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?