An Expert 10×10 Puzzle Solved with Forced Chain in 47 Steps

10×10 · expert · 47 steps

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This 10×10 board rates as expert (difficulty score 2118). Solving it from scratch takes 47 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 remaining candidate in this region touches E7, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
Next, marking A1 would force B3, B4, B5, and 3 more cleared (every open cell left in this region sits in column B, so its marker has to land there), then C3, C4, C7, and 3 more cleared (every open cell left in this region sits in column C, so its marker has to land there), then D7 (the last open cell in region 8), and the chain continues, and then region 7 would be left with no open cell for its marker. So A1 can't be the marker there.
Then, if B1 were the marker, it would force A3, A4, A5, and 3 more cleared (every open cell left in this region sits in column A, so its marker has to land there), then C3, C4, C7, and 3 more cleared (every open cell left in this region sits in column C, so its marker has to land there), then D7 (the last open cell in region 8), 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.
After that, every remaining candidate in this region touches D2, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
Now, that one can't be the marker: at C1 it would force A9, B9, G9, and 3 more cleared (every open cell left in this region sits in row 9, so its marker has to land there), then D10, E10, F10, and 4 more cleared (every open cell left in this region sits in row 10, so its marker has to land there), then A7, A8, B3, and 2 more cleared (together, two regions have open cells only in columns A and B), and the chain continues, and region 7 would be left with no open cell for its marker.
From there, marking J1 would force G9, I9, D10, and 4 more cleared (every open cell in column H belongs to the same region, so that region's marker has to be somewhere in this column), then I3 (the last open cell in column I), then A2 and B2 cleared (every open cell left in this region sits in row 2, so its marker has to land there), and the chain continues, and then column D would be left with no open cell for its marker. So J1 can't be the marker there.
Following that, if C2 were the marker, it would force A9, B9, G9, and 3 more cleared (every open cell left in this region sits in row 9, so its marker has to land there), then D10, E10, F10, and 4 more cleared (every open cell left in this region sits in row 10, so its marker has to land there), then F3, G3, I3, 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 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.
First, that one can't be the marker: at F1 it would force E3 (the last open cell in region 1), then D5, D7, D9, and 1 more cleared (every open cell left in this region sits in column D, so its marker has to land there), then C4, C5, C6, and 1 more cleared (every open cell left in this region sits in column C, so its marker has to land there), and the chain continues, and region 10 would be left with no open cell for its marker.
Next, if F2 were the marker, it would force D1 (the last open cell in region 1), then E4, E5, E9, and 1 more cleared (every open cell left in this region sits in column E, so its marker has to land there), then C3, C4, C5, and 2 more cleared (every open cell left in this region sits in column C, so its marker has to land there), 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.
Then, marking H2 would force E3 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 J7, J8, G9, and 6 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 then column J would be left with no open cell for its marker. So H2 can't be the marker there.
After that, that one can't be the marker: at J2 it would force H6, H7, H8, and 7 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 H1 (the last open cell in column H), then E3 (the last open cell in region 1), and the chain continues, and column C would be left with no open cell for its marker.
Now, that one can't be the marker: at B3 it would force A5, A6, A7, and 1 more cleared (every open cell left in this region sits in column A, so its marker has to land there), then C7, C8, C9, and 1 more cleared (every open cell left in this region sits in column C, so its marker has to land there), then D7 (the last open cell in region 8), and the chain continues, and region 6 would be left with no open cell for its marker.
From there, marking C3 would force A9, B9, G9, and 3 more cleared (every open cell left in this region sits in row 9, so its marker has to land there), then D10, E10, F10, and 4 more cleared (every open cell left in this region sits in row 10, so its marker has to land there), then H1, I1 and I2 cleared (every open cell in column G belongs to the same region, so that region's marker has to be somewhere in this column), and the chain continues, and then column J would be left with no open cell for its marker. So C3 can't be the marker there.
Following that, if D3 were the marker, it would force E1 (the last open cell in region 1), then F9 and F10 cleared (every open cell left in this region sits in column F, so its marker has to land there), then C5, C6 and C7 cleared (every open cell left in this region sits in column C, so its marker has to land there), 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.
First, that one can't be the marker: at E3 it would force G2, I2 and G4 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 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 A7, A8, B7, and 1 more cleared (together, two regions have open cells only in columns A and B), and the chain continues, and row 6 would be left with no open cell for its marker.
Next, marking F3 would force A4, B4, A5, 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 D4, D5, D7, and 5 more cleared (together, two regions have open cells only in columns D and E), then C4 and C7 cleared (every open cell left in this region sits in column C, so its marker has to land there), and the chain continues, and then column J would be left with no open cell for its marker. So F3 can't be the marker there.
Then, if G3 were the marker, it would force E2 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 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 A7, A8, B7, and 1 more cleared (together, two regions have open cells only in columns A and B), and the chain continues, and column H 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 H3 it would force D1 and E1 cleared (every open cell left in this region sits in row 1, so its marker has to land there), then E2 (the last open cell in region 1), then G9, I9, D10, and 3 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 column C would be left with no open cell for its marker.
Now, marking I3 would force E2 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 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 D5, E5, F5, 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), and the chain continues, and then column J would be left with no open cell for its marker. So I3 can't be the marker there.
From there, that one can't be the marker: at A2 it would force J3 (the last open cell in row 3), then G1, H1 and I1 cleared (every open cell left in this region sits in row 1, so its marker has to land there), then G4 (the last open cell in region 2), and the chain continues, and column I would be left with no open cell for its marker.
Following that, marking B2 would force J3 (the last open cell in row 3), then G1, H1 and I1 cleared (every open cell left in this region sits in row 1, so its marker has to land there), then G4 (the last open cell in region 2), and the chain continues, and then column I would be left with no open cell for its marker. So B2 can't be the marker there.
First, marking A4 would force J3 (the last open cell in row 3), then B7 and B8 cleared (every open cell left in this region sits in column B, so its marker has to land there), then F7, G7 and H7 cleared (every open cell left in this region sits in row 7, so its marker has to land there), and the chain continues, and then row 6 would be left with no open cell for its marker. So A4 can't be the marker there.
Next, if B4 were the marker, it would force J3 (the last open cell in row 3), then A7 and A8 cleared (every open cell left in this region sits in column A, so its marker has to land there), then F7, G7 and H7 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 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 C4 it would force A9, B9, G9, and 3 more cleared (every open cell left in this region sits in row 9, so its marker has to land there), then D10, E10, F10, and 4 more cleared (every open cell left in this region sits in row 10, so its marker has to land there), then H1, I1 and I2 cleared (every open cell in column G belongs to the same region, so that region's marker has to be somewhere in this column), and the chain continues, and region 1 would be left with no open cell for its marker.
After that, marking G1 would force E2 (the last open cell in region 1), then D7, D9, D10, and 2 more cleared (together, two regions have open cells only in columns D and F), then C5, C6 and C7 cleared (every open cell left in this region sits in column C, so its marker has to land there), and the chain continues, and then region 10 would be left with no open cell for its marker. So G1 can't be the marker there.
Now, marking D4 would force E6, E8, E9, and 1 more cleared (every open cell left in this region sits in column E, so its marker has to land there), then F9 and F10 cleared (every open cell left in this region sits in column F, so its marker has to land there), then C6 and C7 cleared (every open cell left in this region sits in column C, so its marker has to land there), and the chain continues, and then region 10 would be left with no open cell for its marker. So D4 can't be the marker there.
From there, if E4 were the marker, it would force D1 (the last open cell in region 1), then F9 and F10 cleared (every open cell left in this region sits in column F, so its marker has to land there), then C5, C6 and C7 cleared (every open cell left in this region sits in column C, so its marker has to land there), 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.
Following that, that one can't be the marker: at F4 it would force D7, D9, D10, and 2 more cleared (together, two regions have open cells only in columns D and E), then C5, C6 and C7 cleared (every open cell left in this region sits in column C, so its marker has to land there), then A9, A10, B9, and 1 more cleared (together, two regions have open cells only in columns A and B), and region 10 would be left with no open cell for its marker.
First, if E1 were the marker, it would force G4 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 J3, H5, I5, 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 A3 (the last open cell in row 3), 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.
Next, if H1 were the marker, it would force E2 (the last open cell in region 1), then J3, I5, J5, 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 A3 (the last open cell in row 3), 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.
Then, that one can't be the marker: at I1 it would force E2 (the last open cell in region 1), then J3, H5, J5, 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), then A3 (the last open cell in row 3), and the chain continues, and row 6 would be left with no open cell for its marker.
After that, place a marker at D1. Row 1 has exactly one open cell left. Placing here clears the rest of row 1, column D, and its 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 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, 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.
First, every open cell left in this region sits in column B, so its marker has to land there; that clears every other open cell in the column.
Next, place a marker at C7. Region has exactly one open cell left. Placing here clears the rest of row 7, column C, and its region.
Then, 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.
After that, place a marker at B10. Region has exactly one open cell left. Placing here clears the rest of row 10, column B, and its region.
Now, 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.
From there, together, two regions have open cells only in columns E and F; 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.
Following that, place a marker at G5. Region has exactly one open cell left. Placing here clears the rest of row 5, column G, and its region. Placing here also clears its 3 touching neighbours.
First, place a marker at I2. Region has exactly one open cell left. Placing here clears the rest of row 2, column I, and its region.
Next, place a marker at J4. Region has exactly one open cell left. Placing here clears the rest of row 4, column J, and its region.
Then, place a marker at H8. Region has exactly one open cell left. Placing here clears the rest of row 8, column H, and its region.
After that, place a marker at E6. Region has exactly one open cell left. Placing here clears the rest of row 6, column E, and its region.
Finally, place a marker at F9. Region has exactly one open cell left.

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