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The defining subgroup gives the exact alias relations

Statement

Let $F$ be the regular fraction determined by $p$ independent generator equations $x_{G_{i}} = s_{i}$ on the full $2^{k}$ sign design, and let $H$ be its signed defining subgroup. Then $∣ F ∣ = 2^{k - p}$ , $∣ H ∣ = 2^{p}$ , and $H$ consists of all signed words $(ϵ, C)$ for which $ϵ x_{C} = + 1$ on every run in $F$ . For all factor subsets $A, B \subseteq [k]$ and signs $σ \in {- 1, + 1}$ ,

$x_{A} = σ x_{B} on F ⟺ (σ, A △ B) \in H .$

In particular, the columns agree exactly when $(+ 1, A △ B) \in H$ ; a relation with sign $- 1$ means the columns are opposites.

Facts & Assumptions

Given: Integers $k \geq 1$ and $1 \leq p \leq k$ , $p$ subsets $G_{i} \subseteq [k]$ with linearly independent incidence vectors over $F_{2}$ , signs $s_{i} \in {- 1, + 1}$ , the resulting regular fraction $F$ , and its signed defining subgroup $H$ .

[F1]

Regular two-level fractions, signed defining relations, and resolution: $p$ independent factor subsets $G_{i}$ and signs $s_{i}$ specify the fraction by the equations $x_{G_{i}} = s_{i}$ .

[F2]

Regular two-level fractions, signed defining relations, and resolution: the signed generators are $h_{i} = (s_{i}, G_{i})$ , $H = ⟨ h_{1}, \dots, h_{p} ⟩$ , and signed-word multiplication uses symmetric difference of factor subsets.

[A1]

The incidence vectors $1_{G_{i}} \in F_{2}^{k}$ are linearly independent; this is part of the stated generator hypothesis.

Proof

technique · encode signs by bits and row-reduce the independent generator equations

1.1givenF1A1choosealgebra

Write each run coordinate as $x_{j} = (- 1)^{z_{j}}$ and each generator sign as $s_{i} = (- 1)^{b_{i}}$ , with $z \in F_{2}^{k}$ and $b \in F_{2}^{p}$ . Let $M$ have rows $1_{G_{i}}$ . The equations for $F$ become $M z = b$ . By [A1], $M$ has rank $p$ ; row reduction gives $p$ pivot coordinates and $k - p$ free coordinates, and every assignment of the free coordinates determines exactly one pivot assignment. Therefore $∣ F ∣ = 2^{k - p}$ .

2.1givenF1A1step 1.1algebra

Row-reduce $M$ to $R = U M$ , where $U$ is invertible, and let $D = {d_{1}, \dots, d_{p}}$ be the pivot coordinates; write $R_{i}$ for row $i$ of $R$ . For a subset $C$ , put $c = 1_{C}$ ; on a solution $z$ , $x_{C} = (- 1)^{c \cdot z}$ . Substitution of the pivot equations gives $c \cdot z = \sum_{i} c_{d_{i}} b_{i}^{'} + \sum_{j \notin D} (c_{j} + \sum_{i} c_{d_{i}} R_{i j}) z_{j}$ , where $b^{'} = U b$ . This exponent is constant over all free coordinates exactly when every displayed free-coordinate coefficient is zero. Since the pivot columns of $R$ are the identity, those equations say precisely that $c = \sum_{i} c_{d_{i}} R_{i}$ , so $c$ lies in the row space of $M$ .

3.1givenF1F2A1step 1.1step 2.1algebra

If $c$ is outside the row space of $M$ , step 2.1 shows that $x_{C}$ is nonconstant on $F$ , so neither sign $ϵ$ makes $ϵ x_{C} = + 1$ on every run. If $c$ is in the row space, then $c = \sum_{i} λ_{i} 1_{G_{i}}$ for a unique $λ \in F_{2}^{p}$ , and on $F$ we have $x_{C} = (- 1)^{λ \cdot b} = \prod_{i} s_{i}^{λ_{i}}$ . Exactly one sign $ϵ = \prod_{i} s_{i}^{λ_{i}}$ makes $ϵ x_{C} = + 1$ on all of the nonempty set $F$ , and $(ϵ, C) = \prod_{i : λ_{i} = 1} (s_{i}, G_{i})$ . Conversely, every such product evaluates to $+ 1$ . Independence makes the $2^{p}$ choices of $λ$ distinct, proving that $H$ is precisely the set of constant- $+ 1$ signed words and $∣ H ∣ = 2^{p}$ .

4.1F1F2step 3.1algebra∎

For any $A, B \subseteq [k]$ , cancellation of repeated factors gives $x_{A} x_{B} = x_{A △ B}$ . Therefore $x_{A} = σ x_{B}$ on $F$ exactly when $σ x_{A △ B} = + 1$ there. By step 3.1 and the definition of $H$ in [F2], this is equivalent to $(σ, A △ B) \in H$ . The case $σ = + 1$ is equality of columns, and $σ = - 1$ is opposition.

Sources

Oehlert §18.2 and NIST/SEMATECH §§5.3.3.4.2–5.3.3.4.4 present signed generators, defining relations, and alias multiplication. The exact row-space characterization and both directions of the column criterion are proved here.

Depends on

Regular two-level fractions, signed defining relations, and resolution

Used by

Cited to discharge well-definedness by Regular two-level fractions, signed defining relations, and resolution.

Dependency tree · two levels

2 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

Gary W. Oehlert, A First Course in Design and Analysis of Experiments, Chapter 18, §18.2 (standard reference, not scraped)
NIST/SEMATECH e-Handbook of Statistical Methods, §§5.3.3.4.2–5.3.3.4.4 (standard reference, not scraped)