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Suppose we have $8 \times 8$ chessboard such that two squares are adjacent iff they share a common side. So we can imagine this board as an $8 \times 8$ grid. There is a pawn in the upper left corner square and he can move only to adjacent squares.

The question is whether it can complete a Hamiltonian path between the upper left corner (where he starts) and bottom right corner, i.e. can he come from the upper left to the bottom right using every square only once and end up using them all?

I've tried to simplify the problem and tried finding the solution on $4 \times 4$ grid, but without success, so my intuition tells me there is no solution to the $8 \times 8$ case, but I was unable to prove so, as I'm not familiar with any operable necessary conditions for a graph to have Hamiltonian path.

Any help would be welcome! Thanks in advance!

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  • $\begingroup$ Note: What you're asking for is actually a Hamiltonian path. (A Hamiltonian cycle would end at the same place it began.) $\endgroup$ – paw88789 Feb 1 '17 at 18:02
  • $\begingroup$ @paw88789 you're right of course, edited $\endgroup$ – windircurse Feb 1 '17 at 18:10
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Hint in the Form of Leading Questions. What color is the upper left corner square? What color is the lower right corner square? What happens to the color of the pawn's square on each step? How many steps must the pawn take?

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  • $\begingroup$ That's ingenious, thank you very much. $\endgroup$ – windircurse Feb 1 '17 at 18:47

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