ref: 05c536e50d0c3114e1de5283eac97ff22bad0fc7
dir: /tents.c/
/*
* tents.c: Puzzle involving placing tents next to trees subject to
* some confusing conditions.
*
* TODO:
*
* - it might be nice to make setter-provided tent/nontent clues
* inviolable?
* * on the other hand, this would introduce considerable extra
* complexity and size into the game state; also inviolable
* clues would have to be marked as such somehow, in an
* intrusive and annoying manner. Since they're never
* generated by _my_ generator, I'm currently more inclined
* not to bother.
*
* - more difficult levels at the top end?
* * for example, sometimes we can deduce that two BLANKs in
* the same row are each adjacent to the same unattached tree
* and to nothing else, implying that they can't both be
* tents; this enables us to rule out some extra combinations
* in the row-based deduction loop, and hence deduce more
* from the number in that row than we could otherwise do.
* * that by itself doesn't seem worth implementing a new
* difficulty level for, but if I can find a few more things
* like that then it might become worthwhile.
* * I wonder if there's a sensible heuristic for where to
* guess which would make a recursive solver viable?
*/
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <assert.h>
#include <ctype.h>
#include <limits.h>
#include <math.h>
#include "puzzles.h"
#include "matching.h"
/*
* Design discussion
* -----------------
*
* The rules of this puzzle as available on the WWW are poorly
* specified. The bits about tents having to be orthogonally
* adjacent to trees, tents not being even diagonally adjacent to
* one another, and the number of tents in each row and column
* being given are simple enough; the difficult bit is the
* tent-to-tree matching.
*
* Some sources use simplistic wordings such as `each tree is
* exactly connected to only one tent', which is extremely unclear:
* it's easy to read erroneously as `each tree is _orthogonally
* adjacent_ to exactly one tent', which is definitely incorrect.
* Even the most coherent sources I've found don't do a much better
* job of stating the rule.
*
* A more precise statement of the rule is that it must be possible
* to find a bijection f between tents and trees such that each
* tree T is orthogonally adjacent to the tent f(T), but that a
* tent is permitted to be adjacent to other trees in addition to
* its own. This slightly non-obvious criterion is what gives this
* puzzle most of its subtlety.
*
* However, there's a particularly subtle ambiguity left over. Is
* the bijection between tents and trees required to be _unique_?
* In other words, is that bijection conceptually something the
* player should be able to exhibit as part of the solution (even
* if they aren't actually required to do so)? Or is it sufficient
* to have a unique _placement_ of the tents which gives rise to at
* least one suitable bijection?
*
* The puzzle shown to the right of this .T. 2 *T* 2
* paragraph illustrates the problem. There T.T 0 -> T-T 0
* are two distinct bijections available. .T. 2 *T* 2
* The answer to the above question will
* determine whether it's a valid puzzle. 202 202
*
* This is an important question, because it affects both the
* player and the generator. Eventually I found all the instances
* of this puzzle I could Google up, solved them all by hand, and
* verified that in all cases the tree/tent matching was uniquely
* determined given the tree and tent positions. Therefore, the
* puzzle as implemented in this source file takes the following
* policy:
*
* - When checking a user-supplied solution for correctness, only
* verify that there exists _at least_ one matching.
* - When generating a puzzle, enforce that there must be
* _exactly_ one.
*
* Algorithmic implications
* ------------------------
*
* Another way of phrasing the tree/tent matching criterion is to
* say that the bipartite adjacency graph between trees and tents
* has a perfect matching. That is, if you construct a graph which
* has a vertex per tree and a vertex per tent, and an edge between
* any tree and tent which are orthogonally adjacent, it is
* possible to find a set of N edges of that graph (where N is the
* number of trees and also the number of tents) which between them
* connect every tree to every tent.
*
* The most efficient known algorithms for finding such a matching
* given a graph, as far as I'm aware, are the Munkres assignment
* algorithm (also known as the Hungarian algorithm) and the
* Ford-Fulkerson algorithm (for finding optimal flows in
* networks). Each of these takes O(N^3) running time; so we're
* talking O(N^3) time to verify any candidate solution to this
* puzzle. That's just about OK if you're doing it once per mouse
* click (and in fact not even that, since the sensible thing to do
* is check all the _other_ puzzle criteria and only wade into this
* quagmire if none are violated); but if the solver had to keep
* doing N^3 work internally, then it would probably end up with
* more like N^5 or N^6 running time, and grid generation would
* become very clunky.
*
* Fortunately, I've been able to prove a very useful property of
* _unique_ perfect matchings, by adapting the proof of Hall's
* Marriage Theorem. For those unaware of Hall's Theorem, I'll
* recap it and its proof: it states that a bipartite graph
* contains a perfect matching iff every set of vertices on the
* left side of the graph have a neighbourhood _at least_ as big on
* the right.
*
* This condition is obviously satisfied if a perfect matching does
* exist; each left-side node has a distinct right-side node which
* is the one assigned to it by the matching, and thus any set of n
* left vertices must have a combined neighbourhood containing at
* least the n corresponding right vertices, and possibly others
* too. Alternatively, imagine if you had (say) three left-side
* nodes all of which were connected to only two right-side nodes
* between them: any perfect matching would have to assign one of
* those two right nodes to each of the three left nodes, and still
* give the three left nodes a different right node each. This is
* of course impossible.
*
* To prove the converse (that if every subset of left vertices
* satisfies the Hall condition then a perfect matching exists),
* consider trying to find a proper subset of the left vertices
* which _exactly_ satisfies the Hall condition: that is, its right
* neighbourhood is precisely the same size as it. If we can find
* such a subset, then we can split the bipartite graph into two
* smaller ones: one consisting of the left subset and its right
* neighbourhood, the other consisting of everything else. Edges
* from the left side of the former graph to the right side of the
* latter do not exist, by construction; edges from the right side
* of the former to the left of the latter cannot be part of any
* perfect matching because otherwise the left subset would not be
* left with enough distinct right vertices to connect to (this is
* exactly the same deduction used in Solo's set analysis). You can
* then prove (left as an exercise) that both these smaller graphs
* still satisfy the Hall condition, and therefore the proof will
* follow by induction.
*
* There's one other possibility, which is the case where _no_
* proper subset of the left vertices has a right neighbourhood of
* exactly the same size. That is, every left subset has a strictly
* _larger_ right neighbourhood. In this situation, we can simply
* remove an _arbitrary_ edge from the graph. This cannot reduce
* the size of any left subset's right neighbourhood by more than
* one, so if all neighbourhoods were strictly bigger than they
* needed to be initially, they must now still be _at least as big_
* as they need to be. So we can keep throwing out arbitrary edges
* until we find a set which exactly satisfies the Hall condition,
* and then proceed as above. []
*
* That's Hall's theorem. I now build on this by examining the
* circumstances in which a bipartite graph can have a _unique_
* perfect matching. It is clear that in the second case, where no
* left subset exactly satisfies the Hall condition and so we can
* remove an arbitrary edge, there cannot be a unique perfect
* matching: given one perfect matching, we choose our arbitrary
* removed edge to be one of those contained in it, and then we can
* still find a perfect matching in the remaining graph, which will
* be a distinct perfect matching in the original.
*
* So it is a necessary condition for a unique perfect matching
* that there must be at least one proper left subset which
* _exactly_ satisfies the Hall condition. But now consider the
* smaller graph constructed by taking that left subset and its
* neighbourhood: if the graph as a whole had a unique perfect
* matching, then so must this smaller one, which means we can find
* a proper left subset _again_, and so on. Repeating this process
* must eventually reduce us to a graph with only one left-side
* vertex (so there are no proper subsets at all); this vertex must
* be connected to only one right-side vertex, and hence must be so
* in the original graph as well (by construction). So we can
* discard this vertex pair from the graph, and any other edges
* that involved it (which will by construction be from other left
* vertices only), and the resulting smaller graph still has a
* unique perfect matching which means we can do the same thing
* again.
*
* In other words, given any bipartite graph with a unique perfect
* matching, we can find that matching by the following extremely
* simple algorithm:
*
* - Find a left-side vertex which is only connected to one
* right-side vertex.
* - Assign those vertices to one another, and therefore discard
* any other edges connecting to that right vertex.
* - Repeat until all vertices have been matched.
*
* This algorithm can be run in O(V+E) time (where V is the number
* of vertices and E is the number of edges in the graph), and the
* only way it can fail is if there is not a unique perfect
* matching (either because there is no matching at all, or because
* it isn't unique; but it can't distinguish those cases).
*
* Thus, the internal solver in this source file can be confident
* that if the tree/tent matching is uniquely determined by the
* tree and tent positions, it can find it using only this kind of
* obvious and simple operation: assign a tree to a tent if it
* cannot possibly belong to any other tent, and vice versa. If the
* solver were _only_ trying to determine the matching, even that
* `vice versa' wouldn't be required; but it can come in handy when
* not all the tents have been placed yet. I can therefore be
* reasonably confident that as long as my solver doesn't need to
* cope with grids that have a non-unique matching, it will also
* not need to do anything complicated like set analysis between
* trees and tents.
*/
/*
* 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
/*
* 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(TRICKY,Tricky,t)
#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 tents_diffnames[] = { DIFFLIST(TITLE) };
static char const tents_diffchars[] = DIFFLIST(ENCODE);
#define DIFFCONFIG DIFFLIST(CONFIG)
enum {
COL_BACKGROUND,
COL_GRID,
COL_GRASS,
COL_TREETRUNK,
COL_TREELEAF,
COL_TENT,
COL_ERROR,
COL_ERRTEXT,
COL_ERRTRUNK,
NCOLOURS
};
enum { BLANK, TREE, TENT, NONTENT, MAGIC };
struct game_params {
int w, h;
int diff;
};
struct numbers {
int refcount;
int *numbers;
};
struct game_state {
game_params p;
char *grid;
struct numbers *numbers;
bool completed, used_solve;
};
static game_params *default_params(void)
{
game_params *ret = snew(game_params);
ret->w = ret->h = 8;
ret->diff = DIFF_EASY;
return ret;
}
static const struct game_params tents_presets[] = {
{8, 8, DIFF_EASY},
{8, 8, DIFF_TRICKY},
{10, 10, DIFF_EASY},
{10, 10, DIFF_TRICKY},
{15, 15, DIFF_EASY},
{15, 15, DIFF_TRICKY},
};
static bool game_fetch_preset(int i, char **name, game_params **params)
{
game_params *ret;
char str[80];
if (i < 0 || i >= lenof(tents_presets))
return false;
ret = snew(game_params);
*ret = tents_presets[i];
sprintf(str, "%dx%d %s", ret->w, ret->h, tents_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)
{
params->w = params->h = atoi(string);
while (*string && isdigit((unsigned char)*string)) string++;
if (*string == 'x') {
string++;
params->h = atoi(string);
while (*string && isdigit((unsigned char)*string)) string++;
}
if (*string == 'd') {
int i;
string++;
for (i = 0; i < DIFFCOUNT; i++)
if (*string == tents_diffchars[i])
params->diff = i;
if (*string) string++;
}
}
static char *encode_params(const game_params *params, bool full)
{
char buf[120];
sprintf(buf, "%dx%d", params->w, params->h);
if (full)
sprintf(buf + strlen(buf), "d%c",
tents_diffchars[params->diff]);
return dupstr(buf);
}
static config_item *game_configure(const game_params *params)
{
config_item *ret;
char buf[80];
ret = snewn(4, 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 = "Difficulty";
ret[2].type = C_CHOICES;
ret[2].u.choices.choicenames = DIFFCONFIG;
ret[2].u.choices.selected = params->diff;
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->h = atoi(cfg[1].u.string.sval);
ret->diff = cfg[2].u.choices.selected;
return ret;
}
static const char *validate_params(const game_params *params, bool full)
{
/*
* Generating anything under 4x4 runs into trouble of one kind
* or another.
*/
if (params->w < 4 || params->h < 4)
return "Width and height must both be at least four";
if (params->w > (INT_MAX - 1) / params->h)
return "Width times height must not be unreasonably large";
return NULL;
}
/*
* Scratch space for solver.
*/
enum { N, U, L, R, D, MAXDIR }; /* link directions */
#define dx(d) ( ((d)==R) - ((d)==L) )
#define dy(d) ( ((d)==D) - ((d)==U) )
#define F(d) ( U + D - (d) )
struct solver_scratch {
char *links; /* mapping between trees and tents */
int *locs;
char *place, *mrows, *trows;
};
static struct solver_scratch *new_scratch(int w, int h)
{
struct solver_scratch *ret = snew(struct solver_scratch);
ret->links = snewn(w*h, char);
ret->locs = snewn(max(w, h), int);
ret->place = snewn(max(w, h), char);
ret->mrows = snewn(3 * max(w, h), char);
ret->trows = snewn(3 * max(w, h), char);
return ret;
}
static void free_scratch(struct solver_scratch *sc)
{
sfree(sc->trows);
sfree(sc->mrows);
sfree(sc->place);
sfree(sc->locs);
sfree(sc->links);
sfree(sc);
}
/*
* Solver. Returns 0 for impossibility, 1 for success, 2 for
* ambiguity or failure to converge.
*/
static int tents_solve(int w, int h, const char *grid, int *numbers,
char *soln, struct solver_scratch *sc, int diff)
{
int x, y, d, i, j;
char *mrow, *trow, *trow1, *trow2;
/*
* Set up solver data.
*/
memset(sc->links, N, w*h);
/*
* Set up solution array.
*/
memcpy(soln, grid, w*h);
/*
* Main solver loop.
*/
while (1) {
bool done_something = false;
/*
* Any tent which has only one unattached tree adjacent to
* it can be tied to that tree.
*/
for (y = 0; y < h; y++)
for (x = 0; x < w; x++)
if (soln[y*w+x] == TENT && !sc->links[y*w+x]) {
int linkd = 0;
for (d = 1; d < MAXDIR; d++) {
int x2 = x + dx(d), y2 = y + dy(d);
if (x2 >= 0 && x2 < w && y2 >= 0 && y2 < h &&
soln[y2*w+x2] == TREE &&
!sc->links[y2*w+x2]) {
if (linkd)
break; /* found more than one */
else
linkd = d;
}
}
if (d == MAXDIR && linkd == 0) {
#ifdef SOLVER_DIAGNOSTICS
if (verbose)
printf("tent at %d,%d cannot link to anything\n",
x, y);
#endif
return 0; /* no solution exists */
} else if (d == MAXDIR) {
int x2 = x + dx(linkd), y2 = y + dy(linkd);
#ifdef SOLVER_DIAGNOSTICS
if (verbose)
printf("tent at %d,%d can only link to tree at"
" %d,%d\n", x, y, x2, y2);
#endif
sc->links[y*w+x] = linkd;
sc->links[y2*w+x2] = F(linkd);
done_something = true;
}
}
if (done_something)
continue;
if (diff < 0)
break; /* don't do anything else! */
/*
* Mark a blank square as NONTENT if it is not orthogonally
* adjacent to any unmatched tree.
*/
for (y = 0; y < h; y++)
for (x = 0; x < w; x++)
if (soln[y*w+x] == BLANK) {
bool can_be_tent = false;
for (d = 1; d < MAXDIR; d++) {
int x2 = x + dx(d), y2 = y + dy(d);
if (x2 >= 0 && x2 < w && y2 >= 0 && y2 < h &&
soln[y2*w+x2] == TREE &&
!sc->links[y2*w+x2])
can_be_tent = true;
}
if (!can_be_tent) {
#ifdef SOLVER_DIAGNOSTICS
if (verbose)
printf("%d,%d cannot be a tent (no adjacent"
" unmatched tree)\n", x, y);
#endif
soln[y*w+x] = NONTENT;
done_something = true;
}
}
if (done_something)
continue;
/*
* Mark a blank square as NONTENT if it is (perhaps
* diagonally) adjacent to any other tent.
*/
for (y = 0; y < h; y++)
for (x = 0; x < w; x++)
if (soln[y*w+x] == BLANK) {
int dx, dy;
bool imposs = false;
for (dy = -1; dy <= +1; dy++)
for (dx = -1; dx <= +1; dx++)
if (dy || dx) {
int x2 = x + dx, y2 = y + dy;
if (x2 >= 0 && x2 < w && y2 >= 0 && y2 < h &&
soln[y2*w+x2] == TENT)
imposs = true;
}
if (imposs) {
#ifdef SOLVER_DIAGNOSTICS
if (verbose)
printf("%d,%d cannot be a tent (adjacent tent)\n",
x, y);
#endif
soln[y*w+x] = NONTENT;
done_something = true;
}
}
if (done_something)
continue;
/*
* Any tree which has exactly one {unattached tent, BLANK}
* adjacent to it must have its tent in that square.
*/
for (y = 0; y < h; y++)
for (x = 0; x < w; x++)
if (soln[y*w+x] == TREE && !sc->links[y*w+x]) {
int linkd = 0, linkd2 = 0, nd = 0;
for (d = 1; d < MAXDIR; d++) {
int x2 = x + dx(d), y2 = y + dy(d);
if (!(x2 >= 0 && x2 < w && y2 >= 0 && y2 < h))
continue;
if (soln[y2*w+x2] == BLANK ||
(soln[y2*w+x2] == TENT && !sc->links[y2*w+x2])) {
if (linkd)
linkd2 = d;
else
linkd = d;
nd++;
}
}
if (nd == 0) {
#ifdef SOLVER_DIAGNOSTICS
if (verbose)
printf("tree at %d,%d cannot link to anything\n",
x, y);
#endif
return 0; /* no solution exists */
} else if (nd == 1) {
int x2 = x + dx(linkd), y2 = y + dy(linkd);
#ifdef SOLVER_DIAGNOSTICS
if (verbose)
printf("tree at %d,%d can only link to tent at"
" %d,%d\n", x, y, x2, y2);
#endif
soln[y2*w+x2] = TENT;
sc->links[y*w+x] = linkd;
sc->links[y2*w+x2] = F(linkd);
done_something = true;
} else if (nd == 2 && (!dx(linkd) != !dx(linkd2)) &&
diff >= DIFF_TRICKY) {
/*
* If there are two possible places where
* this tree's tent can go, and they are
* diagonally separated rather than being
* on opposite sides of the tree, then the
* square (other than the tree square)
* which is adjacent to both of them must
* be a non-tent.
*/
int x2 = x + dx(linkd) + dx(linkd2);
int y2 = y + dy(linkd) + dy(linkd2);
assert(x2 >= 0 && x2 < w && y2 >= 0 && y2 < h);
if (soln[y2*w+x2] == BLANK) {
#ifdef SOLVER_DIAGNOSTICS
if (verbose)
printf("possible tent locations for tree at"
" %d,%d rule out tent at %d,%d\n",
x, y, x2, y2);
#endif
soln[y2*w+x2] = NONTENT;
done_something = true;
}
}
}
if (done_something)
continue;
/*
* If localised deductions about the trees and tents
* themselves haven't helped us, it's time to resort to the
* numbers round the grid edge. For each row and column, we
* go through all possible combinations of locations for
* the unplaced tents, rule out any which have adjacent
* tents, and spot any square which is given the same state
* by all remaining combinations.
*/
for (i = 0; i < w+h; i++) {
int start, step, len, start1, start2, n, k;
if (i < w) {
/*
* This is the number for a column.
*/
start = i;
step = w;
len = h;
if (i > 0)
start1 = start - 1;
else
start1 = -1;
if (i+1 < w)
start2 = start + 1;
else
start2 = -1;
} else {
/*
* This is the number for a row.
*/
start = (i-w)*w;
step = 1;
len = w;
if (i > w)
start1 = start - w;
else
start1 = -1;
if (i+1 < w+h)
start2 = start + w;
else
start2 = -1;
}
if (diff < DIFF_TRICKY) {
/*
* In Easy mode, we don't look at the effect of one
* row on the next (i.e. ruling out a square if all
* possibilities for an adjacent row place a tent
* next to it).
*/
start1 = start2 = -1;
}
k = numbers[i];
/*
* Count and store the locations of the free squares,
* and also count the number of tents already placed.
*/
n = 0;
for (j = 0; j < len; j++) {
if (soln[start+j*step] == TENT)
k--; /* one fewer tent to place */
else if (soln[start+j*step] == BLANK)
sc->locs[n++] = j;
}
if (n == 0)
continue; /* nothing left to do here */
/*
* Now we know we're placing k tents in n squares. Set
* up the first possibility.
*/
for (j = 0; j < n; j++)
sc->place[j] = (j < k ? TENT : NONTENT);
/*
* We're aiming to find squares in this row which are
* invariant over all valid possibilities. Thus, we
* maintain the current state of that invariance. We
* start everything off at MAGIC to indicate that it
* hasn't been set up yet.
*/
mrow = sc->mrows;
trow = sc->trows;
trow1 = sc->trows + len;
trow2 = sc->trows + 2*len;
memset(mrow, MAGIC, 3*len);
/*
* And iterate over all possibilities.
*/
while (1) {
int p;
bool valid;
/*
* See if this possibility is valid. The only way
* it can fail to be valid is if it contains two
* adjacent tents. (Other forms of invalidity, such
* as containing a tent adjacent to one already
* placed, will have been dealt with already by
* other parts of the solver.)
*/
valid = true;
for (j = 0; j+1 < n; j++)
if (sc->place[j] == TENT &&
sc->place[j+1] == TENT &&
sc->locs[j+1] == sc->locs[j]+1) {
valid = false;
break;
}
if (valid) {
/*
* Merge this valid combination into mrow.
*/
memset(trow, MAGIC, len);
memset(trow+len, BLANK, 2*len);
for (j = 0; j < n; j++) {
trow[sc->locs[j]] = sc->place[j];
if (sc->place[j] == TENT) {
int jj;
for (jj = sc->locs[j]-1; jj <= sc->locs[j]+1; jj++)
if (jj >= 0 && jj < len)
trow1[jj] = trow2[jj] = NONTENT;
}
}
for (j = 0; j < 3*len; j++) {
if (trow[j] == MAGIC)
continue;
if (mrow[j] == MAGIC || mrow[j] == trow[j]) {
/*
* Either this is the first valid
* placement we've found at all, or
* this square's contents are
* consistent with every previous valid
* combination.
*/
mrow[j] = trow[j];
} else {
/*
* This square's contents fail to match
* what they were in a different
* combination, so we cannot deduce
* anything about this square.
*/
mrow[j] = BLANK;
}
}
}
/*
* Find the next combination of k choices from n.
* We do this by finding the rightmost tent which
* can be moved one place right, doing so, and
* shunting all tents to the right of that as far
* left as they can go.
*/
p = 0;
for (j = n-1; j > 0; j--) {
if (sc->place[j] == TENT)
p++;
if (sc->place[j] == NONTENT && sc->place[j-1] == TENT) {
sc->place[j-1] = NONTENT;
sc->place[j] = TENT;
while (p--)
sc->place[++j] = TENT;
while (++j < n)
sc->place[j] = NONTENT;
break;
}
}
if (j <= 0)
break; /* we've finished */
}
/*
* It's just possible that _no_ placement was valid, in
* which case we have an internally inconsistent
* puzzle.
*/
if (mrow[sc->locs[0]] == MAGIC)
return 0; /* inconsistent */
/*
* Now go through mrow and see if there's anything
* we've deduced which wasn't already mentioned in soln.
*/
for (j = 0; j < len; j++) {
int whichrow;
for (whichrow = 0; whichrow < 3; whichrow++) {
char *mthis = mrow + whichrow * len;
int tstart = (whichrow == 0 ? start :
whichrow == 1 ? start1 : start2);
if (tstart >= 0 &&
mthis[j] != MAGIC && mthis[j] != BLANK &&
soln[tstart+j*step] == BLANK) {
int pos = tstart+j*step;
#ifdef SOLVER_DIAGNOSTICS
if (verbose)
printf("%s %d forces %s at %d,%d\n",
step==1 ? "row" : "column",
step==1 ? start/w : start,
mthis[j] == TENT ? "tent" : "non-tent",
pos % w, pos / w);
#endif
soln[pos] = mthis[j];
done_something = true;
}
}
}
}
if (done_something)
continue;
if (!done_something)
break;
}
/*
* The solver has nothing further it can do. Return 1 if both
* soln and sc->links are completely filled in, or 2 otherwise.
*/
for (y = 0; y < h; y++)
for (x = 0; x < w; x++) {
if (soln[y*w+x] == BLANK)
return 2;
if (soln[y*w+x] != NONTENT && sc->links[y*w+x] == 0)
return 2;
}
return 1;
}
static char *new_game_desc(const game_params *params_in, random_state *rs,
char **aux, bool interactive)
{
game_params params_copy = *params_in; /* structure copy */
game_params *params = ¶ms_copy;
int w = params->w, h = params->h;
int ntrees = w * h / 5;
char *grid = snewn(w*h, char);
char *puzzle = snewn(w*h, char);
int *numbers = snewn(w+h, int);
char *soln = snewn(w*h, char);
int *order = snewn(w*h, int);
int *treemap = snewn(w*h, int);
int maxedges = ntrees*4 + w*h;
int *adjdata = snewn(maxedges, int);
int **adjlists = snewn(ntrees, int *);
int *adjsizes = snewn(ntrees, int);
int *outr = snewn(4*ntrees, int);
struct solver_scratch *sc = new_scratch(w, h);
char *ret, *p;
int i, j, nl, nr;
int *adjptr;
/*
* Since this puzzle has many global deductions and doesn't
* permit limited clue sets, generating grids for this puzzle
* is hard enough that I see no better option than to simply
* generate a solution and see if it's unique and has the
* required difficulty. This turns out to be computationally
* plausible as well.
*
* We chose our tree count (hence also tent count) by dividing
* the total grid area by five above. Why five? Well, w*h/4 is
* the maximum number of tents you can _possibly_ fit into the
* grid without violating the separation criterion, and to
* achieve that you are constrained to a very small set of
* possible layouts (the obvious one with a tent at every
* (even,even) coordinate, and trivial variations thereon). So
* if we reduce the tent count a bit more, we enable more
* random-looking placement; 5 turns out to be a plausible
* figure which yields sensible puzzles. Increasing the tent
* count would give puzzles whose solutions were too regimented
* and could be solved by the use of that knowledge (and would
* also take longer to find a viable placement); decreasing it
* would make the grids emptier and more boring.
*
* Actually generating a grid is a matter of first placing the
* tents, and then placing the trees by the use of matching.c
* (finding a distinct square adjacent to every tent). We do it
* this way round because otherwise satisfying the tent
* separation condition would become onerous: most randomly
* chosen tent layouts do not satisfy this condition, so we'd
* have gone to a lot of work before finding that a candidate
* layout was unusable. Instead, we place the tents first and
* ensure they meet the separation criterion _before_ doing
* lots of computation; this works much better.
*
* This generation strategy can fail at many points, including
* as early as tent placement (if you get a bad random order in
* which to greedily try the grid squares, you won't even
* manage to find enough mutually non-adjacent squares to put
* the tents in). Then it can fail if matching.c doesn't manage
* to find a good enough matching (i.e. the tent placements don't
* admit any adequate tree placements); and finally it can fail
* if the solver finds that the problem has the wrong
* difficulty (including being actually non-unique). All of
* these, however, are insufficiently frequent to cause
* trouble.
*/
if (params->diff > DIFF_EASY && params->w <= 4 && params->h <= 4)
params->diff = DIFF_EASY; /* downgrade to prevent tight loop */
while (1) {
/*
* Make a list of grid squares which we'll permute as we pick
* the tent locations.
*
* We'll also need to index all the potential tree squares,
* i.e. the ones adjacent to the tents.
*/
for (i = 0; i < w*h; i++) {
order[i] = i;
treemap[i] = -1;
}
/*
* Place tents at random without making any two adjacent.
*/
memset(grid, BLANK, w*h);
j = ntrees;
nr = 0;
/* Loop end condition: either j==0 (we've placed all the
* tents), or the number of grid squares we have yet to try
* is too few to fit the remaining tents into. */
for (i = 0; j > 0 && i+j <= w*h; i++) {
int which, x, y, d, tmp;
int dy, dx;
bool ok = true;
which = i + random_upto(rs, w*h - i);
tmp = order[which];
order[which] = order[i];
order[i] = tmp;
x = order[i] % w;
y = order[i] / w;
for (dy = -1; dy <= +1; dy++)
for (dx = -1; dx <= +1; dx++)
if (x+dx >= 0 && x+dx < w &&
y+dy >= 0 && y+dy < h &&
grid[(y+dy)*w+(x+dx)] == TENT)
ok = false;
if (ok) {
grid[order[i]] = TENT;
for (d = 1; d < MAXDIR; d++) {
int x2 = x + dx(d), y2 = y + dy(d);
if (x2 >= 0 && x2 < w && y2 >= 0 && y2 < h &&
treemap[y2*w+x2] == -1) {
treemap[y2*w+x2] = nr++;
}
}
j--;
}
}
if (j > 0)
continue; /* couldn't place all the tents */
/*
* Build up the graph for matching.c.
*/
adjptr = adjdata;
nl = 0;
for (i = 0; i < w*h; i++) {
if (grid[i] == TENT) {
int d, x = i % w, y = i / w;
adjlists[nl] = adjptr;
for (d = 1; d < MAXDIR; d++) {
int x2 = x + dx(d), y2 = y + dy(d);
if (x2 >= 0 && x2 < w && y2 >= 0 && y2 < h) {
assert(treemap[y2*w+x2] != -1);
*adjptr++ = treemap[y2*w+x2];
}
}
adjsizes[nl] = adjptr - adjlists[nl];
nl++;
}
}
/*
* Call the matching algorithm to actually place the trees.
*/
j = matching(ntrees, nr, adjlists, adjsizes, rs, NULL, outr);
if (j < ntrees)
continue; /* couldn't place all the trees */
/*
* Fill in the trees in the grid, by cross-referencing treemap
* (which maps a grid square to its index as known to
* matching()) against the output from matching().
*
* Note that for these purposes we don't actually care _which_
* tent each potential tree square is assigned to - we only
* care whether it was assigned to any tent at all, in order
* to decide whether to put a tree in it.
*/
for (i = 0; i < w*h; i++)
if (treemap[i] != -1 && outr[treemap[i]] != -1)
grid[i] = TREE;
/*
* I think it looks ugly if there isn't at least one of
* _something_ (tent or tree) in each row and each column
* of the grid. This doesn't give any information away
* since a completely empty row/column is instantly obvious
* from the clues (it has no trees and a zero).
*/
for (i = 0; i < w; i++) {
for (j = 0; j < h; j++) {
if (grid[j*w+i] != BLANK)
break; /* found something in this column */
}
if (j == h)
break; /* found empty column */
}
if (i < w)
continue; /* a column was empty */
for (j = 0; j < h; j++) {
for (i = 0; i < w; i++) {
if (grid[j*w+i] != BLANK)
break; /* found something in this row */
}
if (i == w)
break; /* found empty row */
}
if (j < h)
continue; /* a row was empty */
/*
* Now set up the numbers round the edge.
*/
for (i = 0; i < w; i++) {
int n = 0;
for (j = 0; j < h; j++)
if (grid[j*w+i] == TENT)
n++;
numbers[i] = n;
}
for (i = 0; i < h; i++) {
int n = 0;
for (j = 0; j < w; j++)
if (grid[i*w+j] == TENT)
n++;
numbers[w+i] = n;
}
/*
* And now actually solve the puzzle, to see whether it's
* unique and has the required difficulty.
*/
for (i = 0; i < w*h; i++)
puzzle[i] = grid[i] == TREE ? TREE : BLANK;
i = tents_solve(w, h, puzzle, numbers, soln, sc, params->diff-1);
j = tents_solve(w, h, puzzle, numbers, soln, sc, params->diff);
/*
* We expect solving with difficulty params->diff to have
* succeeded (otherwise the problem is too hard), and
* solving with diff-1 to have failed (otherwise it's too
* easy).
*/
if (i == 2 && j == 1)
break;
}
/*
* That's it. Encode as a game ID.
*/
ret = snewn((w+h)*40 + ntrees + (w*h)/26 + 1, char);
p = ret;
j = 0;
for (i = 0; i <= w*h; i++) {
bool c = (i < w*h ? grid[i] == TREE : true);
if (c) {
*p++ = (j == 0 ? '_' : j-1 + 'a');
j = 0;
} else {
j++;
while (j > 25) {
*p++ = 'z';
j -= 25;
}
}
}
for (i = 0; i < w+h; i++)
p += sprintf(p, ",%d", numbers[i]);
*p++ = '\0';
ret = sresize(ret, p - ret, char);
/*
* And encode the solution as an aux_info.
*/
*aux = snewn(ntrees * 40, char);
p = *aux;
*p++ = 'S';
for (i = 0; i < w*h; i++)
if (grid[i] == TENT)
p += sprintf(p, ";T%d,%d", i%w, i/w);
*p++ = '\0';
*aux = sresize(*aux, p - *aux, char);
free_scratch(sc);
sfree(outr);
sfree(adjdata);
sfree(adjlists);
sfree(adjsizes);
sfree(treemap);
sfree(order);
sfree(soln);
sfree(numbers);
sfree(puzzle);
sfree(grid);
return ret;
}
/*
* Grid description format:
*
* _ = tree
* a = 1 BLANK then TREE
* ...
* y = 25 BLANKs then TREE
* z = 25 BLANKs
* ! = set previous square to TENT
* - = set previous square to NONTENT
*
* Last character must be one that would insert a tree as the first
* square after the grid.
*/
static const char *validate_desc(const game_params *params, const char *desc)
{
int w = params->w, h = params->h;
int area, i;
area = 0;
while (*desc && *desc != ',') {
if (*desc == '_')
area++;
else if (*desc >= 'a' && *desc < 'z')
area += *desc - 'a' + 2;
else if (*desc == 'z')
area += 25;
else if (*desc == '!' || *desc == '-') {
if (area == 0 || area > w * h)
return "Tent or non-tent placed off the grid";
} else
return "Invalid character in grid specification";
desc++;
}
if (area < w * h + 1)
return "Not enough data to fill grid";
else if (area > w * h + 1)
return "Too much data to fill grid";
for (i = 0; i < w+h; i++) {
if (!*desc)
return "Not enough numbers given after grid specification";
else if (*desc != ',')
return "Invalid character in number list";
desc++;
while (*desc && isdigit((unsigned char)*desc)) desc++;
}
if (*desc)
return "Unexpected additional data at end of game description";
return NULL;
}
static game_state *new_game(midend *me, const game_params *params,
const char *desc)
{
int w = params->w, h = params->h;
game_state *state = snew(game_state);
int i;
state->p = *params; /* structure copy */
state->grid = snewn(w*h, char);
state->numbers = snew(struct numbers);
state->numbers->refcount = 1;
state->numbers->numbers = snewn(w+h, int);
state->completed = state->used_solve = false;
i = 0;
memset(state->grid, BLANK, w*h);
while (*desc) {
int run, type;
type = TREE;
if (*desc == '_')
run = 0;
else if (*desc >= 'a' && *desc < 'z')
run = *desc - ('a'-1);
else if (*desc == 'z') {
run = 25;
type = BLANK;
} else {
assert(*desc == '!' || *desc == '-');
run = -1;
type = (*desc == '!' ? TENT : NONTENT);
}
desc++;
i += run;
assert(i >= 0 && i <= w*h);
if (i == w*h) {
assert(type == TREE);
break;
} else {
if (type != BLANK)
state->grid[i++] = type;
}
}
for (i = 0; i < w+h; i++) {
assert(*desc == ',');
desc++;
state->numbers->numbers[i] = atoi(desc);
while (*desc && isdigit((unsigned char)*desc)) desc++;
}
assert(!*desc);
return state;
}
static game_state *dup_game(const game_state *state)
{
int w = state->p.w, h = state->p.h;
game_state *ret = snew(game_state);
ret->p = state->p; /* structure copy */
ret->grid = snewn(w*h, char);
memcpy(ret->grid, state->grid, w*h);
ret->numbers = state->numbers;
state->numbers->refcount++;
ret->completed = state->completed;
ret->used_solve = state->used_solve;
return ret;
}
static void free_game(game_state *state)
{
if (--state->numbers->refcount <= 0) {
sfree(state->numbers->numbers);
sfree(state->numbers);
}
sfree(state->grid);
sfree(state);
}
static char *solve_game(const game_state *state, const game_state *currstate,
const char *aux, const char **error)
{
int w = state->p.w, h = state->p.h;
if (aux) {
/*
* If we already have the solution, save ourselves some
* time.
*/
return dupstr(aux);
} else {
struct solver_scratch *sc = new_scratch(w, h);
char *soln;
int ret;
char *move, *p;
int i;
soln = snewn(w*h, char);
ret = tents_solve(w, h, state->grid, state->numbers->numbers,
soln, sc, DIFFCOUNT-1);
free_scratch(sc);
if (ret != 1) {
sfree(soln);
if (ret == 0)
*error = "This puzzle is not self-consistent";
else
*error = "Unable to find a unique solution for this puzzle";
return NULL;
}
/*
* Construct a move string which turns the current state
* into the solved state.
*/
move = snewn(w*h * 40, char);
p = move;
*p++ = 'S';
for (i = 0; i < w*h; i++)
if (soln[i] == TENT)
p += sprintf(p, ";T%d,%d", i%w, i/w);
*p++ = '\0';
move = sresize(move, p - move, char);
sfree(soln);
return move;
}
}
static bool game_can_format_as_text_now(const game_params *params)
{
return params->w <= 1998 && params->h <= 1998; /* 999 tents */
}
static char *game_text_format(const game_state *state)
{
int w = state->p.w, h = state->p.h, r, c;
int cw = 4, ch = 2, gw = (w+1)*cw + 2, gh = (h+1)*ch + 1, len = gw * gh;
char *board = snewn(len + 1, char);
sprintf(board, "%*s\n", len - 2, "");
for (r = 0; r <= h; ++r) {
for (c = 0; c <= w; ++c) {
int cell = r*ch*gw + cw*c, center = cell + gw*ch/2 + cw/2;
int i = r*w + c, n = 1000;
if (r == h && c == w) /* NOP */;
else if (c == w) n = state->numbers->numbers[w + r];
else if (r == h) n = state->numbers->numbers[c];
else switch (state->grid[i]) {
case BLANK: board[center] = '.'; break;
case TREE: board[center] = 'T'; break;
case TENT: memcpy(board + center - 1, "//\\", 3); break;
case NONTENT: break;
default: memcpy(board + center - 1, "wtf", 3);
}
if (n < 100) {
board[center] = '0' + n % 10;
if (n >= 10) board[center - 1] = '0' + n / 10;
} else if (n < 1000) {
board[center + 1] = '0' + n % 10;
board[center] = '0' + n / 10 % 10;
board[center - 1] = '0' + n / 100;
}
board[cell] = '+';
memset(board + cell + 1, '-', cw - 1);
for (i = 1; i < ch; ++i) board[cell + i*gw] = '|';
}
for (c = 0; c < ch; ++c) {
board[(r*ch+c)*gw + gw - 2] =
c == 0 ? '+' : r < h ? '|' : ' ';
board[(r*ch+c)*gw + gw - 1] = '\n';
}
}
memset(board + len - gw, '-', gw - 2 - cw);
for (c = 0; c <= w; ++c) board[len - gw + cw*c] = '+';
return board;
}
struct game_ui {
int dsx, dsy; /* coords of drag start */
int dex, dey; /* coords of drag end */
int drag_button; /* -1 for none, or a button code */
bool drag_ok; /* dragged off the window, to cancel */
int cx, cy; /* cursor position. */
bool cdisp; /* is cursor displayed? */
};
static game_ui *new_ui(const game_state *state)
{
game_ui *ui = snew(game_ui);
ui->dsx = ui->dsy = -1;
ui->dex = ui->dey = -1;
ui->drag_button = -1;
ui->drag_ok = false;
ui->cx = ui->cy = 0;
ui->cdisp = 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)
{
}
static const char *current_key_label(const game_ui *ui,
const game_state *state, int button)
{
int w = state->p.w;
int v = state->grid[ui->cy*w+ui->cx];
if (IS_CURSOR_SELECT(button) && ui->cdisp) {
switch (v) {
case BLANK:
return button == CURSOR_SELECT ? "Tent" : "Green";
case TENT: case NONTENT: return "Clear";
}
}
return "";
}
struct game_drawstate {
int tilesize;
bool started;
game_params p;
int *drawn, *numbersdrawn;
int cx, cy; /* last-drawn cursor pos, or (-1,-1) if absent. */
};
#define PREFERRED_TILESIZE 32
#define TILESIZE (ds->tilesize)
#define TLBORDER (TILESIZE/2)
#define BRBORDER (TILESIZE*3/2)
#define COORD(x) ( (x) * TILESIZE + TLBORDER )
#define FROMCOORD(x) ( ((x) - TLBORDER + TILESIZE) / TILESIZE - 1 )
#define FLASH_TIME 0.30F
static int drag_xform(const game_ui *ui, int x, int y, int v)
{
int xmin, ymin, xmax, ymax;
xmin = min(ui->dsx, ui->dex);
xmax = max(ui->dsx, ui->dex);
ymin = min(ui->dsy, ui->dey);
ymax = max(ui->dsy, ui->dey);
#ifndef STYLUS_BASED
/*
* Left-dragging has no effect, so we treat a left-drag as a
* single click on dsx,dsy.
*/
if (ui->drag_button == LEFT_BUTTON) {
xmin = xmax = ui->dsx;
ymin = ymax = ui->dsy;
}
#endif
if (x < xmin || x > xmax || y < ymin || y > ymax)
return v; /* no change outside drag area */
if (v == TREE)
return v; /* trees are inviolate always */
if (xmin == xmax && ymin == ymax) {
/*
* Results of a simple click. Left button sets blanks to
* tents; right button sets blanks to non-tents; either
* button clears a non-blank square.
* If stylus-based however, it loops instead.
*/
if (ui->drag_button == LEFT_BUTTON)
#ifdef STYLUS_BASED
v = (v == BLANK ? TENT : (v == TENT ? NONTENT : BLANK));
else
v = (v == BLANK ? NONTENT : (v == NONTENT ? TENT : BLANK));
#else
v = (v == BLANK ? TENT : BLANK);
else
v = (v == BLANK ? NONTENT : BLANK);
#endif
} else {
/*
* Results of a drag. Left-dragging has no effect.
* Right-dragging sets all blank squares to non-tents and
* has no effect on anything else.
*/
if (ui->drag_button == RIGHT_BUTTON)
v = (v == BLANK ? NONTENT : v);
else
#ifdef STYLUS_BASED
v = (v == BLANK ? NONTENT : v);
#else
/* do nothing */;
#endif
}
return v;
}
static char *interpret_move(const game_state *state, game_ui *ui,
const game_drawstate *ds,
int x, int y, int button)
{
int w = state->p.w, h = state->p.h;
char tmpbuf[80];
bool shift = button & MOD_SHFT, control = button & MOD_CTRL;
button &= ~MOD_MASK;
if (button == LEFT_BUTTON || button == RIGHT_BUTTON) {
x = FROMCOORD(x);
y = FROMCOORD(y);
if (x < 0 || y < 0 || x >= w || y >= h)
return NULL;
ui->drag_button = button;
ui->dsx = ui->dex = x;
ui->dsy = ui->dey = y;
ui->drag_ok = true;
ui->cdisp = false;
return UI_UPDATE;
}
if ((IS_MOUSE_DRAG(button) || IS_MOUSE_RELEASE(button)) &&
ui->drag_button > 0) {
int xmin, ymin, xmax, ymax;
char *buf;
const char *sep;
int buflen, bufsize, tmplen;
x = FROMCOORD(x);
y = FROMCOORD(y);
if (x < 0 || y < 0 || x >= w || y >= h) {
ui->drag_ok = false;
} else {
/*
* Drags are limited to one row or column. Hence, we
* work out which coordinate is closer to the drag
* start, and move it _to_ the drag start.
*/
if (abs(x - ui->dsx) < abs(y - ui->dsy))
x = ui->dsx;
else
y = ui->dsy;
ui->dex = x;
ui->dey = y;
ui->drag_ok = true;
}
if (IS_MOUSE_DRAG(button))
return UI_UPDATE;
/*
* The drag has been released. Enact it.
*/
if (!ui->drag_ok) {
ui->drag_button = -1;
return UI_UPDATE; /* drag was just cancelled */
}
xmin = min(ui->dsx, ui->dex);
xmax = max(ui->dsx, ui->dex);
ymin = min(ui->dsy, ui->dey);
ymax = max(ui->dsy, ui->dey);
assert(0 <= xmin && xmin <= xmax && xmax < w);
assert(0 <= ymin && ymin <= ymax && ymax < h);
buflen = 0;
bufsize = 256;
buf = snewn(bufsize, char);
sep = "";
for (y = ymin; y <= ymax; y++)
for (x = xmin; x <= xmax; x++) {
int v = drag_xform(ui, x, y, state->grid[y*w+x]);
if (state->grid[y*w+x] != v) {
tmplen = sprintf(tmpbuf, "%s%c%d,%d", sep,
(int)(v == BLANK ? 'B' :
v == TENT ? 'T' : 'N'),
x, y);
sep = ";";
if (buflen + tmplen >= bufsize) {
bufsize = buflen + tmplen + 256;
buf = sresize(buf, bufsize, char);
}
strcpy(buf+buflen, tmpbuf);
buflen += tmplen;
}
}
ui->drag_button = -1; /* drag is terminated */
if (buflen == 0) {
sfree(buf);
return UI_UPDATE; /* drag was terminated */
} else {
buf[buflen] = '\0';
return buf;
}
}
if (IS_CURSOR_MOVE(button)) {
ui->cdisp = true;
if (shift || control) {
int len = 0, i, indices[2];
indices[0] = ui->cx + w * ui->cy;
move_cursor(button, &ui->cx, &ui->cy, w, h, false);
indices[1] = ui->cx + w * ui->cy;
/* NONTENTify all unique traversed eligible squares */
for (i = 0; i <= (indices[0] != indices[1]); ++i)
if (state->grid[indices[i]] == BLANK ||
(control && state->grid[indices[i]] == TENT)) {
len += sprintf(tmpbuf + len, "%sN%d,%d", len ? ";" : "",
indices[i] % w, indices[i] / w);
assert(len < lenof(tmpbuf));
}
tmpbuf[len] = '\0';
if (len) return dupstr(tmpbuf);
} else
move_cursor(button, &ui->cx, &ui->cy, w, h, false);
return UI_UPDATE;
}
if (ui->cdisp) {
char rep = 0;
int v = state->grid[ui->cy*w+ui->cx];
if (v != TREE) {
#ifdef SINGLE_CURSOR_SELECT
if (button == CURSOR_SELECT)
/* SELECT cycles T, N, B */
rep = v == BLANK ? 'T' : v == TENT ? 'N' : 'B';
#else
if (button == CURSOR_SELECT)
rep = v == BLANK ? 'T' : 'B';
else if (button == CURSOR_SELECT2)
rep = v == BLANK ? 'N' : 'B';
else if (button == 'T' || button == 'N' || button == 'B')
rep = (char)button;
#endif
}
if (rep) {
sprintf(tmpbuf, "%c%d,%d", (int)rep, ui->cx, ui->cy);
return dupstr(tmpbuf);
}
} else if (IS_CURSOR_SELECT(button)) {
ui->cdisp = true;
return UI_UPDATE;
}
return NULL;
}
static game_state *execute_move(const game_state *state, const char *move)
{
int w = state->p.w, h = state->p.h;
char c;
int x, y, m, n, i, j;
game_state *ret = dup_game(state);
while (*move) {
c = *move;
if (c == 'S') {
int i;
ret->used_solve = true;
/*
* Set all non-tree squares to NONTENT. The rest of the
* solve move will fill the tents in over the top.
*/
for (i = 0; i < w*h; i++)
if (ret->grid[i] != TREE)
ret->grid[i] = NONTENT;
move++;
} else if (c == 'B' || c == 'T' || c == 'N') {
move++;
if (sscanf(move, "%d,%d%n", &x, &y, &n) != 2 ||
x < 0 || y < 0 || x >= w || y >= h) {
free_game(ret);
return NULL;
}
if (ret->grid[y*w+x] == TREE) {
free_game(ret);
return NULL;
}
ret->grid[y*w+x] = (c == 'B' ? BLANK : c == 'T' ? TENT : NONTENT);
move += n;
} else {
free_game(ret);
return NULL;
}
if (*move == ';')
move++;
else if (*move) {
free_game(ret);
return NULL;
}
}
/*
* Check for completion.
*/
for (i = n = m = 0; i < w*h; i++) {
if (ret->grid[i] == TENT)
n++;
else if (ret->grid[i] == TREE)
m++;
}
if (n == m) {
int *gridids, *adjdata, **adjlists, *adjsizes, *adjptr;
/*
* We have the right number of tents, which is a
* precondition for the game being complete. Now check that
* the numbers add up.
*/
for (i = 0; i < w; i++) {
n = 0;
for (j = 0; j < h; j++)
if (ret->grid[j*w+i] == TENT)
n++;
if (ret->numbers->numbers[i] != n)
goto completion_check_done;
}
for (i = 0; i < h; i++) {
n = 0;
for (j = 0; j < w; j++)
if (ret->grid[i*w+j] == TENT)
n++;
if (ret->numbers->numbers[w+i] != n)
goto completion_check_done;
}
/*
* Also, check that no two tents are adjacent.
*/
for (y = 0; y < h; y++)
for (x = 0; x < w; x++) {
if (x+1 < w &&
ret->grid[y*w+x] == TENT && ret->grid[y*w+x+1] == TENT)
goto completion_check_done;
if (y+1 < h &&
ret->grid[y*w+x] == TENT && ret->grid[(y+1)*w+x] == TENT)
goto completion_check_done;
if (x+1 < w && y+1 < h) {
if (ret->grid[y*w+x] == TENT &&
ret->grid[(y+1)*w+(x+1)] == TENT)
goto completion_check_done;
if (ret->grid[(y+1)*w+x] == TENT &&
ret->grid[y*w+(x+1)] == TENT)
goto completion_check_done;
}
}
/*
* OK; we have the right number of tents, they match the
* numeric clues, and they satisfy the non-adjacency
* criterion. Finally, we need to verify that they can be
* placed in a one-to-one matching with the trees such that
* every tent is orthogonally adjacent to its tree.
*
* This bit is where the hard work comes in: we have to do
* it by finding such a matching using matching.c.
*/
gridids = snewn(w*h, int);
adjdata = snewn(m*4, int);
adjlists = snewn(m, int *);
adjsizes = snewn(m, int);
/* Assign each tent and tree a consecutive vertex id for
* matching(). */
for (i = n = 0; i < w*h; i++) {
if (ret->grid[i] == TENT)
gridids[i] = n++;
}
assert(n == m);
for (i = n = 0; i < w*h; i++) {
if (ret->grid[i] == TREE)
gridids[i] = n++;
}
assert(n == m);
/* Build the vertices' adjacency lists. */
adjptr = adjdata;
for (y = 0; y < h; y++)
for (x = 0; x < w; x++)
if (ret->grid[y*w+x] == TREE) {
int d, treeid = gridids[y*w+x];
adjlists[treeid] = adjptr;
/*
* Here we use the direction enum declared for
* the solver. We make use of the fact that the
* directions are declared in the order
* U,L,R,D, meaning that we go through the four
* neighbours of any square in numerically
* increasing order.
*/
for (d = 1; d < MAXDIR; d++) {
int x2 = x + dx(d), y2 = y + dy(d);
if (x2 >= 0 && x2 < w && y2 >= 0 && y2 < h &&
ret->grid[y2*w+x2] == TENT) {
*adjptr++ = gridids[y2*w+x2];
}
}
adjsizes[treeid] = adjptr - adjlists[treeid];
}
n = matching(m, m, adjlists, adjsizes, NULL, NULL, NULL);
sfree(gridids);
sfree(adjdata);
sfree(adjlists);
sfree(adjsizes);
if (n != m)
goto completion_check_done;
/*
* We haven't managed to fault the grid on any count. Score!
*/
ret->completed = true;
}
completion_check_done:
return ret;
}
/* ----------------------------------------------------------------------
* Drawing routines.
*/
static void game_compute_size(const game_params *params, int tilesize,
int *x, int *y)
{
/* fool the macros */
struct dummy { int tilesize; } dummy, *ds = &dummy;
dummy.tilesize = tilesize;
*x = TLBORDER + BRBORDER + TILESIZE * params->w;
*y = TLBORDER + BRBORDER + TILESIZE * params->h;
}
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_GRASS * 3 + 0] = 0.7F;
ret[COL_GRASS * 3 + 1] = 1.0F;
ret[COL_GRASS * 3 + 2] = 0.5F;
ret[COL_TREETRUNK * 3 + 0] = 0.6F;
ret[COL_TREETRUNK * 3 + 1] = 0.4F;
ret[COL_TREETRUNK * 3 + 2] = 0.0F;
ret[COL_TREELEAF * 3 + 0] = 0.0F;
ret[COL_TREELEAF * 3 + 1] = 0.7F;
ret[COL_TREELEAF * 3 + 2] = 0.0F;
ret[COL_TENT * 3 + 0] = 0.8F;
ret[COL_TENT * 3 + 1] = 0.7F;
ret[COL_TENT * 3 + 2] = 0.0F;
ret[COL_ERROR * 3 + 0] = 1.0F;
ret[COL_ERROR * 3 + 1] = 0.0F;
ret[COL_ERROR * 3 + 2] = 0.0F;
ret[COL_ERRTEXT * 3 + 0] = 1.0F;
ret[COL_ERRTEXT * 3 + 1] = 1.0F;
ret[COL_ERRTEXT * 3 + 2] = 1.0F;
ret[COL_ERRTRUNK * 3 + 0] = 0.6F;
ret[COL_ERRTRUNK * 3 + 1] = 0.0F;
ret[COL_ERRTRUNK * 3 + 2] = 0.0F;
*ncolours = NCOLOURS;
return ret;
}
static game_drawstate *game_new_drawstate(drawing *dr, const game_state *state)
{
int w = state->p.w, h = state->p.h;
struct game_drawstate *ds = snew(struct game_drawstate);
int i;
ds->tilesize = 0;
ds->started = false;
ds->p = state->p; /* structure copy */
ds->drawn = snewn(w*h, int);
for (i = 0; i < w*h; i++)
ds->drawn[i] = MAGIC;
ds->numbersdrawn = snewn(w+h, int);
for (i = 0; i < w+h; i++)
ds->numbersdrawn[i] = 2;
ds->cx = ds->cy = -1;
return ds;
}
static void game_free_drawstate(drawing *dr, game_drawstate *ds)
{
sfree(ds->drawn);
sfree(ds->numbersdrawn);
sfree(ds);
}
enum {
ERR_ADJ_TOPLEFT = 4,
ERR_ADJ_TOP,
ERR_ADJ_TOPRIGHT,
ERR_ADJ_LEFT,
ERR_ADJ_RIGHT,
ERR_ADJ_BOTLEFT,
ERR_ADJ_BOT,
ERR_ADJ_BOTRIGHT,
ERR_OVERCOMMITTED
};
static int *find_errors(const game_state *state, char *grid)
{
int w = state->p.w, h = state->p.h;
int *ret = snewn(w*h + w + h, int);
int *tmp = snewn(w*h*2, int), *dsf = tmp + w*h;
int x, y;
/*
* This function goes through a grid and works out where to
* highlight play errors in red. The aim is that it should
* produce at least one error highlight for any complete grid
* (or complete piece of grid) violating a puzzle constraint, so
* that a grid containing no BLANK squares is either a win or is
* marked up in some way that indicates why not.
*
* So it's easy enough to highlight errors in the numeric clues
* - just light up any row or column number which is not
* fulfilled - and it's just as easy to highlight adjacent
* tents. The difficult bit is highlighting failures in the
* tent/tree matching criterion.
*
* A natural approach would seem to be to apply the matching.c
* algorithm to find the tent/tree matching; if this fails, it
* could be made to produce as a by-product some set of trees
* which have too few tents between them (or vice versa). However,
* it's bad for localising errors, because it's not easy to make
* the algorithm narrow down to the _smallest_ such set of trees:
* if trees A and B have only one tent between them, for instance,
* it might perfectly well highlight not only A and B but also
* trees C and D which are correctly matched on the far side of
* the grid, on the grounds that those four trees between them
* have only three tents.
*
* Also, that approach fares badly when you introduce the
* additional requirement that incomplete grids should have
* errors highlighted only when they can be proved to be errors
* - so that trees should not be marked as having too few tents
* if there are enough BLANK squares remaining around them that
* could be turned into the missing tents (to do so would be
* patronising, since the overwhelming likelihood is not that
* the player has forgotten to put a tree there but that they
* have merely not put one there _yet_). However, tents with too
* few trees can be marked immediately, since those are
* definitely player error.
*
* So I adopt an alternative approach, which is to consider the
* bipartite adjacency graph between trees and tents
* ('bipartite' in the sense that for these purposes I
* deliberately ignore two adjacent trees or two adjacent
* tents), divide that graph up into its connected components
* using a dsf, and look for components which contain different
* numbers of trees and tents. This allows me to highlight
* groups of tents with too few trees between them immediately,
* and then in order to find groups of trees with too few tents
* I redo the same process but counting BLANKs as potential
* tents (so that the only trees highlighted are those
* surrounded by enough NONTENTs to make it impossible to give
* them enough tents).
*
* However, this technique is incomplete: it is not a sufficient
* condition for the existence of a perfect matching that every
* connected component of the graph has the same number of tents
* and trees. An example of a graph which satisfies the latter
* condition but still has no perfect matching is
*
* A B C
* | / ,/|
* | / ,'/ |
* | / ,' / |
* |/,' / |
* 1 2 3
*
* which can be realised in Tents as
*
* B
* A 1 C 2
* 3
*
* The matching-error highlighter described above will not mark
* this construction as erroneous. However, something else will:
* the three tents in the above diagram (let us suppose A,B,C
* are the tents, though it doesn't matter which) contain two
* diagonally adjacent pairs. So there will be _an_ error
* highlighted for the above layout, even though not all types
* of error will be highlighted.
*
* And in fact we can prove that this will always be the case:
* that the shortcomings of the matching-error highlighter will
* always be made up for by the easy tent adjacency highlighter.
*
* Lemma: Let G be a bipartite graph between n trees and n
* tents, which is connected, and in which no tree has degree
* more than two (but a tent may). Then G has a perfect matching.
*
* (Note: in the statement and proof of the Lemma I will
* consistently use 'tree' to indicate a type of graph vertex as
* opposed to a tent, and not to indicate a tree in the graph-
* theoretic sense.)
*
* Proof:
*
* If we can find a tent of degree 1 joined to a tree of degree
* 2, then any perfect matching must pair that tent with that
* tree. Hence, we can remove both, leaving a smaller graph G'
* which still satisfies all the conditions of the Lemma, and
* which has a perfect matching iff G does.
*
* So, wlog, we may assume G contains no tent of degree 1 joined
* to a tree of degree 2; if it does, we can reduce it as above.
*
* If G has no tent of degree 1 at all, then every tent has
* degree at least two, so there are at least 2n edges in the
* graph. But every tree has degree at most two, so there are at
* most 2n edges. Hence there must be exactly 2n edges, so every
* tree and every tent must have degree exactly two, which means
* that the whole graph consists of a single loop (by
* connectedness), and therefore certainly has a perfect
* matching.
*
* Alternatively, if G does have a tent of degree 1 but it is
* not connected to a tree of degree 2, then the tree it is
* connected to must have degree 1 - and, by connectedness, that
* must mean that that tent and that tree between them form the
* entire graph. This trivial graph has a trivial perfect
* matching. []
*
* That proves the lemma. Hence, in any case where the matching-
* error highlighter fails to highlight an erroneous component
* (because it has the same number of tents as trees, but they
* cannot be matched up), the above lemma tells us that there
* must be a tree with degree more than 2, i.e. a tree
* orthogonally adjacent to at least three tents. But in that
* case, there must be some pair of those three tents which are
* diagonally adjacent to each other, so the tent-adjacency
* highlighter will necessarily show an error. So any filled
* layout in Tents which is not a correct solution to the puzzle
* must have _some_ error highlighted by the subroutine below.
*
* (Of course it would be nicer if we could highlight all
* errors: in the above example layout, we would like to
* highlight tents A,B as having too few trees between them, and
* trees 2,3 as having too few tents, in addition to marking the
* adjacency problems. But I can't immediately think of any way
* to find the smallest sets of such tents and trees without an
* O(2^N) loop over all subsets of a given component.)
*/
/*
* ret[0] through to ret[w*h-1] give error markers for the grid
* squares. After that, ret[w*h] to ret[w*h+w-1] give error
* markers for the column numbers, and ret[w*h+w] to
* ret[w*h+w+h-1] for the row numbers.
*/
/*
* Spot tent-adjacency violations.
*/
for (x = 0; x < w*h; x++)
ret[x] = 0;
for (y = 0; y < h; y++) {
for (x = 0; x < w; x++) {
if (y+1 < h && x+1 < w &&
((grid[y*w+x] == TENT &&
grid[(y+1)*w+(x+1)] == TENT) ||
(grid[(y+1)*w+x] == TENT &&
grid[y*w+(x+1)] == TENT))) {
ret[y*w+x] |= 1 << ERR_ADJ_BOTRIGHT;
ret[(y+1)*w+x] |= 1 << ERR_ADJ_TOPRIGHT;
ret[y*w+(x+1)] |= 1 << ERR_ADJ_BOTLEFT;
ret[(y+1)*w+(x+1)] |= 1 << ERR_ADJ_TOPLEFT;
}
if (y+1 < h &&
grid[y*w+x] == TENT &&
grid[(y+1)*w+x] == TENT) {
ret[y*w+x] |= 1 << ERR_ADJ_BOT;
ret[(y+1)*w+x] |= 1 << ERR_ADJ_TOP;
}
if (x+1 < w &&
grid[y*w+x] == TENT &&
grid[y*w+(x+1)] == TENT) {
ret[y*w+x] |= 1 << ERR_ADJ_RIGHT;
ret[y*w+(x+1)] |= 1 << ERR_ADJ_LEFT;
}
}
}
/*
* Spot numeric clue violations.
*/
for (x = 0; x < w; x++) {
int tents = 0, maybetents = 0;
for (y = 0; y < h; y++) {
if (grid[y*w+x] == TENT)
tents++;
else if (grid[y*w+x] == BLANK)
maybetents++;
}
ret[w*h+x] = (tents > state->numbers->numbers[x] ||
tents + maybetents < state->numbers->numbers[x]);
}
for (y = 0; y < h; y++) {
int tents = 0, maybetents = 0;
for (x = 0; x < w; x++) {
if (grid[y*w+x] == TENT)
tents++;
else if (grid[y*w+x] == BLANK)
maybetents++;
}
ret[w*h+w+y] = (tents > state->numbers->numbers[w+y] ||
tents + maybetents < state->numbers->numbers[w+y]);
}
/*
* Identify groups of tents with too few trees between them,
* which we do by constructing the connected components of the
* bipartite adjacency graph between tents and trees
* ('bipartite' in the sense that we deliberately ignore
* adjacency between tents or between trees), and highlighting
* all the tents in any component which has a smaller tree
* count.
*/
dsf_init(dsf, w*h);
/* Construct the equivalence classes. */
for (y = 0; y < h; y++) {
for (x = 0; x < w-1; x++) {
if ((grid[y*w+x] == TREE && grid[y*w+x+1] == TENT) ||
(grid[y*w+x] == TENT && grid[y*w+x+1] == TREE))
dsf_merge(dsf, y*w+x, y*w+x+1);
}
}
for (y = 0; y < h-1; y++) {
for (x = 0; x < w; x++) {
if ((grid[y*w+x] == TREE && grid[(y+1)*w+x] == TENT) ||
(grid[y*w+x] == TENT && grid[(y+1)*w+x] == TREE))
dsf_merge(dsf, y*w+x, (y+1)*w+x);
}
}
/* Count up the tent/tree difference in each one. */
for (x = 0; x < w*h; x++)
tmp[x] = 0;
for (x = 0; x < w*h; x++) {
y = dsf_canonify(dsf, x);
if (grid[x] == TREE)
tmp[y]++;
else if (grid[x] == TENT)
tmp[y]--;
}
/* And highlight any tent belonging to an equivalence class with
* a score less than zero. */
for (x = 0; x < w*h; x++) {
y = dsf_canonify(dsf, x);
if (grid[x] == TENT && tmp[y] < 0)
ret[x] |= 1 << ERR_OVERCOMMITTED;
}
/*
* Identify groups of trees with too few tents between them.
* This is done similarly, except that we now count BLANK as
* equivalent to TENT, i.e. we only highlight such trees when
* the user hasn't even left _room_ to provide tents for them
* all. (Otherwise, we'd highlight all trees red right at the
* start of the game, before the user had done anything wrong!)
*/
#define TENT(x) ((x)==TENT || (x)==BLANK)
dsf_init(dsf, w*h);
/* Construct the equivalence classes. */
for (y = 0; y < h; y++) {
for (x = 0; x < w-1; x++) {
if ((grid[y*w+x] == TREE && TENT(grid[y*w+x+1])) ||
(TENT(grid[y*w+x]) && grid[y*w+x+1] == TREE))
dsf_merge(dsf, y*w+x, y*w+x+1);
}
}
for (y = 0; y < h-1; y++) {
for (x = 0; x < w; x++) {
if ((grid[y*w+x] == TREE && TENT(grid[(y+1)*w+x])) ||
(TENT(grid[y*w+x]) && grid[(y+1)*w+x] == TREE))
dsf_merge(dsf, y*w+x, (y+1)*w+x);
}
}
/* Count up the tent/tree difference in each one. */
for (x = 0; x < w*h; x++)
tmp[x] = 0;
for (x = 0; x < w*h; x++) {
y = dsf_canonify(dsf, x);
if (grid[x] == TREE)
tmp[y]++;
else if (TENT(grid[x]))
tmp[y]--;
}
/* And highlight any tree belonging to an equivalence class with
* a score more than zero. */
for (x = 0; x < w*h; x++) {
y = dsf_canonify(dsf, x);
if (grid[x] == TREE && tmp[y] > 0)
ret[x] |= 1 << ERR_OVERCOMMITTED;
}
#undef TENT
sfree(tmp);
return ret;
}
static void draw_err_adj(drawing *dr, game_drawstate *ds, int x, int y)
{
int coords[8];
int yext, xext;
/*
* Draw a diamond.
*/
coords[0] = x - TILESIZE*2/5;
coords[1] = y;
coords[2] = x;
coords[3] = y - TILESIZE*2/5;
coords[4] = x + TILESIZE*2/5;
coords[5] = y;
coords[6] = x;
coords[7] = y + TILESIZE*2/5;
draw_polygon(dr, coords, 4, COL_ERROR, COL_GRID);
/*
* Draw an exclamation mark in the diamond. This turns out to
* look unpleasantly off-centre if done via draw_text, so I do
* it by hand on the basis that exclamation marks aren't that
* difficult to draw...
*/
xext = TILESIZE/16;
yext = TILESIZE*2/5 - (xext*2+2);
draw_rect(dr, x-xext, y-yext, xext*2+1, yext*2+1 - (xext*3),
COL_ERRTEXT);
draw_rect(dr, x-xext, y+yext-xext*2+1, xext*2+1, xext*2, COL_ERRTEXT);
}
static void draw_tile(drawing *dr, game_drawstate *ds,
int x, int y, int v, bool cur, bool printing)
{
int err;
int tx = COORD(x), ty = COORD(y);
int cx = tx + TILESIZE/2, cy = ty + TILESIZE/2;
err = v & ~15;
v &= 15;
clip(dr, tx, ty, TILESIZE, TILESIZE);
if (!printing) {
draw_rect(dr, tx, ty, TILESIZE, TILESIZE, COL_GRID);
draw_rect(dr, tx+1, ty+1, TILESIZE-1, TILESIZE-1,
(v == BLANK ? COL_BACKGROUND : COL_GRASS));
}
if (v == TREE) {
int i;
(printing ? draw_rect_outline : draw_rect)
(dr, cx-TILESIZE/15, ty+TILESIZE*3/10,
2*(TILESIZE/15)+1, (TILESIZE*9/10 - TILESIZE*3/10),
(err & (1<<ERR_OVERCOMMITTED) ? COL_ERRTRUNK : COL_TREETRUNK));
for (i = 0; i < (printing ? 2 : 1); i++) {
int col = (i == 1 ? COL_BACKGROUND :
(err & (1<<ERR_OVERCOMMITTED) ? COL_ERROR :
COL_TREELEAF));
int sub = i * (TILESIZE/32);
draw_circle(dr, cx, ty+TILESIZE*4/10, TILESIZE/4 - sub,
col, col);
draw_circle(dr, cx+TILESIZE/5, ty+TILESIZE/4, TILESIZE/8 - sub,
col, col);
draw_circle(dr, cx-TILESIZE/5, ty+TILESIZE/4, TILESIZE/8 - sub,
col, col);
draw_circle(dr, cx+TILESIZE/4, ty+TILESIZE*6/13, TILESIZE/8 - sub,
col, col);
draw_circle(dr, cx-TILESIZE/4, ty+TILESIZE*6/13, TILESIZE/8 - sub,
col, col);
}
} else if (v == TENT) {
int coords[6];
int col;
coords[0] = cx - TILESIZE/3;
coords[1] = cy + TILESIZE/3;
coords[2] = cx + TILESIZE/3;
coords[3] = cy + TILESIZE/3;
coords[4] = cx;
coords[5] = cy - TILESIZE/3;
col = (err & (1<<ERR_OVERCOMMITTED) ? COL_ERROR : COL_TENT);
draw_polygon(dr, coords, 3, (printing ? -1 : col), col);
}
if (err & (1 << ERR_ADJ_TOPLEFT))
draw_err_adj(dr, ds, tx, ty);
if (err & (1 << ERR_ADJ_TOP))
draw_err_adj(dr, ds, tx+TILESIZE/2, ty);
if (err & (1 << ERR_ADJ_TOPRIGHT))
draw_err_adj(dr, ds, tx+TILESIZE, ty);
if (err & (1 << ERR_ADJ_LEFT))
draw_err_adj(dr, ds, tx, ty+TILESIZE/2);
if (err & (1 << ERR_ADJ_RIGHT))
draw_err_adj(dr, ds, tx+TILESIZE, ty+TILESIZE/2);
if (err & (1 << ERR_ADJ_BOTLEFT))
draw_err_adj(dr, ds, tx, ty+TILESIZE);
if (err & (1 << ERR_ADJ_BOT))
draw_err_adj(dr, ds, tx+TILESIZE/2, ty+TILESIZE);
if (err & (1 << ERR_ADJ_BOTRIGHT))
draw_err_adj(dr, ds, tx+TILESIZE, ty+TILESIZE);
if (cur) {
int coff = TILESIZE/8;
draw_rect_outline(dr, tx + coff, ty + coff,
TILESIZE - coff*2 + 1, TILESIZE - coff*2 + 1,
COL_GRID);
}
unclip(dr);
draw_update(dr, tx+1, ty+1, TILESIZE-1, TILESIZE-1);
}
/*
* Internal redraw function, used for printing as well as drawing.
*/
static void int_redraw(drawing *dr, game_drawstate *ds,
const game_state *oldstate, const game_state *state,
int dir, const game_ui *ui,
float animtime, float flashtime, bool printing)
{
int w = state->p.w, h = state->p.h;
int x, y;
bool flashing;
int cx = -1, cy = -1;
bool cmoved = false;
char *tmpgrid;
int *errors;
if (ui) {
if (ui->cdisp) { cx = ui->cx; cy = ui->cy; }
if (cx != ds->cx || cy != ds->cy) cmoved = true;
}
if (printing || !ds->started) {
if (printing)
print_line_width(dr, TILESIZE/64);
/*
* Draw the grid.
*/
for (y = 0; y <= h; y++)
draw_line(dr, COORD(0), COORD(y), COORD(w), COORD(y), COL_GRID);
for (x = 0; x <= w; x++)
draw_line(dr, COORD(x), COORD(0), COORD(x), COORD(h), COL_GRID);
}
if (flashtime > 0)
flashing = (int)(flashtime * 3 / FLASH_TIME) != 1;
else
flashing = false;
/*
* Find errors. For this we use _part_ of the information from a
* currently active drag: we transform dsx,dsy but not anything
* else. (This seems to strike a good compromise between having
* the error highlights respond instantly to single clicks, but
* not giving constant feedback during a right-drag.)
*/
if (ui && ui->drag_button >= 0) {
tmpgrid = snewn(w*h, char);
memcpy(tmpgrid, state->grid, w*h);
tmpgrid[ui->dsy * w + ui->dsx] =
drag_xform(ui, ui->dsx, ui->dsy, tmpgrid[ui->dsy * w + ui->dsx]);
errors = find_errors(state, tmpgrid);
sfree(tmpgrid);
} else {
errors = find_errors(state, state->grid);
}
/*
* Draw the grid.
*/
for (y = 0; y < h; y++) {
for (x = 0; x < w; x++) {
int v = state->grid[y*w+x];
bool credraw = false;
/*
* We deliberately do not take drag_ok into account
* here, because user feedback suggests that it's
* marginally nicer not to have the drag effects
* flickering on and off disconcertingly.
*/
if (ui && ui->drag_button >= 0)
v = drag_xform(ui, x, y, v);
if (flashing && (v == TREE || v == TENT))
v = NONTENT;
if (cmoved) {
if ((x == cx && y == cy) ||
(x == ds->cx && y == ds->cy)) credraw = true;
}
v |= errors[y*w+x];
if (printing || ds->drawn[y*w+x] != v || credraw) {
draw_tile(dr, ds, x, y, v, (x == cx && y == cy), printing);
if (!printing)
ds->drawn[y*w+x] = v;
}
}
}
/*
* Draw (or redraw, if their error-highlighted state has
* changed) the numbers.
*/
for (x = 0; x < w; x++) {
if (printing || ds->numbersdrawn[x] != errors[w*h+x]) {
char buf[80];
draw_rect(dr, COORD(x), COORD(h)+1, TILESIZE, BRBORDER-1,
COL_BACKGROUND);
sprintf(buf, "%d", state->numbers->numbers[x]);
draw_text(dr, COORD(x) + TILESIZE/2, COORD(h+1),
FONT_VARIABLE, TILESIZE/2, ALIGN_HCENTRE|ALIGN_VNORMAL,
(errors[w*h+x] ? COL_ERROR : COL_GRID), buf);
draw_update(dr, COORD(x), COORD(h)+1, TILESIZE, BRBORDER-1);
if (!printing)
ds->numbersdrawn[x] = errors[w*h+x];
}
}
for (y = 0; y < h; y++) {
if (printing || ds->numbersdrawn[w+y] != errors[w*h+w+y]) {
char buf[80];
draw_rect(dr, COORD(w)+1, COORD(y), BRBORDER-1, TILESIZE,
COL_BACKGROUND);
sprintf(buf, "%d", state->numbers->numbers[w+y]);
draw_text(dr, COORD(w+1), COORD(y) + TILESIZE/2,
FONT_VARIABLE, TILESIZE/2, ALIGN_HRIGHT|ALIGN_VCENTRE,
(errors[w*h+w+y] ? COL_ERROR : COL_GRID), buf);
draw_update(dr, COORD(w)+1, COORD(y), BRBORDER-1, TILESIZE);
if (!printing)
ds->numbersdrawn[w+y] = errors[w*h+w+y];
}
}
if (cmoved) {
ds->cx = cx;
ds->cy = cy;
}
sfree(errors);
}
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_redraw(dr, ds, oldstate, state, dir, ui, animtime, flashtime, false);
}
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->used_solve && !newstate->used_solve)
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->cdisp) {
*x = COORD(ui->cx);
*y = COORD(ui->cy);
*w = *h = TILESIZE;
}
}
static int game_status(const game_state *state)
{
return state->completed ? +1 : 0;
}
static void game_print_size(const game_params *params, float *x, float *y)
{
int pw, ph;
/*
* I'll use 6mm squares by default.
*/
game_compute_size(params, 600, &pw, &ph);
*x = pw / 100.0F;
*y = ph / 100.0F;
}
static void game_print(drawing *dr, const game_state *state, int tilesize)
{
int c;
/* Ick: fake up `ds->tilesize' for macro expansion purposes */
game_drawstate ads, *ds = &ads;
game_set_size(dr, ds, NULL, tilesize);
c = print_mono_colour(dr, 1); assert(c == COL_BACKGROUND);
c = print_mono_colour(dr, 0); assert(c == COL_GRID);
c = print_mono_colour(dr, 1); assert(c == COL_GRASS);
c = print_mono_colour(dr, 0); assert(c == COL_TREETRUNK);
c = print_mono_colour(dr, 0); assert(c == COL_TREELEAF);
c = print_mono_colour(dr, 0); assert(c == COL_TENT);
int_redraw(dr, ds, NULL, state, +1, NULL, 0.0F, 0.0F, true);
}
#ifdef COMBINED
#define thegame tents
#endif
const struct game thegame = {
"Tents", "games.tents", "tents",
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,
true, game_can_format_as_text_now, game_text_format,
new_ui,
free_ui,
encode_ui,
decode_ui,
NULL, /* game_request_keys */
game_changed_state,
current_key_label,
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, NULL, /* timing_state */
REQUIRE_RBUTTON, /* flags */
};
#ifdef STANDALONE_SOLVER
#include <stdarg.h>
int main(int argc, char **argv)
{
game_params *p;
game_state *s, *s2;
char *id = NULL, *desc;
const char *err;
bool grade = false;
int ret, diff;
bool really_verbose = false;
struct solver_scratch *sc;
while (--argc > 0) {
char *p = *++argv;
if (!strcmp(p, "-v")) {
really_verbose = 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);
s2 = new_game(NULL, p, desc);
sc = new_scratch(p->w, p->h);
/*
* When solving an Easy puzzle, we don't want to bother the
* user with Hard-level deductions. For this reason, we grade
* the puzzle internally before doing anything else.
*/
ret = -1; /* placate optimiser */
for (diff = 0; diff < DIFFCOUNT; diff++) {
ret = tents_solve(p->w, p->h, s->grid, s->numbers->numbers,
s2->grid, sc, diff);
if (ret < 2)
break;
}
if (diff == DIFFCOUNT) {
if (grade)
printf("Difficulty rating: too hard to solve internally\n");
else
printf("Unable to find a unique solution\n");
} else {
if (grade) {
if (ret == 0)
printf("Difficulty rating: impossible (no solution exists)\n");
else if (ret == 1)
printf("Difficulty rating: %s\n", tents_diffnames[diff]);
} else {
verbose = really_verbose;
ret = tents_solve(p->w, p->h, s->grid, s->numbers->numbers,
s2->grid, sc, diff);
if (ret == 0)
printf("Puzzle is inconsistent\n");
else
fputs(game_text_format(s2), stdout);
}
}
return 0;
}
#endif
/* vim: set shiftwidth=4 tabstop=8: */