Corrupting files

[Tom Thekathyil]

Hi Robert,

Thanks for your response: that was for a trivial case :)

Now let's try a curveball.  We substitute lines 9 to 12 for the
equivalent _somewhere else_ in the code, so it won't be a simple
transform.  This is based on a rule that a message sent on the 12th
day of June would have certain properties, so no memorizing is
required.

8   JuNi0jiIA6
9             nS1MSGrUoLv0VInSrfTKpEJtHCN7aksVxIOuiYgJySp6nWM0o8zpVL
10  1g5g8ipqHD45e5cDQOB2bRxqPLF+oUPHE0daaGtzUiccUGlKmuikOPjGlZKpqHQx
11  zVkrE/uEQil6UJMM/lhGXLI+pg4FzleotlWz0Dhc2lLqjqMHGTzt7uxcR6IFsqJT
12  HNkl21JswgxN0DlZaWLhBQeoAKKFbZWpZz4kbN9vYjTsqGhsMnNplH
13                                                        GZvEnQ2oGy
14  qGlhUpW75BKVXgp2SWVqIkWJkws5VUofMQrblF19Pma1rKiK4GXUBK20k36sOj5y

Let's consider another scenario where lines 9 to 12 are meaningless
code inserted into the message. B has the rule to dispose of it but
no one else would know the location and length of corruption.

My gut feeling is that the human element throws a spanner into the
algorithm.

Regards, Tom

(Haven't had time to consider the other responses, but many thanks
- lots to learn here :) )


On Sun, 2006-06-11 at 23:41 -0500, Robert J. Hansen wrote:
> > Since no one apart from A & B knows how the encrypted file has been
> > corrupted, this seems to be a method of increasing security.
> 
> There are some serious problems you'd have to hurdle.  Let's assume
> they're all hurdled, though, and that it works pretty much as you'd
> expect.  If we make that assumption, then we can talk about this scheme
> in the best-case scenario.
> 
> A reasonably long text message might be 3000 characters long.
> Transposing the case of one of these letters gives us 3000 permutations.
>  Two gives us roughly nine million.  Three gives us about 25 billion.
> It's doubtful that you'd want to transpose more than three letters, due
> to the difficulty of someone remembering "was I supposed to transpose
> letters 1442, 1991 and 2047, or 1442, 1991 and 2074?"
> 
> Log-2 of 25 billion is about 35.  You've just added a factor of 2^35
> difficulty to breaking the message... but that's an _addition_, not a
> multiplication.  You're going to recover enough plaintext at the
> beginning of the message to make it clear when you have the right key or
> not.
> 
> If you're going to posit the existence of an adversary who can do 2^128
> work to break your key, do you really think you're gaining anything by
> _adding_ 2^35 work?  2^128 + 2^35 is so close to 2^128 as makes
> absolutely no difference whatsoever.
> 
> > Question: Is there in theory any way of breaking the corrupted
> > encryption through brute force?
> 
> Yes.  As shown above, the additional work factor you're introducing is
> trivial compared to the work in recovering the key in the first place.
> 
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