Generic selectors
Exact matches only
Search in title
Search in content
Search in posts
Search in pages
Filter by Categories
nmims post
Objective Type Set
Online MCQ Assignment
Question Solution
Solved Question
Uncategorized

1. On adding the two points P (4,2) and Q (10, 6) in the elliptic curve E11(1,1) we get
a) (9,3)
b) (6,4)
c) (7,5)
d) (2,8)

View Answer

Answer: b [Reason:] Apply ECC to obtain P+Q=(6,4).

2. If P = (1,4) in the elliptic curve E13(1, 1) , then 4P is
a) (4, 2)
b) (7, 0)
c) (5, 1)
d) (8, 1)

View Answer

Answer: d [Reason:] Apply ECC via adding P+P=2P then, 4P=2P+2P.

3. Multiply the point P=(8, 1) by a constant 3, thus find 3P, in the elliptic curve E13(1, 1)
a) (10,7)
b) (12,6)
c) (11,1)
d) (9,8)

View Answer

Answer: a [Reason:] P+P=2P then, 3P=2P+P Thus we get Q=3P = (10, 7).

4. Bob selects E67(2, 3) as the elliptic curve over GF(p). He selects e1 = (2, 22) and d = 4.
Then he calculates e2 = d × e1 .What is the value of e2?
a) (23,49)
b) (16,55)
c) (12,19)
d) (13,45)

View Answer

Answer: d [Reason:] e2 = d × e1 ; e2 =(13, 45).

5. Bob selects E67(2, 3) as the elliptic curve over GF(p). He selects e1 = (2, 22) and d = 4.
Then he calculates e2 = d × e1 and the publicly announces the tuple (E, e1, e2). Now, Alice wants to send the plaintext P = (24, 26) to Bob and she selects r = 2. What are C1 and C2?
a) C1=(35,1) ; C2 =(21,44)
b) C1=(44,21) ; C2 =(1,35)
c) C1=(44,21) ; C2 =(44,21)
d) C1=(21,44); C2 =(35,1)

View Answer

Answer: a [Reason:] Alice finds the points C1= r × e1 ; C1= (35, 1), C2=P + r × e2 ; C2= (21, 44).

6. P = C1 – (d x C2)
Is this above stated formula true with respect to ECC?
a) True
b) False

View Answer

Answer: b [Reason:] P = C2 – (d x C1).

7. For the point P (11, 2) defined in the curve E13(1, 1). What is –P?
a) (12,4)
b) (10,7)
c) (11,11)
d) (11,12)

View Answer

Answer: c [Reason:] The inverse of P(11,2) is (11,11) or (11,-2).

8. For the point P (7, 0) defined in the curve E13(1, 1). What is –P?
a) (7,1)
b) (8,12)
c) (8,1)
d) (7,0)

View Answer

Answer: d [Reason:] The inverse of P(11,2) is (11,11) or (11,-2).