An Expert 9×9 Puzzle Solved with Forced Chain in 42 Steps

9×9 · expert · 42 steps

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This 9×9 board rates as expert (difficulty score 1843). Solving it from scratch takes 42 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.

First, 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.
Next, marking A1 would immediately leave region 4 with no open cell for its marker, so it can't be the marker there.
Then, marking D1 would force C3 (the last open cell in region 1), then A5, A6, A7, and 2 more cleared (every open cell left in this region sits in column A, so its marker has to land there), then B5 (the last open cell in region 6), and the chain continues, and then region 8 would be left with no open cell for its marker. So D1 can't be the marker there.
After that, marking B2 would immediately leave region 1 with no open cell for its marker, so it can't be the marker there.
Now, 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.
From there, marking E2 would force F4 (the last open cell in region 5), then A3 (the last open cell in region 4), then G1 (the last open cell in region 2), and then region 3 would be left with no open cell for its marker. So E2 can't be the marker there.
Following that, if F2 were the marker, it would force C3, D3, I3, and 5 more cleared (together, two regions have open cells only in rows 3 and 4), then H1 and I1 cleared (every open cell left in this region sits in row 1, so its marker has to land there), and region 3 would end up with no open cell for its marker, which can't happen. So it can't go there.
First, marking H2 would force I4 (the last open cell in region 3), then A3 (the last open cell in region 4), then E7, E8 and E9 cleared (every open cell left in this region sits in column E, 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 H2 can't be the marker there.
Next, if D3 were the marker, it would force E1, F1, G1, and 2 more cleared (every open cell left in this region sits in row 1, so its marker has to land there), then G5, G6, G7, and 2 more cleared (every open cell left in this region sits in column G, so its marker has to land there), then I5, I6, I7, and 2 more cleared (every open cell left in this region sits in column I, so its marker has to land there), and the chain continues, and column I would end up with no open cell for its marker, which can't happen. So it can't go there.
Then, that one can't be the marker: place it at E3, and region 5 would be left with no open cell for its marker.
After that, 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.
Now, marking F3 would force B1, C1, H1, and 1 more cleared (every open cell left in this region sits in row 1, so its marker has to land there), then A2 and I2 cleared (every open cell left in this region sits in row 2, so its marker has to land there), then A4 (the last open cell in region 4), and then region 3 would be left with no open cell for its marker. So F3 can't be the marker there.
From there, every remaining candidate in this region touches F5, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
Following that, if E1 were the marker, it would force F4 (the last open cell in region 5), then C6, C7, C8, and 1 more cleared (every open cell left in this region sits in column C, so its marker has to land there), then I5, I6, I7, and 2 more cleared (every open cell left in this region sits in column I, so its marker has to land there), and the chain continues, and column I would end up with no open cell for its marker, which can't happen. So it can't go there.
First, that one can't be the marker: at H3 it would force I1 (the last open cell in region 3), then A2 cleared (every open cell left in this region sits in row 2, so its marker has to land there), then A4 (the last open cell in region 4), and the chain continues, and column E would be left with no open cell for its marker.
Next, marking A4 would force D2, D7, D8, and 1 more cleared (every open cell left in this region sits in column D, so its marker has to land there), then E7, E8 and E9 cleared (every open cell left in this region sits in column E, so its marker has to land there), then G5, H5, I5, and 6 more cleared (together, two regions have open cells only in rows 5 and 6), and the chain continues, and then column E would be left with no open cell for its marker. So A4 can't be the marker there.
Then, if B1 were the marker, it would force A3 (the last open cell in region 4), then G5, G6, G7, and 2 more cleared (every open cell left in this region sits in column G, so its marker has to land there), then I5, I6, I7, and 2 more cleared (every open cell left in this region sits in column I, so its marker has to land there), and the chain continues, and row 6 would end up with no open cell for its marker, which can't happen. So it can't go there.
After that, marking G1 would force I5, I6, I7, and 2 more cleared (every open cell left in this region sits in column I, so its marker has to land there), then I2 and I3 cleared (together, two regions have open cells only in rows 2 and 3), then I4 (the last open cell in region 3), and the chain continues, and then row 6 would be left with no open cell for its marker. So G1 can't be the marker there.
Now, if H1 were the marker, it would force G5, G6, G7, and 2 more cleared (every open cell left in this region sits in column G, so its marker has to land there), then G3 cleared (together, two regions have open cells only in rows 2 and 3), then G4 (the last open cell in region 2), and the chain continues, and row 6 would end up with no open cell for its marker, which can't happen. So it can't go there.
From there, every open cell left in this region sits in column I, so its marker has to land there; that clears every other open cell in the column.
Following that, that one can't be the marker: at I1 it would force G5, G6, G7, and 2 more cleared (every open cell left in this region sits in column G, so its marker has to land there), then G2 and G3 cleared (together, two regions have open cells only in rows 2 and 3), then G4 (the last open cell in region 2), and the chain continues, and row 6 would be left with no open cell for its marker.
First, if C2 were the marker, it would force A3 (the last open cell in region 4), then I4 (the last open cell in region 3), then F1 (the last open cell in region 2), and the chain continues, and column E would end up with no open cell for its marker, which can't happen. So it can't go there.
Next, that one can't be the marker: at G2 it would force A3 (the last open cell in region 4), then C1 (the last open cell in region 1), then I4 (the last open cell in region 3), and the chain continues, and row 6 would be left with no open cell for its marker.
Then, if I2 were the marker, it would force A3 (the last open cell in region 4), then C1 (the last open cell in region 1), then G4 (the last open cell in region 2), and the chain continues, and row 6 would end up with no open cell for its marker, which can't happen. So it can't go there.
After that, every remaining candidate in this region touches H4, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
Now, every remaining candidate in this region touches G6, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
From there, that one can't be the marker: at C1 it would force A2 (the last open cell in row 2), then G5, G7, G8, and 1 more cleared (every open cell left in this region sits in column G, so its marker has to land there), then B4, D4, E4, and 1 more cleared (together, two regions have open cells only in rows 3 and 4), and the chain continues, and row 6 would be left with no open cell for its marker.
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.
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.
Next, together, two regions have open cells only in rows 2 and 3; 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.
Then, place a marker at I4. Region has exactly one open cell left. Placing here clears the rest of row 4, column I, and its region. Placing here also clears its 1 touching neighbour.
After that, every open cell left in this region sits in column G, so its marker has to land there; that clears every other open cell in the column.
Now, every open cell in column H 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.
From there, 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.
Following that, place a marker at G7. Region has exactly one open cell left. Placing here clears the rest of row 7, column G, and its region. Placing here also clears its 1 touching neighbour.
First, place a marker at H9. Region has exactly one open cell left. Placing here clears the rest of row 9, column H, and its region.
Next, every remaining candidate in this region touches D5 and D6, so whichever candidate ends up marked will clear them, and they can be ruled out now regardless of which one wins.
Then, place a marker at B5. Region has exactly one open cell left. Placing here clears the rest of row 5, column B, and its region.
After that, place a marker at E6. Region has exactly one open cell left.
Now, place a marker at C8. Region has exactly one open cell left. Placing here clears the rest of row 8, column C, and its region.
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.
Finally, place a marker at A3. Region has exactly one open cell left.

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