[Paper Review] Erd\H os-Ko-Rado theorem for $\{0,\pm 1\}$-vectors
This paper extends the classical Erdös-Ko-Rado theorem to {0,±1}-vectors with exactly k +1s and one -1, determining the maximum size of intersecting families under the condition that no two vectors have a scalar product of -2. The key result reveals a phase transition at n = k², where the standard Erdös-Ko-Rado construction becomes suboptimal, and provides exact formulas for m(n,k,1) in both the range 2k ≤ n ≤ k² and n > k² using extremal set theory and shifting techniques.
The main object of this paper is to determine the maximum number of $\{0,\pm 1\}$-vectors subject to the following condition. All vectors have length $n$, exactly $k$ of the coordinates are $+1$ and one is $-1$, $n \geq 2k$. Moreover, there are no two vectors whose scalar product equals the possible minimum, $-2$. Thus, this problem may be seen as an extension of the classical Erd\H os-Ko-Rado theorem. Rather surprisingly there is a phase transition in the behaviour of the maximum at $n=k^2$. Nevertheless, our solution is complete. The main tools are from extremal set theory and some of them might be of independent interest.
Motivation & Objective
- To extend the classical ErdÖs-Ko-Rado theorem to families of {0,±1}-vectors with exactly k coordinates +1 and one coordinate -1.
- To determine the maximum size m(n,k,1) of intersecting families where no two vectors have scalar product -2, the theoretical minimum.
- To identify the threshold at n = k² where the standard ErdÖs-Ko-Rado construction ceases to be optimal.
- To provide a complete characterization of m(n,k,1) for all n ≥ 2k using advanced extremal set theory tools.
Proposed method
- Define m(n,k,1) as the maximum size of a family of {0,±1}-vectors with k +1s and one -1, such that no two vectors have scalar product -2.
- Use shifting techniques to transform families into more structured forms while preserving intersecting properties and vector counts.
- Apply combinatorial bounds on shadow sizes and cross-intersecting families to control the growth of families under constraints.
- Establish recursive inequalities via the construction of extended families in higher dimensions, leading to the recurrence m(n+1,k,1) ≥ m(n,k,1) + (n choose k).
- Prove that for n ≥ k², equality holds: m(n+1,k,1) = m(n,k,1) + (n choose k), implying a linear growth phase.
- Use case analysis based on the size of the family of vectors with -1 in a fixed coordinate, applying bounds from extremal set theory to derive contradictions for larger-than-optimal families.
Experimental results
Research questions
- RQ1What is the maximum size of a family of {0,±1}-vectors with exactly k +1s and one -1, such that no two vectors have scalar product -2, for n ≥ 2k?
- RQ2Does the classical ErdÖs-Ko-Rado construction remain optimal for m(n,k,1) when n > k²?
- RQ3How does the extremal family size m(n,k,1) behave asymptotically for n > k², and what is the exact formula?
- RQ4What structural phase transition occurs at n = k² in the behavior of extremal intersecting families of {0,±1}-vectors?
- RQ5Can the problem be solved completely for l=1 using shifting and shadow techniques in extremal set theory?
Key findings
- For 2k ≤ n ≤ k², the maximum family size is exactly m(n,k,1) = k × (n-1 choose k), achieved by fixing one coordinate to +1.
- For n > k², the ErdÖs-Ko-Rado construction is no longer optimal, and the maximum size grows linearly with n.
- The recurrence m(n+1,k,1) = m(n,k,1) + (n choose k) holds for all n ≥ k², establishing a linear growth phase.
- The exact formula for n > k² is m(n,k,1) = m(k²,k,1) + ∑_{i=k²}^{n-1} (i choose k), derived from the recurrence.
- The phase transition at n = k² is confirmed by the change in extremal structure: from a star-like construction to a linearly growing family.
- The proof relies on advanced extremal set theory, including shifting, shadow bounds, and case analysis on the size of families with -1 in a fixed coordinate.
Better researchstarts right now
From reading papers to final review, dramatically reduce your research time.
No credit card · Free plan available
This review was created by AI and reviewed by human editors.