Mini Puzzle

Yes

No. I've heard it before so I won't give the game away. Clever solution.

Codey

I give up. What's the secret? ffeJ

Place the dominoes in diagonal fashion starting from top left to bottom right and remain on the edge. In effect place dominoes only on top and right edge of the diagram.This will require 13 dominoes.

Once you do this you will notice a perfect square of side of six boxes.

This will leave 36 squares which will hold 18 dominoes.

Cheers.

Randall is right, it can NOT be solved.

If you have a solution please post the diagram. Following your directions the vacant square end up either one row up or one row left of the corner square which is occupied by 1/2 of the last dominoe.

OK, those of you who've tried it probably know now that it is impossible. So the real puzzle is to prove it.

This puzzle is impossible. I have devised a fast proof, which I post here for revision. I don't claim is the best or even correct, but I'm quite busy to spent too much time on it. So here it is:

Let v_i be the dominoes positioned in a vertical position, starting from row i and extending down to row i+1.

It is clear that the number of dominoes that are in a vertical position in the first row with the 7 squares width, must be odd, in order to allow the rest horizontal ones to fill the row. Therefore, v1 is odd. The dominoes now that are vertical starting in row 1 (which has a width of 8 squares), are V₁+v₂ (because V₁ ones have their "feet" and V₂ have their "heads" in row 2 respectively) must be even, so V₂ must be odd. Continuing in the same manner, we see that all vertical dominoes starting from all rows 1-7 downwards are odd. (Obviously row 8 has no vertical dominoes starting from it.) So the total number of dominoes positioned vertically must be odd, as V₁+v₂+...V₇ is odd (sum of odd number of odd numbers!) Consequently the dominoes positioned horizontally must be even, as the total number of dominoes is 62/2=31.

Now turn the figure 90 degrees to the left or right, and repeat the above analysis. You will reach to the conclusion that the vertical ones must be odd again. But these ones used to be horizontal before and should have been even. Oops!...

Brilliant,

I'm pretty sure your proof is sound, but, there is a much simpler one.

As a clue: I threw a bit of a curved ball by supplying the template instead of letting people cover two opposite corners of a real chess board.

SOLUTION

First: clearly the answer is no.

To prove it: simply observe that the diagonally opposite corner of a chess board are both the same colour; wherever you place a domino it covers one white and one black square; when you have place 30 dominoes you will always have two squares left of the opposite colour to those removed, and, these clearly cannot be adjacent.

Just in case someone besides me is wondering if you could get the dominoes to fit if you started with a 9 x 9 grid instead of 8 x 8:

... it still doesn't work. From this we can see that no matter what the original dimensions of the grid are, if you remove 2 squares from diagonally opposite corners, you cannot cover the grid with dominoes.