Port of Simon Tatham's Puzzle Collection

Original revision: 5123b1bf68777ffa86e651f178046b26a87cf2d9

MIT Licensed. Some games still crash and others are unplayable due to
issues with controls. Still need a "real" polygon filling algorithm.

Currently builds one plugin per puzzle (about 40 in total, around 100K
each on ARM), but can easily be made to build a single monolithic
overlay (800K or so on ARM).

The following games are at least partially broken for various reasons,
and have been disabled on this commit:

Cube:     failed assertion with "Icosahedron" setting
Keen:     input issues
Mines:    weird stuff happens on target
Palisade: input issues
Solo:     input issues, occasional crash on target
Towers:   input issues
Undead:   input issues
Unequal:  input and drawing issues (concave polys)
Untangle: input issues

Features left to do:
 - In-game help system
 - Figure out the weird bugs

Change-Id: I7c69b6860ab115f973c8d76799502e9bb3d52368

1 files changed, 3688 insertions, 0 deletions

diff --git a/apps/plugins/puzzles/loopy.c b/apps/plugins/puzzles/loopy.c
new file mode 100644
index 0000000000..a931b31c37
--- /dev/null
+++ b/apps/plugins/puzzles/loopy.c
@@ -0,0 +1,3688 @@
+/*
+ * loopy.c:
+ *
+ * An implementation of the Nikoli game 'Loop the loop'.
+ * (c) Mike Pinna, 2005, 2006
+ * Substantially rewritten to allowing for more general types of grid.
+ * (c) Lambros Lambrou 2008
+ *
+ * vim: set shiftwidth=4 :set textwidth=80:
+ */
+
+/*
+ * Possible future solver enhancements:
+ * 
+ *  - There's an interesting deductive technique which makes use
+ *    of topology rather than just graph theory. Each _face_ in
+ *    the grid is either inside or outside the loop; you can tell
+ *    that two faces are on the same side of the loop if they're
+ *    separated by a LINE_NO (or, more generally, by a path
+ *    crossing no LINE_UNKNOWNs and an even number of LINE_YESes),
+ *    and on the opposite side of the loop if they're separated by
+ *    a LINE_YES (or an odd number of LINE_YESes and no
+ *    LINE_UNKNOWNs). Oh, and any face separated from the outside
+ *    of the grid by a LINE_YES or a LINE_NO is on the inside or
+ *    outside respectively. So if you can track this for all
+ *    faces, you figure out the state of the line between a pair
+ *    once their relative insideness is known.
+ *     + The way I envisage this working is simply to keep an edsf
+ *       of all _faces_, which indicates whether they're on
+ *       opposite sides of the loop from one another. We also
+ *       include a special entry in the edsf for the infinite
+ *       exterior "face".
+ *     + So, the simple way to do this is to just go through the
+ *       edges: every time we see an edge in a state other than
+ *       LINE_UNKNOWN which separates two faces that aren't in the
+ *       same edsf class, we can rectify that by merging the
+ *       classes. Then, conversely, an edge in LINE_UNKNOWN state
+ *       which separates two faces that _are_ in the same edsf
+ *       class can immediately have its state determined.
+ *     + But you can go one better, if you're prepared to loop
+ *       over all _pairs_ of edges. Suppose we have edges A and B,
+ *       which respectively separate faces A1,A2 and B1,B2.
+ *       Suppose that A,B are in the same edge-edsf class and that
+ *       A1,B1 (wlog) are in the same face-edsf class; then we can
+ *       immediately place A2,B2 into the same face-edsf class (as
+ *       each other, not as A1 and A2) one way round or the other.
+ *       And conversely again, if A1,B1 are in the same face-edsf
+ *       class and so are A2,B2, then we can put A,B into the same
+ *       face-edsf class.
+ *        * Of course, this deduction requires a quadratic-time
+ *          loop over all pairs of edges in the grid, so it should
+ *          be reserved until there's nothing easier left to be
+ *          done.
+ * 
+ *  - The generalised grid support has made me (SGT) notice a
+ *    possible extension to the loop-avoidance code. When you have
+ *    a path of connected edges such that no other edges at all
+ *    are incident on any vertex in the middle of the path - or,
+ *    alternatively, such that any such edges are already known to
+ *    be LINE_NO - then you know those edges are either all
+ *    LINE_YES or all LINE_NO. Hence you can mentally merge the
+ *    entire path into a single long curly edge for the purposes
+ *    of loop avoidance, and look directly at whether or not the
+ *    extreme endpoints of the path are connected by some other
+ *    route. I find this coming up fairly often when I play on the
+ *    octagonal grid setting, so it might be worth implementing in
+ *    the solver.
+ *
+ *  - (Just a speed optimisation.)  Consider some todo list queue where every
+ *    time we modify something we mark it for consideration by other bits of
+ *    the solver, to save iteration over things that have already been done.
+ */
+
+#include <stdio.h>
+#include <stdlib.h>
+#include <stddef.h>
+#include <string.h>
+#include "rbassert.h"
+#include <ctype.h>
+#include <math.h>
+
+#include "puzzles.h"
+#include "tree234.h"
+#include "grid.h"
+#include "loopgen.h"
+
+/* Debugging options */
+
+/*
+#define DEBUG_CACHES
+#define SHOW_WORKING
+#define DEBUG_DLINES
+*/
+
+/* ----------------------------------------------------------------------
+ * Struct, enum and function declarations
+ */
+
+enum {
+    COL_BACKGROUND,
+    COL_FOREGROUND,
+    COL_LINEUNKNOWN,
+    COL_HIGHLIGHT,
+    COL_MISTAKE,
+    COL_SATISFIED,
+    COL_FAINT,
+    NCOLOURS
+};
+
+struct game_state {
+    grid *game_grid; /* ref-counted (internally) */
+
+    /* Put -1 in a face that doesn't get a clue */
+    signed char *clues;
+
+    /* Array of line states, to store whether each line is
+     * YES, NO or UNKNOWN */
+    char *lines;
+
+    unsigned char *line_errors;
+    int exactly_one_loop;
+
+    int solved;
+    int cheated;
+
+    /* Used in game_text_format(), so that it knows what type of
+     * grid it's trying to render as ASCII text. */
+    int grid_type;
+};
+
+enum solver_status {
+    SOLVER_SOLVED,    /* This is the only solution the solver could find */
+    SOLVER_MISTAKE,   /* This is definitely not a solution */
+    SOLVER_AMBIGUOUS, /* This _might_ be an ambiguous solution */
+    SOLVER_INCOMPLETE /* This may be a partial solution */
+};
+
+/* ------ Solver state ------ */
+typedef struct solver_state {
+    game_state *state;
+    enum solver_status solver_status;
+    /* NB looplen is the number of dots that are joined together at a point, ie a
+     * looplen of 1 means there are no lines to a particular dot */
+    int *looplen;
+
+    /* Difficulty level of solver.  Used by solver functions that want to
+     * vary their behaviour depending on the requested difficulty level. */
+    int diff;
+
+    /* caches */
+    char *dot_yes_count;
+    char *dot_no_count;
+    char *face_yes_count;
+    char *face_no_count;
+    char *dot_solved, *face_solved;
+    int *dotdsf;
+
+    /* Information for Normal level deductions:
+     * For each dline, store a bitmask for whether we know:
+     * (bit 0) at least one is YES
+     * (bit 1) at most one is YES */
+    char *dlines;
+
+    /* Hard level information */
+    int *linedsf;
+} solver_state;
+
+/*
+ * Difficulty levels. I do some macro ickery here to ensure that my
+ * enum and the various forms of my name list always match up.
+ */
+
+#define DIFFLIST(A) \
+    A(EASY,Easy,e) \
+    A(NORMAL,Normal,n) \
+    A(TRICKY,Tricky,t) \
+    A(HARD,Hard,h)
+#define ENUM(upper,title,lower) DIFF_ ## upper,
+#define TITLE(upper,title,lower) #title,
+#define ENCODE(upper,title,lower) #lower
+#define CONFIG(upper,title,lower) ":" #title
+enum { DIFFLIST(ENUM) DIFF_MAX };
+static char const *const diffnames[] = { DIFFLIST(TITLE) };
+static char const diffchars[] = DIFFLIST(ENCODE);
+#define DIFFCONFIG DIFFLIST(CONFIG)
+
+/*
+ * Solver routines, sorted roughly in order of computational cost.
+ * The solver will run the faster deductions first, and slower deductions are
+ * only invoked when the faster deductions are unable to make progress.
+ * Each function is associated with a difficulty level, so that the generated
+ * puzzles are solvable by applying only the functions with the chosen
+ * difficulty level or lower.
+ */
+#define SOLVERLIST(A) \
+    A(trivial_deductions, DIFF_EASY) \
+    A(dline_deductions, DIFF_NORMAL) \
+    A(linedsf_deductions, DIFF_HARD) \
+    A(loop_deductions, DIFF_EASY)
+#define SOLVER_FN_DECL(fn,diff) static int fn(solver_state *);
+#define SOLVER_FN(fn,diff) &fn,
+#define SOLVER_DIFF(fn,diff) diff,
+SOLVERLIST(SOLVER_FN_DECL)
+static int (*(solver_fns[]))(solver_state *) = { SOLVERLIST(SOLVER_FN) };
+static int const solver_diffs[] = { SOLVERLIST(SOLVER_DIFF) };
+static const int NUM_SOLVERS = sizeof(solver_diffs)/sizeof(*solver_diffs);
+
+struct game_params {
+    int w, h;
+    int diff;
+    int type;
+};
+
+/* line_drawstate is the same as line_state, but with the extra ERROR
+ * possibility.  The drawing code copies line_state to line_drawstate,
+ * except in the case that the line is an error. */
+enum line_state { LINE_YES, LINE_UNKNOWN, LINE_NO };
+enum line_drawstate { DS_LINE_YES, DS_LINE_UNKNOWN,
+                      DS_LINE_NO, DS_LINE_ERROR };
+
+#define OPP(line_state) \
+    (2 - line_state)
+
+
+struct game_drawstate {
+    int started;
+    int tilesize;
+    int flashing;
+    int *textx, *texty;
+    char *lines;
+    char *clue_error;
+    char *clue_satisfied;
+};
+
+static char *validate_desc(const game_params *params, const char *desc);
+static int dot_order(const game_state* state, int i, char line_type);
+static int face_order(const game_state* state, int i, char line_type);
+static solver_state *solve_game_rec(const solver_state *sstate);
+
+#ifdef DEBUG_CACHES
+static void check_caches(const solver_state* sstate);
+#else
+#define check_caches(s)
+#endif
+
+/* ------- List of grid generators ------- */
+#define GRIDLIST(A) \
+    A(Squares,GRID_SQUARE,3,3) \
+    A(Triangular,GRID_TRIANGULAR,3,3) \
+    A(Honeycomb,GRID_HONEYCOMB,3,3) \
+    A(Snub-Square,GRID_SNUBSQUARE,3,3) \
+    A(Cairo,GRID_CAIRO,3,4) \
+    A(Great-Hexagonal,GRID_GREATHEXAGONAL,3,3) \
+    A(Octagonal,GRID_OCTAGONAL,3,3) \
+    A(Kites,GRID_KITE,3,3) \
+    A(Floret,GRID_FLORET,1,2) \
+    A(Dodecagonal,GRID_DODECAGONAL,2,2) \
+    A(Great-Dodecagonal,GRID_GREATDODECAGONAL,2,2) \
+    A(Penrose (kite/dart),GRID_PENROSE_P2,3,3) \
+    A(Penrose (rhombs),GRID_PENROSE_P3,3,3)
+
+#define GRID_NAME(title,type,amin,omin) #title,
+#define GRID_CONFIG(title,type,amin,omin) ":" #title
+#define GRID_TYPE(title,type,amin,omin) type,
+#define GRID_SIZES(title,type,amin,omin) \
+    {amin, omin, \
+     "Width and height for this grid type must both be at least " #amin, \
+     "At least one of width and height for this grid type must be at least " #omin,},
+static char const *const gridnames[] = { GRIDLIST(GRID_NAME) };
+#define GRID_CONFIGS GRIDLIST(GRID_CONFIG)
+static grid_type grid_types[] = { GRIDLIST(GRID_TYPE) };
+#define NUM_GRID_TYPES (sizeof(grid_types) / sizeof(grid_types[0]))
+static const struct {
+    int amin, omin;
+    char *aerr, *oerr;
+} grid_size_limits[] = { GRIDLIST(GRID_SIZES) };
+
+/* Generates a (dynamically allocated) new grid, according to the
+ * type and size requested in params.  Does nothing if the grid is already
+ * generated. */
+static grid *loopy_generate_grid(const game_params *params,
+                                 const char *grid_desc)
+{
+    return grid_new(grid_types[params->type], params->w, params->h, grid_desc);
+}
+
+/* ----------------------------------------------------------------------
+ * Preprocessor magic
+ */
+
+/* General constants */
+#define PREFERRED_TILE_SIZE 32
+#define BORDER(tilesize) ((tilesize) / 2)
+#define FLASH_TIME 0.5F
+
+#define BIT_SET(field, bit) ((field) & (1<<(bit)))
+
+#define SET_BIT(field, bit)  (BIT_SET(field, bit) ? FALSE : \
+                              ((field) |= (1<<(bit)), TRUE))
+
+#define CLEAR_BIT(field, bit) (BIT_SET(field, bit) ? \
+                               ((field) &= ~(1<<(bit)), TRUE) : FALSE)
+
+#define CLUE2CHAR(c) \
+    ((c < 0) ? ' ' : c < 10 ? c + '0' : c - 10 + 'A')
+
+/* ----------------------------------------------------------------------
+ * General struct manipulation and other straightforward code
+ */
+
+static game_state *dup_game(const game_state *state)
+{
+    game_state *ret = snew(game_state);
+
+    ret->game_grid = state->game_grid;
+    ret->game_grid->refcount++;
+
+    ret->solved = state->solved;
+    ret->cheated = state->cheated;
+
+    ret->clues = snewn(state->game_grid->num_faces, signed char);
+    memcpy(ret->clues, state->clues, state->game_grid->num_faces);
+
+    ret->lines = snewn(state->game_grid->num_edges, char);
+    memcpy(ret->lines, state->lines, state->game_grid->num_edges);
+
+    ret->line_errors = snewn(state->game_grid->num_edges, unsigned char);
+    memcpy(ret->line_errors, state->line_errors, state->game_grid->num_edges);
+    ret->exactly_one_loop = state->exactly_one_loop;
+
+    ret->grid_type = state->grid_type;
+    return ret;
+}
+
+static void free_game(game_state *state)
+{
+    if (state) {
+        grid_free(state->game_grid);
+        sfree(state->clues);
+        sfree(state->lines);
+        sfree(state->line_errors);
+        sfree(state);
+    }
+}
+
+static solver_state *new_solver_state(const game_state *state, int diff) {
+    int i;
+    int num_dots = state->game_grid->num_dots;
+    int num_faces = state->game_grid->num_faces;
+    int num_edges = state->game_grid->num_edges;
+    solver_state *ret = snew(solver_state);
+
+    ret->state = dup_game(state);
+
+    ret->solver_status = SOLVER_INCOMPLETE;
+    ret->diff = diff;
+
+    ret->dotdsf = snew_dsf(num_dots);
+    ret->looplen = snewn(num_dots, int);
+
+    for (i = 0; i < num_dots; i++) {
+        ret->looplen[i] = 1;
+    }
+
+    ret->dot_solved = snewn(num_dots, char);
+    ret->face_solved = snewn(num_faces, char);
+    memset(ret->dot_solved, FALSE, num_dots);
+    memset(ret->face_solved, FALSE, num_faces);
+
+    ret->dot_yes_count = snewn(num_dots, char);
+    memset(ret->dot_yes_count, 0, num_dots);
+    ret->dot_no_count = snewn(num_dots, char);
+    memset(ret->dot_no_count, 0, num_dots);
+    ret->face_yes_count = snewn(num_faces, char);
+    memset(ret->face_yes_count, 0, num_faces);
+    ret->face_no_count = snewn(num_faces, char);
+    memset(ret->face_no_count, 0, num_faces);
+
+    if (diff < DIFF_NORMAL) {
+        ret->dlines = NULL;
+    } else {
+        ret->dlines = snewn(2*num_edges, char);
+        memset(ret->dlines, 0, 2*num_edges);
+    }
+
+    if (diff < DIFF_HARD) {
+        ret->linedsf = NULL;
+    } else {
+        ret->linedsf = snew_dsf(state->game_grid->num_edges);
+    }
+
+    return ret;
+}
+
+static void free_solver_state(solver_state *sstate) {
+    if (sstate) {
+        free_game(sstate->state);
+        sfree(sstate->dotdsf);
+        sfree(sstate->looplen);
+        sfree(sstate->dot_solved);
+        sfree(sstate->face_solved);
+        sfree(sstate->dot_yes_count);
+        sfree(sstate->dot_no_count);
+        sfree(sstate->face_yes_count);
+        sfree(sstate->face_no_count);
+
+        /* OK, because sfree(NULL) is a no-op */
+        sfree(sstate->dlines);
+        sfree(sstate->linedsf);
+
+        sfree(sstate);
+    }
+}
+
+static solver_state *dup_solver_state(const solver_state *sstate) {
+    game_state *state = sstate->state;
+    int num_dots = state->game_grid->num_dots;
+    int num_faces = state->game_grid->num_faces;
+    int num_edges = state->game_grid->num_edges;
+    solver_state *ret = snew(solver_state);
+
+    ret->state = state = dup_game(sstate->state);
+
+    ret->solver_status = sstate->solver_status;
+    ret->diff = sstate->diff;
+
+    ret->dotdsf = snewn(num_dots, int);
+    ret->looplen = snewn(num_dots, int);
+    memcpy(ret->dotdsf, sstate->dotdsf,
+           num_dots * sizeof(int));
+    memcpy(ret->looplen, sstate->looplen,
+           num_dots * sizeof(int));
+
+    ret->dot_solved = snewn(num_dots, char);
+    ret->face_solved = snewn(num_faces, char);
+    memcpy(ret->dot_solved, sstate->dot_solved, num_dots);
+    memcpy(ret->face_solved, sstate->face_solved, num_faces);
+
+    ret->dot_yes_count = snewn(num_dots, char);
+    memcpy(ret->dot_yes_count, sstate->dot_yes_count, num_dots);
+    ret->dot_no_count = snewn(num_dots, char);
+    memcpy(ret->dot_no_count, sstate->dot_no_count, num_dots);
+
+    ret->face_yes_count = snewn(num_faces, char);
+    memcpy(ret->face_yes_count, sstate->face_yes_count, num_faces);
+    ret->face_no_count = snewn(num_faces, char);
+    memcpy(ret->face_no_count, sstate->face_no_count, num_faces);
+
+    if (sstate->dlines) {
+        ret->dlines = snewn(2*num_edges, char);
+        memcpy(ret->dlines, sstate->dlines,
+               2*num_edges);
+    } else {
+        ret->dlines = NULL;
+    }
+
+    if (sstate->linedsf) {
+        ret->linedsf = snewn(num_edges, int);
+        memcpy(ret->linedsf, sstate->linedsf,
+               num_edges * sizeof(int));
+    } else {
+        ret->linedsf = NULL;
+    }
+
+    return ret;
+}
+
+static game_params *default_params(void)
+{
+    game_params *ret = snew(game_params);
+
+#ifdef SLOW_SYSTEM
+    ret->h = 7;
+    ret->w = 7;
+#else
+    ret->h = 10;
+    ret->w = 10;
+#endif
+    ret->diff = DIFF_EASY;
+    ret->type = 0;
+
+    return ret;
+}
+
+static game_params *dup_params(const game_params *params)
+{
+    game_params *ret = snew(game_params);
+
+    *ret = *params;                       /* structure copy */
+    return ret;
+}
+
+static const game_params presets[] = {
+#ifdef SMALL_SCREEN
+    {  7,  7, DIFF_EASY, 0 },
+    {  7,  7, DIFF_NORMAL, 0 },
+    {  7,  7, DIFF_HARD, 0 },
+    {  7,  7, DIFF_HARD, 1 },
+    {  7,  7, DIFF_HARD, 2 },
+    {  5,  5, DIFF_HARD, 3 },
+    {  7,  7, DIFF_HARD, 4 },
+    {  5,  4, DIFF_HARD, 5 },
+    {  5,  5, DIFF_HARD, 6 },
+    {  5,  5, DIFF_HARD, 7 },
+    {  3,  3, DIFF_HARD, 8 },
+    {  3,  3, DIFF_HARD, 9 },
+    {  3,  3, DIFF_HARD, 10 },
+    {  6,  6, DIFF_HARD, 11 },
+    {  6,  6, DIFF_HARD, 12 },
+#else
+    {  7,  7, DIFF_EASY, 0 },
+    {  10,  10, DIFF_EASY, 0 },
+    {  7,  7, DIFF_NORMAL, 0 },
+    {  10,  10, DIFF_NORMAL, 0 },
+    {  7,  7, DIFF_HARD, 0 },
+    {  10,  10, DIFF_HARD, 0 },
+    {  10,  10, DIFF_HARD, 1 },
+    {  12,  10, DIFF_HARD, 2 },
+    {  7,  7, DIFF_HARD, 3 },
+    {  9,  9, DIFF_HARD, 4 },
+    {  5,  4, DIFF_HARD, 5 },
+    {  7,  7, DIFF_HARD, 6 },
+    {  5,  5, DIFF_HARD, 7 },
+    {  5,  5, DIFF_HARD, 8 },
+    {  5,  4, DIFF_HARD, 9 },
+    {  5,  4, DIFF_HARD, 10 },
+    {  10, 10, DIFF_HARD, 11 },
+    {  10, 10, DIFF_HARD, 12 }
+#endif
+};
+
+static int game_fetch_preset(int i, char **name, game_params **params)
+{
+    game_params *tmppar;
+    char buf[80];
+
+    if (i < 0 || i >= lenof(presets))
+        return FALSE;
+
+    tmppar = snew(game_params);
+    *tmppar = presets[i];
+    *params = tmppar;
+    sprintf(buf, "%dx%d %s - %s", tmppar->h, tmppar->w,
+            gridnames[tmppar->type], diffnames[tmppar->diff]);
+    *name = dupstr(buf);
+
+    return TRUE;
+}
+
+static void free_params(game_params *params)
+{
+    sfree(params);
+}
+
+static void decode_params(game_params *params, char const *string)
+{
+    params->h = params->w = atoi(string);
+    params->diff = DIFF_EASY;
+    while (*string && isdigit((unsigned char)*string)) string++;
+    if (*string == 'x') {
+        string++;
+        params->h = atoi(string);
+        while (*string && isdigit((unsigned char)*string)) string++;
+    }
+    if (*string == 't') {
+        string++;
+        params->type = atoi(string);
+        while (*string && isdigit((unsigned char)*string)) string++;
+    }
+    if (*string == 'd') {
+        int i;
+        string++;
+        for (i = 0; i < DIFF_MAX; i++)
+            if (*string == diffchars[i])
+                params->diff = i;
+        if (*string) string++;
+    }
+}
+
+static char *encode_params(const game_params *params, int full)
+{
+    char str[80];
+    sprintf(str, "%dx%dt%d", params->w, params->h, params->type);
+    if (full)
+        sprintf(str + strlen(str), "d%c", diffchars[params->diff]);
+    return dupstr(str);
+}
+
+static config_item *game_configure(const 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 = "Grid type";
+    ret[2].type = C_CHOICES;
+    ret[2].sval = GRID_CONFIGS;
+    ret[2].ival = params->type;
+
+    ret[3].name = "Difficulty";
+    ret[3].type = C_CHOICES;
+    ret[3].sval = DIFFCONFIG;
+    ret[3].ival = params->diff;
+
+    ret[4].name = NULL;
+    ret[4].type = C_END;
+    ret[4].sval = NULL;
+    ret[4].ival = 0;
+
+    return ret;
+}
+
+static game_params *custom_params(const config_item *cfg)
+{
+    game_params *ret = snew(game_params);
+
+    ret->w = atoi(cfg[0].sval);
+    ret->h = atoi(cfg[1].sval);
+    ret->type = cfg[2].ival;
+    ret->diff = cfg[3].ival;
+
+    return ret;
+}
+
+static char *validate_params(const game_params *params, int full)
+{
+    if (params->type < 0 || params->type >= NUM_GRID_TYPES)
+        return "Illegal grid type";
+    if (params->w < grid_size_limits[params->type].amin ||
+        params->h < grid_size_limits[params->type].amin)
+        return grid_size_limits[params->type].aerr;
+    if (params->w < grid_size_limits[params->type].omin &&
+        params->h < grid_size_limits[params->type].omin)
+        return grid_size_limits[params->type].oerr;
+
+    /*
+     * This shouldn't be able to happen at all, since decode_params
+     * and custom_params will never generate anything that isn't
+     * within range.
+     */
+    assert(params->diff < DIFF_MAX);
+
+    return NULL;
+}
+
+/* Returns a newly allocated string describing the current puzzle */
+static char *state_to_text(const game_state *state)
+{
+    grid *g = state->game_grid;
+    char *retval;
+    int num_faces = g->num_faces;
+    char *description = snewn(num_faces + 1, char);
+    char *dp = description;
+    int empty_count = 0;
+    int i;
+
+    for (i = 0; i < num_faces; i++) {
+        if (state->clues[i] < 0) {
+            if (empty_count > 25) {
+                dp += sprintf(dp, "%c", (int)(empty_count + 'a' - 1));
+                empty_count = 0;
+            }
+            empty_count++;
+        } else {
+            if (empty_count) {
+                dp += sprintf(dp, "%c", (int)(empty_count + 'a' - 1));
+                empty_count = 0;
+            }
+            dp += sprintf(dp, "%c", (int)CLUE2CHAR(state->clues[i]));
+        }
+    }
+
+    if (empty_count)
+        dp += sprintf(dp, "%c", (int)(empty_count + 'a' - 1));
+
+    retval = dupstr(description);
+    sfree(description);
+
+    return retval;
+}
+
+#define GRID_DESC_SEP '_'
+
+/* Splits up a (optional) grid_desc from the game desc. Returns the
+ * grid_desc (which needs freeing) and updates the desc pointer to
+ * start of real desc, or returns NULL if no desc. */
+static char *extract_grid_desc(const char **desc)
+{
+    char *sep = strchr(*desc, GRID_DESC_SEP), *gd;
+    int gd_len;
+
+    if (!sep) return NULL;
+
+    gd_len = sep - (*desc);
+    gd = snewn(gd_len+1, char);
+    memcpy(gd, *desc, gd_len);
+    gd[gd_len] = '\0';
+
+    *desc = sep+1;
+
+    return gd;
+}
+
+/* We require that the params pass the test in validate_params and that the
+ * description fills the entire game area */
+static char *validate_desc(const game_params *params, const char *desc)
+{
+    int count = 0;
+    grid *g;
+    char *grid_desc, *ret;
+
+    /* It's pretty inefficient to do this just for validation. All we need to
+     * know is the precise number of faces. */
+    grid_desc = extract_grid_desc(&desc);
+    ret = grid_validate_desc(grid_types[params->type], params->w, params->h, grid_desc);
+    if (ret) return ret;
+
+    g = loopy_generate_grid(params, grid_desc);
+    if (grid_desc) sfree(grid_desc);
+
+    for (; *desc; ++desc) {
+        if ((*desc >= '0' && *desc <= '9') || (*desc >= 'A' && *desc <= 'Z')) {
+            count++;
+            continue;
+        }
+        if (*desc >= 'a') {
+            count += *desc - 'a' + 1;
+            continue;
+        }
+        return "Unknown character in description";
+    }
+
+    if (count < g->num_faces)
+        return "Description too short for board size";
+    if (count > g->num_faces)
+        return "Description too long for board size";
+
+    grid_free(g);
+
+    return NULL;
+}
+
+/* Sums the lengths of the numbers in range [0,n) */
+/* See equivalent function in solo.c for justification of this. */
+static int len_0_to_n(int n)
+{
+    int len = 1; /* Counting 0 as a bit of a special case */
+    int i;
+
+    for (i = 1; i < n; i *= 10) {
+        len += max(n - i, 0);
+    }
+
+    return len;
+}
+
+static char *encode_solve_move(const game_state *state)
+{
+    int len;
+    char *ret, *p;
+    int i;
+    int num_edges = state->game_grid->num_edges;
+
+    /* This is going to return a string representing the moves needed to set
+     * every line in a grid to be the same as the ones in 'state'.  The exact
+     * length of this string is predictable. */
+
+    len = 1;  /* Count the 'S' prefix */
+    /* Numbers in all lines */
+    len += len_0_to_n(num_edges);
+    /* For each line we also have a letter */
+    len += num_edges;
+
+    ret = snewn(len + 1, char);
+    p = ret;
+
+    p += sprintf(p, "S");
+
+    for (i = 0; i < num_edges; i++) {
+        switch (state->lines[i]) {
+          case LINE_YES:
+            p += sprintf(p, "%dy", i);
+            break;
+          case LINE_NO:
+            p += sprintf(p, "%dn", i);
+            break;
+        }
+    }
+
+    /* No point in doing sums like that if they're going to be wrong */
+    assert(strlen(ret) <= (size_t)len);
+    return ret;
+}
+
+static game_ui *new_ui(const game_state *state)
+{
+    return NULL;
+}
+
+static void free_ui(game_ui *ui)
+{
+}
+
+static char *encode_ui(const game_ui *ui)
+{
+    return NULL;
+}
+
+static void decode_ui(game_ui *ui, const char *encoding)
+{
+}
+
+static void game_changed_state(game_ui *ui, const game_state *oldstate,
+                               const game_state *newstate)
+{
+}
+
+static void game_compute_size(const game_params *params, int tilesize,
+                              int *x, int *y)
+{
+    int grid_width, grid_height, rendered_width, rendered_height;
+    int g_tilesize;
+
+    grid_compute_size(grid_types[params->type], params->w, params->h,
+                      &g_tilesize, &grid_width, &grid_height);
+
+    /* multiply first to minimise rounding error on integer division */
+    rendered_width = grid_width * tilesize / g_tilesize;
+    rendered_height = grid_height * tilesize / g_tilesize;
+    *x = rendered_width + 2 * BORDER(tilesize) + 1;
+    *y = rendered_height + 2 * BORDER(tilesize) + 1;
+}
+
+static void game_set_size(drawing *dr, game_drawstate *ds,
+                          const game_params *params, int tilesize)
+{
+    ds->tilesize = tilesize;
+}
+
+static float *game_colours(frontend *fe, int *ncolours)
+{
+    float *ret = snewn(3 * NCOLOURS, float);
+
+    frontend_default_colour(fe, &ret[COL_BACKGROUND * 3]);
+
+    ret[COL_FOREGROUND * 3 + 0] = 0.0F;
+    ret[COL_FOREGROUND * 3 + 1] = 0.0F;
+    ret[COL_FOREGROUND * 3 + 2] = 0.0F;
+
+    /*
+     * We want COL_LINEUNKNOWN to be a yellow which is a bit darker
+     * than the background. (I previously set it to 0.8,0.8,0, but
+     * found that this went badly with the 0.8,0.8,0.8 favoured as a
+     * background by the Java frontend.)
+     */
+    ret[COL_LINEUNKNOWN * 3 + 0] = ret[COL_BACKGROUND * 3 + 0] * 0.9F;
+    ret[COL_LINEUNKNOWN * 3 + 1] = ret[COL_BACKGROUND * 3 + 1] * 0.9F;
+    ret[COL_LINEUNKNOWN * 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_MISTAKE * 3 + 0] = 1.0F;
+    ret[COL_MISTAKE * 3 + 1] = 0.0F;
+    ret[COL_MISTAKE * 3 + 2] = 0.0F;
+
+    ret[COL_SATISFIED * 3 + 0] = 0.0F;
+    ret[COL_SATISFIED * 3 + 1] = 0.0F;
+    ret[COL_SATISFIED * 3 + 2] = 0.0F;
+
+    /* We want the faint lines to be a bit darker than the background.
+     * Except if the background is pretty dark already; then it ought to be a
+     * bit lighter.  Oy vey.
+     */
+    ret[COL_FAINT * 3 + 0] = ret[COL_BACKGROUND * 3 + 0] * 0.9F;
+    ret[COL_FAINT * 3 + 1] = ret[COL_BACKGROUND * 3 + 1] * 0.9F;
+    ret[COL_FAINT * 3 + 2] = ret[COL_BACKGROUND * 3 + 2] * 0.9F;
+
+    *ncolours = NCOLOURS;
+    return ret;
+}
+
+static game_drawstate *game_new_drawstate(drawing *dr, const game_state *state)
+{
+    struct game_drawstate *ds = snew(struct game_drawstate);
+    int num_faces = state->game_grid->num_faces;
+    int num_edges = state->game_grid->num_edges;
+    int i;
+
+    ds->tilesize = 0;
+    ds->started = 0;
+    ds->lines = snewn(num_edges, char);
+    ds->clue_error = snewn(num_faces, char);
+    ds->clue_satisfied = snewn(num_faces, char);
+    ds->textx = snewn(num_faces, int);
+    ds->texty = snewn(num_faces, int);
+    ds->flashing = 0;
+
+    memset(ds->lines, LINE_UNKNOWN, num_edges);
+    memset(ds->clue_error, 0, num_faces);
+    memset(ds->clue_satisfied, 0, num_faces);
+    for (i = 0; i < num_faces; i++)
+        ds->textx[i] = ds->texty[i] = -1;
+
+    return ds;
+}
+
+static void game_free_drawstate(drawing *dr, game_drawstate *ds)
+{
+    sfree(ds->textx);
+    sfree(ds->texty);
+    sfree(ds->clue_error);
+    sfree(ds->clue_satisfied);
+    sfree(ds->lines);
+    sfree(ds);
+}
+
+static int game_timing_state(const game_state *state, game_ui *ui)
+{
+    return TRUE;
+}
+
+static float game_anim_length(const game_state *oldstate,
+                              const game_state *newstate, int dir, game_ui *ui)
+{
+    return 0.0F;
+}
+
+static int game_can_format_as_text_now(const game_params *params)
+{
+    if (params->type != 0)
+        return FALSE;
+    return TRUE;
+}
+
+static char *game_text_format(const game_state *state)
+{
+    int w, h, W, H;
+    int x, y, i;
+    int cell_size;
+    char *ret;
+    grid *g = state->game_grid;
+    grid_face *f;
+
+    assert(state->grid_type == 0);
+
+    /* Work out the basic size unit */
+    f = g->faces; /* first face */
+    assert(f->order == 4);
+    /* The dots are ordered clockwise, so the two opposite
+     * corners are guaranteed to span the square */
+    cell_size = abs(f->dots[0]->x - f->dots[2]->x);
+
+    w = (g->highest_x - g->lowest_x) / cell_size;
+    h = (g->highest_y - g->lowest_y) / cell_size;
+
+    /* Create a blank "canvas" to "draw" on */
+    W = 2 * w + 2;
+    H = 2 * h + 1;
+    ret = snewn(W * H + 1, char);
+    for (y = 0; y < H; y++) {
+        for (x = 0; x < W-1; x++) {
+            ret[y*W + x] = ' ';
+        }
+        ret[y*W + W-1] = '\n';
+    }
+    ret[H*W] = '\0';
+
+    /* Fill in edge info */
+    for (i = 0; i < g->num_edges; i++) {
+        grid_edge *e = g->edges + i;
+        /* Cell coordinates, from (0,0) to (w-1,h-1) */
+        int x1 = (e->dot1->x - g->lowest_x) / cell_size;
+        int x2 = (e->dot2->x - g->lowest_x) / cell_size;
+        int y1 = (e->dot1->y - g->lowest_y) / cell_size;
+        int y2 = (e->dot2->y - g->lowest_y) / cell_size;
+        /* Midpoint, in canvas coordinates (canvas coordinates are just twice
+         * cell coordinates) */
+        x = x1 + x2;
+        y = y1 + y2;
+        switch (state->lines[i]) {
+          case LINE_YES:
+            ret[y*W + x] = (y1 == y2) ? '-' : '|';
+            break;
+          case LINE_NO:
+            ret[y*W + x] = 'x';
+            break;
+          case LINE_UNKNOWN:
+            break; /* already a space */
+          default:
+            assert(!"Illegal line state");
+        }
+    }
+
+    /* Fill in clues */
+    for (i = 0; i < g->num_faces; i++) {
+        int x1, x2, y1, y2;
+
+        f = g->faces + i;
+        assert(f->order == 4);
+        /* Cell coordinates, from (0,0) to (w-1,h-1) */
+        x1 = (f->dots[0]->x - g->lowest_x) / cell_size;
+        x2 = (f->dots[2]->x - g->lowest_x) / cell_size;
+        y1 = (f->dots[0]->y - g->lowest_y) / cell_size;
+        y2 = (f->dots[2]->y - g->lowest_y) / cell_size;
+        /* Midpoint, in canvas coordinates */
+        x = x1 + x2;
+        y = y1 + y2;
+        ret[y*W + x] = CLUE2CHAR(state->clues[i]);
+    }
+    return ret;
+}
+
+/* ----------------------------------------------------------------------
+ * Debug code
+ */
+
+#ifdef DEBUG_CACHES
+static void check_caches(const solver_state* sstate)
+{
+    int i;
+    const game_state *state = sstate->state;
+    const grid *g = state->game_grid;
+
+    for (i = 0; i < g->num_dots; i++) {
+        assert(dot_order(state, i, LINE_YES) == sstate->dot_yes_count[i]);
+        assert(dot_order(state, i, LINE_NO) == sstate->dot_no_count[i]);
+    }
+
+    for (i = 0; i < g->num_faces; i++) {
+        assert(face_order(state, i, LINE_YES) == sstate->face_yes_count[i]);
+        assert(face_order(state, i, LINE_NO) == sstate->face_no_count[i]);
+    }
+}
+
+#if 0
+#define check_caches(s) \
+    do { \
+        fprintf(stderr, "check_caches at line %d\n", __LINE__); \
+        check_caches(s); \
+    } while (0)
+#endif
+#endif /* DEBUG_CACHES */
+
+/* ----------------------------------------------------------------------
+ * Solver utility functions
+ */
+
+/* Sets the line (with index i) to the new state 'line_new', and updates
+ * the cached counts of any affected faces and dots.
+ * Returns TRUE if this actually changed the line's state. */
+static int solver_set_line(solver_state *sstate, int i,
+                           enum line_state line_new
+#ifdef SHOW_WORKING
+                           , const char *reason
+#endif
+                           )
+{
+    game_state *state = sstate->state;
+    grid *g;
+    grid_edge *e;
+
+    assert(line_new != LINE_UNKNOWN);
+
+    check_caches(sstate);
+
+    if (state->lines[i] == line_new) {
+        return FALSE; /* nothing changed */
+    }
+    state->lines[i] = line_new;
+
+#ifdef SHOW_WORKING
+    fprintf(stderr, "solver: set line [%d] to %s (%s)\n",
+            i, line_new == LINE_YES ? "YES" : "NO",
+            reason);
+#endif
+
+    g = state->game_grid;
+    e = g->edges + i;
+
+    /* Update the cache for both dots and both faces affected by this. */
+    if (line_new == LINE_YES) {
+        sstate->dot_yes_count[e->dot1 - g->dots]++;
+        sstate->dot_yes_count[e->dot2 - g->dots]++;
+        if (e->face1) {
+            sstate->face_yes_count[e->face1 - g->faces]++;
+        }
+        if (e->face2) {
+            sstate->face_yes_count[e->face2 - g->faces]++;
+        }
+    } else {
+        sstate->dot_no_count[e->dot1 - g->dots]++;
+        sstate->dot_no_count[e->dot2 - g->dots]++;
+        if (e->face1) {
+            sstate->face_no_count[e->face1 - g->faces]++;
+        }
+        if (e->face2) {
+            sstate->face_no_count[e->face2 - g->faces]++;
+        }
+    }
+
+    check_caches(sstate);
+    return TRUE;
+}
+
+#ifdef SHOW_WORKING
+#define solver_set_line(a, b, c) \
+    solver_set_line(a, b, c, __FUNCTION__)
+#endif
+
+/*
+ * Merge two dots due to the existence of an edge between them.
+ * Updates the dsf tracking equivalence classes, and keeps track of
+ * the length of path each dot is currently a part of.
+ * Returns TRUE if the dots were already linked, ie if they are part of a
+ * closed loop, and false otherwise.
+ */
+static int merge_dots(solver_state *sstate, int edge_index)
+{
+    int i, j, len;
+    grid *g = sstate->state->game_grid;
+    grid_edge *e = g->edges + edge_index;
+
+    i = e->dot1 - g->dots;
+    j = e->dot2 - g->dots;
+
+    i = dsf_canonify(sstate->dotdsf, i);
+    j = dsf_canonify(sstate->dotdsf, j);
+
+    if (i == j) {
+        return TRUE;
+    } else {
+        len = sstate->looplen[i] + sstate->looplen[j];
+        dsf_merge(sstate->dotdsf, i, j);
+        i = dsf_canonify(sstate->dotdsf, i);
+        sstate->looplen[i] = len;
+        return FALSE;
+    }
+}
+
+/* Merge two lines because the solver has deduced that they must be either
+ * identical or opposite.   Returns TRUE if this is new information, otherwise
+ * FALSE. */
+static int merge_lines(solver_state *sstate, int i, int j, int inverse
+#ifdef SHOW_WORKING
+                       , const char *reason
+#endif
+                       )
+{
+    int inv_tmp;
+
+    assert(i < sstate->state->game_grid->num_edges);
+    assert(j < sstate->state->game_grid->num_edges);
+
+    i = edsf_canonify(sstate->linedsf, i, &inv_tmp);
+    inverse ^= inv_tmp;
+    j = edsf_canonify(sstate->linedsf, j, &inv_tmp);
+    inverse ^= inv_tmp;
+
+    edsf_merge(sstate->linedsf, i, j, inverse);
+
+#ifdef SHOW_WORKING
+    if (i != j) {
+        fprintf(stderr, "%s [%d] [%d] %s(%s)\n",
+                __FUNCTION__, i, j,
+                inverse ? "inverse " : "", reason);
+    }
+#endif
+    return (i != j);
+}
+
+#ifdef SHOW_WORKING
+#define merge_lines(a, b, c, d) \
+    merge_lines(a, b, c, d, __FUNCTION__)
+#endif
+
+/* Count the number of lines of a particular type currently going into the
+ * given dot. */
+static int dot_order(const game_state* state, int dot, char line_type)
+{
+    int n = 0;
+    grid *g = state->game_grid;
+    grid_dot *d = g->dots + dot;
+    int i;
+
+    for (i = 0; i < d->order; i++) {
+        grid_edge *e = d->edges[i];
+        if (state->lines[e - g->edges] == line_type)
+            ++n;
+    }
+    return n;
+}
+
+/* Count the number of lines of a particular type currently surrounding the
+ * given face */
+static int face_order(const game_state* state, int face, char line_type)
+{
+    int n = 0;
+    grid *g = state->game_grid;
+    grid_face *f = g->faces + face;
+    int i;
+
+    for (i = 0; i < f->order; i++) {
+        grid_edge *e = f->edges[i];
+        if (state->lines[e - g->edges] == line_type)
+            ++n;
+    }
+    return n;
+}
+
+/* Set all lines bordering a dot of type old_type to type new_type
+ * Return value tells caller whether this function actually did anything */
+static int dot_setall(solver_state *sstate, int dot,
+                      char old_type, char new_type)
+{
+    int retval = FALSE, r;
+    game_state *state = sstate->state;
+    grid *g;
+    grid_dot *d;
+    int i;
+
+    if (old_type == new_type)
+        return FALSE;
+
+    g = state->game_grid;
+    d = g->dots + dot;
+
+    for (i = 0; i < d->order; i++) {
+        int line_index = d->edges[i] - g->edges;
+        if (state->lines[line_index] == old_type) {
+            r = solver_set_line(sstate, line_index, new_type);
+            assert(r == TRUE);
+            retval = TRUE;
+        }
+    }
+    return retval;
+}
+
+/* Set all lines bordering a face of type old_type to type new_type */
+static int face_setall(solver_state *sstate, int face,
+                       char old_type, char new_type)
+{
+    int retval = FALSE, r;
+    game_state *state = sstate->state;
+    grid *g;
+    grid_face *f;
+    int i;
+
+    if (old_type == new_type)
+        return FALSE;
+
+    g = state->game_grid;
+    f = g->faces + face;
+
+    for (i = 0; i < f->order; i++) {
+        int line_index = f->edges[i] - g->edges;
+        if (state->lines[line_index] == old_type) {
+            r = solver_set_line(sstate, line_index, new_type);
+            assert(r == TRUE);
+            retval = TRUE;
+        }
+    }
+    return retval;
+}
+
+/* ----------------------------------------------------------------------
+ * Loop generation and clue removal
+ */
+
+static void add_full_clues(game_state *state, random_state *rs)
+{
+    signed char *clues = state->clues;
+    grid *g = state->game_grid;
+    char *board = snewn(g->num_faces, char);
+    int i;
+
+    generate_loop(g, board, rs, NULL, NULL);
+
+    /* Fill out all the clues by initialising to 0, then iterating over
+     * all edges and incrementing each clue as we find edges that border
+     * between BLACK/WHITE faces.  While we're at it, we verify that the
+     * algorithm does work, and there aren't any GREY faces still there. */
+    memset(clues, 0, g->num_faces);
+    for (i = 0; i < g->num_edges; i++) {
+        grid_edge *e = g->edges + i;
+        grid_face *f1 = e->face1;
+        grid_face *f2 = e->face2;
+        enum face_colour c1 = FACE_COLOUR(f1);
+        enum face_colour c2 = FACE_COLOUR(f2);
+        assert(c1 != FACE_GREY);
+        assert(c2 != FACE_GREY);
+        if (c1 != c2) {
+            if (f1) clues[f1 - g->faces]++;
+            if (f2) clues[f2 - g->faces]++;
+        }
+    }
+    sfree(board);
+}
+
+
+static int game_has_unique_soln(const game_state *state, int diff)
+{
+    int ret;
+    solver_state *sstate_new;
+    solver_state *sstate = new_solver_state((game_state *)state, diff);
+
+    sstate_new = solve_game_rec(sstate);
+
+    assert(sstate_new->solver_status != SOLVER_MISTAKE);
+    ret = (sstate_new->solver_status == SOLVER_SOLVED);
+
+    free_solver_state(sstate_new);
+    free_solver_state(sstate);
+
+    return ret;
+}
+
+
+/* Remove clues one at a time at random. */
+static game_state *remove_clues(game_state *state, random_state *rs,
+                                int diff)
+{
+    int *face_list;
+    int num_faces = state->game_grid->num_faces;
+    game_state *ret = dup_game(state), *saved_ret;
+    int n;
+
+    /* We need to remove some clues.  We'll do this by forming a list of all
+     * available clues, shuffling it, then going along one at a
+     * time clearing each clue in turn for which doing so doesn't render the
+     * board unsolvable. */
+    face_list = snewn(num_faces, int);
+    for (n = 0; n < num_faces; ++n) {
+        face_list[n] = n;
+    }
+
+    shuffle(face_list, num_faces, sizeof(int), rs);
+
+    for (n = 0; n < num_faces; ++n) {
+        saved_ret = dup_game(ret);
+        ret->clues[face_list[n]] = -1;
+
+        if (game_has_unique_soln(ret, diff)) {
+            free_game(saved_ret);
+        } else {
+            free_game(ret);
+            ret = saved_ret;
+        }
+    }
+    sfree(face_list);
+
+    return ret;
+}
+
+
+static char *new_game_desc(const game_params *params, random_state *rs,
+                           char **aux, int interactive)
+{
+    /* solution and description both use run-length encoding in obvious ways */
+    char *retval, *game_desc, *grid_desc;
+    grid *g;
+    game_state *state = snew(game_state);
+    game_state *state_new;
+
+    grid_desc = grid_new_desc(grid_types[params->type], params->w, params->h, rs);
+    state->game_grid = g = loopy_generate_grid(params, grid_desc);
+
+    state->clues = snewn(g->num_faces, signed char);
+    state->lines = snewn(g->num_edges, char);
+    state->line_errors = snewn(g->num_edges, unsigned char);
+    state->exactly_one_loop = FALSE;
+
+    state->grid_type = params->type;
+
+    newboard_please:
+
+    memset(state->lines, LINE_UNKNOWN, g->num_edges);
+    memset(state->line_errors, 0, g->num_edges);
+
+    state->solved = state->cheated = FALSE;
+
+    /* Get a new random solvable board with all its clues filled in.  Yes, this
+     * can loop for ever if the params are suitably unfavourable, but
+     * preventing games smaller than 4x4 seems to stop this happening */
+    do {
+        add_full_clues(state, rs);
+    } while (!game_has_unique_soln(state, params->diff));
+
+    state_new = remove_clues(state, rs, params->diff);
+    free_game(state);
+    state = state_new;
+
+
+    if (params->diff > 0 && game_has_unique_soln(state, params->diff-1)) {
+#ifdef SHOW_WORKING
+        fprintf(stderr, "Rejecting board, it is too easy\n");
+#endif
+        goto newboard_please;
+    }
+
+    game_desc = state_to_text(state);
+
+    free_game(state);
+
+    if (grid_desc) {
+        retval = snewn(strlen(grid_desc) + 1 + strlen(game_desc) + 1, char);
+        sprintf(retval, "%s%c%s", grid_desc, (int)GRID_DESC_SEP, game_desc);
+        sfree(grid_desc);
+        sfree(game_desc);
+    } else {
+        retval = game_desc;
+    }
+
+    assert(!validate_desc(params, retval));
+
+    return retval;
+}
+
+static game_state *new_game(midend *me, const game_params *params,
+                            const char *desc)
+{
+    int i;
+    game_state *state = snew(game_state);
+    int empties_to_make = 0;
+    int n,n2;
+    const char *dp;
+    char *grid_desc;
+    grid *g;
+    int num_faces, num_edges;
+
+    grid_desc = extract_grid_desc(&desc);
+    state->game_grid = g = loopy_generate_grid(params, grid_desc);
+    if (grid_desc) sfree(grid_desc);
+
+    dp = desc;
+
+    num_faces = g->num_faces;
+    num_edges = g->num_edges;
+
+    state->clues = snewn(num_faces, signed char);
+    state->lines = snewn(num_edges, char);
+    state->line_errors = snewn(num_edges, unsigned char);
+    state->exactly_one_loop = FALSE;
+
+    state->solved = state->cheated = FALSE;
+
+    state->grid_type = params->type;
+
+    for (i = 0; i < num_faces; i++) {
+        if (empties_to_make) {
+            empties_to_make--;
+            state->clues[i] = -1;
+            continue;
+        }
+
+        assert(*dp);
+        n = *dp - '0';
+        n2 = *dp - 'A' + 10;
+        if (n >= 0 && n < 10) {
+            state->clues[i] = n;
+        } else if (n2 >= 10 && n2 < 36) {
+            state->clues[i] = n2;
+        } else {
+            n = *dp - 'a' + 1;
+            assert(n > 0);
+            state->clues[i] = -1;
+            empties_to_make = n - 1;
+        }
+        ++dp;
+    }
+
+    memset(state->lines, LINE_UNKNOWN, num_edges);
+    memset(state->line_errors, 0, num_edges);
+    return state;
+}
+
+/* Calculates the line_errors data, and checks if the current state is a
+ * solution */
+static int check_completion(game_state *state)
+{
+    grid *g = state->game_grid;
+    int i, ret;
+    int *dsf, *component_state;
+    int nsilly, nloop, npath, largest_comp, largest_size, total_pathsize;
+    enum { COMP_NONE, COMP_LOOP, COMP_PATH, COMP_SILLY, COMP_EMPTY };
+
+    memset(state->line_errors, 0, g->num_edges);
+
+    /*
+     * Find loops in the grid, and determine whether the puzzle is
+     * solved.
+     *
+     * Loopy is a bit more complicated than most puzzles that care
+     * about loop detection. In most of them, loops are simply
+     * _forbidden_; so the obviously right way to do
+     * error-highlighting during play is to light up a graph edge red
+     * iff it is part of a loop, which is exactly what the centralised
+     * findloop.c makes easy.
+     *
+     * But Loopy is unusual in that you're _supposed_ to be making a
+     * loop - and yet _some_ loops are not the right loop. So we need
+     * to be more discriminating, by identifying loops one by one and
+     * then thinking about which ones to highlight, and so findloop.c
+     * isn't quite the right tool for the job in this case.
+     *
+     * Worse still, consider situations in which the grid contains a
+     * loop and also some non-loop edges: there are some cases like
+     * this in which the user's intuitive expectation would be to
+     * highlight the loop (if you're only about half way through the
+     * puzzle and have accidentally made a little loop in some corner
+     * of the grid), and others in which they'd be more likely to
+     * expect you to highlight the non-loop edges (if you've just
+     * closed off a whole loop that you thought was the entire
+     * solution, but forgot some disconnected edges in a corner
+     * somewhere). So while it's easy enough to check whether the
+     * solution is _right_, highlighting the wrong parts is a tricky
+     * problem for this puzzle!
+     *
+     * I'd quite like, in some situations, to identify the largest
+     * loop among the player's YES edges, and then light up everything
+     * other than that. But finding the longest cycle in a graph is an
+     * NP-complete problem (because, in particular, it must return a
+     * Hamilton cycle if one exists).
+     *
+     * However, I think we can make the problem tractable by
+     * exercising the Puzzles principle that it isn't absolutely
+     * necessary to highlight _all_ errors: the key point is that by
+     * the time the user has filled in the whole grid, they should
+     * either have seen a completion flash, or have _some_ error
+     * highlight showing them why the solution isn't right. So in
+     * principle it would be *just about* good enough to highlight
+     * just one error in the whole grid, if there was really no better
+     * way. But we'd like to highlight as many errors as possible.
+     *
+     * In this case, I think the simple approach is to make use of the
+     * fact that no vertex may have degree > 2, and that's really
+     * simple to detect. So the plan goes like this:
+     *
+     *  - Form the dsf of connected components of the graph vertices.
+     *
+     *  - Highlight an error at any vertex with degree > 2. (It so
+     *    happens that we do this by lighting up all the edges
+     *    incident to that vertex, but that's an output detail.)
+     *
+     *  - Any component that contains such a vertex is now excluded
+     *    from further consideration, because it already has a
+     *    highlight.
+     *
+     *  - The remaining components have no vertex with degree > 2, and
+     *    hence they all consist of either a simple loop, or a simple
+     *    path with two endpoints.
+     *
+     *  - For these purposes, group together all the paths and imagine
+     *    them to be a single component (because in most normal
+     *    situations the player will gradually build up the solution
+     *    _not_ all in one connected segment, but as lots of separate
+     *    little path pieces that gradually connect to each other).
+     *
+     *  - After doing that, if there is exactly one (sensible)
+     *    component - be it a collection of paths or a loop - then
+     *    highlight no further edge errors. (The former case is normal
+     *    during play, and the latter is a potentially solved puzzle.)
+     *
+     *  - Otherwise, find the largest of the sensible components,
+     *    leave that one unhighlighted, and light the rest up in red.
+     */
+
+    dsf = snew_dsf(g->num_dots);
+
+    /* Build the dsf. */
+    for (i = 0; i < g->num_edges; i++) {
+        if (state->lines[i] == LINE_YES) {
+            grid_edge *e = g->edges + i;
+            int d1 = e->dot1 - g->dots, d2 = e->dot2 - g->dots;
+            dsf_merge(dsf, d1, d2);
+        }
+    }
+
+    /* Initialise a state variable for each connected component. */
+    component_state = snewn(g->num_dots, int);
+    for (i = 0; i < g->num_dots; i++) {
+        if (dsf_canonify(dsf, i) == i)
+            component_state[i] = COMP_LOOP;
+        else
+            component_state[i] = COMP_NONE;
+    }
+
+    /* Check for dots with degree > 3. Here we also spot dots of
+     * degree 1 in which the user has marked all the non-edges as
+     * LINE_NO, because those are also clear vertex-level errors, so
+     * we give them the same treatment of excluding their connected
+     * component from the subsequent loop analysis. */
+    for (i = 0; i < g->num_dots; i++) {
+        int comp = dsf_canonify(dsf, i);
+        int yes = dot_order(state, i, LINE_YES);
+        int unknown = dot_order(state, i, LINE_UNKNOWN);
+        if ((yes == 1 && unknown == 0) || (yes >= 3)) {
+            /* violation, so mark all YES edges as errors */
+            grid_dot *d = g->dots + i;
+            int j;
+            for (j = 0; j < d->order; j++) {
+                int e = d->edges[j] - g->edges;
+                if (state->lines[e] == LINE_YES)
+                    state->line_errors[e] = TRUE;
+            }
+            /* And mark this component as not worthy of further
+             * consideration. */
+            component_state[comp] = COMP_SILLY;
+
+        } else if (yes == 0) {
+            /* A completely isolated dot must also be excluded it from
+             * the subsequent loop highlighting pass, but we tag it
+             * with a different enum value to avoid it counting
+             * towards the components that inhibit returning a win
+             * status. */
+            component_state[comp] = COMP_EMPTY;
+        } else if (yes == 1) {
+            /* A dot with degree 1 that didn't fall into the 'clearly
+             * erroneous' case above indicates that this connected
+             * component will be a path rather than a loop - unless
+             * something worse elsewhere in the component has
+             * classified it as silly. */
+            if (component_state[comp] != COMP_SILLY)
+                component_state[comp] = COMP_PATH;
+        }
+    }
+
+    /* Count up the components. Also, find the largest sensible
+     * component. (Tie-breaking condition is derived from the order of
+     * vertices in the grid data structure, which is fairly arbitrary
+     * but at least stays stable throughout the game.) */
+    nsilly = nloop = npath = 0;
+    total_pathsize = 0;
+    largest_comp = largest_size = -1;
+    for (i = 0; i < g->num_dots; i++) {
+        if (component_state[i] == COMP_SILLY) {
+            nsilly++;
+        } else if (component_state[i] == COMP_PATH) {
+            total_pathsize += dsf_size(dsf, i);
+            npath = 1;
+        } else if (component_state[i] == COMP_LOOP) {
+            int this_size;
+
+            nloop++;
+
+            if ((this_size = dsf_size(dsf, i)) > largest_size) {
+                largest_comp = i;
+                largest_size = this_size;
+            }
+        }
+    }
+    if (largest_size < total_pathsize) {
+        largest_comp = -1;             /* means the paths */
+        largest_size = total_pathsize;
+    }
+
+    if (nloop > 0 && nloop + npath > 1) {
+        /*
+         * If there are at least two sensible components including at
+         * least one loop, highlight all edges in every sensible
+         * component that is not the largest one.
+         */
+        for (i = 0; i < g->num_edges; i++) {
+            if (state->lines[i] == LINE_YES) {
+                grid_edge *e = g->edges + i;
+                int d1 = e->dot1 - g->dots; /* either endpoint is good enough */
+                int comp = dsf_canonify(dsf, d1);
+                if ((component_state[comp] == COMP_PATH &&
+                     -1 != largest_comp) ||
+                    (component_state[comp] == COMP_LOOP &&
+                     comp != largest_comp))
+                    state->line_errors[i] = TRUE;
+            }
+        }
+    }
+
+    if (nloop == 1 && npath == 0 && nsilly == 0) {
+        /*
+         * If there is exactly one component and it is a loop, then
+         * the puzzle is potentially complete, so check the clues.
+         */
+        ret = TRUE;
+
+        for (i = 0; i < g->num_faces; i++) {
+            int c = state->clues[i];
+            if (c >= 0 && face_order(state, i, LINE_YES) != c) {
+                ret = FALSE;
+                break;
+            }
+        }
+
+        /*
+         * Also, whether or not the puzzle is actually complete, set
+         * the flag that says this game_state has exactly one loop and
+         * nothing else, which will be used to vary the semantics of
+         * clue highlighting at display time.
+         */
+        state->exactly_one_loop = TRUE;
+    } else {
+        ret = FALSE;
+        state->exactly_one_loop = FALSE;
+    }
+
+    sfree(component_state);
+    sfree(dsf);
+
+    return ret;
+}
+
+/* ----------------------------------------------------------------------
+ * Solver logic
+ *
+ * Our solver modes operate as follows.  Each mode also uses the modes above it.
+ *
+ *   Easy Mode
+ *   Just implement the rules of the game.
+ *
+ *   Normal and Tricky Modes
+ *   For each (adjacent) pair of lines through each dot we store a bit for
+ *   whether at least one of them is on and whether at most one is on.  (If we
+ *   know both or neither is on that's already stored more directly.)
+ *
+ *   Advanced Mode
+ *   Use edsf data structure to make equivalence classes of lines that are
+ *   known identical to or opposite to one another.
+ */
+
+
+/* DLines:
+ * For general grids, we consider "dlines" to be pairs of lines joined
+ * at a dot.  The lines must be adjacent around the dot, so we can think of
+ * a dline as being a dot+face combination.  Or, a dot+edge combination where
+ * the second edge is taken to be the next clockwise edge from the dot.
+ * Original loopy code didn't have this extra restriction of the lines being
+ * adjacent.  From my tests with square grids, this extra restriction seems to
+ * take little, if anything, away from the quality of the puzzles.
+ * A dline can be uniquely identified by an edge/dot combination, given that
+ * a dline-pair always goes clockwise around its common dot.  The edge/dot
+ * combination can be represented by an edge/bool combination - if bool is
+ * TRUE, use edge->dot1 else use edge->dot2.  So the total number of dlines is
+ * exactly twice the number of edges in the grid - although the dlines
+ * spanning the infinite face are not all that useful to the solver.
+ * Note that, by convention, a dline goes clockwise around its common dot,
+ * which means the dline goes anti-clockwise around its common face.
+ */
+
+/* Helper functions for obtaining an index into an array of dlines, given
+ * various information.  We assume the grid layout conventions about how
+ * the various lists are interleaved - see grid_make_consistent() for
+ * details. */
+
+/* i points to the first edge of the dline pair, reading clockwise around
+ * the dot. */
+static int dline_index_from_dot(grid *g, grid_dot *d, int i)
+{
+    grid_edge *e = d->edges[i];
+    int ret;
+#ifdef DEBUG_DLINES
+    grid_edge *e2;
+    int i2 = i+1;
+    if (i2 == d->order) i2 = 0;
+    e2 = d->edges[i2];
+#endif
+    ret = 2 * (e - g->edges) + ((e->dot1 == d) ? 1 : 0);
+#ifdef DEBUG_DLINES
+    printf("dline_index_from_dot: d=%d,i=%d, edges [%d,%d] - %d\n",
+           (int)(d - g->dots), i, (int)(e - g->edges),
+           (int)(e2 - g->edges), ret);
+#endif
+    return ret;
+}
+/* i points to the second edge of the dline pair, reading clockwise around
+ * the face.  That is, the edges of the dline, starting at edge{i}, read
+ * anti-clockwise around the face.  By layout conventions, the common dot
+ * of the dline will be f->dots[i] */
+static int dline_index_from_face(grid *g, grid_face *f, int i)
+{
+    grid_edge *e = f->edges[i];
+    grid_dot *d = f->dots[i];
+    int ret;
+#ifdef DEBUG_DLINES
+    grid_edge *e2;
+    int i2 = i - 1;
+    if (i2 < 0) i2 += f->order;
+    e2 = f->edges[i2];
+#endif
+    ret = 2 * (e - g->edges) + ((e->dot1 == d) ? 1 : 0);
+#ifdef DEBUG_DLINES
+    printf("dline_index_from_face: f=%d,i=%d, edges [%d,%d] - %d\n",
+           (int)(f - g->faces), i, (int)(e - g->edges),
+           (int)(e2 - g->edges), ret);
+#endif
+    return ret;
+}
+static int is_atleastone(const char *dline_array, int index)
+{
+    return BIT_SET(dline_array[index], 0);
+}
+static int set_atleastone(char *dline_array, int index)
+{
+    return SET_BIT(dline_array[index], 0);
+}
+static int is_atmostone(const char *dline_array, int index)
+{
+    return BIT_SET(dline_array[index], 1);
+}
+static int set_atmostone(char *dline_array, int index)
+{
+    return SET_BIT(dline_array[index], 1);
+}
+
+static void array_setall(char *array, char from, char to, int len)
+{
+    char *p = array, *p_old = p;
+    int len_remaining = len;
+
+    while ((p = memchr(p, from, len_remaining))) {
+        *p = to;
+        len_remaining -= p - p_old;
+        p_old = p;
+    }
+}
+
+/* Helper, called when doing dline dot deductions, in the case where we
+ * have 4 UNKNOWNs, and two of them (adjacent) have *exactly* one YES between
+ * them (because of dline atmostone/atleastone).
+ * On entry, edge points to the first of these two UNKNOWNs.  This function
+ * will find the opposite UNKNOWNS (if they are adjacent to one another)
+ * and set their corresponding dline to atleastone.  (Setting atmostone
+ * already happens in earlier dline deductions) */
+static int dline_set_opp_atleastone(solver_state *sstate,
+                                    grid_dot *d, int edge)
+{
+    game_state *state = sstate->state;
+    grid *g = state->game_grid;
+    int N = d->order;
+    int opp, opp2;
+    for (opp = 0; opp < N; opp++) {
+        int opp_dline_index;
+        if (opp == edge || opp == edge+1 || opp == edge-1)
+            continue;
+        if (opp == 0 && edge == N-1)
+            continue;
+        if (opp == N-1 && edge == 0)
+            continue;
+        opp2 = opp + 1;
+        if (opp2 == N) opp2 = 0;
+        /* Check if opp, opp2 point to LINE_UNKNOWNs */
+        if (state->lines[d->edges[opp] - g->edges] != LINE_UNKNOWN)
+            continue;
+        if (state->lines[d->edges[opp2] - g->edges] != LINE_UNKNOWN)
+            continue;
+        /* Found opposite UNKNOWNS and they're next to each other */
+        opp_dline_index = dline_index_from_dot(g, d, opp);
+        return set_atleastone(sstate->dlines, opp_dline_index);
+    }
+    return FALSE;
+}
+
+
+/* Set pairs of lines around this face which are known to be identical, to
+ * the given line_state */
+static int face_setall_identical(solver_state *sstate, int face_index,
+                                 enum line_state line_new)
+{
+    /* can[dir] contains the canonical line associated with the line in
+     * direction dir from the square in question.  Similarly inv[dir] is
+     * whether or not the line in question is inverse to its canonical
+     * element. */
+    int retval = FALSE;
+    game_state *state = sstate->state;
+    grid *g = state->game_grid;
+    grid_face *f = g->faces + face_index;
+    int N = f->order;
+    int i, j;
+    int can1, can2, inv1, inv2;
+
+    for (i = 0; i < N; i++) {
+        int line1_index = f->edges[i] - g->edges;
+        if (state->lines[line1_index] != LINE_UNKNOWN)
+            continue;
+        for (j = i + 1; j < N; j++) {
+            int line2_index = f->edges[j] - g->edges;
+            if (state->lines[line2_index] != LINE_UNKNOWN)
+                continue;
+
+            /* Found two UNKNOWNS */
+            can1 = edsf_canonify(sstate->linedsf, line1_index, &inv1);
+            can2 = edsf_canonify(sstate->linedsf, line2_index, &inv2);
+            if (can1 == can2 && inv1 == inv2) {
+                solver_set_line(sstate, line1_index, line_new);
+                solver_set_line(sstate, line2_index, line_new);
+            }
+        }
+    }
+    return retval;
+}
+
+/* Given a dot or face, and a count of LINE_UNKNOWNs, find them and
+ * return the edge indices into e. */
+static void find_unknowns(game_state *state,
+    grid_edge **edge_list, /* Edge list to search (from a face or a dot) */
+    int expected_count, /* Number of UNKNOWNs (comes from solver's cache) */
+    int *e /* Returned edge indices */)
+{
+    int c = 0;
+    grid *g = state->game_grid;
+    while (c < expected_count) {
+        int line_index = *edge_list - g->edges;
+        if (state->lines[line_index] == LINE_UNKNOWN) {
+            e[c] = line_index;
+            c++;
+        }
+        ++edge_list;
+    }
+}
+
+/* If we have a list of edges, and we know whether the number of YESs should
+ * be odd or even, and there are only a few UNKNOWNs, we can do some simple
+ * linedsf deductions.  This can be used for both face and dot deductions.
+ * Returns the difficulty level of the next solver that should be used,
+ * or DIFF_MAX if no progress was made. */
+static int parity_deductions(solver_state *sstate,
+    grid_edge **edge_list, /* Edge list (from a face or a dot) */
+    int total_parity, /* Expected number of YESs modulo 2 (either 0 or 1) */
+    int unknown_count)
+{
+    game_state *state = sstate->state;
+    int diff = DIFF_MAX;
+    int *linedsf = sstate->linedsf;
+
+    if (unknown_count == 2) {
+        /* Lines are known alike/opposite, depending on inv. */
+        int e[2];
+        find_unknowns(state, edge_list, 2, e);
+        if (merge_lines(sstate, e[0], e[1], total_parity))
+            diff = min(diff, DIFF_HARD);
+    } else if (unknown_count == 3) {
+        int e[3];
+        int can[3]; /* canonical edges */
+        int inv[3]; /* whether can[x] is inverse to e[x] */
+        find_unknowns(state, edge_list, 3, e);
+        can[0] = edsf_canonify(linedsf, e[0], inv);
+        can[1] = edsf_canonify(linedsf, e[1], inv+1);
+        can[2] = edsf_canonify(linedsf, e[2], inv+2);
+        if (can[0] == can[1]) {
+            if (solver_set_line(sstate, e[2], (total_parity^inv[0]^inv[1]) ?
+                                LINE_YES : LINE_NO))
+                diff = min(diff, DIFF_EASY);
+        }
+        if (can[0] == can[2]) {
+            if (solver_set_line(sstate, e[1], (total_parity^inv[0]^inv[2]) ?
+                                LINE_YES : LINE_NO))
+                diff = min(diff, DIFF_EASY);
+        }
+        if (can[1] == can[2]) {
+            if (solver_set_line(sstate, e[0], (total_parity^inv[1]^inv[2]) ?
+                                LINE_YES : LINE_NO))
+                diff = min(diff, DIFF_EASY);
+        }
+    } else if (unknown_count == 4) {
+        int e[4];
+        int can[4]; /* canonical edges */
+        int inv[4]; /* whether can[x] is inverse to e[x] */
+        find_unknowns(state, edge_list, 4, e);
+        can[0] = edsf_canonify(linedsf, e[0], inv);
+        can[1] = edsf_canonify(linedsf, e[1], inv+1);
+        can[2] = edsf_canonify(linedsf, e[2], inv+2);
+        can[3] = edsf_canonify(linedsf, e[3], inv+3);
+        if (can[0] == can[1]) {
+            if (merge_lines(sstate, e[2], e[3], total_parity^inv[0]^inv[1]))
+                diff = min(diff, DIFF_HARD);
+        } else if (can[0] == can[2]) {
+            if (merge_lines(sstate, e[1], e[3], total_parity^inv[0]^inv[2]))
+                diff = min(diff, DIFF_HARD);
+        } else if (can[0] == can[3]) {
+            if (merge_lines(sstate, e[1], e[2], total_parity^inv[0]^inv[3]))
+                diff = min(diff, DIFF_HARD);
+        } else if (can[1] == can[2]) {
+            if (merge_lines(sstate, e[0], e[3], total_parity^inv[1]^inv[2]))
+                diff = min(diff, DIFF_HARD);
+        } else if (can[1] == can[3]) {
+            if (merge_lines(sstate, e[0], e[2], total_parity^inv[1]^inv[3]))
+                diff = min(diff, DIFF_HARD);
+        } else if (can[2] == can[3]) {
+            if (merge_lines(sstate, e[0], e[1], total_parity^inv[2]^inv[3]))
+                diff = min(diff, DIFF_HARD);
+        }
+    }
+    return diff;
+}
+
+
+/*
+ * These are the main solver functions.
+ *
+ * Their return values are diff values corresponding to the lowest mode solver
+ * that would notice the work that they have done.  For example if the normal
+ * mode solver adds actual lines or crosses, it will return DIFF_EASY as the
+ * easy mode solver might be able to make progress using that.  It doesn't make
+ * sense for one of them to return a diff value higher than that of the
+ * function itself.
+ *
+ * Each function returns the lowest value it can, as early as possible, in
+ * order to try and pass as much work as possible back to the lower level
+ * solvers which progress more quickly.
+ */
+
+/* PROPOSED NEW DESIGN:
+ * We have a work queue consisting of 'events' notifying us that something has
+ * happened that a particular solver mode might be interested in.  For example
+ * the hard mode solver might do something that helps the normal mode solver at
+ * dot [x,y] in which case it will enqueue an event recording this fact.  Then
+ * we pull events off the work queue, and hand each in turn to the solver that
+ * is interested in them.  If a solver reports that it failed we pass the same
+ * event on to progressively more advanced solvers and the loop detector.  Once
+ * we've exhausted an event, or it has helped us progress, we drop it and
+ * continue to the next one.  The events are sorted first in order of solver
+ * complexity (easy first) then order of insertion (oldest first).
+ * Once we run out of events we loop over each permitted solver in turn
+ * (easiest first) until either a deduction is made (and an event therefore
+ * emerges) or no further deductions can be made (in which case we've failed).
+ *
+ * QUESTIONS:
+ *    * How do we 'loop over' a solver when both dots and squares are concerned.
+ *      Answer: first all squares then all dots.
+ */
+
+static int trivial_deductions(solver_state *sstate)
+{
+    int i, current_yes, current_no;
+    game_state *state = sstate->state;
+    grid *g = state->game_grid;
+    int diff = DIFF_MAX;
+
+    /* Per-face deductions */
+    for (i = 0; i < g->num_faces; i++) {
+        grid_face *f = g->faces + i;
+
+        if (sstate->face_solved[i])
+            continue;
+
+        current_yes = sstate->face_yes_count[i];
+        current_no  = sstate->face_no_count[i];
+
+        if (current_yes + current_no == f->order)  {
+            sstate->face_solved[i] = TRUE;
+            continue;
+        }
+
+        if (state->clues[i] < 0)
+            continue;
+
+        /*
+         * This code checks whether the numeric clue on a face is so
+         * large as to permit all its remaining LINE_UNKNOWNs to be
+         * filled in as LINE_YES, or alternatively so small as to
+         * permit them all to be filled in as LINE_NO.
+         */
+
+        if (state->clues[i] < current_yes) {
+            sstate->solver_status = SOLVER_MISTAKE;
+            return DIFF_EASY;
+        }
+        if (state->clues[i] == current_yes) {
+            if (face_setall(sstate, i, LINE_UNKNOWN, LINE_NO))
+                diff = min(diff, DIFF_EASY);
+            sstate->face_solved[i] = TRUE;
+            continue;
+        }
+
+        if (f->order - state->clues[i] < current_no) {
+            sstate->solver_status = SOLVER_MISTAKE;
+            return DIFF_EASY;
+        }
+        if (f->order - state->clues[i] == current_no) {
+            if (face_setall(sstate, i, LINE_UNKNOWN, LINE_YES))
+                diff = min(diff, DIFF_EASY);
+            sstate->face_solved[i] = TRUE;
+            continue;
+        }
+
+        if (f->order - state->clues[i] == current_no + 1 &&
+            f->order - current_yes - current_no > 2) {
+            /*
+             * One small refinement to the above: we also look for any
+             * adjacent pair of LINE_UNKNOWNs around the face with
+             * some LINE_YES incident on it from elsewhere. If we find
+             * one, then we know that pair of LINE_UNKNOWNs can't
+             * _both_ be LINE_YES, and hence that pushes us one line
+             * closer to being able to determine all the rest.
+             */
+            int j, k, e1, e2, e, d;
+
+            for (j = 0; j < f->order; j++) {
+                e1 = f->edges[j] - g->edges;
+                e2 = f->edges[j+1 < f->order ? j+1 : 0] - g->edges;
+
+                if (g->edges[e1].dot1 == g->edges[e2].dot1 ||
+                    g->edges[e1].dot1 == g->edges[e2].dot2) {
+                    d = g->edges[e1].dot1 - g->dots;
+                } else {
+                    assert(g->edges[e1].dot2 == g->edges[e2].dot1 ||
+                           g->edges[e1].dot2 == g->edges[e2].dot2);
+                    d = g->edges[e1].dot2 - g->dots;
+                }
+
+                if (state->lines[e1] == LINE_UNKNOWN &&
+                    state->lines[e2] == LINE_UNKNOWN) {
+                    for (k = 0; k < g->dots[d].order; k++) {
+                        int e = g->dots[d].edges[k] - g->edges;
+                        if (state->lines[e] == LINE_YES)
+                            goto found;    /* multi-level break */
+                    }
+                }
+            }
+            continue;
+
+          found:
+            /*
+             * If we get here, we've found such a pair of edges, and
+             * they're e1 and e2.
+             */
+            for (j = 0; j < f->order; j++) {
+                e = f->edges[j] - g->edges;
+                if (state->lines[e] == LINE_UNKNOWN && e != e1 && e != e2) {
+                    int r = solver_set_line(sstate, e, LINE_YES);
+                    assert(r);
+                    diff = min(diff, DIFF_EASY);
+                }
+            }
+        }
+    }
+
+    check_caches(sstate);
+
+    /* Per-dot deductions */
+    for (i = 0; i < g->num_dots; i++) {
+        grid_dot *d = g->dots + i;
+        int yes, no, unknown;
+
+        if (sstate->dot_solved[i])
+            continue;
+
+        yes = sstate->dot_yes_count[i];
+        no = sstate->dot_no_count[i];
+        unknown = d->order - yes - no;
+
+        if (yes == 0) {
+            if (unknown == 0) {
+                sstate->dot_solved[i] = TRUE;
+            } else if (unknown == 1) {
+                dot_setall(sstate, i, LINE_UNKNOWN, LINE_NO);
+                diff = min(diff, DIFF_EASY);
+                sstate->dot_solved[i] = TRUE;
+            }
+        } else if (yes == 1) {
+            if (unknown == 0) {
+                sstate->solver_status = SOLVER_MISTAKE;
+                return DIFF_EASY;
+            } else if (unknown == 1) {
+                dot_setall(sstate, i, LINE_UNKNOWN, LINE_YES);
+                diff = min(diff, DIFF_EASY);
+            }
+        } else if (yes == 2) {
+            if (unknown > 0) {
+                dot_setall(sstate, i, LINE_UNKNOWN, LINE_NO);
+                diff = min(diff, DIFF_EASY);
+            }
+            sstate->dot_solved[i] = TRUE;
+        } else {
+            sstate->solver_status = SOLVER_MISTAKE;
+            return DIFF_EASY;
+        }
+    }
+
+    check_caches(sstate);
+
+    return diff;
+}
+
+static int dline_deductions(solver_state *sstate)
+{
+    game_state *state = sstate->state;
+    grid *g = state->game_grid;
+    char *dlines = sstate->dlines;
+    int i;
+    int diff = DIFF_MAX;
+
+    /* ------ Face deductions ------ */
+
+    /* Given a set of dline atmostone/atleastone constraints, need to figure
+     * out if we can deduce any further info.  For more general faces than
+     * squares, this turns out to be a tricky problem.
+     * The approach taken here is to define (per face) NxN matrices:
+     * "maxs" and "mins".
+     * The entries maxs(j,k) and mins(j,k) define the upper and lower limits
+     * for the possible number of edges that are YES between positions j and k
+     * going clockwise around the face.  Can think of j and k as marking dots
+     * around the face (recall the labelling scheme: edge0 joins dot0 to dot1,
+     * edge1 joins dot1 to dot2 etc).
+     * Trivially, mins(j,j) = maxs(j,j) = 0, and we don't even bother storing
+     * these.  mins(j,j+1) and maxs(j,j+1) are determined by whether edge{j}
+     * is YES, NO or UNKNOWN.  mins(j,j+2) and maxs(j,j+2) are related to
+     * the dline atmostone/atleastone status for edges j and j+1.
+     *
+     * Then we calculate the remaining entries recursively.  We definitely
+     * know that
+     * mins(j,k) >= { mins(j,u) + mins(u,k) } for any u between j and k.
+     * This is because any valid placement of YESs between j and k must give
+     * a valid placement between j and u, and also between u and k.
+     * I believe it's sufficient to use just the two values of u:
+     * j+1 and j+2.  Seems to work well in practice - the bounds we compute
+     * are rigorous, even if they might not be best-possible.
+     *
+     * Once we have maxs and mins calculated, we can make inferences about
+     * each dline{j,j+1} by looking at the possible complementary edge-counts
+     * mins(j+2,j) and maxs(j+2,j) and comparing these with the face clue.
+     * As well as dlines, we can make similar inferences about single edges.
+     * For example, consider a pentagon with clue 3, and we know at most one
+     * of (edge0, edge1) is YES, and at most one of (edge2, edge3) is YES.
+     * We could then deduce edge4 is YES, because maxs(0,4) would be 2, so
+     * that final edge would have to be YES to make the count up to 3.
+     */
+
+    /* Much quicker to allocate arrays on the stack than the heap, so
+     * define the largest possible face size, and base our array allocations
+     * on that.  We check this with an assertion, in case someone decides to
+     * make a grid which has larger faces than this.  Note, this algorithm
+     * could get quite expensive if there are many large faces. */
+#define MAX_FACE_SIZE 12
+
+    for (i = 0; i < g->num_faces; i++) {
+        int maxs[MAX_FACE_SIZE][MAX_FACE_SIZE];
+        int mins[MAX_FACE_SIZE][MAX_FACE_SIZE];
+        grid_face *f = g->faces + i;
+        int N = f->order;
+        int j,m;
+        int clue = state->clues[i];
+        assert(N <= MAX_FACE_SIZE);
+        if (sstate->face_solved[i])
+            continue;
+        if (clue < 0) continue;
+
+        /* Calculate the (j,j+1) entries */
+        for (j = 0; j < N; j++) {
+            int edge_index = f->edges[j] - g->edges;
+            int dline_index;
+            enum line_state line1 = state->lines[edge_index];
+            enum line_state line2;
+            int tmp;
+            int k = j + 1;
+            if (k >= N) k = 0;
+            maxs[j][k] = (line1 == LINE_NO) ? 0 : 1;
+            mins[j][k] = (line1 == LINE_YES) ? 1 : 0;
+            /* Calculate the (j,j+2) entries */
+            dline_index = dline_index_from_face(g, f, k);
+            edge_index = f->edges[k] - g->edges;
+            line2 = state->lines[edge_index];
+            k++;
+            if (k >= N) k = 0;
+
+            /* max */
+            tmp = 2;
+            if (line1 == LINE_NO) tmp--;
+            if (line2 == LINE_NO) tmp--;
+            if (tmp == 2 && is_atmostone(dlines, dline_index))
+                tmp = 1;
+            maxs[j][k] = tmp;
+
+            /* min */
+            tmp = 0;
+            if (line1 == LINE_YES) tmp++;
+            if (line2 == LINE_YES) tmp++;
+            if (tmp == 0 && is_atleastone(dlines, dline_index))
+                tmp = 1;
+            mins[j][k] = tmp;
+        }
+
+        /* Calculate the (j,j+m) entries for m between 3 and N-1 */
+        for (m = 3; m < N; m++) {
+            for (j = 0; j < N; j++) {
+                int k = j + m;
+                int u = j + 1;
+                int v = j + 2;
+                int tmp;
+                if (k >= N) k -= N;
+                if (u >= N) u -= N;
+                if (v >= N) v -= N;
+                maxs[j][k] = maxs[j][u] + maxs[u][k];
+                mins[j][k] = mins[j][u] + mins[u][k];
+                tmp = maxs[j][v] + maxs[v][k];
+                maxs[j][k] = min(maxs[j][k], tmp);
+                tmp = mins[j][v] + mins[v][k];
+                mins[j][k] = max(mins[j][k], tmp);
+            }
+        }
+
+        /* See if we can make any deductions */
+        for (j = 0; j < N; j++) {
+            int k;
+            grid_edge *e = f->edges[j];
+            int line_index = e - g->edges;
+            int dline_index;
+
+            if (state->lines[line_index] != LINE_UNKNOWN)
+                continue;
+            k = j + 1;
+            if (k >= N) k = 0;
+
+            /* minimum YESs in the complement of this edge */
+            if (mins[k][j] > clue) {
+                sstate->solver_status = SOLVER_MISTAKE;
+                return DIFF_EASY;
+            }
+            if (mins[k][j] == clue) {
+                /* setting this edge to YES would make at least
+                 * (clue+1) edges - contradiction */
+                solver_set_line(sstate, line_index, LINE_NO);
+                diff = min(diff, DIFF_EASY);
+            }
+            if (maxs[k][j] < clue - 1) {
+                sstate->solver_status = SOLVER_MISTAKE;
+                return DIFF_EASY;
+            }
+            if (maxs[k][j] == clue - 1) {
+                /* Only way to satisfy the clue is to set edge{j} as YES */
+                solver_set_line(sstate, line_index, LINE_YES);
+                diff = min(diff, DIFF_EASY);
+            }
+
+            /* More advanced deduction that allows propagation along diagonal
+             * chains of faces connected by dots, for example, 3-2-...-2-3
+             * in square grids. */
+            if (sstate->diff >= DIFF_TRICKY) {
+                /* Now see if we can make dline deduction for edges{j,j+1} */
+                e = f->edges[k];
+                if (state->lines[e - g->edges] != LINE_UNKNOWN)
+                    /* Only worth doing this for an UNKNOWN,UNKNOWN pair.
+                     * Dlines where one of the edges is known, are handled in the
+                     * dot-deductions */
+                    continue;
+    
+                dline_index = dline_index_from_face(g, f, k);
+                k++;
+                if (k >= N) k = 0;
+    
+                /* minimum YESs in the complement of this dline */
+                if (mins[k][j] > clue - 2) {
+                    /* Adding 2 YESs would break the clue */
+                    if (set_atmostone(dlines, dline_index))
+                        diff = min(diff, DIFF_NORMAL);
+                }
+                /* maximum YESs in the complement of this dline */
+                if (maxs[k][j] < clue) {
+                    /* Adding 2 NOs would mean not enough YESs */
+                    if (set_atleastone(dlines, dline_index))
+                        diff = min(diff, DIFF_NORMAL);
+                }
+            }
+        }
+    }
+
+    if (diff < DIFF_NORMAL)
+        return diff;
+
+    /* ------ Dot deductions ------ */
+
+    for (i = 0; i < g->num_dots; i++) {
+        grid_dot *d = g->dots + i;
+        int N = d->order;
+        int yes, no, unknown;
+        int j;
+        if (sstate->dot_solved[i])
+            continue;
+        yes = sstate->dot_yes_count[i];
+        no = sstate->dot_no_count[i];
+        unknown = N - yes - no;
+
+        for (j = 0; j < N; j++) {
+            int k;
+            int dline_index;
+            int line1_index, line2_index;
+            enum line_state line1, line2;
+            k = j + 1;
+            if (k >= N) k = 0;
+            dline_index = dline_index_from_dot(g, d, j);
+            line1_index = d->edges[j] - g->edges;
+            line2_index = d->edges[k] - g->edges;
+            line1 = state->lines[line1_index];
+            line2 = state->lines[line2_index];
+
+            /* Infer dline state from line state */
+            if (line1 == LINE_NO || line2 == LINE_NO) {
+                if (set_atmostone(dlines, dline_index))
+                    diff = min(diff, DIFF_NORMAL);
+            }
+            if (line1 == LINE_YES || line2 == LINE_YES) {
+                if (set_atleastone(dlines, dline_index))
+                    diff = min(diff, DIFF_NORMAL);
+            }
+            /* Infer line state from dline state */
+            if (is_atmostone(dlines, dline_index)) {
+                if (line1 == LINE_YES && line2 == LINE_UNKNOWN) {
+                    solver_set_line(sstate, line2_index, LINE_NO);
+                    diff = min(diff, DIFF_EASY);
+                }
+                if (line2 == LINE_YES && line1 == LINE_UNKNOWN) {
+                    solver_set_line(sstate, line1_index, LINE_NO);
+                    diff = min(diff, DIFF_EASY);
+                }
+            }
+            if (is_atleastone(dlines, dline_index)) {
+                if (line1 == LINE_NO && line2 == LINE_UNKNOWN) {
+                    solver_set_line(sstate, line2_index, LINE_YES);
+                    diff = min(diff, DIFF_EASY);
+                }
+                if (line2 == LINE_NO && line1 == LINE_UNKNOWN) {
+                    solver_set_line(sstate, line1_index, LINE_YES);
+                    diff = min(diff, DIFF_EASY);
+                }
+            }
+            /* Deductions that depend on the numbers of lines.
+             * Only bother if both lines are UNKNOWN, otherwise the
+             * easy-mode solver (or deductions above) would have taken
+             * care of it. */
+            if (line1 != LINE_UNKNOWN || line2 != LINE_UNKNOWN)
+                continue;
+
+            if (yes == 0 && unknown == 2) {
+                /* Both these unknowns must be identical.  If we know
+                 * atmostone or atleastone, we can make progress. */
+                if (is_atmostone(dlines, dline_index)) {
+                    solver_set_line(sstate, line1_index, LINE_NO);
+                    solver_set_line(sstate, line2_index, LINE_NO);
+                    diff = min(diff, DIFF_EASY);
+                }
+                if (is_atleastone(dlines, dline_index)) {
+                    solver_set_line(sstate, line1_index, LINE_YES);
+                    solver_set_line(sstate, line2_index, LINE_YES);
+                    diff = min(diff, DIFF_EASY);
+                }
+            }
+            if (yes == 1) {
+                if (set_atmostone(dlines, dline_index))
+                    diff = min(diff, DIFF_NORMAL);
+                if (unknown == 2) {
+                    if (set_atleastone(dlines, dline_index))
+                        diff = min(diff, DIFF_NORMAL);
+                }
+            }
+
+            /* More advanced deduction that allows propagation along diagonal
+             * chains of faces connected by dots, for example: 3-2-...-2-3
+             * in square grids. */
+            if (sstate->diff >= DIFF_TRICKY) {
+                /* If we have atleastone set for this dline, infer
+                 * atmostone for each "opposite" dline (that is, each
+                 * dline without edges in common with this one).
+                 * Again, this test is only worth doing if both these
+                 * lines are UNKNOWN.  For if one of these lines were YES,
+                 * the (yes == 1) test above would kick in instead. */
+                if (is_atleastone(dlines, dline_index)) {
+                    int opp;
+                    for (opp = 0; opp < N; opp++) {
+                        int opp_dline_index;
+                        if (opp == j || opp == j+1 || opp == j-1)
+                            continue;
+                        if (j == 0 && opp == N-1)
+                            continue;
+                        if (j == N-1 && opp == 0)
+                            continue;
+                        opp_dline_index = dline_index_from_dot(g, d, opp);
+                        if (set_atmostone(dlines, opp_dline_index))
+                            diff = min(diff, DIFF_NORMAL);
+                    }
+                    if (yes == 0 && is_atmostone(dlines, dline_index)) {
+                        /* This dline has *exactly* one YES and there are no
+                         * other YESs.  This allows more deductions. */
+                        if (unknown == 3) {
+                            /* Third unknown must be YES */
+                            for (opp = 0; opp < N; opp++) {
+                                int opp_index;
+                                if (opp == j || opp == k)
+                                    continue;
+                                opp_index = d->edges[opp] - g->edges;
+                                if (state->lines[opp_index] == LINE_UNKNOWN) {
+                                    solver_set_line(sstate, opp_index,
+                                                    LINE_YES);
+                                    diff = min(diff, DIFF_EASY);
+                                }
+                            }
+                        } else if (unknown == 4) {
+                            /* Exactly one of opposite UNKNOWNS is YES.  We've
+                             * already set atmostone, so set atleastone as
+                             * well.
+                             */
+                            if (dline_set_opp_atleastone(sstate, d, j))
+                                diff = min(diff, DIFF_NORMAL);
+                        }
+                    }
+                }
+            }
+        }
+    }
+    return diff;
+}
+
+static int linedsf_deductions(solver_state *sstate)
+{
+    game_state *state = sstate->state;
+    grid *g = state->game_grid;
+    char *dlines = sstate->dlines;
+    int i;
+    int diff = DIFF_MAX;
+    int diff_tmp;
+
+    /* ------ Face deductions ------ */
+
+    /* A fully-general linedsf deduction seems overly complicated
+     * (I suspect the problem is NP-complete, though in practice it might just
+     * be doable because faces are limited in size).
+     * For simplicity, we only consider *pairs* of LINE_UNKNOWNS that are
+     * known to be identical.  If setting them both to YES (or NO) would break
+     * the clue, set them to NO (or YES). */
+
+    for (i = 0; i < g->num_faces; i++) {
+        int N, yes, no, unknown;
+        int clue;
+
+        if (sstate->face_solved[i])
+            continue;
+        clue = state->clues[i];
+        if (clue < 0)
+            continue;
+
+        N = g->faces[i].order;
+        yes = sstate->face_yes_count[i];
+        if (yes + 1 == clue) {
+            if (face_setall_identical(sstate, i, LINE_NO))
+                diff = min(diff, DIFF_EASY);
+        }
+        no = sstate->face_no_count[i];
+        if (no + 1 == N - clue) {
+            if (face_setall_identical(sstate, i, LINE_YES))
+                diff = min(diff, DIFF_EASY);
+        }
+
+        /* Reload YES count, it might have changed */
+        yes = sstate->face_yes_count[i];
+        unknown = N - no - yes;
+
+        /* Deductions with small number of LINE_UNKNOWNs, based on overall
+         * parity of lines. */
+        diff_tmp = parity_deductions(sstate, g->faces[i].edges,
+                                     (clue - yes) % 2, unknown);
+        diff = min(diff, diff_tmp);
+    }
+
+    /* ------ Dot deductions ------ */
+    for (i = 0; i < g->num_dots; i++) {
+        grid_dot *d = g->dots + i;
+        int N = d->order;
+        int j;
+        int yes, no, unknown;
+        /* Go through dlines, and do any dline<->linedsf deductions wherever
+         * we find two UNKNOWNS. */
+        for (j = 0; j < N; j++) {
+            int dline_index = dline_index_from_dot(g, d, j);
+            int line1_index;
+            int line2_index;
+            int can1, can2, inv1, inv2;
+            int j2;
+            line1_index = d->edges[j] - g->edges;
+            if (state->lines[line1_index] != LINE_UNKNOWN)
+                continue;
+            j2 = j + 1;
+            if (j2 == N) j2 = 0;
+            line2_index = d->edges[j2] - g->edges;
+            if (state->lines[line2_index] != LINE_UNKNOWN)
+                continue;
+            /* Infer dline flags from linedsf */
+            can1 = edsf_canonify(sstate->linedsf, line1_index, &inv1);
+            can2 = edsf_canonify(sstate->linedsf, line2_index, &inv2);
+            if (can1 == can2 && inv1 != inv2) {
+                /* These are opposites, so set dline atmostone/atleastone */
+                if (set_atmostone(dlines, dline_index))
+                    diff = min(diff, DIFF_NORMAL);
+                if (set_atleastone(dlines, dline_index))
+                    diff = min(diff, DIFF_NORMAL);
+                continue;
+            }
+            /* Infer linedsf from dline flags */
+            if (is_atmostone(dlines, dline_index)
+                && is_atleastone(dlines, dline_index)) {
+                if (merge_lines(sstate, line1_index, line2_index, 1))
+                    diff = min(diff, DIFF_HARD);
+            }
+        }
+
+        /* Deductions with small number of LINE_UNKNOWNs, based on overall
+         * parity of lines. */
+        yes = sstate->dot_yes_count[i];
+        no = sstate->dot_no_count[i];
+        unknown = N - yes - no;
+        diff_tmp = parity_deductions(sstate, d->edges,
+                                     yes % 2, unknown);
+        diff = min(diff, diff_tmp);
+    }
+
+    /* ------ Edge dsf deductions ------ */
+
+    /* If the state of a line is known, deduce the state of its canonical line
+     * too, and vice versa. */
+    for (i = 0; i < g->num_edges; i++) {
+        int can, inv;
+        enum line_state s;
+        can = edsf_canonify(sstate->linedsf, i, &inv);
+        if (can == i)
+            continue;
+        s = sstate->state->lines[can];
+        if (s != LINE_UNKNOWN) {
+            if (solver_set_line(sstate, i, inv ? OPP(s) : s))
+                diff = min(diff, DIFF_EASY);
+        } else {
+            s = sstate->state->lines[i];
+            if (s != LINE_UNKNOWN) {
+                if (solver_set_line(sstate, can, inv ? OPP(s) : s))
+                    diff = min(diff, DIFF_EASY);
+            }
+        }
+    }
+
+    return diff;
+}
+
+static int loop_deductions(solver_state *sstate)
+{
+    int edgecount = 0, clues = 0, satclues = 0, sm1clues = 0;
+    game_state *state = sstate->state;
+    grid *g = state->game_grid;
+    int shortest_chainlen = g->num_dots;
+    int loop_found = FALSE;
+    int dots_connected;
+    int progress = FALSE;
+    int i;
+
+    /*
+     * Go through the grid and update for all the new edges.
+     * Since merge_dots() is idempotent, the simplest way to
+     * do this is just to update for _all_ the edges.
+     * Also, while we're here, we count the edges.
+     */
+    for (i = 0; i < g->num_edges; i++) {
+        if (state->lines[i] == LINE_YES) {
+            loop_found |= merge_dots(sstate, i);
+            edgecount++;
+        }
+    }
+
+    /*
+     * Count the clues, count the satisfied clues, and count the
+     * satisfied-minus-one clues.
+     */
+    for (i = 0; i < g->num_faces; i++) {
+        int c = state->clues[i];
+        if (c >= 0) {
+            int o = sstate->face_yes_count[i];
+            if (o == c)
+                satclues++;
+            else if (o == c-1)
+                sm1clues++;
+            clues++;
+        }
+    }
+
+    for (i = 0; i < g->num_dots; ++i) {
+        dots_connected =
+            sstate->looplen[dsf_canonify(sstate->dotdsf, i)];
+        if (dots_connected > 1)
+            shortest_chainlen = min(shortest_chainlen, dots_connected);
+    }
+
+    assert(sstate->solver_status == SOLVER_INCOMPLETE);
+
+    if (satclues == clues && shortest_chainlen == edgecount) {
+        sstate->solver_status = SOLVER_SOLVED;
+        /* This discovery clearly counts as progress, even if we haven't
+         * just added any lines or anything */
+        progress = TRUE;
+        goto finished_loop_deductionsing;
+    }
+
+    /*
+     * Now go through looking for LINE_UNKNOWN edges which
+     * connect two dots that are already in the same
+     * equivalence class. If we find one, test to see if the
+     * loop it would create is a solution.
+     */
+    for (i = 0; i < g->num_edges; i++) {
+        grid_edge *e = g->edges + i;
+        int d1 = e->dot1 - g->dots;
+        int d2 = e->dot2 - g->dots;
+        int eqclass, val;
+        if (state->lines[i] != LINE_UNKNOWN)
+            continue;
+
+        eqclass = dsf_canonify(sstate->dotdsf, d1);
+        if (eqclass != dsf_canonify(sstate->dotdsf, d2))
+            continue;
+
+        val = LINE_NO;  /* loop is bad until proven otherwise */
+
+        /*
+         * This edge would form a loop. Next
+         * question: how long would the loop be?
+         * Would it equal the total number of edges
+         * (plus the one we'd be adding if we added
+         * it)?
+         */
+        if (sstate->looplen[eqclass] == edgecount + 1) {
+            int sm1_nearby;
+
+            /*
+             * This edge would form a loop which
+             * took in all the edges in the entire
+             * grid. So now we need to work out
+             * whether it would be a valid solution
+             * to the puzzle, which means we have to
+             * check if it satisfies all the clues.
+             * This means that every clue must be
+             * either satisfied or satisfied-minus-
+             * 1, and also that the number of
+             * satisfied-minus-1 clues must be at
+             * most two and they must lie on either
+             * side of this edge.
+             */
+            sm1_nearby = 0;
+            if (e->face1) {
+                int f = e->face1 - g->faces;
+                int c = state->clues[f];
+                if (c >= 0 && sstate->face_yes_count[f] == c - 1)
+                    sm1_nearby++;
+            }
+            if (e->face2) {
+                int f = e->face2 - g->faces;
+                int c = state->clues[f];
+                if (c >= 0 && sstate->face_yes_count[f] == c - 1)
+                    sm1_nearby++;
+            }
+            if (sm1clues == sm1_nearby &&
+                sm1clues + satclues == clues) {
+                val = LINE_YES;  /* loop is good! */
+            }
+        }
+
+        /*
+         * Right. Now we know that adding this edge
+         * would form a loop, and we know whether
+         * that loop would be a viable solution or
+         * not.
+         *
+         * If adding this edge produces a solution,
+         * then we know we've found _a_ solution but
+         * we don't know that it's _the_ solution -
+         * if it were provably the solution then
+         * we'd have deduced this edge some time ago
+         * without the need to do loop detection. So
+         * in this state we return SOLVER_AMBIGUOUS,
+         * which has the effect that hitting Solve
+         * on a user-provided puzzle will fill in a
+         * solution but using the solver to
+         * construct new puzzles won't consider this
+         * a reasonable deduction for the user to
+         * make.
+         */
+        progress = solver_set_line(sstate, i, val);
+        assert(progress == TRUE);
+        if (val == LINE_YES) {
+            sstate->solver_status = SOLVER_AMBIGUOUS;
+            goto finished_loop_deductionsing;
+        }
+    }
+
+    finished_loop_deductionsing:
+    return progress ? DIFF_EASY : DIFF_MAX;
+}
+
+/* This will return a dynamically allocated solver_state containing the (more)
+ * solved grid */
+static solver_state *solve_game_rec(const solver_state *sstate_start)
+{
+    solver_state *sstate;
+
+    /* Index of the solver we should call next. */
+    int i = 0;
+    
+    /* As a speed-optimisation, we avoid re-running solvers that we know
+     * won't make any progress.  This happens when a high-difficulty
+     * solver makes a deduction that can only help other high-difficulty
+     * solvers.
+     * For example: if a new 'dline' flag is set by dline_deductions, the
+     * trivial_deductions solver cannot do anything with this information.
+     * If we've already run the trivial_deductions solver (because it's
+     * earlier in the list), there's no point running it again.
+     *
+     * Therefore: if a solver is earlier in the list than "threshold_index",
+     * we don't bother running it if it's difficulty level is less than
+     * "threshold_diff".
+     */
+    int threshold_diff = 0;
+    int threshold_index = 0;
+    
+    sstate = dup_solver_state(sstate_start);
+
+    check_caches(sstate);
+
+    while (i < NUM_SOLVERS) {
+        if (sstate->solver_status == SOLVER_MISTAKE)
+            return sstate;
+        if (sstate->solver_status == SOLVER_SOLVED ||
+            sstate->solver_status == SOLVER_AMBIGUOUS) {
+            /* solver finished */
+            break;
+        }
+
+        if ((solver_diffs[i] >= threshold_diff || i >= threshold_index)
+            && solver_diffs[i] <= sstate->diff) {
+            /* current_solver is eligible, so use it */
+            int next_diff = solver_fns[i](sstate);
+            if (next_diff != DIFF_MAX) {
+                /* solver made progress, so use new thresholds and
+                * start again at top of list. */
+                threshold_diff = next_diff;
+                threshold_index = i;
+                i = 0;
+                continue;
+            }
+        }
+        /* current_solver is ineligible, or failed to make progress, so
+         * go to the next solver in the list */
+        i++;
+    }
+
+    if (sstate->solver_status == SOLVER_SOLVED ||
+        sstate->solver_status == SOLVER_AMBIGUOUS) {
+        /* s/LINE_UNKNOWN/LINE_NO/g */
+        array_setall(sstate->state->lines, LINE_UNKNOWN, LINE_NO,
+                     sstate->state->game_grid->num_edges);
+        return sstate;
+    }
+
+    return sstate;
+}
+
+static char *solve_game(const game_state *state, const game_state *currstate,
+                        const char *aux, char **error)
+{
+    char *soln = NULL;
+    solver_state *sstate, *new_sstate;
+
+    sstate = new_solver_state(state, DIFF_MAX);
+    new_sstate = solve_game_rec(sstate);
+
+    if (new_sstate->solver_status == SOLVER_SOLVED) {
+        soln = encode_solve_move(new_sstate->state);
+    } else if (new_sstate->solver_status == SOLVER_AMBIGUOUS) {
+        soln = encode_solve_move(new_sstate->state);
+        /**error = "Solver found ambiguous solutions"; */
+    } else {
+        soln = encode_solve_move(new_sstate->state);
+        /**error = "Solver failed"; */
+    }
+
+    free_solver_state(new_sstate);
+    free_solver_state(sstate);
+
+    return soln;
+}
+
+/* ----------------------------------------------------------------------
+ * Drawing and mouse-handling
+ */
+
+static char *interpret_move(const game_state *state, game_ui *ui,
+                            const game_drawstate *ds,
+                            int x, int y, int button)
+{
+    grid *g = state->game_grid;
+    grid_edge *e;
+    int i;
+    char *ret, buf[80];
+    char button_char = ' ';
+    enum line_state old_state;
+
+    button &= ~MOD_MASK;
+
+    /* Convert mouse-click (x,y) to grid coordinates */
+    x -= BORDER(ds->tilesize);
+    y -= BORDER(ds->tilesize);
+    x = x * g->tilesize / ds->tilesize;
+    y = y * g->tilesize / ds->tilesize;
+    x += g->lowest_x;
+    y += g->lowest_y;
+
+    e = grid_nearest_edge(g, x, y);
+    if (e == NULL)
+        return NULL;
+
+    i = e - g->edges;
+
+    /* I think it's only possible to play this game with mouse clicks, sorry */
+    /* Maybe will add mouse drag support some time */
+    old_state = state->lines[i];
+
+    switch (button) {
+      case LEFT_BUTTON:
+        switch (old_state) {
+          case LINE_UNKNOWN:
+            button_char = 'y';
+            break;
+          case LINE_YES:
+#ifdef STYLUS_BASED
+            button_char = 'n';
+            break;
+#endif
+          case LINE_NO:
+            button_char = 'u';
+            break;
+        }
+        break;
+      case MIDDLE_BUTTON:
+        button_char = 'u';
+        break;
+      case RIGHT_BUTTON:
+        switch (old_state) {
+          case LINE_UNKNOWN:
+            button_char = 'n';
+            break;
+          case LINE_NO:
+#ifdef STYLUS_BASED
+            button_char = 'y';
+            break;
+#endif
+          case LINE_YES:
+            button_char = 'u';
+            break;
+        }
+        break;
+      default:
+        return NULL;
+    }
+
+
+    sprintf(buf, "%d%c", i, (int)button_char);
+    ret = dupstr(buf);
+
+    return ret;
+}
+
+static game_state *execute_move(const game_state *state, const char *move)
+{
+    int i;
+    game_state *newstate = dup_game(state);
+
+    if (move[0] == 'S') {
+        move++;
+        newstate->cheated = TRUE;
+    }
+
+    while (*move) {
+        i = atoi(move);
+        if (i < 0 || i >= newstate->game_grid->num_edges)
+            goto fail;
+        move += strspn(move, "1234567890");
+        switch (*(move++)) {
+          case 'y':
+            newstate->lines[i] = LINE_YES;
+            break;
+          case 'n':
+            newstate->lines[i] = LINE_NO;
+            break;
+          case 'u':
+            newstate->lines[i] = LINE_UNKNOWN;
+            break;
+          default:
+            goto fail;
+        }
+    }
+
+    /*
+     * Check for completion.
+     */
+    if (check_completion(newstate))
+        newstate->solved = TRUE;
+
+    return newstate;
+
+    fail:
+    free_game(newstate);
+    return NULL;
+}
+
+/* ----------------------------------------------------------------------
+ * Drawing routines.
+ */
+
+/* Convert from grid coordinates to screen coordinates */
+static void grid_to_screen(const game_drawstate *ds, const grid *g,
+                           int grid_x, int grid_y, int *x, int *y)
+{
+    *x = grid_x - g->lowest_x;
+    *y = grid_y - g->lowest_y;
+    *x = *x * ds->tilesize / g->tilesize;
+    *y = *y * ds->tilesize / g->tilesize;
+    *x += BORDER(ds->tilesize);
+    *y += BORDER(ds->tilesize);
+}
+
+/* Returns (into x,y) position of centre of face for rendering the text clue.
+ */
+static void face_text_pos(const game_drawstate *ds, const grid *g,
+                          grid_face *f, int *xret, int *yret)
+{
+    int faceindex = f - g->faces;
+
+    /*
+     * Return the cached position for this face, if we've already
+     * worked it out.
+     */
+    if (ds->textx[faceindex] >= 0) {
+        *xret = ds->textx[faceindex];
+        *yret = ds->texty[faceindex];
+        return;
+    }
+
+    /*
+     * Otherwise, use the incentre computed by grid.c and convert it
+     * to screen coordinates.
+     */
+    grid_find_incentre(f);
+    grid_to_screen(ds, g, f->ix, f->iy,
+                   &ds->textx[faceindex], &ds->texty[faceindex]);
+
+    *xret = ds->textx[faceindex];
+    *yret = ds->texty[faceindex];
+}
+
+static void face_text_bbox(game_drawstate *ds, grid *g, grid_face *f,
+                           int *x, int *y, int *w, int *h)
+{
+    int xx, yy;
+    face_text_pos(ds, g, f, &xx, &yy);
+
+    /* There seems to be a certain amount of trial-and-error involved
+     * in working out the correct bounding-box for the text. */
+
+    *x = xx - ds->tilesize/4 - 1;
+    *y = yy - ds->tilesize/4 - 3;
+    *w = ds->tilesize/2 + 2;
+    *h = ds->tilesize/2 + 5;
+}
+
+static void game_redraw_clue(drawing *dr, game_drawstate *ds,
+                             const game_state *state, int i)
+{
+    grid *g = state->game_grid;
+    grid_face *f = g->faces + i;
+    int x, y;
+    char c[20];
+
+    sprintf(c, "%d", state->clues[i]);
+
+    face_text_pos(ds, g, f, &x, &y);
+    draw_text(dr, x, y,
+              FONT_VARIABLE, ds->tilesize/2,
+              ALIGN_VCENTRE | ALIGN_HCENTRE,
+              ds->clue_error[i] ? COL_MISTAKE :
+              ds->clue_satisfied[i] ? COL_SATISFIED : COL_FOREGROUND, c);
+}
+
+static void edge_bbox(game_drawstate *ds, grid *g, grid_edge *e,
+                      int *x, int *y, int *w, int *h)
+{
+    int x1 = e->dot1->x;
+    int y1 = e->dot1->y;
+    int x2 = e->dot2->x;
+    int y2 = e->dot2->y;
+    int xmin, xmax, ymin, ymax;
+
+    grid_to_screen(ds, g, x1, y1, &x1, &y1);
+    grid_to_screen(ds, g, x2, y2, &x2, &y2);
+    /* Allow extra margin for dots, and thickness of lines */
+    xmin = min(x1, x2) - 2;
+    xmax = max(x1, x2) + 2;
+    ymin = min(y1, y2) - 2;
+    ymax = max(y1, y2) + 2;
+
+    *x = xmin;
+    *y = ymin;
+    *w = xmax - xmin + 1;
+    *h = ymax - ymin + 1;
+}
+
+static void dot_bbox(game_drawstate *ds, grid *g, grid_dot *d,
+                     int *x, int *y, int *w, int *h)
+{
+    int x1, y1;
+
+    grid_to_screen(ds, g, d->x, d->y, &x1, &y1);
+
+    *x = x1 - 2;
+    *y = y1 - 2;
+    *w = 5;
+    *h = 5;
+}
+
+static const int loopy_line_redraw_phases[] = {
+    COL_FAINT, COL_LINEUNKNOWN, COL_FOREGROUND, COL_HIGHLIGHT, COL_MISTAKE
+};
+#define NPHASES lenof(loopy_line_redraw_phases)
+
+static void game_redraw_line(drawing *dr, game_drawstate *ds,
+                             const game_state *state, int i, int phase)
+{
+    grid *g = state->game_grid;
+    grid_edge *e = g->edges + i;
+    int x1, x2, y1, y2;
+    int line_colour;
+
+    if (state->line_errors[i])
+        line_colour = COL_MISTAKE;
+    else if (state->lines[i] == LINE_UNKNOWN)
+        line_colour = COL_LINEUNKNOWN;
+    else if (state->lines[i] == LINE_NO)
+        line_colour = COL_FAINT;
+    else if (ds->flashing)
+        line_colour = COL_HIGHLIGHT;
+    else
+        line_colour = COL_FOREGROUND;
+    if (line_colour != loopy_line_redraw_phases[phase])
+        return;
+
+    /* Convert from grid to screen coordinates */
+    grid_to_screen(ds, g, e->dot1->x, e->dot1->y, &x1, &y1);
+    grid_to_screen(ds, g, e->dot2->x, e->dot2->y, &x2, &y2);
+
+    if (line_colour == COL_FAINT) {
+        static int draw_faint_lines = -1;
+        if (draw_faint_lines < 0) {
+            char *env = getenv("LOOPY_FAINT_LINES");
+            draw_faint_lines = (!env || (env[0] == 'y' ||
+                                         env[0] == 'Y'));
+        }
+        if (draw_faint_lines)
+            draw_line(dr, x1, y1, x2, y2, line_colour);
+    } else {
+        draw_thick_line(dr, 3.0,
+                        x1 + 0.5, y1 + 0.5,
+                        x2 + 0.5, y2 + 0.5,
+                        line_colour);
+    }
+}
+
+static void game_redraw_dot(drawing *dr, game_drawstate *ds,
+                            const game_state *state, int i)
+{
+    grid *g = state->game_grid;
+    grid_dot *d = g->dots + i;
+    int x, y;
+
+    grid_to_screen(ds, g, d->x, d->y, &x, &y);
+    draw_circle(dr, x, y, 2, COL_FOREGROUND, COL_FOREGROUND);
+}
+
+static int boxes_intersect(int x0, int y0, int w0, int h0,
+                           int x1, int y1, int w1, int h1)
+{
+    /*
+     * Two intervals intersect iff neither is wholly on one side of
+     * the other. Two boxes intersect iff their horizontal and
+     * vertical intervals both intersect.
+     */
+    return (x0 < x1+w1 && x1 < x0+w0 && y0 < y1+h1 && y1 < y0+h0);
+}
+
+static void game_redraw_in_rect(drawing *dr, game_drawstate *ds,
+                                const game_state *state,
+                                int x, int y, int w, int h)
+{
+    grid *g = state->game_grid;
+    int i, phase;
+    int bx, by, bw, bh;
+
+    clip(dr, x, y, w, h);
+    draw_rect(dr, x, y, w, h, COL_BACKGROUND);
+
+    for (i = 0; i < g->num_faces; i++) {
+        if (state->clues[i] >= 0) {
+            face_text_bbox(ds, g, &g->faces[i], &bx, &by, &bw, &bh);
+            if (boxes_intersect(x, y, w, h, bx, by, bw, bh))
+                game_redraw_clue(dr, ds, state, i);
+        }
+    }
+    for (phase = 0; phase < NPHASES; phase++) {
+        for (i = 0; i < g->num_edges; i++) {
+            edge_bbox(ds, g, &g->edges[i], &bx, &by, &bw, &bh);
+            if (boxes_intersect(x, y, w, h, bx, by, bw, bh))
+                game_redraw_line(dr, ds, state, i, phase);
+        }
+    }
+    for (i = 0; i < g->num_dots; i++) {
+        dot_bbox(ds, g, &g->dots[i], &bx, &by, &bw, &bh);
+        if (boxes_intersect(x, y, w, h, bx, by, bw, bh))
+            game_redraw_dot(dr, ds, state, i);
+    }
+
+    unclip(dr);
+    draw_update(dr, x, y, w, h);
+}
+
+static void game_redraw(drawing *dr, game_drawstate *ds,
+                        const game_state *oldstate, const game_state *state,
+                        int dir, const game_ui *ui,
+                        float animtime, float flashtime)
+{
+#define REDRAW_OBJECTS_LIMIT 16         /* Somewhat arbitrary tradeoff */
+
+    grid *g = state->game_grid;
+    int border = BORDER(ds->tilesize);
+    int i;
+    int flash_changed;
+    int redraw_everything = FALSE;
+
+    int edges[REDRAW_OBJECTS_LIMIT], nedges = 0;
+    int faces[REDRAW_OBJECTS_LIMIT], nfaces = 0;
+
+    /* Redrawing is somewhat involved.
+     *
+     * An update can theoretically affect an arbitrary number of edges
+     * (consider, for example, completing or breaking a cycle which doesn't
+     * satisfy all the clues -- we'll switch many edges between error and
+     * normal states).  On the other hand, redrawing the whole grid takes a
+     * while, making the game feel sluggish, and many updates are actually
+     * quite well localized.
+     *
+     * This redraw algorithm attempts to cope with both situations gracefully
+     * and correctly.  For localized changes, we set a clip rectangle, fill
+     * it with background, and then redraw (a plausible but conservative
+     * guess at) the objects which intersect the rectangle; if several
+     * objects need redrawing, we'll do them individually.  However, if lots
+     * of objects are affected, we'll just redraw everything.
+     *
+     * The reason for all of this is that it's just not safe to do the redraw
+     * piecemeal.  If you try to draw an antialiased diagonal line over
+     * itself, you get a slightly thicker antialiased diagonal line, which
+     * looks rather ugly after a while.
+     *
+     * So, we take two passes over the grid.  The first attempts to work out
+     * what needs doing, and the second actually does it.
+     */
+
+    if (!ds->started) {
+        redraw_everything = TRUE;
+        /*
+         * But we must still go through the upcoming loops, so that we
+         * set up stuff in ds correctly for the initial redraw.
+         */
+    }
+
+    /* First, trundle through the faces. */
+    for (i = 0; i < g->num_faces; i++) {
+        grid_face *f = g->faces + i;
+        int sides = f->order;
+        int yes_order, no_order;
+        int clue_mistake;
+        int clue_satisfied;
+        int n = state->clues[i];
+        if (n < 0)
+            continue;
+
+        yes_order = face_order(state, i, LINE_YES);
+        if (state->exactly_one_loop) {
+            /*
+             * Special case: if the set of LINE_YES edges in the grid
+             * consists of exactly one loop and nothing else, then we
+             * switch to treating LINE_UNKNOWN the same as LINE_NO for
+             * purposes of clue checking.
+             *
+             * This is because some people like to play Loopy without
+             * using the right-click, i.e. never setting anything to
+             * LINE_NO. Without this special case, if a person playing
+             * in that style fills in what they think is a correct
+             * solution loop but in fact it has an underfilled clue,
+             * then we will display no victory flash and also no error
+             * highlight explaining why not. With this special case,
+             * we light up underfilled clues at the instant the loop
+             * is closed. (Of course, *overfilled* clues are fine
+             * either way.)
+             *
+             * (It might still be considered unfortunate that we can't
+             * warn this style of player any earlier, if they make a
+             * mistake very near the beginning which doesn't show up
+             * until they close the last edge of the loop. One other
+             * thing we _could_ do here is to treat any LINE_UNKNOWN
+             * as LINE_NO if either of its endpoints has yes-degree 2,
+             * reflecting the fact that setting that line to YES would
+             * be an obvious error. But I don't think even that could
+             * catch _all_ clue errors in a timely manner; I think
+             * there are some that won't be displayed until the loop
+             * is filled in, even so, and there's no way to avoid that
+             * with complete reliability except to switch to being a
+             * player who sets things to LINE_NO.)
+             */
+            no_order = sides - yes_order;
+        } else {
+            no_order = face_order(state, i, LINE_NO);
+        }
+
+        clue_mistake = (yes_order > n || no_order > (sides-n));
+        clue_satisfied = (yes_order == n && no_order == (sides-n));
+
+        if (clue_mistake != ds->clue_error[i] ||
+            clue_satisfied != ds->clue_satisfied[i]) {
+            ds->clue_error[i] = clue_mistake;
+            ds->clue_satisfied[i] = clue_satisfied;
+            if (nfaces == REDRAW_OBJECTS_LIMIT)
+                redraw_everything = TRUE;
+            else
+                faces[nfaces++] = i;
+        }
+    }
+
+    /* Work out what the flash state needs to be. */
+    if (flashtime > 0 &&
+        (flashtime <= FLASH_TIME/3 ||
+         flashtime >= FLASH_TIME*2/3)) {
+        flash_changed = !ds->flashing;
+        ds->flashing = TRUE;
+    } else {
+        flash_changed = ds->flashing;
+        ds->flashing = FALSE;
+    }
+
+    /* Now, trundle through the edges. */
+    for (i = 0; i < g->num_edges; i++) {
+        char new_ds =
+            state->line_errors[i] ? DS_LINE_ERROR : state->lines[i];
+        if (new_ds != ds->lines[i] ||
+            (flash_changed && state->lines[i] == LINE_YES)) {
+            ds->lines[i] = new_ds;
+            if (nedges == REDRAW_OBJECTS_LIMIT)
+                redraw_everything = TRUE;
+            else
+                edges[nedges++] = i;
+        }
+    }
+
+    /* Pass one is now done.  Now we do the actual drawing. */
+    if (redraw_everything) {
+        int grid_width = g->highest_x - g->lowest_x;
+        int grid_height = g->highest_y - g->lowest_y;
+        int w = grid_width * ds->tilesize / g->tilesize;
+        int h = grid_height * ds->tilesize / g->tilesize;
+
+        game_redraw_in_rect(dr, ds, state,
+                            0, 0, w + 2*border + 1, h + 2*border + 1);
+    } else {
+
+        /* Right.  Now we roll up our sleeves. */
+
+        for (i = 0; i < nfaces; i++) {
+            grid_face *f = g->faces + faces[i];
+            int x, y, w, h;
+
+            face_text_bbox(ds, g, f, &x, &y, &w, &h);
+            game_redraw_in_rect(dr, ds, state, x, y, w, h);
+        }
+
+        for (i = 0; i < nedges; i++) {
+            grid_edge *e = g->edges + edges[i];
+            int x, y, w, h;
+
+            edge_bbox(ds, g, e, &x, &y, &w, &h);
+            game_redraw_in_rect(dr, ds, state, x, y, w, h);
+        }
+    }
+
+    ds->started = TRUE;
+}
+
+static float game_flash_length(const game_state *oldstate,
+                               const game_state *newstate, int dir, game_ui *ui)
+{
+    if (!oldstate->solved  &&  newstate->solved &&
+        !oldstate->cheated && !newstate->cheated) {
+        return FLASH_TIME;
+    }
+
+    return 0.0F;
+}
+
+static int game_status(const game_state *state)
+{
+    return state->solved ? +1 : 0;
+}
+
+static void game_print_size(const game_params *params, float *x, float *y)
+{
+    int pw, ph;
+
+    /*
+     * I'll use 7mm "squares" by default.
+     */
+    game_compute_size(params, 700, &pw, &ph);
+    *x = pw / 100.0F;
+    *y = ph / 100.0F;
+}
+
+static void game_print(drawing *dr, const game_state *state, int tilesize)
+{
+    int ink = print_mono_colour(dr, 0);
+    int i;
+    game_drawstate ads, *ds = &ads;
+    grid *g = state->game_grid;
+
+    ds->tilesize = tilesize;
+    ds->textx = snewn(g->num_faces, int);
+    ds->texty = snewn(g->num_faces, int);
+    for (i = 0; i < g->num_faces; i++)
+        ds->textx[i] = ds->texty[i] = -1;
+
+    for (i = 0; i < g->num_dots; i++) {
+        int x, y;
+        grid_to_screen(ds, g, g->dots[i].x, g->dots[i].y, &x, &y);
+        draw_circle(dr, x, y, ds->tilesize / 15, ink, ink);
+    }
+
+    /*
+     * Clues.
+     */
+    for (i = 0; i < g->num_faces; i++) {
+        grid_face *f = g->faces + i;
+        int clue = state->clues[i];
+        if (clue >= 0) {
+            char c[20];
+            int x, y;
+            sprintf(c, "%d", state->clues[i]);
+            face_text_pos(ds, g, f, &x, &y);
+            draw_text(dr, x, y,
+                      FONT_VARIABLE, ds->tilesize / 2,
+                      ALIGN_VCENTRE | ALIGN_HCENTRE, ink, c);
+        }
+    }
+
+    /*
+     * Lines.
+     */
+    for (i = 0; i < g->num_edges; i++) {
+        int thickness = (state->lines[i] == LINE_YES) ? 30 : 150;
+        grid_edge *e = g->edges + i;
+        int x1, y1, x2, y2;
+        grid_to_screen(ds, g, e->dot1->x, e->dot1->y, &x1, &y1);
+        grid_to_screen(ds, g, e->dot2->x, e->dot2->y, &x2, &y2);
+        if (state->lines[i] == LINE_YES)
+        {
+            /* (dx, dy) points from (x1, y1) to (x2, y2).
+             * The line is then "fattened" in a perpendicular
+             * direction to create a thin rectangle. */
+            double d = sqrt(SQ((double)x1 - x2) + SQ((double)y1 - y2));
+            double dx = (x2 - x1) / d;
+            double dy = (y2 - y1) / d;
+            int points[8];
+
+            dx = (dx * ds->tilesize) / thickness;
+            dy = (dy * ds->tilesize) / thickness;
+            points[0] = x1 + (int)dy;
+            points[1] = y1 - (int)dx;
+            points[2] = x1 - (int)dy;
+            points[3] = y1 + (int)dx;
+            points[4] = x2 - (int)dy;
+            points[5] = y2 + (int)dx;
+            points[6] = x2 + (int)dy;
+            points[7] = y2 - (int)dx;
+            draw_polygon(dr, points, 4, ink, ink);
+        }
+        else
+        {
+            /* Draw a dotted line */
+            int divisions = 6;
+            int j;
+            for (j = 1; j < divisions; j++) {
+                /* Weighted average */
+                int x = (x1 * (divisions -j) + x2 * j) / divisions;
+                int y = (y1 * (divisions -j) + y2 * j) / divisions;
+                draw_circle(dr, x, y, ds->tilesize / thickness, ink, ink);
+            }
+        }
+    }
+
+    sfree(ds->textx);
+    sfree(ds->texty);
+}
+
+#ifdef COMBINED
+#define thegame loopy
+#endif
+
+const struct game thegame = {
+    "Loopy", "games.loopy", "loopy",
+    default_params,
+    game_fetch_preset,
+    decode_params,
+    encode_params,
+    free_params,
+    dup_params,
+    TRUE, game_configure, custom_params,
+    validate_params,
+    new_game_desc,
+    validate_desc,
+    new_game,
+    dup_game,
+    free_game,
+    1, solve_game,
+    TRUE, game_can_format_as_text_now, game_text_format,
+    new_ui,
+    free_ui,
+    encode_ui,
+    decode_ui,
+    game_changed_state,
+    interpret_move,
+    execute_move,
+    PREFERRED_TILE_SIZE, game_compute_size, game_set_size,
+    game_colours,
+    game_new_drawstate,
+    game_free_drawstate,
+    game_redraw,
+    game_anim_length,
+    game_flash_length,
+    game_status,
+    TRUE, FALSE, game_print_size, game_print,
+    FALSE /* wants_statusbar */,
+    FALSE, game_timing_state,
+    0,                                       /* mouse_priorities */
+};
+
+#ifdef STANDALONE_SOLVER
+
+/*
+ * Half-hearted standalone solver. It can't output the solution to
+ * anything but a square puzzle, and it can't log the deductions
+ * it makes either. But it can solve square puzzles, and more
+ * importantly it can use its solver to grade the difficulty of
+ * any puzzle you give it.
+ */
+
+#include <stdarg.h>
+
+int main(int argc, char **argv)
+{
+    game_params *p;
+    game_state *s;
+    char *id = NULL, *desc, *err;
+    int grade = FALSE;
+    int ret, diff;
+#if 0 /* verbose solver not supported here (yet) */
+    int really_verbose = FALSE;
+#endif
+
+    while (--argc > 0) {
+        char *p = *++argv;
+#if 0 /* verbose solver not supported here (yet) */
+        if (!strcmp(p, "-v")) {
+            really_verbose = TRUE;
+        } else
+#endif
+        if (!strcmp(p, "-g")) {
+            grade = TRUE;
+        } else if (*p == '-') {
+            fprintf(stderr, "%s: unrecognised option `%s'\n", argv[0], p);
+            return 1;
+        } else {
+            id = p;
+        }
+    }
+
+    if (!id) {
+        fprintf(stderr, "usage: %s [-g | -v] <game_id>\n", argv[0]);
+        return 1;
+    }
+
+    desc = strchr(id, ':');
+    if (!desc) {
+        fprintf(stderr, "%s: game id expects a colon in it\n", argv[0]);
+        return 1;
+    }
+    *desc++ = '\0';
+
+    p = default_params();
+    decode_params(p, id);
+    err = validate_desc(p, desc);
+    if (err) {
+        fprintf(stderr, "%s: %s\n", argv[0], err);
+        return 1;
+    }
+    s = new_game(NULL, p, desc);
+
+    /*
+     * When solving an Easy puzzle, we don't want to bother the
+     * user with Hard-level deductions. For this reason, we grade
+     * the puzzle internally before doing anything else.
+     */
+    ret = -1;                          /* placate optimiser */
+    for (diff = 0; diff < DIFF_MAX; diff++) {
+        solver_state *sstate_new;
+        solver_state *sstate = new_solver_state((game_state *)s, diff);
+
+        sstate_new = solve_game_rec(sstate);
+
+        if (sstate_new->solver_status == SOLVER_MISTAKE)
+            ret = 0;
+        else if (sstate_new->solver_status == SOLVER_SOLVED)
+            ret = 1;
+        else
+            ret = 2;
+
+        free_solver_state(sstate_new);
+        free_solver_state(sstate);
+
+        if (ret < 2)
+            break;
+    }
+
+    if (diff == DIFF_MAX) {
+        if (grade)
+            printf("Difficulty rating: harder than Hard, or ambiguous\n");
+        else
+            printf("Unable to find a unique solution\n");
+    } else {
+        if (grade) {
+            if (ret == 0)
+                printf("Difficulty rating: impossible (no solution exists)\n");
+            else if (ret == 1)
+                printf("Difficulty rating: %s\n", diffnames[diff]);
+        } else {
+            solver_state *sstate_new;
+            solver_state *sstate = new_solver_state((game_state *)s, diff);
+
+            /* If we supported a verbose solver, we'd set verbosity here */
+
+            sstate_new = solve_game_rec(sstate);
+
+            if (sstate_new->solver_status == SOLVER_MISTAKE)
+                printf("Puzzle is inconsistent\n");
+            else {
+                assert(sstate_new->solver_status == SOLVER_SOLVED);
+                if (s->grid_type == 0) {
+                    fputs(game_text_format(sstate_new->state), stdout);
+                } else {
+                    printf("Unable to output non-square grids\n");
+                }
+            }
+
+            free_solver_state(sstate_new);
+            free_solver_state(sstate);
+        }
+    }
+
+    return 0;
+}
+
+#endif
+
+/* vim: set shiftwidth=4 tabstop=8: */