ref: 4408476b7584b9a316466fe1bd10867103cddf57
dir: /keen.c/
/*
* keen.c: an implementation of the Times's 'KenKen' puzzle, and
* also of Nikoli's very similar 'Inshi No Heya' puzzle.
*/
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <assert.h>
#include <ctype.h>
#include <math.h>
#include "puzzles.h"
#include "latin.h"
/*
* Difficulty levels. I do some macro ickery here to ensure that my
* enum and the various forms of my name list always match up.
*/
#define DIFFLIST(A) \
A(EASY,Easy,solver_easy,e) \
A(NORMAL,Normal,solver_normal,n) \
A(HARD,Hard,solver_hard,h) \
A(EXTREME,Extreme,NULL,x) \
A(UNREASONABLE,Unreasonable,NULL,u)
#define ENUM(upper,title,func,lower) DIFF_ ## upper,
#define TITLE(upper,title,func,lower) #title,
#define ENCODE(upper,title,func,lower) #lower
#define CONFIG(upper,title,func,lower) ":" #title
enum { DIFFLIST(ENUM) DIFFCOUNT };
static char const *const keen_diffnames[] = { DIFFLIST(TITLE) };
static char const keen_diffchars[] = DIFFLIST(ENCODE);
#define DIFFCONFIG DIFFLIST(CONFIG)
/*
* Clue notation. Important here that ADD and MUL come before SUB
* and DIV, and that DIV comes last.
*/
#define C_ADD 0x00000000L
#define C_MUL 0x20000000L
#define C_SUB 0x40000000L
#define C_DIV 0x60000000L
#define CMASK 0x60000000L
#define CUNIT 0x20000000L
/*
* Maximum size of any clue block. Very large ones are annoying in UI
* terms (if they're multiplicative you end up with too many digits to
* fit in the square) and also in solver terms (too many possibilities
* to iterate over).
*/
#define MAXBLK 6
enum {
COL_BACKGROUND,
COL_GRID,
COL_USER,
COL_HIGHLIGHT,
COL_ERROR,
COL_PENCIL,
NCOLOURS
};
struct game_params {
int w, diff, multiplication_only;
};
struct clues {
int refcount;
int w;
int *dsf;
long *clues;
};
struct game_state {
game_params par;
struct clues *clues;
digit *grid;
int *pencil; /* bitmaps using bits 1<<1..1<<n */
int completed, cheated;
};
static game_params *default_params(void)
{
game_params *ret = snew(game_params);
ret->w = 6;
ret->diff = DIFF_NORMAL;
ret->multiplication_only = FALSE;
return ret;
}
const static struct game_params keen_presets[] = {
{ 4, DIFF_EASY, FALSE },
{ 5, DIFF_EASY, FALSE },
{ 5, DIFF_EASY, TRUE },
{ 6, DIFF_EASY, FALSE },
{ 6, DIFF_NORMAL, FALSE },
{ 6, DIFF_NORMAL, TRUE },
{ 6, DIFF_HARD, FALSE },
{ 6, DIFF_EXTREME, FALSE },
{ 6, DIFF_UNREASONABLE, FALSE },
{ 9, DIFF_NORMAL, FALSE },
};
static int game_fetch_preset(int i, char **name, game_params **params)
{
game_params *ret;
char buf[80];
if (i < 0 || i >= lenof(keen_presets))
return FALSE;
ret = snew(game_params);
*ret = keen_presets[i]; /* structure copy */
sprintf(buf, "%dx%d %s%s", ret->w, ret->w, keen_diffnames[ret->diff],
ret->multiplication_only ? ", multiplication only" : "");
*name = dupstr(buf);
*params = ret;
return TRUE;
}
static void free_params(game_params *params)
{
sfree(params);
}
static game_params *dup_params(const game_params *params)
{
game_params *ret = snew(game_params);
*ret = *params; /* structure copy */
return ret;
}
static void decode_params(game_params *params, char const *string)
{
char const *p = string;
params->w = atoi(p);
while (*p && isdigit((unsigned char)*p)) p++;
if (*p == 'd') {
int i;
p++;
params->diff = DIFFCOUNT+1; /* ...which is invalid */
if (*p) {
for (i = 0; i < DIFFCOUNT; i++) {
if (*p == keen_diffchars[i])
params->diff = i;
}
p++;
}
}
if (*p == 'm') {
p++;
params->multiplication_only = TRUE;
}
}
static char *encode_params(const game_params *params, int full)
{
char ret[80];
sprintf(ret, "%d", params->w);
if (full)
sprintf(ret + strlen(ret), "d%c%s", keen_diffchars[params->diff],
params->multiplication_only ? "m" : "");
return dupstr(ret);
}
static config_item *game_configure(const game_params *params)
{
config_item *ret;
char buf[80];
ret = snewn(4, config_item);
ret[0].name = "Grid size";
ret[0].type = C_STRING;
sprintf(buf, "%d", params->w);
ret[0].u.string.sval = dupstr(buf);
ret[1].name = "Difficulty";
ret[1].type = C_CHOICES;
ret[1].u.choices.choicenames = DIFFCONFIG;
ret[1].u.choices.selected = params->diff;
ret[2].name = "Multiplication only";
ret[2].type = C_BOOLEAN;
ret[2].u.boolean.bval = params->multiplication_only;
ret[3].name = NULL;
ret[3].type = C_END;
return ret;
}
static game_params *custom_params(const config_item *cfg)
{
game_params *ret = snew(game_params);
ret->w = atoi(cfg[0].u.string.sval);
ret->diff = cfg[1].u.choices.selected;
ret->multiplication_only = cfg[2].u.boolean.bval;
return ret;
}
static const char *validate_params(const game_params *params, int full)
{
if (params->w < 3 || params->w > 9)
return "Grid size must be between 3 and 9";
if (params->diff >= DIFFCOUNT)
return "Unknown difficulty rating";
return NULL;
}
/* ----------------------------------------------------------------------
* Solver.
*/
struct solver_ctx {
int w, diff;
int nboxes;
int *boxes, *boxlist, *whichbox;
long *clues;
digit *soln;
digit *dscratch;
int *iscratch;
};
static void solver_clue_candidate(struct solver_ctx *ctx, int diff, int box)
{
int w = ctx->w;
int n = ctx->boxes[box+1] - ctx->boxes[box];
int j;
/*
* This function is called from the main clue-based solver
* routine when we discover a candidate layout for a given clue
* box consistent with everything we currently know about the
* digit constraints in that box. We expect to find the digits
* of the candidate layout in ctx->dscratch, and we update
* ctx->iscratch as appropriate.
*
* The contents of ctx->iscratch are completely different
* depending on whether diff == DIFF_HARD or not. This function
* uses iscratch completely differently between the two cases, and
* the code in solver_common() which consumes the result must
* likewise have an if statement with completely different
* branches for the two cases.
*
* In DIFF_EASY and DIFF_NORMAL modes, the valid entries in
* ctx->iscratch are 0,...,n-1, and each of those entries
* ctx->iscratch[i] gives a bitmap of the possible digits in the
* ith square of the clue box currently under consideration. So
* each entry of iscratch starts off as an empty bitmap, and we
* set bits in it as possible layouts for the clue box are
* considered (and the difference between DIFF_EASY and
* DIFF_NORMAL is just that in DIFF_EASY mode we deliberately set
* more bits than absolutely necessary, hence restricting our own
* knowledge).
*
* But in DIFF_HARD mode, the valid entries are 0,...,2*w-1 (at
* least outside *this* function - inside this function, we also
* use 2*w,...,4*w-1 as scratch space in the loop below); the
* first w of those give the possible digits in the intersection
* of the current clue box with each column of the puzzle, and the
* next w do the same for each row. In this mode, each iscratch
* entry starts off as a _full_ bitmap, and in this function we
* _clear_ bits for digits that are absent from a given row or
* column in each candidate layout, so that the only bits which
* remain set are those for digits which have to appear in a given
* row/column no matter how the clue box is laid out.
*/
if (diff == DIFF_EASY) {
unsigned mask = 0;
/*
* Easy-mode clue deductions: we do not record information
* about which squares take which values, so we amalgamate
* all the values in dscratch and OR them all into
* everywhere.
*/
for (j = 0; j < n; j++)
mask |= 1 << ctx->dscratch[j];
for (j = 0; j < n; j++)
ctx->iscratch[j] |= mask;
} else if (diff == DIFF_NORMAL) {
/*
* Normal-mode deductions: we process the information in
* dscratch in the obvious way.
*/
for (j = 0; j < n; j++)
ctx->iscratch[j] |= 1 << ctx->dscratch[j];
} else if (diff == DIFF_HARD) {
/*
* Hard-mode deductions: instead of ruling things out
* _inside_ the clue box, we look for numbers which occur in
* a given row or column in all candidate layouts, and rule
* them out of all squares in that row or column that
* _aren't_ part of this clue box.
*/
int *sq = ctx->boxlist + ctx->boxes[box];
for (j = 0; j < 2*w; j++)
ctx->iscratch[2*w+j] = 0;
for (j = 0; j < n; j++) {
int x = sq[j] / w, y = sq[j] % w;
ctx->iscratch[2*w+x] |= 1 << ctx->dscratch[j];
ctx->iscratch[3*w+y] |= 1 << ctx->dscratch[j];
}
for (j = 0; j < 2*w; j++)
ctx->iscratch[j] &= ctx->iscratch[2*w+j];
}
}
static int solver_common(struct latin_solver *solver, void *vctx, int diff)
{
struct solver_ctx *ctx = (struct solver_ctx *)vctx;
int w = ctx->w;
int box, i, j, k;
int ret = 0, total;
/*
* Iterate over each clue box and deduce what we can.
*/
for (box = 0; box < ctx->nboxes; box++) {
int *sq = ctx->boxlist + ctx->boxes[box];
int n = ctx->boxes[box+1] - ctx->boxes[box];
long value = ctx->clues[box] & ~CMASK;
long op = ctx->clues[box] & CMASK;
/*
* Initialise ctx->iscratch for this clue box. At different
* difficulty levels we must initialise a different amount of
* it to different things; see the comments in
* solver_clue_candidate explaining what each version does.
*/
if (diff == DIFF_HARD) {
for (i = 0; i < 2*w; i++)
ctx->iscratch[i] = (1 << (w+1)) - (1 << 1);
} else {
for (i = 0; i < n; i++)
ctx->iscratch[i] = 0;
}
switch (op) {
case C_SUB:
case C_DIV:
/*
* These two clue types must always apply to a box of
* area 2. Also, the two digits in these boxes can never
* be the same (because any domino must have its two
* squares in either the same row or the same column).
* So we simply iterate over all possibilities for the
* two squares (both ways round), rule out any which are
* inconsistent with the digit constraints we already
* have, and update the digit constraints with any new
* information thus garnered.
*/
assert(n == 2);
for (i = 1; i <= w; i++) {
j = (op == C_SUB ? i + value : i * value);
if (j > w) break;
/* (i,j) is a valid digit pair. Try it both ways round. */
if (solver->cube[sq[0]*w+i-1] &&
solver->cube[sq[1]*w+j-1]) {
ctx->dscratch[0] = i;
ctx->dscratch[1] = j;
solver_clue_candidate(ctx, diff, box);
}
if (solver->cube[sq[0]*w+j-1] &&
solver->cube[sq[1]*w+i-1]) {
ctx->dscratch[0] = j;
ctx->dscratch[1] = i;
solver_clue_candidate(ctx, diff, box);
}
}
break;
case C_ADD:
case C_MUL:
/*
* For these clue types, I have no alternative but to go
* through all possible number combinations.
*
* Instead of a tedious physical recursion, I iterate in
* the scratch array through all possibilities. At any
* given moment, i indexes the element of the box that
* will next be incremented.
*/
i = 0;
ctx->dscratch[i] = 0;
total = value; /* start with the identity */
while (1) {
if (i < n) {
/*
* Find the next valid value for cell i.
*/
for (j = ctx->dscratch[i] + 1; j <= w; j++) {
if (op == C_ADD ? (total < j) : (total % j != 0))
continue; /* this one won't fit */
if (!solver->cube[sq[i]*w+j-1])
continue; /* this one is ruled out already */
for (k = 0; k < i; k++)
if (ctx->dscratch[k] == j &&
(sq[k] % w == sq[i] % w ||
sq[k] / w == sq[i] / w))
break; /* clashes with another row/col */
if (k < i)
continue;
/* Found one. */
break;
}
if (j > w) {
/* No valid values left; drop back. */
i--;
if (i < 0)
break; /* overall iteration is finished */
if (op == C_ADD)
total += ctx->dscratch[i];
else
total *= ctx->dscratch[i];
} else {
/* Got a valid value; store it and move on. */
ctx->dscratch[i++] = j;
if (op == C_ADD)
total -= j;
else
total /= j;
ctx->dscratch[i] = 0;
}
} else {
if (total == (op == C_ADD ? 0 : 1))
solver_clue_candidate(ctx, diff, box);
i--;
if (op == C_ADD)
total += ctx->dscratch[i];
else
total *= ctx->dscratch[i];
}
}
break;
}
/*
* Do deductions based on the information we've now
* accumulated in ctx->iscratch. See the comments above in
* solver_clue_candidate explaining what data is left in here,
* and how it differs between DIFF_HARD and lower difficulty
* levels (hence the big if statement here).
*/
if (diff < DIFF_HARD) {
#ifdef STANDALONE_SOLVER
char prefix[256];
if (solver_show_working)
sprintf(prefix, "%*susing clue at (%d,%d):\n",
solver_recurse_depth*4, "",
sq[0]/w+1, sq[0]%w+1);
else
prefix[0] = '\0'; /* placate optimiser */
#endif
for (i = 0; i < n; i++)
for (j = 1; j <= w; j++) {
if (solver->cube[sq[i]*w+j-1] &&
!(ctx->iscratch[i] & (1 << j))) {
#ifdef STANDALONE_SOLVER
if (solver_show_working) {
printf("%s%*s ruling out %d at (%d,%d)\n",
prefix, solver_recurse_depth*4, "",
j, sq[i]/w+1, sq[i]%w+1);
prefix[0] = '\0';
}
#endif
solver->cube[sq[i]*w+j-1] = 0;
ret = 1;
}
}
} else {
#ifdef STANDALONE_SOLVER
char prefix[256];
if (solver_show_working)
sprintf(prefix, "%*susing clue at (%d,%d):\n",
solver_recurse_depth*4, "",
sq[0]/w+1, sq[0]%w+1);
else
prefix[0] = '\0'; /* placate optimiser */
#endif
for (i = 0; i < 2*w; i++) {
int start = (i < w ? i*w : i-w);
int step = (i < w ? 1 : w);
for (j = 1; j <= w; j++) if (ctx->iscratch[i] & (1 << j)) {
#ifdef STANDALONE_SOLVER
char prefix2[256];
if (solver_show_working)
sprintf(prefix2, "%*s this clue requires %d in"
" %s %d:\n", solver_recurse_depth*4, "",
j, i < w ? "column" : "row", i%w+1);
else
prefix2[0] = '\0'; /* placate optimiser */
#endif
for (k = 0; k < w; k++) {
int pos = start + k*step;
if (ctx->whichbox[pos] != box &&
solver->cube[pos*w+j-1]) {
#ifdef STANDALONE_SOLVER
if (solver_show_working) {
printf("%s%s%*s ruling out %d at (%d,%d)\n",
prefix, prefix2,
solver_recurse_depth*4, "",
j, pos/w+1, pos%w+1);
prefix[0] = prefix2[0] = '\0';
}
#endif
solver->cube[pos*w+j-1] = 0;
ret = 1;
}
}
}
}
/*
* Once we find one block we can do something with in
* this way, revert to trying easier deductions, so as
* not to generate solver diagnostics that make the
* problem look harder than it is. (We have to do this
* for the Hard deductions but not the Easy/Normal ones,
* because only the Hard deductions are cross-box.)
*/
if (ret)
return ret;
}
}
return ret;
}
static int solver_easy(struct latin_solver *solver, void *vctx)
{
/*
* Omit the EASY deductions when solving at NORMAL level, since
* the NORMAL deductions are a superset of them anyway and it
* saves on time and confusing solver diagnostics.
*
* Note that this breaks the natural semantics of the return
* value of latin_solver. Without this hack, you could determine
* a puzzle's difficulty in one go by trying to solve it at
* maximum difficulty and seeing what difficulty value was
* returned; but with this hack, solving an Easy puzzle on
* Normal difficulty will typically return Normal. Hence the
* uses of the solver to determine difficulty are all arranged
* so as to double-check by re-solving at the next difficulty
* level down and making sure it failed.
*/
struct solver_ctx *ctx = (struct solver_ctx *)vctx;
if (ctx->diff > DIFF_EASY)
return 0;
return solver_common(solver, vctx, DIFF_EASY);
}
static int solver_normal(struct latin_solver *solver, void *vctx)
{
return solver_common(solver, vctx, DIFF_NORMAL);
}
static int solver_hard(struct latin_solver *solver, void *vctx)
{
return solver_common(solver, vctx, DIFF_HARD);
}
#define SOLVER(upper,title,func,lower) func,
static usersolver_t const keen_solvers[] = { DIFFLIST(SOLVER) };
static int solver(int w, int *dsf, long *clues, digit *soln, int maxdiff)
{
int a = w*w;
struct solver_ctx ctx;
int ret;
int i, j, n, m;
ctx.w = w;
ctx.soln = soln;
ctx.diff = maxdiff;
/*
* Transform the dsf-formatted clue list into one over which we
* can iterate more easily.
*
* Also transpose the x- and y-coordinates at this point,
* because the 'cube' array in the general Latin square solver
* puts x first (oops).
*/
for (ctx.nboxes = i = 0; i < a; i++)
if (dsf_canonify(dsf, i) == i)
ctx.nboxes++;
ctx.boxlist = snewn(a, int);
ctx.boxes = snewn(ctx.nboxes+1, int);
ctx.clues = snewn(ctx.nboxes, long);
ctx.whichbox = snewn(a, int);
for (n = m = i = 0; i < a; i++)
if (dsf_canonify(dsf, i) == i) {
ctx.clues[n] = clues[i];
ctx.boxes[n] = m;
for (j = 0; j < a; j++)
if (dsf_canonify(dsf, j) == i) {
ctx.boxlist[m++] = (j % w) * w + (j / w); /* transpose */
ctx.whichbox[ctx.boxlist[m-1]] = n;
}
n++;
}
assert(n == ctx.nboxes);
assert(m == a);
ctx.boxes[n] = m;
ctx.dscratch = snewn(a+1, digit);
ctx.iscratch = snewn(max(a+1, 4*w), int);
ret = latin_solver(soln, w, maxdiff,
DIFF_EASY, DIFF_HARD, DIFF_EXTREME,
DIFF_EXTREME, DIFF_UNREASONABLE,
keen_solvers, &ctx, NULL, NULL);
sfree(ctx.dscratch);
sfree(ctx.iscratch);
sfree(ctx.whichbox);
sfree(ctx.boxlist);
sfree(ctx.boxes);
sfree(ctx.clues);
return ret;
}
/* ----------------------------------------------------------------------
* Grid generation.
*/
static char *encode_block_structure(char *p, int w, int *dsf)
{
int i, currrun = 0;
char *orig, *q, *r, c;
orig = p;
/*
* Encode the block structure. We do this by encoding the
* pattern of dividing lines: first we iterate over the w*(w-1)
* internal vertical grid lines in ordinary reading order, then
* over the w*(w-1) internal horizontal ones in transposed
* reading order.
*
* We encode the number of non-lines between the lines; _ means
* zero (two adjacent divisions), a means 1, ..., y means 25,
* and z means 25 non-lines _and no following line_ (so that za
* means 26, zb 27 etc).
*/
for (i = 0; i <= 2*w*(w-1); i++) {
int x, y, p0, p1, edge;
if (i == 2*w*(w-1)) {
edge = TRUE; /* terminating virtual edge */
} else {
if (i < w*(w-1)) {
y = i/(w-1);
x = i%(w-1);
p0 = y*w+x;
p1 = y*w+x+1;
} else {
x = i/(w-1) - w;
y = i%(w-1);
p0 = y*w+x;
p1 = (y+1)*w+x;
}
edge = (dsf_canonify(dsf, p0) != dsf_canonify(dsf, p1));
}
if (edge) {
while (currrun > 25)
*p++ = 'z', currrun -= 25;
if (currrun)
*p++ = 'a'-1 + currrun;
else
*p++ = '_';
currrun = 0;
} else
currrun++;
}
/*
* Now go through and compress the string by replacing runs of
* the same letter with a single copy of that letter followed by
* a repeat count, where that makes it shorter. (This puzzle
* seems to generate enough long strings of _ to make this a
* worthwhile step.)
*/
for (q = r = orig; r < p ;) {
*q++ = c = *r;
for (i = 0; r+i < p && r[i] == c; i++);
r += i;
if (i == 2) {
*q++ = c;
} else if (i > 2) {
q += sprintf(q, "%d", i);
}
}
return q;
}
static const char *parse_block_structure(const char **p, int w, int *dsf)
{
int a = w*w;
int pos = 0;
int repc = 0, repn = 0;
dsf_init(dsf, a);
while (**p && (repn > 0 || **p != ',')) {
int c, adv;
if (repn > 0) {
repn--;
c = repc;
} else if (**p == '_' || (**p >= 'a' && **p <= 'z')) {
c = (**p == '_' ? 0 : **p - 'a' + 1);
(*p)++;
if (**p && isdigit((unsigned char)**p)) {
repc = c;
repn = atoi(*p)-1;
while (**p && isdigit((unsigned char)**p)) (*p)++;
}
} else
return "Invalid character in game description";
adv = (c != 25); /* 'z' is a special case */
while (c-- > 0) {
int p0, p1;
/*
* Non-edge; merge the two dsf classes on either
* side of it.
*/
if (pos >= 2*w*(w-1))
return "Too much data in block structure specification";
if (pos < w*(w-1)) {
int y = pos/(w-1);
int x = pos%(w-1);
p0 = y*w+x;
p1 = y*w+x+1;
} else {
int x = pos/(w-1) - w;
int y = pos%(w-1);
p0 = y*w+x;
p1 = (y+1)*w+x;
}
dsf_merge(dsf, p0, p1);
pos++;
}
if (adv) {
pos++;
if (pos > 2*w*(w-1)+1)
return "Too much data in block structure specification";
}
}
/*
* When desc is exhausted, we expect to have gone exactly
* one space _past_ the end of the grid, due to the dummy
* edge at the end.
*/
if (pos != 2*w*(w-1)+1)
return "Not enough data in block structure specification";
return NULL;
}
static char *new_game_desc(const game_params *params, random_state *rs,
char **aux, int interactive)
{
int w = params->w, a = w*w;
digit *grid, *soln;
int *order, *revorder, *singletons, *dsf;
long *clues, *cluevals;
int i, j, k, n, x, y, ret;
int diff = params->diff;
char *desc, *p;
/*
* Difficulty exceptions: 3x3 puzzles at difficulty Hard or
* higher are currently not generable - the generator will spin
* forever looking for puzzles of the appropriate difficulty. We
* dial each of these down to the next lower difficulty.
*
* Remember to re-test this whenever a change is made to the
* solver logic!
*
* I tested it using the following shell command:
for d in e n h x u; do
for i in {3..9}; do
echo ./keen --generate 1 ${i}d${d}
perl -e 'alarm 30; exec @ARGV' ./keen --generate 5 ${i}d${d} >/dev/null \
|| echo broken
done
done
* Of course, it's better to do that after taking the exceptions
* _out_, so as to detect exceptions that should be removed as
* well as those which should be added.
*/
if (w == 3 && diff > DIFF_NORMAL)
diff = DIFF_NORMAL;
grid = NULL;
order = snewn(a, int);
revorder = snewn(a, int);
singletons = snewn(a, int);
dsf = snew_dsf(a);
clues = snewn(a, long);
cluevals = snewn(a, long);
soln = snewn(a, digit);
while (1) {
/*
* First construct a latin square to be the solution.
*/
sfree(grid);
grid = latin_generate(w, rs);
/*
* Divide the grid into arbitrarily sized blocks, but so as
* to arrange plenty of dominoes which can be SUB/DIV clues.
* We do this by first placing dominoes at random for a
* while, then tying the remaining singletons one by one
* into neighbouring blocks.
*/
for (i = 0; i < a; i++)
order[i] = i;
shuffle(order, a, sizeof(*order), rs);
for (i = 0; i < a; i++)
revorder[order[i]] = i;
for (i = 0; i < a; i++)
singletons[i] = TRUE;
dsf_init(dsf, a);
/* Place dominoes. */
for (i = 0; i < a; i++) {
if (singletons[i]) {
int best = -1;
x = i % w;
y = i / w;
if (x > 0 && singletons[i-1] &&
(best == -1 || revorder[i-1] < revorder[best]))
best = i-1;
if (x+1 < w && singletons[i+1] &&
(best == -1 || revorder[i+1] < revorder[best]))
best = i+1;
if (y > 0 && singletons[i-w] &&
(best == -1 || revorder[i-w] < revorder[best]))
best = i-w;
if (y+1 < w && singletons[i+w] &&
(best == -1 || revorder[i+w] < revorder[best]))
best = i+w;
/*
* When we find a potential domino, we place it with
* probability 3/4, which seems to strike a decent
* balance between plenty of dominoes and leaving
* enough singletons to make interesting larger
* shapes.
*/
if (best >= 0 && random_upto(rs, 4)) {
singletons[i] = singletons[best] = FALSE;
dsf_merge(dsf, i, best);
}
}
}
/* Fold in singletons. */
for (i = 0; i < a; i++) {
if (singletons[i]) {
int best = -1;
x = i % w;
y = i / w;
if (x > 0 && dsf_size(dsf, i-1) < MAXBLK &&
(best == -1 || revorder[i-1] < revorder[best]))
best = i-1;
if (x+1 < w && dsf_size(dsf, i+1) < MAXBLK &&
(best == -1 || revorder[i+1] < revorder[best]))
best = i+1;
if (y > 0 && dsf_size(dsf, i-w) < MAXBLK &&
(best == -1 || revorder[i-w] < revorder[best]))
best = i-w;
if (y+1 < w && dsf_size(dsf, i+w) < MAXBLK &&
(best == -1 || revorder[i+w] < revorder[best]))
best = i+w;
if (best >= 0) {
singletons[i] = singletons[best] = FALSE;
dsf_merge(dsf, i, best);
}
}
}
/* Quit and start again if we have any singletons left over
* which we weren't able to do anything at all with. */
for (i = 0; i < a; i++)
if (singletons[i])
break;
if (i < a)
continue;
/*
* Decide what would be acceptable clues for each block.
*
* Blocks larger than 2 have free choice of ADD or MUL;
* blocks of size 2 can be anything in principle (except
* that they can only be DIV if the two numbers have an
* integer quotient, of course), but we rule out (or try to
* avoid) some clues because they're of low quality.
*
* Hence, we iterate once over the grid, stopping at the
* canonical element of every >2 block and the _non_-
* canonical element of every 2-block; the latter means that
* we can make our decision about a 2-block in the knowledge
* of both numbers in it.
*
* We reuse the 'singletons' array (finished with in the
* above loop) to hold information about which blocks are
* suitable for what.
*/
#define F_ADD 0x01
#define F_SUB 0x02
#define F_MUL 0x04
#define F_DIV 0x08
#define BAD_SHIFT 4
for (i = 0; i < a; i++) {
singletons[i] = 0;
j = dsf_canonify(dsf, i);
k = dsf_size(dsf, j);
if (params->multiplication_only)
singletons[j] = F_MUL;
else if (j == i && k > 2) {
singletons[j] |= F_ADD | F_MUL;
} else if (j != i && k == 2) {
/* Fetch the two numbers and sort them into order. */
int p = grid[j], q = grid[i], v;
if (p < q) {
int t = p; p = q; q = t;
}
/*
* Addition clues are always allowed, but we try to
* avoid sums of 3, 4, (2w-1) and (2w-2) if we can,
* because they're too easy - they only leave one
* option for the pair of numbers involved.
*/
v = p + q;
if (v > 4 && v < 2*w-2)
singletons[j] |= F_ADD;
else
singletons[j] |= F_ADD << BAD_SHIFT;
/*
* Multiplication clues: above Normal difficulty, we
* prefer (but don't absolutely insist on) clues of
* this type which leave multiple options open.
*/
v = p * q;
n = 0;
for (k = 1; k <= w; k++)
if (v % k == 0 && v / k <= w && v / k != k)
n++;
if (n <= 2 && diff > DIFF_NORMAL)
singletons[j] |= F_MUL << BAD_SHIFT;
else
singletons[j] |= F_MUL;
/*
* Subtraction: we completely avoid a difference of
* w-1.
*/
v = p - q;
if (v < w-1)
singletons[j] |= F_SUB;
/*
* Division: for a start, the quotient must be an
* integer or the clue type is impossible. Also, we
* never use quotients strictly greater than w/2,
* because they're not only too easy but also
* inelegant.
*/
if (p % q == 0 && 2 * (p / q) <= w)
singletons[j] |= F_DIV;
}
}
/*
* Actually choose a clue for each block, trying to keep the
* numbers of each type even, and starting with the
* preferred candidates for each type where possible.
*
* I'm sure there should be a faster algorithm for doing
* this, but I can't be bothered: O(N^2) is good enough when
* N is at most the number of dominoes that fits into a 9x9
* square.
*/
shuffle(order, a, sizeof(*order), rs);
for (i = 0; i < a; i++)
clues[i] = 0;
while (1) {
int done_something = FALSE;
for (k = 0; k < 4; k++) {
long clue;
int good, bad;
switch (k) {
case 0: clue = C_DIV; good = F_DIV; break;
case 1: clue = C_SUB; good = F_SUB; break;
case 2: clue = C_MUL; good = F_MUL; break;
default /* case 3 */ : clue = C_ADD; good = F_ADD; break;
}
for (i = 0; i < a; i++) {
j = order[i];
if (singletons[j] & good) {
clues[j] = clue;
singletons[j] = 0;
break;
}
}
if (i == a) {
/* didn't find a nice one, use a nasty one */
bad = good << BAD_SHIFT;
for (i = 0; i < a; i++) {
j = order[i];
if (singletons[j] & bad) {
clues[j] = clue;
singletons[j] = 0;
break;
}
}
}
if (i < a)
done_something = TRUE;
}
if (!done_something)
break;
}
#undef F_ADD
#undef F_SUB
#undef F_MUL
#undef F_DIV
#undef BAD_SHIFT
/*
* Having chosen the clue types, calculate the clue values.
*/
for (i = 0; i < a; i++) {
j = dsf_canonify(dsf, i);
if (j == i) {
cluevals[j] = grid[i];
} else {
switch (clues[j]) {
case C_ADD:
cluevals[j] += grid[i];
break;
case C_MUL:
cluevals[j] *= grid[i];
break;
case C_SUB:
cluevals[j] = abs(cluevals[j] - grid[i]);
break;
case C_DIV:
{
int d1 = cluevals[j], d2 = grid[i];
if (d1 == 0 || d2 == 0)
cluevals[j] = 0;
else
cluevals[j] = d2/d1 + d1/d2;/* one is 0 :-) */
}
break;
}
}
}
for (i = 0; i < a; i++) {
j = dsf_canonify(dsf, i);
if (j == i) {
clues[j] |= cluevals[j];
}
}
/*
* See if the game can be solved at the specified difficulty
* level, but not at the one below.
*/
if (diff > 0) {
memset(soln, 0, a);
ret = solver(w, dsf, clues, soln, diff-1);
if (ret <= diff-1)
continue;
}
memset(soln, 0, a);
ret = solver(w, dsf, clues, soln, diff);
if (ret != diff)
continue; /* go round again */
/*
* I wondered if at this point it would be worth trying to
* merge adjacent blocks together, to make the puzzle
* gradually more difficult if it's currently easier than
* specced, increasing the chance of a given generation run
* being successful.
*
* It doesn't seem to be critical for the generation speed,
* though, so for the moment I'm leaving it out.
*/
/*
* We've got a usable puzzle!
*/
break;
}
/*
* Encode the puzzle description.
*/
desc = snewn(40*a, char);
p = desc;
p = encode_block_structure(p, w, dsf);
*p++ = ',';
for (i = 0; i < a; i++) {
j = dsf_canonify(dsf, i);
if (j == i) {
switch (clues[j] & CMASK) {
case C_ADD: *p++ = 'a'; break;
case C_SUB: *p++ = 's'; break;
case C_MUL: *p++ = 'm'; break;
case C_DIV: *p++ = 'd'; break;
}
p += sprintf(p, "%ld", clues[j] & ~CMASK);
}
}
*p++ = '\0';
desc = sresize(desc, p - desc, char);
/*
* Encode the solution.
*/
assert(memcmp(soln, grid, a) == 0);
*aux = snewn(a+2, char);
(*aux)[0] = 'S';
for (i = 0; i < a; i++)
(*aux)[i+1] = '0' + soln[i];
(*aux)[a+1] = '\0';
sfree(grid);
sfree(order);
sfree(revorder);
sfree(singletons);
sfree(dsf);
sfree(clues);
sfree(cluevals);
sfree(soln);
return desc;
}
/* ----------------------------------------------------------------------
* Gameplay.
*/
static const char *validate_desc(const game_params *params, const char *desc)
{
int w = params->w, a = w*w;
int *dsf;
const char *ret;
const char *p = desc;
int i;
/*
* Verify that the block structure makes sense.
*/
dsf = snew_dsf(a);
ret = parse_block_structure(&p, w, dsf);
if (ret) {
sfree(dsf);
return ret;
}
if (*p != ',')
return "Expected ',' after block structure description";
p++;
/*
* Verify that the right number of clues are given, and that SUB
* and DIV clues don't apply to blocks of the wrong size.
*/
for (i = 0; i < a; i++) {
if (dsf_canonify(dsf, i) == i) {
if (*p == 'a' || *p == 'm') {
/* these clues need no validation */
} else if (*p == 'd' || *p == 's') {
if (dsf_size(dsf, i) != 2)
return "Subtraction and division blocks must have area 2";
} else if (!*p) {
return "Too few clues for block structure";
} else {
return "Unrecognised clue type";
}
p++;
while (*p && isdigit((unsigned char)*p)) p++;
}
}
if (*p)
return "Too many clues for block structure";
return NULL;
}
static game_state *new_game(midend *me, const game_params *params,
const char *desc)
{
int w = params->w, a = w*w;
game_state *state = snew(game_state);
const char *p = desc;
int i;
state->par = *params; /* structure copy */
state->clues = snew(struct clues);
state->clues->refcount = 1;
state->clues->w = w;
state->clues->dsf = snew_dsf(a);
parse_block_structure(&p, w, state->clues->dsf);
assert(*p == ',');
p++;
state->clues->clues = snewn(a, long);
for (i = 0; i < a; i++) {
if (dsf_canonify(state->clues->dsf, i) == i) {
long clue = 0;
switch (*p) {
case 'a':
clue = C_ADD;
break;
case 'm':
clue = C_MUL;
break;
case 's':
clue = C_SUB;
assert(dsf_size(state->clues->dsf, i) == 2);
break;
case 'd':
clue = C_DIV;
assert(dsf_size(state->clues->dsf, i) == 2);
break;
default:
assert(!"Bad description in new_game");
}
p++;
clue |= atol(p);
while (*p && isdigit((unsigned char)*p)) p++;
state->clues->clues[i] = clue;
} else
state->clues->clues[i] = 0;
}
state->grid = snewn(a, digit);
state->pencil = snewn(a, int);
for (i = 0; i < a; i++) {
state->grid[i] = 0;
state->pencil[i] = 0;
}
state->completed = state->cheated = FALSE;
return state;
}
static game_state *dup_game(const game_state *state)
{
int w = state->par.w, a = w*w;
game_state *ret = snew(game_state);
ret->par = state->par; /* structure copy */
ret->clues = state->clues;
ret->clues->refcount++;
ret->grid = snewn(a, digit);
ret->pencil = snewn(a, int);
memcpy(ret->grid, state->grid, a*sizeof(digit));
memcpy(ret->pencil, state->pencil, a*sizeof(int));
ret->completed = state->completed;
ret->cheated = state->cheated;
return ret;
}
static void free_game(game_state *state)
{
sfree(state->grid);
sfree(state->pencil);
if (--state->clues->refcount <= 0) {
sfree(state->clues->dsf);
sfree(state->clues->clues);
sfree(state->clues);
}
sfree(state);
}
static char *solve_game(const game_state *state, const game_state *currstate,
const char *aux, const char **error)
{
int w = state->par.w, a = w*w;
int i, ret;
digit *soln;
char *out;
if (aux)
return dupstr(aux);
soln = snewn(a, digit);
memset(soln, 0, a);
ret = solver(w, state->clues->dsf, state->clues->clues,
soln, DIFFCOUNT-1);
if (ret == diff_impossible) {
*error = "No solution exists for this puzzle";
out = NULL;
} else if (ret == diff_ambiguous) {
*error = "Multiple solutions exist for this puzzle";
out = NULL;
} else {
out = snewn(a+2, char);
out[0] = 'S';
for (i = 0; i < a; i++)
out[i+1] = '0' + soln[i];
out[a+1] = '\0';
}
sfree(soln);
return out;
}
static int game_can_format_as_text_now(const game_params *params)
{
return TRUE;
}
static char *game_text_format(const game_state *state)
{
return NULL;
}
struct game_ui {
/*
* These are the coordinates of the currently highlighted
* square on the grid, if hshow = 1.
*/
int hx, hy;
/*
* This indicates whether the current highlight is a
* pencil-mark one or a real one.
*/
int hpencil;
/*
* This indicates whether or not we're showing the highlight
* (used to be hx = hy = -1); important so that when we're
* using the cursor keys it doesn't keep coming back at a
* fixed position. When hshow = 1, pressing a valid number
* or letter key or Space will enter that number or letter in the grid.
*/
int hshow;
/*
* This indicates whether we're using the highlight as a cursor;
* it means that it doesn't vanish on a keypress, and that it is
* allowed on immutable squares.
*/
int hcursor;
};
static game_ui *new_ui(const game_state *state)
{
game_ui *ui = snew(game_ui);
ui->hx = ui->hy = 0;
ui->hpencil = ui->hshow = ui->hcursor = 0;
return ui;
}
static void free_ui(game_ui *ui)
{
sfree(ui);
}
static char *encode_ui(const game_ui *ui)
{
return NULL;
}
static void decode_ui(game_ui *ui, const char *encoding)
{
}
static void game_changed_state(game_ui *ui, const game_state *oldstate,
const game_state *newstate)
{
int w = newstate->par.w;
/*
* We prevent pencil-mode highlighting of a filled square, unless
* we're using the cursor keys. So if the user has just filled in
* a square which we had a pencil-mode highlight in (by Undo, or
* by Redo, or by Solve), then we cancel the highlight.
*/
if (ui->hshow && ui->hpencil && !ui->hcursor &&
newstate->grid[ui->hy * w + ui->hx] != 0) {
ui->hshow = 0;
}
}
#define PREFERRED_TILESIZE 48
#define TILESIZE (ds->tilesize)
#define BORDER (TILESIZE / 2)
#define GRIDEXTRA max((TILESIZE / 32),1)
#define COORD(x) ((x)*TILESIZE + BORDER)
#define FROMCOORD(x) (((x)+(TILESIZE-BORDER)) / TILESIZE - 1)
#define FLASH_TIME 0.4F
#define DF_PENCIL_SHIFT 16
#define DF_ERR_LATIN 0x8000
#define DF_ERR_CLUE 0x4000
#define DF_HIGHLIGHT 0x2000
#define DF_HIGHLIGHT_PENCIL 0x1000
#define DF_DIGIT_MASK 0x000F
struct game_drawstate {
int tilesize;
int started;
long *tiles;
long *errors;
char *minus_sign, *times_sign, *divide_sign;
};
static int check_errors(const game_state *state, long *errors)
{
int w = state->par.w, a = w*w;
int i, j, x, y, errs = FALSE;
long *cluevals;
int *full;
cluevals = snewn(a, long);
full = snewn(a, int);
if (errors)
for (i = 0; i < a; i++) {
errors[i] = 0;
full[i] = TRUE;
}
for (i = 0; i < a; i++) {
long clue;
j = dsf_canonify(state->clues->dsf, i);
if (j == i) {
cluevals[i] = state->grid[i];
} else {
clue = state->clues->clues[j] & CMASK;
switch (clue) {
case C_ADD:
cluevals[j] += state->grid[i];
break;
case C_MUL:
cluevals[j] *= state->grid[i];
break;
case C_SUB:
cluevals[j] = abs(cluevals[j] - state->grid[i]);
break;
case C_DIV:
{
int d1 = min(cluevals[j], state->grid[i]);
int d2 = max(cluevals[j], state->grid[i]);
if (d1 == 0 || d2 % d1 != 0)
cluevals[j] = 0;
else
cluevals[j] = d2 / d1;
}
break;
}
}
if (!state->grid[i])
full[j] = FALSE;
}
for (i = 0; i < a; i++) {
j = dsf_canonify(state->clues->dsf, i);
if (j == i) {
if ((state->clues->clues[j] & ~CMASK) != cluevals[i]) {
errs = TRUE;
if (errors && full[j])
errors[j] |= DF_ERR_CLUE;
}
}
}
sfree(cluevals);
sfree(full);
for (y = 0; y < w; y++) {
int mask = 0, errmask = 0;
for (x = 0; x < w; x++) {
int bit = 1 << state->grid[y*w+x];
errmask |= (mask & bit);
mask |= bit;
}
if (mask != (1 << (w+1)) - (1 << 1)) {
errs = TRUE;
errmask &= ~1;
if (errors) {
for (x = 0; x < w; x++)
if (errmask & (1 << state->grid[y*w+x]))
errors[y*w+x] |= DF_ERR_LATIN;
}
}
}
for (x = 0; x < w; x++) {
int mask = 0, errmask = 0;
for (y = 0; y < w; y++) {
int bit = 1 << state->grid[y*w+x];
errmask |= (mask & bit);
mask |= bit;
}
if (mask != (1 << (w+1)) - (1 << 1)) {
errs = TRUE;
errmask &= ~1;
if (errors) {
for (y = 0; y < w; y++)
if (errmask & (1 << state->grid[y*w+x]))
errors[y*w+x] |= DF_ERR_LATIN;
}
}
}
return errs;
}
static char *interpret_move(const game_state *state, game_ui *ui,
const game_drawstate *ds,
int x, int y, int button)
{
int w = state->par.w;
int tx, ty;
char buf[80];
button &= ~MOD_MASK;
tx = FROMCOORD(x);
ty = FROMCOORD(y);
if (tx >= 0 && tx < w && ty >= 0 && ty < w) {
if (button == LEFT_BUTTON) {
if (tx == ui->hx && ty == ui->hy &&
ui->hshow && ui->hpencil == 0) {
ui->hshow = 0;
} else {
ui->hx = tx;
ui->hy = ty;
ui->hshow = 1;
ui->hpencil = 0;
}
ui->hcursor = 0;
return UI_UPDATE;
}
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_UPDATE;
}
}
if (IS_CURSOR_MOVE(button)) {
move_cursor(button, &ui->hx, &ui->hy, w, w, 0);
ui->hshow = ui->hcursor = 1;
return UI_UPDATE;
}
if (ui->hshow &&
(button == CURSOR_SELECT)) {
ui->hpencil = 1 - ui->hpencil;
ui->hcursor = 1;
return UI_UPDATE;
}
if (ui->hshow &&
((button >= '0' && button <= '9' && button - '0' <= w) ||
button == CURSOR_SELECT2 || button == '\b')) {
int n = button - '0';
if (button == CURSOR_SELECT2 || button == '\b')
n = 0;
/*
* Can't make pencil marks in a filled square. This can only
* become highlighted if we're using cursor keys.
*/
if (ui->hpencil && state->grid[ui->hy*w+ui->hx])
return NULL;
sprintf(buf, "%c%d,%d,%d",
(char)(ui->hpencil && n > 0 ? 'P' : 'R'), ui->hx, ui->hy, n);
if (!ui->hcursor) ui->hshow = 0;
return dupstr(buf);
}
if (button == 'M' || button == 'm')
return dupstr("M");
return NULL;
}
static game_state *execute_move(const game_state *from, const char *move)
{
int w = from->par.w, a = w*w;
game_state *ret;
int x, y, i, n;
if (move[0] == 'S') {
ret = dup_game(from);
ret->completed = ret->cheated = TRUE;
for (i = 0; i < a; i++) {
if (move[i+1] < '1' || move[i+1] > '0'+w) {
free_game(ret);
return NULL;
}
ret->grid[i] = move[i+1] - '0';
ret->pencil[i] = 0;
}
if (move[a+1] != '\0') {
free_game(ret);
return NULL;
}
return ret;
} else if ((move[0] == 'P' || move[0] == 'R') &&
sscanf(move+1, "%d,%d,%d", &x, &y, &n) == 3 &&
x >= 0 && x < w && y >= 0 && y < w && n >= 0 && n <= w) {
ret = dup_game(from);
if (move[0] == 'P' && n > 0) {
ret->pencil[y*w+x] ^= 1 << n;
} else {
ret->grid[y*w+x] = n;
ret->pencil[y*w+x] = 0;
if (!ret->completed && !check_errors(ret, NULL))
ret->completed = TRUE;
}
return ret;
} else if (move[0] == 'M') {
/*
* Fill in absolutely all pencil marks everywhere. (I
* wouldn't use this for actual play, but it's a handy
* starting point when following through a set of
* diagnostics output by the standalone solver.)
*/
ret = dup_game(from);
for (i = 0; i < a; i++) {
if (!ret->grid[i])
ret->pencil[i] = (1 << (w+1)) - (1 << 1);
}
return ret;
} else
return NULL; /* couldn't parse move string */
}
/* ----------------------------------------------------------------------
* Drawing routines.
*/
#define SIZE(w) ((w) * TILESIZE + 2*BORDER)
static void game_compute_size(const game_params *params, int tilesize,
int *x, int *y)
{
/* Ick: fake up `ds->tilesize' for macro expansion purposes */
struct { int tilesize; } ads, *ds = &ads;
ads.tilesize = tilesize;
*x = *y = SIZE(params->w);
}
static void game_set_size(drawing *dr, game_drawstate *ds,
const game_params *params, int tilesize)
{
ds->tilesize = tilesize;
}
static float *game_colours(frontend *fe, int *ncolours)
{
float *ret = snewn(3 * NCOLOURS, float);
frontend_default_colour(fe, &ret[COL_BACKGROUND * 3]);
ret[COL_GRID * 3 + 0] = 0.0F;
ret[COL_GRID * 3 + 1] = 0.0F;
ret[COL_GRID * 3 + 2] = 0.0F;
ret[COL_USER * 3 + 0] = 0.0F;
ret[COL_USER * 3 + 1] = 0.6F * ret[COL_BACKGROUND * 3 + 1];
ret[COL_USER * 3 + 2] = 0.0F;
ret[COL_HIGHLIGHT * 3 + 0] = 0.78F * ret[COL_BACKGROUND * 3 + 0];
ret[COL_HIGHLIGHT * 3 + 1] = 0.78F * ret[COL_BACKGROUND * 3 + 1];
ret[COL_HIGHLIGHT * 3 + 2] = 0.78F * ret[COL_BACKGROUND * 3 + 2];
ret[COL_ERROR * 3 + 0] = 1.0F;
ret[COL_ERROR * 3 + 1] = 0.0F;
ret[COL_ERROR * 3 + 2] = 0.0F;
ret[COL_PENCIL * 3 + 0] = 0.5F * ret[COL_BACKGROUND * 3 + 0];
ret[COL_PENCIL * 3 + 1] = 0.5F * ret[COL_BACKGROUND * 3 + 1];
ret[COL_PENCIL * 3 + 2] = ret[COL_BACKGROUND * 3 + 2];
*ncolours = NCOLOURS;
return ret;
}
static const char *const minus_signs[] = { "\xE2\x88\x92", "-" };
static const char *const times_signs[] = { "\xC3\x97", "*" };
static const char *const divide_signs[] = { "\xC3\xB7", "/" };
static game_drawstate *game_new_drawstate(drawing *dr, const game_state *state)
{
int w = state->par.w, a = w*w;
struct game_drawstate *ds = snew(struct game_drawstate);
int i;
ds->tilesize = 0;
ds->started = FALSE;
ds->tiles = snewn(a, long);
for (i = 0; i < a; i++)
ds->tiles[i] = -1;
ds->errors = snewn(a, long);
ds->minus_sign = text_fallback(dr, minus_signs, lenof(minus_signs));
ds->times_sign = text_fallback(dr, times_signs, lenof(times_signs));
ds->divide_sign = text_fallback(dr, divide_signs, lenof(divide_signs));
return ds;
}
static void game_free_drawstate(drawing *dr, game_drawstate *ds)
{
sfree(ds->tiles);
sfree(ds->errors);
sfree(ds->minus_sign);
sfree(ds->times_sign);
sfree(ds->divide_sign);
sfree(ds);
}
static void draw_tile(drawing *dr, game_drawstate *ds, struct clues *clues,
int x, int y, long tile, int only_one_op)
{
int w = clues->w /* , a = w*w */;
int tx, ty, tw, th;
int cx, cy, cw, ch;
char str[64];
tx = BORDER + x * TILESIZE + 1 + GRIDEXTRA;
ty = BORDER + y * TILESIZE + 1 + GRIDEXTRA;
cx = tx;
cy = ty;
cw = tw = TILESIZE-1-2*GRIDEXTRA;
ch = th = TILESIZE-1-2*GRIDEXTRA;
if (x > 0 && dsf_canonify(clues->dsf, y*w+x) == dsf_canonify(clues->dsf, y*w+x-1))
cx -= GRIDEXTRA, cw += GRIDEXTRA;
if (x+1 < w && dsf_canonify(clues->dsf, y*w+x) == dsf_canonify(clues->dsf, y*w+x+1))
cw += GRIDEXTRA;
if (y > 0 && dsf_canonify(clues->dsf, y*w+x) == dsf_canonify(clues->dsf, (y-1)*w+x))
cy -= GRIDEXTRA, ch += GRIDEXTRA;
if (y+1 < w && dsf_canonify(clues->dsf, y*w+x) == dsf_canonify(clues->dsf, (y+1)*w+x))
ch += GRIDEXTRA;
clip(dr, cx, cy, cw, ch);
/* background needs erasing */
draw_rect(dr, cx, cy, cw, ch,
(tile & DF_HIGHLIGHT) ? COL_HIGHLIGHT : COL_BACKGROUND);
/* pencil-mode highlight */
if (tile & DF_HIGHLIGHT_PENCIL) {
int coords[6];
coords[0] = cx;
coords[1] = cy;
coords[2] = cx+cw/2;
coords[3] = cy;
coords[4] = cx;
coords[5] = cy+ch/2;
draw_polygon(dr, coords, 3, COL_HIGHLIGHT, COL_HIGHLIGHT);
}
/*
* Draw the corners of thick lines in corner-adjacent squares,
* which jut into this square by one pixel.
*/
if (x > 0 && y > 0 && dsf_canonify(clues->dsf, y*w+x) != dsf_canonify(clues->dsf, (y-1)*w+x-1))
draw_rect(dr, tx-GRIDEXTRA, ty-GRIDEXTRA, GRIDEXTRA, GRIDEXTRA, COL_GRID);
if (x+1 < w && y > 0 && dsf_canonify(clues->dsf, y*w+x) != dsf_canonify(clues->dsf, (y-1)*w+x+1))
draw_rect(dr, tx+TILESIZE-1-2*GRIDEXTRA, ty-GRIDEXTRA, GRIDEXTRA, GRIDEXTRA, COL_GRID);
if (x > 0 && y+1 < w && dsf_canonify(clues->dsf, y*w+x) != dsf_canonify(clues->dsf, (y+1)*w+x-1))
draw_rect(dr, tx-GRIDEXTRA, ty+TILESIZE-1-2*GRIDEXTRA, GRIDEXTRA, GRIDEXTRA, COL_GRID);
if (x+1 < w && y+1 < w && dsf_canonify(clues->dsf, y*w+x) != dsf_canonify(clues->dsf, (y+1)*w+x+1))
draw_rect(dr, tx+TILESIZE-1-2*GRIDEXTRA, ty+TILESIZE-1-2*GRIDEXTRA, GRIDEXTRA, GRIDEXTRA, COL_GRID);
/* Draw the box clue. */
if (dsf_canonify(clues->dsf, y*w+x) == y*w+x) {
long clue = clues->clues[y*w+x];
long cluetype = clue & CMASK, clueval = clue & ~CMASK;
int size = dsf_size(clues->dsf, y*w+x);
/*
* Special case of clue-drawing: a box with only one square
* is written as just the number, with no operation, because
* it doesn't matter whether the operation is ADD or MUL.
* The generation code above should never produce puzzles
* containing such a thing - I think they're inelegant - but
* it's possible to type in game IDs from elsewhere, so I
* want to display them right if so.
*/
sprintf (str, "%ld%s", clueval,
(size == 1 || only_one_op ? "" :
cluetype == C_ADD ? "+" :
cluetype == C_SUB ? ds->minus_sign :
cluetype == C_MUL ? ds->times_sign :
/* cluetype == C_DIV ? */ ds->divide_sign));
draw_text(dr, tx + GRIDEXTRA * 2, ty + GRIDEXTRA * 2 + TILESIZE/4,
FONT_VARIABLE, TILESIZE/4, ALIGN_VNORMAL | ALIGN_HLEFT,
(tile & DF_ERR_CLUE ? COL_ERROR : COL_GRID), str);
}
/* new number needs drawing? */
if (tile & DF_DIGIT_MASK) {
str[1] = '\0';
str[0] = (tile & DF_DIGIT_MASK) + '0';
draw_text(dr, tx + TILESIZE/2, ty + TILESIZE/2,
FONT_VARIABLE, TILESIZE/2, ALIGN_VCENTRE | ALIGN_HCENTRE,
(tile & DF_ERR_LATIN) ? COL_ERROR : COL_USER, str);
} else {
int i, j, npencil;
int pl, pr, pt, pb;
float bestsize;
int pw, ph, minph, pbest, fontsize;
/* Count the pencil marks required. */
for (i = 1, npencil = 0; i <= w; i++)
if (tile & (1L << (i + DF_PENCIL_SHIFT)))
npencil++;
if (npencil) {
minph = 2;
/*
* Determine the bounding rectangle within which we're going
* to put the pencil marks.
*/
/* Start with the whole square */
pl = tx + GRIDEXTRA;
pr = pl + TILESIZE - GRIDEXTRA;
pt = ty + GRIDEXTRA;
pb = pt + TILESIZE - GRIDEXTRA;
if (dsf_canonify(clues->dsf, y*w+x) == y*w+x) {
/*
* Make space for the clue text.
*/
pt += TILESIZE/4;
/* minph--; */
}
/*
* We arrange our pencil marks in a grid layout, with
* the number of rows and columns adjusted to allow the
* maximum font size.
*
* So now we work out what the grid size ought to be.
*/
bestsize = 0.0;
pbest = 0;
/* Minimum */
for (pw = 3; pw < max(npencil,4); pw++) {
float fw, fh, fs;
ph = (npencil + pw - 1) / pw;
ph = max(ph, minph);
fw = (pr - pl) / (float)pw;
fh = (pb - pt) / (float)ph;
fs = min(fw, fh);
if (fs > bestsize) {
bestsize = fs;
pbest = pw;
}
}
assert(pbest > 0);
pw = pbest;
ph = (npencil + pw - 1) / pw;
ph = max(ph, minph);
/*
* Now we've got our grid dimensions, work out the pixel
* size of a grid element, and round it to the nearest
* pixel. (We don't want rounding errors to make the
* grid look uneven at low pixel sizes.)
*/
fontsize = min((pr - pl) / pw, (pb - pt) / ph);
/*
* Centre the resulting figure in the square.
*/
pl = tx + (TILESIZE - fontsize * pw) / 2;
pt = ty + (TILESIZE - fontsize * ph) / 2;
/*
* And move it down a bit if it's collided with some
* clue text.
*/
if (dsf_canonify(clues->dsf, y*w+x) == y*w+x) {
pt = max(pt, ty + GRIDEXTRA * 3 + TILESIZE/4);
}
/*
* Now actually draw the pencil marks.
*/
for (i = 1, j = 0; i <= w; i++)
if (tile & (1L << (i + DF_PENCIL_SHIFT))) {
int dx = j % pw, dy = j / pw;
str[1] = '\0';
str[0] = i + '0';
draw_text(dr, pl + fontsize * (2*dx+1) / 2,
pt + fontsize * (2*dy+1) / 2,
FONT_VARIABLE, fontsize,
ALIGN_VCENTRE | ALIGN_HCENTRE, COL_PENCIL, str);
j++;
}
}
}
unclip(dr);
draw_update(dr, cx, cy, cw, ch);
}
static void game_redraw(drawing *dr, game_drawstate *ds,
const game_state *oldstate, const game_state *state,
int dir, const game_ui *ui,
float animtime, float flashtime)
{
int w = state->par.w /*, a = w*w */;
int x, y;
if (!ds->started) {
/*
* The initial contents of the window are not guaranteed and
* can vary with front ends. To be on the safe side, all
* games should start by drawing a big background-colour
* rectangle covering the whole window.
*/
draw_rect(dr, 0, 0, SIZE(w), SIZE(w), COL_BACKGROUND);
/*
* Big containing rectangle.
*/
draw_rect(dr, COORD(0) - GRIDEXTRA, COORD(0) - GRIDEXTRA,
w*TILESIZE+1+GRIDEXTRA*2, w*TILESIZE+1+GRIDEXTRA*2,
COL_GRID);
draw_update(dr, 0, 0, SIZE(w), SIZE(w));
ds->started = TRUE;
}
check_errors(state, ds->errors);
for (y = 0; y < w; y++) {
for (x = 0; x < w; x++) {
long tile = 0L;
if (state->grid[y*w+x])
tile = state->grid[y*w+x];
else
tile = (long)state->pencil[y*w+x] << DF_PENCIL_SHIFT;
if (ui->hshow && ui->hx == x && ui->hy == y)
tile |= (ui->hpencil ? DF_HIGHLIGHT_PENCIL : DF_HIGHLIGHT);
if (flashtime > 0 &&
(flashtime <= FLASH_TIME/3 ||
flashtime >= FLASH_TIME*2/3))
tile |= DF_HIGHLIGHT; /* completion flash */
tile |= ds->errors[y*w+x];
if (ds->tiles[y*w+x] != tile) {
ds->tiles[y*w+x] = tile;
draw_tile(dr, ds, state->clues, x, y, tile,
state->par.multiplication_only);
}
}
}
}
static float game_anim_length(const game_state *oldstate,
const game_state *newstate, int dir, game_ui *ui)
{
return 0.0F;
}
static float game_flash_length(const game_state *oldstate,
const game_state *newstate, int dir, game_ui *ui)
{
if (!oldstate->completed && newstate->completed &&
!oldstate->cheated && !newstate->cheated)
return FLASH_TIME;
return 0.0F;
}
static int game_status(const game_state *state)
{
return state->completed ? +1 : 0;
}
static int game_timing_state(const game_state *state, game_ui *ui)
{
if (state->completed)
return FALSE;
return TRUE;
}
static void game_print_size(const game_params *params, float *x, float *y)
{
int pw, ph;
/*
* We use 9mm squares by default, like Solo.
*/
game_compute_size(params, 900, &pw, &ph);
*x = pw / 100.0F;
*y = ph / 100.0F;
}
/*
* Subfunction to draw the thick lines between cells. In order to do
* this using the line-drawing rather than rectangle-drawing API (so
* as to get line thicknesses to scale correctly) and yet have
* correctly mitred joins between lines, we must do this by tracing
* the boundary of each sub-block and drawing it in one go as a
* single polygon.
*/
static void outline_block_structure(drawing *dr, game_drawstate *ds,
int w, int *dsf, int ink)
{
int a = w*w;
int *coords;
int i, n;
int x, y, dx, dy, sx, sy, sdx, sdy;
coords = snewn(4*a, int);
/*
* Iterate over all the blocks.
*/
for (i = 0; i < a; i++) {
if (dsf_canonify(dsf, i) != i)
continue;
/*
* For each block, we need a starting square within it which
* has a boundary at the left. Conveniently, we have one
* right here, by construction.
*/
x = i % w;
y = i / w;
dx = -1;
dy = 0;
/*
* Now begin tracing round the perimeter. At all
* times, (x,y) describes some square within the
* block, and (x+dx,y+dy) is some adjacent square
* outside it; so the edge between those two squares
* is always an edge of the block.
*/
sx = x, sy = y, sdx = dx, sdy = dy; /* save starting position */
n = 0;
do {
int cx, cy, tx, ty, nin;
/*
* Advance to the next edge, by looking at the two
* squares beyond it. If they're both outside the block,
* we turn right (by leaving x,y the same and rotating
* dx,dy clockwise); if they're both inside, we turn
* left (by rotating dx,dy anticlockwise and contriving
* to leave x+dx,y+dy unchanged); if one of each, we go
* straight on (and may enforce by assertion that
* they're one of each the _right_ way round).
*/
nin = 0;
tx = x - dy + dx;
ty = y + dx + dy;
nin += (tx >= 0 && tx < w && ty >= 0 && ty < w &&
dsf_canonify(dsf, ty*w+tx) == i);
tx = x - dy;
ty = y + dx;
nin += (tx >= 0 && tx < w && ty >= 0 && ty < w &&
dsf_canonify(dsf, ty*w+tx) == i);
if (nin == 0) {
/*
* Turn right.
*/
int tmp;
tmp = dx;
dx = -dy;
dy = tmp;
} else if (nin == 2) {
/*
* Turn left.
*/
int tmp;
x += dx;
y += dy;
tmp = dx;
dx = dy;
dy = -tmp;
x -= dx;
y -= dy;
} else {
/*
* Go straight on.
*/
x -= dy;
y += dx;
}
/*
* Now enforce by assertion that we ended up
* somewhere sensible.
*/
assert(x >= 0 && x < w && y >= 0 && y < w &&
dsf_canonify(dsf, y*w+x) == i);
assert(x+dx < 0 || x+dx >= w || y+dy < 0 || y+dy >= w ||
dsf_canonify(dsf, (y+dy)*w+(x+dx)) != i);
/*
* Record the point we just went past at one end of the
* edge. To do this, we translate (x,y) down and right
* by half a unit (so they're describing a point in the
* _centre_ of the square) and then translate back again
* in a manner rotated by dy and dx.
*/
assert(n < 2*w+2);
cx = ((2*x+1) + dy + dx) / 2;
cy = ((2*y+1) - dx + dy) / 2;
coords[2*n+0] = BORDER + cx * TILESIZE;
coords[2*n+1] = BORDER + cy * TILESIZE;
n++;
} while (x != sx || y != sy || dx != sdx || dy != sdy);
/*
* That's our polygon; now draw it.
*/
draw_polygon(dr, coords, n, -1, ink);
}
sfree(coords);
}
static void game_print(drawing *dr, const game_state *state, int tilesize)
{
int w = state->par.w;
int ink = print_mono_colour(dr, 0);
int x, y;
char *minus_sign, *times_sign, *divide_sign;
/* Ick: fake up `ds->tilesize' for macro expansion purposes */
game_drawstate ads, *ds = &ads;
game_set_size(dr, ds, NULL, tilesize);
minus_sign = text_fallback(dr, minus_signs, lenof(minus_signs));
times_sign = text_fallback(dr, times_signs, lenof(times_signs));
divide_sign = text_fallback(dr, divide_signs, lenof(divide_signs));
/*
* Border.
*/
print_line_width(dr, 3 * TILESIZE / 40);
draw_rect_outline(dr, BORDER, BORDER, w*TILESIZE, w*TILESIZE, ink);
/*
* Main grid.
*/
for (x = 1; x < w; x++) {
print_line_width(dr, TILESIZE / 40);
draw_line(dr, BORDER+x*TILESIZE, BORDER,
BORDER+x*TILESIZE, BORDER+w*TILESIZE, ink);
}
for (y = 1; y < w; y++) {
print_line_width(dr, TILESIZE / 40);
draw_line(dr, BORDER, BORDER+y*TILESIZE,
BORDER+w*TILESIZE, BORDER+y*TILESIZE, ink);
}
/*
* Thick lines between cells.
*/
print_line_width(dr, 3 * TILESIZE / 40);
outline_block_structure(dr, ds, w, state->clues->dsf, ink);
/*
* Clues.
*/
for (y = 0; y < w; y++)
for (x = 0; x < w; x++)
if (dsf_canonify(state->clues->dsf, y*w+x) == y*w+x) {
long clue = state->clues->clues[y*w+x];
long cluetype = clue & CMASK, clueval = clue & ~CMASK;
int size = dsf_size(state->clues->dsf, y*w+x);
char str[64];
/*
* As in the drawing code, we omit the operator for
* blocks of area 1.
*/
sprintf (str, "%ld%s", clueval,
(size == 1 ? "" :
cluetype == C_ADD ? "+" :
cluetype == C_SUB ? minus_sign :
cluetype == C_MUL ? times_sign :
/* cluetype == C_DIV ? */ divide_sign));
draw_text(dr,
BORDER+x*TILESIZE + 5*TILESIZE/80,
BORDER+y*TILESIZE + 20*TILESIZE/80,
FONT_VARIABLE, TILESIZE/4,
ALIGN_VNORMAL | ALIGN_HLEFT,
ink, str);
}
/*
* Numbers for the solution, if any.
*/
for (y = 0; y < w; y++)
for (x = 0; x < w; x++)
if (state->grid[y*w+x]) {
char str[2];
str[1] = '\0';
str[0] = state->grid[y*w+x] + '0';
draw_text(dr, BORDER + x*TILESIZE + TILESIZE/2,
BORDER + y*TILESIZE + TILESIZE/2,
FONT_VARIABLE, TILESIZE/2,
ALIGN_VCENTRE | ALIGN_HCENTRE, ink, str);
}
sfree(minus_sign);
sfree(times_sign);
sfree(divide_sign);
}
#ifdef COMBINED
#define thegame keen
#endif
const struct game thegame = {
"Keen", "games.keen", "keen",
default_params,
game_fetch_preset, NULL,
decode_params,
encode_params,
free_params,
dup_params,
TRUE, game_configure, custom_params,
validate_params,
new_game_desc,
validate_desc,
new_game,
dup_game,
free_game,
TRUE, solve_game,
FALSE, game_can_format_as_text_now, game_text_format,
new_ui,
free_ui,
encode_ui,
decode_ui,
game_changed_state,
interpret_move,
execute_move,
PREFERRED_TILESIZE, game_compute_size, game_set_size,
game_colours,
game_new_drawstate,
game_free_drawstate,
game_redraw,
game_anim_length,
game_flash_length,
game_status,
TRUE, FALSE, game_print_size, game_print,
FALSE, /* wants_statusbar */
FALSE, game_timing_state,
REQUIRE_RBUTTON | REQUIRE_NUMPAD, /* flags */
};
#ifdef STANDALONE_SOLVER
#include <stdarg.h>
int main(int argc, char **argv)
{
game_params *p;
game_state *s;
char *id = NULL, *desc;
const char *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: */