
return SET_BIT(dline_array[index], 1);
|
|
}
|
|
|
|
static void array_setall(char *array, char from, char to, int len)
|
|
{
|
|
char *p = array, *p_old = p;
|
|
int len_remaining = len;
|
|
|
|
while ((p = memchr(p, from, len_remaining))) {
|
|
*p = to;
|
|
len_remaining -= p - p_old;
|
|
p_old = p;
|
|
}
|
|
}
|
|
|
|
/* Helper, called when doing dline dot deductions, in the case where we
|
|
* have 4 UNKNOWNs, and two of them (adjacent) have *exactly* one YES between
|
|
* them (because of dline atmostone/atleastone).
|
|
* On entry, edge points to the first of these two UNKNOWNs. This function
|
|
* will find the opposite UNKNOWNS (if they are adjacent to one another)
|
|
* and set their corresponding dline to atleastone. (Setting atmostone
|
|
* already happens in earlier dline deductions) */
|
|
static int dline_set_opp_atleastone(solver_state *sstate,
|
|
grid_dot *d, int edge)
|
|
{
|
|
game_state *state = sstate->state;
|
|
grid *g = state->game_grid;
|
|
int N = d->order;
|
|
int opp, opp2;
|
|
for (opp = 0; opp < N; opp++) {
|
|
int opp_dline_index;
|
|
if (opp == edge || opp == edge+1 || opp == edge-1)
|
|
continue;
|
|
if (opp == 0 && edge == N-1)
|
|
continue;
|
|
if (opp == N-1 && edge == 0)
|
|
continue;
|
|
opp2 = opp + 1;
|
|
if (opp2 == N) opp2 = 0;
|
|
/* Check if opp, opp2 point to LINE_UNKNOWNs */
|
|
if (state->lines[d->edges[opp] - g->edges] != LINE_UNKNOWN)
|
|
continue;
|
|
if (state->lines[d->edges[opp2] - g->edges] != LINE_UNKNOWN)
|
|
continue;
|
|
/* Found opposite UNKNOWNS and they're next to each other */
|
|
opp_dline_index = dline_index_from_dot(g, d, opp);
|
|
return set_atleastone(sstate->dlines, opp_dline_index);
|
|
}
|
|
return FALSE;
|
|
}
|
|
|
|
|
|
/* Set pairs of lines around this face which are known to be identical, to
|
|
* the given line_state */
|
|
static int face_setall_identical(solver_state *sstate, int face_index,
|
|
enum line_state line_new)
|
|
{
|
|
/* can[dir] contains the canonical line associated with the line in
|
|
* direction dir from the square in question. Similarly inv[dir] is
|
|
* whether or not the line in question is inverse to its canonical
|
|
* element. */
|
|
int retval = FALSE;
|
|
game_state *state = sstate->state;
|
|
grid *g = state->game_grid;
|
|
grid_face *f = g->faces + face_index;
|
|
int N = f->order;
|
|
int i, j;
|
|
int can1, can2, inv1, inv2;
|
|
|
|
for (i = 0; i < N; i++) {
|
|
int line1_index = f->edges[i] - g->edges;
|
|
if (state->lines[line1_index] != LINE_UNKNOWN)
|
|
continue;
|
|
for (j = i + 1; j < N; j++) {
|
|
int line2_index = f->edges[j] - g->edges;
|
|
if (state->lines[line2_index] != LINE_UNKNOWN)
|
|
continue;
|
|
|
|
/* Found two UNKNOWNS */
|
|
can1 = edsf_canonify(sstate->linedsf, line1_index, &inv1);
|
|
can2 = edsf_canonify(sstate->linedsf, line2_index, &inv2);
|
|
if (can1 == can2 && inv1 == inv2) {
|
|
solver_set_line(sstate, line1_index, line_new);
|
|
solver_set_line(sstate, line2_index, line_new);
|
|
}
|
|
}
|
|
}
|
|
return retval;
|
|
}
|
|
|
|
/* Given a dot or face, and a count of LINE_UNKNOWNs, find them and
|
|
* return the edge indices into e. */
|
|
static void find_unknowns(game_state *state,
|
|
grid_edge **edge_list, /* Edge list to search (from a face or a dot) */
|
|
int expected_count, /* Number of UNKNOWNs (comes from solver's cache) */
|
|
int *e /* Returned edge indices */)
|
|
{
|
|
int c = 0;
|
|
grid *g = state->game_grid;
|
|
while (c < expected_count) {
|
|
int line_index = *edge_list - g->edges;
|
|
if (state->lines[line_index] == LINE_UNKNOWN) {
|
|
e[c] = line_index;
|
|
c++;
|
|
}
|
|
++edge_list;
|
|
}
|
|
}
|
|
|
|
/* If we have a list of edges, and we know whether the number of YESs should
|
|
* be odd or even, and there are only a few UNKNOWNs, we can do some simple
|
|
* linedsf deductions. This can be used for both face and dot deductions.
|
|
* Returns the difficulty level of the next solver that should be used,
|
|
* or DIFF_MAX if no progress was made. */
|
|
static int parity_deductions(solver_state *sstate,
|
|
grid_edge **edge_list, /* Edge list (from a face or a dot) */
|
|
int total_parity, /* Expected number of YESs modulo 2 (either 0 or 1) */
|
|
int unknown_count)
|
|
{
|
|
game_state *state = sstate->state;
|
|
int diff = DIFF_MAX;
|
|
int *linedsf = sstate->linedsf;
|
|
|
|
if (unknown_count == 2) {
|
|
/* Lines are known alike/opposite, depending on inv. */
|
|
int e[2];
|
|
find_unknowns(state, edge_list, 2, e);
|
|
if (merge_lines(sstate, e[0], e[1], total_parity))
|
|
diff = min(diff, DIFF_HARD);
|
|
} else if (unknown_count == 3) {
|
|
int e[3];
|
|
int can[3]; /* canonical edges */
|
|
int inv[3]; /* whether can[x] is inverse to e[x] */
|
|
find_unknowns(state, edge_list, 3, e);
|
|
can[0] = edsf_canonify(linedsf, e[0], inv);
|
|
can[1] = edsf_canonify(linedsf, e[1], inv+1);
|
|
can[2] = edsf_canonify(linedsf, e[2], inv+2);
|
|
if (can[0] == can[1]) {
|
|
if (solver_set_line(sstate, e[2], (total_parity^inv[0]^inv[1]) ?
|
|
LINE_YES : LINE_NO))
|
|
diff = min(diff, DIFF_EASY);
|
|
}
|
|
if (can[0] == can[2]) {
|
|
if (solver_set_line(sstate, e[1], (total_parity^inv[0]^inv[2]) ?
|
|
LINE_YES : LINE_NO))
|
|
diff = min(diff, DIFF_EASY);
|
|
}
|
|
if (can[1] == can[2]) {
|
|
if (solver_set_line(sstate, e[0], (total_parity^inv[1]^inv[2]) ?
|
|
LINE_YES : LINE_NO))
|
|
diff = min(diff, DIFF_EASY);
|
|
}
|
|
} else if (unknown_count == 4) {
|
|
int e[4];
|
|
int can[4]; /* canonical edges */
|
|
int inv[4]; /* whether can[x] is inverse to e[x] */
|
|
find_unknowns(state, edge_list, 4, e);
|
|
can[0] = edsf_canonify(linedsf, e[0], inv);
|
|
can[1] = edsf_canonify(linedsf, e[1], inv+1);
|
|
can[2] = edsf_canonify(linedsf, e[2], inv+2);
|
|
can[3] = edsf_canonify(linedsf, e[3], inv+3);
|
|
if (can[0] == can[1]) {
|
|
if (merge_lines(sstate, e[2], e[3], total_parity^inv[0]^inv[1]))
|
|
diff = min(diff, DIFF_HARD);
|
|
} else if (can[0] == can[2]) {
|
|
if (merge_lines(sstate, e[1], e[3], total_parity^inv[0]^inv[2]))
|
|
diff = min(diff, DIFF_HARD);
|
|
} else if (can[0] == can[3]) {
|
|
if (merge_lines(sstate, e[1], e[2], total_parity^inv[0]^inv[3]))
|
|
diff = min(diff, DIFF_HARD);
|
|
} else if (can[1] == can[2]) {
|
|
if (merge_lines(sstate, e[0], e[3], total_parity^inv[1]^inv[2]))
|
|
diff = min(diff, DIFF_HARD);
|
|
} else if (can[1] == can[3]) {
|
|
if (merge_lines(sstate, e[0], e[2], total_parity^inv[1]^inv[3]))
|
|
diff = min(diff, DIFF_HARD);
|
|
} else if (can[2] == can[3]) {
|
|
if (merge_lines(sstate, e[0], e[1], total_parity^inv[2]^inv[3]))
|
|
diff = min(diff, DIFF_HARD);
|
|
}
|
|
}
|
|
return diff;
|
|
}
|
|
|
|
|
|
/*
|
|
* These are the main solver functions.
|
|
*
|
|
* Their return values are diff values corresponding to the lowest mode solver
|
|
* that would notice the work that they have done. For example if the normal
|
|
* mode solver adds actual lines or crosses, it will return DIFF_EASY as the
|
|
* easy mode solver might be able to make progress using that. It doesn't make
|
|
* sense for one of them to return a diff value higher than that of the
|
|
* function itself.
|
|
*
|
|
* Each function returns the lowest value it can, as early as possible, in
|
|
* order to try and pass as much work as possible back to the lower level
|
|
* solvers which progress more quickly.
|
|
*/
|
|
|
|
/* PROPOSED NEW DESIGN:
|
|
* We have a work queue consisting of 'events' notifying us that something has
|
|
* happened that a particular solver mode might be interested in. For example
|
|
* the hard mode solver might do something that helps the normal mode solver at
|
|
* dot [x,y] in which case it will enqueue an event recording this fact. Then
|
|
* we pull events off the work queue, and hand each in turn to the solver that
|
|
* is interested in them. If a solver reports that it failed we pass the same
|
|
* event on to progressively more advanced solvers and the loop detector. Once
|
|
* we've exhausted an event, or it has helped us progress, we drop it and
|
|
* continue to the next one. The events are sorted first in order of solver
|
|
* complexity (easy first) then order of insertion (oldest first).
|
|
* Once we run out of events we loop over each permitted solver in turn
|
|
* (easiest first) until either a deduction is made (and an event therefore
|
|
* emerges) or no further deductions can be made (in which case we've failed).
|
|
*
|
|
* QUESTIONS:
|
|
* * How do we 'loop over' a solver when both dots and squares are concerned.
|
|
* Answer: first all squares then all dots.
|
|
*/
|
|
|
|
static int trivial_deductions(solver_state *sstate)
|
|
{
|
|
int i, current_yes, current_no;
|
|
game_state *state = sstate->state;
|
|
grid *g = state->game_grid;
|
|
int diff = DIFF_MAX;
|
|
|
|
/* Per-face deductions */
|
|
for (i = 0; i < g->num_faces; i++) {
|
|
grid_face *f = g->faces + i;
|
|
|
|
if (sstate->face_solved[i])
|
|
continue;
|
|
|
|
current_yes = sstate->face_yes_count[i];
|
|
current_no = sstate->face_no_count[i];
|
|
|
|
if (current_yes + current_no == f->order) {
|
|
sstate->face_solved[i] = TRUE;
|
|
continue;
|
|
}
|
|
|
|
if (state->clues[i] < 0)
|
|
continue;
|
|
|
|
if (state->clues[i] < current_yes) {
|
|
sstate->solver_status = SOLVER_MISTAKE;
|
|
return DIFF_EASY;
|
|
}
|
|
if (state->clues[i] == current_yes) {
|
|
if (face_setall(sstate, i, LINE_UNKNOWN, LINE_NO))
|
|
diff = min(diff, DIFF_EASY);
|
|
sstate->face_solved[i] = TRUE;
|
|
continue;
|
|
}
|
|
|
|
if (f->order - state->clues[i] < current_no) {
|
|
sstate->solver_status = SOLVER_MISTAKE;
|
|
return DIFF_EASY;
|
|
}
|
|
if (f->order - state->clues[i] == current_no) {
|
|
if (face_setall(sstate, i, LINE_UNKNOWN, LINE_YES))
|
|
diff = min(diff, DIFF_EASY);
|
|
sstate->face_solved[i] = TRUE;
|
|
continue;
|
|
}
|
|
}
|
|
|
|
check_caches(sstate);
|
|
|
|
/* Per-dot deductions */
|
|
for (i = 0; i < g->num_dots; i++) {
|
|
grid_dot *d = g->dots + i;
|
|
int yes, no, unknown;
|
|
|
|
if (sstate->dot_solved[i])
|
|
continue;
|
|
|
|
yes = sstate->dot_yes_count[i];
|
|
no = sstate->dot_no_count[i];
|
|
unknown = d->order - yes - no;
|
|
|
|
if (yes == 0) {
|
|
if (unknown == 0) {
|
|
sstate->dot_solved[i] = TRUE;
|
|
} else if (unknown == 1) {
|
|
dot_setall(sstate, i, LINE_UNKNOWN, LINE_NO);
|
|
diff = min(diff, DIFF_EASY);
|
|
sstate->dot_solved[i] = TRUE;
|
|
}
|
|
} else if (yes == 1) {
|
|
if (unknown == 0) {
|
|
sstate->solver_status = SOLVER_MISTAKE;
|
|
return DIFF_EASY;
|
|
} else if (unknown == 1) {
|
|
dot_setall(sstate, i, LINE_UNKNOWN, LINE_YES);
|
|
diff = min(diff, DIFF_EASY);
|
|
}
|
|
} else if (yes == 2) {
|
|
if (unknown > 0) {
|
|
dot_setall(sstate, i, LINE_UNKNOWN, LINE_NO);
|
|
diff = min(diff, DIFF_EASY);
|
|
}
|
|
sstate->dot_solved[i] = TRUE;
|
|
} else {
|
|
sstate->solver_status = SOLVER_MISTAKE;
|
|
return DIFF_EASY;
|
|
}
|
|
}
|
|
|
|
check_caches(sstate);
|
|
|
|
return diff;
|
|
}
|
|
|
|
static int dline_deductions(solver_state *sstate)
|
|
{
|
|
game_state *state = sstate->state;
|
|
grid *g = state->game_grid;
|
|
char *dlines = sstate->dlines;
|
|
int i;
|
|
int diff = DIFF_MAX;
|
|
|
|
/* ------ Face deductions ------ */
|
|
|
|
/* Given a set of dline atmostone/atleastone constraints, need to figure
|
|
* out if we can deduce any further info. For more general faces than
|
|
* squares, this turns out to be a tricky problem.
|
|
* The approach taken here is to define (per face) NxN matrices:
|
|
* "maxs" and "mins".
|
|
* The entries maxs(j,k) and mins(j,k) define the upper and lower limits
|
|
* for the possible number of edges that are YES between positions j and k
|
|
* going clockwise around the face. Can think of j and k as marking dots
|
|
* around the face (recall the labelling scheme: edge0 joins dot0 to dot1,
|
|
* edge1 joins dot1 to dot2 etc).
|
|
* Trivially, mins(j,j) = maxs(j,j) = 0, and we don't even bother storing
|
|
* these. mins(j,j+1) and maxs(j,j+1) are determined by whether edge{j}
|
|
* is YES, NO or UNKNOWN. mins(j,j+2) and maxs(j,j+2) are related to
|
|
* the dline atmostone/atleastone status for edges j and j+1.
|
|
*
|
|
* Then we calculate the remaining entries recursively. We definitely
|
|
* know that
|
|
* mins(j,k) >= { mins(j,u) + mins(u,k) } for any u between j and k.
|
|
* This is because any valid placement of YESs between j and k must give
|
|
* a valid placement between j and u, and also between u and k.
|
|
* I believe it's sufficient to use just the two values of u:
|
|
* j+1 and j+2. Seems to work well in practice - the bounds we compute
|
|
* are rigorous, even if they might not be best-possible.
|
|
*
|
|
* Once we have maxs and mins calculated, we can make inferences about
|
|
* each dline{j,j+1} by looking at the possible complementary edge-counts
|
|
* mins(j+2,j) and maxs(j+2,j) and comparing these with the face clue.
|
|
* As well as dlines, we can make similar inferences about single edges.
|
|
* For example, consider a pentagon with clue 3, and we know at most one
|
|
* of (edge0, edge1) is YES, and at most one of (edge2, edge3) is YES.
|
|
* We could then deduce edge4 is YES, because maxs(0,4) would be 2, so
|
|
* that final edge would have to be YES to make the count up to 3.
|
|
*/
|
|
|
|
/* Much quicker to allocate arrays on the stack than the heap, so
|
|
* define the largest possible face size, and base our array allocations
|
|
* on that. We check this with an assertion, in case someone decides to
|
|
* make a grid which has larger faces than this. Note, this algorithm
|
|
* could get quite expensive if there are many large faces. */
|
|
#define MAX_FACE_SIZE 8
|
|
|
|
for (i = 0; i < g->num_faces; i++) {
|
|
int maxs[MAX_FACE_SIZE][MAX_FACE_SIZE];
|
|
int mins[MAX_FACE_SIZE][MAX_FACE_SIZE];
|
|
grid_face *f = g->faces + i;
|
|
int N = f->order;
|
|
int j,m;
|
|
int clue = state->clues[i];
|
|
assert(N <= MAX_FACE_SIZE);
|
|
if (sstate->face_solved[i])
|
|
continue;
|
|
if (clue < 0) continue;
|
|
|
|
/* Calculate the (j,j+1) entries */
|
|
for (j = 0; j < N; j++) {
|
|
int edge_index = f->edges[j] - g->edges;
|
|
int dline_index;
|
|
enum line_state line1 = state->lines[edge_index];
|
|
enum line_state line2;
|
|
int tmp;
|
|
int k = j + 1;
|
|
if (k >= N) k = 0;
|
|
maxs[j][k] = (line1 == LINE_NO) ? 0 : 1;
|
|
mins[j][k] = (line1 == LINE_YES) ? 1 : 0;
|
|
/* Calculate the (j,j+2) entries */
|
|
dline_index = dline_index_from_face(g, f, k);
|
|
edge_index = f->edges[k] - g->edges;
|
|
line2 = state->lines[edge_index];
|
|
k++;
|
|
if (k >= N) k = 0;
|
|
|
|
/* max */
|
|
tmp = 2;
|
|
if (line1 == LINE_NO) tmp--;
|
|
if (line2 == LINE_NO) tmp--;
|
|
if (tmp == 2 && is_atmostone(dlines, dline_index))
|
|
tmp = 1;
|
|
maxs[j][k] = tmp;
|
|
|
|
/* min */
|
|
tmp = 0;
|
|
if (line1 == LINE_YES) tmp++;
|
|
if (line2 == LINE_YES) tmp++;
|
|
if (tmp == 0 && is_atleastone(dlines, dline_index))
|
|
tmp = 1;
|
|
mins[j][k] = tmp;
|
|
}
|
|
|
|
/* Calculate the (j,j+m) entries for m between 3 and N-1 */
|
|
for (m = 3; m < N; m++) {
|
|
for (j = 0; j < N; j++) {
|
|
int k = j + m;
|
|
int u = j + 1;
|
|
int v = j + 2;
|
|
int tmp;
|
|
if (k >= N) k -= N;
|
|
if (u >= N) u -= N;
|
|
if (v >= N) v -= N;
|
|
maxs[j][k] = maxs[j][u] + maxs[u][k];
|
|
mins[j][k] = mins[j][u] + mins[u][k];
|
|
tmp = maxs[j][v] + maxs[v][k];
|
|
maxs[j][k] = min(maxs[j][k], tmp);
|
|
tmp = mins[j][v] + mins[v][k];
|
|
mins[j][k] = max(mins[j][k], tmp);
|
|
}
|
|
}
|
|
|
|
/* See if we can make any deductions */
|
|
for (j = 0; j < N; j++) {
|
|
int k;
|
|
grid_edge *e = f->edges[j];
|
|
int line_index = e - g->edges;
|
|
int dline_index;
|
|
|
|
if (state->lines[line_index] != LINE_UNKNOWN)
|
|
continue;
|
|
k = j + 1;
|
|
if (k >= N) k = 0;
|
|
|
|
/* minimum YESs in the complement of this edge */
|
|
if (mins[k][j] > clue) {
|
|
sstate->solver_status = SOLVER_MISTAKE;
|
|
return DIFF_EASY;
|
|
}
|
|
if (mins[k][j] == clue) {
|
|
/* setting this edge to YES would make at least
|
|
* (clue+1) edges - contradiction */
|
|
solver_set_line(sstate, line_index, LINE_NO);
|
|
diff = min(diff, DIFF_EASY);
|
|
}
|
|
if (maxs[k][j] < clue - 1) {
|
|
sstate->solver_status = SOLVER_MISTAKE;
|
|
return DIFF_EASY;
|
|
}
|
|
if (maxs[k][j] == clue - 1) {
|
|
/* Only way to satisfy the clue is to set edge{j} as YES */
|
|
solver_set_line(sstate, line_index, LINE_YES);
|
|
diff = min(diff, DIFF_EASY);
|
|
}
|
|
|
|
/* More advanced deduction that allows propagation along diagonal
|
|
* chains of faces connected by dots, for example, 3-2-...-2-3
|
|
* in square grids. */
|
|
if (sstate->diff >= DIFF_TRICKY) {
|
|
/* Now see if we can make dline deduction for edges{j,j+1} */
|
|
e = f->edges[k];
|
|
if (state->lines[e - g->edges] != LINE_UNKNOWN)
|
|
/* Only worth doing this for an UNKNOWN,UNKNOWN pair.
|
|
* Dlines where one of the edges is known, are handled in the
|
|
* dot-deductions */
|
|
continue;
|
|
|
|
dline_index = dline_index_from_face(g, f, k);
|
|
k++;
|
|
if (k >= N) k = 0;
|
|
|
|
/* minimum YESs in the complement of this dline */
|
|
if (mins[k][j] > clue - 2) {
|
|
/* Adding 2 YESs would break the clue */
|
|
if (set_atmostone(dlines, dline_index))
|
|
diff = min(diff, DIFF_NORMAL);
|
|
}
|
|
/* maximum YESs in the complement of this dline */
|
|
if (maxs[k][j] < clue) {
|
|
/* Adding 2 NOs would mean not enough YESs */
|
|
if (set_atleastone(dlines, dline_index))
|
|
diff = min(diff, DIFF_NORMAL);
|
|
}
|
|
}
|
|
}
|
|
}
|
|
|
|
if (diff < DIFF_NORMAL)
|
|
return diff;
|
|
|
|
/* ------ Dot deductions ------ */
|
|
|
|
for (i = 0; i < g->num_dots; i++) {
|
|
grid_dot *d = g->dots + i;
|
|
int N = d->order;
|
|
int yes, no, unknown;
|
|
int j;
|
|
if (sstate->dot_solved[i])
|
|
continue;
|
|
yes = sstate->dot_yes_count[i];
|
|
no = sstate->dot_no_count[i];
|
|
unknown = N - yes - no;
|
|
|
|
for (j = 0; j < N; j++) {
|
|
int k;
|
|
int dline_index;
|
|
int line1_index, line2_index;
|
|
enum line_state line1, line2;
|
|
k = j + 1;
|
|
if (k >= N) k = 0;
|
|
dline_index = dline_index_from_dot(g, d, j);
|
|
line1_index = d->edges[j] - g->edges;
|
|
line2_index = d->edges[k] - g->edges;
|
|
line1 = state->lines[line1_index];
|
|
line2 = state->lines[line2_index];
|
|
|
|
/* Infer dline state from line state */
|
|
if (line1 == LINE_NO || line2 == LINE_NO) {
|
|
if (set_atmostone(dlines, dline_index))
|
|
diff = min(diff, DIFF_NORMAL);
|
|
}
|
|
if (line1 == LINE_YES || line2 == LINE_YES) {
|
|
if (set_atleastone(dlines, dline_index))
|
|
diff = min(diff, DIFF_NORMAL);
|
|
}
|
|
/* Infer line state from dline state */
|
|
if (is_atmostone(dlines, dline_index)) {
|
|
if (line1 == LINE_YES && line2 == LINE_UNKNOWN) {
|
|
solver_set_line(sstate, line2_index, LINE_NO);
|
|
diff = min(diff, DIFF_EASY);
|
|
}
|
|
if (line2 == LINE_YES && line1 == LINE_UNKNOWN) {
|
|
solver_set_line(sstate, line1_index, LINE_NO);
|
|
diff = min(diff, DIFF_EASY);
|
|
}
|
|
}
|
|
if (is_atleastone(dlines, dline_index)) {
|
|
if (line1 == LINE_NO && line2 == LINE_UNKNOWN) {
|
|
solver_set_line(sstate, line2_index, LINE_YES);
|
|
diff = min(diff, DIFF_EASY);
|
|
}
|
|
if (line2 == LINE_NO && line1 == LINE_UNKNOWN) {
|
|
solver_set_line(sstate, line1_index, LINE_YES);
|
|
diff = min(diff, DIFF_EASY);
|
|
}
|
|
}
|
|
/* Deductions that depend on the numbers of lines.
|
|
* Only bother if both lines are UNKNOWN, otherwise the
|
|
* easy-mode solver (or deductions above) would have taken
|
|
* care of it. */
|
|
if (line1 != LINE_UNKNOWN || line2 != LINE_UNKNOWN)
|
|
continue;
|
|
|
|
if (yes == 0 && unknown == 2) {
|
|
/* Both these unknowns must be identical. If we know
|
|
* atmostone or atleastone, we can make progress. */
|
|
if (is_atmostone(dlines, dline_index)) {
|
|
solver_set_line(sstate, line1_index, LINE_NO);
|
|
solver_set_line(sstate, line2_index, LINE_NO);
|
|
diff = min(diff, DIFF_EASY);
|
|
}
|
|
if (is_atleastone(dlines, dline_index)) {
|
|
solver_set_line(sstate, line1_index, LINE_YES);
|
|
solver_set_line(sstate, line2_index, LINE_YES);
|
|
diff = min(diff, DIFF_EASY);
|
|
}
|
|
}
|
|
if (yes == 1) {
|
|
if (set_atmostone(dlines, dline_index))
|
|
diff = min(diff, DIFF_NORMAL);
|
|
if (unknown == 2) {
|
|
if (set_atleastone(dlines, dline_index))
|
|
diff = min(diff, DIFF_NORMAL);
|
|
}
|
|
}
|
|
|
|
/* More advanced deduction that allows propagation along diagonal
|
|
* chains of faces connected by dots, for example: 3-2-...-2-3
|
|
* in square grids. */
|
|
if (sstate->diff >= DIFF_TRICKY) {
|
|
/* If we have atleastone set for this dline, infer
|
|
* atmostone for each "opposite" dline (that is, each
|
|
* dline without edges in common with this one).
|
|
* Again, this test is only worth doing if both these
|
|
* lines are UNKNOWN. For if one of these lines were YES,
|
|
* the (yes == 1) test above would kick in instead. */
|
|
if (is_atleastone(dlines, dline_index)) {
|
|
int opp;
|
|
for (opp = 0; opp < N; opp++) {
|
|
int opp_dline_index;
|
|
if (opp == j || opp == j+1 || opp == j-1)
|
|
continue;
|
|
if (j == 0 && opp == N-1)
|
|
continue;
|
|
if (j == N-1 && opp == 0)
|
|
continue;
|
|
opp_dline_index = dline_index_from_dot(g, d, opp);
|
|
if (set_atmostone(dlines, opp_dline_index))
|
|
diff = min(diff, DIFF_NORMAL);
|
|
}
|
|
if (yes == 0 && is_atmostone(dlines, dline_index)) {
|
|
/* This dline has *exactly* one YES and there are no
|
|
* other YESs. This allows more deductions. */
|
|
if (unknown == 3) {
|
|
/* Third unknown must be YES */
|
|
for (opp = 0; opp < N; opp++) {
|
|
int opp_index;
|
|
if (opp == j || opp == k)
|
|
continue;
|
|
opp_index = d->edges[opp] - g->edges;
|
|
if (state->lines[opp_index] == LINE_UNKNOWN) {
|
|
solver_set_line(sstate, opp_index,
|
|
LINE_YES);
|
|
diff = min(diff, DIFF_EASY);
|
|
}
|
|
}
|
|
} else if (unknown == 4) {
|
|
/* Exactly one of opposite UNKNOWNS is YES. We've
|
|
* already set atmostone, so set atleastone as
|
|
* well.
|
|
*/
|
|
if (dline_set_opp_atleastone(sstate, d, j))
|
|
diff = min(diff, DIFF_NORMAL);
|
|
}
|
|
}
|
|
}
|
|
}
|
|
}
|
|
}
|
|
return diff;
|
|
}
|
|
|
|
static int linedsf_deductions(solver_state *sstate)
|
|
{
|
|
game_state *state = sstate->state;
|
|
grid *g = state->game_grid;
|
|
char *dlines = sstate->dlines;
|
|
int i;
|
|
int diff = DIFF_MAX;
|
|
int diff_tmp;
|
|
|
|
/* ------ Face deductions ------ */
|
|
|
|
/* A fully-general linedsf deduction seems overly complicated
|
|
* (I suspect the problem is NP-complete, though in practice it might just
|
|
* be doable because faces are limited in size).
|
|
* For simplicity, we only consider *pairs* of LINE_UNKNOWNS that are
|
|
* known to be identical. If setting them both to YES (or NO) would break
|
|
* the clue, set them to NO (or YES). */
|
|
|
|
for (i = 0; i < g->num_faces; i++) {
|
|
int N, yes, no, unknown;
|
|
int clue;
|
|
|
|
if (sstate->face_solved[i])
|
|
continue;
|
|
clue = state->clues[i];
|
|
if (clue < 0)
|
|
continue;
|
|
|
|
N = g->faces[i].order;
|
|
yes = sstate->face_yes_count[i];
|
|
if (yes + 1 == clue) {
|
|
if (face_setall_identical(sstate, i, LINE_NO))
|
|
diff = min(diff, DIFF_EASY);
|
|
}
|
|
no = sstate->face_no_count[i];
|
|
if (no + 1 == N - clue) {
|
|
if (face_setall_identical(sstate, i, LINE_YES))
|
|
diff = min(diff, DIFF_EASY);
|
|
}
|
|
|
|
/* Reload YES count, it might have changed */
|
|
yes = sstate->face_yes_count[i];
|
|
unknown = N - no - yes;
|
|
|
|
/* Deductions with small number of LINE_UNKNOWNs, based on overall
|
|
* parity of lines. */
|
|
diff_tmp = parity_deductions(sstate, g->faces[i].edges,
|
|
(clue - yes) % 2, unknown);
|
|
diff = min(diff, diff_tmp);
|
|
}
|
|
|
|
/* ------ Dot deductions ------ */
|
|
for (i = 0; i < g->num_dots; i++) {
|
|
grid_dot *d = g->dots + i;
|
|
int N = d->order;
|
|
int j;
|
|
int yes, no, unknown;
|
|
/* Go through dlines, and do any dline<->linedsf deductions wherever
|
|
* we find two UNKNOWNS. */
|
|
for (j = 0; j < N; j++) {
|
|
int dline_index = dline_index_from_dot(g, d, j);
|
|
int line1_index;
|
|
int line2_index;
|
|
int can1, can2, inv1, inv2;
|
|
int j2;
|
|
line1_index = d->edges[j] - g->edges;
|
|
if (state->lines[line1_index] != LINE_UNKNOWN)
|
|
continue;
|
|
j2 = j + 1;
|
|
if (j2 == N) j2 = 0;
|
|
line2_index = d->edges[j2] - g->edges;
|
|
if (state->lines[line2_index] != LINE_UNKNOWN)
|
|
continue;
|
|
/* Infer dline flags from linedsf */
|
|
can1 = edsf_canonify(sstate->linedsf, line1_index, &inv1);
|
|
can2 = edsf_canonify(sstate->linedsf, line2_index, &inv2);
|
|
if (can1 == can2 && inv1 != inv2) {
|
|
/* These are opposites, so set dline atmostone/atleastone */
|
|
if (set_atmostone(dlines, dline_index))
|
|
diff = min(diff, DIFF_NORMAL);
|
|
if (set_atleastone(dlines, dline_index))
|
|
diff = min(diff, DIFF_NORMAL);
|
|
continue;
|
|
}
|
|
/* Infer linedsf from dline flags */
|
|
if (is_atmostone(dlines, dline_index)
|
|
&& is_atleastone(dlines, dline_index)) {
|
|
if (merge_lines(sstate, line1_index, line2_index, 1))
|
|
diff = min(diff, DIFF_HARD);
|
|
}
|
|
}
|
|
|
|
/* Deductions with small number of LINE_UNKNOWNs, based on overall
|
|
* parity of lines. */
|
|
yes = sstate->dot_yes_count[i];
|
|
no = sstate->dot_no_count[i];
|
|
unknown = N - yes - no;
|
|
diff_tmp = parity_deductions(sstate, d->edges,
|
|
yes % 2, unknown);
|
|
diff = min(diff, diff_tmp);
|
|
}
|
|
|
|
/* ------ Edge dsf deductions ------ */
|
|
|
|
/* If the state of a line is known, deduce the state of its canonical line
|
|
* too, and vice versa. */
|
|
for (i = 0; i < g->num_edges; i++) {
|
|
int can, inv;
|
|
enum line_state s;
|
|
can = edsf_canonify(sstate->linedsf, i, &inv);
|
|
if (can == i)
|
|
continue;
|
|
s = sstate->state->lines[can];
|
|
if (s != LINE_UNKNOWN) {
|
|
if (solver_set_line(sstate, i, inv ? OPP(s) : s))
|
|
diff = min(diff, DIFF_EASY);
|
|
} else {
|
|
s = sstate->state->lines[i];
|
|
if (s != LINE_UNKNOWN) {
|
|
if (solver_set_line(sstate, can, inv ? OPP(s) : s))
|
|
diff = min(diff, DIFF_EASY);
|
|
}
|
|
}
|
|
}
|
|
|
|
return diff;
|
|
}
|
|
|
|
static int loop_deductions(solver_state *sstate)
|
|
{
|
|
int edgecount = 0, clues = 0, satclues = 0, sm1clues = 0;
|
|
game_state *state = sstate->state;
|
|
grid *g = state->game_grid;
|
|
int shortest_chainlen = g->num_dots;
|
|
int loop_found = FALSE;
|
|
int dots_connected;
|
|
int progress = FALSE;
|
|
int i;
|
|
|
|
/*
|
|
* Go through the grid and update for all the new edges.
|
|
* Since merge_dots() is idempotent, the simplest way to
|
|
* do this is just to update for _all_ the edges.
|
|
* Also, while we're here, we count the edges.
|
|
*/
|
|
for (i = 0; i < g->num_edges; i++) {
|
|
if (state->lines[i] == LINE_YES) {
|
|
loop_found |= merge_dots(sstate, i);
|
|
edgecount++;
|
|
}
|
|
}
|
|
|
|
/*
|
|
* Count the clues, count the satisfied clues, and count the
|
|
* satisfied-minus-one clues.
|
|
*/
|
|
for (i = 0; i < g->num_faces; i++) {
|
|
int c = state->clues[i];
|
|
if (c >= 0) {
|
|
int o = sstate->face_yes_count[i];
|
|
if (o == c)
|
|
satclues++;
|
|
else if (o == c-1)
|
|
sm1clues++;
|
|
clues++;
|
|
}
|
|
}
|
|
|
|
for (i = 0; i < g->num_dots; ++i) {
|
|
dots_connected =
|
|
sstate->looplen[dsf_canonify(sstate->dotdsf, i)];
|
|
if (dots_connected > 1)
|
|
shortest_chainlen = min(shortest_chainlen, dots_connected);
|
|
}
|
|
|
|
assert(sstate->solver_status == SOLVER_INCOMPLETE);
|
|
|
|
if (satclues == clues && shortest_chainlen == edgecount) {
|
|
sstate->solver_status = SOLVER_SOLVED;
|
|
/* This discovery clearly counts as progress, even if we haven't
|
|
* just added any lines or anything */
|
|
progress = TRUE;
|
|
goto finished_loop_deductionsing;
|
|
}
|
|
|
|
/*
|
|
* Now go through looking for LINE_UNKNOWN edges which
|
|
* connect two dots that are already in the same
|
|
* equivalence class. If we find one, test to see if the
|
|
* loop it would create is a solution.
|
|
*/
|
|
for (i = 0; i < g->num_edges; i++) {
|
|
grid_edge *e = g->edges + i;
|
|
int d1 = e->dot1 - g->dots;
|
|
int d2 = e->dot2 - g->dots;
|
|
int eqclass, val;
|
|
if (state->lines[i] != LINE_UNKNOWN)
|
|
continue;
|
|
|
|
eqclass = dsf_canonify(sstate->dotdsf, d1);
|
|
if (eqclass != dsf_canonify(sstate->dotdsf, d2))
|
|
continue;
|
|
|
|
val = LINE_NO; /* loop is bad until proven otherwise */
|
|
|
|
/*
|
|
* This edge would form a loop. Next
|
|
* question: how long would the loop be?
|
|
* Would it equal the total number of edges
|
|
* (plus the one we'd be adding if we added
|
|
* it)?
|
|
*/
|
|
if (sstate->looplen[eqclass] == edgecount + 1) {
|
|
int sm1_nearby;
|
|
|
|
/*
|
|
* This edge would form a loop which
|
|
* took in all the edges in the entire
|
|
* grid. So now we need to work out
|
|
* whether it would be a valid solution
|
|
* to the puzzle, which means we have to
|
|
* check if it satisfies all the clues.
|
|
* This means that every clue must be
|
|
* either satisfied or satisfied-minus-
|
|
* 1, and also that the number of
|
|
* satisfied-minus-1 clues must be at
|
|
* most two and they must lie on either
|
|
* side of this edge.
|
|
*/
|
|
sm1_nearby = 0;
|
|
if (e->face1) {
|
|
int f = e->face1 - g->faces;
|
|
int c = state->clues[f];
|
|
if (c >= 0 && sstate->face_yes_count[f] == c - 1)
|
|
sm1_nearby++;
|
|
}
|
|
if (e->face2) {
|
|
int f = e->face2 - g->faces;
|
|
int c = state->clues[f];
|
|
if (c >= 0 && sstate->face_yes_count[f] == c - 1)
|
|
sm1_nearby++;
|
|
}
|
|
if (sm1clues == sm1_nearby &&
|
|
sm1clues + satclues == clues) {
|
|
val = LINE_YES; /* loop is good! */
|
|
}
|
|
}
|
|
|
|
/*
|
|
* Right. Now we know that adding this edge
|
|
* would form a loop, and we know whether
|
|
* that loop would be a viable solution or
|
|
* not.
|
|
*
|
|
* If adding this edge produces a solution,
|
|
* then we know we've found _a_ solution but
|
|
* we don't know that it's _the_ solution -
|
|
* if it were provably the solution then
|
|
* we'd have deduced this edge some time ago
|
|
* without the need to do loop detection. So
|
|
* in this state we return SOLVER_AMBIGUOUS,
|
|
* which has the effect that hitting Solve
|
|
* on a user-provided puzzle will fill in a
|
|
* solution but using the solver to
|
|
* construct new puzzles won't consider this
|
|
* a reasonable deduction for the user to
|
|
* make.
|
|
*/
|
|
progress = solver_set_line(sstate, i, val);
|
|
assert(progress == TRUE);
|
|
if (val == LINE_YES) {
|
|
sstate->solver_status = SOLVER_AMBIGUOUS;
|
|
goto finished_loop_deductionsing;
|
|
}
|
|
}
|
|
|
|
finished_loop_deductionsing:
|
|
return progress ? DIFF_EASY : DIFF_MAX;
|
|
}
|
|
|
|
/* This will return a dynamically allocated solver_state containing the (more)
|
|
* solved grid */
|
|
static solver_state *solve_game_rec(const solver_state *sstate_start)
|
|
{
|
|
solver_state *sstate;
|
|
|
|
/* Index of the solver we should call next. */
|
|
int i = 0;
|
|
|
|
/* As a speed-optimisation, we avoid re-running solvers that we know
|
|
* won't make any progress. This happens when a high-difficulty
|
|
* solver makes a deduction that can only help other high-difficulty
|
|
* solvers.
|
|
* For example: if a new 'dline' flag is set by dline_deductions, the
|
|
* trivial_deductions solver cannot do anything with this information.
|
|
* If we've already run the trivial_deductions solver (because it's
|
|
* earlier in the list), there's no point running it again.
|
|
*
|
|
* Therefore: if a solver is earlier in the list than "threshold_index",
|
|
* we don't bother running it if it's difficulty level is less than
|
|
* "threshold_diff".
|
|
*/
|
|
int threshold_diff = 0;
|
|
int threshold_index = 0;
|
|
|
|
sstate = dup_solver_state(sstate_start);
|
|
|
|
check_caches(sstate);
|
|
|
|
while (i < NUM_SOLVERS) {
|
|
if (sstate->solver_status == SOLVER_MISTAKE)
|
|
return sstate;
|
|
if (sstate->solver_status == SOLVER_SOLVED ||
|
|
sstate->solver_status == SOLVER_AMBIGUOUS) {
|
|
/* solver finished */
|
|
break;
|
|
}
|
|
|
|
if ((solver_diffs[i] >= threshold_diff || i >= threshold_index)
|
|
&& solver_diffs[i] <= sstate->diff) {
|
|
/* current_solver is eligible, so use it */
|
|
int next_diff = solver_fns[i](sstate);
|
|
if (next_diff != DIFF_MAX) {
|
|
/* solver made progress, so use new thresholds and
|
|
* start again at top of list. */
|
|
threshold_diff = next_diff;
|
|
threshold_index = i;
|
|
i = 0;
|
|
continue;
|
|
}
|
|
}
|
|
/* current_solver is ineligible, or failed to make progress, so
|
|
* go to the next solver in the list */
|
|
i++;
|
|
}
|
|
|
|
if (sstate->solver_status == SOLVER_SOLVED ||
|
|
sstate->solver_status == SOLVER_AMBIGUOUS) {
|
|
/* s/LINE_UNKNOWN/LINE_NO/g */
|
|
array_setall(sstate->state->lines, LINE_UNKNOWN, LINE_NO,
|
|
sstate->state->game_grid->num_edges);
|
|
return sstate;
|
|
}
|
|
|
|
return sstate;
|
|
}
|
|
|
|
static char *solve_game(game_state *state, game_state *currstate,
|
|
char *aux, char **error)
|
|
{
|
|
char *soln = NULL;
|
|
solver_state *sstate, *new_sstate;
|
|
|
|
sstate = new_solver_state(state, DIFF_MAX);
|
|
new_sstate = solve_game_rec(sstate);
|
|
|
|
if (new_sstate->solver_status == SOLVER_SOLVED) {
|
|
soln = encode_solve_move(new_sstate->state);
|
|
} else if (new_sstate->solver_status == SOLVER_AMBIGUOUS) {
|
|
soln = encode_solve_move(new_sstate->state);
|
|
/**error = "Solver found ambiguous solutions"; */
|
|
} else {
|
|
soln = encode_solve_move(new_sstate->state);
|
|
/**error = "Solver failed"; */
|
|
}
|
|
|
|
free_solver_state(new_sstate);
|
|
free_solver_state(sstate);
|
|
|
|
return soln;
|
|
}
|
|
|
|
/* ----------------------------------------------------------------------
|
|
* Drawing and mouse-handling
|
|
*/
|
|
|
|
static char *interpret_move(game_state *state, game_ui *ui, game_drawstate *ds,
|
|
int x, int y, int button)
|
|
{
|
|
grid *g = state->game_grid;
|
|
grid_edge *e;
|
|
int i;
|
|
char *ret, buf[80];
|
|
char button_char = ' ';
|
|
enum line_state old_state;
|
|
|
|
button &= ~MOD_MASK;
|
|
|
|
/* Convert mouse-click (x,y) to grid coordinates */
|
|
x -= BORDER(ds->tilesize);
|
|
y -= BORDER(ds->tilesize);
|
|
x = x * g->tilesize / ds->tilesize;
|
|
y = y * g->tilesize / ds->tilesize;
|
|
x += g->lowest_x;
|
|
y += g->lowest_y;
|
|
|
|
e = grid_nearest_edge(g, x, y);
|
|
if (e == NULL)
|
|
return NULL;
|
|
|
|
i = e - g->edges;
|
|
|
|
/* I think it's only possible to play this game with mouse clicks, sorry */
|
|
/* Maybe will add mouse drag support some time */
|
|
old_state = state->lines[i];
|
|
|
|
switch (button) {
|
|
case LEFT_BUTTON:
|
|
switch (old_state) {
|
|
case LINE_UNKNOWN:
|
|
button_char = 'y';
|
|
break;
|
|
case LINE_YES:
|
|
case LINE_NO:
|
|
button_char = 'u';
|
|
break;
|
|
}
|
|
break;
|
|
case MIDDLE_BUTTON:
|
|
button_char = 'u';
|
|
break;
|
|
case RIGHT_BUTTON:
|
|
switch (old_state) {
|
|
case LINE_UNKNOWN:
|
|
button_char = 'n';
|
|
break;
|
|
case LINE_NO:
|
|
case LINE_YES:
|
|
button_char = 'u';
|
|
break;
|
|
}
|
|
break;
|
|
default:
|
|
return NULL;
|
|
}
|
|
|
|
|
|
sprintf(buf, "%d%c", i, (int)button_char);
|
|
ret = dupstr(buf);
|
|
|
|
return ret;
|
|
}
|
|
|
|
static game_state *execute_move(game_state *state, char *move)
|
|
{
|
|
int i;
|
|
game_state *newstate = dup_game(state);
|
|
|
|
if (move[0] == 'S') {
|
|
move++;
|
|
newstate->cheated = TRUE;
|
|
}
|
|
|
|
while (*move) {
|
|
i = atoi(move);
|
|
move += strspn(move, "1234567890");
|
|
switch (*(move++)) {
|
|
case 'y':
|
|
newstate->lines[i] = LINE_YES;
|
|
break;
|
|
case 'n':
|
|
newstate->lines[i] = LINE_NO;
|
|
break;
|
|
case 'u':
|
|
newstate->lines[i] = LINE_UNKNOWN;
|
|
break;
|
|
default:
|
|
goto fail;
|
|
}
|
|
}
|
|
|
|
/*
|
|
* Check for completion.
|
|
*/
|
|
if (check_completion(newstate))
|
|
newstate->solved = TRUE;
|
|
|
|
return newstate;
|
|
|
|
fail:
|
|
free_game(newstate);
|
|
return NULL;
|
|
}
|
|
|
|
/* ----------------------------------------------------------------------
|
|
* Drawing routines.
|
|
*/
|
|
|
|
/* Convert from grid coordinates to screen coordinates */
|
|
static void grid_to_screen(const game_drawstate *ds, const grid *g,
|
|
int grid_x, int grid_y, int *x, int *y)
|
|
{
|
|
*x = grid_x - g->lowest_x;
|
|
*y = grid_y - g->lowest_y;
|
|
*x = *x * ds->tilesize / g->tilesize;
|
|
*y = *y * ds->tilesize / g->tilesize;
|
|
*x += BORDER(ds->tilesize);
|
|
*y += BORDER(ds->tilesize);
|
|
}
|
|
|
|
/* Returns (into x,y) position of centre of face for rendering the text clue.
|
|
*/
|
|
static void face_text_pos(const game_drawstate *ds, const grid *g,
|
|
const grid_face *f, int *x, int *y)
|
|
{
|
|
int i;
|
|
|
|
/* Simplest solution is the centroid. Might not work in some cases. */
|
|
|
|
/* Another algorithm to look into:
|
|
* Find the midpoints of the sides, find the bounding-box,
|
|
* then take the centre of that. */
|
|
|
|
/* Best solution probably involves incentres (inscribed circles) */
|
|
|
|
int sx = 0, sy = 0; /* sums */
|
|
for (i = 0; i < f->order; i++) {
|
|
grid_dot *d = f->dots[i];
|
|
sx += d->x;
|
|
sy += d->y;
|
|
}
|
|
sx /= f->order;
|
|
sy /= f->order;
|
|
|
|
/* convert to screen coordinates */
|
|
grid_to_screen(ds, g, sx, sy, x, y);
|
|
}
|
|
|
|
static void game_redraw(drawing *dr, game_drawstate *ds, game_state *oldstate,
|
|
game_state *state, int dir, game_ui *ui,
|
|
float animtime, float flashtime)
|
|
{
|
|
grid *g = state->game_grid;
|
|
int border = BORDER(ds->tilesize);
|
|
int i, n;
|
|
char c[2];
|
|
int line_colour, flash_changed;
|
|
int clue_mistake;
|
|
int clue_satisfied;
|
|
|
|
if (!ds->started) {
|
|
/*
|
|
* 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.
|
|
*/
|
|
int grid_width = g->highest_x - g->lowest_x;
|
|
int grid_height = g->highest_y - g->lowest_y;
|
|
int w = grid_width * ds->tilesize / g->tilesize;
|
|
int h = grid_height * ds->tilesize / g->tilesize;
|
|
draw_rect(dr, 0, 0, w + 2 * border + 1, h + 2 * border + 1,
|
|
COL_BACKGROUND);
|
|
|
|
/* Draw clues */
|
|
for (i = 0; i < g->num_faces; i++) {
|
|
grid_face *f;
|
|
int x, y;
|
|
|
|
c[0] = CLUE2CHAR(state->clues[i]);
|
|
c[1] = '\0';
|
|
f = g->faces + i;
|
|
face_text_pos(ds, g, f, &x, &y);
|
|
draw_text(dr, x, y, FONT_VARIABLE, ds->tilesize/2,
|
|
ALIGN_VCENTRE | ALIGN_HCENTRE, COL_FOREGROUND, c);
|
|
}
|
|
draw_update(dr, 0, 0, w + 2 * border, h + 2 * border);
|
|
}
|
|
|
|
if (flashtime > 0 &&
|
|
(flashtime <= FLASH_TIME/3 ||
|
|
flashtime >= FLASH_TIME*2/3)) {
|
|
flash_changed = !ds->flashing;
|
|
ds->flashing = TRUE;
|
|
} else {
|
|
flash_changed = ds->flashing;
|
|
ds->flashing = FALSE;
|
|
}
|
|
|
|
/* Some platforms may perform anti-aliasing, which may prevent clean
|
|
* repainting of lines when the colour is changed.
|
|
* If a line needs to be over-drawn in a different colour, erase a
|
|
* bounding-box around the line, then flag all nearby objects for redraw.
|
|
*/
|
|
if (ds->started) {
|
|
const char redraw_flag = (char)(1<<7);
|
|
for (i = 0; i < g->num_edges; i++) {
|
|
char prev_ds = (ds->lines[i] & ~redraw_flag);
|
|
char new_ds = state->lines[i];
|
|
if (state->line_errors[i])
|
|
new_ds = DS_LINE_ERROR;
|
|
|
|
/* If we're changing state, AND
|
|
* the previous state was a coloured line */
|
|
if ((prev_ds != new_ds) && (prev_ds != LINE_NO)) {
|
|
grid_edge *e = g->edges + i;
|
|
int x1 = e->dot1->x;
|
|
int y1 = e->dot1->y;
|
|
int x2 = e->dot2->x;
|
|
int y2 = e->dot2->y;
|
|
int xmin, xmax, ymin, ymax;
|
|
int j;
|
|
grid_to_screen(ds, g, x1, y1, &x1, &y1);
|
|
grid_to_screen(ds, g, x2, y2, &x2, &y2);
|
|
/* Allow extra margin for dots, and thickness of lines */
|
|
xmin = min(x1, x2) - 2;
|
|
xmax = max(x1, x2) + 2;
|
|
ymin = min(y1, y2) - 2;
|
|
ymax = max(y1, y2) + 2;
|
|
/* For testing, I find it helpful to change COL_BACKGROUND
|
|
* to COL_SATISFIED here. */
|
|
draw_rect(dr, xmin, ymin, xmax - xmin + 1, ymax - ymin + 1,
|
|
COL_BACKGROUND);
|
|
draw_update(dr, xmin, ymin, xmax - xmin + 1, ymax - ymin + 1);
|
|
|
|
/* Mark nearby lines for redraw */
|
|
for (j = 0; j < e->dot1->order; j++)
|
|
ds->lines[e->dot1->edges[j] - g->edges] |= redraw_flag;
|
|
for (j = 0; j < e->dot2->order; j++)
|
|
ds->lines[e->dot2->edges[j] - g->edges] |= redraw_flag;
|
|
/* Mark nearby clues for redraw. Use a value that is
|
|
* neither TRUE nor FALSE for this. */
|
|
if (e->face1)
|
|
ds->clue_error[e->face1 - g->faces] = 2;
|
|
if (e->face2)
|
|
ds->clue_error[e->face2 - g->faces] = 2;
|
|
}
|
|
}
|
|
}
|
|
|
|
/* Redraw clue colours if necessary */
|
|
for (i = 0; i < g->num_faces; i++) {
|
|
grid_face *f = g->faces + i;
|
|
int sides = f->order;
|
|
int j;
|
|
n = state->clues[i];
|
|
if (n < 0)
|
|
continue;
|
|
|
|
c[0] = CLUE2CHAR(n);
|
|
c[1] = '\0';
|
|
|
|
clue_mistake = (face_order(state, i, LINE_YES) > n ||
|
|
face_order(state, i, LINE_NO ) > (sides-n));
|
|
|
|
clue_satisfied = (face_order(state, i, LINE_YES) == n &&
|
|
face_order(state, i, LINE_NO ) == (sides-n));
|
|
|
|
if (clue_mistake != ds->clue_error[i]
|
|
|| clue_satisfied != ds->clue_satisfied[i]) {
|
|
int x, y;
|
|
face_text_pos(ds, g, f, &x, &y);
|
|
/* There seems to be a certain amount of trial-and-error
|
|
* involved in working out the correct bounding-box for
|
|
* the text. */
|
|
draw_rect(dr, x - ds->tilesize/4 - 1, y - ds->tilesize/4 - 3,
|
|
ds->tilesize/2 + 2, ds->tilesize/2 + 5,
|
|
COL_BACKGROUND);
|
|
draw_text(dr, x, y,
|
|
FONT_VARIABLE, ds->tilesize/2,
|
|
ALIGN_VCENTRE | ALIGN_HCENTRE,
|
|
clue_mistake ? COL_MISTAKE :
|
|
clue_satisfied ? COL_SATISFIED : COL_FOREGROUND, c);
|
|
draw_update(dr, x - ds->tilesize/4 - 1, y - ds->tilesize/4 - 3,
|
|
ds->tilesize/2 + 2, ds->tilesize/2 + 5);
|
|
|
|
ds->clue_error[i] = clue_mistake;
|
|
ds->clue_satisfied[i] = clue_satisfied;
|
|
|
|
/* Sometimes, the bounding-box encroaches into the surrounding
|
|
* lines (particularly if the window is resized fairly small).
|
|
* So redraw them. */
|
|
for (j = 0; j < f->order; j++)
|
|
ds->lines[f->edges[j] - g->edges] = -1;
|
|
}
|
|
}
|
|
|
|
/* Lines */
|
|
for (i = 0; i < g->num_edges; i++) {
|
|
grid_edge *e = g->edges + i;
|
|
int x1, x2, y1, y2;
|
|
int xmin, ymin, xmax, ymax;
|
|
char new_ds, need_draw;
|
|
new_ds = state->lines[i];
|
|
if (state->line_errors[i])
|
|
new_ds = DS_LINE_ERROR;
|
|
need_draw = (new_ds != ds->lines[i]) ? TRUE : FALSE;
|
|
if (flash_changed && (state->lines[i] == LINE_YES))
|
|
need_draw = TRUE;
|
|
if (!ds->started)
|
|
need_draw = TRUE; /* draw everything at the start */
|
|
ds->lines[i] = new_ds;
|
|
if (!need_draw)
|
|
continue;
|
|
if (state->line_errors[i])
|
|
line_colour = COL_MISTAKE;
|
|
else if (state->lines[i] == LINE_UNKNOWN)
|
|
line_colour = COL_LINEUNKNOWN;
|
|
else if (state->lines[i] == LINE_NO)
|
|
line_colour = COL_BACKGROUND;
|
|
else if (ds->flashing)
|
|
line_colour = COL_HIGHLIGHT;
|
|
else
|
|
line_colour = COL_FOREGROUND;
|
|
|
|
/* Convert from grid to screen coordinates */
|
|
grid_to_screen(ds, g, e->dot1->x, e->dot1->y, &x1, &y1);
|
|
grid_to_screen(ds, g, e->dot2->x, e->dot2->y, &x2, &y2);
|
|
|
|
xmin = min(x1, x2);
|
|
xmax = max(x1, x2);
|
|
ymin = min(y1, y2);
|
|
ymax = max(y1, y2);
|
|
|
|
if (line_colour != COL_BACKGROUND) {
|
|
/* (dx, dy) points roughly from (x1, y1) to (x2, y2).
|
|
* The line is then "fattened" in a (roughly) perpendicular
|
|
* direction to create a thin rectangle. */
|
|
int dx = (x1 > x2) ? -1 : ((x1 < x2) ? 1 : 0);
|
|
int dy = (y1 > y2) ? -1 : ((y1 < y2) ? 1 : 0);
|
|
int points[8];
|
|
points[0] = x1 + dy;
|
|
points[1] = y1 - dx;
|
|
points[2] = x1 - dy;
|
|
points[3] = y1 + dx;
|
|
points[4] = x2 - dy;
|
|
points[5] = y2 + dx;
|
|
points[6] = x2 + dy;
|
|
points[7] = y2 - dx;
|
|
draw_polygon(dr, points, 4, line_colour, line_colour);
|
|
}
|
|
if (ds->started) {
|
|
/* Draw dots at ends of the line */
|
|
draw_circle(dr, x1, y1, 2, COL_FOREGROUND, COL_FOREGROUND);
|
|
draw_circle(dr, x2, y2, 2, COL_FOREGROUND, COL_FOREGROUND);
|
|
}
|
|
draw_update(dr, xmin-2, ymin-2, xmax - xmin + 4, ymax - ymin + 4);
|
|
}
|
|
|
|
/* Draw dots */
|
|
if (!ds->started) {
|
|
for (i = 0; i < g->num_dots; i++) {
|
|
grid_dot *d = g->dots + i;
|
|
int x, y;
|
|
grid_to_screen(ds, g, d->x, d->y, &x, &y);
|
|
draw_circle(dr, x, y, 2, COL_FOREGROUND, COL_FOREGROUND);
|
|
}
|
|
}
|
|
ds->started = TRUE;
|
|
}
|
|
|
|
static float game_flash_length(game_state *oldstate, game_state *newstate,
|
|
int dir, game_ui *ui)
|
|
{
|
|
if (!oldstate->solved && newstate->solved &&
|
|
!oldstate->cheated && !newstate->cheated) {
|
|
return FLASH_TIME;
|
|
}
|
|
|
|
return 0.0F;
|
|
}
|
|
|
|
static void game_print_size(game_params *params, float *x, float *y)
|
|
{
|
|
int pw, ph;
|
|
|
|
/*
|
|
* I'll use 7mm "squares" by default.
|
|
*/
|
|
game_compute_size(params, 700, &pw, &ph);
|
|
*x = pw / 100.0F;
|
|
*y = ph / 100.0F;
|
|
}
|
|
|
|
static void game_print(drawing *dr, game_state *state, int tilesize)
|
|
{
|
|
int ink = print_mono_colour(dr, 0);
|
|
int i;
|
|
game_drawstate ads, *ds = &ads;
|
|
grid *g = state->game_grid;
|
|
|
|
game_set_size(dr, ds, NULL, tilesize);
|
|
|
|
for (i = 0; i < g->num_dots; i++) {
|
|
int x, y;
|
|
grid_to_screen(ds, g, g->dots[i].x, g->dots[i].y, &x, &y);
|
|
draw_circle(dr, x, y, ds->tilesize / 15, ink, ink);
|
|
}
|
|
|
|
/*
|
|
* Clues.
|
|
*/
|
|
for (i = 0; i < g->num_faces; i++) {
|
|
grid_face *f = g->faces + i;
|
|
int clue = state->clues[i];
|
|
if (clue >= 0) {
|
|
char c[2];
|
|
int x, y;
|
|
c[0] = CLUE2CHAR(clue);
|
|
c[1] = '\0';
|
|
face_text_pos(ds, g, f, &x, &y);
|
|
draw_text(dr, x, y,
|
|
FONT_VARIABLE, ds->tilesize / 2,
|
|
ALIGN_VCENTRE | ALIGN_HCENTRE, ink, c);
|
|
}
|
|
}
|
|
|
|
/*
|
|
* Lines.
|
|
*/
|
|
for (i = 0; i < g->num_edges; i++) {
|
|
int thickness = (state->lines[i] == LINE_YES) ? 30 : 150;
|
|
grid_edge *e = g->edges + i;
|
|
int x1, y1, x2, y2;
|
|
grid_to_screen(ds, g, e->dot1->x, e->dot1->y, &x1, &y1);
|
|
grid_to_screen(ds, g, e->dot2->x, e->dot2->y, &x2, &y2);
|
|
if (state->lines[i] == LINE_YES)
|
|
{
|
|
/* (dx, dy) points from (x1, y1) to (x2, y2).
|
|
* The line is then "fattened" in a perpendicular
|
|
* direction to create a thin rectangle. */
|
|
double d = sqrt(SQ((double)x1 - x2) + SQ((double)y1 - y2));
|
|
double dx = (x2 - x1) / d;
|
|
double dy = (y2 - y1) / d;
|
|
int points[8];
|
|
|
|
dx = (dx * ds->tilesize) / thickness;
|
|
dy = (dy * ds->tilesize) / thickness;
|
|
points[0] = x1 + (int)dy;
|
|
points[1] = y1 - (int)dx;
|
|
points[2] = x1 - (int)dy;
|
|
points[3] = y1 + (int)dx;
|
|
points[4] = x2 - (int)dy;
|
|
points[5] = y2 + (int)dx;
|
|
points[6] = x2 + (int)dy;
|
|
points[7] = y2 - (int)dx;
|
|
draw_polygon(dr, points, 4, ink, ink);
|
|
}
|
|
else
|
|
{
|
|
/* Draw a dotted line */
|
|
int divisions = 6;
|
|
int j;
|
|
for (j = 1; j < divisions; j++) {
|
|
/* Weighted average */
|
|
int x = (x1 * (divisions -j) + x2 * j) / divisions;
|
|
int y = (y1 * (divisions -j) + y2 * j) / divisions;
|
|
draw_circle(dr, x, y, ds->tilesize / thickness, ink, ink);
|
|
}
|
|
}
|
|
}
|
|
}
|
|
|
|
#ifdef COMBINED
|
|
#define thegame loopy
|
|
#endif
|
|
|
|
const struct game thegame = {
|
|
"Loopy", "games.loopy", "loopy",
|
|
default_params,
|
|
game_fetch_preset,
|
|
decode_params,
|
|
encode_params,
|
|
free_params,
|
|
dup_params,
|
|
TRUE, game_configure, custom_params,
|
|
validate_params,
|
|
new_game_desc,
|
|
validate_desc,
|
|
new_game,
|
|
dup_game,
|
|
free_game,
|
|
1, solve_game,
|
|
TRUE, game_can_format_as_text_now, game_text_format,
|
|
new_ui,
|
|
free_ui,
|
|
encode_ui,
|
|
decode_ui,
|
|
game_changed_state,
|
|
interpret_move,
|
|
execute_move,
|
|
PREFERRED_TILE_SIZE, game_compute_size, game_set_size,
|
|
game_colours,
|
|
game_new_drawstate,
|
|
game_free_drawstate,
|
|
game_redraw,
|
|
game_anim_length,
|
|
game_flash_length,
|
|
TRUE, FALSE, game_print_size, game_print,
|
|
FALSE /* wants_statusbar */,
|
|
FALSE, game_timing_state,
|
|
0, /* mouse_priorities */
|
|
};
|
|
|
|
#ifdef STANDALONE_SOLVER
|
|
|
|
/*
|
|
* Half-hearted standalone solver. It can't output the solution to
|
|
* anything but a square puzzle, and it can't log the deductions
|
|
* it makes either. But it can solve square puzzles, and more
|
|
* importantly it can use its solver to grade the difficulty of
|
|
* any puzzle you give it.
|
|
*/
|
|
|
|
#include <stdarg.h>
|
|
|
|
int main(int argc, char **argv)
|
|
{
|
|
game_params *p;
|
|
game_state *s;
|
|
char *id = NULL, *desc, *err;
|
|
int grade = FALSE;
|
|
int ret, diff;
|
|
#if 0 /* verbose solver not supported here (yet) */
|
|
int really_verbose = FALSE;
|
|
#endif
|
|
|
|
while (--argc > 0) {
|
|
char *p = *++argv;
|
|
#if 0 /* verbose solver not supported here (yet) */
|
|
if (!strcmp(p, "-v")) {
|
|
really_verbose = TRUE;
|
|
} else
|
|
#endif
|
|
if (!strcmp(p, "-g")) {
|
|
grade = TRUE;
|
|
} else if (*p == '-') {
|
|
fprintf(stderr, "%s: unrecognised option `%s'\n", argv[0], p);
|
|
return 1;
|
|
} else {
|
|
id = p;
|
|
}
|
|
}
|
|
|
|
if (!id) {
|
|
fprintf(stderr, "usage: %s [-g | -v] <game_id>\n", argv[0]);
|
|
return 1;
|
|
}
|
|
|
|
desc = strchr(id, ':');
|
|
if (!desc) {
|
|
fprintf(stderr, "%s: game id expects a colon in it\n", argv[0]);
|
|
return 1;
|
|
}
|
|
*desc++ = '\0';
|
|
|
|
p = default_params();
|
|
decode_params(p, id);
|
|
err = validate_desc(p, desc);
|
|
if (err) {
|
|
fprintf(stderr, "%s: %s\n", argv[0], err);
|
|
return 1;
|
|
}
|
|
s = new_game(NULL, p, desc);
|
|
|
|
/*
|
|
* When solving an Easy puzzle, we don't want to bother the
|
|
* user with Hard-level deductions. For this reason, we grade
|
|
* the puzzle internally before doing anything else.
|
|
*/
|
|
ret = -1; /* placate optimiser */
|
|
for (diff = 0; diff < DIFF_MAX; diff++) {
|
|
solver_state *sstate_new;
|
|
solver_state *sstate = new_solver_state((game_state *)s, diff);
|
|
|
|
sstate_new = solve_game_rec(sstate);
|
|
|
|
if (sstate_new->solver_status == SOLVER_MISTAKE)
|
|
ret = 0;
|
|
else if (sstate_new->solver_status == SOLVER_SOLVED)
|
|
ret = 1;
|
|
else
|
|
ret = 2;
|
|
|
|
free_solver_state(sstate_new);
|
|
free_solver_state(sstate);
|
|
|
|
if (ret < 2)
|
|
break;
|
|
}
|
|
|
|
if (diff == DIFF_MAX) {
|
|
if (grade)
|
|
printf("Difficulty rating: harder than Hard, or ambiguous\n");
|
|
else
|
|
printf("Unable to find a unique solution\n");
|
|
} else {
|
|
if (grade) {
|
|
if (ret == 0)
|
|
printf("Difficulty rating: impossible (no solution exists)\n");
|
|
else if (ret == 1)
|
|
printf("Difficulty rating: %s\n", diffnames[diff]);
|
|
} else {
|
|
solver_state *sstate_new;
|
|
solver_state *sstate = new_solver_state((game_state *)s, diff);
|
|
|
|
/* If we supported a verbose solver, we'd set verbosity here */
|
|
|
|
sstate_new = solve_game_rec(sstate);
|
|
|
|
if (sstate_new->solver_status == SOLVER_MISTAKE)
|
|
printf("Puzzle is inconsistent\n");
|
|
else {
|
|
assert(sstate_new->solver_status == SOLVER_SOLVED);
|
|
if (s->grid_type == 0) {
|
|
fputs(game_text_format(sstate_new->state), stdout);
|
|
} else {
|
|
printf("Unable to output non-square grids\n");
|
|
}
|
|
}
|
|
|
|
free_solver_state(sstate_new);
|
|
free_solver_state(sstate);
|
|
}
|
|
}
|
|
|
|
return 0;
|
|
}
|
|
|
|
#endif
|