DECODING OF CYCLIC CODES OVER THE RING F2hu[tui]
Karim Samei∗and Mohammad Reza Alimoradi∗∗
∗Department of Mathematics, Bu-Ali Sina university, Hamedan, Iran
∗∗Department of Mathematics, Faculty of Mathematical Sciences,
University of Malayer, Malayer, Iran
e-mails: [email protected]; [email protected]
(Received 17 October 2016; accepted 14 March 2018)
In this paper we resolve an open problem about decoding cyclic codes over the ringF2+uF2with u2 = 0.This problem was first proposed by AbuAlrub et al. in (Des Codes Crypt 42: 273-287, 2007). Also we extend this decoding procedure for cyclic codes of arbitrary length over the ringe
F2[u]
huti =F2+uF2+u2F2+· · ·+ut−1F2, whereut= 0.
Key words : Cyclic codes, Hamming distance, decoding, torsion codes.
1. INTRODUCTION
Codes over finite rings have received much attention recently after Hammons et al. [4] found that
some of the best non-linear codes such as the Kerdock, Preparata and Goethal codes can be viewed
as linear codes overZ4 via the Gray map fromZ4n toF22n. Note that codes over finite rings have been studied in different contexts by numerous authors [1, 2, 4, 6, 7, 10]. Among codes over finite
rings the class of cyclic codes is a significant class from both theoretical and practical point of view.
Also cyclic codes are the powerful error-correcting codes and can be efficiently encoded and decoded.
Recall that ringsZ4,F2+uF2, withu2 = 0andF2×F2are three non-isomorphic ring of order four.
Since some of the binary codes with good error-correcting capability are Gray images of cyclic codes
over these rings [1, 2, 4, 10], then the study of codes over these rings is significant. Unlike the binary
field, there is a little work in the literature on the decoding of codes over finite rings. Recently a few
papers have been published, such as decoding of negacyclic codes over the ringZ4,[3]and decoding
of cyclic codes of odd-length over the ringF2+uF2,[9]. In this paper we suggest a simple procedure
ut−1F
2,withut = 0by using of the torsion codes, which are codes over residue field associated to a chain ring. Also we consider Hamming distance in order to estimate the error correction capability
of codes over this ring.
LetRbe a commutative ring with identity, a linear codeC of lengthnis aR-submodule ofRn. For any ringR, a cyclic shift onRnis a permutationσsuch that
σ(c0, c1, . . . , cn−2, cn−1) = (cn−1, c0, c1, . . . , cn−2)
A linear codeC over the ringR that is invariant under the cyclic shift is called a cyclic code.
Recall that with the following correspondence:
Rn→ R[x] hxn−1i
(c0, c1, . . . , cn−1)→c0+c1x+. . .+cn−1xn−1
a cyclic codeCof lengthnoverRis an ideal ofRn= hxRn[−x]1i. The Hamming weight of a codeword
c = (c0, c1, . . . , cn−1) ∈ C is defined by wH(c) = |{i : ci 6= 0}|and the minimum Hamming
distance of a codeCis defined bydH(C) =min{wH(c) : 06=c∈C}.We denote F2hu[tui]byRtand Rt[x]
hxn−1i byRt,n.A finite ringRis called a chain ring if its ideals ordered by inclusion. Clearly a finite
chain ring is a local ring. Also a finite ring is a finite chain ring if and only if it is a local ring and its
maximal ideal is a principal ideal(see proposition2.1in[7]). Lethaibe a unique maximal ideal of
the finite chain ringR, whereais a nilpotent element ofRwith nilpotency indexeand let denote by kthe field hRai. Now assume thatCis a code of lengthnover the chain ringR.Fori= 1,2, . . . , e−1
the projections of(C : ai) over the fieldkare denoted by(C :ai) and are called the torsion codes associated to the codeC, where(C :ai)is defined as(C :ai) ={x∈Rn:xai ∈C}.
2. MAINRESULTS
Udaya and Bonnecaze[9] gave a decoding algorithm for cyclic codes of odd-length over the ring F2+uF2by using a suitable Gray map and ahu,u+viconstruction. They associated to each code Cover the ringF2+uF2two binary codes
Res(C) ={a(x)∈ F2[x]
hxn−1i| ∃b(x), a(x) +ub(x)∈C}
and
T or1(C) ={k(x)∈ F2[x]
In addition the decoding procedure is done in Galois extension ofF2+uF2. In this section we present
a decoding procedure for cyclic code over the ringF2hu[tui]by using of torsion codes. At first we explain
this procedure for cyclic codes over both ringsF2+uF2andF2+uF2+u2F2. Now letCbe a code
of lengthnover the ringF2+uF2. Note thatF2+uF2is a chain ring with maximal idealhui, where uis a nilpotent element with nilpotency index two. Now we associate to the codeCtwo binary codes.
The torsion codeT or1(C)is defined as:
T or1(C) ={k(x)∈F2,n : uk(x)∈C}
and the codeCu is defined as:
Cu={k(x)∈F2,n: t(x) +uk(x)∈C, for some, t(x)∈F2,n}.
Similarly the ringF2+uF2 +u2F2 is a chain ring with unique maximal idealhuiand residue
fieldF2. Now letCbe a cyclic code of lengthnoverF2+uF2+u2F2, we associate to the codeC
two binary codesT or2(C)andCu
2
, which the codeCu2 is defined as:
Cu2 ={c2(x)∈F2,n:c0(x) +uc1(x) +u2c2(x)∈Cfor somec0(x), c1(x)∈F2,n}.
([1],Theorem 1)Let C be a cyclic code of length n over the ring F2 +uF2, then C = hg(x) + up(x), ua(x)i, whereg(x), p(x), a(x)are binary polynomials witha(x)|g(x)andg(x)|(xn−1)in F2, and a(x) |p(x)xgn(x−)1 anddeg(p(x)) < deg(a(x)). ([7],Theorem 3.6).LetC be a cyclic code
overRt, thenC is an ideal inRt,nthat can be generated byC =hg+up1+· · ·+ut−1pt−1, ua1+ u2q
1 +· · · +ut−1qt−2, u2a2 +u3r1 +· · · +ut−1rt−3, . . . , ut−2at−2 +ut−1s1, ut−1at−1i, with at−1(x)|at−2(x)|. . . , |a2(x)|a1(x)|g(x)|(xn−1)inF2.Whennis an odd number, thenC =
hg(x) +ua1(x) +· · ·+ut−1at−1(x)i. Cbe a cyclic code of odd-lengthnover the ringF2+uF2,
thenC=hg(x) +ua(x)i.
PROOF : Since n is an odd number, then xn −1 factors uniquely into a product of distinct irreducible polynomials. Thengcd(a(x),xgn(x−)1) = 1,soa(x)|p(x). Now we know thatdeg p(x)< deg a(x),thenp(x) = 0. SoC = hg(x), ua(x)i. Leth(x) = g(x) +ua(x). Sincenis relatively
prime to two, thenR2,n is a reduced and0-dimensional ring. So by Lemma1in[5],there exist an
idempotente(x) inR2,n such that hg(x)i = he(x)i. Thuse(x) = r(x)g(x) for some polynomial
r(x) ∈ R2,n. Then e(x) = r2(x)g2(x). Ash2(x) = g2(x), then e(x) ∈ hh(x)i.This implies that
g(x)∈ he(x)i ⊆ hh(x)i. Sohg(x), ua(x)i=hh(x)i.
PROOF: Letk(x)∈T or1(C), thenuk(x)∈C.So there existr0(x) +ur1(x), s0(x) +us1(x)∈ R2,nsuch that
uk(x) = (r0(x) +ur1(x))(g(x) +up(x)) + (s0(x) +us1(x))ua(x)
But we know thata(x)|g(x), thenk(x) ∈ ha(x)i.Converselyua(x) ∈C, implies thata(x) ∈
T or1(C).Now Theorem4.2of[6]implies thatdH(C) =dH(T or1(C)).
Lemma 2 — LetCbe a cyclic code of odd-lengthnoverF2+uF2, thenCu=T or1(C).
PROOF : Let c2(x) ∈ T or1(C), then there exist c1(x) ∈ R2,n such that c1(x) + uc2(x) ∈ C=hg(x) +ua(x)i. Soc1(x) +uc2(x) = (h1(x) +uh2(x))(g(x) +ua(x)), for some polynomials h1(x), h2(x)∈F2,n. This implies thatc2(x) =h1(x)a(x)+h2(x)g(x). As we know thata(x)|g(x),
thenc2(x)∈ ha(x)i, soCu⊆T or1(C). Conversely ifc(x)∈T or1(C), thenuc(x)∈C, this implies thatc(x)∈Cu. SoCu=T or1(C).
Theorem 1 — LetC=hg(x)+up(x), ua(x)ibe a cyclic code of lengthnoverF2+uF2, w(x) =
w1(x)+uw2(x)be a received word with an error polynomiale(x) =e1(x)+ue2(x)andwH(ei(x))≤
bdH(T or1(C)−1)
2 c, fori= 1,2. Thenwi(x)will be decoded in binary codeT or1(C).
PROOF: Case(i)Letnbe an odd number andw(x) =c(x) +e(x), wherec(x) =c1(x) +uc2(x)
is a codeword inC. Sinceuc(x) = uc1(x) ∈ C, thenc1(x) ∈ T or1(C). Asuc1(x) =u(w1(x)− e1(x))andT or1(C) is a binary cyclic code, then we can determinee1(x) with using of decoding
algorithms in F2. Since dH(w1, c1) = wH(e1) ≤ bdH(T or12(C)−1)c, then the word w1(x) will be decoded uniquely toc1(x).Similarlyw2(x) = c2(x) +e2(x). Sincec2(x) ∈ Cu = T or1(C)and dH(w2, c2) =wH(e2)≤ bdH(T or12(C)−1)c, thenw2(x)will be decoded uniquely toc2(x).
Case(ii) : Letnbe an even number and w(x) = c(x) +e(x), wherec(x) = c1(x) +uc2(x)
is a codeword inC. Sinceuc(x) = uc1(x) ∈ C, thenc1(x) ∈ T or1(C). Similar to caseiwe can
determinee1(x)in binary codeT or1(C)andw1(x)will be decoded toc1(x). Also by the structure
of codeCthere exist binary polynomialsr1(x), r2(x), s(x) ∈ F2,n such thatc2(x) = r1(x)p(x) + r2(x)g(x) +s(x)a(x). Now we know thata(x)|g(x). Sow2(x)−r1(x)p(x)−e2(x)∈ ha(x)i. Let w´2(x) =w2(x)−r1(x)p(x), thenw´2(x)−e2(x) =ha(x)i=T or1(C). So we can determinee2(x)
in binary codeT or1(C)with using of decoding algorithms inF2.
Lemma 3 — LetC =hg(x) +up1(x) +u2p2(x), ua1(x) +u2q1(x), u2a2(x)ibe a cyclic code
of lengthnoverF2+uF2+u2F2, thenT or2(C) =ha2(x)iand alsodH(C) =dH(T or2(C)).
uh1(x) +u2h2(x), whereh0(x), h1(x), h2(x)are binary polynomials. Thenu2h(x) =u2h0(x) = u2k(x). So T or
2(C) = {k(x) ∈ F2,n : u2k(x) ∈ C}. Then by theorem 4.2 of [6], we have
dH(C) =dH(T or2(C)).Sinceu2a2(x)∈C, thena2(x)∈T or2(C).Soha2(x)i ⊆T or2(C).
Conversely let k(x) ∈ T or2(C), then u2k(x) ∈ C, therefore u2k(x) = (r0(x) +ur1(x) + u2r2(x))(g(x) +up1(x) +u2p2(x)) + (s0(x) +us1(x) +u2s2(x))(ua1(x) +u2q1(x)) + (t0(x) + ut1(x) +u2t2(x))(u2a2(x)).Sok(x) =r2(x)g(x) +s1(x)a1(x) +t0(x)a2(x).But we know that a2(x)|a1(x)|g(x),thenk(x)∈ ha2(x)i.
Lemma 4 — LetCbe a cyclic code of odd-lengthnover the ringF2+uF2+u2F2, thenCu
2
= T or2(C).
PROOF: Letc2 ∈Cu
2
, then there existc0, c1 ∈F2,nsuch that
c0(x) +uc1(x) +u2c2(x)∈C=hg(x) +ua1(x) +u2a2(x)i
So c0(x) + uc1(x) + u2c2(x) = (h0(x) + uh1(x) + u2h2(x))(g(x) + ua1(x) + u2a2(x)).
Thenc2(x) =h0(x)a2(x) +h1(x)a1(x) +h2(x)g(x). Now, we know thata2(x)|a1(x)|g(x), then c2(x) ∈ ha2(x)i.So Cu
2
⊆ T or2(C). Conversely ifc(x) ∈ T or2(C), then u2c(x) ∈ C. This
implies thatc(x)∈Cu2. SoCu2 =T or2(C).
For any cyclic codeCof lengthnoverF2+uF2+u2F2, we define binary codeCuas following:
Cu ={c1∈F2,n:c0+uc1+u2c2∈Cfor somec0, c2 ∈F2,n}.
Lemma 5 — LetC be a cyclic code of odd-lengthnover the ring F2 +uF2+u2, thenCu = T or1(C).
PROOF: At first we show thatT or1(C) = ha1(x)i, whereC = hg(x) +ua1(x) +u2a2(x)i.
ClearlyT or1(C) ={k(x) ∈F2,n :uk(x) +u2t(x) ∈C,∃t(x)∈ F2,n}. Sinceua1(x) ∈ C,then a1(x)∈T or1(C).Conversely ifk(x)∈T or1(C), thenuk(x) +u2t(x)∈C, for somet(x)∈F2,n.
uk(x) +u2t(x) = (h0(x) +uh1(x) +u2h2(x))(g(x) +ua1(x) +u2a2(x))
Sok(x) =h0(x)a1(x) +h1(x)g(x).Thenk(x)∈ ha1(x)i.Sinceua1(x)∈C, thena1(x)∈Cu. Soha1(x)i ⊆Cu. Letc1(x)∈Cu, then there existc0(x), c2(x)∈F2,n,such thatc0(x) +uc1(x) + u2c2(x)∈C. So
c0(x) +uc1(x) +u2c2(x) = (h0(x) +uh1(x) +u2h2(x))(g(x) +ua1(x) +u2a2(x))
Theorem 2 — LetC=hg(x) +up1(x) +u2p2(x), ua1(x) +u2q1(x), u2a2(x)ibe a cyclic code
of lengthnover the ringF2+uF2+u2F2. Ifw(x) =w0(x) +uw1(x) +u2w2(x)be a received word
with an error polynomialse(x) =e0(x)+ue1(x)+u2e2(x). AlsowH(ei(x))≤ b(dH(T or22(C)−1)cfor
i= 0,2.andwH(e1(x))≤ b(dH(T or12(C)−1)c, thenw0(x), w2(x)will be decoded in codeT or2(C)
andw1(x)will be decoded in codeT or1(C).
PROOF : Case (i). Suppose n is an odd number and w(x) = c(x) +e(x), where c(x) = c0(x) + uc1(x) +u2c2(x) is a code-word in C. Since u2c(x) = u2c0(x) ∈ C and u2c0(x) = u2(w
0(x)−e0(x)), this implies thatw0(x)−e0(x) ∈ T or2(C). SinceT or2(C) is a binary code,
we can decodew0 inT or2(C). Sincec1(x) ∈ Cu = T or1(C)andc1(x) = (w1(x)−e1(x)). As T or1(C)is a binary code, then we can decodew1inT or1(C). In this approach we can decodew2(x)
in binary codeCu2 =T or2(C).
Case(ii). Letnbe an even number andw(x) =c(x) +e(x), wherec(x) =c0(x) +uc1(x) + u2c
2(x) is a codeword in C. Similar to case i, we can decode w0(x) in binary code T or2(C).
By using of the structure of code C, there exist binary polynomials r0, r1, r2, s0, s1, t0 such that c1(x) =r0(x)p1(x) +r1(x)g(x) +s0(x)a1(x)and
c2(x) =r0(x)p2(x) +r1(x)p1(x) +s0(x)q1(x) +r2(x)g(x) +s1(x)a1(x) +t0(x)a2(x)
Letw´1(x) =w1(x)−r0(x)p1(x), now we know thata1(x)|g(x). Sow´1(x)−e1(x)∈ ha1(x)i= T or1(C). Sow´1(x)will be decoded in binary codeT or1(C). Thenw´1(x) = d1(x)a1(x) +e1(x)
for some binary polynomial d1(x). Therefore r1(x), s0(x) completly determined by dividing
bi-nary polynomialsd1(x) to b1(x).Let w´2(x) = w2(x)−r0(x)p2(x) −r1(x)p1(x) −s0(x)q1(x)
Asa2(x)|a1(x)|g(x), thenw´2(x)−e2(x) ∈ ha2(x)i = T or2(C).Sow´2(x)and thereforew2(x)
will be decoded in binary codeT or2(C).
We work an example of this decoding procedure.
Let C = h(x −1)7 +u(x −1)5 +u2(x −1)4, u(x−1)6 +u2(x−1)3, u2(x −1)5i be a
cyclic code of length8 overF2 +uF2 +u2F2. Now we know thatdH(C) = dH(T or2(C))and T or2(C) = ha2(x)i =h(x−1)5i, we must computedH(T or2(C)). Since5has a2-adic length1
nonzero expansion, thendH(C) = 4. (see Lemma 10 in [1]). SoC is a1-error-correcting code. If
w(x) =x6+x5+x4+x3+x2+x+1+u(x7+x6+x5+x2+x)+u2(x7+x6+x5+x2+1)be a received
word, thenw0(x) = xa2(x) + (x4 +x3+ 1). So by using of decoding algorithm for cyclic codes
over binary field (see Page 148 in [8]), we obtaine0(x) =x7. Thereforec0(x) = (x−1)7 =g(x). So r0(x) = 1, thenw´1(x) = (x+1)a1(x)+x5+x3+x.Similarly using of decoding algorithm for cyclic
Sor1(x) = (x−1)7, s0(x) = 1. Therefore we obtainw´2(x)−e2(x) = x2a2(x) +x4 +x2 + 1,
thene2(x) = x6.If we correct these errors in the received polynomial, then the vectorw(x)will be
decoded toc(x) =x7+x6+x5+x4+x3+x2+x+ 1 +u(x6+x4+x2+ 1) +u2(x7+x5+ x4+x3+x+ 1).In continue of this section we will describe decoding procedure for cyclic codes
over F2hu[tui] fort ≥ 3. Now let C be a cyclic code over the ringRt, for anyi = 1,2, . . . , t−1 we
define codesCui asCui = {ci(x) ∈ F2,n : c0(x) +uc1(x) +· · ·+ut−1ct−1(x) ∈ C}, for some c0(x), c1(x), . . . , ci−1(x), ci+1(x), . . . , ct−1(x)∈F2,n.
Clearly T ori(C) = {k(x) ∈ F2,n : uik(x) ∈ C} and Cu i
are binary cyclic codes fori = 1,2, . . . , t−1.
Lemma 6 — LetC=hg+up1+· · ·+ut−1pt−1, ua1+· · ·+ut−1qt−2, u2a2+· · ·+ut−1rt−3, . . . , ut−2a
t−2 +ut−1s1, ut−1at−1i be a cyclic code of length noverRt, then T ori(C) = hai(x)i for
i= 1,2, . . . , t−1anddH(C) =dH(T ort−1(C)).
PROOF: Similar to Lemma4.
Lemma 7 — LetC =hg+ua1+u2a2+ut−2at−2+ut−1at−1ibe a cyclic code of odd-length nover the ringRt, then fori= 1,2, . . . , t−1, the relationCu
i
=T ori(C) =hai(x)idoes hold.
Proof : Let k(x) ∈ T ori(C),thenuik(x) ∈ C. Sok(x) ∈ Cu i
.Conversely let ci(x) ∈ Cu i
,
thenc0(x) +uc1(x) +· · ·+ut−1ct−1(x) ∈ Cfor some polynomialsc0(x), c1(x), . . . , ct−1(x).So c0(x) +uc1(x) +· · ·+ut−1ct−1(x) = (r0(x) +ur1(x) +· · ·+ut−1rt−1(x))(g(x) +ua1(x) + u2a
2(x) +ut−2at−2(x) +ut−1at−1(x)).Thenci(x) =g(x)ri(x) +a1(x)ri−1(x) +· · ·+ai(x)r0(x) for anyi= 1,2, . . . , t−1. But we know thatai(x)|ai−1(x)|. . .|a2(x)|a1(x)|g(x), thenci(x)∈
hai(x)i=T ori(C).
Theorem 3 — LetChg+up1+· · ·+ut−1pt−1, ua1+· · ·+ut−1qt−2, u2a2+· · ·+ut−1rt−3, . . . , ut−2 at−2 +ut−1s1, ut−1at−1i be a cyclic code of length n over the ring Rt. If w(x) = w0(x) + uw1(x) +· · ·+ut−1wt−1(x)be a received word with an error polynomialse(x) =e0(x) +ue1(x) +
· · ·+ut−1e
t−1(x), wH(e0(x)) ≤ b(dH(T ort−21(C)−1)c andwH(ei(x)) ≤ b(dH(T or2i(C)−1)c for i =
1,2, . . . , t−1.Thenw0(x)will be decoded in binary codeT ort−1(C)andwi(x)will be decoded in
binary codesT ori(C).fori= 1,2, . . . , t−1.
PROOF: Similar to Theorem2.
3. CONCLUSION
is an arbitrary number with using of torsion codes associated to code, which these codes are binary
codes. A natural open problem is to extend this work to cyclic codes over chain rings, which residue
field of chain ring is of characteristic two. We also expect to present a decoding algorithm for cyclic
codes over this ring with considering the Lee weight.
ACKNOWLEDGEMENT
The authors are thankful to the anonymous referees for their careful reading of the paper and valuable
comments.
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