Equal blocks

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The later large-block theorem extends the three-quarter bound to unequal block sizes, provided every block has size at least the uniformity.

Theorem. Let the ground set of a nonempty $k$-uniform family $\mathcal H$ be partitioned into $m$ blocks, each of size $k$, and suppose the family is invariant under arbitrary permutations within every block. If $k\ge62$ and every at-most-seven edges have a two-point transversal, then

$$\tau(\mathcal H)\le\frac34k+15m. \tag{1}$$

In particular, these families satisfy the conjectured upper bound whenever $m=o(k)$, without a restriction on their total ground-set size. This extends Equal three blocks, but does not address arbitrary families.

Continuous profiles

Let $S$ be a finite nonempty subset of

$$\{a\in[0,c]^m:\textstyle\sum_j a_j=1\}.$$

Write $T(S;c)$ for the infimum of $\sum_jd_j$ over vectors $0\le d_j\le c$ such that no $a\in S$ satisfies $a\le c\mathbf1-d$ coordinatewise.

Lemma. If

$$39/44\le c\le1, \qquad T(S;c)>\frac34+m(1-c), \tag{2}$$

then seven profiles from $S$, with repetitions allowed, have a realization within $m$ blocks of capacity $c$ whose point spectra are subsets of Fano caps. Consequently no pair of points covers all seven rows.

By the Fano-cap covering formula, demands $q_1,\ldots,q_7$ in one block can be realized in this way if

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

Before trimming excess incidences, at most seven cap classes are needed in each block.

Suppose first that some $B\in S$ satisfies $B_j\le2c/3$ for all $j$. The deletion $3B/4$ has total mass $3/4$, so (2) supplies $A\in S$ with $A\le c\mathbf1-3B/4$. Put $B$ on one Fano line and $A$ on its complementary cap. In block $j$, give this cap weight $\max(A_j-3B_j/4,0)$ and each of the other six caps weight $B_j/4$. These weights cover the prescribed demands and have total

$$\max(A_j+3B_j/4,3B_j/2)\le c.$$

Trim excess incidences.

Otherwise every profile has a coordinate greater than $2c/3$. This coordinate is unique, since $4c/3>1$; call it the profile's type. For each represented type $i$, choose a profile $A^i$ minimizing its own coordinate $h_i=A^i_i$, and set

$$e_i=c-h_i,\qquad \epsilon=1-c.$$

Deleting slightly more than $e_i$ in each represented coordinate removes all profiles. Therefore

$$T(S;c)\le\sum_{i\text{ represented}}e_i, \qquad 0\le e_i<c/3. \tag{4}$$

One or two represented types contradict (2), since $2c/3\le2/3<3/4$.

If exactly three types occur, the three-type construction in Equal three blocks applies to their three coordinates. Its proof uses only

$$h_1\ge h_2\ge h_3>2c/3, \qquad h_1+h_2+h_3<3c-3/4,$$

and the fact that each off-type coordinate of $A^i$ is at most $1-h_i$; it does not require all the profile mass to lie in these three coordinates. These hypotheses follow from (4). Use the same type classes $\{1\},\{2,3,4\},\{5,6,7\}$ on the Fano rows. The cited calculations verify (3) in the three type coordinates. In every other coordinate each demand is less than $1-2c/3$, so every line sum is less than $3-2c\le2c$ and the total is less than $7-14c/3\le4c$. Thus (3) holds there as well.

It remains to handle at least four represented types. Relabel them so that

$$e_1\ge e_2\ge e_3\ge e_4\ge\cdots.$$

If $e_2<\epsilon$, then (4) gives

$$T(S;c)<c/3+(m-1)\epsilon<3/4+m\epsilon,$$

contradicting (2). Hence $e_1,e_2\ge\epsilon$.

Use the Fano lines

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

Put $A^1$ on rows $1,2,4$, $A^2$ on rows $3,5$, $A^3$ on row $6$, and $A^4$ on row $7$. Write

$$\delta_i=1-h_i=\epsilon+e_i.$$

Then $\delta_1\ge\delta_2\ge\delta_3\ge\delta_4$, and each off-type coordinate of $A^i$ is at most $\delta_i$. Also $\delta_1<1-2c/3$.

The three type-1 rows are noncollinear. A line containing two of them has its third row of type 2 or 3, so its demand in coordinate 1 is at most

$$2h_1+\delta_2\le2c,$$

because $\epsilon+e_2\le2e_1$. The line through both type-2 rows is $356$, and its demand in coordinate 2 is at most

$$2h_2+\delta_3\le2c,$$

because $\epsilon+e_3\le2e_2$. Every other line-coordinate check involves zero or one own-type row, and follows from

$$3\delta_1<3-2c\le2c, \qquad c+2\delta_1<2-c/3\le2c.$$

For total demands, the first two coordinates satisfy

$$\begin{aligned} Q_1&\le3h_1+2\delta_2+\delta_3+\delta_4 \le3+\delta_1<4-2c/3\le4c,\\ Q_2&\le2h_2+3\delta_1+\delta_3+\delta_4 \le2+3\delta_1<5-2c\le4c. \end{aligned}$$

Each of coordinates 3 and 4 has total demand at most

$$c+6\delta_1<6-3c\le4c.$$

Every other coordinate has total demand at most

$$7\delta_1<7-14c/3\le4c.$$

All these estimates hold for $c\ge39/44$ (indeed $c\ge6/7$ suffices for this four-type construction). Individual demands are already at most $c$. This proves (3) and the lemma.

Integer realization

Suppose (1) fails. Reserve seven points in each block and let $\mathcal H'$ consist of the edges avoiding the reserved $7m$ points. Then

$$\tau(\mathcal H')\ge\tau(\mathcal H)-7m.$$

In particular, $\mathcal H'$ is nonempty. Normalize its integer profiles by $k$ and put $c=(k-7)/k$. Any real deletion removing all these profiles can be rounded upward coordinatewise to an integer deletion, increasing its size by less than $m$. Consequently

$$kT(S;c)\ge\tau(\mathcal H')-m \ge\tau(\mathcal H)-8m >3k/4+7m.$$

Since $k\ge62$, we have $c\ge39/44$, and this is precisely (2).

Apply the lemma and multiply its cap weights by $k$. In each block their total is at most $k-7$. Round the at-most-seven weights upward to integers; the reserved points absorb the increase. Each row's block degree is at least its prescribed integer value. Trim excess incidences separately in each row and block to restore all degrees exactly.

Full block invariance ensures that every resulting row is an edge of $\mathcal H$. Every point spectrum is still a subset of a Fano cap, so the rows have no two-point transversal. Removing repetitions leaves a bad subfamily of at most seven distinct edges, a contradiction.

The exactly-seven convention gives the same result: a family violating (1) has more than seven edges, so a smaller bad subfamily can be extended to seven distinct edges.

Scope

The proof requires complete freedom to realize any allowed profile within every block. It does not provide such freedom in an arbitrary family. With singleton blocks it is not applicable, since its blocks must each have size equal to the uniformity. Neither assertion of Problem 644 is resolved.