An Expert 9×9 Puzzle Solved with Forced Chain in 33 Steps

9×9 · expert · 33 steps

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This 9×9 board rates as expert (difficulty score 1688). Solving it from scratch takes 33 logical steps, using 6 techniques: Single cell, Row/column exclusion, Region confinement, Line confinement, Shape squeeze, Forced chain. Every step below is forced: nothing here is a guess.

First, every remaining candidate in this region touches H6, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
Next, if B1 were the marker, it would force A6 and A7 cleared (every open cell left in this region sits in column A, so its marker has to land there), then C8, D8, E8, and 6 more cleared (together, two regions have open cells only in rows 8 and 9), then C7, F7, G7, and 2 more cleared (every open cell left in this region sits in row 7, so its marker has to land there), and the chain continues, and column C would end up with no open cell for its marker, which can't happen. So it can't go there.
Then, if E1 were the marker, it would force F6, G6, B7, and 2 more cleared (together, two regions have open cells only in rows 6 and 7), then C8, D8, F8, and 4 more cleared (together, two regions have open cells only in rows 8 and 9), and region 8 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 I2, I3, I4, and 3 more cleared (together, two regions have open cells only in columns I and H), then F3, F8 and F9 cleared (every open cell left in this region sits in column F, so its marker has to land there), then A6, B6, C6, and 7 more cleared (together, two regions have open cells only in rows 6 and 7), 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.
Now, if H1 were the marker, it would force I3, I4, I5, and 3 more cleared (together, two regions have open cells only in columns I and G), then F3, F8 and F9 cleared (every open cell left in this region sits in column F, so its marker has to land there), then A6, B6, C6, and 7 more cleared (together, two regions have open cells only in rows 6 and 7), and the chain continues, and column G 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 I1 it would force A7, B7, C7, and 2 more cleared (every open cell left in this region sits in row 7, so its marker has to land there), then E5 cleared (every open cell in row 6 belongs to the same region, so that region's marker has to be somewhere in this row), then G2, G3, G4, and 2 more cleared (together, two regions have open cells only in columns G and H), and the chain continues, and region 8 would be left with no open cell for its marker.
Following that, marking B2 would force F1 (the last open cell in row 1), then A6 and A7 cleared (every open cell left in this region sits in column A, so its marker has to land there), then C8, D8, E8, and 4 more cleared (together, two regions have open cells only in rows 8 and 9), and the chain continues, and then column C would be left with no open cell for its marker. So B2 can't be the marker there.
First, marking E2 would force A3, B3, C3, and 6 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 F6, G6, B7, and 2 more cleared (together, two regions have open cells only in rows 6 and 7), then C8, D8, F8, and 4 more cleared (together, two regions have open cells only in rows 8 and 9), and then region 8 would be left with no open cell for its marker. So E2 can't be the marker there.
Next, that one can't be the marker: at G2 it would force A3, B3, C3, and 6 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 I3, I4, I5, and 2 more cleared (together, two regions have open cells only in columns I and H), then D4 (the last open cell in row 4), and row 3 would be left with no open cell for its marker.
Then, marking D1 would force G3, H3 and I3 cleared (every open cell left in this region sits in row 3, so its marker has to land there), then F2, E4 and F4 cleared (every remaining candidate in this region touches F2 and E4 and F4), then H4, I4, G5, and 5 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), and the chain continues, and then row 5 would be left with no open cell for its marker. So D1 can't be the marker there.
After that, marking H2 would force G4 and G9 cleared (together, two regions have open cells only in columns I and G), then E5 cleared (every remaining candidate in this region touches E5), then B7, D7 and E7 cleared (together, two regions have open cells only in rows 6 and 7), and the chain continues, and then region 8 would be left with no open cell for its marker. So H2 can't be the marker there.
Now, if I2 were the marker, it would force A7, B7, C7, and 2 more cleared (every open cell left in this region sits in row 7, so its marker has to land there), then E5 cleared (every open cell in row 6 belongs to the same region, so that region's marker has to be somewhere in this row), then G3, G4 and G9 cleared (together, two regions have open cells only in columns G and H), and the chain continues, and region 8 would end up with no open cell for its marker, which can't happen. So it can't go there.
From there, if A3 were the marker, it would force F1 (the last open cell in row 1), and row 2 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 B3 it would force F1 (the last open cell in row 1), and row 2 would be left with no open cell for its marker.
First, marking C3 would force F1 (the last open cell in row 1), and then row 2 would be left with no open cell for its marker. So C3 can't be the marker there.
Next, that one can't be the marker: at E3 it would force A2, C2, A4, and 5 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), and row 2 would be left with no open cell for its marker.
Then, marking F3 would immediately leave region 3 with no open cell for its marker, so it can't be the marker there.
After that, if G3 were the marker, it would force A1, C1, A4, and 5 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 F1 (the last open cell in row 1), then I4 (the last open cell in row 4), 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.
Now, that one can't be the marker: at H3 it would force A1, C1, A4, and 5 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 F1 (the last open cell in row 1), and row 4 would be left with no open cell for its marker.
From there, that one can't be the marker: at F1 it would force I3 (the last open cell in row 3), then A4, B4 and C4 cleared (every open cell left in this region sits in row 4, so its marker has to land there), then A7, B7, C7, and 2 more cleared (every open cell left in this region sits in row 7, so its marker has to land there), and the chain continues, and row 6 would be left with no open cell for its marker.
Following that, every open cell left in this region sits in column D, so its marker has to land there; that clears every other open cell in the column.
First, 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.
Next, place a marker at F2. Row 2 has exactly one open cell left. Placing here clears the rest of row 2, column F, and its region.
Then, marking I3 would force D4 (the last open cell in row 4), and then row 5 would be left with no open cell for its marker. So I3 can't be the marker there.
After that, place a marker at D3. Row 3 has exactly one open cell left. Placing here clears the rest of row 3, column D, and its region.
Now, 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.
From there, 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.
Following that, place a marker at I6. Row 6 has exactly one open cell left. Placing here clears the rest of row 6, column I, and its region.
First, place a marker at H4. Region has exactly one open cell left. Placing here clears the rest of row 4, column H, and its region.
Next, place a marker at G8. Region has exactly one open cell left. Placing here clears the rest of row 8, column G, and its region.
Then, place a marker at C9. Region has exactly one open cell left. Placing here clears the rest of row 9, column C, and its region.
After that, place a marker at A1. Region has exactly one open cell left.
Finally, place a marker at B7. Region has exactly one open cell left.

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