Peterson Graph Non-Hamiltonian Proof Explanation I'm working on graph theory and I'm trying to find a generalised elegant proof to non-Hamiltonian graphs. I stumbled onto this proof from D. West, which is simple, but I'm having trouble understanding how it works.
From Wolfram:
If there is a 10-cycle C, then the graph consists of C plus five chords. If each chord joins vertices opposite on C, then there is a 4-cycle. Hence some chord e joins vertices at distance 4 along C. Now no chord incident to a vertex opposite an endpoint of e on C can be added without creating a cycle with at most four vertices. Therefore, the Petersen graph is nonhamiltonian. In fact, it is also the smallest hypohamiltonian graph.
In the following illustration, my interpretation of the above proof is that by connecting opposite vertices in graph C, you are creating 4-cycles, which connected together make a Hamiltonian cycle, thus showing that in order for the graph to be Hamiltonian, you must have more edges. Therefore, the Petersen graph must be non-Hamiltonian.
Furthermore, this can be generalised to any graph as: any graph that does not form any 4-cycles with chords is non-Hamiltonian and by adding chords to create 4-cycles, they will become Hamiltonian.
Is this interpretation correct? If not, can someone please explain this proof in detail?

 A: The starting intuition for this proof is the idea that if a graph $G$ is Hamiltonian, we can draw a diagram of $G$ in which all the vertices are arranged in a circle, and consecutive vertices are adjacent.
For example, the cube graph (below on the left) has a Hamiltonian cycle, and a diagram of the cube graph with the $8$-cycle clearly visible is shown below on the right.

Now let's imagine that the Petersen graph has such a diagram. We don't know yet what the diagram looks like. However, the Petersen graph is $3$-regular, so we know that each vertex we draw must have one additional edge out of it: the "chords", if you will.
Furthermore, one additional property of the Petersen graph is that it does not have any $3$-cycles or $4$-cycles. So the chord out of each vertex must go to one of the three vertices furthest away: anything else would create a $3$-cycle and $4$-cycle. In the diagram below on the right, you see the three possible chords out of the top vertex.

Now, there's two cases. The first is that every chord joins a vertex to the opposite vertex. That gets us the second diagram above, and this is not the Petersen graph: for one thing, it again has $4$-cycles.
So there must be a chord that joins two non-opposite vertices. Such a chord is drawn in the third diagram above. The three red dashed lines are the possible chords out of the bottom vertex, and we see that none of them are possible! Two of them would create $4$-cycles, and one would do worse and create a degree-$4$ vertex.
So our assumption that the Petersen graph had a Hamiltonian cycle must have been wrong.

You ask: can this be generalized? Not meaningfully.
This argument relied on the fact that in a $10$-vertex cycle, if we want to add a chord and not create any $3$-cycles or $4$-cycles, the options are very limited and we can go through all of them and rule them out. In a graph with more vertices, there would be many more possible chords to draw that don't create any $3$-cycles or $4$-cycles.
Already with $12$ vertices, there is a $3$-regular graph with no $3$-cycles or $4$-cycles that has a Hamiltonian cycle:

So all we can say is "If $G$ is a $3$-regular graph on at most $10$ vertices with no $3$-cycles or $4$-cycles, then $G$ is not Hamiltonian", and the Petersen graph is the only graph that matches this description.
