+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 maxflow
+ * algorithm to find the tent/tree matching; if this fails, it
+ * must necessarily terminate with a min-cut which can be
+ * reinterpreted as 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);
+}
+