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Added on 2023-01-17
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Question 1
a) Lower case letters = 26
upper case letters = 26
digits = 10
Total character = 26+26+10
=62
Password lenght = 6
Possible Combination=Possible Number of CharacterPassword Length
= 626
=5.680023558 x 1010
Rate of password generation = 107 / s
Time To Crack = 5.680023558 x 1010 x 10−7
=5680.0236 s
~ 94 minutes
~ 1.5778 hours
a) Lower case letters = 26
upper case letters = 26
digits = 10
Total character = 26+26+10
=62
Password lenght = 6
Possible Combination=Possible Number of CharacterPassword Length
= 626
=5.680023558 x 1010
Rate of password generation = 107 / s
Time To Crack = 5.680023558 x 1010 x 10−7
=5680.0236 s
~ 94 minutes
~ 1.5778 hours
Let password length be N
In binary
log2(6) ~ 2 bits
Password strenght S = 16 * 2
=32bits
Password space = 232
= 4,294,967,296 passwords
Rate = 10,000,000 passwords/ s
Total Time T = 4,294,967,296 *10−6
= 429.5 s
1
b) In case of offline brute force, this protocol is very much insecure as the computational
complexity of the protocol is less hence someone with access the to encrypted text can easily
guess the password and decrypt the ciphertext
1 Ting-Yi Chang, Min-Shiang Hwang & Chou-Chen Yang, ‘Password Authenticated Key Exchange and
Protected Password Change Protocols’, in Symmetry (20738994), vol. 9, 2017, 134.
In binary
log2(6) ~ 2 bits
Password strenght S = 16 * 2
=32bits
Password space = 232
= 4,294,967,296 passwords
Rate = 10,000,000 passwords/ s
Total Time T = 4,294,967,296 *10−6
= 429.5 s
1
b) In case of offline brute force, this protocol is very much insecure as the computational
complexity of the protocol is less hence someone with access the to encrypted text can easily
guess the password and decrypt the ciphertext
1 Ting-Yi Chang, Min-Shiang Hwang & Chou-Chen Yang, ‘Password Authenticated Key Exchange and
Protected Password Change Protocols’, in Symmetry (20738994), vol. 9, 2017, 134.
Question 2
Question 3
Algorithm
GA ist:=0;
initialize P(t);
evaluate P(t);
while not terminate P(t)
dot:=t+1;
P(t):=select P(t−1);
recombine P(t);
mutate P(t);
evaluate P(t);
end whileend GA.
Question 3
Algorithm
GA ist:=0;
initialize P(t);
evaluate P(t);
while not terminate P(t)
dot:=t+1;
P(t):=select P(t−1);
recombine P(t);
mutate P(t);
evaluate P(t);
end whileend GA.
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