/* * 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 <stdio.h> #include <stdlib.h> #include <string.h> #include <assert.h> #include <ctype.h> #include <math.h> #ifdef STANDALONE_SOLVER #include <stdarg.h> 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 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_GRID, COL_CLUE, COL_USER, COL_HIGHLIGHT, COL_ERROR, COL_PENCIL, NCOLOURS }; struct game_params { int c, r, symm, diff; }; struct game_state { int c, r; 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->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 } }, { "2x3 Basic", { 2, 3, SYMM_ROT2, DIFF_SIMPLE } }, { "3x3 Trivial", { 3, 3, SYMM_ROT2, DIFF_BLOCK } }, { "3x3 Basic", { 3, 3, SYMM_ROT2, DIFF_SIMPLE } }, { "3x3 Intermediate", { 3, 3, SYMM_ROT2, DIFF_INTERSECT } }, { "3x3 Advanced", { 3, 3, SYMM_ROT2, DIFF_SET } }, { "3x3 Extreme", { 3, 3, SYMM_ROT2, DIFF_EXTREME } }, { "3x3 Unreasonable", { 3, 3, SYMM_ROT2, DIFF_RECURSIVE } }, #ifndef SLOW_SYSTEM { "3x4 Basic", { 3, 4, SYMM_ROT2, DIFF_SIMPLE } }, { "4x4 Basic", { 4, 4, SYMM_ROT2, DIFF_SIMPLE } }, #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) { ret->c = ret->r = atoi(string); while (*string && isdigit((unsigned char)*string)) string++; if (*string == 'x') { string++; ret->r = atoi(string); while (*string && isdigit((unsigned char)*string)) string++; } while (*string) { if (*string == 'r' || *string == 'm' || *string == 'a') { int sn, sc, sd; sc = *string++; if (*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]; sprintf(str, "%dx%d", params->c, params->r); 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(5, 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 = "Symmetry"; ret[2].type = C_CHOICES; ret[2].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[2].ival = params->symm; ret[3].name = "Difficulty"; ret[3].type = C_CHOICES; ret[3].sval = ":Trivial:Basic:Intermediate:Advanced:Extreme:Unreasonable"; ret[3].ival = params->diff; ret[4].name = NULL; ret[4].type = C_END; ret[4].sval = NULL; ret[4].ival = 0; return ret; } static game_params *custom_params(config_item *cfg) { game_params *ret = snew(game_params); ret->c = atoi(cfg[0].sval); ret->r = atoi(cfg[1].sval); ret->symm = cfg[2].ival; ret->diff = cfg[3].ival; return ret; } static char *validate_params(game_params *params, int full) { if (params->c < 2 || params->r < 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) > 36) return "Unable to support more than 36 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). * * - Mutual neighbour elimination: find two squares A,B and a * number N in the possible set of A, such that putting N in A * would rule out enough possibilities from the mutual * neighbours of A and B that there would be no possibilities * left for B. Thereby rule out N in A. * + The simplest case of this is if B has two possibilities * (wlog {1,2}), and there are two mutual neighbours of A and * B which have possibilities {1,3} and {2,3}. Thus, if A * were to be 3, then those neighbours would contain 1 and 2, * and hence there would be nothing left which could go in B. * + There can be more complex cases of it too: if A and B are * in the same column of large blocks, then they can have * more than two mutual neighbours, some of which can also be * neighbours of one another. Suppose, for example, that B * has possibilities {1,2,3}; there's one square P in the * same column as B and the same block as A, with * possibilities {1,4}; and there are _two_ squares Q,R in * the same column as A and the same block as B with * possibilities {2,3,4}. Then if A contained 4, P would * contain 1, and Q and R would have to contain 2 and 3 in * _some_ order; therefore, once again, B would have no * remaining possibilities. * * - 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. */ /* * Within this solver, I'm going to transform all y-coordinates by * inverting the significance of the block number and the position * within the block. That is, we will start with the top row of * each block in order, then the second row of each block in order, * etc. * * This transformation has the enormous advantage that it means * every row, column _and_ block is described by an arithmetic * progression of coordinates within the cubic array, so that I can * use the same very simple function to do blockwise, row-wise and * column-wise elimination. */ #define YTRANS(y) (((y)%c)*r+(y)/c) #define YUNTRANS(y) (((y)%r)*c+(y)/r) struct solver_usage { int c, r, cr; /* * 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[(x*cr+y)*cr+n-1]. * y-coordinates in here are transformed. */ 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[(y*c+x)*cr+n-1] TRUE if digit n has been placed in block (x,y) */ unsigned char *blk; }; #define cubepos(x,y,n) (((x)*usage->cr+(y))*usage->cr+(n)-1) #define cube(x,y,n) (usage->cube[cubepos(x,y,n)]) /* * 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 c = usage->c, r = usage->r, cr = usage->cr; int i, j, bx, by; 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. */ bx = (x/r)*r; by = y % r; for (i = 0; i < r; i++) for (j = 0; j < c; j++) if (bx+i != x || by+j*r != y) cube(bx+i,by+j*r,n) = FALSE; /* * Enter the number in the result grid. */ usage->grid[YUNTRANS(y)*cr+x] = 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[((y%r)*c+(x/r))*cr+n-1] = TRUE; } static int solver_elim(struct solver_usage *usage, int start, int step #ifdef STANDALONE_SOLVER , char *fmt, ... #endif ) { int c = usage->c, r = usage->r, cr = c*r; 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[start+i*step]) { fpos = start+i*step; m++; } if (m == 1) { int x, y, n; assert(fpos >= 0); n = 1 + fpos % cr; y = fpos / cr; x = y / cr; y %= cr; if (!usage->grid[YUNTRANS(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+YUNTRANS(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 start1, int step1, int start2, int step2 #ifdef STANDALONE_SOLVER , char *fmt, ... #endif ) { int c = usage->c, r = usage->r, cr = c*r; int ret, i; /* * Loop over the first domain and see if there's any set bit * not also in the second. */ for (i = 0; i < cr; i++) { int p = start1+i*step1; if (usage->cube[p] && !(p >= start2 && p < start2+cr*step2 && (p - start2) % step2 == 0)) 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 = 0; i < cr; i++) { int p = start2+i*step2; if (usage->cube[p] && !(p >= start1 && p < start1+cr*step1 && (p - start1) % step1 == 0)) { #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; py = p / cr; px = py / cr; py %= cr; printf("%*s ruling out %d at (%d,%d)\n", solver_recurse_depth*4, "", pn, 1+px, 1+YUNTRANS(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; #ifdef STANDALONE_SOLVER int *bfsprev; #endif }; static int solver_set(struct solver_usage *usage, struct solver_scratch *scratch, int start, int step1, int step2 #ifdef STANDALONE_SOLVER , char *fmt, ... #endif ) { int c = usage->c, r = usage->r, cr = c*r; 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[start+i*step1+j*step2]) 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[start+rowidx[i]*step1+colidx[j]*step2]; /* * 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 = (start+rowidx[i]*step1+ colidx[j]*step2); #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; py = fpos / cr; px = py / cr; py %= cr; printf("%*s ruling out %d at (%d,%d)\n", solver_recurse_depth*4, "", pn, 1+px, 1+YUNTRANS(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; } /* * Try to find a number in the possible set of (x1,y1) which can be * ruled out because it would leave no possibilities for (x2,y2). */ static int solver_mne(struct solver_usage *usage, struct solver_scratch *scratch, int x1, int y1, int x2, int y2) { int c = usage->c, r = usage->r, cr = c*r; int *nb[2]; unsigned char *set = scratch->set; unsigned char *numbers = scratch->rowidx; unsigned char *numbersleft = scratch->colidx; int nnb, count; int i, j, n, nbi; nb[0] = scratch->neighbours; nb[1] = scratch->neighbours + cr; /* * First, work out the mutual neighbour squares of the two. We * can assert that they're not actually in the same block, * which leaves two possibilities: they're in different block * rows _and_ different block columns (thus their mutual * neighbours are precisely the other two corners of the * rectangle), or they're in the same row (WLOG) and different * columns, in which case their mutual neighbours are the * column of each block aligned with the other square. * * We divide the mutual neighbours into two separate subsets * nb[0] and nb[1]; squares in the same subset are not only * adjacent to both our key squares, but are also always * adjacent to one another. */ if (x1 / r != x2 / r && y1 % r != y2 % r) { /* Corners of the rectangle. */ nnb = 1; nb[0][0] = cubepos(x2, y1, 1); nb[1][0] = cubepos(x1, y2, 1); } else if (x1 / r != x2 / r) { /* Same row of blocks; different blocks within that row. */ int x1b = x1 - (x1 % r); int x2b = x2 - (x2 % r); nnb = r; for (i = 0; i < r; i++) { nb[0][i] = cubepos(x2b+i, y1, 1); nb[1][i] = cubepos(x1b+i, y2, 1); } } else { /* Same column of blocks; different blocks within that column. */ int y1b = y1 % r; int y2b = y2 % r; assert(y1 % r != y2 % r); nnb = c; for (i = 0; i < c; i++) { nb[0][i] = cubepos(x2, y1b+i*r, 1); nb[1][i] = cubepos(x1, y2b+i*r, 1); } } /* * Right. Now loop over each possible number. */ for (n = 1; n <= cr; n++) { if (!cube(x1, y1, n)) continue; for (j = 0; j < cr; j++) numbersleft[j] = cube(x2, y2, j+1); /* * Go over every possible subset of each neighbour list, * and see if its union of possible numbers minus n has the * same size as the subset. If so, add the numbers in that * subset to the set of things which would be ruled out * from (x2,y2) if n were placed at (x1,y1). */ memset(set, 0, nnb); count = 0; while (1) { /* * Binary increment: change the rightmost 0 to a 1, and * change all 1s to the right of it to 0s. */ i = nnb; while (i > 0 && set[i-1]) set[--i] = 0, count--; if (i > 0) set[--i] = 1, count++; else break; /* done */ /* * Examine this subset of each neighbour set. */ for (nbi = 0; nbi < 2; nbi++) { int *nbs = nb[nbi]; memset(numbers, 0, cr); for (i = 0; i < nnb; i++) if (set[i]) for (j = 0; j < cr; j++) if (j != n-1 && usage->cube[nbs[i] + j]) numbers[j] = 1; for (i = j = 0; j < cr; j++) i += numbers[j]; if (i == count) { /* * Got one. This subset of nbs, in the absence * of n, would definitely contain all the * numbers listed in `numbers'. Rule them out * of `numbersleft'. */ for (j = 0; j < cr; j++) if (numbers[j]) numbersleft[j] = 0; } } } /* * If we've got nothing left in `numbersleft', we have a * successful mutual neighbour elimination. */ for (j = 0; j < cr; j++) if (numbersleft[j]) break; if (j == cr) { #ifdef STANDALONE_SOLVER if (solver_show_working) { printf("%*smutual neighbour elimination, (%d,%d) vs (%d,%d):\n", solver_recurse_depth*4, "", 1+x1, 1+YUNTRANS(y1), 1+x2, 1+YUNTRANS(y2)); printf("%*s ruling out %d at (%d,%d)\n", solver_recurse_depth*4, "", n, 1+x1, 1+YUNTRANS(y1)); } #endif cube(x1, y1, n) = FALSE; return +1; } } return 0; /* nothing found */ } /* * 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 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 c = usage->c, r = usage->r, cr = c*r; 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, xblk, 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; xblk = xx - (xx % r); for (yt = yy % r; yt < cr; yt += r) for (xt = xblk; xt < xblk+r; xt++) neighbours[nneighbours++] = yt*cr+xt; /* * 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 || (xt / r == x / r && yt % r == y % r))) { #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+YUNTRANS(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+YUNTRANS(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(3*cr, int); scratch->bfsqueue = snewn(cr*cr, int); #ifdef STANDALONE_SOLVER scratch->bfsprev = snewn(cr*cr, int); #endif 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); } static int solver(int c, int r, digit *grid, int maxdiff) { struct solver_usage *usage; struct solver_scratch *scratch; int cr = c*r; int x, y, x2, y2, n, ret; int diff = DIFF_BLOCK; /* * Set up a usage structure as a clean slate (everything * possible). */ usage = snew(struct solver_usage); usage->c = c; usage->r = r; usage->cr = cr; 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); 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, YTRANS(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 (x = 0; x < cr; x += r) for (y = 0; y < r; y++) for (n = 1; n <= cr; n++) if (!usage->blk[(y*c+(x/r))*cr+n-1]) { ret = solver_elim(usage, cubepos(x,y,n), r*cr #ifdef STANDALONE_SOLVER , "positional elimination," " %d in block (%d,%d)", n, 1+x/r, 1+y #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]) { ret = solver_elim(usage, cubepos(0,y,n), cr*cr #ifdef STANDALONE_SOLVER , "positional elimination," " %d in row %d", n, 1+YUNTRANS(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]) { ret = solver_elim(usage, cubepos(x,0,n), cr #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; } } /* * Numeric elimination. */ for (x = 0; x < cr; x++) for (y = 0; y < cr; y++) if (!usage->grid[YUNTRANS(y)*cr+x]) { ret = solver_elim(usage, cubepos(x,y,1), 1 #ifdef STANDALONE_SOLVER , "numeric elimination at (%d,%d)", 1+x, 1+YUNTRANS(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 (x = 0; x < cr; x += r) for (n = 1; n <= cr; n++) /* * solver_intersect() never returns -1. */ if (!usage->row[y*cr+n-1] && !usage->blk[((y%r)*c+(x/r))*cr+n-1] && (solver_intersect(usage, cubepos(0,y,n), cr*cr, cubepos(x,y%r,n), r*cr #ifdef STANDALONE_SOLVER , "intersectional analysis," " %d in row %d vs block (%d,%d)", n, 1+YUNTRANS(y), 1+x/r, 1+y%r #endif ) || solver_intersect(usage, cubepos(x,y%r,n), r*cr, cubepos(0,y,n), cr*cr #ifdef STANDALONE_SOLVER , "intersectional analysis," " %d in block (%d,%d) vs row %d", n, 1+x/r, 1+y%r, 1+YUNTRANS(y) #endif ))) { diff = max(diff, DIFF_INTERSECT); goto cont; } /* * Intersectional analysis, columns vs blocks. */ for (x = 0; x < cr; x++) for (y = 0; y < r; y++) for (n = 1; n <= cr; n++) if (!usage->col[x*cr+n-1] && !usage->blk[(y*c+(x/r))*cr+n-1] && (solver_intersect(usage, cubepos(x,0,n), cr, cubepos((x/r)*r,y,n), r*cr #ifdef STANDALONE_SOLVER , "intersectional analysis," " %d in column %d vs block (%d,%d)", n, 1+x, 1+x/r, 1+y #endif ) || solver_intersect(usage, cubepos((x/r)*r,y,n), r*cr, cubepos(x,0,n), cr #ifdef STANDALONE_SOLVER , "intersectional analysis," " %d in block (%d,%d) vs column %d", n, 1+x/r, 1+y, 1+x #endif ))) { diff = max(diff, DIFF_INTERSECT); goto cont; } if (maxdiff <= DIFF_INTERSECT) break; /* * Blockwise set elimination. */ for (x = 0; x < cr; x += r) for (y = 0; y < r; y++) { ret = solver_set(usage, scratch, cubepos(x,y,1), r*cr, 1 #ifdef STANDALONE_SOLVER , "set elimination, block (%d,%d)", 1+x/r, 1+y #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++) { ret = solver_set(usage, scratch, cubepos(0,y,1), cr*cr, 1 #ifdef STANDALONE_SOLVER , "set elimination, row %d", 1+YUNTRANS(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++) { ret = solver_set(usage, scratch, cubepos(x,0,1), cr, 1 #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; } } /* * Row-vs-column set elimination on a single number. */ for (n = 1; n <= cr; n++) { ret = solver_set(usage, scratch, cubepos(0,0,n), cr*cr, cr #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; } } /* * Mutual neighbour elimination. */ for (y = 0; y+1 < cr; y++) { for (x = 0; x+1 < cr; x++) { for (y2 = y+1; y2 < cr; y2++) { for (x2 = x+1; x2 < cr; x2++) { /* * Can't do mutual neighbour elimination * between elements of the same actual * block. */ if (x/r == x2/r && y%r == y2%r) continue; /* * Otherwise, try (x,y) vs (x2,y2) in both * directions, and likewise (x2,y) vs * (x,y2). */ if (!usage->grid[YUNTRANS(y)*cr+x] && !usage->grid[YUNTRANS(y2)*cr+x2] && (solver_mne(usage, scratch, x, y, x2, y2) || solver_mne(usage, scratch, x2, y2, x, y))) { diff = max(diff, DIFF_EXTREME); goto cont; } if (!usage->grid[YUNTRANS(y)*cr+x2] && !usage->grid[YUNTRANS(y2)*cr+x] && (solver_mne(usage, scratch, x2, y, x, y2) || solver_mne(usage, scratch, x, y2, x2, y))) { 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,YTRANS(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,YTRANS(y),n)) list[j++] = n; #ifdef STANDALONE_SOLVER if (solver_show_working) { char *sep = ""; printf("%*srecursing on (%d,%d) [", solver_recurse_depth*4, "", x, y); 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, y); solver_recurse_depth++; #endif ret = solver(c, r, 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, y); } #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 c, r, cr; /* cr == c*r */ /* 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; /* 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; }; /* * The real recursive step in the generating function. */ static int gridgen_real(struct gridgen_usage *usage, digit *grid) { int c = usage->c, r = usage->r, 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) { memcpy(grid, usage->grid, cr * cr); return TRUE; } /* * 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[((y/c)*c+(x/r))*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[((sy/c)*c+(sx/r))*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. */ usage->row[sy*cr+n-1] = usage->col[sx*cr+n-1] = usage->blk[((sy/c)*c+(sx/r))*cr+n-1] = TRUE; usage->grid[sy*cr+sx] = n; usage->nspaces--; /* Call the solver recursively. Stop when we find a solution. */ if (gridgen_real(usage, grid)) ret = TRUE; /* Revert the usage structure. */ usage->row[sy*cr+n-1] = usage->col[sx*cr+n-1] = usage->blk[((sy/c)*c+(sx/r))*cr+n-1] = FALSE; usage->grid[sy*cr+sx] = 0; usage->nspaces++; if (ret) break; } sfree(digits); return ret; } /* * Entry point to generator. You give it dimensions and a starting * grid, which is simply an array of cr*cr digits. */ static void gridgen(int c, int r, digit *grid, random_state *rs) { struct gridgen_usage *usage; int x, y, cr = c*r; /* * Clear the grid to start with. */ memset(grid, 0, cr*cr); /* * Create a gridgen_usage structure. */ usage = snew(struct gridgen_usage); usage->c = c; usage->r = r; usage->cr = cr; usage->grid = snewn(cr * cr, digit); memcpy(usage->grid, grid, 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); usage->spaces = snewn(cr * cr, struct gridgen_coord); usage->nspaces = 0; usage->rs = rs; /* * Initialise the list of grid spaces. */ for (y = 0; 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. */ gridgen_real(usage, grid); /* * Clean up the usage structure now we have our answer. */ sfree(usage->spaces); sfree(usage->blk); sfree(usage->col); sfree(usage->row); sfree(usage->grid); sfree(usage); } /* ---------------------------------------------------------------------- * End of grid generator code. */ /* * Check whether a grid contains a valid complete puzzle. */ static int check_valid(int c, int r, digit *grid) { int cr = c*r; unsigned char *used; int x, y, 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 (x = 0; x < cr; x += r) { for (y = 0; y < cr; y += c) { int xx, yy; memset(used, FALSE, cr); for (xx = x; xx < x+r; xx++) for (yy = 0; yy < y+c; yy++) if (grid[yy*cr+xx] > 0 && grid[yy*cr+xx] <= cr) used[grid[yy*cr+xx]-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; 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); /* * 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. */ do { /* * Generate a random solved state. */ gridgen(c, r, grid, rs); assert(check_valid(c, r, 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(c, r, 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); } while (solver(c, r, grid2, maxdiff) < maxdiff); 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(5 * 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; } } assert(p - desc < 5 * area); *p++ = '\0'; desc = sresize(desc, p - desc, char); } sfree(grid); return desc; } static char *validate_desc(game_params *params, char *desc) { int area = params->r * params->r * params->c * params->c; int squares = 0; while (*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"; 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->c = params->c; state->r = params->r; 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->completed = state->cheated = FALSE; i = 0; while (*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); return state; } static game_state *dup_game(game_state *state) { game_state *ret = snew(game_state); int c = state->c, r = state->r, cr = c*r, area = cr * cr; ret->c = state->c; ret->r = state->r; 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) { 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 c = state->c, r = state->r, cr = c*r; 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(c, r, 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 c, int r, digit *grid) { int cr = c*r; int x, y; int maxlen; char *ret, *p; /* * There are cr lines of digits, plus r-1 lines of block * separators. Each line contains cr digits, cr-1 separating * spaces, and c-1 two-character block separators. Thus, the * total length of a line is 2*cr+2*c-3 (not counting the * newline), and there are cr+r-1 of them. */ maxlen = (cr+r-1) * (2*cr+2*c-2); ret = snewn(maxlen+1, char); p = ret; for (y = 0; y < cr; y++) { for (x = 0; x < cr; x++) { int ch = grid[y * cr + x]; if (ch == 0) ch = '.'; else if (ch <= 9) ch = '0' + ch; else ch = 'a' + ch-10; *p++ = ch; if (x+1 < cr) { *p++ = ' '; if ((x+1) % r == 0) { *p++ = '|'; *p++ = ' '; } } } *p++ = '\n'; if (y+1 < cr && (y+1) % c == 0) { for (x = 0; x < cr; x++) { *p++ = '-'; if (x+1 < cr) { *p++ = '-'; if ((x+1) % r == 0) { *p++ = '+'; *p++ = '-'; } } } *p++ = '\n'; } } assert(p - ret == maxlen); *p = '\0'; return ret; } static char *game_text_format(game_state *state) { return grid_text_format(state->c, state->r, 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 c = newstate->c, r = newstate->r, cr = c*r; /* * 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 c, r, cr; 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 c = state->c, r = state->r, cr = c*r; 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 c = from->c, r = from->r, cr = c*r; 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(c, r, 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_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] = 0.85F * ret[COL_BACKGROUND * 3 + 0]; ret[COL_HIGHLIGHT * 3 + 1] = 0.85F * ret[COL_BACKGROUND * 3 + 1]; ret[COL_HIGHLIGHT * 3 + 2] = 0.85F * 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 c = state->c, r = state->r, cr = c*r; ds->started = FALSE; ds->c = c; ds->r = r; ds->cr = cr; ds->grid = snewn(cr*cr, digit); memset(ds->grid, 0, 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 c = state->c, r = state->r, cr = c*r; 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 + 2; ty = BORDER + y * TILE_SIZE + 2; cx = tx; cy = ty; cw = TILE_SIZE-3; ch = TILE_SIZE-3; if (x % r) cx--, cw++; if ((x+1) % r) cw++; if (y % c) cy--, ch++; if ((y+1) % c) ch++; clip(dr, cx, cy, cw, ch); /* background needs erasing */ draw_rect(dr, cx, cy, cw, ch, (hl & 15) == 1 ? COL_HIGHLIGHT : COL_BACKGROUND); /* 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 c = state->c, r = state->r, cr = c*r; 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. */ for (x = 0; x <= cr; x++) { int thick = (x % r ? 0 : 1); draw_rect(dr, BORDER + x*TILE_SIZE - thick, BORDER-1, 1+2*thick, cr*TILE_SIZE+3, COL_GRID); } for (y = 0; y <= cr; y++) { int thick = (y % c ? 0 : 1); draw_rect(dr, BORDER-1, BORDER + y*TILE_SIZE - thick, cr*TILE_SIZE+3, 1+2*thick, 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 = (x/r)+(y/c)*c; 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; } } /* * 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[((x/r)+(y/c)*c)*cr+d-1] & 32))) 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 c = state->c, r = state->r, cr = c*r; 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); /* * Grid. */ for (x = 1; x < cr; x++) { print_line_width(dr, (x % r ? 1 : 3) * 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, (y % c ? 1 : 3) * TILE_SIZE / 40); draw_line(dr, BORDER, BORDER+y*TILE_SIZE, BORDER+cr*TILE_SIZE, BORDER+y*TILE_SIZE, ink); } /* * 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", 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, 0, /* flags */ }; #ifdef STANDALONE_SOLVER int main(int argc, char **argv) { game_params *p; game_state *s; char *id = NULL, *desc, *err; int grade = FALSE; int ret; while (--argc > 0) { char *p = *++argv; if (!strcmp(p, "-v")) { solver_show_working = TRUE; } else if (!strcmp(p, "-g")) { grade = TRUE; } else if (*p == '-') { fprintf(stderr, "%s: unrecognised option `%s'\n", argv[0], p); return 1; } else { id = p; } } if (!id) { fprintf(stderr, "usage: %s [-g | -v] <game_id>\n", argv[0]); return 1; } desc = strchr(id, ':'); if (!desc) { fprintf(stderr, "%s: game id expects a colon in it\n", argv[0]); return 1; } *desc++ = '\0'; p = default_params(); decode_params(p, id); err = validate_desc(p, desc); if (err) { fprintf(stderr, "%s: %s\n", argv[0], err); return 1; } s = new_game(NULL, p, desc); ret = solver(p->c, p->r, 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(p->c, p->r, s->grid)); } return 0; } #endif

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