Multiple choice: d, d, c, e, b True/False: F, T, T, T, F .
1. 2. 3. 4.
(d): HM = 8 H(0.75) + 8 H(0.4) ≈ 0.904 13 13
(d): The second least likely codewords have probability 4 and length ⌈log 125 ⌉ = 5. 125 24
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(c): H(B|A) = i P(ai)H(B|ai) = 37H(45) + 47H(58) = 37H(15) + 47H(38).
(e): You can use Euler’s Theorem but it’s easier just to note that 103 ≡ −1 (mod 1001):
MATH3411 INFORMATION, CODES & CIPHERS Test 3 Session 2 2018 SOLUTIONS
101001 ≡ 103333 × 102 ≡ (−1)333 × 100 ≡ −100 ≡ 901 (b) Neither 12 nor 18 are coprime to 28, unlike 3 and 9.
(mod 1001) .
Consider a = 3: Since 33 ≡ −1 (mod 28),
327 ≡339 ≡(−1)9 ≡−1̸≡1
We see that n = 28 is not pseudo-prime to base 3.
Consider a = 9:
We see that n = 28 is pseudo-prime to base 9.
927 ≡3318 ≡(−1)18≡1≡1 6. (i) False: φ(22) = φ(2)φ(11) = 10.
(i) Here,α2 =−α−2=2α+1:
α2 + 1 2α + 2 α3
(ii) α3+α4 =2α+1=α2 =α
α1 = α α5 = 2α α2 =2α+1 α6 =α+2 α3 =2α+2 α7 =α+1 α4 = 2 α8 = 1
(ii) True: x3 +x+1 has no roots in Z2 so it has no linear factor.
Since its degree is 3, it is irreducible. Therefore, Z2[x]/⟨x3 + x + 1⟩ is a field.
16 16 (iii)True:gcd(3,17)=1,3 ≡1(mod17)and32 ≡−1̸≡1(mod17).
(iv) True: 11 = 55 (mod 18) and gcd(5, φ(18)) = 1 (here φ(18) = 5).
(v) False: There are φ(φ(125)) = φ(100) = 40 primitive elements in U125.
(iii) m2(x)=(x−α2)(x−α6)=x2 −(α2 +α6)x+1=x2 +1
Multiple choice: e, c, c, d, a True/False: F, T, T, F, F .
(e): HM = 52 H (0.7) + 53 H (0.2) ≈ 0.786
(c): H(A,B)=H(A)+H(B)−I(A,B)=0.93+0.76−0.56=1.13 (c): By Euler’s Theorem, 5φ(2018) ≡ 51008 ≡ 1 (mod 2018), so
52018 ≡510082×52 ≡12×25≡25 (mod2018).
(d) 6 ≡ 53 (mod 17) and gcd(3,16) = 1; also, 10 ≡ 57 (mod 17) and gcd(7,16) = 1. Therefore, both 6 and 10 are primitive elements in Z17.
The second most likely codewords have probability 50 and length ⌈log 343 ⌉ = 2. 343 3 50
5. (a): 6. (i)
Since its degree is 3, it is irreducible.
(iii) True: There are φ(φ(31)) = φ(30) = 8 primitive elements in U31.
(iv) False: gcd(5, 65) ̸= 1
(v) False: gcd(2, 61) = 1 and 260 ≡ 1 (mod 61); however,fortheprimepowersp=3,5ofn−1=60,2p
Here,α2 =−2α−2=α+1:
α4 α5 α2 α7
α1 = α α5 = 2α
α2 =α+1 α6 =2α+2 α3 =2α+1 α7 =α+2 α4 = 2 α8 = 1
False: φ(48) = φ(3)φ(24 ) = 16.
(ii) True: x3 + x2 + 1 has no roots in Z2 so it has no linear factor.
≡1 (mod31).
2 R1 = α4R1 1 α 1 R2=R2−R1 1 α −−−−−−→ −−−−−−−→
α3 R2=α6R2 1 α5 α 0 R1=R1−R2 1
0 2α+2R2=α−1R2 1 0 α6 α α7 −−−−−−−→ 0 1 α6
−−−−−−−→ 0
(iv) m5(x)=(x−α5)(x−α7)=x2 −(α7 +α5)x+α7α5 =x2 −2x+2=x2 +x+2
sox=y=α6 =2α+2.
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