Domino Puzzle

I want to remove one of the squares from a chess board, and tile the rest of the board with 1x3 dominos.

For which squares on the board is it possible to do this?

How many different ways can you place 21 3*1 dominoes on a 64 square board?

Same question, reworded. Solution in progress.

(all dominoes are identical)

Never mind, I was overthinking it. I get only four squares - c3, c6, f3, and f6.

That's different, I am asking the much more simple question of which square(s) can be the one left out.

Prove it!

It is actually not different - if you have 21 tiles and use them all, you will cover 63 out of 64 squares, leaving one blank.

I did it out on paper - there is only "one" pattern that works, but you can rotate it three times, so you have the four squares.

You have the right answer, and I believe you when you say you found a pattern that works, but to prove that it is the right answer you have to prove that it doesn't work for any other square.

It's very difficult to prove that something doesn't exist; I'm not really sure how to do that. There is probably a way to reduce 21 tiles on 64 squares to a math problem and prove by induction or so, but I don't know how to do it.

I won't give any hints beyond that the proof is elegant.

this is one of the more chess-related posts on the forums

It's at least as chess related as the squares on the chess board thread, so I don't think it belongs in off topic.

I doubt anyone is going to see this anymore so here is the solution:

If you colour the board like this:

Then any domino will cover one red, one blue , and one green square.

Since there are 21 blues, 22 reds, and 21 greens, only a red square can be left out at the end.

However, if you colour similarly in the opposite diagonal direction,

Then again only red squares can be the one left out.

The only squares which are red in both cases are c3,c6,f3,f6.

hi

Those are not domino configurations, it is merely a different colouring of the squares than the normal checkerboard black/white pattern. The argument is that if you paint the squares this way, no matter where you place a 1x3 domino, it will lay overtop one red, one blue, and one green square.

This is a fairly standard kind of proof for this kind of problem (though elegantly explained, FancyKnight).

LaskerFan, imagine that the dominos are transparent when you place them, and are coloured only after they are placed on the board.  Depending where you place the domino, the color will be BRG, RGB, or GBR.  The point is that, if you have a valid placement of your dominos, each domino will cover one of each colour.  The order is irrelevant.  So 21 dominos will have covered 21 reds, 21 blues and 21 greens.  So if you've managed to get them all down, the remaining square must be red.  It's accounting!

Since that argument works for both top and bottom diagrams, the remaining squares must be squares where the top and bottom diagrams are *both* red.

The above arguments show that it's impossible to get a valid setup and leave any other squares (apart from c3, c6, f3 and f6).  Your diagram shows that it's possible to leave those 4 squares, and so together you have your proof!

Try Martin Gardener, or Sam Lloyd. Or some of the mensa puzzle books. Mathematical Digest from the University of Cape Town is also good.

The proof is of the form: if there were a solution is would have to look like A. And here is an example of a solution that looks like A. So A is the only solution to the problem.

A similar proof can be used to show that there are only 5 platonic solids... Left as an exercise

Wow a 2014 forum