ref: bbe866a3819c6a754a5b1d8c5bc5d0701796acfb
dir: /keen.c/
/* * keen.c: an implementation of the Times's 'KenKen' puzzle, and * also of Nikoli's very similar 'Inshi No Heya' puzzle. */ #include <stdio.h> #include <stdlib.h> #include <string.h> #include <assert.h> #include <ctype.h> #include <math.h> #include "puzzles.h" #include "latin.h" /* * Difficulty levels. I do some macro ickery here to ensure that my * enum and the various forms of my name list always match up. */ #define DIFFLIST(A) \ A(EASY,Easy,solver_easy,e) \ A(NORMAL,Normal,solver_normal,n) \ A(HARD,Hard,solver_hard,h) \ A(EXTREME,Extreme,NULL,x) \ A(UNREASONABLE,Unreasonable,NULL,u) #define ENUM(upper,title,func,lower) DIFF_ ## upper, #define TITLE(upper,title,func,lower) #title, #define ENCODE(upper,title,func,lower) #lower #define CONFIG(upper,title,func,lower) ":" #title enum { DIFFLIST(ENUM) DIFFCOUNT }; static char const *const keen_diffnames[] = { DIFFLIST(TITLE) }; static char const keen_diffchars[] = DIFFLIST(ENCODE); #define DIFFCONFIG DIFFLIST(CONFIG) /* * Clue notation. Important here that ADD and MUL come before SUB * and DIV, and that DIV comes last. */ #define C_ADD 0x00000000L #define C_MUL 0x20000000L #define C_SUB 0x40000000L #define C_DIV 0x60000000L #define CMASK 0x60000000L #define CUNIT 0x20000000L /* * Maximum size of any clue block. Very large ones are annoying in UI * terms (if they're multiplicative you end up with too many digits to * fit in the square) and also in solver terms (too many possibilities * to iterate over). */ #define MAXBLK 6 enum { COL_BACKGROUND, COL_GRID, COL_USER, COL_HIGHLIGHT, COL_ERROR, COL_PENCIL, NCOLOURS }; struct game_params { int w, diff; bool multiplication_only; }; struct clues { int refcount; int w; int *dsf; long *clues; }; struct game_state { game_params par; struct clues *clues; digit *grid; int *pencil; /* bitmaps using bits 1<<1..1<<n */ bool completed, cheated; }; static game_params *default_params(void) { game_params *ret = snew(game_params); ret->w = 6; ret->diff = DIFF_NORMAL; ret->multiplication_only = false; return ret; } static const struct game_params keen_presets[] = { { 4, DIFF_EASY, false }, { 5, DIFF_EASY, false }, { 5, DIFF_EASY, true }, { 6, DIFF_EASY, false }, { 6, DIFF_NORMAL, false }, { 6, DIFF_NORMAL, true }, { 6, DIFF_HARD, false }, { 6, DIFF_EXTREME, false }, { 6, DIFF_UNREASONABLE, false }, { 9, DIFF_NORMAL, false }, }; static bool game_fetch_preset(int i, char **name, game_params **params) { game_params *ret; char buf[80]; if (i < 0 || i >= lenof(keen_presets)) return false; ret = snew(game_params); *ret = keen_presets[i]; /* structure copy */ sprintf(buf, "%dx%d %s%s", ret->w, ret->w, keen_diffnames[ret->diff], ret->multiplication_only ? ", multiplication only" : ""); *name = dupstr(buf); *params = ret; return true; } static void free_params(game_params *params) { sfree(params); } static game_params *dup_params(const game_params *params) { game_params *ret = snew(game_params); *ret = *params; /* structure copy */ return ret; } static void decode_params(game_params *params, char const *string) { char const *p = string; params->w = atoi(p); while (*p && isdigit((unsigned char)*p)) p++; if (*p == 'd') { int i; p++; params->diff = DIFFCOUNT+1; /* ...which is invalid */ if (*p) { for (i = 0; i < DIFFCOUNT; i++) { if (*p == keen_diffchars[i]) params->diff = i; } p++; } } if (*p == 'm') { p++; params->multiplication_only = true; } } static char *encode_params(const game_params *params, bool full) { char ret[80]; sprintf(ret, "%d", params->w); if (full) sprintf(ret + strlen(ret), "d%c%s", keen_diffchars[params->diff], params->multiplication_only ? "m" : ""); return dupstr(ret); } static config_item *game_configure(const game_params *params) { config_item *ret; char buf[80]; ret = snewn(4, config_item); ret[0].name = "Grid size"; ret[0].type = C_STRING; sprintf(buf, "%d", params->w); ret[0].u.string.sval = dupstr(buf); ret[1].name = "Difficulty"; ret[1].type = C_CHOICES; ret[1].u.choices.choicenames = DIFFCONFIG; ret[1].u.choices.selected = params->diff; ret[2].name = "Multiplication only"; ret[2].type = C_BOOLEAN; ret[2].u.boolean.bval = params->multiplication_only; ret[3].name = NULL; ret[3].type = C_END; return ret; } static game_params *custom_params(const config_item *cfg) { game_params *ret = snew(game_params); ret->w = atoi(cfg[0].u.string.sval); ret->diff = cfg[1].u.choices.selected; ret->multiplication_only = cfg[2].u.boolean.bval; return ret; } static const char *validate_params(const game_params *params, bool full) { if (params->w < 3 || params->w > 9) return "Grid size must be between 3 and 9"; if (params->diff >= DIFFCOUNT) return "Unknown difficulty rating"; return NULL; } /* ---------------------------------------------------------------------- * Solver. */ struct solver_ctx { int w, diff; int nboxes; int *boxes, *boxlist, *whichbox; long *clues; digit *soln; digit *dscratch; int *iscratch; }; static void solver_clue_candidate(struct solver_ctx *ctx, int diff, int box) { int w = ctx->w; int n = ctx->boxes[box+1] - ctx->boxes[box]; int j; /* * This function is called from the main clue-based solver * routine when we discover a candidate layout for a given clue * box consistent with everything we currently know about the * digit constraints in that box. We expect to find the digits * of the candidate layout in ctx->dscratch, and we update * ctx->iscratch as appropriate. * * The contents of ctx->iscratch are completely different * depending on whether diff == DIFF_HARD or not. This function * uses iscratch completely differently between the two cases, and * the code in solver_common() which consumes the result must * likewise have an if statement with completely different * branches for the two cases. * * In DIFF_EASY and DIFF_NORMAL modes, the valid entries in * ctx->iscratch are 0,...,n-1, and each of those entries * ctx->iscratch[i] gives a bitmap of the possible digits in the * ith square of the clue box currently under consideration. So * each entry of iscratch starts off as an empty bitmap, and we * set bits in it as possible layouts for the clue box are * considered (and the difference between DIFF_EASY and * DIFF_NORMAL is just that in DIFF_EASY mode we deliberately set * more bits than absolutely necessary, hence restricting our own * knowledge). * * But in DIFF_HARD mode, the valid entries are 0,...,2*w-1 (at * least outside *this* function - inside this function, we also * use 2*w,...,4*w-1 as scratch space in the loop below); the * first w of those give the possible digits in the intersection * of the current clue box with each column of the puzzle, and the * next w do the same for each row. In this mode, each iscratch * entry starts off as a _full_ bitmap, and in this function we * _clear_ bits for digits that are absent from a given row or * column in each candidate layout, so that the only bits which * remain set are those for digits which have to appear in a given * row/column no matter how the clue box is laid out. */ if (diff == DIFF_EASY) { unsigned mask = 0; /* * Easy-mode clue deductions: we do not record information * about which squares take which values, so we amalgamate * all the values in dscratch and OR them all into * everywhere. */ for (j = 0; j < n; j++) mask |= 1 << ctx->dscratch[j]; for (j = 0; j < n; j++) ctx->iscratch[j] |= mask; } else if (diff == DIFF_NORMAL) { /* * Normal-mode deductions: we process the information in * dscratch in the obvious way. */ for (j = 0; j < n; j++) ctx->iscratch[j] |= 1 << ctx->dscratch[j]; } else if (diff == DIFF_HARD) { /* * Hard-mode deductions: instead of ruling things out * _inside_ the clue box, we look for numbers which occur in * a given row or column in all candidate layouts, and rule * them out of all squares in that row or column that * _aren't_ part of this clue box. */ int *sq = ctx->boxlist + ctx->boxes[box]; for (j = 0; j < 2*w; j++) ctx->iscratch[2*w+j] = 0; for (j = 0; j < n; j++) { int x = sq[j] / w, y = sq[j] % w; ctx->iscratch[2*w+x] |= 1 << ctx->dscratch[j]; ctx->iscratch[3*w+y] |= 1 << ctx->dscratch[j]; } for (j = 0; j < 2*w; j++) ctx->iscratch[j] &= ctx->iscratch[2*w+j]; } } static int solver_common(struct latin_solver *solver, void *vctx, int diff) { struct solver_ctx *ctx = (struct solver_ctx *)vctx; int w = ctx->w; int box, i, j, k; int ret = 0, total; /* * Iterate over each clue box and deduce what we can. */ for (box = 0; box < ctx->nboxes; box++) { int *sq = ctx->boxlist + ctx->boxes[box]; int n = ctx->boxes[box+1] - ctx->boxes[box]; long value = ctx->clues[box] & ~CMASK; long op = ctx->clues[box] & CMASK; /* * Initialise ctx->iscratch for this clue box. At different * difficulty levels we must initialise a different amount of * it to different things; see the comments in * solver_clue_candidate explaining what each version does. */ if (diff == DIFF_HARD) { for (i = 0; i < 2*w; i++) ctx->iscratch[i] = (1 << (w+1)) - (1 << 1); } else { for (i = 0; i < n; i++) ctx->iscratch[i] = 0; } switch (op) { case C_SUB: case C_DIV: /* * These two clue types must always apply to a box of * area 2. Also, the two digits in these boxes can never * be the same (because any domino must have its two * squares in either the same row or the same column). * So we simply iterate over all possibilities for the * two squares (both ways round), rule out any which are * inconsistent with the digit constraints we already * have, and update the digit constraints with any new * information thus garnered. */ assert(n == 2); for (i = 1; i <= w; i++) { j = (op == C_SUB ? i + value : i * value); if (j > w) break; /* (i,j) is a valid digit pair. Try it both ways round. */ if (solver->cube[sq[0]*w+i-1] && solver->cube[sq[1]*w+j-1]) { ctx->dscratch[0] = i; ctx->dscratch[1] = j; solver_clue_candidate(ctx, diff, box); } if (solver->cube[sq[0]*w+j-1] && solver->cube[sq[1]*w+i-1]) { ctx->dscratch[0] = j; ctx->dscratch[1] = i; solver_clue_candidate(ctx, diff, box); } } break; case C_ADD: case C_MUL: /* * For these clue types, I have no alternative but to go * through all possible number combinations. * * Instead of a tedious physical recursion, I iterate in * the scratch array through all possibilities. At any * given moment, i indexes the element of the box that * will next be incremented. */ i = 0; ctx->dscratch[i] = 0; total = value; /* start with the identity */ while (1) { if (i < n) { /* * Find the next valid value for cell i. */ for (j = ctx->dscratch[i] + 1; j <= w; j++) { if (op == C_ADD ? (total < j) : (total % j != 0)) continue; /* this one won't fit */ if (!solver->cube[sq[i]*w+j-1]) continue; /* this one is ruled out already */ for (k = 0; k < i; k++) if (ctx->dscratch[k] == j && (sq[k] % w == sq[i] % w || sq[k] / w == sq[i] / w)) break; /* clashes with another row/col */ if (k < i) continue; /* Found one. */ break; } if (j > w) { /* No valid values left; drop back. */ i--; if (i < 0) break; /* overall iteration is finished */ if (op == C_ADD) total += ctx->dscratch[i]; else total *= ctx->dscratch[i]; } else { /* Got a valid value; store it and move on. */ ctx->dscratch[i++] = j; if (op == C_ADD) total -= j; else total /= j; ctx->dscratch[i] = 0; } } else { if (total == (op == C_ADD ? 0 : 1)) solver_clue_candidate(ctx, diff, box); i--; if (op == C_ADD) total += ctx->dscratch[i]; else total *= ctx->dscratch[i]; } } break; } /* * Do deductions based on the information we've now * accumulated in ctx->iscratch. See the comments above in * solver_clue_candidate explaining what data is left in here, * and how it differs between DIFF_HARD and lower difficulty * levels (hence the big if statement here). */ if (diff < DIFF_HARD) { #ifdef STANDALONE_SOLVER char prefix[256]; if (solver_show_working) sprintf(prefix, "%*susing clue at (%d,%d):\n", solver_recurse_depth*4, "", sq[0]/w+1, sq[0]%w+1); else prefix[0] = '\0'; /* placate optimiser */ #endif for (i = 0; i < n; i++) for (j = 1; j <= w; j++) { if (solver->cube[sq[i]*w+j-1] && !(ctx->iscratch[i] & (1 << j))) { #ifdef STANDALONE_SOLVER if (solver_show_working) { printf("%s%*s ruling out %d at (%d,%d)\n", prefix, solver_recurse_depth*4, "", j, sq[i]/w+1, sq[i]%w+1); prefix[0] = '\0'; } #endif solver->cube[sq[i]*w+j-1] = 0; ret = 1; } } } else { #ifdef STANDALONE_SOLVER char prefix[256]; if (solver_show_working) sprintf(prefix, "%*susing clue at (%d,%d):\n", solver_recurse_depth*4, "", sq[0]/w+1, sq[0]%w+1); else prefix[0] = '\0'; /* placate optimiser */ #endif for (i = 0; i < 2*w; i++) { int start = (i < w ? i*w : i-w); int step = (i < w ? 1 : w); for (j = 1; j <= w; j++) if (ctx->iscratch[i] & (1 << j)) { #ifdef STANDALONE_SOLVER char prefix2[256]; if (solver_show_working) sprintf(prefix2, "%*s this clue requires %d in" " %s %d:\n", solver_recurse_depth*4, "", j, i < w ? "column" : "row", i%w+1); else prefix2[0] = '\0'; /* placate optimiser */ #endif for (k = 0; k < w; k++) { int pos = start + k*step; if (ctx->whichbox[pos] != box && solver->cube[pos*w+j-1]) { #ifdef STANDALONE_SOLVER if (solver_show_working) { printf("%s%s%*s ruling out %d at (%d,%d)\n", prefix, prefix2, solver_recurse_depth*4, "", j, pos/w+1, pos%w+1); prefix[0] = prefix2[0] = '\0'; } #endif solver->cube[pos*w+j-1] = 0; ret = 1; } } } } /* * Once we find one block we can do something with in * this way, revert to trying easier deductions, so as * not to generate solver diagnostics that make the * problem look harder than it is. (We have to do this * for the Hard deductions but not the Easy/Normal ones, * because only the Hard deductions are cross-box.) */ if (ret) return ret; } } return ret; } static int solver_easy(struct latin_solver *solver, void *vctx) { /* * Omit the EASY deductions when solving at NORMAL level, since * the NORMAL deductions are a superset of them anyway and it * saves on time and confusing solver diagnostics. * * Note that this breaks the natural semantics of the return * value of latin_solver. Without this hack, you could determine * a puzzle's difficulty in one go by trying to solve it at * maximum difficulty and seeing what difficulty value was * returned; but with this hack, solving an Easy puzzle on * Normal difficulty will typically return Normal. Hence the * uses of the solver to determine difficulty are all arranged * so as to double-check by re-solving at the next difficulty * level down and making sure it failed. */ struct solver_ctx *ctx = (struct solver_ctx *)vctx; if (ctx->diff > DIFF_EASY) return 0; return solver_common(solver, vctx, DIFF_EASY); } static int solver_normal(struct latin_solver *solver, void *vctx) { return solver_common(solver, vctx, DIFF_NORMAL); } static int solver_hard(struct latin_solver *solver, void *vctx) { return solver_common(solver, vctx, DIFF_HARD); } #define SOLVER(upper,title,func,lower) func, static usersolver_t const keen_solvers[] = { DIFFLIST(SOLVER) }; static int transpose(int index, int w) { return (index % w) * w + (index / w); } static bool keen_valid(struct latin_solver *solver, void *vctx) { struct solver_ctx *ctx = (struct solver_ctx *)vctx; int w = ctx->w; int box, i; /* * Iterate over each clue box and check it's satisfied. */ for (box = 0; box < ctx->nboxes; box++) { int *sq = ctx->boxlist + ctx->boxes[box]; int n = ctx->boxes[box+1] - ctx->boxes[box]; long value = ctx->clues[box] & ~CMASK; long op = ctx->clues[box] & CMASK; bool fail = false; switch (op) { case C_ADD: { long sum = 0; for (i = 0; i < n; i++) sum += solver->grid[transpose(sq[i], w)]; fail = (sum != value); break; } case C_MUL: { long remaining = value; for (i = 0; i < n; i++) { if (remaining % solver->grid[transpose(sq[i], w)]) { fail = true; break; } remaining /= solver->grid[transpose(sq[i], w)]; } if (remaining != 1) fail = true; break; } case C_SUB: assert(n == 2); if (value != labs(solver->grid[transpose(sq[0], w)] - solver->grid[transpose(sq[1], w)])) fail = true; break; case C_DIV: { int num, den; assert(n == 2); num = max(solver->grid[transpose(sq[0], w)], solver->grid[transpose(sq[1], w)]); den = min(solver->grid[transpose(sq[0], w)], solver->grid[transpose(sq[1], w)]); if (den * value != num) fail = true; break; } } if (fail) { #ifdef STANDALONE_SOLVER if (solver_show_working) { printf("%*sclue at (%d,%d) is violated\n", solver_recurse_depth*4, "", sq[0]/w+1, sq[0]%w+1); printf("%*s (%s clue with target %ld containing [", solver_recurse_depth*4, "", (op == C_ADD ? "addition" : op == C_SUB ? "subtraction": op == C_MUL ? "multiplication" : "division"), value); for (i = 0; i < n; i++) printf(" %d", (int)solver->grid[transpose(sq[i], w)]); printf(" ]\n"); } #endif return false; } } return true; } static int solver(int w, int *dsf, long *clues, digit *soln, int maxdiff) { int a = w*w; struct solver_ctx ctx; int ret; int i, j, n, m; ctx.w = w; ctx.soln = soln; ctx.diff = maxdiff; /* * Transform the dsf-formatted clue list into one over which we * can iterate more easily. * * Also transpose the x- and y-coordinates at this point, * because the 'cube' array in the general Latin square solver * puts x first (oops). */ for (ctx.nboxes = i = 0; i < a; i++) if (dsf_canonify(dsf, i) == i) ctx.nboxes++; ctx.boxlist = snewn(a, int); ctx.boxes = snewn(ctx.nboxes+1, int); ctx.clues = snewn(ctx.nboxes, long); ctx.whichbox = snewn(a, int); for (n = m = i = 0; i < a; i++) if (dsf_canonify(dsf, i) == i) { ctx.clues[n] = clues[i]; ctx.boxes[n] = m; for (j = 0; j < a; j++) if (dsf_canonify(dsf, j) == i) { ctx.boxlist[m++] = (j % w) * w + (j / w); /* transpose */ ctx.whichbox[ctx.boxlist[m-1]] = n; } n++; } assert(n == ctx.nboxes); assert(m == a); ctx.boxes[n] = m; ctx.dscratch = snewn(a+1, digit); ctx.iscratch = snewn(max(a+1, 4*w), int); ret = latin_solver(soln, w, maxdiff, DIFF_EASY, DIFF_HARD, DIFF_EXTREME, DIFF_EXTREME, DIFF_UNREASONABLE, keen_solvers, keen_valid, &ctx, NULL, NULL); sfree(ctx.dscratch); sfree(ctx.iscratch); sfree(ctx.whichbox); sfree(ctx.boxlist); sfree(ctx.boxes); sfree(ctx.clues); return ret; } /* ---------------------------------------------------------------------- * Grid generation. */ static char *encode_block_structure(char *p, int w, int *dsf) { int i, currrun = 0; char *orig, *q, *r, c; orig = p; /* * Encode the block structure. We do this by encoding the * pattern of dividing lines: first we iterate over the w*(w-1) * internal vertical grid lines in ordinary reading order, then * over the w*(w-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*w*(w-1); i++) { int x, y, p0, p1; bool edge; if (i == 2*w*(w-1)) { edge = true; /* terminating virtual edge */ } else { if (i < w*(w-1)) { y = i/(w-1); x = i%(w-1); p0 = y*w+x; p1 = y*w+x+1; } else { x = i/(w-1) - w; y = i%(w-1); p0 = y*w+x; p1 = (y+1)*w+x; } edge = (dsf_canonify(dsf, p0) != dsf_canonify(dsf, p1)); } if (edge) { while (currrun > 25) *p++ = 'z', currrun -= 25; if (currrun) *p++ = 'a'-1 + currrun; else *p++ = '_'; currrun = 0; } else currrun++; } /* * Now go through and compress the string by replacing runs of * the same letter with a single copy of that letter followed by * a repeat count, where that makes it shorter. (This puzzle * seems to generate enough long strings of _ to make this a * worthwhile step.) */ for (q = r = orig; r < p ;) { *q++ = c = *r; for (i = 0; r+i < p && r[i] == c; i++); r += i; if (i == 2) { *q++ = c; } else if (i > 2) { q += sprintf(q, "%d", i); } } return q; } static const char *parse_block_structure(const char **p, int w, int *dsf) { int a = w*w; int pos = 0; int repc = 0, repn = 0; dsf_init(dsf, a); while (**p && (repn > 0 || **p != ',')) { int c; bool adv; if (repn > 0) { repn--; c = repc; } else if (**p == '_' || (**p >= 'a' && **p <= 'z')) { c = (**p == '_' ? 0 : **p - 'a' + 1); (*p)++; if (**p && isdigit((unsigned char)**p)) { repc = c; repn = atoi(*p)-1; while (**p && isdigit((unsigned char)**p)) (*p)++; } } else return "Invalid character in game description"; 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*w*(w-1)) return "Too much data in block structure specification"; if (pos < w*(w-1)) { int y = pos/(w-1); int x = pos%(w-1); p0 = y*w+x; p1 = y*w+x+1; } else { int x = pos/(w-1) - w; int y = pos%(w-1); p0 = y*w+x; p1 = (y+1)*w+x; } dsf_merge(dsf, p0, p1); pos++; } if (adv) { pos++; if (pos > 2*w*(w-1)+1) return "Too much data in block structure specification"; } } /* * 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*w*(w-1)+1) return "Not enough data in block structure specification"; return NULL; } static char *new_game_desc(const game_params *params, random_state *rs, char **aux, bool interactive) { int w = params->w, a = w*w; digit *grid, *soln; int *order, *revorder, *singletons, *dsf; long *clues, *cluevals; int i, j, k, n, x, y, ret; int diff = params->diff; char *desc, *p; /* * Difficulty exceptions: 3x3 puzzles at difficulty Hard or * higher are currently not generable - the generator will spin * forever looking for puzzles of the appropriate difficulty. We * dial each of these down to the next lower difficulty. * * Remember to re-test this whenever a change is made to the * solver logic! * * I tested it using the following shell command: for d in e n h x u; do for i in {3..9}; do echo ./keen --generate 1 ${i}d${d} perl -e 'alarm 30; exec @ARGV' ./keen --generate 5 ${i}d${d} >/dev/null \ || echo broken done done * Of course, it's better to do that after taking the exceptions * _out_, so as to detect exceptions that should be removed as * well as those which should be added. */ if (w == 3 && diff > DIFF_NORMAL) diff = DIFF_NORMAL; grid = NULL; order = snewn(a, int); revorder = snewn(a, int); singletons = snewn(a, int); dsf = snew_dsf(a); clues = snewn(a, long); cluevals = snewn(a, long); soln = snewn(a, digit); while (1) { /* * First construct a latin square to be the solution. */ sfree(grid); grid = latin_generate(w, rs); /* * Divide the grid into arbitrarily sized blocks, but so as * to arrange plenty of dominoes which can be SUB/DIV clues. * We do this by first placing dominoes at random for a * while, then tying the remaining singletons one by one * into neighbouring blocks. */ for (i = 0; i < a; i++) order[i] = i; shuffle(order, a, sizeof(*order), rs); for (i = 0; i < a; i++) revorder[order[i]] = i; for (i = 0; i < a; i++) singletons[i] = true; dsf_init(dsf, a); /* Place dominoes. */ for (i = 0; i < a; i++) { if (singletons[i]) { int best = -1; x = i % w; y = i / w; if (x > 0 && singletons[i-1] && (best == -1 || revorder[i-1] < revorder[best])) best = i-1; if (x+1 < w && singletons[i+1] && (best == -1 || revorder[i+1] < revorder[best])) best = i+1; if (y > 0 && singletons[i-w] && (best == -1 || revorder[i-w] < revorder[best])) best = i-w; if (y+1 < w && singletons[i+w] && (best == -1 || revorder[i+w] < revorder[best])) best = i+w; /* * When we find a potential domino, we place it with * probability 3/4, which seems to strike a decent * balance between plenty of dominoes and leaving * enough singletons to make interesting larger * shapes. */ if (best >= 0 && random_upto(rs, 4)) { singletons[i] = singletons[best] = false; dsf_merge(dsf, i, best); } } } /* Fold in singletons. */ for (i = 0; i < a; i++) { if (singletons[i]) { int best = -1; x = i % w; y = i / w; if (x > 0 && dsf_size(dsf, i-1) < MAXBLK && (best == -1 || revorder[i-1] < revorder[best])) best = i-1; if (x+1 < w && dsf_size(dsf, i+1) < MAXBLK && (best == -1 || revorder[i+1] < revorder[best])) best = i+1; if (y > 0 && dsf_size(dsf, i-w) < MAXBLK && (best == -1 || revorder[i-w] < revorder[best])) best = i-w; if (y+1 < w && dsf_size(dsf, i+w) < MAXBLK && (best == -1 || revorder[i+w] < revorder[best])) best = i+w; if (best >= 0) { singletons[i] = singletons[best] = false; dsf_merge(dsf, i, best); } } } /* Quit and start again if we have any singletons left over * which we weren't able to do anything at all with. */ for (i = 0; i < a; i++) if (singletons[i]) break; if (i < a) continue; /* * Decide what would be acceptable clues for each block. * * Blocks larger than 2 have free choice of ADD or MUL; * blocks of size 2 can be anything in principle (except * that they can only be DIV if the two numbers have an * integer quotient, of course), but we rule out (or try to * avoid) some clues because they're of low quality. * * Hence, we iterate once over the grid, stopping at the * canonical element of every >2 block and the _non_- * canonical element of every 2-block; the latter means that * we can make our decision about a 2-block in the knowledge * of both numbers in it. * * We reuse the 'singletons' array (finished with in the * above loop) to hold information about which blocks are * suitable for what. */ #define F_ADD 0x01 #define F_SUB 0x02 #define F_MUL 0x04 #define F_DIV 0x08 #define BAD_SHIFT 4 for (i = 0; i < a; i++) { singletons[i] = 0; j = dsf_canonify(dsf, i); k = dsf_size(dsf, j); if (params->multiplication_only) singletons[j] = F_MUL; else if (j == i && k > 2) { singletons[j] |= F_ADD | F_MUL; } else if (j != i && k == 2) { /* Fetch the two numbers and sort them into order. */ int p = grid[j], q = grid[i], v; if (p < q) { int t = p; p = q; q = t; } /* * Addition clues are always allowed, but we try to * avoid sums of 3, 4, (2w-1) and (2w-2) if we can, * because they're too easy - they only leave one * option for the pair of numbers involved. */ v = p + q; if (v > 4 && v < 2*w-2) singletons[j] |= F_ADD; else singletons[j] |= F_ADD << BAD_SHIFT; /* * Multiplication clues: above Normal difficulty, we * prefer (but don't absolutely insist on) clues of * this type which leave multiple options open. */ v = p * q; n = 0; for (k = 1; k <= w; k++) if (v % k == 0 && v / k <= w && v / k != k) n++; if (n <= 2 && diff > DIFF_NORMAL) singletons[j] |= F_MUL << BAD_SHIFT; else singletons[j] |= F_MUL; /* * Subtraction: we completely avoid a difference of * w-1. */ v = p - q; if (v < w-1) singletons[j] |= F_SUB; /* * Division: for a start, the quotient must be an * integer or the clue type is impossible. Also, we * never use quotients strictly greater than w/2, * because they're not only too easy but also * inelegant. */ if (p % q == 0 && 2 * (p / q) <= w) singletons[j] |= F_DIV; } } /* * Actually choose a clue for each block, trying to keep the * numbers of each type even, and starting with the * preferred candidates for each type where possible. * * I'm sure there should be a faster algorithm for doing * this, but I can't be bothered: O(N^2) is good enough when * N is at most the number of dominoes that fits into a 9x9 * square. */ shuffle(order, a, sizeof(*order), rs); for (i = 0; i < a; i++) clues[i] = 0; while (1) { bool done_something = false; for (k = 0; k < 4; k++) { long clue; int good, bad; switch (k) { case 0: clue = C_DIV; good = F_DIV; break; case 1: clue = C_SUB; good = F_SUB; break; case 2: clue = C_MUL; good = F_MUL; break; default /* case 3 */ : clue = C_ADD; good = F_ADD; break; } for (i = 0; i < a; i++) { j = order[i]; if (singletons[j] & good) { clues[j] = clue; singletons[j] = 0; break; } } if (i == a) { /* didn't find a nice one, use a nasty one */ bad = good << BAD_SHIFT; for (i = 0; i < a; i++) { j = order[i]; if (singletons[j] & bad) { clues[j] = clue; singletons[j] = 0; break; } } } if (i < a) done_something = true; } if (!done_something) break; } #undef F_ADD #undef F_SUB #undef F_MUL #undef F_DIV #undef BAD_SHIFT /* * Having chosen the clue types, calculate the clue values. */ for (i = 0; i < a; i++) { j = dsf_canonify(dsf, i); if (j == i) { cluevals[j] = grid[i]; } else { switch (clues[j]) { case C_ADD: cluevals[j] += grid[i]; break; case C_MUL: cluevals[j] *= grid[i]; break; case C_SUB: cluevals[j] = labs(cluevals[j] - grid[i]); break; case C_DIV: { int d1 = cluevals[j], d2 = grid[i]; if (d1 == 0 || d2 == 0) cluevals[j] = 0; else cluevals[j] = d2/d1 + d1/d2;/* one is 0 :-) */ } break; } } } for (i = 0; i < a; i++) { j = dsf_canonify(dsf, i); if (j == i) { clues[j] |= cluevals[j]; } } /* * See if the game can be solved at the specified difficulty * level, but not at the one below. */ if (diff > 0) { memset(soln, 0, a); ret = solver(w, dsf, clues, soln, diff-1); if (ret <= diff-1) continue; } memset(soln, 0, a); ret = solver(w, dsf, clues, soln, diff); if (ret != diff) continue; /* go round again */ /* * I wondered if at this point it would be worth trying to * merge adjacent blocks together, to make the puzzle * gradually more difficult if it's currently easier than * specced, increasing the chance of a given generation run * being successful. * * It doesn't seem to be critical for the generation speed, * though, so for the moment I'm leaving it out. */ /* * We've got a usable puzzle! */ break; } /* * Encode the puzzle description. */ desc = snewn(40*a, char); p = desc; p = encode_block_structure(p, w, dsf); *p++ = ','; for (i = 0; i < a; i++) { j = dsf_canonify(dsf, i); if (j == i) { switch (clues[j] & CMASK) { case C_ADD: *p++ = 'a'; break; case C_SUB: *p++ = 's'; break; case C_MUL: *p++ = 'm'; break; case C_DIV: *p++ = 'd'; break; } p += sprintf(p, "%ld", clues[j] & ~CMASK); } } *p++ = '\0'; desc = sresize(desc, p - desc, char); /* * Encode the solution. */ assert(memcmp(soln, grid, a) == 0); *aux = snewn(a+2, char); (*aux)[0] = 'S'; for (i = 0; i < a; i++) (*aux)[i+1] = '0' + soln[i]; (*aux)[a+1] = '\0'; sfree(grid); sfree(order); sfree(revorder); sfree(singletons); sfree(dsf); sfree(clues); sfree(cluevals); sfree(soln); return desc; } /* ---------------------------------------------------------------------- * Gameplay. */ static const char *validate_desc(const game_params *params, const char *desc) { int w = params->w, a = w*w; int *dsf; const char *ret; const char *p = desc; int i; /* * Verify that the block structure makes sense. */ dsf = snew_dsf(a); ret = parse_block_structure(&p, w, dsf); if (ret) { sfree(dsf); return ret; } if (*p != ',') { sfree(dsf); return "Expected ',' after block structure description"; } p++; /* * Verify that the right number of clues are given, and that SUB * and DIV clues don't apply to blocks of the wrong size. */ for (i = 0; i < a; i++) { if (dsf_canonify(dsf, i) == i) { if (*p == 'a' || *p == 'm') { /* these clues need no validation */ } else if (*p == 'd' || *p == 's') { if (dsf_size(dsf, i) != 2) { sfree(dsf); return "Subtraction and division blocks must have area 2"; } } else if (!*p) { sfree(dsf); return "Too few clues for block structure"; } else { sfree(dsf); return "Unrecognised clue type"; } p++; while (*p && isdigit((unsigned char)*p)) p++; } } sfree(dsf); if (*p) return "Too many clues for block structure"; return NULL; } static key_label *game_request_keys(const game_params *params, int *nkeys) { int i; int w = params->w; key_label *keys = snewn(w+1, key_label); *nkeys = w + 1; for (i = 0; i < w; i++) { if (i<9) keys[i].button = '1' + i; else keys[i].button = 'a' + i - 9; keys[i].label = NULL; } keys[w].button = '\b'; keys[w].label = NULL; return keys; } static game_state *new_game(midend *me, const game_params *params, const char *desc) { int w = params->w, a = w*w; game_state *state = snew(game_state); const char *p = desc; int i; state->par = *params; /* structure copy */ state->clues = snew(struct clues); state->clues->refcount = 1; state->clues->w = w; state->clues->dsf = snew_dsf(a); parse_block_structure(&p, w, state->clues->dsf); assert(*p == ','); p++; state->clues->clues = snewn(a, long); for (i = 0; i < a; i++) { if (dsf_canonify(state->clues->dsf, i) == i) { long clue = 0; switch (*p) { case 'a': clue = C_ADD; break; case 'm': clue = C_MUL; break; case 's': clue = C_SUB; assert(dsf_size(state->clues->dsf, i) == 2); break; case 'd': clue = C_DIV; assert(dsf_size(state->clues->dsf, i) == 2); break; default: assert(!"Bad description in new_game"); } p++; clue |= atol(p); while (*p && isdigit((unsigned char)*p)) p++; state->clues->clues[i] = clue; } else state->clues->clues[i] = 0; } state->grid = snewn(a, digit); state->pencil = snewn(a, int); for (i = 0; i < a; i++) { state->grid[i] = 0; state->pencil[i] = 0; } state->completed = false; state->cheated = false; return state; } static game_state *dup_game(const game_state *state) { int w = state->par.w, a = w*w; game_state *ret = snew(game_state); ret->par = state->par; /* structure copy */ ret->clues = state->clues; ret->clues->refcount++; ret->grid = snewn(a, digit); ret->pencil = snewn(a, int); memcpy(ret->grid, state->grid, a*sizeof(digit)); memcpy(ret->pencil, state->pencil, a*sizeof(int)); ret->completed = state->completed; ret->cheated = state->cheated; return ret; } static void free_game(game_state *state) { sfree(state->grid); sfree(state->pencil); if (--state->clues->refcount <= 0) { sfree(state->clues->dsf); sfree(state->clues->clues); sfree(state->clues); } sfree(state); } static char *solve_game(const game_state *state, const game_state *currstate, const char *aux, const char **error) { int w = state->par.w, a = w*w; int i, ret; digit *soln; char *out; if (aux) return dupstr(aux); soln = snewn(a, digit); memset(soln, 0, a); ret = solver(w, state->clues->dsf, state->clues->clues, soln, DIFFCOUNT-1); if (ret == diff_impossible) { *error = "No solution exists for this puzzle"; out = NULL; } else if (ret == diff_ambiguous) { *error = "Multiple solutions exist for this puzzle"; out = NULL; } else { out = snewn(a+2, char); out[0] = 'S'; for (i = 0; i < a; i++) out[i+1] = '0' + soln[i]; out[a+1] = '\0'; } sfree(soln); return out; } struct game_ui { /* * These are the coordinates of the currently highlighted * square on the grid, if hshow is true. */ int hx, hy; /* * This indicates whether the current highlight is a * pencil-mark one or a real one. */ bool hpencil; /* * This indicates whether or not we're showing the highlight * (used to be hx = hy = -1); important so that when we're * using the cursor keys it doesn't keep coming back at a * fixed position. When true, pressing a valid number or letter * key or Space will enter that number or letter in the grid. */ bool hshow; /* * This indicates whether we're using the highlight as a cursor; * it means that it doesn't vanish on a keypress, and that it is * allowed on immutable squares. */ bool hcursor; }; static game_ui *new_ui(const game_state *state) { game_ui *ui = snew(game_ui); ui->hx = ui->hy = 0; ui->hpencil = false; ui->hshow = false; ui->hcursor = false; return ui; } static void free_ui(game_ui *ui) { sfree(ui); } static char *encode_ui(const game_ui *ui) { return NULL; } static void decode_ui(game_ui *ui, const char *encoding) { } static void game_changed_state(game_ui *ui, const game_state *oldstate, const game_state *newstate) { int w = newstate->par.w; /* * We prevent pencil-mode highlighting of a filled square, unless * we're using the cursor keys. 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->hshow && ui->hpencil && !ui->hcursor && newstate->grid[ui->hy * w + ui->hx] != 0) { ui->hshow = false; } } static const char *current_key_label(const game_ui *ui, const game_state *state, int button) { if (ui->hshow && (button == CURSOR_SELECT)) return ui->hpencil ? "Ink" : "Pencil"; return ""; } #define PREFERRED_TILESIZE 48 #define TILESIZE (ds->tilesize) #define BORDER (TILESIZE / 2) #define GRIDEXTRA max((TILESIZE / 32),1) #define COORD(x) ((x)*TILESIZE + BORDER) #define FROMCOORD(x) (((x)+(TILESIZE-BORDER)) / TILESIZE - 1) #define FLASH_TIME 0.4F #define DF_PENCIL_SHIFT 16 #define DF_ERR_LATIN 0x8000 #define DF_ERR_CLUE 0x4000 #define DF_HIGHLIGHT 0x2000 #define DF_HIGHLIGHT_PENCIL 0x1000 #define DF_DIGIT_MASK 0x000F struct game_drawstate { int tilesize; bool started; long *tiles; long *errors; char *minus_sign, *times_sign, *divide_sign; }; static bool check_errors(const game_state *state, long *errors) { int w = state->par.w, a = w*w; int i, j, x, y; bool errs = false; long *cluevals; bool *full; cluevals = snewn(a, long); full = snewn(a, bool); if (errors) for (i = 0; i < a; i++) { errors[i] = 0; full[i] = true; } for (i = 0; i < a; i++) { long clue; j = dsf_canonify(state->clues->dsf, i); if (j == i) { cluevals[i] = state->grid[i]; } else { clue = state->clues->clues[j] & CMASK; switch (clue) { case C_ADD: cluevals[j] += state->grid[i]; break; case C_MUL: cluevals[j] *= state->grid[i]; break; case C_SUB: cluevals[j] = labs(cluevals[j] - state->grid[i]); break; case C_DIV: { int d1 = min(cluevals[j], state->grid[i]); int d2 = max(cluevals[j], state->grid[i]); if (d1 == 0 || d2 % d1 != 0) cluevals[j] = 0; else cluevals[j] = d2 / d1; } break; } } if (!state->grid[i]) full[j] = false; } for (i = 0; i < a; i++) { j = dsf_canonify(state->clues->dsf, i); if (j == i) { if ((state->clues->clues[j] & ~CMASK) != cluevals[i]) { errs = true; if (errors && full[j]) errors[j] |= DF_ERR_CLUE; } } } sfree(cluevals); sfree(full); for (y = 0; y < w; y++) { int mask = 0, errmask = 0; for (x = 0; x < w; x++) { int bit = 1 << state->grid[y*w+x]; errmask |= (mask & bit); mask |= bit; } if (mask != (1 << (w+1)) - (1 << 1)) { errs = true; errmask &= ~1; if (errors) { for (x = 0; x < w; x++) if (errmask & (1 << state->grid[y*w+x])) errors[y*w+x] |= DF_ERR_LATIN; } } } for (x = 0; x < w; x++) { int mask = 0, errmask = 0; for (y = 0; y < w; y++) { int bit = 1 << state->grid[y*w+x]; errmask |= (mask & bit); mask |= bit; } if (mask != (1 << (w+1)) - (1 << 1)) { errs = true; errmask &= ~1; if (errors) { for (y = 0; y < w; y++) if (errmask & (1 << state->grid[y*w+x])) errors[y*w+x] |= DF_ERR_LATIN; } } } return errs; } static char *interpret_move(const game_state *state, game_ui *ui, const game_drawstate *ds, int x, int y, int button) { int w = state->par.w; int tx, ty; char buf[80]; button &= ~MOD_MASK; tx = FROMCOORD(x); ty = FROMCOORD(y); if (tx >= 0 && tx < w && ty >= 0 && ty < w) { if (button == LEFT_BUTTON) { if (tx == ui->hx && ty == ui->hy && ui->hshow && !ui->hpencil) { ui->hshow = false; } else { ui->hx = tx; ui->hy = ty; ui->hshow = true; ui->hpencil = false; } ui->hcursor = false; return UI_UPDATE; } if (button == RIGHT_BUTTON) { /* * Pencil-mode highlighting for non filled squares. */ if (state->grid[ty*w+tx] == 0) { if (tx == ui->hx && ty == ui->hy && ui->hshow && ui->hpencil) { ui->hshow = false; } else { ui->hpencil = true; ui->hx = tx; ui->hy = ty; ui->hshow = true; } } else { ui->hshow = false; } ui->hcursor = false; return UI_UPDATE; } } if (IS_CURSOR_MOVE(button)) { move_cursor(button, &ui->hx, &ui->hy, w, w, false); ui->hshow = true; ui->hcursor = true; return UI_UPDATE; } if (ui->hshow && (button == CURSOR_SELECT)) { ui->hpencil ^= 1; ui->hcursor = true; return UI_UPDATE; } if (ui->hshow && ((button >= '0' && button <= '9' && button - '0' <= w) || button == CURSOR_SELECT2 || button == '\b')) { int n = button - '0'; if (button == CURSOR_SELECT2 || button == '\b') n = 0; /* * Can't make pencil marks in a filled square. This can only * become highlighted if we're using cursor keys. */ if (ui->hpencil && state->grid[ui->hy*w+ui->hx]) return NULL; /* * If you ask to fill a square with what it already contains, * or blank it when it's already empty, that has no effect... */ if ((!ui->hpencil || n == 0) && state->grid[ui->hy*w+ui->hx] == n && state->pencil[ui->hy*w+ui->hx] == 0) { /* ... expect to remove the cursor in mouse mode. */ if (!ui->hcursor) { ui->hshow = false; return UI_UPDATE; } return NULL; } sprintf(buf, "%c%d,%d,%d", (char)(ui->hpencil && n > 0 ? 'P' : 'R'), ui->hx, ui->hy, n); if (!ui->hcursor) ui->hshow = false; return dupstr(buf); } if (button == 'M' || button == 'm') return dupstr("M"); return NULL; } static game_state *execute_move(const game_state *from, const char *move) { int w = from->par.w, a = w*w; game_state *ret; int x, y, i, n; if (move[0] == 'S') { ret = dup_game(from); ret->completed = ret->cheated = true; for (i = 0; i < a; i++) { if (move[i+1] < '1' || move[i+1] > '0'+w) { free_game(ret); return NULL; } ret->grid[i] = move[i+1] - '0'; ret->pencil[i] = 0; } if (move[a+1] != '\0') { free_game(ret); return NULL; } return ret; } else if ((move[0] == 'P' || move[0] == 'R') && sscanf(move+1, "%d,%d,%d", &x, &y, &n) == 3 && x >= 0 && x < w && y >= 0 && y < w && n >= 0 && n <= w) { ret = dup_game(from); if (move[0] == 'P' && n > 0) { ret->pencil[y*w+x] ^= 1 << n; } else { ret->grid[y*w+x] = n; ret->pencil[y*w+x] = 0; if (!ret->completed && !check_errors(ret, NULL)) ret->completed = true; } return ret; } else if (move[0] == 'M') { /* * Fill in absolutely all pencil marks everywhere. (I * wouldn't use this for actual play, but it's a handy * starting point when following through a set of * diagnostics output by the standalone solver.) */ ret = dup_game(from); for (i = 0; i < a; i++) { if (!ret->grid[i]) ret->pencil[i] = (1 << (w+1)) - (1 << 1); } return ret; } else return NULL; /* couldn't parse move string */ } /* ---------------------------------------------------------------------- * Drawing routines. */ #define SIZE(w) ((w) * TILESIZE + 2*BORDER) static void game_compute_size(const 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 = *y = SIZE(params->w); } static void game_set_size(drawing *dr, game_drawstate *ds, const 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_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 const char *const minus_signs[] = { "\xE2\x88\x92", "-" }; static const char *const times_signs[] = { "\xC3\x97", "*" }; static const char *const divide_signs[] = { "\xC3\xB7", "/" }; static game_drawstate *game_new_drawstate(drawing *dr, const game_state *state) { int w = state->par.w, a = w*w; struct game_drawstate *ds = snew(struct game_drawstate); int i; ds->tilesize = 0; ds->started = false; ds->tiles = snewn(a, long); for (i = 0; i < a; i++) ds->tiles[i] = -1; ds->errors = snewn(a, long); ds->minus_sign = text_fallback(dr, minus_signs, lenof(minus_signs)); ds->times_sign = text_fallback(dr, times_signs, lenof(times_signs)); ds->divide_sign = text_fallback(dr, divide_signs, lenof(divide_signs)); return ds; } static void game_free_drawstate(drawing *dr, game_drawstate *ds) { sfree(ds->tiles); sfree(ds->errors); sfree(ds->minus_sign); sfree(ds->times_sign); sfree(ds->divide_sign); sfree(ds); } static void draw_tile(drawing *dr, game_drawstate *ds, struct clues *clues, int x, int y, long tile, bool only_one_op) { int w = clues->w /* , a = w*w */; int tx, ty, tw, th; int cx, cy, cw, ch; char str[64]; tx = BORDER + x * TILESIZE + 1 + GRIDEXTRA; ty = BORDER + y * TILESIZE + 1 + GRIDEXTRA; cx = tx; cy = ty; cw = tw = TILESIZE-1-2*GRIDEXTRA; ch = th = TILESIZE-1-2*GRIDEXTRA; if (x > 0 && dsf_canonify(clues->dsf, y*w+x) == dsf_canonify(clues->dsf, y*w+x-1)) cx -= GRIDEXTRA, cw += GRIDEXTRA; if (x+1 < w && dsf_canonify(clues->dsf, y*w+x) == dsf_canonify(clues->dsf, y*w+x+1)) cw += GRIDEXTRA; if (y > 0 && dsf_canonify(clues->dsf, y*w+x) == dsf_canonify(clues->dsf, (y-1)*w+x)) cy -= GRIDEXTRA, ch += GRIDEXTRA; if (y+1 < w && dsf_canonify(clues->dsf, y*w+x) == dsf_canonify(clues->dsf, (y+1)*w+x)) ch += GRIDEXTRA; clip(dr, cx, cy, cw, ch); /* background needs erasing */ draw_rect(dr, cx, cy, cw, ch, (tile & DF_HIGHLIGHT) ? COL_HIGHLIGHT : COL_BACKGROUND); /* pencil-mode highlight */ if (tile & DF_HIGHLIGHT_PENCIL) { 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); } /* * Draw the corners of thick lines in corner-adjacent squares, * which jut into this square by one pixel. */ if (x > 0 && y > 0 && dsf_canonify(clues->dsf, y*w+x) != dsf_canonify(clues->dsf, (y-1)*w+x-1)) draw_rect(dr, tx-GRIDEXTRA, ty-GRIDEXTRA, GRIDEXTRA, GRIDEXTRA, COL_GRID); if (x+1 < w && y > 0 && dsf_canonify(clues->dsf, y*w+x) != dsf_canonify(clues->dsf, (y-1)*w+x+1)) draw_rect(dr, tx+TILESIZE-1-2*GRIDEXTRA, ty-GRIDEXTRA, GRIDEXTRA, GRIDEXTRA, COL_GRID); if (x > 0 && y+1 < w && dsf_canonify(clues->dsf, y*w+x) != dsf_canonify(clues->dsf, (y+1)*w+x-1)) draw_rect(dr, tx-GRIDEXTRA, ty+TILESIZE-1-2*GRIDEXTRA, GRIDEXTRA, GRIDEXTRA, COL_GRID); if (x+1 < w && y+1 < w && dsf_canonify(clues->dsf, y*w+x) != dsf_canonify(clues->dsf, (y+1)*w+x+1)) draw_rect(dr, tx+TILESIZE-1-2*GRIDEXTRA, ty+TILESIZE-1-2*GRIDEXTRA, GRIDEXTRA, GRIDEXTRA, COL_GRID); /* Draw the box clue. */ if (dsf_canonify(clues->dsf, y*w+x) == y*w+x) { long clue = clues->clues[y*w+x]; long cluetype = clue & CMASK, clueval = clue & ~CMASK; int size = dsf_size(clues->dsf, y*w+x); /* * Special case of clue-drawing: a box with only one square * is written as just the number, with no operation, because * it doesn't matter whether the operation is ADD or MUL. * The generation code above should never produce puzzles * containing such a thing - I think they're inelegant - but * it's possible to type in game IDs from elsewhere, so I * want to display them right if so. */ sprintf (str, "%ld%s", clueval, (size == 1 || only_one_op ? "" : cluetype == C_ADD ? "+" : cluetype == C_SUB ? ds->minus_sign : cluetype == C_MUL ? ds->times_sign : /* cluetype == C_DIV ? */ ds->divide_sign)); draw_text(dr, tx + GRIDEXTRA * 2, ty + GRIDEXTRA * 2 + TILESIZE/4, FONT_VARIABLE, TILESIZE/4, ALIGN_VNORMAL | ALIGN_HLEFT, (tile & DF_ERR_CLUE ? COL_ERROR : COL_GRID), str); } /* new number needs drawing? */ if (tile & DF_DIGIT_MASK) { str[1] = '\0'; str[0] = (tile & DF_DIGIT_MASK) + '0'; draw_text(dr, tx + TILESIZE/2, ty + TILESIZE/2, FONT_VARIABLE, TILESIZE/2, ALIGN_VCENTRE | ALIGN_HCENTRE, (tile & DF_ERR_LATIN) ? COL_ERROR : COL_USER, str); } else { int i, j, npencil; int pl, pr, pt, pb; float bestsize; int pw, ph, minph, pbest, fontsize; /* Count the pencil marks required. */ for (i = 1, npencil = 0; i <= w; i++) if (tile & (1L << (i + DF_PENCIL_SHIFT))) npencil++; if (npencil) { minph = 2; /* * Determine the bounding rectangle within which we're going * to put the pencil marks. */ /* Start with the whole square */ pl = tx + GRIDEXTRA; pr = pl + TILESIZE - GRIDEXTRA; pt = ty + GRIDEXTRA; pb = pt + TILESIZE - GRIDEXTRA; if (dsf_canonify(clues->dsf, y*w+x) == y*w+x) { /* * Make space for the clue text. */ pt += TILESIZE/4; /* minph--; */ } /* * We arrange our pencil marks in a grid layout, with * the number of rows and columns adjusted to allow the * maximum font size. * * So now we work out what the grid size ought to be. */ bestsize = 0.0; pbest = 0; /* Minimum */ for (pw = 3; pw < max(npencil,4); pw++) { float fw, fh, fs; ph = (npencil + pw - 1) / pw; ph = max(ph, minph); fw = (pr - pl) / (float)pw; fh = (pb - pt) / (float)ph; fs = min(fw, fh); if (fs > bestsize) { bestsize = fs; pbest = pw; } } assert(pbest > 0); pw = pbest; ph = (npencil + pw - 1) / pw; ph = max(ph, minph); /* * Now we've got our grid dimensions, work out the pixel * size of a grid element, and round it to the nearest * pixel. (We don't want rounding errors to make the * grid look uneven at low pixel sizes.) */ fontsize = min((pr - pl) / pw, (pb - pt) / ph); /* * Centre the resulting figure in the square. */ pl = tx + (TILESIZE - fontsize * pw) / 2; pt = ty + (TILESIZE - fontsize * ph) / 2; /* * And move it down a bit if it's collided with some * clue text. */ if (dsf_canonify(clues->dsf, y*w+x) == y*w+x) { pt = max(pt, ty + GRIDEXTRA * 3 + TILESIZE/4); } /* * Now actually draw the pencil marks. */ for (i = 1, j = 0; i <= w; i++) if (tile & (1L << (i + DF_PENCIL_SHIFT))) { int dx = j % pw, dy = j / pw; str[1] = '\0'; str[0] = i + '0'; draw_text(dr, pl + fontsize * (2*dx+1) / 2, pt + fontsize * (2*dy+1) / 2, FONT_VARIABLE, fontsize, ALIGN_VCENTRE | ALIGN_HCENTRE, COL_PENCIL, str); j++; } } } unclip(dr); draw_update(dr, cx, cy, cw, ch); } static void game_redraw(drawing *dr, game_drawstate *ds, const game_state *oldstate, const game_state *state, int dir, const game_ui *ui, float animtime, float flashtime) { int w = state->par.w /*, a = w*w */; int x, y; if (!ds->started) { /* * Big containing rectangle. */ draw_rect(dr, COORD(0) - GRIDEXTRA, COORD(0) - GRIDEXTRA, w*TILESIZE+1+GRIDEXTRA*2, w*TILESIZE+1+GRIDEXTRA*2, COL_GRID); draw_update(dr, 0, 0, SIZE(w), SIZE(w)); ds->started = true; } check_errors(state, ds->errors); for (y = 0; y < w; y++) { for (x = 0; x < w; x++) { long tile = 0L; if (state->grid[y*w+x]) tile = state->grid[y*w+x]; else tile = (long)state->pencil[y*w+x] << DF_PENCIL_SHIFT; if (ui->hshow && ui->hx == x && ui->hy == y) tile |= (ui->hpencil ? DF_HIGHLIGHT_PENCIL : DF_HIGHLIGHT); if (flashtime > 0 && (flashtime <= FLASH_TIME/3 || flashtime >= FLASH_TIME*2/3)) tile |= DF_HIGHLIGHT; /* completion flash */ tile |= ds->errors[y*w+x]; if (ds->tiles[y*w+x] != tile) { ds->tiles[y*w+x] = tile; draw_tile(dr, ds, state->clues, x, y, tile, state->par.multiplication_only); } } } } static float game_anim_length(const game_state *oldstate, const game_state *newstate, int dir, game_ui *ui) { return 0.0F; } static float game_flash_length(const game_state *oldstate, const game_state *newstate, int dir, game_ui *ui) { if (!oldstate->completed && newstate->completed && !oldstate->cheated && !newstate->cheated) return FLASH_TIME; return 0.0F; } static void game_get_cursor_location(const game_ui *ui, const game_drawstate *ds, const game_state *state, const game_params *params, int *x, int *y, int *w, int *h) { if(ui->hshow) { *x = BORDER + ui->hx * TILESIZE + 1 + GRIDEXTRA; *y = BORDER + ui->hy * TILESIZE + 1 + GRIDEXTRA; *w = *h = TILESIZE-1-2*GRIDEXTRA; } } static int game_status(const game_state *state) { return state->completed ? +1 : 0; } static void game_print_size(const game_params *params, float *x, float *y) { int pw, ph; /* * We use 9mm squares by default, like Solo. */ game_compute_size(params, 900, &pw, &ph); *x = pw / 100.0F; *y = ph / 100.0F; } /* * Subfunction to draw the 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. */ static void outline_block_structure(drawing *dr, game_drawstate *ds, int w, int *dsf, int ink) { int a = w*w; int *coords; int i, n; int x, y, dx, dy, sx, sy, sdx, sdy; coords = snewn(4*a, int); /* * Iterate over all the blocks. */ for (i = 0; i < a; i++) { if (dsf_canonify(dsf, i) != i) continue; /* * For each block, we need a starting square within it which * has a boundary at the left. Conveniently, we have one * right here, by construction. */ x = i % w; y = i / w; 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; /* * 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 < w && ty >= 0 && ty < w && dsf_canonify(dsf, ty*w+tx) == i); tx = x - dy; ty = y + dx; nin += (tx >= 0 && tx < w && ty >= 0 && ty < w && dsf_canonify(dsf, ty*w+tx) == i); 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 < w && y >= 0 && y < w && dsf_canonify(dsf, y*w+x) == i); assert(x+dx < 0 || x+dx >= w || y+dy < 0 || y+dy >= w || dsf_canonify(dsf, (y+dy)*w+(x+dx)) != i); /* * Record the point we just went past 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*w+2); cx = ((2*x+1) + dy + dx) / 2; cy = ((2*y+1) - dx + dy) / 2; coords[2*n+0] = BORDER + cx * TILESIZE; coords[2*n+1] = BORDER + cy * TILESIZE; n++; } 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); } static void game_print(drawing *dr, const game_state *state, int tilesize) { int w = state->par.w; int ink = print_mono_colour(dr, 0); int x, y; char *minus_sign, *times_sign, *divide_sign; /* Ick: fake up `ds->tilesize' for macro expansion purposes */ game_drawstate ads, *ds = &ads; game_set_size(dr, ds, NULL, tilesize); minus_sign = text_fallback(dr, minus_signs, lenof(minus_signs)); times_sign = text_fallback(dr, times_signs, lenof(times_signs)); divide_sign = text_fallback(dr, divide_signs, lenof(divide_signs)); /* * Border. */ print_line_width(dr, 3 * TILESIZE / 40); draw_rect_outline(dr, BORDER, BORDER, w*TILESIZE, w*TILESIZE, ink); /* * Main grid. */ for (x = 1; x < w; x++) { print_line_width(dr, TILESIZE / 40); draw_line(dr, BORDER+x*TILESIZE, BORDER, BORDER+x*TILESIZE, BORDER+w*TILESIZE, ink); } for (y = 1; y < w; y++) { print_line_width(dr, TILESIZE / 40); draw_line(dr, BORDER, BORDER+y*TILESIZE, BORDER+w*TILESIZE, BORDER+y*TILESIZE, ink); } /* * Thick lines between cells. */ print_line_width(dr, 3 * TILESIZE / 40); outline_block_structure(dr, ds, w, state->clues->dsf, ink); /* * Clues. */ for (y = 0; y < w; y++) for (x = 0; x < w; x++) if (dsf_canonify(state->clues->dsf, y*w+x) == y*w+x) { long clue = state->clues->clues[y*w+x]; long cluetype = clue & CMASK, clueval = clue & ~CMASK; int size = dsf_size(state->clues->dsf, y*w+x); char str[64]; /* * As in the drawing code, we omit the operator for * blocks of area 1. */ sprintf (str, "%ld%s", clueval, (size == 1 ? "" : cluetype == C_ADD ? "+" : cluetype == C_SUB ? minus_sign : cluetype == C_MUL ? times_sign : /* cluetype == C_DIV ? */ divide_sign)); draw_text(dr, BORDER+x*TILESIZE + 5*TILESIZE/80, BORDER+y*TILESIZE + 20*TILESIZE/80, FONT_VARIABLE, TILESIZE/4, ALIGN_VNORMAL | ALIGN_HLEFT, ink, str); } /* * Numbers for the solution, if any. */ for (y = 0; y < w; y++) for (x = 0; x < w; x++) if (state->grid[y*w+x]) { char str[2]; str[1] = '\0'; str[0] = state->grid[y*w+x] + '0'; draw_text(dr, BORDER + x*TILESIZE + TILESIZE/2, BORDER + y*TILESIZE + TILESIZE/2, FONT_VARIABLE, TILESIZE/2, ALIGN_VCENTRE | ALIGN_HCENTRE, ink, str); } sfree(minus_sign); sfree(times_sign); sfree(divide_sign); } #ifdef COMBINED #define thegame keen #endif const struct game thegame = { "Keen", "games.keen", "keen", default_params, game_fetch_preset, NULL, 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, false, NULL, NULL, /* can_format_as_text_now, text_format */ new_ui, free_ui, encode_ui, decode_ui, game_request_keys, game_changed_state, current_key_label, interpret_move, execute_move, PREFERRED_TILESIZE, game_compute_size, game_set_size, game_colours, game_new_drawstate, game_free_drawstate, game_redraw, game_anim_length, game_flash_length, game_get_cursor_location, game_status, true, false, game_print_size, game_print, false, /* wants_statusbar */ false, NULL, /* timing_state */ REQUIRE_RBUTTON | REQUIRE_NUMPAD, /* flags */ }; #ifdef STANDALONE_SOLVER #include <stdarg.h> int main(int argc, char **argv) { game_params *p; game_state *s; char *id = NULL, *desc; const char *err; bool grade = false; int ret, diff; bool really_show_working = false; while (--argc > 0) { char *p = *++argv; if (!strcmp(p, "-v")) { really_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); /* * When solving an Easy puzzle, we don't want to bother the * user with Hard-level deductions. For this reason, we grade * the puzzle internally before doing anything else. */ ret = -1; /* placate optimiser */ solver_show_working = 0; for (diff = 0; diff < DIFFCOUNT; diff++) { memset(s->grid, 0, p->w * p->w); ret = solver(p->w, s->clues->dsf, s->clues->clues, s->grid, diff); if (ret <= diff) break; } if (diff == DIFFCOUNT) { if (grade) printf("Difficulty rating: ambiguous\n"); else printf("Unable to find a unique solution\n"); } else { if (grade) { if (ret == diff_impossible) printf("Difficulty rating: impossible (no solution exists)\n"); else printf("Difficulty rating: %s\n", keen_diffnames[ret]); } else { solver_show_working = really_show_working ? 1 : 0; memset(s->grid, 0, p->w * p->w); ret = solver(p->w, s->clues->dsf, s->clues->clues, s->grid, diff); if (ret != diff) printf("Puzzle is inconsistent\n"); else { /* * We don't have a game_text_format for this game, * so we have to output the solution manually. */ int x, y; for (y = 0; y < p->w; y++) { for (x = 0; x < p->w; x++) { printf("%s%c", x>0?" ":"", '0' + s->grid[y*p->w+x]); } putchar('\n'); } } } } return 0; } #endif /* vim: set shiftwidth=4 tabstop=8: */