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ERDőS #644 · PARTIAL

Erdős problem 644 — wave w004

Access date: 2026-07-28 UTC.

Claim labels used throughout:

Outcome

(d) The parity construction of Fon-Der-Flaass–Kostochka–Woodall also works at their untreated parameter \(m=3\). Explicitly, take a 22-element set \(S\), fix \(X\subset S\) with \(|X|=12\), and let

\[ \mathcal B=\{A\in {S\choose 12}: |A\cap X|\ \text{is odd}\}. \]

The retained exhaustive spectrum computation proves that every seven members of \(\mathcal B\) have a two-point transversal.

(a) The same family has \(\tau(\mathcal B)=10\). Consequently,

\[ \boxed{f(12,7)\ge 10}. \tag{1} \]

(b) The published upper bound of Fon-Der-Flaass–Kostochka–Woodall gives

\[ f(12,7)\le \left\lceil\frac{7\cdot12}{8}\right\rceil=11. \]

Thus the concrete verified regime is

\[ \boxed{10\le f(12,7)\le 11}. \tag{2} \]

This does not settle either asymptotic question on the live page.

Step 0: authoritative live page

I loaded problem 644, its live LaTeX view, and its discussion thread through the Bright Data browser route.

(b, modulo the live page) At access time the status was OPEN. The page listed 0 claimed proofs, Currently working on this problem: None, and Interested in collaborating: None. The mandatory stop condition therefore did not trigger.

The exact statement copied from the live LaTeX view is:

Let $f(k,r)$ be minimal such that if $A_1,A_2,\ldots$ is a family of sets, all of size $k$, such that for every collection of $r$ of the $A_is$ there is some pair $\{x,y\}$ which intersects all of the $A_j$, then there is some set of size $f(k,r)$ which intersects all of the sets $A_i$. Is it true that\[f(k,7)=(1+o(1))\frac{3}{4}k?\]Is it true that for any $r\geq 3$ there exists some constant $c_r$ such that\[f(k,r)=(1+o(1))c_rk?\]

(b, modulo the live page) Its listed known result is that Erdős, Fon-Der-Flaass, Kostochka, and Tuza proved

\[ f(k,3)=2k,\qquad f(k,4)=\lfloor3k/2\rfloor,\qquad f(k,5)=\lfloor5k/4\rfloor,\qquad f(k,6)=k. \]

The only visible comment was by williamwkcook, dated 11 September 2025. (c, as posted) It matches those four formulas to OEIS sequences and speculates that the constants \(c_r\) may be rational and arise from modular constructions. The comment itself says these are possible connections, not exact formulations; it contains no claimed proof or worker marker.

Notation and primary-source check

Write \(\tau(\mathcal H)\) for the minimum size of a transversal of a hypergraph. The page's \(f(k,7)\) is the parameter denoted \(f(k,7,2)\) in the older three-argument literature: uniformity \(k\), local sample size \(7\), and local cover size \(2\).

I verified the following sources and what is used from each:

\[ f(q,7)\le\left\lceil\frac{7q}{8}\right\rceil. \tag{3} \] Its Theorem 1 proves that the parity family on \(7m+1\) vertices, with uniformity and \(|X|\) both \(4m\), has property \((7,2)\) and transversal number \(3m+1\) for \(m\ge10\). A printed remark says the argument can be elaborated to \(m\ge4\), and explicitly says this family fails for \(m=2\). It does not treat \(m=3\).

(c, search miss rather than a theorem) Exact searches for f(12,7,2), f(12; 7; 2), the 1999 title and citations, the author publication list, and the later primary papers found no source reporting the \(m=3\) parity case or a later improvement to (3). This supports treating the computation below as unreported progress, but is not proof of novelty.

The explicit construction and its transversal number

Let

\[ S=\{0,\ldots,21\},\qquad X=\{0,\ldots,11\},\qquad Y=S\setminus X, \]

and define

\[ \mathcal B=\{A\subset S:|A|=12,\ |A\cap X|\equiv1\pmod2\}. \tag{4} \]

(a) Every \(A\in\mathcal B\) meets \(Y\). Indeed, the only 12-subset contained in \(X\) is \(X\) itself, and \(|X\cap X|=12\) is even. Therefore \(Y\), of size 10, is a transversal and

\[ \tau(\mathcal B)\le10. \tag{5} \]

(a) No 9-set \(T\subset S\) is a transversal. Since

\[ |T|=9<|X|=12\quad\text{and}\quad |T|=9<|Y|=10, \]

there are

\[ a\in (S\setminus T)\cap X,\qquad b\in(S\setminus T)\cap Y. \]

Both

\[ (S\setminus T)\setminus\{a\},\qquad (S\setminus T)\setminus\{b\} \]

are 12-subsets disjoint from \(T\). Their intersection sizes with \(X\) differ by one, so exactly one has odd \(X\)-intersection and belongs to \(\mathcal B\). Hence \(T\) misses a member of \(\mathcal B\). Together with (5),

\[ \boxed{\tau(\mathcal B)=10}. \tag{6} \]

(a) The retained script also recomputes

\[ |\mathcal B| =\sum_{\substack{j\ \mathrm{odd}\\0\le12-j\le10}} {12\choose j}{10\choose12-j} =323344. \]

Exact finite reduction of the local property

The only non-elementary part of (1) is showing that every seven members of \(\mathcal B\) have a two-point transversal. Here is the complete reduction certified by the program.

Assume for contradiction that \(A_0,\ldots,A_6\in\mathcal B\) have no two-point transversal. For each ground vertex \(v\in S\), define its miss spectrum

\[ M(v)=\{i\in[7]:v\notin A_i\}. \tag{7} \]

The following deductions are elementary.

  1. (a) Every \(M(v)\) is nonempty; otherwise \(v\) alone hits all seven edges.
  2. (a) The 22 spectra are pairwise intersecting. If \(M(u)\cap M(v)=\varnothing\), then every \(A_i\) contains \(u\) or \(v\), so \(\{u,v\}\) is a transversal.
  3. (a) Every spectrum has size at least three. If \(|M(v)|\le2\), then \(v\) hits at least five edges. Any two 12-subsets of \(S\) intersect because \(12+12>22\); a point in the intersection of the at most two remaining edges, together with \(v\), hits all seven.
  4. (a) Each edge index occurs in exactly ten spectra:

\[ \sum_{v\in S}\mathbf1_{\{i\in M(v)\}}=22-|A_i|=10. \tag{8} \]

  1. (a) Each edge index occurs an odd number of times among the 12 spectra belonging to vertices of \(X\). This is because

\[ |\{v\in X:i\in M(v)\}|=12-|A_i\cap X| \] and \(|A_i\cap X|\) is odd.

Call a spectrum of size three standard. Its support is a pairwise-intersecting family of triples on \([7]\), hence is contained in a maximal such family \(F\).

For a fixed \(F\), introduce nonnegative integer variables \(x_M,y_M\): the multiplicities of spectrum \(M\) among vertices in \(X,Y\), respectively. The exact constraints are

\[ \begin{aligned} \sum_Mx_M&=12,& \sum_My_M&=10,\\ \sum_{M\ni i}(x_M+y_M)&=10 &&(i\in[7]),\\ \sum_{M\ni i}x_M&\equiv1\pmod2 &&(i\in[7]). \tag{9} \end{aligned} \]

The allowed spectra are the triples in \(F\) and all subsets of \([7]\) of size at least four. Two sets of size at least four always intersect on seven points. Since \(F\) is intersecting, the only remaining forbidden disjoint pair is a triple \(T\in F\) and its four-set complement \([7]\setminus T\). Binary support variables enforce that the two cannot both have positive multiplicity.

Thus (9), plus those triple/complement exclusions, is an exact bounded integer model for a bad seven-edge configuration once \(F\) is fixed. Conversely, every alleged bad configuration maps into one of these models after extending its standard spectra to a maximal \(F\) and relabelling the seven edge indices.

The model deliberately does not require the seven rows \(A_i\) to be distinct. This is a relaxation: proving even this larger system infeasible is sufficient for the usual distinct-edge interpretation of “collection.”

Exhaustive computation

The standalone verifier is erdos644_wavew004_reverify.py. Run:

python3 runs/erdos644_wavew004_reverify.py

It requires ortools (tested with version 9.15.6755) and reads no generated data.

(d) From scratch, the script:

  1. directly checks (6) against all \({22\choose9}=497420\) nine-sets;
  2. independently checks an explicit bad \(m=2\) spectrum configuration, agreeing with the 1999 paper;
  3. constructs the intersection graph on all \({7\choose3}=35\) triples;
  4. uses an exact Bron–Kerbosch traversal to enumerate all 6127 labelled maximal intersecting triple families;
  5. verifies pairwise intersection and maximality of every enumerated family;
  6. reduces them under all \(7!=5040\) relabellings to 15 orbits;
  7. rebuilds (9) for every orbit and obtains INFEASIBLE independently from the integer CP-SAT engine and the SCIP mixed-integer engine.

The labelled maximal-family size distribution is

\[ \begin{array}{c|rrrrrr} |F|&7&10&11&12&13&15\\ \hline \#F&30&3185&2100&630&175&7, \end{array} \]

and the 15 orbit representatives have size distribution

\[ \begin{array}{c|rrrrrr} |F|&7&10&11&12&13&15\\ \hline \#\text{ orbits}&1&7&3&1&2&1. \end{array} \]

Their deterministic representative-list digest is 70a0bc2234c4d12434177bc223b88b97cf3076ad514ce6496858b51140357549.

Recorded summary:

PASS explicit m=2 bad seven-edge spectrum configuration
PASS parity family arithmetic: |B|=323344, tau(B)=10 (497420 nine-sets defeated)
PASS maximal intersecting triple families: 6127 labelled, 15 S7-orbits, sha256=70a0bc2234c4d12434177bc223b88b97cf3076ad514ce6496858b51140357549
PASS all 15 exact multiplicity systems infeasible with both engines: CP-SAT branches=21385, conflicts=3112; SCIP nodes=6251
ALL CHECKS PASSED in 14.487 seconds

(d) Therefore the assumed bad seven-edge configuration cannot exist, so the family (4) has property \((7,2)\). Combining this computational fact with the elementary equality (6) proves the lower bound (1), with the computational qualification stated explicitly.

What remains and the precise wall

For the first asymptotic question, the 1999 construction gives

\[ \frac{f(4m,7)}{4m}\ge\frac{3m+1}{4m} \]

for its proved range, while (3) gives

\[ \frac{f(k,7)}k\le\frac78+O(1/k). \]

The missing uniform lemma is exactly an improvement of (3) to

\[ \tau(\mathcal H)\le(3/4+o(1))k \]

for every \(k\)-uniform family with property \((7,2)\), or else a construction asymptotically exceeding \(3k/4\). A computation confined to the 22-vertex parity family cannot provide that uniform step.

Even the concrete value \(f(12,7)\) remains either 10 or 11. Proving it is 10 requires excluding all 12-uniform property-\((7,2)\) families with transversal number 11, not merely the highly symmetric parity family; finding one such family would prove it is 11. The finite spectrum reduction above relies crucially on the prescribed 22-vertex ground set and parity condition and therefore does not extend to arbitrary families.

The second question asks for existence of a limiting linear coefficient for every fixed local sample size. Neither the 1999 fixed-\(7\) argument nor the later general estimates supply an approximate subadditivity or other limit principle in the uniformity parameter. That missing uniformity/limit lemma is logically separate from this small-case computation.

PARTIAL: The explicit 12-uniform parity family on 22 vertices has computationally certified property (7,2) and elementary transversal number 10, proving 10 <= f(12,7) <= 11; both asymptotic questions remain open.

This is the AI working report, labelled by outcome — not an independently verified claim unless marked PROVED. ← ledger