A Sharp 10×10 Puzzle Solved with Forced Chain in 17 Steps

This 10×10 board rates as sharp (difficulty score 972). Solving it from scratch takes 17 logical steps, using 8 techniques: Single cell, Row/column exclusion, Adjacency clear, Region confinement, Line confinement, Paired regions, 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 row 2, so its marker has to land there; that clears every other open cell in the row.
2. Next, together, two regions have open cells only in rows 1 and 3; 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.
3. Then, every open cell in column J belongs to the same region, so that region's marker has to be somewhere in this column; that clears every other open cell in the region.
4. After that, every remaining candidate in this region touches H2, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
5. Now, place a marker at I2. Region has exactly one open cell left. Placing here clears the rest of row 2, column I, and its region. Placing here also clears its 1 touching neighbour.
6. From there, every open cell left in this region sits in column G, so its marker has to land there; that clears every other open cell in the column.
7. Following that, place a marker at H4. Region has exactly one open cell left. Placing here clears the rest of row 4, column H, and its region. Placing here also clears its 1 touching neighbour.
8. First, place a marker at G1. Region has exactly one open cell left. Placing here clears the rest of row 1, column G, and its region.
9. Next, place a marker at A5. Region has exactly one open cell left. Placing here clears the rest of row 5, column A, and its region. Placing here also clears its 1 touching neighbour.
10. Then, marking C3 would force B9 (the last open cell in region 8), and then region 10 would be left with no open cell for its marker. So C3 can't be the marker there.
11. After that, marking F3 would force B10, C10 and J10 cleared (every open cell left in this region sits in row 10, so its marker has to land there), then B9, E9 and J9 cleared (every open cell left in this region sits in row 9, so its marker has to land there), then C6 (the last open cell in region 8), and the chain continues, and then row 10 would be left with no open cell for its marker. So F3 can't be the marker there.
12. Now, place a marker at B3. Region has exactly one open cell left. Placing here clears the rest of row 3, column B, and its region.
13. From there, place a marker at C6. Region has exactly one open cell left. Placing here clears the rest of row 6, column C, and its region. Placing here also clears its 1 touching neighbour.
14. Following that, place a marker at D9. Region has exactly one open cell left. Placing here clears the rest of row 9, column D, and its region. Placing here also clears its 2 touching neighbours.
15. First, place a marker at E7. Column 5 has exactly one open cell left. Placing here clears the rest of row 7, column E, and its region.
16. Next, place a marker at F10. Region has exactly one open cell left. Placing here clears the rest of row 10, column F, and its region.
17. Finally, place a marker at J8. Region has exactly one open cell left.

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