7249561381684235979357186245..3..81..4..8175..81.7.24..13....72...1...85.5...7.61

This puzzle is an exercise in Empty Rectanges, which makes it all new to me. Yahoo!

How to go about it? As I understand it, the only Boxes here that could have ERs in them would be Boxes 4578, and looking at each of these, I think they could only have ERs for candidates 269 (also 4 in Box8). However, the options sharply reduce when you look for the outside Rows or Columns they would need to work with. For Candidate 2 there are Columns 4&6 and Row4, and for 9 there is only c2 and r5. So we end up with a short-list of 5 possible arrangements for ERs, only for candidates 2 and 9 and only working with these units.

The 2ER in Box4 has it’s possible 2s in r4c3 and r5c13, and the intersecting point of the two lines these 2-candidate cells are in is r4c3. This cell can see one of the two instances of candidate 2 in c4. This eliminates 2 from r9c3, but then what?

Yeah, I missed it. Eliminating 2 from r9c3 solves that cell as a 9 and that solves the whole puzzle. There were then only three clues in Box7 and I did not realise that r9c3 was bi-valued.

I went looking through all the other possibles in my short-list, none of which came to anything that would solve the puzzle. So I will have to be more careful. Bring on the ERs!