Talk:Characteristics of Eulerian Graph

Error in Sufficiency Proof
"Otherwise, remove C1 from the graph, leaving the graph G0. The remaining vertices are still even, and since G is connected there is some vertex u in both G0 and C1."

The remaining vertices may not be still even.

Let C1 be a five cycle. choose two neighbors, x and y

add two vertices, a and b, and add edges to make a four cycle x y a b

add two new vertices, c and d, and add edges to make a four cycle x y c d

Now you have a graph with all degree 2 verts, except for x and y, which are degree 4.

Remove C1, and there are two components - a and b with an edge between them, and c and d with an edge between them. All four odd verts.


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 * After you have added those two 4-cycles and removed C1, you are left with a 6-cycle: x, a, b, y, d, c. No odd vertices in sight. --prime mover 17:50, 17 February 2012 (EST)


 * This looks to be a problem of an ambiguous definition. A circuit is an alternating series of vertices and edges. Removing it could mean either removing all the vertices (and hence the edges) or all the edges. Scshunt 00:03, 18 February 2012 (EST)


 * Should no longer be ambiguous, but there's still something that remains unproven on this page, I've just noticed. --prime mover 07:22, 18 February 2012 (EST)