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Files
83121fb826
insist that a variable should be initialised in all branches of an if, instead of just all the non-assertion-failing ones. [originally from svn r7989]
4024 lines
113 KiB
C
4024 lines
113 KiB
C
/*
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* solo.c: the number-placing puzzle most popularly known as `Sudoku'.
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*
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* TODO:
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*
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* - reports from users are that `Trivial'-mode puzzles are still
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* rather hard compared to newspapers' easy ones, so some better
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* low-end difficulty grading would be nice
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* + it's possible that really easy puzzles always have
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* _several_ things you can do, so don't make you hunt too
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* hard for the one deduction you can currently make
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* + it's also possible that easy puzzles require fewer
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* cross-eliminations: perhaps there's a higher incidence of
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* things you can deduce by looking only at (say) rows,
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* rather than things you have to check both rows and columns
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* for
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* + but really, what I need to do is find some really easy
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* puzzles and _play_ them, to see what's actually easy about
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* them
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* + while I'm revamping this area, filling in the _last_
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* number in a nearly-full row or column should certainly be
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* permitted even at the lowest difficulty level.
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* + also Owen noticed that `Basic' grids requiring numeric
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* elimination are actually very hard, so I wonder if a
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* difficulty gradation between that and positional-
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* elimination-only might be in order
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* + but it's not good to have _too_ many difficulty levels, or
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* it'll take too long to randomly generate a given level.
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*
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* - it might still be nice to do some prioritisation on the
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* removal of numbers from the grid
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* + one possibility is to try to minimise the maximum number
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* of filled squares in any block, which in particular ought
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* to enforce never leaving a completely filled block in the
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* puzzle as presented.
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*
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* - alternative interface modes
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* + sudoku.com's Windows program has a palette of possible
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* entries; you select a palette entry first and then click
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* on the square you want it to go in, thus enabling
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* mouse-only play. Useful for PDAs! I don't think it's
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* actually incompatible with the current highlight-then-type
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* approach: you _either_ highlight a palette entry and then
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* click, _or_ you highlight a square and then type. At most
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* one thing is ever highlighted at a time, so there's no way
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* to confuse the two.
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* + then again, I don't actually like sudoku.com's interface;
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* it's too much like a paint package whereas I prefer to
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* think of Solo as a text editor.
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* + another PDA-friendly possibility is a drag interface:
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* _drag_ numbers from the palette into the grid squares.
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* Thought experiments suggest I'd prefer that to the
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* sudoku.com approach, but I haven't actually tried it.
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*/
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/*
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* Solo puzzles need to be square overall (since each row and each
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* column must contain one of every digit), but they need not be
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* subdivided the same way internally. I am going to adopt a
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* convention whereby I _always_ refer to `r' as the number of rows
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* of _big_ divisions, and `c' as the number of columns of _big_
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* divisions. Thus, a 2c by 3r puzzle looks something like this:
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*
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* 4 5 1 | 2 6 3
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* 6 3 2 | 5 4 1
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* ------+------ (Of course, you can't subdivide it the other way
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* 1 4 5 | 6 3 2 or you'll get clashes; observe that the 4 in the
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* 3 2 6 | 4 1 5 top left would conflict with the 4 in the second
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* ------+------ box down on the left-hand side.)
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* 5 1 4 | 3 2 6
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* 2 6 3 | 1 5 4
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*
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* The need for a strong naming convention should now be clear:
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* each small box is two rows of digits by three columns, while the
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* overall puzzle has three rows of small boxes by two columns. So
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* I will (hopefully) consistently use `r' to denote the number of
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* rows _of small boxes_ (here 3), which is also the number of
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* columns of digits in each small box; and `c' vice versa (here
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* 2).
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*
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* I'm also going to choose arbitrarily to list c first wherever
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* possible: the above is a 2x3 puzzle, not a 3x2 one.
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*/
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#include <stdio.h>
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#include <stdlib.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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#ifdef STANDALONE_SOLVER
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#include <stdarg.h>
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int solver_show_working, solver_recurse_depth;
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#endif
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#include "puzzles.h"
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/*
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* To save space, I store digits internally as unsigned char. This
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* imposes a hard limit of 255 on the order of the puzzle. Since
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* even a 5x5 takes unacceptably long to generate, I don't see this
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* as a serious limitation unless something _really_ impressive
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* happens in computing technology; but here's a typedef anyway for
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* general good practice.
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*/
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typedef unsigned char digit;
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#define ORDER_MAX 255
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#define PREFERRED_TILE_SIZE 32
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#define TILE_SIZE (ds->tilesize)
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#define BORDER (TILE_SIZE / 2)
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#define GRIDEXTRA (TILE_SIZE / 32)
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#define FLASH_TIME 0.4F
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enum { SYMM_NONE, SYMM_ROT2, SYMM_ROT4, SYMM_REF2, SYMM_REF2D, SYMM_REF4,
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SYMM_REF4D, SYMM_REF8 };
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enum { DIFF_BLOCK, DIFF_SIMPLE, DIFF_INTERSECT, DIFF_SET, DIFF_EXTREME,
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DIFF_RECURSIVE, DIFF_AMBIGUOUS, DIFF_IMPOSSIBLE };
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enum {
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COL_BACKGROUND,
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COL_XDIAGONALS,
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COL_GRID,
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COL_CLUE,
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COL_USER,
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COL_HIGHLIGHT,
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COL_ERROR,
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COL_PENCIL,
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NCOLOURS
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};
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struct game_params {
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/*
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* For a square puzzle, `c' and `r' indicate the puzzle
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* parameters as described above.
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*
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* A jigsaw-style puzzle is indicated by r==1, in which case c
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* can be whatever it likes (there is no constraint on
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* compositeness - a 7x7 jigsaw sudoku makes perfect sense).
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*/
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int c, r, symm, diff;
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int xtype; /* require all digits in X-diagonals */
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};
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struct block_structure {
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int refcount;
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/*
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* For text formatting, we do need c and r here.
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*/
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int c, r;
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/*
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* For any square index, whichblock[i] gives its block index.
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*
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* For 0 <= b,i < cr, blocks[b][i] gives the index of the ith
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* square in block b.
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*
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* whichblock and blocks are each dynamically allocated in
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* their own right, but the subarrays in blocks are appended
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* to the whichblock array, so shouldn't be freed
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* individually.
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*/
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int *whichblock, **blocks;
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#ifdef STANDALONE_SOLVER
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/*
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* Textual descriptions of each block. For normal Sudoku these
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* are of the form "(1,3)"; for jigsaw they are "starting at
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* (5,7)". So the sensible usage in both cases is to say
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* "elimination within block %s" with one of these strings.
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*
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* Only blocknames itself needs individually freeing; it's all
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* one block.
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*/
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char **blocknames;
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#endif
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};
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struct game_state {
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/*
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* For historical reasons, I use `cr' to denote the overall
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* width/height of the puzzle. It was a natural notation when
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* all puzzles were divided into blocks in a grid, but doesn't
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* really make much sense given jigsaw puzzles. However, the
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* obvious `n' is heavily used in the solver to describe the
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* index of a number being placed, so `cr' will have to stay.
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*/
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int cr;
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struct block_structure *blocks;
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int xtype;
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digit *grid;
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unsigned char *pencil; /* c*r*c*r elements */
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unsigned char *immutable; /* marks which digits are clues */
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int completed, cheated;
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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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ret->c = ret->r = 3;
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ret->xtype = FALSE;
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ret->symm = SYMM_ROT2; /* a plausible default */
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ret->diff = DIFF_BLOCK; /* so is this */
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return ret;
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}
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static void free_params(game_params *params)
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{
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sfree(params);
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}
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static game_params *dup_params(game_params *params)
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{
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game_params *ret = snew(game_params);
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*ret = *params; /* structure copy */
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return ret;
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}
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static int game_fetch_preset(int i, char **name, game_params **params)
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{
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static struct {
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char *title;
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game_params params;
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} presets[] = {
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{ "2x2 Trivial", { 2, 2, SYMM_ROT2, DIFF_BLOCK, FALSE } },
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{ "2x3 Basic", { 2, 3, SYMM_ROT2, DIFF_SIMPLE, FALSE } },
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{ "3x3 Trivial", { 3, 3, SYMM_ROT2, DIFF_BLOCK, FALSE } },
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{ "3x3 Basic", { 3, 3, SYMM_ROT2, DIFF_SIMPLE, FALSE } },
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{ "3x3 Basic X", { 3, 3, SYMM_ROT2, DIFF_SIMPLE, TRUE } },
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{ "3x3 Intermediate", { 3, 3, SYMM_ROT2, DIFF_INTERSECT, FALSE } },
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{ "3x3 Advanced", { 3, 3, SYMM_ROT2, DIFF_SET, FALSE } },
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{ "3x3 Advanced X", { 3, 3, SYMM_ROT2, DIFF_SET, TRUE } },
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{ "3x3 Extreme", { 3, 3, SYMM_ROT2, DIFF_EXTREME, FALSE } },
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{ "3x3 Unreasonable", { 3, 3, SYMM_ROT2, DIFF_RECURSIVE, FALSE } },
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{ "9 Jigsaw Basic", { 9, 1, SYMM_ROT2, DIFF_SIMPLE, FALSE } },
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{ "9 Jigsaw Basic X", { 9, 1, SYMM_ROT2, DIFF_SIMPLE, TRUE } },
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{ "9 Jigsaw Advanced", { 9, 1, SYMM_ROT2, DIFF_SET, FALSE } },
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#ifndef SLOW_SYSTEM
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{ "3x4 Basic", { 3, 4, SYMM_ROT2, DIFF_SIMPLE, FALSE } },
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{ "4x4 Basic", { 4, 4, SYMM_ROT2, DIFF_SIMPLE, FALSE } },
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#endif
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};
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if (i < 0 || i >= lenof(presets))
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return FALSE;
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*name = dupstr(presets[i].title);
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*params = dup_params(&presets[i].params);
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return TRUE;
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}
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static void decode_params(game_params *ret, char const *string)
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{
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int seen_r = FALSE;
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ret->c = ret->r = atoi(string);
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ret->xtype = FALSE;
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while (*string && isdigit((unsigned char)*string)) string++;
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if (*string == 'x') {
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string++;
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ret->r = atoi(string);
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seen_r = TRUE;
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while (*string && isdigit((unsigned char)*string)) string++;
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}
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while (*string) {
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if (*string == 'j') {
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string++;
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if (seen_r)
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ret->c *= ret->r;
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ret->r = 1;
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} else if (*string == 'x') {
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string++;
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ret->xtype = TRUE;
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} else if (*string == 'r' || *string == 'm' || *string == 'a') {
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int sn, sc, sd;
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sc = *string++;
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if (sc == 'm' && *string == 'd') {
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sd = TRUE;
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string++;
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} else {
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sd = FALSE;
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}
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sn = atoi(string);
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while (*string && isdigit((unsigned char)*string)) string++;
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if (sc == 'm' && sn == 8)
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ret->symm = SYMM_REF8;
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if (sc == 'm' && sn == 4)
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ret->symm = sd ? SYMM_REF4D : SYMM_REF4;
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if (sc == 'm' && sn == 2)
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ret->symm = sd ? SYMM_REF2D : SYMM_REF2;
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if (sc == 'r' && sn == 4)
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ret->symm = SYMM_ROT4;
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if (sc == 'r' && sn == 2)
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ret->symm = SYMM_ROT2;
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if (sc == 'a')
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ret->symm = SYMM_NONE;
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} else if (*string == 'd') {
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string++;
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if (*string == 't') /* trivial */
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string++, ret->diff = DIFF_BLOCK;
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else if (*string == 'b') /* basic */
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string++, ret->diff = DIFF_SIMPLE;
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else if (*string == 'i') /* intermediate */
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string++, ret->diff = DIFF_INTERSECT;
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else if (*string == 'a') /* advanced */
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string++, ret->diff = DIFF_SET;
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else if (*string == 'e') /* extreme */
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string++, ret->diff = DIFF_EXTREME;
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else if (*string == 'u') /* unreasonable */
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string++, ret->diff = DIFF_RECURSIVE;
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} else
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string++; /* eat unknown character */
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}
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}
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static char *encode_params(game_params *params, int full)
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{
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char str[80];
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if (params->r > 1)
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sprintf(str, "%dx%d", params->c, params->r);
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else
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sprintf(str, "%dj", params->c);
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if (params->xtype)
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strcat(str, "x");
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if (full) {
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switch (params->symm) {
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case SYMM_REF8: strcat(str, "m8"); break;
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case SYMM_REF4: strcat(str, "m4"); break;
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case SYMM_REF4D: strcat(str, "md4"); break;
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case SYMM_REF2: strcat(str, "m2"); break;
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case SYMM_REF2D: strcat(str, "md2"); break;
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case SYMM_ROT4: strcat(str, "r4"); break;
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/* case SYMM_ROT2: strcat(str, "r2"); break; [default] */
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case SYMM_NONE: strcat(str, "a"); break;
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}
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switch (params->diff) {
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/* case DIFF_BLOCK: strcat(str, "dt"); break; [default] */
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case DIFF_SIMPLE: strcat(str, "db"); break;
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case DIFF_INTERSECT: strcat(str, "di"); break;
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case DIFF_SET: strcat(str, "da"); break;
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case DIFF_EXTREME: strcat(str, "de"); break;
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case DIFF_RECURSIVE: strcat(str, "du"); break;
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}
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}
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return dupstr(str);
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}
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static config_item *game_configure(game_params *params)
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{
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config_item *ret;
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char buf[80];
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ret = snewn(7, config_item);
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ret[0].name = "Columns of sub-blocks";
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ret[0].type = C_STRING;
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sprintf(buf, "%d", params->c);
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ret[0].sval = dupstr(buf);
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ret[0].ival = 0;
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ret[1].name = "Rows of sub-blocks";
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ret[1].type = C_STRING;
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sprintf(buf, "%d", params->r);
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ret[1].sval = dupstr(buf);
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ret[1].ival = 0;
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ret[2].name = "\"X\" (require every number in each main diagonal)";
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ret[2].type = C_BOOLEAN;
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ret[2].sval = NULL;
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ret[2].ival = params->xtype;
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ret[3].name = "Jigsaw (irregularly shaped sub-blocks)";
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ret[3].type = C_BOOLEAN;
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ret[3].sval = NULL;
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ret[3].ival = (params->r == 1);
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ret[4].name = "Symmetry";
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ret[4].type = C_CHOICES;
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ret[4].sval = ":None:2-way rotation:4-way rotation:2-way mirror:"
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"2-way diagonal mirror:4-way mirror:4-way diagonal mirror:"
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"8-way mirror";
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ret[4].ival = params->symm;
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ret[5].name = "Difficulty";
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ret[5].type = C_CHOICES;
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ret[5].sval = ":Trivial:Basic:Intermediate:Advanced:Extreme:Unreasonable";
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ret[5].ival = params->diff;
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ret[6].name = NULL;
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ret[6].type = C_END;
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ret[6].sval = NULL;
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ret[6].ival = 0;
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return ret;
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}
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static game_params *custom_params(config_item *cfg)
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{
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game_params *ret = snew(game_params);
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ret->c = atoi(cfg[0].sval);
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ret->r = atoi(cfg[1].sval);
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ret->xtype = cfg[2].ival;
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if (cfg[3].ival) {
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ret->c *= ret->r;
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ret->r = 1;
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}
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ret->symm = cfg[4].ival;
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ret->diff = cfg[5].ival;
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return ret;
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}
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static char *validate_params(game_params *params, int full)
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{
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if (params->c < 2)
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return "Both dimensions must be at least 2";
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if (params->c > ORDER_MAX || params->r > ORDER_MAX)
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return "Dimensions greater than "STR(ORDER_MAX)" are not supported";
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if ((params->c * params->r) > 35)
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return "Unable to support more than 35 distinct symbols in a puzzle";
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return NULL;
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}
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/* ----------------------------------------------------------------------
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* Solver.
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*
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* This solver is used for two purposes:
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* + to check solubility of a grid as we gradually remove numbers
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* from it
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* + to solve an externally generated puzzle when the user selects
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* `Solve'.
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*
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* It supports a variety of specific modes of reasoning. By
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* enabling or disabling subsets of these modes we can arrange a
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* range of difficulty levels.
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*/
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/*
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* Modes of reasoning currently supported:
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*
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* - Positional elimination: a number must go in a particular
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* square because all the other empty squares in a given
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* row/col/blk are ruled out.
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*
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* - Numeric elimination: a square must have a particular number
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* in because all the other numbers that could go in it are
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* ruled out.
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*
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* - Intersectional analysis: given two domains which overlap
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* (hence one must be a block, and the other can be a row or
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* col), if the possible locations for a particular number in
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* one of the domains can be narrowed down to the overlap, then
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* that number can be ruled out everywhere but the overlap in
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* the other domain too.
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*
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* - Set elimination: if there is a subset of the empty squares
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* within a domain such that the union of the possible numbers
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* in that subset has the same size as the subset itself, then
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* those numbers can be ruled out everywhere else in the domain.
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* (For example, if there are five empty squares and the
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* possible numbers in each are 12, 23, 13, 134 and 1345, then
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* the first three empty squares form such a subset: the numbers
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* 1, 2 and 3 _must_ be in those three squares in some
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* permutation, and hence we can deduce none of them can be in
|
|
* the fourth or fifth squares.)
|
|
* + You can also see this the other way round, concentrating
|
|
* on numbers rather than squares: if there is a subset of
|
|
* the unplaced numbers within a domain such that the union
|
|
* of all their possible positions has the same size as the
|
|
* subset itself, then all other numbers can be ruled out for
|
|
* those positions. However, it turns out that this is
|
|
* exactly equivalent to the first formulation at all times:
|
|
* there is a 1-1 correspondence between suitable subsets of
|
|
* the unplaced numbers and suitable subsets of the unfilled
|
|
* places, found by taking the _complement_ of the union of
|
|
* the numbers' possible positions (or the spaces' possible
|
|
* contents).
|
|
*
|
|
* - Forcing chains (see comment for solver_forcing().)
|
|
*
|
|
* - Recursion. If all else fails, we pick one of the currently
|
|
* most constrained empty squares and take a random guess at its
|
|
* contents, then continue solving on that basis and see if we
|
|
* get any further.
|
|
*/
|
|
|
|
struct solver_usage {
|
|
int cr;
|
|
struct block_structure *blocks;
|
|
/*
|
|
* We set up a cubic array, indexed by x, y and digit; each
|
|
* element of this array is TRUE or FALSE according to whether
|
|
* or not that digit _could_ in principle go in that position.
|
|
*
|
|
* The way to index this array is cube[(y*cr+x)*cr+n-1]; there
|
|
* are macros below to help with this.
|
|
*/
|
|
unsigned char *cube;
|
|
/*
|
|
* This is the grid in which we write down our final
|
|
* deductions. y-coordinates in here are _not_ transformed.
|
|
*/
|
|
digit *grid;
|
|
/*
|
|
* Now we keep track, at a slightly higher level, of what we
|
|
* have yet to work out, to prevent doing the same deduction
|
|
* many times.
|
|
*/
|
|
/* row[y*cr+n-1] TRUE if digit n has been placed in row y */
|
|
unsigned char *row;
|
|
/* col[x*cr+n-1] TRUE if digit n has been placed in row x */
|
|
unsigned char *col;
|
|
/* blk[i*cr+n-1] TRUE if digit n has been placed in block i */
|
|
unsigned char *blk;
|
|
/* diag[i*cr+n-1] TRUE if digit n has been placed in diagonal i */
|
|
unsigned char *diag; /* diag 0 is \, 1 is / */
|
|
};
|
|
#define cubepos2(xy,n) ((xy)*usage->cr+(n)-1)
|
|
#define cubepos(x,y,n) cubepos2((y)*usage->cr+(x),n)
|
|
#define cube(x,y,n) (usage->cube[cubepos(x,y,n)])
|
|
#define cube2(xy,n) (usage->cube[cubepos2(xy,n)])
|
|
|
|
#define ondiag0(xy) ((xy) % (cr+1) == 0)
|
|
#define ondiag1(xy) ((xy) % (cr-1) == 0 && (xy) > 0 && (xy) < cr*cr-1)
|
|
#define diag0(i) ((i) * (cr+1))
|
|
#define diag1(i) ((i+1) * (cr-1))
|
|
|
|
/*
|
|
* Function called when we are certain that a particular square has
|
|
* a particular number in it. The y-coordinate passed in here is
|
|
* transformed.
|
|
*/
|
|
static void solver_place(struct solver_usage *usage, int x, int y, int n)
|
|
{
|
|
int cr = usage->cr;
|
|
int sqindex = y*cr+x;
|
|
int i, bi;
|
|
|
|
assert(cube(x,y,n));
|
|
|
|
/*
|
|
* Rule out all other numbers in this square.
|
|
*/
|
|
for (i = 1; i <= cr; i++)
|
|
if (i != n)
|
|
cube(x,y,i) = FALSE;
|
|
|
|
/*
|
|
* Rule out this number in all other positions in the row.
|
|
*/
|
|
for (i = 0; i < cr; i++)
|
|
if (i != y)
|
|
cube(x,i,n) = FALSE;
|
|
|
|
/*
|
|
* Rule out this number in all other positions in the column.
|
|
*/
|
|
for (i = 0; i < cr; i++)
|
|
if (i != x)
|
|
cube(i,y,n) = FALSE;
|
|
|
|
/*
|
|
* Rule out this number in all other positions in the block.
|
|
*/
|
|
bi = usage->blocks->whichblock[sqindex];
|
|
for (i = 0; i < cr; i++) {
|
|
int bp = usage->blocks->blocks[bi][i];
|
|
if (bp != sqindex)
|
|
cube2(bp,n) = FALSE;
|
|
}
|
|
|
|
/*
|
|
* Enter the number in the result grid.
|
|
*/
|
|
usage->grid[sqindex] = n;
|
|
|
|
/*
|
|
* Cross out this number from the list of numbers left to place
|
|
* in its row, its column and its block.
|
|
*/
|
|
usage->row[y*cr+n-1] = usage->col[x*cr+n-1] =
|
|
usage->blk[bi*cr+n-1] = TRUE;
|
|
|
|
if (usage->diag) {
|
|
if (ondiag0(sqindex)) {
|
|
for (i = 0; i < cr; i++)
|
|
if (diag0(i) != sqindex)
|
|
cube2(diag0(i),n) = FALSE;
|
|
usage->diag[n-1] = TRUE;
|
|
}
|
|
if (ondiag1(sqindex)) {
|
|
for (i = 0; i < cr; i++)
|
|
if (diag1(i) != sqindex)
|
|
cube2(diag1(i),n) = FALSE;
|
|
usage->diag[cr+n-1] = TRUE;
|
|
}
|
|
}
|
|
}
|
|
|
|
static int solver_elim(struct solver_usage *usage, int *indices
|
|
#ifdef STANDALONE_SOLVER
|
|
, char *fmt, ...
|
|
#endif
|
|
)
|
|
{
|
|
int cr = usage->cr;
|
|
int fpos, m, i;
|
|
|
|
/*
|
|
* Count the number of set bits within this section of the
|
|
* cube.
|
|
*/
|
|
m = 0;
|
|
fpos = -1;
|
|
for (i = 0; i < cr; i++)
|
|
if (usage->cube[indices[i]]) {
|
|
fpos = indices[i];
|
|
m++;
|
|
}
|
|
|
|
if (m == 1) {
|
|
int x, y, n;
|
|
assert(fpos >= 0);
|
|
|
|
n = 1 + fpos % cr;
|
|
x = fpos / cr;
|
|
y = x / cr;
|
|
x %= cr;
|
|
|
|
if (!usage->grid[y*cr+x]) {
|
|
#ifdef STANDALONE_SOLVER
|
|
if (solver_show_working) {
|
|
va_list ap;
|
|
printf("%*s", solver_recurse_depth*4, "");
|
|
va_start(ap, fmt);
|
|
vprintf(fmt, ap);
|
|
va_end(ap);
|
|
printf(":\n%*s placing %d at (%d,%d)\n",
|
|
solver_recurse_depth*4, "", n, 1+x, 1+y);
|
|
}
|
|
#endif
|
|
solver_place(usage, x, y, n);
|
|
return +1;
|
|
}
|
|
} else if (m == 0) {
|
|
#ifdef STANDALONE_SOLVER
|
|
if (solver_show_working) {
|
|
va_list ap;
|
|
printf("%*s", solver_recurse_depth*4, "");
|
|
va_start(ap, fmt);
|
|
vprintf(fmt, ap);
|
|
va_end(ap);
|
|
printf(":\n%*s no possibilities available\n",
|
|
solver_recurse_depth*4, "");
|
|
}
|
|
#endif
|
|
return -1;
|
|
}
|
|
|
|
return 0;
|
|
}
|
|
|
|
static int solver_intersect(struct solver_usage *usage,
|
|
int *indices1, int *indices2
|
|
#ifdef STANDALONE_SOLVER
|
|
, char *fmt, ...
|
|
#endif
|
|
)
|
|
{
|
|
int cr = usage->cr;
|
|
int ret, i, j;
|
|
|
|
/*
|
|
* Loop over the first domain and see if there's any set bit
|
|
* not also in the second.
|
|
*/
|
|
for (i = j = 0; i < cr; i++) {
|
|
int p = indices1[i];
|
|
while (j < cr && indices2[j] < p)
|
|
j++;
|
|
if (usage->cube[p]) {
|
|
if (j < cr && indices2[j] == p)
|
|
continue; /* both domains contain this index */
|
|
else
|
|
return 0; /* there is, so we can't deduce */
|
|
}
|
|
}
|
|
|
|
/*
|
|
* We have determined that all set bits in the first domain are
|
|
* within its overlap with the second. So loop over the second
|
|
* domain and remove all set bits that aren't also in that
|
|
* overlap; return +1 iff we actually _did_ anything.
|
|
*/
|
|
ret = 0;
|
|
for (i = j = 0; i < cr; i++) {
|
|
int p = indices2[i];
|
|
while (j < cr && indices1[j] < p)
|
|
j++;
|
|
if (usage->cube[p] && (j >= cr || indices1[j] != p)) {
|
|
#ifdef STANDALONE_SOLVER
|
|
if (solver_show_working) {
|
|
int px, py, pn;
|
|
|
|
if (!ret) {
|
|
va_list ap;
|
|
printf("%*s", solver_recurse_depth*4, "");
|
|
va_start(ap, fmt);
|
|
vprintf(fmt, ap);
|
|
va_end(ap);
|
|
printf(":\n");
|
|
}
|
|
|
|
pn = 1 + p % cr;
|
|
px = p / cr;
|
|
py = px / cr;
|
|
px %= cr;
|
|
|
|
printf("%*s ruling out %d at (%d,%d)\n",
|
|
solver_recurse_depth*4, "", pn, 1+px, 1+py);
|
|
}
|
|
#endif
|
|
ret = +1; /* we did something */
|
|
usage->cube[p] = 0;
|
|
}
|
|
}
|
|
|
|
return ret;
|
|
}
|
|
|
|
struct solver_scratch {
|
|
unsigned char *grid, *rowidx, *colidx, *set;
|
|
int *neighbours, *bfsqueue;
|
|
int *indexlist, *indexlist2;
|
|
#ifdef STANDALONE_SOLVER
|
|
int *bfsprev;
|
|
#endif
|
|
};
|
|
|
|
static int solver_set(struct solver_usage *usage,
|
|
struct solver_scratch *scratch,
|
|
int *indices
|
|
#ifdef STANDALONE_SOLVER
|
|
, char *fmt, ...
|
|
#endif
|
|
)
|
|
{
|
|
int cr = usage->cr;
|
|
int i, j, n, count;
|
|
unsigned char *grid = scratch->grid;
|
|
unsigned char *rowidx = scratch->rowidx;
|
|
unsigned char *colidx = scratch->colidx;
|
|
unsigned char *set = scratch->set;
|
|
|
|
/*
|
|
* We are passed a cr-by-cr matrix of booleans. Our first job
|
|
* is to winnow it by finding any definite placements - i.e.
|
|
* any row with a solitary 1 - and discarding that row and the
|
|
* column containing the 1.
|
|
*/
|
|
memset(rowidx, TRUE, cr);
|
|
memset(colidx, TRUE, cr);
|
|
for (i = 0; i < cr; i++) {
|
|
int count = 0, first = -1;
|
|
for (j = 0; j < cr; j++)
|
|
if (usage->cube[indices[i*cr+j]])
|
|
first = j, count++;
|
|
|
|
/*
|
|
* If count == 0, then there's a row with no 1s at all and
|
|
* the puzzle is internally inconsistent. However, we ought
|
|
* to have caught this already during the simpler reasoning
|
|
* methods, so we can safely fail an assertion if we reach
|
|
* this point here.
|
|
*/
|
|
assert(count > 0);
|
|
if (count == 1)
|
|
rowidx[i] = colidx[first] = FALSE;
|
|
}
|
|
|
|
/*
|
|
* Convert each of rowidx/colidx from a list of 0s and 1s to a
|
|
* list of the indices of the 1s.
|
|
*/
|
|
for (i = j = 0; i < cr; i++)
|
|
if (rowidx[i])
|
|
rowidx[j++] = i;
|
|
n = j;
|
|
for (i = j = 0; i < cr; i++)
|
|
if (colidx[i])
|
|
colidx[j++] = i;
|
|
assert(n == j);
|
|
|
|
/*
|
|
* And create the smaller matrix.
|
|
*/
|
|
for (i = 0; i < n; i++)
|
|
for (j = 0; j < n; j++)
|
|
grid[i*cr+j] = usage->cube[indices[rowidx[i]*cr+colidx[j]]];
|
|
|
|
/*
|
|
* Having done that, we now have a matrix in which every row
|
|
* has at least two 1s in. Now we search to see if we can find
|
|
* a rectangle of zeroes (in the set-theoretic sense of
|
|
* `rectangle', i.e. a subset of rows crossed with a subset of
|
|
* columns) whose width and height add up to n.
|
|
*/
|
|
|
|
memset(set, 0, n);
|
|
count = 0;
|
|
while (1) {
|
|
/*
|
|
* We have a candidate set. If its size is <=1 or >=n-1
|
|
* then we move on immediately.
|
|
*/
|
|
if (count > 1 && count < n-1) {
|
|
/*
|
|
* The number of rows we need is n-count. See if we can
|
|
* find that many rows which each have a zero in all
|
|
* the positions listed in `set'.
|
|
*/
|
|
int rows = 0;
|
|
for (i = 0; i < n; i++) {
|
|
int ok = TRUE;
|
|
for (j = 0; j < n; j++)
|
|
if (set[j] && grid[i*cr+j]) {
|
|
ok = FALSE;
|
|
break;
|
|
}
|
|
if (ok)
|
|
rows++;
|
|
}
|
|
|
|
/*
|
|
* We expect never to be able to get _more_ than
|
|
* n-count suitable rows: this would imply that (for
|
|
* example) there are four numbers which between them
|
|
* have at most three possible positions, and hence it
|
|
* indicates a faulty deduction before this point or
|
|
* even a bogus clue.
|
|
*/
|
|
if (rows > n - count) {
|
|
#ifdef STANDALONE_SOLVER
|
|
if (solver_show_working) {
|
|
va_list ap;
|
|
printf("%*s", solver_recurse_depth*4,
|
|
"");
|
|
va_start(ap, fmt);
|
|
vprintf(fmt, ap);
|
|
va_end(ap);
|
|
printf(":\n%*s contradiction reached\n",
|
|
solver_recurse_depth*4, "");
|
|
}
|
|
#endif
|
|
return -1;
|
|
}
|
|
|
|
if (rows >= n - count) {
|
|
int progress = FALSE;
|
|
|
|
/*
|
|
* We've got one! Now, for each row which _doesn't_
|
|
* satisfy the criterion, eliminate all its set
|
|
* bits in the positions _not_ listed in `set'.
|
|
* Return +1 (meaning progress has been made) if we
|
|
* successfully eliminated anything at all.
|
|
*
|
|
* This involves referring back through
|
|
* rowidx/colidx in order to work out which actual
|
|
* positions in the cube to meddle with.
|
|
*/
|
|
for (i = 0; i < n; i++) {
|
|
int ok = TRUE;
|
|
for (j = 0; j < n; j++)
|
|
if (set[j] && grid[i*cr+j]) {
|
|
ok = FALSE;
|
|
break;
|
|
}
|
|
if (!ok) {
|
|
for (j = 0; j < n; j++)
|
|
if (!set[j] && grid[i*cr+j]) {
|
|
int fpos = indices[rowidx[i]*cr+colidx[j]];
|
|
#ifdef STANDALONE_SOLVER
|
|
if (solver_show_working) {
|
|
int px, py, pn;
|
|
|
|
if (!progress) {
|
|
va_list ap;
|
|
printf("%*s", solver_recurse_depth*4,
|
|
"");
|
|
va_start(ap, fmt);
|
|
vprintf(fmt, ap);
|
|
va_end(ap);
|
|
printf(":\n");
|
|
}
|
|
|
|
pn = 1 + fpos % cr;
|
|
px = fpos / cr;
|
|
py = px / cr;
|
|
px %= cr;
|
|
|
|
printf("%*s ruling out %d at (%d,%d)\n",
|
|
solver_recurse_depth*4, "",
|
|
pn, 1+px, 1+py);
|
|
}
|
|
#endif
|
|
progress = TRUE;
|
|
usage->cube[fpos] = FALSE;
|
|
}
|
|
}
|
|
}
|
|
|
|
if (progress) {
|
|
return +1;
|
|
}
|
|
}
|
|
}
|
|
|
|
/*
|
|
* Binary increment: change the rightmost 0 to a 1, and
|
|
* change all 1s to the right of it to 0s.
|
|
*/
|
|
i = n;
|
|
while (i > 0 && set[i-1])
|
|
set[--i] = 0, count--;
|
|
if (i > 0)
|
|
set[--i] = 1, count++;
|
|
else
|
|
break; /* done */
|
|
}
|
|
|
|
return 0;
|
|
}
|
|
|
|
/*
|
|
* Look for forcing chains. A forcing chain is a path of
|
|
* pairwise-exclusive squares (i.e. each pair of adjacent squares
|
|
* in the path are in the same row, column or block) with the
|
|
* following properties:
|
|
*
|
|
* (a) Each square on the path has precisely two possible numbers.
|
|
*
|
|
* (b) Each pair of squares which are adjacent on the path share
|
|
* at least one possible number in common.
|
|
*
|
|
* (c) Each square in the middle of the path shares _both_ of its
|
|
* numbers with at least one of its neighbours (not the same
|
|
* one with both neighbours).
|
|
*
|
|
* These together imply that at least one of the possible number
|
|
* choices at one end of the path forces _all_ the rest of the
|
|
* numbers along the path. In order to make real use of this, we
|
|
* need further properties:
|
|
*
|
|
* (c) Ruling out some number N from the square at one end of the
|
|
* path forces the square at the other end to take the same
|
|
* number N.
|
|
*
|
|
* (d) The two end squares are both in line with some third
|
|
* square.
|
|
*
|
|
* (e) That third square currently has N as a possibility.
|
|
*
|
|
* If we can find all of that lot, we can deduce that at least one
|
|
* of the two ends of the forcing chain has number N, and that
|
|
* therefore the mutually adjacent third square does not.
|
|
*
|
|
* To find forcing chains, we're going to start a bfs at each
|
|
* suitable square, once for each of its two possible numbers.
|
|
*/
|
|
static int solver_forcing(struct solver_usage *usage,
|
|
struct solver_scratch *scratch)
|
|
{
|
|
int cr = usage->cr;
|
|
int *bfsqueue = scratch->bfsqueue;
|
|
#ifdef STANDALONE_SOLVER
|
|
int *bfsprev = scratch->bfsprev;
|
|
#endif
|
|
unsigned char *number = scratch->grid;
|
|
int *neighbours = scratch->neighbours;
|
|
int x, y;
|
|
|
|
for (y = 0; y < cr; y++)
|
|
for (x = 0; x < cr; x++) {
|
|
int count, t, n;
|
|
|
|
/*
|
|
* If this square doesn't have exactly two candidate
|
|
* numbers, don't try it.
|
|
*
|
|
* In this loop we also sum the candidate numbers,
|
|
* which is a nasty hack to allow us to quickly find
|
|
* `the other one' (since we will shortly know there
|
|
* are exactly two).
|
|
*/
|
|
for (count = t = 0, n = 1; n <= cr; n++)
|
|
if (cube(x, y, n))
|
|
count++, t += n;
|
|
if (count != 2)
|
|
continue;
|
|
|
|
/*
|
|
* Now attempt a bfs for each candidate.
|
|
*/
|
|
for (n = 1; n <= cr; n++)
|
|
if (cube(x, y, n)) {
|
|
int orign, currn, head, tail;
|
|
|
|
/*
|
|
* Begin a bfs.
|
|
*/
|
|
orign = n;
|
|
|
|
memset(number, cr+1, cr*cr);
|
|
head = tail = 0;
|
|
bfsqueue[tail++] = y*cr+x;
|
|
#ifdef STANDALONE_SOLVER
|
|
bfsprev[y*cr+x] = -1;
|
|
#endif
|
|
number[y*cr+x] = t - n;
|
|
|
|
while (head < tail) {
|
|
int xx, yy, nneighbours, xt, yt, i;
|
|
|
|
xx = bfsqueue[head++];
|
|
yy = xx / cr;
|
|
xx %= cr;
|
|
|
|
currn = number[yy*cr+xx];
|
|
|
|
/*
|
|
* Find neighbours of yy,xx.
|
|
*/
|
|
nneighbours = 0;
|
|
for (yt = 0; yt < cr; yt++)
|
|
neighbours[nneighbours++] = yt*cr+xx;
|
|
for (xt = 0; xt < cr; xt++)
|
|
neighbours[nneighbours++] = yy*cr+xt;
|
|
xt = usage->blocks->whichblock[yy*cr+xx];
|
|
for (yt = 0; yt < cr; yt++)
|
|
neighbours[nneighbours++] = usage->blocks->blocks[xt][yt];
|
|
if (usage->diag) {
|
|
int sqindex = yy*cr+xx;
|
|
if (ondiag0(sqindex)) {
|
|
for (i = 0; i < cr; i++)
|
|
neighbours[nneighbours++] = diag0(i);
|
|
}
|
|
if (ondiag1(sqindex)) {
|
|
for (i = 0; i < cr; i++)
|
|
neighbours[nneighbours++] = diag1(i);
|
|
}
|
|
}
|
|
|
|
/*
|
|
* Try visiting each of those neighbours.
|
|
*/
|
|
for (i = 0; i < nneighbours; i++) {
|
|
int cc, tt, nn;
|
|
|
|
xt = neighbours[i] % cr;
|
|
yt = neighbours[i] / cr;
|
|
|
|
/*
|
|
* We need this square to not be
|
|
* already visited, and to include
|
|
* currn as a possible number.
|
|
*/
|
|
if (number[yt*cr+xt] <= cr)
|
|
continue;
|
|
if (!cube(xt, yt, currn))
|
|
continue;
|
|
|
|
/*
|
|
* Don't visit _this_ square a second
|
|
* time!
|
|
*/
|
|
if (xt == xx && yt == yy)
|
|
continue;
|
|
|
|
/*
|
|
* To continue with the bfs, we need
|
|
* this square to have exactly two
|
|
* possible numbers.
|
|
*/
|
|
for (cc = tt = 0, nn = 1; nn <= cr; nn++)
|
|
if (cube(xt, yt, nn))
|
|
cc++, tt += nn;
|
|
if (cc == 2) {
|
|
bfsqueue[tail++] = yt*cr+xt;
|
|
#ifdef STANDALONE_SOLVER
|
|
bfsprev[yt*cr+xt] = yy*cr+xx;
|
|
#endif
|
|
number[yt*cr+xt] = tt - currn;
|
|
}
|
|
|
|
/*
|
|
* One other possibility is that this
|
|
* might be the square in which we can
|
|
* make a real deduction: if it's
|
|
* adjacent to x,y, and currn is equal
|
|
* to the original number we ruled out.
|
|
*/
|
|
if (currn == orign &&
|
|
(xt == x || yt == y ||
|
|
(usage->blocks->whichblock[yt*cr+xt] == usage->blocks->whichblock[y*cr+x]) ||
|
|
(usage->diag && ((ondiag0(yt*cr+xt) && ondiag0(y*cr+x)) ||
|
|
(ondiag1(yt*cr+xt) && ondiag1(y*cr+x)))))) {
|
|
#ifdef STANDALONE_SOLVER
|
|
if (solver_show_working) {
|
|
char *sep = "";
|
|
int xl, yl;
|
|
printf("%*sforcing chain, %d at ends of ",
|
|
solver_recurse_depth*4, "", orign);
|
|
xl = xx;
|
|
yl = yy;
|
|
while (1) {
|
|
printf("%s(%d,%d)", sep, 1+xl,
|
|
1+yl);
|
|
xl = bfsprev[yl*cr+xl];
|
|
if (xl < 0)
|
|
break;
|
|
yl = xl / cr;
|
|
xl %= cr;
|
|
sep = "-";
|
|
}
|
|
printf("\n%*s ruling out %d at (%d,%d)\n",
|
|
solver_recurse_depth*4, "",
|
|
orign, 1+xt, 1+yt);
|
|
}
|
|
#endif
|
|
cube(xt, yt, orign) = FALSE;
|
|
return 1;
|
|
}
|
|
}
|
|
}
|
|
}
|
|
}
|
|
|
|
return 0;
|
|
}
|
|
|
|
static struct solver_scratch *solver_new_scratch(struct solver_usage *usage)
|
|
{
|
|
struct solver_scratch *scratch = snew(struct solver_scratch);
|
|
int cr = usage->cr;
|
|
scratch->grid = snewn(cr*cr, unsigned char);
|
|
scratch->rowidx = snewn(cr, unsigned char);
|
|
scratch->colidx = snewn(cr, unsigned char);
|
|
scratch->set = snewn(cr, unsigned char);
|
|
scratch->neighbours = snewn(5*cr, int);
|
|
scratch->bfsqueue = snewn(cr*cr, int);
|
|
#ifdef STANDALONE_SOLVER
|
|
scratch->bfsprev = snewn(cr*cr, int);
|
|
#endif
|
|
scratch->indexlist = snewn(cr*cr, int); /* used for set elimination */
|
|
scratch->indexlist2 = snewn(cr, int); /* only used for intersect() */
|
|
return scratch;
|
|
}
|
|
|
|
static void solver_free_scratch(struct solver_scratch *scratch)
|
|
{
|
|
#ifdef STANDALONE_SOLVER
|
|
sfree(scratch->bfsprev);
|
|
#endif
|
|
sfree(scratch->bfsqueue);
|
|
sfree(scratch->neighbours);
|
|
sfree(scratch->set);
|
|
sfree(scratch->colidx);
|
|
sfree(scratch->rowidx);
|
|
sfree(scratch->grid);
|
|
sfree(scratch->indexlist);
|
|
sfree(scratch->indexlist2);
|
|
sfree(scratch);
|
|
}
|
|
|
|
static int solver(int cr, struct block_structure *blocks, int xtype,
|
|
digit *grid, int maxdiff)
|
|
{
|
|
struct solver_usage *usage;
|
|
struct solver_scratch *scratch;
|
|
int x, y, b, i, n, ret;
|
|
int diff = DIFF_BLOCK;
|
|
|
|
/*
|
|
* Set up a usage structure as a clean slate (everything
|
|
* possible).
|
|
*/
|
|
usage = snew(struct solver_usage);
|
|
usage->cr = cr;
|
|
usage->blocks = blocks;
|
|
usage->cube = snewn(cr*cr*cr, unsigned char);
|
|
usage->grid = grid; /* write straight back to the input */
|
|
memset(usage->cube, TRUE, cr*cr*cr);
|
|
|
|
usage->row = snewn(cr * cr, unsigned char);
|
|
usage->col = snewn(cr * cr, unsigned char);
|
|
usage->blk = snewn(cr * cr, unsigned char);
|
|
memset(usage->row, FALSE, cr * cr);
|
|
memset(usage->col, FALSE, cr * cr);
|
|
memset(usage->blk, FALSE, cr * cr);
|
|
|
|
if (xtype) {
|
|
usage->diag = snewn(cr * 2, unsigned char);
|
|
memset(usage->diag, FALSE, cr * 2);
|
|
} else
|
|
usage->diag = NULL;
|
|
|
|
scratch = solver_new_scratch(usage);
|
|
|
|
/*
|
|
* Place all the clue numbers we are given.
|
|
*/
|
|
for (x = 0; x < cr; x++)
|
|
for (y = 0; y < cr; y++)
|
|
if (grid[y*cr+x])
|
|
solver_place(usage, x, y, grid[y*cr+x]);
|
|
|
|
/*
|
|
* Now loop over the grid repeatedly trying all permitted modes
|
|
* of reasoning. The loop terminates if we complete an
|
|
* iteration without making any progress; we then return
|
|
* failure or success depending on whether the grid is full or
|
|
* not.
|
|
*/
|
|
while (1) {
|
|
/*
|
|
* I'd like to write `continue;' inside each of the
|
|
* following loops, so that the solver returns here after
|
|
* making some progress. However, I can't specify that I
|
|
* want to continue an outer loop rather than the innermost
|
|
* one, so I'm apologetically resorting to a goto.
|
|
*/
|
|
cont:
|
|
|
|
/*
|
|
* Blockwise positional elimination.
|
|
*/
|
|
for (b = 0; b < cr; b++)
|
|
for (n = 1; n <= cr; n++)
|
|
if (!usage->blk[b*cr+n-1]) {
|
|
for (i = 0; i < cr; i++)
|
|
scratch->indexlist[i] = cubepos2(usage->blocks->blocks[b][i],n);
|
|
ret = solver_elim(usage, scratch->indexlist
|
|
#ifdef STANDALONE_SOLVER
|
|
, "positional elimination,"
|
|
" %d in block %s", n,
|
|
usage->blocks->blocknames[b]
|
|
#endif
|
|
);
|
|
if (ret < 0) {
|
|
diff = DIFF_IMPOSSIBLE;
|
|
goto got_result;
|
|
} else if (ret > 0) {
|
|
diff = max(diff, DIFF_BLOCK);
|
|
goto cont;
|
|
}
|
|
}
|
|
|
|
if (maxdiff <= DIFF_BLOCK)
|
|
break;
|
|
|
|
/*
|
|
* Row-wise positional elimination.
|
|
*/
|
|
for (y = 0; y < cr; y++)
|
|
for (n = 1; n <= cr; n++)
|
|
if (!usage->row[y*cr+n-1]) {
|
|
for (x = 0; x < cr; x++)
|
|
scratch->indexlist[x] = cubepos(x, y, n);
|
|
ret = solver_elim(usage, scratch->indexlist
|
|
#ifdef STANDALONE_SOLVER
|
|
, "positional elimination,"
|
|
" %d in row %d", n, 1+y
|
|
#endif
|
|
);
|
|
if (ret < 0) {
|
|
diff = DIFF_IMPOSSIBLE;
|
|
goto got_result;
|
|
} else if (ret > 0) {
|
|
diff = max(diff, DIFF_SIMPLE);
|
|
goto cont;
|
|
}
|
|
}
|
|
/*
|
|
* Column-wise positional elimination.
|
|
*/
|
|
for (x = 0; x < cr; x++)
|
|
for (n = 1; n <= cr; n++)
|
|
if (!usage->col[x*cr+n-1]) {
|
|
for (y = 0; y < cr; y++)
|
|
scratch->indexlist[y] = cubepos(x, y, n);
|
|
ret = solver_elim(usage, scratch->indexlist
|
|
#ifdef STANDALONE_SOLVER
|
|
, "positional elimination,"
|
|
" %d in column %d", n, 1+x
|
|
#endif
|
|
);
|
|
if (ret < 0) {
|
|
diff = DIFF_IMPOSSIBLE;
|
|
goto got_result;
|
|
} else if (ret > 0) {
|
|
diff = max(diff, DIFF_SIMPLE);
|
|
goto cont;
|
|
}
|
|
}
|
|
|
|
/*
|
|
* X-diagonal positional elimination.
|
|
*/
|
|
if (usage->diag) {
|
|
for (n = 1; n <= cr; n++)
|
|
if (!usage->diag[n-1]) {
|
|
for (i = 0; i < cr; i++)
|
|
scratch->indexlist[i] = cubepos2(diag0(i), n);
|
|
ret = solver_elim(usage, scratch->indexlist
|
|
#ifdef STANDALONE_SOLVER
|
|
, "positional elimination,"
|
|
" %d in \\-diagonal", n
|
|
#endif
|
|
);
|
|
if (ret < 0) {
|
|
diff = DIFF_IMPOSSIBLE;
|
|
goto got_result;
|
|
} else if (ret > 0) {
|
|
diff = max(diff, DIFF_SIMPLE);
|
|
goto cont;
|
|
}
|
|
}
|
|
for (n = 1; n <= cr; n++)
|
|
if (!usage->diag[cr+n-1]) {
|
|
for (i = 0; i < cr; i++)
|
|
scratch->indexlist[i] = cubepos2(diag1(i), n);
|
|
ret = solver_elim(usage, scratch->indexlist
|
|
#ifdef STANDALONE_SOLVER
|
|
, "positional elimination,"
|
|
" %d in /-diagonal", n
|
|
#endif
|
|
);
|
|
if (ret < 0) {
|
|
diff = DIFF_IMPOSSIBLE;
|
|
goto got_result;
|
|
} else if (ret > 0) {
|
|
diff = max(diff, DIFF_SIMPLE);
|
|
goto cont;
|
|
}
|
|
}
|
|
}
|
|
|
|
/*
|
|
* Numeric elimination.
|
|
*/
|
|
for (x = 0; x < cr; x++)
|
|
for (y = 0; y < cr; y++)
|
|
if (!usage->grid[y*cr+x]) {
|
|
for (n = 1; n <= cr; n++)
|
|
scratch->indexlist[n-1] = cubepos(x, y, n);
|
|
ret = solver_elim(usage, scratch->indexlist
|
|
#ifdef STANDALONE_SOLVER
|
|
, "numeric elimination at (%d,%d)",
|
|
1+x, 1+y
|
|
#endif
|
|
);
|
|
if (ret < 0) {
|
|
diff = DIFF_IMPOSSIBLE;
|
|
goto got_result;
|
|
} else if (ret > 0) {
|
|
diff = max(diff, DIFF_SIMPLE);
|
|
goto cont;
|
|
}
|
|
}
|
|
|
|
if (maxdiff <= DIFF_SIMPLE)
|
|
break;
|
|
|
|
/*
|
|
* Intersectional analysis, rows vs blocks.
|
|
*/
|
|
for (y = 0; y < cr; y++)
|
|
for (b = 0; b < cr; b++)
|
|
for (n = 1; n <= cr; n++) {
|
|
if (usage->row[y*cr+n-1] ||
|
|
usage->blk[b*cr+n-1])
|
|
continue;
|
|
for (i = 0; i < cr; i++) {
|
|
scratch->indexlist[i] = cubepos(i, y, n);
|
|
scratch->indexlist2[i] = cubepos2(usage->blocks->blocks[b][i], n);
|
|
}
|
|
/*
|
|
* solver_intersect() never returns -1.
|
|
*/
|
|
if (solver_intersect(usage, scratch->indexlist,
|
|
scratch->indexlist2
|
|
#ifdef STANDALONE_SOLVER
|
|
, "intersectional analysis,"
|
|
" %d in row %d vs block %s",
|
|
n, 1+y, usage->blocks->blocknames[b]
|
|
#endif
|
|
) ||
|
|
solver_intersect(usage, scratch->indexlist2,
|
|
scratch->indexlist
|
|
#ifdef STANDALONE_SOLVER
|
|
, "intersectional analysis,"
|
|
" %d in block %s vs row %d",
|
|
n, usage->blocks->blocknames[b], 1+y
|
|
#endif
|
|
)) {
|
|
diff = max(diff, DIFF_INTERSECT);
|
|
goto cont;
|
|
}
|
|
}
|
|
|
|
/*
|
|
* Intersectional analysis, columns vs blocks.
|
|
*/
|
|
for (x = 0; x < cr; x++)
|
|
for (b = 0; b < cr; b++)
|
|
for (n = 1; n <= cr; n++) {
|
|
if (usage->col[x*cr+n-1] ||
|
|
usage->blk[b*cr+n-1])
|
|
continue;
|
|
for (i = 0; i < cr; i++) {
|
|
scratch->indexlist[i] = cubepos(x, i, n);
|
|
scratch->indexlist2[i] = cubepos2(usage->blocks->blocks[b][i], n);
|
|
}
|
|
if (solver_intersect(usage, scratch->indexlist,
|
|
scratch->indexlist2
|
|
#ifdef STANDALONE_SOLVER
|
|
, "intersectional analysis,"
|
|
" %d in column %d vs block %s",
|
|
n, 1+x, usage->blocks->blocknames[b]
|
|
#endif
|
|
) ||
|
|
solver_intersect(usage, scratch->indexlist2,
|
|
scratch->indexlist
|
|
#ifdef STANDALONE_SOLVER
|
|
, "intersectional analysis,"
|
|
" %d in block %s vs column %d",
|
|
n, usage->blocks->blocknames[b], 1+x
|
|
#endif
|
|
)) {
|
|
diff = max(diff, DIFF_INTERSECT);
|
|
goto cont;
|
|
}
|
|
}
|
|
|
|
if (usage->diag) {
|
|
/*
|
|
* Intersectional analysis, \-diagonal vs blocks.
|
|
*/
|
|
for (b = 0; b < cr; b++)
|
|
for (n = 1; n <= cr; n++) {
|
|
if (usage->diag[n-1] ||
|
|
usage->blk[b*cr+n-1])
|
|
continue;
|
|
for (i = 0; i < cr; i++) {
|
|
scratch->indexlist[i] = cubepos2(diag0(i), n);
|
|
scratch->indexlist2[i] = cubepos2(usage->blocks->blocks[b][i], n);
|
|
}
|
|
if (solver_intersect(usage, scratch->indexlist,
|
|
scratch->indexlist2
|
|
#ifdef STANDALONE_SOLVER
|
|
, "intersectional analysis,"
|
|
" %d in \\-diagonal vs block %s",
|
|
n, 1+x, usage->blocks->blocknames[b]
|
|
#endif
|
|
) ||
|
|
solver_intersect(usage, scratch->indexlist2,
|
|
scratch->indexlist
|
|
#ifdef STANDALONE_SOLVER
|
|
, "intersectional analysis,"
|
|
" %d in block %s vs \\-diagonal",
|
|
n, usage->blocks->blocknames[b], 1+x
|
|
#endif
|
|
)) {
|
|
diff = max(diff, DIFF_INTERSECT);
|
|
goto cont;
|
|
}
|
|
}
|
|
|
|
/*
|
|
* Intersectional analysis, /-diagonal vs blocks.
|
|
*/
|
|
for (b = 0; b < cr; b++)
|
|
for (n = 1; n <= cr; n++) {
|
|
if (usage->diag[cr+n-1] ||
|
|
usage->blk[b*cr+n-1])
|
|
continue;
|
|
for (i = 0; i < cr; i++) {
|
|
scratch->indexlist[i] = cubepos2(diag1(i), n);
|
|
scratch->indexlist2[i] = cubepos2(usage->blocks->blocks[b][i], n);
|
|
}
|
|
if (solver_intersect(usage, scratch->indexlist,
|
|
scratch->indexlist2
|
|
#ifdef STANDALONE_SOLVER
|
|
, "intersectional analysis,"
|
|
" %d in /-diagonal vs block %s",
|
|
n, 1+x, usage->blocks->blocknames[b]
|
|
#endif
|
|
) ||
|
|
solver_intersect(usage, scratch->indexlist2,
|
|
scratch->indexlist
|
|
#ifdef STANDALONE_SOLVER
|
|
, "intersectional analysis,"
|
|
" %d in block %s vs /-diagonal",
|
|
n, usage->blocks->blocknames[b], 1+x
|
|
#endif
|
|
)) {
|
|
diff = max(diff, DIFF_INTERSECT);
|
|
goto cont;
|
|
}
|
|
}
|
|
}
|
|
|
|
if (maxdiff <= DIFF_INTERSECT)
|
|
break;
|
|
|
|
/*
|
|
* Blockwise set elimination.
|
|
*/
|
|
for (b = 0; b < cr; b++) {
|
|
for (i = 0; i < cr; i++)
|
|
for (n = 1; n <= cr; n++)
|
|
scratch->indexlist[i*cr+n-1] = cubepos2(usage->blocks->blocks[b][i], n);
|
|
ret = solver_set(usage, scratch, scratch->indexlist
|
|
#ifdef STANDALONE_SOLVER
|
|
, "set elimination, block %s",
|
|
usage->blocks->blocknames[b]
|
|
#endif
|
|
);
|
|
if (ret < 0) {
|
|
diff = DIFF_IMPOSSIBLE;
|
|
goto got_result;
|
|
} else if (ret > 0) {
|
|
diff = max(diff, DIFF_SET);
|
|
goto cont;
|
|
}
|
|
}
|
|
|
|
/*
|
|
* Row-wise set elimination.
|
|
*/
|
|
for (y = 0; y < cr; y++) {
|
|
for (x = 0; x < cr; x++)
|
|
for (n = 1; n <= cr; n++)
|
|
scratch->indexlist[x*cr+n-1] = cubepos(x, y, n);
|
|
ret = solver_set(usage, scratch, scratch->indexlist
|
|
#ifdef STANDALONE_SOLVER
|
|
, "set elimination, row %d", 1+y
|
|
#endif
|
|
);
|
|
if (ret < 0) {
|
|
diff = DIFF_IMPOSSIBLE;
|
|
goto got_result;
|
|
} else if (ret > 0) {
|
|
diff = max(diff, DIFF_SET);
|
|
goto cont;
|
|
}
|
|
}
|
|
|
|
/*
|
|
* Column-wise set elimination.
|
|
*/
|
|
for (x = 0; x < cr; x++) {
|
|
for (y = 0; y < cr; y++)
|
|
for (n = 1; n <= cr; n++)
|
|
scratch->indexlist[y*cr+n-1] = cubepos(x, y, n);
|
|
ret = solver_set(usage, scratch, scratch->indexlist
|
|
#ifdef STANDALONE_SOLVER
|
|
, "set elimination, column %d", 1+x
|
|
#endif
|
|
);
|
|
if (ret < 0) {
|
|
diff = DIFF_IMPOSSIBLE;
|
|
goto got_result;
|
|
} else if (ret > 0) {
|
|
diff = max(diff, DIFF_SET);
|
|
goto cont;
|
|
}
|
|
}
|
|
|
|
if (usage->diag) {
|
|
/*
|
|
* \-diagonal set elimination.
|
|
*/
|
|
for (i = 0; i < cr; i++)
|
|
for (n = 1; n <= cr; n++)
|
|
scratch->indexlist[i*cr+n-1] = cubepos2(diag0(i), n);
|
|
ret = solver_set(usage, scratch, scratch->indexlist
|
|
#ifdef STANDALONE_SOLVER
|
|
, "set elimination, \\-diagonal"
|
|
#endif
|
|
);
|
|
if (ret < 0) {
|
|
diff = DIFF_IMPOSSIBLE;
|
|
goto got_result;
|
|
} else if (ret > 0) {
|
|
diff = max(diff, DIFF_SET);
|
|
goto cont;
|
|
}
|
|
|
|
/*
|
|
* /-diagonal set elimination.
|
|
*/
|
|
for (i = 0; i < cr; i++)
|
|
for (n = 1; n <= cr; n++)
|
|
scratch->indexlist[i*cr+n-1] = cubepos2(diag1(i), n);
|
|
ret = solver_set(usage, scratch, scratch->indexlist
|
|
#ifdef STANDALONE_SOLVER
|
|
, "set elimination, \\-diagonal"
|
|
#endif
|
|
);
|
|
if (ret < 0) {
|
|
diff = DIFF_IMPOSSIBLE;
|
|
goto got_result;
|
|
} else if (ret > 0) {
|
|
diff = max(diff, DIFF_SET);
|
|
goto cont;
|
|
}
|
|
}
|
|
|
|
if (maxdiff <= DIFF_SET)
|
|
break;
|
|
|
|
/*
|
|
* Row-vs-column set elimination on a single number.
|
|
*/
|
|
for (n = 1; n <= cr; n++) {
|
|
for (y = 0; y < cr; y++)
|
|
for (x = 0; x < cr; x++)
|
|
scratch->indexlist[y*cr+x] = cubepos(x, y, n);
|
|
ret = solver_set(usage, scratch, scratch->indexlist
|
|
#ifdef STANDALONE_SOLVER
|
|
, "positional set elimination, number %d", n
|
|
#endif
|
|
);
|
|
if (ret < 0) {
|
|
diff = DIFF_IMPOSSIBLE;
|
|
goto got_result;
|
|
} else if (ret > 0) {
|
|
diff = max(diff, DIFF_EXTREME);
|
|
goto cont;
|
|
}
|
|
}
|
|
|
|
/*
|
|
* Forcing chains.
|
|
*/
|
|
if (solver_forcing(usage, scratch)) {
|
|
diff = max(diff, DIFF_EXTREME);
|
|
goto cont;
|
|
}
|
|
|
|
/*
|
|
* If we reach here, we have made no deductions in this
|
|
* iteration, so the algorithm terminates.
|
|
*/
|
|
break;
|
|
}
|
|
|
|
/*
|
|
* Last chance: if we haven't fully solved the puzzle yet, try
|
|
* recursing based on guesses for a particular square. We pick
|
|
* one of the most constrained empty squares we can find, which
|
|
* has the effect of pruning the search tree as much as
|
|
* possible.
|
|
*/
|
|
if (maxdiff >= DIFF_RECURSIVE) {
|
|
int best, bestcount;
|
|
|
|
best = -1;
|
|
bestcount = cr+1;
|
|
|
|
for (y = 0; y < cr; y++)
|
|
for (x = 0; x < cr; x++)
|
|
if (!grid[y*cr+x]) {
|
|
int count;
|
|
|
|
/*
|
|
* An unfilled square. Count the number of
|
|
* possible digits in it.
|
|
*/
|
|
count = 0;
|
|
for (n = 1; n <= cr; n++)
|
|
if (cube(x,y,n))
|
|
count++;
|
|
|
|
/*
|
|
* We should have found any impossibilities
|
|
* already, so this can safely be an assert.
|
|
*/
|
|
assert(count > 1);
|
|
|
|
if (count < bestcount) {
|
|
bestcount = count;
|
|
best = y*cr+x;
|
|
}
|
|
}
|
|
|
|
if (best != -1) {
|
|
int i, j;
|
|
digit *list, *ingrid, *outgrid;
|
|
|
|
diff = DIFF_IMPOSSIBLE; /* no solution found yet */
|
|
|
|
/*
|
|
* Attempt recursion.
|
|
*/
|
|
y = best / cr;
|
|
x = best % cr;
|
|
|
|
list = snewn(cr, digit);
|
|
ingrid = snewn(cr * cr, digit);
|
|
outgrid = snewn(cr * cr, digit);
|
|
memcpy(ingrid, grid, cr * cr);
|
|
|
|
/* Make a list of the possible digits. */
|
|
for (j = 0, n = 1; n <= cr; n++)
|
|
if (cube(x,y,n))
|
|
list[j++] = n;
|
|
|
|
#ifdef STANDALONE_SOLVER
|
|
if (solver_show_working) {
|
|
char *sep = "";
|
|
printf("%*srecursing on (%d,%d) [",
|
|
solver_recurse_depth*4, "", x + 1, y + 1);
|
|
for (i = 0; i < j; i++) {
|
|
printf("%s%d", sep, list[i]);
|
|
sep = " or ";
|
|
}
|
|
printf("]\n");
|
|
}
|
|
#endif
|
|
|
|
/*
|
|
* And step along the list, recursing back into the
|
|
* main solver at every stage.
|
|
*/
|
|
for (i = 0; i < j; i++) {
|
|
int ret;
|
|
|
|
memcpy(outgrid, ingrid, cr * cr);
|
|
outgrid[y*cr+x] = list[i];
|
|
|
|
#ifdef STANDALONE_SOLVER
|
|
if (solver_show_working)
|
|
printf("%*sguessing %d at (%d,%d)\n",
|
|
solver_recurse_depth*4, "", list[i], x + 1, y + 1);
|
|
solver_recurse_depth++;
|
|
#endif
|
|
|
|
ret = solver(cr, blocks, xtype, outgrid, maxdiff);
|
|
|
|
#ifdef STANDALONE_SOLVER
|
|
solver_recurse_depth--;
|
|
if (solver_show_working) {
|
|
printf("%*sretracting %d at (%d,%d)\n",
|
|
solver_recurse_depth*4, "", list[i], x + 1, y + 1);
|
|
}
|
|
#endif
|
|
|
|
/*
|
|
* If we have our first solution, copy it into the
|
|
* grid we will return.
|
|
*/
|
|
if (diff == DIFF_IMPOSSIBLE && ret != DIFF_IMPOSSIBLE)
|
|
memcpy(grid, outgrid, cr*cr);
|
|
|
|
if (ret == DIFF_AMBIGUOUS)
|
|
diff = DIFF_AMBIGUOUS;
|
|
else if (ret == DIFF_IMPOSSIBLE)
|
|
/* do not change our return value */;
|
|
else {
|
|
/* the recursion turned up exactly one solution */
|
|
if (diff == DIFF_IMPOSSIBLE)
|
|
diff = DIFF_RECURSIVE;
|
|
else
|
|
diff = DIFF_AMBIGUOUS;
|
|
}
|
|
|
|
/*
|
|
* As soon as we've found more than one solution,
|
|
* give up immediately.
|
|
*/
|
|
if (diff == DIFF_AMBIGUOUS)
|
|
break;
|
|
}
|
|
|
|
sfree(outgrid);
|
|
sfree(ingrid);
|
|
sfree(list);
|
|
}
|
|
|
|
} else {
|
|
/*
|
|
* We're forbidden to use recursion, so we just see whether
|
|
* our grid is fully solved, and return DIFF_IMPOSSIBLE
|
|
* otherwise.
|
|
*/
|
|
for (y = 0; y < cr; y++)
|
|
for (x = 0; x < cr; x++)
|
|
if (!grid[y*cr+x])
|
|
diff = DIFF_IMPOSSIBLE;
|
|
}
|
|
|
|
got_result:;
|
|
|
|
#ifdef STANDALONE_SOLVER
|
|
if (solver_show_working)
|
|
printf("%*s%s found\n",
|
|
solver_recurse_depth*4, "",
|
|
diff == DIFF_IMPOSSIBLE ? "no solution" :
|
|
diff == DIFF_AMBIGUOUS ? "multiple solutions" :
|
|
"one solution");
|
|
#endif
|
|
|
|
sfree(usage->cube);
|
|
sfree(usage->row);
|
|
sfree(usage->col);
|
|
sfree(usage->blk);
|
|
sfree(usage);
|
|
|
|
solver_free_scratch(scratch);
|
|
|
|
return diff;
|
|
}
|
|
|
|
/* ----------------------------------------------------------------------
|
|
* End of solver code.
|
|
*/
|
|
|
|
/* ----------------------------------------------------------------------
|
|
* Solo filled-grid generator.
|
|
*
|
|
* This grid generator works by essentially trying to solve a grid
|
|
* starting from no clues, and not worrying that there's more than
|
|
* one possible solution. Unfortunately, it isn't computationally
|
|
* feasible to do this by calling the above solver with an empty
|
|
* grid, because that one needs to allocate a lot of scratch space
|
|
* at every recursion level. Instead, I have a much simpler
|
|
* algorithm which I shamelessly copied from a Python solver
|
|
* written by Andrew Wilkinson (which is GPLed, but I've reused
|
|
* only ideas and no code). It mostly just does the obvious
|
|
* recursive thing: pick an empty square, put one of the possible
|
|
* digits in it, recurse until all squares are filled, backtrack
|
|
* and change some choices if necessary.
|
|
*
|
|
* The clever bit is that every time it chooses which square to
|
|
* fill in next, it does so by counting the number of _possible_
|
|
* numbers that can go in each square, and it prioritises so that
|
|
* it picks a square with the _lowest_ number of possibilities. The
|
|
* idea is that filling in lots of the obvious bits (particularly
|
|
* any squares with only one possibility) will cut down on the list
|
|
* of possibilities for other squares and hence reduce the enormous
|
|
* search space as much as possible as early as possible.
|
|
*/
|
|
|
|
/*
|
|
* Internal data structure used in gridgen to keep track of
|
|
* progress.
|
|
*/
|
|
struct gridgen_coord { int x, y, r; };
|
|
struct gridgen_usage {
|
|
int cr;
|
|
struct block_structure *blocks;
|
|
/* grid is a copy of the input grid, modified as we go along */
|
|
digit *grid;
|
|
/* row[y*cr+n-1] TRUE if digit n has been placed in row y */
|
|
unsigned char *row;
|
|
/* col[x*cr+n-1] TRUE if digit n has been placed in row x */
|
|
unsigned char *col;
|
|
/* blk[(y*c+x)*cr+n-1] TRUE if digit n has been placed in block (x,y) */
|
|
unsigned char *blk;
|
|
/* diag[i*cr+n-1] TRUE if digit n has been placed in diagonal i */
|
|
unsigned char *diag;
|
|
/* This lists all the empty spaces remaining in the grid. */
|
|
struct gridgen_coord *spaces;
|
|
int nspaces;
|
|
/* If we need randomisation in the solve, this is our random state. */
|
|
random_state *rs;
|
|
};
|
|
|
|
static void gridgen_place(struct gridgen_usage *usage, int x, int y, digit n,
|
|
int placing)
|
|
{
|
|
int cr = usage->cr;
|
|
usage->row[y*cr+n-1] = usage->col[x*cr+n-1] =
|
|
usage->blk[usage->blocks->whichblock[y*cr+x]*cr+n-1] = placing;
|
|
if (usage->diag) {
|
|
if (ondiag0(y*cr+x))
|
|
usage->diag[n-1] = placing;
|
|
if (ondiag1(y*cr+x))
|
|
usage->diag[cr+n-1] = placing;
|
|
}
|
|
usage->grid[y*cr+x] = placing ? n : 0;
|
|
}
|
|
|
|
/*
|
|
* The real recursive step in the generating function.
|
|
*
|
|
* Return values: 1 means solution found, 0 means no solution
|
|
* found on this branch.
|
|
*/
|
|
static int gridgen_real(struct gridgen_usage *usage, digit *grid, int *steps)
|
|
{
|
|
int cr = usage->cr;
|
|
int i, j, n, sx, sy, bestm, bestr, ret;
|
|
int *digits;
|
|
|
|
/*
|
|
* Firstly, check for completion! If there are no spaces left
|
|
* in the grid, we have a solution.
|
|
*/
|
|
if (usage->nspaces == 0)
|
|
return TRUE;
|
|
|
|
/*
|
|
* Next, abandon generation if we went over our steps limit.
|
|
*/
|
|
if (*steps <= 0)
|
|
return FALSE;
|
|
(*steps)--;
|
|
|
|
/*
|
|
* Otherwise, there must be at least one space. Find the most
|
|
* constrained space, using the `r' field as a tie-breaker.
|
|
*/
|
|
bestm = cr+1; /* so that any space will beat it */
|
|
bestr = 0;
|
|
i = sx = sy = -1;
|
|
for (j = 0; j < usage->nspaces; j++) {
|
|
int x = usage->spaces[j].x, y = usage->spaces[j].y;
|
|
int m;
|
|
|
|
/*
|
|
* Find the number of digits that could go in this space.
|
|
*/
|
|
m = 0;
|
|
for (n = 0; n < cr; n++)
|
|
if (!usage->row[y*cr+n] && !usage->col[x*cr+n] &&
|
|
!usage->blk[usage->blocks->whichblock[y*cr+x]*cr+n] &&
|
|
(!usage->diag || ((!ondiag0(y*cr+x) || !usage->diag[n]) &&
|
|
(!ondiag1(y*cr+x) || !usage->diag[cr+n]))))
|
|
m++;
|
|
|
|
if (m < bestm || (m == bestm && usage->spaces[j].r < bestr)) {
|
|
bestm = m;
|
|
bestr = usage->spaces[j].r;
|
|
sx = x;
|
|
sy = y;
|
|
i = j;
|
|
}
|
|
}
|
|
|
|
/*
|
|
* Swap that square into the final place in the spaces array,
|
|
* so that decrementing nspaces will remove it from the list.
|
|
*/
|
|
if (i != usage->nspaces-1) {
|
|
struct gridgen_coord t;
|
|
t = usage->spaces[usage->nspaces-1];
|
|
usage->spaces[usage->nspaces-1] = usage->spaces[i];
|
|
usage->spaces[i] = t;
|
|
}
|
|
|
|
/*
|
|
* Now we've decided which square to start our recursion at,
|
|
* simply go through all possible values, shuffling them
|
|
* randomly first if necessary.
|
|
*/
|
|
digits = snewn(bestm, int);
|
|
j = 0;
|
|
for (n = 0; n < cr; n++)
|
|
if (!usage->row[sy*cr+n] && !usage->col[sx*cr+n] &&
|
|
!usage->blk[usage->blocks->whichblock[sy*cr+sx]*cr+n] &&
|
|
(!usage->diag || ((!ondiag0(sy*cr+sx) || !usage->diag[n]) &&
|
|
(!ondiag1(sy*cr+sx) || !usage->diag[cr+n])))) {
|
|
digits[j++] = n+1;
|
|
}
|
|
|
|
if (usage->rs)
|
|
shuffle(digits, j, sizeof(*digits), usage->rs);
|
|
|
|
/* And finally, go through the digit list and actually recurse. */
|
|
ret = FALSE;
|
|
for (i = 0; i < j; i++) {
|
|
n = digits[i];
|
|
|
|
/* Update the usage structure to reflect the placing of this digit. */
|
|
gridgen_place(usage, sx, sy, n, TRUE);
|
|
usage->nspaces--;
|
|
|
|
/* Call the solver recursively. Stop when we find a solution. */
|
|
if (gridgen_real(usage, grid, steps)) {
|
|
ret = TRUE;
|
|
break;
|
|
}
|
|
|
|
/* Revert the usage structure. */
|
|
gridgen_place(usage, sx, sy, n, FALSE);
|
|
usage->nspaces++;
|
|
}
|
|
|
|
sfree(digits);
|
|
return ret;
|
|
}
|
|
|
|
/*
|
|
* Entry point to generator. You give it parameters and a starting
|
|
* grid, which is simply an array of cr*cr digits.
|
|
*/
|
|
static int gridgen(int cr, struct block_structure *blocks, int xtype,
|
|
digit *grid, random_state *rs, int maxsteps)
|
|
{
|
|
struct gridgen_usage *usage;
|
|
int x, y, ret;
|
|
|
|
/*
|
|
* Clear the grid to start with.
|
|
*/
|
|
memset(grid, 0, cr*cr);
|
|
|
|
/*
|
|
* Create a gridgen_usage structure.
|
|
*/
|
|
usage = snew(struct gridgen_usage);
|
|
|
|
usage->cr = cr;
|
|
usage->blocks = blocks;
|
|
|
|
usage->grid = grid;
|
|
|
|
usage->row = snewn(cr * cr, unsigned char);
|
|
usage->col = snewn(cr * cr, unsigned char);
|
|
usage->blk = snewn(cr * cr, unsigned char);
|
|
memset(usage->row, FALSE, cr * cr);
|
|
memset(usage->col, FALSE, cr * cr);
|
|
memset(usage->blk, FALSE, cr * cr);
|
|
|
|
if (xtype) {
|
|
usage->diag = snewn(2 * cr, unsigned char);
|
|
memset(usage->diag, FALSE, 2 * cr);
|
|
} else {
|
|
usage->diag = NULL;
|
|
}
|
|
|
|
/*
|
|
* Begin by filling in the whole top row with randomly chosen
|
|
* numbers. This cannot introduce any bias or restriction on
|
|
* the available grids, since we already know those numbers
|
|
* are all distinct so all we're doing is choosing their
|
|
* labels.
|
|
*/
|
|
for (x = 0; x < cr; x++)
|
|
grid[x] = x+1;
|
|
shuffle(grid, cr, sizeof(*grid), rs);
|
|
for (x = 0; x < cr; x++)
|
|
gridgen_place(usage, x, 0, grid[x], TRUE);
|
|
|
|
usage->spaces = snewn(cr * cr, struct gridgen_coord);
|
|
usage->nspaces = 0;
|
|
|
|
usage->rs = rs;
|
|
|
|
/*
|
|
* Initialise the list of grid spaces, taking care to leave
|
|
* out the row I've already filled in above.
|
|
*/
|
|
for (y = 1; y < cr; y++) {
|
|
for (x = 0; x < cr; x++) {
|
|
usage->spaces[usage->nspaces].x = x;
|
|
usage->spaces[usage->nspaces].y = y;
|
|
usage->spaces[usage->nspaces].r = random_bits(rs, 31);
|
|
usage->nspaces++;
|
|
}
|
|
}
|
|
|
|
/*
|
|
* Run the real generator function.
|
|
*/
|
|
ret = gridgen_real(usage, grid, &maxsteps);
|
|
|
|
/*
|
|
* Clean up the usage structure now we have our answer.
|
|
*/
|
|
sfree(usage->spaces);
|
|
sfree(usage->blk);
|
|
sfree(usage->col);
|
|
sfree(usage->row);
|
|
sfree(usage);
|
|
|
|
return ret;
|
|
}
|
|
|
|
/* ----------------------------------------------------------------------
|
|
* End of grid generator code.
|
|
*/
|
|
|
|
/*
|
|
* Check whether a grid contains a valid complete puzzle.
|
|
*/
|
|
static int check_valid(int cr, struct block_structure *blocks, int xtype,
|
|
digit *grid)
|
|
{
|
|
unsigned char *used;
|
|
int x, y, i, j, n;
|
|
|
|
used = snewn(cr, unsigned char);
|
|
|
|
/*
|
|
* Check that each row contains precisely one of everything.
|
|
*/
|
|
for (y = 0; y < cr; y++) {
|
|
memset(used, FALSE, cr);
|
|
for (x = 0; x < cr; x++)
|
|
if (grid[y*cr+x] > 0 && grid[y*cr+x] <= cr)
|
|
used[grid[y*cr+x]-1] = TRUE;
|
|
for (n = 0; n < cr; n++)
|
|
if (!used[n]) {
|
|
sfree(used);
|
|
return FALSE;
|
|
}
|
|
}
|
|
|
|
/*
|
|
* Check that each column contains precisely one of everything.
|
|
*/
|
|
for (x = 0; x < cr; x++) {
|
|
memset(used, FALSE, cr);
|
|
for (y = 0; y < cr; y++)
|
|
if (grid[y*cr+x] > 0 && grid[y*cr+x] <= cr)
|
|
used[grid[y*cr+x]-1] = TRUE;
|
|
for (n = 0; n < cr; n++)
|
|
if (!used[n]) {
|
|
sfree(used);
|
|
return FALSE;
|
|
}
|
|
}
|
|
|
|
/*
|
|
* Check that each block contains precisely one of everything.
|
|
*/
|
|
for (i = 0; i < cr; i++) {
|
|
memset(used, FALSE, cr);
|
|
for (j = 0; j < cr; j++)
|
|
if (grid[blocks->blocks[i][j]] > 0 &&
|
|
grid[blocks->blocks[i][j]] <= cr)
|
|
used[grid[blocks->blocks[i][j]]-1] = TRUE;
|
|
for (n = 0; n < cr; n++)
|
|
if (!used[n]) {
|
|
sfree(used);
|
|
return FALSE;
|
|
}
|
|
}
|
|
|
|
/*
|
|
* Check that each diagonal contains precisely one of everything.
|
|
*/
|
|
if (xtype) {
|
|
memset(used, FALSE, cr);
|
|
for (i = 0; i < cr; i++)
|
|
if (grid[diag0(i)] > 0 && grid[diag0(i)] <= cr)
|
|
used[grid[diag0(i)]-1] = TRUE;
|
|
for (n = 0; n < cr; n++)
|
|
if (!used[n]) {
|
|
sfree(used);
|
|
return FALSE;
|
|
}
|
|
for (i = 0; i < cr; i++)
|
|
if (grid[diag1(i)] > 0 && grid[diag1(i)] <= cr)
|
|
used[grid[diag1(i)]-1] = TRUE;
|
|
for (n = 0; n < cr; n++)
|
|
if (!used[n]) {
|
|
sfree(used);
|
|
return FALSE;
|
|
}
|
|
}
|
|
|
|
sfree(used);
|
|
return TRUE;
|
|
}
|
|
|
|
static int symmetries(game_params *params, int x, int y, int *output, int s)
|
|
{
|
|
int c = params->c, r = params->r, cr = c*r;
|
|
int i = 0;
|
|
|
|
#define ADD(x,y) (*output++ = (x), *output++ = (y), i++)
|
|
|
|
ADD(x, y);
|
|
|
|
switch (s) {
|
|
case SYMM_NONE:
|
|
break; /* just x,y is all we need */
|
|
case SYMM_ROT2:
|
|
ADD(cr - 1 - x, cr - 1 - y);
|
|
break;
|
|
case SYMM_ROT4:
|
|
ADD(cr - 1 - y, x);
|
|
ADD(y, cr - 1 - x);
|
|
ADD(cr - 1 - x, cr - 1 - y);
|
|
break;
|
|
case SYMM_REF2:
|
|
ADD(cr - 1 - x, y);
|
|
break;
|
|
case SYMM_REF2D:
|
|
ADD(y, x);
|
|
break;
|
|
case SYMM_REF4:
|
|
ADD(cr - 1 - x, y);
|
|
ADD(x, cr - 1 - y);
|
|
ADD(cr - 1 - x, cr - 1 - y);
|
|
break;
|
|
case SYMM_REF4D:
|
|
ADD(y, x);
|
|
ADD(cr - 1 - x, cr - 1 - y);
|
|
ADD(cr - 1 - y, cr - 1 - x);
|
|
break;
|
|
case SYMM_REF8:
|
|
ADD(cr - 1 - x, y);
|
|
ADD(x, cr - 1 - y);
|
|
ADD(cr - 1 - x, cr - 1 - y);
|
|
ADD(y, x);
|
|
ADD(y, cr - 1 - x);
|
|
ADD(cr - 1 - y, x);
|
|
ADD(cr - 1 - y, cr - 1 - x);
|
|
break;
|
|
}
|
|
|
|
#undef ADD
|
|
|
|
return i;
|
|
}
|
|
|
|
static char *encode_solve_move(int cr, digit *grid)
|
|
{
|
|
int i, len;
|
|
char *ret, *p, *sep;
|
|
|
|
/*
|
|
* It's surprisingly easy to work out _exactly_ how long this
|
|
* string needs to be. To decimal-encode all the numbers from 1
|
|
* to n:
|
|
*
|
|
* - every number has a units digit; total is n.
|
|
* - all numbers above 9 have a tens digit; total is max(n-9,0).
|
|
* - all numbers above 99 have a hundreds digit; total is max(n-99,0).
|
|
* - and so on.
|
|
*/
|
|
len = 0;
|
|
for (i = 1; i <= cr; i *= 10)
|
|
len += max(cr - i + 1, 0);
|
|
len += cr; /* don't forget the commas */
|
|
len *= cr; /* there are cr rows of these */
|
|
|
|
/*
|
|
* Now len is one bigger than the total size of the
|
|
* comma-separated numbers (because we counted an
|
|
* additional leading comma). We need to have a leading S
|
|
* and a trailing NUL, so we're off by one in total.
|
|
*/
|
|
len++;
|
|
|
|
ret = snewn(len, char);
|
|
p = ret;
|
|
*p++ = 'S';
|
|
sep = "";
|
|
for (i = 0; i < cr*cr; i++) {
|
|
p += sprintf(p, "%s%d", sep, grid[i]);
|
|
sep = ",";
|
|
}
|
|
*p++ = '\0';
|
|
assert(p - ret == len);
|
|
|
|
return ret;
|
|
}
|
|
|
|
static char *new_game_desc(game_params *params, random_state *rs,
|
|
char **aux, int interactive)
|
|
{
|
|
int c = params->c, r = params->r, cr = c*r;
|
|
int area = cr*cr;
|
|
struct block_structure *blocks;
|
|
digit *grid, *grid2;
|
|
struct xy { int x, y; } *locs;
|
|
int nlocs;
|
|
char *desc;
|
|
int coords[16], ncoords;
|
|
int maxdiff;
|
|
int x, y, i, j;
|
|
|
|
/*
|
|
* Adjust the maximum difficulty level to be consistent with
|
|
* the puzzle size: all 2x2 puzzles appear to be Trivial
|
|
* (DIFF_BLOCK) so we cannot hold out for even a Basic
|
|
* (DIFF_SIMPLE) one.
|
|
*/
|
|
maxdiff = params->diff;
|
|
if (c == 2 && r == 2)
|
|
maxdiff = DIFF_BLOCK;
|
|
|
|
grid = snewn(area, digit);
|
|
locs = snewn(area, struct xy);
|
|
grid2 = snewn(area, digit);
|
|
|
|
blocks = snew(struct block_structure);
|
|
blocks->c = params->c; blocks->r = params->r;
|
|
blocks->whichblock = snewn(area*2, int);
|
|
blocks->blocks = snewn(cr, int *);
|
|
for (i = 0; i < cr; i++)
|
|
blocks->blocks[i] = blocks->whichblock + area + i*cr;
|
|
#ifdef STANDALONE_SOLVER
|
|
assert(!"This should never happen, so we don't need to create blocknames");
|
|
#endif
|
|
|
|
/*
|
|
* Loop until we get a grid of the required difficulty. This is
|
|
* nasty, but it seems to be unpleasantly hard to generate
|
|
* difficult grids otherwise.
|
|
*/
|
|
while (1) {
|
|
/*
|
|
* Generate a random solved state, starting by
|
|
* constructing the block structure.
|
|
*/
|
|
if (r == 1) { /* jigsaw mode */
|
|
int *dsf = divvy_rectangle(cr, cr, cr, rs);
|
|
int nb = 0;
|
|
|
|
for (i = 0; i < area; i++)
|
|
blocks->whichblock[i] = -1;
|
|
for (i = 0; i < area; i++) {
|
|
int j = dsf_canonify(dsf, i);
|
|
if (blocks->whichblock[j] < 0)
|
|
blocks->whichblock[j] = nb++;
|
|
blocks->whichblock[i] = blocks->whichblock[j];
|
|
}
|
|
assert(nb == cr);
|
|
|
|
sfree(dsf);
|
|
} else { /* basic Sudoku mode */
|
|
for (y = 0; y < cr; y++)
|
|
for (x = 0; x < cr; x++)
|
|
blocks->whichblock[y*cr+x] = (y/c) * c + (x/r);
|
|
}
|
|
for (i = 0; i < cr; i++)
|
|
blocks->blocks[i][cr-1] = 0;
|
|
for (i = 0; i < area; i++) {
|
|
int b = blocks->whichblock[i];
|
|
j = blocks->blocks[b][cr-1]++;
|
|
assert(j < cr);
|
|
blocks->blocks[b][j] = i;
|
|
}
|
|
|
|
if (!gridgen(cr, blocks, params->xtype, grid, rs, area*area))
|
|
continue;
|
|
assert(check_valid(cr, blocks, params->xtype, grid));
|
|
|
|
/*
|
|
* Save the solved grid in aux.
|
|
*/
|
|
{
|
|
/*
|
|
* We might already have written *aux the last time we
|
|
* went round this loop, in which case we should free
|
|
* the old aux before overwriting it with the new one.
|
|
*/
|
|
if (*aux) {
|
|
sfree(*aux);
|
|
}
|
|
|
|
*aux = encode_solve_move(cr, grid);
|
|
}
|
|
|
|
/*
|
|
* Now we have a solved grid, start removing things from it
|
|
* while preserving solubility.
|
|
*/
|
|
|
|
/*
|
|
* Find the set of equivalence classes of squares permitted
|
|
* by the selected symmetry. We do this by enumerating all
|
|
* the grid squares which have no symmetric companion
|
|
* sorting lower than themselves.
|
|
*/
|
|
nlocs = 0;
|
|
for (y = 0; y < cr; y++)
|
|
for (x = 0; x < cr; x++) {
|
|
int i = y*cr+x;
|
|
int j;
|
|
|
|
ncoords = symmetries(params, x, y, coords, params->symm);
|
|
for (j = 0; j < ncoords; j++)
|
|
if (coords[2*j+1]*cr+coords[2*j] < i)
|
|
break;
|
|
if (j == ncoords) {
|
|
locs[nlocs].x = x;
|
|
locs[nlocs].y = y;
|
|
nlocs++;
|
|
}
|
|
}
|
|
|
|
/*
|
|
* Now shuffle that list.
|
|
*/
|
|
shuffle(locs, nlocs, sizeof(*locs), rs);
|
|
|
|
/*
|
|
* Now loop over the shuffled list and, for each element,
|
|
* see whether removing that element (and its reflections)
|
|
* from the grid will still leave the grid soluble.
|
|
*/
|
|
for (i = 0; i < nlocs; i++) {
|
|
int ret;
|
|
|
|
x = locs[i].x;
|
|
y = locs[i].y;
|
|
|
|
memcpy(grid2, grid, area);
|
|
ncoords = symmetries(params, x, y, coords, params->symm);
|
|
for (j = 0; j < ncoords; j++)
|
|
grid2[coords[2*j+1]*cr+coords[2*j]] = 0;
|
|
|
|
ret = solver(cr, blocks, params->xtype, grid2, maxdiff);
|
|
if (ret <= maxdiff) {
|
|
for (j = 0; j < ncoords; j++)
|
|
grid[coords[2*j+1]*cr+coords[2*j]] = 0;
|
|
}
|
|
}
|
|
|
|
memcpy(grid2, grid, area);
|
|
|
|
if (solver(cr, blocks, params->xtype, grid2, maxdiff) == maxdiff)
|
|
break; /* found one! */
|
|
}
|
|
|
|
sfree(grid2);
|
|
sfree(locs);
|
|
|
|
/*
|
|
* Now we have the grid as it will be presented to the user.
|
|
* Encode it in a game desc.
|
|
*/
|
|
{
|
|
char *p;
|
|
int run, i;
|
|
|
|
desc = snewn(7 * area, char);
|
|
p = desc;
|
|
run = 0;
|
|
for (i = 0; i <= area; i++) {
|
|
int n = (i < area ? grid[i] : -1);
|
|
|
|
if (!n)
|
|
run++;
|
|
else {
|
|
if (run) {
|
|
while (run > 0) {
|
|
int c = 'a' - 1 + run;
|
|
if (run > 26)
|
|
c = 'z';
|
|
*p++ = c;
|
|
run -= c - ('a' - 1);
|
|
}
|
|
} else {
|
|
/*
|
|
* If there's a number in the very top left or
|
|
* bottom right, there's no point putting an
|
|
* unnecessary _ before or after it.
|
|
*/
|
|
if (p > desc && n > 0)
|
|
*p++ = '_';
|
|
}
|
|
if (n > 0)
|
|
p += sprintf(p, "%d", n);
|
|
run = 0;
|
|
}
|
|
}
|
|
|
|
if (r == 1) {
|
|
int currrun = 0;
|
|
|
|
*p++ = ',';
|
|
|
|
/*
|
|
* Encode the block structure. We do this by encoding
|
|
* the pattern of dividing lines: first we iterate
|
|
* over the cr*(cr-1) internal vertical grid lines in
|
|
* ordinary reading order, then over the cr*(cr-1)
|
|
* internal horizontal ones in transposed reading
|
|
* order.
|
|
*
|
|
* We encode the number of non-lines between the
|
|
* lines; _ means zero (two adjacent divisions), a
|
|
* means 1, ..., y means 25, and z means 25 non-lines
|
|
* _and no following line_ (so that za means 26, zb 27
|
|
* etc).
|
|
*/
|
|
for (i = 0; i <= 2*cr*(cr-1); i++) {
|
|
int p0, p1, edge;
|
|
|
|
if (i == 2*cr*(cr-1)) {
|
|
edge = TRUE; /* terminating virtual edge */
|
|
} else {
|
|
if (i < cr*(cr-1)) {
|
|
y = i/(cr-1);
|
|
x = i%(cr-1);
|
|
p0 = y*cr+x;
|
|
p1 = y*cr+x+1;
|
|
} else {
|
|
x = i/(cr-1) - cr;
|
|
y = i%(cr-1);
|
|
p0 = y*cr+x;
|
|
p1 = (y+1)*cr+x;
|
|
}
|
|
edge = (blocks->whichblock[p0] != blocks->whichblock[p1]);
|
|
}
|
|
|
|
if (edge) {
|
|
while (currrun > 25)
|
|
*p++ = 'z', currrun -= 25;
|
|
if (currrun)
|
|
*p++ = 'a'-1 + currrun;
|
|
else
|
|
*p++ = '_';
|
|
currrun = 0;
|
|
} else
|
|
currrun++;
|
|
}
|
|
}
|
|
|
|
assert(p - desc < 7 * area);
|
|
*p++ = '\0';
|
|
desc = sresize(desc, p - desc, char);
|
|
}
|
|
|
|
sfree(grid);
|
|
|
|
return desc;
|
|
}
|
|
|
|
static char *validate_desc(game_params *params, char *desc)
|
|
{
|
|
int cr = params->c * params->r, area = cr*cr;
|
|
int squares = 0;
|
|
int *dsf;
|
|
|
|
while (*desc && *desc != ',') {
|
|
int n = *desc++;
|
|
if (n >= 'a' && n <= 'z') {
|
|
squares += n - 'a' + 1;
|
|
} else if (n == '_') {
|
|
/* do nothing */;
|
|
} else if (n > '0' && n <= '9') {
|
|
int val = atoi(desc-1);
|
|
if (val < 1 || val > params->c * params->r)
|
|
return "Out-of-range number in game description";
|
|
squares++;
|
|
while (*desc >= '0' && *desc <= '9')
|
|
desc++;
|
|
} else
|
|
return "Invalid character in game description";
|
|
}
|
|
|
|
if (squares < area)
|
|
return "Not enough data to fill grid";
|
|
|
|
if (squares > area)
|
|
return "Too much data to fit in grid";
|
|
|
|
if (params->r == 1) {
|
|
int pos;
|
|
|
|
/*
|
|
* Now we expect a suffix giving the jigsaw block
|
|
* structure. Parse it and validate that it divides the
|
|
* grid into the right number of regions which are the
|
|
* right size.
|
|
*/
|
|
if (*desc != ',')
|
|
return "Expected jigsaw block structure in game description";
|
|
pos = 0;
|
|
|
|
dsf = snew_dsf(area);
|
|
desc++;
|
|
|
|
while (*desc) {
|
|
int c, adv;
|
|
|
|
if (*desc == '_')
|
|
c = 0;
|
|
else if (*desc >= 'a' && *desc <= 'z')
|
|
c = *desc - 'a' + 1;
|
|
else {
|
|
sfree(dsf);
|
|
return "Invalid character in game description";
|
|
}
|
|
desc++;
|
|
|
|
adv = (c != 25); /* 'z' is a special case */
|
|
|
|
while (c-- > 0) {
|
|
int p0, p1;
|
|
|
|
/*
|
|
* Non-edge; merge the two dsf classes on either
|
|
* side of it.
|
|
*/
|
|
if (pos >= 2*cr*(cr-1)) {
|
|
sfree(dsf);
|
|
return "Too much data in block structure specification";
|
|
} else if (pos < cr*(cr-1)) {
|
|
int y = pos/(cr-1);
|
|
int x = pos%(cr-1);
|
|
p0 = y*cr+x;
|
|
p1 = y*cr+x+1;
|
|
} else {
|
|
int x = pos/(cr-1) - cr;
|
|
int y = pos%(cr-1);
|
|
p0 = y*cr+x;
|
|
p1 = (y+1)*cr+x;
|
|
}
|
|
dsf_merge(dsf, p0, p1);
|
|
|
|
pos++;
|
|
}
|
|
if (adv)
|
|
pos++;
|
|
}
|
|
|
|
/*
|
|
* When desc is exhausted, we expect to have gone exactly
|
|
* one space _past_ the end of the grid, due to the dummy
|
|
* edge at the end.
|
|
*/
|
|
if (pos != 2*cr*(cr-1)+1) {
|
|
sfree(dsf);
|
|
return "Not enough data in block structure specification";
|
|
}
|
|
|
|
/*
|
|
* Now we've got our dsf. Verify that it matches
|
|
* expectations.
|
|
*/
|
|
{
|
|
int *canons, *counts;
|
|
int i, j, c, ncanons = 0;
|
|
|
|
canons = snewn(cr, int);
|
|
counts = snewn(cr, int);
|
|
|
|
for (i = 0; i < area; i++) {
|
|
j = dsf_canonify(dsf, i);
|
|
|
|
for (c = 0; c < ncanons; c++)
|
|
if (canons[c] == j) {
|
|
counts[c]++;
|
|
if (counts[c] > cr) {
|
|
sfree(dsf);
|
|
sfree(canons);
|
|
sfree(counts);
|
|
return "A jigsaw block is too big";
|
|
}
|
|
break;
|
|
}
|
|
|
|
if (c == ncanons) {
|
|
if (ncanons >= cr) {
|
|
sfree(dsf);
|
|
sfree(canons);
|
|
sfree(counts);
|
|
return "Too many distinct jigsaw blocks";
|
|
}
|
|
canons[ncanons] = j;
|
|
counts[ncanons] = 1;
|
|
ncanons++;
|
|
}
|
|
}
|
|
|
|
/*
|
|
* If we've managed to get through that loop without
|
|
* tripping either of the error conditions, then we
|
|
* must have partitioned the entire grid into at most
|
|
* cr blocks of at most cr squares each; therefore we
|
|
* must have _exactly_ cr blocks of _exactly_ cr
|
|
* squares each. I'll verify that by assertion just in
|
|
* case something has gone horribly wrong, but it
|
|
* shouldn't have been able to happen by duff input,
|
|
* only by a bug in the above code.
|
|
*/
|
|
assert(ncanons == cr);
|
|
for (c = 0; c < ncanons; c++)
|
|
assert(counts[c] == cr);
|
|
|
|
sfree(canons);
|
|
sfree(counts);
|
|
}
|
|
|
|
sfree(dsf);
|
|
} else {
|
|
if (*desc)
|
|
return "Unexpected jigsaw block structure in game description";
|
|
}
|
|
|
|
return NULL;
|
|
}
|
|
|
|
static game_state *new_game(midend *me, game_params *params, char *desc)
|
|
{
|
|
game_state *state = snew(game_state);
|
|
int c = params->c, r = params->r, cr = c*r, area = cr * cr;
|
|
int i;
|
|
|
|
state->cr = cr;
|
|
state->xtype = params->xtype;
|
|
|
|
state->grid = snewn(area, digit);
|
|
state->pencil = snewn(area * cr, unsigned char);
|
|
memset(state->pencil, 0, area * cr);
|
|
state->immutable = snewn(area, unsigned char);
|
|
memset(state->immutable, FALSE, area);
|
|
|
|
state->blocks = snew(struct block_structure);
|
|
state->blocks->c = c; state->blocks->r = r;
|
|
state->blocks->refcount = 1;
|
|
state->blocks->whichblock = snewn(area*2, int);
|
|
state->blocks->blocks = snewn(cr, int *);
|
|
for (i = 0; i < cr; i++)
|
|
state->blocks->blocks[i] = state->blocks->whichblock + area + i*cr;
|
|
#ifdef STANDALONE_SOLVER
|
|
state->blocks->blocknames = (char **)smalloc(cr*(sizeof(char *)+80));
|
|
#endif
|
|
|
|
state->completed = state->cheated = FALSE;
|
|
|
|
i = 0;
|
|
while (*desc && *desc != ',') {
|
|
int n = *desc++;
|
|
if (n >= 'a' && n <= 'z') {
|
|
int run = n - 'a' + 1;
|
|
assert(i + run <= area);
|
|
while (run-- > 0)
|
|
state->grid[i++] = 0;
|
|
} else if (n == '_') {
|
|
/* do nothing */;
|
|
} else if (n > '0' && n <= '9') {
|
|
assert(i < area);
|
|
state->immutable[i] = TRUE;
|
|
state->grid[i++] = atoi(desc-1);
|
|
while (*desc >= '0' && *desc <= '9')
|
|
desc++;
|
|
} else {
|
|
assert(!"We can't get here");
|
|
}
|
|
}
|
|
assert(i == area);
|
|
|
|
if (r == 1) {
|
|
int pos = 0;
|
|
int *dsf;
|
|
int nb;
|
|
|
|
assert(*desc == ',');
|
|
|
|
dsf = snew_dsf(area);
|
|
desc++;
|
|
|
|
while (*desc) {
|
|
int c, adv;
|
|
|
|
if (*desc == '_')
|
|
c = 0;
|
|
else {
|
|
assert(*desc >= 'a' && *desc <= 'z');
|
|
c = *desc - 'a' + 1;
|
|
}
|
|
desc++;
|
|
|
|
adv = (c != 25); /* 'z' is a special case */
|
|
|
|
while (c-- > 0) {
|
|
int p0, p1;
|
|
|
|
/*
|
|
* Non-edge; merge the two dsf classes on either
|
|
* side of it.
|
|
*/
|
|
assert(pos < 2*cr*(cr-1));
|
|
if (pos < cr*(cr-1)) {
|
|
int y = pos/(cr-1);
|
|
int x = pos%(cr-1);
|
|
p0 = y*cr+x;
|
|
p1 = y*cr+x+1;
|
|
} else {
|
|
int x = pos/(cr-1) - cr;
|
|
int y = pos%(cr-1);
|
|
p0 = y*cr+x;
|
|
p1 = (y+1)*cr+x;
|
|
}
|
|
dsf_merge(dsf, p0, p1);
|
|
|
|
pos++;
|
|
}
|
|
if (adv)
|
|
pos++;
|
|
}
|
|
|
|
/*
|
|
* When desc is exhausted, we expect to have gone exactly
|
|
* one space _past_ the end of the grid, due to the dummy
|
|
* edge at the end.
|
|
*/
|
|
assert(pos == 2*cr*(cr-1)+1);
|
|
|
|
/*
|
|
* Now we've got our dsf. Translate it into a block
|
|
* structure.
|
|
*/
|
|
nb = 0;
|
|
for (i = 0; i < area; i++)
|
|
state->blocks->whichblock[i] = -1;
|
|
for (i = 0; i < area; i++) {
|
|
int j = dsf_canonify(dsf, i);
|
|
if (state->blocks->whichblock[j] < 0)
|
|
state->blocks->whichblock[j] = nb++;
|
|
state->blocks->whichblock[i] = state->blocks->whichblock[j];
|
|
}
|
|
assert(nb == cr);
|
|
|
|
sfree(dsf);
|
|
} else {
|
|
int x, y;
|
|
|
|
assert(!*desc);
|
|
|
|
for (y = 0; y < cr; y++)
|
|
for (x = 0; x < cr; x++)
|
|
state->blocks->whichblock[y*cr+x] = (y/c) * c + (x/r);
|
|
}
|
|
|
|
/*
|
|
* Having sorted out whichblock[], set up the block index arrays.
|
|
*/
|
|
for (i = 0; i < cr; i++)
|
|
state->blocks->blocks[i][cr-1] = 0;
|
|
for (i = 0; i < area; i++) {
|
|
int b = state->blocks->whichblock[i];
|
|
int j = state->blocks->blocks[b][cr-1]++;
|
|
assert(j < cr);
|
|
state->blocks->blocks[b][j] = i;
|
|
}
|
|
|
|
#ifdef STANDALONE_SOLVER
|
|
/*
|
|
* Set up the block names for solver diagnostic output.
|
|
*/
|
|
{
|
|
char *p = (char *)(state->blocks->blocknames + cr);
|
|
|
|
if (r == 1) {
|
|
for (i = 0; i < cr; i++)
|
|
state->blocks->blocknames[i] = NULL;
|
|
|
|
for (i = 0; i < area; i++) {
|
|
int j = state->blocks->whichblock[i];
|
|
if (!state->blocks->blocknames[j]) {
|
|
state->blocks->blocknames[j] = p;
|
|
p += 1 + sprintf(p, "starting at (%d,%d)",
|
|
1 + i%cr, 1 + i/cr);
|
|
}
|
|
}
|
|
} else {
|
|
int bx, by;
|
|
for (by = 0; by < r; by++)
|
|
for (bx = 0; bx < c; bx++) {
|
|
state->blocks->blocknames[by*c+bx] = p;
|
|
p += 1 + sprintf(p, "(%d,%d)", bx+1, by+1);
|
|
}
|
|
}
|
|
assert(p - (char *)state->blocks->blocknames < cr*(sizeof(char *)+80));
|
|
for (i = 0; i < cr; i++)
|
|
assert(state->blocks->blocknames[i]);
|
|
}
|
|
#endif
|
|
|
|
return state;
|
|
}
|
|
|
|
static game_state *dup_game(game_state *state)
|
|
{
|
|
game_state *ret = snew(game_state);
|
|
int cr = state->cr, area = cr * cr;
|
|
|
|
ret->cr = state->cr;
|
|
ret->xtype = state->xtype;
|
|
|
|
ret->blocks = state->blocks;
|
|
ret->blocks->refcount++;
|
|
|
|
ret->grid = snewn(area, digit);
|
|
memcpy(ret->grid, state->grid, area);
|
|
|
|
ret->pencil = snewn(area * cr, unsigned char);
|
|
memcpy(ret->pencil, state->pencil, area * cr);
|
|
|
|
ret->immutable = snewn(area, unsigned char);
|
|
memcpy(ret->immutable, state->immutable, area);
|
|
|
|
ret->completed = state->completed;
|
|
ret->cheated = state->cheated;
|
|
|
|
return ret;
|
|
}
|
|
|
|
static void free_game(game_state *state)
|
|
{
|
|
if (--state->blocks->refcount == 0) {
|
|
sfree(state->blocks->whichblock);
|
|
sfree(state->blocks->blocks);
|
|
#ifdef STANDALONE_SOLVER
|
|
sfree(state->blocks->blocknames);
|
|
#endif
|
|
sfree(state->blocks);
|
|
}
|
|
sfree(state->immutable);
|
|
sfree(state->pencil);
|
|
sfree(state->grid);
|
|
sfree(state);
|
|
}
|
|
|
|
static char *solve_game(game_state *state, game_state *currstate,
|
|
char *ai, char **error)
|
|
{
|
|
int cr = state->cr;
|
|
char *ret;
|
|
digit *grid;
|
|
int solve_ret;
|
|
|
|
/*
|
|
* If we already have the solution in ai, save ourselves some
|
|
* time.
|
|
*/
|
|
if (ai)
|
|
return dupstr(ai);
|
|
|
|
grid = snewn(cr*cr, digit);
|
|
memcpy(grid, state->grid, cr*cr);
|
|
solve_ret = solver(cr, state->blocks, state->xtype, grid, DIFF_RECURSIVE);
|
|
|
|
*error = NULL;
|
|
|
|
if (solve_ret == DIFF_IMPOSSIBLE)
|
|
*error = "No solution exists for this puzzle";
|
|
else if (solve_ret == DIFF_AMBIGUOUS)
|
|
*error = "Multiple solutions exist for this puzzle";
|
|
|
|
if (*error) {
|
|
sfree(grid);
|
|
return NULL;
|
|
}
|
|
|
|
ret = encode_solve_move(cr, grid);
|
|
|
|
sfree(grid);
|
|
|
|
return ret;
|
|
}
|
|
|
|
static char *grid_text_format(int cr, struct block_structure *blocks,
|
|
int xtype, digit *grid)
|
|
{
|
|
int vmod, hmod;
|
|
int x, y;
|
|
int totallen, linelen, nlines;
|
|
char *ret, *p, ch;
|
|
|
|
/*
|
|
* For non-jigsaw Sudoku, we format in the way we always have,
|
|
* by having the digits unevenly spaced so that the dividing
|
|
* lines can fit in:
|
|
*
|
|
* . . | . .
|
|
* . . | . .
|
|
* ----+----
|
|
* . . | . .
|
|
* . . | . .
|
|
*
|
|
* For jigsaw puzzles, however, we must leave space between
|
|
* _all_ pairs of digits for an optional dividing line, so we
|
|
* have to move to the rather ugly
|
|
*
|
|
* . . . .
|
|
* ------+------
|
|
* . . | . .
|
|
* +---+
|
|
* . . | . | .
|
|
* ------+ |
|
|
* . . . | .
|
|
*
|
|
* We deal with both cases using the same formatting code; we
|
|
* simply invent a vmod value such that there's a vertical
|
|
* dividing line before column i iff i is divisible by vmod
|
|
* (so it's r in the first case and 1 in the second), and hmod
|
|
* likewise for horizontal dividing lines.
|
|
*/
|
|
|
|
if (blocks->r != 1) {
|
|
vmod = blocks->r;
|
|
hmod = blocks->c;
|
|
} else {
|
|
vmod = hmod = 1;
|
|
}
|
|
|
|
/*
|
|
* Line length: we have cr digits, each with a space after it,
|
|
* and (cr-1)/vmod dividing lines, each with a space after it.
|
|
* The final space is replaced by a newline, but that doesn't
|
|
* affect the length.
|
|
*/
|
|
linelen = 2*(cr + (cr-1)/vmod);
|
|
|
|
/*
|
|
* Number of lines: we have cr rows of digits, and (cr-1)/hmod
|
|
* dividing rows.
|
|
*/
|
|
nlines = cr + (cr-1)/hmod;
|
|
|
|
/*
|
|
* Allocate the space.
|
|
*/
|
|
totallen = linelen * nlines;
|
|
ret = snewn(totallen+1, char); /* leave room for terminating NUL */
|
|
|
|
/*
|
|
* Write the text.
|
|
*/
|
|
p = ret;
|
|
for (y = 0; y < cr; y++) {
|
|
/*
|
|
* Row of digits.
|
|
*/
|
|
for (x = 0; x < cr; x++) {
|
|
/*
|
|
* Digit.
|
|
*/
|
|
digit d = grid[y*cr+x];
|
|
|
|
if (d == 0) {
|
|
/*
|
|
* Empty space: we usually write a dot, but we'll
|
|
* highlight spaces on the X-diagonals (in X mode)
|
|
* by using underscores instead.
|
|
*/
|
|
if (xtype && (ondiag0(y*cr+x) || ondiag1(y*cr+x)))
|
|
ch = '_';
|
|
else
|
|
ch = '.';
|
|
} else if (d <= 9) {
|
|
ch = '0' + d;
|
|
} else {
|
|
ch = 'a' + d-10;
|
|
}
|
|
|
|
*p++ = ch;
|
|
if (x == cr-1) {
|
|
*p++ = '\n';
|
|
continue;
|
|
}
|
|
*p++ = ' ';
|
|
|
|
if ((x+1) % vmod)
|
|
continue;
|
|
|
|
/*
|
|
* Optional dividing line.
|
|
*/
|
|
if (blocks->whichblock[y*cr+x] != blocks->whichblock[y*cr+x+1])
|
|
ch = '|';
|
|
else
|
|
ch = ' ';
|
|
*p++ = ch;
|
|
*p++ = ' ';
|
|
}
|
|
if (y == cr-1 || (y+1) % hmod)
|
|
continue;
|
|
|
|
/*
|
|
* Dividing row.
|
|
*/
|
|
for (x = 0; x < cr; x++) {
|
|
int dwid;
|
|
int tl, tr, bl, br;
|
|
|
|
/*
|
|
* Division between two squares. This varies
|
|
* complicatedly in length.
|
|
*/
|
|
dwid = 2; /* digit and its following space */
|
|
if (x == cr-1)
|
|
dwid--; /* no following space at end of line */
|
|
if (x > 0 && x % vmod == 0)
|
|
dwid++; /* preceding space after a divider */
|
|
|
|
if (blocks->whichblock[y*cr+x] != blocks->whichblock[(y+1)*cr+x])
|
|
ch = '-';
|
|
else
|
|
ch = ' ';
|
|
|
|
while (dwid-- > 0)
|
|
*p++ = ch;
|
|
|
|
if (x == cr-1) {
|
|
*p++ = '\n';
|
|
break;
|
|
}
|
|
|
|
if ((x+1) % vmod)
|
|
continue;
|
|
|
|
/*
|
|
* Corner square. This is:
|
|
* - a space if all four surrounding squares are in
|
|
* the same block
|
|
* - a vertical line if the two left ones are in one
|
|
* block and the two right in another
|
|
* - a horizontal line if the two top ones are in one
|
|
* block and the two bottom in another
|
|
* - a plus sign in all other cases. (If we had a
|
|
* richer character set available we could break
|
|
* this case up further by doing fun things with
|
|
* line-drawing T-pieces.)
|
|
*/
|
|
tl = blocks->whichblock[y*cr+x];
|
|
tr = blocks->whichblock[y*cr+x+1];
|
|
bl = blocks->whichblock[(y+1)*cr+x];
|
|
br = blocks->whichblock[(y+1)*cr+x+1];
|
|
|
|
if (tl == tr && tr == bl && bl == br)
|
|
ch = ' ';
|
|
else if (tl == bl && tr == br)
|
|
ch = '|';
|
|
else if (tl == tr && bl == br)
|
|
ch = '-';
|
|
else
|
|
ch = '+';
|
|
|
|
*p++ = ch;
|
|
}
|
|
}
|
|
|
|
assert(p - ret == totallen);
|
|
*p = '\0';
|
|
return ret;
|
|
}
|
|
|
|
static char *game_text_format(game_state *state)
|
|
{
|
|
return grid_text_format(state->cr, state->blocks, state->xtype,
|
|
state->grid);
|
|
}
|
|
|
|
struct game_ui {
|
|
/*
|
|
* These are the coordinates of the currently highlighted
|
|
* square on the grid, or -1,-1 if there isn't one. When there
|
|
* is, pressing a valid number or letter key or Space will
|
|
* enter that number or letter in the grid.
|
|
*/
|
|
int hx, hy;
|
|
/*
|
|
* This indicates whether the current highlight is a
|
|
* pencil-mark one or a real one.
|
|
*/
|
|
int hpencil;
|
|
};
|
|
|
|
static game_ui *new_ui(game_state *state)
|
|
{
|
|
game_ui *ui = snew(game_ui);
|
|
|
|
ui->hx = ui->hy = -1;
|
|
ui->hpencil = 0;
|
|
|
|
return ui;
|
|
}
|
|
|
|
static void free_ui(game_ui *ui)
|
|
{
|
|
sfree(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)
|
|
{
|
|
int cr = newstate->cr;
|
|
/*
|
|
* We prevent pencil-mode highlighting of a filled square. So
|
|
* if the user has just filled in a square which we had a
|
|
* pencil-mode highlight in (by Undo, or by Redo, or by Solve),
|
|
* then we cancel the highlight.
|
|
*/
|
|
if (ui->hx >= 0 && ui->hy >= 0 && ui->hpencil &&
|
|
newstate->grid[ui->hy * cr + ui->hx] != 0) {
|
|
ui->hx = ui->hy = -1;
|
|
}
|
|
}
|
|
|
|
struct game_drawstate {
|
|
int started;
|
|
int cr, xtype;
|
|
int tilesize;
|
|
digit *grid;
|
|
unsigned char *pencil;
|
|
unsigned char *hl;
|
|
/* This is scratch space used within a single call to game_redraw. */
|
|
int *entered_items;
|
|
};
|
|
|
|
static char *interpret_move(game_state *state, game_ui *ui, game_drawstate *ds,
|
|
int x, int y, int button)
|
|
{
|
|
int cr = state->cr;
|
|
int tx, ty;
|
|
char buf[80];
|
|
|
|
button &= ~MOD_MASK;
|
|
|
|
tx = (x + TILE_SIZE - BORDER) / TILE_SIZE - 1;
|
|
ty = (y + TILE_SIZE - BORDER) / TILE_SIZE - 1;
|
|
|
|
if (tx >= 0 && tx < cr && ty >= 0 && ty < cr) {
|
|
if (button == LEFT_BUTTON) {
|
|
if (state->immutable[ty*cr+tx]) {
|
|
ui->hx = ui->hy = -1;
|
|
} else if (tx == ui->hx && ty == ui->hy && ui->hpencil == 0) {
|
|
ui->hx = ui->hy = -1;
|
|
} else {
|
|
ui->hx = tx;
|
|
ui->hy = ty;
|
|
ui->hpencil = 0;
|
|
}
|
|
return ""; /* UI activity occurred */
|
|
}
|
|
if (button == RIGHT_BUTTON) {
|
|
/*
|
|
* Pencil-mode highlighting for non filled squares.
|
|
*/
|
|
if (state->grid[ty*cr+tx] == 0) {
|
|
if (tx == ui->hx && ty == ui->hy && ui->hpencil) {
|
|
ui->hx = ui->hy = -1;
|
|
} else {
|
|
ui->hpencil = 1;
|
|
ui->hx = tx;
|
|
ui->hy = ty;
|
|
}
|
|
} else {
|
|
ui->hx = ui->hy = -1;
|
|
}
|
|
return ""; /* UI activity occurred */
|
|
}
|
|
}
|
|
|
|
if (ui->hx != -1 && ui->hy != -1 &&
|
|
((button >= '1' && button <= '9' && button - '0' <= cr) ||
|
|
(button >= 'a' && button <= 'z' && button - 'a' + 10 <= cr) ||
|
|
(button >= 'A' && button <= 'Z' && button - 'A' + 10 <= cr) ||
|
|
button == ' ' || button == '\010' || button == '\177')) {
|
|
int n = button - '0';
|
|
if (button >= 'A' && button <= 'Z')
|
|
n = button - 'A' + 10;
|
|
if (button >= 'a' && button <= 'z')
|
|
n = button - 'a' + 10;
|
|
if (button == ' ' || button == '\010' || button == '\177')
|
|
n = 0;
|
|
|
|
/*
|
|
* Can't overwrite this square. In principle this shouldn't
|
|
* happen anyway because we should never have even been
|
|
* able to highlight the square, but it never hurts to be
|
|
* careful.
|
|
*/
|
|
if (state->immutable[ui->hy*cr+ui->hx])
|
|
return NULL;
|
|
|
|
/*
|
|
* Can't make pencil marks in a filled square. In principle
|
|
* this shouldn't happen anyway because we should never
|
|
* have even been able to pencil-highlight the square, but
|
|
* it never hurts to be careful.
|
|
*/
|
|
if (ui->hpencil && state->grid[ui->hy*cr+ui->hx])
|
|
return NULL;
|
|
|
|
sprintf(buf, "%c%d,%d,%d",
|
|
(char)(ui->hpencil && n > 0 ? 'P' : 'R'), ui->hx, ui->hy, n);
|
|
|
|
ui->hx = ui->hy = -1;
|
|
|
|
return dupstr(buf);
|
|
}
|
|
|
|
return NULL;
|
|
}
|
|
|
|
static game_state *execute_move(game_state *from, char *move)
|
|
{
|
|
int cr = from->cr;
|
|
game_state *ret;
|
|
int x, y, n;
|
|
|
|
if (move[0] == 'S') {
|
|
char *p;
|
|
|
|
ret = dup_game(from);
|
|
ret->completed = ret->cheated = TRUE;
|
|
|
|
p = move+1;
|
|
for (n = 0; n < cr*cr; n++) {
|
|
ret->grid[n] = atoi(p);
|
|
|
|
if (!*p || ret->grid[n] < 1 || ret->grid[n] > cr) {
|
|
free_game(ret);
|
|
return NULL;
|
|
}
|
|
|
|
while (*p && isdigit((unsigned char)*p)) p++;
|
|
if (*p == ',') p++;
|
|
}
|
|
|
|
return ret;
|
|
} else if ((move[0] == 'P' || move[0] == 'R') &&
|
|
sscanf(move+1, "%d,%d,%d", &x, &y, &n) == 3 &&
|
|
x >= 0 && x < cr && y >= 0 && y < cr && n >= 0 && n <= cr) {
|
|
|
|
ret = dup_game(from);
|
|
if (move[0] == 'P' && n > 0) {
|
|
int index = (y*cr+x) * cr + (n-1);
|
|
ret->pencil[index] = !ret->pencil[index];
|
|
} else {
|
|
ret->grid[y*cr+x] = n;
|
|
memset(ret->pencil + (y*cr+x)*cr, 0, cr);
|
|
|
|
/*
|
|
* We've made a real change to the grid. Check to see
|
|
* if the game has been completed.
|
|
*/
|
|
if (!ret->completed && check_valid(cr, ret->blocks, ret->xtype,
|
|
ret->grid)) {
|
|
ret->completed = TRUE;
|
|
}
|
|
}
|
|
return ret;
|
|
} else
|
|
return NULL; /* couldn't parse move string */
|
|
}
|
|
|
|
/* ----------------------------------------------------------------------
|
|
* Drawing routines.
|
|
*/
|
|
|
|
#define SIZE(cr) ((cr) * TILE_SIZE + 2*BORDER + 1)
|
|
#define GETTILESIZE(cr, w) ( (double)(w-1) / (double)(cr+1) )
|
|
|
|
static void game_compute_size(game_params *params, int tilesize,
|
|
int *x, int *y)
|
|
{
|
|
/* Ick: fake up `ds->tilesize' for macro expansion purposes */
|
|
struct { int tilesize; } ads, *ds = &ads;
|
|
ads.tilesize = tilesize;
|
|
|
|
*x = SIZE(params->c * params->r);
|
|
*y = SIZE(params->c * params->r);
|
|
}
|
|
|
|
static void game_set_size(drawing *dr, game_drawstate *ds,
|
|
game_params *params, int tilesize)
|
|
{
|
|
ds->tilesize = tilesize;
|
|
}
|
|
|
|
static float *game_colours(frontend *fe, int *ncolours)
|
|
{
|
|
float *ret = snewn(3 * NCOLOURS, float);
|
|
|
|
frontend_default_colour(fe, &ret[COL_BACKGROUND * 3]);
|
|
|
|
ret[COL_XDIAGONALS * 3 + 0] = 0.9F * ret[COL_BACKGROUND * 3 + 0];
|
|
ret[COL_XDIAGONALS * 3 + 1] = 0.9F * ret[COL_BACKGROUND * 3 + 1];
|
|
ret[COL_XDIAGONALS * 3 + 2] = 0.9F * ret[COL_BACKGROUND * 3 + 2];
|
|
|
|
ret[COL_GRID * 3 + 0] = 0.0F;
|
|
ret[COL_GRID * 3 + 1] = 0.0F;
|
|
ret[COL_GRID * 3 + 2] = 0.0F;
|
|
|
|
ret[COL_CLUE * 3 + 0] = 0.0F;
|
|
ret[COL_CLUE * 3 + 1] = 0.0F;
|
|
ret[COL_CLUE * 3 + 2] = 0.0F;
|
|
|
|
ret[COL_USER * 3 + 0] = 0.0F;
|
|
ret[COL_USER * 3 + 1] = 0.6F * ret[COL_BACKGROUND * 3 + 1];
|
|
ret[COL_USER * 3 + 2] = 0.0F;
|
|
|
|
ret[COL_HIGHLIGHT * 3 + 0] = 0.78F * ret[COL_BACKGROUND * 3 + 0];
|
|
ret[COL_HIGHLIGHT * 3 + 1] = 0.78F * ret[COL_BACKGROUND * 3 + 1];
|
|
ret[COL_HIGHLIGHT * 3 + 2] = 0.78F * ret[COL_BACKGROUND * 3 + 2];
|
|
|
|
ret[COL_ERROR * 3 + 0] = 1.0F;
|
|
ret[COL_ERROR * 3 + 1] = 0.0F;
|
|
ret[COL_ERROR * 3 + 2] = 0.0F;
|
|
|
|
ret[COL_PENCIL * 3 + 0] = 0.5F * ret[COL_BACKGROUND * 3 + 0];
|
|
ret[COL_PENCIL * 3 + 1] = 0.5F * ret[COL_BACKGROUND * 3 + 1];
|
|
ret[COL_PENCIL * 3 + 2] = ret[COL_BACKGROUND * 3 + 2];
|
|
|
|
*ncolours = NCOLOURS;
|
|
return ret;
|
|
}
|
|
|
|
static game_drawstate *game_new_drawstate(drawing *dr, game_state *state)
|
|
{
|
|
struct game_drawstate *ds = snew(struct game_drawstate);
|
|
int cr = state->cr;
|
|
|
|
ds->started = FALSE;
|
|
ds->cr = cr;
|
|
ds->xtype = state->xtype;
|
|
ds->grid = snewn(cr*cr, digit);
|
|
memset(ds->grid, cr+2, cr*cr);
|
|
ds->pencil = snewn(cr*cr*cr, digit);
|
|
memset(ds->pencil, 0, cr*cr*cr);
|
|
ds->hl = snewn(cr*cr, unsigned char);
|
|
memset(ds->hl, 0, cr*cr);
|
|
ds->entered_items = snewn(cr*cr, int);
|
|
ds->tilesize = 0; /* not decided yet */
|
|
return ds;
|
|
}
|
|
|
|
static void game_free_drawstate(drawing *dr, game_drawstate *ds)
|
|
{
|
|
sfree(ds->hl);
|
|
sfree(ds->pencil);
|
|
sfree(ds->grid);
|
|
sfree(ds->entered_items);
|
|
sfree(ds);
|
|
}
|
|
|
|
static void draw_number(drawing *dr, game_drawstate *ds, game_state *state,
|
|
int x, int y, int hl)
|
|
{
|
|
int cr = state->cr;
|
|
int tx, ty;
|
|
int cx, cy, cw, ch;
|
|
char str[2];
|
|
|
|
if (ds->grid[y*cr+x] == state->grid[y*cr+x] &&
|
|
ds->hl[y*cr+x] == hl &&
|
|
!memcmp(ds->pencil+(y*cr+x)*cr, state->pencil+(y*cr+x)*cr, cr))
|
|
return; /* no change required */
|
|
|
|
tx = BORDER + x * TILE_SIZE + 1 + GRIDEXTRA;
|
|
ty = BORDER + y * TILE_SIZE + 1 + GRIDEXTRA;
|
|
|
|
cx = tx;
|
|
cy = ty;
|
|
cw = TILE_SIZE-1-2*GRIDEXTRA;
|
|
ch = TILE_SIZE-1-2*GRIDEXTRA;
|
|
|
|
if (x > 0 && state->blocks->whichblock[y*cr+x] == state->blocks->whichblock[y*cr+x-1])
|
|
cx -= GRIDEXTRA, cw += GRIDEXTRA;
|
|
if (x+1 < cr && state->blocks->whichblock[y*cr+x] == state->blocks->whichblock[y*cr+x+1])
|
|
cw += GRIDEXTRA;
|
|
if (y > 0 && state->blocks->whichblock[y*cr+x] == state->blocks->whichblock[(y-1)*cr+x])
|
|
cy -= GRIDEXTRA, ch += GRIDEXTRA;
|
|
if (y+1 < cr && state->blocks->whichblock[y*cr+x] == state->blocks->whichblock[(y+1)*cr+x])
|
|
ch += GRIDEXTRA;
|
|
|
|
clip(dr, cx, cy, cw, ch);
|
|
|
|
/* background needs erasing */
|
|
draw_rect(dr, cx, cy, cw, ch,
|
|
((hl & 15) == 1 ? COL_HIGHLIGHT :
|
|
(ds->xtype && (ondiag0(y*cr+x) || ondiag1(y*cr+x))) ? COL_XDIAGONALS :
|
|
COL_BACKGROUND));
|
|
|
|
/*
|
|
* Draw the corners of thick lines in corner-adjacent squares,
|
|
* which jut into this square by one pixel.
|
|
*/
|
|
if (x > 0 && y > 0 && state->blocks->whichblock[y*cr+x] != state->blocks->whichblock[(y-1)*cr+x-1])
|
|
draw_rect(dr, tx-GRIDEXTRA, ty-GRIDEXTRA, GRIDEXTRA, GRIDEXTRA, COL_GRID);
|
|
if (x+1 < cr && y > 0 && state->blocks->whichblock[y*cr+x] != state->blocks->whichblock[(y-1)*cr+x+1])
|
|
draw_rect(dr, tx+TILE_SIZE-1-2*GRIDEXTRA, ty-GRIDEXTRA, GRIDEXTRA, GRIDEXTRA, COL_GRID);
|
|
if (x > 0 && y+1 < cr && state->blocks->whichblock[y*cr+x] != state->blocks->whichblock[(y+1)*cr+x-1])
|
|
draw_rect(dr, tx-GRIDEXTRA, ty+TILE_SIZE-1-2*GRIDEXTRA, GRIDEXTRA, GRIDEXTRA, COL_GRID);
|
|
if (x+1 < cr && y+1 < cr && state->blocks->whichblock[y*cr+x] != state->blocks->whichblock[(y+1)*cr+x+1])
|
|
draw_rect(dr, tx+TILE_SIZE-1-2*GRIDEXTRA, ty+TILE_SIZE-1-2*GRIDEXTRA, GRIDEXTRA, GRIDEXTRA, COL_GRID);
|
|
|
|
/* pencil-mode highlight */
|
|
if ((hl & 15) == 2) {
|
|
int coords[6];
|
|
coords[0] = cx;
|
|
coords[1] = cy;
|
|
coords[2] = cx+cw/2;
|
|
coords[3] = cy;
|
|
coords[4] = cx;
|
|
coords[5] = cy+ch/2;
|
|
draw_polygon(dr, coords, 3, COL_HIGHLIGHT, COL_HIGHLIGHT);
|
|
}
|
|
|
|
/* new number needs drawing? */
|
|
if (state->grid[y*cr+x]) {
|
|
str[1] = '\0';
|
|
str[0] = state->grid[y*cr+x] + '0';
|
|
if (str[0] > '9')
|
|
str[0] += 'a' - ('9'+1);
|
|
draw_text(dr, tx + TILE_SIZE/2, ty + TILE_SIZE/2,
|
|
FONT_VARIABLE, TILE_SIZE/2, ALIGN_VCENTRE | ALIGN_HCENTRE,
|
|
state->immutable[y*cr+x] ? COL_CLUE : (hl & 16) ? COL_ERROR : COL_USER, str);
|
|
} else {
|
|
int i, j, npencil;
|
|
int pw, ph, pmax, fontsize;
|
|
|
|
/* count the pencil marks required */
|
|
for (i = npencil = 0; i < cr; i++)
|
|
if (state->pencil[(y*cr+x)*cr+i])
|
|
npencil++;
|
|
|
|
/*
|
|
* It's not sensible to arrange pencil marks in the same
|
|
* layout as the squares within a block, because this leads
|
|
* to the font being too small. Instead, we arrange pencil
|
|
* marks in the nearest thing we can to a square layout,
|
|
* and we adjust the square layout depending on the number
|
|
* of pencil marks in the square.
|
|
*/
|
|
for (pw = 1; pw * pw < npencil; pw++);
|
|
if (pw < 3) pw = 3; /* otherwise it just looks _silly_ */
|
|
ph = (npencil + pw - 1) / pw;
|
|
if (ph < 2) ph = 2; /* likewise */
|
|
pmax = max(pw, ph);
|
|
fontsize = TILE_SIZE/(pmax*(11-pmax)/8);
|
|
|
|
for (i = j = 0; i < cr; i++)
|
|
if (state->pencil[(y*cr+x)*cr+i]) {
|
|
int dx = j % pw, dy = j / pw;
|
|
|
|
str[1] = '\0';
|
|
str[0] = i + '1';
|
|
if (str[0] > '9')
|
|
str[0] += 'a' - ('9'+1);
|
|
draw_text(dr, tx + (4*dx+3) * TILE_SIZE / (4*pw+2),
|
|
ty + (4*dy+3) * TILE_SIZE / (4*ph+2),
|
|
FONT_VARIABLE, fontsize,
|
|
ALIGN_VCENTRE | ALIGN_HCENTRE, COL_PENCIL, str);
|
|
j++;
|
|
}
|
|
}
|
|
|
|
unclip(dr);
|
|
|
|
draw_update(dr, cx, cy, cw, ch);
|
|
|
|
ds->grid[y*cr+x] = state->grid[y*cr+x];
|
|
memcpy(ds->pencil+(y*cr+x)*cr, state->pencil+(y*cr+x)*cr, cr);
|
|
ds->hl[y*cr+x] = hl;
|
|
}
|
|
|
|
static void game_redraw(drawing *dr, game_drawstate *ds, game_state *oldstate,
|
|
game_state *state, int dir, game_ui *ui,
|
|
float animtime, float flashtime)
|
|
{
|
|
int cr = state->cr;
|
|
int x, y;
|
|
|
|
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.
|
|
*/
|
|
draw_rect(dr, 0, 0, SIZE(cr), SIZE(cr), COL_BACKGROUND);
|
|
|
|
/*
|
|
* Draw the grid. We draw it as a big thick rectangle of
|
|
* COL_GRID initially; individual calls to draw_number()
|
|
* will poke the right-shaped holes in it.
|
|
*/
|
|
draw_rect(dr, BORDER-GRIDEXTRA, BORDER-GRIDEXTRA,
|
|
cr*TILE_SIZE+1+2*GRIDEXTRA, cr*TILE_SIZE+1+2*GRIDEXTRA,
|
|
COL_GRID);
|
|
}
|
|
|
|
/*
|
|
* This array is used to keep track of rows, columns and boxes
|
|
* which contain a number more than once.
|
|
*/
|
|
for (x = 0; x < cr * cr; x++)
|
|
ds->entered_items[x] = 0;
|
|
for (x = 0; x < cr; x++)
|
|
for (y = 0; y < cr; y++) {
|
|
digit d = state->grid[y*cr+x];
|
|
if (d) {
|
|
int box = state->blocks->whichblock[y*cr+x];
|
|
ds->entered_items[x*cr+d-1] |= ((ds->entered_items[x*cr+d-1] & 1) << 1) | 1;
|
|
ds->entered_items[y*cr+d-1] |= ((ds->entered_items[y*cr+d-1] & 4) << 1) | 4;
|
|
ds->entered_items[box*cr+d-1] |= ((ds->entered_items[box*cr+d-1] & 16) << 1) | 16;
|
|
if (ds->xtype) {
|
|
if (ondiag0(y*cr+x))
|
|
ds->entered_items[d-1] |= ((ds->entered_items[d-1] & 64) << 1) | 64;
|
|
if (ondiag1(y*cr+x))
|
|
ds->entered_items[cr+d-1] |= ((ds->entered_items[cr+d-1] & 64) << 1) | 64;
|
|
}
|
|
}
|
|
}
|
|
|
|
/*
|
|
* Draw any numbers which need redrawing.
|
|
*/
|
|
for (x = 0; x < cr; x++) {
|
|
for (y = 0; y < cr; y++) {
|
|
int highlight = 0;
|
|
digit d = state->grid[y*cr+x];
|
|
|
|
if (flashtime > 0 &&
|
|
(flashtime <= FLASH_TIME/3 ||
|
|
flashtime >= FLASH_TIME*2/3))
|
|
highlight = 1;
|
|
|
|
/* Highlight active input areas. */
|
|
if (x == ui->hx && y == ui->hy)
|
|
highlight = ui->hpencil ? 2 : 1;
|
|
|
|
/* Mark obvious errors (ie, numbers which occur more than once
|
|
* in a single row, column, or box). */
|
|
if (d && ((ds->entered_items[x*cr+d-1] & 2) ||
|
|
(ds->entered_items[y*cr+d-1] & 8) ||
|
|
(ds->entered_items[state->blocks->whichblock[y*cr+x]*cr+d-1] & 32) ||
|
|
(ds->xtype && ((ondiag0(y*cr+x) && (ds->entered_items[d-1] & 128)) ||
|
|
(ondiag1(y*cr+x) && (ds->entered_items[cr+d-1] & 128))))))
|
|
highlight |= 16;
|
|
|
|
draw_number(dr, ds, state, x, y, highlight);
|
|
}
|
|
}
|
|
|
|
/*
|
|
* Update the _entire_ grid if necessary.
|
|
*/
|
|
if (!ds->started) {
|
|
draw_update(dr, 0, 0, SIZE(cr), SIZE(cr));
|
|
ds->started = TRUE;
|
|
}
|
|
}
|
|
|
|
static float game_anim_length(game_state *oldstate, game_state *newstate,
|
|
int dir, game_ui *ui)
|
|
{
|
|
return 0.0F;
|
|
}
|
|
|
|
static float game_flash_length(game_state *oldstate, game_state *newstate,
|
|
int dir, game_ui *ui)
|
|
{
|
|
if (!oldstate->completed && newstate->completed &&
|
|
!oldstate->cheated && !newstate->cheated)
|
|
return FLASH_TIME;
|
|
return 0.0F;
|
|
}
|
|
|
|
static int game_timing_state(game_state *state, game_ui *ui)
|
|
{
|
|
return TRUE;
|
|
}
|
|
|
|
static void game_print_size(game_params *params, float *x, float *y)
|
|
{
|
|
int pw, ph;
|
|
|
|
/*
|
|
* I'll use 9mm squares by default. They should be quite big
|
|
* for this game, because players will want to jot down no end
|
|
* of pencil marks in the squares.
|
|
*/
|
|
game_compute_size(params, 900, &pw, &ph);
|
|
*x = pw / 100.0;
|
|
*y = ph / 100.0;
|
|
}
|
|
|
|
static void game_print(drawing *dr, game_state *state, int tilesize)
|
|
{
|
|
int cr = state->cr;
|
|
int ink = print_mono_colour(dr, 0);
|
|
int x, y;
|
|
|
|
/* Ick: fake up `ds->tilesize' for macro expansion purposes */
|
|
game_drawstate ads, *ds = &ads;
|
|
game_set_size(dr, ds, NULL, tilesize);
|
|
|
|
/*
|
|
* Border.
|
|
*/
|
|
print_line_width(dr, 3 * TILE_SIZE / 40);
|
|
draw_rect_outline(dr, BORDER, BORDER, cr*TILE_SIZE, cr*TILE_SIZE, ink);
|
|
|
|
/*
|
|
* Highlight X-diagonal squares.
|
|
*/
|
|
if (state->xtype) {
|
|
int i;
|
|
int xhighlight = print_grey_colour(dr, 0.90F);
|
|
|
|
for (i = 0; i < cr; i++)
|
|
draw_rect(dr, BORDER + i*TILE_SIZE, BORDER + i*TILE_SIZE,
|
|
TILE_SIZE, TILE_SIZE, xhighlight);
|
|
for (i = 0; i < cr; i++)
|
|
if (i*2 != cr-1) /* avoid redoing centre square, just for fun */
|
|
draw_rect(dr, BORDER + i*TILE_SIZE,
|
|
BORDER + (cr-1-i)*TILE_SIZE,
|
|
TILE_SIZE, TILE_SIZE, xhighlight);
|
|
}
|
|
|
|
/*
|
|
* Main grid.
|
|
*/
|
|
for (x = 1; x < cr; x++) {
|
|
print_line_width(dr, TILE_SIZE / 40);
|
|
draw_line(dr, BORDER+x*TILE_SIZE, BORDER,
|
|
BORDER+x*TILE_SIZE, BORDER+cr*TILE_SIZE, ink);
|
|
}
|
|
for (y = 1; y < cr; y++) {
|
|
print_line_width(dr, TILE_SIZE / 40);
|
|
draw_line(dr, BORDER, BORDER+y*TILE_SIZE,
|
|
BORDER+cr*TILE_SIZE, BORDER+y*TILE_SIZE, ink);
|
|
}
|
|
|
|
/*
|
|
* Thick lines between cells. In order to do this using the
|
|
* line-drawing rather than rectangle-drawing API (so as to
|
|
* get line thicknesses to scale correctly) and yet have
|
|
* correctly mitred joins between lines, we must do this by
|
|
* tracing the boundary of each sub-block and drawing it in
|
|
* one go as a single polygon.
|
|
*/
|
|
{
|
|
int *coords;
|
|
int bi, i, n;
|
|
int x, y, dx, dy, sx, sy, sdx, sdy;
|
|
|
|
print_line_width(dr, 3 * TILE_SIZE / 40);
|
|
|
|
/*
|
|
* Maximum perimeter of a k-omino is 2k+2. (Proof: start
|
|
* with k unconnected squares, with total perimeter 4k.
|
|
* Now repeatedly join two disconnected components
|
|
* together into a larger one; every time you do so you
|
|
* remove at least two unit edges, and you require k-1 of
|
|
* these operations to create a single connected piece, so
|
|
* you must have at most 4k-2(k-1) = 2k+2 unit edges left
|
|
* afterwards.)
|
|
*/
|
|
coords = snewn(4*cr+4, int); /* 2k+2 points, 2 coords per point */
|
|
|
|
/*
|
|
* Iterate over all the blocks.
|
|
*/
|
|
for (bi = 0; bi < cr; bi++) {
|
|
|
|
/*
|
|
* For each block, find a starting square within it
|
|
* which has a boundary at the left.
|
|
*/
|
|
for (i = 0; i < cr; i++) {
|
|
int j = state->blocks->blocks[bi][i];
|
|
if (j % cr == 0 || state->blocks->whichblock[j-1] != bi)
|
|
break;
|
|
}
|
|
assert(i < cr); /* every block must have _some_ leftmost square */
|
|
x = state->blocks->blocks[bi][i] % cr;
|
|
y = state->blocks->blocks[bi][i] / cr;
|
|
dx = -1;
|
|
dy = 0;
|
|
|
|
/*
|
|
* Now begin tracing round the perimeter. At all
|
|
* times, (x,y) describes some square within the
|
|
* block, and (x+dx,y+dy) is some adjacent square
|
|
* outside it; so the edge between those two squares
|
|
* is always an edge of the block.
|
|
*/
|
|
sx = x, sy = y, sdx = dx, sdy = dy; /* save starting position */
|
|
n = 0;
|
|
do {
|
|
int cx, cy, tx, ty, nin;
|
|
|
|
/*
|
|
* To begin with, record the point at one end of
|
|
* the edge. To do this, we translate (x,y) down
|
|
* and right by half a unit (so they're describing
|
|
* a point in the _centre_ of the square) and then
|
|
* translate back again in a manner rotated by dy
|
|
* and dx.
|
|
*/
|
|
assert(n < 2*cr+2);
|
|
cx = ((2*x+1) + dy + dx) / 2;
|
|
cy = ((2*y+1) - dx + dy) / 2;
|
|
coords[2*n+0] = BORDER + cx * TILE_SIZE;
|
|
coords[2*n+1] = BORDER + cy * TILE_SIZE;
|
|
n++;
|
|
|
|
/*
|
|
* Now advance to the next edge, by looking at the
|
|
* two squares beyond it. If they're both outside
|
|
* the block, we turn right (by leaving x,y the
|
|
* same and rotating dx,dy clockwise); if they're
|
|
* both inside, we turn left (by rotating dx,dy
|
|
* anticlockwise and contriving to leave x+dx,y+dy
|
|
* unchanged); if one of each, we go straight on
|
|
* (and may enforce by assertion that they're one
|
|
* of each the _right_ way round).
|
|
*/
|
|
nin = 0;
|
|
tx = x - dy + dx;
|
|
ty = y + dx + dy;
|
|
nin += (tx >= 0 && tx < cr && ty >= 0 && ty < cr &&
|
|
state->blocks->whichblock[ty*cr+tx] == bi);
|
|
tx = x - dy;
|
|
ty = y + dx;
|
|
nin += (tx >= 0 && tx < cr && ty >= 0 && ty < cr &&
|
|
state->blocks->whichblock[ty*cr+tx] == bi);
|
|
if (nin == 0) {
|
|
/*
|
|
* Turn right.
|
|
*/
|
|
int tmp;
|
|
tmp = dx;
|
|
dx = -dy;
|
|
dy = tmp;
|
|
} else if (nin == 2) {
|
|
/*
|
|
* Turn left.
|
|
*/
|
|
int tmp;
|
|
|
|
x += dx;
|
|
y += dy;
|
|
|
|
tmp = dx;
|
|
dx = dy;
|
|
dy = -tmp;
|
|
|
|
x -= dx;
|
|
y -= dy;
|
|
} else {
|
|
/*
|
|
* Go straight on.
|
|
*/
|
|
x -= dy;
|
|
y += dx;
|
|
}
|
|
|
|
/*
|
|
* Now enforce by assertion that we ended up
|
|
* somewhere sensible.
|
|
*/
|
|
assert(x >= 0 && x < cr && y >= 0 && y < cr &&
|
|
state->blocks->whichblock[y*cr+x] == bi);
|
|
assert(x+dx < 0 || x+dx >= cr || y+dy < 0 || y+dy >= cr ||
|
|
state->blocks->whichblock[(y+dy)*cr+(x+dx)] != bi);
|
|
|
|
} while (x != sx || y != sy || dx != sdx || dy != sdy);
|
|
|
|
/*
|
|
* That's our polygon; now draw it.
|
|
*/
|
|
draw_polygon(dr, coords, n, -1, ink);
|
|
}
|
|
|
|
sfree(coords);
|
|
}
|
|
|
|
/*
|
|
* Numbers.
|
|
*/
|
|
for (y = 0; y < cr; y++)
|
|
for (x = 0; x < cr; x++)
|
|
if (state->grid[y*cr+x]) {
|
|
char str[2];
|
|
str[1] = '\0';
|
|
str[0] = state->grid[y*cr+x] + '0';
|
|
if (str[0] > '9')
|
|
str[0] += 'a' - ('9'+1);
|
|
draw_text(dr, BORDER + x*TILE_SIZE + TILE_SIZE/2,
|
|
BORDER + y*TILE_SIZE + TILE_SIZE/2,
|
|
FONT_VARIABLE, TILE_SIZE/2,
|
|
ALIGN_VCENTRE | ALIGN_HCENTRE, ink, str);
|
|
}
|
|
}
|
|
|
|
#ifdef COMBINED
|
|
#define thegame solo
|
|
#endif
|
|
|
|
const struct game thegame = {
|
|
"Solo", "games.solo", "solo",
|
|
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,
|
|
TRUE, solve_game,
|
|
TRUE, 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,
|
|
REQUIRE_RBUTTON | REQUIRE_NUMPAD, /* flags */
|
|
};
|
|
|
|
#ifdef STANDALONE_SOLVER
|
|
|
|
int main(int argc, char **argv)
|
|
{
|
|
game_params *p;
|
|
game_state *s;
|
|
char *id = NULL, *desc, *err;
|
|
int grade = FALSE;
|
|
int ret;
|
|
|
|
while (--argc > 0) {
|
|
char *p = *++argv;
|
|
if (!strcmp(p, "-v")) {
|
|
solver_show_working = TRUE;
|
|
} else 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);
|
|
|
|
ret = solver(s->cr, s->blocks, s->xtype, s->grid, DIFF_RECURSIVE);
|
|
if (grade) {
|
|
printf("Difficulty rating: %s\n",
|
|
ret==DIFF_BLOCK ? "Trivial (blockwise positional elimination only)":
|
|
ret==DIFF_SIMPLE ? "Basic (row/column/number elimination required)":
|
|
ret==DIFF_INTERSECT ? "Intermediate (intersectional analysis required)":
|
|
ret==DIFF_SET ? "Advanced (set elimination required)":
|
|
ret==DIFF_EXTREME ? "Extreme (complex non-recursive techniques required)":
|
|
ret==DIFF_RECURSIVE ? "Unreasonable (guesswork and backtracking required)":
|
|
ret==DIFF_AMBIGUOUS ? "Ambiguous (multiple solutions exist)":
|
|
ret==DIFF_IMPOSSIBLE ? "Impossible (no solution exists)":
|
|
"INTERNAL ERROR: unrecognised difficulty code");
|
|
} else {
|
|
printf("%s\n", grid_text_format(s->cr, s->blocks, s->xtype, s->grid));
|
|
}
|
|
|
|
return 0;
|
|
}
|
|
|
|
#endif
|