Here’s a first (Ja)TaHoKu, which I made to prepare for the Logic Masters. Mostly an exercise in the interaction between the Tapa rules and the equal number of cells per row/column/region; that part seems to have potential as a Tapa variation. Would you have thought that even without rooms, the one Tapa clue implies that the mirrored cell has to be shaded?
Rules Place numbers from 1 to 4 into some cells, such that each row, column and region contains each number exactly once. Clues within the grid are Tapa clues; the numbered cells form a valid Tapa solution with respect to these. Clues along the bottom and right edges are skyscraper clues.
I´m not sure, but have that (Ja)Ta(Ho)Ku just one solution? If yes, then I’m surprised! If no, still incredible puzzle :-).
Yes, pretty sure it’s correct.
I found this out some years ago when playing around with Hungarian Tapa. A valid Tapa wall with the same number of cells in every row and colums is already very restricted and there are only a few grid sizes and number of tapa cells that work well together. You can try my Hungarian Tapa in the puzzle portal. If I remember correctly the Tapa part is already unique if you use ? instead of the clue numbers and its surprising how you get uniqueness in parts of the grid far away from every clue just by connectivity. And thats wirhout regions.
Ah yes, i remember solving that one, http://www.logic-masters.de/Raetselportal/Raetsel/zeigen.php?id=0001SG. The restrictions get a little less bad when the grid size grows. 9×9 with 5 shaded cells per row seems a lot easier to work with than this size.
Actually, it seems that the Tapa is already unique with just the one clue and the “4 shaded cells per row/column” restriction. No rooms required.