MATH3411 INFORMATION, CODES & CIPHERS Test 2 Session 2 2016 SOLUTIONS
Multiple choice: e, b, a, d, b True/False: F, T, T, T, F .
1.(e):1.a, 2.b, 3.ba, 4.bab, 5.baba
(d): φ(2016) = φ(25)φ(32)φ(7) = (25 − 24)(32 − 3)6 = 576, so by Euler’s Theorem,
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51155 ≡(5576)2×53 ≡12×125≡125 (mod2016). (b)gcd(3,121)=1and315 ≡1 (mod121)(here,121=23∗15+1).
False: Encode the message bb• : begin
subinterval start
0 0.3 0.3+0.3×0.4 = 0.42 0.42+0.7×0.16=0.532
1 0.4 0.4×0.4 = 0.16 0.16×0.3=0.048
so the message encodes as a number in the interval [0.532, 0.58).
(ii) True: The ternary entropy is approximately 0.966 .
(iii) True: t = 37 and s = 6, so a = 31 and b = 43.
(iv) True: The longest two codeword lengths are 7 and 8.
(v) False: There are φ(125 − 1) = 60 primitive elements in GF(125).
Here,wehavethatα2 =α+1:
α α6 1 R2=α−1R2 1 α5 α7 R1=R1−αR2 1 0 α7
x α7 α + 2 so y = 0 = 0 .
α1 = α α5 = 2α
α2 =α+1 α6 =2α+2 α3 =2α+1 α7 =α+2 α4 = 2 α8 = 1
α3 α4 α2R1=α5R1 1 α α7R2=R2−R1 1
α α7 1 α α7
(iii) {α5, α15 = α7, α21 = α5, . . .} = {α5, α7}, so the minimal polynomial of α5 is
(x−α5)(x−α7)=x2 −(α5 +α7)x+α5α7 =x2 −(2α+α+2)x+α4
= x2 + x + 2 .
Multiple choice: e, d, c, e, d True/False: T, F, T, F, T .
(e): 1.a, 2.b, 3.ba, 4.bb, 5.bbb (d):
s1 HuffE 1
s2 Huff(1) 1
s1 Huff(2) 0 s3 Huff(1) 01 s1 Huff(3) 11
so this is encoded as
(e): φ(123) = φ(3)φ(41) = 2 × 40 = 80, so by Euler’s Theorem,
22016 ≡(280)25×216 ≡125×(27)2×22 ≡1282×4≡52×4≡100×4 (d)gcd(7,25)=1and721×3 ≡−1 (mod25)(here25=23×3+1).
True: Encode the message ab• : begin
subinterval start
0 0.4 0.4 0.4+0.8×0.16=0.528
(mod2016).
1 0.4 0.4×0.4 = 0.16 0.2×0.16=0.032
code to use
encoded symbol
so the message encodes as a number in the interval [0.528, 0.56).
(ii) False: The ternary entropy is approximately 0.921 .
(iii) True: t = 49, s = 12, a = 37, b = 61.
(iv) False: 4
(v) True: gcd(3, 16) = 1 and 53 ≡ 6 (mod 17).
Here,wehavethatα2 =2α+1:
α1 = α α5 = 2α α2 =2α+1 α6 =α+2 α3 =2α+2 α7 =α+1 α4 = 2 α8 = 1
α7 1 α α7
x α4 2 so y = α5 = 2α .
(iii) {α2, α6, α18 = α2, . . .} = {α2, α6}, so the minimal polynomial of α5 is
(x−α2)(x−α6)=x2 −(α2 +α6)x+α8
=x2 −(2α+1+α+2)x+1
= x2 + 1 . 2
α3 α4 α2R1=α5R1 1 α α7R2=R2−R1 1
−−−−−−→ −−−−−−−→
0 α2−α α−α7
R2=αR2 1 α α7 R1=R1−αR2 1 0 α7−2α2 1 0 2
α4 α6 α5 R2=α4R2 1 α2 α
−−−−−→ 0 1 2α −−−−−−−−→ 0 1 2α =
0 α7 2 0 1 2α
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