and a − b = 32.
(i) True: Encode the message bb• :
subinterval start
0 0.3 0.3+0.3×0.4 = 0.42 0.42+0.7×0.16=0.532
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1 0.4 0.4×0.4 = 0.16 0.16×0.3=0.048
MATH3411 INFORMATION, CODES & CIPHERS Test 2 Session 2 2017 SOLUTIONS
Multiple choice: a, b, a, e, b True/False: T, F, F, T, F .
1.(e):1.b, 2.a, 3.aa, 4.aab, 5.aaa
(b): HM = 14 H (0.7) + 43 H (0.4) ≈ 0.904
(a): I(A, B) = H(B) − H(B|A) = H(0.2 + 0.7p) − (0.4p + 0.1),
so d I(A,B)=0.7log ( 1 −1)−0.4. dp 2 0.2+0.7p
d 10.4 −1 SettingdpI(A,B)=0yieldsp=0.7 (20.7 +1) −0.2 ≈0.29.
(e): φ(3411) = φ(32)φ(379) = (32 − 3)378 = 2268, so by Euler’s Theorem,
52272 ≡52268×54 ≡1×625≡625 (b)Forn=1113,⌈√n⌉=34,so
(mod3411).
t 2t + 1 34 69 35 71 36 73 37 75
s2 = t2 − n 43
sot=37ands= 256=16,anda=t+s=53andb=t−s=21;hence,1113=n=ab=53×21
112 × 183 × 256
(iii) False: The second shortest codeword lengths are 4.
(iv) True: There are φ(125 − 1) = φ(22)φ(31) = 60 primitive elements in GF(125).
(v) False: The pseudo-random numbers generated are 7, 2, 9, 6, 0, 5, 15, 1. (i) Here,wehavethatα2 =−1=2:
so the message encodes as a number in the interval [0.532, 0.58). (ii) False: The binary entropy H(S) is approximately 1.53.
(iii)γ4+γ= α =γ6=γ(=α+1)
γ1 =α+1 γ5 =2α+2 γ2 = 2α γ6 = α
γ3 =2α+1 γ7 =α+2
(ii) γ,γ3,γ5,γ7
γ4 + α 2 + α γ7
Multiple choice: d, a, a, d, e True/False: F, T, T, T, T .
(d): Encode the message bb• :
subinterval start
a 0 b 0.5 × 0.5 = 0.25 • 0.25+0.9×0.2=0.43
1 0.5 0.4 × 0.5 = 0.2 0.1×0.02=0.45
so the message encodes as a number in the interval [0.43, 0.45). (a): HM = 35 H (0.7) + 25 H (0.4) ≈ 0.917
(a): I(A, B) = H(B) − H(B|A) = H(0.3 + 0.7p) − (0.5p + 0.1), so d I(A,B)=0.7log ( 1 −1)−0.5.
dp 2 0.3+0.7p
d 10.5 −1
SettingdpI(A,B)=0yieldsp=0.7 (20.7 +1) −0.3 ≈0.11. (d)Forn=1215,⌈√n⌉=35,so
t 2t + 1 35 71 36 73
s2 = t2 − n 10
s ∈ Z? ×
sot=36ands= 81=9,anda=t+s=45andb=t−s=27;
hence, 1215 = n = ab = 45 × 27 and a − b = 18.
(e): φ(99) = φ(32)φ(11) = 6 × 10 = 60, so by Euler’s Theorem,
566 ≡560×56 ≡160×(53)2 ≡1252 ≡262 ≡676≡82
(i)False:1.a,2.b,3.ba,4.bb,5.bba
(ii) True: The ternary entropy is approximately 0.817 .
(iii) True: gcd(8, 65) = 1 and 864 ≡ 1 (mod 65).
(iv) True: Codeword lengths: 1, 3, 3, 4; codewords: 0, 100, 101, 1100.
(v) True: 20 ≡ 55 mod 23 and gcd(5, 22) = 1. (i) Here,wehavethatα3 =α+1.
α5 =α2+α+1 α6 =α2+1
(ii) α3k =(α+1)k =α2+α+1=α5 =α12,so3k≡12 (mod7);hence,k=4. (iii) {α6,α12 = α5,α10 = α3,α6 …} = {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 .
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