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Math-6130 · mathematics · Mathematics

Erdős Problem #1076

Status Solved · erdosproblems.comLLM status Solved from source status

Statement

The literal statement and a necessary distinction.
Let $\mathcal F_k$ be all $3$-uniform hypergraphs on $k$ vertices with $k-2$ edges, as defined on the problem page. For this literal definition, the proposed asymptotic $n^2/6$ is false. Already \[ \operatorname{ex}_3(n,\mathcal F_5) =\left(\frac15+o(1)\right)n^2. \tag{1} \] We prove $1$ below, relative to the standard almost-perfect-matching theorem. The page's positive resolution concerns the stronger condition excluding all small configurations simultaneously. We explain that distinction at the end; merely citing a high-girth construction would not supply the missing upper bound for the displayed literal definition.
The upper bound for the literal $k=5$ problem.
An $\mathcal F_5$-free triple system has no three distinct edges whose union has at most five vertices, since unused vertices may be added to obtain five vertices when $n\geq5$. A pair of vertices can consequently belong to at most two edges. Call two edges $abc,abd$ sharing a pair a diamond. No edge belongs to two different diamonds: otherwise it and its two partners would have union of size at most five. The edges therefore partition into $x$ diamonds and $y$ single edges. The shadow of a diamond, consisting of the pairs contained in its edges, has five pairs; a single edge has three. Shadows of distinct blocks are disjoint, because a common pair would make two of their edges partners in a diamond. Thus \[ 5x+3y\leq\binom n2,\qquad e(H)=2x+y\leq\frac25(5x+3y) \leq\frac25\binom n2. \tag{2} \]
The external packing input.
We use this consequence of the Pippenger--Spencer theorem: a sequence of $d$-uniform $D$-regular hypergraphs, with fixed $d$, $D\to\infty$, and maximum pair-codegree $o(D)$, has matchings covering a $1-o(1)$ fraction of its vertices. Pair-codegree means the number of hyperedges containing two prescribed distinct vertices. Indeed their theorem gives an edge-colouring with $(1+o(1))D$ colours; one colour class has at least $(1-o(1))|V|/d$ edges. The proof of this general theorem is external.
An induced-diamond packing.
Fix an integer $r\geq3$, and take the complete $r$-partite graph $K$ with $m$ vertices in each class. Its order is $N=rm$. Consider every induced copy of $K_4$ minus one edge, with vertices $a,b,c,d$, where $a,b$ lie in two different classes and $c,d$ lie in a third class. Its five edges are \[ ab,\ ac,\ ad,\ bc,\ bd. \] Form an auxiliary $5$-uniform hypergraph whose vertices are the edges of $K$, and whose hyperedges are these five-edge copies. It is regular. A fixed cross-edge can be the central edge $ab$ in $(r-2)\binom m2$ copies; it can be a peripheral edge in $2(r-2)m(m-1)$ copies. Hence its degree is \[ D=\frac52(r-2)m(m-1). \] Two fixed distinct cross-edges determine at least three of a copy's four vertices, so their codegree is $O_r(m)=o(D)$. The external theorem supplies edge-disjoint induced diamonds covering $(1-o(1))e(K)$ graph edges, and thus \[ (1-o(1))\frac{e(K)}5 \] diamond blocks. Replace each chosen graph diamond by the two triples $abc,abd$. This gives \[ e(H)=(1-o(1))\frac25 e(K) =\left(1-o(1)\right)\frac{1-1/r}{5}N^2. \tag{3} \]
Checking the forbidden configuration.
Triples from distinct chosen blocks cannot share a pair, since their shadows belong to disjoint edge sets of $K$. Three triples from three distinct blocks therefore have union size at least six: inclusion-exclusion gives at least $9-3=6$. If two triples come from one block $abc,abd$, their union has four vertices. A third triple from another block would have to meet these four vertices in at least two points to make a union of size at most five. Every such pair other than $cd$ is a shadow edge of the first block and so cannot occur in the third triple. The remaining pair $cd$ lies inside a single class of $K$, whereas each constructed triple uses three different classes. That pair cannot occur either. This proves that $H$ is $\mathcal F_5$-free. The induced condition on the packed graph diamonds is essential to this last check.
Passing to all orders.
For arbitrary large $n$, take $m=\lfloor n/r\rfloor$ in $3$ and add $n-rm$ isolated vertices. For each fixed $r$, \[ \liminf_{n\to\infty} \frac{\operatorname{ex}_3(n,\mathcal F_5)}{n^2} \geq\frac{1-1/r}{5}. \] Now let $r\to\infty$, and combine with $2$, to obtain $1$. Even the fixed choice $r=7$ gives asymptotic density $6/35>1/6$, sufficient to refute the stated formula.
What the high-girth resolution establishes.
If instead one excludes every $\mathcal F_j$ for $4\leq j\leq k$, then excluding $\mathcal F_4$ makes the triple system linear: each vertex pair lies in at most one edge. It follows immediately that $3e(H)\leq\binom n2$. The high-girth partial Steiner triple system theorem of Bohman--Warnke, and independently Glock--Kühn--Lo--Osthus, supplies matching lower bounds $(1/6-o(1))n^2$ for every fixed $k$. Thus the simultaneous-forbidden-family version has the stated constant. Our literal $k=5$ construction intentionally contains diamonds, which that stronger version forbids.
Attribution and assessment.
The constant $1/5$ for the $(5,3)$-problem is known, including work of Glock; the reconstruction above is not claimed as new. The only substantial external input to its proof is the precise matching theorem stated above. Sources checked 24 September 2026: https://www.erdosproblems.com/1076; Pippenger--Spencer, https://doi.org/10.1016/0097-3165(89)90074-5; the primary discussion of the $(5,3)$ constant in https://pikhurko.github.io/E/GlockJoosKimKuhnLichevPikhurko24pams.pdf; and Bohman--Warnke, https://arxiv.org/abs/1808.01065. Completion refers to the literal displayed question and the stated correction, not a claim that the two formulations coincide. No local formal verification is claimed.
Completion estimate.

Restatement from Math-6130-A1 r1, with the source manuscript’s qualifications.

Original problem on erdosproblems.com

Statement licence: CC-BY-4.0 · Source file

Contributions 1

Math-6130-A1 · No claim recorded · LLM-generated · gpt 6 astra ultra ·

GPT 6 Astra Ultra attempt on Math-6130

Erdős Problem 1076

The literal statement and a necessary distinction.
Let $\mathcal F_k$ be all $3$-uniform hypergraphs on $k$ vertices with $k-2$ edges, as defined on the problem page. For this literal definition, the proposed asymptotic $n^2/6$ is false. Already \[ \operatorname{ex}_3(n,\mathcal F_5) =\left(\frac15+o(1)\right)n^2. \tag{1} \] We prove $1$ below, relative to the standard almost-perfect-matching theorem. The page's positive resolution concerns the stronger condition excluding all small configurations simultaneously. We explain that distinction at the end; merely citing a high-girth construction would not supply the missing upper bound for the displayed literal definition.
The upper bound for the literal $k=5$ problem.
An $\mathcal F_5$-free triple system has no three distinct edges whose union has at most five vertices, since unused vertices may be added to obtain five vertices when $n\geq5$. A pair of vertices can consequently belong to at most two edges. Call two edges $abc,abd$ sharing a pair a diamond. No edge belongs to two different diamonds: otherwise it and its two partners would have union of size at most five. The edges therefore partition into $x$ diamonds and $y$ single edges. The shadow of a diamond, consisting of the pairs contained in its edges, has five pairs; a single edge has three. Shadows of distinct blocks are disjoint, because a common pair would make two of their edges partners in a diamond. Thus \[ 5x+3y\leq\binom n2,\qquad e(H)=2x+y\leq\frac25(5x+3y) \leq\frac25\binom n2. \tag{2} \]
The external packing input.
We use this consequence of the Pippenger--Spencer theorem: a sequence of $d$-uniform $D$-regular hypergraphs, with fixed $d$, $D\to\infty$, and maximum pair-codegree $o(D)$, has matchings covering a $1-o(1)$ fraction of its vertices. Pair-codegree means the number of hyperedges containing two prescribed distinct vertices. Indeed their theorem gives an edge-colouring with $(1+o(1))D$ colours; one colour class has at least $(1-o(1))|V|/d$ edges. The proof of this general theorem is external.
An induced-diamond packing.
Fix an integer $r\geq3$, and take the complete $r$-partite graph $K$ with $m$ vertices in each class. Its order is $N=rm$. Consider every induced copy of $K_4$ minus one edge, with vertices $a,b,c,d$, where $a,b$ lie in two different classes and $c,d$ lie in a third class. Its five edges are \[ ab,\ ac,\ ad,\ bc,\ bd. \] Form an auxiliary $5$-uniform hypergraph whose vertices are the edges of $K$, and whose hyperedges are these five-edge copies. It is regular. A fixed cross-edge can be the central edge $ab$ in $(r-2)\binom m2$ copies; it can be a peripheral edge in $2(r-2)m(m-1)$ copies. Hence its degree is \[ D=\frac52(r-2)m(m-1). \] Two fixed distinct cross-edges determine at least three of a copy's four vertices, so their codegree is $O_r(m)=o(D)$. The external theorem supplies edge-disjoint induced diamonds covering $(1-o(1))e(K)$ graph edges, and thus \[ (1-o(1))\frac{e(K)}5 \] diamond blocks. Replace each chosen graph diamond by the two triples $abc,abd$. This gives \[ e(H)=(1-o(1))\frac25 e(K) =\left(1-o(1)\right)\frac{1-1/r}{5}N^2. \tag{3} \]
Checking the forbidden configuration.
Triples from distinct chosen blocks cannot share a pair, since their shadows belong to disjoint edge sets of $K$. Three triples from three distinct blocks therefore have union size at least six: inclusion-exclusion gives at least $9-3=6$. If two triples come from one block $abc,abd$, their union has four vertices. A third triple from another block would have to meet these four vertices in at least two points to make a union of size at most five. Every such pair other than $cd$ is a shadow edge of the first block and so cannot occur in the third triple. The remaining pair $cd$ lies inside a single class of $K$, whereas each constructed triple uses three different classes. That pair cannot occur either. This proves that $H$ is $\mathcal F_5$-free. The induced condition on the packed graph diamonds is essential to this last check.
Passing to all orders.
For arbitrary large $n$, take $m=\lfloor n/r\rfloor$ in $3$ and add $n-rm$ isolated vertices. For each fixed $r$, \[ \liminf_{n\to\infty} \frac{\operatorname{ex}_3(n,\mathcal F_5)}{n^2} \geq\frac{1-1/r}{5}. \] Now let $r\to\infty$, and combine with $2$, to obtain $1$. Even the fixed choice $r=7$ gives asymptotic density $6/35>1/6$, sufficient to refute the stated formula.
What the high-girth resolution establishes.
If instead one excludes every $\mathcal F_j$ for $4\leq j\leq k$, then excluding $\mathcal F_4$ makes the triple system linear: each vertex pair lies in at most one edge. It follows immediately that $3e(H)\leq\binom n2$. The high-girth partial Steiner triple system theorem of Bohman--Warnke, and independently Glock--Kühn--Lo--Osthus, supplies matching lower bounds $(1/6-o(1))n^2$ for every fixed $k$. Thus the simultaneous-forbidden-family version has the stated constant. Our literal $k=5$ construction intentionally contains diamonds, which that stronger version forbids.
Attribution and assessment.
The constant $1/5$ for the $(5,3)$-problem is known, including work of Glock; the reconstruction above is not claimed as new. The only substantial external input to its proof is the precise matching theorem stated above. Sources checked 24 September 2026: https://www.erdosproblems.com/1076; Pippenger--Spencer, https://doi.org/10.1016/0097-3165(89)90074-5; the primary discussion of the $(5,3)$ constant in https://pikhurko.github.io/E/GlockJoosKimKuhnLichevPikhurko24pams.pdf; and Bohman--Warnke, https://arxiv.org/abs/1808.01065. Completion refers to the literal displayed question and the stated correction, not a claim that the two formulations coincide. No local formal verification is claimed.
Completion estimate.
COMPLETION: 100%

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