Codes over a subset of Octonion Integers Cristina Flaut 4 1 0 Abstract. InthispaperwedefinecodesoversomeOctonionintegers. Weprove 2 that in some conditions these codes can correct up to two errors for a transmitted vectorandthecoderateofthecodesisgraterthanthecoderateofthecodesdefined n a on some subset of Quaternion integers. J 0 3 Keywords. Block codes, Cyclic codes, Integer codes, Mannheim distance. ] AMS Classification. 94B15, 94B05. T I . s c [ 0. Preliminaries 1 v Coding Theory is a mathematical domain with many applications in Infor- 8 mation Theory. Various type of codes with their parameters were intensively 2 studied. As one of the important parameters of a code, the distance associated 8 (as Hamming, Lee, Mannheim, etc.) was also studied for many types of codes, 7 . finding values for the minimum or maximum values for such a distances ( see 1 [Ne, In,Fa, El, Pa; 01], [Sa, Na, Re, Sa; 13] ). 0 4 One of the type of codes with a great development in the last years are Inte- 1 ger Codes. Integer codes are codes defined over finite rings of integers modulo : m,m ∈ Z and have some advantages over the traditional block codes. One of v i them is that integer codes are capable to correct limited number of error pat- X ternswhichoccurmostfrequentlywhiletheconventionalcodesintendtocorrect r all possible error patterns. Integer codes have low encoding and decoding com- a plexityandaresuitableforapplicationinrealcommunicationsystems(see[Ko, Mo, Ii, Ha, Ma;10]). There are some other codes similar to the integer codes, as for example codes over Gaussian integers ([Hu; 94], [Gh, Fr; 10], [Ne, In,Fa, El, Pa; 01], [Ri; 95]), codes over Eisenstein–Jacobi integers ([Ne, In,Fa, El, Pa; 01]), a class of error correcting codes based on a generalized Lee distance ([Ni, Hi; 08]), codes over Hurwitz integers ([Gu; 13]), which are intensively studied in the last time. QAM is usedin many digitaldata radio communicationsand data commu- nications applications. The most common errors which appear are those which change a point with its nearest neighbor. The Hamming distance and the Lee distance are not capable to correct these errors in a QAM signal. To improve thissituation,in[Hu; 94],Huber constructedcodesoverGaussianintegerswith 1 a new distance, calledMannheimdistance. He provedthat these codes areable to correctMannheim errorof weight1 andused this new distance to find prop- ertiesofthesecodes(see (see[Mo,Ha,Ko;04])forotherdetails). Nevertheless, in [Ni, Hi; 08] the authors introduced a new distance which generalized the Lee distance and constructed codes capable to correcterrorsof generalizedLee weight one or two. In [Gu; 13], the author generalized some results from [Ne, In,Fa, El, Pa; 01] constructing codes over Hurwitz integers. Due to the structure of the real Octonion algebra, a nonassociative and a noncommutative algebra, in the presented paper, we generalize these results to aspecialsubsetofOctonionintegers,comparingwiththesomeresultsobtained until now. 1. Introduction The octoniondivisionalgebraoverR, denotedby O(R), is a nonassociative unital algebra. This algebra is power-associative (i.e. the subalgebra < x > of O(R), generated by any element x ∈ O(R), is associative), flexible (i.e. x(yx)=(xy)x=xyx, for all x,y ∈O(R)) and has the following properties: 1) O(R) is a free R−module with the basis {1,e ,e ,e ,e ,e ,e ,e }. 2 3 4 5 6 7 8 2) 1 is the unity in O(R). 3) e2 = e2 = e2 = e2 = e2 = e2 = e2 = −1 and e e = −e e = e ,i 6= 2 3 4 5 6 7 8 i j j i k j,i,j ∈{2,...,8},wherek =i⊗j,where⊗is”x-or”fori,j wroteinthedecimal basis (see [Ba; 09]). If x=x +x e +x e +x e +x e +x e +x e +x e ∈O(R), then its 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 conjugate istheoctonionx=x −x e +x e +x e +x e +x e +x e +x e 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 and the norm of the octonion x is N(x)=xx=xx=x2+x2+x2+x2+x2+x2+x2+x2. 1 2 3 4 5 6 7 8 Remark 1.1. The octonionic norm N is multiplicative, therefore for x,y ∈ O(R), we have N(xy)=N(x)N(y). The real part of the octonion x is x and the its vector part is x e +x e + 1 2 2 3 3 x e +x e +x e +x e +x e ∈O(R). 4 4 5 5 6 6 7 7 8 8 In [Co, Sm; 03] p. 55, was described Hurwitz integers or Hurwitz Integral Quaternions,denotedbyH,aselementsoftheformq =x +x e +x e +x e 1 2 2 3 3 4 4 where x ,x ,x ,x are in Z or in Z+ 1. In the same book, p. 99-105, was 1 2 3 4 2 defined Octavian Integers or Octonion Integers to be the set of the elements spanned by i ,i ,i ,i over O(Z), where 1587 2457 2685 2378 1 i = (e +e +e +e ). abcd a b c d 2 2 We will denote this ring with O. O(Z) is also called the set of Gravesian Octo- nion integers, the octonions with all coordinates in Z. 8 Let w = 1 1+ e ∈ O, be an integer octonion and let V={a + 2(cid:18) i(cid:19) iP=2 bw / a,b∈Z}. We remark that N(w)=2 and w2−w+2=0. Since octonion algebrais a power associativealgebra,it results that V is an associative and a commutative ring, V⊂ O. Remark 1.2. For x∈V the following properties are equivalent: i) x is invertible in the algebra V. ii) N(x)=1. iii) x∈{±1,±e ±e ,±e ,±e ,±e ,±e ,±e }. 2 3 4 5 6 7 8 Definition 1.3. The octonion x∈V is prime in V if x is not an invertible element in V and if x=ab, then a or b is an invertible element in V. Proposition 1.4. i) If x,y ∈ V, then there are z,v ∈ V such that x = zy+v, with N(v) < N(y). ii) With the above notation, we have that the remainder v has the formula xy v =x− y, (1.1) (cid:20)yy(cid:21) where the symbol [ ,] is the rounding to the closest integer. For the octonions, the rounding of an octonion integer can be found by rounding the coefficients of the basis, separately, to the closest integer. Proof. i)See [Co, Sm; 03],Theorem2, p. 109,[Da, Sa,Va; 03]Proposition 2.6.2, p. 60 and Lemma 2.6.3., p.61. 8 ii) From the above, we have v = x−zy with w = xy = w e ,w ∈ Z, i i i wi = azi+ri, 0 ≤ ri < a,i ∈ {1,...,8}. Therefore, zi = wai −iPra=i1= wyyi − yryi = (xyyy)i − yryi. Rounding to the closest integer, we have yryi =0, since ri <a.(cid:3) h i Remark 1.5. Let x=a+bw ∈V. We have that N(x)=xx=(a+bw)(a+bw)=a2+ab+2b2 = a+ b 2+7b2 =A2+7B2. 2 4 Proposition 1.6. [Co; 89],[Sa; 14] Let p ∈ N(cid:0)be a p(cid:1)rime number. There are integers a,b such that p=a2+ab+2b2 if p=7k+1,k∈Z.(cid:3) Definition 1.7. With the above notations, let π = x+yw be a prime integer in V and v ,v be two elements in V. If there is v ∈ V, such that 1 2 v − v = vπ, then v ,v are called congruent modulo π and it is denoted 1 2 1 2 v ≡v mod π. 1 2 Proposition 1.8. i) The above relation is an equivalence relation on V. The set of equivalence class is denoted by V and is called the residue class of V modulo π. π 3 ii) The modulo function µ:V→V is µ(x)=v modπ =x− xy y,where π yy h i x=zπ+v, with N(π)<N(y). iii) V is a field isomorphic with Z ,p=N(π),p a prime number. π p Proof. i) Ifv ≡v mod π and v ≡v mod π then there arev,v′ ∈V such 1 2 2 3 ′ ′ that v −v =vπ and v −v =v π. It results v −v =(v+v )π,therefore the 1 2 2 3 1 3 transitivity. We will denote with bold the elements from V . π ii) We define v +v = (v +v )mod π and v ·v = (v v )mod π. These 1 2 1 2 1 2 1 2 ′ ′ multiplications are well defined. Indeed, if v ≡ v mod π and v ≡ v mod π, 1 1 2 2 itresultsv −v′ =uπ,v −v′ =u′π,u,u′ ∈V,therefore(v +v )−(v′ +v′)= 1 1 2 2 1 2 1 2 ′ ′ ′ ′ (u+u)π.FromProposition1.4andsincev =v +uπ,v =v +uπ, itresults 1 1 2 2 ′ ′ v v =v v +M , with M a multiple of π. 1 2 1 2 π π Let f be the map f :Z →V , f(m)=µ(m+π)=(m+π) mod π. (1.2) p π ′ ′ Themapf iswelldefined,sinceifm≡m mod pwehave(m+π)−(m +π)= m−m′ =pq =ππq,q ∈Z, therefore (m+π)≡(m′+π) mod π. Since 1 = v π +v π, if f(m) = v,v = (m+π) mod π ∈ V , we define 1 2 π f−1(v) = m(v π) + m(v π). Indeed, for m ∈ Z , it results m=Mπ + m 1 2 p and m=m = Mπ + m, therefore m(v π) + m(v π) = (m−Mπ)(v π) + 1 2 1 m−Mπ (v π)=m(v π+v π) mod π =m mod π. 2 1 2 ′ (cid:0) The m(cid:1)ap f is a ring morphism. Indeed, f(m)+f(m) = (m+π)mod ′ ′ ′ ′ π+(m +π)mod π = (m+m +π)mod π = f(m+m) and f(m)f(m) = ′ (m+π)(m +π)modπ = = mm′+(m+m′)π+π2 mod π = (mm′+π)mod π.We obtain that V is π iso(cid:0)morphic with Zp.(cid:3) (cid:1) Remark 1.9. 1)ThefieldV hasthepropertythatifx,y ∈V , thenthere π π are z,v ∈ V such that x = zy+v, with N(v) < N(y) and V has N(π) = p π π elements. 2) If π = a+bw ∈ V, then in V two conjugate elements are not in the π same coset. Indeed, if x = m+nw ∈ V ,m,n ∈ Z, is in the same coset as π x = m−nw, therefore π / (x−x), then π / 2nw. We have 2nw = πq,q ∈ V, then N(π) / n, false, since N(x)=(m+ n)2+7n2 >N(π)2 >N(π). 2 4 Remark 1.10. 1) O(Z) has N4(π) elements. ([Ma, Be, Ga; 09]). π 2) From Proposition 1.8 and from Remark 1.9, we have that for v ,v ∈ i j V ,i,j ∈ {1,2,...,p− 1}, v + v = v if and only if k = i + j mod p and π i j k v ·v =v if and only if k =i·j mod p. From here, with the above notations, i j k we have the following labelling procedure: 1)Letπ∈Vbeaprime,withN(π)=p,paprimenumber,π =a+bw,a,b∈ Z. 4 2) Let s ∈ Z be the only solution of the equation a + bx mod p, x ∈ {0,1,2,...,p−1}. 3) The element k ∈ Z is the label of the element v = m + nw ∈ V if p m+ns=k mod p and N(v) is minimum. The aboveRemark generalizesto octonions Theorem1 and Labeling proce- dure from [Ne, In,Fa, El, Pa; 01]. Example 1.11. Using notations from Proposition 1.4, let p = 29 and π =−1+4w, with N(π)=29. We compute the field V . From the above, we π have U ={(m+π) mod π / m∈Z }. It results: π p V ={0,1,2,3,−3−w,−2−w,−1−w,−w,1−w,2−w, π 3−w,4−w,−2w−2,2w−2,−2w,−2w+1,−2w+2, 2+2w,w−4,w−3,w−2,w−1,w,1+w,2+w,3+w,−3,−2,−1}.Using the above labeling procedure, we have f(0)=0,f(1)=1,f(2)=2,f(3)=3, f(4)=−3−w, f(5)=−2−w,f(6)=−1−w,f(7)=−w,f(8)=1−w, f(9)=2−w,f(10)=3−w,f(11)=4−w,f(12)=−2w−2, f(13)=2w−2,f(14)=−2w,f(15)=−2w+1,f(16)=−2w+2, f(17)=2+2w,f(18)=−4+w,f(19)=−3+w,f(20)=−2+w, f(21)=−1+w,f(22)=w,f(23)=1+w,f(24)=2+w, f(25)=3+w,f(26)=−3,f(27)=−2,f(28)=−1. 2. Codes over V π Generalizingthe Hurwitz weightfrom[Gu; 13],wedefine the Cyley-Dickson weight, denotedd . Letπ be aprimeinV,π =a+bw. Letx∈V,x=a +b w. C 0 0 The Cyley-Dickson weight of x is defined as w (x) = |a |+|b |,where x = C 0 0 a +b w mod π, with |a |+|b | minimum. 0 0 0 0 The Cyley-Dickson distance between x,y ∈V is defined as π d (x,y)=w (x−y). C C We will prove that d is a metric. Indeed, for x,y,z three elements in V , we C π have d (x,y)=w (α )=|a |+|b |, where α =x−y =a +b mod π is an C C 1 1 1 1 1 1 element in V and |a |+|b | is minimum. π 1 1 d (y,z)=w (α )=|a |+|b |,whereα =x−y =a +b mod π isanelement C C 2 2 2 2 2 2 in V and |a |+|b | is minimum. π 2 2 d (x,z)=w (α )=|a |+|b |,whereα =x−y =a +b mod πisanelement C C 3 3 3 3 3 3 in V and |a |+|b | is minimum. π 3 3 We have x − y = α + α mod π. It results w (α +α ) ≥ w (α ) since 2 3 C 2 3 C 1 w (α )=|a |+|b | is minimum. C 1 1 1 5 Remark 2.1. The maximum Cayley-Dickson distance d between any Cmax twoelements fromV has the property thatd ≤|a|+|b|, with π =a+bw. π Cmax Remark 2.2. Since the Octonion algebra is alternative, due to Artin’s Theorem (see [Sc; 66]), each two nonzero different elements generate an asso- ciative algebra. From here, for x,y ∈ O we have that xm(xny) = xm+ny, for all m,n∈Z. In the following, we will consider π a prime in V of the norm N(π)=ππ = p,p=7l+1,l∈Z. Let α ,α be two primitive elements (of order p−1) in V , 1 2 π p−1 p−1 such that α 7 = w,α 7 = −w. Let α ∈ {α ,α }. We will consider codes of 1 2 1 2 length n= p−1. 7 Let C be the code defined by the parity-check matrix H, 1 α α2 ... αn−1  1 α8 α16 ... α8(n−1)  H = , (2.1) ... ... ... ... ...  1 α7k+1 α2(7k+1) ... α(7k+1)(n−1)    with k < n. We know that c is a codeword in C if and only if HcT = 0. n−1 From here, if we consider the associate code polynomial c(x) = c xi, we i iP=0 have that c α7l+1 = 0,l ∈ {0,1,...,k}. We consider the polynomial g(x) = (x−α) x−(cid:0)α8 ...(cid:1)x−α7k+1 .Sincetheelementsα,α8,...,α7k+1 aredistinct, from [Li(cid:0), Xi; 04(cid:1)], L(cid:0)emma 8.1.6(cid:1), we have that c(x) is divisible by the generator polynomial g(x). Since g(x) / ( xn±w), therefore g(x) is the generator poly- nomial for the code C. It results that C is a principal ideal in the ring V / π ((xn±w)). Theorem 2.3. Let C be the code defined on V by the parity check matrix π H = 1 α α2 ... αn−1 . (2.2) (cid:0) (cid:1) The code C is able to correct all errors pattern of the form e(x) = e xt, with t 0≤w (e )≤1 and any errors pattern of the form e(x)=e xt, with w (e )= C t t C t 3,e =±w2. t Proof. Let r(x) = c(x)+e(x) be the received polynomial, with c(x) the codeword polynomial and e(x) = e xt denotes the error polynomial with 0 ≤ t w (e )≤1. Since αn =w, or αn =−w andw2 =w−2,d w2 =3, it results C t C et =αnl.Wehavethesyndromes1 =αt+nl =αL,witht,L∈(cid:0)Z,0(cid:1)≤t,L≤n−1. If we reduce L modulo n, we obtain t, the location of the error,and from here, l= L−t, the value of the error.(cid:3) n Example 2.4. Withtheabovenotation,letπ =7+2w,p=71,n=10,α= 2−2w,w =α10 and the parity check matrix H = 1 α α2 α3 α4 α5 α6 α7 α8 α9 . (cid:0) (cid:1) 6 Supposing that the received vector is r = (w,1,w−1,1,1,0,0,0,1,1). Using MAPLE software,we compute the syndrome. It results S =Hrt =−5−2w= α14 mod π. We get L = 14, therefore the location of the error is l = L mod 14=4 mod 10. The value is w = α14−4 = α10 mod π. Therefore the corrected vector is c=r−(0,0,0,0,w,0,0,0,0,0)=(w,1,w−1,1,1−w,0,0,0,1,1) mod π. Theorem 2.5. Let C be a code defined by the parity-check matrix 1 α α2 ... αn−1 H = . (2.3) (cid:18) 1 α8 α16 ... α8(n−1) (cid:19) Then C can corrects any error pattern of the form e(x)= e xi, 0≤i ≤n−1, i with e ∈V . i π Proof. Let r(x) = c(x) + e(x) be the received polynomial, with c(x) the codeword polynomial and e(x) = e xi denotes the error polynomial with i e ∈ V . Then the corresponding vector of the polynomial r(x) is r = c+e i π and we will compute the syndrome S of r. We have e = αq,0 ≤ q ≤ 7n−1. i Therefore the syndrome is S=Hrt= s1 =αi+q =αM1 . (cid:18) s8 =α8i+q =αM2 (cid:19) We obtain ai+q−M1 = 1, with i+q = M1 mod(p−1) and α8i+q−M2 = 1, with 8i+q = M mod(p−1). We get 7i = (M −M ) mod(p−1), therefore the 2 2 1 uniquesolutionofthesystemisi= M2−M1 mod nandq =(M −i)mod(p−1). 7 1 In this way, we found the location and the value of the error.(cid:3) Example 2.6. Letπ =−1+4w,p=29,n=4,α=1−w,−w =α4 mod π, and the parity check matrix 1 α α2 α3 H = . (cid:18) 1 α8 α16 α24 (cid:19) Supposingthatthereceivedvectorisr = α,α2,1,α3 =(1−w,2−2w,1,−3+w).Using MAPLE software, we compute the syndr(cid:0)ome. It resu(cid:1)lts s =α7 S =Hrt = 1 . (cid:18) s8 =α7 (cid:19) The location of the error is i = 7−7 = 0 mod 4 and the value of the error is 7 α7−0 =α7 =17=(2+2w) mod π. Therefore the corrected vector is c=r−(2+2w,0,0,0)=(−1−3w,2−2w,1,−3+w)mod π = =(w−2,2−2w,1,−3+w). Theorem 2.7. Let C be a code defined by the parity-check matrix 1 α α2 ... αn−1 H = 1 α8 α16 ... α8(n−1) . (2.4) 1 α15 α30 ... α15(n−1)   7 Then C can find location and/or can correct some errors pattern of the form e(x)=e xi, 0≤i≤n−1, with e ∈V . i i π Proof. Using notations from the above Theorem, we have e =αq,0≤q ≤ i 7n−1. Therefore the syndrome is: s1 =αi+q =αM1 S=Hrt= s8 =α8i+q =αM2 . s15 =α15i+q =αM3   Since the rankofthe matrix(2.4)is 3,thenthis systemhasalwayssolution. We obtain ai+q−M1 = 1, with i+q = M1 mod(p−1), α8i+q−M2 = 1, with 8i+q = M2 mod(p−1), α15i+q−M3 = 1, with 15i+q = M3 mod(p−1). We canfindthelocationofthe errorif7i=(M −M )mod(p−1),7i=(M −M ) 2 1 3 2 mod(p−1), or, equivalently, i= M2−M1 mod n= M3−M2 mod n and the value of the error q if 7 7 (M −i) mod(p−1)=(M −8i) mod(p−1)=(M −15i)mod(p−1)(=q). (cid:3) 1 2 3 Example 2.8. 1) Let π = −1+4w,p = 29,n = 4,α = 1−w,−w = α4 mod π, and the parity check matrix 1 α α2 α3 H = 1 α8 α16 α24 . 1 α15 α30 α45   Supposing that the received vector is r = (1−w,2−2w,1,−3+w). Using MAPLE software, we compute the syndrome. It results s =α7 =αi+q 1 S =Hrt = s8 =α7 =α8i+q . s =α27 =α15i+q 15   The location of the error is i = 7−7 = 27−7 = 0 mod 4. We can’t find the 7 7 value of the error since α7−0 = α7 = 17 = (2+2w) mod π is different from α27−0 =α27 =11=(4−w) mod π. 2) In the same conditions, supposing that the received vector is r = 1,α3,1,α2 =(1,−3+w,1,−1−w).Again,usingMAPLE,thesyndrome is (cid:0) (cid:1) s =α21 =αi+q 1 S =Hrt = s8 =α11 =α8i+q . s =α18 =α15i+q 15   We can’t find the location and the value of the error, since 2 = 11−21 mod 46= 18−11 mod π =1. 7 7 3)Ifwesupposethatthereceivedvectorisr=(5,0,0,0)=(−2−w,0,0,0), the syndrome is s =α26 1 S =Hrt = s8 =α26 . s =α26 15   8 The location of the error is 0 and the value of the error is 5. Therefore the corrected vector is (0,0,0,0). Theorem 2.9. Let C be a code defined by the parity-check matrix 1 α α2 ... αn−1  1 α8 α16 ... α8(n−1)  H = . (2.5) 1 α15 α30 ... α15(n−1)  1 α22 α44 ... α22(n−1)    Then C can correct some errors pattern of the form e(x) = e xi +e xj, 0 ≤ i j i,j ≤n−1, with e ,e ∈V . i j π Proof. We will prove the general case, when we have two errors. We have e =αq 6=0 and e =αt 6=0,q,t∈Z. We obtain the syndrome: i j s =αi+q +αj+t 1 S=Hrt= s8 =α8i+q +α8j+t . s =α15i+q +α15j+t  15   s =α22i+q +α22j+t  22   Denoting αi+q =A and αj+t =B, it results s =A+B 1 S=Hrt= s8 =α8iA+α8jB  (2.6) s =α15iA+α15jB 15    s22 =α22iA+α22jB    If the system (2.6) admits only one solution, then the code C can correct two errors. First, we will prove the following Lemma. Lemma. With the above notations, if we have two errors, we obtain α7i 6= α7j,0≤i,j ≤n−1 and s s 6=s2. 1 15 8 Proof. If α7i = α7j,then α7(i−j) = 1 and 7n / 7(i−j), false. Supposing that s s −s2 = 0, we have s s = s2. For x = αi+q, it results α14is x+ 1 15 8 1 15 8 1 α14js2 − α14js x = α7i−α7j 2x2 + α14js2 + 2α7j α7i−α7j s x. We get 1 1 1 1 α7i−α7j 2x2 +2α7(cid:0)i+7js x−(cid:1)α14is x−α14js x = 0(cid:0). From he(cid:1)re, x = 0 or 1 1 1 (cid:0)x = −2α7i+(cid:1)7js1+α14is1+α14js1 = s . If x = αi+q = 0, false. If x = αi+q = s (α7i−α7j)2 1 1 implies αj+t =0,false. Now, we return to the proof of the Theorem and we are under conditions α7i 6=α7j,0≤i,j ≤n−1 and s s 6=s2. For B =s −A, it results: 1 15 8 1 A α7i−α7j =s −s α7j 8 1 A(cid:0)α14i−α14(cid:1)j =s −s α14j 15 1 A(cid:0)α21i−α21j(cid:1)=s −s α21j. We obtain: 22 1 (cid:0) (cid:1) s −s α14j = s −s α7j α7i+α7j 15 1 8 1 (cid:0) (cid:1)(cid:0) (cid:1) 9 and s −s α21j = s −s α7j α14i+α7iα7j +α14j . 22 1 8 1 (cid:0) (cid:1)(cid:0) (cid:1) Denoting α7i+α7j =s and α7iα7j =p , we have 7 7 s −s s +p s =0 15 8 7 7 1 and s −s α7j s2−p =s −s α21j. 8 1 7 7 22 1 (cid:0) (cid:1)(cid:0) (cid:1) It results s s −s 8 7 15 p = 7 s 1 and s (s s −s2)=s s −s s . 7 1 15 8 1 22 8 15 We obtain s s −s s 1 22 8 15 s = 7 s s −s2 1 15 8 and for p we get 7 s s −s2 p = 8 22 15. 7 s s −s2 1 15 8 Fromhere,solvingthe equationx2−s x+p =0, we willfind the locations 7 7 and the values of the errors. (cid:3) Example 2.10. 1) Let π = −1+4w,p = 29,n = 4,α = 1−w,−w = α4 mod π, and the parity check matrix 1 α α2 α3  1 α8 α16 α24  H = . 1 α15 α30 α45    1 α22 α44 α66    Supposing that that the received vector is r = 1,α3,1,α2 =(1,−3+w,1,−1−w). Using againMAPLE,the syndrome is (cid:0) (cid:1) s =α21 1 S=Hrt= s8 =α11 . s =α18  15   s22 =α20    We obtain s s −s s 1 22 8 15 s = =(2+w) mod π 7 s s −s2 1 15 8 and 10