The final solution is: >! [![enter image description here][1]][1] Here are the steps: First of all, focus on the top-left corner: >! [![enter image description here][2]][2] >! Here the only possibility for the vertical 23-run is `9 8 6`. But 9 or 8 in d would render the 12-run unsolvable, so d=6. Then a+b+c=6, so they are 1,2,3 in some order. But there's no possible combination for 27 with 1 or 2, so a=3. Then the 27-run must be `3 7 8 9` in some order; but g can't be 9 or 8, so g=7, f=1; so b$\ne$1$\implies $b=2,c=1,e=2. Since the horizontal 23-run is `9 8 6`, and we need 12 more for the 15-run, the only possibilities for the rest of the 15-run are `4 8` or `3 9`. But in the first case, we'd need another 16 for the 22-run, forcing a 7 between a and g, which is bad since g=7. So the part is something like: >! [![enter image description here][3]][3] Now we'll solve four very similar regions. Focus on this part: >! [![enter image description here][4]][4] >! Clearly the only possibilities for the 12-run are `7 5` or `8 4`. But the first case would imply d=1, b=7=a, which is bad. So (a,b,c,d)=(8,6,2,4). Similar strategy can be applied to the other 2x2 parts, giving: >! [![enter image description here][5]][5] >! Now that we know the 23-run considered in the first paragraph is `6 8 9`, we can do some usual sum-chasing to arrive at the following: >! [![enter image description here][6]][6] Now, >! we apply some reasoning along the line "`S` has the only combination `a1 a2 a3`, but that cell can't have `a1` or `a2`, so it's `a3`", to arrive at the following: >! [![enter image description here][7]][7] Next, consider the top-right corner. >! The only possibility for the vertical 24-run is `9 8 7` in some order, and for 30, it's `9 8 7 6` in some order. So their intersection must be 9 or 8. If it's 9, then the 16-run must be `7 9`, and the 24-run would be `9 8`, which means we need to complete the `19` run from 8, 9 and 2 more numbers, which is impossible. So the only possibility is: `8 9 * *` and `9 7 * *`. Also, since the last two asterisks are `1 2` in some order, and there's no combination for 4-cell 26-run with a 1, we get that these two rows are `8 9 7 6` and `9 7 2 1`. Now some sum-chasing leads to this: >! [![enter image description here][8]][8] Now let's try the bottom-left corner. >! The horizontal 23-run has to be `9 6 8`, or `6 9 8`; but the first possibility would mean we need to get a sum of 4 from 3 cells in the 13-run, so it has to be `6 9 8`. The other three cells in the 13-run have to be `1 2 4` in some order. How `1` or `2` in that 12 run isn't possible, so the next row is `4 8`. >! [![enter image description here][9]][9] >! Now focus on the remaining 2x4 grid. By summing by rows and then by column, we conclude that the yellow cells must add up to 13; but the only possible combination for the 11-run is `1 2 3 5` in some order, so these yellow cells must be `5 8`. Now the remaining cells in the bottom row are `1 2 4` in some order, and `4` fits only in the second cell. Now sum chasing yields the following: >! [![enter image description here][10]][10] To finish things off, look at the bottom right: >! it's simple to get that the 23-run is `6 8 9` and the 7-run begins with `4`. Now look at the 22-run. It's first element is `1` or `2` and the third element is among `1 2 3`. So the other two cells must sum up to at least 22-(2+3)=17; but that's the maximum for two cells, so equality holds, and the 22-run is precisely `2 9 3 8`. Now some quick casework yields the following: >! [![enter image description here][11]][11] Now it's solved completely! [1]: https://i.sstatic.net/zTKdF.png [2]: https://i.sstatic.net/Wkt1j.png [3]: https://i.sstatic.net/e0p0B.png [4]: https://i.sstatic.net/AqlEw.png [5]: https://i.sstatic.net/UxqO1.png [6]: https://i.sstatic.net/leWps.png [7]: https://i.sstatic.net/dhy9M.png [8]: https://i.sstatic.net/b55yu.png [9]: https://i.sstatic.net/T4EzF.png [10]: https://i.sstatic.net/JtSux.png [11]: https://i.sstatic.net/oVVki.png