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

This 7×7 board rates as hard (difficulty score 985). 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 open cell left in this region sits in row 1, so its marker has to land there; that clears every other open cell in the row.
2. Next, every open cell left in this region sits in row 2, so its marker has to land there; that clears every other open cell in the row.
3. Then, 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.
4. After that, 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.
5. Now, every remaining candidate in this region touches E2, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
6. 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 3 touching neighbours.
7. Following that, place a marker at F1. Region has exactly one open cell left. Placing here clears the rest of row 1, column F, and its region.
8. First, 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.
9. Next, 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.
10. Then, if G3 were the marker, it would force C4 and E4 cleared (every open cell left in this region sits in row 4, so its marker has to land there), then C5 (the last open cell in region 5), then A4 (the last open cell in region 1), and row 6 would end up with no open cell for its marker, which can't happen. So it can't go there.
11. After that, 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.
12. Now, every open cell left in this region sits in row 6, so its marker has to land there; that clears every other open cell in the row.
13. From there, every remaining candidate in this region touches C5 and B7 and C7, so whichever candidate ends up marked will clear them, and they can be ruled out now regardless of which one wins.
14. Following that, 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.
15. First, 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.
16. Next, place a marker at G5. Region has exactly one open cell left.
17. 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. ∎