A Hard 10×10 Puzzle Solved with Forced Chain in 19 Steps

This 10×10 board rates as hard (difficulty score 996). Solving it from scratch takes 19 logical steps, using 7 techniques: Single cell, Row/column exclusion, Adjacency clear, Region confinement, Line confinement, Shape squeeze, Forced chain. Every step below is forced: nothing here is a guess.

1. First, every open cell left in this region sits in column A, so its marker has to land there; that clears every other open cell in the column.
2. Next, every open cell left in this region sits in column B, so its marker has to land there; that clears every other open cell in the column.
3. Then, every open cell left in this region sits in column C, so its marker has to land there; that clears every other open cell in the column.
4. After that, place a marker at D5. Region has exactly one open cell left. Placing here clears the rest of row 5, column D, and its region. Placing here also clears its 3 touching neighbours.
5. Now, every open cell in row 10 belongs to the same region, so that region's marker has to be somewhere in this row; that clears every other open cell in the region.
6. From there, place a marker at G9. Row 9 has exactly one open cell left. Placing here clears the rest of row 9, column G, and its region. Placing here also clears its 2 touching neighbours.
7. Following that, every open cell in row 8 belongs to the same region, so that region's marker has to be somewhere in this row; that clears every other open cell in the region.
8. First, place a marker at H7. Row 7 has exactly one open cell left. Placing here clears the rest of row 7, column H, and its region. Placing here also clears its 1 touching neighbour.
9. Next, place a marker at J8. Region has exactly one open cell left. Placing here clears the rest of row 8, column J, and its region.
10. Then, place a marker at A6. Row 6 has exactly one open cell left. Placing here clears the rest of row 6, column A, and its region.
11. After that, every remaining candidate in this region touches B2, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
12. Now, every remaining candidate in this region touches C3, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
13. From there, that one can't be the marker: at C1 it would force I10 (the last open cell in column I), then F4 cleared (every open cell left in this region sits in column F, so its marker has to land there), then B4 (the last open cell in row 4), and the chain continues, and column F would be left with no open cell for its marker.
14. Following that, place a marker at C2. Region has exactly one open cell left. Placing here clears the rest of row 2, column C, and its region. Placing here also clears its 1 touching neighbour.
15. First, place a marker at B4. Region has exactly one open cell left. Placing here clears the rest of row 4, column B, and its region.
16. Next, that one can't be the marker: place it at F1, and row 3 would be left with no open cell for its marker.
17. Then, place a marker at F3. Column 6 has exactly one open cell left. Placing here clears the rest of row 3, column F, and its region.
18. After that, place a marker at E1. Region has exactly one open cell left. Placing here clears the rest of row 1, column E, and its region.
19. Finally, place a marker at I10. Region has exactly one open cell left.

The key move here was Forced chain. Once you can spot that, this board falls quickly. ∎