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Tracks: add reverse neighbour deduction in hard mode.

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This is the contrapositive of the deduction introduced in the previous
commit. Previously I said: a square A can have some edges blocked in
such a way that you know it can't be filled without a particular one
of its neighbours B also being filled. And then, if you know the row
containing A and B only has one filled square left to find, then it
can't be A.

This commit adds the obvious followup: if you know the row only has
one _empty_ square left, then for the same reason, it can't be B!

I'm putting this in at the new Hard difficulty, mostly out of
guesswork rather than rigorous play-testing, because I don't remember
ever having _observed_ myself making this deduction in the past. I'm
open to changing the settings if someone has a good argument for it.
This commit is contained in:
Simon Tatham
2020-02-26 06:26:36 +00:00
parent d724e13666
commit 5e9dc42e54
1 changed files with 47 additions and 19 deletions
Download Patch File Download Diff File Expand all files Collapse all files

66
tracks.c
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@ -1200,21 +1200,27 @@ static int solve_check_loose_ends(game_state *state)
return did;
}
static bool solve_check_neighbours_count(
game_state *state, int start, int step, int n, int clueindex)
static void solve_check_neighbours_count(
game_state *state, int start, int step, int n, int clueindex,
bool *onefill, bool *oneempty)
{
int to_fill = state->numbers->numbers[clueindex];
int to_empty = n - to_fill;
int i;
for (i = 0; i < n; i++) {
int p = start + i*step;
if (state->sflags[p] & S_TRACK)
to_fill--;
if (state->sflags[p] & S_NOTRACK)
to_empty--;
}
return (to_fill == 1);
*onefill = (to_fill == 1);
*oneempty = (to_empty == 1);
}
static int solve_check_neighbours_try(game_state *state, int x, int y,
int X, int Y, unsigned dir,
int X, int Y, bool onefill,
bool oneempty, unsigned dir,
const char *what)
{
int w = state->p.w, p = y*w+x, P = Y*w+X;
@ -1230,7 +1236,8 @@ static int solve_check_neighbours_try(game_state *state, int x, int y,
* row/column says that there's only one track square left to
* place, it can't be p, because P would have to be one too,
* violating the clue. So in that situation we can mark p as
* unfilled.
* unfilled. Conversely, if there's only one _non_-track square
* left to place, it can't be P, so we can mark P as filled.
*/
if ((state->sflags[p] | state->sflags[P]) & (S_TRACK | S_NOTRACK))
@ -1241,36 +1248,57 @@ static int solve_check_neighbours_try(game_state *state, int x, int y,
if (possible_exits_except_dir >= 2)
return 0; /* square p need not connect to P, even if it is filled */
/* OK, now we know that if p is filled, P must be filled too.
* But at most one of them can be filled, so it can't be p. */
state->sflags[p] |= S_NOTRACK;
solverdebug(("square (%d,%d) -> NOTRACK: otherwise, that and (%d,%d) "
"would make too many TRACK in %s", x, y, X, Y, what));
return 1;
/* OK, now we know that if p is filled, P must be filled too. */
int did = 0;
if (onefill) {
/* But at most one of them can be filled, so it can't be p. */
state->sflags[p] |= S_NOTRACK;
solverdebug(("square (%d,%d) -> NOTRACK: otherwise, that and (%d,%d) "
"would make too many TRACK in %s", x, y, X, Y, what));
did++;
}
if (oneempty) {
/* Alternatively, at least one of them _must_ be filled, so P
* must be. */
state->sflags[P] |= S_TRACK;
solverdebug(("square (%d,%d) -> TRACK: otherwise, that and (%d,%d) "
"would make too many NOTRACK in %s", X, Y, x, y, what));
did++;
}
return did;
}
static int solve_check_neighbours(game_state *state)
static int solve_check_neighbours(game_state *state, bool both_ways)
{
int w = state->p.w, h = state->p.h, x, y, did = 0;
bool onefill, oneempty;
for (x = 0; x < w; x++) {
if (!solve_check_neighbours_count(state, x, w, h, x))
solve_check_neighbours_count(state, x, w, h, x, &onefill, &oneempty);
if (!both_ways)
oneempty = false; /* disable the harder version of the deduction */
if (!onefill && !oneempty)
continue;
for (y = 0; y+1 < h; y++) {
did += solve_check_neighbours_try(state, x, y, x, y+1,
D, "column");
onefill, oneempty, D, "column");
did += solve_check_neighbours_try(state, x, y+1, x, y,
U, "column");
onefill, oneempty, U, "column");
}
}
for (y = 0; y < h; y++) {
if (!solve_check_neighbours_count(state, y*w, 1, w, w+y))
solve_check_neighbours_count(state, y*w, 1, w, w+y,
&onefill, &oneempty);
if (!both_ways)
oneempty = false; /* disable the harder version of the deduction */
if (!onefill && !oneempty)
continue;
for (x = 0; x+1 < w; x++) {
did += solve_check_neighbours_try(state, x, y, x+1, y,
R, "row");
onefill, oneempty, R, "row");
did += solve_check_neighbours_try(state, x+1, y, x, y,
L, "row");
onefill, oneempty, L, "row");
}
}
return did;
@ -1556,7 +1584,7 @@ static int tracks_solve(game_state *state, int diff, int *max_diff_out)
TRY(DIFF_TRICKY, solve_check_single(state));
TRY(DIFF_TRICKY, solve_check_loose_ends(state));
TRY(DIFF_TRICKY, solve_check_neighbours(state));
TRY(DIFF_TRICKY, solve_check_neighbours(state, false));
TRY(DIFF_HARD, solve_check_neighbours(state, true));
TRY(DIFF_HARD, solve_check_bridge_parity(state, sc));