Linear Equivalence and Involutions
- From: "J.D." <degolyer181@xxxxxxxxx>
- Date: Thu, 11 Mar 2010 10:14:16 -0800 (PST)
First, a couple of clarifying definitions so we are on the same page:
An involution is a function, f, such that for all x in the domain of
the function, f(f(x))=x.
Two permutations (e.g. s-boxes), A and B, where A, B : F(n)^m -->
F(n)^m, are linearly equivalent if there are bijective linear
mappings, P and Q, and constants, p and q, such that A(x)= Q(B(P(x)+p))
OK, so I have a couple of questions:
1) If a permutation that is not itself an involution, but that is
linearly equivalent to its own inverse (i.e. S(x)^-1=Q(S(P(x)+p))+q),
then does that mean there is some permutation T that is linearly
equivalent to S but that _is_ an involution (i.e. T(T(x)=x)?
2) If so, is there some way of finding T, or of constructing it from
S, that is better than a brute force search of prospective affine
I would be grateful for any answers or pointers to papers that might
give an answer.
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