Previous research on regular and nonregular designs has largely been based on orthogonal parameterization. In recent years, baseline designs based on baseline parameterization have garnered significant attention due to their wide applicability. Below, we will provide specific examples to explain what baseline parameterization and orthogonal parameterization are.

For an orthogonal design with N rows and m columns, where the factor levels are 0 and 1, select any s , 2 ≤ s ≤ m columns g 1 , ⋯ , g s , g 1 , ⋯ , g s ∈ { 1 , ⋯ , m } from this design to form a set S . Let ϕ ( S ) denote the vector obtained by summing the corresponding elements of the columns in S ( mod 2 ) . Let φ ( S ) denote the sum of the elements in the vector ϕ ( S ) . Let J s ( S ) = | 2 φ ( S ) − N | . It is important to emphasize that when the factor levels of the design under discussion are −1 and 1, the definition of J s ( S ) undergoes the following corresponding changes: J s ( S ) = | ∑ i = 1 N j i g 1 j i g 2 ⋯ j i g s | . Where j i o represents the element in the i -th row and o , o ∈ { g 1 , ⋯ , g s } -th column of the design. Regardless of whether the factor levels in the orthogonal design are 0, 1, or 1, −1. If J s ( S ) = 0 , then the s columns are said to be completely orthogonal. If J s ( S ) = N , then the s columns are said to be completely confounded. If 0 < J s ( S ) < N , then the s columns are said to be partially confounded. What has been introduced above is orthogonal parameterization. The specific details can be found in Deng and Tang (1999) [1]. The left table in Table 1 represents an orthogonal design with factor levels of 0 and 1, while the middle table in Table 1 represents an orthogonal design with factor levels of −1 and 1. The first and second columns of both tables are completely orthogonal, while the first, second, and fourth columns are completely confounded. The difference between baseline parameterization and orthogonal parameterization lies in the fact that, for a baseline design with N rows and m columns, the m factors do not exhibit orthogonality or confounding. Furthermore, in a baseline design, the factor levels are restricted to 0 and 1. When selecting s factors from the m factors in the baseline design, the interaction effects among these s factors can be represented by the product of the corresponding elements in the s columns. The specific details can be found in Mukerjee and Tang (2012) [2]. In Table 1, columns 1 to 5 in the right-hand table represent the design matrix of the baseline design, while column 6 represents the interaction between columns 1 and 2, and column 7 represents the interaction between columns 1, 2, and 3. Baseline design is a design where factor levels are restricted to 0 and 1, and the number of 0s and 1s in its design matrix can vary arbitrarily. In contrast, orthogonal design under orthogonal parameterization, the factor levels can be 0, 1, or −1, 1 and the design matrix must satisfy orthogonality between any two columns.

Table 1. The table on the left and the table in the middle represent orthogonal designs, while the table on the right represents the baseline design.

We use D to denote a baseline design with N rows and m columns. The matrix D is a matrix with elements 0 and 1, 0 represents the baseline level, while 1 denotes the test level. Let Ω s ( D ) be the set of all N × s submatrices of D . Unless otherwise specified, we will denote this set by Ω s . Here is an example of a baseline design with 8 rows and 5 columns.

Mukerjee and Tang (2012) [2] proposed a theory related to the optimality of baseline designs. They focused on main effects designs and proved that designs with strength 2 have universal optimality in estimating main effects. When estimating main effects, the presence of active interaction effects can cause bias in the estimation of the main effects. Consider the baseline design D , where the m factors of D are denoted as F 1 , ⋯ , F m . Let g 1 , ⋯ , g s , 1 ≤ g 1 < ⋯ < g s ≤ m be any s numbers selected from 1 , ⋯ , m . We denote θ g 1 , ⋯ , g s as the interaction effects of the s factors F g 1 , ⋯ , F g s . Let α ( g 1 , ⋯ , g s ) denote the number of occurrences of the row vector ( 1 , 1 , ⋯ , 1 ) in the submatrix of D formed by the g 1 , ⋯ , g s -th columns. Here, we denote μ ( g 1 , ⋯ , g s ) as an m × 1 column vector. We then define the i -th element of the vector μ ( g 1 , ⋯ , g s ) as follows. When i ∈ { g 1 , ⋯ , g s } , let the i -th element of μ ( g 1 , ⋯ , g s ) be α ( g 1 ⋯ g s ) ; when i ∉ { g 1 , ⋯ , g s } , let the i -th element of μ ( g 1 , ⋯ , g s ) be α ( 〈 i g 1 ⋯ g s 〉 ) , where 〈 i g 1 ... g s 〉 represents the ascending order of i , g 1 , ⋯ , g s . Mukerjee and Tang(2012) [2] pointed out that if there are interaction effects among F g 1 , ⋯ , F g s , these interaction effects will contribute a bias of magnitude ξ ( g 1 , ⋯ , g s ) θ g 1 , ⋯ , g s when estimating the main effects, where ξ ( g 1 , ⋯ , g s ) = ( 2 / N ) ( 2 μ ( g 1 , ⋯ , g s ) − α ( g 1 , ⋯ , g s ) 1 m ) . Here, 1 m represents a column vector of size m × 1 , where all elements are ones. According to the principle that interaction effects of the same order are of equal importance, they provided that K s = ∑ Ω s ξ ( g 1 , ⋯ , g s ) ′ ξ ( g 1 , ⋯ , g s ) . It is evident that K s measures the bias effect of all s -th order interaction effects on the estimation of the main effects. Since lower-order interaction effects are more important than higher-order interaction effects, we aim to find a design that sequentially minimizes ( K 2 , K 3 , ⋯ ) . This is precisely the minimum K -aberration criterion proposed by Mukerjee and Tang (2012) [2]. The minimum K -aberration criterion is defined as follows in Definition 1.

Definition 1. Consider two baseline designs D 1 and D 2 with the same number of rows and columns, both of which are orthogonal designs with strength 2. Let j ′ be the smallest integer at which the values of K j for D 1 and D 2 first differ. If the value of K j ′ for D 1 is smaller than that for D 2 , then D 1 is saidto have a smaller K -aberration than D 2 . A minimum K -aberration design is one in which no design exists with a smaller K -aberration.

Mukerjee and Tang (2012) [2] proved the following expression, where s = 2 , ⋯ , m − 1 , and let α ( ω ) denote the number of rows in ω ∈ Ω s in which all elements are equal to 1.

where T 1 = ∑ ω ∈ Ω s ( α ( ω ) ) 2 and T 2 = ∑ ω * ∈ Ω s + 1 ∑ ω ∘ ∈ Ω s ( ω * ) ( 2 α ( ω * ) − α ( ω ∘ ) ) 2 . The derivation details of this sequence can be found in the literature by Mukerjee and Tang (2012) [2].

In order to reduce the computational effort required to find optimal baseline designs, Mukerjee and Tang (2016) [3] provided an equivalent transformation form Z s for K s , 2 ≤ s ≤ m based on the original approach, and referred to Z s as a moment confounding of a design. The paper points out that the sequential minimization of a design’s K 2 , ⋯ , K m is equivalent to the sequential minimization of the design’s Z 2 , ⋯ , Z m . The method of finding an optimal baseline design by minimizing the sequence of Z 2 , ⋯ , Z m is applicable to both regular and nonregular designs. The definition of Z s is Z s = N − 2 t r ( Z Z ′ ) [ s ] W W ′ . The matrix Z = ( z u j ) , 1 ≤ u ≤ N , 1 ≤ j ≤ m is a baseline design matrix of size N × m with elements 0 and 1. W = J N m − 2 Z , where J N m is an N × m matrix with all elements equal to 1.

This paper starts with the Plackett-Burman design to identify designs with good K -aberration properties. Chapter 2 introduces the relevant symbols and definitions of the Plackett-Burman design. Chapter 3 introduces the properties of baseline designs when studying them starting from the Plackett-Burman design. Chapter 4 presents the application of the theory, where the minimum K -aberration subdesign with 7 columns and 24 rows was identified for a Plackett-Burman design with 24 rows and 23 columns.

In this section, we introduce some symbols and definitions related to Plackett-Burman design that will be used in the derivations of subsequent lemmas or theorems.

We use P to denote a Plackett-Burman design of size N × m , where the elements are either 0 or 1. Plackett-Burman design is typically generated by a special cyclic method [4]. Let the first row of P be α 1 . α 1 is a row vector containing only 0 s and 1 s, with at least one occurrence of both 0 and 1. We represent α 1 as ( a 1 , ⋯ , a m ) , The element a i , i ∈ { 1 , ⋯ , m } refers to the i -th element in the first row of P . Let us assume that α 2 represents the second row of P . Then, the 2^nd to the m -th elements of α 2 are respectively equal to the 1^st to the ( m − 1 ) -th elements of α 1 . The first element of α 2 is equal to the m -th element of α 1 . That is, α 2 = ( a m , a 1 , ⋯ , a m − 1 ) . Let us assume that α 3 represents the third row of P . Then, the 2^nd to the m -th elements of α 3 are respectively equal to the 1^st to the ( m − 1 ) -th elements of α 2 . The first element of α 3 is equal to the m -th element of α 2 . That is, α 3 = ( a m − 1 , a m , a 1 , ⋯ , a m − 2 ) . By the same logic, we can derive α 4 , ⋯ , α m from α 1 . Let α m + 1 represent the ( m + 1 ) -th row of P . Here, we define α m + 1 = ( 0 , 0 , ⋯ , 0 ) or ( 1 , 1 , ⋯ , 1 ) (This depends on the first line of the Plackett-Burman design). Therefore, P can be expressed as follows.

Here is an example of Plackett-Burman design. Consider an N × m Plackett-Burman design, where its first row is

According to the cyclic generation method of Plackett-Burman design, we can obtain its design matrix as follows.

Select an arbitrary s -column submatrix ( w 1 , ⋯ , w s ) ( 1 ≤ w 1 < ⋯ < w s ≤ m ) from an N × m Plackett-Burman design. We let d i represents the number of columns between the w i -th column and the w i + 1 -th column within the m columns of Plackett-Burman design, in this case, i = 1 , ⋯ , s − 1 , it is evident that d i = w i + 1 − w i − 1 . Specifically, let d s represent the number of columns between the w s -th column and the w 1 -th column within the m columns of Plackett-Burman design. Specifically, d s is the sum of the number of columns after the w s -th column and the number of columns before the w 1 -th column within the m columns of Plackett-Burman design. It is evident that d s = m − w s + w 1 − 1 . ( d 1 , ⋯ , d s ) is referred to as the distance vector of ( w 1 , ⋯ , w s ) . Let the set of distance vectors ( d 1 , ⋯ , d s ) , ( d 2 , ⋯ , d s , d 1 ) , ⋯ , ( d s , d 1 , ⋯ , d s − 1 ) be denoted as D s .

To ensure the smooth progress of the proofs of the lemma and theorem in this section, we now provide the definitions of function a i , j and function C i , respectively. The definition of function a i , j can be found in Equation (2), and the definition of function C i can be found in Equation (3).

The Function a i , j is a binary function, with independent variables i and j , where i ∈ { 1 , ⋯ , m } , j ∈ { 0 , ⋯ , 2 m + x − 1 } . In Equation (2), a i − j , a m − j + i and a 2 m − j + i represent the ( i − j ) -th, ( m − j + i ) -th and ( 2 m − j + i ) -th elements in the first row of Plackett-Burman design, respectively.

The function C i is a unary function, with i as its independent variable, where i ∈ { 1 , ⋯ , 2 N − 2 } .

Lemma 1.Select two s -column submatrices, ( g 1 , ⋯ , g s ) and ( g ′ 1 , ⋯ , g ′ s ) , from an N × m Plackett-Burman design. Let the distance vectors of the two s -column submatrices be denoted as ( d 1 , ⋯ , d s ) and ( d ′ 1 , ⋯ , d ′ s ) , respectively. If d i = d ′ i ( i = 1 , ⋯ , s ) , then ( g 1 , ⋯ , g s ) and ( g ′ 1 , ⋯ , g ′ s ) have the same α ( ω ) .

Proof. Select two s -column submatrices, ( g 1 , ⋯ , g s ) and ( g ′ 1 , ⋯ , g ′ s ) , from an N × m Plackett-Burman design. Let the distance vectors of the two s -column submatrices be denoted as ( d 1 , ⋯ , d s ) and ( d ′ 1 , ⋯ , d ′ s ) , respectively. Now, let d i = d ′ i ( i = 1 , ⋯ , s ) . By considering the distance vector of ( g 1 , ⋯ , g s ) , g 1 , ⋯ , g s can be further expressed in the form of Equation (4).

Similarly, by considering the distance vector of ( g ′ 1 , ⋯ , g ′ s ) , g ′ 1 , ⋯ , g ′ s can be further expressed in the form of Equation (5).

(i) g 1 = g ′ 1 . From Equations (4) and (5), it can be observed that when g 1 = g ′ 1 , ( g 1 , ⋯ , g s ) and ( g ′ 1 , ⋯ , g ′ s ) are two identical s -column submatrices selected from an N × m Plackett-Burman design. Therefore, it is evident that ( g 1 , ⋯ , g s ) and ( g ′ 1 , ⋯ , g ′ s ) have the same α ( ω ) .

(ii) g 1 ≠ g ′ 1 . Without loss of generality, let g 1 < g ′ 1 . Let the number of columns between the g 1 -th and g ′ 1 -th columns in the m columns of Plackett-Burman design be d , thus, we have g ′ 1 = g 1 + d + 1 . From this, g ′ 1 , ⋯ , g ′ s can be further expressed as g ′ i = g i + d + 1 ( i = 1 ) , g ′ i = g 1 + d + 1 + ∑ j = 1 i − 1 d j + i − 1 ( i = 2 , ⋯ , s ) . The first row of ( g 1 , ⋯ , g s ) can be represented as

By considering the Plackett-Burman design generation method and the definition of Equation (2), it can be concluded that the ( v + 1 ) ( v = 0 , ⋯ , N − 2 ) -th row of ( g 1 , ⋯ , g s ) can be represented as

The first row of ( g ′ 1 , ⋯ , g ′ s ) can be represented as

By considering the Plackett-Burman design generation method and the definitions of Equations (2) and (3), the C v + d + 2 -th row of ( g ′ 1 , ⋯ , g ′ s ) can be represented as

The element at the z -th position in the C v + d + 2 -th row of ( g ′ 1 , ⋯ , g ′ s ) is a g 1 + d + 1 + ∑ j = 1 z − 1 d j + z − 1 , v + d + 1 , ( z = 1 , ⋯ , s ) . From Equation (2), we can deduce that a g 1 + ∑ j = 1 z − 1 d j + z − 1 , v = a g 1 + d + 1 + ∑ j = 1 z − 1 d j + z − 1 , v + d + 1 . Since a g 1 + ∑ j = 1 z − 1 d j + z − 1 , v is the element in the ( v + 1 ) -th row and z -th column of ( g 1 , ⋯ , g s ) , it follows that the element in the C v + d + 2 -th row and z -th column of ( g ′ 1 , ⋯ , g ′ s ) is equal to the element in the ( v + 1 ) -th row and z -th column of ( g 1 , ⋯ , g s ) .

By considering all values of z , it follows that the C v + d + 2 -th row of ( g ′ 1 , ⋯ , g ′ s ) is identical to the ( v + 1 ) -th row of ( g 1 , ⋯ , g s ) . Since the last row of the Plackett-Burman design consists entirely of zeros or ones, by traversing v , it can be concluded that ( g 1 , ⋯ , g s ) and ( g ′ 1 , ⋯ , g ′ s ) have the same α ( ω ) . ◻

Lemma 2. Select two s -column submatrices, ( g 1 , ⋯ , g s ) and ( g ′ 1 , ⋯ , g ′ s ) , from an N × m Plackett-Burman design. Let the distance vectors of the two s -column submatrices be denoted as ( p 1 , ⋯ , p s ) and ( p ′ 1 , ⋯ , p ′ s ) , respectively. If both ( p 1 , ⋯ , p s ) and ( p ′ 1 , ⋯ , p ′ s ) belong to the set D s , then ( g 1 , ⋯ , g s ) and ( g ′ 1 , ⋯ , g ′ s ) have the same α ( ω ) .

Proof. Select two s -column submatrices, ( g 1 , ⋯ , g s ) and ( g ′ 1 , ⋯ , g ′ s ) , from an N × m Plackett-Burman design. Let the distance vectors of the two s -column submatrices be denoted as ( p 1 , ⋯ , p s ) and ( p ′ 1 , ⋯ , p ′ s ) , respectively. Let both ( p 1 , ⋯ , p s ) and ( p ′ 1 , ⋯ , p ′ s ) belong to D s . Similar to the discussion in Lemma 1, ( g 1 , ⋯ , g s ) can be further expressed as g i = g i ( i = 1 ) , g i = g 1 + ∑ j = 1 i − 1 p j + i − 1 ( i = 2 , ⋯ , s ) . g ′ 1 , ⋯ , g ′ s can be further expressed as g ′ i = g ′ i ( i = 1 ) , g ′ i = g ′ 1 + ∑ j = 1 i − 1 p j + i − 1 ( i = 2 , ⋯ , s ) .

(i) ( p 1 , ⋯ , p s ) = ( p ′ 1 , ⋯ , p ′ s ) . In this case, ( g 1 , ⋯ , g s ) and ( g ′ 1 , ⋯ , g ′ s ) have identical distance vectors. By Lemma 1, it follows that, at this point, ( g 1 , ⋯ , g s ) and ( g ′ 1 , ⋯ , g ′ s ) have the same α ( ω ) .

(ii) ( p 1 , ⋯ , p s ) ≠ ( p ′ 1 , ⋯ , p ′ s ) . Let ( p 1 , ⋯ , p s ) = ( d 1 , ⋯ , d s ) and ( p ′ 1 , ⋯ , p ′ s ) = ( d 2 , ⋯ , d s , d 1 ) . Discuss the values of α ( ω ) for ( g 1 , ⋯ , g s ) and ( g ′ 1 , ⋯ , g ′ s ) under these two distance vectors, respectively. As stated in Lemma 1, we only need to consider the case where both g 1 and g ′ 1 are equal to 1. Now, let both g 1 and g ′ 1 be equal to 1, so g 2 , ⋯ , g s can be further represented as g i = ∑ j = 1 i − 1 d j + i ( i = 2 , ⋯ , s ) . g ′ 2 , ⋯ , g ′ s can be further represented as g ′ i = ∑ j = 2 i d j + i ( i = 2 , ⋯ , s ) . The first row of ( g 1 , ⋯ , g s ) is

Based on the Plackett-Burman design generation method and the definition of Equation (2), it can be concluded that the ( v + 1 ) -th row of ( g 1 , ⋯ , g s ) can be represented as

The first row of ( g ′ 1 , ⋯ , g ′ s ) is

Based on the generation method of Plackett-Burman design and the definitions of Equations (2) and (3), we can deduce that the C v + s + ∑ j = 2 s d j -th row of ( g ′ 1 , ⋯ , g ′ s ) can be represented as

The ( c − 1 ) -th element in the C v + s + ∑ j = 2 s d j -th row of ( g ′ 1 , ⋯ , g ′ s ) is a c − 1 + ∑ j = 2 c − 1 d j , v + s + ∑ j = 2 s d j , ( 3 ≤ c ≤ s ) . From Equation (2), we can deduce that a c − 1 + ∑ j = 2 c − 1 d j , v + s + ∑ j = 2 s d j = a c + ∑ j = 1 c − 1 d j , v . That is, the ( c − 1 ) -th element in the C v + s + ∑ j = 2 s d j -th row of ( g ′ 1 , ⋯ , g ′ s ) is equal to the c -th element in the ( v + 1 ) -th row of ( g 1 , ⋯ , g s ) . In particular, the s -th element in the C v + s + ∑ j = 2 s d j -th row of ( g ′ 1 , ⋯ , g ′ s ) is equal to a s + ∑ j = 2 s d j , v + s + ∑ j = 2 s d j − 1 . From Equation (2), we can deduce that a s + ∑ j = 2 s d j , v + s + ∑ j = 2 s d j − 1 = a 1 , v . That is, the s -th element in the C v + s + ∑ j = 2 s d j -th row of ( g ′ 1 , ⋯ , g ′ s ) is equal to the first element in the v + 1 -th row of ( g 1 , ⋯ , g s ) . The first element in the C v + s + ∑ j = 2 s d j -th row of ( g ′ 1 , ⋯ , g ′ s ) is denoted as a 1 , v + s + ∑ j = 2 s d j − 1 . From Equation (2), we can deduce that a 1 , v + s + ∑ j = 2 s d j − 1 = a 2 + d 1 , v . That is, the first element in the C v + s + ∑ j = 2 s d j -th row of ( g ′ 1 , ⋯ , g ′ s ) is equal to the second element in the ( v + 1 ) -th row of ( g 1 , ⋯ , g s ) .

By iterating over all possible values of v , it can be observed that the ( v + 1 ) -th row of ( g 1 , ⋯ , g s ) corresponds to the C v + s + ∑ j = 2 s d j -th row of ( g ′ 1 , ⋯ , g ′ s ) , with both rows containing the same number of 0 s and 1 s. Therefore, when the distance vectors of ( g 1 , ⋯ , g s ) and ( g ′ 1 , ⋯ , g ′ s ) are equal to ( d 1 , ⋯ , d s ) and ( d 2 , ⋯ , d s , d 1 ) respectively, the α ( ω ) of ( g 1 , ⋯ , g s ) is identical to that of ( g ′ 1 , ⋯ , g ′ s ) . From the scenario where ( p 1 , ⋯ , p s ) = ( d 2 , ⋯ , d s , d 1 ) and ( p ′ 1 , ⋯ , p ′ s ) = ( d 3 , ⋯ , d s , d 1 , d 2 ) , extending all the way to ( p 1 , ⋯ , p s ) = ( d s , d 1 , ⋯ , d s − 1 ) and ( p ′ 1 , ⋯ , p ′ s ) = ( d 1 , ⋯ , d s ) , the same conclusion holds as in the case where ( p 1 , ⋯ , p s ) = ( d 1 , ⋯ , d s ) , ( p ′ 1 , ⋯ , p ′ s ) = ( d 2 , ⋯ , d s , d 1 ) . By the principle of transitivity, if both ( p 1 , ⋯ , p s ) and ( p ′ 1 , ⋯ , p ′ s ) are derived from D s , then the α ( ω ) of ( g 1 , ⋯ , g s ) is identical to that of ( g ′ 1 , ⋯ , g ′ s ) . ◻

Theorem 1. Select two s -column submatrices, ( g 1 , ⋯ , g s ) and ( g ′ 1 , ⋯ , g ′ s ) , from an N × m Plackett-Burman design. Let the distance vectors of the two s -column submatrices be denoted as ( p 1 , ⋯ , p s ) and ( p ′ 1 , ⋯ , p ′ s ) , respectively. If both ( p 1 , ⋯ , p s ) and ( p ′ 1 , ⋯ , p ′ s ) belong to the set D s , then, the s -column sub-design constructed from ( g 1 , ⋯ , g s ) and the s -column sub-design constructed from ( g ′ 1 , ⋯ , g ′ s ) possess the same K -aberration sequence.

Proof. Select two s -column submatrices, ( g 1 , ⋯ , g s ) and ( g ′ 1 , ⋯ , g ′ s ) , from an N × m Plackett-Burman design. Let the distance vectors of the two s -column submatrices be denoted as ( p 1 , ⋯ , p s ) and ( p ′ 1 , ⋯ , p ′ s ) , respectively. Select any u columns from g 1 , ⋯ , g s . Let the selected u columns be denoted as h 1 , ⋯ , h u , and it holds that h 1 < ⋯ < h u . Let the e -th column h e ( e = 1 , ⋯ , u ) among the u columns h 1 , ⋯ , h u be the c e , c e ∈ { 1 , ⋯ , s } -th column among the s columns g 1 , ... , g s . Let ( n 1 , ⋯ , n u ) denote the distance vector of ( h 1 , ⋯ , h u ) . For the selected columns h 1 , ⋯ , h u , we choose a corresponding set of u specific columns h ′ 1 , ⋯ , h ′ u from g ′ 1 , ⋯ , g ′ s . Let h ′ e ( e = 1 , ⋯ , u ) be the c ′ e , c ′ e ∈ { 1 , ⋯ , s } -th column among the s columns g ′ 1 , ⋯ , g ′ s . Let ( n ′ 1 , ⋯ , n ′ u ) denote the distance vector of ( h ′ 1 , ⋯ , h ′ u ) .

(i) ( p 1 , ⋯ , p s ) = ( p ′ 1 , ⋯ , p ′ s ) .

Without loss of generality, let ( p 1 , ⋯ , p s ) = ( p ′ 1 , ⋯ , p ′ s ) = ( d 1 , ⋯ , d s ) . When 1 ≤ e ≤ u − 1 , it follows that n e = ∑ f = c e c e + 1 − 1 d f + c e + 1 − c e − 1 . In particular, n u = ∑ f = c u s d f + ∑ r = 1 c 1 − 1 d r + s − c u + c 1 − 1 . By setting c ′ e = c e , we obtain ( n ′ 1 , ⋯ , n ′ u ) = ( n 1 , ⋯ , n u ) . From Lemma 1, it follows that the α ( ω ) of ( h 1 , ⋯ , h u ) and ( h ′ 1 , ⋯ , h ′ u ) is identical under this condition. Therefore, when ( p 1 , ⋯ , p s ) = ( p ′ 1 , ⋯ , p ′ s ) , for every ( h 1 , ⋯ , h u ) , there exists a corresponding ( h ′ 1 , ⋯ , h ′ u ) such that the α ( ω ) of ( h 1 , ⋯ , h u ) and ( h ′ 1 , ⋯ , h ′ u ) is identical.

(ii) ( p 1 , ⋯ , p s ) ≠ ( p ′ 1 , ⋯ , p ′ s ) .

Let ( p 1 , ⋯ , p s ) = ( d 1 , ⋯ , d s ) and ( p ′ 1 , ⋯ , p ′ s ) = ( d 2 , ⋯ , d s , d 1 ) . When 1 ≤ e ≤ u − 1 , it follows that n e = ∑ f = c e c e + 1 − 1 d f + c e + 1 − c e − 1 . In particular, n u = ∑ f = c u s d f + ∑ r = 1 c 1 − 1 d r + s − c u + c 1 − 1 . When 1 ≤ e ≤ u − 2 , let c ′ e = c e + 1 − 1 and let n ′ e = ∑ f = c e + 1 c e + 2 − 1 d f + c e + 2 − c e + 1 . When e = u and c 1 > 1 , let c ′ u = c 1 − 1 and n ′ u = ∑ f = c 1 c 2 − 1 d f + c 2 − c 1 − 1 . When e = u and c 1 = 1 , let c ′ u = s and n ′ u = ∑ f = 1 c 2 − 1 d f + c 2 − 2 . We can discover that ( h ′ 1 , ⋯ , h ′ u ) precisely corresponds to a u -column submatrix of ( g ′ 1 , ⋯ , g ′ s ) , with the distance vector ( n ′ 1 , ⋯ , n ′ u ) of ( h ′ 1 , ⋯ , h ′ u ) being equal to ( n 2 , ⋯ , n u , n 1 ) . From Lemma 2, it follows that the α ( ω ) of ( h 1 , ⋯ , h u ) and ( h ′ 1 , ⋯ , h ′ u ) is identical under this condition. Therefore, when ( p 1 , ⋯ , p s ) = ( d 1 , ⋯ , d s ) and ( p ′ 1 , ⋯ , p ′ s ) = ( d 2 , ⋯ , d s , d 1 ) , for every ( h 1 , ⋯ , h u ) , there exists a corresponding ( h ′ 1 , ⋯ , h ′ u ) such that the α ( ω ) of ( h 1 , ⋯ , h u ) and ( h ′ 1 , ⋯ , h ′ u ) is identical.