Just to clarify the picture: It's a vertex with one node and a loop

@Semiclassical yes

I'd agree, that's technically a cycle instead of a path. It should be easy to modify the fit the definition exactly though: try the same idea with three nodes, two connecting edges, and a loop.

@D.ZackGarza thank you! So we could draw:

Is that a graph with euler path but not hamiltonian path?

So what's drawn is an Euler path that is not a Hamiltonian path, but the graph does still have a Hamiltonian path.

Namely, just doing $v_1 \to e_1 \to v_2 \to e_3 \to v_3$. So now you just need to modify it a bit so that visiting every vertex somehow requires you to hit a vertex twice

Hint: think about adding a second "branch" off of $v_1$.

@D.ZackGarza how do you go from $e_3$ to $v_3$? It should not be $v_1 \to e_1 \to v_2 \to e_{\color{red}{2}} \to v_3$?

Right, that should be $e_2$. Any path visiting each vertex exactly once will do.

@D.ZackGarza ^^^^^^ it has Euler path (even though it repeats vertex) but I am not sure if it has a Hamiltonian path

It does -- by definition, any path that visits each vertex exactly once will be a Hamiltonian path. So e.g. take $v_1 \to v_2 \to v_3$.

(Note that you don't need to use every edge.)

A modification with a "dead end" might help.

@D.ZackGarza that's right; it has Hamiltonian path

@D.ZackGarza hmmm... I am not sure if I understand it correctly

If I delete $e_4$ I still have a HP (since $v_1\to v_2\to v_3$), so I should do something with $e_4$ but I do not know what

Consider adding a vertex $v_4$ and an edge $e': v_2 \to v_4$, for example.

(Note: this won't actually be a complete solution, since something else goes wrong. But try to check it for Hamiltonian paths.)

@D.ZackGarza this? ^^^^^^^^ Oh

Ok so with this graph we don't have Hamiltonian path since each path must pass trough $v_2$ more than once. Right?

@D.ZackGarza However I am wondering if it has Euler path

Right, exactly. The problem is that now you also can't have an Euler path.

Or, well, can you?

I think I see a few, as long as you're allowed to start/end anywhere.

E.g. try starting at $v_4$ or $v_2$.

So it has Euler path because we have a path $v_4\to e'\to v_2\to e_2\to v_3\to e_4\to v_1\to e_1\to v_2\to e_3\to v_2$?

Yep, looks good. And you only need to produce one.

And once you've shown that no Hamiltonian path can exist, you're good to go.

@D.ZackGarza Yes, exactly. Thank you so much @D.ZackGarza! Well, the proof of no existence of HP must include an exhaustive proof, isn't it? Because our graph is finite? Or can we prove using general methods?

@manooooh An exhaustive proof would certainly work here, but it usually suffices to say some words about why one couldn't exist.

And as it turns out, that graph does actually have a Hamiltonian path, so some slight modification is needed.

I think you'd just need to delete $e_4$. Then the proof that it's not Hamiltonian is definitely related to what you said: any path must pass through $v_2$.

@D.ZackGarza oh you are right: we could have the path $v_4\to e'\to v_2\to e_1\to v_1\to e_4\to v_3$

But once that's fixed (so there are no cycles), the odd valence of $v_2$ and $v_4$ play a major role in the proof.

@D.ZackGarza yeah, we have $gr(v_2)=5$ and $gr(v_4)=1$

Now the other part of the exercise! Draw a graph with Hamiltonian path but not Euler path

This should be easy...

Yep, so you now have a graph that can't admit any Hamiltonian paths. You just need to modify it a bit to allow an Eulerian one.