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I have two independent authentication methods A and B with entropy a and b respectively. say compromise of A does not provide any information on B. Is entropy of two-factor authentication of A and B is a+b?

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    Is this a homework / assignment question? I'm asking because it looks like it could be one of those question establishing confidence in one's newly learned knowledge. If that's the case and we answer it for you, you won't gain that ever so needed confidence, will you?
    – TildalWave
    Sep 21, 2014 at 3:31
  • i will try to research more on it before posting here...its not an assignment though.... Sep 21, 2014 at 3:43

2 Answers 2

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It depends. If you guess A correctly but get B incorrect, does the system tell you that A is correct? If yes, then it's A + B and the average number of guesses it takes is (A + B) / 2. If you don't see which one was wrong, the entropy is A * B and the average number of guesses is A * B / 2.

To show why, consider this example: A=4, B=2. If we try the options sequentially and know which is correct, it would go like this:

A=0, B=0 A ERR; B ERR
A=1, B=1 A ERR; B ERR
A=2, B=2 A ERR; B OK
A=3, B=2 A ERR; B OK
A=4, B=2 A OK; B OK
(5 attempts)

If we don't know which is correct, it would go like this:

A=0, B=0 ERR
A=1, B=0 ERR
A=2, B=0 ERR
A=3, B=0 ERR
A=4, B=0 ERR
A=5, B=0 ERR
A=6, B=0 ERR
A=7, B=0 ERR
A=8, B=0 ERR
A=9, B=0 ERR
A=0, B=1 ERR
A=1, B=1 ERR
A=2, B=1 ERR
A=3, B=1 ERR
A=4, B=1 ERR
A=5, B=1 ERR
A=6, B=1 ERR
A=7, B=1 ERR
A=8, B=1 ERR
A=9, B=1 ERR
A=0, B=2 ERR
A=1, B=2 ERR
A=2, B=2 ERR
A=3, B=2 ERR
A=4, B=2 OK
(24 attempts)
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We measure the entropy of the authentication system in bits. If the entropy of a system S is s bits, this means that after exploring 2^s possibilities the system is certainly broken.

Let us assume entropy of the method A is a=30 bits and the entropy of method B is b=40 bits. Let us also assume that the authentication method A is followed by the method B (Order doesn't matter). If the system notifies the user the failure is due to method A or method B, then the the entropy of the system is max(a,b) = max(30,40) = 40 bits. This is because the attacker can break method A first, which requires 2^30 trials. After breaking A, the attacker can try breaking B by exploring 2^40 possibilities (or vice versa). Since 2^40 >> 2^30, the entropy of the entire system is 2^40.

However, if the authentication failure due to method A or method B is not notified to the attacker, then the total entropy of the system is a+b = 30+40 = 70 bits. This is because for every possibility of method A, the attacker has to try all possibilities of method B i.e. 2^30 * 2^40 = 2^70 possibilities.

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