A Hard 7×7 Puzzle Solved with Forced Chain in 18 Steps

This 7×7 board rates as hard (difficulty score 1092). Solving it from scratch takes 18 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 E, so its marker has to land there; that clears every other open cell in the column.
2. Next, 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.
3. Then, together, two regions have open cells only in rows 4 and 5; 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.
4. After that, every open cell left in this region sits in row 3, so its marker has to land there; that clears every other open cell in the row.
5. Now, 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.
6. From there, every remaining candidate in this region touches B2 and C2 and B4, so whichever candidate ends up marked will clear them, and they can be ruled out now regardless of which one wins.
7. Following that, every remaining candidate in this region touches D7, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
8. First, marking A1 would force G2 (the last open cell in region 2), then B5 (the last open cell in region 4), and then column F would be left with no open cell for its marker. So A1 can't be the marker there.
9. Next, if B1 were the marker, it would force G2 (the last open cell in region 2), then C3 (the last open cell in region 3), then A4 (the last open cell in row 4), and region 6 would end up with no open cell for its marker, which can't happen. So it can't go there.
10. Then, that one can't be the marker: at C1 it would force G2 (the last open cell in region 2), then B3 (the last open cell in region 3), then A5 (the last open cell in region 4), and column F would be left with no open cell for its marker.
11. After that, that one can't be the marker: at F1 it would force G5 (the last open cell in column G), then A4 (the last open cell in region 4), then D2 (the last open cell in region 1), and row 3 would be left with no open cell for its marker.
12. Now, place a marker at F5. Column 6 has exactly one open cell left. Placing here clears the rest of row 5, column F, and its region. Placing here also clears its 1 touching neighbour.
13. From there, place a marker at A4. Region has exactly one open cell left. Placing here clears the rest of row 4, column A, and its region. Placing here also clears its 1 touching neighbour.
14. Following that, place a marker at C3. Region has exactly one open cell left. Placing here clears the rest of row 3, column C, and its region. Placing here also clears its 1 touching neighbour.
15. First, place a marker at D1. Region has exactly one open cell left. Placing here clears the rest of row 1, column D, and its region.
16. Next, place a marker at G2. Region has exactly one open cell left.
17. Then, 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.
18. Finally, place a marker at B6. Region has exactly one open cell left.

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