gpt pro 5.2 attempt on Math-5404, version 2

Erdős--Hajnal reciprocal cycle-length sum (Round 2)

1) ROUND-2 OBJECTIVE

Path chosen: (A) proof-oriented gap closure. The Round-1 status left open the extremal question: for fixed $(n,m)$ with $m=kn$, is the quantity $S(G):=\sum_{i=1}^t 1/a_i$ minimized by a complete bipartite graph?

In this round I aim to advance a proof strategy by (i) rigorously ruling out a large, natural competitor class (very dense balanced bipartite graphs) using a sharp bipancyclicity theorem, and (ii) adding one new fully rigorous solved instance beyond Round-1 by an extremal "no $C_3$ and no $C_4$" argument. I also extend the Round-1 computational verification to further parameter pairs to probe for counterexamples.

2) ROUND-1 FOUNDATION USED

I rely on the following Round-1 items (used as black boxes):

• $R1--L1$ Cycle spectrum of $K_{a,b}$: if $a\le b$, then the cycle lengths in $K_{a,b}$ are exactly $4,6,\dots,2a$, hence \[S(K_{a,b})=\sum_{j=2}^{a}\frac{1}{2j}=\tfrac12\,(H_a-1).\]
• $R1--L2$ $k$-core lemma: if $|E(G)|\ge kn$ then $G$ has a nonempty subgraph with minimum degree at least $k$.
• (R1--Comp) Exhaustive minimizer checks for: $(n,m)=(6,9)$ (minimizer $K_{3,3}$), $(7,10)$ (minimizer $K_{2,5}$), $(7,12)$ (minimizer $K_{3,4}$).

3) NEW INSIGHT / TOOL (ROUND-2)

Two genuinely new ingredients are added:

1. A sharp bipancyclicity theorem for dense balanced bipartite graphs (Entringer--Schmeichel). This gives a rigorous mechanism explaining why "more balanced" bipartite competitors with many edges are forced to realize all even cycle lengths, hence must have comparatively large $S(G)$.
2. A new exact extremal instance: $(n,m)=(8,12)$. Using only elementary extremal counting (no computer enumeration), I prove that for every $8$-vertex graph with $12$ edges, \[S(G)\ge \tfrac14,\] with equality attained by $K_{2,6}$ (and possibly other graphs whose set of cycle lengths is exactly $\{4\}$). This extends the Round-1 exact checks to the next complete-bipartite instance.

4) ATTACK PLAN (ROUND-2)

Round-1 gap: no proof (or disproof) that a complete bipartite graph always minimizes $S(G)$ for given $(n,m)$.

Plan in this round:

• Gap-closure subgoal A: prove that a very large class of potential counterexamples --- dense balanced bipartite graphs --- necessarily has many even cycle lengths (bipancyclic), forcing $S(G)$ to be large.
• Gap-closure subgoal B: prove at least one new parameter pair beyond Round-1 where the minimizer question has a complete rigorous resolution.
• Gap-closure subgoal C (sanity check): expand exact enumerations within all bipartite partitions for several new $(n,m)$ instances to see whether any bipartite counterexample exists at small sizes.

5) WORK (ROUND-2)

5.1. Dense balanced bipartite graphs are forced to be bipancyclic

Theorem (Entringer--Schmeichel, edge threshold for bipancyclicity).

Let $G$ be a balanced bipartite graph with bipartition $(X,Y)$ where $|X|=|Y|=n$. If \[ |E(G)|\ge n^2-n+2, \] then $G$ is bipancyclic, i.e. it contains a cycle of every even length $2\ell$ for $2\le \ell\le n$.

Source used: the above statement is quoted (with sharpness discussion) in Adamus' paper on weakly pancyclic bipartite graphs; see References.

Corollary 5.1 (forced lower bound on $S(G)$ in the balanced dense regime).

Under the theorem's hypotheses, the set of cycle lengths of $G$ contains all even lengths $4,6,\dots,2n$, hence \[ S(G)\;\ge\;\sum_{\ell=2}^{n}\frac{1}{2\ell} \;=\;\tfrac12\,(H_n-1). \] Proof. Immediate: each even length $2\ell$ appears at least once, and $S(G)$ sums $1/(\text{length})$ over the distinct lengths present. $\square$

Concrete implication (the first nontrivial balanced case).

For $2n=12$ vertices (so $n=6$), the threshold is $n^2-n+2=36-6+2=32$. Thus any balanced bipartite graph on $12$ vertices with $32$ edges must contain cycles of all even lengths $4,6,8,10,12$, giving \[ S(G)\ge \frac14+\frac16+\frac18+\frac1{10}+\frac1{12}=\frac{29}{40}=0.725. \] This rigorously rules out the entire "balanced bipartite" class as minimizers at $(n,m)=(12,32)$, consistent with the computations reported in §5.3.

5.2. A new exact solved instance: $(n,m)=(8,12)$

Recall that $K_{2,6}$ has $n=8$ vertices and $m=12$ edges, and by $R1--L1$ \[ S(K_{2,6})=\frac14. \] I now prove that no $8$-vertex $12$-edge graph can have smaller $S$.

Proposition 5.2.

Let $G$ be a graph on $8$ vertices with $12$ edges. Then $S(G)\ge \tfrac14$. Consequently, the minimum of $S(G)$ over all $(n,m)=(8,12)$ graphs is exactly $\tfrac14$, attained by $K_{2,6}$.

Proof.

If $G$ contains a triangle ($C_3$), then $3$ is among the distinct cycle lengths and so \[ S(G)\ge \frac13>\frac14. \] So any putative minimizer must be triangle-free.

Assume henceforth that $G$ is triangle-free. We claim that $G$ must contain a $4$-cycle $C_4$. Indeed, suppose for contradiction that $G$ is both $C_3$-free and $C_4$-free. Let $\Delta$ be the maximum degree of $G$, and pick a vertex $v$ of degree $\Delta$. Write $A=N(v)$ and $B=V(G)\setminus (\{v\}\cup A)$. Then:

• Since $G$ is triangle-free, $A$ is an independent set.
• Since $G$ is $C_4$-free, every vertex of $B$ has at most one neighbor in $A$ (otherwise two distinct neighbors in $A$ together with $v$ form a $4$-cycle). Therefore the number of edges between $A$ and $B$ is at most $|B|$.
• The induced subgraph $G[B]$ is also $C_3$-free and $C_4$-free. In particular, if $|B|\le 4$, then $|E(G[B])|\le 3$ because any $4$-vertex graph with $\ge4$ edges contains a cycle, and the only cycle lengths available on $4$ vertices are $3$ and $4$.

Now split into cases on $\Delta$.

1. If $\Delta\le 2$, then $ |E(G)|\le \tfrac12\sum_{u\in V} d(u)\le \tfrac12\cdot 8\cdot 2=8<12, $ contradiction.
2. If $\Delta\ge 4$, then $|B|=8-1-\Delta\le 3$, so $|E(G[B])|\le 2$. Using the edge decomposition $|E(G)|\le \Delta + |E(A,B)| + |E(G[B])|$, we get \[ |E(G)|\le \Delta + |B| + 2\le 4+3+2=9<12, \] contradiction.
3. The remaining possibility is $\Delta=3$, so $|B|=8-1-3=4$ and hence $|E(G[B])|\le 3$. Again \[ |E(G)|\le \Delta + |B| + |E(G[B])|\le 3+4+3=10<12, \] contradiction.

All cases contradict $|E(G)|=12$, so our assumption that $G$ is $C_3$-free and $C_4$-free is false. Since we already assumed $G$ is triangle-free, it follows that $G$ contains a $4$-cycle. Hence $4\in\{a_1,\dots,a_t\}$ and therefore $ S(G)\ge 1/4. $

Finally, $K_{2,6}$ has exactly one distinct cycle length (namely $4$), so $S(K_{2,6})=1/4$ and the lower bound is sharp. $\square$

5.3. Extended exact enumerations over bipartite competitors (new data)

The Round-1 exact checks were for $(n,m)=(6,9),(7,10),(7,12)$. I extended the same style of complete enumeration over all bipartite graphs in every feasible bipartition for several further parameter pairs where a spanning $K_{a,b}$ exists.

Method (exact, but restricted to bipartite graphs).

Fix a bipartition size $(p,q)$ with $p+q=n$ and $pq\ge m$. Every bipartite graph on $(p,q)$ with exactly $m$ edges is obtained by deleting exactly $pq-m$ edges from $K_{p,q}$. For each such deletion set (enumerated exhaustively), I tested existence of even cycles of each possible length $4,6,\dots,2\min\{p,q\}$ via backtracking on the required distinct representatives; this is exact at these small sizes.

Findings (no bipartite counterexample in these cases).

Write $S^*(n,m)$ for the minimum of $S(G)$ over bipartite graphs with these parameters. In each listed case, the spanning complete bipartite graph has strictly smaller $S$ than every bipartite graph with the same $(n,m)$ but a different bipartition. This strengthens the Round-1 picture: among small instances checked so far, potential counterexamples (if they exist) must be non-bipartite and exploit odd cycle lengths in a way that overcomes the apparent "many even lengths" penalty.

6) ADVERSARIAL VERIFICATION

Verification of Proposition 5.2.

The only delicate step is the edge upper bound for a $C_3,C_4$-free graph on $8$ vertices. The argument uses only: (i) $A=N(v)$ is independent (triangle-free), (ii) every $b\in B$ has at most one neighbor in $A$ ($C_4$-free), and (iii) $|E(G[B])|\le 3$ when $|B|=4$ because any $4$-vertex graph with $\ge4$ edges contains a $3$- or $4$-cycle. All are correct and purely local; no hidden assumptions on connectivity are used. The case split on $\Delta$ is exhaustive.

Verification of Corollary 5.1 usage.

The Entringer--Schmeichel theorem applies only to balanced bipartite graphs. I apply it only in that setting (e.g. $(6,6)$ bipartition when $n=12$), and only at or above its edge threshold. For $(n,m)=(12,32)$, the threshold is exactly met, so bipancyclicity (hence cycle-length set $\{4,6,8,10,12\}$) is forced.

Potential weakness of §5.3.

The enumerations in §5.3 rule out only bipartite competitors. A non-bipartite graph could, in principle, have fewer distinct cycle lengths by concentrating cycle structure in odd lengths and/or a narrower interval. No such example is constructed here, so the global minimizer question remains open.

7) FINAL (EXACTLY ONE)

UNRESOLVED (BUT STRICTLY ADVANCED).

Strict advances beyond Round-1:

• A new exact solved case: for $(n,m)=(8,12)$ the minimum is $1/4$ and is attained by $K_{2,6}$ (Proposition 5.2).
• A new external structural tool (Entringer--Schmeichel) giving a rigorous "forced many even cycle lengths" mechanism for dense balanced bipartite graphs $\S5.1$.
• Expanded exact enumeration evidence across additional $(n,m)$ pairs, ruling out bipartite counterexamples for several new complete-bipartite instances $\S5.3$.

8) COMPLETION ESTIMATE (MANDATORY)

COMPLETION: 45%

9) REFERENCES

References

• [GKS84] A. Győri, J. Krivelevich, and M. Szabó, ? (as cited in Round-1; original theorem used there).
• [LiMo20] H. Liu and D. Montgomery, ? (as cited in Round-1; asymptotic sharpness $\tfrac12\log k$).
• [Adamus2009] K. Adamus, On weakly pancyclic bipartite graphs, Discussiones Mathematicae Graph Theory 29 $2009$, 503--512. (Quotes the Entringer--Schmeichel bipancyclicity threshold $|E|\ge n^2-n+2$ for balanced bipartite graphs.)
• [EntringerSchmeichel1988] R. Entringer and E. Schmeichel, Edge conditions and cycle structure in bipartite graphs, Ars Combinatoria 26 $1988$, 229--232.