Suppose we are given the setup in the title. Two squares are adjacent if and only if they share a common edge. I want to find out whether the obtained graph considering squares as nodes would be Hamiltonian or not? I am out of ideas as to how to solve this? The degree condition that sum of every non adjacent node is greater than number of nodes doesn't hold, neither do I see any set of vertices which I can remove to get more components than the vertices in the removed set. Kindly help. Thank you!
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Edges of the graph are adjacent squares? – Aig May 08 '24 at 20:31
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@Aig no, two nodes are adjacent if they are adjacent squares – Sj2704 May 08 '24 at 20:32
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2Consider the chessboard coloring. The squares in the Hamiltonian path must alternate colors. – Aig May 08 '24 at 20:33
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@Aig And by adjacent, I mean any form of adjacency on the board, right left up or down, just not diagonally – Sj2704 May 08 '24 at 20:33
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Yes, that’s exactly what I wrote. I guess, the word “edge” has two meanings here: edge of the square and edge of the graph. – Aig May 08 '24 at 20:34
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@Aig i am unable to see a Hamiltonian path or a cycle, since the two squares are removed. – Sj2704 May 08 '24 at 20:36
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Two removed squares are of the same color. Hence there are, say, $32$ black and $30$ white squares left. Let there be a Hamiltonian cycle. The color of squares alternates in this cycle, so there must be equally many black and white squares. Contradiction. So there is no Hamiltonian cycle.
Aig
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oh ok got it! It essentially form $K_{m,n}$ where m and n are not equal! Thank you so much – Sj2704 May 08 '24 at 20:43
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1@Sj2704 $K_{m,n}$ usually stands for complete bipartite graph. This is just a bipartite graph. But yes, this is the idea. – Aig May 08 '24 at 20:45
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