Can a graph with 2 nodes of degree 5 have an Euler path?

[cobalt hill]

ik that

For a graph to have an Euler path:
    - Every vertex must have an **even** degree
    - Have exactly **zero** or **two** vertices with an odd degree
``` and all examples I have seen use degree 3 for the odd ones, but not sure if all odd numbers (5, 7, 9...) also work

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[naive trout]

Yes, for every oddnit also works:

This is for two vertices of degree 5

But you can add triangles to make 2 vertices of odd degree (can be different)

[cobalt hill]

ty :)

[zealous storm]

@cobalt hill has given 1 rep to @naive trout

[astral trail]

What would happen if there were 4 vertices of an odd degree? Is an Eulerian path impossible then?

[cobalt hill]

no, its only possible if and only if there are 0 or 2 odd degree vertices

[viral igloo]

If I recall correctly, shouldn't one odd-degree vertex also be possible?
If there are no odd-degree vertices, you can start from and end anywhere.
If there is one odd-degree vertex, you have to start and end on it.
If there are two odd-degree vertices, you have to start on one and end on the other.
And you can't do it for more than two.

Though, now that I think about it, I can't think of any graphs with just one odd-degree vertex...

[naive trout]

No 1 odd degree isn't possible because if you start on it you can't end on it and there are no other odd degree vertices

[astral trail]

As far as simple graphs(not digraphs) are concerned, I think there is a basic theorem like there must an even number of vertices with an odd number. So that 1 vertex with an odd degree graph wouldn't exist, forget about the Eulerian Path.

[viral igloo]

Ohh, I see. Thanks!