Here,wehavethatα3 =α+1.
α5 =α2+α+1 α6 =α2+1
MATH3411 INFORMATION, CODES & CIPHERS Test 2 Session 2 2014 SOLUTIONS
Multiple choice: e, d, c, e, c True/False: T, F, F, T, F .
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1. (e): code number rescaled 0.69
in interval
[0.4, 0.9) [0.4, 0.9) [0, 0.4) [0.9, 1)
decoded symbol
s2 s2 s1 •
(e): φ(2014)=φ(2×19×53)=φ(2)φ(19)φ(53)=1×18×52=936,
so by Euler’s Theorem,
3940 ≡3936×34 ≡1936×81≡81 (mod2014).
(c) gcd(4,15) = 1 and 414 ≡ 1 (mod 15).
(ii) False: The binary entropy is approximately 1.38 .
(iii) False: a = 61 and b = 103, so 2a + b = 267.
(iv) True: The codewords are 0, 10, 110, 1110 .
(v) False: There are φ(26) = 12 primitive elements in GF(27).
The decoded message is then s2s2s1• .
(d): MH = 8 H(0.75) + 5 H(0.4) ≈ 0.873.
(0.69 − 0.4)/.5 = 0.58 (0.58 − 0.4)/.5 = 0.36 0.36/.4 = 0.90
(ii) α3k =(α+1)k =α2+α+1=α5 =α12,so3k≡12 (mod7);hence,k=4.
(iii) {α3,α6,α12 = α5,α10 = α3,…} = {α3,α5,α6},
so the minimal polynomial of α3 is
(x−α3)(x−α5)(x−α6)
=x3 −(α3 +α5 +α6)x2 +(α3α5 +α3α6 +α5α6)x−α3α5α6 =x3 +(α+1+α2 +α+1+α2 +1)x2 +(α+α2 +α4)x+1 =x3 +x2 +(α+α2 +α2 +α)x+1
= x3 + x2 + 1 .
Multiple choice: a, b, a, d, b True/False: T, T, F, T, F .
1. (a): code number rescaled 0.35
in interval
[0, 0.4) [0.4, 0.9) [0.9, 1)
decoded symbol
(b): MH = 25H(0.7) + 53H(0.2) ≈ 0.786. (a)
0.35/.4 = 0.875 (0.875 − 0.4)/.5 = 0.95
(d): φ(123)=φ(3×41)=φ(3)φ(41)=2×40=80, so by Euler’s Theorem,
22014 =(280)25×214 ≡125×(27)2 ≡1282 ≡52 ≡25 (mod123). (b)33 =10and35 =5inZ17 andgcd(3,16)=gcd(5,16)=1.
(ii) True: The binary entropy is approximately 1.495 and by Shannon’s Theorem,
we can get arbitrarily close to this.
(iii) False: a = 61 and b = 103, so 2a + b = 225. (iv) True: The codewords are 0, 100, 101, 1100 .
(v) False: The numbers xi generated are 1,7,2,9,6,0,5,15 . (i)Here,wehavethatα3=α2+1.
α4 =α2+α+1 α5 =α+1
(ii) α3+α4 =α2+1+α2+α+1= α =α2
(iii) {α5,α10 = α3,α6,α12 = α5,…} = {α3,α5,α6},
so the minimal polynomial of α5 is
(x−α3)(x−α5)(x−α6)
=x3 −(α3 +α5 +α6)x2 +(α3α5 +α3α6 +α5α6)x−α3α5α6 =x3 +(α2 +1+α+1+α2 +α)x2 +(α+α2 +α2 +α+1)x+1 = x3 + x + 1 .
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