gpt pro 5.4 attempt on Math-694

Erdős Problem #151 (Round 2): a new obstruction via common neighbourhoods

1) ROUND-2 OBJECTIVE

For Problem #151 I pursue path (C): I do not have a full proof or a counterexample, so the correct Round-2 goal is to isolate a sharp structural obstruction that any counterexample would have to satisfy.

Round 1 already proved the general inequality \[ \tau(G)\le n-\alpha(G), \] where $\alpha(G)$ is the independence number. The new Round-2 goal is to add a second, genuinely different mechanism that also forces small clique transversals.

2) Round-1 FOUNDATION USED

I use the following Round-1 facts.

1. The Round-1 lemma \[ \tau(G)\le n-\alpha(G). \]
2. The Round-1 proof that the conjecture holds trivially for triangle-free graphs.
3. The Round-1 observation that the general bound $\tau(G)\le n-\sqrt n$ is easy but much weaker than the target $\tau(G)\le n-H(n)$.

3) NEW INSIGHT / TOOL (ROUND 2)

The new tool is a codegree lemma.

For an edge $uv\in E(G)$, let \[ \mathrm{codeg}(u,v):=|N(u)\cap N(v)|. \] The new observation is that \[ \tau(G)\le n-\mathrm{codeg}(u,v) \] for every edge $uv$.

This is different in nature from the Round-1 bound involving $\alpha(G)$: it uses a dense local configuration (a large common neighbourhood of one edge), not a sparse global configuration (a large independent set).

4) ATTACK PLAN (ROUND 2)

The exact gap after Round 1 was that we only had one broad sufficient condition for the conjecture: \[ \alpha(G)\ge H(n)\quad\Longrightarrow\quad \tau(G)\le n-H(n). \] I will prove a second broad sufficient condition: \[ \exists\, uv\in E(G)\mbox{ with }|N(u)\cap N(v)|\ge H(n) \quad\Longrightarrow\quad \tau(G)\le n-H(n). \]

Combining the two will give a structural obstruction: every counterexample must satisfy simultaneously \[ \alpha(G)<H(n) \quad\mbox{and}\quad |N(u)\cap N(v)|<H(n)\ \mbox{for every edge }uv. \] In particular, every counterexample must also satisfy \[ \omega(G)\le H(n)+1, \] because any clique of size $r$ gives an edge with at least $r-2$ common neighbours.

5) WORK (ROUND 2)

Lemma (common-neighbourhood transversal).

For every graph $G$ on $n$ vertices and every edge $uv\in E(G)$, \[ \tau(G)\le n-|N(u)\cap N(v)|. \]

Proof.

Fix an edge $uv$. Let \[ S:=N(u)\cap N(v), \qquad T:=V(G)\setminus S. \] Then \[ |T|=n-|S|=n-|N(u)\cap N(v)|. \] I claim that $T$ is a clique transversal.

Suppose not. Then there exists a maximal clique $C$ of size at least $2$ with \[ C\cap T=\emptyset. \] So $C\subseteq S=N(u)\cap N(v)$. In particular, every vertex of $C$ is adjacent to $u$. Therefore \[ C\cup\{u\} \] is a clique strictly larger than $C$, contradicting the maximality of $C$.

So $T$ is indeed a clique transversal, and hence \[ \tau(G)\le |T|=n-|N(u)\cap N(v)|. \]

Corollary 1.

If $G$ has an edge $uv$ such that \[ |N(u)\cap N(v)|\ge H(n), \] then \[ \tau(G)\le n-H(n). \]

Proof.

Apply the lemma to that edge.

Corollary 2.

If \[ \max\left\{\alpha(G),\ \max_{uv\in E(G)} |N(u)\cap N(v)|\right\}\ge H(n), \] then \[ \tau(G)\le n-H(n). \]

Proof.

If $\alpha(G)\ge H(n)$, use the Round-1 bound $\tau(G)\le n-\alpha(G)$. Otherwise, if some edge has common neighbourhood size at least $H(n)$, use Corollary 1.

Corollary 3 (clique obstruction).

If $G$ is a counterexample to the conjectured inequality \[ \tau(G)\le n-H(n), \] then \[ \omega(G)\le H(n)+1. \]

Proof.

Suppose $\omega(G)\ge H(n)+2$, and let $K$ be a clique of that size. Pick any edge $uv$ of $K$. Then every vertex of $K\setminus\{u,v\}$ lies in $N(u)\cap N(v)$, so \[ |N(u)\cap N(v)|\ge |K|-2\ge H(n). \] By Corollary 1 the conjectured inequality would then hold, contradiction.

Corollary 4 (triangle-count version).

Let $t(G)$ denote the number of triangles of $G$ and $e(G)$ the number of edges. Then \[ \tau(G)\le n-\left\lceil \frac{3t(G)}{e(G)}\right\rceil \] whenever $e(G)>0$.

Proof.

The average value of $|N(u)\cap N(v)|$ over all edges $uv$ equals \[ \frac{1}{e(G)}\sum_{uv\in E(G)} |N(u)\cap N(v)| = \frac{3t(G)}{e(G)}, \] since each triangle contributes $1$ to the codegree of each of its $3$ edges. Therefore some edge $uv$ satisfies \[ |N(u)\cap N(v)|\ge \left\lceil \frac{3t(G)}{e(G)}\right\rceil. \] Now apply the lemma.

Round-2 structural obstruction.

Combining Round 1 with Corollary 2, any counterexample $G$ on $n$ vertices must satisfy \[ \alpha(G)<H(n) \quad\mbox{and}\quad |N(u)\cap N(v)|<H(n)\ \mbox{for every edge }uv. \] Equivalently, every counterexample must simultaneously have:

1. no independent set of size $H(n)$, and
2. no edge lying in $H(n)$ or more triangles.

This sharply narrows the possible shape of a counterexample.

6) ADVERSARIAL VERIFICATION

I check the possible pitfalls.

1. Why does a clique contained in $N(u)\cap N(v)$ contradict maximality? Because every vertex of that clique is adjacent to $u$, so adjoining $u$ enlarges the clique. There is no need to use both $u$ and $v$.
2. Does the lemma fail for isolated edges or sparse graphs? No. If $N(u)\cap N(v)=\emptyset$, the lemma simply gives the trivial estimate $\tau(G)\le n$.
3. Does Corollary 3 use more than is proved? No. A clique of size $r$ gives an edge with at least $r-2$ common neighbours, and that is exactly what the lemma needs.
4. Is the triangle-count version correct? Yes. Each triangle contributes to the common-neighbour count of exactly its three edges, so the averaging identity is exact.

7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED).

Round 2 does not prove the Erdős--Gallai--Tuza conjecture, but it adds a genuinely new sufficient condition and a sharp obstruction:

1. Round 1 showed the conjecture whenever $\alpha(G)\ge H(n)$.
2. Round 2 now also shows the conjecture whenever some edge has at least $H(n)$ common neighbours.
3. Therefore any counterexample must have both \[ \alpha(G)<H(n) \quad\mbox{and}\quad \max_{uv\in E(G)}|N(u)\cap N(v)|<H(n), \] and in particular \[ \omega(G)\le H(n)+1. \]

This is a real narrowing of the search space for counterexamples and a clear structural advance beyond Round 1.

8) COMPLETION ESTIMATE

COMPLETION: 58%

9) REFERENCES

1. Erdős Problems, Problem #151, checked on 7 March 2026.
2. Erdős Problems, Problem #610 (for the currently recorded general bound $\tau(G)\le n-\sqrt{2n}+O(1)$), checked on 7 March 2026.