Is a graph consisting of a single node complete in addition to being simple? What about a node with a self loop:it's not simple but is it complete ?
Node with loop graph completion
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You're totally missing the point here.
First, you should be aware that some questions only make sense when they are bundled with definitions, unless you're using the conventional ones.
But, assuming this is just a naive question from a naive guy (and there's absolutely nothing wrong with it :) ), here you can read on the first line:
In the mathematical field of graph theory, a complete graph is a simple undirected graph
Simple => No loops. So no, the node+self loop isn't complete.
Furthermore, and this is the funniest part of your question, is that you cannot apply the definition of complete graph to a single node graph. This happens because "technically", a Graph $G$ is defined as a pair of two sets $(V,E)$, $V$ for the vertexes and $E$ for the edges. So if you were to follow the same page definition, which i will quote:
(...) in which every pair of distinct vertices is connected by (...)
So if we are analysing the single node graph, our set will look like $SingleNode=(\{v \},\{ \})$, and we can't pick a pair of distinct vertices, not even a pair of vertices.
So yeah, the $SingleNode$ vertex is as complete as my hair is purple (I'm bald, so...). But for your purpose I guess is better to take it as if he isn't, actually, i think that's the convention.
In the simple case, we have the complete graph $K_1$.
In the non-simple case, it's trickier. E.g., we could be working with graphs that may have more than one loop per vertex, and arbitrarily many loops. So usually it's not meaningful to think of complete graphs in contexts where loops are permissible.
However, one could imagine contexts where at most one loop per vertex could be present (e.g. transitions in symmetric Markov chains). In such a context, the single vertex with a loop would be complete (and the single vertex graph without the loop would not be).