MATH3302 – Sample Mid-Semester Examination Solutions

Part A

1. (c) 13

2. (b) C

3. (c) EO

4. (a) CYKLU

5. (e) V XGADF

6. (b) A modified shift cipher with keyspace {1, 2, . . . , 25} in which all 25 keys areequally likely.

(d) A shift cipher in which all even keys are used with probability 3/52 and allodd keys are used with probability 1/52.

7. (c) 0011

8. (d) 0011

9. (a) 0010

10. (a) (2 marks) The study of the methods of transforming an intelligible message into onethat is unintelligible and back again.

(b) (5 marks) See page 3 of the notes.

11. (5 marks) 9× 3− 4× 1 = 23 so we need the multiplicative inverse of 23.

26 = 23 + 3 1 = 3− 223 = 7(3) + 2 = 3− (23− 7(3))3 = 2 + 1 = 8(26− 23)− 23

= 8(26)− 9(23)

So 23−1 = −9 = 17 (mod 26).

K−1 = 17

[3 −4−1 9

]=

[25 109 23

]To decrypt AB [

0 1] [

25 109 23

]=

[9 23

]Thus the plaintext is jx.

12. (a) (3 marks) A polyalphabetic substitution cipher which uses a keyword of length m, sayk1, k2, . . . , km. To encrypt the plaintext x1, x2, . . . the key is added to the plaintext,that is, e(xi) = (xi+ki (mod m)) (mod 26) = yi. To decrypt the ciphertext y1, y2, . . . thekey is subtracted from the ciphertext, that is, d(yi) = (yi− ki (mod m)) (mod 26) = xi.

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MATH3302 – Sample Mid-Semester Examination Solutions

(b) (i) (2 marks) If it were then the IC for m = 1 would be close to 0.065. But form = 1 in the above tables we have IC = 0.048 which is closer to the value of0.38 for random text.

(ii) (2 marks) Some of the distances between repeated substrings are 60, 24, 40, 56,156, 184, 28, 101, 80. All of these are divisible by 4 except for 101, so we guessthat the most probable keyword length is m = 4.

(iii) (2 marks) Since the row m = 4 in the second table gives IC values that areclosest to 0.065, Friedman’s test tells us that m = 4 is the most probable keywordlength.

13. (a) (2 marks) A good choice of linear recurrence relation would give a keystream withperiod 24− 1 for any non-zero initial key. For li+4 = li+1 + li+3 with initial key 1000we have the keystream 10000000 . . . . Thus this is not a good choice for a linearrecurrence relation.

(b) (5 marks)

plaintext 101010101010ciphertext 001001011111keystream 100011110101

Hence we have the equations: l5 = 1 = c0, l6 = 1 = c3, l7 = 1 = c2 + c3, l8 = 1 =c1 + c2 + c3. Solving these gives c0 = 1, c3 = 1, c2 = 0 and c1 = 0. Thus the linearrecurrence relation is li+4 = li + li+3.

14. (a) (3 marks) pC(A) = pP(c)pK(k2) + pP(b)pK(k3)

=1

3× 1

3+

1

3× 1

6

=1

6

(b) (3 marks) pP(b|A) =pP(b)pC(A|b)

pC(A)

=1/3× 1/6

1/6

=1

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15. (a) (3 marks) A substitution-permutation cipher which takes as input a plaintext oflength 2n, where the first half is L0 and the second half is R0, and a key K andapplies r rounds of processing according to the rule

Li = Ri−1

Ri = Li−1 ⊕ f(Ri−1, Ki)

The ciphertext is then written as RrLr.

(b) (5 marks) The 48-bit string is written as 8 6-bit strings. The ith 6-bit string isreplaced by a 4-bit string based on the ith S-box as follows. Let the 6-bit string beb1b2b3b4b5b6. Look up the entry in row b1b6 and column b2b3b4b5 of the S-box (where

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MATH3302 – Sample Mid-Semester Examination Solutions

rows are indexed by 00, 01, 10, 11 and columns are indexed by the 4-bit binaryrepresentations of the integers from 0 to 15). The entry in the S-box will be aninteger between 0 and 15, and its 4-bit binary representation is used to replace the6-bit string. Thus a string of length 48 = 8×6 becomes a string of length 32 = 8×4.

(c) (2 marks) The keyspace of size 256 is not big enough to be secure. The secrecybehind the development of the S-boxes reduces user confidence in the system.

16. (a) (2 marks) 01001110 and x6 + x3 + x2 + x

(b) (4 marks) (polynomial method)

(x6 + x1)(x4 + x2) = x10 + x8 + x5 + x3 + x4 + x2

= x2(x4 + x3 + x + 1) + x4 + x3 + x + 1 + x5 + x3 + x4 + x2

= x6 + x3 + x + 1

(table method) {43} ∗ {14} = {03}{bd}+{34} = {03}{f1} = {4b}Thus the answer is 01001011.

17. (5 marks) See proof on page 111 of the notes.

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