250+ TOP MCQs on Cryptography – Decryption and Answers

Discrete Mathematics Multiple Choice Questions on “Cryptography – Decryption”.

1. Suppose that there are two primes, P1 = 229 and p2 = 61. Find the value of z and Φ.
a) 13969, 13680
b) 5853, 23452
c) 7793, 34565
d) 17146, 69262

Answer: a
Clarification: We know that, z = p1*p2 = 229*61 = 13969 and Φ = (p1 – 1)(p2 – 1) = (229 – 1)(61 – 1) = 228*60 = 13680.

2. ________ can decrypt traffic to make it available to all other network security functions such as web proxies.
a) SSL visibility appliances
b) RSA appliances
c) Rodriguez cipher system
d) Standard cipher system

Answer: a
Clarification: In the data loss prevention systems, Web proxies and antivirus network security functions, SSL visibility appliances decrypt traffic to make it available for all networks.

3. The ROT13 caesar cipher system has an offset of ___________
a) 13
b) 45
c) 71
d) 37

Answer: a
Clarification: The ROT13 Caesar cipher system has an offset of 13 and it is one of the comprehensive cipher scheme. However, the Vigenere cipher employs Caesar cipher as one element of the encryption process.

4. In a public key system, the cipher text received is C = 10 if RSA encryption used with a public key(e = 11, n = 77) to deduce the plain text. Determine the value of ϕ(n)?
a) 49
b) 60
c) 123
d) 70

Answer: b
Clarification: Given n = 77, that means p and q must be 7 and 11 that is they must be co-prime to each other. Now we know that ϕ(n) = (p – 1) (q – 1)
ϕ(n) = (7 – 1) (11 – 1)
ϕ(n) = 6*10
ϕ(n) = 60.

5. To encrypt a message _______ is used on the character’s positions.
a) boolean algebra
b) bijective function
c) inverse function
d) surjective function

Answer: b
Clarification: We have a mathematical notion that a bijective function can be used on the characters’ positions to encrypt a message and an inverse function is used to decrypt the message.

6. The public key of given user, in an RSA encryption system is e = 57 and n = 3901. What is the value of Euler’s totient function ϕ(n) for calculating the private key of the user?
a) 4369
b) 3772
c) 871
d) 7892

Answer: b
Clarification: Given that n=3901 and e=31. We know that n = p∗q where p and q are prime numbers, which gives 3901 = 47*83. Now, ϕ(n) is Euler’s totient function i.e., ϕ(n) = (p−1)∗(q−1)
ϕ(n) = (47−1)∗(83−1)
ϕ(n) = 46*82 = 3772.

7. Using RSA algorithm what is the value of cipher test c if the plain text e = 7 and P = 5, q = 16 & n = 832. Determine the Euler’s totient function for the plain text?
a) 47
b) 584
c) 428
d) 60

Answer: d
Clarification: Given plain text (m) = 7, P = 5, Q = 16, where P and Q are two prime integer
n = P * Q ⇒ n = 5*16 = 80 ⇒ Z = (P-1)*(Q-1) ⇒ Z = (5-1)*(16-1) = 4*15 = 60.

8. There are 67 people in a company where they are using secret key encryption and decryption system for privacy purpose. Determine the number of secret keys required for this purpose?
a) 887
b) 6529
c) 2211
d) 834

Answer: c
Clarification: Since every two employee have their own secret key encryption and decryption. Both users have to agree on a secret key to communicate using symmetric cryptography. After that, each message is encrypted with that key it is transmitted and decrypted with the same key. Here, key distribution must be secret. For n = 67 we would need (frac{n(n-1)}{2} = frac{67(67-1)}{2}) = 2211 keys.

9. In a transposition cipher, the plaintext is constructed by the ________ of the ciphertext.
a) permutation
b) combination
c) sequence
d) series

Answer: a
Clarification: In cryptography, a method of encryption where the positions of plaintext held by units are shifted according to a regular system so that the ciphertext constructs a permutation of the plaintext is termed as transposition cipher.

10. How many bits of message does the Secure Hash Algorithm produce?
a) 160 bits
b) 1035 bits
c) 621 bits
d) 3761 bits

Answer: a
Clarification: The Secure Hash Algorithm or SHA is based on MD4 encryption system. This algorithm gives an output of a longer 160-bit message that is why it is harder to construct another message that yields the same resultant message.

250+ TOP MCQs on Advanced Counting Techniques – Recurrence Relation

Discrete Mathematics Multiple Choice Questions on “Advanced Counting Techniques – Recurrence Relation”.

1. Consider the recurrence relation a1=4, an=5n+an-1. The value of a64 is _________
a) 10399
b) 23760
c) 75100
d) 53700

Answer: a
Clarification: an=5n+an-1
= 5n + 5(n-1) + … + an-2
= 5n + 5(n-1) + 5(n − 2) +…+ a1
= 5n + 5(n-1) + 5(n − 2) +…+ 4 [since, a1=4]
= 5n + 5(n-1) + 5(n − 2) +…+ 5.1 – 1
= 5(n + (n − 1)+…+2 + 1) – 1
= 5 * n(n+1)/ 2 – 1
an = 5 * n(n+1)/ 2 – 1
Now, n=64 so the answer is a64 = 10399.

2. Determine the solution of the recurrence relation Fn=20Fn-1 − 25Fn-2 where F0=4 and F1=14.
a) an = 14*5n-1
b) an = 7/2*2n−1/2*6n
c) an = 7/2*2n−3/4*6n+1
d) an = 3*2n−1/2*3n

Answer: b
Clarification: The characteristic equation of the recurrence relation is → x2−20x+36=0
So, (x-2)(x-18)=0. Hence, there are two real roots x1=2 and x2=18. Therefore the solution to the recurrence relation will have the form: an=a2n+b18n. To find a and b, set n=0 and n=1 to get a system of two equations with two unknowns: 4=a20+b180=a+b and 3=a21+b61=2a+6b. Solving this system gives b=-1/2 and a=7/2. So the solution to the recurrence relation is,
an = 7/2*2n−1/2*6n.

3. What is the recurrence relation for 1, 7, 31, 127, 499?
a) bn+1=5bn-1+3
b) bn=4bn+7!
c) bn=4bn-1+3
d) bn=bn-1+1

Answer: c
Clarification: Look at the differences between terms: 1, 7, 31, 124,…. and these are growing by a factor of 4. So, 1⋅4=4, 7⋅4=28, 31⋅4=124, and so on. Note that we always end up with 3 less than the next term. So, bn=4bn-1+3 is the recurrence relation and the initial condition is b0=1.

4. If Sn=4Sn-1+12n, where S0=6 and S1=7, find the solution for the recurrence relation.
a) an=7(2n)−29/6n6n
b) an=6(6n)+6/7n6n
c) an=6(3n+1)−5n
d) an=nn−2/6n6n

Answer: b
Clarification: The characteristic equation of the recurrence relation is → x2−4x-12=0
So, (x-6)(x+2)=0. Only the characteristic root is 6. Therefore the solution to the recurrence relation will have the form: an=a.6n+b.n.6n. To find a and b, set n=0 and n=1 to get a system of two equations with two unknowns: 6=a60+b.0.60=a and 7=a61+b.1.61=2a+6b. Solving this system gives a=6 and b=6/7. So the solution to the recurrence relation is, an=6(6n)−6/7n6n.

5. Find the value of a4 for the recurrence relation an=2an-1+3, with a0=6.
a) 320
b) 221
c) 141
d) 65

Answer: c
Clarification: When n=1, a1=2a0+3, Now a2=2a1+3. By substitution, we get a2=2(2a0+3)+3.
Regrouping the terms, we get a4=141, where a0=6.

6. The solution to the recurrence relation an=an-1+2n, with initial term a0=2 are _________
a) 4n+7
b) 2(1+n)
c) 3n2
d) 5*(n+1)/2

Answer: b
Clarification: When n=1, a1=a0+2. By substitution we get, a2=a1+2 ⇒ a2=(a0+2)+2 and so on. So the solution to the recurrence relation, subject to the initial condition should be an=2+2n=2(1+n).

7. Determine the solution for the recurrence relation bn=8bn-1−12bn-2 with b0=3 and b1=4.
a) 7/2*2n−1/2*6n
b) 2/3*7n-5*4n
c) 4!*6n
d) 2/8n

Answer: a
Clarification: Rewrite the recurrence relation bn-8bn-1+12bn-2=0. Now from the characteristic equation: x2−8x+12=0 we have x: (x−2)(x−6)=0, so x=2 and x=6 are the characteristic roots. Therefore the solution to the recurrence relation will have the form: bn=b2n+c6n. To find b and c, set n=0 and n=1 to get a system of two equations with two unknowns: 3=b20+c60=b+c, and 4=b21+c61=2b+6c. Solving this system gives c=-1/2 and b=7/2. So the solution to the recurrence relation is, bn=7/2*2n−1/2*6n.

8. What is the solution to the recurrence relation an=5an-1+6an-2?
a) 2n2
b) 6n
c) (3/2)n
d) n!*3

Answer: b
Clarification: Check for the left side of the equation with all the options into the recurrence relation. Then, we get that 6n is the required solution to the recurrence relation an=5an-1 + 6an-2.

9. Determine the value of a2 for the recurrence relation an = 17an-1 + 30n with a0=3.
a) 4387
b) 5484
c) 238
d) 1437

Answer: d
Clarification: When n=1, a1=17a0+30, Now a2=17a1+30*2. By substitution, we get a2=17(17a0+30)+60. Then regrouping the terms, we get a2=1437, where a0=3.

10. Determine the solution for the recurrence relation an = 6an-1−8an-2 provided initial conditions a0=3 and a1=5.
a) an = 4 * 2n – 3n
b) an = 3 * 7n – 5*3n
c) an = 5 * 7n
d) an = 3! * 5n

Answer: b
Clarification: The characteristic polynomial is x2−6x+8. By solving the characteristic equation, x2−6x+8=0 we get x=2 and x=4, these are the characteristic roots. Therefore we know that the solution to the recurrence relation has the form an=a*2n+b*4n, for some constants a and b. Now, by using the initial conditions a0 and a1 we have: a=7/2 and b=-1/2. Therefore the solution to the recurrence relation is: an = 4 * 2n – 1*3n = 7/2 * 2n – 1/2*3n.

250+ TOP MCQs on Graphs – Diagraph and Answers

Discrete Mathematics Multiple Choice Questions on “Graphs – Diagraph”.

1. A directed graph or digraph can have directed cycle in which ______
a) starting node and ending node are different
b) starting node and ending node are same
c) minimum four vertices can be there
d) ending node does not exist

Answer: b
Clarification: If the start node and end node are same in the path of a graph then it is termed as directed cycle i.e, c0 = cn. For instance, a c b a is a simple cycle in which start and end nodes are same(a). But, a c b b a is not a simple cycle as there is a loop <b,b>.

2. Let, D = <A, R> be a directed graph or digraph,then D’ = <A’, R’> is a subgraph if ___________
a) A’ ⊂ A and R’ = R ∩ (A’ x A’)
b) A’ ⊂ A and R ⊂ R’ ∩ (A’ x A’)
c) R’ = R ∩ (A’ x A’)
d) A’ ⊆ A and R ⊆ R’ ∩ (A’ x A’)

Answer: a
Clarification: A directed graph or digraph is an ordered pair D<A, R> where A(is a set of nodes of D) is a set and R(the elements of R are the arcs of D) is a binary relation on A. The relation R is called the incidence relation on D. Now, a digraph is a subgraph of D if i)A’ ⊂ A and ii)R’ = R ∩ (A’ x A’). If D’ D, D’ is a proper subgraph of D.

3. The graph representing universal relation is called _______
a) complete digraph
b) partial digraph
c) empty graph
d) partial subgraph

Answer: a
Clarification: Consider, A is a graph with vertices {a, b, c, d} and the universal relation is A x A. The graph representing universal relation is called a complete graph and all ordered pairs are present there.

4. What is a complete digraph?
a) connection of nodes without containing any cycle
b) connecting nodes to make at least three complete cycles
c) start node and end node in a graph are same having a cycle
d) connection of every node with every other node including itself in a digraph

Answer: d
Clarification: Every node should be connected to every other node including itself in a digraph is the complete digraph. Now, graphs are connected, strongly connected and disconnected

5. Disconnected components can be created in case of ___________
a) undirected graphs
b) partial subgraphs
c) disconnected graphs
d) complete graphs

Answer: c
Clarification: By the deletion of one edge from either connected or strongly connected graphs the graph obtained is termed as a disconnected graph. It can have connected components separated by the deletion of the edges. The edge that has to be deleted called cut edge.

6. A simple graph can have _______
a) multiple edges
b) self loops
c) parallel edges
d) no multiple edges, self-loops and parallel edges

Answer: d
Clarification: If a graph say G = <V, E> has no parallel or multiple edges and no self loops contained in it is called a simple graph. An undirected graph may have multiple edges and self-loops.

7. Degree of a graph with 12 vertices is _______
a) 25
b) 56
c) 24
d) 212

Answer: c
Clarification: Number of edges incident on a graph is known as degree of a vertex. Sum of degrees of each vertex is called total degree of the graph. Total degree = 2 * number of vertices. So, if there are 24 vertices then total degree is 24.

8. In a finite graph the number of vertices of odd degree is always ______
a) even
b) odd
c) even or odd
d) infinite

Answer: a
Clarification: In any finite graph, sum of degree of all the vertices = 2 * number of edges.
Sum of degree of all the vertices with even degree + sum of degree of all the vertices with odd degree = 2 * number of edges. Now, even number + sum of degree of all the vertices with odd degree = even number. It is possible if and only if number of odd degree vertices are even.

9. An undirected graph has 8 vertices labelled 1, 2, …,8 and 31 edges. Vertices 1, 3, 5, 7 have degree 8 and vertices 2, 4, 6, 8 have degree 7. What is the degree of vertex 8?
a) 15
b) 8
c) 5
d) 23

Answer: b
Clarification: Vertices 1, 3, 5, 7 have degree 8 and vertices 2, 4, 6, 8 have degree 7. By definition, sum of degree= 2 * No of edges
Let x = degree of vertex 8
8 + 7 + 8 + 7 + 8 + 7 + 8 + x = 2 * 31
53 + x = 61
x = 8
Hence, degree of vertex 8 is 8.

10. G is an undirected graph with n vertices and 26 edges such that each vertex of G has a degree at least 4. Then the maximum possible value of n is ___________
a) 7
b) 43
c) 13
d) 10

Answer: c
Clarification: Let m be min degree and M be a max degree of a graph, then m ≤ 2E/V ≤ M. Here, m=4, E=26, v=?
So, 4 ≤ (2*26)/V
V ≤ (52/4)
V ≤ 13 ⇒ V = 13.

250+ TOP MCQs on Boolean Functions and Answers

Discrete Mathematics Multiple Choice Questions on “Boolean Functions”.

1. What is the use of Boolean identities?
a) Minimizing the Boolean expression
b) Maximizing the Boolean expression
c) To evaluate a logical identity
d) Searching of an algebraic expression

Answer: a
Clarification: Boolean identities are used for minimizing the Boolean expression and transforming into an equivalent expression.

2. _________ is used to implement the Boolean functions.
a) Logical notations
b) Arithmetic logics
c) Logic gates
d) Expressions

Answer: c
Clarification: To implement a Boolean function logic gates are used. Basic logic gates are AND, OR and NOT.

3. Inversion of single bit input to a single bit output using _________
a) NOT gate
b) NOR gate
c) AND gate
d) NAND gate

Answer: a
Clarification: A NOT gate is used to invert a single bit input (say A) to a single bit of output (~A).

4. There are _________ numbers of Boolean functions of degree n.
a) n
b) 2(2*n)
c) n3
d) n(n*2)

Answer: b
Clarification: There are 2n different n-tuples of 0’s and 1’s. A Boolean function is an assignment of 0’s or 1’s to each of these 2 n different n-tuples. Hence, there are 2(2*n) different Boolean functions.

5. A _________ is a Boolean variable.
a) Literal
b) String
c) Keyword
d) Identifier

Answer: a
Clarification: A literal is a Boolean variable or its complement. A maxterm is a sum of n literals and a minterm is a product of n literals.

6. Minimization of function F(A,B,C) = A*B*(B+C) is _________
a) AC
b) B+C
c) B`
d) AB

Answer: d
Clarification: AB(B+C)
= ABB + ABC [Applying distributive rule]
= AB + ABC [Applying Idempotent law]
= AB (1+C)
= AB*1 [As, 1+C=1]
= AB.

7. The set for which the Boolean function is functionally complete is __________
a) {*, %, /}
b) {., +, -}
c) {^, +, -}
d) {%, +, *}

Answer: b
Clarification: A Boolean function is represented by using three operators ., +, -. We can find a smaller set of functionally complete operators if one of the three operators of this set can be expressed in terms of the other two.

8. (X+Y`)(X+Z) can be represented by _____
a) (X+Y`Z)
b) (Y+X`)
c) XY`
d) (X+Z`)

Answer: a
Clarification: (X+Y`) (X+Z)
= XX + XZ + XY`+ Y`Z
= X + XZ + XY`+ Y`Z
= X (1+Z) + XY`+ Y`Z
= X.1 + XY`+ Y`Z
= X (1+Y`) + Y`Z
= X + Y`Z.

9. __________ is a disjunctive normal form.
a) product-of-sums
b) product-of-subtractions
c) sum-of-products
d) sum-of-subtractions

Answer: c
Clarification: The sum of minterms that represents the function is called the sum-of-products expansion or the disjunctive normal form. A Boolean sum of minterms has the value 1 when exactly one of the minterms in the sum has the value 1. It has the value 0 for all other combinations of values of the variables.

10. a ⊕ b = ________
a) (a+b)(a`+b`)
b) (a+b`)
c) b`
d) a` + b`

Answer: a
Clarification: a ⊕ b
= a`b + ab`
= a`b+aa` + bb` + ab` [As, a*a` = 0 and b*b` = 0]
= a`(a+b) + b`(a+b)
= (a+b)(a`+b`).

250+ TOP MCQs on Logics – Logical Equivalences and Answers

Discrete Mathematics Multiple Choice Questions on “Logics – Logical Equivalences”.

1. The compound propositions p and q are called logically equivalent if ________ is a tautology.
a) p ↔ q
b) p → q
c) ¬ (p ∨ q)
d) ¬p ∨ ¬q

Answer: a
Clarification: Definition of logical equivalence.

2. p → q is logically equivalent to ________
a) ¬p ∨ ¬q
b) p ∨ ¬q
c) ¬p ∨ q
d) ¬p ∧ q

Answer: c
Clarification: (p → q) ↔ (¬p ∨ q) is tautology.

3. p ∨ q is logically equivalent to ________
a) ¬q → ¬p
b) q → p
c) ¬p → ¬q
d) ¬p → q

Answer: d
Clarification: (p ∨ q) ↔ (¬p → q) is tautology.

4. ¬ (p ↔ q) is logically equivalent to ________
a) q↔p
b) p↔¬q
c) ¬p↔¬q
d) ¬q↔¬p

Answer: b
Clarification: ¬(p↔q)↔(p↔¬q) is tautology.

5. p ∧ q is logically equivalent to ________
a) ¬ (p → ¬q)
b) (p → ¬q)
c) (¬p → ¬q)
d) (¬p → q)

Answer: a
Clarification: (p ∧ q) ↔ (¬(p → ¬q)) is tautology.

6. Which of the following statement is correct?
a) p ∨ q ≡ q ∨ p
b) ¬(p ∧ q) ≡ ¬p ∨ ¬q
c) (p ∨ q) ∨ r ≡ p ∨ (q ∨ r)
d) All of mentioned

Answer: d
Clarification: Verify using truth table, all are correct.

7. p ↔ q is logically equivalent to ________
a) (p → q) → (q → p)
b) (p → q) ∨ (q → p)
c) (p → q) ∧ (q → p)
d) (p ∧ q) → (q ∧ p)

Answer: c
Clarification: (p ↔ q) ↔ ((p → q) ∧ (q → p)) is tautology.

8. (p → q) ∧ (p → r) is logically equivalent to ________
a) p → (q ∧ r)
b) p → (q ∨ r)
c) p ∧ (q ∨ r)
d) p ∨ (q ∧ r)

Answer: a
Clarification: ((p → q) ∧ (p → r)) ↔ (p → (q ∧ r)) is tautology.

9. (p → r) ∨ (q → r) is logically equivalent to ________
a) (p ∧ q) ∨ r
b) (p ∨ q) → r
c) (p ∧ q) → r
d) (p → q) → r

Answer: c
Clarification: ((p → r) ∨ (q → r)) ↔ ((p ∧ q) → r) is tautology.

10. ¬ (p ↔ q) is logically equivalent to ________
a) p ↔ ¬q
b) ¬p ↔ q
c) ¬p ↔ ¬q
d) ¬q ↔ ¬p

Answer: a
Clarification: (¬ (p ↔ q)) ↔ (p ↔ ¬q) is tautology.

250+ TOP MCQs on Floor and Ceiling Function and Answers

Discrete Mathematics Multiple Choice Questions on “Floor and Ceiling Function”.

1. A floor function map a real number to ___________
a) smallest previous integer
b) greatest previous integer
c) smallest following integer
d) none of the mentioned

Answer: b
Clarification: Floor function f(x) is the largest integer not greater than x.

2. A ceil function map a real number to __________
a) smallest previous integer
b) greatest previous integer
c) smallest following integer
d) none of the mentioned

Answer: c
Clarification: Ceil function f(x) is the smallest integer not less than x.

3. A function f(x) is defined as f(x) = x – [x], where [.] represents GIF then __________
a) f(x) will be intergral part of x
b) f(x) will be fractional part of x
c) f(x) will always be 0
d) none of the mentioned

Answer: b
Clarification: A integral part of a number is subtracted from that number we are left with the fractional part of that number.

4. Floor(2.4) + Ceil(2.9) is equal to __________
a) 4
b) 6
c) 5
d) none of the mentioned

Answer: c
Clarification: Floor(2.4) = 2, Ceil(2.9) = 3, 2 + 3 = 5.

5. For some integer n such that x < n < x + 1, ceil(x) < n.
a) True
b) False

Answer: b
Clarification: If x < n < x + 1 then ceil(x) = n.

6. For some number x, Floor(x) <= x <= Ceil(x).
a) True
b) False

Answer: a
Clarification: Floor function f(x) is the largest integer not greater than x and ceil function f(x) is the smallest integer not less than x.

7. If x, and y are positive numbers both are less than one, then maximum value of floor(x + y) is?
a) 0
b) 1
c) 2
d) -1

Answer: b
Clarification: Since x < 1 and y < 1 this implies x + y < 2 which means maximium value of floor(x + y) is 1.

8. If x, and y are positive numbers both are less than one, then maximum value of ceil(x + y) is?
a) 0
b) 1
c) 2
d) -1

Answer: c
Clarification: Since x < 1 and y < 1 this implies x + y < 2 which means maximum value of ceil(x + y) is 2.

9. If X = Floor(X) = Ceil(X) then __________
a) X is a fractional number
b) X is a Integer
c) X is less than 1
d) none of the mentioned

Answer: b
Clarification: Only in case of integers X = Floor(X) = Ceil(X) holds good.

10. Let n be some integer greater than 1,then floor((n-1)/n) is 1.
a) True
b) False

Answer: b
Clarification: Since (n-1)/n will always be less than one thus f floor((n-1)/n) is 0.