3 4 7. 8/9 2 8/9 1 6. 5

2 5 7. 8/9 3 8/9 1 6. 4.

If you put 34 on the 7 arrow you run into a double 56 pair with r1c4 r1c6 and r5c4 r5c6 which forms a deadly pattern. Sadly this is using uniqueness rather than the usual logic you see in ctc but got me through that part.

I could not find an easier way to prove that spoiler. But I also didn't exhaustively try to run into a contradiction with the incorrect path.

The given grid should be derivable from the start.




At this point it should be noted that both of r6c23 cannot be big, otherwise there are too many big numbers in the row.

If both of them are small, then r6c78 are 56. In box 7 the 6 is confined to row 8, removing it from r8c7. The blue or green squares in box 9 contribute two of 789, as does the thermometer – too many big numbers. So r6c23 are a big and a little number.

The renban within box 7 must contain both a 4 and a 5. The small number from r6c23 means the renban cannot have a 9. Also, there must be at least one big number in r7c23, and at least one small number in r8c23. There are at least two big numbers on the renban, so it cannot have 1. It has 34567 and either 2 or 8.





In box 8, one of the 239 cells must be a 9, which provides some eliminations.

R9c7 cannot be 4 (either green is big, or it has a 78 on one side). Similarly 3 is removed from r8c46.





General whispers, thermo, and sudoku can be done for a bit.

Now it borders on bifurcation, but I think Simon would accept this:

If r8c4 were a 9, then the blue squares are big. This means the big number in r6c23 is a 6, so 6 is not on the renban inside box 7, putting it in the corner. This leaves 2457 on the renban in box 7, leaving r9c2 as the only spot for the 1. 1 is then forced into r5c1 (unique in box) and into r5c7 (23 pair and 123). So r8c4 is a 2.





After this, it solves pretty easily, I think. Let me know how much over the bi-line you think I have gone here.

extending it twice could really skip a lot of steps to figure out parity as it will automatically eliminate 6 and 7 from R6C7 which isn't my goal. Only after your can find that at most 2 big can go on the thermo can you find the bulb has be less than 5. Only when you figure out it can't be 6 or 5 should you break both the rings with parity is my goal.

using set but not getting the squeeze on the 4 2x2s yet