A prominent candidate for a one-way function is exponentiation in a suitable cyc
ID: 3677312 • Letter: A
Question
A prominent candidate for a one-way function is exponentiation in a suitable cyclic group. The underlying assumption is that the discrete logarithm problem cannot be solved efficiently in certain groups. While the size of the group in question obviously plays a role for the difficulty of the discrete logarithm problem, the particular group representation is of crucial importance—even for large groups, exponentiation can be trivial to invert. Let p=220+7 (which is prime) and g=5.
a) Find the smallest natural number x such that gx = 2 mod p (this is a discrete logarithm problem in the multiplicative group generated by g).
b) Find the smallest natural number y such that y g =2 mod p (this is a discrete logarithm problem in the additive group generated by g).
Explanation / Answer
a Ans:
Given that
p=220+7=227(prime)
g=5
Discrete logarithm in multiplicative group function is: gx=2 mod p
where n=2,g=5, p=227 and x=?
log gx=n mod p
x=g^n (mod p)
x=5^2 (mod 227)
=25
The smallest value of x is: 25.
b Ans:
Given that g=5,p=220+7, ,n=2
In discrete logarithm in additive group: y=g^order of n
y.g=n mod p
y=5^1 mod(220+7)
=5
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