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[Gast]

Hello friend , although almost all the algoritmos code blocks of 8 octetos or multiples that does not signify that if we want alone to code 9 octets , is very simple

 

 

 

We have data like: 11 22 33 44 55 66 77 88 99

 

 

 

We divide it into two blocks:

 

 

 

- Block 1: 11 22 33 44 55 66 77 88 - Block 2: 99 00 00 00 00 00 00 00

 

 

 

And figures each block with the key , the should be something like this:

 

 

 

- Block 1: 75 74 32 12 58 09 99 11 - Block 2: 33 55 00 91 E1 A0 11 88

 

 

 

As we know that the data of entrance are 9 we know that the last 7 bytes are garbage.

 

 

 

Since Italy is said that the last this in clear, for my is a wrong theory, we go although they have 10000000 ECM with data can be removed coded by means of an algoritmo.

 

 

 

My theory upon the algorithem is that it will be possible , that is to say the relation among blocks, before they were independent, now I am almost sure that there is relation among the blocks coded, you agree of what I said you of the CFB mode that for me is the present one.

 

 

 

In the principles of S*CA2 capture some 60 ECM and I put them on a page, in a pattern to see if there was some coincidence, we did this like an x-files and the results went negative.

 

 

 

 

 

 

Greetings.

 

 

parisino

 

 

jullie zien er is nog heel weinig van de algo bekent

Gerard