1to9 Dissect
(Eingestellt am 13. Oktober 2010, 23:05 Uhr von zhergan)
Dissect the diagram into fields (outlined areas) of size 9, in a way, that every digit from 1 to 9 occurs in every part exactly once. Keep in mind the following rule: If two horizontally or vertically adjacent digits sum up to 9 then both cells must be placed inside the same outlined area. The given line segments must be satisfied in your solution.
Lösungscode: Give the total number of different outlined areas on row 2, row 6, column 4 and column 9 consecutively.
Kommentare
am 19. Oktober 2010, 21:43 Uhr von zhergan
Hi to all. Name and the puzzle description is changed. I also removed the wrong sudoku label...
am 19. Oktober 2010, 11:08 Uhr von CHalb
This is a variation of ABC dissection (http://wiki.logic-masters.de/index.php?title=ABC_dissection/en).
@CHalb: Hi Christian. You're right. I forgot to look at Puzzlewiki or other sources on the net. Then I may use a name such as "1to9 Dissection" or "1-9 Dissection". What do you think also?
am 19. Oktober 2010, 10:20 Uhr von zhergan
@zuzanina: Actually you're right. It does not fit completely to the rules of the chaos sudoku since there exist duplicating digits in rows and columns. But it is very similar somehow and I hadn't any other idea for the title so I decided to use the title "Inverse Chaos Sudoku". Do you or others have some suggestions about the title? I don't know if I try to design a new one (since the current one had lots of problems), but if I decide to do I may use more appropriate title for that one and also change the title of the current one.
Regards,
Zafer
am 19. Oktober 2010, 10:09 Uhr von zhergan
Hi to all! The puzzle figure has upgraded. I mentioned to give a second line segment. The older version had two different solutions (at least I could obtain two), but luckily both solutions had the same solution code. But if I'm mot missing any other point this new version has a unique solution. I want to apologize one more time from the earlier solvers since it is getting much more easy with all these additions. Sorry for the inconvenience...
Zafer
am 18. Oktober 2010, 14:51 Uhr von derwolf23
@ibag - Danke, so hab ich es auch verstanden, wird aber nicht akzeptiert.
@derwolf23: Hi Wolfgang, sorry for writing late. I added a new line segment to the puzzle. I don't know your proposal satisfy this line segment also. If it also satisfies this new line segment can you send it to me in some format for checking (maybe via PM, in text format by assigning different letters to different areas). Or if it doesn't satisfy the new line segment you may give it another try. Sorry for the trouble...
Zafer
am 18. Oktober 2010, 14:11 Uhr von ibag
@derwolf23: Die Anzahl verschiedener fettumrandeter Gebiete in den angegebenen Zeilen und Spalten.
am 18. Oktober 2010, 14:10 Uhr von Saskia
@Zafer: Wonderful. I understand. Trying again ... :-)
am 18. Oktober 2010, 14:02 Uhr von derwolf23
Kann mir mal jemand schreiben was der Lösungscode (auf deutsch) sein soll. Irgendetwas mache ich bei der Eingabe falsch.
am 17. Oktober 2010, 13:56 Uhr von Saskia
I don't understand: if there is e.g. 151 in a row cannot be both 15 pair in one outlined area?!
@Saskia: Hi! Each outlined area must include digits 1 to 9 once (without duplication). Additional constraint can be explained like this:
If there exist two neighbour digits whose sum is not equal to 9 may be inside the same outlined area or they may be placed in different outlined areas. There is no constraint about them.
But if there exist two neighbour digits whose sum is equal to 9 e.g. 4 and 5, then these two digits MUST be placed inside the same outlined area.
Regards,
Zafer
am 14. Oktober 2010, 08:55 Uhr von zhergan
Hi to all. The figure has changed. I decided to give an additional line...
am 14. Oktober 2010, 08:16 Uhr von zhergan
A typo has been corrected...
am 14. Oktober 2010, 08:13 Uhr von zhergan
Hi to all. I forgot to give the additional constraint to avoid multiple solutions. It is corrected now. You may try again to solve this one. Sorry for the inconvenience...
am 14. Oktober 2010, 01:21 Uhr von uvo
This puzzle seems to have lots of solutions.
