/* * solo.c: the number-placing puzzle most popularly known as `Sudoku'. * * TODO: * * - reports from users are that `Trivial'-mode puzzles are still * rather hard compared to newspapers' easy ones, so some better * low-end difficulty grading would be nice * + it's possible that really easy puzzles always have * _several_ things you can do, so don't make you hunt too * hard for the one deduction you can currently make * + it's also possible that easy puzzles require fewer * cross-eliminations: perhaps there's a higher incidence of * things you can deduce by looking only at (say) rows, * rather than things you have to check both rows and columns * for * + but really, what I need to do is find some really easy * puzzles and _play_ them, to see what's actually easy about * them * + while I'm revamping this area, filling in the _last_ * number in a nearly-full row or column should certainly be * permitted even at the lowest difficulty level. * + also Owen noticed that `Basic' grids requiring numeric * elimination are actually very hard, so I wonder if a * difficulty gradation between that and positional- * elimination-only might be in order * + but it's not good to have _too_ many difficulty levels, or * it'll take too long to randomly generate a given level. * * - it might still be nice to do some prioritisation on the * removal of numbers from the grid * + one possibility is to try to minimise the maximum number * of filled squares in any block, which in particular ought * to enforce never leaving a completely filled block in the * puzzle as presented. * * - alternative interface modes * + sudoku.com's Windows program has a palette of possible * entries; you select a palette entry first and then click * on the square you want it to go in, thus enabling * mouse-only play. Useful for PDAs! I don't think it's * actually incompatible with the current highlight-then-type * approach: you _either_ highlight a palette entry and then * click, _or_ you highlight a square and then type. At most * one thing is ever highlighted at a time, so there's no way * to confuse the two. * + then again, I don't actually like sudoku.com's interface; * it's too much like a paint package whereas I prefer to * think of Solo as a text editor. * + another PDA-friendly possibility is a drag interface: * _drag_ numbers from the palette into the grid squares. * Thought experiments suggest I'd prefer that to the * sudoku.com approach, but I haven't actually tried it. */ /* * Solo puzzles need to be square overall (since each row and each * column must contain one of every digit), but they need not be * subdivided the same way internally. I am going to adopt a * convention whereby I _always_ refer to `r' as the number of rows * of _big_ divisions, and `c' as the number of columns of _big_ * divisions. Thus, a 2c by 3r puzzle looks something like this: * * 4 5 1 | 2 6 3 * 6 3 2 | 5 4 1 * ------+------ (Of course, you can't subdivide it the other way * 1 4 5 | 6 3 2 or you'll get clashes; observe that the 4 in the * 3 2 6 | 4 1 5 top left would conflict with the 4 in the second * ------+------ box down on the left-hand side.) * 5 1 4 | 3 2 6 * 2 6 3 | 1 5 4 * * The need for a strong naming convention should now be clear: * each small box is two rows of digits by three columns, while the * overall puzzle has three rows of small boxes by two columns. So * I will (hopefully) consistently use `r' to denote the number of * rows _of small boxes_ (here 3), which is also the number of * columns of digits in each small box; and `c' vice versa (here * 2). * * I'm also going to choose arbitrarily to list c first wherever * possible: the above is a 2x3 puzzle, not a 3x2 one. */ #include #include #include #include #include #include #ifdef STANDALONE_SOLVER #include int solver_show_working, solver_recurse_depth; #endif #include "puzzles.h" /* * To save space, I store digits internally as unsigned char. This * imposes a hard limit of 255 on the order of the puzzle. Since * even a 5x5 takes unacceptably long to generate, I don't see this * as a serious limitation unless something _really_ impressive * happens in computing technology; but here's a typedef anyway for * general good practice. */ typedef unsigned char digit; #define ORDER_MAX 255 #define PREFERRED_TILE_SIZE 32 #define TILE_SIZE (ds->tilesize) #define BORDER (TILE_SIZE / 2) #define GRIDEXTRA (TILE_SIZE / 32) #define FLASH_TIME 0.4F enum { SYMM_NONE, SYMM_ROT2, SYMM_ROT4, SYMM_REF2, SYMM_REF2D, SYMM_REF4, SYMM_REF4D, SYMM_REF8 }; enum { DIFF_BLOCK, DIFF_SIMPLE, DIFF_INTERSECT, DIFF_SET, DIFF_EXTREME, DIFF_RECURSIVE, DIFF_AMBIGUOUS, DIFF_IMPOSSIBLE }; enum { COL_BACKGROUND, COL_XDIAGONALS, COL_GRID, COL_CLUE, COL_USER, COL_HIGHLIGHT, COL_ERROR, COL_PENCIL, NCOLOURS }; struct game_params { /* * For a square puzzle, `c' and `r' indicate the puzzle * parameters as described above. * * A jigsaw-style puzzle is indicated by r==1, in which case c * can be whatever it likes (there is no constraint on * compositeness - a 7x7 jigsaw sudoku makes perfect sense). */ int c, r, symm, diff; int xtype; /* require all digits in X-diagonals */ }; struct block_structure { int refcount; /* * For text formatting, we do need c and r here. */ int c, r; /* * For any square index, whichblock[i] gives its block index. * * For 0 <= b,i < cr, blocks[b][i] gives the index of the ith * square in block b. * * whichblock and blocks are each dynamically allocated in * their own right, but the subarrays in blocks are appended * to the whichblock array, so shouldn't be freed * individually. */ int *whichblock, **blocks; #ifdef STANDALONE_SOLVER /* * Textual descriptions of each block. For normal Sudoku these * are of the form "(1,3)"; for jigsaw they are "starting at * (5,7)". So the sensible usage in both cases is to say * "elimination within block %s" with one of these strings. * * Only blocknames itself needs individually freeing; it's all * one block. */ char **blocknames; #endif }; struct game_state { /* * For historical reasons, I use `cr' to denote the overall * width/height of the puzzle. It was a natural notation when * all puzzles were divided into blocks in a grid, but doesn't * really make much sense given jigsaw puzzles. However, the * obvious `n' is heavily used in the solver to describe the * index of a number being placed, so `cr' will have to stay. */ int cr; struct block_structure *blocks; int xtype; digit *grid; unsigned char *pencil; /* c*r*c*r elements */ unsigned char *immutable; /* marks which digits are clues */ int completed, cheated; }; static game_params *default_params(void) { game_params *ret = snew(game_params); ret->c = ret->r = 3; ret->xtype = FALSE; ret->symm = SYMM_ROT2; /* a plausible default */ ret->diff = DIFF_BLOCK; /* so is this */ return ret; } static void free_params(game_params *params) { sfree(params); } static game_params *dup_params(game_params *params) { game_params *ret = snew(game_params); *ret = *params; /* structure copy */ return ret; } static int game_fetch_preset(int i, char **name, game_params **params) { static struct { char *title; game_params params; } presets[] = { { "2x2 Trivial", { 2, 2, SYMM_ROT2, DIFF_BLOCK, FALSE } }, { "2x3 Basic", { 2, 3, SYMM_ROT2, DIFF_SIMPLE, FALSE } }, { "3x3 Trivial", { 3, 3, SYMM_ROT2, DIFF_BLOCK, FALSE } }, { "3x3 Basic", { 3, 3, SYMM_ROT2, DIFF_SIMPLE, FALSE } }, { "3x3 Basic X", { 3, 3, SYMM_ROT2, DIFF_SIMPLE, TRUE } }, { "3x3 Intermediate", { 3, 3, SYMM_ROT2, DIFF_INTERSECT, FALSE } }, { "3x3 Advanced", { 3, 3, SYMM_ROT2, DIFF_SET, FALSE } }, { "3x3 Advanced X", { 3, 3, SYMM_ROT2, DIFF_SET, TRUE } }, { "3x3 Extreme", { 3, 3, SYMM_ROT2, DIFF_EXTREME, FALSE } }, { "3x3 Unreasonable", { 3, 3, SYMM_ROT2, DIFF_RECURSIVE, FALSE } }, { "9 Jigsaw Basic", { 9, 1, SYMM_ROT2, DIFF_SIMPLE, FALSE } }, { "9 Jigsaw Basic X", { 9, 1, SYMM_ROT2, DIFF_SIMPLE, TRUE } }, { "9 Jigsaw Advanced", { 9, 1, SYMM_ROT2, DIFF_SET, FALSE } }, #ifndef SLOW_SYSTEM { "3x4 Basic", { 3, 4, SYMM_ROT2, DIFF_SIMPLE, FALSE } }, { "4x4 Basic", { 4, 4, SYMM_ROT2, DIFF_SIMPLE, FALSE } }, #endif }; if (i < 0 || i >= lenof(presets)) return FALSE; *name = dupstr(presets[i].title); *params = dup_params(&presets[i].params); return TRUE; } static void decode_params(game_params *ret, char const *string) { int seen_r = FALSE; ret->c = ret->r = atoi(string); ret->xtype = FALSE; while (*string && isdigit((unsigned char)*string)) string++; if (*string == 'x') { string++; ret->r = atoi(string); seen_r = TRUE; while (*string && isdigit((unsigned char)*string)) string++; } while (*string) { if (*string == 'j') { string++; if (seen_r) ret->c *= ret->r; ret->r = 1; } else if (*string == 'x') { string++; ret->xtype = TRUE; } else if (*string == 'r' || *string == 'm' || *string == 'a') { int sn, sc, sd; sc = *string++; if (sc == 'm' && *string == 'd') { sd = TRUE; string++; } else { sd = FALSE; } sn = atoi(string); while (*string && isdigit((unsigned char)*string)) string++; if (sc == 'm' && sn == 8) ret->symm = SYMM_REF8; if (sc == 'm' && sn == 4) ret->symm = sd ? SYMM_REF4D : SYMM_REF4; if (sc == 'm' && sn == 2) ret->symm = sd ? SYMM_REF2D : SYMM_REF2; if (sc == 'r' && sn == 4) ret->symm = SYMM_ROT4; if (sc == 'r' && sn == 2) ret->symm = SYMM_ROT2; if (sc == 'a') ret->symm = SYMM_NONE; } else if (*string == 'd') { string++; if (*string == 't') /* trivial */ string++, ret->diff = DIFF_BLOCK; else if (*string == 'b') /* basic */ string++, ret->diff = DIFF_SIMPLE; else if (*string == 'i') /* intermediate */ string++, ret->diff = DIFF_INTERSECT; else if (*string == 'a') /* advanced */ string++, ret->diff = DIFF_SET; else if (*string == 'e') /* extreme */ string++, ret->diff = DIFF_EXTREME; else if (*string == 'u') /* unreasonable */ string++, ret->diff = DIFF_RECURSIVE; } else string++; /* eat unknown character */ } } static char *encode_params(game_params *params, int full) { char str[80]; if (params->r > 1) sprintf(str, "%dx%d", params->c, params->r); else sprintf(str, "%dj", params->c); if (params->xtype) strcat(str, "x"); if (full) { switch (params->symm) { case SYMM_REF8: strcat(str, "m8"); break; case SYMM_REF4: strcat(str, "m4"); break; case SYMM_REF4D: strcat(str, "md4"); break; case SYMM_REF2: strcat(str, "m2"); break; case SYMM_REF2D: strcat(str, "md2"); break; case SYMM_ROT4: strcat(str, "r4"); break; /* case SYMM_ROT2: strcat(str, "r2"); break; [default] */ case SYMM_NONE: strcat(str, "a"); break; } switch (params->diff) { /* case DIFF_BLOCK: strcat(str, "dt"); break; [default] */ case DIFF_SIMPLE: strcat(str, "db"); break; case DIFF_INTERSECT: strcat(str, "di"); break; case DIFF_SET: strcat(str, "da"); break; case DIFF_EXTREME: strcat(str, "de"); break; case DIFF_RECURSIVE: strcat(str, "du"); break; } } return dupstr(str); } static config_item *game_configure(game_params *params) { config_item *ret; char buf[80]; ret = snewn(7, config_item); ret[0].name = "Columns of sub-blocks"; ret[0].type = C_STRING; sprintf(buf, "%d", params->c); ret[0].sval = dupstr(buf); ret[0].ival = 0; ret[1].name = "Rows of sub-blocks"; ret[1].type = C_STRING; sprintf(buf, "%d", params->r); ret[1].sval = dupstr(buf); ret[1].ival = 0; ret[2].name = "\"X\" (require every number in each main diagonal)"; ret[2].type = C_BOOLEAN; ret[2].sval = NULL; ret[2].ival = params->xtype; ret[3].name = "Jigsaw (irregularly shaped sub-blocks)"; ret[3].type = C_BOOLEAN; ret[3].sval = NULL; ret[3].ival = (params->r == 1); ret[4].name = "Symmetry"; ret[4].type = C_CHOICES; ret[4].sval = ":None:2-way rotation:4-way rotation:2-way mirror:" "2-way diagonal mirror:4-way mirror:4-way diagonal mirror:" "8-way mirror"; ret[4].ival = params->symm; ret[5].name = "Difficulty"; ret[5].type = C_CHOICES; ret[5].sval = ":Trivial:Basic:Intermediate:Advanced:Extreme:Unreasonable"; ret[5].ival = params->diff; ret[6].name = NULL; ret[6].type = C_END; ret[6].sval = NULL; ret[6].ival = 0; return ret; } static game_params *custom_params(config_item *cfg) { game_params *ret = snew(game_params); ret->c = atoi(cfg[0].sval); ret->r = atoi(cfg[1].sval); ret->xtype = cfg[2].ival; if (cfg[3].ival) { ret->c *= ret->r; ret->r = 1; } ret->symm = cfg[4].ival; ret->diff = cfg[5].ival; return ret; } static char *validate_params(game_params *params, int full) { if (params->c < 2) return "Both dimensions must be at least 2"; if (params->c > ORDER_MAX || params->r > ORDER_MAX) return "Dimensions greater than "STR(ORDER_MAX)" are not supported"; if ((params->c * params->r) > 35) return "Unable to support more than 35 distinct symbols in a puzzle"; return NULL; } /* ---------------------------------------------------------------------- * Solver. * * This solver is used for two purposes: * + to check solubility of a grid as we gradually remove numbers * from it * + to solve an externally generated puzzle when the user selects * `Solve'. * * It supports a variety of specific modes of reasoning. By * enabling or disabling subsets of these modes we can arrange a * range of difficulty levels. */ /* * Modes of reasoning currently supported: * * - Positional elimination: a number must go in a particular * square because all the other empty squares in a given * row/col/blk are ruled out. * * - Numeric elimination: a square must have a particular number * in because all the other numbers that could go in it are * ruled out. * * - Intersectional analysis: given two domains which overlap * (hence one must be a block, and the other can be a row or * col), if the possible locations for a particular number in * one of the domains can be narrowed down to the overlap, then * that number can be ruled out everywhere but the overlap in * the other domain too. * * - Set elimination: if there is a subset of the empty squares * within a domain such that the union of the possible numbers * in that subset has the same size as the subset itself, then * those numbers can be ruled out everywhere else in the domain. * (For example, if there are five empty squares and the * possible numbers in each are 12, 23, 13, 134 and 1345, then * the first three empty squares form such a subset: the numbers * 1, 2 and 3 _must_ be in those three squares in some * 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 if (*desc >= 'a' && *desc <= 'z') c = *desc - 'a' + 1; else assert(!"Shouldn't get here"); 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] \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