I've proved that a grid dimension of 2 is capable of irretrievably

[sgt-puzzles.git] / mines.c

/*
 * mines.c: Minesweeper clone with sophisticated grid generation.
 * 
 * Still TODO:
 *
 *  - think about configurably supporting question marks. Once,
 *    that is, we've thought about configurability in general!
 */

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

#include "tree234.h"
#include "puzzles.h"

enum {
    COL_BACKGROUND, COL_BACKGROUND2,
    COL_1, COL_2, COL_3, COL_4, COL_5, COL_6, COL_7, COL_8,
    COL_MINE, COL_BANG, COL_CROSS, COL_FLAG, COL_FLAGBASE, COL_QUERY,
    COL_HIGHLIGHT, COL_LOWLIGHT,
    NCOLOURS
};

#define TILE_SIZE 20
#define BORDER (TILE_SIZE * 3 / 2)
#define HIGHLIGHT_WIDTH 2
#define OUTER_HIGHLIGHT_WIDTH 3
#define COORD(x)  ( (x) * TILE_SIZE + BORDER )
#define FROMCOORD(x)  ( ((x) - BORDER + TILE_SIZE) / TILE_SIZE - 1 )

#define FLASH_FRAME 0.13F

struct game_params {
    int w, h, n;
    int unique;
};

struct mine_layout {
    /*
     * This structure is shared between all the game_states for a
     * given instance of the puzzle, so we reference-count it.
     */
    int refcount;
    char *mines;
    /*
     * If we haven't yet actually generated the mine layout, here's
     * all the data we will need to do so.
     */
    int n, unique;
    random_state *rs;
    midend_data *me;                   /* to give back the new game desc */
};

struct game_state {
    int w, h, n, dead, won;
    int used_solve, just_used_solve;
    struct mine_layout *layout;        /* real mine positions */
    signed char *grid;                         /* player knowledge */
    /*
     * Each item in the `grid' array is one of the following values:
     * 
     *  - 0 to 8 mean the square is open and has a surrounding mine
     *    count.
     * 
     *  - -1 means the square is marked as a mine.
     * 
     *  - -2 means the square is unknown.
     * 
     *  - -3 means the square is marked with a question mark
     *    (FIXME: do we even want to bother with this?).
     * 
     *  - 64 means the square has had a mine revealed when the game
     *    was lost.
     * 
     *  - 65 means the square had a mine revealed and this was the
     *    one the player hits.
     * 
     *  - 66 means the square has a crossed-out mine because the
     *    player had incorrectly marked it.
     */
};

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

    ret->w = ret->h = 9;
    ret->n = 10;
    ret->unique = TRUE;

    return ret;
}

static int game_fetch_preset(int i, char **name, game_params **params)
{
    game_params *ret;
    char str[80];
    static const struct { int w, h, n; } values[] = {
        {9, 9, 10},
        {16, 16, 40},
        {30, 16, 99},
    };

    if (i < 0 || i >= lenof(values))
        return FALSE;

    ret = snew(game_params);
    ret->w = values[i].w;
    ret->h = values[i].h;
    ret->n = values[i].n;
    ret->unique = TRUE;

    sprintf(str, "%dx%d, %d mines", ret->w, ret->h, ret->n);

    *name = dupstr(str);
    *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 == 'x') {
        p++;
        params->h = atoi(p);
        while (*p && isdigit((unsigned char)*p)) p++;
    } else {
        params->h = params->w;
    }
    if (*p == 'n') {
        p++;
        params->n = atoi(p);
        while (*p && (*p == '.' || isdigit((unsigned char)*p))) p++;
    } else {
        params->n = params->w * params->h / 10;
    }

    while (*p) {
        if (*p == 'a') {
            p++;
            params->unique = FALSE;
        } else
            p++;                       /* skip any other gunk */
    }
}

static char *encode_params(game_params *params, int full)
{
    char ret[400];
    int len;

    len = sprintf(ret, "%dx%d", params->w, params->h);
    /*
     * Mine count is a generation-time parameter, since it can be
     * deduced from the mine bitmap!
     */
    if (full)
        len += sprintf(ret+len, "n%d", params->n);
    if (full && !params->unique)
        ret[len++] = 'a';
    assert(len < lenof(ret));
    ret[len] = '\0';

    return dupstr(ret);
}

static config_item *game_configure(game_params *params)
{
    config_item *ret;
    char buf[80];

    ret = snewn(5, config_item);

    ret[0].name = "Width";
    ret[0].type = C_STRING;
    sprintf(buf, "%d", params->w);
    ret[0].sval = dupstr(buf);
    ret[0].ival = 0;

    ret[1].name = "Height";
    ret[1].type = C_STRING;
    sprintf(buf, "%d", params->h);
    ret[1].sval = dupstr(buf);
    ret[1].ival = 0;

    ret[2].name = "Mines";
    ret[2].type = C_STRING;
    sprintf(buf, "%d", params->n);
    ret[2].sval = dupstr(buf);
    ret[2].ival = 0;

    ret[3].name = "Ensure solubility";
    ret[3].type = C_BOOLEAN;
    ret[3].sval = NULL;
    ret[3].ival = params->unique;

    ret[4].name = NULL;
    ret[4].type = C_END;
    ret[4].sval = NULL;
    ret[4].ival = 0;

    return ret;
}

static game_params *custom_params(config_item *cfg)
{
    game_params *ret = snew(game_params);

    ret->w = atoi(cfg[0].sval);
    ret->h = atoi(cfg[1].sval);
    ret->n = atoi(cfg[2].sval);
    if (strchr(cfg[2].sval, '%'))
        ret->n = ret->n * (ret->w * ret->h) / 100;
    ret->unique = cfg[3].ival;

    return ret;
}

static char *validate_params(game_params *params)
{
    /*
     * Lower limit on grid size: each dimension must be at least 3.
     * 1 is theoretically workable if rather boring, but 2 is a
     * real problem: there is often _no_ way to generate a uniquely
     * solvable 2xn Mines grid. You either run into two mines
     * blocking the way and no idea what's behind them, or one mine
     * and no way to know which of the two rows it's in. If the
     * mine count is even you can create a soluble grid by packing
     * all the mines at one end (so what when you hit a two-mine
     * wall there are only as many covered squares left as there
     * are mines); but if it's odd, you are doomed, because you
     * _have_ to have a gap somewhere which you can't determine the
     * position of.
     */
    if (params->w <= 2 || params->h <= 2)
        return "Width and height must both be greater than two";
    if (params->n > params->w * params->h - 9)
        return "Too many mines for grid size";

    /*
     * FIXME: Need more constraints here. Not sure what the
     * sensible limits for Minesweeper actually are. The limits
     * probably ought to change, however, depending on uniqueness.
     */

    return NULL;
}

/* ----------------------------------------------------------------------
 * Minesweeper solver, used to ensure the generated grids are
 * solvable without having to take risks.
 */

/*
 * Count the bits in a word. Only needs to cope with 16 bits.
 */
static int bitcount16(int word)
{
    word = ((word & 0xAAAA) >> 1) + (word & 0x5555);
    word = ((word & 0xCCCC) >> 2) + (word & 0x3333);
    word = ((word & 0xF0F0) >> 4) + (word & 0x0F0F);
    word = ((word & 0xFF00) >> 8) + (word & 0x00FF);

    return word;
}

/*
 * We use a tree234 to store a large number of small localised
 * sets, each with a mine count. We also keep some of those sets
 * linked together into a to-do list.
 */
struct set {
    short x, y, mask, mines;
    int todo;
    struct set *prev, *next;
};

static int setcmp(void *av, void *bv)
{
    struct set *a = (struct set *)av;
    struct set *b = (struct set *)bv;

    if (a->y < b->y)
        return -1;
    else if (a->y > b->y)
        return +1;
    else if (a->x < b->x)
        return -1;
    else if (a->x > b->x)
        return +1;
    else if (a->mask < b->mask)
        return -1;
    else if (a->mask > b->mask)
        return +1;
    else
        return 0;
}

struct setstore {
    tree234 *sets;
    struct set *todo_head, *todo_tail;
};

static struct setstore *ss_new(void)
{
    struct setstore *ss = snew(struct setstore);
    ss->sets = newtree234(setcmp);
    ss->todo_head = ss->todo_tail = NULL;
    return ss;
}

/*
 * Take two input sets, in the form (x,y,mask). Munge the first by
 * taking either its intersection with the second or its difference
 * with the second. Return the new mask part of the first set.
 */
static int setmunge(int x1, int y1, int mask1, int x2, int y2, int mask2,
                    int diff)
{
    /*
     * Adjust the second set so that it has the same x,y
     * coordinates as the first.
     */
    if (abs(x2-x1) >= 3 || abs(y2-y1) >= 3) {
        mask2 = 0;
    } else {
        while (x2 > x1) {
            mask2 &= ~(4|32|256);
            mask2 <<= 1;
            x2--;
        }
        while (x2 < x1) {
            mask2 &= ~(1|8|64);
            mask2 >>= 1;
            x2++;
        }
        while (y2 > y1) {
            mask2 &= ~(64|128|256);
            mask2 <<= 3;
            y2--;
        }
        while (y2 < y1) {
            mask2 &= ~(1|2|4);
            mask2 >>= 3;
            y2++;
        }
    }

    /*
     * Invert the second set if `diff' is set (we're after A &~ B
     * rather than A & B).
     */
    if (diff)
        mask2 ^= 511;

    /*
     * Now all that's left is a logical AND.
     */
    return mask1 & mask2;
}

static void ss_add_todo(struct setstore *ss, struct set *s)
{
    if (s->todo)
        return;                        /* already on it */

#ifdef SOLVER_DIAGNOSTICS
    printf("adding set on todo list: %d,%d %03x %d\n",
           s->x, s->y, s->mask, s->mines);
#endif

    s->prev = ss->todo_tail;
    if (s->prev)
        s->prev->next = s;
    else
        ss->todo_head = s;
    ss->todo_tail = s;
    s->next = NULL;
    s->todo = TRUE;
}

static void ss_add(struct setstore *ss, int x, int y, int mask, int mines)
{
    struct set *s;

    assert(mask != 0);

    /*
     * Normalise so that x and y are genuinely the bounding
     * rectangle.
     */
    while (!(mask & (1|8|64)))
        mask >>= 1, x++;
    while (!(mask & (1|2|4)))
        mask >>= 3, y++;

    /*
     * Create a set structure and add it to the tree.
     */
    s = snew(struct set);
    s->x = x;
    s->y = y;
    s->mask = mask;
    s->mines = mines;
    s->todo = FALSE;
    if (add234(ss->sets, s) != s) {
        /*
         * This set already existed! Free it and return.
         */
        sfree(s);
        return;
    }

    /*
     * We've added a new set to the tree, so put it on the todo
     * list.
     */
    ss_add_todo(ss, s);
}

static void ss_remove(struct setstore *ss, struct set *s)
{
    struct set *next = s->next, *prev = s->prev;

#ifdef SOLVER_DIAGNOSTICS
    printf("removing set %d,%d %03x\n", s->x, s->y, s->mask);
#endif
    /*
     * Remove s from the todo list.
     */
    if (prev)
        prev->next = next;
    else if (s == ss->todo_head)
        ss->todo_head = next;

    if (next)
        next->prev = prev;
    else if (s == ss->todo_tail)
        ss->todo_tail = prev;

    s->todo = FALSE;

    /*
     * Remove s from the tree.
     */
    del234(ss->sets, s);

    /*
     * Destroy the actual set structure.
     */
    sfree(s);
}

/*
 * Return a dynamically allocated list of all the sets which
 * overlap a provided input set.
 */
static struct set **ss_overlap(struct setstore *ss, int x, int y, int mask)
{
    struct set **ret = NULL;
    int nret = 0, retsize = 0;
    int xx, yy;

    for (xx = x-3; xx < x+3; xx++)
        for (yy = y-3; yy < y+3; yy++) {
            struct set stmp, *s;
            int pos;

            /*
             * Find the first set with these top left coordinates.
             */
            stmp.x = xx;
            stmp.y = yy;
            stmp.mask = 0;

            if (findrelpos234(ss->sets, &stmp, NULL, REL234_GE, &pos)) {
                while ((s = index234(ss->sets, pos)) != NULL &&
                       s->x == xx && s->y == yy) {
                    /*
                     * This set potentially overlaps the input one.
                     * Compute the intersection to see if they
                     * really overlap, and add it to the list if
                     * so.
                     */
                    if (setmunge(x, y, mask, s->x, s->y, s->mask, FALSE)) {
                        /*
                         * There's an overlap.
                         */
                        if (nret >= retsize) {
                            retsize = nret + 32;
                            ret = sresize(ret, retsize, struct set *);
                        }
                        ret[nret++] = s;
                    }

                    pos++;
                }
            }
        }

    ret = sresize(ret, nret+1, struct set *);
    ret[nret] = NULL;

    return ret;
}

/*
 * Get an element from the head of the set todo list.
 */
static struct set *ss_todo(struct setstore *ss)
{
    if (ss->todo_head) {
        struct set *ret = ss->todo_head;
        ss->todo_head = ret->next;
        if (ss->todo_head)
            ss->todo_head->prev = NULL;
        else
            ss->todo_tail = NULL;
        ret->next = ret->prev = NULL;
        ret->todo = FALSE;
        return ret;
    } else {
        return NULL;
    }
}

struct squaretodo {
    int *next;
    int head, tail;
};

static void std_add(struct squaretodo *std, int i)
{
    if (std->tail >= 0)
        std->next[std->tail] = i;
    else
        std->head = i;
    std->tail = i;
    std->next[i] = -1;
}

static void known_squares(int w, int h, struct squaretodo *std,
                          signed char *grid,
                          int (*open)(void *ctx, int x, int y), void *openctx,
                          int x, int y, int mask, int mine)
{
    int xx, yy, bit;

    bit = 1;

    for (yy = 0; yy < 3; yy++)
        for (xx = 0; xx < 3; xx++) {
            if (mask & bit) {
                int i = (y + yy) * w + (x + xx);

                /*
                 * It's possible that this square is _already_
                 * known, in which case we don't try to add it to
                 * the list twice.
                 */
                if (grid[i] == -2) {

                    if (mine) {
                        grid[i] = -1;   /* and don't open it! */
                    } else {
                        grid[i] = open(openctx, x + xx, y + yy);
                        assert(grid[i] != -1);   /* *bang* */
                    }
                    std_add(std, i);

                }
            }
            bit <<= 1;
        }
}

/*
 * This is data returned from the `perturb' function. It details
 * which squares have become mines and which have become clear. The
 * solver is (of course) expected to honourably not use that
 * knowledge directly, but to efficently adjust its internal data
 * structures and proceed based on only the information it
 * legitimately has.
 */
struct perturbation {
    int x, y;
    int delta;                         /* +1 == become a mine; -1 == cleared */
};
struct perturbations {
    int n;
    struct perturbation *changes;
};

/*
 * Main solver entry point. You give it a grid of existing
 * knowledge (-1 for a square known to be a mine, 0-8 for empty
 * squares with a given number of neighbours, -2 for completely
 * unknown), plus a function which you can call to open new squares
 * once you're confident of them. It fills in as much more of the
 * grid as it can.
 * 
 * Return value is:
 * 
 *  - -1 means deduction stalled and nothing could be done
 *  - 0 means deduction succeeded fully
 *  - >0 means deduction succeeded but some number of perturbation
 *    steps were required; the exact return value is the number of
 *    perturb calls.
 */
static int minesolve(int w, int h, int n, signed char *grid,
                     int (*open)(void *ctx, int x, int y),
                     struct perturbations *(*perturb)(void *ctx,
                                                      signed char *grid,
                                                      int x, int y, int mask),
                     void *ctx, random_state *rs)
{
    struct setstore *ss = ss_new();
    struct set **list;
    struct squaretodo astd, *std = &astd;
    int x, y, i, j;
    int nperturbs = 0;

    /*
     * Set up a linked list of squares with known contents, so that
     * we can process them one by one.
     */
    std->next = snewn(w*h, int);
    std->head = std->tail = -1;

    /*
     * Initialise that list with all known squares in the input
     * grid.
     */
    for (y = 0; y < h; y++) {
        for (x = 0; x < w; x++) {
            i = y*w+x;
            if (grid[i] != -2)
                std_add(std, i);
        }
    }

    /*
     * Main deductive loop.
     */
    while (1) {
        int done_something = FALSE;
        struct set *s;

        /*
         * If there are any known squares on the todo list, process
         * them and construct a set for each.
         */
        while (std->head != -1) {
            i = std->head;
#ifdef SOLVER_DIAGNOSTICS
            printf("known square at %d,%d [%d]\n", i%w, i/w, grid[i]);
#endif
            std->head = std->next[i];
            if (std->head == -1)
                std->tail = -1;

            x = i % w;
            y = i / w;

            if (grid[i] >= 0) {
                int dx, dy, mines, bit, val;
#ifdef SOLVER_DIAGNOSTICS
                printf("creating set around this square\n");
#endif
                /*
                 * Empty square. Construct the set of non-known squares
                 * around this one, and determine its mine count.
                 */
                mines = grid[i];
                bit = 1;
                val = 0;
                for (dy = -1; dy <= +1; dy++) {
                    for (dx = -1; dx <= +1; dx++) {
#ifdef SOLVER_DIAGNOSTICS
                        printf("grid %d,%d = %d\n", x+dx, y+dy, grid[i+dy*w+dx]);
#endif
                        if (x+dx < 0 || x+dx >= w || y+dy < 0 || y+dy >= h)
                            /* ignore this one */;
                        else if (grid[i+dy*w+dx] == -1)
                            mines--;
                        else if (grid[i+dy*w+dx] == -2)
                            val |= bit;
                        bit <<= 1;
                    }
                }
                if (val)
                    ss_add(ss, x-1, y-1, val, mines);
            }

            /*
             * Now, whether the square is empty or full, we must
             * find any set which contains it and replace it with
             * one which does not.
             */
            {
#ifdef SOLVER_DIAGNOSTICS
                printf("finding sets containing known square %d,%d\n", x, y);
#endif
                list = ss_overlap(ss, x, y, 1);

                for (j = 0; list[j]; j++) {
                    int newmask, newmines;

                    s = list[j];

                    /*
                     * Compute the mask for this set minus the
                     * newly known square.
                     */
                    newmask = setmunge(s->x, s->y, s->mask, x, y, 1, TRUE);

                    /*
                     * Compute the new mine count.
                     */
                    newmines = s->mines - (grid[i] == -1);

                    /*
                     * Insert the new set into the collection,
                     * unless it's been whittled right down to
                     * nothing.
                     */
                    if (newmask)
                        ss_add(ss, s->x, s->y, newmask, newmines);

                    /*
                     * Destroy the old one; it is actually obsolete.
                     */
                    ss_remove(ss, s);
                }

                sfree(list);
            }

            /*
             * Marking a fresh square as known certainly counts as
             * doing something.
             */
            done_something = TRUE;
        }

        /*
         * Now pick a set off the to-do list and attempt deductions
         * based on it.
         */
        if ((s = ss_todo(ss)) != NULL) {

#ifdef SOLVER_DIAGNOSTICS
            printf("set to do: %d,%d %03x %d\n", s->x, s->y, s->mask, s->mines);
#endif
            /*
             * Firstly, see if this set has a mine count of zero or
             * of its own cardinality.
             */
            if (s->mines == 0 || s->mines == bitcount16(s->mask)) {
                /*
                 * If so, we can immediately mark all the squares
                 * in the set as known.
                 */
#ifdef SOLVER_DIAGNOSTICS
                printf("easy\n");
#endif
                known_squares(w, h, std, grid, open, ctx,
                              s->x, s->y, s->mask, (s->mines != 0));

                /*
                 * Having done that, we need do nothing further
                 * with this set; marking all the squares in it as
                 * known will eventually eliminate it, and will
                 * also permit further deductions about anything
                 * that overlaps it.
                 */
                continue;
            }

            /*
             * Failing that, we now search through all the sets
             * which overlap this one.
             */
            list = ss_overlap(ss, s->x, s->y, s->mask);

            for (j = 0; list[j]; j++) {
                struct set *s2 = list[j];
                int swing, s2wing, swc, s2wc;

                /*
                 * Find the non-overlapping parts s2-s and s-s2,
                 * and their cardinalities.
                 * 
                 * I'm going to refer to these parts as `wings'
                 * surrounding the central part common to both
                 * sets. The `s wing' is s-s2; the `s2 wing' is
                 * s2-s.
                 */
                swing = setmunge(s->x, s->y, s->mask, s2->x, s2->y, s2->mask,
                                 TRUE);
                s2wing = setmunge(s2->x, s2->y, s2->mask, s->x, s->y, s->mask,
                                 TRUE);
                swc = bitcount16(swing);
                s2wc = bitcount16(s2wing);

                /*
                 * If one set has more mines than the other, and
                 * the number of extra mines is equal to the
                 * cardinality of that set's wing, then we can mark
                 * every square in the wing as a known mine, and
                 * every square in the other wing as known clear.
                 */
                if (swc == s->mines - s2->mines ||
                    s2wc == s2->mines - s->mines) {
                    known_squares(w, h, std, grid, open, ctx,
                                  s->x, s->y, swing,
                                  (swc == s->mines - s2->mines));
                    known_squares(w, h, std, grid, open, ctx,
                                  s2->x, s2->y, s2wing,
                                  (s2wc == s2->mines - s->mines));
                    continue;
                }

                /*
                 * Failing that, see if one set is a subset of the
                 * other. If so, we can divide up the mine count of
                 * the larger set between the smaller set and its
                 * complement, even if neither smaller set ends up
                 * being immediately clearable.
                 */
                if (swc == 0 && s2wc != 0) {
                    /* s is a subset of s2. */
                    assert(s2->mines > s->mines);
                    ss_add(ss, s2->x, s2->y, s2wing, s2->mines - s->mines);
                } else if (s2wc == 0 && swc != 0) {
                    /* s2 is a subset of s. */
                    assert(s->mines > s2->mines);
                    ss_add(ss, s->x, s->y, swing, s->mines - s2->mines);
                }
            }

            sfree(list);

            /*
             * In this situation we have definitely done
             * _something_, even if it's only reducing the size of
             * our to-do list.
             */
            done_something = TRUE;
        } else if (n >= 0) {
            /*
             * We have nothing left on our todo list, which means
             * all localised deductions have failed. Our next step
             * is to resort to global deduction based on the total
             * mine count. This is computationally expensive
             * compared to any of the above deductions, which is
             * why we only ever do it when all else fails, so that
             * hopefully it won't have to happen too often.
             * 
             * If you pass n<0 into this solver, that informs it
             * that you do not know the total mine count, so it
             * won't even attempt these deductions.
             */

            int minesleft, squaresleft;
            int nsets, setused[10], cursor;

            /*
             * Start by scanning the current grid state to work out
             * how many unknown squares we still have, and how many
             * mines are to be placed in them.
             */
            squaresleft = 0;
            minesleft = n;
            for (i = 0; i < w*h; i++) {
                if (grid[i] == -1)
                    minesleft--;
                else if (grid[i] == -2)
                    squaresleft++;
            }

#ifdef SOLVER_DIAGNOSTICS
            printf("global deduction time: squaresleft=%d minesleft=%d\n",
                   squaresleft, minesleft);
            for (y = 0; y < h; y++) {
                for (x = 0; x < w; x++) {
                    int v = grid[y*w+x];
                    if (v == -1)
                        putchar('*');
                    else if (v == -2)
                        putchar('?');
                    else if (v == 0)
                        putchar('-');
                    else
                        putchar('0' + v);
                }
                putchar('\n');
            }
#endif

            /*
             * If there _are_ no unknown squares, we have actually
             * finished.
             */
            if (squaresleft == 0) {
                assert(minesleft == 0);
                break;
            }

            /*
             * First really simple case: if there are no more mines
             * left, or if there are exactly as many mines left as
             * squares to play them in, then it's all easy.
             */
            if (minesleft == 0 || minesleft == squaresleft) {
                for (i = 0; i < w*h; i++)
                    if (grid[i] == -2)
                        known_squares(w, h, std, grid, open, ctx,
                                      i % w, i / w, 1, minesleft != 0);
                continue;              /* now go back to main deductive loop */
            }

            /*
             * Failing that, we have to do some _real_ work.
             * Ideally what we do here is to try every single
             * combination of the currently available sets, in an
             * attempt to find a disjoint union (i.e. a set of
             * squares with a known mine count between them) such
             * that the remaining unknown squares _not_ contained
             * in that union either contain no mines or are all
             * mines.
             * 
             * Actually enumerating all 2^n possibilities will get
             * a bit slow for large n, so I artificially cap this
             * recursion at n=10 to avoid too much pain.
             */
            nsets = count234(ss->sets);
            if (nsets <= lenof(setused)) {
                /*
                 * Doing this with actual recursive function calls
                 * would get fiddly because a load of local
                 * variables from this function would have to be
                 * passed down through the recursion. So instead
                 * I'm going to use a virtual recursion within this
                 * function. The way this works is:
                 * 
                 *  - we have an array `setused', such that
                 *    setused[n] is 0 or 1 depending on whether set
                 *    n is currently in the union we are
                 *    considering.
                 * 
                 *  - we have a value `cursor' which indicates how
                 *    much of `setused' we have so far filled in.
                 *    It's conceptually the recursion depth.
                 * 
                 * We begin by setting `cursor' to zero. Then:
                 * 
                 *  - if cursor can advance, we advance it by one.
                 *    We set the value in `setused' that it went
                 *    past to 1 if that set is disjoint from
                 *    anything else currently in `setused', or to 0
                 *    otherwise.
                 * 
                 *  - If cursor cannot advance because it has
                 *    reached the end of the setused list, then we
                 *    have a maximal disjoint union. Check to see
                 *    whether its mine count has any useful
                 *    properties. If so, mark all the squares not
                 *    in the union as known and terminate.
                 * 
                 *  - If cursor has reached the end of setused and
                 *    the algorithm _hasn't_ terminated, back
                 *    cursor up to the nearest 1, turn it into a 0
                 *    and advance cursor just past it.
                 * 
                 *  - If we attempt to back up to the nearest 1 and
                 *    there isn't one at all, then we have gone
                 *    through all disjoint unions of sets in the
                 *    list and none of them has been helpful, so we
                 *    give up.
                 */
                struct set *sets[lenof(setused)];
                for (i = 0; i < nsets; i++)
                    sets[i] = index234(ss->sets, i);

                cursor = 0;
                while (1) {

                    if (cursor < nsets) {
                        int ok = TRUE;

                        /* See if any existing set overlaps this one. */
                        for (i = 0; i < cursor; i++)
                            if (setused[i] &&
                                setmunge(sets[cursor]->x,
                                         sets[cursor]->y,
                                         sets[cursor]->mask,
                                         sets[i]->x, sets[i]->y, sets[i]->mask,
                                         FALSE)) {
                                ok = FALSE;
                                break;
                            }

                        if (ok) {
                            /*
                             * We're adding this set to our union,
                             * so adjust minesleft and squaresleft
                             * appropriately.
                             */
                            minesleft -= sets[cursor]->mines;
                            squaresleft -= bitcount16(sets[cursor]->mask);
                        }

                        setused[cursor++] = ok;
                    } else {
#ifdef SOLVER_DIAGNOSTICS
                        printf("trying a set combination with %d %d\n",
                               squaresleft, minesleft);
#endif /* SOLVER_DIAGNOSTICS */

                        /*
                         * We've reached the end. See if we've got
                         * anything interesting.
                         */
                        if (squaresleft > 0 &&
                            (minesleft == 0 || minesleft == squaresleft)) {
                            /*
                             * We have! There is at least one
                             * square not contained within the set
                             * union we've just found, and we can
                             * deduce that either all such squares
                             * are mines or all are not (depending
                             * on whether minesleft==0). So now all
                             * we have to do is actually go through
                             * the grid, find those squares, and
                             * mark them.
                             */
                            for (i = 0; i < w*h; i++)
                                if (grid[i] == -2) {
                                    int outside = TRUE;
                                    y = i / w;
                                    x = i % w;
                                    for (j = 0; j < nsets; j++)
                                        if (setused[j] &&
                                            setmunge(sets[j]->x, sets[j]->y,
                                                     sets[j]->mask, x, y, 1,
                                                     FALSE)) {
                                            outside = FALSE;
                                            break;
                                        }
                                    if (outside)
                                        known_squares(w, h, std, grid,
                                                      open, ctx,
                                                      x, y, 1, minesleft != 0);
                                }

                            done_something = TRUE;
                            break;     /* return to main deductive loop */
                        }

                        /*
                         * If we reach here, then this union hasn't
                         * done us any good, so move on to the
                         * next. Backtrack cursor to the nearest 1,
                         * change it to a 0 and continue.
                         */
                        while (--cursor >= 0 && !setused[cursor]);
                        if (cursor >= 0) {
                            assert(setused[cursor]);

                            /*
                             * We're removing this set from our
                             * union, so re-increment minesleft and
                             * squaresleft.
                             */
                            minesleft += sets[cursor]->mines;
                            squaresleft += bitcount16(sets[cursor]->mask);

                            setused[cursor++] = 0;
                        } else {
                            /*
                             * We've backtracked all the way to the
                             * start without finding a single 1,
                             * which means that our virtual
                             * recursion is complete and nothing
                             * helped.
                             */
                            break;
                        }
                    }

                }

            }
        }

        if (done_something)
            continue;

#ifdef SOLVER_DIAGNOSTICS
        /*
         * Dump the current known state of the grid.
         */
        printf("solver ran out of steam, ret=%d, grid:\n", nperturbs);
        for (y = 0; y < h; y++) {
            for (x = 0; x < w; x++) {
                int v = grid[y*w+x];
                if (v == -1)
                    putchar('*');
                else if (v == -2)
                    putchar('?');
                else if (v == 0)
                    putchar('-');
                else
                    putchar('0' + v);
            }
            putchar('\n');
        }

        {
            struct set *s;

            for (i = 0; (s = index234(ss->sets, i)) != NULL; i++)
                printf("remaining set: %d,%d %03x %d\n", s->x, s->y, s->mask, s->mines);
        }
#endif

        /*
         * Now we really are at our wits' end as far as solving
         * this grid goes. Our only remaining option is to call
         * a perturb function and ask it to modify the grid to
         * make it easier.
         */
        if (perturb) {
            struct perturbations *ret;
            struct set *s;

            nperturbs++;

            /*
             * Choose a set at random from the current selection,
             * and ask the perturb function to either fill or empty
             * it.
             * 
             * If we have no sets at all, we must give up.
             */
            if (count234(ss->sets) == 0) {
#ifdef SOLVER_DIAGNOSTICS
                printf("perturbing on entire unknown set\n");
#endif
                ret = perturb(ctx, grid, 0, 0, 0);
            } else {
                s = index234(ss->sets, random_upto(rs, count234(ss->sets)));
#ifdef SOLVER_DIAGNOSTICS
                printf("perturbing on set %d,%d %03x\n", s->x, s->y, s->mask);
#endif
                ret = perturb(ctx, grid, s->x, s->y, s->mask);
            }

            if (ret) {
                assert(ret->n > 0);    /* otherwise should have been NULL */

                /*
                 * A number of squares have been fiddled with, and
                 * the returned structure tells us which. Adjust
                 * the mine count in any set which overlaps one of
                 * those squares, and put them back on the to-do
                 * list. Also, if the square itself is marked as a
                 * known non-mine, put it back on the squares-to-do
                 * list.
                 */
                for (i = 0; i < ret->n; i++) {
#ifdef SOLVER_DIAGNOSTICS
                    printf("perturbation %s mine at %d,%d\n",
                           ret->changes[i].delta > 0 ? "added" : "removed",
                           ret->changes[i].x, ret->changes[i].y);
#endif

                    if (ret->changes[i].delta < 0 &&
                        grid[ret->changes[i].y*w+ret->changes[i].x] != -2) {
                        std_add(std, ret->changes[i].y*w+ret->changes[i].x);
                    }

                    list = ss_overlap(ss,
                                      ret->changes[i].x, ret->changes[i].y, 1);

                    for (j = 0; list[j]; j++) {
                        list[j]->mines += ret->changes[i].delta;
                        ss_add_todo(ss, list[j]);
                    }

                    sfree(list);
                }

                /*
                 * Now free the returned data.
                 */
                sfree(ret->changes);
                sfree(ret);

#ifdef SOLVER_DIAGNOSTICS
                /*
                 * Dump the current known state of the grid.
                 */
                printf("state after perturbation:\n");
                for (y = 0; y < h; y++) {
                    for (x = 0; x < w; x++) {
                        int v = grid[y*w+x];
                        if (v == -1)
                            putchar('*');
                        else if (v == -2)
                            putchar('?');
                        else if (v == 0)
                            putchar('-');
                        else
                            putchar('0' + v);
                    }
                    putchar('\n');
                }

                {
                    struct set *s;

                    for (i = 0; (s = index234(ss->sets, i)) != NULL; i++)
                        printf("remaining set: %d,%d %03x %d\n", s->x, s->y, s->mask, s->mines);
                }
#endif

                /*
                 * And now we can go back round the deductive loop.
                 */
                continue;
            }
        }

        /*
         * If we get here, even that didn't work (either we didn't
         * have a perturb function or it returned failure), so we
         * give up entirely.
         */
        break;
    }

    /*
     * See if we've got any unknown squares left.
     */
    for (y = 0; y < h; y++)
        for (x = 0; x < w; x++)
            if (grid[y*w+x] == -2) {
                nperturbs = -1;        /* failed to complete */
                break;
            }

    /*
     * Free the set list and square-todo list.
     */
    {
        struct set *s;
        while ((s = delpos234(ss->sets, 0)) != NULL)
            sfree(s);
        freetree234(ss->sets);
        sfree(ss);
        sfree(std->next);
    }

    return nperturbs;
}

/* ----------------------------------------------------------------------
 * Grid generator which uses the above solver.
 */

struct minectx {
    signed char *grid;
    int w, h;
    int sx, sy;
    int allow_big_perturbs;
    random_state *rs;
};

static int mineopen(void *vctx, int x, int y)
{
    struct minectx *ctx = (struct minectx *)vctx;
    int i, j, n;

    assert(x >= 0 && x < ctx->w && y >= 0 && y < ctx->h);
    if (ctx->grid[y * ctx->w + x])
        return -1;                     /* *bang* */

    n = 0;
    for (i = -1; i <= +1; i++) {
        if (x + i < 0 || x + i >= ctx->w)
            continue;
        for (j = -1; j <= +1; j++) {
            if (y + j < 0 || y + j >= ctx->h)
                continue;
            if (i == 0 && j == 0)
                continue;
            if (ctx->grid[(y+j) * ctx->w + (x+i)])
                n++;
        }
    }

    return n;
}

/* Structure used internally to mineperturb(). */
struct square {
    int x, y, type, random;
};
static int squarecmp(const void *av, const void *bv)
{
    const struct square *a = (const struct square *)av;
    const struct square *b = (const struct square *)bv;
    if (a->type < b->type)
        return -1;
    else if (a->type > b->type)
        return +1;
    else if (a->random < b->random)
        return -1;
    else if (a->random > b->random)
        return +1;
    else if (a->y < b->y)
        return -1;
    else if (a->y > b->y)
        return +1;
    else if (a->x < b->x)
        return -1;
    else if (a->x > b->x)
        return +1;
    return 0;
}

/*
 * Normally this function is passed an (x,y,mask) set description.
 * On occasions, though, there is no _localised_ set being used,
 * and the set being perturbed is supposed to be the entirety of
 * the unreachable area. This is signified by the special case
 * mask==0: in this case, anything labelled -2 in the grid is part
 * of the set.
 * 
 * Allowing perturbation in this special case appears to make it
 * guaranteeably possible to generate a workable grid for any mine
 * density, but they tend to be a bit boring, with mines packed
 * densely into far corners of the grid and the remainder being
 * less dense than one might like. Therefore, to improve overall
 * grid quality I disable this feature for the first few attempts,
 * and fall back to it after no useful grid has been generated.
 */
static struct perturbations *mineperturb(void *vctx, signed char *grid,
                                         int setx, int sety, int mask)
{
    struct minectx *ctx = (struct minectx *)vctx;
    struct square *sqlist;
    int x, y, dx, dy, i, n, nfull, nempty;
    struct square **tofill, **toempty, **todo;
    int ntofill, ntoempty, ntodo, dtodo, dset;
    struct perturbations *ret;
    int *setlist;

    if (!mask && !ctx->allow_big_perturbs)
        return NULL;

    /*
     * Make a list of all the squares in the grid which we can
     * possibly use. This list should be in preference order, which
     * means
     * 
     *  - first, unknown squares on the boundary of known space
     *  - next, unknown squares beyond that boundary
     *  - as a very last resort, known squares, but not within one
     *    square of the starting position.
     * 
     * Each of these sections needs to be shuffled independently.
     * We do this by preparing list of all squares and then sorting
     * it with a random secondary key.
     */
    sqlist = snewn(ctx->w * ctx->h, struct square);
    n = 0;
    for (y = 0; y < ctx->h; y++)
        for (x = 0; x < ctx->w; x++) {
            /*
             * If this square is too near the starting position,
             * don't put it on the list at all.
             */
            if (abs(y - ctx->sy) <= 1 && abs(x - ctx->sx) <= 1)
                continue;

            /*
             * If this square is in the input set, also don't put
             * it on the list!
             */
            if ((mask == 0 && grid[y*ctx->w+x] == -2) ||
                (x >= setx && x < setx + 3 &&
                 y >= sety && y < sety + 3 &&
                 mask & (1 << ((y-sety)*3+(x-setx)))))
                continue;

            sqlist[n].x = x;
            sqlist[n].y = y;

            if (grid[y*ctx->w+x] != -2) {
                sqlist[n].type = 3;    /* known square */
            } else {
                /*
                 * Unknown square. Examine everything around it and
                 * see if it borders on any known squares. If it
                 * does, it's class 1, otherwise it's 2.
                 */

                sqlist[n].type = 2;

                for (dy = -1; dy <= +1; dy++)
                    for (dx = -1; dx <= +1; dx++)
                        if (x+dx >= 0 && x+dx < ctx->w &&
                            y+dy >= 0 && y+dy < ctx->h &&
                            grid[(y+dy)*ctx->w+(x+dx)] != -2) {
                            sqlist[n].type = 1;
                            break;
                        }
            }

            /*
             * Finally, a random number to cause qsort to
             * shuffle within each group.
             */
            sqlist[n].random = random_bits(ctx->rs, 31);

            n++;
        }

    qsort(sqlist, n, sizeof(struct square), squarecmp);

    /*
     * Now count up the number of full and empty squares in the set
     * we've been provided.
     */
    nfull = nempty = 0;
    if (mask) {
        for (dy = 0; dy < 3; dy++)
            for (dx = 0; dx < 3; dx++)
                if (mask & (1 << (dy*3+dx))) {
                    assert(setx+dx <= ctx->w);
                    assert(sety+dy <= ctx->h);
                    if (ctx->grid[(sety+dy)*ctx->w+(setx+dx)])
                        nfull++;
                    else
                        nempty++;
                }
    } else {
        for (y = 0; y < ctx->h; y++)
            for (x = 0; x < ctx->w; x++)
                if (grid[y*ctx->w+x] == -2) {
                    if (ctx->grid[y*ctx->w+x])
                        nfull++;
                    else
                        nempty++;
                }
    }

    /*
     * Now go through our sorted list until we find either `nfull'
     * empty squares, or `nempty' full squares; these will be
     * swapped with the appropriate squares in the set to either
     * fill or empty the set while keeping the same number of mines
     * overall.
     */
    ntofill = ntoempty = 0;
    if (mask) {
        tofill = snewn(9, struct square *);
        toempty = snewn(9, struct square *);
    } else {
        tofill = snewn(ctx->w * ctx->h, struct square *);
        toempty = snewn(ctx->w * ctx->h, struct square *);
    }
    for (i = 0; i < n; i++) {
        struct square *sq = &sqlist[i];
        if (ctx->grid[sq->y * ctx->w + sq->x])
            toempty[ntoempty++] = sq;
        else
            tofill[ntofill++] = sq;
        if (ntofill == nfull || ntoempty == nempty)
            break;
    }

    /*
     * If we haven't found enough empty squares outside the set to
     * empty it into _or_ enough full squares outside it to fill it
     * up with, we'll have to settle for doing only a partial job.
     * In this case we choose to always _fill_ the set (because
     * this case will tend to crop up when we're working with very
     * high mine densities and the only way to get a solvable grid
     * is going to be to pack most of the mines solidly around the
     * edges). So now our job is to make a list of the empty
     * squares in the set, and shuffle that list so that we fill a
     * random selection of them.
     */
    if (ntofill != nfull && ntoempty != nempty) {
        int k;

        assert(ntoempty != 0);

        setlist = snewn(ctx->w * ctx->h, int);
        i = 0;
        if (mask) {
            for (dy = 0; dy < 3; dy++)
                for (dx = 0; dx < 3; dx++)
                    if (mask & (1 << (dy*3+dx))) {
                        assert(setx+dx <= ctx->w);
                        assert(sety+dy <= ctx->h);
                        if (!ctx->grid[(sety+dy)*ctx->w+(setx+dx)])
                            setlist[i++] = (sety+dy)*ctx->w+(setx+dx);
                    }
        } else {
            for (y = 0; y < ctx->h; y++)
                for (x = 0; x < ctx->w; x++)
                    if (grid[y*ctx->w+x] == -2) {
                        if (!ctx->grid[y*ctx->w+x])
                            setlist[i++] = y*ctx->w+x;
                    }
        }
        assert(i > ntoempty);
        /*
         * Now pick `ntoempty' items at random from the list.
         */
        for (k = 0; k < ntoempty; k++) {
            int index = k + random_upto(ctx->rs, i - k);
            int tmp;

            tmp = setlist[k];
            setlist[k] = setlist[index];
            setlist[index] = tmp;
        }
    } else
        setlist = NULL;

    /*
     * Now we're pretty much there. We need to either
     *  (a) put a mine in each of the empty squares in the set, and
     *      take one out of each square in `toempty'
     *  (b) take a mine out of each of the full squares in the set,
     *      and put one in each square in `tofill'
     * depending on which one we've found enough squares to do.
     * 
     * So we start by constructing our list of changes to return to
     * the solver, so that it can update its data structures
     * efficiently rather than having to rescan the whole grid.
     */
    ret = snew(struct perturbations);
    if (ntofill == nfull) {
        todo = tofill;
        ntodo = ntofill;
        dtodo = +1;
        dset = -1;
        sfree(toempty);
    } else {
        /*
         * (We also fall into this case if we've constructed a
         * setlist.)
         */
        todo = toempty;
        ntodo = ntoempty;
        dtodo = -1;
        dset = +1;
        sfree(tofill);
    }
    ret->n = 2 * ntodo;
    ret->changes = snewn(ret->n, struct perturbation);
    for (i = 0; i < ntodo; i++) {
        ret->changes[i].x = todo[i]->x;
        ret->changes[i].y = todo[i]->y;
        ret->changes[i].delta = dtodo;
    }
    /* now i == ntodo */
    if (setlist) {
        int j;
        assert(todo == toempty);
        for (j = 0; j < ntoempty; j++) {
            ret->changes[i].x = setlist[j] % ctx->w;
            ret->changes[i].y = setlist[j] / ctx->w;
            ret->changes[i].delta = dset;
            i++;
        }
        sfree(setlist);
    } else if (mask) {
        for (dy = 0; dy < 3; dy++)
            for (dx = 0; dx < 3; dx++)
                if (mask & (1 << (dy*3+dx))) {
                    int currval = (ctx->grid[(sety+dy)*ctx->w+(setx+dx)] ? +1 : -1);
                    if (dset == -currval) {
                        ret->changes[i].x = setx + dx;
                        ret->changes[i].y = sety + dy;
                        ret->changes[i].delta = dset;
                        i++;
                    }
                }
    } else {
        for (y = 0; y < ctx->h; y++)
            for (x = 0; x < ctx->w; x++)
                if (grid[y*ctx->w+x] == -2) {
                    int currval = (ctx->grid[y*ctx->w+x] ? +1 : -1);
                    if (dset == -currval) {
                        ret->changes[i].x = x;
                        ret->changes[i].y = y;
                        ret->changes[i].delta = dset;
                        i++;
                    }
                }
    }
    assert(i == ret->n);

    sfree(sqlist);
    sfree(todo);

    /*
     * Having set up the precise list of changes we're going to
     * make, we now simply make them and return.
     */
    for (i = 0; i < ret->n; i++) {
        int delta;

        x = ret->changes[i].x;
        y = ret->changes[i].y;
        delta = ret->changes[i].delta;

        /*
         * Check we're not trying to add an existing mine or remove
         * an absent one.
         */
        assert((delta < 0) ^ (ctx->grid[y*ctx->w+x] == 0));

        /*
         * Actually make the change.
         */
        ctx->grid[y*ctx->w+x] = (delta > 0);

        /*
         * Update any numbers already present in the grid.
         */
        for (dy = -1; dy <= +1; dy++)
            for (dx = -1; dx <= +1; dx++)
                if (x+dx >= 0 && x+dx < ctx->w &&
                    y+dy >= 0 && y+dy < ctx->h &&
                    grid[(y+dy)*ctx->w+(x+dx)] != -2) {
                    if (dx == 0 && dy == 0) {
                        /*
                         * The square itself is marked as known in
                         * the grid. Mark it as a mine if it's a
                         * mine, or else work out its number.
                         */
                        if (delta > 0) {
                            grid[y*ctx->w+x] = -1;
                        } else {
                            int dx2, dy2, minecount = 0;
                            for (dy2 = -1; dy2 <= +1; dy2++)
                                for (dx2 = -1; dx2 <= +1; dx2++)
                                    if (x+dx2 >= 0 && x+dx2 < ctx->w &&
                                        y+dy2 >= 0 && y+dy2 < ctx->h &&
                                        ctx->grid[(y+dy2)*ctx->w+(x+dx2)])
                                        minecount++;
                            grid[y*ctx->w+x] = minecount;
                        }
                    } else {
                        if (grid[(y+dy)*ctx->w+(x+dx)] >= 0)
                            grid[(y+dy)*ctx->w+(x+dx)] += delta;
                    }
                }
    }

#ifdef GENERATION_DIAGNOSTICS
    {
        int yy, xx;
        printf("grid after perturbing:\n");
        for (yy = 0; yy < ctx->h; yy++) {
            for (xx = 0; xx < ctx->w; xx++) {
                int v = ctx->grid[yy*ctx->w+xx];
                if (yy == ctx->sy && xx == ctx->sx) {
                    assert(!v);
                    putchar('S');
                } else if (v) {
                    putchar('*');
                } else {
                    putchar('-');
                }
            }
            putchar('\n');
        }
        printf("\n");
    }
#endif

    return ret;
}

static char *minegen(int w, int h, int n, int x, int y, int unique,
                     random_state *rs)
{
    char *ret = snewn(w*h, char);
    int success;
    int ntries = 0;

    do {
        success = FALSE;
        ntries++;

        memset(ret, 0, w*h);

        /*
         * Start by placing n mines, none of which is at x,y or within
         * one square of it.
         */
        {
            int *tmp = snewn(w*h, int);
            int i, j, k, nn;

            /*
             * Write down the list of possible mine locations.
             */
            k = 0;
            for (i = 0; i < h; i++)
                for (j = 0; j < w; j++)
                    if (abs(i - y) > 1 || abs(j - x) > 1)
                        tmp[k++] = i*w+j;

            /*
             * Now pick n off the list at random.
             */
            nn = n;
            while (nn-- > 0) {
                i = random_upto(rs, k);
                ret[tmp[i]] = 1;
                tmp[i] = tmp[--k];
            }

            sfree(tmp);
        }

#ifdef GENERATION_DIAGNOSTICS
        {
            int yy, xx;
            printf("grid after initial generation:\n");
            for (yy = 0; yy < h; yy++) {
                for (xx = 0; xx < w; xx++) {
                    int v = ret[yy*w+xx];
                    if (yy == y && xx == x) {
                        assert(!v);
                        putchar('S');
                    } else if (v) {
                        putchar('*');
                    } else {
                        putchar('-');
                    }
                }
                putchar('\n');
            }
            printf("\n");
        }
#endif

        /*
         * Now set up a results grid to run the solver in, and a
         * context for the solver to open squares. Then run the solver
         * repeatedly; if the number of perturb steps ever goes up or
         * it ever returns -1, give up completely.
         *
         * We bypass this bit if we're not after a unique grid.
         */
        if (unique) {
            signed char *solvegrid = snewn(w*h, char);
            struct minectx actx, *ctx = &actx;
            int solveret, prevret = -2;

            ctx->grid = ret;
            ctx->w = w;
            ctx->h = h;
            ctx->sx = x;
            ctx->sy = y;
            ctx->rs = rs;
            ctx->allow_big_perturbs = (ntries > 100);

            while (1) {
                memset(solvegrid, -2, w*h);
                solvegrid[y*w+x] = mineopen(ctx, x, y);
                assert(solvegrid[y*w+x] == 0); /* by deliberate arrangement */

                solveret =
                    minesolve(w, h, n, solvegrid, mineopen, mineperturb, ctx, rs);
                if (solveret < 0 || (prevret >= 0 && solveret >= prevret)) {
                    success = FALSE;
                    break;
                } else if (solveret == 0) {
                    success = TRUE;
                    break;
                }
            }

            sfree(solvegrid);
        } else {
            success = TRUE;
        }

    } while (!success);

    return ret;
}

/*
 * The Mines game descriptions contain the location of every mine,
 * and can therefore be used to cheat.
 * 
 * It would be pointless to attempt to _prevent_ this form of
 * cheating by encrypting the description, since Mines is
 * open-source so anyone can find out the encryption key. However,
 * I think it is worth doing a bit of gentle obfuscation to prevent
 * _accidental_ spoilers: if you happened to note that the game ID
 * starts with an F, for example, you might be unable to put the
 * knowledge of those mines out of your mind while playing. So,
 * just as discussions of film endings are rot13ed to avoid
 * spoiling it for people who don't want to be told, we apply a
 * keyless, reversible, but visually completely obfuscatory masking
 * function to the mine bitmap.
 */
static void obfuscate_bitmap(unsigned char *bmp, int bits, int decode)
{
    int bytes, firsthalf, secondhalf;
    struct step {
        unsigned char *seedstart;
        int seedlen;
        unsigned char *targetstart;
        int targetlen;
    } steps[2];
    int i, j;

    /*
     * My obfuscation algorithm is similar in concept to the OAEP
     * encoding used in some forms of RSA. Here's a specification
     * of it:
     * 
     *  + We have a `masking function' which constructs a stream of
     *    pseudorandom bytes from a seed of some number of input
     *    bytes.
     * 
     *  + We pad out our input bit stream to a whole number of
     *    bytes by adding up to 7 zero bits on the end. (In fact
     *    the bitmap passed as input to this function will already
     *    have had this done in practice.)
     * 
     *  + We divide the _byte_ stream exactly in half, rounding the
     *    half-way position _down_. So an 81-bit input string, for
     *    example, rounds up to 88 bits or 11 bytes, and then
     *    dividing by two gives 5 bytes in the first half and 6 in
     *    the second half.
     * 
     *  + We generate a mask from the second half of the bytes, and
     *    XOR it over the first half.
     * 
     *  + We generate a mask from the (encoded) first half of the
     *    bytes, and XOR it over the second half. Any null bits at
     *    the end which were added as padding are cleared back to
     *    zero even if this operation would have made them nonzero.
     * 
     * To de-obfuscate, the steps are precisely the same except
     * that the final two are reversed.
     * 
     * Finally, our masking function. Given an input seed string of
     * bytes, the output mask consists of concatenating the SHA-1
     * hashes of the seed string and successive decimal integers,
     * starting from 0.
     */

    bytes = (bits + 7) / 8;
    firsthalf = bytes / 2;
    secondhalf = bytes - firsthalf;

    steps[decode ? 1 : 0].seedstart = bmp + firsthalf;
    steps[decode ? 1 : 0].seedlen = secondhalf;
    steps[decode ? 1 : 0].targetstart = bmp;
    steps[decode ? 1 : 0].targetlen = firsthalf;

    steps[decode ? 0 : 1].seedstart = bmp;
    steps[decode ? 0 : 1].seedlen = firsthalf;
    steps[decode ? 0 : 1].targetstart = bmp + firsthalf;
    steps[decode ? 0 : 1].targetlen = secondhalf;

    for (i = 0; i < 2; i++) {
        SHA_State base, final;
        unsigned char digest[20];
        char numberbuf[80];
        int digestpos = 20, counter = 0;

        SHA_Init(&base);
        SHA_Bytes(&base, steps[i].seedstart, steps[i].seedlen);

        for (j = 0; j < steps[i].targetlen; j++) {
            if (digestpos >= 20) {
                sprintf(numberbuf, "%d", counter++);
                final = base;
                SHA_Bytes(&final, numberbuf, strlen(numberbuf));
                SHA_Final(&final, digest);
                digestpos = 0;
            }
            steps[i].targetstart[j] ^= digest[digestpos++];
        }

        /*
         * Mask off the pad bits in the final byte after both steps.
         */
        if (bits % 8)
            bmp[bits / 8] &= 0xFF & (0xFF00 >> (bits % 8));
    }
}

static char *new_mine_layout(int w, int h, int n, int x, int y, int unique,
                             random_state *rs, char **game_desc)
{
    signed char *grid, *ret, *p;
    unsigned char *bmp;
    int i, area;

#ifdef TEST_OBFUSCATION
    static int tested_obfuscation = FALSE;
    if (!tested_obfuscation) {
        /*
         * A few simple test vectors for the obfuscator.
         * 
         * First test: the 28-bit stream 1234567. This divides up
         * into 1234 and 567[0]. The SHA of 56 70 30 (appending
         * "0") is 15ce8ab946640340bbb99f3f48fd2c45d1a31d30. Thus,
         * we XOR the 16-bit string 15CE into the input 1234 to get
         * 07FA. Next, we SHA that with "0": the SHA of 07 FA 30 is
         * 3370135c5e3da4fed937adc004a79533962b6391. So we XOR the
         * 12-bit string 337 into the input 567 to get 650. Thus
         * our output is 07FA650.
         */
        {
            unsigned char bmp1[] = "\x12\x34\x56\x70";
            obfuscate_bitmap(bmp1, 28, FALSE);
            printf("test 1 encode: %s\n",
                   memcmp(bmp1, "\x07\xfa\x65\x00", 4) ? "failed" : "passed");
            obfuscate_bitmap(bmp1, 28, TRUE);
            printf("test 1 decode: %s\n",
                   memcmp(bmp1, "\x12\x34\x56\x70", 4) ? "failed" : "passed");
        }
        /*
         * Second test: a long string to make sure we switch from
         * one SHA to the next correctly. My input string this time
         * is simply fifty bytes of zeroes.
         */
        {
            unsigned char bmp2[50];
            unsigned char bmp2a[50];
            memset(bmp2, 0, 50);
            memset(bmp2a, 0, 50);
            obfuscate_bitmap(bmp2, 50 * 8, FALSE);
            /*
             * SHA of twenty-five zero bytes plus "0" is
             * b202c07b990c01f6ff2d544707f60e506019b671. SHA of
             * twenty-five zero bytes plus "1" is
             * fcb1d8b5a2f6b592fe6780b36aa9d65dd7aa6db9. Thus our
             * first half becomes
             * b202c07b990c01f6ff2d544707f60e506019b671fcb1d8b5a2.
             * 
             * SHA of that lot plus "0" is
             * 10b0af913db85d37ca27f52a9f78bba3a80030db. SHA of the
             * same string plus "1" is
             * 3d01d8df78e76d382b8106f480135a1bc751d725. So the
             * second half becomes
             * 10b0af913db85d37ca27f52a9f78bba3a80030db3d01d8df78.
             */
            printf("test 2 encode: %s\n",
                   memcmp(bmp2, "\xb2\x02\xc0\x7b\x99\x0c\x01\xf6\xff\x2d\x54"
                          "\x47\x07\xf6\x0e\x50\x60\x19\xb6\x71\xfc\xb1\xd8"
                          "\xb5\xa2\x10\xb0\xaf\x91\x3d\xb8\x5d\x37\xca\x27"
                          "\xf5\x2a\x9f\x78\xbb\xa3\xa8\x00\x30\xdb\x3d\x01"
                          "\xd8\xdf\x78", 50) ? "failed" : "passed");
            obfuscate_bitmap(bmp2, 50 * 8, TRUE);
            printf("test 2 decode: %s\n",
                   memcmp(bmp2, bmp2a, 50) ? "failed" : "passed");
        }
    }
#endif

    grid = minegen(w, h, n, x, y, unique, rs);

    if (game_desc) {
        /*
         * Set up the mine bitmap and obfuscate it.
         */
        area = w * h;
        bmp = snewn((area + 7) / 8, unsigned char);
        memset(bmp, 0, (area + 7) / 8);
        for (i = 0; i < area; i++) {
            if (grid[i])
                bmp[i / 8] |= 0x80 >> (i % 8);
        }
        obfuscate_bitmap(bmp, area, FALSE);

        /*
         * Now encode the resulting bitmap in hex. We can work to
         * nibble rather than byte granularity, since the obfuscation
         * function guarantees to return a bit string of the same
         * length as its input.
         */
        ret = snewn((area+3)/4 + 100, char);
        p = ret + sprintf(ret, "%d,%d,m", x, y);   /* 'm' == masked */
        for (i = 0; i < (area+3)/4; i++) {
            int v = bmp[i/2];
            if (i % 2 == 0)
                v >>= 4;
            *p++ = "0123456789abcdef"[v & 0xF];
        }
        *p = '\0';

        sfree(bmp);

        *game_desc = ret;
    }   

    return grid;
}

static char *new_game_desc(game_params *params, random_state *rs,
                           game_aux_info **aux, int interactive)
{
    if (!interactive) {
        /*
         * For batch-generated grids, pre-open one square.
         */
        int x = random_upto(rs, params->w);
        int y = random_upto(rs, params->h);
        signed char *grid;
        char *desc;

        grid = new_mine_layout(params->w, params->h, params->n,
                               x, y, params->unique, rs, &desc);
        sfree(grid);
        return desc;
    } else {
        char *rsdesc, *desc;

        rsdesc = random_state_encode(rs);
        desc = snewn(strlen(rsdesc) + 100, char);
        sprintf(desc, "r%d,%c,%s", params->n, params->unique ? 'u' : 'a', rsdesc);
        sfree(rsdesc);
        return desc;
    }
}

static void game_free_aux_info(game_aux_info *aux)
{
    assert(!"Shouldn't happen");
}

static char *validate_desc(game_params *params, char *desc)
{
    int wh = params->w * params->h;
    int x, y;

    if (*desc == 'r') {
        if (!*desc || !isdigit((unsigned char)*desc))
            return "No initial mine count in game description";
        while (*desc && isdigit((unsigned char)*desc))
            desc++;                    /* skip over mine count */
        if (*desc != ',')
            return "No ',' after initial x-coordinate in game description";
        desc++;
        if (*desc != 'u' && *desc != 'a')
            return "No uniqueness specifier in game description";
        desc++;
        if (*desc != ',')
            return "No ',' after uniqueness specifier in game description";
        /* now ignore the rest */
    } else {
        if (!*desc || !isdigit((unsigned char)*desc))
            return "No initial x-coordinate in game description";
        x = atoi(desc);
        if (x < 0 || x >= params->w)
            return "Initial x-coordinate was out of range";
        while (*desc && isdigit((unsigned char)*desc))
            desc++;                    /* skip over x coordinate */
        if (*desc != ',')
            return "No ',' after initial x-coordinate in game description";
        desc++;                        /* eat comma */
        if (!*desc || !isdigit((unsigned char)*desc))
            return "No initial y-coordinate in game description";
        y = atoi(desc);
        if (y < 0 || y >= params->h)
            return "Initial y-coordinate was out of range";
        while (*desc && isdigit((unsigned char)*desc))
            desc++;                    /* skip over y coordinate */
        if (*desc != ',')
            return "No ',' after initial y-coordinate in game description";
        desc++;                        /* eat comma */
        /* eat `m', meaning `masked', if present */
        if (*desc == 'm')
            desc++;
        /* now just check length of remainder */
        if (strlen(desc) != (wh+3)/4)
            return "Game description is wrong length";
    }

    return NULL;
}

static int open_square(game_state *state, int x, int y)
{
    int w = state->w, h = state->h;
    int xx, yy, nmines, ncovered;

    if (!state->layout->mines) {
        /*
         * We have a preliminary game in which the mine layout
         * hasn't been generated yet. Generate it based on the
         * initial click location.
         */
        char *desc;
        state->layout->mines = new_mine_layout(w, h, state->layout->n,
                                               x, y, state->layout->unique,
                                               state->layout->rs,
                                               &desc);
        midend_supersede_game_desc(state->layout->me, desc);
        sfree(desc);
        random_free(state->layout->rs);
        state->layout->rs = NULL;
    }

    if (state->layout->mines[y*w+x]) {
        /*
         * The player has landed on a mine. Bad luck. Expose the
         * mine that killed them, but not the rest (in case they
         * want to Undo and carry on playing).
         */
        state->dead = TRUE;
        state->grid[y*w+x] = 65;
        return -1;
    }

    /*
     * Otherwise, the player has opened a safe square. Mark it to-do.
     */
    state->grid[y*w+x] = -10;          /* `todo' value internal to this func */

    /*
     * Now go through the grid finding all `todo' values and
     * opening them. Every time one of them turns out to have no
     * neighbouring mines, we add all its unopened neighbours to
     * the list as well.
     * 
     * FIXME: We really ought to be able to do this better than
     * using repeated N^2 scans of the grid.
     */
    while (1) {
        int done_something = FALSE;

        for (yy = 0; yy < h; yy++)
            for (xx = 0; xx < w; xx++)
                if (state->grid[yy*w+xx] == -10) {
                    int dx, dy, v;

                    assert(!state->layout->mines[yy*w+xx]);

                    v = 0;

                    for (dx = -1; dx <= +1; dx++)
                        for (dy = -1; dy <= +1; dy++)
                            if (xx+dx >= 0 && xx+dx < state->w &&
                                yy+dy >= 0 && yy+dy < state->h &&
                                state->layout->mines[(yy+dy)*w+(xx+dx)])
                                v++;

                    state->grid[yy*w+xx] = v;

                    if (v == 0) {
                        for (dx = -1; dx <= +1; dx++)
                            for (dy = -1; dy <= +1; dy++)
                                if (xx+dx >= 0 && xx+dx < state->w &&
                                    yy+dy >= 0 && yy+dy < state->h &&
                                    state->grid[(yy+dy)*w+(xx+dx)] == -2)
                                    state->grid[(yy+dy)*w+(xx+dx)] = -10;
                    }

                    done_something = TRUE;
                }

        if (!done_something)
            break;
    }

    /*
     * Finally, scan the grid and see if exactly as many squares
     * are still covered as there are mines. If so, set the `won'
     * flag and fill in mine markers on all covered squares.
     */
    nmines = ncovered = 0;
    for (yy = 0; yy < h; yy++)
        for (xx = 0; xx < w; xx++) {
            if (state->grid[yy*w+xx] < 0)
                ncovered++;
            if (state->layout->mines[yy*w+xx])
                nmines++;
        }
    assert(ncovered >= nmines);
    if (ncovered == nmines) {
        for (yy = 0; yy < h; yy++)
            for (xx = 0; xx < w; xx++) {
                if (state->grid[yy*w+xx] < 0)
                    state->grid[yy*w+xx] = -1;
        }
        state->won = TRUE;
    }

    return 0;
}

static game_state *new_game(midend_data *me, game_params *params, char *desc)
{
    game_state *state = snew(game_state);
    int i, wh, x, y, ret, masked;
    unsigned char *bmp;

    state->w = params->w;
    state->h = params->h;
    state->n = params->n;
    state->dead = state->won = FALSE;
    state->used_solve = state->just_used_solve = FALSE;

    wh = state->w * state->h;

    state->layout = snew(struct mine_layout);
    state->layout->refcount = 1;

    state->grid = snewn(wh, char);
    memset(state->grid, -2, wh);

    if (*desc == 'r') {
        desc++;
        state->layout->n = atoi(desc);
        while (*desc && isdigit((unsigned char)*desc))
            desc++;                    /* skip over mine count */
        if (*desc) desc++;             /* eat comma */
        if (*desc == 'a')
            state->layout->unique = FALSE;
        else
            state->layout->unique = TRUE;
        desc++;
        if (*desc) desc++;             /* eat comma */

        state->layout->mines = NULL;
        state->layout->rs = random_state_decode(desc);
        state->layout->me = me;

    } else {
        state->layout->rs = NULL;
        state->layout->me = NULL;

        state->layout->mines = snewn(wh, char);
        x = atoi(desc);
        while (*desc && isdigit((unsigned char)*desc))
            desc++;                    /* skip over x coordinate */
        if (*desc) desc++;             /* eat comma */
        y = atoi(desc);
        while (*desc && isdigit((unsigned char)*desc))
            desc++;                    /* skip over y coordinate */
        if (*desc) desc++;             /* eat comma */

        if (*desc == 'm') {
            masked = TRUE;
            desc++;
        } else {
            /*
             * We permit game IDs to be entered by hand without the
             * masking transformation.
             */
            masked = FALSE;
        }

        bmp = snewn((wh + 7) / 8, unsigned char);
        memset(bmp, 0, (wh + 7) / 8);
        for (i = 0; i < (wh+3)/4; i++) {
            int c = desc[i];
            int v;

            assert(c != 0);            /* validate_desc should have caught */
            if (c >= '0' && c <= '9')
                v = c - '0';
            else if (c >= 'a' && c <= 'f')
                v = c - 'a' + 10;
            else if (c >= 'A' && c <= 'F')
                v = c - 'A' + 10;
            else
                v = 0;

            bmp[i / 2] |= v << (4 * (1 - (i % 2)));
        }

        if (masked)
            obfuscate_bitmap(bmp, wh, TRUE);

        memset(state->layout->mines, 0, wh);
        for (i = 0; i < wh; i++) {
            if (bmp[i / 8] & (0x80 >> (i % 8)))
                state->layout->mines[i] = 1;
        }

        ret = open_square(state, x, y);
    }

    return state;
}

static game_state *dup_game(game_state *state)
{
    game_state *ret = snew(game_state);

    ret->w = state->w;
    ret->h = state->h;
    ret->n = state->n;
    ret->dead = state->dead;
    ret->won = state->won;
    ret->used_solve = state->used_solve;
    ret->just_used_solve = state->just_used_solve;
    ret->layout = state->layout;
    ret->layout->refcount++;
    ret->grid = snewn(ret->w * ret->h, char);
    memcpy(ret->grid, state->grid, ret->w * ret->h);

    return ret;
}

static void free_game(game_state *state)
{
    if (--state->layout->refcount <= 0) {
        sfree(state->layout->mines);
        if (state->layout->rs)
            random_free(state->layout->rs);
        sfree(state->layout);
    }
    sfree(state->grid);
    sfree(state);
}

static game_state *solve_game(game_state *state, game_aux_info *aux,
                              char **error)
{
    /*
     * Simply expose the entire grid as if it were a completed
     * solution.
     */
    game_state *ret;
    int yy, xx;

    if (!state->layout->mines) {
        *error = "Game has not been started yet";
        return NULL;
    }

    ret = dup_game(state);
    for (yy = 0; yy < ret->h; yy++)
        for (xx = 0; xx < ret->w; xx++) {

            if (ret->layout->mines[yy*ret->w+xx]) {
                ret->grid[yy*ret->w+xx] = -1;
            } else {
                int dx, dy, v;

                v = 0;

                for (dx = -1; dx <= +1; dx++)
                    for (dy = -1; dy <= +1; dy++)
                        if (xx+dx >= 0 && xx+dx < ret->w &&
                            yy+dy >= 0 && yy+dy < ret->h &&
                            ret->layout->mines[(yy+dy)*ret->w+(xx+dx)])
                            v++;

                ret->grid[yy*ret->w+xx] = v;
            }
        }
    ret->used_solve = ret->just_used_solve = TRUE;
    ret->won = TRUE;

    return ret;
}

static char *game_text_format(game_state *state)
{
    char *ret;
    int x, y;

    ret = snewn((state->w + 1) * state->h + 1, char);
    for (y = 0; y < state->h; y++) {
        for (x = 0; x < state->w; x++) {
            int v = state->grid[y*state->w+x];
            if (v == 0)
                v = '-';
            else if (v >= 1 && v <= 8)
                v = '0' + v;
            else if (v == -1)
                v = '*';
            else if (v == -2 || v == -3)
                v = '?';
            else if (v >= 64)
                v = '!';
            ret[y * (state->w+1) + x] = v;
        }
        ret[y * (state->w+1) + state->w] = '\n';
    }
    ret[(state->w + 1) * state->h] = '\0';

    return ret;
}

struct game_ui {
    int hx, hy, hradius;               /* for mouse-down highlights */
    int flash_is_death;
    int deaths;
};

static game_ui *new_ui(game_state *state)
{
    game_ui *ui = snew(game_ui);
    ui->hx = ui->hy = -1;
    ui->hradius = 0;
    ui->deaths = 0;
    ui->flash_is_death = FALSE;        /* *shrug* */
    return ui;
}

static void free_ui(game_ui *ui)
{
    sfree(ui);
}

static game_state *make_move(game_state *from, game_ui *ui, game_drawstate *ds,
                             int x, int y, int button)
{
    game_state *ret;
    int cx, cy;

    if (from->dead || from->won)
        return NULL;                   /* no further moves permitted */

    if (!IS_MOUSE_DOWN(button) && !IS_MOUSE_DRAG(button) &&
        !IS_MOUSE_RELEASE(button))
        return NULL;

    cx = FROMCOORD(x);
    cy = FROMCOORD(y);
    if (cx < 0 || cx >= from->w || cy < 0 || cy >= from->h)
        return NULL;

    if (button == LEFT_BUTTON || button == LEFT_DRAG ||
        button == MIDDLE_BUTTON || button == MIDDLE_DRAG) {
        /*
         * Mouse-downs and mouse-drags just cause highlighting
         * updates.
         */
        ui->hx = cx;
        ui->hy = cy;
        ui->hradius = (from->grid[cy*from->w+cx] >= 0 ? 1 : 0);
        return from;
    }

    if (button == RIGHT_BUTTON) {
        /*
         * Right-clicking only works on a covered square, and it
         * toggles between -1 (marked as mine) and -2 (not marked
         * as mine).
         *
         * FIXME: question marks.
         */
        if (from->grid[cy * from->w + cx] != -2 &&
            from->grid[cy * from->w + cx] != -1)
            return NULL;

        ret = dup_game(from);
        ret->just_used_solve = FALSE;
        ret->grid[cy * from->w + cx] ^= (-2 ^ -1);

        return ret;
    }

    if (button == LEFT_RELEASE || button == MIDDLE_RELEASE) {
        ui->hx = ui->hy = -1;
        ui->hradius = 0;

        /*
         * At this stage we must never return NULL: we have adjusted
         * the ui, so at worst we return `from'.
         */

        /*
         * Left-clicking on a covered square opens a tile. Not
         * permitted if the tile is marked as a mine, for safety.
         * (Unmark it and _then_ open it.)
         */
        if (button == LEFT_RELEASE &&
            (from->grid[cy * from->w + cx] == -2 ||
             from->grid[cy * from->w + cx] == -3)) {
            ret = dup_game(from);
            ret->just_used_solve = FALSE;
            open_square(ret, cx, cy);
            if (ret->dead)
                ui->deaths++;
            return ret;
        }

        /*
         * Left-clicking or middle-clicking on an uncovered tile:
         * first we check to see if the number of mine markers
         * surrounding the tile is equal to its mine count, and if
         * so then we open all other surrounding squares.
         */
        if (from->grid[cy * from->w + cx] > 0) {
            int dy, dx, n;

            /* Count mine markers. */
            n = 0;
            for (dy = -1; dy <= +1; dy++)
                for (dx = -1; dx <= +1; dx++)
                    if (cx+dx >= 0 && cx+dx < from->w &&
                        cy+dy >= 0 && cy+dy < from->h) {
                        if (from->grid[(cy+dy)*from->w+(cx+dx)] == -1)
                            n++;
                    }

            if (n == from->grid[cy * from->w + cx]) {
                ret = dup_game(from);
                ret->just_used_solve = FALSE;
                for (dy = -1; dy <= +1; dy++)
                    for (dx = -1; dx <= +1; dx++)
                        if (cx+dx >= 0 && cx+dx < ret->w &&
                            cy+dy >= 0 && cy+dy < ret->h &&
                            (ret->grid[(cy+dy)*ret->w+(cx+dx)] == -2 ||
                             ret->grid[(cy+dy)*ret->w+(cx+dx)] == -3))
                            open_square(ret, cx+dx, cy+dy);
                if (ret->dead)
                    ui->deaths++;
                return ret;
            }
        }

        return from;
    }

    return NULL;
}

/* ----------------------------------------------------------------------
 * Drawing routines.
 */

struct game_drawstate {
    int w, h, started;
    signed char *grid;
    /*
     * Items in this `grid' array have all the same values as in
     * the game_state grid, and in addition:
     * 
     *  - -10 means the tile was drawn `specially' as a result of a
     *    flash, so it will always need redrawing.
     * 
     *  - -22 and -23 mean the tile is highlighted for a possible
     *    click.
     */
};

static void game_size(game_params *params, int *x, int *y)
{
    *x = BORDER * 2 + TILE_SIZE * params->w;
    *y = BORDER * 2 + TILE_SIZE * params->h;
}

static float *game_colours(frontend *fe, game_state *state, int *ncolours)
{
    float *ret = snewn(3 * NCOLOURS, float);

    frontend_default_colour(fe, &ret[COL_BACKGROUND * 3]);

    ret[COL_BACKGROUND2 * 3 + 0] = ret[COL_BACKGROUND * 3 + 0] * 19.0 / 20.0;
    ret[COL_BACKGROUND2 * 3 + 1] = ret[COL_BACKGROUND * 3 + 1] * 19.0 / 20.0;
    ret[COL_BACKGROUND2 * 3 + 2] = ret[COL_BACKGROUND * 3 + 2] * 19.0 / 20.0;

    ret[COL_1 * 3 + 0] = 0.0F;
    ret[COL_1 * 3 + 1] = 0.0F;
    ret[COL_1 * 3 + 2] = 1.0F;

    ret[COL_2 * 3 + 0] = 0.0F;
    ret[COL_2 * 3 + 1] = 0.5F;
    ret[COL_2 * 3 + 2] = 0.0F;

    ret[COL_3 * 3 + 0] = 1.0F;
    ret[COL_3 * 3 + 1] = 0.0F;
    ret[COL_3 * 3 + 2] = 0.0F;

    ret[COL_4 * 3 + 0] = 0.0F;
    ret[COL_4 * 3 + 1] = 0.0F;
    ret[COL_4 * 3 + 2] = 0.5F;

    ret[COL_5 * 3 + 0] = 0.5F;
    ret[COL_5 * 3 + 1] = 0.0F;
    ret[COL_5 * 3 + 2] = 0.0F;

    ret[COL_6 * 3 + 0] = 0.0F;
    ret[COL_6 * 3 + 1] = 0.5F;
    ret[COL_6 * 3 + 2] = 0.5F;

    ret[COL_7 * 3 + 0] = 0.0F;
    ret[COL_7 * 3 + 1] = 0.0F;
    ret[COL_7 * 3 + 2] = 0.0F;

    ret[COL_8 * 3 + 0] = 0.5F;
    ret[COL_8 * 3 + 1] = 0.5F;
    ret[COL_8 * 3 + 2] = 0.5F;

    ret[COL_MINE * 3 + 0] = 0.0F;
    ret[COL_MINE * 3 + 1] = 0.0F;
    ret[COL_MINE * 3 + 2] = 0.0F;

    ret[COL_BANG * 3 + 0] = 1.0F;
    ret[COL_BANG * 3 + 1] = 0.0F;
    ret[COL_BANG * 3 + 2] = 0.0F;

    ret[COL_CROSS * 3 + 0] = 1.0F;
    ret[COL_CROSS * 3 + 1] = 0.0F;
    ret[COL_CROSS * 3 + 2] = 0.0F;

    ret[COL_FLAG * 3 + 0] = 1.0F;
    ret[COL_FLAG * 3 + 1] = 0.0F;
    ret[COL_FLAG * 3 + 2] = 0.0F;

    ret[COL_FLAGBASE * 3 + 0] = 0.0F;
    ret[COL_FLAGBASE * 3 + 1] = 0.0F;
    ret[COL_FLAGBASE * 3 + 2] = 0.0F;

    ret[COL_QUERY * 3 + 0] = 0.0F;
    ret[COL_QUERY * 3 + 1] = 0.0F;
    ret[COL_QUERY * 3 + 2] = 0.0F;

    ret[COL_HIGHLIGHT * 3 + 0] = 1.0F;
    ret[COL_HIGHLIGHT * 3 + 1] = 1.0F;
    ret[COL_HIGHLIGHT * 3 + 2] = 1.0F;

    ret[COL_LOWLIGHT * 3 + 0] = ret[COL_BACKGROUND * 3 + 0] * 2.0 / 3.0;
    ret[COL_LOWLIGHT * 3 + 1] = ret[COL_BACKGROUND * 3 + 1] * 2.0 / 3.0;
    ret[COL_LOWLIGHT * 3 + 2] = ret[COL_BACKGROUND * 3 + 2] * 2.0 / 3.0;

    *ncolours = NCOLOURS;
    return ret;
}

static game_drawstate *game_new_drawstate(game_state *state)
{
    struct game_drawstate *ds = snew(struct game_drawstate);

    ds->w = state->w;
    ds->h = state->h;
    ds->started = FALSE;
    ds->grid = snewn(ds->w * ds->h, char);

    memset(ds->grid, -99, ds->w * ds->h);

    return ds;
}

static void game_free_drawstate(game_drawstate *ds)
{
    sfree(ds->grid);
    sfree(ds);
}

static void draw_tile(frontend *fe, int x, int y, int v, int bg)
{
    if (v < 0) {
        int coords[12];
        int hl = 0;

        if (v == -22 || v == -23) {
            v += 20;

            /*
             * Omit the highlights in this case.
             */
            draw_rect(fe, x, y, TILE_SIZE, TILE_SIZE,
                      bg == COL_BACKGROUND ? COL_BACKGROUND2 : bg);
            draw_line(fe, x, y, x + TILE_SIZE - 1, y, COL_LOWLIGHT);
            draw_line(fe, x, y, x, y + TILE_SIZE - 1, COL_LOWLIGHT);
        } else {
            /*
             * Draw highlights to indicate the square is covered.
             */
            coords[0] = x + TILE_SIZE - 1;
            coords[1] = y + TILE_SIZE - 1;
            coords[2] = x + TILE_SIZE - 1;
            coords[3] = y;
            coords[4] = x;
            coords[5] = y + TILE_SIZE - 1;
            draw_polygon(fe, coords, 3, TRUE, COL_LOWLIGHT ^ hl);
            draw_polygon(fe, coords, 3, FALSE, COL_LOWLIGHT ^ hl);

            coords[0] = x;
            coords[1] = y;
            draw_polygon(fe, coords, 3, TRUE, COL_HIGHLIGHT ^ hl);
            draw_polygon(fe, coords, 3, FALSE, COL_HIGHLIGHT ^ hl);

            draw_rect(fe, x + HIGHLIGHT_WIDTH, y + HIGHLIGHT_WIDTH,
                      TILE_SIZE - 2*HIGHLIGHT_WIDTH, TILE_SIZE - 2*HIGHLIGHT_WIDTH,
                      bg);
        }

        if (v == -1) {
            /*
             * Draw a flag.
             */
#define SETCOORD(n, dx, dy) do { \
    coords[(n)*2+0] = x + TILE_SIZE * (dx); \
    coords[(n)*2+1] = y + TILE_SIZE * (dy); \
} while (0)
            SETCOORD(0, 0.6, 0.35);
            SETCOORD(1, 0.6, 0.7);
            SETCOORD(2, 0.8, 0.8);
            SETCOORD(3, 0.25, 0.8);
            SETCOORD(4, 0.55, 0.7);
            SETCOORD(5, 0.55, 0.35);
            draw_polygon(fe, coords, 6, TRUE, COL_FLAGBASE);
            draw_polygon(fe, coords, 6, FALSE, COL_FLAGBASE);

            SETCOORD(0, 0.6, 0.2);
            SETCOORD(1, 0.6, 0.5);
            SETCOORD(2, 0.2, 0.35);
            draw_polygon(fe, coords, 3, TRUE, COL_FLAG);
            draw_polygon(fe, coords, 3, FALSE, COL_FLAG);
#undef SETCOORD

        } else if (v == -3) {
            /*
             * Draw a question mark.
             */
            draw_text(fe, x + TILE_SIZE / 2, y + TILE_SIZE / 2,
                      FONT_VARIABLE, TILE_SIZE * 6 / 8,
                      ALIGN_VCENTRE | ALIGN_HCENTRE,
                      COL_QUERY, "?");
        }
    } else {
        /*
         * Clear the square to the background colour, and draw thin
         * grid lines along the top and left.
         * 
         * Exception is that for value 65 (mine we've just trodden
         * on), we clear the square to COL_BANG.
         */
        draw_rect(fe, x, y, TILE_SIZE, TILE_SIZE,
                  (v == 65 ? COL_BANG :
                   bg == COL_BACKGROUND ? COL_BACKGROUND2 : bg));
        draw_line(fe, x, y, x + TILE_SIZE - 1, y, COL_LOWLIGHT);
        draw_line(fe, x, y, x, y + TILE_SIZE - 1, COL_LOWLIGHT);

        if (v > 0 && v <= 8) {
            /*
             * Mark a number.
             */
            char str[2];
            str[0] = v + '0';
            str[1] = '\0';
            draw_text(fe, x + TILE_SIZE / 2, y + TILE_SIZE / 2,
                      FONT_VARIABLE, TILE_SIZE * 7 / 8,
                      ALIGN_VCENTRE | ALIGN_HCENTRE,
                      (COL_1 - 1) + v, str);

        } else if (v >= 64) {
            /*
             * Mark a mine.
             * 
             * FIXME: this could be done better!
             */
#if 0
            draw_text(fe, x + TILE_SIZE / 2, y + TILE_SIZE / 2,
                      FONT_VARIABLE, TILE_SIZE * 7 / 8,
                      ALIGN_VCENTRE | ALIGN_HCENTRE,
                      COL_MINE, "*");
#else
            {
                int cx = x + TILE_SIZE / 2;
                int cy = y + TILE_SIZE / 2;
                int r = TILE_SIZE / 2 - 3;
                int coords[4*5*2];
                int xdx = 1, xdy = 0, ydx = 0, ydy = 1;
                int tdx, tdy, i;

                for (i = 0; i < 4*5*2; i += 5*2) {
                    coords[i+2*0+0] = cx - r/6*xdx + r*4/5*ydx;
                    coords[i+2*0+1] = cy - r/6*xdy + r*4/5*ydy;
                    coords[i+2*1+0] = cx - r/6*xdx + r*ydx;
                    coords[i+2*1+1] = cy - r/6*xdy + r*ydy;
                    coords[i+2*2+0] = cx + r/6*xdx + r*ydx;
                    coords[i+2*2+1] = cy + r/6*xdy + r*ydy;
                    coords[i+2*3+0] = cx + r/6*xdx + r*4/5*ydx;
                    coords[i+2*3+1] = cy + r/6*xdy + r*4/5*ydy;
                    coords[i+2*4+0] = cx + r*3/5*xdx + r*3/5*ydx;
                    coords[i+2*4+1] = cy + r*3/5*xdy + r*3/5*ydy;

                    tdx = ydx;
                    tdy = ydy;
                    ydx = xdx;
                    ydy = xdy;
                    xdx = -tdx;
                    xdy = -tdy;
                }

                draw_polygon(fe, coords, 5*4, TRUE, COL_MINE);
                draw_polygon(fe, coords, 5*4, FALSE, COL_MINE);

                draw_rect(fe, cx-r/3, cy-r/3, r/3, r/4, COL_HIGHLIGHT);
            }
#endif

            if (v == 66) {
                /*
                 * Cross through the mine.
                 */
                int dx;
                for (dx = -1; dx <= +1; dx++) {
                    draw_line(fe, x + 3 + dx, y + 2,
                              x + TILE_SIZE - 3 + dx,
                              y + TILE_SIZE - 2, COL_CROSS);
                    draw_line(fe, x + TILE_SIZE - 3 + dx, y + 2,
                              x + 3 + dx, y + TILE_SIZE - 2,
                              COL_CROSS);
                }
            }
        }
    }

    draw_update(fe, x, y, TILE_SIZE, TILE_SIZE);
}

static void game_redraw(frontend *fe, game_drawstate *ds, game_state *oldstate,
                        game_state *state, int dir, game_ui *ui,
                        float animtime, float flashtime)
{
    int x, y;
    int mines, markers, bg;

    if (flashtime) {
        int frame = (flashtime / FLASH_FRAME);
        if (frame % 2)
            bg = (ui->flash_is_death ? COL_BACKGROUND : COL_LOWLIGHT);
        else
            bg = (ui->flash_is_death ? COL_BANG : COL_HIGHLIGHT);
    } else
        bg = COL_BACKGROUND;

    if (!ds->started) {
        int coords[10];

        draw_rect(fe, 0, 0,
                  TILE_SIZE * state->w + 2 * BORDER,
                  TILE_SIZE * state->h + 2 * BORDER, COL_BACKGROUND);
        draw_update(fe, 0, 0,
                    TILE_SIZE * state->w + 2 * BORDER,
                    TILE_SIZE * state->h + 2 * BORDER);

        /*
         * Recessed area containing the whole puzzle.
         */
        coords[0] = COORD(state->w) + OUTER_HIGHLIGHT_WIDTH - 1;
        coords[1] = COORD(state->h) + OUTER_HIGHLIGHT_WIDTH - 1;
        coords[2] = COORD(state->w) + OUTER_HIGHLIGHT_WIDTH - 1;
        coords[3] = COORD(0) - OUTER_HIGHLIGHT_WIDTH;
        coords[4] = coords[2] - TILE_SIZE;
        coords[5] = coords[3] + TILE_SIZE;
        coords[8] = COORD(0) - OUTER_HIGHLIGHT_WIDTH;
        coords[9] = COORD(state->h) + OUTER_HIGHLIGHT_WIDTH - 1;
        coords[6] = coords[8] + TILE_SIZE;
        coords[7] = coords[9] - TILE_SIZE;
        draw_polygon(fe, coords, 5, TRUE, COL_HIGHLIGHT);
        draw_polygon(fe, coords, 5, FALSE, COL_HIGHLIGHT);

        coords[1] = COORD(0) - OUTER_HIGHLIGHT_WIDTH;
        coords[0] = COORD(0) - OUTER_HIGHLIGHT_WIDTH;
        draw_polygon(fe, coords, 5, TRUE, COL_LOWLIGHT);
        draw_polygon(fe, coords, 5, FALSE, COL_LOWLIGHT);

        ds->started = TRUE;
    }

    /*
     * Now draw the tiles. Also in this loop, count up the number
     * of mines and mine markers.
     */
    mines = markers = 0;
    for (y = 0; y < ds->h; y++)
        for (x = 0; x < ds->w; x++) {
            int v = state->grid[y*ds->w+x];

            if (v == -1)
                markers++;
            if (state->layout->mines && state->layout->mines[y*ds->w+x])
                mines++;

            if ((v == -2 || v == -3) &&
                (abs(x-ui->hx) <= ui->hradius && abs(y-ui->hy) <= ui->hradius))
                v -= 20;

            if (ds->grid[y*ds->w+x] != v || bg != COL_BACKGROUND) {
                draw_tile(fe, COORD(x), COORD(y), v, bg);
                ds->grid[y*ds->w+x] = (bg == COL_BACKGROUND ? v : -10);
            }
        }

    if (!state->layout->mines)
        mines = state->layout->n;

    /*
     * Update the status bar.
     */
    {
        char statusbar[512];
        if (state->dead) {
            sprintf(statusbar, "DEAD!");
        } else if (state->won) {
            if (state->used_solve)
                sprintf(statusbar, "Auto-solved.");
            else
                sprintf(statusbar, "COMPLETED!");
        } else {
            sprintf(statusbar, "Marked: %d / %d", markers, mines);
        }
        if (ui->deaths)
            sprintf(statusbar + strlen(statusbar),
                    "  Deaths: %d", ui->deaths);
        status_bar(fe, statusbar);
    }
}

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->used_solve || newstate->used_solve)
        return 0.0F;

    if (dir > 0 && !oldstate->dead && !oldstate->won) {
        if (newstate->dead) {
            ui->flash_is_death = TRUE;
            return 3 * FLASH_FRAME;
        }
        if (newstate->won) {
            ui->flash_is_death = FALSE;
            return 2 * FLASH_FRAME;
        }
    }
    return 0.0F;
}

static int game_wants_statusbar(void)
{
    return TRUE;
}

static int game_timing_state(game_state *state)
{
    if (state->dead || state->won || !state->layout->mines)
        return FALSE;
    return TRUE;
}

#ifdef COMBINED
#define thegame mines
#endif

const struct game thegame = {
    "Mines", "games.mines",
    default_params,
    game_fetch_preset,
    decode_params,
    encode_params,
    free_params,
    dup_params,
    TRUE, game_configure, custom_params,
    validate_params,
    new_game_desc,
    game_free_aux_info,
    validate_desc,
    new_game,
    dup_game,
    free_game,
    TRUE, solve_game,
    TRUE, game_text_format,
    new_ui,
    free_ui,
    make_move,
    game_size,
    game_colours,
    game_new_drawstate,
    game_free_drawstate,
    game_redraw,
    game_anim_length,
    game_flash_length,
    game_wants_statusbar,
    TRUE, game_timing_state,
    BUTTON_BEATS(LEFT_BUTTON, RIGHT_BUTTON),
};