A Hard 7×7 Puzzle Solved with Forced Chain in 17 Steps

This 7×7 board rates as hard (difficulty score 1062). Solving it from scratch takes 17 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 A, so its marker has to land there; that clears every other open cell in the column.
2. Next, every remaining candidate in this region touches C2, 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 F2, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
4. After that, every remaining candidate in this region touches B3 and B4, so whichever candidate ends up marked will clear them, and they can be ruled out now regardless of which one wins.
5. Now, every remaining candidate in this region touches G5, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
6. From there, together, two regions have open cells only in rows 6 and 7; 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.
7. Following that, that one can't be the marker: at C1 it would force B7 (the last open cell in region 7), then G6 (the last open cell in region 6), then F3 (the last open cell in region 2), and row 2 would be left with no open cell for its marker.
8. First, marking D1 would force G2 (the last open cell in row 2), then C6 (the last open cell in row 6), and then column B would be left with no open cell for its marker. So D1 can't be the marker there.
9. Next, if E1 were the marker, it would force B2 (the last open cell in row 2), and row 3 would end up with no open cell for its marker, which can't happen. So it can't go there.
10. Then, together, two regions have open cells only in columns F and G; 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, place a marker at C6. Row 6 has exactly one open cell left. Placing here clears the rest of row 6, column C, and its region. Placing here also clears its 3 touching neighbours.
12. Now, place a marker at E7. Region has exactly one open cell left. Placing here clears the rest of row 7, column E, and its region.
13. From there, place a marker at D4. Region has exactly one open cell left. Placing here clears the rest of row 4, column D, and its region.
14. Following that, 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. Placing here also clears its 1 touching neighbour.
15. First, place a marker at B1. Region has exactly one open cell left. Placing here clears the rest of row 1, column B, and its region.
16. Next, place a marker at G2. Region has exactly one open cell left.
17. Finally, place a marker at F5. Region has exactly one open cell left.

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