Fully logical and without using temps. Nice result and resolution, well done and thank you flover.
Two technical ways can be used to solve the hanjie at the very end (in R15C8-R17C13). Choose the one you prefer ;)
- With smile-logic: Exactly one cell is still to be blacken in each column C8-13. We can thus use smile-logic (https://webpbn.com/index.cgi?page=solving.html)
- With wolog hypothesis (without loss of generality): We know that the 4 from R17 pass by C8 or C13. By symmetry of the problem, we can assume that the 4 passes by C8 and complete (activate temporaries). As the picture obtained is symmetrical, we can conclude that we would obtained the same if we assumed the 4 passes by C13 (validate hypothesis, desactivate temporaries).
Two technical ways can be used to solve the hanjie at the very end (in R15C8-R17C13). Choose the one you prefer ;)
- With smile-logic: Exactly one cell is still to be blacken in each column C8-13. We can thus use smile-logic (https://webpbn.com/index.cgi?page=solving.html)
- With wolog hypothesis (without loss of generality): We know that the 4 from R17 pass by C8 or C13. By symmetry of the problem, we can assume that the 4 passes by C8 and complete (activate temporaries). As the picture obtained is symmetrical, we can conclude that we would obtained the same if we assumed the 4 passes by C13 (validate hypothesis, desactivate temporaries).