An Expert 10×10 Puzzle Solved with Forced Chain in 41 Steps

10×10 · expert · 41 steps

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This 10×10 board rates as expert (difficulty score 1826). Solving it from scratch takes 41 logical steps, using 8 techniques: Single cell, Row/column exclusion, Adjacency clear, Region confinement, Line confinement, Paired regions, Shape squeeze, Forced chain. Every step below is forced: nothing here is a guess.

First, every open cell left in this region sits in row 10, so its marker has to land there; that clears every other open cell in the row.
Next, every remaining candidate in this region touches F8, 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 I9, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
After that, that one can't be the marker: at C1 it would force B7, B8 and 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 B6 (the last open cell in column B), then G4, F5, G5, and 3 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 region 9 would be left with no open cell for its marker.
Now, marking D1 would force B7, B8, C8, and 2 more 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 B6 (the last open cell in column B), and then column C 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 B7, B8, C8, and 2 more 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 B6 (the last open cell in column B), and column C 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 F1 it would force G4, G5, G6, and 1 more cleared (every open cell left in this region sits in column G, so its marker has to land there), then H8 and H9 cleared (every remaining candidate in this region touches H8 and H9), then J10, I2, I3, and 4 more cleared (together, two regions have open cells only in columns J and I), and the chain continues, and region 4 would be left with no open cell for its marker.
First, if H1 were the marker, it would force I3, I5, I6, and 1 more cleared (together, two regions have open cells only in columns J and I), then G3, G4, G5, and 5 more cleared (together, two regions have open cells only in columns G and F), then E3, E5, E6, and 3 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 row 5 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 H3, H4, H5, and 2 more cleared (together, two regions have open cells only in columns J and H), then G2, G3, G4, and 6 more cleared (together, two regions have open cells only in columns G and F), then E3, E5, E6, and 3 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 row 5 would be left with no open cell for its marker.
Then, marking J1 would force A8, B8, C8, and 7 more cleared (together, two regions have open cells only in rows 8 and 9), then H2, H3, H4, and 7 more cleared (together, two regions have open cells only in columns H and I), then G2, G3, G4, and 6 more cleared (together, two regions have open cells only in columns G and F), and the chain continues, and then row 5 would be left with no open cell for its marker. So J1 can't be the marker there.
After that, that one can't be the marker: at A2 it would force G1 (the last open cell in row 1), then F9 (the last open cell in region 9), then C7, D7, E7, and 3 more cleared (together, two regions have open cells only in rows 7 and 8), and the chain continues, and column E would be left with no open cell for its marker.
Now, marking B2 would force G1 (the last open cell in row 1), then F9 (the last open cell in region 9), then C8 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 then column E 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 G1 (the last open cell in row 1), then F9 (the last open cell in region 9), then B7 and B8 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 D 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 D2 it would force G1 (the last open cell in row 1), then F9 (the last open cell in region 9), then E5 cleared (every open cell left in this region sits in column E, so its marker has to land there), and the chain continues, and column C would be left with no open cell for its marker.
First, if F2 were the marker, it would force G4, G5, G6, and 1 more cleared (every open cell left in this region sits in column G, so its marker has to land there), then A3, B3, C3, 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 H8 and H9 cleared (every remaining candidate in this region touches H8 and H9), and the chain continues, and region 4 would end up with no open cell for its marker, which can't happen. So it can't go there.
Next, marking H2 would force A3, B3, C3, 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 J10 and I10 cleared (together, two regions have open cells only in columns J and I), and then row 10 would be left with no open cell for its marker. So H2 can't be the marker there.
Then, if I2 were the marker, it would force J7, J8, J9, and 1 more cleared (every open cell left in this region sits in column J, so its marker has to land there), then H10 (the last open cell in region 10), 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 F9 (the last open cell in region 9), then J2 (the last open cell in row 2), 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 row 10 would be left with no open cell for its marker. So G1 can't be the marker there.
Now, 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.
From there, that one can't be the marker: at J2 it would force A8, B8, C8, and 7 more cleared (together, two regions have open cells only in rows 8 and 9), then D3, E3, D4, and 4 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), then D6 (the last open cell in column D), and the chain continues, and column E would be left with no open cell for its marker.
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, if D3 were the marker, it would force G2 (the last open cell in region 2), then F9 (the last open cell in region 9), then C8 (the last open cell in column C), 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.
Next, that one can't be the marker: at E3 it would force G2 (the last open cell in region 2), then F9 (the last open cell in region 9), then C8 (the last open cell in column C), and the chain continues, and row 5 would be left with no open cell for its marker.
Then, that one can't be the marker: at H3 it would force E2 (the last open cell in region 2), then J7, J8, J9, and 1 more cleared (every open cell left in this region sits in column J, so its marker has to land there), then I8 (the last open cell in region 8), and row 10 would be left with no open cell for its marker.
After that, marking I3 would force J7, J8, J9, and 1 more cleared (every open cell left in this region sits in column J, so its marker has to land there), then H10 (the last open cell in region 10), and then region 8 would be left with no open cell for its marker. So I3 can't be the marker there.
Now, place a marker at J3. Row 3 has exactly one open cell left. Placing here clears the rest of row 3, column J, and its region.
From there, 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.
Following that, every open cell in column C belongs to the same region, so that region's marker has to be somewhere in this column; that clears every other open cell in the region.
First, place a marker at D6. Column 4 has exactly one open cell left. Placing here clears the rest of row 6, column D, and its region. Placing here also clears its 2 touching neighbours.
Next, place a marker at C4. Region has exactly one open cell left. Placing here clears the rest of row 4, column C, and its region.
Then, place a marker at E2. Column 5 has exactly one open cell left. Placing here clears the rest of row 2, column E, and its region.
After that, together, two regions have open cells only in columns H and I; 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.
Now, place a marker at A5. Row 5 has exactly one open cell left. Placing here clears the rest of row 5, column A, and its region.
From there, place a marker at B1. Region has exactly one open cell left.
Following 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.
First, 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.
Next, that one can't be the marker: at F7 it would force G9 (the last open cell in region 9), and column H would be left with no open cell for its marker.
Then, place a marker at G7. Region has exactly one open cell left. Placing here clears the rest of row 7, column G, and its region. Placing here also clears its 1 touching neighbour.
After that, place a marker at I8. Region has exactly one open cell left. Placing here clears the rest of row 8, column I, and its region.
Now, place a marker at F9. Region has exactly one open cell left.
Finally, place a marker at H10. Region has exactly one open cell left.

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