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 2128). Solving it from scratch takes 46 logical steps, using 7 techniques: Single cell, Row/column exclusion, Adjacency clear, Region 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 column A, so its marker has to land there; that clears every other open cell in the column.
Next, marking D1 would force B5, B6, B7, and 3 more cleared (every open cell left in this region sits in column B, so its marker has to land there), then C3, 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 E8 (the last open cell in region 10), and the chain continues, and then row 10 would be left with no open cell for its marker. So D1 can't be the marker there.
Then, if E1 were the marker, it would force B5, B6, B7, and 3 more cleared (every open cell left in this region sits in column B, so its marker has to land there), then C8, C9, C10, and 3 more cleared (together, two regions have open cells only in columns C and D), and region 10 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 A2 it would force C1 and J1 cleared (every open cell left in this region sits in row 1, so its marker has to land there), then B4 (the last open cell in region 1), then D3, E3, F3, and 1 more 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 C would be left with no open cell for its marker.
Now, if C2 were the marker, it would force B4 (the last open cell in region 1), then D7 (the last open cell in region 7), and region 10 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 B5, B6, B7, and 3 more cleared (every open cell left in this region sits in column B, so its marker has to land there), then B1 and J1 cleared (every open cell left in this region sits in row 1, so its marker has to land there), then C8, C9 and C10 cleared (every open cell left in this region sits in column C, so its marker has to land there), and the chain continues, and row 10 would be left with no open cell for its marker.
Following that, marking E2 would force B1, C1 and J1 cleared (every open cell left in this region sits in row 1, so its marker has to land there), then B5, B6, B7, and 3 more cleared (every open cell left in this region sits in column B, so its marker has to land there), then A3, G3, A4, and 5 more cleared (together, two regions have open cells only in rows 3 and 4), and the chain continues, and then region 7 would be left with no open cell for its marker. So E2 can't be the marker there.
First, if A3 were the marker, it would force F1, G1, H1, and 2 more cleared (every open cell left in this region sits in row 1, so its marker has to land there), then G2 (the last open cell in region 2), then J4 (the last open cell in region 3), 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.
Next, marking C3 would force B1 (the last open cell in region 1), then G2 (the last open cell in region 2), then J4 (the last open cell in region 3), and the chain continues, and then region 10 would be left with no open cell for its marker. So C3 can't be the marker there.
Then, if D3 were the marker, it would force B9, B10, C8, and 2 more cleared (together, two regions have open cells only in columns B and C), then E8 (the last open cell in region 10), then G7, I7, J7, and 5 more cleared (every open cell in row 9 belongs to the same region, so that region's marker has to be somewhere in this row), and row 10 would end up with no open cell for its marker, which can't happen. So it can't go there.
After that, marking F3 would force B1, C1 and J1 cleared (every open cell left in this region sits in row 1, so its marker has to land there), then B5, B6, B7, and 3 more cleared (every open cell left in this region sits in column B, so its marker has to land there), then D7, D8, D9, and 1 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 then row 10 would be left with no open cell for its marker. So F3 can't be the marker there.
Now, if G3 were the marker, it would force B1, C1 and J1 cleared (every open cell left in this region sits in row 1, so its marker has to land there), then B5, B6, B7, and 3 more cleared (every open cell left in this region sits in column B, so its marker has to land there), then D7, D8, D9, and 1 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 row 10 would end up with no open cell for its marker, which can't happen. So it can't go there.
From there, marking A4 would force J1, F2, H2, and 2 more cleared (together, two regions have open cells only in rows 1 and 2), then E3 cleared (every open cell left in this region sits in row 3, so its marker has to land there), then D7, D8, D9, and 1 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 then region 7 would be left with no open cell for its marker. So A4 can't be the marker there.
Following that, every remaining candidate in this region touches B6, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
First, if B4 were the marker, it would force D6, E6, F6, and 10 more cleared (together, two regions have open cells only in rows 6 and 7), then H5 (the last open cell in region 8), and region 6 would end up with no open cell for its marker, which can't happen. So it can't go there.
Next, marking J1 would force G2 (the last open cell in region 2), then B3 (the last open cell in region 1), then D7, D8, D9, and 1 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 then row 10 would be left with no open cell for its marker. So J1 can't be the marker there.
Then, marking H2 would force F1 (the last open cell in region 2), then B3 (the last open cell in region 1), then D7, D8, D9, and 1 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 then row 10 would be left with no open cell for its marker. So H2 can't be the marker there.
After that, if I2 were the marker, it would force B1 and C1 cleared (every open cell left in this region sits in row 1, so its marker has to land there), then B3 (the last open cell in region 1), then D7, D8, D9, and 1 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 row 10 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 J2 it would force B1 and C1 cleared (every open cell left in this region sits in row 1, so its marker has to land there), then B3 (the last open cell in region 1), then D7, D8, D9, and 1 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 row 10 would be left with no open cell for its marker.
From there, every remaining candidate in this region touches I4, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
Following that, that one can't be the marker: at H3 it would force B1 and C1 cleared (every open cell left in this region sits in row 1, so its marker has to land there), then B2 (the last open cell in region 1), then D7, D8, D9, and 1 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 row 10 would be left with no open cell for its marker.
First, marking I3 would force F2 cleared (together, two regions have open cells only in rows 1 and 2), then D7, D8, D9, and 1 more cleared (every open cell left in this region sits in column D, so its marker has to land there), then B9, B10, C8, and 2 more cleared (together, two regions have open cells only in columns B and C), and the chain continues, and then row 10 would be left with no open cell for its marker. So I3 can't be the marker there.
Next, 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.
Then, if J3 were the marker, it would force F2 cleared (together, two regions have open cells only in rows 1 and 2), then D7, D8, D9, and 1 more cleared (every open cell left in this region sits in column D, so its marker has to land there), then B9, B10, C8, and 2 more cleared (together, two regions have open cells only in columns B and C), and the chain continues, and row 10 would end up with no open cell for its marker, which can't happen. So it can't go there.
After that, 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. Placing here also clears its 1 touching neighbour.
Now, every remaining candidate in this region touches F6, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
From there, if B1 were the marker, it would force G2 (the last open cell in region 2), then E3 (the last open cell in row 3), then F5 (the last open cell in region 6), 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.
Following that, that one can't be the marker: at F1 it would force B2 (the last open cell in row 2), then E3 (the last open cell in row 3), then G5 (the last open cell in region 6), and the chain continues, and region 7 would be left with no open cell for its marker.
First, marking G1 would force B2 (the last open cell in row 2), then E3 (the last open cell in row 3), then F5 (the last open cell in region 6), and the chain continues, and then region 7 would be left with no open cell for its marker. So G1 can't be the marker there.
Next, marking B2 would force E3 (the last open cell in row 3), then A5, C5 and H5 cleared (every open cell left in this region sits in row 5, so its marker has to land there), then C6, C7, D7, and 2 more cleared (together, two regions have open cells only in rows 6 and 7), and then region 7 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 A5 it would force E6 (the last open cell in region 6), then F2 (the last open cell in region 5), then H7 (the last open cell in region 8), and the chain continues, and region 7 would be left with no open cell for its marker.
After that, every remaining candidate in this region touches B7, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
Now, if H1 were the marker, it would force B3 (the last open cell in region 1), then F2 (the last open cell in row 2), then C6, E6, C7, and 3 more cleared (together, two regions have open cells only in rows 6 and 7), 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, marking B5 would force C1 (the last open cell in region 1), then G2 (the last open cell in region 2), then E6 (the last open cell in region 6), and then row 3 would be left with no open cell for its marker. So B5 can't be the marker there.
Following that, if C5 were the marker, it would force B3 (the last open cell in region 1), then F2 (the last open cell in region 5), then I1 (the last open cell in region 2), 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 I1 it would force B3 (the last open cell in region 1), then F2 (the last open cell in row 2), then E6, G6, H6, and 3 more cleared (together, two regions have open cells only in rows 6 and 7), and the chain continues, and region 6 would be left with no open cell for its marker.
Next, place a marker at G2. Region has exactly one open cell left. Placing here clears the rest of row 2, column G, and its region.
Then, place a marker at C1. Row 1 has exactly one open cell left. Placing here clears the rest of row 1, column C, and its region.
After that, place a marker at E3. Row 3 has exactly one open cell left. Placing here clears the rest of row 3, column E, and its region.
Now, place a marker at F5. Region has exactly one open cell left. Placing here clears the rest of row 5, column F, and its region.
From there, 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.
Following that, 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. Placing here also clears its 1 touching neighbour.
First, place a marker at A6. Region has exactly one open cell left. Placing here clears the rest of row 6, column A, and its region.
Next, place a marker at H7. Region has exactly one open cell left. Placing here clears the rest of row 7, column H, and its region.
Then, place a marker at D10. Region has exactly one open cell left. Placing here clears the rest of row 10, column D, 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. ∎