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May 8 at 8:52 comment added Jaap Scherphuis I don't really see how the Klein 4 group applies either. The usual proof involves 3-colouring the holes such that any three-in-a-row contains all three colours. Then every move will increment or decrement the number of pegs in each colour. A simple parity argument then tells you what colour hole the final peg must be in.
May 4 at 12:03 comment added CommunityBot Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer.
May 4 at 11:53 history edited Kan't
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S May 4 at 11:45 history asked 점시맙 CC BY-SA 4.0