An Expert 7×7 Puzzle Solved with Forced Chain in 18 Steps

This 7×7 board rates as expert (difficulty score 1130). Solving it from scratch takes 18 logical steps, using 8 techniques: Single cell, Row/column exclusion, Adjacency clear, Region confinement, Line confinement, Paired regions, Shape squeeze, Forced chain. Every step below is forced: nothing here is a guess.

1. First, place a marker at G2. Region has exactly one open cell left. Placing here clears the rest of row 2, column G, and its region. Placing here also clears its 2 touching neighbours.
2. Next, 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.
3. Then, every open cell in row 1 belongs to the same region, so that region's marker has to be somewhere in this row; that clears every other open cell in the region.
4. After that, every remaining candidate in this region touches D6, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
5. Now, marking A1 would force E3 (the last open cell in row 3), then F7 (the last open cell in region 5), and then column D would be left with no open cell for its marker. So A1 can't be the marker there.
6. From there, if E1 were the marker, it would force A3 (the last open cell in row 3), then B7 and F7 cleared (every open cell left in this region sits in row 7, so its marker has to land there), then F4 (the last open cell in region 5), and the chain continues, and row 7 would end up with no open cell for its marker, which can't happen. So it can't go there.
7. Following that, that one can't be the marker: at E3 it would force F7 (the last open cell in region 5), and region 7 would be left with no open cell for its marker.
8. First, place a marker at A3. Row 3 has exactly one open cell left. Placing here clears the rest of row 3, column A, and its region.
9. Next, marking D1 would force C5 and C6 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 E5 (the last open cell in row 5), and then row 4 would be left with no open cell for its marker. So D1 can't be the marker there.
10. Then, together, two regions have open cells only in columns B and C; 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, marking D4 would immediately leave row 5 with no open cell for its marker, so it can't be the marker there.
12. Now, if B1 were the marker, it would force D5 cleared (every remaining candidate in this region touches D5), then D7 (the last open cell in column D), and row 6 would end up with no open cell for its marker, which can't happen. So it can't go there.
13. From there, place a marker at C1. Region has exactly one open cell left. Placing here clears the rest of row 1, column C, and its region.
14. Following that, every open cell in row 5 belongs to the same region, so that region's marker has to be somewhere in this row; that clears every other open cell in the region.
15. First, place a marker at F4. Row 4 has exactly one open cell left. Placing here clears the rest of row 4, column F, and its region. Placing here also clears its 1 touching neighbour.
16. Next, 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 1 touching neighbour.
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. ∎