1. Sample Solution Final exam: Cryptology Design Fundamentals
Grundlagen des kryptographischen Systementwurfs
Module ID: ET-IDA-28
Short questions (closed book test)
Kurzfragenteil (ohne Hilfsmittel)
Zeit: 30 Minuten
Please write your answer on the same question page.
Bitte schreiben Sie die Lösungen auf die Aufgabenblätter.
Name: ……………………………………………..
Matr. Nr.: ………………………….………………
Sample Solution
Contribution of : Martina Georgieva
Ivanov Stoyan
Prof. W. Adi

2. Q1: Compute gcd(627,494).  gcd ( 627 , 494 ) = 19n1 n2 q r
627
494 1
133 3 95 38 2 19
 gcd ( 627 , 494 ) = 19
Q2: Compute gcd( , ).
(1 P)
gcd ( – 1 , 2 21 – 1 ) = 2 gcd ( 49, 21 ) – 1 = 2 7 – 1 = 127
MH: Unterscheidet sich der Font auf dieser Folie absichtlich von den anderen? n1 n2 q r 49 21 2 7 3
 gcd ( 49 , 21 ) = 7
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3. 1. Diffie-Hellman key exchange system
Q3: On which claimed one-way function is the security of the following cryptosystems based on?
1. Diffie-Hellman key exchange system
Discrete logarithm problem
2. Rabin lock Integer factoring problem
As computing the square root in a ring modulo m=p.q is only possible if p and q are known.
(Rabin lock encrypts the clear text X to the cipher text Y as Y = x2 mod m, m=p.q p and q
are two secret primes)
3. „Elliptic-Curve“ Cryptosystem
Based on the difficulty of division in an additive group defined on elliptic curve over finite fields.
( no computationally feasible division algorithm is known in the group)
(3 P)

4. it is  ( 27 ) =  ( 3 3 ) = 3 3 ( 1 – 1/3 ) = 3 3 . 2/3 = 18 elements
Q4: Compute the number of integers less than 27, which have a multiplicative inverse modulu 27.
it is  ( 27 ) =  ( 3 3 ) = 3 3 ( 1 – 1/3 ) = /3 = 18 elements
Q5: How many elements are there in the group of units Z*m , if m = 52· 33 . Compute the highest multiplicative order for a unit in Z*m?
Number of elements in Z*m =  ( ) = ( 1 – 1/5 )( 1 – 1/3 ) = 360
The highest multiplicative order is  ( ) called Carmichael’s function  ( m ).
 ( ) = lcm [  ( 5 2 ) ,  ( 3 3 ) ]
= lcm [ ( 5 – 1 )5 2 – 1 , ( 3 – 1 )3 3 – 1 ]
= lcm ( 20, 18 ) = ( ) / gcd ( ) = ( ) / 2 = 180
(1 P)
(4 P)
MH: Unterscheidet sich der Font auf dieser Folie absichtlich von den anderen?
Name:__________________ Matr.Nr.:____________________
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5. Q6: Reduce the following expressions to the smallest positive integers: 1. R41 ( – ( 83 ) 2 · 43 2 ) = R41 [ – ( +1 ) 2 · ( 2 ) 2 ] = R41 ( 1 – 4)
= R41 ( - 3 ) = - 3 = 41 – 3 = 38
2. ( 3 – 2x 2 ) ( 1 + 2x2 + 4x4 ) over GF(5) =
= 3 + 6x x 4 – 2x 2 – 4x 4 – 8x 6
= 3 + 4x 2 + 8x 4 – 8x 6
= 3 + 4x 2 + 3x 4 + 2x 6
(3 P)

6. Q7: What is a cyclic group?
Is a group which can be completely generated by exponentiation of one of its elements.
Q8: What is a mathematical „Involution“ function?
Is a function which is equal to its inverse function. ( F= F-1 )
Q9: Sketch one unconditionally secure cipher and set its operation conditions.
Conditions:
1) The length of the key Z should be equal to the length of message M.
or at least [H(Z) > H(M)]
2) No key or a part of it is allowed to be used more than one time.
(1 P)
(1 P)
(2 P)
Y = M  Z M + + M
(Vernam Cipher 1926) Z Z
MH: Unterscheidet sich der Font auf dieser Folie absichtlich von den anderen?
Name:__________________ Matr.Nr.:____________________
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7. Which multiplicative orders are possible in GF(43)?
Q10: in GF(43).
Which multiplicative orders are possible in GF(43)?
Possible orders are the divisors of  ( 43 ) i.e. the divisors of ( ) = 42
these are : 1, 2, 3, 6, 7, 14, 21, 42
Compute the number of primitive elements in GF(43).
# of primitive elements =  [  ( 43 ) ] =  ( 42 ) =  ( )
= (2 – 1)(3 – 1)(7 – 1) = 12
3. Which are the minimum tests required to find out weather an element β from GF(43) is a primitive one?
If all the following tests are true β ≠ 1, β 2 ≠ 1, β 3 ≠ 1, β 6 ≠ 1, β 7 ≠ 1, β 14 ≠ 1 ,
and β 21 ≠ 1 , then β is a primitive element!
(6 P)
MH: Unterscheidet sich der Font auf dieser Folie absichtlich von den anderen?

8. 2 7 = 42 = -1 ≠ 1, 2 14 = ( 2 7 ) 2 = 1  the order of 2 is 14
Compute the multiplicative order of 2 in GF(43). 2 ≠ 1, = 4 ≠ 1, = 8 ≠ 1, = 64 = 21 ≠ 1,
2 7 = 42 = -1 ≠ 1, = ( 2 7 ) 2 =  the order of 2 is 14
Compute the smallest positive integer t for which 2-1 = 2t holds.
As the order of 2 is 14, mod 14 = =  t = 13
(As the order of 2 is 14 the smallest modulus in the exponent is 14).
( 42 can also be used as a modulus in the exponent, however would not deliver the smallest t )
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9. Compute a8 and give the corresponding binary vector for a8 .
Q11: GF(26) is generated by the irreducible and primitive polynom P(x)= x6 + x + 1. The element a = = x + 1 is selected from GF(26).
Compute a8 and give the corresponding binary vector for a8 .
a = (x +1), a 2 = (x 2 + 1), a 4 = (x 2 + 1) 2 = x 4 + 1,
a 8 = (x 4 + 1) 2 = x = x 3 + x =
as x 6 + x + 1 = 0  x 6 = x  x 7 = x 2 + x  x8 = x 3 + x 2
2. Compute the multiplicative order of a2 (Hint: a= 1+x = x6 )
a = (x +1) as 1 + x = x 6  a = x 6 a 2 = x 12
As P(x) is primitive, the order of x is 26-1=63
ord (a 2 ) = ord (x 12) = (ord x) / gcd (ord x , 12) = (2 6 – 1) / gcd (2 6 – 1 , 12) = 63/3 = 21
3. Compute the smallest positive integer t for which a-1 = at holds.
The modulus in the exponent for a is the order of a
ord (a) = ord (x6) = (ord x) / gcd (ord x , 6) = (2 6 – 1) / gcd (2 6 – 1 , 6) = 63/3 = 21
 a -1 mod 21 = a = a 20  t = 20
(10 P)
MH: Unterscheidet sich der Font auf dieser Folie absichtlich von den anderen?

10. Carmicheal´s function (m) :
Annex:
Euler Function (m)
For m = p1 p2 p pt
e1 e2 e et
(m) = m ( ) ( ) …… P1 1 P2 1
Carmicheal´s function (m) :
 (2)= 1, (22) = 2, (2e) = 2e for e  3:
(pe)= (pe) = (p - 1)pe-1 for p odd prim.
for m = p1e1 p2e2 p3e pnen
(m) = lcm [ (p1e1 ), (p2e2 ), … (pnen ) ]
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