axis wrote:Okay, I'm still lost. I understand disjoint subsets but I'm not too sure on reducing candidates from the cells that hold the unique subset. I believe in this puzzles case, the 5,7,9 in block 3 was the triple candidates for the cells r1c8, r2c7, r2c8. However, what confuses me is why this works when r1c8 doesn't hold all three just two of them?

HI!

Failing a response from my elders and betters, let's have a shot at saying how I tackled this:

I actually started with the 1,2,4,8 set because I couldn't, at first, see how the 5,7,9 set worked. There are 4 numerals so 4 locations need to be part of this set --- and further locations with one or more of the constituent numbers can have those numbers deleted. The 4 locations are r1c9, r3c7, r3c8,r3c9. So the other 2's in that square and, crucially, the 8 in r1c8 can all be removed.

That leaves as possibles 5 & 9 in r1c8, and 5,7,9 in r2c7 and r2c8: the grouping Animator was talking about.

After that repeated application of the same concept elsewhere led to the solution.

Hope that helps...