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

This 7×7 board rates as hard (difficulty score 1111). Solving it from scratch takes 17 logical steps, using 7 techniques: Single cell, Row/column exclusion, Adjacency clear, Region confinement, Line confinement, Shape squeeze, Forced chain. Every step below is forced: nothing here is a guess.

1. First, every remaining candidate in this region touches F3, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
2. Next, marking A1 would force B3 (the last open cell in region 4), then D2 and E2 cleared (every open cell left in this region sits in row 2, so its marker has to land there), then D4 (the last open cell in region 2), and then region 5 would be left with no open cell for its marker. So A1 can't be the marker there.
3. Then, every open cell in column A 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, if B1 were the marker, it would force D5, E5, D6, and 4 more cleared (every open cell in column C belongs to the same region, so that region's marker has to be somewhere in this column), then E2 cleared (every open cell in column D belongs to the same region, so that region's marker has to be somewhere in this column), then E3 cleared (every remaining candidate in this region touches E3), and the chain continues, and region 3 would end up with no open cell for its marker, which can't happen. So it can't go there.
5. Now, that one can't be the marker: at C1 it would force D5, E5, D6, and 4 more cleared (every open cell in column B belongs to the same region, so that region's marker has to be somewhere in this column), then E2 cleared (every open cell in column D belongs to the same region, so that region's marker has to be somewhere in this column), then A3, E3 and A4 cleared (together, two regions have open cells only in rows 3 and 4), and the chain continues, and row 3 would be left with no open cell for its marker.
6. From there, marking D1 would immediately leave region 2 with no open cell for its marker, so it can't be the marker there.
7. Following that, if E1 were the marker, it would force G2 (the last open cell in region 3), then F4 (the last open cell in region 5), and region 7 would end up with no open cell for its marker, which can't happen. So it can't go there.
8. First, that one can't be the marker: at F1 it would force E3 (the last open cell in region 3), then G4 (the last open cell in region 5), and region 7 would be left with no open cell for its marker.
9. Next, place a marker at G1. Row 1 has exactly one open cell left. Placing here clears the rest of row 1, column G, 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, every open cell left in this region sits in column F, so its marker has to land there; that clears every other open cell in the column.
12. Now, place a marker at E4. Region has exactly one open cell left. Placing here clears the rest of row 4, column E, and its region. Placing here also clears its 3 touching neighbours.
13. From there, place a marker at D2. Region has exactly one open cell left. Placing here clears the rest of row 2, column D, and its region. Placing here also clears its 1 touching neighbour.
14. Following that, place a marker at B5. Region has exactly one open cell left. Placing here clears the rest of row 5, column B, and its region. Placing here also clears its 2 touching neighbours.
15. First, place a marker at F6. Region has exactly one open cell left.
16. Next, place a marker at C7. Region has exactly one open cell left. Placing here clears the rest of row 7, column C, and its region.
17. Finally, place a marker at A3. Region has exactly one open cell left.

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