A Hard 10×10 Puzzle Solved with Forced Chain in 18 Steps

This 10×10 board rates as hard (difficulty score 991). Solving it from scratch takes 18 logical steps, using 7 techniques: Single cell, Row/column exclusion, Adjacency clear, Region confinement, Paired regions, Shape squeeze, Forced chain. Every step below is forced: nothing here is a guess.

1. First, place a marker at J1. Region has exactly one open cell left. Placing here clears the rest of row 1, column J, and its region. Placing here also clears its 1 touching neighbour.
2. Next, place a marker at H2. Region has exactly one open cell left. Placing here clears the rest of row 2, column H, and its region. Placing here also clears its 2 touching neighbours.
3. Then, every open cell left in this region sits in column I, so its marker has to land there; that clears every other open cell in the column.
4. After that, place a marker at G5. Region has exactly one open cell left. Placing here clears the rest of row 5, column G, and its region. Placing here also clears its 2 touching neighbours.
5. Now, together, two regions have open cells only in rows 9 and 10; with one marker each, those two markers have to fill exactly those two rows, so every other region's open cells there can be cleared.
6. From there, place a marker at I6. Region has exactly one open cell left. Placing here clears the rest of row 6, column I, and its region.
7. Following that, together, two regions have open cells only in rows 7 and 8; with one marker each, those two markers have to fill exactly those two rows, so every other region's open cells there can be cleared.
8. First, that one can't be the marker: at B3 it would force E4 (the last open cell in region 2), then C9 and C10 cleared (every open cell left in this region sits in column C, so its marker has to land there), then F7 and F8 cleared (every open cell left in this region sits in column F, so its marker has to land there), and the chain continues, and column C would be left with no open cell for its marker.
9. Next, marking C3 would force A4 (the last open cell in region 1), then B8 (the last open cell in region 6), then D10 (the last open cell in region 10), and then column E would be left with no open cell for its marker. So C3 can't be the marker there.
10. Then, together, two regions have open cells only in columns F and E; with one marker each, those two markers have to fill exactly those two columns, so every other region's open cells there can be cleared.
11. After that, every open cell left in this region sits in column D, so its marker has to land there; that clears every other open cell in the column.
12. Now, every remaining candidate in this region touches C7 and C8, so whichever candidate ends up marked will clear them, and they can be ruled out now regardless of which one wins.
13. From there, place a marker at B8. Region has exactly one open cell left. Placing here clears the rest of row 8, column B, and its region. Placing here also clears its 2 touching neighbours.
14. Following that, place a marker at D7. Region has exactly one open cell left.
15. First, place a marker at F9. Row 9 has exactly one open cell left. Placing here clears the rest of row 9, column F, and its region.
16. Next, place a marker at E4. Region has exactly one open cell left. Placing here clears the rest of row 4, column E, and its region.
17. Then, place a marker at A3. Region has exactly one open cell left. Placing here clears the rest of row 3, column A, and its region.
18. Finally, place a marker at C10. Region has exactly one open cell left.

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