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I did all the 11 graphs but how can i prove it? I did all the grade of the vertices but im not sure how to proof that there are only 11

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    "how to proof that there are only 11" I would do a case-by-case analysis. Split into cases that are smaller and easier to analyse than "all graphs on 4 vertices", and make absolutely sure that you have all graphs in each case. – Arthur Sep 28 '20 at 18:29
  • Related: https://math.stackexchange.com/questions/599675/why-there-are-11-non-isomorphic-graphs-of-order-4 – Barry Cipra Sep 28 '20 at 19:12
  • thank you so much! –  Sep 28 '20 at 19:22
  • By the way, in this context, "grado" should be translated as "degree", not "grade." – saulspatz Sep 28 '20 at 19:25

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While I think that Arthur's comment is the way to go, if that doesn't convince you, you can cross-check against the number of labeled graphs. That is, label the vertices A,B,C,D say, and consider all ways of joining them.

The adjacency matrix matrix is $4\times4$, and is completely determined by the $6$ values in the upper triangle, so there are $2^6$ labeled graphs on $4$ vertices. Count how many labeled graphs correspond to each of your graphs. If the total comes to $64$, you know you've got them all.

For example, one of your graphs has one edge and two isolated vertices. There are $\binom42=6$ ways to choose the adjacent vertices, so this graph corresponds to $6$ labeled graphs.

saulspatz
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