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Department of Mathemati s, Mahidol University Kit Tyabandha, PhD Midterm Examination Coding Theory th 27 January 2006 Time: 2 hours (12:30{2:30pm) 1. Our task is to design a Hamming ode in the ase where there are4 he k digits. First (cid:12)nd the binary-representative matrix M to be used for the purpose.[1℄y Find the length of the ode word [1℄ and that of a message word [1℄. What is the redundan y of this ode?[1℄ Then onstru t the ode.[5℄ And then ode the message words 10101010101[3℄ and 00111100101[3℄. 4 4 Solution.Wehave r =4, n=2 (cid:0)1=15 and m=2 (cid:0)4(cid:0)1=11. Ea h ode word has 15 digits and 4 he k digits. # Ea h message The redundan y of the ode is 4. # he k digits b1 b2 b4 b8 " " " " ode word b1 b2 b3 b4 b5 b6 b7 b8 b9 b10 b11 b12 b13 b14 b15 # # # # # # # # # # # message word a1 a2 a3 a4 a5 a6 a7 a8 a9 a10 a11 r The (2 (cid:0)1)(cid:2)r matrix M is then 0 1 0 0 0 1 B0 0 1 0C B C B0 0 1 1C B C B0 1 0 0C B C B0 1 0 1C B C B0 1 1 0C B C B0 1 1 1C B C M =B1 0 0 0C B C B1 0 0 1C B C B1 0 1 0C B C B1 0 1 1C B C B1 1 0 0C B C B1 1 0 1C A 1 1 1 0 1 1 1 1 Then form the matrix equation (b1 b2 (cid:1)(cid:1)(cid:1) b15)M =0 whi h gives four equations in four unknown, b8+b9+b10+b11+b12+b13+b14+b15 =0 b4+b5+b6+b7+b12+b13+b14+b15 =0 b2+b3+b6+b7+b10+b11+b14+b15 =0 b1+b3+b5+b7+b9+b11+b13+b15 =0 The message word 10101010101be omes the ode word b1 b2 1 b4 0 1 0 b8 1 0 1 0 1 0 1 y Numbers pla ed between square bra kets are marks. th Coding theory, Midterm Exam {1{ From 27 Jan 05, as of 27 January, 2006 Department of Mathemati s, Mahidol University Kit Tyabandha, PhD Hen e, b8 =0, b4 =1, b2 =0 and b1 =1, and the ode word is 101101001010101. # The message 00111100101be omes the ode word b1 b2 0 b4 0 1 1 b8 1 1 0 0 1 0 1 whi h gives us b8 =0, b4 =0, b2 =0 and b1 =1. Therefore the ode word is 100001101100101. # (cid:10) (cid:11) 3 2 3 2 2. Show that the polynomial x +x +1 is irredu ible [4℄ and the element (cid:11)=x+ x +x +1 (cid:10) (cid:11) 3 2 ofF[x℄= x +x +1 primitive[4℄. Then usethis to onstru tabinaryBCH odeoflength7and minimum distan e 3.[5℄ And then (cid:12)nd the ode word for the message 1101.[2℄ 3 2 Solution.Suppose x +x +1 is redu ible, then it must have either x or x+1 as a fa tor, then 3 2 x=0 or 1 is a root of x +x +1. But 2 x +x 3 2 x x +x +1 3 x 2 x +1 2 x 1 2 x 3 2 x+1 x +x +1 3 2 x +x 1 3 2 3 2 from whi h we an see that both xjx +x +1 and x+1jx +x +1 give a remainder 1. 3 2 Therefore neither x nor x+1 divides x +x +1, hen e the latter is irredu ible. # pn(cid:0)1 k Weknowthatf(x) in Fp[x℄ ofdegreenisprimitiveif f(x)jx (cid:0)1and f(x)6jx (cid:0)1forany n 23(cid:0)1 7 k <p (cid:0)1. We have x (cid:0)1=x (cid:0)1. Then 4 3 2 x +x +x +1 3 2 7 x +x +1 x (cid:0)1 7 6 4 x +x +x 6 4 x +x +1 6 5 3 x +x +x 5 4 3 x +x +x +1 5 4 2 x +x +x 3 2 3 2 7 x +x +1 !0 ) x +x +1jx (cid:0)1 For k <7; if k =6; 3 2 x +x +x 3 2 6 x +x +1 x (cid:0)1 6 5 3 x +x +x 5 3 x +x +1 5 4 2 x +x +x 4 3 2 x +x +x +1 4 3 x +x +x 2 3 2 6 x +x+1 6=0 ) x +x +16jx (cid:0)1 If k =5; th Coding theory, Midterm Exam {2{ From 27 Jan 05, as of 27 January, 2006 Department of Mathemati s, Mahidol University Kit Tyabandha, PhD 2 x +x+1 3 2 5 x +x +1 x (cid:0)1 5 4 2 x +x +x 4 2 x +x +1 4 3 x +x +x 3 2 x +x +x+1 3 2 x +x +1 3 2 5 x 6=0 ) x +x +16jx (cid:0)1 If k =4; x+1 3 2 4 x +x +1 x (cid:0)1 4 3 x +x +x 3 x +x+1 3 2 x +x +1 2 3 2 4 x +4 6=0 ) x +x +16jx (cid:0)1 (cid:10) (cid:11) 3 2 3 3 2 When k =3, x +x +16jx (cid:0)1 is obvious. Therefore(cid:11)=x+ x +x +1 is aprimitive of (cid:10) (cid:11) 3 2 F[x℄= x +x +1 . 3 2 n Then, (cid:11) satis(cid:12)es (cid:11) +(cid:11) +1=0. For F a (cid:12)nite (cid:12)eld of order p with k as its prime sub(cid:12)eld, p we know that (cid:11) and (cid:11) have the same minimum polynomial over k for every (cid:11) 2 F. Here p =1; 2 3 therefore (cid:11) and (cid:11) have the same minimum polynomial, hen e the generating polynomial is x + 2 2 3 x +1. Letthemessagebea0a1a2a3. Thenthemessag(cid:0)epolynomia(cid:1)lisa(x)=a0+a1x+a2x +a3x . 3 2 The orresponding ode polynomial is therefore a(x) x +x +1 . In other words, 2 3 4 5 6 a0+a1x+(a0+a2)x +(a0+a1+a3)x +(a1+a2)x +(a2+a3)x +a3x The ode word is thus (a0;a1;(a0+a2);(a0+a1+a3);(a1+a2);(a2+a3);a3) # For the message 1101, the ode word is 1111111. # th Coding theory, Midterm Exam {3{ From 27 Jan 05, as of 27 January, 2006