An Expert 11×11 Puzzle Solved with Forced Chain in 61 Steps

11×11 · expert · 61 steps

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This 11×11 board rates as expert (difficulty score 2629). Solving it from scratch takes 61 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, marking A1 would force C2, D2, E2, and 4 more cleared (every open cell left in this region sits in row 2, so its marker has to land there), then K3 (the last open cell in region 4), and then region 2 would be left with no open cell for its marker. So A1 can't be the marker there.
Next, if B1 were the marker, it would force D2, E2, F2, and 3 more cleared (every open cell left in this region sits in row 2, so its marker has to land there), then K3 (the last open cell in region 4), and region 2 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 C1 it would force E2, F2, I2, and 2 more cleared (every open cell left in this region sits in row 2, so its marker has to land there), then K3 (the last open cell in region 4), and region 2 would be left with no open cell for its marker.
After that, marking D1 would force A2, I2, J2, and 1 more cleared (every open cell left in this region sits in row 2, so its marker has to land there), then K3 (the last open cell in region 4), then J5 (the last open cell in region 5), and the chain continues, and then region 9 would be left with no open cell for its marker. So D1 can't be the marker there.
Now, if H1 were the marker, it would force A2, A3, B3, and 4 more cleared (together, two regions have open cells only in rows 2 and 3), then J6, J7, J8, and 9 more cleared (together, two regions have open cells only in columns J and K), then I8 (the last open cell in region 10), 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.
From there, that one can't be the marker: at A2 it would force I1, J1 and K1 cleared (every open cell left in this region sits in row 1, so its marker has to land there), then K3 (the last open cell in region 4), and region 2 would be left with no open cell for its marker.
Following that, marking B2 would force I1, J1 and K1 cleared (every open cell left in this region sits in row 1, so its marker has to land there), then K3 (the last open cell in region 4), then J5 (the last open cell in region 5), and the chain continues, and then row 10 would be left with no open cell for its marker. So B2 can't be the marker there.
First, if F2 were the marker, region 3 would end up with no open cell for its marker, which can't happen. So it can't go there.
Next, marking H2 would force A3, B3, C3, and 3 more cleared (every open cell left in this region sits in row 3, so its marker has to land there), then J6, J7, J8, and 9 more cleared (together, two regions have open cells only in columns J and K), then I8 (the last open cell in region 10), and the chain continues, and then region 9 would be left with no open cell for its marker. So H2 can't be the marker there.
Then, if I2 were the marker, it would force K6, K7, K8, and 3 more cleared (every open cell left in this region sits in column K, so its marker has to land there), then K1 cleared (every open cell left in this region sits in row 1, so its marker has to land there), then K3 (the last open cell in region 4), and region 2 would end up with no open cell for its marker, which can't happen. So it can't go there.
After that, if A3 were the marker, it would force G2, J2 and K2 cleared (every open cell left in this region sits in row 2, so its marker has to land there), then E1, F1 and G1 cleared (every open cell left in this region sits in row 1, so its marker has to land there), and region 3 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 B3 it would force G2, J2 and K2 cleared (every open cell left in this region sits in row 2, so its marker has to land there), then E1, F1 and G1 cleared (every open cell left in this region sits in row 1, so its marker has to land there), and region 3 would be left with no open cell for its marker.
From there, marking C3 would force E2 (the last open cell in region 2), then G1 (the last open cell in region 3), and then region 4 would be left with no open cell for its marker. So C3 can't be the marker there.
Following that, that one can't be the marker: at E3 it would force C2 (the last open cell in region 2), then F1 and G1 cleared (every open cell left in this region sits in row 1, so its marker has to land there), and region 3 would be left with no open cell for its marker.
First, that one can't be the marker: at H3 it would force I1, J1 and K1 cleared (every open cell left in this region sits in row 1, so its marker has to land there), then C2, D2 and E2 cleared (every open cell left in this region sits in row 2, so its marker has to land there), and region 2 would be left with no open cell for its marker.
Next, marking I3 would force G2 and K2 cleared (every open cell left in this region sits in row 2, so its marker has to land there), then E1, F1 and G1 cleared (every open cell left in this region sits in row 1, so its marker has to land there), and then region 3 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 E1, F1 and G1 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 3), and region 2 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 C5, D5, E5, and 4 more cleared (every open cell left in this region sits in row 5, so its marker has to land there), then H6 (the last open cell in region 6), then C7, J7, K7, and 5 more cleared (together, two regions have open cells only in rows 7 and 8), and the chain continues, and then row 8 would be left with no open cell for its marker. So A4 can't be the marker there.
Now, if C2 were the marker, it would force K3 (the last open cell in row 3), then J5 (the last open cell in region 5), then A8, B8, D8, and 3 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 region 6 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 K3 (the last open cell in row 3), then J5 (the last open cell in region 5), then A8, B8, C8, and 3 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 region 6 would be left with no open cell for its marker.
Following that, marking E2 would force G1 (the last open cell in region 3), then K3 (the last open cell in region 4), then J5 (the last open cell in region 5), and the chain continues, and then region 6 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 row 3, so its marker has to land there; that clears every other open cell in the row.
Next, that one can't be the marker: at C4 it would force A5, E5, F5, and 1 more cleared (every open cell left in this region sits in row 5, so its marker has to land there), then B10 and B11 cleared (together, two regions have open cells only in columns A and B), then G2 cleared (every remaining candidate in this region touches G2), and the chain continues, and region 9 would be left with no open cell for its marker.
Then, marking D4 would force A5, F5 and I5 cleared (every open cell left in this region sits in row 5, so its marker has to land there), then G2 cleared (every remaining candidate in this region touches G2), then I1, J1 and K1 cleared (every open cell left in this region sits in row 1, 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.
After that, if E4 were the marker, it would force G3 (the last open cell in region 2), then F1 (the last open cell in region 3), then A5, C5 and I5 cleared (every open cell left in this region sits in row 5, so its marker has to land there), and the chain continues, and column D would end up with no open cell for its marker, which can't happen. So it can't go there.
Now, marking G4 would force D3 (the last open cell in region 2), then I1, J1 and K1 cleared (every open cell left in this region sits in row 1, so its marker has to land there), then A5, C5, E5, 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 G4 can't be the marker there.
From there, if H4 were the marker, it would force B5 (the last open cell in region 6), and region 5 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 K5 (the last open cell in region 5), then H6 (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), and region 10 would be left with no open cell for its marker.
First, that one can't be the marker: at F1 it would force J6, J7, J8, and 9 more cleared (together, two regions have open cells only in columns J and K), then A8, B8, C8, and 3 more cleared (every open cell left in this region sits in row 8, so its marker has to land there), then B6 cleared (every remaining candidate in this region touches B6), and the chain continues, and region 9 would be left with no open cell for its marker.
Next, marking F3 would force I1, J1 and K1 cleared (every open cell left in this region sits in row 1, so its marker has to land there), then J6, J7, J8, and 9 more cleared (together, two regions have open cells only in columns J and K), then A8, B8, C8, and 3 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 then region 9 would be left with no open cell for its marker. So F3 can't be the marker there.
Then, that one can't be the marker: at A5 it would force F4 cleared (every open cell left in this region sits in row 4, so its marker has to land there), then H6 (the last open cell in region 6), then C7, J7, K7, and 5 more cleared (together, two regions have open cells only in rows 7 and 8), and the chain continues, and row 8 would be left with no open cell for its marker.
After that, if K1 were the marker, it would force G2 (the last open cell in region 3), then D3 (the last open cell in region 2), then J6, J7, J8, and 3 more cleared (every open cell left in this region sits in column J, 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.
Now, marking K2 would force J6, 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 A8, B8, C8, and 4 more cleared (every open cell left in this region sits in row 8, so its marker has to land there), then I5 cleared (every remaining candidate in this region touches I5), and the chain continues, and then column J would be left with no open cell for its marker. So K2 can't be the marker there.
From there, marking B5 would force A9, A10 and A11 cleared (every open cell left in this region sits in column A, so its marker has to land there), then C9, C10 and C11 cleared (every open cell left in this region sits in column C, so its marker has to land there), then D7, E7, F7, and 13 more cleared (together, two regions have open cells only in rows 7 and 8), and the chain continues, and then region 9 would be left with no open cell for its marker. So B5 can't be the marker there.
Following that, marking J4 would force I1 (the last open cell in region 4), then G2 (the last open cell in region 3), then D3 (the last open cell in region 2), and the chain continues, and then region 9 would be left with no open cell for its marker. So J4 can't be the marker there.
First, if C5 were the marker, it would force K4 (the last open cell in region 5), then H6 (the last open cell in region 6), then D9, E9, F9, and 8 more cleared (every open cell in row 10 belongs to the same region, so that region's marker has to be somewhere in this row), 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.
Next, that one can't be the marker: at D5 it would force G3 (the last open cell in region 2), then E1 (the last open cell in region 3), then J2 (the last open cell in region 4), and the chain continues, and region 9 would be left with no open cell for its marker.
Then, marking E5 would force K4 (the last open cell in region 5), then H6 (the last open cell in region 6), then G3, G8, G9, and 2 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 then region 9 would be left with no open cell for its marker. So E5 can't be the marker there.
After that, if F5 were the marker, it would force K4 (the last open cell in region 5), then H6 (the last open cell in region 6), then G8 (the last open cell in region 7), 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.
Now, if E1 were the marker, it would force J2 (the last open cell in region 4), then K6, K7, K8, and 3 more cleared (every open cell left in this region sits in column K, so its marker has to land there), then A8, B8, C8, and 3 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 region 6 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, place a marker at D3. Region has exactly one open cell left. Placing here clears the rest of row 3, column D, and its region.
First, every remaining candidate in this region touches F7, 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 H6, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
Then, together, two regions have open cells only in rows 4 and 5; 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.
After that, marking G1 would force J2 (the last open cell in region 4), then K6, K7, K8, and 3 more cleared (every open cell left in this region sits in column K, so its marker has to land there), then A8, B8, C8, 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 then region 9 would be left with no open cell for its marker. So G1 can't be the marker there.
Now, 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.
From there, marking J1 would force K6, K7, K8, and 3 more cleared (every open cell left in this region sits in column K, so its marker has to land there), then A8, B8, C8, and 2 more cleared (every open cell left in this region sits in row 8, so its marker has to land there), then B6, C6, E6, and 3 more cleared (together, two regions have open cells only in rows 6 and 7), and then region 9 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.
First, 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.
Next, every open cell left in this region sits in row 4, so its marker has to land there; that clears every other open cell in the row.
Then, 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.
After that, together, two regions have open cells only in columns J and K; 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.
Now, together, two regions have open cells only in rows 6 and 8; 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.
From there, place a marker at B9. Region has exactly one open cell left. Placing here clears the rest of row 9, column B, and its region. Placing here also clears its 2 touching neighbours.
Following that, place a marker at F4. Region has exactly one open cell left. Placing here clears the rest of row 4, column F, and its region.
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 E8. Region has exactly one open cell left. Placing here clears the rest of row 8, column E, and its region.
Then, 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.
After that, place a marker at K10. Region has exactly one open cell left. Placing here clears the rest of row 10, column K, and its region.
Finally, place a marker at J5. Region has exactly one open cell left.

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