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Another chessboard riddle (slightly easier than the previous one)

Started by Yoni, October 03, 2004, 04:48 PM

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Yoni

Given is an 8x8 chessboard with two of the diagonally opposite corners removed.


Is it possible to cover the entire board with domino tiles?
A domino tile is a rectangular 1x2 (or 2x1) tile.
You can't stack more than one tile on the same square.

If so, show how.
If not, prove it.

hismajesty

Are we allowed to physically change the domino tile in any way? :P

Yoni


Meh

If you want to know my solution, read the black!!
Its impossible, with maths it doesnt work out as far as I can see, and I also tried doing it by using Paint to show which squares had been used.

Yoni

Quote from: Meh on October 06, 2004, 08:40 AM
If you want to know my solution, read the black!!
Its impossible, with maths it doesnt work out as far as I can see, and I also tried doing it by using Paint to show which squares had been used.
That != solution.

Meh

My solution, sorry Misread I had to prove it.
If there are 62 squares half is 31. I divided it by 2 as the dominoes are 1*2 meaning they will ocupy 2 squares either way they go. If you do one half the board you are left with a vacant squre as there has already been 1 square taken away in the corner. Same for the other side. This means that the 1 square left on either side are not together and in most cases are diagonal from each other so the dominoes cant be placed on it.. Take the 1 away from each half of the board leaving you with 2 blank squares that cant be filled as they are not together.

http://rabidgamerz.co.uk/images/proof.PNG As an example.

Edit: More in depth.

Yoni

Quote from: Meh on October 06, 2004, 10:10 AM
My solution, sorry Misread I had to prove it.
If there are 62 squares half is 31. I divided it by 2 as the dominoes are 1*2 meaning they will ocupy 2 squares either way they go. If you do one half the board you are left with a vacant squre as there has already been 1 square taken away in the corner. Same for the other side. This means that the 1 square left on either side are not together and in most cases are diagonal from each other so the dominoes cant be placed on it.. Take the 1 away from each half of the board leaving you with 2 blank squares that cant be filled as they are not together.

http://rabidgamerz.co.uk/images/proof.PNG As an example.

Edit: More in depth.

My comments on your solution in black.

I talked to you and you said that by "half", you mean the right half vs. the left half.
Observe:
http://israel.valhallalegends.com/images/counterexample.png

In this image I took away 2 different corners.
If you split the board in 2 halves in the same way as your proof, then you still have 31 on each half!

So, according to your proof, the board in counterexample.png should not be solvable. But it is - see the red markings.
Therefore, your proof is incorrect.


Nice try! You've come closer than anybody in this thread, at least.

Meh

Thanks, I dint think you could change the board, If you work it out wih the missing corners diagonally oposite It wont work. to my knowledge.

Edit: Forgot 1 word it needed to mkae true my proof!

Yoni

No, I showed you different conditions under which your proof accepts the conditions but fails. Therefore your proof is wrong. A small addition/change won't make it right.

iago

<black>
Each domino must cover a white tile and a black tile (I dont think I need to prove that, it's obvious).  Therefore, to work, we have to have the same number of white and black.  We removed 2 white tiles, and therefore we have 30 white and 32 black, and thus it's impossible.
</black>

I haven't read the rest of the thread, just the first post, but do I win anything? :)

<edit> it's actually pretty easy and obvious, if you read the solution to your last chess problem.
This'll make an interesting test for broken AV:
QuoteX5O!P%@AP[4\PZX54(P^)7CC)7}$EICAR-STANDARD-ANTIVIRUS-TEST-FILE!$H+H*


Meh

Thats kinda what I meant I just didnt include colours.
I meant that there were 2 odd tiles left out and couldnt be filled.


I am not very good at explanations  :P

iago

But you need to prove that they're left over, not just say they will be.

Continues waiting for his prize
This'll make an interesting test for broken AV:
QuoteX5O!P%@AP[4\PZX54(P^)7CC)7}$EICAR-STANDARD-ANTIVIRUS-TEST-FILE!$H+H*


Yoni

iago wins the prize! :D
(Click image for mp3 goodness)


Quote from: iago on October 06, 2004, 11:11 AM
<edit> it's actually pretty easy and obvious, if you read the solution to your last chess problem.
True! :)

Meh: Your explanation was understandable, but the idea was wrong. You'll get over it.

Meh

Damn I wanted to win. LOL! I could never explain theorys. I wanted a badge  :'(

iago

This'll make an interesting test for broken AV:
QuoteX5O!P%@AP[4\PZX54(P^)7CC)7}$EICAR-STANDARD-ANTIVIRUS-TEST-FILE!$H+H*