shithub: puzzles

Download patch

ref: 514bd502be62be1b645e13488740fd34f8674d47
parent: a79ab3494119f2544f5ad05637e629ec703be4c5
author: Simon Tatham <anakin@pobox.com>
date: Sun Dec 27 05:01:23 EST 2009 

New puzzle! 'Keen', a clone of KenKen.

[originally from svn r8796]


--- a/dsf.c
+++ b/dsf.c
@@ -165,7 +165,7 @@
 	 * We always make the smaller of v1 and v2 the new canonical
 	 * element. This ensures that the canonical element of any
 	 * class in this structure is always the first element in
-	 * it.
+	 * it. 'Keen' depends critically on this property.
 	 *
 	 * (Jonas Koelker previously had this code choosing which
 	 * way round to connect the trees by examining the sizes of
--- a/icons/Makefile
+++ b/icons/Makefile
@@ -1,8 +1,8 @@
 # Makefile for Puzzles icons.
 
 PUZZLES = blackbox bridges cube dominosa fifteen filling flip galaxies guess \
-	  inertia lightup loopy map mines net netslide pattern pegs rect \
-	  samegame sixteen slant solo tents twiddle unequal untangle
+	  inertia keen lightup loopy map mines net netslide pattern pegs \
+	  rect samegame sixteen slant solo tents twiddle unequal untangle
 
 BASE = $(patsubst %,%-base.png,$(PUZZLES))
 WEB = $(patsubst %,%-web.png,$(PUZZLES))
@@ -60,6 +60,7 @@
 galaxies-ibase.png : override CROP=288x288 165x165+0+0
 guess-ibase.png : override CROP=263x420 178x178+75+17
 inertia-ibase.png : override CROP=321x321 128x128+193+0
+keen-ibase.png : override CROP=288x288 96x96+24+120
 lightup-ibase.png : override CROP=256x256 112x112+144+0
 loopy-ibase.png : override CROP=257x257 113x113+0+0
 mines-ibase.png : override CROP=240x240 110x110+130+130
--- /dev/null
+++ b/icons/keen.sav
@@ -1,0 +1,62 @@
+SAVEFILE:41:Simon Tatham's Portable Puzzle Collection
+VERSION :1:1
+GAME    :4:Keen
+PARAMS  :3:5de
+CPARAMS :3:5de
+SEED    :15:846699649745236
+DESC    :48:a__a_3a_5a_a_a_3b_bac_,a5m15a7a10s2s2d2s3m40m2s2
+AUXINFO :52:6105af67c6ebc8de056b59ebc9a463aa54e75f647055c0a6c1bd
+NSTATES :2:53
+STATEPOS:2:39
+MOVE    :6:P0,4,2
+MOVE    :6:P0,4,4
+MOVE    :6:P0,4,5
+MOVE    :6:P1,4,2
+MOVE    :6:P1,4,4
+MOVE    :6:P1,4,5
+MOVE    :6:P1,3,2
+MOVE    :6:P1,3,4
+MOVE    :6:P1,3,5
+MOVE    :6:R2,2,4
+MOVE    :6:R1,2,2
+MOVE    :6:P1,3,2
+MOVE    :6:P1,4,2
+MOVE    :6:R0,4,2
+MOVE    :6:R2,3,2
+MOVE    :6:R2,4,1
+MOVE    :6:R1,4,4
+MOVE    :6:R1,3,5
+MOVE    :6:P3,4,3
+MOVE    :6:P3,4,5
+MOVE    :6:P4,4,3
+MOVE    :6:P4,4,5
+MOVE    :6:R4,4,5
+MOVE    :6:R3,4,3
+MOVE    :6:P3,1,2
+MOVE    :6:P3,1,5
+MOVE    :6:P3,0,2
+MOVE    :6:P3,0,5
+MOVE    :6:R3,2,1
+MOVE    :6:R3,3,4
+MOVE    :6:P2,0,3
+MOVE    :6:P2,0,5
+MOVE    :6:P2,1,3
+MOVE    :6:P2,1,5
+MOVE    :6:P0,1,1
+MOVE    :6:P0,1,3
+MOVE    :6:P1,1,1
+MOVE    :6:P1,1,3
+MOVE    :6:R2,0,3
+MOVE    :6:R2,1,5
+MOVE    :6:R3,0,5
+MOVE    :6:R3,1,2
+MOVE    :6:R4,1,4
+MOVE    :6:R4,2,3
+MOVE    :6:R4,0,2
+MOVE    :6:R4,3,1
+MOVE    :6:R0,2,5
+MOVE    :6:R0,3,3
+MOVE    :6:R1,1,3
+MOVE    :6:R0,1,1
+MOVE    :6:R1,0,1
+MOVE    :6:R0,0,4
--- /dev/null
+++ b/keen.R
@@ -1,0 +1,25 @@
+# -*- makefile -*-
+
+KEEN_LATIN_EXTRA = tree234 maxflow dsf
+KEEN_EXTRA = latin KEEN_LATIN_EXTRA
+
+keen    : [X] GTK COMMON keen KEEN_EXTRA keen-icon|no-icon
+
+keen    : [G] WINDOWS COMMON keen KEEN_EXTRA keen.res|noicon.res
+
+keensolver : [U] keen[STANDALONE_SOLVER] latin[STANDALONE_SOLVER] KEEN_LATIN_EXTRA STANDALONE
+keensolver : [C] keen[STANDALONE_SOLVER] latin[STANDALONE_SOLVER] KEEN_LATIN_EXTRA STANDALONE
+
+ALL += keen[COMBINED] KEEN_EXTRA
+
+!begin gtk
+GAMES += keen
+!end
+
+!begin >list.c
+    A(keen) \
+!end
+
+!begin >wingames.lst
+keen.exe:Keen
+!end
--- /dev/null
+++ b/keen.c
@@ -1,0 +1,2387 @@
+/*
+ * keen.c: an implementation of the Times's 'KenKen' puzzle.
+ */
+
+#include <stdio.h>
+#include <stdlib.h>
+#include <string.h>
+#include <assert.h>
+#include <ctype.h>
+#include <math.h>
+
+#include "puzzles.h"
+#include "latin.h"
+
+/*
+ * Difficulty levels. I do some macro ickery here to ensure that my
+ * enum and the various forms of my name list always match up.
+ */
+#define DIFFLIST(A) \
+    A(EASY,Easy,solver_easy,e) \
+    A(NORMAL,Normal,solver_normal,n) \
+    A(HARD,Hard,solver_hard,h) \
+    A(EXTREME,Extreme,NULL,x) \
+    A(UNREASONABLE,Unreasonable,NULL,u)
+#define ENUM(upper,title,func,lower) DIFF_ ## upper,
+#define TITLE(upper,title,func,lower) #title,
+#define ENCODE(upper,title,func,lower) #lower
+#define CONFIG(upper,title,func,lower) ":" #title
+enum { DIFFLIST(ENUM) DIFFCOUNT };
+static char const *const keen_diffnames[] = { DIFFLIST(TITLE) };
+static char const keen_diffchars[] = DIFFLIST(ENCODE);
+#define DIFFCONFIG DIFFLIST(CONFIG)
+
+/*
+ * Clue notation. Important here that ADD and MUL come before SUB
+ * and DIV, and that DIV comes last. 
+ */
+#define C_ADD 0x00000000L
+#define C_MUL 0x20000000L
+#define C_SUB 0x40000000L
+#define C_DIV 0x60000000L
+#define CMASK 0x60000000L
+#define CUNIT 0x20000000L
+
+enum {
+    COL_BACKGROUND,
+    COL_GRID,
+    COL_USER,
+    COL_HIGHLIGHT,
+    COL_ERROR,
+    COL_PENCIL,
+    NCOLOURS
+};
+
+struct game_params {
+    int w, diff;
+};
+
+struct clues {
+    int refcount;
+    int w;
+    int *dsf;
+    long *clues;
+};
+
+struct game_state {
+    game_params par;
+    struct clues *clues;
+    digit *grid;
+    int *pencil;		       /* bitmaps using bits 1<<1..1<<n */
+    int completed, cheated;
+};
+
+static game_params *default_params(void)
+{
+    game_params *ret = snew(game_params);
+
+    ret->w = 6;
+    ret->diff = DIFF_NORMAL;
+
+    return ret;
+}
+
+const static struct game_params keen_presets[] = {
+    {  4, DIFF_EASY         },
+    {  5, DIFF_EASY         },
+    {  6, DIFF_EASY         },
+    {  6, DIFF_NORMAL       },
+    {  6, DIFF_HARD         },
+    {  6, DIFF_EXTREME      },
+    {  6, DIFF_UNREASONABLE },
+    {  9, DIFF_NORMAL       },
+};
+
+static int game_fetch_preset(int i, char **name, game_params **params)
+{
+    game_params *ret;
+    char buf[80];
+
+    if (i < 0 || i >= lenof(keen_presets))
+        return FALSE;
+
+    ret = snew(game_params);
+    *ret = keen_presets[i]; /* structure copy */
+
+    sprintf(buf, "%dx%d %s", ret->w, ret->w, keen_diffnames[ret->diff]);
+
+    *name = dupstr(buf);
+    *params = ret;
+    return TRUE;
+}
+
+static void free_params(game_params *params)
+{
+    sfree(params);
+}
+
+static game_params *dup_params(game_params *params)
+{
+    game_params *ret = snew(game_params);
+    *ret = *params;		       /* structure copy */
+    return ret;
+}
+
+static void decode_params(game_params *params, char const *string)
+{
+    char const *p = string;
+
+    params->w = atoi(p);
+    while (*p && isdigit((unsigned char)*p)) p++;
+
+    if (*p == 'd') {
+        int i;
+        p++;
+        params->diff = DIFFCOUNT+1; /* ...which is invalid */
+        if (*p) {
+            for (i = 0; i < DIFFCOUNT; i++) {
+                if (*p == keen_diffchars[i])
+                    params->diff = i;
+            }
+            p++;
+        }
+    }
+}
+
+static char *encode_params(game_params *params, int full)
+{
+    char ret[80];
+
+    sprintf(ret, "%d", params->w);
+    if (full)
+        sprintf(ret + strlen(ret), "d%c", keen_diffchars[params->diff]);
+
+    return dupstr(ret);
+}
+
+static config_item *game_configure(game_params *params)
+{
+    config_item *ret;
+    char buf[80];
+
+    ret = snewn(3, config_item);
+
+    ret[0].name = "Grid size";
+    ret[0].type = C_STRING;
+    sprintf(buf, "%d", params->w);
+    ret[0].sval = dupstr(buf);
+    ret[0].ival = 0;
+
+    ret[1].name = "Difficulty";
+    ret[1].type = C_CHOICES;
+    ret[1].sval = DIFFCONFIG;
+    ret[1].ival = params->diff;
+
+    ret[2].name = NULL;
+    ret[2].type = C_END;
+    ret[2].sval = NULL;
+    ret[2].ival = 0;
+
+    return ret;
+}
+
+static game_params *custom_params(config_item *cfg)
+{
+    game_params *ret = snew(game_params);
+
+    ret->w = atoi(cfg[0].sval);
+    ret->diff = cfg[1].ival;
+
+    return ret;
+}
+
+static char *validate_params(game_params *params, int full)
+{
+    if (params->w < 3 || params->w > 9)
+        return "Grid size must be between 3 and 9";
+    if (params->diff >= DIFFCOUNT)
+        return "Unknown difficulty rating";
+    return NULL;
+}
+
+/* ----------------------------------------------------------------------
+ * Solver.
+ */
+
+struct solver_ctx {
+    int w, diff;
+    int nboxes;
+    int *boxes, *boxlist, *whichbox;
+    long *clues;
+    digit *soln;
+    digit *dscratch;
+    int *iscratch;
+};
+
+static void solver_clue_candidate(struct solver_ctx *ctx, int diff, int box)
+{
+    int w = ctx->w;
+    int n = ctx->boxes[box+1] - ctx->boxes[box];
+    int j;
+
+    /*
+     * This function is called from the main clue-based solver
+     * routine when we discover a candidate layout for a given clue
+     * box consistent with everything we currently know about the
+     * digit constraints in that box. We expect to find the digits
+     * of the candidate layout in ctx->dscratch, and we update
+     * ctx->iscratch as appropriate.
+     */
+    if (diff == DIFF_EASY) {
+	unsigned mask = 0;
+	/*
+	 * Easy-mode clue deductions: we do not record information
+	 * about which squares take which values, so we amalgamate
+	 * all the values in dscratch and OR them all into
+	 * everywhere.
+	 */
+	for (j = 0; j < n; j++)
+	    mask |= 1 << ctx->dscratch[j];
+	for (j = 0; j < n; j++)
+	    ctx->iscratch[j] |= mask;
+    } else if (diff == DIFF_NORMAL) {
+	/*
+	 * Normal-mode deductions: we process the information in
+	 * dscratch in the obvious way.
+	 */
+	for (j = 0; j < n; j++)
+	    ctx->iscratch[j] |= 1 << ctx->dscratch[j];
+    } else if (diff == DIFF_HARD) {
+	/*
+	 * Hard-mode deductions: instead of ruling things out
+	 * _inside_ the clue box, we look for numbers which occur in
+	 * a given row or column in all candidate layouts, and rule
+	 * them out of all squares in that row or column that
+	 * _aren't_ part of this clue box.
+	 */
+	int *sq = ctx->boxlist + ctx->boxes[box];
+
+	for (j = 0; j < 2*w; j++)
+	    ctx->iscratch[2*w+j] = 0;
+	for (j = 0; j < n; j++) {
+	    int x = sq[j] / w, y = sq[j] % w;
+	    ctx->iscratch[2*w+x] |= 1 << ctx->dscratch[j];
+	    ctx->iscratch[3*w+y] |= 1 << ctx->dscratch[j];
+	}
+	for (j = 0; j < 2*w; j++)
+	    ctx->iscratch[j] &= ctx->iscratch[2*w+j];
+    }
+}
+
+static int solver_common(struct latin_solver *solver, void *vctx, int diff)
+{
+    struct solver_ctx *ctx = (struct solver_ctx *)vctx;
+    int w = ctx->w;
+    int box, i, j, k;
+    int ret = 0, total;
+
+    /*
+     * Iterate over each clue box and deduce what we can.
+     */
+    for (box = 0; box < ctx->nboxes; box++) {
+	int *sq = ctx->boxlist + ctx->boxes[box];
+	int n = ctx->boxes[box+1] - ctx->boxes[box];
+	int value = ctx->clues[box] & ~CMASK;
+	int op = ctx->clues[box] & CMASK;
+
+	if (diff == DIFF_HARD) {
+	    for (i = 0; i < n; i++)
+		ctx->iscratch[i] = (1 << (w+1)) - (1 << 1);
+	} else {
+	    for (i = 0; i < n; i++)
+		ctx->iscratch[i] = 0;
+	}
+
+	switch (op) {
+	  case C_SUB:
+	  case C_DIV:
+	    /*
+	     * These two clue types must always apply to a box of
+	     * area 2. Also, the two digits in these boxes can never
+	     * be the same (because any domino must have its two
+	     * squares in either the same row or the same column).
+	     * So we simply iterate over all possibilities for the
+	     * two squares (both ways round), rule out any which are
+	     * inconsistent with the digit constraints we already
+	     * have, and update the digit constraints with any new
+	     * information thus garnered.
+	     */
+	    assert(n == 2);
+
+	    for (i = 1; i <= w; i++) {
+		j = (op == C_SUB ? i + value : i * value);
+		if (j > w) break;
+
+		/* (i,j) is a valid digit pair. Try it both ways round. */
+
+		if (solver->cube[sq[0]*w+i-1] &&
+		    solver->cube[sq[1]*w+j-1]) {
+		    ctx->dscratch[0] = i;
+		    ctx->dscratch[1] = j;
+		    solver_clue_candidate(ctx, diff, box);
+		}
+
+		if (solver->cube[sq[0]*w+j-1] &&
+		    solver->cube[sq[1]*w+i-1]) {
+		    ctx->dscratch[0] = j;
+		    ctx->dscratch[1] = i;
+		    solver_clue_candidate(ctx, diff, box);
+		}
+	    }
+
+	    break;
+
+	  case C_ADD:
+	  case C_MUL:
+	    /*
+	     * For these clue types, I have no alternative but to go
+	     * through all possible number combinations.
+	     *
+	     * Instead of a tedious physical recursion, I iterate in
+	     * the scratch array through all possibilities. At any
+	     * given moment, i indexes the element of the box that
+	     * will next be incremented.
+	     */
+	    i = 0;
+	    ctx->dscratch[i] = 0;
+	    total = value;	       /* start with the identity */
+	    while (1) {
+		if (i < n) {
+		    /*
+		     * Find the next valid value for cell i.
+		     */
+		    for (j = ctx->dscratch[i] + 1; j <= w; j++) {
+			if (op == C_ADD ? (total < j) : (total % j != 0))
+			    continue;  /* this one won't fit */
+			if (!solver->cube[sq[i]*w+j-1])
+			    continue;  /* this one is ruled out already */
+			for (k = 0; k < i; k++)
+			    if (ctx->dscratch[k] == j &&
+				(sq[k] % w == sq[i] % w ||
+				 sq[k] / w == sq[i] / w))
+				break; /* clashes with another row/col */
+			if (k < i)
+			    continue;
+
+			/* Found one. */
+			break;
+		    }
+
+		    if (j > w) {
+			/* No valid values left; drop back. */
+			i--;
+			if (i < 0)
+			    break;     /* overall iteration is finished */
+			if (op == C_ADD)
+			    total += ctx->dscratch[i];
+			else
+			    total *= ctx->dscratch[i];
+		    } else {
+			/* Got a valid value; store it and move on. */
+			ctx->dscratch[i++] = j;
+			if (op == C_ADD)
+			    total -= j;
+			else
+			    total /= j;
+			ctx->dscratch[i] = 0;
+		    }
+		} else {
+		    if (total == (op == C_ADD ? 0 : 1))
+			solver_clue_candidate(ctx, diff, box);
+		    i--;
+		    if (op == C_ADD)
+			total += ctx->dscratch[i];
+		    else
+			total *= ctx->dscratch[i];
+		}
+	    }
+
+	    break;
+	}
+
+	if (diff < DIFF_HARD) {
+#ifdef STANDALONE_SOLVER
+	    char prefix[256];
+
+	    if (solver_show_working)
+		sprintf(prefix, "%*susing clue at (%d,%d):\n",
+			solver_recurse_depth*4, "",
+			sq[0]/w+1, sq[0]%w+1);
+	    else
+		prefix[0] = '\0';	       /* placate optimiser */
+#endif
+
+	    for (i = 0; i < n; i++)
+		for (j = 1; j <= w; j++) {
+		    if (solver->cube[sq[i]*w+j-1] &&
+			!(ctx->iscratch[i] & (1 << j))) {
+#ifdef STANDALONE_SOLVER
+			if (solver_show_working) {
+			    printf("%s%*s  ruling out %d at (%d,%d)\n",
+				   prefix, solver_recurse_depth*4, "",
+				   j, sq[i]/w+1, sq[i]%w+1);
+			    prefix[0] = '\0';
+			}
+#endif
+			solver->cube[sq[i]*w+j-1] = 0;
+			ret = 1;
+		    }
+		}
+	} else {
+#ifdef STANDALONE_SOLVER
+	    char prefix[256];
+
+	    if (solver_show_working)
+		sprintf(prefix, "%*susing clue at (%d,%d):\n",
+			solver_recurse_depth*4, "",
+			sq[0]/w+1, sq[0]%w+1);
+	    else
+		prefix[0] = '\0';	       /* placate optimiser */
+#endif
+
+	    for (i = 0; i < 2*w; i++) {
+		int start = (i < w ? i*w : i-w);
+		int step = (i < w ? 1 : w);
+		for (j = 1; j <= w; j++) if (ctx->iscratch[i] & (1 << j)) {
+#ifdef STANDALONE_SOLVER
+		    char prefix2[256];
+
+		    if (solver_show_working)
+			sprintf(prefix2, "%*s  this clue requires %d in"
+				" %s %d:\n", solver_recurse_depth*4, "",
+				j, i < w ? "column" : "row", i%w+1);
+		    else
+			prefix2[0] = '\0';   /* placate optimiser */
+#endif
+
+		    for (k = 0; k < w; k++) {
+			int pos = start + k*step;
+			if (ctx->whichbox[pos] != box &&
+			    solver->cube[pos*w+j-1]) {
+#ifdef STANDALONE_SOLVER
+			    if (solver_show_working) {
+				printf("%s%s%*s   ruling out %d at (%d,%d)\n",
+				       prefix, prefix2,
+				       solver_recurse_depth*4, "",
+				       j, pos/w+1, pos%w+1);
+				prefix[0] = prefix2[0] = '\0';
+			    }
+#endif
+			    solver->cube[pos*w+j-1] = 0;
+			    ret = 1;
+			}
+		    }
+		}
+	    }
+
+	    /*
+	     * Once we find one block we can do something with in
+	     * this way, revert to trying easier deductions, so as
+	     * not to generate solver diagnostics that make the
+	     * problem look harder than it is. (We have to do this
+	     * for the Hard deductions but not the Easy/Normal ones,
+	     * because only the Hard deductions are cross-box.)
+	     */
+	    if (ret)
+		return ret;
+	}
+    }
+
+    return ret;
+}
+
+static int solver_easy(struct latin_solver *solver, void *vctx)
+{
+    /*
+     * Omit the EASY deductions when solving at NORMAL level, since
+     * the NORMAL deductions are a superset of them anyway and it
+     * saves on time and confusing solver diagnostics.
+     *
+     * Note that this breaks the natural semantics of the return
+     * value of latin_solver. Without this hack, you could determine
+     * a puzzle's difficulty in one go by trying to solve it at
+     * maximum difficulty and seeing what difficulty value was
+     * returned; but with this hack, solving an Easy puzzle on
+     * Normal difficulty will typically return Normal. Hence the
+     * uses of the solver to determine difficulty are all arranged
+     * so as to double-check by re-solving at the next difficulty
+     * level down and making sure it failed.
+     */
+    struct solver_ctx *ctx = (struct solver_ctx *)vctx;
+    if (ctx->diff > DIFF_EASY)
+	return 0;
+    return solver_common(solver, vctx, DIFF_EASY);
+}
+
+static int solver_normal(struct latin_solver *solver, void *vctx)
+{
+    return solver_common(solver, vctx, DIFF_NORMAL);
+}
+
+static int solver_hard(struct latin_solver *solver, void *vctx)
+{
+    return solver_common(solver, vctx, DIFF_HARD);
+}
+
+#define SOLVER(upper,title,func,lower) func,
+static usersolver_t const keen_solvers[] = { DIFFLIST(SOLVER) };
+
+static int solver(int w, int *dsf, long *clues, digit *soln, int maxdiff)
+{
+    int a = w*w;
+    struct solver_ctx ctx;
+    int ret;
+    int i, j, n, m;
+    
+    ctx.w = w;
+    ctx.soln = soln;
+    ctx.diff = maxdiff;
+
+    /*
+     * Transform the dsf-formatted clue list into one over which we
+     * can iterate more easily.
+     *
+     * Also transpose the x- and y-coordinates at this point,
+     * because the 'cube' array in the general Latin square solver
+     * puts x first (oops).
+     */
+    for (ctx.nboxes = i = 0; i < a; i++)
+	if (dsf_canonify(dsf, i) == i)
+	    ctx.nboxes++;
+    ctx.boxlist = snewn(a, int);
+    ctx.boxes = snewn(ctx.nboxes+1, int);
+    ctx.clues = snewn(ctx.nboxes, long);
+    ctx.whichbox = snewn(a, int);
+    for (n = m = i = 0; i < a; i++)
+	if (dsf_canonify(dsf, i) == i) {
+	    ctx.clues[n] = clues[i];
+	    ctx.boxes[n] = m;
+	    for (j = 0; j < a; j++)
+		if (dsf_canonify(dsf, j) == i) {
+		    ctx.boxlist[m++] = (j % w) * w + (j / w);   /* transpose */
+		    ctx.whichbox[ctx.boxlist[m-1]] = n;
+		}
+	    n++;
+	}
+    assert(n == ctx.nboxes);
+    assert(m == a);
+    ctx.boxes[n] = m;
+
+    ctx.dscratch = snewn(a+1, digit);
+    ctx.iscratch = snewn(max(a+1, 4*w), int);
+
+    ret = latin_solver(soln, w, maxdiff,
+		       DIFF_EASY, DIFF_HARD, DIFF_EXTREME,
+		       DIFF_EXTREME, DIFF_UNREASONABLE,
+		       keen_solvers, &ctx, NULL, NULL);
+
+    sfree(ctx.dscratch);
+    sfree(ctx.iscratch);
+    sfree(ctx.whichbox);
+    sfree(ctx.boxlist);
+    sfree(ctx.boxes);
+    sfree(ctx.clues);
+
+    return ret;
+}
+
+/* ----------------------------------------------------------------------
+ * Grid generation.
+ */
+
+static char *encode_block_structure(char *p, int w, int *dsf)
+{
+    int i, currrun = 0;
+    char *orig, *q, *r, c;
+
+    orig = p;
+
+    /*
+     * Encode the block structure. We do this by encoding the
+     * pattern of dividing lines: first we iterate over the w*(w-1)
+     * internal vertical grid lines in ordinary reading order, then
+     * over the w*(w-1) internal horizontal ones in transposed
+     * reading order.
+     *
+     * We encode the number of non-lines between the lines; _ means
+     * zero (two adjacent divisions), a means 1, ..., y means 25,
+     * and z means 25 non-lines _and no following line_ (so that za
+     * means 26, zb 27 etc).
+     */
+    for (i = 0; i <= 2*w*(w-1); i++) {
+	int x, y, p0, p1, edge;
+
+	if (i == 2*w*(w-1)) {
+	    edge = TRUE;       /* terminating virtual edge */
+	} else {
+	    if (i < w*(w-1)) {
+		y = i/(w-1);
+		x = i%(w-1);
+		p0 = y*w+x;
+		p1 = y*w+x+1;
+	    } else {
+		x = i/(w-1) - w;
+		y = i%(w-1);
+		p0 = y*w+x;
+		p1 = (y+1)*w+x;
+	    }
+	    edge = (dsf_canonify(dsf, p0) != dsf_canonify(dsf, p1));
+	}
+
+	if (edge) {
+	    while (currrun > 25)
+		*p++ = 'z', currrun -= 25;
+	    if (currrun)
+		*p++ = 'a'-1 + currrun;
+	    else
+		*p++ = '_';
+	    currrun = 0;
+	} else
+	    currrun++;
+    }
+
+    /*
+     * Now go through and compress the string by replacing runs of
+     * the same letter with a single copy of that letter followed by
+     * a repeat count, where that makes it shorter. (This puzzle
+     * seems to generate enough long strings of _ to make this a
+     * worthwhile step.)
+     */
+    for (q = r = orig; r < p ;) {
+	*q++ = c = *r;
+
+	for (i = 0; r+i < p && r[i] == c; i++);
+	r += i;
+
+	if (i == 2) {
+	    *q++ = c;
+	} else if (i > 2) {
+	    q += sprintf(q, "%d", i);
+	}
+    }
+    
+    return q;
+}
+
+static char *parse_block_structure(const char **p, int w, int *dsf)
+{
+    int a = w*w;
+    int pos = 0;
+    int repc = 0, repn = 0;
+
+    dsf_init(dsf, a);
+
+    while (**p && (repn > 0 || **p != ',')) {
+	int c, adv;
+
+	if (repn > 0) {
+	    repn--;
+	    c = repc;
+	} else if (**p == '_' || (**p >= 'a' && **p <= 'z')) {
+	    c = (**p == '_' ? 0 : **p - 'a' + 1);
+	    (*p)++;
+	    if (**p && isdigit((unsigned char)**p)) {
+		repc = c;
+		repn = atoi(*p)-1;
+		while (**p && isdigit((unsigned char)**p)) (*p)++;
+	    }
+	} else
+	    return "Invalid character in game description";
+
+	adv = (c != 25);	       /* 'z' is a special case */
+
+	while (c-- > 0) {
+	    int p0, p1;
+
+	    /*
+	     * Non-edge; merge the two dsf classes on either
+	     * side of it.
+	     */
+	    if (pos >= 2*w*(w-1))
+		return "Too much data in block structure specification";
+	    if (pos < w*(w-1)) {
+		int y = pos/(w-1);
+		int x = pos%(w-1);
+		p0 = y*w+x;
+		p1 = y*w+x+1;
+	    } else {
+		int x = pos/(w-1) - w;
+		int y = pos%(w-1);
+		p0 = y*w+x;
+		p1 = (y+1)*w+x;
+	    }
+	    dsf_merge(dsf, p0, p1);
+
+	    pos++;
+	}
+	if (adv) {
+	    pos++;
+	    if (pos > 2*w*(w-1)+1)
+		return "Too much data in block structure specification";
+	}
+    }
+
+    /*
+     * When desc is exhausted, we expect to have gone exactly
+     * one space _past_ the end of the grid, due to the dummy
+     * edge at the end.
+     */
+    if (pos != 2*w*(w-1)+1)
+	return "Not enough data in block structure specification";
+
+    return NULL;
+}
+
+static char *new_game_desc(game_params *params, random_state *rs,
+			   char **aux, int interactive)
+{
+    int w = params->w, a = w*w;
+    digit *grid, *soln;
+    int *order, *revorder, *singletons, *dsf;
+    long *clues, *cluevals;
+    int i, j, k, n, x, y, ret;
+    int diff = params->diff;
+    char *desc, *p;
+
+    /*
+     * Difficulty exceptions: 3x3 puzzles at difficulty Hard or
+     * higher are currently not generable - the generator will spin
+     * forever looking for puzzles of the appropriate difficulty. We
+     * dial each of these down to the next lower difficulty.
+     *
+     * Remember to re-test this whenever a change is made to the
+     * solver logic!
+     *
+     * I tested it using the following shell command:
+
+for d in e n h x u; do
+  for i in {3..9}; do
+    echo ./keen --generate 1 ${i}d${d}
+    perl -e 'alarm 30; exec @ARGV' ./keen --generate 5 ${i}d${d} >/dev/null \
+      || echo broken
+  done
+done
+
+     * Of course, it's better to do that after taking the exceptions
+     * _out_, so as to detect exceptions that should be removed as
+     * well as those which should be added.
+     */
+    if (w == 3 && diff > DIFF_NORMAL)
+	diff = DIFF_NORMAL;
+
+    grid = NULL;
+
+    order = snewn(a, int);
+    revorder = snewn(a, int);
+    singletons = snewn(a, int);
+    dsf = snew_dsf(a);
+    clues = snewn(a, long);
+    cluevals = snewn(a, long);
+    soln = snewn(a, digit);
+
+    while (1) {
+	/*
+	 * First construct a latin square to be the solution.
+	 */
+	sfree(grid);
+	grid = latin_generate(w, rs);
+
+	/*
+	 * Divide the grid into arbitrarily sized blocks, but so as
+	 * to arrange plenty of dominoes which can be SUB/DIV clues.
+	 * We do this by first placing dominoes at random for a
+	 * while, then tying the remaining singletons one by one
+	 * into neighbouring blocks.
+	 */
+	for (i = 0; i < a; i++)
+	    order[i] = i;
+	shuffle(order, a, sizeof(*order), rs);
+	for (i = 0; i < a; i++)
+	    revorder[order[i]] = i;
+
+	for (i = 0; i < a; i++)
+	    singletons[i] = TRUE;
+
+	dsf_init(dsf, a);
+
+	/* Place dominoes. */
+	for (i = 0; i < a; i++) {
+	    if (singletons[i]) {
+		int best = -1;
+
+		x = i % w;
+		y = i / w;
+
+		if (x > 0 && singletons[i-1] &&
+		    (best == -1 || revorder[i-1] < revorder[best]))
+		    best = i-1;
+		if (x+1 < w && singletons[i+1] &&
+		    (best == -1 || revorder[i+1] < revorder[best]))
+		    best = i+1;
+		if (y > 0 && singletons[i-w] &&
+		    (best == -1 || revorder[i-w] < revorder[best]))
+		    best = i-w;
+		if (y+1 < w && singletons[i+w] &&
+		    (best == -1 || revorder[i+w] < revorder[best]))
+		    best = i+w;
+
+		/*
+		 * When we find a potential domino, we place it with
+		 * probability 3/4, which seems to strike a decent
+		 * balance between plenty of dominoes and leaving
+		 * enough singletons to make interesting larger
+		 * shapes.
+		 */
+		if (best >= 0 && random_upto(rs, 4)) {
+		    singletons[i] = singletons[best] = FALSE;
+		    dsf_merge(dsf, i, best);
+		}
+	    }
+	}
+
+	/* Fold in singletons. */
+	for (i = 0; i < a; i++) {
+	    if (singletons[i]) {
+		int best = -1;
+
+		x = i % w;
+		y = i / w;
+
+		if (x > 0 &&
+		    (best == -1 || revorder[i-1] < revorder[best]))
+		    best = i-1;
+		if (x+1 < w &&
+		    (best == -1 || revorder[i+1] < revorder[best]))
+		    best = i+1;
+		if (y > 0 &&
+		    (best == -1 || revorder[i-w] < revorder[best]))
+		    best = i-w;
+		if (y+1 < w &&
+		    (best == -1 || revorder[i+w] < revorder[best]))
+		    best = i+w;
+
+		if (best >= 0) {
+		    singletons[i] = FALSE;
+		    dsf_merge(dsf, i, best);
+		}
+	    }
+	}
+
+	/*
+	 * Decide what would be acceptable clues for each block.
+	 *
+	 * Blocks larger than 2 have free choice of ADD or MUL;
+	 * blocks of size 2 can be anything in principle (except
+	 * that they can only be DIV if the two numbers have an
+	 * integer quotient, of course), but we rule out (or try to
+	 * avoid) some clues because they're of low quality.
+	 *
+	 * Hence, we iterate once over the grid, stopping at the
+	 * canonical element of every >2 block and the _non_-
+	 * canonical element of every 2-block; the latter means that
+	 * we can make our decision about a 2-block in the knowledge
+	 * of both numbers in it.
+	 *
+	 * We reuse the 'singletons' array (finished with in the
+	 * above loop) to hold information about which blocks are
+	 * suitable for what.
+	 */
+#define F_ADD     0x01
+#define F_ADD_BAD 0x02
+#define F_SUB     0x04
+#define F_SUB_BAD 0x08
+#define F_MUL     0x10
+#define F_MUL_BAD 0x20
+#define F_DIV     0x40
+#define F_DIV_BAD 0x80
+	for (i = 0; i < a; i++) {
+	    singletons[i] = 0;
+	    j = dsf_canonify(dsf, i);
+	    k = dsf_size(dsf, j);
+	    if (j == i && k > 2) {
+		singletons[j] |= F_ADD | F_MUL;
+	    } else if (j != i && k == 2) {
+		/* Fetch the two numbers and sort them into order. */
+		int p = grid[j], q = grid[i], v;
+		if (p < q) {
+		    int t = p; p = q; q = t;
+		}
+
+		/*
+		 * Addition clues are always allowed, but we try to
+		 * avoid sums of 3, 4, (2w-1) and (2w-2) if we can,
+		 * because they're too easy - they only leave one
+		 * option for the pair of numbers involved.
+		 */
+		v = p + q;
+		if (v > 4 && v < 2*w-2)
+		    singletons[j] |= F_ADD;
+		else
+		    singletons[j] |= F_ADD_BAD;
+
+		/*
+		 * Multiplication clues: similarly, we prefer clues
+		 * of this type which leave multiple options open.
+		 * We can't rule out all the others, though, because
+		 * there are very very few 2-square multiplication
+		 * clues that _don't_ leave only one option.
+		 */
+		v = p * q;
+		n = 0;
+		for (k = 1; k <= w; k++)
+		    if (v % k == 0 && v / k <= w && v / k != k)
+			n++;
+		if (n > 1)
+		    singletons[j] |= F_MUL;
+		else
+		    singletons[j] |= F_MUL_BAD;
+
+		/*
+		 * Subtraction: we completely avoid a difference of
+		 * w-1.
+		 */
+		v = p - q;
+		if (v < w-1)
+		    singletons[j] |= F_SUB;
+
+		/*
+		 * Division: for a start, the quotient must be an
+		 * integer or the clue type is impossible. Also, we
+		 * never use quotients strictly greater than w/2,
+		 * because they're not only too easy but also
+		 * inelegant.
+		 */
+		if (p % q == 0 && 2 * (p / q) <= w)
+		    singletons[j] |= F_DIV;
+	    }
+	}
+
+	/*
+	 * Actually choose a clue for each block, trying to keep the
+	 * numbers of each type even, and starting with the
+	 * preferred candidates for each type where possible.
+	 *
+	 * I'm sure there should be a faster algorithm for doing
+	 * this, but I can't be bothered: O(N^2) is good enough when
+	 * N is at most the number of dominoes that fits into a 9x9
+	 * square.
+	 */
+	shuffle(order, a, sizeof(*order), rs);
+	for (i = 0; i < a; i++)
+	    clues[i] = 0;
+	while (1) {
+	    int done_something = FALSE;
+
+	    for (k = 0; k < 4; k++) {
+		long clue;
+		int good, bad;
+		switch (k) {
+		  case 0: clue = C_DIV; good = F_DIV; bad = F_DIV_BAD; break;
+		  case 1: clue = C_SUB; good = F_SUB; bad = F_SUB_BAD; break;
+		  case 2: clue = C_MUL; good = F_MUL; bad = F_MUL_BAD; break;
+		  default /* case 3 */ :
+		    clue = C_ADD; good = F_ADD; bad = F_ADD_BAD; break;
+		}
+
+		for (i = 0; i < a; i++) {
+		    j = order[i];
+		    if (singletons[j] & good) {
+			clues[j] = clue;
+			singletons[j] = 0;
+			break;
+		    }
+		}
+		if (i == a) {
+		    /* didn't find a nice one, use a nasty one */
+		    for (i = 0; i < a; i++) {
+			j = order[i];
+			if (singletons[j] & good) {
+			    clues[j] = clue;
+			    singletons[j] = 0;
+			    break;
+			}
+		    }
+		}
+		if (i < a)
+		    done_something = TRUE;
+	    }
+
+	    if (!done_something)
+		break;
+	}
+#undef F_ADD
+#undef F_ADD_BAD
+#undef F_SUB
+#undef F_SUB_BAD
+#undef F_MUL
+#undef F_MUL_BAD
+#undef F_DIV
+#undef F_DIV_BAD
+
+	/*
+	 * Having chosen the clue types, calculate the clue values.
+	 */
+	for (i = 0; i < a; i++) {
+	    j = dsf_canonify(dsf, i);
+	    if (j == i) {
+		cluevals[j] = grid[i];
+	    } else {
+		switch (clues[j]) {
+		  case C_ADD:
+		    cluevals[j] += grid[i];
+		    break;
+		  case C_MUL:
+		    cluevals[j] *= grid[i];
+		    break;
+		  case C_SUB:
+		    cluevals[j] = abs(cluevals[j] - grid[i]);
+		    break;
+		  case C_DIV:
+		    {
+			int d1 = cluevals[j], d2 = grid[i];
+			if (d1 == 0 || d2 == 0)
+			    cluevals[j] = 0;
+			else
+			    cluevals[j] = d2/d1 + d1/d2;/* one is 0 :-) */
+		    }
+		    break;
+		}
+	    }
+	}
+
+	for (i = 0; i < a; i++) {
+	    j = dsf_canonify(dsf, i);
+	    if (j == i) {
+		clues[j] |= cluevals[j];
+	    }
+	}
+
+	/*
+	 * See if the game can be solved at the specified difficulty
+	 * level, but not at the one below.
+	 */
+	if (diff > 0) {
+	    memset(soln, 0, a);
+	    ret = solver(w, dsf, clues, soln, diff-1);
+	    if (ret <= diff-1)
+		continue;
+	}
+	memset(soln, 0, a);
+	ret = solver(w, dsf, clues, soln, diff);
+	if (ret != diff)
+	    continue;		       /* go round again */
+
+	/*
+	 * I wondered if at this point it would be worth trying to
+	 * merge adjacent blocks together, to make the puzzle
+	 * gradually more difficult if it's currently easier than
+	 * specced, increasing the chance of a given generation run
+	 * being successful.
+	 *
+	 * It doesn't seem to be critical for the generation speed,
+	 * though, so for the moment I'm leaving it out.
+	 */
+
+	/*
+	 * We've got a usable puzzle!
+	 */
+	break;
+    }
+
+    /*
+     * Encode the puzzle description.
+     */
+    desc = snewn(40*a, char);
+    p = desc;
+    p = encode_block_structure(p, w, dsf);
+    *p++ = ',';
+    for (i = 0; i < a; i++) {
+	j = dsf_canonify(dsf, i);
+	if (j == i) {
+	    switch (clues[j] & CMASK) {
+	      case C_ADD: *p++ = 'a'; break;
+	      case C_SUB: *p++ = 's'; break;
+	      case C_MUL: *p++ = 'm'; break;
+	      case C_DIV: *p++ = 'd'; break;
+	    }
+	    p += sprintf(p, "%ld", clues[j] & ~CMASK);
+	}
+    }
+    *p++ = '\0';
+    desc = sresize(desc, p - desc, char);
+
+    /*
+     * Encode the solution.
+     */
+    assert(memcmp(soln, grid, a) == 0);
+    *aux = snewn(a+2, char);
+    (*aux)[0] = 'S';
+    for (i = 0; i < a; i++)
+	(*aux)[i+1] = '0' + soln[i];
+    (*aux)[a+1] = '\0';
+
+    sfree(grid);
+    sfree(order);
+    sfree(revorder);
+    sfree(singletons);
+    sfree(dsf);
+    sfree(clues);
+    sfree(cluevals);
+    sfree(soln);
+
+    return desc;
+}
+
+/* ----------------------------------------------------------------------
+ * Gameplay.
+ */
+
+static char *validate_desc(game_params *params, char *desc)
+{
+    int w = params->w, a = w*w;
+    int *dsf;
+    char *ret;
+    const char *p = desc;
+    int i;
+
+    /*
+     * Verify that the block structure makes sense.
+     */
+    dsf = snew_dsf(a);
+    ret = parse_block_structure(&p, w, dsf);
+    if (ret) {
+	sfree(dsf);
+	return ret;
+    }
+
+    if (*p != ',')
+	return "Expected ',' after block structure description";
+    p++;
+
+    /*
+     * Verify that the right number of clues are given, and that SUB
+     * and DIV clues don't apply to blocks of the wrong size.
+     */
+    for (i = 0; i < a; i++) {
+	if (dsf_canonify(dsf, i) == i) {
+	    if (*p == 'a' || *p == 'm') {
+		/* these clues need no validation */
+	    } else if (*p == 'd' || *p == 's') {
+		if (dsf_size(dsf, i) != 2)
+		    return "Subtraction and division blocks must have area 2";
+	    } else if (!*p) {
+		return "Too few clues for block structure";
+	    } else {
+		return "Unrecognised clue type";
+	    }
+	    p++;
+	    while (*p && isdigit((unsigned char)*p)) p++;
+	}
+    }
+    if (*p)
+	return "Too many clues for block structure";
+
+    return NULL;
+}
+
+static game_state *new_game(midend *me, game_params *params, char *desc)
+{
+    int w = params->w, a = w*w;
+    game_state *state = snew(game_state);
+    char *err;
+    const char *p = desc;
+    int i;
+
+    state->par = *params;	       /* structure copy */
+    state->clues = snew(struct clues);
+    state->clues->refcount = 1;
+    state->clues->w = w;
+    state->clues->dsf = snew_dsf(a);
+    err = parse_block_structure(&p, w, state->clues->dsf);
+
+    assert(*p == ',');
+    p++;
+
+    state->clues->clues = snewn(a, long);
+    for (i = 0; i < a; i++) {
+	if (dsf_canonify(state->clues->dsf, i) == i) {
+	    long clue = 0;
+	    switch (*p) {
+	      case 'a':
+		clue = C_ADD;
+		break;
+	      case 'm':
+		clue = C_MUL;
+		break;
+	      case 's':
+		clue = C_SUB;
+		assert(dsf_size(state->clues->dsf, i) == 2);
+		break;
+	      case 'd':
+		clue = C_DIV;
+		assert(dsf_size(state->clues->dsf, i) == 2);
+		break;
+	      default:
+		assert(!"Bad description in new_game");
+	    }
+	    p++;
+	    clue |= atol(p);
+	    while (*p && isdigit((unsigned char)*p)) p++;
+	    state->clues->clues[i] = clue;
+	} else
+	    state->clues->clues[i] = 0;
+    }
+
+    state->grid = snewn(a, digit);
+    state->pencil = snewn(a, int);
+    for (i = 0; i < a; i++) {
+	state->grid[i] = 0;
+	state->pencil[i] = 0;
+    }
+
+    state->completed = state->cheated = FALSE;
+
+    return state;
+}
+
+static game_state *dup_game(game_state *state)
+{
+    int w = state->par.w, a = w*w;
+    game_state *ret = snew(game_state);
+
+    ret->par = state->par;	       /* structure copy */
+
+    ret->clues = state->clues;
+    ret->clues->refcount++;
+
+    ret->grid = snewn(a, digit);
+    ret->pencil = snewn(a, int);
+    memcpy(ret->grid, state->grid, a*sizeof(digit));
+    memcpy(ret->pencil, state->pencil, a*sizeof(int));
+
+    ret->completed = state->completed;
+    ret->cheated = state->cheated;
+
+    return ret;
+}
+
+static void free_game(game_state *state)
+{
+    sfree(state->grid);
+    sfree(state->pencil);
+    if (--state->clues->refcount <= 0) {
+	sfree(state->clues->dsf);
+	sfree(state->clues->clues);
+	sfree(state->clues);
+    }
+    sfree(state);
+}
+
+static char *solve_game(game_state *state, game_state *currstate,
+			char *aux, char **error)
+{
+    int w = state->par.w, a = w*w;
+    int i, ret;
+    digit *soln;
+    char *out;
+
+    if (aux)
+	return dupstr(aux);
+
+    soln = snewn(a, digit);
+    memset(soln, 0, a);
+
+    ret = solver(w, state->clues->dsf, state->clues->clues,
+		 soln, DIFFCOUNT-1);
+
+    if (ret == diff_impossible) {
+	*error = "No solution exists for this puzzle";
+	out = NULL;
+    } else if (ret == diff_ambiguous) {
+	*error = "Multiple solutions exist for this puzzle";
+	out = NULL;
+    } else {
+	out = snewn(a+2, char);
+	out[0] = 'S';
+	for (i = 0; i < a; i++)
+	    out[i+1] = '0' + soln[i];
+	out[a+1] = '\0';
+    }
+
+    sfree(soln);
+    return out;
+}
+
+static int game_can_format_as_text_now(game_params *params)
+{
+    return TRUE;
+}
+
+static char *game_text_format(game_state *state)
+{
+    return NULL;
+}
+
+struct game_ui {
+    /*
+     * These are the coordinates of the currently highlighted
+     * square on the grid, if hshow = 1.
+     */
+    int hx, hy;
+    /*
+     * This indicates whether the current highlight is a
+     * pencil-mark one or a real one.
+     */
+    int hpencil;
+    /*
+     * This indicates whether or not we're showing the highlight
+     * (used to be hx = hy = -1); important so that when we're
+     * using the cursor keys it doesn't keep coming back at a
+     * fixed position. When hshow = 1, pressing a valid number
+     * or letter key or Space will enter that number or letter in the grid.
+     */
+    int hshow;
+    /*
+     * This indicates whether we're using the highlight as a cursor;
+     * it means that it doesn't vanish on a keypress, and that it is
+     * allowed on immutable squares.
+     */
+    int hcursor;
+};
+
+static game_ui *new_ui(game_state *state)
+{
+    game_ui *ui = snew(game_ui);
+
+    ui->hx = ui->hy = 0;
+    ui->hpencil = ui->hshow = ui->hcursor = 0;
+
+    return ui;
+}
+
+static void free_ui(game_ui *ui)
+{
+    sfree(ui);
+}
+
+static char *encode_ui(game_ui *ui)
+{
+    return NULL;
+}
+
+static void decode_ui(game_ui *ui, char *encoding)
+{
+}
+
+static void game_changed_state(game_ui *ui, game_state *oldstate,
+                               game_state *newstate)
+{
+    int w = newstate->par.w;
+    /*
+     * We prevent pencil-mode highlighting of a filled square, unless
+     * we're using the cursor keys. So if the user has just filled in
+     * a square which we had a pencil-mode highlight in (by Undo, or
+     * by Redo, or by Solve), then we cancel the highlight.
+     */
+    if (ui->hshow && ui->hpencil && !ui->hcursor &&
+        newstate->grid[ui->hy * w + ui->hx] != 0) {
+        ui->hshow = 0;
+    }
+}
+
+#define PREFERRED_TILESIZE 48
+#define TILESIZE (ds->tilesize)
+#define BORDER (TILESIZE / 2)
+#define GRIDEXTRA max((TILESIZE / 32),1)
+#define COORD(x) ((x)*TILESIZE + BORDER)
+#define FROMCOORD(x) (((x)+(TILESIZE-BORDER)) / TILESIZE - 1)
+
+#define FLASH_TIME 0.4F
+
+#define DF_PENCIL_SHIFT 16
+#define DF_ERR_LATIN 0x8000
+#define DF_ERR_CLUE 0x4000
+#define DF_HIGHLIGHT 0x2000
+#define DF_HIGHLIGHT_PENCIL 0x1000
+#define DF_DIGIT_MASK 0x000F
+
+struct game_drawstate {
+    int tilesize;
+    int started;
+    long *tiles;
+    long *errors;
+    char *minus_sign, *times_sign, *divide_sign;
+};
+
+static int check_errors(game_state *state, long *errors)
+{
+    int w = state->par.w, a = w*w;
+    int i, j, x, y, errs = FALSE;
+    long *cluevals;
+    int *full;
+
+    cluevals = snewn(a, long);
+    full = snewn(a, int);
+
+    if (errors)
+	for (i = 0; i < a; i++) {
+	    errors[i] = 0;
+	    full[i] = TRUE;
+	}
+
+    for (i = 0; i < a; i++) {
+	long clue;
+
+	j = dsf_canonify(state->clues->dsf, i);
+	if (j == i) {
+	    cluevals[i] = state->grid[i];
+	} else {
+	    clue = state->clues->clues[j] & CMASK;
+
+	    switch (clue) {
+	      case C_ADD:
+		cluevals[j] += state->grid[i];
+		break;
+	      case C_MUL:
+		cluevals[j] *= state->grid[i];
+		break;
+	      case C_SUB:
+		cluevals[j] = abs(cluevals[j] - state->grid[i]);
+		break;
+	      case C_DIV:
+		{
+		    int d1 = cluevals[j], d2 = state->grid[i];
+		    if (d1 == 0 || d2 == 0)
+			cluevals[j] = 0;
+		    else
+			cluevals[j] = d2/d1 + d1/d2;/* one of them is 0 :-) */
+		}
+		break;
+	    }
+	}
+
+	if (!state->grid[i])
+	    full[j] = FALSE;
+    }
+
+    for (i = 0; i < a; i++) {
+	j = dsf_canonify(state->clues->dsf, i);
+	if (j == i) {
+	    if ((state->clues->clues[j] & ~CMASK) != cluevals[i]) {
+		errs = TRUE;
+		if (errors && full[j])
+		    errors[j] |= DF_ERR_CLUE;
+	    }
+	}
+    }
+
+    sfree(cluevals);
+    sfree(full);
+
+    for (y = 0; y < w; y++) {
+	int mask = 0, errmask = 0;
+	for (x = 0; x < w; x++) {
+	    int bit = 1 << state->grid[y*w+x];
+	    errmask |= (mask & bit);
+	    mask |= bit;
+	}
+
+	if (mask != (1 << (w+1)) - (1 << 1)) {
+	    errs = TRUE;
+	    errmask &= ~1;
+	    if (errors) {
+		for (x = 0; x < w; x++)
+		    if (errmask & (1 << state->grid[y*w+x]))
+			errors[y*w+x] |= DF_ERR_LATIN;
+	    }
+	}
+    }
+
+    for (x = 0; x < w; x++) {
+	int mask = 0, errmask = 0;
+	for (y = 0; y < w; y++) {
+	    int bit = 1 << state->grid[y*w+x];
+	    errmask |= (mask & bit);
+	    mask |= bit;
+	}
+
+	if (mask != (1 << (w+1)) - (1 << 1)) {
+	    errs = TRUE;
+	    errmask &= ~1;
+	    if (errors) {
+		for (y = 0; y < w; y++)
+		    if (errmask & (1 << state->grid[y*w+x]))
+			errors[y*w+x] |= DF_ERR_LATIN;
+	    }
+	}
+    }
+
+    return errs;
+}
+
+static char *interpret_move(game_state *state, game_ui *ui, game_drawstate *ds,
+			    int x, int y, int button)
+{
+    int w = state->par.w;
+    int tx, ty;
+    char buf[80];
+
+    button &= ~MOD_MASK;
+
+    tx = FROMCOORD(x);
+    ty = FROMCOORD(y);
+
+    if (tx >= 0 && tx < w && ty >= 0 && ty < w) {
+        if (button == LEFT_BUTTON) {
+	    if (tx == ui->hx && ty == ui->hy &&
+		ui->hshow && ui->hpencil == 0) {
+                ui->hshow = 0;
+            } else {
+                ui->hx = tx;
+                ui->hy = ty;
+                ui->hshow = 1;
+                ui->hpencil = 0;
+            }
+            ui->hcursor = 0;
+            return "";		       /* UI activity occurred */
+        }
+        if (button == RIGHT_BUTTON) {
+            /*
+             * Pencil-mode highlighting for non filled squares.
+             */
+            if (state->grid[ty*w+tx] == 0) {
+                if (tx == ui->hx && ty == ui->hy &&
+                    ui->hshow && ui->hpencil) {
+                    ui->hshow = 0;
+                } else {
+                    ui->hpencil = 1;
+                    ui->hx = tx;
+                    ui->hy = ty;
+                    ui->hshow = 1;
+                }
+            } else {
+                ui->hshow = 0;
+            }
+            ui->hcursor = 0;
+            return "";		       /* UI activity occurred */
+        }
+    }
+    if (IS_CURSOR_MOVE(button)) {
+        move_cursor(button, &ui->hx, &ui->hy, w, w, 0);
+        ui->hshow = ui->hcursor = 1;
+        return "";
+    }
+    if (ui->hshow &&
+        (button == CURSOR_SELECT)) {
+        ui->hpencil = 1 - ui->hpencil;
+        ui->hcursor = 1;
+        return "";
+    }
+
+    if (ui->hshow &&
+	((button >= '0' && button <= '9' && button - '0' <= w) ||
+	 button == CURSOR_SELECT2 || button == '\b')) {
+	int n = button - '0';
+	if (button == CURSOR_SELECT2 || button == '\b')
+	    n = 0;
+
+        /*
+         * Can't make pencil marks in a filled square. This can only
+         * become highlighted if we're using cursor keys.
+         */
+        if (ui->hpencil && state->grid[ui->hy*w+ui->hx])
+            return NULL;
+
+	sprintf(buf, "%c%d,%d,%d",
+		(char)(ui->hpencil && n > 0 ? 'P' : 'R'), ui->hx, ui->hy, n);
+
+        if (!ui->hcursor) ui->hshow = 0;
+
+	return dupstr(buf);
+    }
+
+    if (button == 'M' || button == 'm')
+        return dupstr("M");
+
+    return NULL;
+}
+
+static game_state *execute_move(game_state *from, char *move)
+{
+    int w = from->par.w, a = w*w;
+    game_state *ret;
+    int x, y, i, n;
+
+    if (move[0] == 'S') {
+	ret = dup_game(from);
+	ret->completed = ret->cheated = TRUE;
+
+	for (i = 0; i < a; i++) {
+	    if (move[i+1] < '1' || move[i+1] > '0'+w) {
+		free_game(ret);
+		return NULL;
+	    }
+	    ret->grid[i] = move[i+1] - '0';
+	    ret->pencil[i] = 0;
+	}
+
+	if (move[a+1] != '\0') {
+	    free_game(ret);
+	    return NULL;
+	}
+
+	return ret;
+    } else if ((move[0] == 'P' || move[0] == 'R') &&
+	sscanf(move+1, "%d,%d,%d", &x, &y, &n) == 3 &&
+	x >= 0 && x < w && y >= 0 && y < w && n >= 0 && n <= w) {
+
+	ret = dup_game(from);
+        if (move[0] == 'P' && n > 0) {
+            ret->pencil[y*w+x] ^= 1 << n;
+        } else {
+            ret->grid[y*w+x] = n;
+            ret->pencil[y*w+x] = 0;
+
+            if (!ret->completed && !check_errors(ret, NULL))
+                ret->completed = TRUE;
+        }
+	return ret;
+    } else if (move[0] == 'M') {
+	/*
+	 * Fill in absolutely all pencil marks everywhere. (I
+	 * wouldn't use this for actual play, but it's a handy
+	 * starting point when following through a set of
+	 * diagnostics output by the standalone solver.)
+	 */
+	ret = dup_game(from);
+	for (i = 0; i < a; i++) {
+	    if (!ret->grid[i])
+		ret->pencil[i] = (1 << (w+1)) - (1 << 1);
+	}
+	return ret;
+    } else
+	return NULL;		       /* couldn't parse move string */
+}
+
+/* ----------------------------------------------------------------------
+ * Drawing routines.
+ */
+
+#define SIZE(w) ((w) * TILESIZE + 2*BORDER)
+
+static void game_compute_size(game_params *params, int tilesize,
+			      int *x, int *y)
+{
+    /* Ick: fake up `ds->tilesize' for macro expansion purposes */
+    struct { int tilesize; } ads, *ds = &ads;
+    ads.tilesize = tilesize;
+
+    *x = *y = SIZE(params->w);
+}
+
+static void game_set_size(drawing *dr, game_drawstate *ds,
+			  game_params *params, int tilesize)
+{
+    ds->tilesize = tilesize;
+}
+
+static float *game_colours(frontend *fe, int *ncolours)
+{
+    float *ret = snewn(3 * NCOLOURS, float);
+
+    frontend_default_colour(fe, &ret[COL_BACKGROUND * 3]);
+
+    ret[COL_GRID * 3 + 0] = 0.0F;
+    ret[COL_GRID * 3 + 1] = 0.0F;
+    ret[COL_GRID * 3 + 2] = 0.0F;
+
+    ret[COL_USER * 3 + 0] = 0.0F;
+    ret[COL_USER * 3 + 1] = 0.6F * ret[COL_BACKGROUND * 3 + 1];
+    ret[COL_USER * 3 + 2] = 0.0F;
+
+    ret[COL_HIGHLIGHT * 3 + 0] = 0.78F * ret[COL_BACKGROUND * 3 + 0];
+    ret[COL_HIGHLIGHT * 3 + 1] = 0.78F * ret[COL_BACKGROUND * 3 + 1];
+    ret[COL_HIGHLIGHT * 3 + 2] = 0.78F * ret[COL_BACKGROUND * 3 + 2];
+
+    ret[COL_ERROR * 3 + 0] = 1.0F;
+    ret[COL_ERROR * 3 + 1] = 0.0F;
+    ret[COL_ERROR * 3 + 2] = 0.0F;
+
+    ret[COL_PENCIL * 3 + 0] = 0.5F * ret[COL_BACKGROUND * 3 + 0];
+    ret[COL_PENCIL * 3 + 1] = 0.5F * ret[COL_BACKGROUND * 3 + 1];
+    ret[COL_PENCIL * 3 + 2] = ret[COL_BACKGROUND * 3 + 2];
+
+    *ncolours = NCOLOURS;
+    return ret;
+}
+
+static const char *const minus_signs[] = { "\xE2\x88\x92", "-" };
+static const char *const times_signs[] = { "\xC3\x97", "*" };
+static const char *const divide_signs[] = { "\xC3\xB7", "/" };
+
+static game_drawstate *game_new_drawstate(drawing *dr, game_state *state)
+{
+    int w = state->par.w, a = w*w;
+    struct game_drawstate *ds = snew(struct game_drawstate);
+    int i;
+
+    ds->tilesize = 0;
+    ds->started = FALSE;
+    ds->tiles = snewn(a, long);
+    for (i = 0; i < a; i++)
+	ds->tiles[i] = -1;
+    ds->errors = snewn(a, long);
+    ds->minus_sign = text_fallback(dr, minus_signs, lenof(minus_signs));
+    ds->times_sign = text_fallback(dr, times_signs, lenof(times_signs));
+    ds->divide_sign = text_fallback(dr, divide_signs, lenof(divide_signs));
+
+    return ds;
+}
+
+static void game_free_drawstate(drawing *dr, game_drawstate *ds)
+{
+    sfree(ds->tiles);
+    sfree(ds->errors);
+    sfree(ds->minus_sign);
+    sfree(ds->times_sign);
+    sfree(ds->divide_sign);
+    sfree(ds);
+}
+
+void draw_tile(drawing *dr, game_drawstate *ds, struct clues *clues,
+	       int x, int y, long tile)
+{
+    int w = clues->w /* , a = w*w */;
+    int tx, ty, tw, th;
+    int cx, cy, cw, ch;
+    char str[64];
+
+    tx = BORDER + x * TILESIZE + 1 + GRIDEXTRA;
+    ty = BORDER + y * TILESIZE + 1 + GRIDEXTRA;
+
+    cx = tx;
+    cy = ty;
+    cw = tw = TILESIZE-1-2*GRIDEXTRA;
+    ch = th = TILESIZE-1-2*GRIDEXTRA;
+
+    if (x > 0 && dsf_canonify(clues->dsf, y*w+x) == dsf_canonify(clues->dsf, y*w+x-1))
+	cx -= GRIDEXTRA, cw += GRIDEXTRA;
+    if (x+1 < w && dsf_canonify(clues->dsf, y*w+x) == dsf_canonify(clues->dsf, y*w+x+1))
+	cw += GRIDEXTRA;
+    if (y > 0 && dsf_canonify(clues->dsf, y*w+x) == dsf_canonify(clues->dsf, (y-1)*w+x))
+	cy -= GRIDEXTRA, ch += GRIDEXTRA;
+    if (y+1 < w && dsf_canonify(clues->dsf, y*w+x) == dsf_canonify(clues->dsf, (y+1)*w+x))
+	ch += GRIDEXTRA;
+
+    clip(dr, cx, cy, cw, ch);
+
+    /* background needs erasing */
+    draw_rect(dr, cx, cy, cw, ch,
+	      (tile & DF_HIGHLIGHT) ? COL_HIGHLIGHT : COL_BACKGROUND);
+
+    /*
+     * Draw the corners of thick lines in corner-adjacent squares,
+     * which jut into this square by one pixel.
+     */
+    if (x > 0 && y > 0 && dsf_canonify(clues->dsf, y*w+x) != dsf_canonify(clues->dsf, (y-1)*w+x-1))
+	draw_rect(dr, tx-GRIDEXTRA, ty-GRIDEXTRA, GRIDEXTRA, GRIDEXTRA, COL_GRID);
+    if (x+1 < w && y > 0 && dsf_canonify(clues->dsf, y*w+x) != dsf_canonify(clues->dsf, (y-1)*w+x+1))
+	draw_rect(dr, tx+TILESIZE-1-2*GRIDEXTRA, ty-GRIDEXTRA, GRIDEXTRA, GRIDEXTRA, COL_GRID);
+    if (x > 0 && y+1 < w && dsf_canonify(clues->dsf, y*w+x) != dsf_canonify(clues->dsf, (y+1)*w+x-1))
+	draw_rect(dr, tx-GRIDEXTRA, ty+TILESIZE-1-2*GRIDEXTRA, GRIDEXTRA, GRIDEXTRA, COL_GRID);
+    if (x+1 < w && y+1 < w && dsf_canonify(clues->dsf, y*w+x) != dsf_canonify(clues->dsf, (y+1)*w+x+1))
+	draw_rect(dr, tx+TILESIZE-1-2*GRIDEXTRA, ty+TILESIZE-1-2*GRIDEXTRA, GRIDEXTRA, GRIDEXTRA, COL_GRID);
+
+    /* pencil-mode highlight */
+    if (tile & DF_HIGHLIGHT_PENCIL) {
+        int coords[6];
+        coords[0] = cx;
+        coords[1] = cy;
+        coords[2] = cx+cw/2;
+        coords[3] = cy;
+        coords[4] = cx;
+        coords[5] = cy+ch/2;
+        draw_polygon(dr, coords, 3, COL_HIGHLIGHT, COL_HIGHLIGHT);
+    }
+
+    /* Draw the box clue. */
+    if (dsf_canonify(clues->dsf, y*w+x) == y*w+x) {
+	long clue = clues->clues[y*w+x];
+	long cluetype = clue & CMASK, clueval = clue & ~CMASK;
+	int size = dsf_size(clues->dsf, y*w+x);
+	/*
+	 * Special case of clue-drawing: a box with only one square
+	 * is written as just the number, with no operation, because
+	 * it doesn't matter whether the operation is ADD or MUL.
+	 * The generation code above should never produce puzzles
+	 * containing such a thing - I think they're inelegant - but
+	 * it's possible to type in game IDs from elsewhere, so I
+	 * want to display them right if so.
+	 */
+	sprintf (str, "%ld%s", clueval,
+		 (size == 1 ? "" :
+		  cluetype == C_ADD ? "+" :
+		  cluetype == C_SUB ? ds->minus_sign :
+		  cluetype == C_MUL ? ds->times_sign :
+		  /* cluetype == C_DIV ? */ ds->divide_sign));
+	draw_text(dr, tx + GRIDEXTRA * 2, ty + GRIDEXTRA * 2 + TILESIZE/4,
+		  FONT_VARIABLE, TILESIZE/4, ALIGN_VNORMAL | ALIGN_HLEFT,
+		  (tile & DF_ERR_CLUE ? COL_ERROR : COL_GRID), str);
+    }
+
+    /* new number needs drawing? */
+    if (tile & DF_DIGIT_MASK) {
+	str[1] = '\0';
+	str[0] = (tile & DF_DIGIT_MASK) + '0';
+	draw_text(dr, tx + TILESIZE/2, ty + TILESIZE/2,
+		  FONT_VARIABLE, TILESIZE/2, ALIGN_VCENTRE | ALIGN_HCENTRE,
+		  (tile & DF_ERR_LATIN) ? COL_ERROR : COL_USER, str);
+    } else {
+        int i, j, npencil;
+	int pl, pr, pt, pb;
+	float bestsize;
+	int pw, ph, minph, pbest, fontsize;
+
+        /* Count the pencil marks required. */
+        for (i = 1, npencil = 0; i <= w; i++)
+            if (tile & (1 << (i + DF_PENCIL_SHIFT)))
+		npencil++;
+	if (npencil) {
+
+	    minph = 2;
+
+	    /*
+	     * Determine the bounding rectangle within which we're going
+	     * to put the pencil marks.
+	     */
+	    /* Start with the whole square */
+	    pl = tx + GRIDEXTRA;
+	    pr = pl + TILESIZE - GRIDEXTRA;
+	    pt = ty + GRIDEXTRA;
+	    pb = pt + TILESIZE - GRIDEXTRA;
+
+	    /*
+	     * We arrange our pencil marks in a grid layout, with
+	     * the number of rows and columns adjusted to allow the
+	     * maximum font size.
+	     *
+	     * So now we work out what the grid size ought to be.
+	     */
+	    bestsize = 0.0;
+	    pbest = 0;
+	    /* Minimum */
+	    for (pw = 3; pw < max(npencil,4); pw++) {
+		float fw, fh, fs;
+
+		ph = (npencil + pw - 1) / pw;
+		ph = max(ph, minph);
+		fw = (pr - pl) / (float)pw;
+		fh = (pb - pt) / (float)ph;
+		fs = min(fw, fh);
+		if (fs > bestsize) {
+		    bestsize = fs;
+		    pbest = pw;
+		}
+	    }
+	    assert(pbest > 0);
+	    pw = pbest;
+	    ph = (npencil + pw - 1) / pw;
+	    ph = max(ph, minph);
+
+	    /*
+	     * Now we've got our grid dimensions, work out the pixel
+	     * size of a grid element, and round it to the nearest
+	     * pixel. (We don't want rounding errors to make the
+	     * grid look uneven at low pixel sizes.)
+	     */
+	    fontsize = min((pr - pl) / pw, (pb - pt) / ph);
+
+	    /*
+	     * Centre the resulting figure in the square.
+	     */
+	    pl = tx + (TILESIZE - fontsize * pw) / 2;
+	    pt = ty + (TILESIZE - fontsize * ph) / 2;
+
+	    /*
+	     * And move it down a bit if it's collided with some
+	     * clue text.
+	     */
+	    if (dsf_canonify(clues->dsf, y*w+x) == y*w+x) {
+		pt = max(pt, ty + GRIDEXTRA * 3 + TILESIZE/4);
+	    }
+
+	    /*
+	     * Now actually draw the pencil marks.
+	     */
+	    for (i = 1, j = 0; i <= w; i++)
+		if (tile & (1 << (i + DF_PENCIL_SHIFT))) {
+		    int dx = j % pw, dy = j / pw;
+
+		    str[1] = '\0';
+		    str[0] = i + '0';
+		    draw_text(dr, pl + fontsize * (2*dx+1) / 2,
+			      pt + fontsize * (2*dy+1) / 2,
+			      FONT_VARIABLE, fontsize,
+			      ALIGN_VCENTRE | ALIGN_HCENTRE, COL_PENCIL, str);
+		    j++;
+		}
+	}
+    }
+
+    unclip(dr);
+
+    draw_update(dr, cx, cy, cw, ch);
+}
+
+static void game_redraw(drawing *dr, game_drawstate *ds, game_state *oldstate,
+			game_state *state, int dir, game_ui *ui,
+			float animtime, float flashtime)
+{
+    int w = state->par.w /*, a = w*w */;
+    int x, y;
+
+    if (!ds->started) {
+	/*
+	 * The initial contents of the window are not guaranteed and
+	 * can vary with front ends. To be on the safe side, all
+	 * games should start by drawing a big background-colour
+	 * rectangle covering the whole window.
+	 */
+	draw_rect(dr, 0, 0, SIZE(w), SIZE(w), COL_BACKGROUND);
+
+	/*
+	 * Big containing rectangle.
+	 */
+	draw_rect(dr, COORD(0) - GRIDEXTRA, COORD(0) - GRIDEXTRA,
+		  w*TILESIZE+1+GRIDEXTRA*2, w*TILESIZE+1+GRIDEXTRA*2,
+		  COL_GRID);
+
+	draw_update(dr, 0, 0, SIZE(w), SIZE(w));
+
+	ds->started = TRUE;
+    }
+
+    check_errors(state, ds->errors);
+
+    for (y = 0; y < w; y++) {
+	for (x = 0; x < w; x++) {
+	    long tile = 0L;
+
+	    if (state->grid[y*w+x])
+		tile = state->grid[y*w+x];
+	    else
+		tile = (long)state->pencil[y*w+x] << DF_PENCIL_SHIFT;
+
+	    if (ui->hshow && ui->hx == x && ui->hy == y)
+		tile |= (ui->hpencil ? DF_HIGHLIGHT_PENCIL : DF_HIGHLIGHT);
+
+            if (flashtime > 0 &&
+                (flashtime <= FLASH_TIME/3 ||
+                 flashtime >= FLASH_TIME*2/3))
+                tile |= DF_HIGHLIGHT;  /* completion flash */
+
+	    tile |= ds->errors[y*w+x];
+
+	    if (ds->tiles[y*w+x] != tile) {
+		ds->tiles[y*w+x] = tile;
+		draw_tile(dr, ds, state->clues, x, y, tile);
+	    }
+	}
+    }
+}
+
+static float game_anim_length(game_state *oldstate, game_state *newstate,
+			      int dir, game_ui *ui)
+{
+    return 0.0F;
+}
+
+static float game_flash_length(game_state *oldstate, game_state *newstate,
+			       int dir, game_ui *ui)
+{
+    if (!oldstate->completed && newstate->completed &&
+	!oldstate->cheated && !newstate->cheated)
+        return FLASH_TIME;
+    return 0.0F;
+}
+
+static int game_timing_state(game_state *state, game_ui *ui)
+{
+    if (state->completed)
+	return FALSE;
+    return TRUE;
+}
+
+static void game_print_size(game_params *params, float *x, float *y)
+{
+    int pw, ph;
+
+    /*
+     * We use 9mm squares by default, like Solo.
+     */
+    game_compute_size(params, 900, &pw, &ph);
+    *x = pw / 100.0F;
+    *y = ph / 100.0F;
+}
+
+/*
+ * Subfunction to draw the thick lines between cells. In order to do
+ * this using the line-drawing rather than rectangle-drawing API (so
+ * as to get line thicknesses to scale correctly) and yet have
+ * correctly mitred joins between lines, we must do this by tracing
+ * the boundary of each sub-block and drawing it in one go as a
+ * single polygon.
+ */
+static void outline_block_structure(drawing *dr, game_drawstate *ds,
+				    int w, int *dsf, int ink)
+{
+    int a = w*w;
+    int *coords;
+    int i, n;
+    int x, y, dx, dy, sx, sy, sdx, sdy;
+
+    coords = snewn(4*a, int);
+
+    /*
+     * Iterate over all the blocks.
+     */
+    for (i = 0; i < a; i++) {
+	if (dsf_canonify(dsf, i) != i)
+	    continue;
+
+	/*
+	 * For each block, we need a starting square within it which
+	 * has a boundary at the left. Conveniently, we have one
+	 * right here, by construction.
+	 */
+	x = i % w;
+	y = i / w;
+	dx = -1;
+	dy = 0;
+
+	/*
+	 * Now begin tracing round the perimeter. At all
+	 * times, (x,y) describes some square within the
+	 * block, and (x+dx,y+dy) is some adjacent square
+	 * outside it; so the edge between those two squares
+	 * is always an edge of the block.
+	 */
+	sx = x, sy = y, sdx = dx, sdy = dy;   /* save starting position */
+	n = 0;
+	do {
+	    int cx, cy, tx, ty, nin;
+
+	    /*
+	     * Advance to the next edge, by looking at the two
+	     * squares beyond it. If they're both outside the block,
+	     * we turn right (by leaving x,y the same and rotating
+	     * dx,dy clockwise); if they're both inside, we turn
+	     * left (by rotating dx,dy anticlockwise and contriving
+	     * to leave x+dx,y+dy unchanged); if one of each, we go
+	     * straight on (and may enforce by assertion that
+	     * they're one of each the _right_ way round).
+	     */
+	    nin = 0;
+	    tx = x - dy + dx;
+	    ty = y + dx + dy;
+	    nin += (tx >= 0 && tx < w && ty >= 0 && ty < w &&
+		    dsf_canonify(dsf, ty*w+tx) == i);
+	    tx = x - dy;
+	    ty = y + dx;
+	    nin += (tx >= 0 && tx < w && ty >= 0 && ty < w &&
+		    dsf_canonify(dsf, ty*w+tx) == i);
+	    if (nin == 0) {
+		/*
+		 * Turn right.
+		 */
+		int tmp;
+		tmp = dx;
+		dx = -dy;
+		dy = tmp;
+	    } else if (nin == 2) {
+		/*
+		 * Turn left.
+		 */
+		int tmp;
+
+		x += dx;
+		y += dy;
+
+		tmp = dx;
+		dx = dy;
+		dy = -tmp;
+
+		x -= dx;
+		y -= dy;
+	    } else {
+		/*
+		 * Go straight on.
+		 */
+		x -= dy;
+		y += dx;
+	    }
+
+	    /*
+	     * Now enforce by assertion that we ended up
+	     * somewhere sensible.
+	     */
+	    assert(x >= 0 && x < w && y >= 0 && y < w &&
+		   dsf_canonify(dsf, y*w+x) == i);
+	    assert(x+dx < 0 || x+dx >= w || y+dy < 0 || y+dy >= w ||
+		   dsf_canonify(dsf, (y+dy)*w+(x+dx)) != i);
+
+	    /*
+	     * Record the point we just went past at one end of the
+	     * edge. To do this, we translate (x,y) down and right
+	     * by half a unit (so they're describing a point in the
+	     * _centre_ of the square) and then translate back again
+	     * in a manner rotated by dy and dx.
+	     */
+	    assert(n < 2*w+2);
+	    cx = ((2*x+1) + dy + dx) / 2;
+	    cy = ((2*y+1) - dx + dy) / 2;
+	    coords[2*n+0] = BORDER + cx * TILESIZE;
+	    coords[2*n+1] = BORDER + cy * TILESIZE;
+	    n++;
+
+	} while (x != sx || y != sy || dx != sdx || dy != sdy);
+
+	/*
+	 * That's our polygon; now draw it.
+	 */
+	draw_polygon(dr, coords, n, -1, ink);
+    }
+
+    sfree(coords);
+}
+
+static void game_print(drawing *dr, game_state *state, int tilesize)
+{
+    int w = state->par.w;
+    int ink = print_mono_colour(dr, 0);
+    int x, y;
+    char *minus_sign, *times_sign, *divide_sign;
+
+    /* Ick: fake up `ds->tilesize' for macro expansion purposes */
+    game_drawstate ads, *ds = &ads;
+    game_set_size(dr, ds, NULL, tilesize);
+
+    minus_sign = text_fallback(dr, minus_signs, lenof(minus_signs));
+    times_sign = text_fallback(dr, times_signs, lenof(times_signs));
+    divide_sign = text_fallback(dr, divide_signs, lenof(divide_signs));
+
+    /*
+     * Border.
+     */
+    print_line_width(dr, 3 * TILESIZE / 40);
+    draw_rect_outline(dr, BORDER, BORDER, w*TILESIZE, w*TILESIZE, ink);
+
+    /*
+     * Main grid.
+     */
+    for (x = 1; x < w; x++) {
+	print_line_width(dr, TILESIZE / 40);
+	draw_line(dr, BORDER+x*TILESIZE, BORDER,
+		  BORDER+x*TILESIZE, BORDER+w*TILESIZE, ink);
+    }
+    for (y = 1; y < w; y++) {
+	print_line_width(dr, TILESIZE / 40);
+	draw_line(dr, BORDER, BORDER+y*TILESIZE,
+		  BORDER+w*TILESIZE, BORDER+y*TILESIZE, ink);
+    }
+
+    /*
+     * Thick lines between cells.
+     */
+    print_line_width(dr, 3 * TILESIZE / 40);
+    outline_block_structure(dr, ds, w, state->clues->dsf, ink);
+
+    /*
+     * Clues.
+     */
+    for (y = 0; y < w; y++)
+	for (x = 0; x < w; x++)
+	    if (dsf_canonify(state->clues->dsf, y*w+x) == y*w+x) {
+		long clue = state->clues->clues[y*w+x];
+		long cluetype = clue & CMASK, clueval = clue & ~CMASK;
+		int size = dsf_size(state->clues->dsf, y*w+x);
+		char str[64];
+
+		/*
+		 * As in the drawing code, we omit the operator for
+		 * blocks of area 1.
+		 */
+		sprintf (str, "%ld%s", clueval,
+			 (size == 1 ? "" :
+			  cluetype == C_ADD ? "+" :
+			  cluetype == C_SUB ? minus_sign :
+			  cluetype == C_MUL ? times_sign :
+			  /* cluetype == C_DIV ? */ divide_sign));
+
+		draw_text(dr,
+			  BORDER+x*TILESIZE + 5*TILESIZE/80,
+			  BORDER+y*TILESIZE + 20*TILESIZE/80,
+			  FONT_VARIABLE, TILESIZE/4,
+			  ALIGN_VNORMAL | ALIGN_HLEFT,
+			  ink, str);
+	    }
+
+    /*
+     * Numbers for the solution, if any.
+     */
+    for (y = 0; y < w; y++)
+	for (x = 0; x < w; x++)
+	    if (state->grid[y*w+x]) {
+		char str[2];
+		str[1] = '\0';
+		str[0] = state->grid[y*w+x] + '0';
+		draw_text(dr, BORDER + x*TILESIZE + TILESIZE/2,
+			  BORDER + y*TILESIZE + TILESIZE/2,
+			  FONT_VARIABLE, TILESIZE/2,
+			  ALIGN_VCENTRE | ALIGN_HCENTRE, ink, str);
+	    }
+}
+
+#ifdef COMBINED
+#define thegame keen
+#endif
+
+const struct game thegame = {
+    "Keen", "games.keen", "keen",
+    default_params,
+    game_fetch_preset,
+    decode_params,
+    encode_params,
+    free_params,
+    dup_params,
+    TRUE, game_configure, custom_params,
+    validate_params,
+    new_game_desc,
+    validate_desc,
+    new_game,
+    dup_game,
+    free_game,
+    TRUE, solve_game,
+    FALSE, game_can_format_as_text_now, game_text_format,
+    new_ui,
+    free_ui,
+    encode_ui,
+    decode_ui,
+    game_changed_state,
+    interpret_move,
+    execute_move,
+    PREFERRED_TILESIZE, game_compute_size, game_set_size,
+    game_colours,
+    game_new_drawstate,
+    game_free_drawstate,
+    game_redraw,
+    game_anim_length,
+    game_flash_length,
+    TRUE, FALSE, game_print_size, game_print,
+    FALSE,			       /* wants_statusbar */
+    FALSE, game_timing_state,
+    REQUIRE_RBUTTON | REQUIRE_NUMPAD,  /* flags */
+};
+
+#ifdef STANDALONE_SOLVER
+
+#include <stdarg.h>
+
+int main(int argc, char **argv)
+{
+    game_params *p;
+    game_state *s;
+    char *id = NULL, *desc, *err;
+    int grade = FALSE;
+    int ret, diff, really_show_working = FALSE;
+
+    while (--argc > 0) {
+        char *p = *++argv;
+        if (!strcmp(p, "-v")) {
+            really_show_working = TRUE;
+        } else if (!strcmp(p, "-g")) {
+            grade = TRUE;
+        } else if (*p == '-') {
+            fprintf(stderr, "%s: unrecognised option `%s'\n", argv[0], p);
+            return 1;
+        } else {
+            id = p;
+        }
+    }
+
+    if (!id) {
+        fprintf(stderr, "usage: %s [-g | -v] <game_id>\n", argv[0]);
+        return 1;
+    }
+
+    desc = strchr(id, ':');
+    if (!desc) {
+        fprintf(stderr, "%s: game id expects a colon in it\n", argv[0]);
+        return 1;
+    }
+    *desc++ = '\0';
+
+    p = default_params();
+    decode_params(p, id);
+    err = validate_desc(p, desc);
+    if (err) {
+        fprintf(stderr, "%s: %s\n", argv[0], err);
+        return 1;
+    }
+    s = new_game(NULL, p, desc);
+
+    /*
+     * When solving an Easy puzzle, we don't want to bother the
+     * user with Hard-level deductions. For this reason, we grade
+     * the puzzle internally before doing anything else.
+     */
+    ret = -1;			       /* placate optimiser */
+    solver_show_working = FALSE;
+    for (diff = 0; diff < DIFFCOUNT; diff++) {
+	memset(s->grid, 0, p->w * p->w);
+	ret = solver(p->w, s->clues->dsf, s->clues->clues,
+		     s->grid, diff);
+	if (ret <= diff)
+	    break;
+    }
+
+    if (diff == DIFFCOUNT) {
+	if (grade)
+	    printf("Difficulty rating: ambiguous\n");
+	else
+	    printf("Unable to find a unique solution\n");
+    } else {
+	if (grade) {
+	    if (ret == diff_impossible)
+		printf("Difficulty rating: impossible (no solution exists)\n");
+	    else
+		printf("Difficulty rating: %s\n", keen_diffnames[ret]);
+	} else {
+	    solver_show_working = really_show_working;
+	    memset(s->grid, 0, p->w * p->w);
+	    ret = solver(p->w, s->clues->dsf, s->clues->clues,
+			 s->grid, diff);
+	    if (ret != diff)
+		printf("Puzzle is inconsistent\n");
+	    else {
+		/*
+		 * We don't have a game_text_format for this game,
+		 * so we have to output the solution manually.
+		 */
+		int x, y;
+		for (y = 0; y < p->w; y++) {
+		    for (x = 0; x < p->w; x++) {
+			printf("%s%c", x>0?" ":"", '0' + s->grid[y*p->w+x]);
+		    }
+		    putchar('\n');
+		}
+	    }
+	}
+    }
+
+    return 0;
+}
+
+#endif
+
+/* vim: set shiftwidth=4 tabstop=8: */
--- a/puzzles.but
+++ b/puzzles.but
@@ -22,6 +22,12 @@
 
 \define{dash} \u2013{-}
 
+\define{times} \u00D7{*}
+
+\define{divide} \u00F7{/}
+
+\define{minus} \u2212{-}
+
 This is a collection of small one-player puzzle games.
 
 \copyright This manual is copyright 2004-2009 Simon Tatham. All rights
@@ -2450,6 +2456,110 @@
 grid, through the \q{Type} menu.
 
 
+\C{keen} \i{Keen}
+
+\cfg{winhelp-topic}{games.keen}
+
+You have a square grid; each square may contain a digit from 1 to
+the size of the grid. The grid is divided into blocks of varying
+shape and size, with arithmetic clues written in them. Your aim is
+to fully populate the grid with digits such that:
+
+\b Each row contains only one occurrence of each digit
+
+\b Each column contains only one occurrence of each digit
+
+\b The digits in each block can be combined to form the number
+stated in the clue, using the arithmetic operation given in the
+clue. That is:
+
+\lcont{
+
+\b An addition clue means that the sum of the digits in the block
+must be the given number. For example, \q{15+} means the contents of
+the block adds up to fifteen.
+
+\b A multiplication clue (e.g. \q{60\times}), similarly, means that
+the product of the digits in the block must be the given number.
+
+\b A subtraction clue will always be written in a block of size two,
+and it means that one of the digits in the block is greater than the
+other by the given amount. For example, \q{2\minus} means that one
+of the digits in the block is 2 more than the other, or equivalently
+that one digit minus the other one is 2. The two digits could be
+either way round, though.
+
+\b A division clue (e.g. \q{3\divide}), similarly, is always in a
+block of size two and means that one digit divided by the other is
+equal to the given amount.
+
+Note that a block may contain more than one digit the same (provided
+the identical ones are not in the same row and column). This rule is
+precisely the opposite of the rule in Solo's \q{Killer} mode (see
+\k{solo}).
+
+}
+
+This puzzle appears in the Times under the name \q{KenKen}.
+
+
+\H{keen-controls} \i{Keen controls}
+
+\IM{Keen controls} controls, for Keen
+
+Keen shares much of its control system with Solo (and Unequal).
+
+To play Keen, simply click the mouse in any empty square and then
+type a digit on the keyboard to fill that square. If you make a
+mistake, click the mouse in the incorrect square and press Space to
+clear it again (or use the Undo feature).
+
+If you \e{right}-click in a square and then type a number, that
+number will be entered in the square as a \q{pencil mark}. You can
+have pencil marks for multiple numbers in the same square. Squares
+containing filled-in numbers cannot also contain pencil marks.
+
+The game pays no attention to pencil marks, so exactly what you use
+them for is up to you: you can use them as reminders that a
+particular square needs to be re-examined once you know more about a
+particular number, or you can use them as lists of the possible
+numbers in a given square, or anything else you feel like.
+
+To erase a single pencil mark, right-click in the square and type
+the same number again.
+
+All pencil marks in a square are erased when you left-click and type
+a number, or when you left-click and press space. Right-clicking and
+pressing space will also erase pencil marks.
+
+As for Solo, the cursor keys can be used in conjunction with the
+digit keys to set numbers or pencil marks. Use the cursor keys to
+move a highlight around the grid, and type a digit to enter it in
+the highlighted square. Pressing return toggles the highlight into a
+mode in which you can enter or remove pencil marks.
+
+Pressing M will fill in a full set of pencil marks in every square
+that does not have a main digit in it.
+
+(All the actions described in \k{common-actions} are also available.)
+
+\H{keen-parameters} \I{parameters, for Keen}Keen parameters
+
+These parameters are available from the \q{Custom...} option on the
+\q{Type} menu.
+
+\dt \e{Grid size}
+
+\dd Specifies the size of the grid. Lower limit is 3; upper limit is
+9 (because the user interface would become more difficult with
+\q{digits} bigger than 9!).
+
+\dt \e{Difficulty}
+
+\dd Controls the difficulty of the generated puzzle. At Unreasonable
+level, some backtracking will be required, but the solution should
+still be unique. The remaining levels require increasingly complex
+reasoning to avoid having to backtrack.
 
 \A{licence} \I{MIT licence}\ii{Licence}