--- /dev/null
+/*
+ * keen.c: an implementation of the Times's 'KenKen' 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
+
+enum {
+ COL_BACKGROUND,
+ COL_GRID,
+ COL_USER,
+ COL_HIGHLIGHT,
+ COL_ERROR,
+ COL_PENCIL,
+ NCOLOURS
+};
+
+struct game_params {
+ int w, diff;
+};
+
+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 */
+ int completed, cheated;
+};
+
+static game_params *default_params(void)
+{
+ game_params *ret = snew(game_params);
+
+ ret->w = 6;
+ ret->diff = DIFF_NORMAL;
+
+ return ret;
+}
+
+const static struct game_params keen_presets[] = {
+ { 4, DIFF_EASY },
+ { 5, DIFF_EASY },
+ { 6, DIFF_EASY },
+ { 6, DIFF_NORMAL },
+ { 6, DIFF_HARD },
+ { 6, DIFF_EXTREME },
+ { 6, DIFF_UNREASONABLE },
+ { 9, DIFF_NORMAL },
+};
+
+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", ret->w, ret->w, keen_diffnames[ret->diff]);
+
+ *name = dupstr(buf);
+ *params = ret;
+ return TRUE;
+}
+
+static void free_params(game_params *params)
+{
+ sfree(params);
+}
+
+static game_params *dup_params(game_params *params)
+{
+ game_params *ret = snew(game_params);
+ *ret = *params; /* structure copy */
+ return ret;
+}
+
+static 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++;
+ }
+ }
+}
+
+static char *encode_params(game_params *params, int full)
+{
+ char ret[80];
+
+ sprintf(ret, "%d", params->w);
+ if (full)
+ sprintf(ret + strlen(ret), "d%c", keen_diffchars[params->diff]);
+
+ return dupstr(ret);
+}
+
+static config_item *game_configure(game_params *params)
+{
+ config_item *ret;
+ char buf[80];
+
+ ret = snewn(3, 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 = NULL;
+ ret[2].type = C_END;
+ ret[2].sval = NULL;
+ ret[2].ival = 0;
+
+ return ret;
+}
+
+static game_params *custom_params(config_item *cfg)
+{
+ game_params *ret = snew(game_params);
+
+ ret->w = atoi(cfg[0].sval);
+ ret->diff = cfg[1].ival;
+
+ return ret;
+}
+
+static char *validate_params(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.
+ */
+ 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];
+ int value = ctx->clues[box] & ~CMASK;
+ int op = ctx->clues[box] & CMASK;
+
+ if (diff == DIFF_HARD) {
+ for (i = 0; i < n; 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;
+ }
+
+ 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(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 &&
+ (best == -1 || revorder[i-1] < revorder[best]))
+ best = i-1;
+ if (x+1 < w &&
+ (best == -1 || revorder[i+1] < revorder[best]))
+ best = i+1;
+ if (y > 0 &&
+ (best == -1 || revorder[i-w] < revorder[best]))
+ best = i-w;
+ if (y+1 < w &&
+ (best == -1 || revorder[i+w] < revorder[best]))
+ best = i+w;
+
+ if (best >= 0) {
+ singletons[i] = FALSE;
+ dsf_merge(dsf, i, best);
+ }
+ }
+ }
+
+ /*
+ * 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_ADD_BAD 0x02
+#define F_SUB 0x04
+#define F_SUB_BAD 0x08
+#define F_MUL 0x10
+#define F_MUL_BAD 0x20
+#define F_DIV 0x40
+#define F_DIV_BAD 0x80
+ for (i = 0; i < a; i++) {
+ singletons[i] = 0;
+ j = dsf_canonify(dsf, i);
+ k = dsf_size(dsf, j);
+ 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;
+
+ /*
+ * Multiplication clues: similarly, we prefer clues
+ * of this type which leave multiple options open.
+ * We can't rule out all the others, though, because
+ * there are very very few 2-square multiplication
+ * clues that _don't_ leave only one option.
+ */
+ v = p * q;
+ n = 0;
+ for (k = 1; k <= w; k++)
+ if (v % k == 0 && v / k <= w && v / k != k)
+ n++;
+ if (n > 1)
+ singletons[j] |= F_MUL;
+ else
+ singletons[j] |= F_MUL_BAD;
+
+ /*
+ * 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; bad = F_DIV_BAD; break;
+ case 1: clue = C_SUB; good = F_SUB; bad = F_SUB_BAD; break;
+ case 2: clue = C_MUL; good = F_MUL; bad = F_MUL_BAD; break;
+ default /* case 3 */ :
+ clue = C_ADD; good = F_ADD; bad = F_ADD_BAD; 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 */
+ for (i = 0; i < a; i++) {
+ j = order[i];
+ if (singletons[j] & good) {
+ clues[j] = clue;
+ singletons[j] = 0;
+ break;
+ }
+ }
+ }
+ if (i < a)
+ done_something = TRUE;
+ }
+
+ if (!done_something)
+ break;
+ }
+#undef F_ADD
+#undef F_ADD_BAD
+#undef F_SUB
+#undef F_SUB_BAD
+#undef F_MUL
+#undef F_MUL_BAD
+#undef F_DIV
+#undef F_DIV_BAD
+
+ /*
+ * 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(game_params *params, 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, game_params *params, char *desc)
+{
+ int w = params->w, a = w*w;
+ game_state *state = snew(game_state);
+ char *err;
+ 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);
+ err = 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(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(game_state *state, game_state *currstate,
+ 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(game_params *params)
+{
+ return TRUE;
+}
+
+static char *game_text_format(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(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(game_ui *ui)
+{
+ return NULL;
+}
+
+static void decode_ui(game_ui *ui, char *encoding)
+{
+}
+
+static void game_changed_state(game_ui *ui, game_state *oldstate,
+ game_state *newstate)
+{
+ int 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(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 = cluevals[j], d2 = state->grid[i];
+ if (d1 == 0 || d2 == 0)
+ cluevals[j] = 0;
+ else
+ cluevals[j] = d2/d1 + d1/d2;/* one of them is 0 :-) */
+ }
+ 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(game_state *state, game_ui *ui, 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(game_state *from, 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(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,
+ 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, 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);
+}
+
+void draw_tile(drawing *dr, game_drawstate *ds, struct clues *clues,
+ int x, int y, long tile)
+{
+ 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);
+
+ /*
+ * 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);
+
+ /* 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 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 ? "" :
+ 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 & (1 << (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;
+
+ /*
+ * 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 & (1 << (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, game_state *oldstate,
+ game_state *state, int dir, 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);
+ }
+ }
+ }
+}
+
+static float game_anim_length(game_state *oldstate, game_state *newstate,
+ int dir, game_ui *ui)
+{
+ return 0.0F;
+}
+
+static float game_flash_length(game_state *oldstate, game_state *newstate,
+ int dir, game_ui *ui)
+{
+ if (!oldstate->completed && newstate->completed &&
+ !oldstate->cheated && !newstate->cheated)
+ return FLASH_TIME;
+ return 0.0F;
+}
+
+static int game_timing_state(game_state *state, game_ui *ui)
+{
+ if (state->completed)
+ return FALSE;
+ return TRUE;
+}
+
+static void game_print_size(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, 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);
+ }
+}
+
+#ifdef COMBINED
+#define thegame keen
+#endif
+
+const struct game thegame = {
+ "Keen", "games.keen", "keen",
+ default_params,
+ game_fetch_preset,
+ decode_params,
+ encode_params,
+ free_params,
+ dup_params,
+ TRUE, game_configure, custom_params,
+ validate_params,
+ new_game_desc,
+ validate_desc,
+ new_game,
+ dup_game,
+ free_game,
+ TRUE, solve_game,
+ 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,
+ TRUE, FALSE, game_print_size, game_print,
+ FALSE, /* wants_statusbar */
+ FALSE, game_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, *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] <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 = 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: */
~mdw / sgt / puzzles / commitdiff