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Re: A block encryption processing idea taken from linear algebra

From Mok-Kong Shen <mok-kong.shen@t-online.de>
Newsgroups comp.programming
Subject Re: A block encryption processing idea taken from linear algebra
Date 2013-06-18 08:43 +0200
Organization albasani.net
Message-ID <kpovhs$j6$1@news.albasani.net> (permalink)
References <kpmeiq$5hu$1@news.albasani.net> <n1h0nq2aai5z$.ympp3gs2xcce$.dlg@40tude.net> <kpn4r4$jd1$1@news.albasani.net> <4nv1jdhieo4y.1le3iwvmj42o3.dlg@40tude.net>

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Am 18.06.2013 08:03, schrieb JJ:
> On Mon, 17 Jun 2013 16:01:09 +0200, Mok-Kong Shen wrote:
>> Am 17.06.2013 12:13, schrieb JJ:
>>> On Mon, 17 Jun 2013 09:41:14 +0200, Mok-Kong Shen wrote:
>>>> Using this as a hint, we propose to do for block encryption processing
>>>> of n blocks, x1, x2, ... xn, the follwoing, where the f's are
>>>> invertible non-linear functions, the r's are pseudo-random numbers and
>>>> the assignments are performed sequentially (the f's and the r's are
>>>> (secret) key-dependent and different for different rounds, if more
>>>> then one rounds are used):
>>>>
>>>> x1 := f1(x1 + x2 ... + xn + r1)
>>>> x2 := f2(x1 + x2 ... + xn + r2)
>>>>       ................
>>>> xn := fn(x1 + x2 ... + xn + rn)
>>>
>>> I'm lost. If n is the number of blocks, then xn is the block number n,
>>> right?
>>>
>>> But this:
>>> x1 + x2 ... + xn + r1
>>> suggest to add one block with another then add them with a random number.
>>>
>>> How can a block of data be, I assume, arithmetically be added by another
>>> block and added with a number?
>>>
>>> Also... with that information alone, the encryption doesn't seem to be
>>> decryptable.
>>
>> If the block size is m bits, then the arithmetics is mod 2**m. If the
>> f's are invertible, then decryption is evidently without problems,
>> just like each step of the iterative solution of a system of linear
>> equations could be reversed (though one of course never does that in
>> practice).
>
> Assuming I take out the r1, r2, etc. (the salt) from the equation, the
> result would be:
>
> x1 := f1(x1 + x2 ... + xn)
> x2 := f2(x1 + x2 ... + xn)
>       ................
> xn := fn(x1 + x2 ... + xn)
>
> It would make all block data to be same as the first block. It'll destroy
> the data and keep only the first block.
>
> Something is definitely missing here.

Why would it make all blocks the same? Note that the assigments
are done "sequentially". (Cf. what is generally known as the
Gauss-Seidel (or single-step) method of iterative solutions
of systems of linear equations.) For example, for the 2nd assignment
the x1 in f2() is already the "new" value determined by the 1st
assignment.

M. K. Shen

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Thread

A block encryption processing idea taken from linear algebra Mok-Kong Shen <mok-kong.shen@t-online.de> - 2013-06-17 09:41 +0200
  Re: A block encryption processing idea taken from linear algebra JJ <duh@nah.meh> - 2013-06-17 17:13 +0700
    Re: A block encryption processing idea taken from linear algebra Mok-Kong Shen <mok-kong.shen@t-online.de> - 2013-06-17 16:01 +0200
      Re: A block encryption processing idea taken from linear algebra JJ <duh@nah.meh> - 2013-06-18 13:03 +0700
        Re: A block encryption processing idea taken from linear algebra Mok-Kong Shen <mok-kong.shen@t-online.de> - 2013-06-18 08:43 +0200
          Re: A block encryption processing idea taken from linear algebra JJ <duh@nah.meh> - 2013-06-18 18:12 +0700
            Re: A block encryption processing idea taken from linear algebra Mok-Kong Shen <mok-kong.shen@t-online.de> - 2013-06-18 15:08 +0200
              Re: A block encryption processing idea taken from linear algebra JJ <duh@nah.meh> - 2013-06-19 15:53 +0700

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