An Expert 11×11 Puzzle Solved with Forced Chain in 50 Steps

11×11 · expert · 50 steps

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This 11×11 board rates as expert (difficulty score 2088). 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 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 J2, 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 E10, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
After that, every remaining candidate in this region touches B10, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
Now, if K1 were the marker, it would force A11, B11, C11, and 2 more cleared (every open cell left in this region sits in row 11, so its marker has to land there), then D2, D3, D4, and 4 more cleared (every open cell left in this region sits in column D, so its marker has to land there), then C2, C3, C4, and 4 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 row 9 would end up with no open cell for its marker, which can't happen. So it can't go there.
From there, 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.
Following that, if C2 were the marker, it would force B11 (the last open cell in region 9), then A9 (the last open cell in region 8), then D10 (the last open cell in region 10), and region 11 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 D2 it would force E11 (the last open cell in region 10), then K10 (the last open cell in region 11), then C9 (the last open cell in region 9), and region 8 would be left with no open cell for its marker.
Next, marking K2 would force I1 (the last open cell in region 3), then A11, B11, C11, and 2 more cleared (every open cell left in this region sits in row 11, so its marker has to land there), then D3, D4, D5, and 3 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 9 would be left with no open cell for its marker. So K2 can't be the marker there.
Then, if A3 were the marker, it would force B5, B6, B7, and 4 more cleared (together, two regions have open cells only in columns B and C), then E5 cleared (every remaining candidate in this region touches E5), then C9 cleared (every remaining candidate in this region touches C9), 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, marking C3 would force B11 (the last open cell in region 9), then A9 (the last open cell in region 8), then D10 (the last open cell in region 10), and then region 11 would be left with no open cell for its marker. So C3 can't be the marker there.
Now, if D3 were the marker, it would force E11 (the last open cell in region 10), then K10 (the last open cell in region 11), then C9 (the last open cell in region 9), and region 8 would end up with no open cell for its marker, which can't happen. So it can't go there.
From there, marking I3 would force J1 (the last open cell in region 3), then A2 and B2 cleared (every open cell left in this region sits in row 2, so its marker has to land there), then A6, A7, A8, and 3 more cleared (every open cell left in this region sits in column A, so its marker has to land there), and the chain continues, and then column G would be left with no open cell for its marker. So I3 can't be the marker there.
Following that, that one can't be the marker: at K3 it would force A11, B11, C11, and 2 more cleared (every open cell left in this region sits in row 11, so its marker has to land there), then D4, D5, D6, and 2 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 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 row 9 would be left with no open cell for its marker.
First, marking A4 would force B6, B7, C5, and 2 more cleared (together, two regions have open cells only in columns B and C), then D7, D8, D9, and 2 more cleared (every open cell left in this region sits in column D, so its marker has to land there), then E11 (the last open cell in region 10), and the chain continues, and then column B 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 A2 (the last open cell in region 1), then C8 (the last open cell in region 8), 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 C4 it would force B11 (the last open cell in region 9), then A9 (the last open cell in region 8), and region 1 would be left with no open cell for its marker.
After that, that one can't be the marker: at E3 it would force D5, D6, D7, and 1 more cleared (every open cell left in this region sits in column D, so its marker has to land there), then G2, F5, G5, and 9 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 A5 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 9 would be left with no open cell for its marker.
Now, that one can't be the marker: at G2 it would force B3 (the last open cell in row 3), then D4, E4 and F4 cleared (every open cell left in this region sits in row 4, so its marker has to land there), then C5, C6, C7, 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 column C would be left with no open cell for its marker.
From there, marking D4 would force E11 (the last open cell in region 10), then K10 (the last open cell in region 11), then C9 (the last open cell in region 9), and then region 8 would be left with no open cell for its marker. So D4 can't be the marker there.
Following that, marking F3 would force H2 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 A5 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, B6, C6, and 5 more cleared (every open cell in row 5 belongs to the same region, so that region's marker has to be somewhere in this row), and the chain continues, and then column K would be left with no open cell for its marker. So F3 can't be the marker there.
First, marking G4 would force B3 (the last open cell in row 3), then C5, C6, C7, and 1 more cleared (every open cell left in this region sits in column C, so its marker has to land there), then D5 (the last open cell in row 5), and the chain continues, and then column C would be left with no open cell for its marker. So G4 can't be the marker there.
Next, if H4 were the marker, it would force B3 and A5 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 (the last open cell in row 3), then I1 (the last open cell in region 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.
Then, if J3 were the marker, it would force I1 (the last open cell in region 3), then A2 and B2 cleared (every open cell left in this region sits in row 2, so its marker has to land there), then A5 (the last open cell in region 1), and the chain continues, and column K 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 B3 (the last open cell in row 3), then C5, C6, C7, and 1 more cleared (every open cell left in this region sits in column C, so its marker has to land there), then A6, A7 and A8 cleared (every open cell left in this region sits in column A, so its marker has to land there), and the chain continues, and row 5 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 I4 it would force J1 (the last open cell in region 3), then B3 and A5 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 G3 (the last open cell in row 3), and the chain continues, and row 8 would be left with no open cell for its marker.
From there, if G3 were the marker, it would force E2 (the last open cell in region 2), and row 4 would end up with no open cell for its marker, which can't happen. So it can't go there.
Following that, marking E2 would force B3 (the last open cell in row 3), then D5, D6, D7, and 1 more cleared (every open cell left in this region sits in column D, so its marker has to land there), then C5, C6, C7, 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 then row 7 would be left with no open cell for its marker. So E2 can't be the marker there.
First, every open cell left in this region sits in column H, so its marker has to land there; that clears every other open cell in the column.
Next, marking J4 would force I1 (the last open cell in region 3), then H3 (the last open cell in region 2), then A5 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 then row 6 would be left with no open cell for its marker. So J4 can't be the marker there.
Then, if K4 were the marker, it would force A11, B11, C11, and 2 more cleared (every open cell left in this region sits in row 11, so its marker has to land there), then D5, D6, D7, and 1 more cleared (every open cell left in this region sits in column D, so its marker has to land there), then C5, C6, C7, 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 column J would end up with no open cell for its marker, which can't happen. So it can't go there.
After that, 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.
Now, every remaining candidate in this region touches E5 and F5, so whichever candidate ends up marked will clear them, and they can be ruled out now regardless of which one wins.
From there, marking J1 would force G5, I5, K5, and 1 more cleared (every open cell in row 6 belongs to the same region, so that region's marker has to be somewhere in this row), then A5 (the last open cell in row 5), then H2 (the last open cell in row 2), and then row 3 would be left with no open cell for its marker. So J1 can't be the marker there.
Following that, 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. Placing here also clears its 1 touching neighbour.
First, 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.
Next, 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.
Then, 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.
After that, place a marker at J6. Row 6 has exactly one open cell left. Placing here clears the rest of row 6, column J, and its region. Placing here also clears its 1 touching neighbour.
Now, 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 1 touching neighbour.
From there, 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.
Following that, every open cell left in this region sits in column D, so its marker has to land there; that clears every other open cell in the column.
First, place a marker at F7. Row 7 has exactly one open cell left. Placing here clears the rest of row 7, column F, and its region.
Next, every open cell in row 8 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.
Then, place a marker at A2. Column 1 has exactly one open cell left. Placing here clears the rest of row 2, column A, and its region.
After that, every remaining candidate in this region touches C9, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
Now, place a marker at D9. Row 9 has exactly one open cell left. Placing here clears the rest of row 9, column D, and its region. Placing here also clears its 2 touching neighbours.
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.
Following that, place a marker at C11. Region has exactly one open cell left. Placing here clears the rest of row 11, column C, and its region.
Finally, place a marker at K10. Region has exactly one open cell left.

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