To derive the maximum size $m$ of a graph $G$ with order $n$ and no $\Theta$-subgraphs, let's start with the definitions and assumptions provided.

Definitions and Initial Setup

• A $\Theta$-graph consists of three internally vertex-disjoint paths joining two vertices.
• Graph $G$ has $a$ blocks which are isomorphic to $K_2$ and $b$ blocks which are cycles.
• Let the lengths of these $b$ cycles be $c_1, c_2, \ldots, c_b$.

Expression of $n$ and $m$ in Terms of $a$, $b$, and Cycle Lengths

Each block $K_2$ contributes 2 vertices and 1 edge:

• There are $2a$ vertices contributed by the $a$ $K_2$ blocks.
• There are $a$ edges from these $K_2$ blocks.

Each cycle $C_i$ (of length $c_i$) contributes $c_i$ vertices and $c_i$ edges:

• There are $\sum_{i=1}^{b} c_i$ vertices contributed by the $b$ cycles.
• There are $\sum_{i=1}^{b} c_i$ edges from these cycles.

The total number of vertices $n$ is: $n = 2a + \sum_{i=1}^{b} c_i$

The total number of edges $m$ is: $m = a + \sum_{i=1}^{b} c_i$

Upper Bound on $m$

To obtain an upper bound on $m$, let's consider the constraints imposed by the absence of $\Theta$-subgraphs. A graph without $\Theta$-subgraphs cannot have any two vertices connected by more than two internally vertex-disjoint paths. This limitation affects how we construct the blocks.

Constructing the Graph for Maximum $m$

One way to construct such a graph while maximizing the number of edges is to use cycles and $K_2$ blocks in a specific arrangement.

Case for Odd $n$

For odd $n$, it has been derived that: $m_{\max} = \left\lfloor \frac{3}{2} (n - 1) \right\rfloor$

Case for Even $n$

For even $n$, we need to consider a slightly different construction to avoid $\Theta$-subgraphs while maximizing edges.

If $n$ is even, let's denote $n = 2k$ where $k$ is an integer. The maximum number of edges can be obtained by forming a cycle of length $2k$.

Constructing the Cycle

Consider a single cycle of length $2k$ (i.e., all vertices forming one big cycle). This avoids the $\Theta$-subgraphs and maximizes the number of edges: $m = n$

Therefore, for even $n$: $m_{\max} = n$

Conclusion

For odd $n$, the maximum number of edges in an $n$-vertex graph with no $\Theta$-subgraphs is: $m_{\max} = \left\lfloor \frac{3}{2} (n - 1) \right\rfloor$

For even $n$, the maximum number of edges in an $n$-vertex graph with no $\Theta$-subgraphs is: $m_{\max} = n$

This bound can be achieved by forming a single cycle of length $n$ for even $n$.

Do you want details or have any questions?

Here are 5 numbered questions you could ask:

1. How do you construct a $\Theta$-free graph for odd $n$?
2. Can you explain why cycles avoid $\Theta$-subgraphs?
3. How would the construction change if self-loops were allowed?
4. Can this method be extended to hypergraphs?
5. What are the implications of this result in graph theory?

Tip: When dealing with combinatorial structures, always check for forbidden configurations to simplify problem-solving strategies.