An Expert 7×7 Puzzle Solved with Forced Chain in 28 Steps

This 7×7 board rates as expert (difficulty score 1456). Solving it from scratch takes 28 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, every open cell left in this region sits in row 7, so its marker has to land there; that clears every other open cell in the row.
2. Next, every remaining candidate in this region touches C5, 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 A6 and B6, so whichever candidate ends up marked will clear them, and they can be ruled out now regardless of which one wins.
4. After that, marking A1 would force B7 (the last open cell in region 7), and then region 6 would be left with no open cell for its marker. So A1 can't be the marker there.
5. Now, if B1 were the marker, it would force C6 (the last open cell in region 6), then A7 (the last open cell in region 7), and region 1 would end up with no open cell for its marker, which can't happen. So it can't go there.
6. From there, that one can't be the marker: at C1 it would force B5 (the last open cell in region 6), then A7 (the last open cell in region 7), and region 1 would be left with no open cell for its marker.
7. Following that, if E1 were the marker, it would force A2 and G2 cleared (every open cell left in this region sits in row 2, so its marker has to land there), then F6 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 G3, G4, G5, and 1 more cleared (every open cell in column F belongs to the same region, so that region's marker has to be somewhere in this column), and column G would end up with no open cell for its marker, which can't happen. So it can't go there.
8. First, marking G1 would force A3 (the last open cell in row 3), then B7 (the last open cell in region 7), and then region 6 would be left with no open cell for its marker. So G1 can't be the marker there.
9. Next, every open cell in column G 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.
10. Then, that one can't be the marker: at A2 it would force D1 (the last open cell in region 2), then B7 (the last open cell in region 7), and region 6 would be left with no open cell for its marker.
11. After that, marking B2 would force C6 (the last open cell in region 6), then A7 (the last open cell in region 7), and then region 1 would be left with no open cell for its marker. So B2 can't be the marker there.
12. Now, if C2 were the marker, it would force B5 (the last open cell in region 6), then A3 (the last open cell in region 1), and row 7 would end up with no open cell for its marker, which can't happen. So it can't go there.
13. From there, marking E2 would immediately leave row 1 with no open cell for its marker, so it can't be the marker there.
14. Following that, every open cell left in this region sits in column D, so its marker has to land there; that clears every other open cell in the column.
15. First, together, two regions have open cells only in rows 5 and 6; 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.
16. Next, 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.
17. Then, every remaining candidate in this region touches B3, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
18. After that, if F2 were the marker, it would force D1 (the last open cell in region 2), then G4 (the last open cell in region 4), then A3 (the last open cell in region 1), and the chain continues, and column C would end up with no open cell for its marker, which can't happen. So it can't go there.
19. Now, marking D1 would force G2 (the last open cell in row 2), then F6 (the last open cell in column F), then B5 (the last open cell in region 6), and the chain continues, and then row 7 would be left with no open cell for its marker. So D1 can't be the marker there.
20. 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 2 touching neighbours.
21. Following that, place a marker at F1. Row 1 has exactly one open cell left. Placing here clears the rest of row 1, column F, and its region.
22. First, if A3 were the marker, it would force G4 (the last open cell in region 4), then B7 (the last open cell in region 7), and column C would end up with no open cell for its marker, which can't happen. So it can't go there.
23. Next, place a marker at G3. Row 3 has exactly one open cell left. Placing here clears the rest of row 3, column G, and its region.
24. Then, every remaining candidate in this region touches B5, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
25. After that, place a marker at C6. Region has exactly one open cell left. Placing here clears the rest of row 6, column C, and its region. Placing here also clears its 1 touching neighbour.
26. Now, place a marker at E5. Region has exactly one open cell left.
27. From there, place a marker at A7. Region has exactly one open cell left. Placing here clears the rest of row 7, column A, and its region.
28. Finally, place a marker at B4. Region has exactly one open cell left.

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