Clockhand
(Published on 9. January 2019, 17:56 by Voyager)
At each letter starts a path which goes only orthogonally and on which a small pointer and a big pointer have to be gathered. While entering it's cell each pointer will be fixed at once (that means before a potential bending on its cell) relatively to the movement direction; i.e. if the path bends, then the already collected pointers turn in the appropriate direction of rotation. When all pathes are finished, each outside the grid pictured display must be seen on one of the ends of the pathes by looking in the respectively last taken direction of movement. Each cell has to be used from exactly one path exactly once.
Here is an example:
Here is the puzzle:
Here is an example:
Here is the puzzle:
Solution code: For A - E the index of it's display. Then for A - E the count of the from the path used cells. (For the example the Code would be 3125128.)
Last changed on on 27. July 2021, 16:20
Solved by pokerke, jessica6, Luigi, Joe Average, simorin, Zzzyxas, Nothere, moss, ibag, tuace, ch1983, ffricke, sf2l, dm_litv, zorant, rimodech, NikolaZ
Comments
on 27. July 2021, 16:20 by Voyager
Label Path puzzle set
on 7. February 2019, 16:03 by ibag
Da hab ich eine ganze Weile gebraucht, um den Einstieg zu finden. Schön!
on 9. January 2019, 18:38 by Voyager
Description of solution code corrected.
Beschreibung vom Lösungscode korrigiert.
