Re: One time pad keys recovery from misusage
 From: "James H. Newman" <NewJames@xxxxxxxxxxx>
 Date: Sat, 4 Oct 2008 18:11:27 +0000 (UTC)
On Sat, 04 Oct 2008 18:57:17 +0100, rossum wrote:
On Fri, 3 Oct 2008 19:47:43 +0000 (UTC), "James H. Newman"
<NewJames@xxxxxxxxxxx> wrote:
I understand that if a one time pad key is used more than onceIf K is the key that is used twice then we have:
then recovering this key is almost trivial. Can anybody please explain
how this would be done?
C1 = P1 xor K
C2 = P2 xor K
where C1, C2 are the two cyphertexts and P1, P2 are the two plaintexts.
Now take X = C1 xor C2 = (P1 xor K) xor (P2 xor K) which simplifies to:
X = P1 xor P2
Notice that the key K has disappeared since K xor K = 0. Since there is
no key left then no knowledge of the key is needed to decrypt both
messages. For example, if we know that the language is English then we
could take some likely text like " the " and run it along X, xor'ing as
we go, to see if we get anything sensible as a result.
Thanks very much; that was very useful.
.
 References:
 One time pad keys recovery from misusage
 From: James H. Newman
 Re: One time pad keys recovery from misusage
 From: rossum
 One time pad keys recovery from misusage
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