A Hard 7×7 Puzzle Solved with Forced Chain in 19 Steps

This 7×7 board rates as hard (difficulty score 1113). Solving it from scratch takes 19 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 7, 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 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.
3. Then, every open cell left in this region sits in column A, so its marker has to land there; that clears every other open cell in the column.
4. After that, 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.
5. Now, place a marker at A3. Region has exactly one open cell left. Placing here clears the rest of row 3, column A, and its region. Placing here also clears its 2 touching neighbours.
6. From there, that one can't be the marker: at C1 it would force B7 (the last open cell in column B), then E5 cleared (every remaining candidate in this region touches E5), then F5 (the last open cell in row 5), and the chain continues, and row 6 would be left with no open cell for its marker.
7. Following that, marking D1 would immediately leave row 2 with no open cell for its marker, so it can't be the marker there.
8. First, if E1 were the marker, it would force C2 (the last open cell in region 1), then D6 (the last open cell in region 5), and column G would end up with no open cell for its marker, which can't happen. So it can't go there.
9. Next, that one can't be the marker: at F1 it would force C2 (the last open cell in region 1), then G6 (the last open cell in region 6), then E5 (the last open cell in region 5), and row 4 would be left with no open cell for its marker.
10. Then, marking G1 would force C2 (the last open cell in region 1), then B7 (the last open cell in column B), then F4 cleared (every open cell left in this region sits in column F, so its marker has to land there), and the chain continues, and then column E would be left with no open cell for its marker. So G1 can't be the marker there.
11. After that, place a marker at B1. Row 1 has exactly one open cell left. Placing here clears the rest of row 1, column B, and its region.
12. Now, 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.
13. From there, 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.
14. Following that, every remaining candidate in this region touches F5 and F6, 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 G6. Region has exactly one open cell left. Placing here clears the rest of row 6, column G, and its region.
16. Next, place a marker at E5. Region has exactly one open cell left. Placing here also clears its 2 touching neighbours.
17. Then, place a marker at C4. Region has exactly one open cell left. Placing here clears the rest of row 4, column C, and its region.
18. After that, place a marker at D7. Region has exactly one open cell left. Placing here clears the rest of row 7, column D, and its region.
19. Finally, place a marker at F2. Region has exactly one open cell left.

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