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map.c
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map.c
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/*
* map.c: Game involving four-colouring a map.
*/
/*
* TODO:
*
* - clue marking
* - better four-colouring algorithm?
*/
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <assert.h>
#include <ctype.h>
#include <math.h>
#include "puzzles.h"
/*
* In standalone solver mode, `verbose' is a variable which can be
* set by command-line option; in debugging mode it's simply always
* true.
*/
#if defined STANDALONE_SOLVER
#define SOLVER_DIAGNOSTICS
bool verbose = false;
#elif defined SOLVER_DIAGNOSTICS
#define verbose true
#endif
/*
* I don't seriously anticipate wanting to change the number of
* colours used in this game, but it doesn't cost much to use a
* #define just in case :-)
*/
#define FOUR 4
#define THREE (FOUR-1)
#define FIVE (FOUR+1)
#define SIX (FOUR+2)
/*
* Ghastly run-time configuration option, just for Gareth (again).
*/
static int flash_type = -1;
static float flash_length;
/*
* Difficulty levels. I do some macro ickery here to ensure that my
* enum and the various forms of my name list always match up.
*/
#define DIFFLIST(A) \
A(EASY,Easy,e) \
A(NORMAL,Normal,n) \
A(HARD,Hard,h) \
A(RECURSE,Unreasonable,u)
#define ENUM(upper,title,lower) DIFF_ ## upper,
#define TITLE(upper,title,lower) #title,
#define ENCODE(upper,title,lower) #lower
#define CONFIG(upper,title,lower) ":" #title
enum { DIFFLIST(ENUM) DIFFCOUNT };
static char const *const map_diffnames[] = { DIFFLIST(TITLE) };
static char const map_diffchars[] = DIFFLIST(ENCODE);
#define DIFFCONFIG DIFFLIST(CONFIG)
enum { TE, BE, LE, RE }; /* top/bottom/left/right edges */
enum {
COL_BACKGROUND,
COL_GRID,
COL_0, COL_1, COL_2, COL_3,
COL_ERROR, COL_ERRTEXT,
NCOLOURS
};
struct game_params {
int w, h, n, diff;
};
struct map {
int refcount;
int *map;
int *graph;
int n;
int ngraph;
bool *immutable;
int *edgex, *edgey; /* position of a point on each edge */
int *regionx, *regiony; /* position of a point in each region */
};
struct game_state {
game_params p;
struct map *map;
int *colouring, *pencil;
bool completed, cheated;
};
static game_params *default_params(void)
{
game_params *ret = snew(game_params);
#ifdef PORTRAIT_SCREEN
ret->w = 16;
ret->h = 18;
#else
ret->w = 20;
ret->h = 15;
#endif
ret->n = 30;
ret->diff = DIFF_NORMAL;
return ret;
}
static const struct game_params map_presets[] = {
#ifdef PORTRAIT_SCREEN
{16, 18, 30, DIFF_EASY},
{16, 18, 30, DIFF_NORMAL},
{16, 18, 30, DIFF_HARD},
{16, 18, 30, DIFF_RECURSE},
{25, 30, 75, DIFF_NORMAL},
{25, 30, 75, DIFF_HARD},
#else
{20, 15, 30, DIFF_EASY},
{20, 15, 30, DIFF_NORMAL},
{20, 15, 30, DIFF_HARD},
{20, 15, 30, DIFF_RECURSE},
{30, 25, 75, DIFF_NORMAL},
{30, 25, 75, DIFF_HARD},
#endif
};
static bool game_fetch_preset(int i, char **name, game_params **params)
{
game_params *ret;
char str[80];
if (i < 0 || i >= lenof(map_presets))
return false;
ret = snew(game_params);
*ret = map_presets[i];
sprintf(str, "%dx%d, %d regions, %s", ret->w, ret->h, ret->n,
map_diffnames[ret->diff]);
*name = dupstr(str);
*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 == 'x') {
p++;
params->h = atoi(p);
while (*p && isdigit((unsigned char)*p)) p++;
} else {
params->h = params->w;
}
if (*p == 'n') {
p++;
params->n = atoi(p);
while (*p && (*p == '.' || isdigit((unsigned char)*p))) p++;
} else {
params->n = params->w * params->h / 8;
}
if (*p == 'd') {
int i;
p++;
for (i = 0; i < DIFFCOUNT; i++)
if (*p == map_diffchars[i])
params->diff = i;
if (*p) p++;
}
}
static char *encode_params(const game_params *params, bool full)
{
char ret[400];
sprintf(ret, "%dx%dn%d", params->w, params->h, params->n);
if (full)
sprintf(ret + strlen(ret), "d%c", map_diffchars[params->diff]);
return dupstr(ret);
}
static config_item *game_configure(const game_params *params)
{
config_item *ret;
char buf[80];
ret = snewn(5, config_item);
ret[0].name = "Width";
ret[0].type = C_STRING;
sprintf(buf, "%d", params->w);
ret[0].u.string.sval = dupstr(buf);
ret[1].name = "Height";
ret[1].type = C_STRING;
sprintf(buf, "%d", params->h);
ret[1].u.string.sval = dupstr(buf);
ret[2].name = "Regions";
ret[2].type = C_STRING;
sprintf(buf, "%d", params->n);
ret[2].u.string.sval = dupstr(buf);
ret[3].name = "Difficulty";
ret[3].type = C_CHOICES;
ret[3].u.choices.choicenames = DIFFCONFIG;
ret[3].u.choices.selected = params->diff;
ret[4].name = NULL;
ret[4].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->h = atoi(cfg[1].u.string.sval);
ret->n = atoi(cfg[2].u.string.sval);
ret->diff = cfg[3].u.choices.selected;
return ret;
}
static const char *validate_params(const game_params *params, bool full)
{
if (params->w < 2 || params->h < 2)
return "Width and height must be at least two";
if (params->n < 5)
return "Must have at least five regions";
if (params->n > params->w * params->h)
return "Too many regions to fit in grid";
return NULL;
}
/* ----------------------------------------------------------------------
* Cumulative frequency table functions.
*/
/*
* Initialise a cumulative frequency table. (Hardly worth writing
* this function; all it does is to initialise everything in the
* array to zero.)
*/
static void cf_init(int *table, int n)
{
int i;
for (i = 0; i < n; i++)
table[i] = 0;
}
/*
* Increment the count of symbol `sym' by `count'.
*/
static void cf_add(int *table, int n, int sym, int count)
{
int bit;
bit = 1;
while (sym != 0) {
if (sym & bit) {
table[sym] += count;
sym &= ~bit;
}
bit <<= 1;
}
table[0] += count;
}
/*
* Cumulative frequency lookup: return the total count of symbols
* with value less than `sym'.
*/
static int cf_clookup(int *table, int n, int sym)
{
int bit, index, limit, count;
if (sym == 0)
return 0;
assert(0 < sym && sym <= n);
count = table[0]; /* start with the whole table size */
bit = 1;
while (bit < n)
bit <<= 1;
limit = n;
while (bit > 0) {
/*
* Find the least number with its lowest set bit in this
* position which is greater than or equal to sym.
*/
index = ((sym + bit - 1) &~ (bit * 2 - 1)) + bit;
if (index < limit) {
count -= table[index];
limit = index;
}
bit >>= 1;
}
return count;
}
/*
* Single frequency lookup: return the count of symbol `sym'.
*/
static int cf_slookup(int *table, int n, int sym)
{
int count, bit;
assert(0 <= sym && sym < n);
count = table[sym];
for (bit = 1; sym+bit < n && !(sym & bit); bit <<= 1)
count -= table[sym+bit];
return count;
}
/*
* Return the largest symbol index such that the cumulative
* frequency up to that symbol is less than _or equal to_ count.
*/
static int cf_whichsym(int *table, int n, int count) {
int bit, sym, top;
assert(count >= 0 && count < table[0]);
bit = 1;
while (bit < n)
bit <<= 1;
sym = 0;
top = table[0];
while (bit > 0) {
if (sym+bit < n) {
if (count >= top - table[sym+bit])
sym += bit;
else
top -= table[sym+bit];
}
bit >>= 1;
}
return sym;
}
/* ----------------------------------------------------------------------
* Map generation.
*
* FIXME: this isn't entirely optimal at present, because it
* inherently prioritises growing the largest region since there
* are more squares adjacent to it. This acts as a destabilising
* influence leading to a few large regions and mostly small ones.
* It might be better to do it some other way.
*/
#define WEIGHT_INCREASED 2 /* for increased perimeter */
#define WEIGHT_DECREASED 4 /* for decreased perimeter */
#define WEIGHT_UNCHANGED 3 /* for unchanged perimeter */
/*
* Look at a square and decide which colours can be extended into
* it.
*
* If called with index < 0, it adds together one of
* WEIGHT_INCREASED, WEIGHT_DECREASED or WEIGHT_UNCHANGED for each
* colour that has a valid extension (according to the effect that
* it would have on the perimeter of the region being extended) and
* returns the overall total.
*
* If called with index >= 0, it returns one of the possible
* colours depending on the value of index, in such a way that the
* number of possible inputs which would give rise to a given
* return value correspond to the weight of that value.
*/
static int extend_options(int w, int h, int n, int *map,
int x, int y, int index)
{
int c, i, dx, dy;
int col[8];
int total = 0;
if (map[y*w+x] >= 0) {
assert(index < 0);
return 0; /* can't do this square at all */
}
/*
* Fetch the eight neighbours of this square, in order around
* the square.
*/
for (dy = -1; dy <= +1; dy++)
for (dx = -1; dx <= +1; dx++) {
int index = (dy < 0 ? 6-dx : dy > 0 ? 2+dx : 2*(1+dx));
if (x+dx >= 0 && x+dx < w && y+dy >= 0 && y+dy < h)
col[index] = map[(y+dy)*w+(x+dx)];
else
col[index] = -1;
}
/*
* Iterate over each colour that might be feasible.
*
* FIXME: this routine currently has O(n) running time. We
* could turn it into O(FOUR) by only bothering to iterate over
* the colours mentioned in the four neighbouring squares.
*/
for (c = 0; c < n; c++) {
int count, neighbours, runs;
/*
* One of the even indices of col (representing the
* orthogonal neighbours of this square) must be equal to
* c, or else this square is not adjacent to region c and
* obviously cannot become an extension of it at this time.
*/
neighbours = 0;
for (i = 0; i < 8; i += 2)
if (col[i] == c)
neighbours++;
if (!neighbours)
continue;
/*
* Now we know this square is adjacent to region c. The
* next question is, would extending it cause the region to
* become non-simply-connected? If so, we mustn't do it.
*
* We determine this by looking around col to see if we can
* find more than one separate run of colour c.
*/
runs = 0;
for (i = 0; i < 8; i++)
if (col[i] == c && col[(i+1) & 7] != c)
runs++;
if (runs > 1)
continue;
assert(runs == 1);
/*
* This square is a possibility. Determine its effect on
* the region's perimeter (computed from the number of
* orthogonal neighbours - 1 means a perimeter increase, 3
* a decrease, 2 no change; 4 is impossible because the
* region would already not be simply connected) and we're
* done.
*/
assert(neighbours > 0 && neighbours < 4);
count = (neighbours == 1 ? WEIGHT_INCREASED :
neighbours == 2 ? WEIGHT_UNCHANGED : WEIGHT_DECREASED);
total += count;
if (index >= 0 && index < count)
return c;
else
index -= count;
}
assert(index < 0);
return total;
}
static void genmap(int w, int h, int n, int *map, random_state *rs)
{
int wh = w*h;
int x, y, i, k;
int *tmp;
assert(n <= wh);
tmp = snewn(wh, int);
/*
* Clear the map, and set up `tmp' as a list of grid indices.
*/
for (i = 0; i < wh; i++) {
map[i] = -1;
tmp[i] = i;
}
/*
* Place the region seeds by selecting n members from `tmp'.
*/
k = wh;
for (i = 0; i < n; i++) {
int j = random_upto(rs, k);
map[tmp[j]] = i;
tmp[j] = tmp[--k];
}
/*
* Re-initialise `tmp' as a cumulative frequency table. This
* will store the number of possible region colours we can
* extend into each square.
*/
cf_init(tmp, wh);
/*
* Go through the grid and set up the initial cumulative
* frequencies.
*/
for (y = 0; y < h; y++)
for (x = 0; x < w; x++)
cf_add(tmp, wh, y*w+x,
extend_options(w, h, n, map, x, y, -1));
/*
* Now repeatedly choose a square we can extend a region into,
* and do so.
*/
while (tmp[0] > 0) {
int k = random_upto(rs, tmp[0]);
int sq;
int colour;
int xx, yy;
sq = cf_whichsym(tmp, wh, k);
k -= cf_clookup(tmp, wh, sq);
x = sq % w;
y = sq / w;
colour = extend_options(w, h, n, map, x, y, k);
map[sq] = colour;
/*
* Re-scan the nine cells around the one we've just
* modified.
*/
for (yy = max(y-1, 0); yy < min(y+2, h); yy++)
for (xx = max(x-1, 0); xx < min(x+2, w); xx++) {
cf_add(tmp, wh, yy*w+xx,
-cf_slookup(tmp, wh, yy*w+xx) +
extend_options(w, h, n, map, xx, yy, -1));
}
}
/*
* Finally, go through and normalise the region labels into
* order, meaning that indistinguishable maps are actually
* identical.
*/
for (i = 0; i < n; i++)
tmp[i] = -1;
k = 0;
for (i = 0; i < wh; i++) {
assert(map[i] >= 0);
if (tmp[map[i]] < 0)
tmp[map[i]] = k++;
map[i] = tmp[map[i]];
}
sfree(tmp);
}
/* ----------------------------------------------------------------------
* Functions to handle graphs.
*/
/*
* Having got a map in a square grid, convert it into a graph
* representation.
*/
static int gengraph(int w, int h, int n, int *map, int *graph)
{
int i, j, x, y;
/*
* Start by setting the graph up as an adjacency matrix. We'll
* turn it into a list later.
*/
for (i = 0; i < n*n; i++)
graph[i] = 0;
/*
* Iterate over the map looking for all adjacencies.
*/
for (y = 0; y < h; y++)
for (x = 0; x < w; x++) {
int v, vx, vy;
v = map[y*w+x];
if (x+1 < w && (vx = map[y*w+(x+1)]) != v)
graph[v*n+vx] = graph[vx*n+v] = 1;
if (y+1 < h && (vy = map[(y+1)*w+x]) != v)
graph[v*n+vy] = graph[vy*n+v] = 1;
}
/*
* Turn the matrix into a list.
*/
for (i = j = 0; i < n*n; i++)
if (graph[i])
graph[j++] = i;
return j;
}
static int graph_edge_index(int *graph, int n, int ngraph, int i, int j)
{
int v = i*n+j;
int top, bot, mid;
bot = -1;
top = ngraph;
while (top - bot > 1) {
mid = (top + bot) / 2;
if (graph[mid] == v)
return mid;
else if (graph[mid] < v)
bot = mid;
else
top = mid;
}
return -1;
}
#define graph_adjacent(graph, n, ngraph, i, j) \
(graph_edge_index((graph), (n), (ngraph), (i), (j)) >= 0)
static int graph_vertex_start(int *graph, int n, int ngraph, int i)
{
int v = i*n;
int top, bot, mid;
bot = -1;
top = ngraph;
while (top - bot > 1) {
mid = (top + bot) / 2;
if (graph[mid] < v)
bot = mid;
else
top = mid;
}
return top;
}
/* ----------------------------------------------------------------------
* Generate a four-colouring of a graph.
*
* FIXME: it would be nice if we could convert this recursion into
* pseudo-recursion using some sort of explicit stack array, for
* the sake of the Palm port and its limited stack.
*/
static bool fourcolour_recurse(int *graph, int n, int ngraph,
int *colouring, int *scratch, random_state *rs)
{
int nfree, nvert, start, i, j, k, c, ci;
int cs[FOUR];
/*
* Find the smallest number of free colours in any uncoloured
* vertex, and count the number of such vertices.
*/
nfree = FIVE; /* start off bigger than FOUR! */
nvert = 0;
for (i = 0; i < n; i++)
if (colouring[i] < 0 && scratch[i*FIVE+FOUR] <= nfree) {
if (nfree > scratch[i*FIVE+FOUR]) {
nfree = scratch[i*FIVE+FOUR];
nvert = 0;
}
nvert++;
}
/*
* If there aren't any uncoloured vertices at all, we're done.
*/
if (nvert == 0)
return true; /* we've got a colouring! */
/*
* Pick a random vertex in that set.
*/
j = random_upto(rs, nvert);
for (i = 0; i < n; i++)
if (colouring[i] < 0 && scratch[i*FIVE+FOUR] == nfree)
if (j-- == 0)
break;
assert(i < n);
start = graph_vertex_start(graph, n, ngraph, i);
/*
* Loop over the possible colours for i, and recurse for each
* one.
*/
ci = 0;
for (c = 0; c < FOUR; c++)
if (scratch[i*FIVE+c] == 0)
cs[ci++] = c;
shuffle(cs, ci, sizeof(*cs), rs);
while (ci-- > 0) {
c = cs[ci];
/*
* Fill in this colour.
*/
colouring[i] = c;
/*
* Update the scratch space to reflect a new neighbour
* of this colour for each neighbour of vertex i.
*/
for (j = start; j < ngraph && graph[j] < n*(i+1); j++) {
k = graph[j] - i*n;
if (scratch[k*FIVE+c] == 0)
scratch[k*FIVE+FOUR]--;
scratch[k*FIVE+c]++;
}
/*
* Recurse.
*/
if (fourcolour_recurse(graph, n, ngraph, colouring, scratch, rs))
return true; /* got one! */
/*
* If that didn't work, clean up and try again with a
* different colour.
*/
for (j = start; j < ngraph && graph[j] < n*(i+1); j++) {
k = graph[j] - i*n;
scratch[k*FIVE+c]--;
if (scratch[k*FIVE+c] == 0)
scratch[k*FIVE+FOUR]++;
}
colouring[i] = -1;
}
/*
* If we reach here, we were unable to find a colouring at all.
* (This doesn't necessarily mean the Four Colour Theorem is
* violated; it might just mean we've gone down a dead end and
* need to back up and look somewhere else. It's only an FCT
* violation if we get all the way back up to the top level and
* still fail.)
*/
return false;
}
static void fourcolour(int *graph, int n, int ngraph, int *colouring,
random_state *rs)
{
int *scratch;
int i;
bool retd;
/*
* For each vertex and each colour, we store the number of
* neighbours that have that colour. Also, we store the number
* of free colours for the vertex.
*/
scratch = snewn(n * FIVE, int);
for (i = 0; i < n * FIVE; i++)
scratch[i] = (i % FIVE == FOUR ? FOUR : 0);
/*
* Clear the colouring to start with.
*/
for (i = 0; i < n; i++)
colouring[i] = -1;
retd = fourcolour_recurse(graph, n, ngraph, colouring, scratch, rs);
assert(retd); /* by the Four Colour Theorem :-) */
sfree(scratch);
}
/* ----------------------------------------------------------------------
* Non-recursive solver.
*/
struct solver_scratch {
unsigned char *possible; /* bitmap of colours for each region */
int *graph;
int n;
int ngraph;
int *bfsqueue;
int *bfscolour;
#ifdef SOLVER_DIAGNOSTICS
int *bfsprev;
#endif
int depth;
};
static struct solver_scratch *new_scratch(int *graph, int n, int ngraph)
{
struct solver_scratch *sc;
sc = snew(struct solver_scratch);
sc->graph = graph;
sc->n = n;
sc->ngraph = ngraph;
sc->possible = snewn(n, unsigned char);
sc->depth = 0;
sc->bfsqueue = snewn(n, int);
sc->bfscolour = snewn(n, int);
#ifdef SOLVER_DIAGNOSTICS
sc->bfsprev = snewn(n, int);
#endif
return sc;
}
static void free_scratch(struct solver_scratch *sc)
{
sfree(sc->possible);
sfree(sc->bfsqueue);
sfree(sc->bfscolour);
#ifdef SOLVER_DIAGNOSTICS
sfree(sc->bfsprev);
#endif
sfree(sc);
}
/*
* Count the bits in a word. Only needs to cope with FOUR bits.
*/
static int bitcount(int word)
{
assert(FOUR <= 4); /* or this needs changing */
word = ((word & 0xA) >> 1) + (word & 0x5);
word = ((word & 0xC) >> 2) + (word & 0x3);
return word;
}
#ifdef SOLVER_DIAGNOSTICS
static const char colnames[FOUR] = { 'R', 'Y', 'G', 'B' };
#endif
static bool place_colour(struct solver_scratch *sc,
int *colouring, int index, int colour
#ifdef SOLVER_DIAGNOSTICS
, const char *verb
#endif
)
{
int *graph = sc->graph, n = sc->n, ngraph = sc->ngraph;
int j, k;
if (!(sc->possible[index] & (1 << colour))) {
#ifdef SOLVER_DIAGNOSTICS
if (verbose)
printf("%*scannot place %c in region %d\n", 2*sc->depth, "",
colnames[colour], index);
#endif
return false; /* can't do it */
}
sc->possible[index] = 1 << colour;
colouring[index] = colour;
#ifdef SOLVER_DIAGNOSTICS
if (verbose)
printf("%*s%s %c in region %d\n", 2*sc->depth, "",
verb, colnames[colour], index);
#endif
/*
* Rule out this colour from all the region's neighbours.
*/
for (j = graph_vertex_start(graph, n, ngraph, index);
j < ngraph && graph[j] < n*(index+1); j++) {
k = graph[j] - index*n;
#ifdef SOLVER_DIAGNOSTICS
if (verbose && (sc->possible[k] & (1 << colour)))
printf("%*s ruling out %c in region %d\n", 2*sc->depth, "",
colnames[colour], k);
#endif
sc->possible[k] &= ~(1 << colour);
}
return true;
}
#ifdef SOLVER_DIAGNOSTICS
static char *colourset(char *buf, int set)
{
int i;
char *p = buf;
const char *sep = "";
for (i = 0; i < FOUR; i++)
if (set & (1 << i)) {
p += sprintf(p, "%s%c", sep, colnames[i]);
sep = ",";
}
return buf;
}
#endif
/*
* Returns 0 for impossible, 1 for success, 2 for failure to
* converge (i.e. puzzle is either ambiguous or just too
* difficult).
*/
static int map_solver(struct solver_scratch *sc,
int *graph, int n, int ngraph, int *colouring,
int difficulty)
{
int i;
if (sc->depth == 0) {
/*
* Initialise scratch space.
*/
for (i = 0; i < n; i++)
sc->possible[i] = (1 << FOUR) - 1;
/*
* Place clues.
*/
for (i = 0; i < n; i++)
if (colouring[i] >= 0) {
if (!place_colour(sc, colouring, i, colouring[i]
#ifdef SOLVER_DIAGNOSTICS
, "initial clue:"
#endif
)) {
#ifdef SOLVER_DIAGNOSTICS
if (verbose)
printf("%*sinitial clue set is inconsistent\n",
2*sc->depth, "");
#endif
return 0; /* the clues aren't even consistent! */
}
}
}
/*
* Now repeatedly loop until we find nothing further to do.
*/
while (1) {
bool done_something = false;
if (difficulty < DIFF_EASY)
break; /* can't do anything at all! */
/*
* Simplest possible deduction: find a region with only one
* possible colour.
*/
for (i = 0; i < n; i++) if (colouring[i] < 0) {
int p = sc->possible[i];
if (p == 0) {
#ifdef SOLVER_DIAGNOSTICS
if (verbose)
printf("%*sregion %d has no possible colours left\n",
2*sc->depth, "", i);
#endif
return 0; /* puzzle is inconsistent */
}
if ((p & (p-1)) == 0) { /* p is a power of two */
int c;