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How to compute $x_0$ in Chaotic Logistic map in special method?

cn flag

I know that the Chaotic Logistic map is $x_{n+1}=Rx_n(1−x_n)$. I read some articles about the Chaotic Logistic Map, but some things are not clear. Someone wrote in his article the following: The initial condition $x_0$ for the logistic map is extracted from the string of $256$ bits ($32$ characters) taken in ASCII form and denoted as $K=K_1,K_2,K_3,\cdots,K_{32}$ ($K_i$ denotes the $8$-bit key character in the $i$-th key position). The value of the initial condition for the logistic map is given by.

\begin{equation} x_0=\sum_{i=1}^{32}\text{mod}(K_i\times 10^i,1) \end{equation} Another author wrote the following

enter image description here

My question: How does he calculate $x_0$ in that function, where we know $x_0$ should lie within the interval $[1,0]$.

Daniel S avatar
ru flag
Are you sure that the summands are not $\mathrm{mod}(K_i\times 10^{-i},1)$?
Mhsz avatar
cn flag
Yes I am sure..
Mhsz avatar
cn flag
See for modifying my question.
fgrieu avatar
ng flag
The literature on cryptographic applications of the Chaotic Logistic map is riddled with errors (and the whole subject is seldom treated seriously or published in reputable journals on cryptography). The equation that the question considers uses non-standard notation, and seems stuck to $x_0=0$. If for some reason it's really needed to work on something similar, what about using$$x_0=\frac{\left(\displaystyle\sum_{i=1}^{32}K_i\times 10^{i-1}\right)\bmod10^{32}}{10^{32}}$$which at least uses standard notation and can take $10^{32}$ values all in $[0,1)$?
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