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I was recently studying some literature on chaos theory and non-linear equations . where in various ciphers were created using non- linear equations like Lorenz equation . Are the data generated from these equations truly random , is it true that the cipher created using non -linear equations are much stronger than AES and other ciphers used currently.

PS: I am very naive in maths and I apologies , if this sounds ridiculous.

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IMHO, "random" is the cop-out way of saying "we can't find an obvious pattern". – J. M. Nov 22 '10 at 14:50

"Random" is usually a poorly defined term, so it is impossible to give a clear answer. Many times what is called random behavior in a dynamical system is more appropriately called chaotic. If you plot the output values (however you take them) you notice that certain values are much more common than others, just from one value you can't predict the next. Usually for random numbers the first of many requirements is that all numbers are equally likely.

When you speak of the strength of an encryption algorithm, what you are usually asking is "given certain data (read about cryptology attacks to see what sorts of data are considered), can you learn something about the key or about the plaintext that was encrypted. Encryption algorithms are (should be) carefully designed on this point. Generally speaking you need to scramble the bits in well-defined ways. Dynamical systems are not designed with this in mind and may well have vulnerabilities when used for encryption.

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but what if the data generated by the lorenz data is used for ciphers like vernam cipher (one-time pad) which totally generate avalanche effect ( as far as i know) . wouldnt using non - linear equations for such ciphers compound its effect. – Rahul Nov 22 '10 at 14:29
The clumpiness in Lorentz data that I mention would mean that your pad is not truly random. There may be more 1's than 0's for example, or more 11110000's than there should be, or something like that. For a good one-time pad, this should not be true. – Ross Millikan Nov 22 '10 at 16:14

The random (chaotic) result from the logistic map, $x\leftarrow 4x(1-x)$ (which is truly chaotic if $x$ starts as almost any irrational number between $0$ and $1$), actually ends up reflecting the final bit of rounding error when using any precision floating point representation.

It is still pretty random, but not worth using for any cryptographic uses.

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