CONTACT:  

Ari Dwi Hartanto,    ari@ugm.ac.id       Department of Mathematics, Universitas Gadjah Mada, Sleman, 

D.I. Yogyakarta 55281, Indonesia  

 

       The article can be accessed here. https://doi.org/10.15642/mantik.2021.7.1.1-8  

Binary Cyclic Pearson Codes  
 

 

Ari Dwi Hartanto1*, Al. Sutjijana2 

1,2Department of Mathematics, Universitas Gadjah Mada, Yogyakarta, Indonesia 

 

 

 

Article history: 

Received Sep 2, 2019 

Revised Feb 4, 2021 

Accepted Mar 3, 2021 

 

 

Kata Kunci:  

Jarak Pearson, Kode 

Pearson, Kode siklis 

 

Abstrak. Fenomena gain atau offset yang tidak terduga pada sistem 

komunikasi dan media penyimpan data modern seperti media 

penyimpanan berjenis optik (CD) dan memori non-volatile (flash) 

merupakan gangguan yang serius. Permasalahan ini dapat ditangani 

dengan mengaplikasikan jarak Pearson pada detektor error pada 

sistem tersebut karena jarak Pearson menawarkan kekebalan terhadap 

gain dan offset yang tidak menentu. Jarak ini hanya dapat digunakan 

pada suatu himpunan codewords tertentu, yaitu himpunan 

Pearson/kode Pearson. Salah satu contoh kode Pearson dapat 

ditemukan di kelas kode T-constrained. Dalam paper ini, diberikan 

kode 2-constrained biner dengan sifat siklis. Konstruksi kode ini 

diadopsi dari konstruksi pada kode siklis, akan tetapi kode yang 

dihasilkan tidak dapat dipandang sebagai kode siklik. 

 

Keywords:  

Pearson distance, 

Pearson code, Cyclic 

code 

 

 

Abstract. The phenomena of unknown gain or offset on 

communication systems and modern storage such as optical data 

storage and non-volatile memory (flash) becomes a serious problem. 

This problem can be handled by Pearson distance applied to the 

detector because it offers immunity to gain and offset mismatch. This 

distance can only be used for a specific set of codewords, called 

Pearson codes. An interesting example of Pearson code can be found 

in the T-constrained code class. In this paper, we present binary 2-

constrained codes with the cyclic property. The construction of this 

code is adopted from cyclic codes, but it cannot be considered as 

cyclic codes 

  

 

How to cite: 

A. D. Hartanto and Al. Sutjijana, “Binary Cyclic Pearson Codes”, J. Mat. Mantik, vol. 7, no. 1, 

pp. 1-8, May 2021. 

 

  

Jurnal Matematika MANTIK 

Vol. 7, No. 1, May 2021, pp. 1-8 

 

ISSN: 2527-3159 (print) 2527-3167 (online) 

mailto:ari@ugm.ac.id
https://doi.org/10.15642/mantik.2021.7.1.1-8
http://u.lipi.go.id/1458103791


Jurnal Matematika MANTIK 

Vol. 7, No. 1, May 2021, pp. 1-8 

2 

 

1. Introduction 

Based on [1] and [2], gain and/or offset mismatch often occur on modern storages and 

communication channels. In non-volatile memories, for instance: a flash memory, the data 

is stored in a floating gate. It can leak away from the floating gate and affects a shift of the 

offset. In optical disc media, the signal retrieved by the sensor depends on the dimensions 

of the written features and the quality of the light path. The quality of the light path usually 

occurs because of fingerprints and scratches on the disc surface. These lead to gain and 

offset mismatch of the retrieved signal. Regarding this problem, Weber et.al. [3] presented 

a system code which involves the Person distance. 

Pearson distance is resistant to gain and offset mismatch, so it can be an alternative to 

the Euclidean distance for improving the error performance of noisy channels with 

unknown gain and offset. Unfortunately, Pearson distance cannot be used for arbitrary sets 

of codewords, but it can only be used for a Pearson set/Pearson code, a set of codewords 

with special properties. Weber, Swart, and Immink [3] stated that one of the known Pearson 

code is 𝑇(𝑛, 𝑞). This code is a member of the T-constrained code class which consists of 
sequences in which each of T pre-determined reference symbols appears at least once ([1], 

[3], [4]). 

Let 𝐶 ⊆ 𝑄𝑛  be a q-ary code of length n, where 𝑄 = {0,1, ⋯ , 𝑞 − 1} is the code 
alphabet of size 𝑞 ≥ 2. The alphabet symbols in codebook C will not be treated as elements 
of 𝑍𝑞, but as just integers. The reader should be careful that this situation will be different 

when we discuss linear codes in this paper. 

Weber, Swart, and Immink ([3], [5]) presented two coding procedures, i.e. fixed-to-

fixed (FF) and variable-to-fixed (VF) length coding schemes which are special codes of 

𝑇(𝑛, 𝑞). We present another special code of 𝑇(𝑛, 𝑞), that is Binary Cyclic Pearson Codes. 
We focus on how to construct the Pearson codes C in which the codewords have cyclic 

property, that is, if (𝑎0, 𝑎1, ⋯ , 𝑎𝑛−1) ∈ 𝐶 then (𝑎𝑛−1, 𝑎0, 𝑎1, ⋯ , 𝑎𝑛−2) ∈ 𝐶. Our idea is 
taking the similar way of cyclic code construction (see [6] and [7]) for constructing a cyclic 

Pearson code. Cyclic code is one of the most important classes of linear codes. The cyclic 

codes have a rich algebraic structure and can be efficiently implemented using simple shift 

registers [8]. Although the cyclic codes possess algebraic structure, we do not emphasize 

algebraic structure on the binary cyclic Pearson codes. We only concern on the set of 

codewords that have cyclic property and satisfy the axioms of Pearson code. We investigate 

whether it can be constructed and the simple encoding scheme with a generator matrix can 

be built in a similar way on the simple encoding scheme of cyclic codes. 

The rest of this paper is organized as follows. We start our presentation with a brief 

introduction to Pearson Distance, Pearson codes, and T-constrained codes in Section 2. In 

Section 3, we give a brief exposition of the cyclic codes, and then present binary cyclic 

Pearson codes. Finally, we draw some conclusions in Section 4. 

 

2. Preliminaries 

We first introduce a necessary notation used in this paper. For simplicity, we use the 

shorthand notation 𝛼𝒖 + 𝛽 = (𝛼𝑢1 + 𝛽, 𝛼𝑢2 + 𝛽, ⋯ , 𝛼𝑢𝑛 + 𝛽) as described in [3]. The 
bold letter indicates a vector that can mean either a message word or a codeword. Suppose 

a codeword c is sent, and then a vector 𝒓 = 𝑎(𝒄 + 𝒗) + 𝑏 is received, where 𝑎, 𝑏 ∈ 𝑅, 𝑎 >
0, and 𝑣 = (𝑣1, ⋯ , 𝑣𝑛) , 𝑣𝑖 ∈ 𝑅. The real numbers a and b are called gain and offset, 
respectively, and the entries of vector 𝑣 are noise sample from a zero-mean Gaussian 
distribution. The gain and offset will not affect sent codewords if we use Pearson distance 

detection ([9], [10]). 

The following is the Pearson distance that is a well-known concept in statistics. Let 𝒙 
be a vector in 𝑅𝑛. Define: 



Ari Dwi Hartanto, Al. Sutjijana 

Binary Cyclic Pearson Codes 

3 

𝒙 =
1

𝑛
∑ 𝑥𝑖

𝑛

𝑖=1

, 

and 

𝜎𝑥
2 = ∑(𝑥𝑖 − 𝒙)

2

𝑛

𝑖=1

. 

The Pearson correlation coefficient of x and y is defined by 

𝜌𝑥,𝑦 =
∑ (𝑥𝑖 − 𝒙)

𝑛
𝑖=1 (𝑦𝑖 − 𝒚)

𝜎𝑥 𝜎𝑦
. 

Next, the Pearson distance is defined as follow: 

𝛿(𝒙, 𝒚) = 1 − 𝜌𝑥,𝑦 . 

To know the bound of 𝛿(𝒙, 𝒚), we first note that −1 ≤ 𝜌𝑥,𝑦 ≤ 1. This implies that 0 ≤

𝛿(𝒙, 𝒚) ≤ 2. 
By nearest neighbour decoding, a received vector will be decoded to the codeword 

closest to it, with respect to Pearson distance. If we have a received vector r, the minimum 

Pearson distance detector outputs the codeword: 

𝒄𝟎 =𝑎𝑟𝑔 𝑎𝑟𝑔 𝑚𝑖𝑛
𝑐∈𝑆

 𝛿(𝒓, 𝒄), 

where S is the codebook. 

 

We can observe that for all 𝛼, 𝛽 ∈ 𝑅, 𝛼 > 0, we have 𝜌
𝑥,𝛼𝑦+𝛽

= 𝜌𝑥,𝑦 . The equation 

implies that 𝛿(𝒙, 𝛼𝒚 + 𝛽) = 𝛿(𝒙, 𝒚). It means that the Pearson distance is invariant under 
translation and scale. As a result, the minimum Pearson distance detector is immune to gain 

and offset mismatch. However, this arises a weakness. We cannot use the minimum 

Pearson distance detector for arbitrary codebooks because it cannot distinguish between 

vector x and 𝒚 = 𝛼𝒙 + 𝛽 where 𝛼, 𝛽 ∈ 𝑅, 𝛼 > 0, (𝛼, 𝛽) ≠ (1,0), or mathematically, it can 
be written as 

𝛿(𝒓, 𝒙) = 𝛿(𝒓, 𝒚) 
if 𝒚 = 𝛼𝒙 + 𝛽, for all 𝛼, 𝛽 ∈ 𝑅 with 𝛼 > 0 and (𝛼, 𝛽) ≠ (1,0). Because of this fact, our 

codebook C should has property that if 𝒄 ∈ 𝐶 then 𝛼𝒄 + 𝛽 ∉ 𝐶 for all 𝛼, 𝛽 ∈ 𝑅 with 𝛼 > 0 
and (𝛼, 𝛽) ≠ (1,0). Moreover, another property that has to be satisfied by codebook C is 
that vectors in the form of 𝒙 = (𝑎, 𝑎, ⋯ , 𝑎) is not allowed in the codebook C since the 
vectors will lead to an undefined Pearson correlation coefficient. 

 

Definition 1. Let C be a nonempty subset of 𝑄𝑛. The set C is called Pearson code (or 
Pearson set) if it satisfies the following properties: 

a. If 𝒄 ∈ 𝐶 then 𝛼𝒄 + 𝛽 ∉ 𝐶 for all 𝛼, 𝛽 ∈ 𝑅 with 𝛼 > 0 and (𝛼, 𝛽) ≠ (1,0).  
b. 𝒙 = (𝑎, 𝑎, ⋯ , 𝑎) ∉ 𝐶 for all 𝑎 ∈ 𝑅.  

 

To have an example of Pearson codes, we first give a description of T-constrained codes. 

Let 𝑇 be an integer with 1 ≤ 𝑇 < 𝑞 and reference symbols 𝑎1, ⋯ , 𝑎𝑇  be element of 𝑄.  A 
T-constrained code [1], denoted by 𝑆𝑞,𝑛(𝑎1, ⋯ , 𝑎𝑇 ),  consist of all q-ary codewords of 

length n, (𝑥1, ⋯ , 𝑥𝑛 ), such that 
#{𝑖: 𝑥𝑖 = 𝑗} > 0 𝑓𝑜𝑟 𝑒𝑎𝑐ℎ 𝑗 ∈ {𝑎1, ⋯ , 𝑎𝑇 }. 

Note that not all T-constrained codes are Pearson codes. For 𝑇 = 2, we can see that both 2-
constrained code 𝑆𝑞,𝑛(0,1) and 𝑆𝑞,𝑛(0, 𝑞 − 1) are Pearson codes, but for 𝑞 ≥ 5, 𝑆𝑞,𝑛(0,2) 

is not a Pearson code since it does not meet the first property of Pearson code. 

We now focus to discuss the specific case, that is the T-constrained code 𝑆𝑞,𝑛(0,1). 

This code is Pearson codes, and will be denoted by 𝑇(𝑛, 𝑞) as in [3]. Weber, Swart, and 
Immink [3] proposed easy coding procedures, i.e., fixed-to-fixed (FF) and variable-to-fixed 

(VF) length coding schemes, denoted by 𝑇𝐹𝐹(𝑛, 𝑞) and 𝑇𝑉𝐹 (𝑛, 𝑞), respectively. The simple 



Jurnal Matematika MANTIK 

Vol. 7, No. 1, May 2021, pp. 1-8 

4 

 

FF scheme is to fill the first 𝑛 − 2 positions in the codeword c with information symbols 
and to fill the last two symbols for reference purpose, i.e., 𝑐𝑛−1 = 0 and 𝑐𝑛 = 1. For VF 
schemes, it can be described as follows. 

a. Take 𝑛 − 2 information from the q-ary source, and then set these as (𝑐1, 𝑐2, ⋯ , 𝑐𝑛−2). 
b. If there exists 𝑖, 1 ≤ 𝑖 ≤ 𝑛 − 2, such that 𝑐𝑖 = 0, then choose 𝑐𝑛−1 to be a new 

information symbol, otherwise set 𝑐𝑛−1 = 0. 
c. If there exists 𝑖, 1 ≤ 𝑖 ≤ 𝑛 − 1,  such that 𝑐𝑖 = 1, then choose 𝑐𝑛 to be a new 

information symbol, otherwise set 𝑐𝑛 = 1. 
The resulted codewords from both schemes are in 𝑇(𝑛, 𝑞) since it contains at least one 
symbol 0 and at least one symbol 1. 

 

3. Cyclic Pearson Codes 

In this section, we intend to construct cyclic Pearson codes by using similar ways as 

construction of cyclic codes. Let F be a finite field of order q, and 𝑉𝑛 (𝐹) = {(𝑥1𝑥2 ⋯ 𝑥𝑛 ) ∣
𝑥𝑖 , ⋯ , 𝑥𝑛 ∈ 𝐹} be a vector space over F. A nonempty subset C of 𝑉𝑛(𝐹) is called cyclic 
code if it is a subspace and every (𝑎1𝑎2 ⋯ 𝑎𝑛−1𝑎𝑛) ∈ 𝐶 implies (𝑎𝑛𝑎1𝑎2 ⋯ 𝑎𝑛−1) ∈ 𝐶.  
Cyclic codes can be constructed by ring polynomial approach. 

Clearly, 𝑉𝑛(𝐹) is an abelian group under vector addition, but it is not a ring since we 
have not had a multiplication between any two vectors yet. In order to have multiplication 

operation such that 𝑉𝑛 (𝐹) is a ring, the easiest way is to associate vectors in 𝑉𝑛(𝐹) with 
polynomials in 𝐹[𝑥], that is, if 𝒂 = (𝑎0𝑎1 ⋯ 𝑎𝑛−1) ∈ 𝑉𝑛 (𝐹),  then let 

𝑎(𝑥) = 𝑎0 + 𝑎1 + ⋯ + 𝑎𝑛−1𝑥
𝑛−1. 

We select the polynomial 𝑓(𝑥) = 𝑥𝑛 − 1 ∈ 𝐹[𝑥] to have a quotient ring 𝐹[𝑥]/⟨𝑓(𝑥)⟩.  
For all 𝒂, 𝒃 ∈ 𝑉𝑛(𝐹), let 𝑎(𝑥)𝑏(𝑥) = 𝑣(𝑥) where 𝑣(𝑥) is the polynomial of least degree in 
the equivalence class [𝑎(𝑥)𝑏(𝑥)] of 𝐹[𝑥]/⟨𝑓(𝑥)⟩. Note that 𝑣(𝑥) is the remainder 
polynomial when 𝑎(𝑥)𝑏(𝑥) is divided by 𝑓(𝑥),  and it is a polynomial of degree at most 
𝑛 − 1 over F, which is associated with an element of 𝑉𝑛(𝐹). In conclusion, now we have 
multiplication between any two vectors in 𝑉𝑛(𝐹) which is defined by this way, and with 
the association between the vectors and polynomials, we can essentially think of 𝑉𝑛 (𝐹) and 
𝐹[𝑥]/⟨𝑓(𝑥)⟩ interchangeable. Moreover, we can prove that 𝑉𝑛(𝐹) is a ring. 

To have cyclic property of vectors, the choice of the polynomial 𝑓(𝑥) = 𝑥𝑛 − 1 is 
most suitable. Multiplication a polynomial 𝑣(𝑥) in 𝐹[𝑥]/⟨𝑓(𝑥)⟩ by x corresponds to a 
cyclic shift of v. This results in a fundamental theorem in cyclic codes presented as follows. 

A nonempty subset C of 𝑉𝑛 (𝐹) is a cyclic code if only if the set of polynomials I associated 
with C is an ideal in the ring 𝐹[𝑥]/⟨𝑥𝑛 − 1⟩. This property helps us in constructing cyclic 
codes. In fact, there is a unique monic polynomial of least degree that generates a nonzero 

ideal I of 𝐹[𝑥]/⟨𝑥𝑛 − 1⟩, i.e., a monic polynomial divisor of 𝑓(𝑥) = 𝑥𝑛 − 1.  Thus, there 
is a 1 − 1 correspondence between cyclic codes in 𝑉𝑛 (𝐹) and monic polynomials 𝑔(𝑥) ∈
𝐹[𝑥] that divide 𝑓(𝑥). 

A monic polynomial 𝑔(𝑥) divisor of 𝑓(𝑥) = 𝑥𝑛 − 1 over F having degree 𝑛 − 𝑘 will 
become the generator for a cyclic code C of dimension k in 𝑉𝑛 (𝐹). The encoder encodes 
the message polynomial 𝑎(𝑥) (of degree less than or equal to 𝑘 − 1) to the codeword 
𝑎(𝑥)𝑔(𝑥) ∈ 𝐹[𝑥]/⟨𝑓(𝑥)⟩. Thus, the generator matrix for cyclic code C is given in terms of 
𝑔(𝑥) by 

𝐺 = [

𝑔(𝑥)
𝑥𝑔(𝑥)

⋮
𝑥𝑘−1𝑔(𝑥)

]. 

Note that G is a matrix of 𝑘 × 𝑛 where the 𝑖𝑡ℎ row of G, 2 ≤ 𝑖 ≤ 𝑘, is the 𝑖 − 1 right shifts 
of the first row of G. 



Ari Dwi Hartanto, Al. Sutjijana 

Binary Cyclic Pearson Codes 

5 

We now turn our discussion to Pearson codes. For simplicity, sometimes we will write 

a vector in 𝑉𝑛(𝐹) or in codes C in the form 𝑎0𝑎1 ⋯ 𝑎𝑛 instead of (𝑎0, 𝑎1, ⋯ , 𝑎𝑛). We note 
that the cyclic codes C are not Pearson codes since 00 ⋯ 0 belongs to C. We observe some 
nonempty subsets of cyclic code C such that it is Pearson codes. In case our discussion is 

in Pearson code, we treat the information symbols of codewords as real numbers instead of 

elements of a finite field. 

In order to find a nonempty subset of cyclic codes C such that it is a Pearson code, 

clearly, we should eliminate the zero vector from C since it breaks the second axiom of the 

definition of Pearson codes. We should observe whether there exist other elements having 

form 𝑎𝑎 ⋯ 𝑎 in C. Let us see a (7,4)-cyclic code over 𝑍2 generated by 𝑔(𝑥) = 1 + 𝑥 + 𝑥
3. 

The generator matrix of code C is 

𝐺 = [1 + 𝑥 + 𝑥3 𝑥 + 𝑥2 + 𝑥4 𝑥2 + 𝑥3 + 𝑥5 𝑥3 + 𝑥4 + 𝑥6 ]
= [1 1 0 1 0 0 0 0 1 1 0 1 0 0 0 0 1 1 0 1 0 0 0 0 1 1 0 1 ]. 

The message 𝒎 = 1011 is encoded to be codeword 𝒄 = 𝒎𝐺 = 1111111 ∈ 𝐶. Such 
codeword must be omitted from C. It means that if our observation is in (𝑛, 𝑘)-cyclic codes, 
we have to make attention for messages from which codewords having form 𝑎𝑎 ⋯ 𝑎. We 
note that such messages are the polynomials which are quotient of division ℎ(𝑥) = 𝑟 +
𝑟𝑥 + ⋯ + 𝑟𝑥𝑛−1, where 𝑟 ∈ 𝐹, by the generator polynomial 𝑔(𝑥) in 𝐹[𝑥]/⟨𝑥𝑛 − 1⟩. 

Clearly, if c is a codeword of (𝑛, 𝑘)-cyclic code 𝐶 over finite field F, then 𝛼𝑐 +
(1𝐹, 1𝐹 , ⋯ , 1𝐹) must be element of C, for all 𝛼 ∈ 𝑁. This does not meet the first axiom of 
the definition of Pearson codes. Now we go on a specific case, that is (𝑛, 𝑘)-cyclic code 𝐶 
over finite field 𝑍2. In this case, we have a good result. We note that 𝐶 − {00 ⋯ 0,11 ⋯ 1)} 
is a 2-constrained code 𝑆2,𝑛(0,1) if we treat the information symbols of codewords in C as 

real numbers. In conclusion, we obtain that the code 𝑃 = 𝐶 − {00 ⋯ 0,11 ⋯ 1} is a Pearson 
code, and we call this code the binary cyclic Pearson codes. 

For a simple encoding scheme, we intend to use the generator matrix, but we face a 

problem when the message word is the zero vector. We should give a special treatment for 

this message word. First, we need to limit n to an even positive integer. We correspond the 

zero vector with the codeword 1010 ⋯ 10. Since 1010 ⋯ 10 is a codeword, 0101 ⋯ 01 
must be considered as a codeword. We correspond 0101 ⋯ 01 with a message word 
𝑎0𝑎1 ⋯ 𝑎𝑘−1 such that 𝑎0𝑎1 ⋯ 𝑎𝑘−1𝐺 = 111 ⋯ 1. Unfortunately, we cannot guarantee that 
such a message word will exist. In order to guarantee the existence of such a message word, 

the polynomial ℎ(𝑥) = 1 + 𝑥 + ⋯ + 𝑥𝑛−1 must be divisible by the generator polynomial 
𝑔(𝑥). By this condition, we have 𝑎0 + 𝑎1𝑥 + ⋯ + 𝑎𝑘−1𝑥

𝑘−1 is ℎ(𝑥)/𝑔(𝑥).  Next, we 
should be careful if 1010 ⋯ 10 belongs to (𝑛, 𝑘)-cyclic code C. It should not be allowed 
since we have corresponded 1010 ⋯ 10 with message word 000 ⋯ 0. 

 
Table 1.  Some generator polynomials of binary cyclic Pearson codes 

n k Generator Polynomial 

2 1 𝑔(𝑥) = 1 + 𝑥 

4 1 𝑔(𝑥) = (1 + 𝑥)
3 

6 5 𝑔(𝑥) = 1 + 𝑥  

 3 𝑔(𝑥) = (1 + 𝑥)(1 + 𝑥 + 𝑥
2) 

 1 𝑔(𝑥) = (1 + 𝑥)(1 + 𝑥 + 𝑥
2)2 

8 1 𝑔(𝑥) = (1 + 𝑥)7  

10 9 𝑔(𝑥) = 1 + 𝑥  

 5 𝑔(𝑥) = (1 + 𝑥)(1 + 𝑥 + 𝑥
2 + 𝑥 3 + 𝑥 4) 

 1 𝑔(𝑥) = (1 + 𝑥)(1 + 𝑥 + 𝑥
2 + 𝑥 3 + 𝑥 4)2 



Jurnal Matematika MANTIK 

Vol. 7, No. 1, May 2021, pp. 1-8 

6 

 

 
Table 1.  Some generator polynomials of binary cyclic Pearson codes (continued) 

n k Generator Polynomial 

12 9 𝑔(𝑥) = (1 + 𝑥)3 

 7 𝑔(𝑥) = (1 + 𝑥)3(1 + 𝑥 + 𝑥 2) 

 5 𝑔(𝑥) = (1 + 𝑥)3(1 + 𝑥 + 𝑥 2)2 

 3 𝑔(𝑥) = (1 + 𝑥)3(1 + 𝑥 + 𝑥 2)3 

 1 𝑔(𝑥) = (1 + 𝑥)3(1 + 𝑥 + 𝑥 2)4 

 

Based on the explanation above, not all (𝑛, 𝑘)-cyclic codes over 𝑍2 can be used for 
constructing a binary cyclic Pearson code. As summary, the following conditions must be 

held in order that (𝑛, 𝑘)-cyclic codes over 𝑍2 can be taken for constructing a binary cyclic 
Pearson code: 

a. n is even. 

b. The polynomial ℎ(𝑥) = 1 + 𝑥 + ⋯ + 𝑥𝑛−1 is divisible by the generator polynomial 
𝑔(𝑥). 

c. The polynomial 1 + 𝑥2 + 𝑥4 + ⋯ + 𝑥𝑛−2 is not divisible by the generator polynomial 
𝑔(𝑥).  

 

By using MAGMA (Computational Algebra System) (see [11]), we observe all possibility 

of (𝑛, 𝑘)-cyclic codes over 𝑍2, for 𝑛 = 2,4, ⋯ ,12, that meets three conditions above. Table 
1 gives all generator polynomials of binary cyclic Pearson codes for 𝑛 = 2,4, ⋯ ,12. 
Example 1. Consider 𝑛 = 6, 𝑉6(𝑍2), and 𝑓(𝑥) = 𝑥

6 − 1. We take polynomial 𝑔(𝑥) = 1 +
𝑥 as a generator of the binary cyclic Pearson code, so we have a generator matrix: 

𝐺 = [1 1 0 0 0 0 0 1 1 0 0 0 0 0 1 1 0 0 0 0 0 1 1 0 0 0 0 0 1 1 ]. 
We divide 1 + 𝑥 + 𝑥2 + 𝑥3 + 𝑥4 + 𝑥5 by 𝑔(𝑥), and we have 1 + 𝑥2 + 𝑥4 as a result. 
Then, the message words 00000 and 10101 are encoded to the codewords 101010 and 

010101, respectively. For all 𝒎 ∈ 𝑉5(𝑍2) − {00000,10101},  m is encoded to codeword 
𝒎𝐺. Therefore, the binary cyclic Pearson code of length 𝑛 = 6 and with generator 𝑔(𝑥) =
1 + 𝑥 is 

𝑃 = {𝒎𝐺 ∣  𝒎 ∈ 𝑉5(𝑍2) − {00000,10101}} ∪ 𝑃0, 
where 𝑃0 = {101010,010101}. 
 

4. Conclusions 

A binary cyclic code of even length usually has a small Hamming distance. Some of 

the Hamming distances of cyclic codes can be seen on Table 2. A code with small distance 

can be said as a bad code because it can only detect and correct small errors. However, it 

does not mean that the code is always useless. As we explained above, an even length of 

code become one of the sufficient conditions for constructing a binary cyclic Pearson code. 

Therefore, some of cyclic codes of even length are needed for binary cyclic Pearson codes. 

In this paper we have investigated binary cyclic Pearson codes for 𝑛 = 2,4,6,8,10,12,  
and the result is that the binary cyclic Pearson codes exist for each of these n. We have not 

yet investigated codes of even length generally. Our conjecture is the binary cyclic Pearson 

codes exist for all even positive integer n. 

 
  



Ari Dwi Hartanto, Al. Sutjijana 

Binary Cyclic Pearson Codes 

7 

Table 2.  Some generators of (𝑛, 𝑘, 𝑑)-cyclic codes and its compatibility for binary cyclic Pearson 
codes (BCPC) 

n k Generator Polynomial d 
Compatibility of generator 

for BCPC 

2 1 𝑔(𝑥) = 1 + 𝑥 2 yes 

4 3 𝑔(𝑥) = 1 + 𝑥 2 no 

2 𝑔(𝑥) = (1 + 𝑥)2 2 no 

1 𝑔(𝑥) = (1 + 𝑥)3 4 yes 

6 5 𝑔(𝑥) = 1 + 𝑥 2 yes 

4 𝑔(𝑥) = (1 + 𝑥)2 2 no 

4 𝑔(𝑥) = 1 + 𝑥 + 𝑥 2 2 no 

2 𝑔(𝑥) = (1 + 𝑥 + 𝑥 2)2 3 no 

3 𝑔(𝑥) = (1 + 𝑥)(1 + 𝑥 + 𝑥 2) 2 yes 

2 𝑔(𝑥) = (1 + 𝑥)2(1 + 𝑥 + 𝑥 2) 4 no 

1 𝑔(𝑥) = (1 + 𝑥)(1 + 𝑥 + 𝑥 2)2 6 yes 

8 7 𝑔(𝑥) = (1 + 𝑥) 2 no 

6 𝑔(𝑥) = (1 + 𝑥)2 2 no 

5 𝑔(𝑥) = (1 + 𝑥)3 2 no 

4 𝑔(𝑥) = (1 + 𝑥)4 2 no 

3 𝑔(𝑥) = (1 + 𝑥)5 4 no 

2 𝑔(𝑥) = (1 + 𝑥)6 4 no 

1 𝑔(𝑥) = (1 + 𝑥)7 8 yes 

10 9 𝑔(𝑥) = 1 + 𝑥 2 yes 

8 𝑔(𝑥) = (1 + 𝑥)2 2 no 

6 𝑔(𝑥) = 1 + 𝑥 + 𝑥 2 + 𝑥 3 + 𝑥 4 2 no 

2 𝑔(𝑥) = (1 + 𝑥 + 𝑥 2 + 𝑥 3 + 𝑥 4)2 5 no 

5 𝑔(𝑥) = (1 + 𝑥)(1 + 𝑥 + 𝑥 2 + 𝑥 3 + 𝑥 4) 2 yes 

4 𝑔(𝑥) = (1 + 𝑥)2(1 + 𝑥 + 𝑥 2 + 𝑥 3 + 𝑥 4) 4 no 

1 𝑔(𝑥) = (1 + 𝑥)(1 + 𝑥 + 𝑥 2 + 𝑥 3 + 𝑥 4)2 10 yes 

12 11 𝑔(𝑥) = 1 + 𝑥 2 no 

 10 𝑔(𝑥) = (1 + 𝑥)2 2 no 

9 𝑔(𝑥) = (1 + 𝑥)3 2 yes 

8 𝑔(𝑥) = (1 + 𝑥)4 2 no 

10 𝑔(𝑥) = 1 + 𝑥 + 𝑥 2 2 no 

8 𝑔(𝑥) = (1 + 𝑥 + 𝑥 2)2 2 no 

6 𝑔(𝑥) = (1 + 𝑥 + 𝑥 2)3 3 no 

4 𝑔(𝑥) = (1 + 𝑥 + 𝑥 2)4 3 no 

9 𝑔(𝑥) = (1 + 𝑥)(1 + 𝑥 + 𝑥 2) 2 no 

7 𝑔(𝑥) = (1 + 𝑥)(1 + 𝑥 + 𝑥 2)2 2 no 

5 𝑔(𝑥) = (1 + 𝑥)(1 + 𝑥 + 𝑥 2)3 4 no 

3 𝑔(𝑥) = (1 + 𝑥)(1 + 𝑥 + 𝑥 2)4 6 no 

8 𝑔(𝑥) = (1 + 𝑥)2(1 + 𝑥 + 𝑥 2) 2 no 

6 𝑔(𝑥) = (1 + 𝑥)2(1 + 𝑥 + 𝑥 2)2 2 no 

 

  



Jurnal Matematika MANTIK 

Vol. 7, No. 1, May 2021, pp. 1-8 

8 

 

Table 2.  Some generators of (𝑛, 𝑘, 𝑑)-cyclic codes and its compatibility for binary cyclic Pearson 
codes (BCPC) (continued) 

n k Generator Polynomial d 
Compatibility of generator 

for BCPC 

12 4 𝑔(𝑥) = (1 + 𝑥)2(1 + 𝑥 + 𝑥 2)3 4 no 

 2 𝑔(𝑥) = (1 + 𝑥)2(1 + 𝑥 + 𝑥 2)4 6 no 

7 𝑔(𝑥) = (1 + 𝑥)3(1 + 𝑥 + 𝑥 2) 4 yes 

5 𝑔(𝑥) = (1 + 𝑥)3(1 + 𝑥 + 𝑥 2)2 4 yes 

3 𝑔(𝑥) = (1 + 𝑥)3(1 + 𝑥 + 𝑥 2)3 4 yes 

1 𝑔(𝑥) = (1 + 𝑥)3(1 + 𝑥 + 𝑥 2)4 12 yes 

 

Acknowledgement 

We thanks to Jos Weber from Delft University of Technology (TU Delft) for the 

discussion and his papers shared with us. 

 

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