On linear lexicographic codes: Ninth column construction of the ternary Golay code

On linear lexicographic codes: Ninth col umn cons tr uction of the tern ar y Gola y code Y uki Irie ∗ 1 1 Graduat e F aculty of Inter disciplinar y R esearc h, U niv ersity of Y amanashi, Y amanashi, Japan Abstract . W e characterize linear lexicographic p -ary codes. Using this characterization, when p ⩾ 3 , w e deter mine the dimensions of linear lexicographic codes obtained from se veral bases including the standard basis, ex cept f or those of certain minimum distances. In these e x cluded cases, w e may obtain linear codes of higher dimensions; f or instance, when p = 3 and d = 6 , the ternar y Gola y code is obtained. Ke ywor ds. Le xicog raphic codes, Gola y codes, Solomon-S tiﬄer codes, Griesmer bound Mathematics Subject Classiﬁcations. 94B05 1. Introduction Le xicographic codes were introduced b y Lev enshtein [ 5 ] and independentl y b y Conw a y and Sloane [ 2 ]. Although a binar y le xicographic code is linear , a non-binary lexicographic code is not alw ay s linear , so sev eral algor ithms f or producing linear codes ha v e been proposed [ 1 , 7 ]. In this paper , we use a slightl y diﬀerent approach. F or a prime p , w e inv estig ate linear le xicog raphic p -ary codes produced b y the original algorithm de veloped b y Le venshtein and Con wa y-Sloane. W e ﬁrst giv e a characterization f or linear lexicographic p -ar y codes. Using this characterization, when p > 2 , w e determine the dimensions of linear le xicog raphic codes obtained from se v eral bases including the standard basis, e x cept for cases with speciﬁc minimum distances. In these e x cluded cases, w e may obtain linear codes of higher dimensions; f or ins tance, when p = 3 and d = 6 , we can obtain the ternar y Golay code. 1.1. Lexicographic codes In this paper , we onl y consider le xicographic codes o v er a prime ﬁeld F p . W e identify F p with { 0 , 1 , . . . , p − 1 } . ∗ Supported b y JSPS KAKENHI Grant Number JP20K14277 and JP24K16892. 2 Y uki Irie Let F N p denote the set of [ x i ] i ∈ N such that x i ∈ F p and x i = 0 f or all but only ﬁnitel y many i ∈ N , where N is the set of non-negativ e integers. Let F be an (ordered) basis ( f i ) i ∈ N of F N p . For x ∈ F N p , w e wr ite x = X i ∈ N x [ i ; F ] f i . When no confusion can ar ise, w e simply write x [ i ] instead of x [ i ; F ] . Let E denote the s tandard basis ( e 0 , e 1 , . . . ) of F N p , where e i =  0 · · · 0 1 0 · · ·  with 1 in the i -th coordinate; f or e xample, e 0 =  1 0 0 · · ·  and e 1 =  0 1 0 · · ·  . W e write x i = x [ i,E ] . A basis F of F N p induces a total order on F N p as f ollow s. Let x and y be tw o distinct elements in F N p . Then there e xists N ∈ N such that x [ N ]  = y [ N ] and x [ i ] = y [ i ] f or i > N . W e wr ite x < F y if x [ N ] < y [ N ] . For ex ample, if p = 3 then  0 0 0 · · ·  < E  1 0 0 · · ·  < E  2 0 0 · · ·  < E  0 1 0 · · ·  < E · · · Let d ∈ N with d ⩾ 2 . For a ∈ N , let x F,d ( a ) or simpl y x ( a ) denote the minimum of z ∈ F N p with respect to < F such that d ( z , x ( b )) ⩾ d f or 0 ⩽ b < a , where d ( x , y ) is the Hamming distance betw een x and y , that is, d ( x , y ) = |{ i ∈ N : x i  = y i }| . Deﬁne x i ( a ) and x [ i ] ( a ) b y x ( a ) = X i ∈ N x i ( a ) e i = X i ∈ N x [ i ] ( a ) f i . For k ∈ N , let X k ; F ,d = { x (0) , x (1) , . . . , x ( p k − 1) } . W e call X k ; F ,d a lexicog raphic code . Let x = [ x i ] i ∈ N ∈ F N p . F or a subset S of N , let x   S denote the v ector obtained from x b y deleting coordinates in N \ S , that is, x   S = [ x i ] i ∈ S . Let supp F ( x ) = { i ∈ N : x [ i ; F ]  = 0 } . F or e xample, if x = e 0 + e 2 =  1 0 1 0 0 · · ·  , then supp E ( x ) = { 0 , 2 } and x   { 0 , 1 , 2 } =  1 0 1  . If X is a subset of F N p , then let supp F ( X ) = [ x ∈ X supp F ( x ) and X sp = n x   supp E ( X ) : x ∈ X o . Example 1.1. Let p = 3 , F = E , and d = 2 . Then x (0) =  0 0 0 0 0 · · ·  , x (1) =  1 1 0 0 0 · · ·  , x (2) =  2 2 0 0 0 · · ·  , x (3) =  1 0 1 0 0 · · ·  , x (4) =  0 1 1 0 0 · · ·  , x (5) =  2 0 2 0 0 · · ·  , x (6) =  0 2 2 0 0 · · ·  , x (7) =  1 0 0 1 0 · · ·  , x (8) =  0 1 0 1 0 · · ·  . Theref ore X 1 sp =   0 0  ,  1 1  ,  2 2   , X 2 sp =   0 0 0 0  ,  1 1 0 0  , . . . ,  0 1 0 1   . Note that X 1 sp is a linear code, but X 2 sp is not. 3 1.2. Linearity of lexicographic codes Theorem 1.2 ( [ 2 , 5 ]) . If p = 2 , t hen X k ; F ,d is a linear code. Example 1.3 ( [ 2 ]) . If p = 2 , then X 12; E , 8 sp is the (extended) binar y Gola y code. In contrast to binar y le xicographic codes, a p -ar y le xicographic code may not be linear when p ⩾ 3 as we ha v e seen in Example 1.1 . Our ﬁrs t result giv es the conditions f or X k ; F ,d to be linear . For a ∈ N , let a i denote the i -th digit of a in the p -ar y expansion, that is, a = P a i p i and a i ∈ { 0 , 1 , . . . , p − 1 } . W e deﬁne e x F,d ( a ) as f ollo ws: e x F,d ( a ) = X i ∈ N a i x F,d ( p i ) . Note that e x F,d ( p i ) = x F,d ( p i ) . Theorem 1.4 . The f ollowing thr ee conditions are equiv alent. (1) X k ; F ,d is a linear code. (2) x F,d ( a ) = e x F,d ( a ) f or 0 ⩽ a < p k . (3) x F,d ( a ) = e x F,d ( a ) f or 0 ⩽ a < p k with P i ∈ N a i ⩽ p − 1 . Remar k 1.5 . Theorem 1.4 can be considered as a generalization of Theorem 1.2 because (3) in Theorem 1.4 holds tr iviall y when p = 2 . Indeed, if p = 2 and P a i ⩽ 1 , then a must be a pow er of 2, and hence x F,d ( a ) = e x F,d ( a ) . 1.3. Dimensions of linear lexicographic codes Using Theorem 1.4 , w e will deter mine the dimensions of se v eral linear le xicographic codes. R ecall that when p = 2 , the lexicographic code X 12; E , 8 sp is the Golay code. Ho we ver , when p = 3 , the lexicographic code X 6; E , 6 sp is not the ter nary Golay code, and is not ev en a linear code; in [ 3 ], it is pointed out that a modiﬁed algor ithm produces the ternar y Gola y code (see Example 1.8 ). W e here giv e another constr uction of the ter nary Gola y code. For distinct non-neg ativ e integ ers ξ and η , let F ( ξ , η ) be the basis ( f i ) i ∈ N of F N p deﬁned b y f i = ( e i if i  = ξ , e ξ + e η if i = ξ . The basis F ( ξ , η ) diﬀers from s tandard basis E only in the ξ -th element. Let F = { E } ∪ { F ( ξ , η ) : ξ , η ∈ N , ξ  = η } . For con v enience, let E = F ( ∞ , ∞ ) . For a basis F of F N p , let k ( F , d ) denote the maximum k suc h that X k ; F ,d is a linear code; if X k ; F ,d is a linear code f or all k , then let k ( F , d ) = ∞ . The follo wing theorem sa y s that a slight modiﬁcation of the standard basis allo ws the le xicog raphic code algorithm to produce the ternar y Golay code. 4 Y uki Irie Theorem 1.6 . Let p = 3 and F ∈ F . Then k ( F, 6) =      6 if F = F ( ξ , 9) , 0 ⩽ ξ ⩽ 8 , 3 if F = F ( ξ , η ) , 0 ⩽ ξ ⩽ 8 , η ⩾ 10 , 2 otherwise . In particular , if F = F ( ξ , 9) and 0 ⩽ ξ ⩽ 8 , then X 6; F, 6 sp is the ternar y Golay code. Note that, in Theorem 1.6 , the dimension k ( F, 6) is signiﬁcantly larg er in the case when F = F ( ξ , 9) and 0 ⩽ ξ ⩽ 8 . The f ollo wing theorem sho ws that, e x cept f or certain d , this phenomenon does not occur . Theorem 1.7 . Let F ∈ F . (1) k ( F, d ) = ∞ if p = 2 . (2) k ( F, d ) ⩾ k if p = 3 , d = 3 k − 1 d ′ , and d ′ ⩾ 2 ; mor eov er , if k − 1 is the 3 -adic or der of d and d ′ ⩾ 8 , then k ( F, d ) = k . (3) k ( F, d ) ⩽ 2 if p ⩾ 5 . While proving Theorem 1.7 , we will deter mine, in cer tain cases, the structure of the f ollo wing le xicographic-code v ariants discussed in [ 1 , 3 ]. For a non-negativ e integ er a , deﬁne λ F,d ( a ) , or simpl y λ ( a ) , as f ollo ws. If a is a pow er of p , then λ ( a ) is the minimum of z ∈ F N p such that d ( z , λ ( b )) ⩾ d f or 0 ⩽ b < a ; otherwise, λ ( a ) = P i ∈ N a i λ ( p i ) . For k ∈ N , let L k ; F ,d = { λ (0) , λ (1) , . . . , λ ( p k − 1) } . W e deﬁne λ i ( a ) and λ [ i ] ( a ) in the same wa y as x i ( a ) and x [ i ] ( a ) , that is, x ( a ) = P λ i ( a ) e i = P λ [ i ] ( a ) f i . N ote that, b y Theorem 1.4 , w e see that if X k ; F ,d is linear , then x ( a ) = e x ( a ) = λ ( a ) f or 0 ⩽ a < p k , and X k ; F ,d = L k ; F ,d . Example 1.8 ( [ 3 ]) . If p = 3 , then L 6; E , 6 sp is the ter nary Gola y code. Example 1.9. Let p = 3 . Then L 1; E , 2 sp is g enerated b y  1 1  and L 2; E , 2 sp is g enerated b y  1 1 0  and  1 0 1  . Note that L 1; E , 2 sp = X 1; E , 2 sp and L 2; E , 2 sp  = X 2; E , 2 sp . Theorem 1.10. If d = p k − 1 d ′ with d ′ ⩾ 2 and F ∈ F , then L k ; F ,d sp is a  p k − 1 p − 1 d ′ , k , p k − 1 d ′  code, whic h is obtained by r epeating t he  p k − 1 p − 1 , k , p k − 1  simplex code d ′ times. Theorem 1.11. Let d = p k − 1 ( pq + r ) , wher e 0 < r < p . If q + r − p + 1 ⩾ 2 , then, L k +1; F ,d sp is a  p k − 1 p − 1 ( pq + r ) + q + 1 , k + 1 , d  Solomon-Stiﬄer code. 1.4. Organization This paper is org anized as f ollo ws. In Section 2 , we prov e Theorem 1.4 . After re vie wing the Griesmer bound in Section 3 , w e pro v e Theorem 1.10 in Section 4 . In Section 5 , w e sho w that X ⟨ k − 1 ⟩ ; F ,d is linear if p = 3 and d = 3 k − 1 d ′ with d ′ ⩾ 2 . Section 6 pro v es Theorem 1.11 . Finall y , w e pro v e Theorem 1.7 in Section 7 . 5 2. A characterization of linear lexicographic codes W e ﬁx a prime p , an integer d with d ⩾ 2 , and a basis F of F N p . F or a, b ∈ N , let a ⊕ p b = P i ∈ N ( a i ⊕ p b i ) p i and a ⊖ p b = P i ∈ N ( a i ⊖ p b i ) p i , where a i ⊕ p b i and a i ⊖ p b i are the addition and subtraction of a i and b i in F p , respectiv ely . For ex ample, if p = 3 , then 7 ⊕ 3 13 = (1 + 2 · 3 + 0 · 3 2 ) ⊕ 3 (1 + 1 · 3 + 1 · 3 2 ) = (1 ⊕ 3 1) + (2 ⊕ 3 1) · 3 + (0 ⊕ 1) · 3 2 = 2 + 0 · 3 + 1 · 3 2 = 11 and 7 ⊖ 3 13 = (1 + 2 · 3) ⊖ 3 (1 + 3 + 3 2 ) = 3 + 2 · 3 2 = 21 . Note that e x ( a ⊕ p b ) = X ( a i ⊕ p b i ) x ( p i ) = X a i x ( p i ) + X b i x ( p i ) = e x ( a ) + e x ( b ) , λ ( a ⊕ p b ) = X ( a i ⊕ p b i ) λ ( p i ) = X a i λ ( p i ) + X b i λ ( p i ) = λ ( a ) + λ ( b ) . W e will repeatedly use the f ollo wing simple lemma. Lemma 2.1. Let y ∈ F N p . (1) If d ( y , x ( b )) ⩾ d for 0 ⩽ b < a , t hen y ⩾ F x ( a ) . (2) If d ( y , λ ( b )) ⩾ d for 0 ⩽ b < p k , then y ⩾ F λ ( p k ) . Pr oof. By deﬁnition, x ( a ) is the minimum of z such that d ( z , x ( b )) ⩾ d f or 0 ⩽ b < a . Thus x ( a ) ⩽ F y . The proof of (2) is the same. For i, a ∈ N , w e deﬁne e x [ i ] ( a ) by e x ( a ) = P i e x [ i ] ( a ) f i . The ne xt lemma is a k e y to pro ve Theorem 1.4 . Lemma 2.2. Suppose t hat a is the smallest integ er satisfying x ( a )  = e x ( a ) . Let N = max { i ∈ N : x [ i ] ( a )  = e x [ i ] ( a ) } . If b is a positive integ er suc h that b i ⩽ a i f or i ∈ N , t hen e x [ N ] ( b )  = 0 . (2.1) Pr oof. W e ﬁrst sho w that x ( a ) < F e x ( a ) . Let p k − 1 ⩽ a < p k . Since x ( a )  = e x ( a ) , it f ollo ws that p k − 1 < a . For 0 ⩽ b < a , d ( e x ( a ) , x ( b )) = d ( e x ( a ) , e x ( b )) = d  k − 1 X i =0 a i x ( p i ) , k − 1 X i =0 b i x ( p i )  = d  ( a k − 1 ⊖ p b k − 1 ) x ( p k − 1 ) , k − 2 X i =0 ( b i ⊖ p a i ) x ( p i )  . 6 Y uki Irie Let c = P k − 2 i =0 ( b i ⊖ p a i ) p i . Since c < a , w e see that x ( c ) = e x ( c ) = k − 2 X i =0 ( b i ⊖ p a i ) x ( p i ) . If a k − 1 ⊖ p b k − 1 = 0 , then d ( e x ( a ) , x ( b )) = d ( x (0) , x ( c )) ⩾ d . If a k − 1 ⊖ p b k − 1  = 0 , then d  ( a k − 1 ⊖ p b k − 1 ) x ( p k − 1 ) , k − 2 X i =0 ( b i ⊖ p a i ) x ( p i )  = d  x ( p k − 1 ) , 1 ( a k − 1 ⊖ p b k − 1 ) k − 2 X i =0 ( b i ⊖ p a i ) x ( p i )  ⩾ d. Theref ore d ( e x ( a ) , x ( b )) ⩾ d . Since e x ( a )  = x ( a ) , it f ollo ws from Lemma 2.1 that x ( a ) < F e x ( a ) . In particular , x [ N ] ( a ) < e x [ N ] ( a ) . (2.2) W e no w sho w ( 2.1 ). Suppose that b  = 0 and b [ i ] ⩽ a [ i ] f or i ∈ N . If b = a , then ( 2.1 ) f ollo ws from ( 2.2 ). Suppose that b < a . W e claim that e x ( a ) − x ( b ) < x ( a ) − x ( b ) . (2.3) Assume ( 2.3 ) f or the moment. By the deﬁnition of N , f or i > N , e x [ i ] ( a ) − x [ i ] ( b ) = x [ i ] ( a ) − x [ i ] ( b ) . By ( 2.3 ), e x [ N ] ( a ) − x [ N ] ( b ) < x [ N ] ( a ) − x [ N ] ( b ) . It f ollo ws from ( 2.2 ) that e x [ N ] ( b ) = x [ N ] ( b )  = 0 . W e ﬁnall y show ( 2.3 ). Since b [ i ] ⩽ a [ i ] , w e see that a ⊖ p b = a − b < a . Hence e x ( a ) − x ( b ) = e x ( a ) − e x ( b ) = e x ( a ⊖ p b ) = x ( a ⊖ p b ) . Hence to prov e ( 2.3 ), it suﬃces to sho w that, f or c < a ⊖ p b = a − b , d ( x ( a ) − x ( b ) , x ( c )) ⩾ d. Since c ⊕ b ⩽ c + b < a , we see that x ( c ⊕ b ) = x ( c ) + x ( b ) and d ( x ( a ) − x ( b ) , x ( c )) = d ( x ( a ) , x ( c ) + x ( b )) = d ( x ( a ) , x ( c ⊕ b )) ⩾ d. It f ollo ws from Lemma 2.1 that ( 2.3 ) holds. 7 Lemma 2.3. F or 0 ⩽ h ⩽ k − 1 , le t M h = max { i ∈ N : x [ i ] ( p h )  = 0 } . Then the f ollowing asser tions hold. (1) M 0 < M 1 < · · · < M k − 1 . (2) F or 0 ⩽ h, l < k − 1 , λ [ M h ] ( p l ) = δ h,l = ( 1 if h = l , 0 if h  = l . (3) If X ⟨ k − 1 ⟩ is a linear code, then x ( a ) = e x ( a ) = λ ( a ) f or 0 ⩽ a < p k . Pr oof. W e sho w the lemma b y induction on k . Suppose that k = 1 ; then (1) is ob vious. By the deﬁnition of M 0 , we see that x [ M 0 ] ( p 0 )  = 0 . From the minimality of x ( p 0 ) , w e see that x [ M 0 ] ( p 0 ) = 1 . Thus (2) holds. This implies that e x (1) < e x (2) < · · · < e x ( p − 1) . (2.4) Suppose that X ⟨ 0 ⟩ is a linear code. Then X ⟨ 0 ⟩ = { e x (0) , e x (1) , . . . , e x ( p − 1) } . It f ollo w s from ( 2.4 ) that x ( a ) = e x ( a ) = λ ( a ) f or 0 ⩽ a F λ ( p k − 2 ) , w e see that M k − 1 ⩾ M k − 2 . Let 0 ⩽ h < k − 1 , M = M h , α = λ [ M ] ( p k − 1 ) , and y = λ ( p k − 1 ) − α λ ( p h ) . By the induction hypothesis, λ [ M ] ( p h ) = 1 , so y [ M ] = 0 and y ⩽ F λ ( p h ) . For 0 ⩽ a < p k − 1 , d ( y , λ ( a )) = d  λ ( p k − 1 ) − α λ ( p h ) , λ ( a )  = d  λ ( p k − 1 ) , λ ( a ) + α λ ( p h )  = d  λ ( p k − 1 ) , λ ( a ⊕ p αp h )  ⩾ d. Lemma 2.1 implies that y ⩾ F λ ( p k − 1 ) . Hence y = λ ( p k − 1 ) , that is, x [ M ] ( p k − 1 ) = α = 0 . Theref ore (1) and (2) hold. From (1) and (2), if 0 ⩽ b < a < p k , then e x ( b ) < e x ( a ) . Theref ore (3) holds. pr oof of Theor em 1.4 . By Lemma 2.3 , w e see that (1) implies (2). It is obvious that (2) implies (1) and (3). W e show that (3) implies (2). Assume that a is the smallest integ er satisfying x ( a )  = e x ( a ) and a < p k . Let N be as in Lemma 2.2 and α i = x [ N ] ( p i ) . W e sho w that b 0 α 0 ⊕ p · · · ⊕ p b k − 1 α k − 1 = 0 f or some 0 < b < a satisfying b i ⩽ a i f or i ∈ N , (2.5) which contradicts Lemma 2.2 since e x [ N ] ( b ) = b 0 α 0 ⊕ p · · · ⊕ p b k − 1 α k − 1 . Consider the sequence ( β i ) i =1 , 2 ,...,n = ( α 0 , . . . , α 0 | {z } a 0 , α 1 , . . . , α 1 | {z } a 1 , . . . , α k − 1 , . . . , α k − 1 | {z } a k − 1 ) and let γ j = β 1 ⊕ p β 2 ⊕ p · · · ⊕ p β j . If { γ j : 1 ⩽ j ⩽ n } = F p , then ( 2.5 ) is ob vious. Suppose that { γ j : 1 ⩽ j ⩽ n }  = F p . Since n ⩾ p , it f ollo ws that γ j = γ h f or some j < h , so γ h − γ j = β j +1 ⊕ p · · · ⊕ p β h = 0 , which yields ( 2.5 ). Theref ore e x ( a ) = x ( a ) f or 0 ⩽ a < p k . 8 Y uki Irie Remar k 2.4 . Let p = 3 , and let a and N be the same as in Lemma 2.2 . Theorem 1.4 implies that a = 3 h +3 l f or some h and l . Moreo v er , ( x [ N ] (3 h ) , x [ N ] (3 l )) ∈ { (1 , 1) , (2 , 2) } . Indeed, it f ollo w s from Lemma 2.2 that x [ N ] (3 h ) , x [ N ] (3 l ) , and x [ N ] (3 l ) + x [ N ] (3 h ) are all nonzero. Theref ore ( x [ N ] (3 h ) , x [ N ] (3 l )) ∈ { (1 , 1) , (2 , 2) } . W e introduce some notation before moving on to the ne xt section. F or k ∈ N , let ⟨ k ⟩ = { 0 , 1 , . . . , k } . For a ⟨ k ⟩ × N matr ix X and i ∈ N , let X i denote the i -th column of X . W e wr ite X i ∈ F ⟨ k ⟩× 1 p or X i ∈ F ⟨ k ⟩ p . For v ∈ F ⟨ k ⟩ p , let CI ( X , v ) denote the set of column inde x es i such that the i -th column X i equals v , that is, CI ( X , v ) = { i ∈ N : X i = v } . Let ci ( X , v ) = | CI ( X , v ) | . Example 2.5. Let p = 3 , d = d , and F = E . Let X ⟨ 1 ⟩ =  x (3 0 ) x (3 1 )  =  0 1 2 3 4 1 1 1 0 0 · · · 0 2 1 0 1 0 · · · 1  . Then X ⟨ 1 ⟩ 0 = [ 1 2 ] , X ⟨ 1 ⟩ 1 = [ 1 1 ] , X ⟨ 1 ⟩ 2 = [ 1 0 ] , X ⟨ 1 ⟩ 3 = [ 0 1 ] , CI  X ⟨ 1 ⟩ , [ 1 2 ]  = { 0 } , CI  X ⟨ 1 ⟩ , [ 1 1 ]  = { 1 } , CI  X ⟨ 1 ⟩ , [ 1 0 ]  = { 2 } , CI  X ⟨ 1 ⟩ , [ 0 1 ]  = { 3 } , CI  X ⟨ 1 ⟩ , [ 0 0 ]  = { 4 , 5 , . . . } . 3. Griesmer bound In this section, w e recall the Griesmer bound. Let g p ( k , d ) = k − 1 X i =0 l d p i m Theorem 3.1 (Griesmer Bound [ 4 , 6 ]) . If C is an [ n, k , d ] code ov er F p , then n ⩾ g p ( k , d ) . Corollary 3.2. Let C be an [ n, k , d ] code ov er F p with g enerator matrix X . If C meets the Griesmer bound and d ⩽ d ′ p k − 1 , then ci ( X , v ) ⩽ d ′ f or v ∈ F ⟨ k − 1 ⟩ p . Pr oof. Suppose that ci ( X , w ) ⩾ d ′ + 1 f or some w ∈ F ⟨ k − 1 ⟩ p . Since C meets the Griesmer bound, w  = 0 0 0 . Hence w e ma y assume that X =      1 . . . 1 . . . 0 . . . 0 . . . . . . . . . X ′ 0 . . . 0      d ′ + 1 z }| { . The matrix X ′ g enerates an [ n − d ′ − 1 , k − 1 , d ] code. By the Gr iesmer bound, n − d ′ − 1 ⩾ g p ( k − 1 , d ) . Thus n ⩾ d ′ + 1 + g p ( k − 1 , d ) > g p ( k , d ) , a contradiction. 9 Let V ⟨ k ⟩ = { v ∈ F ⟨ k ⟩ p : v 0 = · · · = v i − 1 = 0 , v i = 1 f or some i } , V ⟨ k ⟩ 0 0 0 = V ⟨ k ⟩ ∪ { 0 0 0 } , e ⟨ k ⟩ = " 0 0 . . . 0 1 # ∈ V ⟨ k ⟩ . For e xample, if p = 3 , then V ⟨ 0 ⟩ = { [1] } , V ⟨ 1 ⟩ = { [ 1 2 ] , [ 1 1 ] , [ 1 0 ] , [ 0 1 ] } . Note that | V ⟨ k − 1 ⟩ | = p k − 1 p − 1 . As w e hav e seen in Ex ample 2.5 , a nonzero column of X ⟨ 1 ⟩ is in V ⟨ 1 ⟩ . W e will sho w this is alw ay s true in Section 4. Corollary 3.3. Let C be an [ n, k , p k − 1 d ′ ] code ov er F p with g enerat or matrix X . Le t V be a subset of F ⟨ k ⟩ p of size p k − 1 p − 1 . Suppose t hat CI ( X , v ) = ∅ f or v ∈ F ⟨ k − 1 ⟩ p \ V . (3.1) Then the f ollowing two conditions ar e eq uivalent. (1) C meets the Griesmer bound. (2) ci ( X , v ) = d ′ f or v ∈ V . Pr oof. By the Gr iesmer bound, n ⩾ g p ( k , p k − 1 d ′ ) = p k − 1 p − 1 d ′ . Suppose that C meets the Gr iesmer bound. Then n = p k − 1 p − 1 d ′ and ci ( X , v ) ⩽ d ′ f or v ∈ V b y Corollar y 3.2 . Since | V | = p k − 1 p − 1 , it f ollow s that ci ( X , v ) = d ′ . Con v ersel y , suppose that ci ( X , v ) = d ′ f or v ∈ V . Then n = p k − 1 p − 1 d ′ , so C meets the Gr iesmer bound. Note that w e can apply Corollar y 3.3 f or V = V ⟨ k − 1 ⟩ since V ⟨ k − 1 ⟩ = p k − 1 p − 1 . 4. Le xicographic codes meeting the Griesmer bound Fix F = F ( ξ , η ) ∈ F and an integer d such that d ⩾ 2 . Let Θ = { ξ , η } . In this section, w e pro v e Theorem 1.10 , whic h s tates that L ⟨ k − 1 ⟩ meets the Gr iesmer bound, b y using Corollary 3.3 . 10 Y uki Irie Let X ⟨ k ⟩ = X ⟨ k ⟩ ; F ,d =      x ( p 0 ) x ( p 1 ) . . . x ( p k )      and X ⟨ k ⟩ sp =      x ( p 0 )   S x ( p 1 )   S . . . x ( p k )   S      , where S = supp E ( X ⟨ k ⟩ ) = S i ∈⟨ k ⟩ supp E ( x ( p i )) . Similar ly , let Λ ⟨ k ⟩ =  λ ( p i )  i ∈⟨ k ⟩ and Λ ⟨ k ⟩ sp =  x ( p i )   S  i ∈⟨ k ⟩ , where S = supp E (Λ ⟨ k ⟩ ) . For a ∈ N and v ∈ F ⟨ k ⟩ p , let a • v = M i ∈⟨ k ⟩ p a i v i ∈ F p . For e xample, if p = 3 , a = 7 = 1 + 2 · 3 , and v = h 1 1 2 i ∈ F ⟨ 2 ⟩ p , then a • v = 1 · 1 ⊕ 3 2 · 1 ⊕ 3 0 · 2 = 0 . W e begin with an easy lemma to calculate distance. Lemma 4.1. Let X be a ⟨ k ⟩ × N matrix. F or a ∈ N , d  x k , X i ∈⟨ k − 1 ⟩ a i x i  = X v ∈ F ⟨ k ⟩ p ,v k  = v • a ci ( X , v ) . (4.1) wher e X =  x 0 . . . x k  . Pr oof. Let y = P i ∈⟨ k − 1 ⟩ a i x i . W e count the number of coordinates in which x k and y agree. Let v = X i , that is, i ∈ CI ( X , v ) . Then the i -th component of x k is v k and that of y is a • v . Theref ore ( 4.1 ) f ollow s. The ne xt lemma enables us to sho w ( 3.1 ) f or Λ ⟨ k ⟩ sp . Lemma 4.2. Let F = F ( ξ , η ) ∈ F . (1) If ξ ∈ supp E ( L ⟨ k ⟩ ) , then η ∈ supp E ( L ⟨ k ⟩ ) . (2) If η ∈ supp E ( L ⟨ k ⟩ ) and ξ < η , then ξ ∈ supp E ( L ⟨ k ⟩ ) . Pr oof. (1) Assume that η ∈ supp E ( L ⟨ k ⟩ ) . Let h be the smallest integer with ξ ∈ supp E ( λ ( p h )) . Then h ⩽ k . Note that λ i ( p h ) = ( λ [ i ] ( p h ) if i  = η , λ [ η ] ( p h ) + λ [ ξ ] ( p h ) if i = η . 11 Since η ∈ supp E ( L ⟨ k ⟩ ) , it f ollow s that 0 = λ η ( p h ) = λ [ η ] ( p h ) + λ [ ξ ] ( p h ) , and hence λ [ η ] ( p h ) = − λ [ ξ ] ( p h )  = 0 . Let y = λ ( p h ) − λ [ η ] ( p h ) f η ; then y < F λ ( p h ) . W e sho w that d ( y , λ ( a )) ⩾ d f or 0 ⩽ a < p h . For i  = η , w e see that y i = λ i ( p h ) . Since y η = λ [ η ] ( p h ) + λ [ ξ ] ( p h ) − λ [ η ] ( p h ) = λ [ ξ ] ( p h )  = 0 and λ η ( p l ) = 0 f or 0 ⩽ l ⩽ k , it f ollo ws that d ( y , λ ( a )) = d ( λ ( p h ) , λ ( a )) + 1 > d f or 0 ⩽ a < p h . Thus Lemma 2.1 implies that y ⩾ F λ ( p h ) , whic h is impossible. Therefore η ∈ supp E ( L ⟨ k ⟩ ) . (2) Assume that ξ ∈ supp E ( L ⟨ k ⟩ ) . Let h be the smallest integer suc h that η ∈ supp E ( λ ( p h )) ; then h ⩽ k and 0  = λ η ( p h ) = λ [ η ] ( p h ) + λ [ ξ ] ( p h ) = λ [ η ] ( p h ) . Let y = λ ( p h ) − f η + f ξ = λ ( p h ) + e ξ . Since ξ < η and λ [ η ] ( p h ) > 0 , it f ollow s that y < F λ ( p h ) . Moreo v er , since y ξ = λ ξ ( p h ) + 1 = 1 and λ ξ ( p l ) = 0 f or 0 ⩽ l ⩽ k , we see that d ( y , λ ( a )) ⩾ d ( λ ( p h ) , λ ( a )) ⩾ d f or 0 ⩽ a < p h . Theref ore y ⩾ F λ ( p h ) , a contradiction. Remar k 4.3 . From Lemma 4.2 , we see that supp F ( L ⟨ k ⟩ ) ⊆ supp E ( L ⟨ k ⟩ ) . Indeed, let N ∈ N \ supp E ( L ⟨ k ⟩ ) . If N  = η , then N ∈ N \ supp F ( L ⟨ k ⟩ ) . By Lemma 4.2 , ξ ∈ N \ supp E ( L ⟨ k ⟩ ) . Hence λ [ η ] ( p i ) = λ η ( p i ) = 0 f or 0 ⩽ i ⩽ k . Theref ore N = η ∈ N \ supp F ( L ⟨ k ⟩ ) . Lemma 4.4. If v ∈ F ⟨ k ⟩ p \ V ⟨ k ⟩ 0 0 0 , then CI  Λ ⟨ k ⟩ , v  = ∅ . In particular , ( 3.1 ) holds f or Λ ⟨ k ⟩ sp . Pr oof. Since v ∈ V ⟨ k ⟩ 0 0 0 , we see that there e xists h such that v h ⩾ 2 and v l = 0 f or 0 ⩽ l < h . Assume that i ∈ CI  Λ ⟨ k ⟩ , v  . Then λ i ( p l ) = v l f or 0 ⩽ l ⩽ k . Case 1 ( i  = ξ ) . Suppose that λ [ i ] ( p h )  = 0 . Let y = λ ( p h ) − f i ; then y < F λ ( p h ) and y i = λ i ( p h ) − 1 = v h − 1  = 0 . Since λ i ( p l ) = v l = 0 f or 0 ⩽ l < h , it f ollo ws that d ( y , λ ( a )) = d  x ( p h ) , λ ( a )  ⩾ d f or 0 ⩽ a < p h , whic h is impossible. Suppose that λ [ i ] ( p h ) = 0 . Since λ i ( p h ) = v h ⩾ 2 , it f ollow s that i = η and λ ξ ( p h ) = λ η ( p h ) ⩾ 2 . Let y = λ ( p h ) − f ξ ; then y < F λ ( p h ) . Since η ∈ supp E ( L h − 1 ) , it f ollo ws from Lemma 4.2 that ξ ∈ supp E ( L ⟨ h − 1 ⟩ ) . Hence λ η ( p l ) = λ ξ ( p l ) = 0 f or 0 ⩽ l < h . Since y η = y ξ = λ ξ ( p h ) − 1  = 0 , it f ollo w s that d ( y , λ ( a )) ⩾ d for 0 ⩽ a η . Let y = λ ( p h ) − f ξ + f η = x ( p h ) − e ξ ; then y < F λ ( p h ) . Since λ ξ ( p l ) = 0 f or 0 ⩽ l < h , it f ollo ws that d ( y , λ ( a )) = d  λ ( p h ) , λ ( a )  ⩾ d f or 0 ⩽ a < h , whic h is impossible. Suppose that ξ < η . Let y = λ ( p h ) − f ξ − λ [ η ] ( p h ) f η ; then y < F λ ( p h ) and y η = y ξ = λ ξ ( p h ) − 1 = v h − 1  = 0 . Since ξ ∈ supp E ( L ⟨ h − 1 ⟩ ) and ξ < η , it f ollow s Lemma 4.2 that η ∈ supp E ( L ⟨ h − 1 ⟩ ) . Theref ore d ( y , λ ( a )) ⩾ d f or 0 ⩽ a < h , a contradiction. 12 Y uki Irie For v ∈ F ⟨ k − 1 ⟩ p and α ∈ F p , let v α =  v α  ∈ F ⟨ k ⟩ p . (4.2) For e xample, if v = [ 0 2 ] and α = 1 , then v α = h 0 2 1 i . Similar l y , f or a ⟨ k − 1 ⟩ × N matr ix X and y ∈ F N p , let X y =  X y  ∈ F ⟨ k ⟩× N p . (4.3) Lemma 4.5 (Order lemma) . Le t M ∈ CI  X ⟨ k ⟩ , v α  \ { ξ } and N ∈ CI  X ⟨ k ⟩ , v β  \ { ξ } , wher e v ∈ F ⟨ k − 1 ⟩ p , and α , β ∈ F p . If α < β and λ [ N ] ( p k ) = λ N ( p k ) (= β ) , t hen M > N . Pr oof. Let y = λ ( p k ) + ( β − α ) f M + ( α − β ) f N = λ ( p k ) + ( β − α ) e M + ( α − β ) e N . Note that if M < N , then y < F λ ( p k ) since y [ N ] = α < β = λ [ n ] ( p k ) . By the deﬁnition of y , CI  Λ ⟨ k − 1 ⟩ , y , v γ  =      CI  Λ ⟨ k ⟩ , v α  ∪ { N } \ { M } if γ = α CI  Λ ⟨ k ⟩ , v β  ∪ { M } \ { N } if γ = β CI  Λ ⟨ k ⟩ , v γ  otherwise . In particular , ci  Λ ⟨ k − 1 ⟩ , y , v γ  = ci  Λ ⟨ k ⟩ , v γ  . It f ollow s from Lemma 4.1 that d ( y , λ ( a )) = d  λ ( p k ) , λ ( a )  ⩾ d f or 0 ⩽ a N . As w e ha v e seen in the proof of Lemma 4.5 , the distance d ( y , x ∗ ( a )) is deter mined b y CI  Λ ⟨ k − 1 ⟩ , y , v γ  . The f ollo wing lemma ensures that there e xists a suitable y . Lemma 4.6. Le t X be a ⟨ k ⟩ × N matrix and v ∈ F ⟨ k − 1 ⟩ p . Let m 0 , m 1 , . . . , m p − 1 be integ ers such that P α ∈ F p m α = ci ( X , v ) and m α ⩾ 2 . Then, f or y ∈ F N p , t here exists ˜ y ∈ F N p satisfying t he f ollowing two conditions: (1) ci  X ˜ y , v α  = m α f or α ∈ F p . (2) supp E ( y − ˜ y ) = supp F ( y − ˜ y ) ⊆ CI ( X , v ) \ Θ . In particular , CI  X ˜ y , w α  = CI ( X y , w α ) f or α ∈ F p and w ∈ F ⟨ k − 1 ⟩ p \ { v } . Pr oof. Let l α = ci ( X y , v α ) and D = P α ∈ F p | l α − m α | . W e sho w the lemma by induction on D . If D = 0 , then y itself satisﬁes the tw o conditions. Suppose that D ⩾ 1 . Since P α ∈ F p l α = ci ( X , v ) = P α ∈ F p m α , it f ollo ws that there e xist β , γ ∈ F p such that l β > m β and l γ < m γ . 13 Since l β > m β ⩾ 2 , the set CI  X y , v β  \ Θ contains an element N . Let z = y + ( γ − β ) f N = y + ( γ − β ) e N and l ′ α = ci ( X z , v α ) . W e see that l ′ α =      l β − 1 if α = β , l γ + 1 if α = γ , l α otherwise . By the induction h ypothesis, there e xists ˜ z ∈ F N p satisfying (1) and (2) f or z , that is, ci  X ˜ z , v α  = m α f or α ∈ F p and supp E ( z − ˜ z ) = supp F ( z − ˜ z ) ⊆ CI ( X , v ) \ Θ . Since supp F ( y − ˜ z ) = supp F ( z + ( α − β ) f N − ˜ z ) ⊆ supp F ( z − ˜ z ) ∪ { i } ⊆ CI ( X , v ) \ Θ , it f ollo ws that supp F ( y − ˜ z ) = supp E ( y − ˜ z ) , and that ˜ z satisﬁes (1) and (2) also f or y . Example 4.7 . Let p = 3 , d = 7 , and F = F (1 , 3) . Then x (1) =  1 1 1 1 1 1 1 0 0 · · ·  and ci  X ⟨ 0 ⟩ , 1  = 7 . Consider m 0 = 3 , m 1 = 2 , m 2 = 2 , and y =  2 2 2 2 2 1 1 0 0 · · ·  . Since X ⟨ 0 ⟩ , y =  0 1 2 3 4 5 6 7 8 ··· 1 1 1 1 1 1 1 0 0 ··· 2 2 2 2 2 1 1 0 0 ···  , w e see that CI  X ⟨ 0 ⟩ , y , [ 1 0 ]  = ∅ , CI  X ⟨ 0 ⟩ , y , [ 1 1 ]  = { 5 , 6 } , CI  X ⟨ 0 ⟩ , y , [ 1 2 ]  = { 0 , 1 , 2 , 3 , 4 } . Let ˜ y =  0 2 0 2 0 1 1 0 0 · · ·  . Then X ⟨ 0 ⟩ , ˜ y =  0 1 2 3 4 5 6 7 8 ··· 1 1 1 1 1 1 1 0 0 ··· 0 2 0 2 0 1 1 0 0 ···  , so CI  X ⟨ 0 ⟩ , ˜ y , [ 1 0 ]  = { 0 , 2 , 4 } , CI  X ⟨ 0 ⟩ , ˜ y , [ 1 1 ]  = { 5 , 6 } , CI  X ⟨ 0 ⟩ , ˜ y , [ 1 2 ]  = { 1 , 4 } . Moreo v er y − ˜ y = 2 e 0 + 2 e 2 + 2 e 4 = 2 f 0 + 2 f 2 + 2 f 4 . Thus ˜ y satisﬁes the tw o conditions (1) and (2). Pr oof of Theor em 1.10 . Let − 1 ⩽ h < k . W e pro v e the theorem b y induction on h . Since the length of L ⟨ h ⟩ sp equals P v ∈ V ⟨ h ⟩ ci  Λ ⟨ h ⟩ , v  and g p ( h, d ) = d ′ ( p k − 1 + p k − 2 + · · · + p k − h − 1 ) , it suﬃces to show that X v ∈ V ⟨ h ⟩ ci  Λ ⟨ h ⟩ , v  = d ′ ( p k − 1 + p k − 2 + · · · + p k − h − 1 ) . If h = − 1 , then both sides equal zero. Suppose that h ⩾ 0 . By the induction h ypothesis, X v ∈ V ⟨ h − 1 ⟩ ci  Λ ⟨ h − 1 ⟩ , v  = d ′ ( p k − 1 + p k − 2 + · · · + p k − h ) . 14 Y uki Irie Hence X v ∈ V ⟨ h ⟩ ci  Λ ⟨ h ⟩ , v  = X v ∈ V ⟨ h − 1 ⟩ ci  Λ ⟨ h − 1 ⟩ , v  + ci  Λ ⟨ h ⟩ , e ⟨ h ⟩  = d ′ ( p k − 1 + p k − 2 + · · · + p k − h ) + ci  Λ ⟨ h ⟩ , e ⟨ h ⟩  . Assume that ci  Λ ⟨ h ⟩ , e ⟨ h ⟩  > d ′ p k − h − 1 . Since d ′ ⩾ 2 , there e xists M ∈ CI  Λ ⟨ h ⟩ , e ⟨ h ⟩  \ Θ . Let y = λ ( p h ) − f M = λ ( p h ) − e M . Then ci  Λ ⟨ h − 1 ⟩ , y , e ⟨ h ⟩  ⩾ d ′ p k − h − 1 and y < F λ ( p h ) because λ [ M ] ( p h ) = λ M ( p h ) = 1 . Since L ⟨ h − 1 ⟩ sp meets the Gr iesmer bound, it f ollo w s from Lemma 4.4 and Corollary 3.3 that ci  Λ ⟨ h − 1 ⟩ , v  = d ′ p k − 1 p h − 1 = d ′ p k − h f or v ∈ V ⟨ h − 1 ⟩ . Note that d ′ p k − h = d ′ p k − h − 1 + · · · + d ′ p k − h − 1 | {z } p . Lemma 4.6 show s that there e xists ˜ y such that ci  Λ ⟨ h − 1 ⟩ , ˜ y , v α  = d ′ p k − h − 1 f or v ∈ V ⟨ h − 1 ⟩ , α ∈ F p (4.4) and supp F ( y − ˜ y ) = supp E ( y − ˜ y ) ⊆ [ v ∈ V ⟨ h − 1 ⟩ CI  Λ ⟨ h − 1 ⟩ , v  \ Θ . Since ci  Λ ⟨ h − 1 ⟩ , ˜ y , e ⟨ h ⟩  = ci  Λ ⟨ h − 1 ⟩ , y , e ⟨ h ⟩  ⩾ d ′ p k − h − 1 and ci  Λ ⟨ h − 1 ⟩ , ˜ y , w  = 0 f or w ∈ V ⟨ h ⟩ , it f ollo ws from ( 4.4 ) that d ( ˜ y , λ ( a )) ⩾ d f or 0 ⩽ a < p h , so ˜ y ⩾ F λ ( p h ) . From Lemma 4.5 , if N ∈ CI  Λ ⟨ h − 1 ⟩ , v  \ Θ , then N < M . Thus max supp F ( ˜ y − λ ( p h )) = M , and hence ˜ y < F λ ( p h ) , a contradiction. 5. T ernary linear lexicographic codes In this section, w e sho w that X ⟨ k − 1 ⟩ is a linear code when p = 3 , d = p k − 1 d ′ , and d ′ ⩾ 2 . 5.1. Strong er v ersions of lemmas in Section 4 Lemma 5.1. Let h ∈ N . Suppose t hat ci  Λ ⟨ h ⟩ , v  ⩾ 2 for v ∈ V ⟨ h ⟩ . (1) If λ ξ ( p h )  = 0 , t hen λ η ( p h )  = 0 . (2) If ξ < η , then λ ξ ( p h ) = λ η ( p h ) ; in par ticular , λ [ η ] ( p h ) = 0 . (3) If ξ > η and λ ξ ( p h )  = λ η ( p h ) , then λ ξ ( p h ) < λ η ( p h ) ; in particular , λ [ η ] ( p h )  = p − 1 when λ ξ ( p h )  = 0 . 15 Pr oof. (1) W e ﬁrst sho w (1) assuming (2) and (3). Suppose that λ ξ ( p h )  = 0 . Since λ η ( p h ) = λ [ η ] ( p h ) + λ [ ξ ] ( p h ) , we ma y assume that λ [ η ] ( p h )  = 0 . It f ollow s from (2) and (3) that ξ > η and λ ξ ( p h ) < λ η ( p h ) . Thus λ η ( p h )  = 0 . (2) Suppose that ξ < η . If ξ ∈ supp E ( L ⟨ h ⟩ ) , then it f ollow s from Lemma 4.2 that η ∈ supp E ( L ⟨ h ⟩ ) , and hence λ η ( p h ) = λ ξ ( p h ) = 0 . Suppose that ξ ∈ supp E ( L ⟨ h ⟩ ) . Lemma 4.2 yields η ∈ supp E ( L ⟨ h ⟩ ) . Let v α = Λ ⟨ h ⟩ ξ , that is, ξ ∈ CI  Λ ⟨ h ⟩ , v α  . W e sho w that η ∈ CI  Λ ⟨ h ⟩ , v α  b y induction on h . If h = 0 , then it f ollo ws from Lemma 4.4 that v α = 1 and η ∈ CI  Λ ⟨ 0 ⟩ , 1  . Suppose that h ⩾ 1 . By the induction h ypothesis, λ ξ ( p l ) = λ η ( p l ) for 0 ⩽ l < h , and hence Λ ⟨ h − 1 ⟩ η = Λ ⟨ h − 1 ⟩ ξ = v . Thus η ∈ CI  Λ ⟨ h − 1 ⟩ , v  . Let β = λ η ( p h ) , that is, η ∈ CI  Λ ⟨ h ⟩ , v β  . Assume that β  = α ( = λ ξ ( p h ) ). Since β = α + λ [ η ] ( p h ) , w e see that λ [ η ] ( p h )  = 0 . Let γ = max { α, β } and ϵ = min { α, β } . Since ci  Λ ⟨ h ⟩ , v γ  ⩾ 2 and   CI  Λ ⟨ h ⟩ , v γ  ∩ Θ   = 1 , there e xists N ∈ CI  Λ ⟨ h ⟩ , v γ  \ Θ . Let y = λ ( p h ) + ( γ − α ) f ξ + ( α − β ) f η + ( ϵ − γ ) f N . ξ η N λ [ i ] ( p h ) α β − α γ λ i ( p h ) α β γ y [ i ] γ 0 ϵ y i γ γ ϵ Since ξ < η and ϵ < γ , it f ollo ws that y < F λ ( p h ) . Because { γ , γ , ϵ } and { α , β , γ } are eq ual as multisets, we see that ci  Λ ⟨ h − 1 ⟩ , y , w  = ci  Λ ⟨ h ⟩ , w  f or w ∈ V ⟨ h ⟩ . It f ollo ws from Lemma 4.1 that d  y , λ ( a )  = d  λ ( p h ) , λ ( a )  ⩾ d . This implies that y > F λ ( p h ) , a contradiction. (3) If ξ ∈ supp E ( L ⟨ h ⟩ ) , then the assertion is obvious. Suppose that ξ ∈ supp E ( L ⟨ h ⟩ ) . Lemma 4.2 sho ws that η ∈ supp E ( L ⟨ h ⟩ ) . Let v α = Λ ⟨ h ⟩ ξ and w β = Λ ⟨ h ⟩ η , that is, ξ ∈ CI  Λ ⟨ h ⟩ , v α  and η ∈ CI  Λ ⟨ h ⟩ , w β  . Assume that w = 0 0 0 . Lemma 4.2 implies that v = 0 0 0 . Since ξ , η ∈ supp E ( L ⟨ h ⟩ ) , it f ollo ws from Lemma 4.4 that α = β = 1 , a contradiction. Thus w  = 0 0 0 , so w ∈ V ⟨ h − 1 ⟩ and w α ∈ V ⟨ h ⟩ . Since ci  Λ ⟨ h ⟩ , w α  ⩾ 2 , there e xists M ∈ CI  Λ ⟨ h ⟩ , w α  \ Θ . Let y = λ ( p h ) + ( α − β ) f η + ( β − α ) f M . η M ξ λ [ i ] ( p h ) β − α α α λ i ( p h ) β α α y [ i ] 0 β α y i α β α It f ollo ws from Lemma 4.1 that d  y , λ ( a )  = d  λ ( p h , λ ( a )  ⩾ d f or 0 ⩽ a F λ ( p h ) . Theref ore M > η and β > α . Lemma 5.2. Let h ∈ N . Suppose t hat ci  Λ ⟨ h ⟩ , w  ⩾ 2 f or w ∈ V ⟨ h ⟩ . Le t M ∈ CI  Λ ⟨ h ⟩ , v α  and N ∈ CI  Λ ⟨ h ⟩ , v β  \ { ξ } . If α < β and λ [ N ] ( p h ) = λ N ( p h ) , then M > N . 16 Y uki Irie Pr oof. When M  = ξ , the lemma f ollo w s from Lemma 4.5 . Suppose that M = ξ . Case 1 ( η < ξ ) . If N = η , then N = η < ξ = M , and hence w e ma y assume that N  = η . Let y = λ ( p h ) + ( β − α ) f ξ + ( α − β ) f η + ( α − β ) f N It f ollo ws from Lemma 4.1 that d  y , x ( a )  ⩾ d f or 0 ⩽ a F x ( p h ) . Theref ore N < ξ = M . N η ξ λ [ i ] ( p h ) β γ α λ i ( p h ) β γ + α α y [ i ] α γ + α − β β y i α γ + α β Case 2 ( η > ξ ) . By Lemma 5.1 , w e see that Λ ⟨ h ⟩ η = Λ ⟨ h ⟩ ξ = v α , and hence η ∈ CI  Λ ⟨ h ⟩ , v α  . Thus N  = η . Since ci  Λ ⟨ h ⟩ , v β  ⩾ 2 , there e xists L ∈ CI  Λ ⟨ h ⟩ , v β  \ { N } . Note that L ∈ Θ . Let y = λ ( p h ) + ( β − α ) f ξ + ( α − β ) f L + ( α − β ) f N . N L ξ η λ [ i ] ( p h ) β β α 0 λ i ( p h ) β β α α y [ i ] α α β 0 y i α α β β It f ollo ws from Lemma 4.1 that d  y , x ( a )  ⩾ d f or 0 ⩽ a F λ ( p h ) . Theref ore N < ξ = M . Lemma 5.3. Le t L < j < M and h ∈ N ∪ { − 1 } . Suppose that ci  Λ ⟨ h ⟩ , v  ⩾ 2 f or v ∈ V ⟨ h ⟩ 0 0 0 , and that L, M ∈ CI  Λ ⟨ h ⟩ , t  f or some t ∈ V ⟨ h ⟩ 0 0 0 , where V ⟨− 1 ⟩ 0 0 0 = { [ ] } and CI  Λ ⟨− 1 ⟩ , [ ]  = N . If one of t he f ollowing tw o conditions holds, then j ∈ CI  Λ ⟨ h ⟩ , t  . (1) j, M ∈ Θ . (2) j, M ∈ Θ . Pr oof. W e sho w the lemma b y induction on h . This is obvious when h = − 1 . Suppose that h ⩾ 0 , and let w α = t , where w ∈ V ⟨ h − 1 ⟩ 0 0 0 and α ∈ F p . By the induction h ypothesis, j ∈ CI  Λ ⟨ h − 1 ⟩ , w  . Let β = λ j ( p h ) ; then j ∈ CI  Λ ⟨ h ⟩ , w β  . (1) Firs t, assume that β < α . Since j ∈ CI  Λ ⟨ h ⟩ , w β  and M ∈ CI  Λ ⟨ h ⟩ , w α  \ Θ , it f ollow s from Lemma 5.2 that j > M , a contradiction. Ne xt, assume that α < β . Since L ∈ CI  Λ ⟨ h ⟩ , w α  and j ∈ CI  Λ ⟨ h ⟩ , w β  \ Θ , it f ollo w s from Lemma 5.2 that L > j , a contradiction. Theref ore α = β . (2) If j = ξ and M = η , then ξ < η ; it f ollo w s from Lemma 5.1 that α = β . Suppose that j = η and M = ξ . Let y = λ ( p h ) + ( α − β ) f j + ( β − α ) f L . 17 L j M λ [ i ] ( p h ) α β − α α λ i ( p h ) α β α y [ i ] β 0 α y i β α α Note that y ⩽ F λ ( p h ) . Since d ( y , λ ( a )) ⩾ d f or 0 ⩽ a N in mos t cases. The follo wing lemma provides necessar y conditions f or M < N . Lemma 5.4. Let h ∈ N . Suppose t hat ci  Λ ⟨ h ⟩ , w  ⩾ 2 f or w ∈ V ⟨ h ⟩ . Le t M ∈ CI  Λ ⟨ h ⟩ , v α  and N ∈ CI  Λ ⟨ h ⟩ , v β  . Suppose t hat α < β and M < N . Then the follo wing stat ements hold. (1) λ [ η ] ( p h ) = 0 and x η ( p h ) = x ξ ( p h ) = β  = 0 . (2) If η < ξ , then N = η and β = λ η ( p h ) = λ ξ ( p h ) . (3) If ξ < η , then N ∈ Θ ; mor eov er , if N = ξ , then α = β − 1 and N = M + 1 . Pr oof. Lemma 5.2 implies that N ∈ Θ . (1) W e sho w λ [ η ] ( p h ) = 0 b y assuming (2). If η < ξ , then β = λ η ( p h ) = λ ξ ( p h ) b y (2), and hence λ [ η ] ( p h ) = 0 . Suppose that ξ < η . Lemma 5.1 sho ws that λ [ η ] ( p h ) = 0 . Hence λ η ( p h ) = λ ξ ( p h ) = λ N ( p h ) = β since N ∈ Θ . (2) Assume that N = ξ . Let y = λ ( p h ) + ( α − β ) f ξ + ( β − α )( f η + f M ) . Note that y < F λ ( p h ) . Moreo v er , d  y , λ ( a )  = d  λ ( p h ) , λ ( a )  ⩾ d f or 0 ⩽ a F λ ( p h ) , a contradiction. M = η N = ξ λ [ i ] ( p h ) α − β β λ i ( p h ) α β y [ i ] β − α α y i β α M η N = ξ λ [ i ] ( p h ) α γ β λ i ( p h ) α γ + β β y [ i ] β γ + β − α α y i β γ + β α Theref ore N = η . Since M < N = η < ξ , it f ollo ws that M  = ξ . W e sho w that there e xists L ∈ CI  Λ ⟨ h ⟩ , v β − 1  \ Θ such L ⩽ M . Indeed, if α = β − 1 , then L = M satisﬁes the condition. If α < β − 1 , then there e xists L ∈ CI  Λ ⟨ h ⟩ , v β − 1  \ Θ , and we see that L < M b y Lemma 5.2 . Let y = λ ( p h ) − f N + f L . L N = η ξ λ [ i ] ( p h ) β − 1 β − γ γ λ i ( p h ) β − 1 β γ y [ i ] β β − γ − 1 γ y i β β − 1 γ 18 Y uki Irie It f ollo ws from Lemma 4.1 that d ( y , λ ( a )) ⩾ d f or 0 ⩽ a F λ ( p h ) and β − γ < β − γ − 1 . Theref ore β = γ = λ ξ ( p h ) . (3) W e ha v e sho wn N ∈ Θ . Suppose that N = ξ . Since ξ < η , it follo ws from Lemma 5.1 that λ [ η ] ( p h ) = 0 and η , ξ ∈ CI  Λ ⟨ h ⟩ , v β  . W e ﬁrst sho w that if L ∈ CI  Λ ⟨ h ⟩ , v α  \ { M } , then L > N . ( ∗ ) Let y = λ ( p h ) + ( β − α )( f M + f L ) + ( α − β ) f ξ . M L N = ξ η λ [ i ] ( p h ) α α β 0 λ i ( p h ) α α β β y [ i ] β β α 0 y i β β α α It f ollow s from Lemma 4.1 that d  y , λ ( a )  f or 0 ⩽ a F λ ( p h ) . Theref ore L > N . W e ne xt sho w that N = M + 1 . Assume that N  = M + 1 . Since M < N , we see that M < M + 1 < N . Let L ∈ CI  Λ ⟨ h ⟩ , v α  \ { M } ; then L > N b y ( ∗ ) . Since M < M + 1 < L and M + 1 , L ∈ Θ , it f ollo ws from Lemma 5.3 that M + 1 ∈ CI  Λ ⟨ h ⟩ , v α  . From ( ∗ ), w e see that M + 1 > N , a contradiction. Theref ore N = M + 1 . Finall y , we sho w that α = β − 1 . Assume that α < β − 1 , and let M ′ ∈ CI  Λ ⟨ h ⟩ , v β − 1  . Since M ′ ∈ Θ , it f ollo ws from Lemma 5.2 that M ′ < M , and hence M ′ < N . Theref ore M ′ satisﬁes the same conditions as M . This implies that M ′ = N − 1 , a contradiction. Example 5.5. (1) Let p = 3 , d = 4 , ξ = 3 , and η = 7 . Then x (1) =  1 1 0 1 0 0 0 1 0 · · ·  . Note that 2 , 4 , 5 , 6 ∈ CI  Λ ⟨ 0 ⟩ , 0  and 0 , 1 , 3 , 7 ∈ CI  Λ ⟨ 0 ⟩ , 1  . Thus, f or e xample, the cases ( M , N ) = (2 , 3) , (4 , 7) cor respond to Lemma 5.4 (3). (2) Let p = 3 , d = 6 , ξ = 6 , and η = 5 . Then x (1) =  1 1 1 1 1 1 0 0 0 · · ·  , x (3) =  2 2 1 0 0 1 1 1 0 · · ·  . Since 3 , 4 ∈ CI  Λ ⟨ 1 ⟩ , [ 1 0 ]  and η = 5 ∈ CI  Λ ⟨ 1 ⟩ , [ 1 1 ]  , w e see that the cases ( M , N ) = (3 , 5) , (4 , 5) correspond to Lemma 5.4 (2). Lemma 5.6. Let h ∈ N . Suppose that ci  Λ ⟨ h ⟩ , w  ⩾ 2 for w ∈ V ⟨ h ⟩ . Le t M ∈ N \ supp E ( L ⟨ h ⟩ ) and N ∈ supp E ( L ⟨ h ⟩ ) . If M < N , then ξ ⩽ M + 1 , ξ < η , M ∈ Θ , and N ∈ Θ . Pr oof. W e may assume that N ∈ CI  Λ ⟨ h ⟩ , e ⟨ h ⟩  and M ∈ CI  Λ ⟨ h ⟩ , 0 0 0  . Since M < N , it f ollow s from Lemma 5.4 that N ∈ Θ and λ η ( p h ) = λ ξ ( p h ) = λ N ( p h ) = 1 . Hence M ∈ Θ since λ M ( p h ) = 0 . W e show that ξ < η . Assume that η < ξ . It f ollo ws from Lemma 5.4 that N = η . Let y = λ ( p h ) − f ξ + f M + f η ; then y < F λ ( p h ) . 19 M N = η ξ λ [ i ] ( p h ) 0 0 1 λ i ( p h ) 0 1 1 y [ i ] 1 1 0 y i 1 1 0 Since M , N ∈ CI  Λ ⟨ h − 1 ⟩ , 0 0 0  , it f ollo ws that λ M ( a ) = λ N ( a ) = 0 f or 0 ⩽ a F λ ( p h ) , a contradiction. Theref ore ξ < η . It remains to sho w that ξ ⩽ M + 1 . If ξ ⩽ M , then it is obvious. Suppose that ξ ⩾ M + 1 . By appl ying Lemma 5.4 f or M and ξ , we see that ξ = M + 1 . Theref ore ξ ⩽ M + 1 . 5.2. Linearity br eaking In this subsection, w e use the f ollowing notation. Let p = 3 and d = 3 k − 1 d ′ with d ′ ⩾ 2 . Let a be the smallest integ er a suc h that x ( a )  = λ ( a ) . It f ollo w s from Theorem 1.4 that a = 3 h + 3 l f or some h and l , where l ⩽ h . Let y = x (3 h + 3 l ) , z = y − x (3 h ) , and N = max supp F ( y − x (3 h ) − x (3 l )) = max  i ∈ N : y [ i ]  = x [ i ] (3 h ) + x [ i ] (3 l )  . By Lemma 2.2 , ( x [ N ] (3 h ) , x [ N ] (3 l )) = (1 , 1) or (2 , 2) . (5.1) Lemma 5.7. Suppose t hat h < k . If N = η , t hen η < ξ . Pr oof. It f ollow s from ( 5.1 ) that x [ N ] (3 h ) = x [ η ] (3 h )  = 0 . By Lemma 5.1 , we see that η < ξ . Lemma 5.8. If h < k , then the minimum distances of L ⟨ h − 1 ⟩ , y and L ⟨ l − 1 ⟩ , z ar e at least d , where L ⟨ h − 1 ⟩ , y and L ⟨ l − 1 ⟩ , z ar e linear codes with g enerator matrices Λ ⟨ h − 1 ⟩ , y and Λ ⟨ l − 1 ⟩ , z , respectiv ely. Pr oof. Let 0 ⩽ a < 3 h + 3 l . Since x ( a ) = λ ( a ) , it f ollow s that d ( y , λ ( a )) = d ( y , x ( a )) ⩾ d f or 0 ⩽ a < 3 h + 3 l . Thus the minimum distance of L ⟨ h − 1 ⟩ , y is at least d . If 0 ⩽ a < 3 l , then a ⊕ 3 3 h < 3 l + 3 h , and hence d ( z , λ ( a )) = d  y − λ (3 h ) , λ ( a )  = d  y , λ ( a ⊕ 3 3 h )  ⩾ d. Theref ore the minimum dis tance of L ⟨ l − 1 ⟩ , z is at least d . Lemma 5.9. If h < k , then supp E ( y ) ⊆ supp E ( L ⟨ h ⟩ ) and supp E ( z ) ⊆ supp E ( L ⟨ l ⟩ ) . Pr oof. Theorem 1.10 sho ws that L ⟨ k − 1 ⟩ meets the Griesmer bound. It f ollo ws from Corollar y 3.3 that ci  Λ ⟨ k − 1 ⟩ , v  = d ′ f or v ∈ V ⟨ k − 1 ⟩ . Hence ci  Λ ⟨ h ⟩ , v  = 3 k − h − 1 d ′ f or v ∈ V ⟨ h ⟩ . Note that Λ ⟨ h ⟩ = X ⟨ h ⟩ and y [ i ] = x [ i ] (3 h ) + x [ i ] (3 l ) , z [ i ] = x [ i ] (3 l ) (5.2) 20 Y uki Irie f or i > N . By ( 5.1 ), ( x [ N ] (3 h ) , x [ N ] (3 l )) = (1 , 1) or (2 , 2) . Thus N ∈ supp F ( L ⟨ l ⟩ ) ⊆ supp E ( L ⟨ l ⟩ ) by R emark 4.3 . No w assume that there e xists M ∈ supp E ( z ) \ supp E ( L ⟨ l ⟩ ) . W e see that M  = N and M ∈ supp F ( L ⟨ l ⟩ ) . Case 1 ( N < M ) . N ote that z M  = 0 . Since M > N , it f ollo ws that z [ M ] = x [ M ] (3 l ) = 0 . Hence M = η and z ξ = z η  = 0 . Since η = M ∈ supp E ( L ⟨ l ⟩ ) it f ollo w s from Lemma 4.2 that ξ ∈ supp E ( L ⟨ l ⟩ ) , so x ξ (3 l ) = x [ ξ ] (3 l ) = 0  = z [ ξ ] . By the deﬁnition of N , w e see that ξ < N . Since ξ ∈ supp E ( L ⟨ l ⟩ ) and N ∈ supp E ( L ⟨ l ⟩ ) , it f ollo ws from Lemma 5.6 that ξ ∈ Θ , a contradiction. Case 2 ( N > M ) . It f ollow s from Lemma 5.6 that ξ ⩽ M + 1 , ξ < η , and N ∈ Θ . By Lemma 5.7 , N = ξ . For w = y , z , w e will compute   supp E ( L ⟨ h − 1 ⟩ , w ) \ supp E ( L ⟨ h − 1 ⟩ )   (5.3) in tw o w ay s, where L ⟨ h − 1 ⟩ , w is the code with generator matr ix Λ ⟨ h − 1 ⟩ , w . In the ﬁrs t w ay , the results will be at least 3 k − h − 1 d ′ , but in the second w ay , one of them will be less than 3 k − h − 1 d ′ . Step 1. W e sho w that l = h and N ∈ CI  X ⟨ h ⟩ , e ⟨ h ⟩  . (5.4) From Lemma 2.3 , it suﬃces to sho w that N = max { i ∈ N : x [ i ] (3 l )  = 0 } . For i > N , w e see that x [ i ] (3 l ) = 0 . Indeed, if i = η , then x [ η ] (3 l ) = 0 since ξ < η . Suppose that i  = η . Since M ∈ supp E ( L ⟨ l ⟩ ) , it follo ws from Lemma 5.6 that supp E ( L ⟨ l ⟩ ) \ Θ ⊆ N ξ = N , it f ollo w s from ( 5.2 ) and Lemma 5.1 that w [ η ] = ( x [ η ] (3 h ) + x [ η ] (3 l ) = 0 if w = y x [ η ] (3 l ) = 0 if w = z . Hence w η = w ξ . If w = y , then w ξ = y N = 0 b y ( 5.9 ). Suppose that w = z . Then y ξ = y N  = 0 . Since y ξ < x ξ (3 h ) + x ξ (3 l ) = 1 + 1 , it f ollo ws that y ξ = 1 , and hence z ξ = 0 . Theref ore w ξ = w η = 0 . By ( 5.8 ), supp E ( L ⟨ h − 1 ⟩ , w ) \ supp E ( L ⟨ h − 1 ⟩ ) ⊆ CI  X ⟨ h ⟩ , e ⟨ h ⟩  ∪ { N − 1 } \ { ξ , η } . From ( 5.4 ), we see that ξ , η ∈ CI  Λ ⟨ h ⟩ , e ⟨ h ⟩  , and hence   supp E ( L ⟨ h − 1 ⟩ , w ) \ supp E ( L ⟨ h − 1 ⟩ )   ⩽ 3 k − h − 1 d ′ + 1 − 2 < 3 k − h − 1 d ′ , contrary to ( 5.5 ). Theref ore we conclude that supp E ( z ) ⊆ supp E ( L ⟨ l ⟩ ) and supp E ( y ) ⊆ supp E ( L ⟨ h ⟩ ) . Corollary 5.10. If h < k , then L ⟨ h − 1 ⟩ , y sp and L ⟨ l − 1 ⟩ , z sp meet the Griesmer bound. 22 Y uki Irie Pr oof. By Lemma 5.8 , the minimum distances of L ⟨ h − 1 ⟩ , y and L ⟨ l − 1 ⟩ , z are at leas t d . Lemma 5.9 sho ws that supp E ( L ⟨ h − 1 ⟩ , y ) ⊆ supp E ( L ⟨ h ⟩ ) and supp E ( L ⟨ l − 1 ⟩ , z ) ⊆ supp E ( L ⟨ l ⟩ ) . Since L ⟨ h ⟩ sp and L ⟨ l ⟩ sp meet the Griesmer bound b y Theorem 1.10 , it f ollo ws that L ⟨ h − 1 ⟩ , y sp and L ⟨ l − 1 ⟩ , z sp also attain the Griesmer bound. Lemma 5.11. Suppose that h < k . Let V ⟨ h − 1 ⟩ , y = ( V ⟨ h ⟩ if l < h, V ⟨ h ⟩ ∪ { 2 e ⟨ h ⟩ } \ { e ⟨ h ⟩ } if l = h, and V ⟨ h − 1 ⟩ , y 0 0 0 = V ⟨ h − 1 ⟩ , y ∪ { 0 0 0 ⟨ h ⟩ } . If w α ∈ F ⟨ h ⟩ p \ V ⟨ h − 1 ⟩ , y 0 0 0 , then CI  X ⟨ h − 1 ⟩ , y , w α  = ∅ . In particular , ci  X ⟨ h − 1 ⟩ , y , v  = 3 k − h − 1 d ′ f or v ∈ V ⟨ h − 1 ⟩ , y . Pr oof. If w ∈ V ⟨ h − 1 ⟩ 0 0 0 , then CI  X ⟨ h − 1 ⟩ , y , w α  ⊆ CI  X ⟨ h − 1 ⟩ , w  = ∅ . Suppose that w = 0 0 0 ∈ F ⟨ h − 1 ⟩ p . Assume that there e xists L ∈ CI  X ⟨ h − 1 ⟩ , y , 0 0 0 α  . Since w α  = 0 0 0 , it f ollo ws that L ∈ supp E ( y ) ⊆ supp E ( L ⟨ h ⟩ ) . Theref ore L ∈ CI  X ⟨ h ⟩ , e ⟨ h ⟩  since L ∈ CI  X ⟨ h − 1 ⟩ , 0 0 0  . Moreo v er , z L = y L − x L ( p h ) = α − 1 . Case 1 ( l < h ) . Since 0 0 0 α ∈ V ⟨ h ⟩ 0 0 0 , we see that α = 2 . Hence z L = α − 1 = 1 and L ∈ supp E ( z ) . Since L ∈ CI  X ⟨ h − 1 ⟩ , 0 0 0  and l < h , it f ollow s that L ∈ supp E ( L ⟨ l ⟩ ) , how ev er , supp E ( z ) ⊆ supp E ( L ⟨ l ⟩ ) , a contradiction. Case 2 ( l = h ) . Since 0 0 0 α ∈ V ⟨ h − 1 ⟩ , y 0 0 0 , w e see that α = 1 . Hence z L = α − 1 = 0 and L ∈ supp E ( z ) . Since l = h and L ∈ CI  X ⟨ h ⟩ , e ⟨ h ⟩  , it f ollo ws that L ∈ supp E ( L ⟨ h − 1 ⟩ , z ) and L ∈ supp E ( L ⟨ h ⟩ ) , contrary to supp E ( L ⟨ h − 1 ⟩ , z ) = supp E ( L ⟨ h ⟩ ) . Theref ore CI  X ⟨ h − 1 ⟩ , y , w α  = ∅ f or w α ∈ F ⟨ h ⟩ p \ V ⟨ h − 1 ⟩ , y 0 0 0 . Since   V ⟨ h − 1 ⟩ , y   =   V ⟨ h ⟩   , it f ollow s from Corollary 3.3 that ci  X ⟨ h − 1 ⟩ , y , v  = 3 k − h − 1 d ′ f or v ∈ V ⟨ h − 1 ⟩ , y . W e will e xamine the components of y and z , ultimately leading to a contradiction. Lemma 5.12. Suppose t hat h < k . (1) Le t 0 ⩽ m ⩽ h and w ∈ V ⟨ m ⟩ 0 0 0 . If CI  X ⟨ m ⟩ , w  ⊆ Z >N , then y i = x i (3 h ) + x i (3 l ) f or i ∈ CI  X ⟨ m ⟩ , w  . (2) If CI  X ⟨ h ⟩ , v α  ⊆ Z >N , then CI  X ⟨ h ⟩ , v α  = CI  X ⟨ h − 1 ⟩ , y , v α + γ  , where γ is the l -th component of v α ; in par ticular , y i  = α + γ for i ∈ CI  X ⟨ h − 1 ⟩ , v  ∩ Z ⩽ N . (3) If CI  X ⟨ l ⟩ , v α  ⊆ Z >N , then CI  X ⟨ l ⟩ , v α  = CI  X ⟨ l − 1 ⟩ , z , v α  ; in particular , z i  = α f or i ∈ CI  X ⟨ l − 1 ⟩ , v  ∩ Z ⩽ N . 23 Pr oof. (1) This is ob vious when i  = η . Suppose that i = η . W e ﬁrst sho w that ξ > N . If ξ > η , then ξ > η = i > N . Suppose that ξ < η . R ecall that X ⟨ h ⟩ = Λ ⟨ h ⟩ . It follo ws from Theorem 1.10 and Corollar y 3.3 that ci  X ⟨ m ⟩ , w  = 3 k − m − 1 d ′ ⩾ 2 . Hence X ⟨ m ⟩ ξ = X ⟨ m ⟩ η = w b y Lemma 5.1 . Thus ξ ∈ CI  X ⟨ m ⟩ , w  ⊆ Z >N and ξ > N . Theref ore y η = y [ η ] + y [ ξ ] = x [ η ] (3 h ) + x [ η ] (3 l ) + x [ ξ ] (3 h ) + x [ ξ ] (3 l ) = x η (3 h ) + x η (3 l ) . (2) Corollary 5.10 sho ws that L ⟨ h − 1 ⟩ , y sp meets the Gr iesmer bound. By (1), we see that y i = x i (3 h ) + x i (3 l ) = α + γ f or i ∈ CI  X ⟨ h ⟩ , v α  . Theref ore i ∈ CI  X ⟨ h − 1 ⟩ , y , v α + γ  , that is, CI  X ⟨ h ⟩ , v α  ⊆ CI  X ⟨ h − 1 ⟩ , y , v α + γ  . Since L ⟨ h − 1 ⟩ , y sp meets the Gr iesmer bound, it f ol- lo w s from Corollar y 3.2 that CI  X ⟨ h ⟩ , v α  = CI  X ⟨ h − 1 ⟩ , y , v α + γ  . In par ticular , if i ∈ CI  X ⟨ h − 1 ⟩ , v  ∩ Z ⩽ N , then i ∈ CI  X ⟨ h − 1 ⟩ , y , v α + γ  , so y i  = α + γ . (3) Corollary 5.10 sho w s that L ⟨ l − 1 ⟩ , z sp meets the Griesmer bound. By (1), w e see that z i = x i (3 l ) = α f or i ∈ CI  X ⟨ l ⟩ , v α  . Theref ore CI  X ⟨ l ⟩ , v α  ⊆ CI  X ⟨ l − 1 ⟩ , z , v α  . Since L ⟨ l − 1 ⟩ , z sp meets the Gr iesmer bound, it f ollo w s that CI  X ⟨ l ⟩ , v α  = CI  X ⟨ l − 1 ⟩ , z , v α  . Lemma 5.13. Suppose that h < k . (1) y N  = x N (3 h ) + x N (3 l ) . (2) If N ∈ CI  X ⟨ h ⟩ , v β  , t hen ther e exist M ∈ N and α ∈ F p satisfying the f ollowing conditions: M ∈ CI  X ⟨ h ⟩ , v α  , α > β , M < N , and y M = x N (3 h ) + x N (3 l ) . Pr oof. (1) This is obvious when N  = η since y [ N ]  = x [ N ] (3 h ) + x [ N ] (3 l ) . Suppose that N = η . Lemma 5.7 show s that N = η < ξ , and hence y ξ = x [ ξ ] (3 h ) + x [ ξ ] (3 l ) . Theref ore y N = y [ η ] + y [ ξ ] = y [ η ] + x [ ξ ] (3 h ) + x [ ξ ] (3 l )  = x [ η ] (3 h ) + x [ η ] (3 l ) + x [ ξ ] (3 h ) + x [ ξ ] (3 l ) = x η (3 h ) + x η (3 l ) . (2) Let γ = x N (3 l ) ; then y N  = β + γ b y (1). N ote that CI  X ⟨ h − 1 ⟩ , v  = [ δ ∈ F 3 CI  X ⟨ h ⟩ , v δ  = [ δ ∈ F 3 CI  X ⟨ h − 1 ⟩ , y , v δ  . Step 1. W e sho w that v β + γ ∈ V ⟨ h − 1 ⟩ , y and ci  X ⟨ h − 1 ⟩ , y , v β + γ  = 3 k − h − 1 d ′ . Suppose that v  = 0 0 0 . Since N ∈ CI  X ⟨ h ⟩ , v β  , it f ollo ws that v ∈ V ⟨ h − 1 ⟩ , and hence v β + γ ∈ V ⟨ h − 1 ⟩ , y . Suppose that v = 0 0 0 . Then N ∈ CI  X ⟨ h ⟩ , 0 0 0 β  . Since x [ N ] (3 l ) ∈ { 1 , 2 } b y ( 5.1 ), it f ollo w s that N ∈ supp F ( L ⟨ l ⟩ ) ⊆ supp E ( L ⟨ l ⟩ ) , and hence β = 1 and l = h . Theref ore γ = x N (3 l ) = 1 and v β + γ = 2 e ⟨ h ⟩ ∈ V ⟨ h − 1 ⟩ , y . Lemma 5.11 sho ws that ci  X ⟨ h − 1 ⟩ , y , v β + γ  = 3 k − h − 1 d ′ . 24 Y uki Irie Step 2. W e sho w that there e xist M ∈ N and α ∈ F p satisfying the f ollo wing three conditions. (i) α  = β . (ii) M ∈ CI  X ⟨ h ⟩ , v α  . (iii) M ∈ CI  X ⟨ h − 1 ⟩ , y , v β + γ  , that is, y M = β + γ . By (1), we see that N ∈ CI  X ⟨ h ⟩ , v β  \ CI  X ⟨ h − 1 ⟩ , y , v β + γ  . Thus there e xists M ∈ CI  X ⟨ h − 1 ⟩ , y , v β + γ  \ CI  X ⟨ h ⟩ , v β  since ci  X ⟨ h − 1 ⟩ , y , v β + γ  = ci  X ⟨ h ⟩ , v β  = 3 k − h − 1 d ′ b y Step 1. Let v α = X ⟨ h ⟩ M . Since M ∈ CI  X ⟨ h ⟩ , v β  , w e see that α  = β . Thus M and α satisfy the conditions (i)–(iii). Note that x M (3 h ) + x M (3 l )  = β + γ . (5.10) Indeed, if l < h , then x M (3 l ) = x N (3 l ) = γ since M , N ∈ CI  X ⟨ h − 1 ⟩ , v  , and hence x M (3 h ) + x M (3 l ) = α + γ  = β + γ . If l = h , then γ = x N (3 l ) = x N (3 h ) = β , and hence x M (3 h ) + x M (3 l ) = 2 α  = 2 β = β + γ . Theref ore ( 5.10 ) holds. Step 3. W e show that if M and α satisfy (i)–(iii) and α < β , then N = ξ , ξ < η , and M = N − 1 . Assume that CI  X ⟨ h ⟩ , v α  ⊆ Z >N . Since M ∈ CI  X ⟨ h ⟩ , v α  , it f ollo ws from Lemma 5.12 that y M = x M (3 h ) + x M (3 l ) . Ho w ev er , y M = β + γ b y (iii), contrar y to ( 5.10 ). Hence CI  X ⟨ h ⟩ , v α  ⊆ Z >N . Thus there e xists L ∈ CI  X ⟨ h ⟩ , v α  such that L < N . Since α < β , it f ollo w s from Lemma 5.4 that N ∈ Θ and x [ η ] (3 h ) = 0 . Since x [ N ] (3 h ) ∈ { 1 , 2 } , w e see that N  = η , and hence N = ξ . Lemmas 5.4 and 5.1 impl y that ξ < η and x η (3 h ) = x ξ (3 h ) = β . W e sho w that M < N . Since x M (3 h ) = α and x η (3 h ) = β , it f ollo w s that M  = η . If M > N , then y M = y [ M ] = x [ M ] (3 h ) + x [ M ] (3 l ) = x M (3 h ) + x M (3 l ) , contrary to ( 5.10 ) since y M = β + γ . Theref ore M < N . Lemma 5.4 sho w s that M = N − 1 . Step 4. W e show that if M and α satisfy (i)–(iii) and α < β , then there e xist M ′ and α ′ satisfying (i)–(iii) and M ′  = M ; in particular α ′ > β . By Step 3, w e see that N = ξ and ξ < η . Lemma 5.1 implies that X ⟨ h ⟩ η = X ⟨ h ⟩ ξ = v β . Since η > ξ = N , it f ollow s that y [ η ] = x [ η ] (3 h ) + x [ η ] (3 l ) = 0 + 0 , and that y η = y ξ = y N  = β + γ b y (1). Hence X ⟨ h − 1 ⟩ , y η = X ⟨ h − 1 ⟩ , y ξ  = v β + γ . Theref ore ξ , η ∈ CI  X ⟨ h ⟩ , v β  and ξ , η ∈ CI  X ⟨ h − 1 ⟩ , y , v β + γ  . This implies that the size of CI  X ⟨ h − 1 ⟩ , y , v β + γ  \ CI  X ⟨ h ⟩ , v β  is at least tw o. Hence there exis ts M ′ ∈ CI  X ⟨ h − 1 ⟩ , y , v β + γ  \ CI  X ⟨ h ⟩ , v β  such that M ′  = M . Let v α ′ = X ⟨ h ⟩ M ′ . It f ollo ws from S tep 3 that α ′ > β since otherwise M ′ = N − 1 = M . From S teps 2–4, w e conclude that there e xist α and M satisfying (i)–(iii) and α > β . Step 5. If M < N , then M and α satisfy the conditions. Suppose that M > N . W e sho w that ξ and α satisfy the conditions. Since M > N , w e see that y [ M ] = x [ M ] (3 h ) + x [ M ] (3 l ) . Hence M = η because y M = β + γ  = x M (3 h ) + x M (3 l ) . Since M ∈ CI  X ⟨ h ⟩ , v α  , N ∈ 25 CI  X ⟨ h ⟩ , v β  , α > β , and M > N , it f ollo w s from Lemma 5.4 that x [ M ] (3 h ) = x [ η ] (3 h ) = 0 . If ξ > N , then y M = y η = y [ η ] + y [ ξ ] = x [ η ] (3 h ) + x [ η ] (3 l ) + x [ ξ ] (3 h ) + x [ ξ ] (3 l ) = x η (3 h ) + x η (3 l ) = x M (3 h ) + x M (3 l )  = β + γ , a contradiction. Thus ξ < N < M = η . Lemma 5.1 sho ws that X ⟨ h ⟩ ξ = X ⟨ h ⟩ η = v α . Hence ξ ∈ CI  X ⟨ h ⟩ , v α  . Since y [ η ] = x [ η ] (3 h ) + x [ η ] (3 l ) = 0 , it f ollo w s that y ξ = y η = y M = β + γ . Theref ore ξ and α satisfy the conditions. Corollary 5.14. Suppose that h < k . Let α and β be as in Lemma 5.13 . Then x [ N ] (3 h ) = x [ N ] (3 l ) = 1 , α = 2 , and β = 1 . Pr oof. W e ﬁrst show that x [ N ] (3 h ) = x [ N ] (3 l ) = 1 . By ( 5.1 ), ( x [ N ] (3 h ) , x [ N ] (3 l )) = (1 , 1) or (2 , 2) . Assume that ( x [ N ] (3 h ) , x [ N ] (3 l )) = (2 , 2) . Since x N (3 h ) = β < α , it f ollo ws that β  = 2 . Hence x N (3 h )  = 2 = x [ N ] (3 h ) . This implies that N = η and x ξ (3 h )  = 0 . By Lemma 5.1 , w e see that x [ η ] (3 h ) = 0 when ξ < η , and x [ η ] (3 h )  = p − 1 = 2 when ξ > η , a contradiction. Theref ore ( x [ N ] (3 h ) , x [ N ] (3 l )) = (1 , 1) . W e ne xt sho w that α = 2 and β = 1 . If N  = η , then β = x N (3 h ) = x [ N ] (3 h ) = 1 , and hence α = 2 since α > β . Suppose that N = η . Assume that β = x η (3 h )  = 1 . Since β < α ⩽ 2 , w e see that x η (3 h ) = 0 . Since x [ η ] (3 h ) = 1  = 0 , it f ollo ws from Lemma 5.1 that ξ > η . This implies that x η (3 h ) = 0 < x ξ (3 h ) , contrar y to Lemma 5.1 . Theref ore x η (3 h ) = β = 1 and α = 2 . 5.3. T ernary linear lexicographic codes Theorem 5.15. Let p = 3 , d = 3 k − 1 d ′ , and d ′ ⩾ 2 . Then X ⟨ k − 1 ⟩ is a linear code. Pr oof. Assume that X ⟨ k − 1 ⟩ is not linear . Let h , l , y , z , N , M , α , and β be as in Lemma 5.13 . Then x [ N ] (3 h ) = x [ N ] (3 l ) = 1 , α = 2 , and β = 1 . Let t = X ⟨ l − 1 ⟩ N and v = X ⟨ h − 1 ⟩ N . Case 1 ( N ∈ Θ ) . Note that y M = x N (3 h ) + x N (3 l ) = 2 and M ∈ CI  X ⟨ h ⟩ , v 2  . This implies that z M = y M − x M (3 h ) = 2 − 2 = 0 . Since N ∈ Θ and N ∈ CI  X ⟨ l ⟩ , t 1  , it f ollo ws from Lemma 5.2 that CI  X ⟨ l ⟩ , t 0  ⊆ Z >N . Since M ∈ CI  X ⟨ l − 1 ⟩ , t  ∩ Z γ . Since N = η , N ∈ CI  X ⟨ l ⟩ , t γ +1  , γ < γ + 1 , and x [ N ] (3 l ) = x N (3 l ) = 1 , it f ollo ws from Lemma 5.2 that CI  X ⟨ l ⟩ , t γ  ⊆ Z >N . Lemma 5.12 sho ws that z L  = γ f or L ∈ CI  X ⟨ l − 1 ⟩ , t  ∩ Z ⩽ N . (5.11) Since y M = x N (3 h ) + x N (3 l ) = 1 + γ + 1 = γ + 2 , it f ollow s that z M = y M − x M (3 h ) = γ + 2 − 2 = γ . Ho we ver M ∈ CI  X ⟨ l − 1 ⟩ , t  and M < N , contrary to ( 5.11 ). 26 Y uki Irie Case 3 ( N = ξ ) . Since y N = β < α , w e see that y N = 0 , 1 . Case 3.1 ( y N = 1 ). W e ﬁrs t sho w that CI  X ⟨ l ⟩ , t 0  \ { N − 1 } ⊆ CI  X ⟨ l − 1 ⟩ , z , t 0  . (5.12) Note that z N = y N − x N (3 h ) = 1 − 1 = 0 . By Lemma 5.12 , we see that CI  X ⟨ l ⟩ , t 0  ⊆ Z >N . Since ξ = N ∈ CI  X ⟨ l ⟩ , t 1  , it f ollo ws from Lemma 5.4 that ξ < η and CI  X ⟨ l ⟩ , t 0  \ { N − 1 } ⊆ Z >N . This implies that if i ∈ CI  X ⟨ l ⟩ , t 0  \ { N − 1 } , then i > N , and hence z i = x i ( p l ) = 0 . Theref ore ( 5.12 ) holds. W e ne xt sho w that { ξ , η } ⊆ CI  X ⟨ l − 1 ⟩ , z , t 0  . (5.13) Since η > ξ , w e see that X ⟨ l ⟩ η = X ⟨ l ⟩ ξ = t 1 , that is, η , ξ ∈ CI  X ⟨ l ⟩ , t 1  . (5.14) Since η > ξ = N , it f ollo ws that that z [ η ] = x [ η ] (3 l ) = 0 . Hence z η = z ξ + z [ η ] = 0 + 0 = 0 . Theref ore ( 5.13 ) holds. From ( 5.12 ) – ( 5.14 ), we conclude that ci  X ⟨ l − 1 ⟩ , z , t 0  ⩾ 2 + ci  X ⟨ l ⟩ , t 0  − 1 = 3 k − l − 1 d ′ + 1 , contrary to Corollar ies 5.10 and 3.2 . Case 3.2 ( y N = 0 ). Step 1. W e sho w that ξ < η and CI  X ⟨ l ⟩ , t 0  \ { N − 1 } ⊆ Z >N . (5.15) Let γ = x M (3 l ) . Since M ∈ CI  X ⟨ h ⟩ , v 2  , w e see that γ = 2 when if l = h . Moreov er , since N ∈ CI  X ⟨ h ⟩ , v 2  , it f ollo ws that if l < h , then γ = x M (3 l ) = x N (3 l ) = 1 . Theref ore γ = ( 2 if l = h 1 if l < h. Note that z M = y M − x M (3 h ) = x N (3 h ) + x N (3 l ) − 2 = 1 + 1 − 2 = 0 . Thus M ∈ CI  X ⟨ l − 1 ⟩ , z , t 0  . It f ollo ws that CI  X ⟨ l − 1 ⟩ , z , t 0   = CI  X ⟨ l ⟩ , t 0  since M ∈ CI  X ⟨ l ⟩ , t 0  . By Lemma 5.12 , we see that CI  X ⟨ l ⟩ , t 0  ⊆ Z >N . Thus there exis ts L ∈ CI  X ⟨ l ⟩ , t 0  such that L < N . Lemma 5.4 sho w s that L = N − 1 and ξ < η . Moreo v er , ( 5.15 ) holds. 27 Step 2. W e sho w that (1) ξ , η ∈ CI  X ⟨ h − 1 ⟩ , y , v 0  and (2) v  = 0 0 0 ⟨ h − 1 ⟩ (1) Since N = ξ < η , w e see that X ⟨ h ⟩ ξ = X ⟨ h ⟩ η = v 1 . Moreo v er , y [ η ] = x [ η ] (3 h ) + x [ η ] (3 l ) = 0 + 0 , and hence y η = y ξ = 0 . Theref ore ξ , η ∈ CI  X ⟨ h − 1 ⟩ , y , v 0  . (2) If v = 0 0 0 ⟨ h − 1 ⟩ , then N ∈ CI  X ⟨ h − 1 ⟩ , y , 0 0 0 h  and N ∈ supp E ( L ⟨ h − 1 ⟩ , y ) , contrary to N ∈ CI  X ⟨ h ⟩ , e ⟨ h ⟩  ⊆ supp E ( L ⟨ h ⟩ ) = supp E ( L ⟨ h − 1 ⟩ , y ) . Theref ore v  = 0 0 0 ⟨ h − 1 ⟩ . Step 3. W e sho w that CI  X ⟨ h ⟩ , v 0  ⊆ Z >N . Assume that there e xists L ′ ∈ CI  X ⟨ h ⟩ , v 0  ∩ Z ⩽ N . Since N ∈ CI  X ⟨ h ⟩ , v 1  , we see that L ′ < N . Lemma 5.4 sho ws that L ′ = N − 1 , and hence L ′ = L ∈ CI  X ⟨ l ⟩ , t 0  . Assume that l < h . Then x L (3 l ) = v l = x N (3 l ) = 1 . Ho we ver , since X ⟨ l ⟩ L = t 0 , w e see that x L (3 l ) = 0 , a contradiction. Thus l = h . Theref ore CI  X ⟨ h ⟩ , v 0  \ { N − 1 } ∪ { ξ , η } ⊆ CI  X ⟨ h − 1 ⟩ , y , v 0  , contrary to Corollar ies 5.10 and 3.2 . Theref ore CI  X ⟨ h ⟩ , v 0  ⊆ Z >N . Step 4. W e sho w that l < h and   CI  X ⟨ h − 1 ⟩ , y , v 2  ∩ CI  X ⟨ h ⟩ , v 2    ⩾ 2 . (5.16) From S tep 3 and Lemma 5.12 , w e see that y N  = γ , where γ is the l -th component of v 0 . Since y N = 0 and the h -th component of v 0 is 0, we see that l < h and γ = x N (3 l ) = 1 . Lemma 5.12 sho ws that CI  X ⟨ h − 1 ⟩ , y , v 1  = CI  X ⟨ h ⟩ , v 0  . Theref ore CI  X ⟨ h − 1 ⟩ , v  \ CI  X ⟨ h − 1 ⟩ , y , v 1  = CI  X ⟨ h − 1 ⟩ , v  \ CI  X ⟨ h ⟩ , v 0  = = CI  X ⟨ h − 1 ⟩ , y , v 0  ∪ CI  X ⟨ h − 1 ⟩ , y , v 2  CI  X ⟨ h ⟩ , v 1  ∪ CI  X ⟨ h ⟩ , v 2  . (5.17) From S tep 2, we know that η , ξ ∈ CI  X ⟨ h − 1 ⟩ , y , v 0  . Since η , ξ ∈ CI  X ⟨ h ⟩ , v 1  , it f ollo w s from ( 5.17 ) that ( 5.16 ) holds. Step 5. W e sho w that X ⟨ k − 1 ⟩ is a linear code. Since X ⟨ h ⟩ M = v 2 and y M = x N (3 h ) + x N (3 l ) = 2 , it f ollo w s that M ∈ CI  X ⟨ h − 1 ⟩ , y , v 2  ∩ CI  X ⟨ h ⟩ , v 2  . By ( 5.16 ), we see that there e xists M ′ such that M ′  = M and M ′ ∈ CI  X ⟨ h ⟩ , v 2  ∩ CI  X ⟨ h − 1 ⟩ , y , v 2  . Note that z M = y M − x M (3 h ) = 2 − 2 = 0 , z M ′ = y M ′ − x M ′ (3 h ) = 2 − 2 = 0 . This implies that M , M ′ ∈ CI  X ⟨ l − 1 ⟩ , z , t 0  . W e sho w that CI  X ⟨ l ⟩ , t 0  \ { N − 1 } ⊆ CI  X ⟨ l − 1 ⟩ , z , t 0  . Let L ∈ CI  X ⟨ l ⟩ , t 0  \ { N − 1 } . By S tep 1, L > N . Since X ⟨ l ⟩ η = X ⟨ l ⟩ ξ = X ⟨ l ⟩ N = t 1 , it f ollo w s that L  = η . Thus 28 Y uki Irie z L = z [ L ] = y [ L ] − x [ L ] (3 h ) = x [ L ] (3 h ) + x [ L ] (3 l ) − x [ L ] (3 h ) = x [ L ] (3 l ) . Since X ⟨ l ⟩ L = t 0 , it f ollow s that x [ L ] (3 l ) = 0 , and that z L = x [ L ] (3 l ) = 0 . Theref ore  CI  X ⟨ l ⟩ , t 0  \ { N − 1 }  ∪ { M , M ′ } ⊆ CI  X ⟨ l − 1 ⟩ , z , t 0  . Since M , M ′ ∈ CI  X ⟨ l ⟩ , t 1  , it f ollo w s that ci  X ⟨ l − 1 ⟩ , z , t 0  ⩾ ci  X ⟨ l ⟩ , t 0  − 1 + 2 = 3 k − l − 1 + 1 , a contradiction. Theref ore w e conclude that X ⟨ k − 1 ⟩ is a linear code. 6. Solomon-Stiﬄer codes Let p = 3 and d = 3 k − 1 d ′ = 3 k − 1 (3 q + r ) , where 0 < r < 3 . In the ne xt section, w e pro v e that x (3 k + 1)  = x (3 k ) + x (1) when q + r ⩾ 2 ; in par ticular , X ⟨ k ⟩ is not a linear code. T o this end, w e study x (3 k ) in this section. Note that x ( a ) = λ ( a ) f or 0 ⩽ a ⩽ 3 k . 6.1. π d -distributed matrices Example 6.1. Let p = 3 , F = E , and d = 15 . Then X ⟨ 2 ⟩ = Λ ⟨ 2 ⟩ =   11111 11111 11111 00000 00000 · · · 22222 11111 00000 11111 00000 · · · 22110 22100 21100 21100 11000 · · ·   . As w e hav e sho wn in Section 5 , ci  X ⟨ 1 ⟩ , v  = 5 f or v ∈ V ⟨ 1 ⟩ . Obser v e that ci  X ⟨ 2 ⟩ , h 1 2 2 i = ci  X ⟨ 2 ⟩ , h 1 2 1 i = 2 and ci  X ⟨ 2 ⟩ , h 1 2 0 i = 1 . A ctually , f or v ∈ V ⟨ 2 ⟩ , ci  X ⟨ 2 ⟩ , v  = ( 1 if v • 1 1 1 = 0 , 2 if v • 1 1 1  = 0 . Based on the abov e e xample, we introduce the f ollo wing notation. Let d = p k − 1 d ′ = p k − 1 ( pq + r ) , where 0 ⩽ r < p . For 0 ⩽ l ⩽ k and v ∈ F ⟨ l ⟩ p \ { 0 0 0 } , we deﬁne π d ( v ) as f ollow s: if v ∈ V ⟨ l ⟩ , then π d ( v ) = 0 ; if v ∈ V ⟨ l ⟩ , then π d ( v ) =     d p l  if l < k or v • 1 1 1  = 0 , q + r − p + 1 if l = k and v • 1 1 1 = 0 . Note that if v ∈ V ⟨ k − 1 ⟩ , then X α ∈ F p π d ( v α ) = q + r − p + 1 + ( p − 1)  d p k  = pq + r = π d ( v ) . 29 Let x i ∈ F N p f or 0 ⩽ i ⩽ l and X =  x 0 x 1 . . . x l  ∈ F ⟨ l ⟩× N p . F or W ⊆ F ⟨ l ⟩ p \ { 0 0 0 } , the matrix X is said to be π d -distribut ed on W if ci ( X , v ) = π d ( v ) f or v ∈ W . If X is π d -distributed on F ⟨ l ⟩ p \ { 0 0 0 } , then X is said to be π d -distributed. Example 6.2 . Let p = 3 and d = 15 = 3 · (3 · 1 + 2) . Then q = 1 and r = 2 , so q + 1 = 2 and q + r − p + 1 = 1 . Thus X ⟨ 2 ⟩ is π 15 -distributed. Our goal is to sho w the ne xt result. Proposition 6.3. Let d = p k − 1 ( pq + r ) . If q + r − p + 1 ⩾ 2 , t hen Λ ⟨ k ⟩ is π d -distribut ed. Note that a π d -distributed matr ix g enerates a code with minimum dis tance d because this code is a Solomon-S tiﬄer code [ 6 ]. Thus the f ollo wing lemma holds. Lemma 6.4. Let d = p k − 1 d ′ = p k − 1 ( pq + r ) , 0 ⩽ l ⩽ k , and X =  x 0 x 1 . . . x l  ∈ F ⟨ l ⟩× N p . Let C be the code g ener ated by the r ow s of X . If X is π d -distribut ed, then the minimum distance of C equals d and wt( x i ) = d f or 0 ⩽ i ⩽ l . 6.2. Restriction For Φ ⊆ N and y ∈ F N p , let y   [Φ; F ] = X i ∈ Φ y [ i ; F ] ( f i )   Φ ∈ F Φ p . For simplicity , we wr ite y   [Φ] instead of y   [Φ; F ] . For h ∈ N , let Φ h = N \ supp E ( x ( p h )) =  i ∈ N : λ i ( p h ) = 0  . (6.1) The ne xt lemma allo ws an inductiv e approach. Lemma 6.5. Let h ∈ N . If ci  Λ ⟨ h ⟩ , v  ⩾ 2 for v ∈ V ⟨ h ⟩ , then y   Φ h = y   [Φ h ] f or y ∈ F N p . Pr oof. Let Φ = Φ h . It suﬃces to sho w that ( y   Φ ) i = ( y   [Φ] ) i f or i ∈ Φ . By deﬁnition, ( y   Φ ) i = y i = ( y [ η ] + y [ ξ ] if i = η , y [ i ] otherwise , ( y   [Φ] ) i = ( y [ η ] + y [ ξ ] if i = η and ξ ∈ Φ , y [ i ] otherwise . Thus w e ma y assume that i = η and ξ ∈ Φ . Since i ∈ Φ and ξ ∈ Φ , it f ollow s that λ η ( p h ) = 0 and λ ξ ( p h )  = 0 . In par ticular , λ η ( p h ) < λ ξ ( p h ) , whic h contradicts Lemma 5.1 . 30 Y uki Irie For Φ ⊆ N , let F   Φ = ( f i   Φ ) i ∈ Φ . Then F   Φ is a basis of F Φ p since, f or i ∈ Φ , f i   Φ = ( e η   Φ + e ξ   Φ if i = η , e i   Φ otherwise . When Φ = { i j : j ∈ N } with i 0 < i 1 < · · · , identifying i j with j allo ws us to treat F   Φ as an element of F . 6.3. Outline of proof of Pr oposition 6.3 W e will pro v e the proposition b y induction on k . By the induction h ypothesis, (D k ) f or 0 ⩽ l ⩽ k − 1 and ˜ F ∈ F , if ord p ( ˜ d ) ⩾ l − 1 , then the matrix Λ ⟨ l ⟩ ; ˜ F, ˜ d is π ˜ d -distributed. For m ⩽ l and w ∈ F ⟨ m ⟩ p , let V ⟨ l ⟩ ( w ) = { v ∈ V ⟨ l ⟩ : v i = w i f or 0 ⩽ i ⩽ m } . For e xample, V ⟨ 1 ⟩ (0) = V ⟨ 1 ⟩ ([0]) =  [ v 0 v 1 ] ∈ V ⟨ 1 ⟩ : v 0 = 0  = { [ 0 1 ] } . Let F ∗ = F   Φ 0 , E ∗ = E   Φ 0 , d ∗ =  d p  . x ∗ ( a ) = x F ∗ ,d ∗ ( a ) and λ ∗ ( a ) = λ F ∗ ,d ∗ ( a ) . W e ﬁrs t sho w that λ ( p l )   Φ 0 = λ ∗ ( p l − 1 ) f or 0 ⩽ l ⩽ k in Lemma 6.8 . This implies that Λ ⟨ k ⟩ is π d -distributed on V ⟨ k ⟩ (0) . By using the restr iction to Φ 1 , w e ne xt sho w that Λ ⟨ k ⟩ is π d -distributed on V ⟨ k ⟩ ([ 1 0 ]) . Using these results, we ﬁnally pro v e that Λ ⟨ k ⟩ is π d -distributed when p = 3 . 6.4. Λ ⟨ k ⟩ is π d -distributed on V ⟨ k ⟩ (0) For W ⊆ F ⟨ m ⟩ p , let V ⟨ l ⟩ ( W ) = [ w ∈ W V ⟨ l ⟩ ( w ) and CI ( X , W ) = [ w ∈ W CI ( X , w ) . Lemma 6.6. Let 1 ⩽ l ⩽ k and X ∈ F ⟨ l ⟩× N p . Suppose that q + r − p + 1 ⩾ 2 . Let W be a subset of V ⟨ l − 1 ⟩ 0 0 0 suc h t hat 0 0 0 ∈ W . If X is π d -distribut ed and X y is π d -distribut ed on V ⟨ l ⟩ ( W ) , then ther e exists ˜ y satisfying the f ollowing two conditions. (1) X ˜ y is π d -distribut ed. 31 (2) supp F ( ˜ y − y ) = supp E ( ˜ y − y ) ⊆ N \ (Θ ∪ CI ( X , W )) ; in par ticular , ˜ y   CI( X, W ) = y   CI( X, W ) . Pr oof. Let n = #  v ∈ F ⟨ l − 1 ⟩ p : v α  = 0 0 0 and ci ( X y , v α )  = π d ( v α ) f or some α ∈ F p  . W e show b y induction on n . If n = 0 , then y satisﬁes the condition. Suppose that n > 0 and ci ( X y , v α )  = π d ( v α ) . W e ﬁrst sho w that v ∈ V ⟨ l − 1 ⟩ . Since v ∈ W , w e see that v  = 0 0 0 . Moreo v er , ci ( X , w ) = 0 f or w ∈ V ⟨ l − 1 ⟩ 0 0 0 since X is π d -distributed, and hence ci ( X y , w α ) = 0 = π d ( w α ) . Theref ore v ∈ V ⟨ l − 1 ⟩ . Since X is π d -distributed, it f ollo w s that ci ( X , v ) = p k − l ( pq + r ) . Note that q + r − p + 1 ⩾ 2 and q + 1 ⩾ 2 . It f ollo ws from Lemma 4.6 that there e xists z satisfying the follo wing tw o conditions: • ci ( X z , v α ) = π d ( v α ) f or α ∈ F p . • supp F ( z − y ) = supp E ( z − y ) ⊆ CI ( X , v ) \ Θ . Since v ∈ W , w e see that z   CI( X, w ) = y   CI( X, w ) f or w ∈ W ; in par ticular , ci ( X z , w α ) = ci ( X y , w α ) = π d ( w α ) . Hence X z is π d -distributed on V ⟨ l ⟩ ( W ∪ { v } ) . By the induction h ypothesis, there exis ts ˜ z such that X ˜ z is π d -distributed and supp F ( ˜ z − z ) = supp E ( ˜ z − z ) ⊆ N \ (Θ ∪ CI ( X , W ∪ { v } )) ⊆ N \ (Θ ∪ CI ( X , W )) . Since v ∈ W , it f ollo w s that CI ( X , v ) ⊆ N \ CI ( X , W ) , and that supp F ( ˜ z − y ) ⊆ supp F ( z − y ) ∪ supp F ( ˜ z − z ) ⊆  CI ( X , v ) \ Θ  ∪  N \ (Θ ∪ CI ( X , W ))  = N \ (Θ ∪ CI ( X, W )) . Thus supp F ( ˜ z − y ) = supp E ( ˜ z − y ) . Theref ore ˜ z satisﬁes the conditions. Lemma 6.7. Let 1 ⩽ l ⩽ k and v ∈ F ⟨ l − 1 ⟩ p \ { 0 0 0 } . Suppose that w is a vect or in F ⟨ l ⟩ p obtained fr om v by inserting 0, t hat is, w = [ v 0 ··· v i − 1 0 v i ··· v l − 1 ] ⊺ f or some i . Then π d ∗ ( v ) = π d ( w ) . 32 Y uki Irie Pr oof. If v ∈ V ⟨ l − 1 ⟩ , then w ∈ V ⟨ l ⟩ , and hence π d ∗ ( v ) = 0 = π d ( w ) . Suppose that v ∈ V ⟨ l − 1 ⟩ . Then w ∈ V ⟨ l ⟩ . Since l ⌈ d p ⌉ p l − 1 m = l d p l m , it f ollo ws that π d ∗ ( v ) =    l d p l m if l − 1 < k − 1 or v • 1 1 1  = 0 q + r − p + 1 otherwise = π d ( w ) . Lemma 6.8 ( Φ 0 lemma) . Suppose that q + r − p + 1 ⩾ 2 . If (D k ) holds, then λ ( p l )   Φ 0 = λ ∗ ( p l − 1 ) f or 0 ⩽ l ⩽ k , wher e λ ∗ ( p − 1 ) = λ ∗ (0) . Moreo v er , Λ ⟨ k ⟩ is π d -distribut ed on V ⟨ k ⟩ (0) . Pr oof. W e sho w the lemma b y induction on l . If l = 0 , then λ ( p 0 )   Φ 0 = 0 0 0 = λ ∗ (0) . Suppose that 1 ⩽ l ⩽ k . Step 1. W e sho w λ ( p l )   Φ 0 ⩾ F ∗ λ ∗ ( p l − 1 ) . Since Λ ⟨ k − 1 ⟩ is π d -distributed, w e see that ci  Λ ⟨ 0 ⟩ , 1  = d , so the weight of λ (1) eq uals d . This sho ws that the code generated by λ ( p 1 )   Φ 0 , . . . , λ ( p l )   Φ 0 , which is the residual code of Λ ⟨ l ⟩ with respect to λ (1) , has minimum distance at leas t d ∗ since d ∗ =  d p  ⩽ d −  ( p − 1) d p  . By the induction hypothesis, λ ( p i )   Φ 0 = λ ∗ ( p i − 1 ) f or 0 ⩽ i ⩽ l − 1 . Since L ⟨ l − 1 ⟩ sp meets the Griesmer bound, it f ollo ws that ci  Λ ⟨ l ⟩ , e ⟨ l ⟩  = | supp E ( L ⟨ l ⟩ ) | − | supp E ( L ⟨ l − 1 ⟩ ) | ⩾ 1 , and hence ci  Λ ⟨ l ⟩ , e ⟨ l ⟩  ⩾ 1 . This implies that λ ( p l +1 )   Φ 0 is not generated b y λ ∗ ( p i ) f or 0 ⩽ i ⩽ l − 2 . Thus λ ( p l )   Φ 0  = λ ∗ ( a ) ( 0 ⩽ a < p l − 1 ). Theref ore λ ( p l )   Φ 0 ⩾ F ∗ λ ∗ ( p l − 1 ) . Let y = X i ∈ Φ 0 λ ∗ [ i ] ( p l − 1 ) f i . Lemma 6.5 show s that y   Φ 0 = y   [Φ 0 ] = λ ∗ ( p l − 1 ) . Step 2. W e sho w that Λ ⟨ l − 1 ⟩ , y is π d -distributed on V ⟨ l ⟩ (0) ; note that this implies that we can appl y Lemma 6.6 to Λ ⟨ l − 1 ⟩ , y since Λ ⟨ l − 1 ⟩ is π d -distributed. Let Λ ∗⟨ l − 1 ⟩ = Λ ⟨ l − 1 ⟩ ; F ∗ ,d ∗ . It f ollo w s from (D k ) that Λ ∗⟨ l − 1 ⟩ is d ∗ -distributed. Thus f or v ∈ V ⟨ l − 1 ⟩ , ci  Λ ∗⟨ l − 1 ⟩ , v  = π d ∗ ( v ) . W e see that CI  Λ ∗⟨ l − 1 ⟩ , v  = CI  Λ ⟨ l − 1 ⟩ , y , [ 0 v ]  . 33 Indeed, CI  Λ ⟨ l − 1 ⟩ , y , [ 0 v ]  = { i ∈ N : [ λ i (1) λ i ( p ) ··· λ i ( p l − 1 ) y i ] = [ 0 v 0 ··· v l − 2 v l − 1 ] } = { i ∈ Φ 0 : [ λ ∗ i (1) ··· λ ∗ i ( p l − 2 ) λ ∗ i ( p l − 1 ) ] = [ v 0 ··· v l − 2 v l − 1 ] } = CI  Λ ∗⟨ l − 1 ⟩ , v  . Lemma 6.7 yields ci  Λ ⟨ l − 1 ⟩ , y , [ 0 v ]  = ci  Λ ∗⟨ l − 1 ⟩ , v  = π d ∗ ( v ) = π d ([ 0 v ]) . Theref ore Λ ⟨ l − 1 ⟩ , y is π d -distributed on V ⟨ l ⟩ (0) . Step 3. W e show that λ ( p l )   Φ 0 = λ ∗ ( p l − 1 ) . By Step 1, λ ( p l )   Φ 0 ⩾ F ∗ λ ∗ ( p l − 1 ) . Assume that λ ( p l )   Φ 0 > F ∗ λ ∗ ( p l − 1 ) . It f ollo w s from S tep 2 and Lemma 6.6 that there exis ts z satisfying the f ollo wing tw o condi- tions: (1) Λ ⟨ l − 1 ⟩ , z is π d -distributed. (2) supp E ( z − y ) = supp F ( z − y ) ⊆ N \ (Φ 0 ∪ Θ) = CI  Λ ⟨ 0 ⟩ , 1  \ Θ . In particular , z   Φ 0 = y   Φ 0 = λ ∗ ( p l − 1 ) . It f ollo ws from Lemma 6.4 that d ( z , λ ( a )) ⩾ d. (0 ⩽ a F λ ( p l ) . Let N = max { i ∈ N : z [ i ]  = λ [ i ] ( p l ) } and M = max { i ∈ Φ 0 : z [ i ]  = λ [ i ] ( p l ) } . Note that z [ N ] > λ [ N ] ( p l ) . Moreo v er , z [ M ] < λ [ M ] ( p l ) since z   Φ 0 = λ ∗ ( p l − 1 ) < F ∗ λ ( p l )   Φ 0 . In par ticular , M < N and N ∈ Φ 0 . Theref ore N ∈ CI  Λ ⟨ 0 ⟩ , 1  . Moreo v er , N ∈ CI  Λ ⟨ 0 ⟩ , 1  \ Θ . Indeed, since supp F ( y ) ⊆ Φ 0 , it f ollo ws that y [ N ] = 0 . Since z [ N ]  = 0 , we see that N ∈ supp F ( z − y ) ⊆ CI  Λ ⟨ 0 ⟩ , 1  \ Θ . Theref ore N ∈ CI  Λ ⟨ 0 ⟩ , 1  \ Θ and M ∈ Φ 0 . It f ollo w s from Lemma 5.2 that M > N , whic h is impossible. Theref ore λ ( p l )   Φ 0 = λ ∗ ( p l − 1 ) . Finall y , since y   Φ 0 = λ ∗ ( p l − 1 ) = λ ( p l )   Φ 0 , it f ollow s from S tep 2 that Λ ⟨ l ⟩ is π d -distributed on V ⟨ l ⟩ (0) . 34 Y uki Irie 6.5. Λ ⟨ k ⟩ is π d -distributed on V ⟨ k ⟩ ([ 1 0 ]) W e introduce the f ollo wing notation. F ♯ = F   Φ 1 , E ♯ = E   Φ 1 , d ♯ = d ∗ =  d p  , x ♯ ( a ) = x F ♯ ,d ♯ ( a ) , λ ♯ ( a ) = λ F ♯ ,d ♯ ( a ) , and X ♯ ⟨ l ⟩ = X ⟨ l ⟩ ; F ♯ ,d ♯ . For simplicity , we wr ite CI (Λ , v ) ins tead of CI  Λ ⟨ l ⟩ , v  f or v ∈ F ⟨ l ⟩ p Lemma 6.9. Let k ⩾ 2 . Suppose t hat q + r − p + 1 ⩾ 2 . If (D k ) holds, then λ ( p 0 ) | Φ 1 = λ ♯ ( p 0 ) . Pr oof. The proof will be divided into f our s teps. Step 1. W e sho w that wt( λ ( p 0 )   Φ 1 ) = d ♯ ; in par ticular , λ ( p 0 )   Φ 1 ⩾ F ♯ λ ♯ ( p 0 ) . Note that supp E ( λ ( p 0 )   Φ 1 ) = CI (Λ , [ 1 0 ]) . It f ollo ws from (D k ) that Λ ⟨ 1 ⟩ is π d -distributed, and hence ci (Λ , [ 1 0 ]) = d ♯ . Theref ore wt( λ ( p 0 )   Φ 1 ) = d ♯ . This yields λ ( p 0 )   Φ 1 ⩾ F ♯ λ ( p 0 ) . Assume that λ ( p 0 )   Φ 1 > F ♯ λ ( p 0 ) . Let y = X i ∈ N \ Φ 1 λ [ i ] ( p 0 ) f i + X i ∈ Φ 1 λ ♯ [ i ] ( p 0 ) f i . Step 2. W e sho w that wt( y ) < d . By Lemma 6.5 , y   Φ 1 = y   [Φ 1 ] = X i ∈ Φ 1 λ ♯ [ i ] ( p 0 ) f i   Φ 1 = λ ♯ ( p 0 ) . Note that supp F ( y − λ ( p 0 )) ⊆ Φ 1 . Since y   Φ 1 = λ ♯ ( p 0 ) < F ♯ λ ( p 0 )   Φ 1 , it f ollo ws that y < F λ ( p 0 ) . Hence wt( y ) < d . Step 3. W e sho w that ξ ∈ Φ 1 , η ∈ N \ Φ 1 , λ [ ξ ] ( p 0 ) = 1 , and λ ♯ [ ξ ] ( p 0 ) = 0 . Note that wt( y ) < d and wt( λ ( p 0 )) = d . From Step 1, wt( λ ( p 0 )   Φ 1 ) = d ♯ . Moreo v er , wt( y   Φ 1 ) = wt( λ ♯ ( p 0 )) = d ♯ . Hence y   N \ Φ 1  = λ ( p 0 )   N \ Φ 1 . Since y   [ N \ Φ 1 ] = λ ( p 0 )   [ N \ Φ 1 ] , it f ollo ws that y   N \ (Φ 1 ∪{ η } ) = λ ( p 0 )   N \ (Φ 1 ∪{ η } ) . Theref ore η ∈ N \ Φ 1 , ξ ∈ Φ 1 , y η = 0 , and λ η ( p 0 ) = 1 . W e see that y η = y [ η ] + y [ ξ ] = λ [ η ] ( p 0 ) + λ ♯ [ ξ ] ( p 0 ) = 0 35 and λ η ( p 0 ) = λ [ η ] ( p 0 ) + λ [ ξ ] ( p 0 ) = 1 . Since y [ η ] = λ [ η ] ( p 0 ) ∈ { 0 , 1 } and λ ♯ [ ξ ] ( p 0 ) , λ [ ξ ] ( p 0 ) ∈ { 0 , 1 } , it f ollo w s that λ [ η ] ( p 0 ) = λ ♯ [ ξ ] ( p 0 ) = 0 and λ [ ξ ] ( p 0 ) = 1 . Step 4. W e ﬁnall y sho w that λ ( p 0 )   Φ 1 = λ ♯ ( p 0 ) . Since λ ξ ( p 0 ) = 1 and ξ ∈ Φ 1 , it f ollo ws that ξ ∈ CI (Λ , 1) ∩ Φ 1 = CI (Λ , [ 1 0 ]) . Thus λ ξ ( p ) = 0 . Since λ η ( p 0 ) = 1 and η ∈ Φ 1 , w e see that η ∈ CI  Λ ,  [ 1 1 ] , . . .  1 p − 1   . Note that ci (Λ , v ) = p k − 2 ( pq + r ) ⩾ 2 f or v ∈ V ⟨ 1 ⟩ since Λ ⟨ 1 ⟩ is π d -distributed. If ξ < η , then Lemma 5.1 sho ws that λ η ( p ) = λ ξ ( p ) = 0 , a contradiction. Thus η < ξ . Let z = y + f η . Then z < F λ ( p 0 ) because z [ ξ ] = y [ ξ ] = 0 , λ [ ξ ] ( p 0 ) = 1 , and ξ > η . Moreov er , z   N \ Φ 1 = λ ( p 0 )   N \ Φ 1 since z   N \ (Φ 1 ∪{ η } ) = y   N \ (Φ 1 ∪{ η } ) = λ ( p 0 )   N \ (Φ 1 ∪{ η } ) and z η = 1 = λ η ( p 0 ) . It f ollow s that wt( z ) = wt( z   N \ Φ 1 ) + wt( z   Φ 1 ) = wt( λ ( p 0 )   N \ Φ 1 ) + wt( y   Φ 1 ) = d − d ♯ + d ♯ = d. Hence z ⩾ F x ( p 0 ) , a contradiction. Theref ore λ ( p 0 )   Φ 1 = λ ♯ ( p 0 ) . Let λ ( a ; F, d ) = λ F,d ( a ) . Lemma 6.10. Let k ⩾ 2 . Suppose t hat q + r − p + 1 ⩾ 2 . If (D k ) holds, then λ ( p l )   Φ 1 = λ ♯ ( p l − ) f or 0 ⩽ l ⩽ k , wher e l − =      0 l = 0 , − 1 l = 1 , l − 1 l ⩾ 2 . Pr oof. W e sho w the lemma b y induction on l . If l = 0 , then this f ollo w s from the pre vious lemma. If l = 1 , then λ ( p 1 )   Φ 1 = 0 0 0 = λ ♯ (0) . Suppose that 2 ⩽ l ⩽ k . Then l − = l − 1 . Step 1. W e sho w that λ ( p l )   Φ 1 ⩾ F ♯ λ ♯ ( p l − 1 ) . By Lemma 6.5 , λ ( p l )   Φ 1 = λ ( p l )   [Φ 1 ] = X i ∈ Φ 1 λ [ i ] ( p l ) f i   Φ 1 . Note that N \ Φ 1 = { i ∈ N : λ i ( p )  = 0 } = supp E ( λ ( p )) . Since Λ ⟨ 1 ⟩ is π d -distributed, it f ollow s from Lemma 6.4 that wt( λ ( p )) = d . Theref ore the code g enerated b y λ ( p 0 )   Φ 1 , λ ( p 2 )   Φ 1 , 36 Y uki Irie λ ( p 3 )   Φ 1 , . . . , λ ( p l − 1 )   Φ 1 , λ ( p l )   Φ 1 has minimum distance at least d ♯ since d ♯ = d p ⩽ d − j ( p − 1) d p k . By the induction h ypothesis, we see that λ ( p 0 )   Φ 1 = λ ♯ ( p 0 ) and λ ( p l )   Φ 1 = λ ♯ ( p i − 1 ) f or 2 ⩽ i ⩽ l − 1 . Since ci  Λ , e ⟨ l ⟩  ⩾ 1 and CI  Λ , e ⟨ l ⟩  ⊆ Φ 1 , it f ollo w s that λ ( p l )   Φ 1  = λ ♯ ( a ) f or 0 ⩽ a < p l − 1 . Theref ore λ ( p l )   Φ 1 ⩾ F ♯ λ ♯ ( p l − 1 ) . Step 2. W e sho w that λ ♯ ( p l − 1 )   CI ( Λ ♯ , 0 ) = λ ∗ ( p l − 1 )   CI(Λ ∗ , 0) (6.2) and λ [ i ] ( p l ) = λ ∗ [ i ] ( p l − 1 ) = λ ♯ [ i ] ( p l − 1 ) f or i ∈ CI (Λ , [ 0 0 ]) . (6.3) Note that d ♯ = p k − 2 ( pq + r ) . Let F ∗ ♯ = F   Φ 0 ∩ Φ 1 = F   CI ( Λ , [ 0 0 ]) and d ∗ ♯ =  d p 2  . Since (D k ) holds, we see that (D k − 1 ) holds. By Lemma 6.8 , λ ♯ ( p l − 1 )   CI ( Λ ♯ , 0 ) = λ  p l − 1 ; F ♯ , d ♯    CI ( Λ ♯ , 0 ) = λ  p l − 2 ; F ♯   CI ( Λ ♯ , 0 ) , d ∗ ♯  . W e claim that F ♯   CI ( Λ ♯ , 0 ) = F ∗ ♯ . Indeed, it f ollo w s from Lemma 6.9 that CI  Λ ♯ , 0  = { i ∈ Φ 1 : λ ♯ i ( p 0 ) = 0 } = { i ∈ N : λ i ( p ) = λ i ( p 0 ) = 0 } = CI (Λ , [ 0 0 ]) . Hence F ♯   CI ( Λ ♯ , 0 ) =  F   Φ 1     CI ( Λ , [ 0 0 ]) = F   CI ( Λ , [ 0 0 ]) = F ∗ ♯ . Theref ore λ ♯ ( p l − 1 )   CI ( Λ ♯ , 0 ) = λ  p l − 2 ; F ∗ ♯ , d ∗ ♯  . W e ne xt sho w that λ  p l − 2 ; F ∗ ♯ , d ∗ ♯  = λ ∗ ( p l − 1 )   CI(Λ ∗ , 0) . It f ollo ws from Lemma 6.8 that CI (Λ ∗ , 0) = { i ∈ Φ 0 : λ ∗ i ( p 0 ) = 0 } = { i ∈ N : λ i ( p 0 ) = λ i ( p 1 ) = 0 } = CI (Λ , [ 0 0 ]) . Ag ain, Lemma 6.8 implies that λ ∗ ( p l − 1 )   CI(Λ ∗ , 0) = λ  p l − 1 ; F ∗ , d ∗    CI(Λ ∗ , 0) = λ  p l − 2 ; F ∗   CI(Λ ∗ , 0) , d ∗ ♯  = λ  p l − 2 ; F ∗ ♯ , d ∗ ♯  . 37 Theref ore ( 6.2 ) holds. Since CI (Λ ∗ , 0) = CI (Λ , [ 0 0 ]) , it f ollo ws from Lemma 6.5 that λ ∗ ( p l − 1 )   CI(Λ ∗ , 0) = λ ∗ ( p l − 1 )   [CI ( Λ , [ 0 0 ]) ] = X i ∈ CI ( Λ , [ 0 0 ]) λ ∗ [ i ] ( p l − 1 )  f i   CI(Λ , 0)     CI ( Λ , [ 0 0 ]) = X i ∈ CI ( Λ , [ 0 0 ]) λ ∗ [ i ] ( p l − 1 ) f i   CI ( Λ , [ 0 0 ]) and λ ♯ ( p l − 1 )   CI ( Λ ♯ , 0 ) = λ ♯ ( p l − 1 )   [CI ( Λ , [ 0 0 ]) ] = X i ∈ CI ( Λ , [ 0 0 ]) λ ♯ [ i ] ( p l − 1 ) f i   CI ( Λ , [ 0 0 ]) . By ( 6.2 ), λ ♯ [ i ] ( p l − 1 ) = λ ∗ [ i ] ( p l − 1 ) f or i ∈ CI (Λ , [ 0 0 ]) . It f ollo ws from Lemmas 6.5 and 6.8 that λ ( p l )   [CI(Λ , 0)] = λ ( p l )   CI(Λ , 0) = λ ∗ ( p l − 1 ) . Theref ore ( 6.3 ) holds. Step 3. Let y = X i ∈ Φ 1 λ ♯ [ i ] ( p l − 1 ) f i + X i ∈ CI ( Λ , [ 0 1 ]) λ [ i ] ( p l ) f i W e sho w the f ollo wing. supp F ( y ) ⊆ Φ 0 ∪ Φ 1 . y   Φ 1 = λ ♯ ( p l − 1 ) . y   Φ 0 = λ ∗ ( p l − 1 ) . By the deﬁnition of y , w e see that supp F ( y ) ⊆ Φ 0 ∪ Φ 1 . Lemma 6.5 sho ws that y   Φ 1 = y   [Φ 1 ] = X i ∈ Φ 1 λ ♯ [ i ] ( p l − 1 ) f i   Φ 1 = λ ♯ ( p l − 1 ) . From S tep 2, y = X i ∈ Φ 1 λ ♯ [ i ] ( p l − 1 ) f i + X i ∈ CI ( Λ , [ 0 1 ]) λ [ i ] ( p l ) f i = X i ∈ CI ( Λ , [ 1 0 ]) λ ♯ [ i ] ( p l − 1 ) f i + X i ∈ CI ( Λ , [ 0 0 ]) λ ♯ [ i ] ( p l − 1 ) f i + X i ∈ CI ( Λ , [ 0 1 ]) λ [ i ] ( p l ) f i = X i ∈ CI ( Λ , [ 1 0 ]) λ ♯ [ i ] ( p l − 1 ) f i + X i ∈ CI ( Λ , [ 0 0 ]) λ [ i ] ( p l ) f i + X i ∈ CI ( Λ , [ 0 1 ]) λ [ i ] ( p l ) f i = X i ∈ CI ( Λ , [ 1 0 ]) λ ♯ [ i ] ( p l − 1 ) f i + X i ∈ Φ 0 λ [ i ] ( p l ) f i . 38 Y uki Irie Theref ore y   Φ 0 = y   [Φ 0 ] = X i ∈ Φ 0 λ [ i ] ( p l ) f i   Φ 0 = λ ( p l )   [Φ 0 ] = λ ( p l )   Φ 0 = λ ∗ ( p l − 1 ) . Step 4. W e sho w that Λ ⟨ l − 1 ⟩ , y is π d -distributed on V ⟨ l ⟩ (0) ∪ V ⟨ l ⟩ ([ 1 0 ]) . W e ﬁrst sho w that Λ ⟨ l − 1 ⟩ , y is π d -distributed on V ⟨ l ⟩ (0) . Lemma 6.8 implies that Λ ⟨ l ⟩ is π d -distributed on V ⟨ l ⟩ (0) . Since y   CI(Λ , 0) = λ ( p l )   CI(Λ , 0) , it f ollo ws that CI  Λ ⟨ l − 1 ⟩ , y , [ 0 v ]  = CI  Λ ⟨ l ⟩ , [ 0 v ]  f or [ 0 v ] ∈ V ⟨ l ⟩ (0) . Theref ore Λ ⟨ l − 1 ⟩ , y is π d -distributed on V ⟨ l ⟩ (0) . W e ne xt sho w that Λ ⟨ l − 1 ⟩ , y is π d -distributed on V ⟨ l ⟩ ([ 1 0 ]) . For w ∈ F ⟨ l − 2 ⟩ p , ci  Λ ⟨ l − 1 ⟩ , y , h 1 0 w i = ci  Λ ♯ ⟨ l − 1 ⟩ , [ 1 w ]  . Indeed, CI  Λ ⟨ l − 1 ⟩ , y , h 1 0 w i = { i ∈ N : [ λ i ( p 0 ) λ i ( p 1 ) λ i ( p 2 ) ··· λ i ( p l − 1 ) y i ] = [ 1 0 w ] } = { i ∈ CI (Λ , [ 1 0 ]) : [ λ i ( p 0 ) λ i ( p 2 ) ··· λ i ( p l − 1 ) y i ] = [ 1 w ] } = { i ∈ CI (Λ , [ 1 0 ]) : [ λ ♯ i ( p 0 ) λ ♯ i ( p 1 ) ··· λ ♯ i ( p l − 2 ) λ ♯ i ( p l − 1 ) ] = [ 1 w ] } = CI  Λ ♯ ⟨ l − 1 ⟩ , [ 1 w ]  . Since Λ ♯ ⟨ l − 1 ⟩ is π d ♯ -distributed, it f ollo ws from Lemma 6.7 that Λ ⟨ l − 1 ⟩ , y is π d -distributed on V ⟨ l ⟩ ([ 1 0 ]) . Step 5. W e ﬁnally sho w that λ ( p l )   Φ 1 = λ ♯ ( p l − 1 ) . From Step 1, λ ( p l )   Φ 1 ⩾ F ♯ λ ♯ ( p l − 1 ) . As- sume that λ ( p l )   Φ 1 > F ♯ λ ♯ ( p l − 1 ) . From Step 4, Λ ⟨ l − 1 ⟩ , y is π d -distributed on V ⟨ l ⟩ (0) ∪ V ⟨ l ⟩ ([ 1 0 ]) . Let W = V ⟨ l − 1 ⟩ 0 0 0 (0) ∪ V ⟨ l − 1 ⟩ ([ 1 0 ]) . Then V ⟨ l ⟩ ( W ) = V ⟨ l ⟩ (0) ∪ V ⟨ l ⟩ ([ 1 0 ]) and CI (Λ , W ) = CI (Λ , 0) ∪ CI (Λ , [ 1 0 ]) . It f ollo ws Lemma 6.6 that there exis ts z satisfying the follo wing tw o conditions. (1) Λ ⟨ l − 1 ⟩ , z is π d -distributed. (2) supp E ( z − y ) = supp F ( z − y ) ⊆ N \ (CI (Λ , W ) ∪ Θ) = CI  Λ ,  [ 1 1 ] , . . . ,  1 p − 1   \ Θ . In particular , 39 z   Φ 1 = y   Φ 1 = λ ♯ ( p l − 1 )  = λ ( p l )   Φ 1 , z   Φ 0 = y   Φ 0 = λ ∗ ( p l − 1 ) = λ ( p l )   Φ 0 . Since Λ ⟨ l − 1 ⟩ , z is π d -distributed, it f ollow s that d ( z , λ ( a )) ⩾ d f or 0 ⩽ a F λ ( p l ) . Let N = max { i ∈ N : z [ i ]  = λ [ i ] ( p l ) } and M = max { i ∈ Φ 1 : z [ i ]  = λ [ i ] ( p l ) } . Since z   [Φ 0 ] = z   Φ 0 = λ ( p l )   Φ 0 = λ ( p l )   [Φ 0 ] , it f ollo w s that N , M ∈ Φ 0 . In par ticular , M ∈ Φ 1 \ Φ 0 = CI (Λ , [ 1 0 ]) . W e sho w that M < N . Since z > F λ ( p l ) , it f ollow s that z [ N ] > λ [ N ] ( p l ) . R ecall that z   Φ 1 = λ ♯ ( p l − 1 ) < F ♯ λ ( p l )   Φ 1 . This implies that z [ M ] < λ [ M ] ( p l ) . Hence M < N . W e next sho w that M > N . Recall that N ∈ Φ 0 . Moreo v er , N ∈ Φ 1 because if N ∈ Φ 1 , then N = M . Since supp F ( y ) ⊆ Φ 0 ∪ Φ 1 , it f ollo ws that y [ N ] = 0 . Thus N ∈ supp F ( z − y ) ⊆ CI  Λ ,  [ 1 1 ] , . . . ,  1 p − 1   \ Θ . Since M ∈ CI (Λ , [ 1 0 ]) , it follo ws from Lemma 5.2 that M > N , a contradiction. Theref ore λ ( p l )   Φ 1 = λ ♯ ( p l − 1 ) . Pr oof of Pr oposition 6.3 . Theorem 1.10 sho w s that the code L ⟨ k − 1 ⟩ sp meets the Gr iesmer bound. It f ollow s from Corollar y 3.3 that Λ ⟨ k − 1 ⟩ is π d -distributed. W e sho w that Λ ⟨ k ⟩ is π d -distributed b y induction on k . Case 1 ( k = 1 ) . Note that d = pq + r . Let n be the length of L ⟨ 1 ⟩ sp . Then n = ci (Λ , 1) + ci (Λ , [ 0 1 ]) = d + ci (Λ , [ 0 1 ]) = pq + r + ci (Λ , [ 0 1 ]) . Step 1. W e ﬁrst show that ci (Λ , [ 0 1 ]) = q + 1 . By the Gr iesmer bound, n ⩾ d + ⌈ d p ⌉ = pq + r + q + 1 , and hence ci (Λ , [ 0 1 ]) ⩾ q + 1 . Assume that ci (Λ , [ 0 1 ]) > q + 1 . Since q + 1 ⩾ 2 , there e xists M ∈ CI (Λ , [ 0 1 ]) \ Θ . Let y = λ ( p ) − f M = λ ( p ) − e M . Note that y [ M ] = λ [ M ] ( p ) − 1 = 0 and ci  Λ ⟨ 0 ⟩ , y , [ 0 1 ]  = ci (Λ , [ 0 1 ]) − 1 ⩾ q + 1 . Since q + 1 , q + r − 2 ⩾ 2 , it f ollo ws from Lemma 4.6 that there e xists z satisfying the f ollo wing conditions. ci  Λ ⟨ 0 ⟩ , z , [ 1 α ]  = π d ([ 1 α ]) = ( q + 1 if α  = p − 1 , q + r − p + 1 if α = p − 1 . supp E ( y − z ) = supp F ( y − z ) ⊆ CI (Λ , 1) \ Θ . W e claim that z < F λ ( p ) . Since M ∈ CI (Λ , [ 0 1 ]) , w e see that z [ M ] = y [ M ] = 0 < 1 = λ [ M ] ( p ) . For i ∈ supp E ( y − z ) ⊆ CI (Λ , 1) \ Θ , w e see that i < M b y Lemma 4.5 . Hence z < F λ ( p ) . Moreo v er , since ci  Λ ⟨ 0 ⟩ , z , [ 0 1 ]  = ci  Λ ⟨ 0 ⟩ , y , [ 0 1 ]  ⩾ q + 1 and Λ ⟨ 0 ⟩ , z is π d -distributed on V ⟨ 1 ⟩ (1) , it follo ws from Lemma 6.4 that d ( z , λ ( a )) ⩾ d f or 0 ⩽ a < p . Hence z ⩾ F λ ( p ) , which is impossible. Theref ore ci (Λ , [ 0 1 ]) = q + 1 . In particular , L ⟨ 1 ⟩ sp meets the Gr iesmer bound. 40 Y uki Irie Step 2. W e ne xt sho w that ci  Λ ,  1 p − 1  = q + r − p + 1 . Assume that ci  Λ ,  1 p − 1  > q + r − p + 1 . Since q + 1 , q + r − p + 1 ⩾ 2 , it f ollo ws from Lemma 4.6 that there e xist y satisfying the f ollo wing conditions. ci  Λ ⟨ 0 ⟩ , y , [ 1 α ]  = π d ([ 1 α ]) = ( q + 1 if α  = p − 1 , q + r − p + 1 if α = p − 1 . supp E ( y − λ ( p )) = supp F ( y − λ ( p )) ⊆ CI (Λ , 1) \ Θ . Since d = pq + r < p ( q + 1) , it f ollo w s from Corollar y 3.2 that ci (Λ , [ 1 0 ]) ⩽ q + 1 . Hence w e ma y assume that supp E ( y − λ ( p )) = supp F ( y − λ ( p )) ⊆ CI  Λ ,  1 p − 1  \ Θ . For i ∈ supp F ( y − λ ( p )) , w e see that λ [ i ] ( p ) = p − 1 , so y < F λ ( p ) . Ho w ev er , d ( y , λ ( a )) ⩾ d f or 0 ⩽ a < p , which is impossible. Theref ore ci  Λ ,  1 p − 1  = q + r − p + 1 and ci (Λ , [ 1 0 ]) = ci (Λ , [ 1 1 ]) = q + 1 because P ci (Λ , [ 1 α ]) = pq + r and ci (Λ , [ 1 α ]) ⩽ q + 1 . Case 2 ( k ⩾ 2 ) . Step 1. W e ﬁrst sho w that Λ ⟨ k ⟩ is π d -distributed on V ⟨ k ⟩ (0) ∪ V ⟨ k ⟩ ([ 1 0 ]) . By the induction hypothesis, (D k ) holds. It f ollo ws from Lemmas 6.8 and 6.10 that Λ ⟨ k ⟩ is π d -distributed on V ⟨ k ⟩ (0) ∪ V ⟨ k ⟩ ([ 1 0 ]) . Step 2. For α , β ∈ F p \ { 0 } , let b ( α, β ) = α + β p + p k − p 2 = α + β p + ( p − 1) p 2 + · · · + ( p − 1) p k − 1 . Let Ω = V ⟨ k − 1 ⟩  [ 1 1 ] , [ 1 2 ] , . . . ,  1 p − 1   . T o pro v e Λ ⟨ k ⟩ is π d -distributed on V ⟨ k ⟩ ([ 1 1 ]) , we sho w that d ( λ ( p k ) , λ ( b ( α , β )) = p k − 2  ( p 2 + p − 1) q + ( p + 1) r − 1  − X v ∈ Ω ci  Λ ⟨ k ⟩ , v v • b ( α,β )  . Let a = b ( α, β ) . By Lemma 4.1 , d ( λ ( p k ) , λ ( a )) = X v ∈ V ⟨ k ⟩ ,v k  = v • a ci  Λ ⟨ k ⟩ , v  = X v ∈ V ⟨ k ⟩ ci  Λ ⟨ k ⟩ , v  − X v ∈ V ⟨ k − 1 ⟩ ci  Λ ⟨ k ⟩ , v v • a  = ( pq + r ) p k − 1 p − 1 + q + 1 − X v ∈ V ⟨ k − 1 ⟩ ci  Λ ⟨ k ⟩ , v v • a  = p k p − 1 ( pq + r ) − q + r − p + 1 p − 1 − X v ∈ V ⟨ k − 1 ⟩ ci  Λ ⟨ k ⟩ , v v • a  . 41 For v ∈ V ⟨ k − 1 ⟩ (0) ∪ V ⟨ k − 1 ⟩ ([ 1 0 ]) , w e calculate ci  Λ ⟨ k ⟩ , v v • a  . From Steps 1 and 2, Λ ⟨ k ⟩ is π d -distributed on V ⟨ k − 1 ⟩ (0) ∪ V ⟨ k − 1 ⟩ ([ 1 0 ]) , and hence ci  Λ ⟨ k ⟩ , v v • a  is determined b y v v • a • 1 1 1 . Since v • a = v 0 α + v 1 β − v 2 − · · · − v k − 1 , it f ollo ws that v v • a • 1 1 1 = (1 + α ) v 0 + (1 + β ) v 1 . Since ( v 0 , v 1 ) ∈ { (0 , 0) , (0 , 1) , (1 , 0) } and α, β ∈ F p \ p − 1 , it f ollo w s that v v • a • 1 1 1 = 0 ⇐ ⇒ v 0 = v 1 = 0 . Theref ore X v ∈ V ⟨ k − 1 ⟩ (0) ∪ V ⟨ k − 1 ⟩ ([ 1 0 ]) ci  Λ ⟨ k ⟩ , v v • a  = ( q + r − p − 1) | V ⟨ k − 1 ⟩ ([ 0 0 ]) | + ( q + 1)  | V ⟨ k − 1 ⟩ ([ 0 1 ] , [ 1 0 ]) |  = ( q + r − p + 1) p k − 2 − 1 p − 1 + ( q + 1)2 p k − 2 = p k − 2 p − 1  2 pq + p + r − q − 1  − q + r − p + 1 p − 1 . It f ollo ws that d ( λ ( p k ) , λ ( a )) = p k p − 1 ( pq + r ) − q + r − p + 1 p − 1 − X v ∈ V ⟨ k − 1 ⟩ ci  Λ ⟨ k ⟩ , v v • a  = p k − 2  ( p 2 + p − 1) q + ( p + 1) r − 1  − X v ∈ Ω ci  Λ ⟨ k ⟩ , v v • a  . Step 3. Let γ ∈ F p \ { 0 } and D = p − 2 X a 0 =0 p − 2 X a 1 =0 d  λ ( p k ) , λ ( b ( a 0 , a 1 ))  + p − 2 X a 0 =0 d  λ ( p k ) , λ ( b ( a 0 , ( a 0 + 1) γ − 1))  . By calculating D , w e sho w that Λ ⟨ k ⟩ is π d -distributed on V ⟨ k ⟩  1 − γ − 1  . T o this end, it suﬃces to calculate ci  Λ ⟨ k ⟩ , v v • a  f or v ∈ Ω b y S tep 3. Let σ = − v 2 − · · · − v k − 1 . Then v • b ( a 0 , a 1 ) = a 0 + v 1 a 1 + σ. Note that n v • b ( a 0 , a 1 ) : a 1 ∈ F p \ { p − 1 } o = F p \ { a 0 − v 1 + σ } . 42 Y uki Irie Hence p − 2 X a 0 =0 p − 2 X a 1 =0 ci  Λ ⟨ k ⟩ , v v • b ( a 0 ,a 1 )  = ( p − 1) ci  Λ ⟨ k − 1 ⟩ , v  − X a 0 ∈{ 0 , 1 ,...,p − 2 } ci  Λ ⟨ k ⟩ , v a 0 − v 1 + σ  = ( p − 2) ci  Λ ⟨ k − 1 ⟩ , v  + ci  Λ ⟨ k ⟩ , v − 1 − v 1 + σ  . Moreo v er , v • b  a 0 , ( a 0 + 1) γ − 1  = a 0 + v 1  ( a 0 + 1) γ − 1  + σ = a 0 (1 + v 1 γ ) + v 1 γ − v 1 + σ. Let ν = − γ − 1 . Note that if v 1  = ν , then 1 + v 1 γ  = 0 , and hence n v • b  a 0 , ( a 0 + 1) γ − 1  : a 0 ∈ F p \ { p − 1 } o = F p \ { − 1 − v 1 + σ } . If v 1 = ν , then n v • b  a 0 , ( a 0 + 1) γ − 1  : a 0 ∈ F p \ { p − 1 } o = { − 1 − v 1 + σ } . Hence p − 2 X a 0 =0 ci  Λ ⟨ k ⟩ , v v • b ( a 0 , ( a 0 +1) γ − 1)  = ( ci  Λ ⟨ k − 1 ⟩ , v  − ci  Λ ⟨ k ⟩ , v − 1 − v 1 + σ  if v 1  = ν, ( p − 1) ci  Λ ⟨ k ⟩ , v − 1 − v 1 + σ  if v 1 = ν = ( ( p − 1)( pq + r ) − ci  Λ ⟨ k ⟩ , v − 1 − v 1 + σ  if v 1  = ν, ( p − 1) ci  Λ ⟨ k ⟩ , v − 1 − v 1 + σ  if v 1 = ν. Theref ore p − 2 X a 0 =0 p − 2 X a 1 =0 ci  Λ ⟨ k ⟩ , v v • b ( a 0 ,a 1 )  + p − 2 X a 0 =0 ci  Λ ⟨ k ⟩ , v v • b ( a 0 , ( a 0 +1) γ − 1)  = ( ( p − 1)( pq + r ) if v 1  = ν, ( p − 2)( pq + r ) + p ci  Λ ⟨ k ⟩ , v − 1 − v 1 + σ  if v 1 = ν. Note that, f or v ∈ Ω , v − 1 − v 1 + σ • 1 1 1 = − 1 + v 1 + v 2 + · · · + v k − 1 − 1 − v 1 + σ = 0 . 43 Hence v − 1 − v 1 + σ is an element w in V ⟨ k ⟩ ([ 1 v 1 ]) with w • 1 1 1 = 0 . Theref ore D = p − 2 X a 0 =0 p − 2 X a 1 =0 d  λ ( p k ) , λ ( b ( a 0 , a 1 ))  + p − 2 X a 0 =0 d  λ ( p k ) , λ ( b ( a 0 , ( a 0 + 1) γ − 1))  = p k − 1 ( p − 1)  ( p 2 + p − 1) q + ( p + 1) r − 1  − X v ∈ Ω ,v 1  = ν ( p − 1)( pq + r ) − X v ∈ Ω ,v 1 = ν ( p − 2)( pq + r ) + p ci  Λ ⟨ k ⟩ , v − 1 − v 1 + σ  = p k − 1 ( p − 1)  ( p 2 + p − 1) q + ( p + 1) r − 1  − p k − 2 ( p − 1)( p − 2)( pq + r ) − p k − 2 ( p − 2)( pq + r ) − p X w ∈ V ⟨ k ⟩ ([ 1 ν ]) , w • 1 1 1=0 ci  Λ ⟨ k ⟩ , w  = p k − 1  ( p 3 − p 2 + 1) q + ( p 2 − p + 1) r p + 1  − p X w ∈ V ⟨ k ⟩ ([ 1 ν ]) , w • 1 1 1=0 ci  Λ ⟨ k ⟩ , w  ⩾ ( p 2 − p ) d = ( p 2 − p )( pq + r ) . This implies that X w ∈ V ⟨ k ⟩ ([ 1 ν ]) , w • 1 1 1=0 ci  Λ ⟨ k ⟩ , w  ⩽ p k − 2 ( q + r − p + 1) . Since ci  Λ ⟨ k ⟩ , w  ⩾ q + r − p + 1 , it f ollo w s that ci  Λ ⟨ k ⟩ , w  = q + r − p + 1 f or w ∈ V ⟨ k ⟩ ([ 1 ν ]) with w • 1 1 1 = 0 . W r ite w = v β . Then ci  Λ ⟨ k ⟩ , v α  = q + 1 f or α  = β . This implies that Λ ⟨ k ⟩ is π d -distributed on V ⟨ k ⟩ ([ 1 ν ]) . Theref ore Λ ⟨ k ⟩ is π d -distributed. 7. Proof of Theorem 1.7 7.1. Case p = 3 W e show that x (3 k + 1)  = x (3 k ) + x (1) . Let y = x (3 k ) + x (1) . Theorem 5.15 implies that x ( a ) = λ ( a ) f or 0 ⩽ a ⩽ 3 k . Let v ∈ V ⟨ k − 1 ⟩ (1) and β = − v 0 − v 1 − · · · − v k − 1 . Note that ci  X ⟨ k − 1 ⟩ , y , v α  = ( δ + ϵ − 2 if α = β + 1 , δ + 1 if α = β , β + 2 . Since δ + 1 , δ + ϵ − 2 ⩾ 2 , it f ollo ws from Lemma 4.6 that there exis ts ˜ y satisfying the f ollowing tw o conditions: (1) ci  X ⟨ k − 1 ⟩ , ˜ y , v α  = ( δ + ϵ − 2 if α = β , δ + 1 if α = β + 1 , β + 2 . (2) supp F ( ˜ y − y ) = supp E ( ˜ y − y ) ⊆ CI  X ⟨ k − 1 ⟩ , y , v β  \ Θ . 44 Y uki Irie Note that CI  X ⟨ k − 1 ⟩ , y , v β  = CI  X ⟨ k ⟩ , v β +2  . CI  X ⟨ k ⟩ , v β  CI  X ⟨ k ⟩ , v β +2  CI  X ⟨ k ⟩ , v β +1  x (3 k ) β δ + ϵ − 2 ( β + 2) δ +1 ( β + 1) δ +1 y ( β + 1) δ + ϵ − 2 β δ +1 ( β + 2) δ +1 ˜ y ( β + 1) δ + ϵ − 2 ( β + 1) 3 − ϵ β δ + ϵ − 2 ( β + 2) δ +1 By repeatedl y using Lemma 4.6 , we see that there e xists z satisfying the f ollo wing tw o condi- tions: (1) ci  X ⟨ k − 1 ⟩ , z , v α  = ( δ + ϵ − 2 if α = β , δ + 1 if α = β + 1 , β + 2 . (2) supp F ( z − y ) = supp E ( z − y ) ⊆ S v ∈ V ⟨ k − 1 ⟩ (1) CI  X ⟨ k ⟩ , v β +2  \ Θ It suﬃces to sho w that z < F y = x (3 k ) + x (1) and d ( z , x ( a )) ⩾ d f or 0 ⩽ a ⩽ 3 k . Since X ⟨ k − 1 ⟩ , z is π d -distributed, it f ollo ws that d ( z , x ( a )) ⩾ d f or 0 ⩽ a < 3 k . Moreo v er , by the deﬁnition of z , w e see that d  z , x (3 k )  = d . Let N = max { i ∈ N : z [ i ]  = x [ i ] (3 k ) + x [ i ] (1) } . Then N ∈ CI  X ⟨ k − 1 ⟩ , [ 1 0 ··· 0 ] ⊺  \ Θ . CI  X ⟨ k ⟩ , v 2  CI  X ⟨ k ⟩ , v 1  CI  X ⟨ k ⟩ , v 0  x (3 k ) 2 δ + ϵ − 2 1 δ +1 0 δ +1 x (3 k ) + x (1) 0 δ + ϵ − 2 2 δ +1 1 δ +1 z 0 δ + ϵ − 2 0 3 − ϵ 2 δ + ϵ − 2 1 δ +1 Hence z N = 0 < 2 = x N (3 k ) + x N (1) and z < F x (3 k ) + x (1) . Theref ore x (3 k + 1)  = x (3 k ) + x (1) . 7.2. Case p ⩾ 5 , d = 2 Proposition 7.1. If d = 2 and ξ ∈ { 0 , 1 } , then k ( F , d ) = 2 . Pr oof. Let { ξ , ι } = { 0 , 1 } and F ′ =  f ⊺ ξ f ⊺ ι f ⊺ 2 f ⊺ 3  ⊺ . Step 1. W e ﬁrst sho w that x (1) =  1 0 0 0  F ′ . If ξ = 0 , then x (1) = f ξ . Suppose that ξ = 1 . Since x (1)  =  0 α 0 0  F ′ , w e see that x (1) =  1 0 0 0  F ′ = f ξ . 45 Step 2. W e ne xt sho w that x ( a 0 + a 1 p ) =  a 0 a 1 a 1 0  F ′ b y induction on a 1 . Let a 1 = 0 . If a 0 = 1 , then it f ollo ws from Step 1 that x (1) =  1 0 0 0  F ′ . Since the distance between  1 α 0 0  F ′ and x (1) equals one, w e see that x (2) =  2 0 0 0  F ′ . In general, x ( a 0 ) =  a 0 0 0 0  F ′ . Suppose that a 1 ⩾ 1 . F or α < a 1 , b y induction h ypothesis, w e see that x ( µ + αp ) =  µ α α 0  F ′ . N ote that the dis tance betw een  λ µ α 0  F ′ and  λ α α 0  F ′ is at mos t one. Theref ore x ( a 0 + a 1 p ) =  λ µ a 1 0  F ′ . F or α < a 1 , the dis tance betw een  λ α a 1 0  F ′ and x ( α + λp ) =  λ α α 0  F ′ is at most one. Theref ore x ( a 1 p ) =  0 a 1 a 1 0  F ′ . W e see that the distance betw een  a 0 a 1 a 1 0  F ′ and x ( α + λp ) equals tw o. Theref ore x ( a 0 + a 1 p ) =  a 0 a 1 a 1 0  F ′ . Step 3. W e sho w that x ( p 2 ) =  0 1 0 1  F ′ . The distance between  λ µ ν 0  F ′ and x ( λ + µp ) =  λ µ µ 0  F ′ is at most one. The distance between  λ 0 0 1  F ′ and x ( λ ) =  λ 0 0 0  F ′ is at most one Theref ore x ( p 2 ) =  0 1 0 1  F ′ . Step 4. W e sho w that if x ( α + p 2 ) = x ( α ) + x ( p 2 ) f or 1 ⩽ α < p , then x ( p + p 2 )  = x ( p ) + x ( p 2 ) . Note that x ( p ) + x ( p 2 ) =  0 2 1 1  F ′ . W e sho w that the dis tance between  0 0 1 1  F ′ and the f ounded w ords is at least tw o. Indeed, the distance betw een  0 0 1 1  F ′ and x ( a 0 + a 1 p ) =  a 0 a 1 a 1 0  F ′ is at least tw o f or a 1 = 0 , 1 . Moreo v er , the distance betw een  0 0 1 1  F ′ and x ( α + p 2 ) =  α 1 0 1  F ′ is at least tw o. Theref ore x ( p + p 2 )  = x ( p ) + x ( p 2 ) . Proposition 7.2 . If d = 2 and ξ = 2 , then k ( F, d ) = 2 . Pr oof. Let F ′ =  f ⊺ 0 f ⊺ 1 f ⊺ 2 f ⊺ 3  ⊺ . W e ﬁrs t sho w that x ( a 1 p + a 0 ) =  a 0 a 0 a 1 0  F ′ . b y induction on a 1 . Let a 1 = 0 . W e see that x (1) =  1 1 0 0  F . Hence x ( a 0 ) =  a 0 a 0 0 0  F . Let a 1 ⩾ 1 . F or α < a 1 , the distance between  λ µ α 0  F and x ( λ + αp ) =  λ λ α 0  F eq uals one. Thus x ( a 0 + a 1 p ) =  λ µ a 1 0  F . Note that F or α < a 0 , d   λ α a 1 0  F ,  a 0 a 0 a 1 0  F  = 1 . d   α µ a 1 0  F ,  a 0 a 0 a 1 0  F  = 1 . Hence x ( a 0 + a 1 p ) =  a 0 a 0 a 1 0  F . W e ne xt sho w that x ( p 2 ) =  1 0 0 1  F . The distance between  λ µ ν 0  F and x ( λ + ν p ) =  λ λ ν 0  F eq uals one. Thus x ( p 2 ) =  1 0 0 1  F . Finall y , w e show that x (1 + p 2 )  = x (1) + x ( p 2 ) . N ote that x (1) + x ( p 2 ) =  2 1 0 1  F . Since d   0 1 0 1  F , x ( a )  ⩾ 2 f or 0 ⩽ a ⩽ p 2 , 46 Y uki Irie it f ollo ws that x (1 + p 2 )  = x (1) + x ( p 2 ) . Proposition 7.3. If d = 2 and ξ ⩾ 3 , t hen k ( F, d ) = 1 . Pr oof. W e see that x ( α ) = α f 0 + α f 1 f or α ∈ F p and x ( p ) = f 0 + f 2 . Moreo ver , f 1 + f 2 < F f 0 + f 1 + f 2 = x ( p ) + x (1) and d ( f 1 + f 2 , x ( a )) ⩾ 2 f or 0 ⩽ a ⩽ p . Hence x ( p + 1)  = x ( p ) + x (1) . 7.3. Case p ⩾ 5 , d ⩾ 3 Let F ∈ F and d ⩾ 3 . Lemma 7.4. (1) ci ( X , 1) = d . (2) ci ( X , [ 0 1 ]) ⩾ max α ∈ F p ci ( X , [ 1 α ]) . Mor eo ver , if CI ( X , [ 0 1 ]) \ Θ  = ∅ , then ci ( X, [ 0 1 ]) = max α ∈ F p ci ( X , [ 1 α ]) . Pr oof. By deﬁnition, X ⟨ 0 ⟩ = Λ ⟨ 0 ⟩ . Hence Theorem 1.10 sho w s that ci ( X , 1) = d . Since d ( x ( p ) , x ( α )) = ci ( X , 1) + ci ( X , [ 0 1 ]) − ci ( X , [ 1 α ]) ⩾ d , it f ollow s that ci ( X , [ 0 1 ]) ⩾ ci ( X , [ 1 α ]) . Suppose that CI ( X , [ 0 1 ]) \ Θ  = ∅ , and let M ∈ CI ( X , [ 0 1 ]) \ Θ and y = x ( p ) − f M . Since y [ M ] = x [ M ] ( p ) − 1 = 0 , w e see that y < F x ( p ) . Assume that ci ( X , [ 0 1 ]) > max α ∈ F p ci ( X , [ 1 α ]) . Then, f or α ∈ F p , d ( y , x ( α )) = ci ( X , 1) + ci ( X , [ 0 1 ]) − ci ( X , [ α 1 ]) − 1 ⩾ ci ( X , 1) = d. This implies that y ⩾ F x ( p ) , a contradiction. Lemma 7.5. Let α, β ∈ F p with α < β and CI  X ,  1 β   = ∅ . If ther e exist γ suc h that α < γ ⩽ β and CI  X ,  1 γ  \ Θ  = ∅ , then ci ( X , [ 1 α ]) ⩾ ci  X ,  1 β  . Pr oof. Assume that ci ( X , [ 1 α ]) < ci  X ,  1 β  . Let M ∈ CI  X ,  1 γ  \ Θ and y = x ( p ) + ( α − γ ) f M = x ( p ) + ( α − γ ) e M . W e see that y [ M ] = x [ M ] ( p ) + α − γ = γ + α − γ = α < γ = x [ M ] ( p ) . This implies that y < F x ( p ) . Note that d ( y , x ( a )) ⩾ d ( x ( p ) , x ( a )) ⩾ d f or 0 ⩽ a ci ( X , [ 1 α ]) , and hence d ( y , x ( α )) ⩾ d . Theref ore y ⩾ F x ( p ) , whic h is impossible. Lemma 7.6. Let α , β ∈ F p with α < β . If CI  X ,  1 β   = ∅ , then CI ( X , [ 1 α ])  = ∅ . 47 Pr oof. When CI  X ,  1 β  \ Θ  = ∅ , it f ollo ws from Lemma 7.5 that ci ( X , [ 1 α ]) ⩾ ci  X ,  1 β  > 0 . Suppose that CI  X ,  1 β  \ Θ = ∅ , and assume that CI ( X , [ 1 α ]) = ∅ . Let M ∈ CI  X ,  1 β  and y = x ( p ) + ( α − β ) f M . Then M = η or ξ . Note that the f ollo wing s tatements hold. (i) If M = η , then α − x [ ξ ] ( p ) > β − x [ ξ ] ( p ) . (ii) If M = ξ , then ci ( X , [ 0 1 ]) = 1 . Indeed, if M = η , then d ( y , x ( a )) ⩾ d f or 0 ⩽ a F x ( p ) , which implies that y [ M ] = α − x [ ξ ] ( p ) > β − x [ ξ ] ( p ) = x [ M ] ( p ) . If M = ξ , then y < F x ( p ) , and hence d ( y , x ( α )) = ci ( X , 1) + ci ( X , [ 0 1 ]) − ci ( X , [ 1 α ]) − 2 = d + ci ( X , [ 0 1 ]) − 2 < d, which yields ci ( X , [ 0 1 ]) = 1 . W e sho w that ξ ∈ CI ( X , 1) . Assume that ξ ∈ CI ( X , 1) . Then M = η . By (i), α − x [ ξ ] ( p ) > β − x [ ξ ] ( p ) . It f ollo ws that ξ ∈ CI ( X, [ 0 1 ]) , and hence α − 1 > β − 1 . Hence α = 0 and ci ( X , [ 1 0 ]) = 0 . It f ollo w s from Lemma 7.5 that CI ( X , 1) ⊆ Θ , and hence ci ( X , 1) = d = 1 , a contradiction. Hence ξ ∈ CI ( X , 1) . Lemma 4.2 implies that η ∈ CI ( X , 1) . Assume that ξ , η ∈ CI  X ,  1 β  . Then ci ( X , [ 0 1 ]) ⩾ ci  X ,  1 β  ⩾ 2 , contrar y to (ii). Theref ore CI  X ,  1 β  = { ξ } or { η } . Case 1 ( CI  X ,  1 β  = { ξ } ) . Let η ∈ CI  X ,  1 γ  . Let z = x ( p ) + ( α − β ) f ξ + (1 + β − γ ) f η . Then z [ ξ ] = x [ ξ ] ( p ) + α − β = α < β = x [ ξ ] ( p ) and z [ η ] = x [ η ] ( p ) + 1 + β − γ = 1 ⩽ γ − β = x [ η ] ( p ) , and hence z < f x ( p ) . Theref ore d ( z , x ( a )) < d f or some 0 ⩽ a 0 . Thus CI ( X , [ 1 α +1 ]) \ Θ  = ∅ . It f ollo ws from Lemma 7.5 that ci ( X , [ 1 α ]) ⩾ ci  X ,  1 β  ⩾ 1 , a contradiction. Case 2 ( CI  X ,  1 β  = { η } ) . Let ξ ∈ CI  X ,  1 γ  . If γ > α , then the proof f ollo w s from Case 1. Suppose that γ < α < β . Since M = η , it f ollo ws from (i) that α − γ > β − γ , whic h is impossible. Lemma 7.7. If ther e exists α, γ ∈ F p satisfying the f ollowing two conditions, then k ( F , d ) = 1 . (1) γ ⩾ 1 , CI  X ,  1 p − γ  \ Θ  = ∅ . (2) 1 ⩽ α ⩽ p − 2 γ − 1 , CI ( X , [ 1 α ]) \ Θ  = ∅ . 48 Y uki Irie Pr oof. It suﬃces to show that if x ( p + β ) = x ( p ) + x ( β ) f or 1 ⩽ β < γ , then x ( p + γ )  = x ( p ) + x ( γ ) . Let N ∈ CI  X ,  1 p − γ  \ Θ and M ∈ CI ( X , [ 1 α ]) \ Θ . Since α N . N ote that x [ N ] ( p ) + x [ N ] ( γ ) = ( p − γ ) ⊕ p γ = 0 and x [ M ] ( p ) + x [ M ] ( γ ) = α + γ . Let y = x ( p ) + x ( γ ) − ( α + γ ) f M + ( α + γ ) f N . W e see that y [ M ] = 0 < α + γ = x [ M ] ( p ) + x [ M ] ( γ ) , and hence y < F x ( p ) + x ( γ ) . Since 1 ⩽ α ⩽ p − 2 γ − 1 , it f ollo ws that α + γ ⩽ p − γ − 1 ; in par ticular , α + γ  = p − γ , p − γ + 1 , . . . , p − 1 , 0 . Hence d ( y , x ( a )) ⩾ d f or 0 ⩽ a < p + γ . Theref ore x ( p + γ )  = x ( p ) + x ( γ ) . N M x [ i ] (1) 1 1 x [ i ] ( p ) p − γ α x [ i ] ( p ) + x [ i ] (1) p − γ + 1 α + 1 x [ i ] ( p ) + x [ i ] (2) p − γ + 2 α + 2 . . . x [ i ] ( p ) + x [ i ] ( γ − 1) p − 1 α + γ − 1 x [ i ] ( p ) + x [ i ] ( γ ) 0 α + γ y [ i ] α + γ 0 Proposition 7.8. Let p ⩾ 7 . If CI  X ,  1 p − 1  \ Θ  = ∅ , then k ( F , d ) = 1 . Pr oof. Lemma 7.6 implies that CI ( X , [ 1 α ])  = ∅ f or 0 ⩽ α ⩽ p − 2 . Since p ⩾ 7 , it f ollo ws that CI ( X , [ 1 α ]) \ Θ  = ∅ f or some α with 1 ⩽ α ⩽ p − 3 . Lemma 7.7 show s that k ( F , d ) = 1 . Proposition 7.9. Let p ⩾ 7 . If CI  X ,  1 p − 1  = Θ , then k ( F , d ) = 1 . Pr oof. Lemma 7.6 implies that CI ( X , [ 1 α ])  = ∅ f or 0 ⩽ α ⩽ p − 2 . Thus CI  X ,  1 p − 2  \ Θ  = ∅ . Since p ⩾ 7 , we see that p − 5 ⩾ 1 and CI  X ,  1 p − 5  \ Θ  = ∅ . Lemma 7.7 show s that k ( F, d ) = 1 . Proposition 7.10 . Let p ⩾ 11 . If CI  X ,  1 p − 1  ⊆ Θ and ci  X ,  1 p − 1  = 1 , then k ( F, d ) = 1 . Pr oof. Suppose that CI  X ,  1 p − 2  \ Θ  = ∅ . Since CI ( X , [ 1 1 ]) \ Θ  = ∅ or CI ( X , [ 1 2 ]) \ Θ  = ∅ , it f ollo ws from Lemma 7.7 that k ( F , d ) = 1 . Suppose that CI  X ,  1 p − 2  \ Θ = ∅ . Then CI  X ,  1 p − 1  ∪ CI  X ,  1 p − 2  = Θ . Since p ⩾ 11 , w e see that p − 7 ⩾ 1 . Moreo v er , CI  X ,  1 p − 3  \ Θ  = ∅ and CI  X ,  1 p − 7  \ Θ  = ∅ . Lemma 7.7 show s that k ( F , d ) = 1 . Lemma 7.11. If one of the f ollowing two conditions holds, then k ( F , d ) = 1 . (1) ci  X ,  1 p − 1  ⩽ ci ( X , [ 0 1 ]) − 1 and CI ( X , [ 1 α ]) \ Θ  = ∅ f or some 1 ⩽ α ⩽ p − 2 . (2) ci  X ,  1 p − 1  ⩽ ci ( X , [ 0 1 ]) − 2 and CI ( X , [ 1 α ]) = Θ f or some 1 ⩽ α ⩽ p − 2 . 49 Pr oof. When (1) holds, let M ∈ CI ( X , [ 1 α ]) \ Θ . When (2) holds, let M = ξ . Let y = x ( p ) + x (1) − (1 + α ) f M . Then x [ M ] ( p ) + x [ M ] (1) = α + 1 > 0 and y [ M ] = 0 . Thus y < F x ( p ) + x (1) . It suﬃces to sho w that d  y , x ( a )  ⩾ d f or 0 ⩽ a ⩽ p . Firs t, suppose that (1) holds. Since y M = 0 and M ∈ CI ( X, [ 1 α ]) , it f ollo ws that d  y , x ( a )  ⩾ d  x ( p ) + x (1) , x ( a )  ⩾ d f or 1 ⩽ a ⩽ p. Moreo v er , since ci  X ,  1 p − 1  ⩽ ci ( X , [ 0 1 ]) − 1 and CI  X ⟨ 0 ⟩ , y , [ 1 0 ]  = CI  X ,  1 p − 1  ∪ { M } , it f ollo ws that d  y , x (0)  = ci ( X , 1) + ci ( X, [ 0 1 ]) − ci  X ,  1 p − 1  − 1 ⩾ d. Thus d  y , x ( a )  ⩾ d f or 0 ⩽ a ⩽ p . Ne xt, suppose that (2) holds. Since y M = y ξ = y η = 0 and M ∈ CI ( X , [ 1 α ]) , it f ollo ws that d  y , x ( a )  ⩾ d f or 1 ⩽ a ⩽ p . Moreo v er , since ci  X ,  1 p − 1  ⩽ ci ( X , [ 0 1 ]) − 2 and CI  X ⟨ 0 ⟩ , y , [ 1 0 ]  = CI  X ,  1 p − 1  ∪ Θ , it f ollo ws that d  y , x (0)  = ci ( X , 1) + ci ( X, [ 0 1 ]) − ci  X ,  1 p − 1  − 2 ⩾ d. Theref ore d  y , x ( a )  ⩾ d f or 0 ⩽ a ⩽ p . Proposition 7.12 . Let p ⩾ 5 . If CI  X ,  1 p − 1  = ∅ , then k ( F , d ) = 1 . Pr oof. If CI ( X , [ 1 α ]) \ Θ  = ∅ f or some α with 1 ⩽ α ⩽ p − 2 , then Lemma 7.11 sho ws that k ( F , d ) = 1 . Suppose that CI ( X , [ 1 α ]) \ Θ = ∅ f or 1 ⩽ α ⩽ p − 2 . W e ﬁrst sho w that ci ( X , [ 1 0 ]) ⩽ 2 . Suppose that ci ( X , [ 1 0 ]) ⩾ 3 . Then there e xists M ∈ CI ( X , [ 1 0 ]) \ Θ . Since p ⩾ 5 , w e see that there exis ts β suc h that CI  X ,  1 β  = ∅ and 1 ⩽ β ⩽ p − 2 . W e also see that there exis ts N ∈ CI ( X , [ 0 1 ]) \ Θ since ci ( X , [ 0 1 ]) ⩾ CI ( X , [ 1 0 ]) ⩾ 3 . Note that N > M . Let y = x ( p ) + β f M − f N . Then y [ N ] = x [ N ] ( p ) − 1 = 0 , and hence y < F x ( p ) . Moreo v er , d  y , x ( β )  = ci ( X , 1) + ci ( X, [ 0 1 ]) − ci  X ,  1 β  − 1 ⩾ d. Thus d ( y , x ( a )) ⩾ d f or 0 ⩽ a < p, a contradiction. Hence ci ( X , [ 1 0 ]) ⩽ 2 . Since CI ( X , [ 1 α ]) ⊆ Θ f or 1 ⩽ α ⊆ p − 1 , it f ollo ws that d = P p − 1 α =0 ci ( X , [ 1 α ]) ⩽ 4 Thus d = 3 or 4 . Case 1 ( d = 4 ) . Since d = P p − 1 α =0 ci ( X , [ 1 α ]) and CI ( X , [ 1 α ]) ⊆ Θ f or α  = 0 , it f ollo w s that ci ( X , [ 1 0 ]) = 2 and CI ( X , [ 1 1 ]) ∪ CI ( X , [ 1 2 ]) = Θ . Let ci ( X , [ 1 0 ]) = { N , M } and y = f M + f N + f ξ . 50 Y uki Irie Then d ( y , x (0)) = 4 . Since ξ ∈ CI ( X , 1) , it f ollo ws that y = x (1) . Recall that CI ( X , [ 1 1 ]) ∪ CI ( X , [ 1 2 ]) = Θ . Assume that ci ( X , [ 1 2 ]) = 1 . Let CI ( X , [ 0 1 ]) = { i 0 , i 1 } and E ′ =  e ⊺ ξ e ⊺ η e ⊺ N e ⊺ M e ⊺ i 0 e ⊺ i 1  ⊺ . Then w e see that x (1) =  1 1 1 1 0 0  E ′ , x ( p ) =  1 2 0 0 1 1  E ′ . Let y = x ( p ) − f i 1 + 3 f M . Then y =  1 2 3 0 1 0  E ′ . Since i 1 > M , w e see that y < F x ( p ) . Moreo v er , d ( y , x ( a )) ⩾ 4 f or 0 ⩽ a < p , a contradiction. Hence ci ( X , [ 1 2 ]) = 0 and CI ( X , [ 1 1 ]) = Θ . Lemma 7.11 sho w s that k ( F , d ) = 1 . Case 2 ( d = 3 ) . Case 2.1. ci ( X , [ 0 1 ]) = 2 . Since CI ( X , 1) ∩ Θ  = ∅ , w e see that CI ( X , [ 0 1 ]) \ Θ  = ∅ . It f ollo ws from Lemma 7.4 that ci ( X , [ 1 α ]) = 2 f or some α . Assume that ci ( X , [ 1 0 ]) = 2 . Since P α ∈ F p ci ( X , [ 1 α ]) = 3 , it f ollo ws from Lemma 7.6 that ci ( X , [ 1 1 ]) = 1 . Moreo ver , there e xists N ∈ CI ( X , [ 1 0 ]) \ Θ and M ∈ CI ( X , [ 0 1 ]) \ Θ . N ote that N < M . Let CI ( X, [ 1 1 ]) = { i 0 } , CI ( X , [ 1 0 ]) = { N , i 1 } , CI ( X , [ 0 1 ]) = { M , i 2 } , and E ′ =  e ⊺ i 0 e ⊺ N e ⊺ i 1 e ⊺ M e ⊺ i 2  ⊺ . Then x (1) =  1 1 1 0 0  E ′ . x ( p ) =  1 0 0 1 1  E ′ . Let y = x ( p ) + 2 f N − f M . Then y =  1 2 0 0 1  E ′ . Since y [ M ] = 0 and N < M , it f ollo ws that y < F x ( p ) . Moreo v er , d ( y , x ( a )) ⩾ 3 f or 0 ⩽ a < p , a contradiction. Theref ore ci ( X , [ 1 0 ]) = 1 . This implies that ci ( X , [ 1 1 ]) = 2 . Thus CI ( X , [ 1 1 ]) = Θ . Theref ore Lemma 7.11 sho ws that k ( F , d ) = 1 . Case 2.2. ci ( X , [ 0 1 ]) = 1 . It suﬃces to sho w that if x ( p + 1) = x ( p ) + x (1) , then x ( p + 2)  = x ( p ) + x (2) . Since d = 3 , it f ollo ws that ci ( X , [ 1 0 ]) = ci ( X , [ 1 1 ]) = ci ( X , [ 1 2 ]) = 1 . Let CI ( X , [ 1 0 ]) = { N } and CI ( X , [ 0 1 ]) = { M } . N ote that  CI ( X , [ 1 1 ]) , CI ( X , [ 1 2 ])  =  { ξ } , { η }  or  { η } , { ξ }  . W e show that  CI ( X , [ 1 1 ]) , CI ( X , [ 1 2 ])  =  { ξ } , { η }  . If  CI ( X , [ 1 1 ]) , CI ( X , [ 1 2 ])  =  { ξ } , { η }  , then x [ ξ ] ( p ) = 1 and x [ η ] ( p ) = 1 . If  CI ( X , [ 1 1 ]) , CI ( X , [ 1 2 ])  =  { η } , { ξ }  , then x [ ξ ] ( p ) = 2 and x [ η ] ( p ) = p − 1 . Thus CI ( X , [ 1 1 ]) = { ξ } and CI ( X , [ 1 2 ]) = { η } . 51 Let F ′ =  f ⊺ ξ f ⊺ η f ⊺ N f ⊺ M  ⊺ , and E ′ =  e ⊺ ξ e ⊺ η e ⊺ N e ⊺ M  ⊺ . Then x (1) =  1 0 1 0  F ′ =  1 1 1 0  E ′ . x ( p ) =  1 1 0 1  F ′ =  1 2 0 1  E ′ . It f ollo ws that x ( p ) + x (1) =  2 1 1 1  F ′ =  2 3 1 1  E ′ . x ( p ) + x (2) =  3 1 2 1  F ′ =  3 4 2 1  E ′ . Let y =  0 1 2 1  F ′ =  0 1 2 1  E ′ . Then y < F x ( p ) + x (2) and d ( y , x ( a )) ⩾ d f or 0 ⩽ a ⩽ p + 1 . Theref ore x ( p + 2)  = x ( p ) + x (2) . Proposition 7.13. If p ⩾ 5 , t hen k ( F, d ) ⩽ 2 . Pr oof. From Propositions 7.8 – 7.12 , if one of the f ollo wing conditions holds, then k ( F , d ) = 1 . (1) p ⩾ 7 and CI  X ,  1 p − 1  \ Θ  = ∅ . (2) p ⩾ 7 and CI  X ,  1 p − 1  = Θ . (3) p ⩾ 11 and CI  X ,  1 p − 1  = { ξ } or { η } . (4) p ⩾ 5 and CI  X ,  1 p − 1  = ∅ . The remaining cases are as f ollo w s. (1) ′ p = 5 , 7 and CI  X ,  1 p − 1  = { ξ } or { η } . (2) ′ p = 5 and CI  X ,  1 p − 1  \ Θ  = ∅ . (3) ′ p = 5 and CI  X ,  1 p − 1  = Θ . (1) ′ p = 5 , 7 and CI  X ,  1 p − 1  = { ξ } or { η } . Lemma 7.6 show s that CI ( X , [ 1 α ])  = ∅ f or 0 ⩽ α ⩽ p − 1 . Since CI  X ,  1 p − 1  ∩ Θ  = ∅ , it f ollo ws that CI ( X , [ 1 1 ]) \ Θ  = ∅ or CI ( X , [ 1 2 ]) \ Θ  = ∅ . If ci ( X , [ 0 1 ]) ⩾ 2 , then Lemma 7.11 sho w s that k ( F , d ) = 1 . Hence w e ma y assume that ci ( X , [ 0 1 ]) = 1 . Theref ore d = p . By computer , w e see that k ( F , d ) = 1 . (2) ′ p = 5 and CI  X ,  1 p − 1  \ Θ  = ∅ . If CI ( X , [ 1 α ]) \ Θ  = ∅ for some 1 ⩽ α ⩽ p − 3 = 2 , then k ( F, d ) = 1 . Hence we ma y assume that CI ( X , [ 1 1 ]) ∪ CI ( X , [ 1 2 ]) = Θ . Since CI ( X , [ 1 4 ])  = ∅ , it f ollo w s from Lemma 7.6 that ci ( X , [ 1 α ])  = 0 , and hence ci ( X , [ 1 1 ]) = ci ( X , [ 1 2 ]) = 1 . Thus CI ( X , [ 1 1 ]) = { ξ } and CI ( X, [ 1 2 ]) = { η } . N ote that x [ η ] ( p ) = 1 and x η ( p ) = 2 . Assume that ci ( X , [ 0 1 ]) ⩾ 2 . Let y = x ( p ) − f η . Then y [ η ] = 0 , and hence y < F x ( p ) . Moreo v er , d  y , x (1)  = ci ( X , 1) + ci ( X, [ 0 1 ]) − 2 ⩾ d. 52 Y uki Irie Thus d  y , x ( a )  ⩾ d f or 0 ⩽ a < p , a contradiction. Hence ci ( X , [ 0 1 ]) = 1 and d = 5 . By computer , w e see that k ( F, 5) = ( 2 if ξ = 2 and η ∈ { 3 , 4 , 5 } , 1 otherwise . (3) ′ p = 5 and CI  X ,  1 p − 1  = Θ . Lemma 7.6 sho ws that CI ( X , [ 1 α ])  = ∅ f or 0 ⩽ α ⩽ 4 . Note that CI ( X , [ 1 4 ]) = Θ and CI ( X , [ 1 3 ]) \ Θ  = ∅ . Lemma 7.5 sho ws that ci ( X , [ 1 1 ]) ⩾ 2 . Let M , M ′ ∈ CI ( X , [ 1 1 ]) . W e sho w that max { M , M ′ } > ξ . Let y = x ( p ) − 3 f ξ + 3( f M + f ′ M ) . W e see that d ( y , x ( a )) ⩾ d f or 0 ⩽ a ξ . ξ η M M ′ x i (1) 1 1 1 1 x i ( p ) 4 4 1 1 y i 1 1 4 4 Let z = x ( p ) + x (1) + 2 f ξ − 2( f M + f M ′ ) . Then z < F x (1) + x ( p ) and d ( z , x ( a )) ⩾ d f or 0 ⩽ a ⩽ p . Theref ore x ( p + 1)  = x ( p ) + x (1) . ξ η M M ′ x i (1) 1 1 1 1 x i ( p ) 4 4 1 1 x i ( p ) + x i (1) 0 0 2 2 z i 2 2 0 0 R ef erences [1] Jeﬀre y T . Bonn. F orcing Linear ity on Greedy Codes. Designs, Codes and Cr yp togr aphy , 9(1):39–49, 1996. [2] J. H. Con wa y and N. J. A. Sloane. Lexicographic Codes: Er ror -cor recting Codes from Game Theory. IEEE T ransactions on Inf or mation Theor y , 32(3):337–348, 1986. [3] J. H. Conw ay , R. T . Cur tis, S. P . N or ton, R. A. Parker , and R. A. Wilson. Atlas of Finit e Gr oups: Maximal Subgr oups and Ordinary Charact ers for Simple Gr oups . Clarendon Press, Oxf ord, 1986. 53 [4] J. H. Gr iesmer . A Bound f or Er ror -Cor recting Codes. IBM Jour nal of Resear c h and Dev el- opment , 4(5):532–542, 1960. [5] V . I. Le v enshtein. A class of sy stematic codes. Soviet Mat hematics. Doklady , 1:368–371, 1960. [6] G. Solomon and J. J. S tiﬄer . Algebraicall y punctured cy clic codes. Information and Control , 8(2):170–179, 1965. [7] A. J. v an Zanten and I. N engah Supar ta. On the Construction of Linear q-ar y Le xicodes. Designs, Codes and Cryptog raphy , 37(1):15–29, 2005.

On linear lexicographic codes: Ninth column construction of the ternary Golay code

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