Money Sudoku
(Eingestellt am 18. Februar 2020, 23:01 Uhr von hepcecob)
Green cells show the combined relationship of the numbers above, below, to the left and right of the cell. 2 or 4 lines overlapping each other cancel each other out. If there are 1 or 3 lines, then a line will stay.
The green clues in this puzzle also show the solved number.
The puzzle is solvable by logic, there is no need to create a program and/or brute force every single combination of numbers to figure this out.
Lösungscode: Enter the first and last row with no spaces (18 characters total)
Kommentare
am 22. Februar 2020, 13:20 Uhr von Rollo
OK, I didn't see that. It was fun anyway :-).
But please hide your comment.
am 22. Februar 2020, 07:24 Uhr von Rollo
Mühsam.
Ich seh übrigens immer noch kein Geld. Kann jemand den Titel erklären?
am 20. Februar 2020, 09:32 Uhr von hepcecob
Copied contents to German version
am 20. Februar 2020, 04:40 Uhr von Nylimb
@hepcecob: As Joe Average and Ragna mentioned, the German version of your puzzle is blank. I had the same thing happen with my first puzzle here, and was told that there's a bug in the way the 2 versions are handled:
If you only create an English version of your puzzle, the website is supposed to show the English version for German speakers, but that's not working. So to make it easier for German speakers to see a puzzle, it's recommended that you copy the English version into the box for the German one.
am 19. Februar 2020, 23:05 Uhr von hepcecob
@Ragna
The whole puzzle is money
am 19. Februar 2020, 21:48 Uhr von Ragna
No money ???
am 19. Februar 2020, 11:29 Uhr von hepcecob
updated description stating that no brute forcing is required. Puzzle is solvable by logic alone.
am 19. Februar 2020, 09:39 Uhr von Joe Average
The puzzle only shows in the English Version, not the German one.
@cdwg2000
Think of all the numbers as seven -segment-numbers on a digital clock. There's three horizontal bars and four vertical ones that are either turned on or off, to form the digits that appear on the display.
At the green cells, you have to sum up how many times each of the seven segments appears in the digits of the orthogonally adjecent cells. If it's an odd number of times, it appears in the green cell, if it's an even number of times, it does not.
Take that 6 in r2c2, for example. It has four adjecent cells. Each of the three horizontal bars needs to appear an odd number of times in those four surrounding cells, so either once or thrice. But the top right vertical segment needs to appaear evently, either twice or four times, for it to be turned off in the 6-digit.
am 19. Februar 2020, 05:31 Uhr von Nylimb
I enjoyed that. I wrote a computer program to figure out the possible combinations of digits that could be used around the green cells, because I wasn't patient enough to try to do that by hand. I did solve the resulting sudoku by hand, though.
am 19. Februar 2020, 00:46 Uhr von cdwg2000
Is there a more detailed rule introduction?
am 18. Februar 2020, 23:05 Uhr von hepcecob
Added text
