TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-26

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Covering ${0, 1}^{n}$ minus the origin by affine hyperplanes avoiding the origin needs at least $n$ of them

Statement

For $1 \leq k \leq m$ , let $a_{k} \in R^{n}$ be nonzero, let $b_{k} \in R$ be nonzero, and put

$H_{k} : = {x \in R^{n} : ⟨ a_{k}, x ⟩ = b_{k}} .$

If every nonzero vertex of the cube ${0, 1}^{n}$ lies on at least one $H_{k}$ , then $m \geq n$ .

Facts & Assumptions

Given: nonzero vectors $a_{k} \in R^{n}$ and nonzero scalars $b_{k}$ , with the hyperplanes $H_{k} = {x : ⟨ a_{k}, x ⟩ = b_{k}}$ covering every nonzero cube vertex.

[L1]

Over a field, if $\deg f = \sum_{i} t_{i}$ , the coefficient of $x_{1}^{t_{1}} \dots x_{n}^{t_{n}}$ is nonzero, and $∣ S_{i} ∣ > t_{i}$ for every $i$ , then $f$ is nonzero at some point of $S_{1} \times \dots \times S_{n}$ (Alon's Combinatorial Nullstellensatz: if $\deg f = \sum_{i} t_{i}$ , the coefficient of $x_{1}^{t_{1}} \dots x_{n}^{t_{n}}$ in $f$ is nonzero, and $∣ S_{i} ∣ > t_{i}$ , then $f (s_{1}, \dots, s_{n}) \neq 0$ for some $s_{i} \in S_{i}$ ).

[F1]

The standard bilinear form is $⟨ a, x ⟩ = \sum_{i < n} a_{i} x_{i}$ (The standard bilinear form $⟨ x, y ⟩ = \sum_{i < n} x_{i} y_{i}$ on $F^{n}$ ).

Proof

technique · contradiction

1.1assume-contraF1construct

Suppose, for contradiction, that $m < n$ , and define $f (x) : = (- 1)^{n + m} b_{1} \dots b_{m} \prod_{i < n} (x_{i} - 1) - \prod_{k = 1}^{m} (⟨ a_{k}, x ⟩ - b_{k}) .$

2.1step 1.1given

The polynomial $f$ vanishes on every vertex of ${0, 1}^{n}$ . At the origin, the two terms are equal by construction, so they cancel. At any nonzero cube vertex, the first product vanishes because some coordinate equals $1$ , and the second vanishes because that vertex lies on one of the hyperplanes.

2.2F1step 1.1

The total degree of $f$ is $n$ , and the coefficient of $\prod_{i < n} x_{i}$ is $\pm b_{1} \dots b_{m} \neq 0$ : the first product contributes that coefficient, while the second product has degree $m < n$ and contributes nothing to that top monomial.

3.1L1step 2.1step 2.2discharge-contradiction∎

Apply [L1] to the $n$ coordinates indexed by $i < n$ , taking $t_{i} = 1$ and $S_{i} = {0, 1}$ for every $i < n$ . Step 2.1 says that $f$ vanishes on the whole grid ${0, 1}^{n}$ , but step 2.2 says its top coefficient is nonzero and the degree hypothesis is exactly the required one. This contradiction proves $m \geq n$ .

Remarks

The constant in the first term is chosen only to force cancellation at the origin. That check is the one place where a sign error can hide.

Depends on

Used by

The n hyperplanes xᵢ=1 cover {0,1}ⁿ except the origin, so the Alon–Füredi bound is tight Example

Dependency tree · two levels

25 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

N. Alon, Combinatorial Nullstellensatz, Theorem 6.3 (standard reference, not scraped)
J. Matousek, Thirty-three Miniatures, Miniature 16 (standard reference, not scraped)

Covering {0,1}n{0,1}n minus the origin by affine hyperplanes avoiding the origin needs at least nn of them