Ramsey number $R(3,7)=23$

Addendum - A proof of the claim that the reduced $H_2$ has no independent set of size 6

Suppose that there is a set $X$ of 6 independent points.

First we prove $X$ is not independent in $H_2$

In view of the symmetry of $H_2$ we can suppose that $p_{12} \in X$. The only points in $H_2$ independent of $p_{12}$ are $p_{13}-p_{24},p_{14}-p_{23},p_{15}-p_{26},p_{16}-p_{25}$ where the dashes indicate some of the edges in $H_2$. We can only have one of each pair of joined points in $X$ and so there can be at most $5$ independent points in $H_2$.

We now know that $X$ contains two adjacent points in the Hamiltonian path

There are 15 possibilities (unless we establish some symmetries amongst the points in the path). We shall consider one possibility; that $X$ contains $p_{35}$ and $p_{16}$.

The only points independent of both $p_{35}$ and $p_{16}$ in the reduced $H_2$ are either adjacent to one of them in the Hamiltonian path or contain both one of the suffices $3,5$ and one of the suffices $1,6$. So the possibilities are $p_{12}-p_{36},p_{13}-p_{56},p_{15}-p_{23}$ where the dashes indicate some of the edges in the reduced $H_2$. Again there can be at most $5$ independent points.

Let's first deal with the fact that there are many triangles and these are in $H_2$.

Think of the points as being in three groups:-

The preferred point $p$

The six points $p_1, ..., p_6$ of $H_1$ which are joined to $p$

The 15 points $p_{ij}$ of $H_2$, each of which is joined to two of the points in $H_1$ and to six of the points in $H_2$.

Now let us look at possibilities for triangles.

(1) Consider the point $p$. This is only joined to points of $H_1$ and no two points of $H_2$ are joined to each other so $p$ is not in a triangle.

(2) Consider the point $p_1$ of $H_1$. This is joined to the five points $p_{12}, ..., p_{16}$ of $H_2$ but no pair of these points are joined and so $p_1$ is not in a triangle. Similarly no point of $H_1$ is in a triangle.

We now know that triangles can only be in $H_2$.

Let's look for just those which include $p_{12}$. They are $p_{12},p_{34},p_{56}$ and $p_{12},p_{35},p_{46}$ and $p_{12},p_{36},p_{45}$ and ...

There are lots!

The next part of the proof starts with the assertion that the reduced $H_2$ (i.e. with the Hamiltonian cycle removed) is a (3,6) graph.

The author does not prove this but you might like to check that you can see this to be true.

Once you have done this it is now obvious that the reduced graph $G$ has no triangles because all the original triangles were in $H_2$. So the crux of the proof is to obtain a contradiction to the supposition:

Suppose G has an independent set of size 7

Since $H_2$ has no independent set of size 6 we would require at least two points from $p,p_1, ..., p_6$. However, $p$ is joined to the points of $H_1$ and so we require at least two points from $H_1$.

If we had three points from $H_1$, say $p_1, p_2, p_3$, then the only points from $H_2$ that we can have are $p_{45},p_{46},p_{56}$ i.e a maximum of 6 points. Similarly, if we had four points from $H_1$, then we can have only 1 point from $H_2$ and if we had five or six points from $H_1$, then we can have no points from $H_2$.

The only case needing study is if we have two points from $H_1$, say $p_1, p_2$.

This is an important part of the author's construction so check that you understand the set up we now have. The only points in $H_2$ independent of $p_1, p_2$ are $$p_{34}-p_{56},p_{35}-p_{46},p_{36}-p_{45}$$ where the dashes indicate edges in the author's original graph. So, for an independent set of size 7 we require at least two of these three edges to be in the Hamiltonian cycle that was removed. A quick check will show you that only $p_{34}-p_{56}$ has been removed and so there is no independent set of size 7 including $p_1, p_2$.

Now we have to do this analysis for any two distinct points $p_x, p_y$ of $H_1$. Therefore we must prove that for any four distinct indices $a,b,c,d$, at most one of the three edges in $$p_{ab}-p_{cd},p_{ac}-p_{bd},p_{ad}-p_{bc}$$ are in the Hamiltonian cycle. This has to be checked; the author does not do so but claims that the check is easy. (At worst it's a mechanical task and you may spot some clever shortcuts.)