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Full and partial foldovers separate aliases by parity

Statement

Let $F$ be a regular fraction with signed defining subgroup $H$ . For a specified set of factors $J \subseteq [k]$ , define $τ_{J} x$ by $(τ_{J} x)_{j} = - x_{j}$ for $j \in J$ and $(τ_{J} x)_{j} = x_{j}$ for $j \notin J$ . The partial foldover on $J$ is the run table formed by stacking the original runs with one transformed copy $τ_{J} (F)$ ; the complete mirror foldover is the case $J = [k]$ . Its residual signed defining subgroup is exactly

$H_{J} = {(ϵ, A) \in H : ∣ A \cap J ∣ is even} .$

Thus an originally signed-aliased pair $A, B$ remains signed-aliased on the augmented table exactly when $∣ (A △ B) \cap J ∣$ is even; if it is odd, the pair is separated. For a complete mirror foldover, a word of order $r = ∣ A △ B ∣$ changes sign by $(- 1)^{r}$ . Therefore aliases between effects of opposite order parity separate, while same-parity relations remain. The displayed formula gives the full residual alias set for every partial foldover; a selected-factor follow-up need not remove other aliases.

Facts & Assumptions

Given: $k \geq 1$ , a regular fraction $F$ , its signed defining subgroup $H$ , a factor subset $J \subseteq [k]$ , and the stacked run table containing both $F$ and $τ_{J} (F)$ .

[F1]

Regular two-level fractions, signed defining relations, and resolution: a signed word $(ϵ, A)$ evaluates as $ϵ x_{A}$ , and $H$ is the defining subgroup of the original fraction.

[F2]

The defining subgroup gives the exact alias relations: $A$ and $B$ are signed aliases with sign $σ$ exactly when $(σ, A △ B) \in H$ .

Proof

technique · evaluate signed monomials on the appended runs and apply the alias criterion

1.1givenF1algebra

For a signed word $(ϵ, A)$ and an original run $x$ , multiplication of its factors gives $ϵ (τ_{J} x)_{A} = (- 1)^{∣ A \cap J ∣} ϵ x_{A}$ . In particular, for $J = [k]$ the multiplier is $(- 1)^{∣ A ∣}$ .

2.1givenF1step 1.1algebra

Every word in $H$ evaluates to $+ 1$ on all original rows. By step 1.1 it evaluates to $(- 1)^{∣ A \cap J ∣}$ on every appended row. Since both blocks are present, it is a constant- $+ 1$ relation on the stacked table exactly when $∣ A \cap J ∣$ is even. Any relation on the stacked table must also hold on its original rows and hence belong to $H$ ; therefore the residual subgroup is exactly $H_{J}$ .

3.1givenF2step 2.1algebra∎

For any two effects, $x_{A} = σ x_{B}$ on a run table exactly when the signed word $(σ, A △ B)$ evaluates to $+ 1$ on every row. By step 2.1, an originally signed-aliased pair retains that relation on the stacked table precisely when $∣ (A △ B) \cap J ∣$ is even; if it is odd, the relation changes sign between blocks and no constant signed alias remains. A pair not signed-aliased on the original fraction cannot become one because the original rows remain in the augmented table. When $J = [k]$ , $∣ A △ B ∣ \equiv ∣ A ∣ + ∣ B ∣ (m o d 2)$ , so exactly opposite-parity pairs separate among the original aliases, and even-word relations remain.

Sources

Oehlert §18.7 develops mirror foldover by reversing every sign. NIST/SEMATECH §§5.3.3.8.1–5.3.3.8.2 covers both mirror and selected-factor follow-ups. The subgroup formula and exact residual condition for arbitrary $J$ are proved here.

Depends on

Used by

Full and targeted foldovers of a resolution III fraction Example

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Sources

Gary W. Oehlert, A First Course in Design and Analysis of Experiments, Chapter 18, §18.7 (standard reference, not scraped)
NIST/SEMATECH e-Handbook of Statistical Methods, §§5.3.3.8.1–5.3.3.8.2, Mirror-Image and Alternative Foldover Designs (standard reference, not scraped)