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I am a beginner in graph theory. For the above Qn, I think yes. I was thinking this when i was thinking about theorem - every complete graph with even number of vertices has perfect matching. Coming to the above QN in title,

My approach -

connected graph has spanning tree subgraph(by removing all the extra edges). Now since there are even number of vertices we have odd numbber of edges in spanning subgraph. Now i remove edge between even order vertices in spanning subgraph. That proves that we have edges subset with vertices having odd degree with these edges subset which are disjoint again. Kindly correct me ifthere are any mistakes in the argument

I found the following article related to this - - Picking edges from a connected graph so that any vertex is incident with an odd number of those edges

Here the author (1st answer - 2nd method), author talks about splitting vertex of even degree into two. I did not understand how splitting of even order vertices done. Kindly any one please explain the process.

Nascimento de Cos
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    No. Consider a star graph with $4$ vertices. (One vertex of degree $3$, three vertices of degree $1$.( – saulspatz Feb 15 '20 at 02:18
  • Oh yes. I was thinking in terms of trees. Sir,Thank you for enligtening me. can you pls have a look at the above link(author talking about splitting vertex of even degree --> what is the meaning of splitting vertex of even degree) – Nascimento de Cos Feb 15 '20 at 02:19
  • He is just talking about pairing off the even degree vertices. I.e., if the even-degree vertices are $A,B,C,D,E, ...$, grouping them together in sets of 2: ${A,D}, {B, Z}, {C, H}, ...$. In his case, he labels the first pair $a_1, b_1$, the second pair $a_2, b_2$, etc. – Paul Sinclair Feb 15 '20 at 17:08

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