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Taking advantage of one-time pad key reuse?
How does one attack a two-time pad (i.e. one time pad with key reuse)?

Denote C is cipher text, m is message, and PRG(k) is a psuedo randomize generator, generate an key.

Because we have :

C1 = m1 xor PRG(k)
C2 = m2 xor PRG(k)

--> C1 xor C2 = m1  xor m2 

So, this is a problem because if m1 and m2 is a real text (a text that is in normal life) and use ASCII, you can predict m1 and m2 if you know m1 xor m2.

I don't understand last statement so much. Why we can predicted m1 and m2 when we just know m1 xor m2 ? I have though some additional information, such as in normal text, E is the most appearance character, after that is T... but I still cannot explain.

Thanks :)

share|improve this question

marked as duplicate by poncho, mikeazo Sep 25 '12 at 17:22

This question was marked as an exact duplicate of an existing question.

There is also this: Taking advantage of one-time pad key reuse? – mikeazo Sep 25 '12 at 17:20
@hqt, take a look at the two questions linked to above. They should answer your question. Therefore I am going to close your question. If after reading them, you feel that they do not address your question, please edit the question, explain why, and flag it to be reopened. – mikeazo Sep 25 '12 at 17:21