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The answer in this question defined how to calculate the probability of rotational cryptanalysis on modulo multiplication $\odot$. This paper defined an algebraic equation of how to calculate the rotational probability of modular addition $\boxplus$ which is

$$ \textbf{P}(\overrightarrow{X\boxplus Y} = \overrightarrow{X}\boxplus\overrightarrow{Y}) = \frac{1}{4}(1+2^{-r} + 2^{r-n}+ 2^{-n}) $$ where $r$ is rotational value and smaller than modular value $n$

and experimental results of 8-bit ($n$) addition shows similar values of

$$ \begin{array}{c|c} r & f \\ \hline 0 & 65536 \\ 1 & 24768 \\ 2 & 20800 \\ 3 & 19008 \\ 4 & 18496 \\ 5 & 19008 \\ 6 & 20800 \\ 7 & 24768 \\ \end{array} $$

where f is the frequency of $ \overrightarrow{X\boxplus Y} = \overrightarrow{X}\boxplus\overrightarrow{Y} $ for each $r$ . the probability of 1 bit rotation is $\frac{24768}{65536}$ is around $.378$ (nearly same as in the table in ). the other thing is you find reflection of frequency values , 1 bit rotation is equal to 7 bit rotation (2 and 6 so on). The following figure shows a flipped bell distribution of rotational modulo addition.

enter image description here

The next step

I conducted the same experiment (8 bit size) but with the following equation (double rotation).

$$ \overrightarrow{X} \odot \overrightarrow{Y} = \overrightarrow{\overrightarrow{(X\odot Y)}} $$ the following table shows the frequency:

$$ \begin{array}{c|c} r & f \\ \hline 0 & 65536 \\ 1 & 4730 \\ 2 & 1082 \\ 3 & 904 \\ 4 & 1664 \\ 5 & 904 \\ 6 & 1082 \\ 7 & 4730 \\ \end{array} $$

enter image description here from the table and figure, i see there is a reflection of frequency values (suggesting a similar equation of rotational cryptanalysis probability is applied as following:

$$ \textbf{P}(\overrightarrow{X} \odot \overrightarrow{Y} = \overrightarrow{\overrightarrow{(X\odot Y)}}) = \frac{1}{t}(1+2^{-e_0} + 2^{e_0-e_1}+ 2^{-e_1}) $$

Question

if my guess is right , how to calculate the $t , e_0 , e_1$ values?

NOTE Iam happy to share the source code in the question.

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    $\begingroup$ Ask a clear question up-front. What does "guess" refer to? How in the world does your "guess" change the fact of how the t,e0,e1 values will be calculated? Which experiment are you referring to? $\endgroup$
    – Patriot
    Commented Oct 8, 2019 at 10:20
  • $\begingroup$ if i understand your comment right , I mapped the equation used in addition expression to multiplication. my guess was t, eo, e1 are only function of rotation constant (r) and size (n) , however , after ploting the tables of rotational addition and multiplication , I found the two plots are different , the addition plot is smooth fliped bell distribution , the multiplication is not (there is a bumb when r =4 f=1644) , I guess the proposed equation is not correct $\endgroup$
    – hardyrama
    Commented Dec 17, 2019 at 9:05

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