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Handshaking lemma states in an undirected graph an even number of vertices must have odd degree.

However 3 people shaking hands with each other, 6 hand shakes, or two a each. So there are no vertices with odd degree.

Does the handshaking lemma hold true because 0 is even and there are zero vertices of odd degree?

I'm not doubting the lemma is true, just thinking I'm missing something really obvious.

Am_I_Helpful
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user1768079
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1 Answers1

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Does the handshaking lemma hold true because 0 is even and there are zero vertices of odd degree?

Yes. Since all 3 vertices are of even-degree, so there are zero vertices of odd-degree.

You're absolutely correct. Same is the case when people = 1.

Am_I_Helpful
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  • Thank you, I saw your typo, but I made a similar typo when I posted my question. – user1768079 Oct 23 '16 at 19:30
  • @user1768079 - I hope it is clear to you logically as well. *`An even number of people must have shaken an odd number of other people's hands`* --- which can be verified thinking logically too. – Am_I_Helpful Oct 23 '16 at 19:31
  • It was clear to me, I just wasn't sure if I was missing something in cases like this. I thought maybe I was misunderstanding the definition of something. All of the examples I was given showed graphs without a cycle that included every vertex or had some other condition that made it different from a graph with odd number of vertices all of equal, even degrees. – user1768079 Oct 23 '16 at 19:34
  • @user1768079 - It is good that you decided to ask it here. It shows you're inquisitive. Keep it up! Good Luck. – Am_I_Helpful Oct 23 '16 at 19:35