We study non-trivial intersection problems for multi-part hypergraphs, excluding the usual extremal examples determined by fixed vertices or fixed coordinates. Our first result determines the exact value of the non-trivial \(t\)-intersection problem in the symmetric product \([n]^r\) for \(1\le t\le r-2\) and all \(n\ge2\). Frankl and Nie proved a two-candidate formula for sufficiently large \(n\) and conjectured it for all \(n\ge 2\); our formula shows that the conjectured expression must be enlarged, in small ranges of \(n\), by additional Ahlswede–Khachatrian ball-type terms.
Our second result concerns intersecting families in general products \(X_1\times\cdots\times X_r\), where \(|X_i|=n_i\), with no common vertex. Let \(m_0(1,n_1,\ldots,n_r)\) denote the largest size of such a family. We show that this number is equal to the maximum of \(\sum_{X\in \mathcal{D}}\prod_{i\in X}(n_i-1)\) over all downsets \(\mathcal{D}\subseteq 2^{[r]}\) such that \(\bigcup_{X\in \mathcal{D}}X=[r]\) and no two members of \(\mathcal{D}\) have union \([r]\). This finite reduction separates the intersection obstruction from the part sizes and yields explicit fully asymmetric formulas for \(r=4,5,6\).
Throughout \([r]=\{1,\ldots,r\}\). Let \(r\ge2\) and let \(X_1,\ldots,X_r\) be finite sets. An \(r\)-partite \(r\)-graph with parts \(X_1,\ldots,X_r\) will be identified with a family \[\mathcal{F}\subseteq X_1\times\cdots\times X_r =\{(x_1,\ldots,x_r):x_i\in X_i,\;1\le i\le r\}.\] The fixed-coordinate set of \(\mathcal{F}\) is defined as \[\bigcap \mathcal{F} := \{i\in [r]: \text{all sequences in } \mathcal{F}\text{ have the same entry in the } i\text{-th coordinate}\}.\] For two vectors \(\mathbf{x}=(x_1,\ldots,x_r)\) and \(\mathbf{y}=(y_1,\ldots,y_r)\) in the same product, let \[\operatorname{agr}(\mathbf{x},\mathbf{y}) := \{i\in[r]:x_i=y_i\}.\] Thus \(\mathbf{x}\) and \(\mathbf{y}\) are disjoint as hypergraph edges if and only if \(\operatorname{agr}(\mathbf{x},\mathbf{y})=\emptyset\). A family \(\mathcal{F}\subseteq[n]^r\) is \(t\)-intersecting if \(|\operatorname{agr}(\mathbf{x},\mathbf{y})|\ge t\) for all \(\mathbf{x},\mathbf{y}\in\mathcal{F}\); when \(t=1\) we simply say intersecting. A \(t\)-intersecting family \(\mathcal{F}\) is non-trivial if \(|\bigcap\mathcal{F}|<t\).
The classical Erdős–Ko–Rado theorem [1] states that if \(\mathcal{F}\subseteq\binom{[n]}{r}\) is intersecting and \(n\ge 2r\), then \(|\mathcal{F}|\le\binom{n-1}{r-1}\). The extremal families are the trivial stars, consisting of all \(r\)-sets containing a fixed vertex. This naturally leads to the corresponding non-trivial problem, in which such stars are excluded. Hilton and Milner [2] resolved this question by showing that any non-trivial intersecting family \(\mathcal{F}\subseteq\binom{[n]}{r}\) with \(n>2r\) satisfies \[|\mathcal{F}| \le \binom{n-1}{r-1}-\binom{n-r-1}{r-1}+1.\] The bound is sharp: The extremal family is called the Hilton–Milner family \[\mathcal{H}_{x,B}(n) = \{B\} \cup \left\{ F\in\binom{[n]}{r}: x\in F,\;F\cap B\neq\emptyset \right\},~\text{ where }~ x\in[n] ~\text{and}~B\in\binom{[n]\setminus\{x\}}{r}.\]
The Hilton–Milner theorem, whose short proof was later given by Frankl and Füredi [3], initiated a line of work in which the common core of an intersecting family is controlled or excluded. For \(t>1\), the corresponding non-trivial problem asks for the largest \(t\)-intersecting family. Frankl [4] determined this maximum for sufficiently large \(n\). The extremal family is \[\mathcal{A}_1(n,r,t) = \left\{ F\in \binom{[n]}{r}: |F\cap [t+2]|\ge t+1 \right\}\] when \(t+1\le r\le 2t+1\), while for \(r>2t+1\) it is the Hilton–Milner type family \[\mathcal{H}_{r,t}(n) = \left\{ [r+1]\setminus\{i\}: i\in [r] \right\} \cup \left\{ F\in \binom{[n]}{r}: [t]\subseteq F,\, F\cap [t+1,r+1]\neq \emptyset \right\},\] where \([t+1, r+1] = \{t+1, t+2, \dots, r+1\}\). Ahlswede and Khachatrian [5] then proved the complete non-trivial intersection theorem, determining the maximum for all relevant \(n,r,t\). Their theorem shows that the extremal value is governed by the complete intersection theorem and hence by the Ahlswede–Khachatrian families \[\mathcal{A}_i(n,r,t) = \left\{ F\in \binom{[n]}{r}: |F\cap [t+2i]|\ge t+i \right\},~1\le i\le \left\lfloor\frac{r-t}{2}\right\rfloor.\] Thus the extremal picture is not merely the Hilton–Milner construction with \(t\) fixed coordinates; depending on the parameter range, ball-type Ahlswede–Khachatrian families also appear.
Turning to multi-part versions, Kwan, Sudakov, and Vieira [6] studied non-trivially intersecting multi-part families, where the ground set is partitioned into parts and each member has a prescribed intersection size with each part. The extremal examples are multi-part Hilton–Milner-type families, constructed by choosing a distinguished part together with possible auxiliary parts.
Our first problem concerns \(t\)-intersecting families in \(r\)-partite \(r\)-graphs. For positive integers \(n,t, r\) with \(n\ge r>t\), define \[\iota_0(t,n;r):= \max\{|\mathcal{F}|:\mathcal{F}\subseteq[n]^r\text{ is non-trivial and }t\text{-intersecting}\}.\] Frankl and Nie [7] determined this maximum for sufficiently large \(n\), proving that \[\iota_0(t,n;r) = \max\left\{ n^{r-t}-(n-1)^{r-t}+t(n-1), (t+2)n^{r-t-1}-(t+1)n^{r-t-2} \right\},\] and conjectured that the same two-candidate formula holds for every \(n\ge2\). We show that this conjecture is asymptotically correct but fails for certain values of \(n\): in small parameter ranges, additional Ahlswede–Khachatrian ball-type examples may achieve larger size.
Theorem 1. For integers \(r\ge3\), \(1\le t\le r-2\), \(n\ge 2\) and \(q:=r-t\), we have \[\iota_0(t,n;r) = \max\left\{ W_{r,t}(n),\, A_i(t,n;r):1\le i\le\left\lfloor\frac{q}{2}\right\rfloor \right\},\] where \[W_{r,t}(n):=n^q-(n-1)^q+t(n-1),\] and \[A_i(t,n;r):= n^{q-2i}\sum_{j=0}^i\binom{t+2i}{j}(n-1)^j.\]
In Theorem 1, the term \(W_{r,t}(n)\) corresponds to a multi-part Hilton–Milner construction, while \(A_i(t,n;r)\) is the product analogue of an Ahlswede–Khachatrian ball. For fixed \(r\) and \(t\), two terms \(W_{r,t}(n)\) and \(A_1(t,n;r)\) have the largest possible degree, namely \(r-t-1\), whereas \(A_i(t,n;r)\) has degree at most \(r-t-2\) for every \(i\ge2\). Hence the formula reduces to the Frankl–Nie two-candidate maximum when \(n\) is sufficiently large, but for small \(n\) the lower-degree ball terms may dominate.
We now turn to a different non-trivial condition, formulated in terms of matching and covering numbers. For a family \(\mathcal{F}\subseteq X_1\times\cdots\times X_r\), let \(\nu(\mathcal{F})\) be the maximum number of pairwise disjoint edges in \(\mathcal{F}\), and let \(\tau(\mathcal{F})\) be the minimum size of a vertex set meeting every edge. For \(r=2\), König’s theorem [8] gives \(\nu(\mathcal{F})=\tau(\mathcal{F})\); for \(r\ge3\) the two parameters differ in an essential way.
Our second result deals with the Hilton–Milner case in general multi-part products. Given \(\mathcal{F}\subseteq X_1\times\cdots\times X_r\), \(\mathcal{F}\) is intersecting means \(\nu(\mathcal{F})\le1\). Meyer [9] proved that every intersecting subhypergraph of \([n]^r\) has at most \(n^{r-1}\) edges. Deza and Frankl [10] proved the asymmetric version: if \(n_1\ge n_2\ge\cdots\ge n_r\), then every intersecting subhypergraph of \([n_1]\times\cdots\times[n_r]\) has at most \(\prod_{\ell=1}^{r-1}n_\ell\) edges. More generally, Frankl [11] showed that \(\nu(\mathcal{F})\le s\) implies \(|\mathcal{F}|\le s\prod_{\ell=1}^{r-1}n_\ell\) in the same ordered setting. Since these extremal examples have small covering number, the natural non-trivial condition is \(\tau(\mathcal{F})>s\); here we treat the case \(s=1\).
Following Frankl and Nie [7], for \(n_i=|X_i|\) we define \(m_0(1,n_1,\ldots,n_r)\) to be the maximum size of a family \(\mathcal{F}\subseteq X_1\times\cdots\times X_r\) satisfying \[\nu(\mathcal{F})\le1<\tau(\mathcal{F}).\] Equivalently, \(\mathcal{F}\) is intersecting but has no common vertex. This agrees with \(\iota_0(1,n;r)\) in the symmetric case \(n_1=\cdots=n_r=n\), although for \(t>1\) the condition \(|\bigcap\mathcal{F}|<t\) differs from the covering condition \(\tau(\mathcal{F})>t\). Frankl and Nie [7] studied this problem and proved, among other results, the symmetric formula \[m_0(1,n,\ldots,n)=n^{r-1}-(n-1)^{r-1}+n-1\] for all \(r\ge 3\) and \(n\ge 2\), and the asymmetric three-part formula \[m_0(1,n_1,n_2,n_3)=n_1+n_2+n_3-2.\] The symmetric formula above is the case \(s=1\) of a conjecture of Lu and Ma [12], which states that, for sufficiently large \(n\), \(r\ge 4\) and \(s<n\), \[m_0(s,n,\ldots,n)=s n^{r-1}-(n-1)^{r-1}+n-s,\] where \(m_0(s,n,\ldots,n)\) denotes the analogous maximum under \(\nu(F)\le s<\tau(F)\). Frankl and Nie [7] later confirmed this conjecture for sufficiently large \(n\), and also pointed out that, already for four parts, no single asymmetric expression governs all parameter ranges. We give a finite exact reduction which accounts for this dependence on the part sizes.
Let \(a_i:=n_i-1\) for \(i\in[r]\). For a subset \(X\subseteq[r]\), set \[a_X:=\prod_{i\in X}a_i, \text{ and } \qquad a_\emptyset:=1.\]
A family \(\mathcal{D}\subseteq 2^{[r]}\) is called available if it satisfies the following
\(\mathcal{D}\) is a downset: if \(X\in \mathcal{D}\) and \(Y\subseteq X\), then \(Y\in \mathcal{D}\);
\(\bigcup_{X\in \mathcal{D}}X=[r]\);
\(X\cup Y\ne[r]\) for all \(X,Y\in \mathcal{D}\).
Let \(\mathfrak D_r\) be the collection of all available families \(\mathcal{D}\subseteq 2^{[r]}\).
Theorem 2. For every \(r\ge3\) and \(n_1,\ldots,n_r\ge2\), \[m_0(1,n_1,\ldots,n_r) = \max_{\mathcal{D}\in\mathfrak D_r}\sum_{X\in \mathcal{D}}a_X.\]
Theorem 2 reduces the multi-part Hilton–Milner problem to a weighted problem on the Boolean lattice. Applying this reduction, we determine the fully asymmetric values for \(r=4,5,6\). For \(r=4\), the optimization reduces to intersecting graphs on four vertices and is governed by the star–triangle dichotomy. For \(r=5\), weighted intersecting graphs still suffice. For \(r=6\), one must also choose among complementary pairs of triples. These cases illustrate the increasing complexity of the asymmetric problem.
The proofs of the two main results use a common shifting philosophy, but in different ways. In both problems we first replace an extremal family by a shifted one without changing its size or the relevant intersection property. This makes it possible to record an edge by the set of coordinates in which it differs from a fixed reference edge, and hence to pass from the multi-part product to a weighted problem on \(2^{[r]}\).
For the symmetric \(t\)-intersection problem, this reduction leads to a weighted non-trivial \(t\)-intersection problem on the Boolean lattice. The weights are determined by the product structure: a coordinate in the support contributes a factor \(n-1\), while the remaining coordinates contribute freely. After this translation, the Ahlswede–Khachatrian complete non-trivial intersection theorem, together with the weighted form of Bey–Engel, identifies the only possible extremal shapes. Evaluating their weights gives the term \(W_{r,t}(n)\) and the ball terms \(A_i(t,n;r)\) in Theorem 1. This is also where the additional small-\(n\) candidates enter.
For the multi-part Hilton–Milner problem, the shifted family is encoded more directly by a downset \(\mathcal{D}\subseteq2^{[r]}\). The condition that the family has no common vertex becomes \(\bigcup_{X\in \mathcal{D}}X=[r]\), while the intersecting condition becomes the finite obstruction \(X\cup Y\ne[r]\) for all \(X,Y\in \mathcal{D}\). Thus the original product problem is converted into the weighted downset optimization in Theorem 2. The explicit formulas for \(r=4,5,6\) are then obtained by solving this finite optimization: graphs suffice for \(r=4\) and \(r=5\), while for \(r=6\) complementary pairs of triples also have to be considered.
The paper is organized as follows. Section \(2\) contains the shifting and weighted intersection tools. Section \(3\) proves Theorem 1. Section \(4\) proves Theorem 2 and derives the explicit formulas for \(r=4,5,6\).
In this section, we collect the notation and preliminary results needed for our proofs. We begin by introducing a multi-part adaptation of the classical shifting operation, due to Kwan, Sudakov, and Vieira [6].
Definition 1. (Shifting) Let \(\mathcal{F} \subseteq X_1 \times \cdots \times X_r\) be an \(r\)-partite \(r\)-graph, and let \(\ell, j\) be integers with \(1\le \ell\le r\) and \(1<j\le |X_\ell|\). Define the \((1\leftarrow j)\)-shift under \([n_{\ell}]\) by \(S_j^{(\ell)}(\mathcal{F})=\{S_j^{(\ell)}(F):F=(a_1,\ldots,a_\ell,\ldots,a_r)\in\mathcal{F}\}\), where \[S_j^{(\ell)}\bigl((a_1,\ldots,a_\ell,\ldots,a_r)\bigr)= \begin{cases} F'=(a_1,\ldots,1,\ldots,a_r) & \text{if } a_\ell=j \text{ and } F'\notin\mathcal{F},\\ (a_1,\ldots,a_r) & \text{otherwise.} \end{cases}\]
We say that \(\mathcal{F}\) is \(\ell\)-shifted if \(S_j^{(\ell)}(\mathcal{F})=\mathcal{F}\) for all \(j\). Further, call \(\mathcal{F}\) coordinate-wise shifted if it is \(\ell\)-shifted for every \(1\le \ell\le r\). The following result, given by Kwan, Sudakov and Vieira [6], shows that shifting operation maintains the intersecting property.
Theorem 3 (Kwan–Sudakov–Vieira [6]). Let \(r\ge 2\) and \(n_1,\ldots,n_r\ge 2\). If there exists a non-trivial intersecting family \(\mathcal{F} \subseteq [n_1] \times \cdots \times [n_r]\), then there exists a maximum-size non-trivial intersecting family \(\mathcal{F}' \subseteq [n_1] \times \cdots \times [n_r]\) which is coordinate-wise shifted.
We remark that the original definition and theorem of Kwan–Sudakov–Vieira are more general than the form stated above. It applies to families of separated sets \[\mathcal{G}\subseteq \prod_{s=1}^p \binom{X_s}{k_s}\] for arbitrary positive integers \(p,k_1,\ldots,k_p\), and uses the usual shifts inside each part. For our purposes, the simplest case stated above (i.e., \(p = r\) and \(k_1 = \cdots = k_r = 1\)) suffices.
For a vector \(\mathbf{x}=(x_1,\ldots,x_r)\), define its projection as \[P(\mathbf{x}):=\{i\in[r]:x_i=1\}.\] For a shifted family of tuples, projections reduce the problem to set systems.
Lemma 1 (Frankl–Nie [7]). For integers \(r\ge 3\) and \(1\le t\le r-2\), if \(\mathcal{F} \subseteq [n_1] \times \cdots \times [n_r]\) is coordinate-wise shifted and non-trivial \(t\)-intersecting, then the family of projections \[P(\mathcal{F}):=\{P(\mathbf{x}):\mathbf{x}\in\mathcal{F}\}\subseteq 2^{[r]}\] is non-trivial \(t\)-intersecting as well.
A major problem with the shifting operation is that it may decrease the covering number \(\tau\); in particular, it might turn a non-trivial family into a trivial one. We say a non-trivial \(t\)-intersecting family \(\mathcal{F}\subseteq X_1\times\dots \times X_r\) is \(\ell\)-shift-resistant if there exists \(x \in X_{\ell}\) such that \(S^{(\ell)}_x (\mathcal{F})\) is trivial, i.e. \(| \cap S^{(\ell)}_x (\mathcal{F})| = t\). Frankl and Nie also proved the following shift-resistant bound, which is used only in the symmetric \(t\)-intersection theorem below.
Lemma 2 (Frankl–Nie [7]). For integers \(n\ge 2\), \(r\ge 3\) and \(r-2\ge t\ge 1\), let \(\mathcal{F}\subseteq [n]^r\) be non-trivial \(t\)-intersecting such that for every \(1\le \ell\le r\), \(\mathcal{F}\) is either \(\ell\)-shifted or \(\ell\)-shift-resistant, and \(\mathcal{F}\) is not coordinate-wise shifted. Then \[|\mathcal{F}|\le (t+2)n^{r-t-1}-(t+1)n^{r-t-2}.\]
In order to prove Theorem 1, we shall use the following \(p\)-biased consequence of the weighted non-trivial \(t\)-intersection theorem of Bey and Engel [13]. We state it in a form adapted to the product measure used in this paper. For \(0<p\le 1\) and \(\mathcal{U}\subseteq 2^{[r]}\), define the \(p\)-biased measure of \(\mathcal{U}\) by \[\mu_p(\mathcal{U}) = \sum_{A\in\mathcal{U}} p^{|A|}(1-p)^{r-|A|}.\]
Theorem 4 (Bey–Engel [13]). Let \(0<p\le 1/2\) and \(1\le t\le r-2\). Let \(\mathcal{U}\subseteq 2^{[r]}\) be non-trivial and \(t\)-intersecting, meaning that \[|A\cap B|\ge t\quad\text{for all }A,B\in\mathcal{U}, \qquad \left|\bigcap_{A\in\mathcal{U}}A\right|<t.\] For \(1\le i\le \lfloor (r-t)/2\rfloor\), define the Ahlswede–Khachatrian families \[\mathcal{S}_i:=\{A\subseteq[r]: |A\cap[t+2i]|\ge t+i\}.\] Define the Hilton–Milner type family \[\mathcal{H}_{r,t}:= \{A\subseteq[r]:[t]\subseteq A,\, A\cap\{t+1,\ldots,r\}\ne\emptyset\} \cup \{[r]\setminus\{h\}:h\in[t]\}.\] Then \[\mu_p(\mathcal{U}) \le \max\left\{ \mu_p(\mathcal{H}_{r,t}),\, \mu_p(\mathcal{S}_i):1\le i\le \left\lfloor\frac{r-t}{2}\right\rfloor \right\}.\]
Remark 5. Theorem 4 is the \(p\)-biased specialization of a more general size-dependent weighted theorem of Bey and Engel. In that setting one fixes non-negative weights \(w_0,\ldots,w_r\) and considers \[W(\mathcal{V}) := \sum_{A\in\mathcal{V}} w_{|A|} \qquad (\mathcal{V}\subseteq 2^{[r]}).\] Bey–Engel’s non-trivial \(t\)-intersection theorem gives the corresponding maximum, under their hypotheses, in terms of their weighted Ahlswede–Khachatrian and Hilton–Milner type candidates. The form used here is obtained by taking the weight \[w_j=\alpha^{-j}, \qquad \alpha=\frac{1-p}{p}.\] Since \(0<p\le 1/2\), we have \(\alpha\ge 1\), and for every \(\mathcal{V}\subseteq 2^{[r]}\), \[\mu_p(\mathcal{V}) = \sum_{A\in\mathcal{V}}p^{|A|}(1-p)^{r-|A|} = (1-p)^r\sum_{A\in\mathcal{V}}\alpha^{-|A|} = (1-p)^r W(\mathcal{V}).\] The factor \((1-p)^r\) is independent of \(\mathcal{V}\). Hence maximizing \(p\)-biased measure is equivalent to maximizing the Bey–Engel size-dependent weight with \(w_j=\alpha^{-j}\). After this normalization and the harmless relabelling of the ground-set size as \(r\), the weighted theorem gives exactly the form stated in Theorem 4.
In this section, we prove Theorem 1 using Theorem 4. Let \(r\ge3\), \(t\le r-2\) and \(n\ge 2\) be positive integers and let \(q=r-t\). We first prove the lower bound. Write \(T=[t]\) and \(Q=\{t+1,\ldots,r\}\). Let \[\mathcal{W}_1:=\{\mathbf{x}\in[n]^r:x_i=1\text{ for all }i\in T,\text{ and }x_s=1\text{ for some }s\in Q\},\] and \[\mathcal{W}_2:= \bigcup_{h\in T}\{\mathbf{x}\in[n]^r:x_i=1\text{ for all }i\ne h,\;x_h\ne1\}.\] Set \(\mathcal{W}:=\mathcal{W}_1\cup\mathcal{W}_2\). Clearly \(\left|\bigcap \mathcal{W}\right|<t\) and \[|\mathcal{W}|=n^q-(n-1)^q+t(n-1)=W_{r,t}(n).\] Now we show \(\mathcal{W}\) is \(t\)-intersecting. Indeed, if two vectors both lie in \(\mathcal{W}_1\), they agree on all \(t\) coordinates of \(T\). If one lies in \(\mathcal{W}_1\) and the other in \(\mathcal{W}_2\), they agree on at least \(t-1\) coordinates of \(T\) and also on one coordinate of \(Q\) where the vector in \(\mathcal{W}_1\) has value \(1\). If both lie in \(\mathcal{W}_2\), they differ from the all-\(1\) vector in at most two coordinates, and therefore agree in at least \(r-2\ge t\) coordinates. Thus \(\mathcal{W}\) is a non-trivial \(t\)-intersecting family, and then \[\iota_0(t,n;r)\ge W_{r,t}(n).\]
For \(1\le i\le\lfloor q/2\rfloor\), define the Hamming-ball type family \[\mathcal{A}_i:= \left\{\mathbf{x}\in[n]^r: \left|\{s\in[t+2i]:x_s\ne1\}\right|\le i \right\}.\] Any two members of \(\mathcal{A}_i\) have at least \[(t+2i)-i-i=t\] coordinates among \([t+2i]\) where both are equal to \(1\). Hence \(\mathcal{A}_i\) is \(t\)-intersecting. Since \(i\ge1\), no coordinate is fixed in all members of \(\mathcal{A}_i\), so it is non-trivial. Counting by the number \(j\) of non-\(1\) coordinates inside \([t+2i]\) gives \[|\mathcal{A}_i|=n^{q-2i}\sum_{j=0}^i\binom{t+2i}{j}(n-1)^j=A_i(t,n;r).\] Thus the lower bound follows.
For the upper bound, let \(\mathcal{F}\subseteq[n]^r\) be non-trivial and \(t\)-intersecting. We iteratively apply coordinate-wise shifts that do not destroy non-triviality until no further such shift is possible. This process terminates because each non-trivial shift strictly decreases the sum of all coordinate values over the family. Since each shift preserves cardinality and \(t\)-intersection, the terminal family \(\mathcal{F}^\ast\) satisfies \[|\mathcal{F}^\ast|=|\mathcal{F}|, \qquad \left|\bigcap \mathcal{F}^\ast\right|<t.\] For each coordinate, \(\mathcal{F}^\ast\) is either shifted in that coordinate or is shift-resistant in the sense of Lemma 2. If \(\mathcal{F}^\ast\) is not coordinate-wise shifted, Lemma 2 gives \[|\mathcal{F}| =|\mathcal{F}^\ast| \le (t+2)n^{q-1}-(t+1)n^{q-2} =A_1(t,n;r),\] which is already at most the claimed maximum.
Otherwise, \(\mathcal{F}^\ast\) is coordinate-wise shifted. Hence, we may assume without loss of generality that \(\mathcal{F}\) is coordinate-wise shifted.
Let \[\mathcal{U}:=P(\mathcal{F})=\{P(\mathbf{x}):\mathbf{x}\in\mathcal{F}\}\subseteq2^{[r]}.\] By Lemma 1, \(\mathcal{U}\) is non-trivial and \(t\)-intersecting. For each \(A\in\mathcal{U}\), there are exactly \((n-1)^{r-|A|}\) vectors in \([n]^r\) having projection \(A\), and then \[|\mathcal{F}|\le\sum_{A\in\mathcal{U}}(n-1)^{r-|A|}.\] Put \(p=1/n\). Since \(n\ge2\), we have \(0<p\le1/2\). It follows from \((n-1)^{r-|A|}=n^r p^{|A|}(1-p)^{r-|A|}\) that \[|\mathcal{F}|\le n^r\mu_p(\mathcal{U}).\] By Theorem 4, \[\mu_p(\mathcal{U}) \le \max\left\{ \mu_p(\mathcal{H}_{r,t}),\, \mu_p(\mathcal{S}_i):1\le i\le\left\lfloor\frac{q}{2}\right\rfloor \right\}.\] Thus, to complete the proof, it remains only to compute these measures.
For \(\mathcal{S}_i=\{A:|A\cap[t+2i]|\ge t+i\}\), write \(j=|[t+2i]\setminus A|\). Then \(0\le j\le i\) and \[\mu_p(\mathcal{S}_i)= \sum_{j=0}^i\binom{t+2i}{j}p^{t+2i-j}(1-p)^j.\] Multiplying by \(n^r\) and using \(p=1/n\) gives \[n^r\mu_p(\mathcal{S}_i) =n^{q-2i}\sum_{j=0}^i\binom{t+2i}{j}(n-1)^j =A_i(t,n;r).\]
For \(\mathcal{H}_{r,t}\), its first part \[\{A:[t]\subseteq A,\;A\cap Q\ne\emptyset\},\qquad Q=\{t+1,\ldots,r\},\] has measure \[p^t\bigl(1-(1-p)^q\bigr),\] so after multiplying by \(n^r\) contributes \[n^q-(n-1)^q.\] The second part consists of the \(t\) sets \([r]\setminus\{h\}\) with \(h\in[t]\). Each has measure \(p^{r-1}(1-p)\), so after multiplying by \(n^r\) contributes \(n-1\). Thus \[n^r\mu_p(\mathcal{H}_{r,t})=n^q-(n-1)^q+t(n-1)=W_{r,t}(n).\] Combining these estimates gives \[|\mathcal{F}| \le \max\left\{W_{r,t}(n),A_i(t,n;r):1\le i\le\left\lfloor\frac{q}{2}\right\rfloor\right\}.\] Together with the lower bound, this proves the theorem.
Proof of Theorem 2. By Theorem 3, there is a maximum non-trivial intersecting family \(\mathcal{F}\subseteq[n_1]\times\cdots\times[n_r]\) that is coordinate-wise shifted. Define \[\mathcal{U}:=P(\mathcal{F})=\{P(\mathbf{x}):\mathbf{x}\in\mathcal{F}\}\subseteq2^{[r]}.\] Then \(\mathcal{U}\) is intersecting and \(\bigcap\mathcal{U}=\emptyset\) by Lemma 1. We next prove projection-saturation. Suppose \(A\in\mathcal{U}\) and \(\mathbf{x}\) is any vector with \(P(\mathbf{x})=A\). If \(\mathbf{x}\notin\mathcal{F}\), then adding \(\mathbf{x}\) to \(\mathcal{F}\) preserves intersection: for every \(\mathbf{y}\in\mathcal{F}\), the sets \(A=P(\mathbf{x})\) and \(P(\mathbf{y})\) intersect, so \(\mathbf{x}\) and \(\mathbf{y}\) agree in a coordinate where both values are \(1\). Adding a vector cannot create a common fixed coordinate. Hence non-triviality is also preserved, contradicting maximality. Therefore \[\mathcal{F}=\{\mathbf{x}:P(\mathbf{x})\in\mathcal{U}\},\] and consequently \[\label{can-calF} |\mathcal{F}|=\sum_{A\in\mathcal{U}}\prod_{i\notin A}(n_i-1).\tag{1}\]
Set \(\mathcal{D}:=\{[r]\setminus A:A\in\mathcal{U}\}\). We claim that \(\mathcal{D}\) is available. The condition \(\bigcap\mathcal{U}=\emptyset\) gives \(\bigcup \mathcal{D}=[r]\). Since \(\mathcal{U}\) is intersecting, no two sets of \(\mathcal{D}\) have union \([r]\). It suffices to show \(\mathcal{D}\) is a downset. Otherwise, there would exist \([r]\setminus A\in \mathcal{D}\) and \([r]\setminus B\subseteq [r]\setminus A\) such that \([r]\setminus B\notin \mathcal{D}\). This means that \(A\subseteq B\), \(A\in\mathcal{U}\) and \(B\notin\mathcal{U}\). Choose \(\mathbf{x}\in\mathcal{F}\) with \(P(\mathbf{x})=A\). Since \(\mathcal{F}\) is coordinate-wise shifted, replacing any coordinate of \(\mathbf{x}\) outside \(A\) by \(1\) keeps the resulting vector in \(\mathcal{F}\); iterating over the coordinates in \(B\setminus A\) gives a member of \(\mathcal{F}\) whose projection is \(B\), a contradiction.
Recall that for \(X\in [r]\), \(a_X=\prod_{i\in X}(n_i-1)\). By 1 , we have that \[|\mathcal{F}|\le \max\limits_{\mathcal{D}\in\mathfrak D_r}\sum\limits_{X\in \mathcal{D}}a_X.\] Conversely, let \(\mathcal{D}\in\mathfrak D_r\) and put \[\mathcal{U}:=\{[r]\setminus X:X\in \mathcal{D}\}.\] Then \(\bigcap \mathcal{U}=\emptyset\), and \(\mathcal{U}\) is intersecting because \(X\cup Y\ne[r]\) for all \(X,Y\in \mathcal{D}\). Define \[\mathcal{F}_\mathcal{D}:=\{\mathbf{x}\in[n_1]\times\cdots\times[n_r]:P(\mathbf{x})\in\mathcal{U}\}.\] Then \(\mathcal{F}_\mathcal{D}\) is intersecting, since any two projection sets in \(\mathcal{U}\) intersect and the corresponding vectors agree with value \(1\) in a common coordinate. It remains to check that \(\mathcal{F}_\mathcal{D}\) has no common vertex. Since \(\mathcal{D}\) is a downset, we have \(\emptyset\in \mathcal{D}\), and hence \([r]\in\mathcal{U}\). Thus \(\mathcal{F}_\mathcal{D}\) contains the all-one vector, so no vertex with value different from \(1\) can belong to every member of \(\mathcal{F}_\mathcal{D}\). On the other hand, since \(\bigcup_{X\in \mathcal{D}}X=[r]\), for each \(i\in[r]\) there exists \(X\in \mathcal{D}\) with \(i\in X\). Then \(A=[r]\setminus X\) belongs to \(\mathcal{U}\) and does not contain \(i\). Hence \(\mathcal{F}_\mathcal{D}\) contains a vector whose \(i\)th coordinate is not equal to \(1\). Therefore the vertex \(1\) of \(X_i\) is not common to all members of \(\mathcal{F}_\mathcal{D}\) either. Since this holds for all \(i\in [r]\), the family \(\mathcal{F}_\mathcal{D}\) has no common vertex. Hence \(\mathcal{F}_\mathcal{D}\) is non-trivial and intersecting, and \[|\mathcal{F}_\mathcal{D}|=\sum_{X\in \mathcal{D}}a_X.\] Taking the maximum over \(\mathcal{D}\) completes the proof of Theorem 2. ◻
Next, we will give the exact values of \(m_0(1,n_1,\ldots,n_r)\) for \(r=4,5,6\). Before that, we define an intersecting graph as a graph where every pair of edges shares a common vertex.
Let \(\mathcal{I}_4\) denote the set of all intersecting graphs on vertex set \([4]\), including the empty graph.
Theorem 6. For \(n_i\ge2\) and \(a_i=n_i-1\) for \(i\in [4]\), we have \[m_0(1,n_1,n_2,n_3,n_4) = 1+\sum_{i=1}^4a_i+ \max_{G\in\mathcal{I}_4}\sum_{ij\in E(G)}a_i a_j.\] In particular, if \(n_1\ge n_2\ge n_3\ge n_4\), then \[m_0(1,n_1,n_2,n_3,n_4) = \max\left\{ n_1(n_2+n_3+n_4-2), \, n_1n_2n_3-(n_1-1)(n_2-1)(n_3-1)+n_4-1 \right\}.\] Equivalently, under this ordering of the part sizes, the final maximum is attained either by the star centered at \(1\) or by the triangle on \(\{1,2,3\}\).
Proof. By Theorem 2, it suffices to optimize over \(\mathcal{D}\in\mathfrak D_4\). Since \(\mathcal{D}\) is a downset and \(\bigcup \mathcal{D}=[4]\), all singletons belong to \(\mathcal{D}\). Hence \(\mathcal{D}\) cannot contain a \(3\)-set, because such a set together with the missing singleton would have union \([4]\). Also \([4]\notin \mathcal{D}\). Thus \(\mathcal{D}\) has only levels \(0,1,2\). Define a graph G on \([4]\) by \[E(G):=\mathcal{D}\cap\binom{[4]}2.\] If \(e,f\in E\) are disjoint, then \(e\cup f=[4]\), contradicting the definition of \(\mathcal{D}\). Hence \(G\in\mathcal{I}_4\). Hence \(E\) is an intersecting graph. Conversely, every intersecting graph \(E\) yields an available family \[\mathcal{D}=\{\emptyset\}\cup\binom{[4]}1\cup E.\] Therefore \[m_0(1,n_1,n_2,n_3,n_4) =1+\sum_{i=1}^4a_i+ \max_{G\in\mathcal{I}_4}\sum_{ij\in E(G)}a_i a_j.\]
Assume now \(n_1\ge n_2\ge n_3\ge n_4\). An intersecting graph on four vertices is contained in a star or in a triangle; since all edge weights are positive, an optimum is a full star or a full triangle. The largest star is centered at vertex \(1\), giving contribution \[a_1a_2+a_1a_3+a_1a_4,\] while the largest triangle is on \(\{1,2,3\}\), giving contribution \[a_1a_2+a_1a_3+a_2a_3.\] Adding \(1+\sum_i a_i\) gives respectively \[(1+a_1)(1+a_2+a_3+a_4)=n_1(n_2+n_3+n_4-2)\] and \[n_1n_2n_3-a_1a_2a_3+a_4 =n_1n_2n_3-(n_1-1)(n_2-1)(n_3-1)+n_4-1.\] The result follows. ◻
Let \(\mathcal{I}_5\) be the set of all intersecting graphs on \([5]\), including the empty graph. For \(e\in\binom{[5]}2\), write \(\overline{e}:=[5]\setminus e\) and \[\Delta_e:=a_{\overline{e}}-a_e.\]
Theorem 7. For \(n_i\ge2\) and \(a_i=n_i-1\) for \(i\in [5]\), we have \[m_0(1,n_1,\ldots,n_5) = 1+\sum_{i=1}^5a_i+ \sum_{1\le i<j\le5}a_i a_j + \max_{G\in\mathcal{I}_5}\sum_{e\in E(G)}\Delta_e.\] Equivalently, the final maximum may be taken over subgraphs of stars and triangles on \([5]\).
Proof. Let \(\mathcal{D}\in\mathfrak D_5\) be optimal. As before, since \(\mathcal{D}\) is a downset and \(\bigcup_{X\in \mathcal{D}}X=[5]\), all singletons belong to \(\mathcal{D}\). Hence \(\mathcal{D}\) contains no \(4\)-set and no \(5\)-set. Thus only levels \(0,1,2,3\) can occur.
For \(e\in\binom{[5]}2\), we define a graph \(G\) on \([5]\) by \[E(G):=\left\{e\in\binom{[5]}2:\bar e\in \mathcal{D}\right\}.\] If \(e,f\in E(G)\) were disjoint, then \[\bar e\cup \bar f=[5],\] contradicting the defining condition of \(\mathfrak D_5\). Hence \(G\in\mathcal{I}_5\).
For every \(e\in\binom{[5]}2\), the condition \(X\cup Y\ne[5]\) forbids both \(e\) and \(\bar e\) from belonging to \(\mathcal{D}\). Conversely, since the original weights \(a_X\) are positive, optimality forbids both \(e\) and \(\bar e\) from being absent. Indeed, if neither \(e\) nor \(\bar e\) belongs to \(\mathcal{D}\), then adding the \(2\)-set \(e\) preserves the downset property, since all singletons already belong to \(\mathcal{D}\). It also preserves the condition \(X\cup Y\ne[5]\), because the only set of size at most \(3\) whose union with \(e\) is \([5]\) is \(\bar e\), which is absent. Thus exactly one of \(e\) and \(\bar e\) belongs to \(\mathcal{D}\). Therefore \[\mathcal{D}= \{\emptyset\}\cup\binom{[5]}1\cup \left(\binom{[5]}2\setminus E(G)\right) \cup\{\bar e:e\in E(G)\}.\] Consequently, \[\sum_{X\in \mathcal{D}}a_X = 1+\sum_{i=1}^5a_i+\sum_{e\in\binom{[5]}2}a_e +\sum_{e\in E(G)}(a_{\bar e}-a_e).\] This gives the upper bound \[m_0(1,n_1,\ldots,n_5) \le 1+\sum_{i=1}^5a_i+\sum_{1\le i<j\le5}a_ia_j +\max_{G\in\mathcal{I}_5}\sum_{e\in E(G)}\Delta_e.\]
Conversely, let \(G\in\mathcal{I}_5\) and define \[\mathcal{D}_G= \{\emptyset\}\cup\binom{[5]}1\cup \left(\binom{[5]}2\setminus E(G)\right) \cup\{\bar e:e\in E(G)\}.\] We claim that \(\mathcal{D}_G\in\mathfrak D_5\). First, \(\mathcal{D}_G\) is a downset. Since \(\mathcal{D}_G\) contains \(\emptyset\) and all singletons, the only nontrivial point is to check the \(2\)-subsets of the \(3\)-sets \(\bar e\) with \(e\in E(G)\). Let \(f\in\binom{[5]}2\) and suppose \(f\subseteq\bar e\) for some \(e\in E(G)\). Then \(f\cap e=\emptyset\). Since \(G\) is pairwise intersecting, we have \(f\notin E(G)\), and hence \[f\in \binom{[5]}2\setminus E(G)\subseteq \mathcal{D}_G.\] Thus \(\mathcal{D}_G\) is a downset.
It remains to verify the admissibility condition. Since \(\mathcal{D}_G\) contains all singletons, we have \[\bigcup_{X\in \mathcal{D}_G}X=[5].\] Moreover, no two members of \(\mathcal{D}_G\) have union \([5]\). Indeed, among sets of sizes at most \(3\), a full union can occur only from a complementary \(2\)–\(3\) pair or from two \(3\)-sets. A complementary \(2\)–\(3\) pair cannot occur, because whenever \(\bar e\in \mathcal{D}_G\), the corresponding \(2\)-set \(e\) has been deleted. If two \(3\)-sets \(\bar e,\bar f\in \mathcal{D}_G\) had union \([5]\), then \(e\cap f=\emptyset\), contradicting the fact that \(G\) is pairwise intersecting. Hence \(\mathcal{D}_G\in\mathfrak D_5\).
Therefore the upper bound is sharp, and \[m_0(1,n_1,\ldots,n_5) = 1+\sum_{i=1}^5a_i+\sum_{1\le i<j\le5}a_ia_j +\max_{G\in\mathcal{I}_5}\sum_{e\in E(G)}\Delta_e.\]
Finally, any intersecting graph on five vertices is contained in a star or in a triangle. The maximum \(\max_{G\in\mathcal{I}_5}\sum_{e\in E(G)}\Delta_e\) may be checked over subgraphs of stars and triangles. Since the weights \(\Delta_e\) may be negative, an optimal graph need not be a full star or a full triangle. ◻
Let \(\mathcal{I}_6\) be the set of all intersecting graphs on \([6]\), including the empty graph. For \(e\in\binom{[6]}2\), write \(\overline{e}=[6]\setminus e\) and \[\Delta_e:=a_{\overline{e}}-a_e.\] Set \[B_2^{(6)}:=1+\sum_{i=1}^6a_i+ \sum_{1\le i<j\le6}a_i a_j.\] For \(G\in\mathcal{I}_6\), define \[\mathcal{A}(G):= \left\{C\in\binom{[6]}3:\text{there is no }e\in E(G)\text{ with }e\subseteq C\right\}.\] The \(3\)-subsets of \([6]\) are paired by complementation. Define \[\Phi(G):= \sum_{\{C,[6]\setminus C\}} \max\left( \{0\}\cup \{a_C:C\in\mathcal{A}(G)\}\cup \{a_{[6]\setminus C}:[6]\setminus C\in\mathcal{A}(G)\} \right),\] where each complementary pair is counted once.
Theorem 8. For \(n_i\ge2\) and \(a_i=n_i-1\) for \(i\in [6]\), we have \[m_0(1,n_1,\ldots,n_6) = B_2^{(6)}+ \max_{G\in\mathcal{I}_6} \left( \sum_{e\in E(G)}\Delta_e+ \Phi(G) \right).\]
Proof. Let \(\mathcal{D}\in\mathfrak D_6\) be optimal. Since \(\mathcal{D}\) contains all singletons, it contains no \(5\)-set and no \(6\)-set. Thus every member of \(\mathcal{D}\) has size at most \(4\).
Define \[E(G):=\{e\in\binom{[6]}2:\overline{e}\in \mathcal{D}\}.\] If \(e,f\in E(G)\) are disjoint, then \(\overline{e}\cup\overline{f}=[6]\), impossible. Thus \(G\in\mathcal{I}_6\). As in the \(r=5\) case, optimality forces \[\mathcal{D}\cap\binom{[6]}2=\binom{[6]}2\setminus E(G),\] for the only obstruction to adding a missing \(2\)-set \(e\) is the presence of its complementary \(4\)-set \(\overline{e}\). Thus \[\mathcal{D}\cap\binom{[6]}4=\{\overline{e}:e\in E(G)\}.\] The total contribution from levels \(0,1,2,4\) is \[1+\sum_{i=1}^6a_i +\sum_{e\in\binom{[6]}2\setminus E(G)}a_e +\sum_{e\in E(G)}a_{\overline{e}} = B_2^{(6)}+\sum_{e\in E(G)}(a_{\overline{e}}-a_e).\] It remains to bound the possible contribution from the third level. Put \[H:=\mathcal{D}\cap\binom{[6]}3.\] We first show that \(H\subseteq\mathcal{A}(G)\), if \(C\in H\) and some \(e\in E(G)\) satisfies \(e\subseteq C\), then \(C\cup\overline{e}=[6]\), contradicting the conditions of \(\mathfrak{D}_6\). Hence \(H\subseteq\mathcal{A}(G)\). Also \(H\) cannot contain both \(C\) and \([6]\setminus C\). Hence the total weight of \(H\) is at most \(\Phi(G)\).
Since \(\Delta_e=a_{\overline{e}}-a_e\), we obtain \[\sum_{X\in \mathcal{D}}a_X \le B_2^{(6)}+\sum_{e\in E(G)}\Delta_e+\Phi(G).\] Taking the maximum over \(G\in\mathcal{I}_6\) gives the upper bound \[m_0(1,n_1,\ldots,n_6) \le B_2^{(6)} + \max_{G\in\mathcal{I}_6} \left( \sum_{e\in E(G)}\Delta_e+\Phi(G) \right).\]
We now prove the reverse inequality. Fix \(G\in\mathcal{I}_6\). For each complementary pair \(\{C, [6]\setminus C\}\) of \(3\)-sets, choose a member of maximum weight among the members of this pair that lie in \(\mathcal{A}(G)\); if neither member lies in \(\mathcal{A}(G)\), choose nothing. Let \(H_G\) be the family of chosen triples. Then \[H_G\subseteq \mathcal{A}(G), \qquad \sum_{C\in H_G}a_C=\Phi(G).\] Define \[\mathcal{D}_G:= \{\emptyset\} \cup\binom{[6]}1 \cup\left(\binom{[6]}2\setminus E(G)\right) \cup H_G \cup\{\overline{e}:e\in E(G)\}.\] We claim that \(\mathcal{D}_G\in\mathfrak D_6\). First, \(\mathcal{D}_G\) is a downset. Since it contains \(\emptyset\) and all singletons, the only nontrivial points to check are the chosen \(3\)-sets and the present \(4\)-sets. Let \(C\in H_G\). Since \(C\in \mathcal{A}(G)\), no edge of \(G\) is contained in \(C\). Therefore every \(2\)-subset of \(C\) lies in \(\binom{[6]}2\setminus E(G)\subseteq \mathcal{D}_G\).
Now let \(\overline{e}\) be a present \(4\)-set, with \(e \in E(G)\). Clearly, all \(2\)-subsets of \(\overline{e}\) belong to \(\binom{[6]}2\setminus E(G)\). Let \(C \subseteq \overline{e}\) be any \(3\)-subset. Since \(C \subseteq \overline{e} = [6]\setminus e\), every \(2\)-subset of \(C\) is disjoint from \(e\). Thus, by the intersection of \(G\), we must have \(C\in\mathcal{A}(G)\). On the other hand, the complementary \(3\)-set \([6]\setminus C\) contains \(e\), and therefore \([6]\setminus C \notin \mathcal{A}(G)\), the construction of \(H_G\) forces \(C\in H_G\). This proves that \(\mathcal{D}_G\) is a downset.
Second, since \(\mathcal{D}_G\) has no member of size larger than \(4\), the only possible ways for two members to have union \([6]\) are of types \(2-4\), \(3-3\), \(3-4\), and \(4-4\). For a \(2-4\) pair, every present \(4\)-set has the form \(\overline{e}\) with \(e\in E(G)\). However, in the construction of \(\mathcal{D}_G\), the set \(e\) has already been deleted. Hence no complementary \(2-4\) pair occurs.
For a \(3-3\) pair, two \(3\)-sets have union \([6]\) exactly when they are complementary. But \(H_G\) contains at most one member from each complementary pair, so this cannot occur.
For a \(3-4\) pair, let \(C\in H_G\) and let \(\overline{e}\) be a present \(4\)-set with \(e\in E(G)\). If \(C\cup\overline{e}=[6]\), then \(e\subseteq C\). This contradicts \(C\in \mathcal{A}(G)\). Hence no \(3-4\) pair has full union.
Finally, consider two present \(4\)-sets \(\overline{e}\) and \(\overline{f}\), where \(e, f\in E(G)\). Their union is \([6]\) if and only if \(e\cap f=\emptyset\), this contradicts the intersection of \(G\). Thus \(X\cup Y\neq [6]\) for all \(X, Y\in \mathcal{D}_G\).
Third, since \(\mathcal{D}_G\) contains all singletons, we have \[\bigcup_{X\in \mathcal{D}_G} X=[6].\] Consequently, \(\mathcal{D}_G\in\mathfrak{D}_6\).
It remains to compute its weight. By construction, \[\begin{align} \sum_{X\in \mathcal{D}_G} a_X &= 1+\sum_{i=1}^{6} a_i +\sum_{e\in \binom{[6]}{2}\setminus E(G)} a_e +\sum_{C\in H_G} a_C +\sum_{e\in E(G)} a_{\bar e}\\ &=B_2^{(6)}+\sum_{e\in E(G)}\Delta_e+\Phi(G). \end{align}\]
Since this construction works for every \(G\in \mathcal{I}_6\), we obtain \[m_0(1,n_1,\ldots,n_6) \ge B_2^{(6)}+\max_{G\in \mathcal{I}_6}\left(\sum_{e\in E(G)}\Delta_e+\Phi(G)\right).\]
Together with the upper bound, this proves \[m_0(1,n_1,\ldots,n_6)=B_2^{(6)}+\max_{G\in \mathcal{I}_6}\left(\sum_{e\in E(G)}\Delta_e+\Phi(G)\right).\] The theorem follows. ◻
The authors used generative AI tools to assist checking proofs, and improving exposition.