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EventHorizon
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Suppose you are dropped onto a random square on an N by M chessboard. You can move in the four compass directions, and you cannot return to a square once you have left it.

Under what conditions can you visit all squares? When is it not possible to visit all squares?

Can you prove your answers? Easily?

Edited by EventHorizon
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If you are dropped in a corner, you will always be able to visit every square. If you are dropped on an edge, you will be able to visit every square if the edge you're dropped on has an even number of squares. Or if the edge you're on has an odd number of squares, you can visit every square if you are an odd number of squares away from both corners of your edge. Not sure about the middle yet...

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I've played all of the zelda games, and most of them have puzzles exactly like this one. theres a different route to take depending on where u land on the board.

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If you are dropped in a corner, you will always be able to visit every square. If you are dropped on an edge, you will be able to visit every square if the edge you're dropped on has an even number of squares. Or if the edge you're on has an odd number of squares, you can visit every square if you are an odd number of squares away from both corners of your edge. Not sure about the middle yet...

Yes, but there is a more simply way of stating it.....you may get it once you figure out the middle ones.

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I wasn't asking how...I was asking when it is possible and when it isn't possible. And for a proof.

Pretty simple, really.

If there are an odd number of squares, there will be more of one color than another. Since you have to alternate light/dark, if you start on the color which has fewer squares, you will never be able to visit all the squares. In any other case you will be able to.

Edited by Duh Puck
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A knight can cover all squares without repetition, can't prove easily - name a square and I'll provide the answer.

would not all otters except bishop, including pawn that can be queened.

I'm missing the question are'nt I?

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Pretty simple, really.

If there are an odd number of squares, there will be more of one color than another. Since you have to alternate light/dark, if you start on the color which has fewer squares, you will never be able to visit all the squares. In any other case you will be able to.

What about for even?

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What about for even?

I kinda figured that fell under the category of "any other case."

Edit: I can't seem to type even the simplest reply without phat-phingering something. I should use preview more.

Edited by Duh Puck
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Pretty simple, really.

If there are an odd number of squares, there will be more of one color than another. Since you have to alternate light/dark, if you start on the color which has fewer squares, you will never be able to visit all the squares. In any other case you will be able to.

Yup....good job.

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