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Fractional factorial aliases and foldover

1 · Prerequisites

2 · Summary

Regular two-level fractions are selected from the coded {−1,+1}k run cube by independent generator equations. Their signed defining subgroup records all exact and opposite-column relations; resolution is the shortest nonidentity defining word. The proofs establish the run count and show that an equal-column alias gives only a combined coefficient in an unrestricted linear model.

A complete mirror foldover reverses every factor sign, so a word of order r changes by (−1)r. It separates opposite-parity aliases while same-parity relations may remain. Reversing a selected set J retains exactly the original defining words with even intersection with J; this formula lists residual aliases for partial follow-ups. Resolution and aliasing are design-matrix properties. Sparsity or hierarchy assumptions used to interpret a combined effect are additional modeling assumptions, not consequences of the design.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

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

Definition

Fix integers k≥1 and 1≤p≤k, and put [k]={1,…,k}. Let F2={0,1} with addition and multiplication modulo 2. For a factor subset G⊆[k], its incidence vector 1G∈F2k has coordinate 1 at j exactly when j∈G. Choose p subsets G1,…,Gp whose incidence vectors are linearly independent over F2, and choose signs si∈{−1,+1}. Using the run space and product columns of Two-level full factorial designs and coded effects, define the regular fraction

F={x∈{−1,+1}[k]:xGi=si for 1≤i≤p},xG=∏j∈Gxj.

The corresponding design is called a regular 2k−p fractional factorial; The defining subgroup gives the exact alias relations ↗ proves that F has exactly 2k−p runs, including the case p=k with one run. The choices si=−1 are alternate fractions and are part of the definition.

A signed word is a pair (ϵ,A) with ϵ∈{−1,+1} and A⊆[k]. It evaluates at x as ϵxA and multiplies by

(ϵ,A)(δ,B)=(ϵδ,A△B),A△B=(A∖B)∪(B∖A).

The identity word is (+1,∅). The signed generator hi=(si,Gi) evaluates to +1 on F. The defining subgroup H is generated by h1,…,hp; the alias theorem proves it is exactly the subgroup of signed words that evaluate to +1 on every run in F.

For A,B⊆[k], their restricted sign columns are equal-column aliases when xA=xB for every x∈F. They are signed aliases with sign σ∈{−1,+1} when xA=σxB on every run; sign +1 is the equal-column case and sign −1 records opposite columns. The alias theorem gives the exact criterion for each sign.

The word length of (ϵ,A) is ∣A∣, the number of factor letters, independent of ϵ. The resolution is the minimum ∣A∣ among nonidentity words (ϵ,A)∈H; it is written as a Roman numeral. Since p≥1 and the generators are independent, H has a nonidentity word of nonzero length.

Remarks

  • Equal or opposite design columns make the corresponding unrestricted linear-model coefficients nonseparable; see Aliased factorial columns prevent separate coefficient identification. A hierarchy or sparsity premise is an additional modeling assumption, not a consequence of the fraction.
  • The resolution records only the shortest defining word. It does not describe the complete alias structure; that is given by the whole subgroup H.

Sources

Oehlert §§18.2 and 18.4 and NIST/SEMATECH §§5.3.3.4.2–5.3.3.4.4 support the generator, signed-relation, and resolution conventions. The run count and exact subgroup characterization are proved locally in the justified-by theorem.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passOpen item page →

The defining subgroup gives the exact alias relations

Statement

Let F be the regular fraction determined by p independent generator equations xGi=si on the full 2k sign design, and let H be its signed defining subgroup. Then ∣F∣=2k−p, ∣H∣=2p, and H consists of all signed words (ϵ,C) for which ϵxC=+1 on every run in F. For all factor subsets A,B⊆[k] and signs σ∈{−1,+1},

xA=σxB on F⟺(σ,A△B)∈H.

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

Facts & Assumptions

Given: Integers k≥1 and 1≤p≤k, p subsets Gi⊆[k] with linearly independent incidence vectors over F2, signs si∈{−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 Gi and signs si specify the fraction by the equations xGi=si.

[F2]

Regular two-level fractions, signed defining relations, and resolution: the signed generators are hi=(si,Gi), H=⟨h1,…,hp⟩, and signed-word multiplication uses symmetric difference of factor subsets.

[A1]

The incidence vectors 1Gi∈F2k 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 xj=(−1)zj and each generator sign as si=(−1)bi, with z∈F2k and b∈F2p. Let M have rows 1Gi. The equations for F become Mz=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∣=2k−p.

2.1givenF1A1step 1.1algebra

Row-reduce M to R=UM, where U is invertible, and let D={d1,…,dp} be the pivot coordinates; write Ri for row i of R. For a subset C, put c=1C; on a solution z, xC=(−1)c⋅z. Substitution of the pivot equations gives c⋅z=∑icdibi′+∑j∉D(cj+∑icdiRij)zj, where b′=Ub. 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=∑icdiRi, 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 xC is nonconstant on F, so neither sign ϵ makes ϵxC=+1 on every run. If c is in the row space, then c=∑iλi1Gi for a unique λ∈F2p, and on F we have xC=(−1)λ⋅b=∏isiλi. Exactly one sign ϵ=∏isiλi makes ϵxC=+1 on all of the nonempty set F, and (ϵ,C)=∏i:λi=1(si,Gi). Conversely, every such product evaluates to +1. Independence makes the 2p choices of λ distinct, proving that H is precisely the set of constant-+1 signed words and ∣H∣=2p.

4.1F1F2step 3.1algebra∎

For any A,B⊆[k], cancellation of repeated factors gives xAxB=xA△B. Therefore xA=σxB on F exactly when σxA△B=+1 there. By step 3.1 and the definition of H in [F2], this is equivalent to (σ,A△B)∈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.

TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (gpt-6-luna)Open item page →

Aliased factorial columns prevent separate coefficient identification

Statement

Let cA,cB∈RF be the columns for distinct factorial effects A,B⊆[k] in a regular fraction, and suppose cA=σcB for a fixed σ∈{−1,+1}. Consider the unrestricted fixed-design linear model Y=Xβ+ϵ, where these columns occur in X and the distribution of the error vector ϵ does not depend on β. A coefficient combination is identifiable if it has the same value for all parameter vectors giving the same distribution of Y. The contribution of these two columns depends only on σβA+βB; neither coefficient is separately identifiable while the parameter space remains unrestricted. For equal columns (σ=+1), the relevant combination is the sum; for opposite columns (σ=−1), it is the difference. Additional runs can remove the ambiguity only if they break the column relation, and hierarchy or sparsity restrictions are extra modeling assumptions.

Facts & Assumptions

Given: A regular fraction with distinct effects A,B, design columns cA,cB satisfying cA=σcB, and the fixed-design model and error-law condition in the Statement.

[F1]

Regular two-level fractions, signed defining relations, and resolution: signed-alias columns satisfy xA=σxB on every run in the fraction; sign +1 is equality and sign −1 is opposition.

Proof

technique · exhibit the coefficient combination in the fitted vector and a parameter direction that leaves it unchanged
1.1givenF1algebra

The two terms in the fitted vector combine as βAcA+βBcB=βA(σcB)+βBcB=(σβA+βB)cB. Thus the fitted vector depends on this pair only through σβA+βB, with all other coefficients fixed.

2.1givenstep 1.1algebra∎

For any t∈R, replace βA by βA+t and βB by βB−σt. Then σ(βA+t)+(βB−σt)=σβA+βB, so the combination in step 1.1 and therefore Xβ are unchanged. Because the error law is independent of β, the resulting response distributions are identical. Taking t≠0 changes both coefficients, proving that neither is separately identifiable in the unrestricted model.

Sources

Oehlert §18.6 and NIST/SEMATECH §5.3.3.4.3 explain the ambiguity left by aliased effects and the extra role of sparsity assumptions. The parameter nonidentifiability argument is given explicitly above.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6-luna)Open item page →

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⊆[k], define τJx by (τJx)j=−xj for j∈J and (τJx)j=xj for j∉J. The partial foldover on J is the run table formed by stacking the original runs with one transformed copy τJ(F); the complete mirror foldover is the case J=[k]. Its residual signed defining subgroup is exactly

HJ={(ϵ,A)∈H:∣A∩J∣ is even}.

Thus an originally signed-aliased pair A,B remains signed-aliased on the augmented table exactly when ∣(A△B)∩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≥1, a regular fraction F, its signed defining subgroup H, a factor subset J⊆[k], and the stacked run table containing both F and τJ(F).

[F1]

Regular two-level fractions, signed defining relations, and resolution: a signed word (ϵ,A) evaluates as ϵxA, 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)∈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(τJx)A=(−1)∣A∩J∣ϵxA. 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∩J∣ on every appended row. Since both blocks are present, it is a constant-+1 relation on the stacked table exactly when ∣A∩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 HJ.

3.1givenF2step 2.1algebra∎

For any two effects, xA=σxB 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)∩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∣≡∣A∣+∣B∣(mod2), 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.

5 · Examples, counterexamples and false statements

None yet.

Sources