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There seem to be two distinctly different ways of approaching induction with respect to graphs, where one of them has been described to me as incorrect. What about this particular method fails in induction?

Deconstructing a graph (correct)

Let $\mathcal{F_n}$ be a family of graphs with property $P$ on $n$ vertices. Consider an arbitrary graph $G$ on $n+1$ vertices, and locate some $v\in V(G)$ such that $G-v = H\in\mathcal{F_n}$ if it exists. Considering this vertex, construct an argument proving that $G\in\mathcal{F_{n+1}}$.

Constructing a graph (incorrect)

Consider an arbitrary graph $G\in\mathcal{F_n}$. Let $v$ be some vertex added to $G$ -- denote this new graph $\hat{G} = G + v$. Considering this vertex and how it must be added to $G$, construct an argument proving that $\hat{G}\in\mathcal{F_{n+1}}$.

My intuition tells me that the second methods fails, as not all graphs may be created from the initially given $G$, but I have not been able to prove this myself. Am I correct here, or is there something more going on?

Example

This question peaked my interest, and I want to prove the inductive step by considering a graph with a spanning path, adding a vertex to the graph and showing that no matter how edges are added to this new graph it must also have a spanning path. This, however, makes use of the second method of induction, and hence I am unsure if this proof is correct.

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    With the second method you can prove that all graphs are paths. – 2017-02-21
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    What about the second method permits that, and how can you show it? – 2017-02-21

1 Answers 1

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Aside from the problem you state, i.e., we don't know all graphs in the family can be constructed this way, there's another problem:

When deleting a vertex, we delete all the edges it is an endpoint of, so $G-v$ is a uniquely defined graph. However, when adding a vertex, $G+v$ is not uniquely defined (unless you add it as an isolated vertex), as it will vary with how edges are added.

We can overcome this problem by checking all possible ways of adding edges in $G+v$ with endpoint $v$, but there could be $2^{|V(G)|}$ ways of doing this, which may be impractical (depending on the family).

However, proofs along the lines of "adding vertices" could work, e.g., if it is already known that some vertex-adding algorithm generates all possible graphs (in fact, we might be studying a family of graphs defined this way), and it might help if the degree was restricted in some way.