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| author | Simon Tatham <anakin@pobox.com> | 2020-02-26 06:07:18 +0000 |
|---|---|---|
| committer | Simon Tatham <anakin@pobox.com> | 2020-02-26 06:32:35 +0000 |
| commit | 4f2f8a9d173f34097a1636a3205ca0d50a39efee (patch) | |
| tree | a91d2871b0e9581d55b553f8e6a33d73cb84db17 /pattern.R | |
| parent | b3098efbc489be685ff644cde8dd6844f0198479 (diff) | |
| download | puzzles-4f2f8a9d173f34097a1636a3205ca0d50a39efee.zip puzzles-4f2f8a9d173f34097a1636a3205ca0d50a39efee.tar.gz puzzles-4f2f8a9d173f34097a1636a3205ca0d50a39efee.tar.bz2 puzzles-4f2f8a9d173f34097a1636a3205ca0d50a39efee.tar.xz | |
Tracks: new neighbour-based deduction.
This is a deduction I've been using in my own head for years: if you
only have one remaining filled square to put in a row, then it can't
be any square that has two adjacent edges blocked, because if that
square contains anything at all then it would have to be a corner
piece, and a corner piece forces the square next to it to be filled as
well.
I ran across a puzzle today that this implementation couldn't solve,
but I solved it fine by hand and found the deduction I was using that
wasn't implemented here. Now it is.
Diffstat (limited to 'pattern.R')
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