Perpendicular Celtic iteration
First fold the real input: construct z' = |xₙ| + iyₙ. Square it: (z')² = (|xₙ|² − yₙ²) + i(2·|xₙ|·yₙ). Then fold the real part of that result: zₙ₊₁ = ||xₙ|² − yₙ²| + i(2·|xₙ|·yₙ) + c.
The double fold reflects the real component twice per iteration: before the square, negative x mirrors to positive; after it, a negative resulting real part mirrors to nonnegative. The imaginary component folds only indirectly, through |xₙ| in the cross term.
Consider c = −0.5. For the Mandelbrot set: z₁ = −0.5, z₂ = −0.25, bounded. For the Perpendicular Celtic: z₁ = ||0|² − 0²| + i(2·|0|·0) − 0.5 = 0 − 0.5 = −0.5, then z₂ = ||−0.5|² − 0²| + i(2·|−0.5|·0) − 0.5 = |0.25| − 0.5 = −0.25, same orbit. But try c = 0.3i: the double fold redirects the orbit in a way that neither the Mandelbrot set nor the Burning Ship reproduces.
