answersLogoWhite

0

Yes.

An example:

_____A---------B________ A connected directly to B and D by one path.

_____|_______/|\________ B connected directly to A and E by one path, and to C by two paths.

_____|______/_|_\_______

_____|_____/___\_|______

_____|__E/_____\|______ E connected directly to B and D by one path.

_____|____\_____C______ C connected directly to B and D by two paths.

_____|_____\____|\_____

_____|______\___|__\___

_____|_______\__|__/___

_____|________\_|_/____

_____|_________\|/_____

_____-------------D_____ D connected directly to A and E by one path, and to C by two paths.

There is an Euler circuit: ABCDEBCDA

But a Hamiltonian circuit is impossible: as part of a circuit A can only be reached by the path BAD, but once BAD has been traversed it is impossible to get to both C and E without returning to B or D first. However there is a Hamiltonian Path: ABCDE.

User Avatar

Wiki User

9y ago

Still curious? Ask our experts.

Chat with our AI personalities

RossRoss
Every question is just a happy little opportunity.
Chat with Ross
ReneRene
Change my mind. I dare you.
Chat with Rene
ViviVivi
Your ride-or-die bestie who's seen you through every high and low.
Chat with Vivi

Add your answer:

Earn +20 pts
Q: Can a graph have an Euler circuit but not a Hamiltonian circuit?
Write your answer...
Submit
Still have questions?
magnify glass
imp