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rockbox/apps/plugins/puzzles/src/keen.c
Franklin Wei 5094aaa4d4 puzzles: Follow cursor in zoom mode and general code cleanup. 
Frontends now have a way to retrieve the backend cursor position with some
changes I've submitted upstream. With this information, we can now follow
the cursor around in "interaction mode" while zoomed in, eliminating (most)
need for mode switching.

Also does some cleanup of the frontend code.

Change-Id: I1ba118f67564a3baed95435f5619b73cfa3ae87a
2020-07-06 23:00:13 -04:00

2608 lines
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/*
* 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;
bool 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 */
bool 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;
}
static const 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 bool 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, bool 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, bool 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 transpose(int index, int w)
{
return (index % w) * w + (index / w);
}
static bool keen_valid(struct latin_solver *solver, void *vctx)
{
struct solver_ctx *ctx = (struct solver_ctx *)vctx;
int w = ctx->w;
int box, i;
/*
* Iterate over each clue box and check it's satisfied.
*/
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;
bool fail = false;
switch (op) {
case C_ADD: {
long sum = 0;
for (i = 0; i < n; i++)
sum += solver->grid[transpose(sq[i], w)];
fail = (sum != value);
break;
}
case C_MUL: {
long remaining = value;
for (i = 0; i < n; i++) {
if (remaining % solver->grid[transpose(sq[i], w)]) {
fail = true;
break;
}
remaining /= solver->grid[transpose(sq[i], w)];
}
if (remaining != 1)
fail = true;
break;
}
case C_SUB:
assert(n == 2);
if (value != labs(solver->grid[transpose(sq[0], w)] -
solver->grid[transpose(sq[1], w)]))
fail = true;
break;
case C_DIV: {
int num, den;
assert(n == 2);
num = max(solver->grid[transpose(sq[0], w)],
solver->grid[transpose(sq[1], w)]);
den = min(solver->grid[transpose(sq[0], w)],
solver->grid[transpose(sq[1], w)]);
if (den * value != num)
fail = true;
break;
}
}
if (fail) {
#ifdef STANDALONE_SOLVER
if (solver_show_working) {
printf("%*sclue at (%d,%d) is violated\n",
solver_recurse_depth*4, "",
sq[0]/w+1, sq[0]%w+1);
printf("%*s (%s clue with target %ld containing [",
solver_recurse_depth*4, "",
(op == C_ADD ? "addition" : op == C_SUB ? "subtraction":
op == C_MUL ? "multiplication" : "division"), value);
for (i = 0; i < n; i++)
printf(" %d", (int)solver->grid[transpose(sq[i], w)]);
printf(" ]\n");
}
#endif
return false;
}
}
return true;
}
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, keen_valid, &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;
bool 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;
bool 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, bool 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) {
bool 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] = labs(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 key_label *game_request_keys(const game_params *params, int *nkeys)
{
int i;
int w = params->w;
key_label *keys = snewn(w+1, key_label);
*nkeys = w + 1;
for (i = 0; i < w; i++) {
if (i<9) keys[i].button = '1' + i;
else keys[i].button = 'a' + i - 9;
keys[i].label = NULL;
}
keys[w].button = '\b';
keys[w].label = NULL;
return keys;
}
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 = false;
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 bool 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 is true.
*/
int hx, hy;
/*
* This indicates whether the current highlight is a
* pencil-mark one or a real one.
*/
bool 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 true, pressing a valid number or letter
* key or Space will enter that number or letter in the grid.
*/
bool 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.
*/
bool hcursor;
};
static game_ui *new_ui(const game_state *state)
{
game_ui *ui = snew(game_ui);
ui->hx = ui->hy = 0;
ui->hpencil = false;
ui->hshow = false;
ui->hcursor = false;
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 = false;
}
}
#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;
bool started;
long *tiles;
long *errors;
char *minus_sign, *times_sign, *divide_sign;
};
static bool check_errors(const game_state *state, long *errors)
{
int w = state->par.w, a = w*w;
int i, j, x, y;
bool errs = false;
long *cluevals;
bool *full;
cluevals = snewn(a, long);
full = snewn(a, bool);
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] = labs(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) {
ui->hshow = false;
} else {
ui->hx = tx;
ui->hy = ty;
ui->hshow = true;
ui->hpencil = false;
}
ui->hcursor = false;
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 = false;
} else {
ui->hpencil = true;
ui->hx = tx;
ui->hy = ty;
ui->hshow = true;
}
} else {
ui->hshow = false;
}
ui->hcursor = false;
return UI_UPDATE;
}
}
if (IS_CURSOR_MOVE(button)) {
move_cursor(button, &ui->hx, &ui->hy, w, w, false);
ui->hshow = true;
ui->hcursor = true;
return UI_UPDATE;
}
if (ui->hshow &&
(button == CURSOR_SELECT)) {
ui->hpencil ^= 1;
ui->hcursor = true;
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 = false;
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, bool 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 void game_get_cursor_location(const game_ui *ui,
const game_drawstate *ds,
const game_state *state,
const game_params *params,
int *x, int *y, int *w, int *h)
{
if(ui->hshow) {
*x = BORDER + ui->hx * TILESIZE + 1 + GRIDEXTRA;
*y = BORDER + ui->hy * TILESIZE + 1 + GRIDEXTRA;
*w = *h = TILESIZE-1-2*GRIDEXTRA;
}
}
static int game_status(const game_state *state)
{
return state->completed ? +1 : 0;
}
static bool 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_request_keys,
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_get_cursor_location,
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;
bool grade = false;
int ret, diff;
bool 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 = 0;
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 ? 1 : 0;
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: */
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