3 ms·
You can see this in the flower image. The target image has a lot of symmetries (reflections, rotations) that also exist in the GoL ruleset. So must be a parent
by remcob 7y ago
You can see this in the flower image. The target image has a lot of symmetries (reflections, rotations) that also exist in the GoL ruleset. So must be a parent that also has these symmetries. Yet the found solution has none of these symmetries.
- OscarCunningham 7y agoConsider the 3×3 block of live cells: ..... .ooo. .ooo. .ooo. ..... It has no parent with the same symmetry, since every neighbour of the centre cell would be the same but a cell can't live or be born with 0 or 4 neighbours. But it does have a parent with less symmetry: ..... ..... ooooo ..... .....
- remcob 7y agoTrue, and this is a counterexample to my claim that the parent would retain the symmetry. I think it can be refined though: The parent still has hor/ver mirror symmetries, and the solution rotated is also a solution. So there is a set of (in this case two) parents that is invariant under the same symmetries. In general, a parent should still be a parent when any of the targets symmetries are applied. (But it wouldn't necessarily be the same parent). This can be captured in a SAT by insisting that any symmetry of a solution is also a solution, which adds a lot of new constraints and no new variables, so should help with finding.