Zombie Outbreak
(Published on 31. October 2023, 14:57 by Nahileon)
The secret experiments of a biotech company have gone horribly wrong. A virus has been released that makes an infected person lose control over their body and gives them an insatiable appetite for human flesh. A tiny scratch from one of the infected is all it takes to become one of them. The number of people controlled by the virus is already staggeringly high and with every minute that passes it is spreading further.
In all this chaos you have received the desperate call of three scientists, who figured out a plan that could stop the virus. They think it might have been engineered with a weakness. A substance that can kill the virus quickly if an infected is treated with it. However they don't know what this substance is and how to make it.
To find out, they want to reach three locations on the company campus. Two research facilities, where they hope to find data on the virus and a laboratory that has everything that is required to create the cure. The only problem is the horde of hungry zombies between them and their destinations. This is why they asked for your help.
Here is what you know:
The zombies can be sorted into nine different weight classes. From 1, very fast but weak, to 9, very strong but slow. A scientist can pass a zombie by either outrunning it or by overpowering it. However, for that to be save, the scientist must be a lot faster or a lot stronger. Your team has gathered some data via drones, that shows the approximate whereabouts of some of the zombies. It isn't much, but you will have to make due with what you have.
Will you be able to guide the scientists savely to their destinations? This might be humanity's best chance of avoiding a full zombie apocalypse. The pressure is high, but there might even be hope for the infected.
Good luck!
Normal sudoku rules apply.
Killer: The digits in a cage add to the number in the top left corner of the cage. Digits may not repeat within a cage.
Draw three orthogonally connected nonbranching paths between the given digits and the three colored cells, such that each of the given digits is connected to exactly one colored cell and vice versa.
Paths may not cross given digits or colored cells, but they may cross each other.
The digits on a path differ from the starting digit by at least 3. (This does not include the colored cells!)
The blue cell (r9c5) contains the number of digits that are larger than the green cell (r5c9) and smaller than the red cell (r5c1).
Solution code: Row 9 of the sudoku.
Comments
on 16. July 2024, 08:23 by QuiltyAsCharged
Cool puzzle! Drawing the three paths is an interesting challenge. Even when you have a basic idea of where they go, it's tricky to get the exact steps down.
Once I had the paths drawn, I had a hard time resolving the rest of the sudoku. This part is not my strength, and I may have missed something that would simplify the process.
Here's a link with the solution check enabled: https://sudokupad.app/zfc903uuah
on 1. November 2023, 16:03 by Nahileon
Clarified the rules
on 1. November 2023, 13:40 by sanabas
Sorry for being a pain again. I'm convinced my logic is sound, and the puzzle is impossible with my reading of the rules. But now I think I've found another option for how I was misreading the rules. I read it as the given digits being the start of the paths, because the rules say 'between the given digits and the three colored cells', rather than between the coloured cells and the given digits. So the path starting at the given 5 can only have 1289 on it, etc.
Now I think it is the coloured squares that are meant to be the start of the paths, and hopefully I can get a solution this way...
...I get multiple solutions that way. It's been fun to attempt, but I'm out of ideas.
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The restriction of a difference of 3 is only for the paths and not for the colored cells themselves. Sorry for causing this confusion.
on 1. November 2023, 04:57 by sanabas
My initial assumption for 'The blue cell (r9c5) contains the number of digits that are larger than the green cell (r5c9) and smaller than the red cell (r5c1).' is that if green is 2 and red is 8, then blue would be 5, as the digits both larger than green and smaller than red are 34567. But I *think* I've proved that is impossible.
So does that mean if green is 2 and red is 8 that blue would need to be 14, because there are 7 digits larger than green + 7 digits smaller than blue? Or does it mean that blue would be 7, and therefore green + red must equal 10?
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It means what you thought it means. Basically Blue = Red - Green -1.
