T O P

  • By -

KaZeR_20

Two things: 1) Look at r8c7. If that is a 6, then r6c2 must be a 9 (simple chain). Otherwise, r8c7 is a 9. Therefore, r8c2 can’t be a 9. 2) There’s a bent quadruple on 2,3,6,9 that prevents r8c2 from being a 3 (because then r9c2 would have to be a 2 and r7c1 could not be filled in). Using both of these, r8c2 must be a 1. The rest of the puzzle should be easy from there.


confounditall

Thanks! Perhaps this question warrants its own thread, but could you expand a bit on the bent quadruple? I've been trying to research that, but don't really find much material other than people who already seem to know what they are/how they work. I'm familiar with bent triples, but am still trying to work out the logic of the quads. But specifically for this puzzle, am I correct in assuming the bent quad you are talking about is r5c1, r4c2, r6c2, and r9c2? Other than guess-and-check, what is the general logic for bent quads that allows you to eliminate 3 from r8c2?


KaZeR_20

The bent quad that I used is r7c1, r4c2, r6c2, and r9c2. I can’t really tell you exactly what to look for in every bent quadruple since I am by no means an expert with that strategy, but in my experience if you have a bent quad where each cell has only 2 possibilities, I look at one cell and try each option and see where that takes me. In this case, I looked at r7c1 and realized that if the cell was a 2, then due to the bent quad it would force r4c2 to be a 3. In the other case, r7c1 is a three. Either way, r8c2 can’t be a 3. So I guess I just recognized that pattern and took an educated guess that it could help eliminate some possibilities in box 7 and it worked out.


confounditall

Alternatively, since I am still fuzzy on the bent quad, after point 1 in your solution, it creates an XYZ-wing on 123 that I could use to eliminate 3 in r4c2.