An Expert 9×9 Puzzle Solved with Forced Chain in 36 Steps

9×9 · expert · 36 steps

Play this board yourself →

1×

2×

4×

5×

7×

8×

9×

This 9×9 board rates as expert (difficulty score 1764). Solving it from scratch takes 36 logical steps, using 7 techniques: Single cell, Row/column exclusion, Adjacency clear, 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 E3, 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 H8, 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 B9 (the last open cell in region 6), then I2, I3, I4, and 2 more cleared (every open cell left in this region sits in column I, so its marker has to land there), then E5, F5, D6, and 2 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 region 8 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 C1 it would force B9 cleared (every open cell in column A belongs to the same region, so that region's marker has to be somewhere in this column), then E5, F5, D6, and 2 more cleared (every open cell in column B belongs to the same region, so that region's marker has to be somewhere in this column), then F8, G8, E9, and 1 more cleared (every open cell in column D belongs to the same region, so that region's marker has to be somewhere in this column), and the chain continues, and column B would be left with no open cell for its marker.
Now, marking D1 would force F5, F6, F7, and 2 more cleared (every open cell left in this region sits in column F, so its marker has to land there), then B9 cleared (every open cell in column A belongs to the same region, so that region's marker has to be somewhere in this column), then E5, C6, E6, and 2 more cleared (every open cell in column B belongs to the same region, so that region's marker has to be somewhere in this column), and the chain continues, and then column B would be left with no open cell for its marker. So D1 can't be the marker there.
From there, if E1 were the marker, it would force F5, F6, F7, and 2 more cleared (every open cell left in this region sits in column F, 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), then A8, B8, C8, and 2 more cleared (together, two regions have open cells only in rows 8 and 9), and the chain continues, and region 5 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 I1 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 A2, A3, A4, and 1 more cleared (every open cell left in this region sits in column A, so its marker has to land there), then G7, G8 and H7 cleared (together, two regions have open cells only in columns G and H), and the chain continues, and row 8 would be left with no open cell for its marker.
First, that one can't be the marker: at A2 it would force B9 (the last open cell in region 6), then F1, F5, F6, and 2 more cleared (every open cell left in this region sits in column F, so its marker has to land there), 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), and the chain continues, and row 4 would be left with no open cell for its marker.
Next, if C2 were the marker, it would force F1, F5, F6, and 3 more cleared (every open cell left in this region sits in column F, so its marker has to land there), then G3, H3 and H4 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 B9 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 column B would end up with no open cell for its marker, which can't happen. So it can't go there.
Then, marking E2 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 A8, B8, C8, and 2 more cleared (together, two regions have open cells only in rows 8 and 9), then A6 (the last open cell in region 6), and the chain continues, and then column C would be left with no open cell for its marker. So E2 can't be the marker there.
After that, if F2 were the marker, it would force H5, H6, H7, and 1 more cleared (every open cell left in this region sits in column H, so its marker has to land there), then G7 and G8 cleared (together, two regions have open cells only in columns I and G), then E4, E5, E6, and 1 more cleared (every open cell left in this region sits in column E, so its marker has to land there), and the chain continues, and region 5 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 F4 (the last open cell in region 3), then B1 (the last open cell in row 1), then I3 (the last open cell in row 3), and row 5 would be left with no open cell for its marker.
From there, marking H2 would force B1 (the last open cell in row 1), then F3 (the last open cell in row 3), then I4 (the last open cell in row 4), and the chain continues, and then column C 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 F1, F5, F6, and 3 more cleared (every open cell left in this region sits in column F, so its marker has to land there), then A9, B9, C9, and 2 more cleared (every open cell left in this region sits in row 9, so its marker has to land there), then A3, A4 and A5 cleared (every open cell left in this region sits in column A, so its marker has to land there), and the chain continues, and region 5 would end up with no open cell for its marker, which can't happen. So it can't go there.
First, if A3 were the marker, it would force B9 (the last open cell in region 6), then D2 (the last open cell in row 2), then I4, I5 and I6 cleared (every open cell left in this region sits in column I, so its marker has to land there), 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.
Next, marking C3 would immediately leave row 2 with no open cell for its marker, so it can't be the marker there.
Then, if D3 were the marker, row 2 would end up with no open cell for its marker, which can't happen. So it can't go there.
After that, marking F3 would force B2 (the last open cell in row 2), then H4 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 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 F3 can't be the marker there.
Now, marking B2 would force F4 (the last open cell in region 3), then H3 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 (the last open cell in row 3), and then row 5 would be left with no open cell for its marker. So B2 can't be the marker there.
From there, place a marker at D2. Row 2 has exactly one open cell left. Placing here clears the rest of row 2, column D, and its region.
Following that, marking G1 would force H7 cleared (together, two regions have open cells only in columns I and H), then F8 cleared (every remaining candidate in this region touches F8), then A9, B9, H9, and 1 more cleared (every open cell left in this region sits in row 9, 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 G1 can't be the marker there.
First, if G3 were the marker, it would force B1 (the last open cell in row 1), then I4 (the last open cell in row 4), then H9 (the last open cell in region 9), 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.
Next, if B1 were the marker, it would force H5, H6, H7, and 1 more cleared (every open cell left in this region sits in column H, so its marker has to land there), then G7 and G8 cleared (together, two regions have open cells only in columns I and G), then I3, G4 and I4 cleared (every remaining candidate in this region touches I3 and G4 and I4), 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.
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, marking I3 would force A9, B9, C9, and 2 more cleared (every open cell left in this region sits in row 9, so its marker has to land there), then A4 and A5 cleared (every open cell left in this region sits in column A, so its marker has to land there), then A8, B8, C8, and 1 more cleared (every open cell left in this region sits in row 8, so its marker has to land there), and the chain continues, and then column C would be left with no open cell for its marker. So I3 can't be the marker there.
Now, place a marker at B3. Row 3 has exactly one open cell left. Placing here clears the rest of row 3, column B, 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 open cell in row 5 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, place a marker at A6. Row 6 has exactly one open cell left. Placing here clears the rest of row 6, column A, and its region.
Next, place a marker at C9. Column 3 has exactly one open cell left. Placing here clears the rest of row 9, column C, and its region.
Then, 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.
After that, place a marker at G4. Region has exactly one open cell left. Placing here clears the rest of row 4, column G, and its region. Placing here also clears its 1 touching neighbour.
Now, place a marker at E5. Region has exactly one open cell left. Placing here clears the rest of row 5, column E, and its region.
From there, place a marker at I8. Row 8 has exactly one open cell left. Placing here clears the rest of row 8, column I, and its region. Placing here also clears its 1 touching neighbour.
Following that, 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 H1. Region has exactly one open cell left.

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