Let's first stare at the board for a moment. There are 26 circles. It is similar to a killer cage, we have to find different digits in 1~9 and sum to 26.

1+2+...+6=21. It means that we must choose at least one number from 7, 8, 9 to make 26.

We can group the circles into 3 groups.

• The green group are all in 4 boxes, a number can only repeat 4 times.

• The blue group are all in r3, c7, a number can only repeat 2 times.

• The red group is a single cell, a number can only show up once.

• Therefore, for any number in the circles, they can only show up 7 times.

• It means that we can only use numbers from 1 to 7.

1+2+...+7=28, the only way to sum to 26 is 1+3+4+5+6+7=26.

Therefore, 2, 8, 9 are never in a circle.

Also, 7 must be in the red cell. (I forgot to put it, don't be surprise when you see it in step4)

In box7, numbers 2, 8, 9 cannot be in circles. But three of cells are occupied by thermos.

We cannot put 9 on any of thermo cells in box7. The only cell to place 9 is r9c2.

Similarly, we cannot place 8 in r8c1 nor r8c3. 8 must be in r7c2.

Now, 2 must be in r8c1 or r8c3. Both of them are on thermo, so 1 must be in r9c1 or r9c3.

Because 1 is in one of the circles now. We have found the only 1 in circles. Therefore, we cannot place 1 in other circles.

box8 is impacted now. We cannot place 1, 2, 8, 9 in circles.

And, we cannot place 1, 2, 8, 9 in r7c4 either, because of the thermo constraint. Therefore, 1289 forms a quadruple in box8. (And we can make them 19 pairs and 28 pairs)

Then we can check box4. Similarly, 1, 2, 8 cannot be in circles.

r4c3 is very limited:

• obviously 1 cannot be there.

• 2 cannot be there either, otherwise we need to place 1 in r5c3, which is a circle.

• 8 cannot be there either, it's too far away from the end of the thermo.

Therefore, 128 forms a triple in box4!

r6c3 must be 8.

r4c1 and r5c2 forms a 12 pair.

Therefore, in box7, we cannot place 2 in c1. It must go to c3.

Now we need to think about 3. We need to place exactly three 3s in circles.

In box5, 3 must be in a circle.

In box4 and box7, the only cell that is not in a circle is on thermos, but both 12 are already used in those boxes.

Therefore, 3 must be in circles in box4, box7 too.

Now we have found all three 3s in circles.

All other circles an only contain 4567.

All other 3s must not be in a circle.

Also, r3c7 cannot be 6. Because we can only place four 6s in the green group of step 1.

And, because 6 and 7 need to show up six times, they must not be in non-circled cells in greep and blue groups.

You can do some sudoku before this step, but you will need to come back here eventually...

Now, let's think about r4c3, r8c1.

Both of them can only be 45.

They cannot both be 5. Otherwise we cannot place five 5s in circles.

Can they both be 4? No

• If both of them are 4, we need to place four 4s using the remaining circles. 

• The best we can get is having four 4s from exactly row3, col7, box5, box8.

• But, box8, row3, col7 can only have two 4s.

Therefore, r4c3 and r8c1 are different.

We can remove 45 from cells that see both of them.

And, because one of them is 4, at least one of r5c3, r9c1 must be 3.

r3c7 must be 4. If it is 5, we cannot place five 5s in circles.

r3c4 can only be 8 now.

This 8 resolves 28 pair in box8 and box5.

And the 8 put pressure on row4, and force the numbers on the thermo in box6.

The rest of the puzzle can be solved by using common Sudoku skills. (remember to respect the pencil marks)