Hello,I am trying to understand Euler circle or not. If a graph has an euler path ,then it has at most 2 vertices with odd degree. (If I understand it right.)
I find some graphs I try solve them and ask you if my answers are right.
On graph 1. it is Eulerian. We have u0,u1,u2,u3. We have 4 vertices. Each vertice has 2 edges max so it is Euler.
On graph 2. u0,u1,u2,u3,u4. I have 5 vertices with 6 edges. I have circle the u2 ,u1 on the image (I have circle and u4 but wrong circle ,only u2 and u3 I wanted) .So, u2,u3 are vertices that have 3 edges. I wasn't sure if it is Euler or not. Cause the most 2 vertiies with odd edges. Maybe it is Euler?
On graph 3. I have 20 vertices (If we count u0 until u19 is 20 vertices).Its not Euler cause it has all those u1,u2,u3,u9,u14,u16,u17,u18 all those vertices have 3 edges each one that makes not Euler. u6 or u7 for example has 4 edges.Not Euler
**On graph 4.**It is u0 until u7 so it is 8 vertices. u0,u1,u6,u7 are have odd degree (3 edges) so it is not euler .
On graph 5. the same it is not Euler example u0,u1,u2 are 3 vertices and have 3 edges each one . Not euler
As you say, the first graph has an Euler circuit, and the third, fourth, and fifth all have more than two vertices of odd degree, so they have neither an Euler circuit nor an Euler path. The second graph is the interesting one. It has some vertices of odd degree, so it does not have an Euler circuit, but it has only two of them, so it does have an Euler path with those vertices as endpoints, for instance $u_2u_3u_0u_1u_4u_2u_1$. However, in order to be Eulerian, a graph must have an Euler circuit; an Euler path is not enough. Thus, only the first graph is Eulerian.