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

This 7×7 board rates as hard (difficulty score 1069). 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 row 1, 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 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.
3. 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.
4. After that, every remaining candidate in this region touches E5, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
5. Now, that one can't be the marker: at C1 it would force E4, E6 and E7 cleared (every open cell left in this region sits in column E, so its marker has to land there), then D6 and D7 cleared (every open cell left in this region sits in column D, so its marker has to land there), then B4 cleared (every open cell left in this region sits in column B, so its marker has to land there), and the chain continues, and column F would be left with no open cell for its marker.
6. From there, marking D1 would force E3 (the last open cell in region 2), and then region 5 would be left with no open cell for its marker. So D1 can't be the marker there.
7. Following that, that one can't be the marker: at D2 it would force F3 (the last open cell in region 3), and region 5 would be left with no open cell for its marker.
8. First, if B1 were the marker, it would force E4, E6 and E7 cleared (every open cell left in this region sits in column E, so its marker has to land there), then D6 and D7 cleared (every open cell left in this region sits in column D, so its marker has to land there), then C4 cleared (every open cell left in this region sits in column C, so its marker has to land there), and the chain continues, and column F would end up with no open cell for its marker, which can't happen. So it can't go there.
9. Next, place a marker at A1. Region has exactly one open cell left. Placing here clears the rest of row 1, column A, and its region.
10. Then, every open cell left in this region sits in row 4, so its marker has to land there; that clears every other open cell in the row.
11. After that, place a marker at D5. Region has exactly one open cell left. Placing here clears the rest of row 5, column D, and its region. Placing here also clears its 3 touching neighbours.
12. Now, place a marker at B4. Region has exactly one open cell left. Placing here clears the rest of row 4, column B, and its region.
13. From there, place a marker at G6. Region has exactly one open cell left. Placing here clears the rest of row 6, column G, and its region.
14. Following that, every remaining candidate in this region touches E2 and E3, so whichever candidate ends up marked will clear them, and they can be ruled out now regardless of which one wins.
15. First, place a marker at C2. Region has exactly one open cell left. Placing here clears the rest of row 2, column C, and its region.
16. Next, place a marker at F3. Region has exactly one open cell left.
17. Finally, place a marker at E7. Region has exactly one open cell left.

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