An Expert 9×9 Puzzle Solved with Forced Chain in 41 Steps

9×9 · expert · 41 steps

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This 9×9 board rates as expert (difficulty score 1940). Solving it from scratch takes 41 logical steps, using 7 techniques: Single cell, Row/column exclusion, Adjacency clear, 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 H8, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
Next, every remaining candidate in this region touches F8, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
Then, marking A1 would force B8 (the last open cell in region 4), then D9, E9, H9, and 1 more cleared (every open cell left in this region sits in row 9, so its marker has to land there), then I7 (the last open cell in region 8), and the chain continues, and then row 3 would be left with no open cell for its marker. So A1 can't be the marker there.
After that, that one can't be the marker: at F1 it would force G3, G4, G5, and 2 more cleared (every open cell left in this region sits in column G, so its marker has to land there), then H3, I3 and I4 cleared (every open cell in row 2 belongs to the same region, so that region's marker has to be somewhere in this row), then B8 cleared (every open cell in column A belongs to the same region, so that region's marker has to be somewhere in this column), and the chain continues, and region 8 would be left with no open cell for its marker.
Now, that one can't be the marker: at I1 it would force H9 (the last open cell in region 8), and region 9 would be left with no open cell for its marker.
From there, that one can't be the marker: at A2 it would force B8 (the last open cell in region 4), then D9, E9, H9, and 1 more cleared (every open cell left in this region sits in row 9, so its marker has to land there), then I7 (the last open cell in region 8), and the chain continues, and row 4 would be left with no open cell for its marker.
Following that, every open cell in column A 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.
First, if F2 were the marker, it would force G6 and G7 cleared (every open cell left in this region sits in column G, so its marker has to land there), then H9, I7, I8, and 1 more cleared (together, two regions have open cells only in columns H and I), and region 8 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 F9 (the last open cell in region 9), then I5, I6, I7, and 1 more cleared (every open cell left in this region sits in column I, so its marker has to land there), and region 8 would be left with no open cell for its marker.
Then, if E1 were the marker, it would force G3, H3, I3, and 1 more cleared (every open cell in row 2 belongs to the same region, so that region's marker has to be somewhere in this row), then H4, H5, H6, and 3 more cleared (together, two regions have open cells only in columns H and I), then G4, G5, F3, and 3 more cleared (together, two regions have open cells only in columns G and F), and the chain continues, and region 6 would end up with no open cell for its marker, which can't happen. So it can't go there.
After that, if I2 were the marker, it would force H9 (the last open cell in region 8), and region 9 would end up with no open cell for its marker, which can't happen. So it can't go there.
Now, marking C3 would force H2 (the last open cell in row 2), and then row 1 would be left with no open cell for its marker. So C3 can't be the marker there.
From there, if D3 were the marker, it would force H2 (the last open cell in row 2), and row 1 would end up with no open cell for its marker, which can't happen. So it can't go there.
Following that, marking F3 would force G1, G6 and G7 cleared (every open cell left in this region sits in column G, so its marker has to land there), then D5, C6, D6, and 6 more cleared (every open cell in column E belongs to the same region, so that region's marker has to be somewhere in this column), then H9, I7, I8, and 1 more cleared (together, two regions have open cells only in columns H and I), and then region 8 would be left with no open cell for its marker. So F3 can't be the marker there.
First, marking B2 would force G3, H3, I3, and 1 more cleared (every open cell in row 1 belongs to the same region, so that region's marker has to be somewhere in this row), then E3 (the last open cell in row 3), then D5, D6, D8, and 1 more cleared (every open cell in column C belongs to the same region, so that region's marker has to be somewhere in this column), and the chain continues, and then column F would be left with no open cell for its marker. So B2 can't be the marker there.
Next, if G3 were the marker, it would force F9 (the last open cell in region 9), then I5 and I6 cleared (every open cell left in this region sits in column I, so its marker has to land there), then C2, D2 and E2 cleared (every open cell in row 1 belongs to the same region, so that region's marker has to be somewhere in this row), and row 2 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: at H3 it would force I5 and I6 cleared (every open cell left in this region sits in column I, so its marker has to land there), then C2, D2 and E2 cleared (every open cell in row 1 belongs to the same region, so that region's marker has to be somewhere in this row), and row 2 would be left with no open cell for its marker.
After that, marking I3 would force H9 (the last open cell in region 8), and then region 9 would be left with no open cell for its marker. So I3 can't be the marker there.
Now, if B4 were the marker, row 3 would end up with no open cell for its marker, which can't happen. So it can't go there.
From there, marking G1 would force F9 (the last open cell in region 9), then I5 and I6 cleared (every open cell left in this region sits in column I, so its marker has to land there), then H4 cleared (every open cell left in this region sits in column H, so its marker has to land there), and the chain continues, and then row 3 would be left with no open cell for its marker. So G1 can't be the marker there.
Following that, together, two regions have open cells only in columns H and I; 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.
First, together, two regions have open cells only in columns G and F; 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.
Next, 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.
Then, marking E2 would force D7 (the last open cell in region 3), then G6 (the last open cell in region 7), then F9 (the last open cell in region 9), and then region 6 would be left with no open cell for its marker. So E2 can't be the marker there.
After that, that one can't be the marker: at B3 it would force I4 (the last open cell in row 4), then H9 (the last open cell in region 8), and region 9 would be left with no open cell for its marker.
Now, marking D1 would force H2 (the last open cell in row 2), then E3, E4, E8, and 1 more cleared (every open cell left in this region sits in column E, so its marker has to land there), then A3 (the last open cell in row 3), and the chain continues, and then column F would be left with no open cell for its marker. So D1 can't be the marker there.
From there, that one can't be the marker: at D2 it would force H1 (the last open cell in row 1), then A3 (the last open cell in row 3), then E4, 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 column F would be left with no open cell for its marker.
Following that, that one can't be the marker: at E3 it would force D7 (the last open cell in region 3), then G6 (the last open cell in region 7), then F9 (the last open cell in region 9), and region 6 would be left with no open cell for its marker.
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.
Next, that one can't be the marker: at C1 it would force H2 (the last open cell in row 2), then B5, B6 and B7 cleared (every open cell in row 4 belongs to the same region, so that region's marker has to be somewhere in this row), then B9 (the last open cell in column B), and the chain continues, and column F would be left with no open cell for its marker.
Then, if H1 were the marker, it would force C2 (the last open cell in region 1), then B5, B6 and B7 cleared (every open cell in row 4 belongs to the same region, so that region's marker has to be somewhere in this row), then B9 (the last open cell in column B), and the chain continues, and column F would end up with no open cell for its marker, which can't happen. So it can't go there.
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.
Now, place a marker at H2. Row 2 has exactly one open cell left. Placing here clears the rest of row 2, column H, and its region.
From there, every open cell in row 4 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.
Following that, every remaining candidate in this region touches D5, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
First, place a marker at E5. Row 5 has exactly one open cell left. Placing here clears the rest of row 5, column E, and its region. Placing here also clears its 1 touching neighbour.
Next, 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.
Then, place a marker at G6. Row 6 has exactly one open cell left. Placing here clears the rest of row 6, column G, and its region.
After that, place a marker at F9. Region has exactly one open cell left. Placing here clears the rest of row 9, column F, and its region.
Now, place a marker at D8. Region has exactly one open cell left. Placing here clears the rest of row 8, column D, and its region.
Finally, place a marker at I7. Region has exactly one open cell left.

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