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Article

Minimal Linear Codes Constructed from Sunflowers

School of Mathematics, Southeast University, Nanjing 210096, China
*
Author to whom correspondence should be addressed.
Entropy 2023, 25(12), 1669; https://doi.org/10.3390/e25121669
Submission received: 19 November 2023 / Revised: 10 December 2023 / Accepted: 14 December 2023 / Published: 18 December 2023
(This article belongs to the Section Information Theory, Probability and Statistics)

Abstract

:
Sunflower in coding theory is a class of important subspace codes and can be used to construct linear codes. In this paper, we study the minimality of linear codes over F q constructed from sunflowers of size s in all cases. For any sunflower, the corresponding linear code is minimal if s q + 1 , and not minimal if 2 s 3 q . In the case where 3 < s q , for some sunflowers, the corresponding linear codes are minimal, whereas for some other sunflowers, the corresponding linear codes are not minimal.
MSC:
94B05; 94A62

1. Introduction

Let F q be the finite field with q elements and F q n the vector space with dimension n over F q . For a vector v = ( v 1 , , v n ) F q n , let Suppt ( v ) : = { 1 i n : v i 0 } be the support of v . The Hamming weight of v is wt ( v ) :=# Suppt ( v ) . For any two vectors u , v F q n , if Suppt ( u ) Suppt ( v ) , we say that v covers u (or u is covered by v ) and write u v . Clearly, a v v for all a F q .
An [ n , m ] q linear code C over F q is an m-dimensional subspace of F q n . A codeword c in a linear code C is called minimal if c covers only the codewords a c for all a F q , but no other codewords in C . If every codeword in C is minimal, then C is said to be a minimal linear code. Minimal linear codes have interesting applications in secret sharing [1,2,3,4,5] and secure two-party computation [6,7], and could be decoded with a minimum distance decoding method [8].
Up to now, there are two approaches to studying minimal linear codes. One is the algebraic method and the other is the geometric method. The algebraic method is based on the Hamming weights of the codewords. In [8], Ashikhmin and Barg gave a sufficient condition for a linear code to be minimal. Many minimal linear codes satisfying the condition w m i n w m a x > q 1 q are obtained from linear codes with few weights; for example [9,10]. Cohen et al. [7] provided an example to show that the condition w m i n w m a x > q 1 q is not necessary for a linear code to be minimal. Ding, Heng, and Zhou [11,12] derived a sufficient and necessary condition on all Hamming weights for a given linear code to be minimal.
When using the algebraic method to prove the minimality of a given linear code, one needs to know all the Hamming weights in the code, which is very difficult in general. Even if all the Hamming weights are known, it is hard to use the algebraic method to prove the minimality. In this paper, we will use the geometric approaches to study the minimality of some linear codes. Based on the geometric approaches (see [13,14,15]) it is easier to construct minimal linear codes or to prove the minimality of some linear codes (see [16,17,18,19,20,21]).
Sunflower in coding theory is a class of important subspace codes and can be used to construct linear codes, see [22]. Let s be the number of the elements in a sunflower. In [23], (Theorem 10), the authors proved that if s p + 1 , then the corresponding linear code over F p is minimal, where p is a prime number.
In this paper, we will use the approach used in [14] to consider the minimality of linear codes over F q constructed from sunflowers for all s. We obtain the following three results: (1) when s q + 1 , for any sunflower, the corresponding linear code is minimal; (2) when 2 s 3 q , for any sunflower, the corresponding linear code is not minimal; (3) when 3 < s q , for some sunflowers, the corresponding linear codes are minimal, wherea for some other sunflowers, the corresponding linear codes are not minimal.
This paper is organized as follows. In Section 2, we introduce some basic knowledge about sunflowers, Euclidean inner product, and minimal linear codes. In Section 3, we consider the linear codes constructed from sunflowers and discuss the minimality of these linear codes in three cases. In Section 4, we conclude this paper.

2. Preliminaries

2.1. Sunflower

Throughout this paper, let k and t 0 be two positive integers, m = 2 k + t 0 and l = k + t 0 . Let 2 s q k + 1 be a positive integer, T 0 F q m be a subspace of F q m , and dim T 0 = t 0 . We denote G q ( l , m ) the set of l-dimensional vector subspaces of F q m . We define
Φ = { E i F q m : dim E i = l , E i E j = T 0 , 1 i j s } .
Then, Φ G q ( l , m ) is a sunflower of F q m and the space T 0 is called the center of the sunflower Φ .
Lemma 1.
Let Φ G q ( l , m ) be a sunflower and T 0 the center of Φ. For any E i , E j Φ with 1 i j s , we have F q m = E i + E j .
Proof. 
Since
dim ( E i + E j ) = dim ( E i ) + dim ( E j ) dim ( E i E j ) = dim ( E i ) + dim ( E j ) dim ( T 0 ) = l + l t 0 = m
and E i + E j F q m , we have F q m = E i + E j . □
Lemma 2.
Let Φ G q ( l , m ) be a sunflower and T 0 the center of Φ. For any E i , E j Φ with 1 i j s , we have E i E j = { 0 } .
Proof. 
Assume that z E i E j . It follows from Lemma 1 that z F q m , which implies z = 0 . □

2.2. Euclidean Inner Product

Let m be a positive integer. For x = ( x 1 , x 2 , , x m ) , y = ( y 1 , y 2 , , y m ) F q m , the Euclidean inner product of x and y is given by
< x , y > : = x y T = i = 1 m x i y i .
For any S F q m , we define
Span ( S ) : = i = 1 r λ i s i | r N , s i S , λ i F q ,
S : = { v F q m | vs T = 0 , for any s S } .
Then, Span ( S ) and S are vector spaces over F q and
dim ( Span ( S ) ) + dim ( S ) = m .

2.3. Minimal Linear Codes

All linear codes can be constructed by the following way. Let m n be two positive integers. Let G : = [ d 1 , , d n ] be an m × n matrix over F q and D : = { d 1 , , d n } be a multiset. Let r ( D ) = r ( G ) denote the rank of G, which is equal to the dimension of the vector space Span ( D ) over F q . Let
C ( D ) : = c ( x ) = x G = ( xd 1 T , , xd n T ) , x F q m .
Then, C ( D ) is an [ n , r ( D ) ] q linear code with generator matrix G. We always study the minimality of C ( D ) by considering some appropriate multisets D.
To present the sufficient and necessary condition for minimal linear codes in [14], some concepts are needed. For any y F q m , we define
H ( y ) : = y = { x F q m xy T = 0 } ,
H ( y , D ) : = D H ( y ) = { x D xy T = 0 } ,
V ( y , D ) : = Span ( H ( y , D ) ) .
It is obvious that H ( y , D ) V ( y , D ) H ( y ) .
Proposition 1
([14]). For any x , y F q m , c ( x ) c ( y ) if and only if H ( y , D ) H ( x , D ) .
Let y F q m { 0 } . The following lemma gives a sufficient and necessary condition for the codeword c ( y ) C ( D ) to be minimal.
Lemma 3
([14] (Theorem 3.1)). Let y F q m { 0 } . Then, the following three conditions are equivalent:
(1)
c ( y ) is minimal in C ( D ) ;
(2)
dim V ( y , D ) = m 1 ;
(3)
V ( y , D ) = H ( y ) .
The following lemma gives a sufficient and necessary condition for linear codes over F q to be minimal.
Lemma 4
([14] (Theorem 3.2)). The following three conditions are equivalent:
(1)
C ( D ) is minimal;
(2)
for any y F q m { 0 } , dim V ( y , D ) = m 1 ;
(3)
for any y F q m { 0 } , V ( y , D ) = H ( y ) .
By the following lemma, we can obtain infinity of many minimal linear codes from any known minimal linear codes.
Lemma 5
([14] (Proposition 4.1)). Let D 1 D 2 be two multisets with elements in F q m and r ( D 1 ) = r ( D 2 ) = m . If C ( D 1 ) is minimal, then C ( D 2 ) is minimal.
The following corollary is trivial.
Corollary 1.
Let D 1 D 2 be two multisets with elements in F q m and r ( D 1 ) = r ( D 2 ) = m . If C ( D 2 ) is not minimal, then C ( D 1 ) is not minimal.
In the following section, we will use the above lemmas to consider the minimality of linear codes constructed from sunflowers.

3. The Minimality of Linear Codes Constructed from Sunflowers

In this section, we consider the linear codes constructed from sunflowers and discuss the minimality of these linear codes.
Let
Φ = { E i F q m : dim E i = l , E i E j = T 0 , 1 i j s } .
be a sunflower of F q m and T 0 the center of Φ .
Let
D : = i = 1 s E i T 0 = i = 1 s ( E i T 0 ) .
It is easy to see that C ( D ) is a [ s ( q l q t 0 ) , m ] q linear code.
The following lemmas are important in the proofs of this section.
Lemma 6
([24] (Lemma 3.1)). For all y F q m { 0 } , E F q m and dim ( E ) = r , we have H ( y , E ) = V ( y , E ) and
dim V ( y , E ) = r , i f y E ; r 1 , i f y E .
By linear algebra, we can obtain the following lemma.
Lemma 7.
Let y F q m { 0 } . If for any E i Φ , y E i , 1 i s . For any E i 0 , E j 0 Φ , E i 0 E j 0 , let D 1 = ( E i 0 E j 0 ) T 0 . We have
rank H ( y , D 1 ) = m 2 , i f y T 0 ; m 1 , i f y T 0 .
Proof. 
Since y E i , it follows from Lemma 6 that dim H ( y , E i 0 ) = dim H ( y , E j 0 ) = l 1 . Note that H ( y , T 0 ) H ( y , E i 0 ) and H ( y , T 0 ) H ( y , E j 0 ) .
If y T 0 , then H ( y , T 0 ) = T 0 . Suppose that
H ( y , T 0 ) = T 0 = Span { γ 1 , γ 2 , , γ t 0 } , H ( y , E i 0 ) = Span { α 1 , α 2 , , α k 1 , γ 1 , γ 2 , , γ t 0 } , H ( y , E j 0 ) = Span { β 1 , β 2 , , β k 1 , γ 1 , γ 2 , , γ t 0 } .
Then, we have
H ( y , E i 0 ) T 0 { α 1 , α 2 , α k 1 , α 1 + γ 1 , α 1 + γ 2 , , α 1 + γ t 0 } , H ( y , E j 0 ) T 0 { β 1 , β 2 , , β k 1 , β 1 + γ 1 , β 1 + γ 2 , , β 1 + γ t 0 } .
Since H ( y , D 1 ) = ( H ( y , E i 0 ) H ( y , E j 0 ) ) T 0 , the above equations lead to
H ( y , D 1 ) { α 1 , α 2 , , α k 1 , β 1 , β 2 , , β k 1 , α 1 + γ 1 , α 1 + γ 2 , , α 1 + γ t 0 } ,
i.e., rank H ( y , D 1 ) = m 2 .
If y T 0 , then dim H ( y , T 0 ) = t 0 1 by Lemma 6. Suppose that
H ( y , T 0 ) = Span { γ 1 , γ 2 , , γ t 0 1 } , T 0 = Span { γ 1 , γ 2 , , γ t 0 1 , γ t 0 } , H ( y , E i 0 ) = Span { α 1 , α 2 , , α k , γ 1 , γ 2 , , γ t 0 1 } , H ( y , E j 0 ) = Span { β 1 , β 2 , , β k , γ 1 , γ 2 , , γ t 0 1 } .
Then, we have
H ( y , E i 0 ) T 0 { α 1 , α 2 , α k , α 1 + γ 1 , α 1 + γ 2 , , α 1 + γ t 0 1 } , H ( y , E j 0 ) T 0 { β 1 , β 2 , , β k , β 1 + γ 1 , β 1 + γ 2 , , β 1 + γ t 0 1 } .
Since H ( y , D 1 ) = ( H ( y , E i 0 ) H ( y , E j 0 ) ) T 0 , the above equations yield
H ( y , D 1 ) { α 1 , α 2 , , α k , β 1 , β 2 , , β k , α 1 + γ 1 , α 1 + γ 2 , , α 1 + γ t 0 1 } ,
i.e., rank H ( y , D 1 ) = m 1 . The proof is completed. □
Now, we consider the minimality of C ( D ) in three cases. First, when s q + 1 , we have
Theorem 1.
Let Φ = { E 1 , , E s } be a sunflower of F q m with center T 0 of dimension t 0 . If s q + 1 , then C ( D ) is an [ s ( q l q t 0 ) , m ] q minimal linear code.
Proof. 
According to Lemma 4, we only need to prove that for any y F q m { 0 } , dim V ( y , D ) = m 1 . By (2), we obtain
H ( y , D ) = D H ( y ) = i = 1 s ( H ( y , E i ) T 0 ) .
There are three cases:
(1) If there exists E i 0 Φ such that y E i 0 , then we have dim H ( y , E i 0 ) = l from Lemma 6. According to Lemma 2, for any E j 0 Φ with E j 0 E i 0 , we have y E j 0 . Then, it follows from Lemma 6 that dim H ( y , E j 0 ) = l 1 . Since y E i 0 T 0 , we have H ( y , T 0 ) = T 0 . We set
H ( y , T 0 ) = T 0 = Span { γ 1 , γ 2 , , γ t 0 } .
When k = 1 , we set
H ( y , E i 0 ) = Span { α 1 , γ 1 , γ 2 , , γ t 0 } .
By (3), we have H ( y , D ) { α 1 , α 1 + γ 1 , α 1 + γ 2 , , α 1 + γ t 0 } , and so dim V ( y , D ) = m 1 .
When k > 1 , we set
H ( y , E i 0 ) = Span { α 1 , α 2 , , α k , γ 1 , γ 2 , , γ t 0 } ,
and
H ( y , E j 0 ) = Span { β 1 , β 2 , , β k 1 , γ 1 , γ 2 , , γ t 0 } .
By (3), we have
H ( y , D ) { α 1 , α 2 , , α k , β 1 , β 2 , , β k 1 , α 1 + γ 1 , α 1 + γ 2 , , α 1 + γ t 0 } .
Since
rank { α 1 , α 2 , , α k , β 1 , β 2 , , β k 1 , α 1 + γ 1 , α 1 + γ 2 , , α 1 + γ t 0 } = m 1 ,
it is easy to obtain dim V ( y , D ) = m 1 .
(2) If for any E i Φ , 1 i s , we have y E i and y T 0 , then dim H ( y , E i 0 ) = dim H ( y , E j 0 ) = l 1 for any E i 0 , E j 0 Φ with E i 0 E j 0 . Since y T 0 , dim H ( y , T 0 ) = t 0 1 . We set
H ( y , T 0 ) = Span { γ 1 , γ 2 , , γ t 0 1 } , T 0 = Span { γ 1 , γ 2 , , γ t 0 } .
When k = 1 , we set
H ( y , E i 0 ) = Span { α 1 , γ 1 , γ 2 , , γ t 0 1 } , H ( y , E j 0 ) = Span { β 1 , γ 1 , γ 2 , , γ t 0 1 } .
Then,
H ( y , D ) { α 1 , β 1 , α 1 + γ 1 , α 1 + γ 2 , , α 1 + γ t 0 1 } .
Since
rank { α 1 , β 1 , α 1 + γ 1 , α 1 + γ 2 , , α 1 + γ t 0 1 } = m 1 ,
it is easy to obtain dim V ( y , D ) = m 1 .
When k > 1 , let D 1 = ( E i 0 E j 0 ) T 0 . By Lemma 7, we have rank ( H ( y , D 1 ) ) = m 1 , thus dim V ( y , D ) = m 1 .
(3) If for any E i Φ , 1 i s , we have y E i and y T 0 ; then, it follows from Lemma 6 that dim H ( y , E i ) = l 1 and dim H ( y , T 0 ) = t 0 .
When k = 1 , we obtain dim T 0 =2 and dim E i =1, 1 i s , then E i is the one-dimensional subspace of T 0 . There are q + 1 one dimensional subspace of T 0 , since s q + 1 , we obtain s = q + 1 . By Lemma 2, for any E i , E j Φ , E i E j , we have E i E j = { 0 } . Thus,
T 0 = i = 1 s E i .
Since y T 0 , by (4), there exists E j Φ , such that y E j , a contradiction. So k 1 .
When k > 1 , we have dim H ( y , E 1 ) =dim H ( y , E 2 ) = l 1 . Let D 1 = ( E 1 E 2 ) T 0 . By Lemma 7, we have rank ( H ( y , D 1 ) ) = m 2 . We set
T 0 = H ( y , T 0 ) = Span { γ 1 , γ 2 , , γ t 0 } .
E 1 = Span { α 1 , α 2 , , α k 1 , α k , γ 1 , γ 2 , , γ t 0 } , H ( y , E 1 ) = Span { α 1 , α 2 , , α k 1 , γ 1 , γ 2 , , γ t 0 } .
E 2 = Span { β 1 , β 2 , , β k 1 , β k , γ 1 , γ 2 , , γ t 0 } , H ( y , E 2 ) = Span { β 1 , β 2 , , β k 1 , γ 1 , γ 2 , , γ t 0 } .
Let
B = { α 1 , α 2 , , α k 1 , β 1 , β 2 , , β k 1 , α 1 + γ 1 , α 1 + γ 2 , , α 1 + γ t 0 } .
Then, rank B = m 2 and B H ( y , D ) . Let V = F q m , W = Span ( B ) and V ¯ = V / W the quotient space of V over W. We have dim V ¯ = 2 and V = Span { α k ¯ , β k ¯ } . Let π be the standard map from V to V ¯ . For any E i Φ , 1 i s , π ( E i ) is a subspace of V ¯ . It is easily seen that dim π ( E i ) = 1 or 2. There are the following two cases.
(i) If there exists E i 0 Φ such that dim π ( E i 0 ) = 2 , then π ( E i 0 ) = V ¯ . There must exist α E i 0 such that
π ( α ) = α k b β k ¯ , where b = ( α k y T ) / ( β k y T ) .
So, α = α k b β k + w , where w W . It is simply checked that α T 0 , α H ( y ) and α W . We obtain
( B { α } ) H ( y , D ) , rank ( B { α } ) = m 1 .
Thus, dim V ( y , D ) = m 1 .
(ii) If for any E i Φ we have dim π ( E i ) = 1 , combining that V = E i + E j for any E i , E j Φ with E i E j in accordance with Lemma 1, we have
V ¯ = π ( V ) = π ( E i ) + π ( E j ) and π ( E i ) π ( E j ) .
Since V ¯ has only q + 1 one-dimensional subspace and s q + 1 , we have s = q + 1 and V ¯ = i = 1 s π ( E i ) . There must exist E j 0 Φ such that
π ( E j 0 ) = Span { α k b β k ¯ } , where b = ( α k y T ) / ( β k y T ) .
Hence, there exists α = α k b β k + w E j 0 , where w W , such that π ( α ) = α k b β k ¯ . One can easily deduce that α T 0 , α H ( y ) and α W . We obtain
( B { α } ) H ( y , D ) , rank ( B { α } ) = m 1 .
Thus, dim V ( ( y ) , D ) = m 1 .
In conclusion, for any y F q m { 0 } , we have dim V ( y , D ) = m 1 , so C ( D ) is a minimal linear code. □
Remark 1.
In Theorem 1, if q = p is a prime number, then it becomes [23] (Theorem 10). So Theorem 1 is a generalization of [23] (Theorem 10). Our method is different from theirs. When s q , our method also can be used to study the minimality of the linear codes, whereas theirs can not.
Example 1.
Let e 1 , , e m be the standard basis of F q m . Let
T 0 = Span ( { e 2 k + 1 , e 2 k + 2 , , e m } ) = { ( 0 , 0 , t ) | t F q t 0 } .
 For any b F q , we define
E b = Span { e 1 + b e k + 1 , e 2 + b e k + 2 , , e k + b e 2 k , e 2 k + 1 , , e m } .
 Suppose that
Φ = { E b | b F q } Span { e k + 1 , e k + 2 , , e 2 k , e 2 k + 1 , , e m }
 and 
D = E i Φ ( E i T 0 ) .
 It is easy to see that Φ is a sunflower of F q m with center T 0 and s = q + 1 . Here, we take q = 4 , k = 3 , and t 0 = 1 . With the help of Magma, we verify that the code C ( D ) is a minimal [ 1260 , 7 ] 4 linear code with minimum distance 768, and
w min w max = 4 5 > 3 4 .
Now, we consider the minimality of C ( D ) when 2 s 3 q . If s = 3 , we have
Theorem 2.
Let Φ = { E 1 , , E s } be a sunflower of F q m with center T 0 of dimension t 0 . If s = 3 q , then C ( D ) is not minimal.
Proof. 
To prove C ( D ) is not minimal, by Lemma 4, we only need to prove there exists y 0 F q m { 0 } such that dim V ( y 0 , D ) m 2 .
When s = 3 , Φ = { E 1 , E 2 , E 3 } . By Lemma 2 we know E 1 E 2 = { 0 } . Then, for any y 1 E 2 { 0 } , we have y 1 E 1 and y 1 T 0 . Thus, dim H ( y 1 , E 1 ) = l 1 , dim H ( y 1 , E 2 ) = l , and dim H ( y 1 , T 0 ) = t 0 . We set
T 0 = H ( y 1 , T 0 ) = Span { γ 1 , γ 2 , , γ t 0 } ,
E 1 = Span { α 1 , α 2 , , α k 1 , α k , γ 1 , γ 2 , , γ t 0 } , H ( y 1 , E 1 ) = Span { α 1 , α 2 , , α k 1 , γ 1 , γ 2 , , γ t 0 } ,
E 2 = H ( y 1 , E 2 ) = Span { β 1 , β 2 , , β k , γ 1 , γ 2 , , γ t 0 } ,
where α k y 1 T = 1 . Let
E 1 = Span { α 1 , α 2 , , α k } , E 2 = Span { β 1 , β 2 , , β k } ,
we have F q m = E 1 E 2 T 0 . For any η E 3 , there exist unique α E 1 , β E 2 , γ T 0 , such that η = α + β + γ . Since α + β = η γ E 3 , for any α E 1 , there exists unique β E 2 such that α + β E 3 . Let φ be a map from E 1 to E 2 satisfying φ ( α ) = β . We can see φ is an isomorphism from E 1 to E 2 and
E 3 = { x + φ ( x ) | x E 1 } T 0 = E 3 T 0 .
Since y 1 E 1 , y 1 T 0 , and E 1 = E 1 T 0 , we have y 1 ( E 1 ) , dim H ( y 1 , E 1 ) = k 1 , dim φ ( H ( y 1 , E 1 ) ) = k 1 , and dim φ ( H ( y 1 , E 1 ) ) = m ( k 1 ) = k + t 0 + 1 . Thus,
dim ( φ ( H ( y 1 , E 1 ) ) E 1 ) = dim φ ( H ( y 1 , E 1 ) ) + dim ( E 1 ) dim ( φ ( H ( y 1 , E 1 ) ) + E 1 ) k + t 0 + 1 + k m = 1 .
Since q 3 , there exists y 2 ( φ ( H ( y 1 , E 1 ) ) E 1 ) { 0 } such that φ ( α k ) y 2 T 1 . It is easy to see y 2 E 2 and y 2 T 0 . Let y 0 = y 1 + y 2 , we obtain y 0 E 1 , y 0 E 2 , and y 0 T 0 . Since α k + φ ( α k ) E 3 and
( α k + φ ( α k ) ) y 0 T = ( α k + φ ( α k ) ) ( y 1 + y 2 ) T = α k y 1 T + α k y 2 T + φ ( α k ) y 1 T + φ ( α k ) y 2 T = 1 + 0 + 0 + φ ( α k ) y 2 T 0 ,
we obtain y 0 E 3 . Thus, y 0 E i , 1 i 3 and dim H ( y 0 , E i ) = l 1 .
(1) When k = 1 , dim H ( y 0 , E i ) = t 0 , since T 0 H ( y 0 , E i ) , we have T 0 = H ( y 0 , E i ) . Thus,
H ( y 0 , D ) = i = 1 3 ( H ( y , E i ) T 0 ) = .
Thus, C ( D ) is not minimal.
(2) When k > 1 , since E i = E i T 0 , we have y 0 ( E i ) and dim H ( y 0 , E i ) = k 1 . Thus,
H ( y 0 , E i ) = H ( y 0 , E i ) T 0 , 1 i 3 .
By Lemma 6, it is easily verified that
H ( y 0 , E 1 ) = H ( y 1 , E 1 ) ,
H ( y 0 , E 2 ) = H ( y 2 , E 2 ) = φ ( H ( y 1 , E 1 ) ) = φ ( H ( y 0 , E 1 ) ) ,
H ( y 0 , E 3 ) = { x + φ ( x ) | x H ( y 0 , E 1 ) } Span ( H ( y 0 , E 1 ) H ( y 0 , E 2 ) ) .
Then, dim V ( y 0 , D ) = m 2 . By Lemma 4, we have that c ( y 0 ) is not minimal. □
Combining Theorem 2 and Corollary 1, we have
Corollary 2.
Let Φ = { E 1 , , E s } be a partial spread of F q m . If 2 s 3 q , then C ( D ) is not minimal.
Now, we consider the minimality of C ( D ) when 4 s q . We recall from (5) that
T 0 = Span ( { e 2 k + 1 , e 2 k + 2 , , e m } ) = { ( 0 , 0 , t ) | t F q t 0 } .
We will show that some sunflowers Φ with center T 0 , C ( D ) are minimal, whereas some other sunflowers Φ with center T 0 , C ( D ) are not minimal.
First, we construct some sunflowers Φ such that C ( D ) are minimal. Let k 2 , f ( x ) be an irreducible polynomial in F q [ x ] of degree k and M F q k × k be a matrix with characteristic polynomial f ( x ) . We define
E 1 = { ( x , 0 , t ) | x F q k , t F q t 0 } , E 2 = { ( 0 , x , t ) | x F q k , t F q t 0 } , E 3 = { ( x , x , t ) | x F q k , t F q t 0 } , E 4 = { ( x , x M , t ) | x F q k , t F q t 0 } ,
and
Φ = { E 1 , E 2 , E 3 , E 4 } .
We can see Φ is a sunflower with center T 0 .
Theorem 3.
For the sunflower Φ defined in (8), the linear code C ( D ) is minimal.
Proof. 
According to Lemma 4, we only need to prove that for any y F q m { 0 } , dim V ( y , D ) = m 1 . There are three cases:
(1)
If there exists E i 0 Φ such that y E i 0 , the proof is similar as that in Theorem 1 (1).
(2)
If for any E i Φ , 1 i s , we have y E i and y T 0 , then the proof is similar to that in Theorem 1 (2).
(3)
If for any E i Φ , 1 i s , we have y E i and y T 0 , the proof is as follows. Let y = ( y 1 , y 2 , y 3 ) where y 1 , y 2 F q k , y 3 F q t 0 . Next, we define two linear transformations φ , ψ from F q k to F q k :
φ ( x ) = x , ψ ( x ) = x M , x F q k .
Then,
E 3 = { ( x , φ ( x ) , t ) | x F q k , t F q t 0 } , E 4 = { ( x , φ ( x ) , t ) | x F q k , t F q t 0 } .
Let
E 1 = { ( x , 0 , 0 ) | x F q k } , E 2 = { ( 0 , x , 0 ) | x F q k } , E 3 = { ( x , φ ( x ) , 0 ) | x F q k } , E 4 = { ( x , ψ ( x ) , 0 ) | x F q k } .
It is easy to verify that
E i = E i T 0 , 1 i 4 .
Let
S : = Span { H ( y , E 1 ) H ( y , E 2 ) } = Span { { H ( y , E 1 ) H ( y , E 2 ) } T 0 } = { ( α , β , 0 ) | α H ( y , E 1 ) , β H ( y , E 2 ) } { ( 0 , 0 , t ) | t F q t 0 } = S T 0 .
By Lemma 7, we have dim.
Now, we prove H ( y , E 3 ) S or H ( y , E 4 ) S . If not, assume that H ( y , E 3 ) S and H ( y , E 4 ) S . By H ( y , E 3 ) S , it is obvious that H ( y , E 3 ) S . Since y E 3 and y T 0 , we have y E 3 , and then dim H ( y , E 3 ) = k 1 . There exists α 1 , , α k 1 H ( y 1 ) , β 1 , , β k 1 H ( y 2 ) such that ( α 1 , β 1 , 0 ) , , ( α k 1 , β k 1 , 0 ) is a basis of H ( y , E 3 ) . Then, (10) yields β i = φ ( α i ) . It is effortlessly demonstrated that α 1 , , α k 1 is a basis of H ( y 1 ) , and β 1 , , β k 1 is a basis of H ( y 2 ) . Thus,
φ ( H ( y 1 ) ) = H ( y 2 ) .
Similarly, by H ( y , E 4 ) S , we obtain
ψ ( H ( y 1 ) ) = H ( y 2 ) .
Then, we have
ψ ( H ( y 1 ) ) = H ( y 2 ) = φ ( H ( y 1 ) ) = H ( y 1 ) .
That is to say, H ( y 1 ) is the ψ - invariant subspace of F q k .
Let α 1 , , α k 1 , α k be a basis of F q k , where α 1 , , α k 1 is a basis of H ( y 1 ) . Then, the matrix of ψ with respect to this basis is
B = B 1 B 2 0 b ,
where B 1 is the matrix of ψ | H ( y 1 ) with respect to α 1 , , α k 1 . Note that M is the matrix of ψ with respect to the standard basis, and thus M and B are similar and have the same characteristic polynomial. So
f ( x ) = | x I B 1 | ( x b ) ,
a contradiction with the irreducibility of f ( x ) . Hence, H ( y , E 3 ) S or H ( y , E 4 ) S . It is easy to see that r ( { H ( y , E 1 ) H ( y , E 2 ) H ( y , E 3 ) } T 0 ) = m 1 or r ( { H ( y , E 1 ) H ( y , E 2 ) H ( y , E 4 ) } T 0 ) = m 1 . So, dim V ( y , D ) = m 1 .
In conclusion, for any y F q m { 0 } , dim V ( y , D ) = m 1 . By Lemma 4, C ( D ) is minimal. □
Combining Theorem 3 and Lemma 5, we have
Corollary 3.
Let s 4 and Φ = { E 1 , , E s } be a sunflower of F q m with center T 0 . If { E 1 , E 2 , E 3 , E 4 } are defined as (7), then C ( D ) is minimal.
Example 2.
Take q = 5 , k = 2 , and t 0 = 1 . Let f ( x ) = x 2 + x + 1 and
M = 0 1 1 1 .
 It is easily checked that f ( x ) F q [ x ] is an irreducible polynomial of degree 2 and the characteristic polynomial of M. Then, the code C ( D ) constructed based on Theorem 3 is a minimal [ 480 , 5 ] 5 linear code with minimum distance 300, and
w min w max = 3 4 < 4 5 .
Now, we construct some sunflowers Φ with center T 0 such that C ( D ) are not minimal. Let us recall from (6) that
E b = Span { e 1 + b e k + 1 , e 2 + b e k + 2 , , e k + b e 2 k , e 2 k + 1 , , e m } .
Let
Φ = { E b | b F q } .
It is easy to see that Φ is a sunflower of F q m with center T 0 .
Theorem 4.
For the sunflower Φ defined in (13), the linear code C ( D ) is not minimal.
Proof. 
Let y 0 = e 1 . Then, for any b F q , we obtain
H ( y 0 , E b ) = Span { e 2 + b e k + 2 , , e k + b e 2 k , e 2 k + 1 , , e m } Span { e 2 , , e k , e k + 2 , , e 2 k , e 2 k + 1 , , e m } .
By (3), we have
H ( y 0 , D ) Span ( { e 2 , , e k , e k + 2 , , e 2 k , e 2 k + 1 , , e m } ) .
Then, dim V ( y 0 , D ) m 2 . By Lemma 4, we have that c ( y 0 ) is not minimal and C ( D ) is not minimal. □
Combining Theorem 4 and Corollary 1, we have
Corollary 4.
Let 3 < s q and S F q where # S = s . Let Φ = { E b | b S } . Then, C ( D ) is not minimal.
Remark 2.
In Theorem 3, Corollary 3, Theorem 4, and Corollary 4, the center of the sunflower Φ is the special subspace T 0 . When the center is a general subspace, we have not yet proved the minimality of C ( D ) .
Example 3.
Take q = 3 , k = 2 , and t 0 = 2 . Then, the code C ( D ) constructed based on Theorem 4 is [ 216 , 6 ] 3 linear code with minimum distance 108, and
w min w max = 2 3 .
According to Magma experiments, there exists y 1 = [ 1 , 0 , 0 , 0 , 0 , 0 ] F 3 6 such that dim V ( y 1 , D ) = 4 . Then, it follows from Lemma 4 that C ( D ) is not minimal.

4. Concluding Remarks

In this paper, we use the approach used in [14] to study the minimality of linear codes constructed from sunflowers in all cases. In [23], the authors proved that if the number s of the elements in a sunflower satisfying s p + 1 , then the corresponding linear code over F p is minimal, where p is a prime number. Our results in this paper generalize [23] (Theorem 10). We discuss the minimality of linear codes constructed from sunflowers for all s. We obtain the following three results: (1) when s q + 1 , for any sunflower, the corresponding linear code is minimal; (2) when 2 s 3 q , for any sunflower, the corresponding linear code is not minimal; (3) when 3 < s q , for some sunflowers, the corresponding linear codes are minimal, whereas for some other sunflowers, the corresponding linear codes are not minimal.

Author Contributions

Writing—original draft preparation, X.W.; writing—review and editing, W.L. All authors have read and agreed to the published version of the manuscript.

Funding

This research was funded by National Natural Science Foundation of China (Grant No. 11971102).

Data Availability Statement

No new data were created or analyzed in this study. Data sharing is not applicable to this article.

Conflicts of Interest

The authors declare no conflict of interest.

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Wu, X.; Lu, W. Minimal Linear Codes Constructed from Sunflowers. Entropy 2023, 25, 1669. https://doi.org/10.3390/e25121669

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Wu, Xia, and Wei Lu. 2023. "Minimal Linear Codes Constructed from Sunflowers" Entropy 25, no. 12: 1669. https://doi.org/10.3390/e25121669

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Wu, X., & Lu, W. (2023). Minimal Linear Codes Constructed from Sunflowers. Entropy, 25(12), 1669. https://doi.org/10.3390/e25121669

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