There has been work like this done before, though I don't have references handy. Check Schneier's "Applied Cryptography". It hasn't been very successful - the output of the system looks very random, but you can predict each value from the last, so known plaintext attack kills it. And even if the mathematics are strong, the implementation can be weak. At 09:18 PM 9/8/97 PDT, Nobuki Nakatuji wrote:

P(i) Plaintext,C(i) Ciphertext,K(j) Key,Ch(i) Chaos signal, L Irrational number

P(),C()-->Manage in byte,Length supposing that n byte. K()-->Character line from ASCII CODE,Length supposing that m.

Ch(n) begin Xn+1=aXn(1.0-Xn) return Xn+1 end

How long are Xn and Ch(n)? Double (64-bit IEEE floating point)? Is 0<Ch(n)<1 ? (I assume yes...) How long is L (since a true irrational is infinitely long)? Is L part of the key, or shared by everybody?

f(K) begin w = Sigmaj strtoul(K(j))j delay=int(w/L) return(double)(w/L-delay) end

How is f(K) used? Initialize Ch(0)? What is strtoul? String to Unsigned Long? If size(w) == ul == 32 bits, you only have a 32-bit key, too weak. If size(w) == 64 bits, maybe you have a chance. Be sure each piece of K(j) is long enough - adding a bunch of short numbers together does not produce a long number. An MD5 hash would be much better.

1.K input 2.delay generato 3.Ch(i) generato 4.P(i) acquire 5.C(i)=P(i) XOR Ch(i+delay) 6.C(i) output

How do you XOR a plaintext byte with (double) Ch(i)? Do you really just use 1 byte of the Ch(i)? Then you may have a chance. If you use all the bits of Ch(i), then known plaintext lets you take C(i) XOR P(i) == Ch(i), which lets you generate Ch(i+1).... You may not know the key, but you don't need to if you know the function.