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A 920-block explicit construction guaranteeing a triple intersection with every 6-subset of [60]

Paulo Henrique Cunha Gomes

TL;DR

The paper presents an explicit construction of $920$ blocks of size $6$ on the ground set $[60]$ so that every $6$-subset intersects some block in at least three elements, i.e., a triple-intersection covering of $J(60,6)$. The construction partitions the ground set into halves, forms two block families by unions of three disjoint pairs within each half, and appends ten fixed consecutive blocks, achieving $|\,\mathcal{B}\,|=920$. A key result is the guaranteed triple intersection for all $|S|=6$ subsets, proved by a case split on the distribution of $S$ across halves and leveraging the pair structure. The work also provides a lower bound on the minimal possible size $M$ of such a family ($97\le M\le 920$) and discusses how different partition viewpoints influence the existence of triple intersections, including an obstruction for certain partition schemes. Overall, the paper delivers a simple, explicit, and deterministic construction and clarifies the role of partition topology in enforcing intersection properties, with open questions remaining on the exact minimal size.

Abstract

We present an explicit family $\mathcal{B}$ of $920$ subsets of size $6$ of $[60]=\{1,\dots,60\}$ with the property that every $6$-subset $S\subset[60]$ intersects at least one block $B\in\mathcal{B}$ in at least three elements, i.e.\ $|S\cap B|\ge 3$. The construction is purely combinatorial, based on a partition of the ground set into pairs and a pigeonhole argument. We also record a simple counting lower bound and discuss how different partitions of the ten base blocks affect the emergence of triple intersections.

A 920-block explicit construction guaranteeing a triple intersection with every 6-subset of [60]

TL;DR

The paper presents an explicit construction of blocks of size on the ground set so that every -subset intersects some block in at least three elements, i.e., a triple-intersection covering of . The construction partitions the ground set into halves, forms two block families by unions of three disjoint pairs within each half, and appends ten fixed consecutive blocks, achieving . A key result is the guaranteed triple intersection for all subsets, proved by a case split on the distribution of across halves and leveraging the pair structure. The work also provides a lower bound on the minimal possible size of such a family () and discusses how different partition viewpoints influence the existence of triple intersections, including an obstruction for certain partition schemes. Overall, the paper delivers a simple, explicit, and deterministic construction and clarifies the role of partition topology in enforcing intersection properties, with open questions remaining on the exact minimal size.

Abstract

We present an explicit family of subsets of size of with the property that every -subset intersects at least one block in at least three elements, i.e.\ . The construction is purely combinatorial, based on a partition of the ground set into pairs and a pigeonhole argument. We also record a simple counting lower bound and discuss how different partitions of the ten base blocks affect the emergence of triple intersections.
Paper Structure (7 sections, 1 theorem, 10 equations)

This paper contains 7 sections, 1 theorem, 10 equations.

Table of Contents

  1. Introduction
  2. Construction
  3. Main result
  4. Bounds and minimality
  5. A (5,5) partition viewpoint
  6. A (3,3,4) partition and an obstruction
  7. Conclusion

Key Result

Theorem 1

Let $\mathcal{B}$ be the family of $920$ blocks constructed above. For every $S\subset[60]$ with $|S|=6$, there exists $B\in\mathcal{B}$ such that $|S\cap B|\ge 3$.

Theorems & Definitions (2)

  • Theorem 1
  • proof