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- Newsgroups: sci.crypt
- Path: sparky!uunet!newsgate.watson.ibm.com!yktnews!admin!wo0z!lwloen
- From: lwloen@rchland.vnet.ibm.com (Larry Loen)
- Subject: Re: perfect secrecy
- Sender: news@rchland.ibm.com
- Message-ID: <1992Dec29.152217.23308@rchland.ibm.com>
- Date: Tue, 29 Dec 1992 15:22:17 GMT
- Reply-To: lwloen@rchland.vnet.ibm.com
- Disclaimer: This posting represents the poster's views, not necessarily those of IBM
- References: <1992Dec29.121656.4726@nuscc.nus.sg>
- Nntp-Posting-Host: wo0z.rchland.ibm.com
- Organization: IBM Rochester
- Lines: 90
-
- In article <1992Dec29.121656.4726@nuscc.nus.sg>, isc40019@nusunix1.nus.sg (THAI LING) writes:
- |> I think this is a neat folder to discuss topics on cyptography.
- |> I am currently doing a course on it and there are some questions
- |> on my mind.Can anyone help?
- |>
- |> Does anyone know how to prove the following :
- |>
- |> Q1 ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
- |> Definition : M = the message space
- |> K = the key space
- |> C = the cytogram space
- |> ! ! means the cardinality of
- |>
- |> Given that !M! = !K! = !C! then perfect secrecy is achieved if
- |>
- |> (1) all the keys are equiprobable
- |> (2) there is exactly one transformation ( i.e one key ) from a given
- |> message to a given cyptogram
- |>
-
- If this is meant for an arbitrary cryptosystem, it is easily _disproven_.
-
- Suppose one has the well-broken cryptosystem that uses a "random" number
- generator of the form x(n+1) = (A * x(n) + C) mod M and that the
- cipher text is formed by exclusive orring successive terms of x with
- the plaintext.
-
- Suppose further that the value of x(0) (that is, the first term) is
- chosen by flipping a perfectly balanced coin to establish each bit
- in x's representation.
-
- Now, if A, M, and C are correctly chosen (all relatively prime to each
- other, if memory serves and this is usual for such a system), each value
- from 0...M-1 will appear exactly once. Now, since the plaintext is
- broken into groups of size M as well, it appears to me that the
- conditions described above have been achieved if one presumes a general
- enough plaintext to admit all 0...M-1 values, which is also usual.
-
- Yet, this system is well-studied and can be broken (for example) if
- A, C, and M are known (usual crypto assumptions) and if two successive
- values of the plaintext can be guessed using ordinary algebra. There
- are also breaks relating to the problem that there may be two messages
- who share values of x.
-
- At best, the conditions you cite are "necessary", but are certainly not
- "sufficient". They probably are not even "necessary".
-
- |> Q2 ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
- |>
- |> how can the key space be increased by using a combination of keys are
- |> encryption instead of a single key replacement?
- |> i.e instead of A ...... B
- |> . .
- |> . .
- |> . .
- |> . .
- |> Z K
- |>
- |> to try AB ....... CD
- |> . .
- |> . .
- |> . .
- |> . .
- |> AZ ...... WE ...etc
- |>
- |> Thanks in advance !!!!!
- |>
- |> Linda
-
- If I understand your question, the answer is very specific to the
- system. For instance, in DES, it can be shown that an "enhanced"
- DES using two keys done like so:
-
- E(k1,E(k2,x))
-
- can be defeated in only about double the number of
- operations rather that 2 to the 110 operations as might be guessed
- if one is granted the luxury of a table of size 2 to the 67th bytes.
-
-
- If you don't mind my asking, what cryptography books do you have available
- to you? These questions should be able to be ferreted out of available
- sources (such as H. F. Gaines' "Cryptanalysis"). On the other hand,
- I have no idea what restrictions one labors under outside of the US.
- (If you choose not to answer this question, I will understand; I really
- _don't_ know what restrictions you labor under and it is sometimes a
- mistake to even ask).
- --
- Larry W. Loen | My Opinions are decidedly my own, so please
- | do not attribute them to my employer
-