A "10" cage must be one of the 4 combinations:

• {1, 9}, {2, 8}, {3, 7}, {4, 6}

"10" cages in box 1 can't contain the same combinations as "10" cages in box 4.

• Therefore, 4 "10" cages in box 1 and box 4 must contain all 4 combinations.

If we forget about the thermo constraints for now, we can label these 8 cells (in 4 "10" cages) as {A, A'}, {B, B'}, {C, C'}, {D, D'}. Each pair sums to 10.

With the thermo inside "10" cages, we know the order of each pair.

We can't use alphabet on f-puzzle. Let's just use {1, 9}, {2, 8}, {3, 7}, {4, 6} for {A, A'}, {B, B'}, {C, C'}, {D, D'}.

• But please remember that these are labels, instead of real digits.

We haven't assigned a label for 5, let's just use 5.

Until the final step, all "digits" in this solution path are actually "labels".

To simplify the board, I also removed all thermo. They are not very useful until the last step.

The "10" cage in box 7 must be {3, 7} or {4, 6}.

• Putting {1, 9} or {2, 8} in this cage would break box 4.

We have a 1289 quadruple in column 3.

The highlighted 46 pair see the highlighted empty cells.

• Therefore, {4, 6} goes to column 3 in box 1.

Also, for column 3, 5 must go to box 1. This is a 456 triple in box 1.

The highlighted empty cells must be 3 and 7.

And the "10" cage in box 7 must be 37 pair.

Yellow "10" cages can't be {3, 7} nor {4, 6}, they must be {1, 9} or {2, 8}

There are 1289 quadruple in row 4 and row 5.

5 in box 4 must go to row 6, we can eliminate 5 from r4c5 and r5c5.

The green cell in box 7 (r7c3) sees all yellow cells in box 4.

This cell can only be 1289. Whatever we put in r7c3 must go to the green cell in box 4 (r6c2).

Therefore, the green cell must be 1 or 9.

• {1, 9} is the intersection of {1, 5, 9} and {1, 2, 8, 9}

In box 4, 5 can only go to r6c1.

The green "10" cage in box 5 (r6c5, r6c6) can't be {1, 9}, {3, 7}, {4, 6}. Therefore, it must be {2, 8}.

• If we put other combinations into this case, one of the yellow cells (r4c5, r5c5, r6c2) will break.

This eliminates 28 from r6c3. r6c2, r6c3 become a 19 pair now.

r4c3, r5c3 become 28 pair now, and 8 in box 1 disambiguate the order.

We can remove 28 from the other "10" cage in box 5 (r4c6, r5c6), and it becomes a 19 pair now.

The 19 pair affects the "10" cages in box 6. The left one (r4c8, r5c8) must be 28 pair, and the right one (r4c9, r5c9) must be 19 pair.

The 19 pair in box 4 eliminate 19 from r8c2, the only candidate in r8c2 is 4.

In box 5, 6 can only go to r4c5 (hidden single).

The green "10" cage in box 8 can't be {2, 8}, {3, 7} and {4, 6}. Due to,

• {2, 8} in box 5

• 4 in box 7

• 37 pair in box 7

• Therefore, it must be 19 pair

The green cell (r7c4) sees all yellow cells (r4c4, r5c4, r6c4, r5c5), therefore, it can't be 3, 4, 5, 7.

The remaining candidates for r7c4 is 2 and 6.

In box 2, 8 can only go to r1c5 or r3c5.

The "10" cage in box 9 (r7c9, r8c9) can't be 19 pair.

We can also remove candidate 2 from r7c9. Therefore, 8 is removed from r8c9.

In box 9, 8 must go to row 7 (r7c7 or r7c9)

Now r7c1 must be "2".

7 in box 5 must be in r5c5, r6c4, they both see r7c6.

7 in box 8 must be in the yellow cells (r7c5, r8c6), the both see r8c3.

r8c3 (the green cell) must be 3.

We can put a bunch of 7 to the grid now.

The green cell (r7c7) is also a 7.

This force the 8 into the "10" cage in box 9. This "10" cage must be {2, 8} pair.

The green cell (r3c9) can only be 4 or 5.

Now r3c6 is a naked single, it must be 6.

r3c8 can only be 1 or 9.

r1c7, r3c7, r3c8 is an 189 triple.

The yellow cells (r3c8, r4c6) are a knight's move apart, therefore, they must contain different digits (1 and 9).

The green cell (r3c5) sees both yellow cells. Therefore, it can't be 1 or 9. Therefore it must be 8.

Similarly, the yellow cells (r3c9, r4c7) are a knight's move apart, they must be different digits (4 and 5).

The green cell (r2c8) sees both yellows cells, so it can't be 4 or 5, and it must be 3.

3 in box 2 can only go to r1c5.

3 in box 8 can only go to r7c6.

3 in box 5 can only go to r6c4.

It should be easy to finish the rest of the grid.

After we finish the entire grid, let's put the thermometers back.

Not at all surprising, the length 4 thermometer is broken.

But we can easily fix it by relabeling the digits:

• {1, 9} ==> {1, 9}

• {2, 8} ==> {3, 7}

• {3, 7} ==> {4, 6}

• {4, 6} ==> {2, 8}

This will give us a grid that satisfied all constraints.

This is the solution to the puzzle.