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It's pretty obvious why it doesn't work using the sequence algorithms.

So the way I thought of it:(n is the number of vertices)

at n = 1: you get 0, which is the graph of just one vertex

at n =2: you get 1, 0 which is two vertices with one having a degree of 1, but the other vertex has no edges so this is impossible.

this continues on for n greater than 2 as well. Not sure if this is enough to answer the question.

Edit: possible new solution, there are n possible degrees for n vertices which is impossible when $n\geq2$

johnson
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  • You mean, why is it impossible for a graph to exist with those degrees? – Mike Pierce Nov 02 '16 at 15:51
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    this is a classic results for simple graphs: see http://math.stackexchange.com/questions/251310/if-n-is-a-natural-number-ge-2-how-do-i-prove-that-any-graph-with-n-vertic – cats Nov 02 '16 at 15:52

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