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Yes.

Example:

....................................................

...A * ...........................................

......|.\ ......................................... eg Euler circuit: ACDCBA

......|...\ ........... --------- .............

......|.....\........./...............\............ The Hamilton circuit is impossible as it has two

......|.......\...../...................\.......... halves (ACD & CD) connected to each other only

......|.........\./.......................\........ at vertex C. Once vertex C has been reached in

......|.......C *........................* D.... one half, it can only be used to start a path in

......|........./.\......................./......... the other half, or complete the cycle in the

......|......./.....\.................../........... current half; or if the path starts at C, it will end

......|...../.........\.............../............. without the other half being visited before C is

......|.../ ........... --------- .............. revisited.

......|./ ...........................................

...B *..............................................

......................................................

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More answers

Yes, a graph can have an Euler circuit (a circuit that visits every edge exactly once) but not have a Hamiltonian circuit (a circuit that visits every vertex exactly once). This can happen when the graph has certain degree requirements that allow for the Euler circuit but prevent the existence of a Hamiltonian circuit.

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Q: Can a graph have a euler circuit but not a hamiltonian circuit?
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