# An Application of thet-Extension of the p-Fibonacci Pascal Matrix in Coding Theory

Authors: Elahe Mehraban; Mansour Hashemi
Journal: Advances in Mathematical Physics
 (2022)

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## Abstract

We consider thet-extension of the p-Fibonacci Pascal matrix. First, we study the k-th power of the t-extension of the p-Fibonacci lower and upper triangular Pascal matrix. Then, we obtain a new code which is named the t-extension of the p-Fibonacci Pascal matrix coding/decoding by using them.


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## Body

## 1. Introduction

The Fibonacci sequence is defined by the recurrence relationfn=fn−1+fn−2,n≥3, with the initial values f1=f2=1. Recently, there are many papers devoted to the study of the Fibonacci and generalized Fibonacci sequences; for example, see [1, 2]. Such the generalization, which will be used in this paper, is the t-extension of the p-Fibonacci sequence.Definition 1.
For constantst≥1 and p≥0, the t-extension of the p-Fibonacci sequence fpt,nn=0∞ is given with the following recurrence relation (see [3, 4]):
(1)fpt,n=0,n<1,1,n=1,tfpt,n−1+fpt,n−p−1,n>1.For example, fort=3 and p=2, we have f23,n=3f23,n−1+f23,n−3 and
(2)f23,n−∞∞=⋯,0,1,3,9⋯.In [4–13], the authors studied the properties of the Pascal matrix. We recall the following definition from [5].Definition 2.
Then×n lower triangular Pascal matrix Ln=lij is defined as
(3)li,j=ij,ifi≥j,0,otherwise,and the n×n upper triangular matrix Un=uij is defined as
(4)ui,j=ij,ifj≥i,0,otherwise.Then×n, t-extension of the p-Fibonacci lower triangular Pascal matrix, denoted by Lpt,n, is defined as follows:
(5)Lpt,n=fi,ji,j=1,⋯,n=fpt,i−j+1,ifi−j+1≥0,0,ifi−j+1<0,and the inverse of the t-extension of the p-Fibonacci lower triangular Pascal matrix, Lpt,n−1=li,j′i,j=1,⋯,n, is given as
(6)li,j′=1,ifi=j,−t,ifj=i−1,−1,ifj=i−p+1,0,otherwise.See [14] for more details.For example, we have(7)L13,4=f3,1000f3,2f3,100f3,3f3,2f3,10f3,4f3,3f3,2f3,1=1000310010310331031,L13,4−1=1000−t100−1−t100−1−t1=1000−3100−1−3100−1−31.Now, we define thet-extension of the p-Fibonacci upper triangular Pascal matrix, denoted by Upt,n, as follows:
(8)Upt,n=mi,ji,j=1,⋯,n=fpt,j−i+1,ifj−i+1≥0,0,ifj−i+1<0.The inverse of thet-extension of the p-Fibonacci upper triangular Pascal matrix denoted by Upt,n−1 is defined as follows:
(9)Upt,n−1=mi,j′i,j=1,⋯,n=1,ifi=j,−t,ifi=j−1,−1,ifi=j−p+1,0,otherwise.An application of the Fibonacci matrix and its generalization is in coding theory (see [15–22]). In this paper, using the t-extension of the p-Fibonacci lower and upper triangular Pascal matrix, we present a new code matrix.The rest of the paper is organized as follows: in Section2, we obtain the k-th power of the t-extension of the p-Fibonacci lower triangular matrix and upper triangular Pascal matrix. Also, we give the inverse of them 2≤n≤4. Section 3 is devoted to obtaining a coding and decoding on the t-extension of the p-Fibonacci Pascal matrix.


## 2. Thek-th Power of the t-Extension of the p-Fibonacci

In this section, we calculate thek-th power of the t-extension of the p-Fibonacci lower triangular Pascal matrix, the t-extension of the p-Fibonacci upper triangular Pascal matrix, and their inverse matrices. First, we denote the k-th power of Lpt,n and Upt,n by Lpt,nk and Upt,nk, respectively.Theorem 3.
Forp=1 and t,k≥1, we have
(10)iL1t,2k=10kt1,L1t,3k=100kt10k2t2+kt2+1kt1,L1t,4k=1000kt100k2t2+kt2+1kt10bkt3+t+kt3+2t+∑i=2ki2t2+it2+1tk2t2+kt2+1kt1,where b1=0, b2=2, and bk=bk−1+k−1 for k≥3.
(11)iiU1t,2k=1kt01,U1t,3k=1ktk2t2+kt2+101kt001,U1t,4k=1tkk2t2+kt2+1bkt3+t+kt3+2t+∑i=2ki2t2+it2+1t01tkk2t2+kt2+1001tk0001,where b1=0, b2=2, and bk=bk−1+k−1,k≥3.Proof.
(i)
Since the proof of the result forn=2, 3, and 4 is similar, we prove the result by induction on k for n=4. Let k=1. Then, by relation (3), we have
(12)L1t,4=f1t,1000f1t,200f1t,3f1t,2f1t,10f1t,4f1t,3f1t,2f1t,1=1000t100t2+1t10t3+2tt2+1t1.
Suppose the result holds fork=l. Therefore, we have
(13)L1t,nl+1=L1t,nl×L1t,n=1000lt100l2t2+lt2+1lt10blt3+t+lt3+2t+∑i=2li2t2+it2+1tl2t2+lt2+1lt1×1000t100t2+1t10t3+2tt2+1t1=1000l+1t100l+12t2+l+1t2+1lt10ul+12t2+l+1t2+1l+1t1,where
(14)u=bl+1t3+t+lt3+2t+∑i=2li2t2+it2+1tl+12t2+l+1t2+1.(ii)
The proof is similar to the proof of the propositioni and then is omittedExample 1.
Forn=4 and k=3, we have
(15)L1t,43=10003t1003t2+3t2+13t104t3+t+3t3+2t+t2+2t2+1t3t2+3t2+13t1.Similar to the proof of Theorem3, we can prove the following theorems.Theorem 4.
Forn=4, p=2, and t≥1, we have
(16)iL2t,nk=1000kt100k2+kt2kt104+∑i=3ki−12+2i−1t3+kk2+kt2+kt2+1kt1,iiU2t,nk=1tkk2+kt24+∑i=3ki−12+2i−1t3+k01tkk2+kt2001tk0001.Theorem 5.
Forn=4, p≥3, and t≥1, we have
(17)iLpt,nk=1000kt100k2+kt2kt104+∑i=3ki−12+2i−1t3k2+kt2+kt2+1kt1,iiUpt,nk=1tkk2+kt24+∑i=3ki−12+2i−1t301tkk2+kt2001tk0001.
We denote the inverse ofLpt,nk and Upt,nk by Lpt,n−k and Upt,n−k, respectively. Now, we have the following theorem.Theorem 6.
Forn=4 and p=1,
(18)iL1t,2−k=10−kt1,L1t,3−k=100−kt10k2t2−k−kt1,L1t,4−k=1000−tk100k2t2−k−tk10ckk2t2−k−tk1,where c1=0, c2=2t, and
(19)ck=ck−1+k−1t−∑i=3ki−12t2−i−1t,for k≥3.
(20)iiU1t,2−k=1−kt01,U1t,3−k=1−tkk2t2−k01−tk001,U1t,n−k=1−tkk2t2−kck01−tkk2t2−k001−tk0001,where c1=0, c2=2t, and
(21)ck=ck−1+k−1t−∑i=3ki−12t2−i−1t,k≥3.Proof.
(i)
It is sufficient to show that the result holds forn=4. For this, we use the induction method on k. Let k=1. Then, by relation (4), we have
(22)L1t,n−k=1000−t100−1−t100−1−t1.
Suppose the statement holds fork=l. Then, we have
(23)L1t,n−l+1=L1t,n−l×L1t,n−1=1000−lt100l2t2−l−lt10cll2t2−l−lt11000−t100−1−t100−1−t1=1000−l+1t100l+12t2−l+1−l+1t10cl+l2t2−lt+ltl+12t2−l−l+1t1.
Since(24)cl−l2t2−lt+lt=cl+lt−∑i=3li−12t2−i−1t=cl+1,we have result.
The proof (ii) is similar to the proof of the proposition (i). So, we omit it.Example 2.
Forn=4 and t=5, we have
(25)L1t,n−5=1000−5t100−10t2−5−5t1011t−t2−2t−3t2−3t−6t2−4t−10t2−5−5t1.
By induction onk, we can prove the following theorems and therefore we will omit their proofs.Theorem 7.
Forn=4 and p=2,
(26)iLpt,4−k=1000−tk100k2t2−tk10−∑i=3ki−12t3−kk2t2−tk1,iiUpt,n−k=1−tkk2t2−∑i=3ki−12t3−k01−tkk2t2001−tk0001.Theorem 8.
Forn=4 and p≥3,
(27)iLpt,n−k=1000−tk100k2t2−tk10−∑i=3ki−12t3k2t2−tk1,iiUpt,n−k=1−tkk2t2−∑i=3ki−12t301−tkk2t2001−tk0001.Lemma 9.
Form,s,p∈ℕ and n=4, we get
(28)iLPt,nm×LPt,ns=LPt,ns×LPt,nm=LPt,nm+s,iiUPt,nm×UPt,ns=UPt,ns×UPt,nm=UPt,nm+s.Proof.
(i)
By Theorem3 for p=1, we have
(29)L1t,nm×L1t,ns=1000mt100m2t2+mt2+1mt10bmt3+t+mt3+2t+∑i=2mi2t2+it2+1tm2t2+mt2+1mt1×1000st100s2t2+st2+1st10bst3+t+st3+2t+∑i=2si2t2+it2+1ts2t2+st2+1kt1=1000m+st100m+s2t2+m+st2+1m+st10abm+st1=L1t,nm+s,where
(30)a=bm+st3+t+m+st3+2t+∑i=2m+si2t2+it2+1t,b=m+s2t2+m+st2+1.In a similar way, one can prove the remaining cases.


## 3. Coding and Decoding onk-th Power the t-Extension of the p-Fibonacci Pascal Matrix

In the present section, first, we introduce a coding and decoding on thet-extension of the p-Fibonacci Pascal matrix and get its error detection and correction.For2≤n≤4 and an initial message Mn×n, we will name a transformation E=Lpt,nk×M×Upt,nk as the t-extension of the p-Fibonacci Pascal coding and a transformation M=Lpt,n−k×E×Upt,n−k as the t-extension of the p-Fibonacci Pascal decoding. Note that the matrix E is as a code matrix and all of the elements of M are positive.Lemma 10.
For2≤n≤4 and k,p∈ℕ, detE=detM.Proof.
SinceE=Lpt,nk×M×Upt,nk, we have
(31)detE=detLpt,nk×M×Upt,nk=detUpt,nk×detM×detLpt,nk=1×detM×1=detM.Now, we obtain a relation for elements of the code matrixE. We study three cases. The rest cases are similar.Case 1.
Letn=2 and p=1. For the matrix M2×2, we have
(32)E=L1t,2k×M×U1t,2k=10tk1m1m2m3m41tk01=m1tkm1+m2tkm1+m3t2k2m1+tkm3+m2+m4.
On the other hand, we get(33)M=L1t,2−kE×U1t,2−k=10−tk1e1e2e3e41−tk01=e1−tke1+e2−tke1+e3t2k2e1−tke2+e3+e4=m1m2m3m4.
We obtain(34)m1=e1≥0,m2=−tke1+e2≥0,m3=−tke1+e3≥0,m4=t2k2e1−tke2+e3+e4≥0.
So, we have(35)e1e2≤1tk,e1e3≤1tk,e1e4≤1t2k2.Case 2.
Suppose thatp=1 and n=3. For M3×3,
(36)M=L1t,3−k×E×L1t,3−k=100−tk10k2t2−k−tk1e1e2e3e4e5e6e7e8e91−tkk2t2−k01−tk001=e1−tke1+e2a−tke1+e4−tke1+e4−tk+−tke2+e5bk2t2−ke1−tke4+e7cd,where
(37)a=k2t2−ke1−tke2+e3,b=k2t2−k−tke1+e4+−tke2+e5−tk+−tke3+e6,c=k2t2−ke1−tke4+e7−tk+k2t2−ke2−tke5+e8,d=k2t2−ke1−tke4+e7k2t2−k+k2t2−ke2−tke5+e8−tk+k2t2−ke3−tke6+e9.
So,(38)m1=e1≥0,m2=−tke1+e2≥0,m3=k2t2−ke1−tke2+e3≥0,m4=−tke1+e4≥0,m5=−tke1+e4−tk+−tke2+e5≥0,m6=k2t2−k−tke1+e4+−tke2+e5−tk+−tke3+e6≥0,m7=k2t2−ke1−tke4+e7≥0,m8=k2t2−ke1−tke4+e7−tk+k2t2−ke2−tke5+e8≥0,m9=k2t2−ke1−tke4+e7k2t2−k+k2t2−ke2−tke5+e8−tk≥0.
Therefore, we get(39)e1e2≤1tk,e1e3≤1t2k2−k2t2−k,(40)e1e4≤1tk,e1e5≤1t2k2,(41)e1e6≤1t3k3−k2kt2−k2t,e1e7≤1t2k2−k2t2−k,(42)e1e8≤1t3k3−k2kt2−k2t,e1e9≤1t2k2+k−k2t22.Case 3.
Letp=2 and n=4, we obtain
(43)M=L2t,4−k×E×Lpt,4−k=1000−tk100k2t2−tk10−∑i=3kk2t3−kk2t2−tk1e1e2e3e4e5e6e7e8e9e10e11e12e13e14e15e161−tkk2t2−∑i=3kk2t3−k01−tkk2t2001−tk0001=e1−tke1+e2k2t2e1−tke2+e3h1−tke1+e5−tke1+e5−tk+−tke2+e6h2h3h4h5h6h7h8h9h10h11,where
(44)h1=−∑i=3kk2t3−ke1+k2t2e2−tke3+e4,h2=−tke1+e5k2t2+−tke2+e6−tk−tke3+e7,h3=−tke1+e5−∑i=3kk2t3−k+−tke2+e6k2t2+−tke3+e7−tk+−tke4+e8,h4=k2t2e1−tke5+e9,h5=k2t2e1−tke5+e9−tk+k2t2e2−tke6+e10,h6=k2t2e1−tke5+e9k2t2+k2t2e2−tke6+e10−tk+k2t2e3−tke7+e11,h7=k2t2e1−tke5+e9−∑i=3kk2t3−k+k2t2e2−tke6+e10k2t2+k2t2e3−tke7+e11−tk+k2t2e4−tke8+e12,h8=−∑i=3kk2t3−ke1+k2t2e5−tke9+e13,h9=−∑i=3kk2t3−ke1+k2t2e5−tke9+e13−tk+−∑i=3kk2t3−ke2+k2t2e6−tke10+e14,h10=−∑i=3kk2t3−ke1+k2t2e5−tke9+e13k2t2+−∑i=3kk2t3−ke2+k2t2e6−tke10+e14−tk+−∑i=3kk2t3−ke3+k2t2e7−tke11+e15,h11=−∑i=3kk2t3−ke1+k2t2e5−tke9+e13−∑i=3kk2t3−k+−∑i=3kk2t3−ke2+k2t2e6−tke10+e14k2t2+−∑i=3kk2t3−ke3+k2t2e7−tke11+e15−tk+−∑i=3kk2t3−ke4+k2t2e8−tke12+e16.
So,(45)m1=e1≥0,m2=−tke1+e2≥0,m3=k2t2−ke1−tke2+e3≥0,m4=−∑i=3kk2t3−ke1+k2t2e2−−tke3+e4≥0,m5=−tke1+e5≥0,m6=−tke1+e5−tk+−tke2+e6≥0,m7=−tke1+e5k2t2+−tke2+e6−tk−tke3+e7≥0,m8=−tke1+e5−∑i=3kk2t3−k+−tke2+e6k2t2+−tke3+e7−tk+−tke4+e8≥0,m9=k2t2e1−tke5+e9≥0,m10=k2t2e1−tke5+e9−tk+k2t2e2−tke6+e10≥0,m11=k2t2e1−tke5+e9k2t2+k2t2e2−tke6+e10−tk+k2t2e3−tke7+e11≥0,m12=k2t2e1−tke5+e9−∑i=3kk2t3−k+k2t2e2−tke6+e10k2t2+k2t2e3−tke7+e11−tk+k2t2e4−tke8+e12≥0,m13=−∑i=3kk2t3−ke1+k2t2e5−tke9+e13≥0,m14=−∑i=3kk2t3−ke1+k2t2e5−tke9+e13−tk+−∑i=3kk2t3−ke2+k2t2e6−tke10+e14≥0,m15=−∑i=3kk2t3−ke1+k2t2e5−tke9+e13k2t2+−∑i=3kk2t3−ke2+k2t2e6−tke10+e14−tk+−∑i=3kk2t3−ke3+k2t2e7−tke11+e15≥0,m16=−∑i=3kk2t3−ke1+k2t2e5−tke9+e13−∑i=3kk2t3−k+−∑i=3kk2t3−ke2+k2t2e6−tke10+e14k2t2+−∑i=3kk2t3−ke3+k2t2e7−tke11+e15−tk+−∑i=3kk2t3−ke4+k2t2e8−tke12+e16≥0.
Therefore, we get(46)e1≥0,e1e2≤1tk,e1e3≤1t2k2−k2t2,e1e4≤1∑i=3kk2t3+k−t3k3,e1e5≤1tk,e2e6≤1tk,e3e7≤1tk,e4e8≤1tk,e1e9≤1t2k2−k2t2,e3e11≤1t2k2−k2t2,e4e8≤1t2k2−k2t2,e1e13≤1∑i=3kk2t3+k−t3k3,e2e14≤1∑i=3kk2t3+k−t3k3,e3e15≤1∑i=3kk2t3+k−t3k3,e4e16≤1∑i=3kk2t3+k−t3k3.
In a similar way, we obtain the following relations forp≥3:
(47)e1≥0,e1e2≤1tk,(48)e1e3≤1t2k2−k2t2,e1e4≤1∑i=3kk2t3−t3k3,(49)e1e5≤1tk,e2e6≤1tk,(50)e3e7≤1tk,e4e8≤1tk,(51)e1e9≤1t2k2−k2t2,e3e11≤1t2k2−k2t2,(52)e4e8≤1t2k2−k2t2,e1e13≤1∑i=3kk2t3−t3k3,(53)e2e14≤1∑i=3kk2t3−t3k3,e3e15≤1∑i=3kk2t3−t3k3,(54)e4e16≤1∑i=3kk2t3−t3k3.We encounter a question about relations (35)–(54). How large should k be to claim that inequalities (35)–(54) can be converted into equalities? We will give an approximate answer to it by two examples.Example 3.
Forp=1,t=5 and an initial message M2×2, we have
(55)M=35820.
We considerk=1; then,
(56)E=L15,21×M×L15,21=1051358201501=32024146.
Here, we have the following:(57)3,320≃0.015,324≃0.125,3146≃0.020.
From (35), we have 1/tk=1/5=0.2 and 1/t2k2=1/25=0.04.
Now, fork=100, we have
(58)E=L15,2100×M×U15,2100=10500135820150001=310051508756520.
In this case,(59)3,31005≃0.0029,31508≃0.0019,3756520≃0.0000039.
Since1/tk=1/500=0.002 and 1/t2k2=1/250000=0.000004, we have the following relations for k=100:
(60)e1e2≃1tk,e1e3≃1tk,e1e4≃1t2k2.Example 4.
Forp=1,t=3 and an initial message as
(61)M=157223489,we consider k=2; then,
(62)E=L13,32×M×U13,32=10061029611572234891629016001=11166880349454352525.
In this case, we have the following values:(63)1,111≃0.09,166≃0.015,18≃0.125,180≃0.0125,1349≃0.0028,145≃0.022,1435≃0.00229,12525≃0.000396.
From (39)–(42), we have
(64)e1e2≤1tk=16≃0.166,e1e3≤1t2k2−k2t2−k=129≃0.0344,e1e4≤1tk=16≃0.166,e1e5≤1t2k2=136≃0.0277,e1e6≤1t3k3−k2kt2−k2t=1186≃0.0053,e1e7≤1t2k2−k2t2−k=129≃0.0344,e1e8≤1t3k3−k2kt2−k2t=1186≃0.0053,e1e9≤1t2k2+k−k2t22=1292≃0.001189.
Now, letk=100. Then, we get
(65)E=L13,3100×M×U13,3100=10020010545020011572234891200545001200001=1205645720241402130270358541198458324847519.
In this case, we obtain the following values:(66)1,1205≃0.0048,16457≃0.00015,1202≃0.0049,141402≃0.000024,11302703≃0.00000076,15854≃0.00017,11198458≃0.00000083,1324847519≃0.000003.
From (39)–(42), we have
(67)e1e2≤1tk=1200≃0.005,e1e3≤1t2k2−k2t2−k=120300≃0.000049,e1e4≤1tk=1200≃0.005,e1e5≤1t2k2=140000≃0.000025,e1e6≤1t3k3−k2kt2−k2t=1204000≃0.0000049,e1e7≤1t2k2−k2t2−k=120300≃0.000049,e1e8≤1t3k3−k2tk2−k2t=12040000≃0.0000049,e1e9≤1t2k2+k−k2t22=1203002≃0.0000024.
Therefore, by consideringt and the matrix M, we can find k=100 is an appropriate value.
Now, we calculate the error detection for thet-extension of the p-Fibonacci Pascal decoding. Error detection and correction of the code message E are the most important aim of the coding theory. First, we should determine the error detection in the code message E. For this, using the property of the determinant of a matrix, we can check the transmitted message E in the communication channel. We have
(68)E=Lpt,nk×M×Upt,nk,detE=detLpt,nk×M×Upt,nk=detM.
Therefore, it is clear that the determinant of the initial messageM is connected with the determinant of the code message E. So, we obtain the determinant of the matrix M. detM treats as a controller of entries of the code matrix E received from the communication channel. After receiving the code matrix E and computing the determinant of M, we will compute the determinant of E. Then, we will compare them together. If detE=detM, this means that the matrix E has passed from the communication channel without error. Otherwise, according to the matrix E of the order n×n1≤n≤4, we have “single,” “double,” ⋯, “n2-fold” errors. Thus, we get
(69)1Cn2+2Cn2+⋯+n2Cn2=2n2−1.Now, we will compute the error correction. Supposep=1 and n=2. Hence, there exists only one error in the matrix E received from the communication channel. The four variants of the single errors in the code matrix E are as follows:
(70)1ae2e3e42e1be3e4,3e1e2ce4,4e1e2e3d,where a,b,c, and d are the destroyed elements. By algebraic equations and Lemma 10, we obtain
(71)a=detM+e2e3e4,(72)b=−detM+e1e4e3,(73)c=−detM+e1e4e2(74)d=detM+e2e3e1.From relations (71)–(74) and (35), we calculate the destroyed element in the code matrix E.In a similar way, we can correct “double” and “triple” errors. For example, we consider the following case of double errors in the matrixE,
(75)ghe3e4,where g and h are the destroyed elements. We have
(76)ge4−he3=detM.Similarly, we compute the destroyed element in the code matrixE. So, there are 222−1=15 errors for n=2,p=1, and we can correct all cases of this method except the fourfold. Thus, the correctable possibility of it is equal to 14/15=0.9333=%93.33.Similarly, there are232−1=511 errors for n=3,p=1. Since detE=detM and from (40), we can correct up to eight cases of this method except the ninefold. Therefore, we get that correctable possibility of the method is equal to 510/511=0.9980=%99.80. And finally, we obtain 216−2/216−1≃1=%100 for n=4,p≥3.Thet-extension of the p-Fibonacci Pascal coding method has a high correction ability in comparison to the classical (algebraic) coding method. The reason is the use of matrix theory to get error-correction codes in this coding method, while in algebraic coding method there are very small information elements and bits and their combinations are the objects of detection and correction. For example, we compare the t-extension of the p-Fibonacci Pascal coding method to the Hamming coding by an example and show the correctable ability of the error is much greater in the t-extension of the p-Fibonacci Pascal coding method. For n=15 and k=10, we consider Hamming code method. We will obtain the correctable ability of the error. There is 210215−210=32505856 error messages and the number 215−210=31744 can just be corrected; then, the correctable ability of the error equals
(77)3174432505856=0.0009765=%0.09765.Besides, the correctable ability of the error in thet-extension of the 3-Fibonacci Pascal coding method is %99.80. Thus, the correctable ability of the k-Fibonacci Pascal is 10000 times more than Hamming one.


## 4. Conclusion

In this paper, we obtained thek-th power of the t-extension of the p-Fibonacci matrix Lpt,nk and the inverse of them Lpt,n−k for 2≤n≤4. Then, using them, we introduced a new coding/decoding method. The t-extension of the p-Fibonacci Pascal coding method is the main application of the matrices Lpt,nk and Lpt,n−k. The t-extension of the p-Fibonacci Pascal matrix coding/decoding was calculated very quickly by computer. Also, the correcting and detection abilities of this coding method were very high in comparison with a classical algebraic coding/decoding method.



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*Source: 4619136-2022-10-10.xml*