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 2033). Solving it from scratch takes 46 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 1, so its marker has to land there; that clears every other open cell in the row.
Next, every remaining candidate in this region touches I2, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
Then, that one can't be the marker: at A2 it would force H3 (the last open cell in region 2), then E4 (the last open cell in region 4), then D6, G6, D7, and 3 more cleared (every open cell in column F 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.
After that, marking B2 would force H3 (the last open cell in region 2), then E4 (the last open cell in region 4), then D6, G6, D7, and 3 more cleared (every open cell in column F belongs to the same region, so that region's marker has to be somewhere in this column), and the chain continues, and then column C would be left with no open cell for its marker. So B2 can't be the marker there.
Now, if C2 were the marker, it would force H3 (the last open cell in region 2), then E4 (the last open cell in region 4), then D6, G6, D7, and 3 more cleared (every open cell in column F belongs to the same region, so that region's marker has to be somewhere in this column), and the chain continues, and column B 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 D2 it would force H3 (the last open cell in region 2), 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 I5, I7, I8, and 5 more cleared (together, two regions have open cells only in columns I and J), and the chain continues, and row 8 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, marking E2 would immediately leave region 4 with no open cell for its marker, so it can't be the marker there.
Next, that one can't be the marker: at G2 it would force I5, I7, I8, and 5 more cleared (together, two regions have open cells only in columns I and J), then H4, H5, H6, and 2 more cleared (every open cell left in this region sits in column H, so its marker has to land there), then F6, F7, F8, and 2 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 A would be left with no open cell for its marker.
Then, marking H2 would force J1 (the last open cell in region 3), then I5, I7, I8, and 2 more cleared (every open cell left in this region sits in column I, so its marker has to land there), then G10 (the last open cell in region 10), and the chain continues, and then row 8 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 H1 (the last open cell in region 3), then I5, I7, I8, and 2 more cleared (every open cell left in this region sits in column I, so its marker has to land there), then G10 (the last open cell in region 10), and the chain continues, and row 8 would be left with no open cell for its marker.
Now, place a marker at F2. Region has exactly one open cell left. Placing here clears the rest of row 2, column F, and its region. Placing here also clears its 2 touching neighbours.
From there, every remaining candidate in this region touches D4, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
Following that, 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.
First, that one can't be the marker: at B3 it would force C5 (the last open cell in region 1), then E4 (the last open cell in region 4), then D8, D9 and D10 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 region 7 would be left with no open cell for its marker.
Next, marking I3 would force E4 (the last open cell in region 4), then C5 (the last open cell in region 1), then H1, H8, H9, and 1 more cleared (every open cell left in this region sits in column H, so its marker has to land there), and the chain continues, and then row 8 would be left with no open cell for its marker. So I3 can't be the marker there.
Then, if J3 were the marker, it would force E4 (the last open cell in region 4), then C5 (the last open cell in region 1), then H8, H9, H10, and 3 more cleared (together, two regions have open cells only in columns H and I), and the chain continues, and row 8 would end up with no open cell for its marker, which can't happen. So it can't go there.
After that, marking A4 would force D3 (the last open cell in region 4), then C5 (the last open cell in region 1), then E8, E9 and E10 cleared (every open cell in column B belongs to the same region, so that region's marker has to be somewhere in this column), and the chain continues, and then row 9 would be left with no open cell for its marker. So A4 can't be the marker there.
Now, that one can't be the marker: place it at C4, and row 3 would be left with no open cell for its marker.
From there, if H4 were the marker, it would force D3 (the last open cell in region 4), then C5 (the last open cell in region 1), then I8, I9, I10, and 3 more cleared (together, two regions have open cells only in columns I and J), and the chain continues, and row 8 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 I4 it would force D3 (the last open cell in region 4), then C5 (the last open cell in region 1), then H1, H8, H9, and 1 more cleared (every open cell left in this region sits in column H, so its marker has to land there), and the chain continues, and row 8 would be left with no open cell for its marker.
First, marking J4 would force D3 (the last open cell in region 4), then C5 (the last open cell in region 1), then H8, H9, H10, and 3 more cleared (together, two regions have open cells only in columns H and I), and the chain continues, and then row 8 would be left with no open cell for its marker. So J4 can't be the marker there.
Next, that one can't be the marker: at A5 it would force C3 (the last open cell in region 1), then E4 (the last open cell in region 4), then D8, D9 and D10 cleared (every open cell in column B belongs to the same region, so that region's marker has to be somewhere in this column), and the chain continues, and row 9 would be left with no open cell for its marker.
Then, marking B5 would force D3 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 4), then D6, G6, D7, and 1 more cleared (together, two regions have open cells only in rows 6 and 7), and the chain continues, and then column C would be left with no open cell for its marker. So B5 can't be the marker there.
After that, that one can't be the marker: place it at D5, and region 4 would be left with no open cell for its marker.
Now, marking E5 would force D3 (the last open cell in region 4), and then region 1 would be left with no open cell for its marker. So E5 can't be the marker there.
From there, marking J1 would force I6 (the last open cell in region 7), then G7, 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 H9 cleared (every open cell left in this region sits in column H, so its marker has to land there), and the chain continues, and then row 9 would be left with no open cell for its marker. So J1 can't be the marker there.
Following that, marking H5 would force I1 (the last open cell in region 3), then E4 (the last open cell in row 4), then J8, J9 and J10 cleared (every open cell left in this region sits in column J, so its marker has to land there), and the chain continues, and then row 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 H1 (the last open cell in region 3), then J7 (the last open cell in region 7), then G10 (the last open cell in region 10), and the chain continues, and row 8 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 D3 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 4), then H8, H9, H10, and 3 more cleared (together, two regions have open cells only in columns H and I), and the chain continues, and row 8 would be left with no open cell for its marker.
Then, every remaining candidate in this region touches I7, 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 I1 it would force J8, J9 and J10 cleared (every open cell left in this region sits in column J, so its marker has to land there), then G6, G7, G8, and 2 more cleared (together, two regions have open cells only in columns G and H), then D8, D9, D10, and 3 more cleared (together, two regions have open cells only in columns D and E), and the chain continues, and row 9 would be left with no open cell for its marker.
Now, place a marker at H1. Region has exactly one open cell left. Placing here clears the rest of row 1, column H, and its region.
From there, every open cell left in this region sits in column G, so its marker has to land there; that clears every other open cell in the column.
Following that, together, two regions have open cells only in columns D and E; 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.
First, together, two regions have open cells only in rows 6 and 7; 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.
Next, every remaining candidate in this region touches B8, 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 B9, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
After that, place a marker at B10. Column 2 has exactly one open cell left. Placing here clears the rest of row 10, column B, and its region. Placing here also clears its 1 touching neighbour.
Now, place a marker at C8. Region has exactly one open cell left. Placing here clears the rest of row 8, column C, and its region. Placing here also clears its 1 touching neighbour.
From there, place a marker at A3. Region has exactly one open cell left. Placing here clears the rest of row 3, column A, and its region.
Following that, place a marker at E4. Region has exactly one open cell left. Placing here clears the rest of row 4, column E, and its region.
First, place a marker at G5. Region has exactly one open cell left.
Next, 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.
Then, 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.
Finally, place a marker at I9. Region has exactly one open cell left.

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