changelog: document last change

[sgt-puzzles.git] / keen.c

/*
 * keen.c: an implementation of the Times's 'KenKen' puzzle, and
 * also of Nikoli's very similar 'Inshi No Heya' puzzle.
 */

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <assert.h>
#include <ctype.h>
#include <math.h>

#include "puzzles.h"
#include "latin.h"

/*
 * Difficulty levels. I do some macro ickery here to ensure that my
 * enum and the various forms of my name list always match up.
 */
#define DIFFLIST(A) \
    A(EASY,Easy,solver_easy,e) \
    A(NORMAL,Normal,solver_normal,n) \
    A(HARD,Hard,solver_hard,h) \
    A(EXTREME,Extreme,NULL,x) \
    A(UNREASONABLE,Unreasonable,NULL,u)
#define ENUM(upper,title,func,lower) DIFF_ ## upper,
#define TITLE(upper,title,func,lower) #title,
#define ENCODE(upper,title,func,lower) #lower
#define CONFIG(upper,title,func,lower) ":" #title
enum { DIFFLIST(ENUM) DIFFCOUNT };
static char const *const keen_diffnames[] = { DIFFLIST(TITLE) };
static char const keen_diffchars[] = DIFFLIST(ENCODE);
#define DIFFCONFIG DIFFLIST(CONFIG)

/*
 * Clue notation. Important here that ADD and MUL come before SUB
 * and DIV, and that DIV comes last. 
 */
#define C_ADD 0x00000000L
#define C_MUL 0x20000000L
#define C_SUB 0x40000000L
#define C_DIV 0x60000000L
#define CMASK 0x60000000L
#define CUNIT 0x20000000L

/*
 * Maximum size of any clue block. Very large ones are annoying in UI
 * terms (if they're multiplicative you end up with too many digits to
 * fit in the square) and also in solver terms (too many possibilities
 * to iterate over).
 */
#define MAXBLK 6

enum {
    COL_BACKGROUND,
    COL_GRID,
    COL_USER,
    COL_HIGHLIGHT,
    COL_ERROR,
    COL_PENCIL,
    NCOLOURS
};

struct game_params {
    int w, diff, multiplication_only;
};

struct clues {
    int refcount;
    int w;
    int *dsf;
    long *clues;
};

struct game_state {
    game_params par;
    struct clues *clues;
    digit *grid;
    int *pencil;                       /* bitmaps using bits 1<<1..1<<n */
    int completed, cheated;
};

static game_params *default_params(void)
{
    game_params *ret = snew(game_params);

    ret->w = 6;
    ret->diff = DIFF_NORMAL;
    ret->multiplication_only = FALSE;

    return ret;
}

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 <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: */