An Expert 10×10 Puzzle Solved with Forced Chain in 46 Steps

10×10 · expert · 46 steps

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This 10×10 board rates as expert (difficulty score 2083). Solving it from scratch takes 46 logical steps, using 6 techniques: Single cell, Row/column exclusion, Adjacency clear, Region confinement, Line confinement, Forced chain. Every step below is forced: nothing here is a guess.

First, marking A1 would force J2 (the last open cell in row 2), then G7, H7, H8, and 1 more cleared (every open cell in column I belongs to the same region, so that region's marker has to be somewhere in this column), then B3, B4, B5, and 7 more cleared (together, two regions have open cells only in columns B and C), and the chain continues, and then row 4 would be left with no open cell for its marker. So A1 can't be the marker there.
Next, if B1 were the marker, it would force J2 (the last open cell in row 2), then G7, H7, H8, and 1 more cleared (every open cell in column I belongs to the same region, so that region's marker has to be somewhere in this column), then A3, A4, A5, and 6 more cleared (together, two regions have open cells only in columns A and C), 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, that one can't be the marker: at C1 it would force A2, A3, A4, and 5 more cleared (together, two regions have open cells only in columns A and B), then D3 (the last open cell in region 3), then J2 (the last open cell in row 2), and row 4 would be left with no open cell for its marker.
After that, if H1 were the marker, it would force G7, J8, J9, and 1 more cleared (every open cell in column I belongs to the same region, so that region's marker has to be somewhere in this column), and column J would end up with no open cell for its marker, which can't happen. So it can't go there.
Now, marking B2 would force I3, J3, I4, 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 A4, A5, C4, and 4 more cleared (together, two regions have open cells only in columns A and C), then D3 (the last open cell in region 3), and then row 4 would be left with no open cell for its marker. So B2 can't be the marker there.
From there, if C2 were the marker, it would force I3, J3, I4, 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 A8, A9, A10, and 2 more cleared (together, two regions have open cells only in columns A and B), 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, if I2 were the marker, row 1 would end up with no open cell for its marker, which can't happen. So it can't go there.
First, if G3 were the marker, it would force J2, I4, J4, 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), then A2 (the last open cell in row 2), then C4, C5, C6, and 2 more cleared (together, two regions have open cells only in columns B and C), 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.
Next, that one can't be the marker: at H3 it would force J2, J4, I5, and 4 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 A2 (the last open cell in row 2), then C4, C5, C6, and 2 more cleared (together, two regions have open cells only in columns B and C), and the chain continues, and row 4 would be left with no open cell for its marker.
Then, marking I3 would force D2, E2, F2, and 4 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 A2 (the last open cell in row 2), then G7, H7, H8, and 1 more cleared (every open cell in column J belongs to the same region, so that region's marker has to be somewhere in this column), and the chain continues, and then row 4 would be left with no open cell for its marker. So I3 can't be the marker there.
After that, if J3 were the marker, it would force D2, E2, F2, 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 A2 (the last open cell in row 2), then G7, H7, H8, and 1 more cleared (every open cell in column I belongs to the same region, so that region's marker has to be somewhere in this column), 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 C4 it would force A8, A9, A10, and 2 more cleared (together, two regions have open cells only in columns A and B), and region 9 would be left with no open cell for its marker.
From there, marking E2 would force I4, J4, I5, and 4 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 A4, B4, A5, and 2 more cleared (every open cell in row 3 belongs to the same region, so that region's marker has to be somewhere in this row), then D4 (the last open cell in row 4), and then row 5 would be left with no open cell for its marker. So E2 can't be the marker there.
Following that, if F2 were the marker, it would force I4, J4, I5, and 4 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 A4, B4, A5, and 2 more cleared (every open cell in row 3 belongs to the same region, so that region's marker has to be somewhere in this row), then E5, G5, G6, and 1 more cleared (every open cell in row 4 belongs to the same region, so that region's marker has to be somewhere in this row), 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.
First, that one can't be the marker: at F4 it would force A6, B6, C6, and 4 more cleared (every open cell left in this region sits in row 6, so its marker has to land there), then J2, I5, J5, 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 A2 (the last open cell in row 2), and row 3 would be left with no open cell for its marker.
Next, marking G4 would force J2, I5, J5, and 3 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 A2 (the last open cell in row 2), then E3 (the last open cell in row 3), and the chain continues, and then row 6 would be left with no open cell for its marker. So G4 can't be the marker there.
Then, if H4 were the marker, it would force J2, J5, I6, and 2 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 A2 (the last open cell in row 2), then E5, F5, F6, and 2 more cleared (every open cell in row 3 belongs to the same region, so that region's marker has to be somewhere in this row), and the chain continues, and row 6 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: at A2 it would force D4, E4, E5, and 6 more cleared (every open cell in row 3 belongs to the same region, so that region's marker has to be somewhere in this row), then I1, J1, I5, and 4 more 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 H5 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 the chain continues, and row 6 would be left with no open cell for its marker.
Now, if D3 were the marker, it would force I1, J1, J2, and 5 more 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 G2, H2 and H5 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.
From there, that one can't be the marker: at I4 it would force D2, G2 and H2 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.
Following that, marking J4 would force D2, G2, H2, 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), and then row 2 would be left with no open cell for its marker. So J4 can't be the marker there.
First, if A3 were the marker, it would force E5, F5, G5, and 4 more 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 C5, C6, C7, and 1 more cleared (together, two regions have open cells only in columns B and C), then D1, D2, D7, and 7 more cleared (together, two regions have open cells only in columns D and E), and the chain continues, and region 10 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 B3 it would force E5, F5, G5, and 4 more 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 C5, C6, C7, and 1 more cleared (together, two regions have open cells only in columns A and C), then D1, D2, D7, and 7 more cleared (together, two regions have open cells only in columns D and E), and the chain continues, and region 10 would be left with no open cell for its marker.
Then, that one can't be the marker: at D2 it would force F3 (the last open cell in row 3), then I5, J5, I6, and 2 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 A5, B5 and B6 cleared (every open cell in row 4 belongs to the same region, so that region's marker has to be somewhere in this row), and the chain continues, and row 6 would be left with no open cell for its marker.
After that, marking C3 would force E4 (the last open cell in row 4), then D6 (the last open cell in region 5), then F9 (the last open cell in region 7), and then region 10 would be left with no open cell for its marker. So C3 can't be the marker there.
Now, every open cell in row 3 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, 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, marking D1 would force J2 (the last open cell in row 2), then C5 (the last open cell in row 5), then A4 (the last open cell in region 3), and then row 6 would be left with no open cell for its marker. So D1 can't be the marker there.
First, if E1 were the marker, it would force F3 (the last open cell in region 4), then J2 (the last open cell in row 2), then C6 and D6 cleared (every open cell in row 5 belongs to the same region, so that region's marker has to be somewhere in this row), and the chain continues, and region 7 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 F1 it would force E3 (the last open cell in region 4), then J2 (the last open cell in row 2), then C6 and D6 cleared (every open cell in row 5 belongs to the same region, so that region's marker has to be somewhere in this row), and the chain continues, and region 7 would be left with no open cell for its marker.
Then, marking G1 would force J2 (the last open cell in row 2), then C6, D6, E6, and 1 more cleared (every open cell in row 5 belongs to the same region, so that region's marker has to be somewhere in this row), then A6 (the last open cell in row 6), and the chain continues, and then column I would be left with no open cell for its marker. So G1 can't be the marker there.
After that, 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.
Now, 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.
From there, 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.
Following that, 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.
First, place a marker at B4. Region has exactly one open cell left. Placing here clears the rest of row 4, column B, and its region. Placing here also clears its 1 touching neighbour.
Next, 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.
Then, place a marker at C10. Region has exactly one open cell left. Placing here clears the rest of row 10, column C, and its region.
After that, every open cell left in this region sits in row 9, so its marker has to land there; that clears every other open cell in the row.
Now, every open cell left in this region sits in row 8, so its marker has to land there; that clears every other open cell in the row.
From there, place a marker at J1. Column 10 has exactly one open cell left. Placing here clears the rest of row 1, column J, and its region.
Following that, place a marker at I7. Column 9 has exactly one open cell left. Placing here clears the rest of row 7, column I, and its region.
First, place a marker at H2. Column 8 has exactly one open cell left. Placing here clears the rest of row 2, column H, and its region.
Next, place a marker at G8. Column 7 has exactly one open cell left. Placing here clears the rest of row 8, column G, and its region. Placing here also clears its 1 touching neighbour.
Then, place a marker at E9. Region has exactly one open cell left. Placing here clears the rest of row 9, column E, and its region.
Finally, place a marker at F3. Region has exactly one open cell left.

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