A Severe 10×10 Puzzle Solved with Forced Chain in 50 Steps

10×10 · severe · 50 steps

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This 10×10 board rates as severe (difficulty score 2295). Solving it from scratch takes 50 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 B4, 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 B9, 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 D5, E5, F5, and 4 more cleared (every open cell left in this region sits in row 5, so its marker has to land there), then B6, B7, B8, and 3 more cleared (together, two regions have open cells only in columns B and C), then F6, G6, H6, 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 region 8 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 A6, A7, C6, and 2 more cleared (together, two regions have open cells only in columns A and C), then F6, G6, H6, and 2 more cleared (every open cell left in this region sits in row 6, so its marker has to land there), then D2, D3, D4, and 8 more cleared (together, two regions have open cells only in columns D and E), 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 A6, A7, B6, and 2 more cleared (together, two regions have open cells only in columns A and B), then F6, G6, H6, and 2 more cleared (every open cell left in this region sits in row 6, so its marker has to land there), then D3, D4, D5, and 7 more cleared (together, two regions have open cells only in columns D and E), and the chain continues, and column E would be left with no open cell for its marker.
From there, that one can't be the marker: at A2 it would force D5, E5, F5, and 4 more cleared (every open cell left in this region sits in row 5, so its marker has to land there), then J3 and J4 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 B6, B7, B8, and 3 more cleared (together, two regions have open cells only in columns B and C), and the chain continues, and region 8 would be left with no open cell for its marker.
Following that, marking B2 would force J3, J4, I5, 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 A6, A7, C6, and 2 more cleared (together, two regions have open cells only in columns A and C), then F6, G6, H6, 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 B2 can't be the marker there.
First, if H1 were the marker, it would force C2 (the last open cell in row 2), then A6, A7, B6, and 2 more cleared (together, two regions have open cells only in columns A and B), then F6, G6, I6, and 1 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.
Next, that one can't be the marker: at I1 it would force C2 (the last open cell in row 2), then A6, A7, B6, and 2 more cleared (together, two regions have open cells only in columns A and B), then F6, G6, H6, and 1 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 be left with no open cell for its marker.
Then, if C2 were the marker, it would force J3, J4, I5, 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 A6, A7, B6, and 2 more cleared (together, two regions have open cells only in columns A and B), then F6, G6, H6, 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.
After that, that one can't be the marker: place it at D2, and row 1 would be left with no open cell for its marker.
Now, marking E2 would immediately leave row 1 with no open cell for its marker, so it can't be the marker there.
From there, that one can't be the marker: at G2 it would force F4 (the last open cell in region 3), then D1 (the last open cell in row 1), and region 4 would be left with no open cell for its marker.
Following that, if A3 were the marker, it would force D5, E5, F5, and 4 more cleared (every open cell left in this region sits in row 5, so its marker has to land there), then F2, J2 and J4 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 F4 and 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), 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.
First, that one can't be the marker: at B3 it would force D5, E5, F5, and 4 more cleared (every open cell left in this region sits in row 5, so its marker has to land there), then F2, J2 and J4 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 F4 and 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), and the chain continues, and region 8 would be left with no open cell for its marker.
Next, if E1 were the marker, it would force C5, C6, C7, and 3 more cleared (every open cell left in this region sits in column C, so its marker has to land there), then F3, G3, H3, and 3 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 A6, A7, B6, and 2 more cleared (together, two regions have open cells only in columns A and B), 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, that one can't be the marker: at F1 it would force C5, C6, C7, and 3 more cleared (every open cell left in this region sits in column C, so its marker has to land there), then G3, H3, G4, and 1 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 A6, A7, B6, and 2 more cleared (together, two regions have open cells only in columns A and B), and the chain continues, and region 8 would be left with no open cell for its marker.
After that, marking G1 would force I2 (the last open cell in row 2), then C5, C6, C7, and 3 more cleared (every open cell left in this region sits in column C, so its marker has to land there), then A6, A7, B6, and 2 more cleared (together, two regions have open cells only in columns A and B), and the chain continues, and then column E would be left with no open cell for its marker. So G1 can't be the marker there.
Now, marking J1 would force H2 (the last open cell in row 2), then C5, C6, C7, and 3 more cleared (every open cell left in this region sits in column C, so its marker has to land there), then A6, A7, B6, and 2 more cleared (together, two regions have open cells only in columns A and B), and the chain continues, and then column E would be left with no open cell for its marker. So J1 can't be the marker there.
From there, 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.
Following that, marking I3 would force F2 (the last open cell in row 2), then G6, G8, G9, and 1 more cleared (every open cell left in this region sits in column G, so its marker has to land there), then E6, E7, E8, and 2 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 region 6 would be left with no open cell for its marker. So I3 can't be the marker there.
First, that one can't be the marker: at F4 it would force G6 (the last open cell in region 4), then J2 (the last open cell in row 2), and row 3 would be left with no open cell for its marker.
Next, if H4 were the marker, it would force E5, F5 and J5 cleared (every open cell left in this region sits in row 5, so its marker has to land there), then E3 cleared (together, two regions have open cells only in rows 2 and 3), then A6, B6, C6, and 1 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 row 3 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 I4 it would force H2 cleared (every open cell left in this region sits in row 2, so its marker has to land there), then E5, F5 and G5 cleared (every open cell left in this region sits in row 5, so its marker has to land there), then E3 cleared (every open cell left in this region sits in row 3, so its marker has to land there), and the chain continues, and column H would be left with no open cell for its marker.
After that, marking E5 would force A4 (the last open cell in region 6), then B10, C9 and C10 cleared (together, two regions have open cells only in columns B and C), and then region 9 would be left with no open cell for its marker. So E5 can't be the marker there.
Now, marking F3 would force G6 (the last open cell in region 4), then J2 (the last open cell in row 2), then A4 (the last open cell in row 4), and then row 5 would be left with no open cell for its marker. So F3 can't be the marker there.
From there, that one can't be the marker: at G5 it would force A4 (the last open cell in region 6), then E3 (the last open cell in region 4), then J2 (the last open cell in region 2), and the chain continues, and region 9 would be left with no open cell for its marker.
Following that, marking H5 would force A4 (the last open cell in region 6), then E3 cleared (together, two regions have open cells only in rows 2 and 3), then F6 (the last open cell in region 4), and the chain continues, and then region 8 would be left with no open cell for its marker. So H5 can't be the marker there.
First, if I5 were the marker, it would force A4 (the last open cell in region 6), then H2 (the last open cell in row 2), then E3 (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.
Next, that one can't be the marker: at J5 it would force A4 (the last open cell in region 6), then G3 and H3 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 E3 (the last open cell in row 3), and the chain continues, and region 9 would be left with no open cell for its marker.
Then, if A6 were the marker, it would force C5 (the last open cell in region 6), then B10 (the last open cell in region 9), then E7 (the last open cell in region 8), and region 4 would end up with no open cell for its marker, which can't happen. So it can't go there.
After that, marking C6 would force A8, A9 and A10 cleared (every open cell left in this region sits in column A, so its marker has to land there), then B10 (the last open cell in region 9), then E7 (the last open cell in region 8), and the chain continues, and then region 2 would be left with no open cell for its marker. So C6 can't be the marker there.
Now, if F2 were the marker, it would force H3 (the last open cell in row 3), then A4 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 E4 (the last open cell in row 4), and the chain continues, and column J would end up with no open cell for its marker, which can't happen. So it can't go there.
From there, every open cell left in this region sits in column J, so its marker has to land there; that clears every other open cell in the column.
Following that, if I2 were the marker, it would force J4 (the last open cell in region 2), then E3 (the last open cell in row 3), then B6 cleared (every remaining candidate in this region touches B6), and the chain continues, and column G 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 A4 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 B6 cleared (every remaining candidate in this region touches B6), then C7, F7, G7, and 2 more cleared (every open cell left in this region sits in row 7, so its marker has to land there), and the chain continues, and column I would be left with no open cell for its marker.
Next, if E4 were the marker, it would force B6 cleared (every remaining candidate in this region touches B6), then C7, F7, G7, and 2 more cleared (every open cell left in this region sits in row 7, so its marker has to land there), then A8, F8, G8, and 2 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 column I 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 J3 (the last open cell in row 3), then A4 (the last open cell in row 4), then F5 (the last open cell in row 5), and the chain continues, and then region 5 would be left with no open cell for its marker. So H2 can't be the marker there.
After that, place a marker at J2. Row 2 has exactly one open cell left. Placing here clears the rest of row 2, column J, and its region.
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, place a marker at A4. Row 4 has exactly one open cell left. Placing here clears the rest of row 4, column A, and its region.
Following that, place a marker at F5. Row 5 has exactly one open cell left. Placing here clears the rest of row 5, column F, and its region. Placing here also clears its 1 touching neighbour.
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, every open cell left in this region sits in column C, so its marker has to land there; that clears every other open cell in the column.
Then, place a marker at E7. Region has exactly one open cell left. Placing here clears the rest of row 7, column E, and its region.
After that, place a marker at B6. Region has exactly one open cell left. Placing here clears the rest of row 6, column B, and its region.
Now, 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. Placing here also clears its 2 touching neighbours.
From there, place a marker at G3. Region has exactly one open cell left. Placing here clears the rest of row 3, column G, and its region.
Following that, place a marker at I10. Region has exactly one open cell left. Placing here clears the rest of row 10, column I, and its region.
Finally, place a marker at C9. Region has exactly one open cell left.

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