A Severe 10×10 Puzzle Solved with Forced Chain in 49 Steps

10×10 · severe · 49 steps

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This 10×10 board rates as severe (difficulty score 2230). Solving it from scratch takes 49 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 H2, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
Next, if E1 were the marker, it would force G3, H3, I3, and 1 more cleared (every open cell left in this region sits in row 3, so its marker has to land there), then G2 (the last open cell in region 3), then J4 (the last open cell in region 4), 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.
Then, if H1 were the marker, it would force G3 (the last open cell in region 3), then I7, I8, I9, and 4 more cleared (together, two regions have open cells only in columns I and J), and region 9 would end up with no open cell for its marker, which can't happen. So it can't go there.
After that, if F2 were the marker, it would force H3 (the last open cell in region 3), then A5, B5, C5, and 2 more cleared (every open cell left in this region sits in row 5, so its marker has to land there), then D1, D7, D8, and 2 more cleared (every open cell left in this region sits in column D, so its marker has to land there), 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, if I2 were the marker, it would force G4, G5, G6, and 4 more cleared (every open cell left in this region sits in column G, so its marker has to land there), then H4, H5 and J5 cleared (together, two regions have open cells only in columns H and J), then F4 (the last open cell in region 5), and the chain continues, and region 6 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 F1 it would force J5, J6, J7, and 3 more cleared (every open cell left in this region sits in column J, so its marker has to land there), then A3, B3, C3, and 2 more cleared (every open cell left in this region sits in row 3, so its marker has to land there), then A6, B6, C6, and 2 more cleared (every open cell left in this region sits in row 6, so its marker has to land there), and the chain continues, and region 1 would be left with no open cell for its marker.
Following that, marking G1 would force J5, J6, J7, and 3 more cleared (every open cell left in this region sits in column J, so its marker has to land there), then A6, B6, C6, and 3 more cleared (every open cell left in this region sits in row 6, so its marker has to land there), then H4, H5, I3, and 2 more cleared (together, two regions have open cells only in columns H and I), and the chain continues, and then region 1 would be left with no open cell for its marker. So G1 can't be the marker there.
First, that one can't be the marker: at A2 it would force D3, E3, F3, and 2 more cleared (every open cell left in this region sits in row 3, so its marker has to land there), then E4 (the last open cell in region 2), then B5 and C5 cleared (every open cell left in this region sits in row 5, so its marker has to land there), and the chain continues, and column F would be left with no open cell for its marker.
Next, marking B2 would force D3, E3, F3, and 2 more cleared (every open cell left in this region sits in row 3, so its marker has to land there), then E4 (the last open cell in region 2), then C5 cleared (every open cell left in this region sits in row 5, so its marker has to land there), and the chain continues, and then column F would be left with no open cell for its marker. So B2 can't be the marker there.
Then, that one can't be the marker: at J2 it would force A3, B3, C3, and 3 more cleared (every open cell left in this region sits in row 3, so its marker has to land there), then E4 (the last open cell in region 2), then A5, B5 and C5 cleared (every open cell left in this region sits in row 5, so its marker has to land there), and the chain continues, and region 9 would be left with no open cell for its marker.
After that, marking D1 would force G2 (the last open cell in row 2), then J5, J6, J7, and 3 more cleared (every open cell left in this region sits in column J, so its marker has to land there), then E5, E6, E7, 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 then column B would be left with no open cell for its marker. So D1 can't be the marker there.
Now, if A3 were the marker, it would force G2 (the last open cell in region 3), then E4 (the last open cell in region 2), then B5 and C5 cleared (every open cell left in this region sits in row 5, so its marker has to land there), and the chain continues, and column F 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 B3 it would force G2 (the last open cell in region 3), then E4 (the last open cell in region 2), then C5 cleared (every open cell left in this region sits in row 5, so its marker has to land there), and the chain continues, and column F would be left with no open cell for its marker.
Following that, marking F3 would immediately leave region 3 with no open cell for its marker, so it can't be the marker there.
First, marking I3 would force J1 (the last open cell in region 4), then G2 (the last open cell in region 3), then E4 (the last open cell in region 2), and the chain continues, and then region 9 would be left with no open cell for its marker. So I3 can't be the marker there.
Next, if J3 were the marker, it would force G2 (the last open cell in region 3), then E4 (the last open cell in region 2), then A5, B5 and C5 cleared (every open cell left in this region sits in row 5, so its marker has to land there), 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, marking A1 would force J4 (the last open cell in region 4), then B5, C5, D5, and 2 more cleared (every open cell left in this region sits in row 5, so its marker has to land there), then B6, C6, D6, and 2 more cleared (every open cell left in this region sits in row 6, so its marker has to land there), and the chain continues, and then column E would be left with no open cell for its marker. So A1 can't be the marker there.
After that, if B1 were the marker, it would force J4 (the last open cell in region 4), then C5, D5, E5, and 1 more cleared (every open cell left in this region sits in row 5, so its marker has to land there), then A6, C6, D6, and 2 more cleared (every open cell left in this region sits in row 6, so its marker has to land there), and the chain continues, and column E 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 C1 it would force J4 (the last open cell in region 4), then B5, D5, E5, and 1 more cleared (every open cell left in this region sits in row 5, so its marker has to land there), then A6, B6, D6, and 2 more cleared (every open cell left in this region sits in row 6, so its marker has to land there), and the chain continues, and region 9 would be left with no open cell for its marker.
From there, 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.
Following that, that one can't be the marker: at D2 it would force B5, C5, E5, and 1 more cleared (together, two regions have open cells only in rows 4 and 5), then F4, F8, F9, and 1 more cleared (every open cell left in this region sits in column F, so its marker has to land there), then A6, B6, C6, and 8 more cleared (together, two regions have open cells only in rows 6 and 7), and the chain continues, and column B would be left with no open cell for its marker.
First, if D3 were the marker, it would force G2 (the last open cell in region 3), then B5, C5, E5, and 1 more cleared (together, two regions have open cells only in rows 4 and 5), then F4, F8, F9, and 1 more cleared (every open cell left in this region sits in column F, so its marker has to land there), and the chain continues, and column E would end up with no open cell for its marker, which can't happen. So it can't go there.
Next, every open cell left in this region sits in column E, so its marker has to land there; that clears every other open cell in the column.
Then, marking A4 would force C5, D5 and F5 cleared (every open cell left in this region sits in row 5, so its marker has to land there), then B6, C6, B7, and 5 more cleared (together, two regions have open cells only in rows 6 and 7), then F8, D9, F9, and 3 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 column F would be left with no open cell for its marker. So A4 can't be the marker there.
After that, if B4 were the marker, it would force D5 and F5 cleared (every open cell left in this region sits in row 5, so its marker has to land there), then A6, C6, A7, and 5 more cleared (together, two regions have open cells only in rows 6 and 7), then F8, D9, F9, and 3 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 column F would end up with no open cell for its marker, which can't happen. So it can't go there.
Now, marking D4 would force E2 (the last open cell in region 2), then A5 (the last open cell in region 1), and then region 5 would be left with no open cell for its marker. So D4 can't be the marker there.
From there, marking G4 would immediately leave region 3 with no open cell for its marker, so it can't be the marker there.
Following that, if H4 were the marker, it would force G2 (the last open cell in region 3), then E3 (the last open cell in region 2), then A5 (the last open cell in region 1), 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 I4 it would force J1 (the last open cell in region 4), then G6, G7, G8, and 2 more cleared (every open cell left in this region sits in column G, so its marker has to land there), then H6 (the last open cell in region 8), and region 9 would be left with no open cell for its marker.
Next, marking B5 would force F4 (the last open cell in region 5), then E2 (the last open cell in region 2), then H3 (the last open cell in region 3), and then region 1 would be left with no open cell for its marker. So B5 can't be the marker there.
Then, if C5 were the marker, region 1 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 D5 it would force F4 (the last open cell in region 5), then E2 (the last open cell in region 2), then H3 (the last open cell in region 3), and region 1 would be left with no open cell for its marker.
Now, marking J1 would force A6, B6, C6, and 2 more cleared (every open cell left in this region sits in row 6, so its marker has to land there), then F4, F8, F9, and 1 more cleared (every open cell left in this region sits in column F, so its marker has to land there), then A5 and F5 cleared (every open cell left in this region sits in row 5, so its marker has to land there), and the chain continues, and then region 9 would be left with no open cell for its marker. So J1 can't be the marker there.
From there, place a marker at I1. Region has exactly one open cell left. Placing here clears the rest of row 1, column I, and its region.
Following that, that one can't be the marker: at G2 it would force H5 and J5 cleared (together, two regions have open cells only in columns H and J), then F4 (the last open cell in region 5), and column E would be left with no open cell for its marker.
First, every open cell left in this region sits in row 3, so its marker has to land there; that clears every other open cell in the row.
Next, if G3 were the marker, it would force A5 and F5 cleared (every open cell left in this region sits in row 5, so its marker has to land there), then C6, C7, C8, and 2 more cleared (every open cell left in this region sits in column C, so its marker has to land there), then H7, H8, H9, and 4 more cleared (together, two regions have open cells only in columns H and J), and region 9 would end up with no open cell for its marker, which can't happen. So it can't go there.
Then, place a marker at H3. Region has exactly one open cell left. Placing here clears the rest of row 3, column H, and its region.
After that, together, two regions have open cells only in columns G and J; 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 F4. Region has exactly one open cell left. Placing here clears the rest of row 4, column F, and its region.
From there, place a marker at E2. Region has exactly one open cell left. Placing here clears the rest of row 2, column E, and its region.
Following that, place a marker at A5. Region has exactly one open cell left. Placing here clears the rest of row 5, column A, and its region. Placing here also clears its 1 touching neighbour.
First, place a marker at D6. Region has exactly one open cell left. Placing here clears the rest of row 6, column D, and its region. Placing here also clears its 1 touching neighbour.
Next, place a marker at J7. Region has exactly one open cell left. Placing here clears the rest of row 7, column J, and its region.
Then, 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.
After that, every remaining candidate in this region touches C8 and C9, so whichever candidate ends up marked will clear them, and they can be ruled out now regardless of which one wins.
Now, 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. Placing here also clears its 1 touching neighbour.
From there, place a marker at B8. Region has exactly one open cell left. Placing here clears the rest of row 8, column B, and its region.
Finally, place a marker at G9. Region has exactly one open cell left.

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