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Fix loophole in Palisade completion checker.

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A user pointed out that if you construct a 'solution' in which no clue
square has too _many_ borders but at least one has too few, and then
bring those clues up to their count by adding extra stray border lines
_inside_ a connected component (avoiding actually dividing any
component completely into two), then the game checker treats that as
solved for victory-flash purposes, on the grounds that (a) the grid is
divided into components of the right size and (b) all clues are
satisfied.

A small example is 4x4n4:22a2b2c33, with the non-solution of dividing
the grid into four 2x2 square blocks and then adding a spurious extra
edge between the two 3 clues. The old Palisade completion check would
flash for victory _at the same time_ as highlighting the spurious edge
in COL_ERROR.

Fixed by enforcing in is_solved() that every border line must separate
two distinct connected components.
This commit is contained in:
Simon Tatham
2015-11-03 06:45:49 +00:00
parent d60e348aae
commit 13b67b9d09
1 changed files with 31 additions and 1 deletions
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32
palisade.c
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@ -511,10 +511,20 @@ static void dfs_dsf(int i, int w, borderflag *border, int *dsf, int black)
static int is_solved(const game_params *params, clue *clues,
borderflag *border)
{
int wh = params->w * params->h, k = params->k, *dsf = snew_dsf(wh), i;
int w = params->w, h = params->h, wh = w*h, k = params->k;
int i, x, y;
int *dsf = snew_dsf(wh);
assert (dsf[0] == UNVISITED); /* check: UNVISITED and dsf.c match up */
/*
* A game is solved if:
*
* - the borders drawn on the grid divide it into connected
* components such that every square is in a component of the
* correct size
* - the borders also satisfy the clue set
*/
for (i = 0; i < wh; ++i) {
if (dsf[i] == UNVISITED) dfs_dsf(i, params->w, border, dsf, TRUE);
if (dsf_size(dsf, i) != k) goto error;
@ -522,6 +532,26 @@ static int is_solved(const game_params *params, clue *clues,
if (clues[i] != bitcount[border[i] & BORDER_MASK]) goto error;
}
/*
* ... and thirdly:
*
* - there are no *stray* borders, in that every border is
* actually part of the division between two components.
* Otherwise you could cheat by finding a subdivision which did
* not *exceed* any clue square's counter, and then adding a
* few extra edges.
*/
for (y = 0; y < h; y++) {
for (x = 0; x < w; x++) {
if (x+1 < w && (border[y*w+x] & BORDER_R) &&
dsf_canonify(dsf, y*w+x) == dsf_canonify(dsf, y*w+(x+1)))
goto error;
if (y+1 < h && (border[y*w+x] & BORDER_D) &&
dsf_canonify(dsf, y*w+x) == dsf_canonify(dsf, (y+1)*w+x))
goto error;
}
}
sfree(dsf);
return TRUE;