An Expert 9×9 Puzzle Solved with Forced Chain in 35 Steps

9×9 · expert · 35 steps

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This 9×9 board rates as expert (difficulty score 1755). Solving it from scratch takes 35 logical steps, using 6 techniques: Single cell, Row/column exclusion, Adjacency clear, Line confinement, Paired regions, Forced chain. Every step below is forced: nothing here is a guess.

First, together, two regions have open cells only in rows 8 and 9; 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.
Next, together, two regions have open cells only in rows 6 and 7; 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, together, two regions have open cells only in rows 4 and 5; 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.
After that, marking A1 would force C9 (the last open cell in column C), then D5 (the last open cell in column D), and then column E would be left with no open cell for its marker. So A1 can't be the marker there.
Now, if B1 were the marker, it would force C9 (the last open cell in column C), then D5 (the last open cell in column D), and column E would end up with no open cell for its marker, which can't happen. So it can't go there.
From there, marking D1 would force C9 (the last open cell in column C), then B6 and B7 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 then column B would be left with no open cell for its marker. So D1 can't be the marker there.
Following that, if E1 were the marker, it would force C9 (the last open cell in column C), then D5 (the last open cell in column D), then I4 (the last open cell in region 5), and the chain continues, and row 3 would end up with no open cell for its marker, which can't happen. So it can't go there.
First, if H1 were the marker, it would force I3 (the last open cell in region 3), then D5 and E5 cleared (every open cell left in this region sits in row 5, so its marker has to land there), then F4, F6 and G6 cleared (every remaining candidate in this region touches F4 and F6 and G6), and the chain continues, and region 9 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 A2 it would force C9 (the last open cell in column C), then D5 (the last open cell in column D), and column E would be left with no open cell for its marker.
Then, marking B2 would force C9 (the last open cell in column C), then D5 (the last open cell in column D), and then column E would be left with no open cell for its marker. So B2 can't be the marker there.
After that, that one can't be the marker: at D2 it would force C9 (the last open cell in column C), then B6 and B7 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 column B would be left with no open cell for its marker.
Now, marking E2 would force C9 (the last open cell in column C), then D5 (the last open cell in column D), then I4 (the last open cell in region 5), and then region 3 would be left with no open cell for its marker. So E2 can't be the marker there.
From there, that one can't be the marker: at G2 it would force I4, I5 and I7 cleared (every open cell left in this region sits in column I, so its marker has to land there), then D5 and E5 cleared (every open cell left in this region sits in row 5, so its marker has to land there), then E3 and F5 cleared (every remaining candidate in this region touches E3 and F5), and the chain continues, and row 9 would be left with no open cell for its marker.
Following that, if A3 were the marker, it would force C9 (the last open cell in column C), then D5 (the last open cell in column D), and column E 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 B3 it would force C9 (the last open cell in column C), then D5 (the last open cell in column D), and column E would be left with no open cell for its marker.
Next, if D3 were the marker, it would force C9 (the last open cell in column C), then B6 and B7 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 column B 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 I1 it would force D5 and E5 cleared (every open cell left in this region sits in row 5, so its marker has to land there), then D9 (the last open cell in column D), then B6 and B7 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 column B would be left with no open cell for its marker.
After that, if C2 were the marker, it would force G3 cleared (every open cell left in this region sits in row 3, so its marker has to land there), then I4 cleared (every remaining candidate in this region touches I4), then D5 and E5 cleared (every open cell left in this region sits in row 5, so its marker has to land there), and the chain continues, and column B would end up with no open cell for its marker, which can't happen. So it can't go there.
Now, marking G1 would force I2 (the last open cell in row 2), then D5 and E5 cleared (every open cell left in this region sits in row 5, so its marker has to land there), then D9 (the last open cell in column D), and the chain continues, and then row 4 would be left with no open cell for its marker. So G1 can't be the marker there.
From there, if F2 were the marker, it would force C1 (the last open cell in row 1), then I4 cleared (every remaining candidate in this region touches I4), then E4 (the last open cell in row 4), and the chain continues, and column B would end up with no open cell for its marker, which can't happen. So it can't go there.
Following that, every open cell in row 2 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.
First, that one can't be the marker: at C1 it would force G3 (the last open cell in region 2), then I2 (the last open cell in region 3), then E4 (the last open cell in row 4), and the chain continues, and column D would be left with no open cell for its marker.
Next, 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.
Then, if I2 were the marker, it would force E4 (the last open cell in row 4), then C3 (the last open cell in region 1), then D9 (the last open cell in column D), and the chain continues, and column B 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 H2. Region has exactly one open cell left. Placing here clears the rest of row 2, column H, and its region.
Now, that one can't be the marker: at E3 it would force D5 (the last open cell in region 4), then I4 (the last open cell in region 5), then C9 (the last open cell in column C), and the chain continues, and region 7 would be left with no open cell for its marker.
From there, place a marker at C3. Region has exactly one open cell left. Placing here clears the rest of row 3, column C, and its region.
Following that, if E4 were the marker, it would force D9 (the last open cell in column D), then G8 (the last open cell in region 9), then I5 (the last open cell in region 5), and region 7 would end up with no open cell for its marker, which can't happen. So it can't go there.
First, place a marker at I4. Row 4 has exactly one open cell left. Placing here clears the rest of row 4, column I, and its region.
Next, together, two regions have open cells only in columns E and G; 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.
Then, place a marker at D5. Region has exactly one open cell left. Placing here clears the rest of row 5, column D, and its region. Placing here also clears its 1 touching neighbour.
After that, place a marker at E9. Column 5 has exactly one open cell left. Placing here clears the rest of row 9, column E, and its region.
Now, place a marker at A8. Region has exactly one open cell left. Placing here clears the rest of row 8, column A, and its region. Placing here also clears its 1 touching neighbour.
From there, place a marker at B6. Region has exactly one open cell left. Placing here clears the rest of row 6, column B, and its region.
Finally, place a marker at G7. Region has exactly one open cell left.

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