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Research article

Some results on ordinary words of standard Reed-Solomon codes

  • Received: 23 July 2019 Accepted: 26 August 2019 Published: 04 September 2019
  • MSC : 11C08, 94B35

  • The Reed-Solomon codes are widely used to establish a reliable channel to transmit information in digital communication which has a strong error correction capability and a variety of efficient decoding algorithm. We usually use the maximum likelihood decoding algorithm (MLD) in the decoding process of Reed-Solomon codes. MLD algorithm lies in determining its error distance. Li, Wan, Hong and Wu et al obtained some results on the error distance. For the Reed-Solomon code RSq(Fq,k), the received word u is called an ordinary word of RSq(Fq,k) if the error distance d(u,RSq(Fq,k))=ndeg(u(x)) with u(x) being the Lagrange interpolation polynomial of u. In this paper, we make use of the polynomial method and particularly, we use the KŚnig-Rados theorem on the number of nonzero solutions of polynomial equation over finite fields to show that if q4,2kq2, then the received word uFq1q of degree q2 is an ordinary word of RSq(Fq,k) if and only if its Lagrange interpolation polynomial u(x) is of the form u(x)=λq2i=kaq2ixi+fk1(x) with a,λFq and fk1(x)Fq[x] being of degree at most k1. This answers partially an open problem proposed by J.Y. Li and D.Q. Wan in [On the subset sum problem over finite fields, Finite Fields Appls. 14 (2008), 911-929].

    Citation: Xiaofan Xu, Yongchao Xu, Shaofang Hong. Some results on ordinary words of standard Reed-Solomon codes[J]. AIMS Mathematics, 2019, 4(5): 1336-1347. doi: 10.3934/math.2019.5.1336

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  • The Reed-Solomon codes are widely used to establish a reliable channel to transmit information in digital communication which has a strong error correction capability and a variety of efficient decoding algorithm. We usually use the maximum likelihood decoding algorithm (MLD) in the decoding process of Reed-Solomon codes. MLD algorithm lies in determining its error distance. Li, Wan, Hong and Wu et al obtained some results on the error distance. For the Reed-Solomon code RSq(Fq,k), the received word u is called an ordinary word of RSq(Fq,k) if the error distance d(u,RSq(Fq,k))=ndeg(u(x)) with u(x) being the Lagrange interpolation polynomial of u. In this paper, we make use of the polynomial method and particularly, we use the KŚnig-Rados theorem on the number of nonzero solutions of polynomial equation over finite fields to show that if q4,2kq2, then the received word uFq1q of degree q2 is an ordinary word of RSq(Fq,k) if and only if its Lagrange interpolation polynomial u(x) is of the form u(x)=λq2i=kaq2ixi+fk1(x) with a,λFq and fk1(x)Fq[x] being of degree at most k1. This answers partially an open problem proposed by J.Y. Li and D.Q. Wan in [On the subset sum problem over finite fields, Finite Fields Appls. 14 (2008), 911-929].


    Let Fq be the finite field of q elements with characteristic p. Let D={x1,,xn} be a subset of Fq, which is called the evaluation set. The generalized Reed-Solomon code RSq(D,k) of length n and dimension k over Fq is defined as follows:

    RSq(D,k):={(f(x1),,f(xn))Fnqf(x)Fq[x],degf(x)k1}.

    If D=Fq, then it is called standard Reed-Solomon code, i.e.,

    RSq(Fq,k):={(f(1),f(α),,f(αq2))Fnqf(x)Fq[x],degf(x)k1}, (1.1)

    where α is a primitive element of Fq. We refer the above definition as the polynomial code version of the standard Reed-Solomon code. If D=Fq, then it is called the extended Reed-Solomon code. For any [n,k]q linear code C, the minimum distance d(C) is defined by

    d(C):=min{d(x,y)xC,yC,xy},

    where d(,) denotes the Hamming distance of two words which is the number of different entries of them and w() denotes the Hamming weight of a word which is the number of its non-zero entries. Since the Reed-Solomon code is a linear code, we have

    d(C)=min0xC{d(x,0)}=min0xC{w(x)}.

    The error distance to code C of a received word uFnq is defined by

    d(u,C):=minvC{d(u,v)}.

    Clearly, d(u,C)=0 if and only if uC. The most important algorithmic problem in coding theory is the maximum likelihood decoding (MLD): Given a received word uFnq, find a codeword vC such that d(u,v)=d(u,C), then we decode u to v [4]. Therefore, it is very crucial to decide d(u,C) for the word u. When decoding the generalized Reed-Solomon code RSq(D,k), for a received word u=(u1,,un)Fnq, we define the Lagrange interpolation polynomial u(x) of u by

    u(x):=ni=1uinj=1jixxjxixjFq[x], (1.2)

    i.e., u(x) is the unique polynomial of degree deg(u(x))n1 such that u(xi)=ui for 1in. For uFnq, we define the degree of u(x) to be the degree of u, i.e., deg(u):=deg(u(x)). It is clear that uRSq(D,k) if and only if d(u,RSq(D,k))=0 if and only if deg(u)k1. Equivalently, uRSq(D,k) if and only if d(u,RSq(D,k))1 if and only if kdeg(u)n1. Evidently, we have the following simple bounds due to Li and Wan [3].

    Theorem 1.1. [3] Let u be a received word such that uRSq(D,k). Then

    ndeg(u)d(u,RSq(D,k))nk.

    Let uFnq. If d(u,RSq(D,k))=nk, then the received word u is called a deep hole of RSq(D,k). In 2007, Cheng and Murray [1] conjectured that a word u is a deep hole of RSq(Fq,k) if and only if u(x)=axk+fk1(x), where u(x) is the Lagrange interpolation polynomial of the received word u and aFq. In 2012, Wu and Hong [9] disproved this conjecture by presenting a new class of deep holes for standard Reed-Solomon codes RSq(Fq,k). In fact, let q4 and 2kq2. They showed that the received word u is a deep hole if its Lagrange interpolation polynomial equals axq2+fk1(x). Later on, the main result of [9] is extended to the generalized Reed-Solomon code in [2]. Recently, some progress on deep holes of generalized projective Reed-Solomon codes are made in [10] and [11].

    On the other hand, the received word u is called an ordinary word of RSq(D,k) if d(u,RSq(D,k))=ndeg(u(x)). If deg(u)=k, then the upper bound is equal to the lower bound which implies that u is a deep hole and also an ordinary word. This immediately gives (q1)qk ordinary words. We call these trivial ordinary words. It is an interesting problem to determine all the ordinary words. In 2008, Li and Wan [3] proposed an open problem to determine all the ordinary words of the standard Reed-Solomon code. In [4], by using Weil's estimate on character sums, the following result is obtained.

    Theorem 1.2. [4] Let uFqq be such that k+1deg(u)q1. Assume that q>max((deg(u))2,(deg(u)k1)2+ilonε) and k>(4ilonε+1)(deg(u)k)+4ilonε+2 for some constant ilonε>0. Then u is an ordinary word of extended Reed-Solomon code RSq(Fq,k).

    Furthermore, using Weil's character sum estimate and Li-Wan sieve for distinct coordinates counting, Zhu and Wan [12] showed the following result.

    Theorem 1.3. [12] Let uFqq be such that k+1deg(u)q1. If there are positive constants c1 and c2 such that deg(u)k<c1q1/2,(deg(u)k+1)log2q<k<c2q, then u is an ordinary word of extended Reed-Solomon code RSq(Fq,k).

    In [5], Li and Zhu proved the following result.

    Theorem 1.4. [5] Let 3k+2q1, and uFqq be represented by polynomial u(x)=xk+2bxk+1+cxk+v(x) with degv(x)k1. If k+2=q1 and b2=c, then u is an ordinary word of extended Reed-Solomon code RSq(Fq,k).

    In this paper, we make use of a well-known result, i.e. the so-called König-Rados theorem, to find all the ordinary words of degree q2 of standard Reed-Solomon code RSq(Fq,k). The main result of this paper can be stated as follows.

    Theorem 1.5.Let q4,2kq2 and uFq1q be a received word with u(x) being its Lagrange interpolation polynomial and degu(x)=q2. Then u is an ordinary word of RSq(Fq,k) if and only if u(x) is of the following form

    u(x)=λq2i=kaq2ixi+fk1(x)

    with a,λFq and fk1(x)Fq[x] being of degree at most k1.

    If one picks k=q2, then the ordinary words given by Theorem 1.5 are just the trivial ones. From Theorem 1.5, the following interesting result follows immediately.

    Proposition 1.6. Let q4,2kq2. Then the number of ordinary words of degree q2 of the standard Reed-Solomon code RSq(Fq,k) is equal to (q1)2qk.

    This paper is organized as follows. First, in Section 2, we show several preliminary lemmas that are needed in the proof of Theorem 1.5. Consequently, we show Theorem 1.5 in Section 3. Finally, we present two examples to illustrate the validity of our main result.

    In this section, our main goal is to prove several lemmas that are needed in the proof of Theorem 1.5. In what follows, we let

    Pk1:={f(x)f(x)Fq[x],degf(x)k1}

    and

    f(Fq):=(f(1),f(α),,f(αq2)),

    where α is a primitive element of Fq. We begin with the following lemma.

    Lemma 2.1. Let u,vFq1q be two words. If u=λv+fk1(Fq), where λFq and fk1(x)Fq[x] is a polynomial of degree at most k1. Then each of the following is true:

    (ⅰ). We have d(u,RSq(Fq,k))=d(v,RSq(Fq,k)).

    (ⅱ). The word u is an ordinary word of RSq(Fq,k) if and only if the word v is an ordinary word of RSq(Fq,k).

    Proof. (ⅰ). Since RSq(Fq,k) is a linear code, we obtain that

    d(u,RSq(Fq,k))=mincRSq(Fq,k){d(u,c)}=mincRSq(Fq,k){d(λv+fk1(Fq),c)}=mincRSq(Fq,k){d(λv+fk1(Fq),c+fk1(Fq)}=minc(x)Pk1#{xFq|λv(x)+fk1(x)c(x)fk1(x)0}=minc(x)Pk1#{xFq|λv(x)c(x)0}=mincRSq(Fq,k){d(λv,c)}=mincRSq(Fq,k){d(λv,λc)}(since λFq)=minc(x)Pk1#{xFq|λv(x)λc(x)0}=minc(x)Pk1#{xFq|v(x)c(x)0}=mincRSq(Fq,k){d(v,c)}=d(v,RSq(Fq,k))

    as desired.

    (ⅱ). Since u=λv+fk1(Fq), one has degu=degv. Hence u is an ordinary word of RSq(Fq,k) if and only if

    d(u,RSq(Fq,k))=q1degu,

    if and only if

    d(u,RSq(Fq,k))=q1degv. (2.1)

    But part (ⅰ) tells us that d(u,RSq(Fq,k))=d(v,RSq(Fq,k)). So (2.1) holds if and only if

    d(v,RSq(Fq,k))=q1degv. (2.2)

    So u is ordinary holds if and only if (2.2) is true. In other words, u is ordinary if and only if v is ordinary.

    This completes the proof of Lemma 2.1.

    Consequently, we give another useful fact.

    Lemma 2.2. Let uFq1q be a received word and u(x) be its Lagrange interpolation polynomial. Then one has

    d(u,RSq(Fq,k))=q1maxv(x)Pk1#{βFq|u(β)=v(β)}.

    Proof. By (1.1), we have

    d(u,RSq(Fq,k))=minvRSq(Fq,k){d(u,v)}=minv(x)Pk1#{1iq1|u(αi1)v(αi1)}=q1maxv(x)Pk1#{1iq1|u(αi1)=v(αi1)}=q1maxv(x)Pk1#{βFq|u(β)=v(β)}

    as required.

    The proof of Lemma 2.2 is complete.

    The following result gives a formula on the number of nonzero solutions of polynomial equation over finite fields and is due to König and Rados (see, for instance, [6,7,8]). It is a key and important ingredient in the proof of our main result.

    Lemma 2.3. (König-Rados) ([6,7,8]) Let f(x)=a0+a1x++aq2xq2Fq[x]. Then the number of nonzero solution of equation f(x)=0 in Fq is equal to q1rank(A), where A is the left (q1)×(q1) circulant matrix defined by

    A:=(a0a1aq3aq2a1a2aq2a0aq2a0aq4aq3).

    In this section, we use the lemmas presented in the previous section to give the proof of Theorem 1.5.

    Proof of Theorem 1.5. First of all, we show the sufficiency part. Let

    u(x)=λq2i=kaq2ixi+fk1(x),

    where a and λFq, fk1(x)Fq[x] is a polynomial of degree at most k1. Define

    uk(x):=q2i=kaq2ixi. (3.1)

    Then u(x)=λuk(x)+fk1(x). Now we pick a primitive element α of Fq and let

    uk:=(uk(1),uk(α),,uk(αq2)).

    By Lemma 2.1, one gets that

    d(u,RSq(Fq,k))=d(uk,RSq(Fq,k)).

    Therefore, in order to show that

    u:=(u(1),u(α),,u(αq2))

    is an ordinary word, it suffices to prove that uk is an ordinary word. Equivalently, we need only to show that

    d(uk,RSq(Fq,k))=q1deguk(x)=1, (3.2)

    since deguk(x)=q2. This will be done in what follows.

    By Lemma 2.2, we have

    d(uk,RSq(Fq,k))=q1maxv(x)Pk1#{βFq|uk(β)=v(β)}. (3.3)

    For any v(x)Pk1, one has degv(x)k1. But deguk(x)=q2k. Hence

    deg(uk(x)v(x))=deguk(x).

    It then follows that for any v(x)Pk1, one has

    #{γFq|uk(γ)=v(γ)}=#{γFq|uk(γ)v(γ)=0}deg(uk(x)v(x))=deguk(x)=q2. (3.4)

    On the other hand, we take

    v0(x):=k1i=0aq2ixi.

    Then v0(x)Pk1. Furthermore, by (3.1) we have

    #{γFq|uk(γ)v0(γ)=0}=#{γFq|q2i=kaq2iγi+k1i=0aq2iγi=0}=#{γFq|q2i=0aq2iγi=0}. (3.5)

    Since

    xq11=q1i=1(xαi)

    and aFq implying that

    xq11=(xa)q2i=0aq2ixi,

    it then follows that

    q2i=0aq2ixi=q1i=1(xαi)xa.

    This infers that

    {γFq|q2i=0aq2iγi=0}=Fq{a},

    from which one can derive that

    #{γFq|q2i=0aq2iγi=0}=q2. (3.6)

    So (3.4) together with (3.5) and (3.6) implies that

    maxv(x)Pk1#{γFq|uk(γ)v(γ)=0}=q2. (3.7)

    Hence (3.2) follows immediately from (3.3) and (3.7). So u is an ordinary word of RSq(Fq,k). This finishes the proof of the sufficiency part.

    Now we turn our attention to the proof of the necessity part. Let u be an ordinary word of RSq(Fq,k) and degu(x)=q2. Then by the definition of ordinary word, we have

    d(u,RSq(Fq,k))=q1(q2)=1.

    Hence by Lemma 2.2, one has

    maxv(x)Pk1#{xiFq|u(x)v(x)=0}=q1d(u,RSq(Fq,k))=q11=q2.

    Notice that for any v(x)Pk1, one has

    #{xiFq|u(x)v(x)=0}deg(u(x)v(x))=q2.

    So there is a polynomial v0(x)Pk1 such that

    #{xFq|u(x)v0(x)=0}=q2.

    Write

    u(x)=q2i=0uixi

    and

    v0(x)=k1i=0vixi.

    Let uq2=λ. Since degu(x)=q2, one has λFq. Then

    u(x)v0(x)=q2i=0uixik1i=0vixi=q2i=kuixi+k1i=0(uivi)xi=λ(q2i=kλ1uixi+k1i=0λ1(uivi)xi) (since λFq):=λq2i=0cixi,

    with ci=λ1ui for all integers i with kiq2 and ci=λ1(uivi) for all integers i with 0ik1. One then deduces that

    #{xFq|q2i=0cixi=0}=q2. (3.8)

    On the other hand, Lemma 2.3 yields that

    #{xFq|q2i=0cixi=0}=q1rank(B), (3.9)

    where B is the left circulant matrix defined by

    B:=(c0c1cq3cq2c1c2cq2c0cq2c0cq4cq3).

    So from (3.8) and (3.9), we derive that rank(B)=1. Since cq2=λ1λ=1, one has

    B=(c0c1cq31c1c21c01c0cq4cq3).

    Assume that cq3=0. Then B holds a nonzero minor of order 2, and so one gets that rank(B)2, which is impossible. Hence we must have cq30. In the following, we let cq3=a. Then aFq. For each integer i with 1iq1, let ri denote the i-th row of the matrix B. Then rank(ri)=1 for each integer i with 1iq1. Since rank(B) = 1, there exists an element aFq such that r1=ar2. Then we deduce that cq2=ac0 and ck=ack+1 for 0kq3. Since cq2=1, one has a=cq3,c0=a1=aq2 and ck=a1ck1 for 1kq2. It then follows that ck=aq2k for 0kq2. This implies that

    u(x)v0(x)=λq2i=0aq2ixi.

    Therefore

    u(x)=λq2i=kaq2ixi+λk1i=0aq2ixi+v0(x)=λq2i=kaq2ixi+fk1(x),

    where

    fk1(x):=λk1i=0aq2ixi+v0(x)Pk1.

    So the necessity part is proved.

    This concludes the proof of Theorem 1.5

    In this last section, we supply two examples to demonstrate the validity of Theorem 1.5.

    Example 4.1. Let q=7,n=q1=6,k=3. Putting α=3 gives us the standard Reed-Solomon code

    RS7(F7,3)={(f(1),f(3),,f(35))F67f(x)F7[x],degf(x)2}.

    Using the MATLAB 2011a programming, we search for the ordinary word and find out all the ordinary words of degree q2=5 of standard Reed-Solomon code RS7(F7,3) that are listed in Table 1. By (1.2), we can get the Lagrange interpolation polynomial u(x) of the ordinary word u of degree 5 of RS7(F7,3) listed also in Table 1. This coincides with Theorem 1.5.

    Table 1.  Ordinary words of degree 5 for RS7(F7,3).} λF7, f=l2e2+l1e+l0, f(x)=l2x2+l1x+l0 with ei=(1,3i,2i,6i,4i,5i) and l0,l1,l2 running over F7, d(u,v)=1.
    Ordinary word u LIP u(x) of u Codeword v LIP v(x) of v
    λ(3,1,0,6,0,4)+f λ(x5+x4+x3)+f(x) λ(4,1,0,6,0,4)+f λ(6x2+6x+6)+f(x)
    λ(0,2,5,4,0,3)+f λ(x5+2x4+4x3)+f(x) λ(0,2,2,4,0,3)+f λ(6x2+5x+3)+f(x)
    λ(6,1,5,0,2,0)+f λ(x5+3x4+2x3)+f(x) λ(6,6,5,0,2,0)+f λ(x2+3x+2)+f(x)
    λ(0,5,0,1,6,2)+f λ(x5+4x4+2x3)+f(x) λ(0,5,0,1,1,2)+f λ(6x2+3x+5)+f(x)
    λ(3,0,4,0,5,2)+f λ(x5+5x4+4x3)+f(x) λ(3,0,4,0,5,5)+f λ(x2+5x+4)+f(x)
    λ(1,0,3,4,6,0)+f λ(x5+6x4+x3)+f(x) λ(1,0,3,3,6,0)+f λ(x2+6x+1)+f(x)

     | Show Table
    DownLoad: CSV

    Suppose that u is an ordinary word of degree 5 of RS7(F7,3). Then

    d(u,RS7(F7,3))=ndegu(x)=65=1.

    On the other hand, one has

    d(u,RS7(F7,3))=minvRS7(F7,3){d(u,v)}.

    So it is sufficient to find a codeword v in RS7(F7,3) such that d(u,v)=1. For the received ordinary word u=λ(3,1,0,6,0,4)+f of degree 5, by (1.2) we compute and get that u(x)=λ5i=3xi+f(x). Furthermore, one can search and find the word v=λ(4,1,0,6,0,4)+fRS7(F7,3) such that d(u,v)=1. For the other ordinary words u of degree 5, one can also find the corresponding codewords v such that d(u,v)=1. We can easily compute the Lagrange interpolation polynomial v(x) of v also listed in Table 1.

    Example 4.2. Let q=11,n=q1=10,k=5. Putting α=2 gives us the standard Reed-Solomon code

    RS11(F11,5)={(f(1),f(2),,f(29))F1011f(x)F11[x],degf(x)5}.

    Using the MATLAB 2011a programming, we search for the ordinary word and find out all the ordinary words of degree q2=9 of standard Reed-Solomon code RS11(F11,5) that are listed in Table 2. By (1.2), one can easily calculate the Lagrange interpolation polynomial u(x) of the ordinary word u of degree 9 of RS11(F11,5) listed also in Table 2. This coincides with Theorem 1.5.

    Table 2.  Ordinary words of degree 9 for RS11(F11,5).} λF11, f=l4e4+l3e3+l2e2+l1e+l0, f(x)=l4x4+l3x3+l2x2+l1x+l0 with ei=(1,2i,4i,8i,5i,10i,9i,7i,3i,6i) and l0,l1,l2,l3,l4 running over F11, d(u,v)=1.
    Ordinary word u LIP u(x) of u Codeword v LIP v(x) of v
    λ(5,2,0,5,0,10,0,4,0,7)+f λ(x9+x8+x7+10x6+x5)+f(x) λ(6,2,0,5,0,10,0,4,0,7)+f λ(10x4+10x3+10x2+10x+10)+f(x)
    λ(9,8,1,0,8,0,5,0,2,0)+f λ(x9+2x8+4x7+8x6+5x5)+f(x) λ(9,3,1,0,8,0,5,0,2,0)+f λ(x4+2x3+4x2+8x+5)+f(x)
    λ(0,9,0,7,0,5,0,6,9,8)+f λ(x9+3x8+9x7+5x6+4x5)+f(x) λ(0,9,0,7,0,5,0,6,2,8)+f λ(10x4+8x3+2x2+6x+7)+f(x)
    λ(0,10,4,6,0,4,0,8,0,1)+f λ(x9+4x8+5x7+9x6+3x5)+f(x) λ(0,10,7,6,0,4,0,8,0,1)+f λ(10x4+7x3+6x2+2x+8)+f(x)
    λ(0,3,0,8,1,7,0,1,0,2)+f λ(x9+5x8+3x7+4x6+9x5)+f(x) λ(0,3,0,8,10,7,0,1,0,2)+f λ(10x4+6x3+8x2+7x+2)+f(x)
    λ(4,0,10,0,9,0,8,0,3,10)+f λ(x9+6x8+3x7+7x6+9x5)+f(x) λ(4,0,10,0,9,0,8,0,3,1)+f λ(x4+6x3+3x2+7x+9)+f(x)
    λ(7,0,3,0,10,0,1,7,5,0)+f λ(x9+7x8+5x7+2x6+3x5)+f(x) λ(7,0,3,0,10,0,1,4,5,0)+f λ(x4+7x3+5x2+2x+3)+f(x)
    λ(6,0,5,2,3,0,2,0,4,0)+f λ(x9+8x8+9x7+6x6+4x5)+f(x) λ(6,0,5,9,3,0,2,0,4,0)+f λ(x4+8x3+9x2+6x+4)+f(x)
    λ(0,6,0,9,0,2,3,10,0,3)+f λ(x9+9x8+4x7+3x6+5x5)+f(x) λ(0,6,0,9,0,2,8,10,0,3)+f λ(10x4+2x3+7x2+8x+6)+f(x)
    λ(1,0,7,0,4,6,9,0,6,0)+f λ(x9+10x8+x7+10x6+x5)+f(x) λ(1,0,7,0,4,5,9,0,6,0)+f λ(x4+10x3+x2+10x+1)+f(x)

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    Suppose that u is an ordinary word of degree 9 of RS11(F11,5). Then

    d(u,RS11(F11,5))=ndegu(x)=109=1.

    On the other hand, one has

    d(u,RS11F11,5))=minvRS11(F11,5){d(u,v)}.

    So it is sufficient to find a codeword v in RS11(F11,5) such that d(u,v)=1. For the received ordinary word u=λ(5,2,0,5,0,10,0,4,0,7)+f of degree 9, by (1.2) we compute and get that u(x)=λ9i=5xi+f(x). Furthermore, one can search and find the codeword v=λ(6,2,0,5,0,10,0,4,0,7)+fRS11(F11,5) such that d(u,v)=1. For the other ordinary words u of degree 9, one can also find the corresponding codewords v such that d(u,v)=1. It is easy to compute the Lagrange interpolation polynomial v(x) of v that is listed in Table 2.

    Remark 4.3. In this paper, we determine all the ordinary words of maximal degree q2 of the standard Reed-Solomon code RSq(Fq,k). In the close future, we will explore the ordinary words of degree no more than q3 of the standard Reed-Solomon code RSq(Fq,k).

    The authors would like to thank the anonymous referees for their careful readings of the manuscript and helpful comments and suggestions that improved the presentation of the paper. This work was supported partially by National Science Foundation of China Grant # 11771304, by the Fundamental Research Funds for the Central Universities and by Foundation of Sichuan Tourism University under Grant # 19SCTUZZ01.

    We declare that we have no conflict of interest.



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