Euler's path same as Hamiltonian cycle

AI Thread Summary
The discussion focuses on identifying graphs with n vertices where the Euler's path coincides with the Hamiltonian cycle. It suggests that regular graphs with a degree greater than n/2, as per Dirac's Theorem, may exhibit this property, particularly when n is even. However, it emphasizes that having both paths does not guarantee they are identical. The key condition is that the graph must contain only the edges of the Hamiltonian cycle without any additional edges. The Cycle Graph is proposed as a potential unique example, and a recommendation is made to use the GraphSolver app for further exploration.
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Homework Statement


Find (classify) all graphs with n vertices who's Euler's path is the same as their Hamiltonian cycle.

The Attempt at a Solution


I'd say that any regular graphs with a degree greater than n/2 (Dirac's Theorem) with and n divisible by 2 has both and Euler's path and a Hamiltonian cycle. But that doesn't mean they're the same path on that graph. Also that probably does not define all graphs where that's the case.
 
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Think about what each of the definitions mean. It seems that it's a pretty strong condition that a path touching each vertex once also covers each edge once. Specifically, it means the graph needs to have only the edges in the hamiltonian cycle, and no others. See if you can go from here to find a characterization of such graphs.
 
I don't think there are other graphs but the Cycle Graph that meets that criteria :\
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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