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 1746). Solving it from scratch takes 35 logical steps, using 6 techniques: Single cell, Row/column exclusion, Adjacency clear, Line confinement, Shape squeeze, Forced chain. Every step below is forced: nothing here is a guess.

First, every remaining candidate in this region touches C2, 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 H7, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
Then, 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.
After that, marking A1 would immediately leave region 2 with no open cell for its marker, so it can't be the marker there.
Now, if E1 were the marker, it would force B2 (the last open cell in region 2), then I3 (the last open cell in row 3), then H6 (the last open cell in region 7), and the chain continues, and row 5 would end up with no open cell for its marker, which can't happen. So it can't go there.
From there, that one can't be the marker: at F1 it would force B2 (the last open cell in region 2), then E6 (the last open cell in region 6), then I3 (the last open cell in row 3), and region 7 would be left with no open cell for its marker.
Following that, marking G1 would force B2 (the last open cell in region 2), then A9, C9, D9, and 2 more cleared (every open cell left in this region sits in row 9, so its marker has to land there), then H5, I3, I4, and 1 more cleared (together, two regions have open cells only in columns H and I), 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.
First, if H1 were the marker, it would force B2 (the last open cell in region 2), then I3, I4, I5, and 1 more cleared (every open cell left in this region sits in column I, so its marker has to land there), then E3 (the last open cell in row 3), and row 4 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 I1 it would force B2 (the last open cell in region 2), then H6 (the last open cell in region 7), then E3 (the last open cell in row 3), and row 4 would be left with no open cell for its marker.
Then, every open cell in row 1 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.
After that, if F2 were the marker, it would force E6 (the last open cell in region 6), then I3, I4, I5, and 1 more cleared (every open cell left in this region sits in column I, so its marker has to land there), then A5, B5 and C5 cleared (every open cell left in this region sits in row 5, so its marker has to land there), and the chain continues, and row 4 would end up with no open cell for its marker, which can't happen. So it can't go there.
Now, that one can't be the marker: at G2 it would force A9, B9, C9, and 3 more cleared (every open cell left in this region sits in row 9, so its marker has to land there), then H5, I3, I4, and 1 more cleared (together, two regions have open cells only in columns H and I), then A5, B5, C5, and 1 more cleared (every open cell left in this region sits in row 5, so its marker has to land there), and the chain continues, and row 4 would be left with no open cell for its marker.
From there, marking H2 would force I4, I5 and I9 cleared (every open cell left in this region sits in column I, so its marker has to land there), then G5 and G6 cleared (every open cell left in this region sits in column G, so its marker has to land there), then A5, B5, C5, and 1 more cleared (every open cell left in this region sits in row 5, so its marker has to land there), and the chain continues, and then row 4 would be left with no open cell for its marker. So H2 can't be the marker there.
Following that, if I2 were the marker, it would force H6 (the last open cell in region 7), then F3 and F4 cleared (every open cell left in this region sits in column F, so its marker has to land there), then G3 and G4 cleared (every open cell left in this region sits in column G, so its marker has to land there), and the chain continues, and row 7 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 F4, G4 and H4 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 I4 (the last open cell in row 4), and row 5 would be left with no open cell for its marker.
Next, marking C3 would force A2 (the last open cell in row 2), then D1 (the last open cell in region 2), then I4 (the last open cell in row 4), and the chain continues, and then row 8 would be left with no open cell for its marker. So C3 can't be the marker there.
Then, if D3 were the marker, it would force A2 (the last open cell in row 2), then C1 (the last open cell in region 2), then I4 (the last open cell in row 4), and the chain continues, and row 8 would end up with no open cell for its marker, which can't happen. So it can't go there.
After that, that one can't be the marker: place it at E3, and row 2 would be left with no open cell for its marker.
Now, marking F3 would force E6 (the last open cell in region 6), then A2 (the last open cell in row 2), then B5 (the last open cell in region 5), and the chain continues, and then region 7 would be left with no open cell for its marker. So F3 can't be the marker there.
From there, if G3 were the marker, it would force A2 (the last open cell in row 2), then I4 (the last open cell in row 4), then H6 (the last open cell in region 7), and region 9 would end up with no open cell for its marker, which can't happen. So it can't go there.
Following that, that one can't be the marker: at H3 it would force A2 (the last open cell in row 2), and row 4 would be left with no open cell for its marker.
First, marking D1 would force A2 (the last open cell in row 2), then I3 (the last open cell in row 3), then H6 (the last open cell in region 7), and the chain continues, and then region 9 would be left with no open cell for its marker. So D1 can't be the marker there.
Next, that one can't be the marker: at A2 it would force C1 (the last open cell in region 2), then I3 (the last open cell in row 3), then H6 (the last open cell in region 7), and the chain continues, and region 9 would be left with no open cell for its marker.
Then, 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.
After that, marking I3 would force H6 (the last open cell in region 7), then A5, C5 and D5 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 B5 (the last open cell in row 5), and the chain continues, and then row 7 would be left with no open cell for its marker. So I3 can't be the marker there.
Now, 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.
From there, 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.
Following that, place a marker at H6. Region has exactly one open cell left. Placing here clears the rest of row 6, column H, and its region. Placing here also clears its 1 touching neighbour.
First, 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.
Next, place a marker at C1. Region has exactly one open cell left. Placing here clears the rest of row 1, column C, and its region. Placing here also clears its 1 touching neighbour.
Then, place a marker at E2. Region has exactly one open cell left. Placing here clears the rest of row 2, column E, and its region.
After that, every remaining candidate in this region touches G8, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
Now, place a marker at G9. Region has exactly one open cell left. Placing here clears the rest of row 9, column G, and its region. Placing here also clears its 1 touching neighbour.
From there, place a marker at F7. Region has exactly one open cell left. Placing here clears the rest of row 7, column F, and its region.
Finally, place a marker at D8. Region has exactly one open cell left.

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