A Hard 10×10 Puzzle Solved with Forced Chain in 21 Steps

This 10×10 board rates as hard (difficulty score 1085). Solving it from scratch takes 21 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, every open cell left in this region sits in column J, 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 I, so its marker has to land there; that clears every other open cell in the column.
3. Then, 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 2 touching neighbours.
4. After that, every open cell left in this region sits in row 6, so its marker has to land there; that clears every other open cell in the row.
5. Now, together, two regions have open cells only in columns G and F; 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.
6. From there, place a marker at E10. Region has exactly one open cell left. Placing here clears the rest of row 10, column E, and its region. Placing here also clears its 1 touching neighbour.
7. Following that, every remaining candidate in this region touches F5, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
8. First, place a marker at G2. Region has exactly one open cell left. Placing here clears the rest of row 2, column G, and its region.
9. Next, place a marker at F6. Region has exactly one open cell left.
10. Then, 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.
11. After that, 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.
12. Now, every remaining candidate in this region touches C8 and C9, so whichever candidate ends up marked will clear them, and they can be ruled out now regardless of which one wins.
13. From there, if B1 were the marker, it would force J3 (the last open cell in region 3), then D8 (the last open cell in region 10), and column C would end up with no open cell for its marker, which can't happen. So it can't go there.
14. Following that, marking D1 would force J3 (the last open cell in region 3), then B9 (the last open cell in region 10), and then row 8 would be left with no open cell for its marker. So D1 can't be the marker there.
15. First, marking J1 would force D3 (the last open cell in region 1), then B9 (the last open cell in region 10), and then row 8 would be left with no open cell for its marker. So J1 can't be the marker there.
16. Next, place a marker at J3. Region has exactly one open cell left. Placing here clears the rest of row 3, column J, and its region.
17. Then, place a marker at C1. Region has exactly one open cell left. Placing here clears the rest of row 1, column C, and its region.
18. After that, marking D7 would immediately leave row 8 with no open cell for its marker, so it can't be the marker there.
19. Now, place a marker at D8. Column 4 has exactly one open cell left. Placing here clears the rest of row 8, column D, and its region.
20. From there, place a marker at B7. Region has exactly one open cell left. Placing here clears the rest of row 7, column B, and its region.
21. Finally, place a marker at I9. Region has exactly one open cell left.

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