
Mutilated chessboard problem - ColinWright
https://en.wikipedia.org/wiki/Mutilated_chessboard_problem
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Grustaf
The problem where one square of each is colour is removed is much more
interesting. It's not obvious to me why it should always be possible to solve
that problem, regardless of which squares are removed, but apparently it is.

I wonder if it can be proved without resorting to looking at classes of cases
(removed squares right next to each other, removed squares with two squares
between them, etc)?

~~~
ColinWright
Yes - there is a simple, clear, elegant way to see that removing one square of
each colour leaves the board coverable. It's even given in the article.

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probably_wrong
I got the correct answer immediately because I've seen a similar problem
before in HN, although with a Tetris lamp instead.

Link: [http://jackm.co.uk/the-simple-proof-of-the-tetris-
lamp/](http://jackm.co.uk/the-simple-proof-of-the-tetris-lamp/)

Discussion:
[https://news.ycombinator.com/item?id=8870479](https://news.ycombinator.com/item?id=8870479)

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eli173
I was shown a similar problem (I believe it was originally a Putnam question)
which asked if you could cover a chessboard missing one corner with L-shaped
triominos, and then asked to prove whether or not you could always cover a
2nx2n board with one corner missing. Neat problem

~~~
wolfgke
Indeed (spoiler!) an easy, but nice induction proof, which easily can be
turned into an algorithm.

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glxc
"here's another interview brain teaser for sociopathic managers"

~~~
wolfgke
Rather a simple exercise for a first-semester exercise sheet.

~~~
danbruc
Only if you already know the trick to attack this class of problems.

~~~
wolfgke
No, it's really that easy: Clearly the complete coverings with dominoes biject
to matchings of the graph with vertices = squares and edges defined by 2-sets
of squares sharing an edge on the chessboard. Clearly this graph is bipartite,
but the two elements of the bipartision are of different cardinality. Thus no
perfect matching can exist.

