From 881746789a489fad85aae8317555f73dbe261556 Mon Sep 17 00:00:00 2001 From: Franklin Wei Date: Sat, 29 Apr 2017 18:21:56 -0400 Subject: puzzles: refactor and resync with upstream This brings puzzles up-to-date with upstream revision 2d333750272c3967cfd5cd3677572cddeaad5932, though certain changes made by me, including cursor-only Untangle and some compilation fixes remain. Upstream code has been moved to its separate subdirectory and future syncs can be done by simply copying over the new sources. Change-Id: Ia6506ca5f78c3627165ea6791d38db414ace0804 --- apps/plugins/puzzles/src/keen.c | 2479 +++++++++++++++++++++++++++++++++++++++ 1 file changed, 2479 insertions(+) create mode 100644 apps/plugins/puzzles/src/keen.c (limited to 'apps/plugins/puzzles/src/keen.c') diff --git a/apps/plugins/puzzles/src/keen.c b/apps/plugins/puzzles/src/keen.c new file mode 100644 index 0000000000..fdaae32e5d --- /dev/null +++ b/apps/plugins/puzzles/src/keen.c @@ -0,0 +1,2479 @@ +/* + * keen.c: an implementation of the Times's 'KenKen' puzzle, and + * also of Nikoli's very similar 'Inshi No Heya' puzzle. + */ + +#include +#include +#include +#include +#include +#include + +#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, 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<w = 6; + ret->diff = DIFF_NORMAL; + ret->multiplication_only = FALSE; + + return ret; +} + +const static 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 int 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, int 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].sval = dupstr(buf); + ret[0].ival = 0; + + ret[1].name = "Difficulty"; + ret[1].type = C_CHOICES; + ret[1].sval = DIFFCONFIG; + ret[1].ival = params->diff; + + ret[2].name = "Multiplication only"; + ret[2].type = C_BOOLEAN; + ret[2].sval = NULL; + ret[2].ival = params->multiplication_only; + + ret[3].name = NULL; + ret[3].type = C_END; + ret[3].sval = NULL; + ret[3].ival = 0; + + return ret; +} + +static game_params *custom_params(const config_item *cfg) +{ + game_params *ret = snew(game_params); + + ret->w = atoi(cfg[0].sval); + ret->diff = cfg[1].ival; + ret->multiplication_only = cfg[2].ival; + + return ret; +} + +static char *validate_params(const game_params *params, int 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 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, &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, 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 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, 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, int 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) { + int 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] = abs(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 char *validate_desc(const game_params *params, const char *desc) +{ + int w = params->w, a = w*w; + int *dsf; + 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 != ',') + 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) + return "Subtraction and division blocks must have area 2"; + } else if (!*p) { + return "Too few clues for block structure"; + } else { + return "Unrecognised clue type"; + } + p++; + while (*p && isdigit((unsigned char)*p)) p++; + } + } + if (*p) + return "Too many clues for block structure"; + + return NULL; +} + +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 = 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, 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; +} + +static int game_can_format_as_text_now(const game_params *params) +{ + return TRUE; +} + +static char *game_text_format(const game_state *state) +{ + return NULL; +} + +struct game_ui { + /* + * These are the coordinates of the currently highlighted + * square on the grid, if hshow = 1. + */ + int hx, hy; + /* + * This indicates whether the current highlight is a + * pencil-mark one or a real one. + */ + int 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 hshow = 1, pressing a valid number + * or letter key or Space will enter that number or letter in the grid. + */ + int 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. + */ + int hcursor; +}; + +static game_ui *new_ui(const game_state *state) +{ + game_ui *ui = snew(game_ui); + + ui->hx = ui->hy = 0; + ui->hpencil = ui->hshow = ui->hcursor = 0; + + 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 = 0; + } +} + +#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; + int started; + long *tiles; + long *errors; + char *minus_sign, *times_sign, *divide_sign; +}; + +static int check_errors(const game_state *state, long *errors) +{ + int w = state->par.w, a = w*w; + int i, j, x, y, errs = FALSE; + long *cluevals; + int *full; + + cluevals = snewn(a, long); + full = snewn(a, int); + + 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] = abs(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 == 0) { + ui->hshow = 0; + } else { + ui->hx = tx; + ui->hy = ty; + ui->hshow = 1; + ui->hpencil = 0; + } + ui->hcursor = 0; + return ""; /* UI activity occurred */ + } + 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 = 0; + } else { + ui->hpencil = 1; + ui->hx = tx; + ui->hy = ty; + ui->hshow = 1; + } + } else { + ui->hshow = 0; + } + ui->hcursor = 0; + return ""; /* UI activity occurred */ + } + } + if (IS_CURSOR_MOVE(button)) { + move_cursor(button, &ui->hx, &ui->hy, w, w, 0); + ui->hshow = ui->hcursor = 1; + return ""; + } + if (ui->hshow && + (button == CURSOR_SELECT)) { + ui->hpencil = 1 - ui->hpencil; + ui->hcursor = 1; + return ""; + } + + 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; + + sprintf(buf, "%c%d,%d,%d", + (char)(ui->hpencil && n > 0 ? 'P' : 'R'), ui->hx, ui->hy, n); + + if (!ui->hcursor) ui->hshow = 0; + + 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, int 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) { + /* + * 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(w), SIZE(w), COL_BACKGROUND); + + /* + * 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 int game_status(const game_state *state) +{ + return state->completed ? +1 : 0; +} + +static int game_timing_state(const game_state *state, game_ui *ui) +{ + if (state->completed) + return FALSE; + return TRUE; +} + +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, game_can_format_as_text_now, game_text_format, + new_ui, + free_ui, + encode_ui, + decode_ui, + game_changed_state, + interpret_move, + execute_move, + PREFERRED_TILESIZE, game_compute_size, game_set_size, + game_colours, + game_new_drawstate, + game_free_drawstate, + game_redraw, + game_anim_length, + game_flash_length, + game_status, + TRUE, FALSE, game_print_size, game_print, + FALSE, /* wants_statusbar */ + FALSE, game_timing_state, + REQUIRE_RBUTTON | REQUIRE_NUMPAD, /* flags */ +}; + +#ifdef STANDALONE_SOLVER + +#include + +int main(int argc, char **argv) +{ + game_params *p; + game_state *s; + char *id = NULL, *desc, *err; + int grade = FALSE; + int ret, diff, 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] \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 = FALSE; + 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; + 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: */ -- cgit v1.2.3