If you found that video interesting, then perhaps this is an interesting picture. [Someone going by the pseudonym shye is responsible for this puzzle, and it even has a name. Imagine that, dearest reader: a sudoku with a name! Rather grandly, this one is called "Valtari"]
If you found that picture interesting, then perhaps these are some interesting words.
- Add rows 1, 3, 5, 7, 9 together with box 5 (in blue, with double weights in R5C4 and R5C6).
- Subtract Columns 3, 5, 7 (in red).
- The blue cells, considered with multiplicity, have the same contents as the red cells plus 3 extra copies of the digits 1-9.
- In the double-weighted blue cells, the only possibilities are 2, 3 and 4.
- There are already exactly two each of 2, 3 and 4 appearing as given digits in the single-weighted blue cells.
- This means that whichever two of the three digits ends up in the double-weighted blue cells will end up bring the multiplicity of both of those digits up to 4 within the blue cells.
- This means that the two empty red cells have the same contents as the two empty double-weighted blue cells.
- It also means that those two digits cannot appear again in any other blue cell.
- More on this to come I'm sure, following this good work by Philip Newman: https://docs.google.com/document/d/1Y7m2tTUDmUTj3BZ0w5viNuQlaiBVv9kmYksSLrm_XLE/edit
UPDATE: as far as I can tell, there is no need to add Box 5 to the equation here - this makes the intersections slightly simpler (no need for double-weights any more) - but it does make it harder to identify where you are supposed to be looking.
If you didn't find that interesting, then I'll leave you with some whimsical musings. I always thought that an obscure and difficult sudoku solving technique going by the name Exocet was named for a powerful missile that could bust open the hardest of puzzles. Cela a plus de sens en français.
Hi, Tom. Great observation
ReplyDeleteJust a few comments:
1. This could be explained through the Fred's theorem. The red cells (intersections of rows 2, 4, 6 & 8 and columns 3, 5 & 7) have the same contents as the blue cells (could be defined as the cells not included in any of rows 2, 4, 6 8 or columns 3, 5, 7) minus 2 full sets 1-9.
2. Your final conclusion should be even stronger: none of the other blue cells may contain ANY of 2, 3 or 4 (regardless of what's in the double-weighted blue cells). This immediately solves the puzzle (no strategies other than the basic ones are then required)
3. This sudoku can be easily solved by using MSLS on 2,3 & 4 in rows 3, 5 & 7:
- 3 must be either in r1c3 or r1c5 in raw 1
- 4 must be either in r3c5 or r3c7 in raw 3
- r4c7 and r6c3 must both contain 2, 3 or 4
- r7c3, r7c5 & r7c7 are a 2-3-4 triple
- 2 in raw 9 must be either in r9c3, r9c5 or r9c7
Then r4c9 is a 6, r1c7 & r3c7 are a 6-8 pair, the only place for 4 in raw 3 is r3c5, etc. The rest is trivial
And sorry for my English - I am not an English speaker
Hi Unknown, no need to apologise for your English, I can understand everything you wrote well. Thanks for your explanation of the technique from yet another perspective and another intersection - I think it's very helpful and is something I missed!
ReplyDeleteOn Fred's theorem - yes this is exactly what I was trying to get at. I got into the habit of calling it intersection theory, and I think there's a comment on this blog from Sam CL explaining the nicer way of thinking about it. Others have also called it Set Equivalence Theory ("SET").
My understanding of this puzzle is that it was specifically designed to show-off the exocet technique; what I was trying to explore is the extent to which different solving techniques can be interpreted through the lens of Fred/intersection theory/SET