Characterization of three-order confounding via consulting sets

Abstract

The aliased effect-number pattern, proposed by Zhang et al. (Stat Sin 18:1689–1705, 2008), is often used to express the overall confounding between factorial effects in two-level regular designs. The confounding relationships of main effects and two-factor interactions have been well revealed in literature, but little is known about three and higher-order interactions. To fill the gaps, this paper aims to study the confounding of three-order interactions in any two-level design. When the factor number of the design is larger, we derive the confounding among lower-order and three-order interactions via the complementary method. Further, confounding formulas can be obtained owing to the consulting sets of the structured designs. These two approaches can cover all designs in the sense of isomorphism. As an application of the previous results, the confounding numbers of three-order interactions are tabulated for 16, 32, and 64-run optimal designs under the general minimum lower-order confounding criterion.

Appendix: Proofs of Lemmas 1–4

Proof of Lemma 1

Let $A_1=\{(d_1,d_2,d_3): d_1\in \overline{D}, d_2,d_3\in D, d_1d_2d_3=\gamma \}$ and $A_2=\{(d_1,d_2,d_3): d_1\in D, d_2,d_3\in \overline{D}, d_1d_2d_3=\gamma \}$. By the definition of $B_3(D,\gamma )$, for any $\gamma \in H_q$, we have

$$\begin{aligned} B_3(H_q,\gamma )= & {} \#\{(d_1,d_2,d_3): d_1,d_2,d_3\in D\cup \overline{D}, d_1d_2d_3=\gamma \}\\= & {} \#\{(d_1,d_2,d_3): d_1,d_2,d_3\in D, d_1d_2d_3=\gamma \}\\{} & {} +\,\#\{(d_1,d_2,d_3): d_1,d_2,d_3\in \overline{D}, d_1d_2d_3=\gamma \}\\{} & {} +\,\#\{(d_1,d_2,d_3): d_1\in \overline{D}, d_2,d_3\in D, d_1d_2d_3=\gamma \}\\{} & {} +\,\#\{(d_1,d_2,d_3): d_1\in D, d_2,d_3\in \overline{D}, d_1d_2d_3=\gamma \}\\= & {} B_3(D,\gamma )+B_3(\overline{D},\gamma )+\#A_1+\#A_2. \end{aligned}$$

From Proposition 1, there are $(N-2)(N-4)/2$ factors $d_i$ satisfying $d_{i_1}d_{i_2}d_{i_3}=\gamma$ without considering the repetition. It implies that for any fix factor in $H_q\backslash \gamma$, say $d_1$, equation $\#\{d_1: d_1d_2d_3=\gamma , d_1,d_2,d_3\in H_q\backslash \gamma \}=2$. Hence, we have

$$\begin{aligned} \sum \limits _{i=1}^2 i\#A_i+3B_3(\overline{D}, \gamma )= {\left\{ \begin{array}{ll} (N-4)(N-n-1)/2, &{} if \ \gamma \in D,\\ (N-4)(N-n-2)/2, &{} if \ \gamma \in \overline{D}. \end{array}\right. } \end{aligned}$$

Through the above equation, it leads to

$$\begin{aligned} B_3(D, \gamma )={\left\{ \begin{array}{ll} 2B_3(\overline{D}, \gamma )+ (N-4)(3n-2N+1)/6+\#A_2, &{} if \ \gamma \in D,\\ 2B_3(\overline{D}, \gamma )+ (N-4)(3n-2N+4)/6+\#A_2, &{} if \ \gamma \in \overline{D}. \end{array}\right. } \end{aligned}$$

(7)

Next, we calculate the value of $\#A_2$. For any factor $\gamma \in D$, it follows that

$$\begin{aligned} \#A_2= & {} \#\{(d_1,d_2,d_3): d_1,d_2\in \overline{D},d_3\in D, d_1d_2d_3=\gamma \}\\= & {} \#\{(d_1,d_2,\gamma d_3): d_1,d_2\in \overline{D}, d_3 \in H_q, d_1d_2d_3=I\}\\{} & {} -3\#\{(d_1,d_2,\gamma d_3): d_1,d_2\in \overline{D}, \gamma d_3 \in \overline{D}, d_1d_2d_3=I\}\\{} & {} -\#\{(d_1,d_2,\gamma d_3): d_1,d_2\in \overline{D}, \gamma d_3= I, d_1d_2d_3=I\}\\= & {} \#\{(d_1,d_2,\gamma d_3): d_1,d_2\in \overline{D}, d_3 \in H_q, d_1d_2d_3=I\}\\{} & {} -3\#\{(d_1,d_2,d_3): d_1,d_2,d_3\in \overline{D}, d_1d_2d_3=\gamma \}\\{} & {} -\#\{(d_1,d_2): d_1,d_2\in \overline{D}, d_1d_2=\gamma \}\\= & {} (N-n-1)(N-n-2)/2-3B_3(\overline{D}, \gamma )-B_2(\overline{D}, \gamma ). \end{aligned}$$

Similar to the above derivation, it can be obtained that $\#A_2=(N-n-2)(N-n-3)/2-3B_3(\overline{D}, \gamma )-B_2(\overline{D}, \gamma )$ for $\gamma \in \overline{D}$. Substituting $\#A_2$ into Eq. (7), the proof is complete. $\square$

Proof of Lemma 2

Let $A_1=\{(d_1,d_2,d_3): d_1\in [H_q\backslash H_r], d_2,d_3\in D\backslash [H_q\backslash H_r], d_1d_2d_3=\gamma \}$ and $A_2=\{(d_1,d_2,d_3): d_1\in D\backslash [H_q\backslash H_r], d_2,d_3\in [H_q\backslash H_r], d_1d_2d_3=\gamma \}$. For any $\gamma \in H_q$, it follows that

$$\begin{aligned} B_3(D,\gamma )= & {} \#\{(d_1,d_2,d_3): d_1,d_2,d_3\in D, d_1d_2d_3=\gamma \}\\= & {} \#\{(d_1,d_2,d_3): d_1,d_2,d_3\in [H_q\backslash H_r], d_1d_2d_3=\gamma \}\\{} & {} +\,\#\{(d_1,d_2,d_3): d_1,d_2,d_3\in D\backslash [H_q\backslash H_r], d_1d_2d_3=\gamma \}\\{} & {} +\,\#\{(d_1,d_2,d_3): d_1\in [H_q\backslash H_r], d_2,d_3\in D\backslash [H_q\backslash H_r], d_1d_2d_3=\gamma \}\\{} & {} +\,\#\{(d_1,d_2,d_3): d_1\in D\backslash [H_q\backslash H_r], d_2,d_3\in [H_q\backslash H_r], d_1d_2d_3=\gamma \}\\= & {} B_3([H_q\backslash H_r], \gamma )+B_3(D\backslash [H_q\backslash H_r],\gamma )+\#A_1+\#A_2. \end{aligned}$$

For any $\gamma \in H_r, \#A_1=0$. Then

$$\begin{aligned} B_3(D, \gamma )= B_3([H_q\backslash H_r], \gamma ) + B_3(D\backslash [H_q\backslash H_r], \gamma ) + \#A_2. \end{aligned}$$

Let $A_2'=\{(d_1,d_2,d_3): d_1\in H_r, d_2,d_3\in [H_q\backslash H_r], d_1d_2d_3=\gamma \}$. For $d_1, \gamma \in H_r$, there must exist $d_2, d_3\in [H_q\backslash H_r]$ such that $d_1d_2d_3=\gamma$. Hence, we have $\#A_2'=(2^r-2)(N-2^r)/2$ for $\gamma \in H_r$. It implies that for any fix factor in the remaining $2^r-2$ factors in $H_r$, say $d_1$, equation $\#\{d_1: d_1d_2d_3=\gamma , d_1,d_2,d_3,\gamma \in H_r\}=(N-2^r)/2$. Then,

$$\begin{aligned} \#A_2={\left\{ \begin{array}{ll} (N-2^r)(n-N+2^r-1)/2, &{} if \ \gamma \in D\backslash [H_q\backslash H_r],\\ (N-2^r)(n-N+2^r)/2, &{} if \ \gamma \in H_q\backslash D. \end{array}\right. } \end{aligned}$$

(8)

For $\gamma \in [H_q\backslash H_r]$, $B_3(D\backslash [H_q\backslash H_r], \gamma )=0$. Thus

$$\begin{aligned} B_3(D, \gamma )= B_3([H_q\backslash H_r], \gamma ) + \#A_1 + \#A_2. \end{aligned}$$

Similar to the derivation of Lemma 1, it can be obtained that $\#A_1=(n-N+2^r)(n-N+2^r-1)/2$ and $\#A_2=(N-2^{r+1})(n-N+2^r)/2$. Therefore, the proof is complete. $\square$

Proof of Lemma 3

By the definition of $B_3(D,\gamma )$ and $\gamma \in [H_q\backslash H_{q-1}]$, we have

$$\begin{aligned} B_3([H_q\backslash H_{q-1}],\gamma )= & {} \#\{(d_1,d_2,d_3): d_1,d_2,d_3\in D\cup [H_q\backslash H_{q-1}]\backslash D, d_1d_2d_3=\gamma \}\\= & {} \#\{(d_1,d_2,d_3): d_1,d_2,d_3\in D, d_1d_2d_3=\gamma \}\\{} & {} +\,\#\{(d_1,d_2,d_3): d_1,d_2,d_3\in [H_q\backslash H_{q-1}]\backslash D, d_1d_2d_3=\gamma \}\\{} & {} +\,\#\{(d_1,d_2,d_3): d_1\in [H_q\backslash H_{q-1}]\backslash D, d_2,d_3\in D, d_1d_2d_3=\gamma \}\\{} & {} +\,\#\{(d_1,d_2,d_3): d_1\in D, d_2,d_3\in [H_q\backslash H_{q-1}]\backslash D, d_1d_2d_3=\gamma \}\\= & {} B_3(D,\gamma )+B_3([H_q\backslash H_{q-1}]\backslash D,\gamma )+\#A_1+\#A_2, \end{aligned}$$

where $A_1=\{(d_1,d_2,d_3): d_1\in [H_q\backslash H_{q-1}]\backslash D, d_2,d_3\in D, d_1d_2d_3=\gamma \}$ and $A_2=\{(d_1,d_2,d_3): d_1\in D, d_2,d_3\in [H_q\backslash H_{q-1}]\backslash D, d_1d_2d_3=\gamma \}$. Obviously, $\sum \limits _{i=1}^2i\#A_i+3B_3([H_q\backslash H_{q-1}]\backslash D,\gamma )$ is the total number of the factors in $[H_q\backslash H_{q-1}]\backslash D$, appearing in the sets: $\{(d_1, d_2, d_3): d_i \in [H_q\backslash H_{q-1}], i=1,2,3, d_1d_2d_3=\gamma \}$. By Proposition 3, there are $(N-2)(N-4)/8$ factors satisfying $d_1 d_2 d_3=\gamma$. Hence,

$$\begin{aligned} \sum \limits _{i=1}^2i\#A_i+3B_3([H_q\backslash H_{q-1}]\backslash D,\gamma )= {\left\{ \begin{array}{ll} (N/2-n)(N-4)/4, &{} if \ \gamma \in D,\\ (N/2-n-1)(N-4)/4, &{} if \ \gamma \in [H_q\backslash H_{q-1}]. \end{array}\right. } \end{aligned}$$

Thus, $B_3([H_q\backslash H_{q-1}],\gamma )=B_3(D,\gamma )+\#A_1+\#A_2+B_3([H_q\backslash H_{q-1}]\backslash D,\gamma )$. Further, we have

$$\begin{aligned} B_3([H_q\backslash H_{q-1}], \gamma )= & {} B_3(D, \gamma )+\sum _{i=1}^2i\#A_i+3B_3([H_q \backslash H_{q-1}]\backslash D,\gamma )\\{} & {} \quad -\#A_2-2B_3([H_q\backslash H_{q-1}]\backslash D, \gamma ). \end{aligned}$$

Based on the equation above, it is vital to calculate the value of $\#A_2$. Similarly to the proofs of Lemmas 1 and 2, we can obtain

$$\begin{aligned} \#A_2= {\left\{ \begin{array}{ll} (N-2n)(N-2n-2)/8-3B_3([H_q\backslash H_{q-1}] \backslash D, \gamma ), &{} if \ \gamma \in D,\\ (N-2n-2)(N-2n-4)/8-3B_3([H_q\backslash H_{q-1}] \backslash D, \gamma ), &{} if \ \gamma \in [H_q\backslash H_{q-1}]\backslash D. \end{array}\right. } \end{aligned}$$

The proof is completed by combining the above conclusions. $\square$

Proof of Lemma 4

By the definition of $B_2(dH_r,\gamma )$ and $\gamma \in H_r$, we have

$$\begin{aligned} B_2(dHr,\gamma )= & {} \#\{(d_1,d_2): d_1,d_2\in dH_r, d_1d_2=\gamma \}\\= & {} \#\{(d_1,d_2): d_1,d_2\in [H_q\backslash H_{q-1}]\backslash D, d_1d_2=\gamma \}\\{} & {} +\,\#\{(d_1,d_2): d_1,d_2\in \widetilde{D}, d_1d_2=\gamma \}\\{} & {} +\,\#\{(d_1,d_2): d_1\in [H_q\backslash H_{q-1}]\backslash D, d_2\in \widetilde{D}, d_1d_2=\gamma \}\\= & {} B_2(\widetilde{D},\gamma )+B_2([H_q\backslash H_{q-1}]\backslash D,\gamma )+\#T_1, \end{aligned}$$

where $T_1=\{(d_1,d_2): d_1\in [H_q\backslash H_{q-1}]\backslash D, d_2\in \widetilde{D}, d_1d_2=\gamma \}$. Thus,

$$\begin{aligned} \#T_1=\#T_1+2B_2([H_q\backslash H_{q-1}]\backslash D, \gamma ) - 2B_2(d \{ [H_q\backslash H_{q-1}]\backslash D\}, \gamma ). \end{aligned}$$

For any $\gamma \in H_r$, we have

$$\begin{aligned} B_2(dH_r, \gamma )=B_2([H_q\backslash H_{q-1}]\backslash D, \gamma )+B_2(\widetilde{D}, \gamma )+\#T_1= (2^r-2)/2. \end{aligned}$$

If $\gamma \in d \{ [H_q\backslash H_{q-1}] \backslash D\}$, then $\#T_1+2B_2([H_q\backslash H_{q-1}]\backslash D, \gamma )= N/2-n-1$ and

$$\begin{aligned} B_2(d \{[H_q\backslash H_{q-1}]\backslash D\}, \gamma )=B_2(H_r \backslash \{[H_q\backslash H_{q-1}]\backslash D\},\gamma )+N/2-n- (2^r-2)/2. \end{aligned}$$

If $\gamma \in H_r \backslash \{[H_q\backslash H_{q-1}]\backslash D\},$ then $\#T_1+2B_2([H_q\backslash H_{q-1}]\backslash D, \gamma )= N/2-n$ and

$$\begin{aligned} B_2([H_q\backslash H_{q-1}]\backslash D, \gamma )=B_2(\widetilde{D},\gamma )+N/2-n- (2^r-2)/2+1. \end{aligned}$$

Therefore,

$$\begin{aligned} B_2([H_q\backslash H_{q-1}] \backslash D, \gamma )= {\left\{ \begin{array}{ll} B_2(\widetilde{D}, \gamma )+N/2-2^{r-1}-n, &{} if \ \gamma \in d \{[H_q\backslash H_{q-1}] \backslash D\},\\ B_2(\widetilde{D}, \gamma )+N/2-2^{r-1}-n+1, &{} if \ \gamma \in H_r \backslash \{[H_q\backslash H_{q-1}]\backslash D\}. \end{array}\right. }\nonumber \\ \end{aligned}$$

(9)

Substituting Eq. (9) into Lemma 3, we complete the proof. $\square$