That’s An Answer, Not A Pencil!

This is an open question. It represents a shortcut to the cryptanalyst between multiple rounds of a cipher algorithm.

Represent the cipher as y = f(c) in two dimensions. 

For superior ciphers, f(c) is NOT A GROUP.

Question: can we devise a y = Ef(c) in three (3) dimensions such that Ef(c) IS A GROUP?

Then we apply some creative calculation f(c) and Ef(c) to derive z1 and z2. Then we attempt to use z2 to compute (for example) Ef(z2) == f(f(f(f(f(z1))))), effectively skipping 4 rounds.

Is there a way to CALCULATE, or DERIVE these formulae?

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