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

This 10×10 board rates as hard (difficulty score 1023). Solving it from scratch takes 19 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 row 1, so its marker has to land there; that clears every other open cell in the row.
2. Next, every remaining candidate in this region touches I2, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
3. Then, every remaining candidate in this region touches I4, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
4. After that, that one can't be the marker: at I1 it would force J5, J6, J7, and 3 more cleared (every open cell left in this region sits in column J, so its marker has to land there), then A9, C9 and H9 cleared (every open cell in row 10 belongs to the same region, so that region's marker has to be somewhere in this row), then F6, B7, C7, and 8 more cleared (every open cell in row 9 belongs to the same region, so that region's marker has to be somewhere in this row), and the chain continues, and row 10 would be left with no open cell for its marker.
5. Now, marking J1 would force I3 (the last open cell in region 5), then G2 (the last open cell in region 3), and then region 2 would be left with no open cell for its marker. So J1 can't be the marker there.
6. From there, place a marker at H1. Region has exactly one open cell left. Placing here clears the rest of row 1, column H, and its region. Placing here also clears its 1 touching neighbour.
7. Following that, together, two regions have open cells only in rows 2 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.
8. First, place a marker at J4. Region has exactly one open cell left. Placing here clears the rest of row 4, column J, and its region. Placing here also clears its 1 touching neighbour.
9. Next, place a marker at G3. Region has exactly one open cell left. Placing here clears the rest of row 3, column G, and its region. Placing here also clears its 1 touching neighbour.
10. Then, place a marker at E2. Region has exactly one open cell left. Placing here clears the rest of row 2, column E, and its region.
11. After that, place a marker at F5. Region has exactly one open cell left. Placing here clears the rest of row 5, column F, and its region.
12. Now, place a marker at D6. Region has exactly one open cell left. Placing here clears the rest of row 6, column D, and its region. Placing here also clears its 1 touching neighbour.
13. From there, place a marker at A7. Region has exactly one open cell left. Placing here clears the rest of row 7, column A, and its region. Placing here also clears its 1 touching neighbour.
14. Following that, every remaining candidate in this region touches C9, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
15. First, every open cell left in this region sits in row 10, so its marker has to land there; that clears every other open cell in the row.
16. Next, every remaining candidate in this region touches B9, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
17. Then, place a marker at C8. Region has exactly one open cell left. Placing here clears the rest of row 8, column C, and its region.
18. After that, place a marker at I9. Region has exactly one open cell left.
19. Finally, place a marker at B10. Region has exactly one open cell left.

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