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Let's suppose I want to modify Kuznyechik block cipher by choosing a random S-box (taken from /dev/random for example).

How can I calculate/generate the inverse S-box?

Does anyone know the formula or algorithm used to do this?

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2 Answers 2

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Sagemath SBox Package is a friend of SBox learners/designers.

For an invertible SBox;

#         0  1  2  3  4  5  6  7       #index
S = SBox([0, 1, 3, 6, 7, 4, 5, 2])     #output
Sinv = S.inverse()
print(Sinv)

outputs

(0, 1, 7, 2, 5, 6, 3, 4)

Actually, the implementation of the inverse is not hard; just reverse the index-output relation. Remember, invertible SBox is just a permutation.


Note that the source code of SageMath SBox is here and as a good library it first controls the SBox is a permutation or not and returns an SBox oject;

        if not self.is_permutation():
            raise TypeError("S-Box must be a permutation")

        cdef Py_ssize_t i
        cdef list L = [self._S_list[i] for i in range(1 << self.m)]

        return SBox([L.index(i) for i in range(1 << self.m)],
                    big_endian=self._big_endian)
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  • $\begingroup$ Thanks for the reply. Do you know some similar package/program in C/C++ that does the same of Sagemath SBox package? $\endgroup$ Mar 21, 2022 at 23:58
  • $\begingroup$ @phantomcraft I'm not aware of it, however, you can use it in Python $\endgroup$
    – kelalaka
    Mar 22, 2022 at 6:40
  • $\begingroup$ I got, thank you. $\endgroup$ Mar 23, 2022 at 1:18
  • $\begingroup$ Sorry, I selected the other question as "useful" because I was interested in a C implementation. $\endgroup$ Mar 23, 2022 at 1:22
  • $\begingroup$ @phantomcraft then you are asking on the wrong Stack Overflow site. This is not a programming site and you asked Does anyone know the formula or algorithm used to do this? so I've given you the easiest way, and the simple algorithm reverse the index-output relation.. Without my answer, this question is more to be off-topic side. Have fun. $\endgroup$
    – kelalaka
    Mar 23, 2022 at 8:47
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Assuming S-boxes are a permutation.

Here is example in Python:

S = (2, 0, 1)
inverse = [0] * len(S)

for i in range(len(S)):
    inverse[S[i]] = i

print(inverse)

Here is example in C:

unsigned int S[256] = {...};
unsigned int inverse[256];

for (int i = 0; i < 256; i++)
{
    inverse[S[i]] = i;
}
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  • $\begingroup$ Quite simple, I didn't realize that it would be that easy. Thanks! $\endgroup$ Mar 23, 2022 at 1:18

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