An Expert 11×11 Puzzle Solved with Forced Chain in 54 Steps

11×11 · expert · 54 steps

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This 11×11 board rates as expert (difficulty score 2362). Solving it from scratch takes 54 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 I5, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
Next, if I2 were the marker, it would force J7, J8, J9, and 2 more cleared (every open cell left in this region sits in column J, so its marker has to land there), then K7, K8, K9, and 2 more cleared (every open cell left in this region sits in column K, so its marker has to land there), then A6, B6, C6, and 5 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 K 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 J2 it would force I4 (the last open cell in region 7), then A5, B5, C5, and 3 more cleared (every open cell left in this region sits in row 5, so its marker has to land there), then K6 (the last open cell in region 5), and region 9 would be left with no open cell for its marker.
After that, that one can't be the marker: at H3 it would force A1, B1, C1, and 4 more cleared (every open cell left in this region sits in row 1, so its marker has to land there), then J1, J7, J8, and 3 more cleared (every open cell left in this region sits in column J, so its marker has to land there), then K7, K8, K9, and 2 more cleared (every open cell left in this region sits in column K, so its marker has to land there), and the chain continues, and row 11 would be left with no open cell for its marker.
Now, that one can't be the marker: at I1 it would force J3, J7, J8, and 3 more cleared (every open cell left in this region sits in column J, so its marker has to land there), then K7, K8, K9, and 2 more cleared (every open cell left in this region sits in column K, so its marker has to land there), then A6, B6, C6, and 5 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 K would be left with no open cell for its marker.
From there, marking I3 would force J7, J8, J9, and 2 more cleared (every open cell left in this region sits in column J, so its marker has to land there), then A5, B5, C5, and 4 more cleared (every open cell left in this region sits in row 5, so its marker has to land there), then J6 (the last open cell in region 7), 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.
Following that, that one can't be the marker: at F1 it would force G5, G6, G7, and 4 more cleared (every open cell left in this region sits in column G, so its marker has to land there), then H6, H7, H8, and 3 more cleared (every open cell left in this region sits in column H, so its marker has to land there), then I6, I7, J3, and 2 more cleared (together, two regions have open cells only in columns I and J), and the chain continues, and region 9 would be left with no open cell for its marker.
First, if J3 were the marker, region 7 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 K, so its marker has to land there; that clears every other open cell in the column.
Then, marking A4 would force J1, J7, J8, and 3 more cleared (every open cell left in this region sits in column J, so its marker has to land there), then H2 cleared (every remaining candidate in this region touches H2), then D5, E5, J5, and 1 more 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 A4 can't be the marker there.
After that, if B4 were the marker, it would force J1, J7, J8, and 3 more cleared (every open cell left in this region sits in column J, so its marker has to land there), then H2 cleared (every remaining candidate in this region touches H2), then J5 and K5 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.
Now, that one can't be the marker: at C4 it would force J1, J7, J8, and 3 more cleared (every open cell left in this region sits in column J, so its marker has to land there), then H2 cleared (every remaining candidate in this region touches H2), then A5, J5 and K5 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.
From there, marking D4 would force J1, J7, J8, and 3 more cleared (every open cell left in this region sits in column J, so its marker has to land there), then H2 cleared (every remaining candidate in this region touches H2), then A5, B5, J5, and 1 more 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 D4 can't be the marker there.
Following that, if E4 were the marker, it would force J1, J7, J8, and 3 more cleared (every open cell left in this region sits in column J, so its marker has to land there), then G6, G7, G8, and 9 more cleared (together, two regions have open cells only in columns G and H), then I8, I9, I10, and 1 more cleared (every open cell left in this region sits in column I, so its marker has to land there), and the chain continues, and row 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 F4 it would force J1, J7, J8, and 3 more cleared (every open cell left in this region sits in column J, so its marker has to land there), then H1, H6, H7, and 4 more cleared (every open cell left in this region sits in column H, so its marker has to land there), then G6, G7, G8, and 3 more cleared (every open cell left in this region sits in column G, so its marker has to land there), and the chain continues, and region 8 would be left with no open cell for its marker.
Next, marking E2 would force C1 (the last open cell in region 3), then G5, G6, G7, and 4 more cleared (every open cell left in this region sits in column G, so its marker has to land there), then H4 cleared (every remaining candidate in this region touches H4), and the chain continues, and then region 10 would be left with no open cell for its marker. So E2 can't be the marker there.
Then, marking G4 would force H2 (the last open cell in region 6), then J7, J8, J9, and 2 more cleared (every open cell left in this region sits in column J, so its marker has to land there), then I8, I9, I10, and 1 more cleared (every open cell left in this region sits in column I, so its marker has to land there), and the chain continues, and then row 11 would be left with no open cell for its marker. So G4 can't be the marker there.
After that, marking H2 would force J1 (the last open cell in region 4), then I4 (the last open cell in region 7), then G6 (the last open cell in region 9), and the chain continues, and then column K would be left with no open cell for its marker. So H2 can't be the marker there.
Now, marking J1 would force I4 (the last open cell in region 7), then A5, B5, C5, and 3 more cleared (every open cell left in this region sits in row 5, so its marker has to land there), then A6, B6, C6, and 4 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 2 would be left with no open cell for its marker. So J1 can't be the marker there.
From there, that one can't be the marker: at D2 it would force F3 (the last open cell in region 3), then A1 and K1 cleared (every open cell left in this region sits in row 1, so its marker has to land there), then G6, G7, G8, and 9 more cleared (together, two regions have open cells only in columns G and H), and the chain continues, and region 8 would be left with no open cell for its marker.
Following that, if F2 were the marker, it would force H1 (the last open cell in region 4), then G5 (the last open cell in region 6), then I8, I9, I10, and 4 more cleared (together, two regions have open cells only in columns I and J), and region 8 would end up with no open cell for its marker, which can't happen. So it can't go there.
First, marking F3 would force A1, B1 and K1 cleared (every open cell left in this region sits in row 1, so its marker has to land there), then G6, G7, G8, and 9 more cleared (together, two regions have open cells only in columns G and H), then I8, I9, I10, and 4 more cleared (together, two regions have open cells only in columns I and J), and then region 8 would be left with no open cell for its marker. So F3 can't be the marker there.
Next, if H4 were the marker, it would force G6, G7, G8, and 3 more cleared (every open cell left in this region sits in column G, so its marker has to land there), then J7, J8, J9, and 2 more cleared (every open cell left in this region sits in column J, so its marker has to land there), then I8, I9, I10, and 1 more cleared (every open cell left in this region sits in column I, so its marker has to land there), and the chain continues, and row 11 would end up with no open cell for its marker, which can't happen. So it can't go there.
Then, every open cell left in this region sits in row 5, so its marker has to land there; that clears every other open cell in the row.
After that, every remaining candidate in this region touches G6, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
Now, marking A1 would force G5, G7, G8, and 3 more cleared (every open cell left in this region sits in column G, so its marker has to land there), then K2, B3, C3, and 2 more cleared (together, two regions have open cells only in rows 2 and 3), then E6 (the last open cell in region 2), and the chain continues, and then column K would be left with no open cell for its marker. So A1 can't be the marker there.
From there, if B1 were the marker, it would force E3 (the last open cell in region 3), then G2 (the last open cell in region 4), then A6, C6, D6, and 3 more cleared (together, two regions have open cells only in rows 4 and 6), and the chain continues, and region 10 would end up with no open cell for its marker, which can't happen. So it can't go there.
Following that, if K1 were the marker, it would force G5, G7, G8, and 3 more cleared (every open cell left in this region sits in column G, so its marker has to land there), then J6 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 A2, B2, A3, and 3 more cleared (together, two regions have open cells only in rows 2 and 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.
First, that one can't be the marker: at A2 it would force C3, D3 and K3 cleared (together, two regions have open cells only in rows 1 and 3), then E6 (the last open cell in region 2), then K4 (the last open cell in region 5), and region 7 would be left with no open cell for its marker.
Next, marking B2 would force K3 cleared (together, two regions have open cells only in rows 1 and 3), then A6, C6, D6, and 3 more cleared (together, two regions have open cells only in rows 4 and 6), then C7 (the last open cell in region 1), and the chain continues, and then region 10 would be left with no open cell for its marker. So B2 can't be the marker there.
Then, if C2 were the marker, it would force E6 (the last open cell in region 2), then K4 cleared (every open cell left in this region sits in row 4, so its marker has to land there), then K3 (the last open cell in region 5), and the chain continues, 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, marking G1 would force E3 (the last open cell in region 3), and then region 2 would be left with no open cell for its marker. So G1 can't be the marker there.
Now, if H1 were the marker, it would force E3 (the last open cell in region 3), and region 2 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 G, so its marker has to land there; that clears every other open cell in the column.
Following that, 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.
First, marking K2 would force G3 (the last open cell in region 4), then E6 (the last open cell in region 2), then H5 (the last open cell in region 6), and the chain continues, and then region 1 would be left with no open cell for its marker. So K2 can't be the marker there.
Next, place a marker at G2. Row 2 has exactly one open cell left. Placing here clears the rest of row 2, column G, and its region.
Then, if A3 were the marker, it would force E6 (the last open cell in region 2), then H5 (the last open cell in region 6), then J4 (the last open cell in region 7), 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, that one can't be the marker: at B3 it would force A6, C6, D6, and 3 more cleared (together, two regions have open cells only in rows 4 and 6), then C7 (the last open cell in region 1), then J8 (the last open cell in region 9), and region 10 would be left with no open cell for its marker.
Now, marking C3 would force A6, B6, D6, and 3 more cleared (together, two regions have open cells only in rows 4 and 6), then B7 (the last open cell in region 1), then J8 (the last open cell in region 9), and then region 10 would be left with no open cell for its marker. So C3 can't be the marker there.
From there, marking D1 would force E6 (the last open cell in region 2), then H5 (the last open cell in region 6), then J4 (the last open cell in region 7), and then row 3 would be left with no open cell for its marker. So D1 can't be the marker there.
Following that, that one can't be the marker: at K3 it would force E6 (the last open cell in region 2), then C1 (the last open cell in region 3), then I4 (the last open cell in region 7), and row 5 would be left with no open cell for its marker.
First, place a marker at D3. Row 3 has exactly one open cell left. Placing here clears the rest of row 3, column D, and its region.
Next, together, two regions have open cells only in rows 4 and 6; 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.
Then, every open cell left in this region sits in row 7, so its marker has to land there; that clears every other open cell in the row.
After that, place a marker at J8. Region has exactly one open cell left. Placing here clears the rest of row 8, column J, and its region. Placing here also clears its 1 touching neighbour.
Now, place a marker at I4. Region has exactly one open cell left. Placing here clears the rest of row 4, column I, and its region. Placing here also clears its 1 touching neighbour.
From there, 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.
Following that, place a marker at K6. Region has exactly one open cell left.
First, place a marker at H9. Region has exactly one open cell left. Placing here clears the rest of row 9, column H, and its region.
Next, 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.
Then, place a marker at E1. Region has exactly one open cell left. Placing here clears the rest of row 1, column E, and its region.
After that, place a marker at B7. Region has exactly one open cell left.
Finally, place a marker at A11. Region has exactly one open cell left.

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