Equal three blocks

---

Theorem. Suppose a $k$-uniform family $\mathcal H$ is invariant under arbitrary permutations within three blocks, each of size $k$. For $k\ge62$, the local two-point property for at most seven edges implies

$$\tau(\mathcal H)\le\frac{3k}{4}+24. \tag{1}$$

The same assertion holds for the property on exactly seven distinct edges: a family violating (1) has more than seven edges, so any smaller bad subfamily extends to seven.

This deals with ground size $3k$, outside the range $n\le9k/4+o(k)$ of the general many-block bound in Block-profile families. It does not settle unrestricted three-block families or either assertion of Problem 644.

A continuous lemma

For a finite nonempty set $S$ of profiles $a=(a_1,a_2,a_3)$ satisfying

$$a_j\in[0,c],\qquad a_1+a_2+a_3=1,$$

write $T(S;c)$ for the infimum of $d_1+d_2+d_3$ over deletions $0\le d_j\le c$ for which no $a\in S$ satisfies $a_j\le c-d_j$ for every $j$.

Lemma. If

$$\frac{39}{44}\le c\le\frac98, \qquad T(S;c)>\frac34, \tag{2}$$

then seven profiles from $S$, allowing repetitions, have a weighted realization within three blocks of capacity $c$ in which no pair of points covers the seven rows. Before deleting excess incidences, this realization uses at most seven support classes per block.

Use the seven Fano caps as allowable point spectra. The Fano-cap covering formula says that demands $q_1,\ldots,q_7$ in one block can be realized within capacity $c$, with at most seven covering classes before trimming, provided

$$\max_iq_i\le c,\qquad \sum_{i\in L}q_i\le2c\quad\text{for every Fano line }L, \qquad \sum_iq_i\le4c. \tag{3}$$

Subsets of caps remain allowable spectra. No two such spectra cover all seven rows.

If some profile $B$ satisfies $B_j\le2c/3$ for every $j$, deletion $3B/4$ has total mass $3/4$. Hence there is a profile $A\le c\mathbf1-3B/4$. Put profile $B$ on one Fano line and $A$ on its complementary cap. In block $j$, weight that cap by $\max(A_j-3B_j/4,0)$, and each of the six other caps by $B_j/4$. These cover the row demands with total mass

$$\max\{A_j+3B_j/4,\ 3B_j/2\}\le c.$$

Trimming excess incidences gives the desired realization.

Otherwise every profile has a coordinate exceeding $2c/3$. Since $c>3/4$, it has exactly one such coordinate; call this its type. For every represented type $i$, let $h_i$ be the minimum $i$th coordinate of a profile of that type, and choose a minimizing profile $A^i$. Deleting just more than $c-h_i$ from each represented type's block removes all profiles, so

$$T(S;c)\le\sum_{i\text{ represented}}(c-h_i). \tag{4}$$

With only one or two represented types, the right side is less than $2c/3\le3/4$. Thus all three types occur. Relabel them so that

$$1\ge h_1\ge h_2\ge h_3>\frac{2c}{3}, \qquad H:=h_1+h_2+h_3<3c-\frac34. \tag{5}$$

Every off-type coordinate of $A^i$ is at most $1-h_i$.

Put profile $A^1$ on one Fano point, and profiles $A^2,A^3$ on two complementary triples of the other six points, neither triple being a line. For instance, with Fano lines

$$123,\ 145,\ 167,\ 246,\ 257,\ 347,\ 356,$$

use the three type classes $\{1\},\{2,3,4\},\{5,6,7\}$. We verify (3) separately in every coordinate.

The maximum individual demand is at most $c$. A line containing no row of a coordinate's own type has demand less than $3-2c\le2c$. A line containing exactly one such row has demand at most

$$h_1+2(1-h_3) <\frac c3+\frac54\le2c,$$

where (5) gives $h_1<3c-3/4-2h_3$. No type class contains a line. A line with two own-type rows can therefore involve only type 2 or 3; its demand is at most

$$2h_2+1-h_3 <\frac{5c}{3}+\frac14\le2c,$$

using $2h_2+h_3\le H<3c-3/4$. These inequalities use only $c\ge3/4$.

The total demands in coordinates 1, 2, and 3, respectively, satisfy

$$\begin{aligned} Q_1&\le h_1+3(1-h_2)+3(1-h_3) <6-3c\le4c,\\ Q_2&\le3h_2+(1-h_1)+3(1-h_3) \le4+2h_2-3h_3 <\frac{13}{4}+\frac c3\le4c,\\ Q_3&\le3h_3+(1-h_1)+3(1-h_2) \le4-h_1<4-\frac{2c}{3}\le4c. \end{aligned}$$

For the first inequality use $h_1\le c$; for the middle one use $2h_2\le H-h_3$. The outer two require only $c\ge6/7$, and the middle one requires $c\ge39/44$. This proves (3), hence the lemma.

In the special case $c=1$, the no-light-profile argument works already under $T(S;1)>2/3$: equation (4) then gives $H<7/3$, and the same line and total estimates apply. The light-profile case above still uses the threshold $3/4$.

Finite rounding

Suppose (1) fails. Reserve seven points in each of the three blocks and let $\mathcal H'$ consist of the edges disjoint from these 21 points. Then

$$\tau(\mathcal H')\ge\tau(\mathcal H)-21>\frac{3k}{4}+3. \tag{6}$$

The family $\mathcal H'$ is invariant within its three remaining blocks, each of size $k-7$.

Let $S$ be its integer profiles divided by $k$, and put $c=(k-7)/k$. Any real deletion in these remaining blocks which removes all allowed profiles can be rounded up separately in each block. The resulting integer deletion still removes every edge, with increase less than three in total size. Consequently

$$kT(S;c)\ge\tau(\mathcal H')-3>\frac{3k}{4}.$$

For $k\ge62$, we have $39/44\le c<1$, so the continuous lemma applies.

Scale its weighted covering construction by $k$. In each block its at most seven cap weights have total at most $k-7$. Round each weight upward to an integer. The resulting at most seven extra points fit in the reserved points of that block. Every row's block degree is now at least its specified integer profile coordinate. Delete excess incidences, separately for each row and block, until all degrees agree with their specified profiles.

Each resulting row is an edge of $\mathcal H$, by block invariance. Every point spectrum is a subset of a Fano cap, so no pair covers the rows. Removing repeated rows, if any, leaves a bad subfamily of at most seven distinct edges, a contradiction.

Scope

The proof uses full invariance within each block and constructs new edges from allowed profiles. It cannot be applied merely because an arbitrary family has three reference blocks. The constants are not optimized. No counterexample among general three-or-more-block profile families has been established.