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Let's start with the most restricted clues.
The minimum value of the highlighted region is (3+4+5+6+7) + 3 + 1, equals to 14+15. Therefore the highlighted region is forced to be minimum values.
The highlighted region is forced too:
The minimum value of Box4 is 4+5+6=15
The minimum value of Box7 is 3+4+5+6+7+8=33
The minimum value of Box8 is 2+3+4=9
The total is 15+33+9=57=14+20+23
In both box1 and box7, digit 3 must be on the diagonal of 14.
In box9, digit 2 and digit 3 must be on row 7.
On the highlighted column, digit 1 can only be in r8c6.
Digit 1 is already placed in box2.
Digit 1 cannot be in the middle of a thermometer.
Digit 1 is already place on row 9.
There is a 123 triple in box9.
Digit 4 in both box8 and box9 are on row8 and row9. Therefore digit 4 in box7 must be on row7.
Digit 7 in column1 can only be in r3c1 or r7c1.
r7c1 is impossible. Because placing digit 7 in r7c1, we need to place digit 4 on the diagonal of 14 as well, which is impossible.
Therefore, digit 7 must be in r3c1.
After we placed digit 7 in r3c1, the board looks like this.
r7c3 cannot be digit 4. Otherwise the sum cannot reach 23.
Therefore, digit 4 of box4 and box7 are locked on column1 and column2. Therefore digit 4 of box1 must be on column3.
We can fill in more digits.
The longest path on the thermometer is highlighted. It's 7 cells long. And it's very restricted, 3 cannot be on the thermometer. It can only use 1, 2, 4, 5, 6, 7, 8, 9.
Now we found a 24 pair in column4, and 13 pair in r5c5, r6c5.
In box5, digit 9 can only be in r6c4.
In box5, digit 2 can only be in r4c4.
And we can get several digits.
The highlighted diagonal is more restricted now:
5+6+8+3+2=24, it only has 1 degree of freedom.
It means that we can remove 5,6 from the candidate of r7c3.
r4c3, r7c3 forms a 78 pair now. And r2c3, r8c3 become a 56 pair.
Because r8c3 can only be 5 or 6, the highlighted diagonal must be 4+7+3=14, instead of 5+6+3=14. Therefore, r7c1=4, r8c2=7.
And we can fill in several digits now.
The rest of the puzzle should be easy.