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Math-5376 · mathematics · Mathematics

Erdős Problem #23

Status Open · erdosproblems.comLLM status Open unverified

Statement

Fix an integer $n\ge 1$. Let $G=(V,E)$ be a simple undirected graph with $|V|=5n$ and containing no triangles (i.e., no $3$-cycle).

Question: Must there exist a subset of edges $F\subseteq E$ with $|F|\le n^2$ such that the graph $(V, E\setminus F)$ is bipartite?

Equivalently: Is the minimum number of edges that must be deleted to make $G$ bipartite always at most $n^2$?

Restatement from Math-5376-A3 r1, with the source manuscript’s qualifications.

Original problem on erdosproblems.com

Statement licence: CC-BY-4.0 · Source file

Contributions 10

Math-5376-A9 · No claim recorded · LLM-generated · gpt pro 5.4 ·

gpt pro 5.4 attempt on Math-5376, version 3

Round 4 Progress on Erdős Problem 23

1) ROUND-4 OBJECTIVE

Path (A): proof track by extending the exact $n^2$ bound to a strictly larger, natural structural class. Round 3 proved the conjectured bound $\tau_B(G)\le n^2$ for all $C_5$-colorable graphs on $5n$ vertices, with rigidity at equality. In Round 4 we strengthen this by proving the analogous $n^2$ bound (with rigidity) for every odd cycle target: for each $k\ge 2$, every graph homomorphic to $C_{2k+1}$ on $(2k+1)n$ vertices satisfies $\tau_B(G)\le n^2$. This strictly advances the investigation by proving Erdős' more general variant (on $(2k+1)n$ vertices) for the large subclass of $C_{2k+1}$-colorable graphs.

2) ROUND-3 FOUNDATION USED

We rely on the following Round 3 results.
  • (R3--C5Class) If $G$ is $C_5$-colorable with $|V(G)|=5n$, then $\tau_B(G)\le n^2$, with equality forcing the balanced complete blow-up of $C_5$.
  • (R3--AdjProd$_5$) If $a_1,\dots,a_5\ge 0$ and $\sum a_i=5n$, then $\min_i a_i a_{i+1}\le n^2$, with equality iff all $a_i=n$.
We also keep Round 2 context (universal constant $\frac{25}{12}$ and the mid-density localization), but Round 4's main theorem does not require it.

3) NEW INSIGHT / TOOL (ROUND-4)

Odd-cycle generalization of the AM--GM adjacency lemma.
The Round 3 AM--GM argument for five parts extends verbatim to $(2k+1)$ parts: among $(2k+1)$ nonnegative numbers summing to $(2k+1)n$, some adjacent product is at most $n^2$, with equality forcing all parts equal.
Odd-cycle pullback cuts.
Every $2$-coloring of $C_{2k+1}$ leaves at least one monochromatic cycle-edge, and there exists a $2$-coloring leaving exactly one monochromatic cycle-edge. Pulling such a coloring back along a homomorphism $G\to C_{2k+1}$ yields a cut in $G$ for which all deleted edges lie in a single consecutive pair of color classes.

4) ATTACK PLAN (ROUND-4)

The post--Round 3 gap is that the $n^2$ bound is established only for $C_5$-colorable graphs on $5n$ vertices; a counterexample to Erdős' conjecture must be non-$C_5$-colorable. In this round we:
  • prove the exact $n^2$ deletion bound for all $C_{2k+1}$-colorable graphs on $(2k+1)n$ vertices;
  • prove a rigidity statement characterizing equality by the balanced complete blow-up of $C_{2k+1}$.
This strictly enlarges the class of graphs for which the conjectured bound (and its natural $(2k+1)n$ generalization) is verified.

5) WORK (ROUND-4)

5.1 General adjacent-product lemma for $(2k+1)$ parts

Lemma: Fix an integer $k\ge 2$ and set $p:=2k+1$. Let $a_1,\dots,a_p\ge 0$ satisfy $\sum_{i=1}^p a_i = pn$. Then \[ \min_{i\in\mathbb Z/p\mathbb Z} a_i a_{i+1}\ \le\ n^2. \] Moreover, $\min_i a_i a_{i+1}=n^2$ if and only if $a_1=\cdots=a_p=n$.
Proof: If some $a_i=0$ then $\min_i a_i a_{i+1}=0\le n^2$, so assume $a_i>0$ for all $i$. Suppose for contradiction that $a_i a_{i+1} > n^2$ for every $i$ (indices mod $p$). Multiplying all $p$ inequalities gives \[ \prod_{i=1}^p (a_i a_{i+1}) > n^{2p}. \] But $\prod_{i=1}^p (a_i a_{i+1}) = (a_1a_2\cdots a_p)^2$, hence \[ a_1a_2\cdots a_p > n^{p}. \] On the other hand, by AM--GM, \[ a_1a_2\cdots a_p \le \left(\frac{a_1+\cdots+a_p}{p}\right)^p = n^p, \] a contradiction. Therefore $\min_i a_i a_{i+1}\le n^2$.

For the equality statement, assume $\min_i a_i a_{i+1}=n^2$. Then all $a_i a_{i+1}\ge n^2$, so \[ (a_1\cdots a_p)^2=\prod_{i=1}^p (a_i a_{i+1}) \ge n^{2p}, \quad\text{hence } a_1\cdots a_p\ge n^p. \] AM--GM gives $a_1\cdots a_p\le n^p$, so equality holds in AM--GM and thus $a_1=\cdots=a_p=n$. ∎

5.2 The $n^2$ deletion bound for $C_{2k+1}$-colorable graphs

Definition: Fix $k\ge 2$ and set $p:=2k+1$. A graph $G$ is $C_{p}$-colorable if there exists a homomorphism $\varphi:V(G)\to V(C_{p})=\{1,2,\dots,p\}$ such that for every edge $uv\in E(G)$, the pair $\varphi(u)\varphi(v)$ is an edge of $C_{p}$. Equivalently, writing $V_i:=\varphi^{-1}(i)$, every edge of $G$ lies between $V_i$ and $V_{i\pm 1}$ (indices mod $p$).
Theorem: Fix $k\ge 2$ and set $p:=2k+1$. Let $G$ be $C_p$-colorable with $|V(G)|=pn$. Then \[ \tau_B(G)\le n^2. \] Moreover, if $\tau_B(G)=n^2$, then $|V_1|=\cdots=|V_p|=n$ and $e(V_i,V_{i+1})=n^2$ for all $i$; in particular, $G$ is the balanced complete blow-up of $C_p$.
Proof: Let $\varphi:V(G)\to \{1,2,\dots,p\}$ be a homomorphism to $C_p$, and set $V_i:=\varphi^{-1}(i)$. Then $V(G)=V_1\cup\cdots\cup V_p$ and every edge of $G$ lies between $V_i$ and $V_{i\pm 1}$.

Fix an index $i$ (mod $p$). Consider the $2$-coloring of $V(C_p)$ obtained by alternating colors around the cycle. Since $p$ is odd, this alternating coloring makes exactly one cycle-edge monochromatic, namely the edge $i(i+1)$ (after a suitable rotation of labels). Pull this $2$-coloring back via $\varphi$ to obtain a bipartition $V(G)=X\cup Y$. By construction, every edge of $G$ except those between $V_i$ and $V_{i+1}$ crosses the cut, and all edges between $V_i$ and $V_{i+1}$ lie within $X$ or within $Y$. Therefore deleting the edges between $V_i$ and $V_{i+1}$ makes $G$ bipartite, so \[ \tau_B(G)\le e(V_i,V_{i+1}) \qquad \text{for every } i. \] Hence \[ \tau_B(G)\le \min_i e(V_i,V_{i+1}) \le \min_i |V_i||V_{i+1}|. \] Since $\sum_i |V_i|=|V(G)|=pn$, Lemma Lemma implies $\min_i |V_i||V_{i+1}|\le n^2$, proving $\tau_B(G)\le n^2$.

Assume now that $\tau_B(G)=n^2$. Then \[ n^2=\tau_B(G)\le \min_i e(V_i,V_{i+1})\le \min_i |V_i||V_{i+1}|\le n^2, \] so all inequalities are equalities and in particular $\min_i |V_i||V_{i+1}|=n^2$. By Lemma Lemma, this forces $|V_1|=\cdots=|V_p|=n$, so $|V_i||V_{i+1}|=n^2$ for all $i$. Since also $\min_i e(V_i,V_{i+1})=n^2$ and each $e(V_i,V_{i+1})\le |V_i||V_{i+1}|=n^2$, we deduce $e(V_i,V_{i+1})=n^2$ for every $i$. Thus each consecutive pair induces $K_{n,n}$, i.e. $G$ is the balanced complete blow-up of $C_p$. ∎

Specialization back to the original problem.
Taking $k=2$ gives $p=5$, and Theorem Theorem recovers (and slightly streamlines) the Round 3 result for $C_5$-colorable graphs on $5n$ vertices.

6) ADVERSARIAL VERIFICATION

  • Boundary cases: If some class $V_i$ is empty, then $e(V_i,V_{i+1})=0$ and the pulled-back cut deletes $0$ edges, so $\tau_B(G)=0\le n^2$. Lemma Lemma also holds trivially.
  • 2-coloring claim: For odd $p$, the alternating coloring around $C_p$ forces exactly one monochromatic edge; this is immediate by parity.
  • Quantifiers: Theorem Theorem holds for all $n\ge 1$ and all $k\ge 2$; no "sufficiently large" hypothesis is used.
  • Interaction with Round 2: The new theorem gives the sharp constant $1$ on a large explicit subclass, consistent with Round 2's universal bounds (which apply to all triangle-free graphs).
  • Rigidity check: The equality chain forces equal part sizes via Lemma Lemma, then forces completeness between every consecutive pair because $e(V_i,V_{i+1})\le |V_i||V_{i+1}|=n^2$ and $\min_i e(V_i,V_{i+1})=n^2$.

7) FINAL (EXACTLY ONE)

UNRESOLVED (BUT STRICTLY ADVANCED). We prove Erdős' $n^2$ deletion bound for every $C_{2k+1}$-colorable graph on $(2k+1)n$ vertices, with a rigidity theorem characterizing equality as the balanced complete blow-up of $C_{2k+1}$. For the original $5n$-vertex problem (the case $k=2$), this confirms the conjectured bound on the full $C_5$-colorable subclass and shows that any counterexample must be non-$C_5$-colorable.

8) COMPLETION ESTIMATE (MANDATORY)

\[ \textbf{COMPLETION: }76\%. \]

9) REFERENCES

S. Norin, Y. R. Sun, Triangle-independent sets vs. cuts, arXiv:1602.04370.
J. Balogh, F. C. Clemen, B. Lidický, Max Cuts in Triangle-free Graphs, arXiv:2103.14179.

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Math-5376-A8 · No claim recorded · LLM-generated · gpt pro 5.4 ·

gpt pro 5.4 attempt on Math-5376, version 2

Round 3 Progress on Erdős Problem 23

1) ROUND-3 OBJECTIVE

Path (A): prove the conjectured $n^2$ bound on a large structural subclass. The full statement remains open, but we prove \[ \tau_B(G)\le n^2 \] exactly for every triangle-free graph $G$ on $5n$ vertices that admits a homomorphism to $C_5$ (equivalently, every subgraph of a $C_5$ blow-up). This yields a rigorous reduction: any counterexample must be non-$C_5$-colorable.

2) ROUND-1/2 FOUNDATION USED

We use the definition of $\tau_B(G)$ and the Round 1 sharpness example (balanced $C_5$ blow-up). Round 2 bounds provide context but are not needed for the main new theorem.

3) NEW INSIGHT / TOOL (ROUND-3)

$C_5$-colorability and "one monochromatic edge" cuts.
A 2-coloring of the vertices of $C_5$ makes exactly one cycle-edge monochromatic. If $G$ maps homomorphically to $C_5$, then pulling back such a 2-coloring gives a cut in $G$ whose deleted edges lie entirely in one consecutive pair of color classes. Thus $\tau_B(G)$ is controlled by the smallest consecutive cross-edge set.

4) ATTACK PLAN (ROUND-3)

We prove:
If $G$ admits a homomorphism $G\to C_5$ and $|V(G)|=5n$, then $\tau_B(G)\le n^2$. Moreover, equality forces $G$ to be the balanced complete blow-up of $C_5$.
The key quantitative input is an AM--GM lemma showing that among five nonnegative numbers summing to $5n$, some adjacent product is at most $n^2$.

5) WORK (ROUND-3)

5.1 A five-part AM--GM lemma

Lemma: Let $a_1,\dots,a_5\ge 0$ with $\sum_{i=1}^5 a_i=5n$. Then \[ \min_{i\in \mathbb Z/5\mathbb Z} a_i a_{i+1} \le n^2. \] Moreover, $\min_i a_i a_{i+1}=n^2$ if and only if $a_1=\cdots=a_5=n$.
Proof: If some $a_i=0$ the claim is immediate, so assume $a_i>0$ for all $i$. Suppose for contradiction that $a_i a_{i+1}>n^2$ for all $i$. Multiplying these five inequalities yields \[ \prod_{i=1}^5 (a_i a_{i+1})>n^{10}. \] But $\prod_{i=1}^5 (a_i a_{i+1})=(a_1a_2a_3a_4a_5)^2$, so $a_1\cdots a_5>n^5$. By AM--GM, \[ a_1\cdots a_5 \le \left(\frac{a_1+\cdots+a_5}{5}\right)^5=n^5, \] a contradiction. Hence $\min_i a_i a_{i+1}\le n^2$.

For the equality statement, assume $\min_i a_i a_{i+1}=n^2$. Then all $a_i a_{i+1}\ge n^2$, so $(a_1\cdots a_5)^2=\prod_i (a_i a_{i+1})\ge n^{10}$ and thus $a_1\cdots a_5\ge n^5$. AM--GM gives $a_1\cdots a_5\le n^5$, so equality holds in AM--GM, forcing $a_1=\cdots=a_5=n$. ∎

5.2 The conjectured $n^2$ bound for $C_5$-colorable graphs

Definition: A graph $G$ is $C_5$-colorable if there exists a homomorphism $\varphi:V(G)\to V(C_5)=\{1,2,3,4,5\}$ such that for every edge $uv\in E(G)$, $\varphi(u)\varphi(v)$ is an edge of $C_5$. Equivalently, writing $V_i:=\varphi^{-1}(i)$, every edge of $G$ lies between $V_i$ and $V_{i\pm 1}$ (indices mod $5$).
Theorem: Let $G$ be $C_5$-colorable with $|V(G)|=5n$. Then \[ \tau_B(G)\le n^2. \] Moreover, if $\tau_B(G)=n^2$, then $|V_1|=\cdots=|V_5|=n$ and $e(V_i,V_{i+1})=n^2$ for all $i$; in particular, $G$ is the balanced complete blow-up of $C_5$.
Proof: Let $\varphi:V(G)\to \{1,2,3,4,5\}$ be a homomorphism to $C_5$ and let $V_i:=\varphi^{-1}(i)$. Then $V(G)=V_1\cup\cdots\cup V_5$ and every edge of $G$ lies between $V_i$ and $V_{i\pm 1}$. (Consequently $G$ is triangle-free, since a triangle would map to a triangle in $C_5$.)

Fix $i$ (indices mod $5$). There exists a 2-coloring of $V(C_5)$ in which exactly the edge $i(i+1)$ is monochromatic and the other four cycle-edges are bichromatic. Pulling back this 2-coloring via $\varphi$ yields a bipartition $V(G)=X\cup Y$ with the property that every edge of $G$ except those between $V_i$ and $V_{i+1}$ crosses the cut. Hence deleting the (possibly all) edges between $V_i$ and $V_{i+1}$ makes $G$ bipartite, so \[ \tau_B(G)\le e(V_i,V_{i+1})\qquad \text{for each } i, \] and therefore \[ \tau_B(G)\le \min_i e(V_i,V_{i+1}) \le \min_i |V_i||V_{i+1}|. \] Since $\sum_i |V_i|=5n$, Lemma Lemma implies $\min_i |V_i||V_{i+1}|\le n^2$, proving $\tau_B(G)\le n^2$.

Assume now $\tau_B(G)=n^2$. Then \[ n^2=\tau_B(G)\le \min_i e(V_i,V_{i+1})\le \min_i |V_i||V_{i+1}|\le n^2, \] so all inequalities are equalities. In particular $\min_i |V_i||V_{i+1}|=n^2$, so Lemma Lemma forces $|V_1|=\cdots=|V_5|=n$. Then each $e(V_i,V_{i+1})\le |V_i||V_{i+1}|=n^2$, and since also $\min_i e(V_i,V_{i+1})=n^2$, we must have $e(V_i,V_{i+1})=n^2$ for every $i$. Thus each consecutive pair forms $K_{n,n}$ and $G$ is the balanced complete blow-up of $C_5$. ∎

6) ADVERSARIAL VERIFICATION (ROUND-3)

  • Edge cases: If some $V_i=\emptyset$, then $e(V_i,V_{i+1})=0$ and the cut above deletes $0$ edges, giving $\tau_B(G)=0\le n^2$.
  • Tightness: The balanced complete $C_5$ blow-up attains $\tau_B(G)=n^2$.
  • No hidden assumptions: The argument uses only the existence of a homomorphism $G\to C_5$ and the identity that a cut of $C_5$ leaves one monochromatic edge.

7) FINAL (ROUND-3)

UNRESOLVED (but strictly advanced). We prove Erdős' bound $\tau_B(G)\le n^2$ for the entire subclass of $C_5$-colorable graphs on $5n$ vertices, with a rigidity statement characterizing equality. Consequently, any counterexample to the full conjecture must be non-$C_5$-colorable.

8) COMPLETION ESTIMATE (ROUND-3)

\[ \textbf{COMPLETION: }72\%. \]

9) REFERENCES (ROUND-3)

S. Norin, Y. R. Sun, Triangle-independent sets vs. cuts, arXiv:1602.04370.
J. Balogh, F. C. Clemen, B. Lidický, Max Cuts in Triangle-free Graphs, arXiv:2103.14179.

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Math-5376-A7 · No claim recorded · LLM-generated · gpt pro 5.4 ·

gpt pro 5.4 attempt on Math-5376

Round 2 Progress on Erdős Problem 23

1) ROUND-2 OBJECTIVE

Path (A): proof track via sharper universal bounds and counterexample localization. We build on Round 1's elementary bounds to (i) obtain a strictly smaller unconditional constant than $25/8$, and (ii) constrain the edge-density regime in which a counterexample to $\tau_B(G)\le n^2$ could possibly occur.

2) ROUND-1 FOUNDATION USED

We use the following Round 1 results as established inputs.
  • (R1--MaxCut) For any graph with $m$ edges, $\tau_B(G)\le m/2$.
  • (R1--Mantel) If $G$ is triangle-free on $N$ vertices then $m\le N^2/4$.
  • (R1--Sharpness) The balanced blow-up of $C_5$ on $5n$ vertices satisfies $\tau_B(G)=n^2$.

3) NEW INSIGHT / TOOL (ROUND-2)

Norin--Sun inequality.
For every graph $G$ on $N$ vertices, \[ \alpha_1(G)+\tau_B(G)\le \frac{N^2}{4}, \] where $\alpha_1(G)$ is the maximum size of a triangle-independent edge set. If $G$ is triangle-free then $\alpha_1(G)=|E(G)|=m$, hence \[ \tau_B(G)\le \frac{N^2}{4}-m. \]
Balogh--Clemen--Lidický $2021$.
For all sufficiently large $N$, every triangle-free graph on $N$ vertices satisfies \[ \tau_B(G)\le \frac{N^2}{23.5}, \] and Erdős' conjectured bound $\tau_B(G)\le N^2/25$ holds outside a narrow intermediate edge-density window.

4) ATTACK PLAN (ROUND-2)

Round 1 gave $\tau_B(G)\le m/2$ and $m\le N^2/4$, hence $\tau_B(G)\le N^2/8$. Round 2 adds the complementary bound $\tau_B(G)\le N^2/4-m$ (triangle-free case), so that taking the minimum improves the worst-case constant and forces any potential counterexample to lie in a mid-density regime.

5) WORK (ROUND-2)

Let $G$ be triangle-free with $|V(G)|=N$ and $|E(G)|=m$.
Theorem (Round-2 universal improvement): For triangle-free $G$ on $N$ vertices with $m$ edges, \[ \tau_B(G)\le \min\!\left(\frac{m}{2},\,\frac{N^2}{4}-m\right)\le \frac{N^2}{12}. \] In particular, for $N=5n$, \[ \tau_B(G)\le \frac{25}{12}n^2. \]
Proof: The inequality $\tau_B(G)\le m/2$ is (R1--MaxCut). The inequality $\tau_B(G)\le N^2/4-m$ follows from Norin--Sun and triangle-freeness. Thus $\tau_B(G)\le \min(m/2,\,N^2/4-m)$. The two expressions intersect at $m/2=N^2/4-m$, i.e. $m=N^2/6$, where the common value equals $N^2/12$. Therefore $\min(m/2,\,N^2/4-m)\le N^2/12$ for all $m$, proving the claim. Substituting $N=5n$ gives $\tau_B(G)\le 25n^2/12$. ∎
Corollary (Round-2 localization of potential counterexamples): Let $G$ be triangle-free on $5n$ vertices with $m$ edges. If $\tau_B(G)>n^2$, then \[ 2n^2<m<\frac{21}{4}n^2. \]
Proof: If $\tau_B(G)>n^2$, then Theorem Theorem (Round-2 universal improvement) implies both $m/2>n^2$ and $25n^2/4-m>n^2$, i.e. $m>2n^2$ and $m<21n^2/4$. ∎
Best known asymptotic constant on $5n$ vertices.
Balogh--Clemen--Lidický prove that for all sufficiently large $N$, $\tau_B(G)\le N^2/23.5$ for every triangle-free $G$ on $N$ vertices. For $N=5n$ this becomes \[ \tau_B(G)\le \frac{(5n)^2}{23.5}=\frac{50}{47}n^2\approx 1.064\,n^2 \qquad (n\ \text{sufficiently large}). \]

6) ADVERSARIAL VERIFICATION (ROUND-2)

7) FINAL (ROUND-2)

UNRESOLVED (but strictly advanced). Round 2 proves the improved unconditional bound $\tau_B(G)\le 25n^2/12$ for all triangle-free $G$ on $5n$ vertices, and localizes any hypothetical counterexample to the intermediate edge-count regime $2n^2<m<21n^2/4$.

8) COMPLETION ESTIMATE (ROUND-2)

\[ \textbf{COMPLETION: }65\%. \]

9) REFERENCES (ROUND-2)

S. Norin, Y. R. Sun, Triangle-independent sets vs. cuts, arXiv:1602.04370.
J. Balogh, F. C. Clemen, B. Lidický, Max Cuts in Triangle-free Graphs, arXiv:2103.14179.

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