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The exercise: Asks for both of Eulerian circuit and path circuit.

Conditions: 1)-Should stop at the same point that started from. 2)- Don't repeat edges. 3)-Should cross all edges

After long time of focusing I found the Eulerian path, I tried so much on the circuit but could not find it. The conclusion that it got is that there is only Eulerian path, Eulerian circuit is not exist here. Is that correct!!,image is imbedded

Thanks in advance

User
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    There is a theorem that says that a connected graph has an Euler circuit if and only if all of its vertices have even degree; $a$ and $d$ have degree $3$, so this graph has no Euler circuit. The theorem also says that a connected graph has an Euler path but no Euler circuit if and only if it has exactly two vertices of odd degree, in which case any Euler path must have those two vertices as endpoints. – Brian M. Scott Nov 06 '20 at 20:03
  • Did your teacher not introduce these concepts with the famed "Bridges of Königsberg" problem and Euler's solution, which is exactly the result Brian Scott mentions, and was also the birth of graph theory? – Paul Sinclair Nov 07 '20 at 00:54
  • Thanks for your helpful comments, @BrianM.Scott: that was fantastic and very beneficial , thanks. Paul : looks like we just jumped to the middle without studying this part, I'll try to understand more about it, looks very important, thanks – User Nov 07 '20 at 06:19
  • @UserUser: You’re very welcome. – Brian M. Scott Nov 07 '20 at 18:22

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