An Expert 11×11 Puzzle Solved with Forced Chain in 77 Steps

11×11 · expert · 77 steps

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This 11×11 board rates as expert (difficulty score 3233). Solving it from scratch takes 77 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 J7, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
Next, that one can't be the marker: at I1 it would force K5, K6, K7, and 4 more cleared (every open cell left in this region sits in column K, so its marker has to land there), then J6 (the last open cell in region 9), and region 6 would be left with no open cell for its marker.
Then, marking J1 would force K5, K6, K7, and 4 more cleared (every open cell left in this region sits in column K, so its marker has to land there), then I6, I7, I8, and 3 more cleared (every open cell left in this region sits in column I, so its marker has to land there), and then region 9 would be left with no open cell for its marker. So J1 can't be the marker there.
After that, that one can't be the marker: at E3 it would force A1, G1, H1, and 1 more cleared (every open cell left in this region sits in row 1, so its marker has to land there), then G2, H2, I2, and 4 more cleared (together, two regions have open cells only in rows 2 and 4), 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), and the chain continues, and row 5 would be left with no open cell for its marker.
Now, if J3 were the marker, it would force K1 (the last open cell in region 4), then I5, I8, I9, and 2 more cleared (every open cell left in this region sits in column I, so its marker has to land there), then E2, G2, H2, and 4 more cleared (together, two regions have open cells only in rows 2 and 4), and the chain continues, and column I would end up with no open cell for its marker, which can't happen. So it can't go there.
From there, if I5 were the marker, region 6 would end up with no open cell for its marker, which can't happen. So it can't go there.
Following that, marking K5 would force J2 (the last open cell in region 4), 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), then F6, G6, H6, and 7 more cleared (together, two regions have open cells only in rows 6 and 7), and the chain continues, and then region 3 would be left with no open cell for its marker. So K5 can't be the marker there.
First, that one can't be the marker: at K6 it would force J2 (the last open cell in region 4), then I4 (the last open cell in region 6), then A5, B5, C5, and 2 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 region 3 would be left with no open cell for its marker.
Next, if B7 were the marker, it would force F8, G8, F9, and 5 more cleared (every open cell in row 11 belongs to the same region, so that region's marker has to be somewhere in this row), then G6, D8, E8, and 7 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.
Then, if K7 were the marker, it would force J2 (the last open cell in region 4), then I4 (the last open cell in region 6), then G1, H1, F3, and 1 more cleared (together, two regions have open cells only in rows 1 and 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.
After that, that one can't be the marker: at A8 it would force F9, K9, F10, and 3 more cleared (every open cell in row 11 belongs to the same region, so that region's marker has to be somewhere in this row), then G6, C7, D7, and 8 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 be left with no open cell for its marker.
Now, marking B8 would force F9, K9, F10, and 3 more cleared (every open cell in row 11 belongs to the same region, so that region's marker has to be somewhere in this row), then G6, D7, E7, and 7 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 then row 10 would be left with no open cell for its marker. So B8 can't be the marker there.
From there, that one can't be the marker: at F7 it would force K8 (the last open cell in row 8), then J2 (the last open cell in region 4), then I4 (the last open cell in region 6), and the chain continues, and row 10 would be left with no open cell for its marker.
Following that, marking G7 would force K8 (the last open cell in row 8), then J2 (the last open cell in region 4), then I4 (the last open cell in region 6), and the chain continues, and then region 3 would be left with no open cell for its marker. So G7 can't be the marker there.
First, marking E8 would force F1, F2, F3, and 2 more cleared (every open cell left in this region sits in column F, so its marker has to land there), then I10, I11, J2, and 2 more cleared (together, two regions have open cells only in columns I and J), then K9, K10 and K11 cleared (every open cell left in this region sits in column K, 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 E8 can't be the marker there.
Next, if F8 were the marker, it would force E1, E6 and E10 cleared (every open cell left in this region sits in column E, so its marker has to land there), then A9, B9, A10, and 2 more cleared (every open cell in row 11 belongs to the same region, so that region's marker has to be somewhere in this row), then G6, C7, D7, and 4 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.
Then, marking E2 would force K8, K9, K10, and 1 more cleared (every open cell left in this region sits in column K, so its marker has to land there), then G8, I10, J10, and 4 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), then G6, C7, D7, and 10 more cleared (every open cell in row 8 belongs to the same region, so that region's marker has to be somewhere in this row), and the chain continues, and then region 2 would be left with no open cell for its marker. So E2 can't be the marker there.
After that, marking A4 would force G5, H5, J5, and 4 more cleared (together, two regions have open cells only in rows 5 and 6), then I7, I8, I9, and 2 more cleared (every open cell left in this region sits in column I, so its marker has to land there), then K8 (the last open cell in region 9), and the chain continues, and then column I would be left with no open cell for its marker. So A4 can't be the marker there.
Now, if K4 were the marker, it would force I6, I7, I8, and 3 more cleared (every open cell left in this region sits in column I, so its marker has to land there), then J6 (the last open cell in region 9), 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), and the chain continues, and region 3 would end up with no open cell for its marker, which can't happen. So it can't go there.
From there, marking F3 would force E5 (the last open cell in region 5), then A1, A2, B2, and 2 more cleared (together, two regions have open cells only in rows 1 and 2), then I4 and J4 cleared (every open cell left in this region sits in row 4, so its marker has to land there), and then region 6 would be left with no open cell for its marker. So F3 can't be the marker there.
Following that, that one can't be the marker: at G5 it would force E4 (the last open cell in region 5), then I6, I7, I8, and 3 more cleared (every open cell left in this region sits in column I, so its marker has to land there), then K9, K10, K11, and 4 more cleared (together, two regions have open cells only in columns K and J), and the chain continues, and region 2 would be left with no open cell for its marker.
First, if G6 were the marker, it would force K8 (the last open cell in row 8), then J2 (the last open cell in region 4), then I4 (the last open cell in region 6), and the chain continues, and region 2 would end up with no open cell for its marker, which can't happen. So it can't go there.
Next, if H7 were the marker, it would force K8 (the last open cell in row 8), then J2 (the last open cell in region 4), then I4 (the last open cell in region 6), and the chain continues, and column G 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 G8 it would force H10 cleared (every open cell left in this region sits in column H, so its marker has to land there), then A9, B9, A10, and 2 more cleared (every open cell in row 11 belongs to the same region, so that region's marker has to be somewhere in this row), then C7, D7, E7, and 2 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 be left with no open cell for its marker.
After that, if F2 were the marker, it would force K8, K9, K10, and 1 more cleared (every open cell left in this region sits in column K, so its marker has to land there), then E6, E7, E9, and 2 more cleared (every open cell left in this region sits in column E, so its marker has to land there), then C7, D7, C9, and 8 more cleared (every open cell in row 8 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.
Now, marking H2 would force K8, K9, K10, and 1 more cleared (every open cell left in this region sits in column K, so its marker has to land there), then C7, D7, E7, and 9 more cleared (every open cell in row 8 belongs to the same region, so that region's marker has to be somewhere in this row), then G11 (the last open cell in column G), 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.
From there, that one can't be the marker: at H3 it would force J2, J6, J8, and 3 more cleared (every open cell left in this region sits in column J, so its marker has to land there), then K8, K9, K10, and 1 more cleared (every open cell left in this region sits in column K, 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 10 would be left with no open cell for its marker.
Following that, if D6 were the marker, it would force I7 (the last open cell in row 7), then C8 (the last open cell in row 8), then A1, G1, H1, and 1 more cleared (every open cell left in this region sits in row 1, 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.
First, that one can't be the marker: at C7 it would force K8 (the last open cell in row 8), then J2 (the last open cell in region 4), then I4 (the last open cell in region 6), and the chain continues, and region 7 would be left with no open cell for its marker.
Next, that one can't be the marker: at E6 it would force F4 (the last open cell in region 5), then I7 (the last open cell in row 7), then J5 (the last open cell in region 6), and the chain continues, and row 10 would be left with no open cell for its marker.
Then, marking D7 would force K8 (the last open cell in row 8), then J2 (the last open cell in region 4), then I4 (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 D7 can't be the marker there.
After that, if E7 were the marker, it would force K8 (the last open cell in row 8), then J2 (the last open cell in region 4), then I4 (the last open cell in region 6), 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, marking A1 would force I7 (the last open cell in row 7), then J2, J9, J10, and 1 more cleared (every open cell left in this region sits in column J, so its marker has to land there), then K9, K10 and K11 cleared (every open cell left in this region sits in column K, so its marker has to land there), and the chain continues, and then region 10 would be left with no open cell for its marker. So A1 can't be the marker there.
From there, marking C3 would force A2 (the last open cell in region 1), then K1 (the last open cell in region 4), then I7 (the last open cell in row 7), 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.
Following that, that one can't be the marker: at A5 it would force I7 (the last open cell in row 7), then J4 (the last open cell in region 6), then F6 (the last open cell in region 5), and the chain continues, and row 10 would be left with no open cell for its marker.
First, marking B5 would force I7 (the last open cell in row 7), then J4 (the last open cell in region 6), then F6 (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 B5 can't be the marker there.
Next, if C5 were the marker, it would force I7 (the last open cell in row 7), then J4 (the last open cell in region 6), then F6 (the last open cell in region 5), and the chain continues, and region 1 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 D5 it would force I7 (the last open cell in row 7), then J4 (the last open cell in region 6), then F6 (the last open cell in region 5), and the chain continues, and region 1 would be left with no open cell for its marker.
After that, that one can't be the marker: at A2 it would force I7 (the last open cell in row 7), then K9, K10 and K11 cleared (every open cell left in this region sits in column K, so its marker has to land there), then J9, J10 and J11 cleared (every open cell left in this region sits in column J, so its marker has to land there), and the chain continues, and region 3 would be left with no open cell for its marker.
Now, if A3 were the marker, it would force I7 (the last open cell in row 7), then J2, J9, J10, and 1 more cleared (every open cell left in this region sits in column J, so its marker has to land there), then K9, K10 and K11 cleared (every open cell left in this region sits in column K, so its marker has to land there), and the chain continues, and region 3 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 B6 it would force I7 (the last open cell in row 7), then C1, C8, C9, and 2 more cleared (every open cell left in this region sits in column C, so its marker has to land there), then D8 (the last open cell in row 8), and the chain continues, and row 10 would be left with no open cell for its marker.
Following that, marking C6 would force I7 (the last open cell in row 7), then D8 (the last open cell in row 8), then G1, H1 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 row 10 would be left with no open cell for its marker. So C6 can't be the marker there.
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, every remaining candidate in this region touches C10, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
Then, marking B2 would force K8, K9, K10, and 1 more cleared (every open cell left in this region sits in column K, so its marker has to land there), then C9, D9, E9, and 7 more cleared (every open cell in row 8 belongs to the same region, so that region's marker has to be somewhere in this row), then F9 (the last open cell in row 9), and the chain continues, and then row 11 would be left with no open cell for its marker. So B2 can't be the marker there.
After that, that one can't be the marker: at D2 it would force K8, K9, K10, and 1 more cleared (every open cell left in this region sits in column K, so its marker has to land there), then C9, E9, G9, and 6 more cleared (every open cell in row 8 belongs to the same region, so that region's marker has to be somewhere in this row), then C8 cleared (together, two regions have open cells only in columns B and C), and the chain continues, and column G would be left with no open cell for its marker.
Now, if E4 were the marker, it would force G1, H1 and K1 cleared (every open cell left in this region sits in row 1, so its marker has to land there), then G2, I2, G3, and 1 more cleared (together, two regions have open cells only in rows 2 and 3), then J5 (the last open cell in region 6), and the chain continues, and region 4 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 J5 it would force K8, K9, K10, and 1 more cleared (every open cell left in this region sits in column K, so its marker has to land there), then I7 (the last open cell in region 9), then A6 (the last open cell in region 7), and the chain continues, and row 10 would be left with no open cell for its marker.
Following that, if B1 were the marker, it would force C4 (the last open cell in region 1), then I6, I7, I8, and 3 more cleared (every open cell left in this region sits in column I, so its marker has to land there), then A7 (the last open cell in row 7), and the chain continues, and row 3 would end up with no open cell for its marker, which can't happen. So it can't go there.
First, if H1 were the marker, it would force D8, D9, D10, and 1 more cleared (every open cell left in this region sits in column D, so its marker has to land there), then F4 and F6 cleared (every open cell in row 5 belongs to the same region, so that region's marker has to be somewhere in this row), then B3, K3, B4, and 1 more cleared (together, two regions have open cells only in rows 3 and 4), and the chain continues, and region 4 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 G2 it would force K8, K9, K10, and 1 more cleared (every open cell left in this region sits in column K, so its marker has to land there), then F4 and F6 cleared (every open cell in row 5 belongs to the same region, so that region's marker has to be somewhere in this row), then C9, D9, E9, and 5 more cleared (every open cell in row 8 belongs to the same region, so that region's marker has to be somewhere in this row), and the chain continues, and region 2 would be left with no open cell for its marker.
Then, if K1 were the marker, it would force D8, D9, D10, and 1 more cleared (every open cell left in this region sits in column D, so its marker has to land there), then C9, E9, G9, and 6 more cleared (every open cell in row 8 belongs to the same region, so that region's marker has to be somewhere in this row), then C8 cleared (together, two regions have open cells only in columns C and B), 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.
After that, if C2 were the marker, it would force K3 (the last open cell in region 4), then I4 (the last open cell in region 6), then J6 (the last open cell in region 9), and the chain continues, and region 2 would end up with no open cell for its marker, which can't happen. So it can't go there.
Now, marking G1 would force D8, D9, D10, and 1 more cleared (every open cell left in this region sits in column D, so its marker has to land there), then F4 and F6 cleared (every open cell in row 5 belongs to the same region, so that region's marker has to be somewhere in this row), then I3, K3, I4, and 1 more cleared (together, two regions have open cells only in rows 3 and 4), and the chain continues, and then region 4 would be left with no open cell for its marker. So G1 can't be the marker there.
From there, 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.
Following that, if I2 were the marker, it would force K3 (the last open cell in region 4), then J6 (the last open cell in region 9), then A7 (the last open cell in region 7), and the chain continues, and region 5 would end up with no open cell for its marker, which can't happen. So it can't go there.
First, 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.
Next, together, two regions have open cells only in rows 3 and 4; 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 column H, 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 rows 5 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.
Now, place a marker at A7. Region has exactly one open cell left. Placing here clears the rest of row 7, column A, and its region.
From there, place a marker at K8. Region has exactly one open cell left. Placing here clears the rest of row 8, column K, and its region. Placing here also clears its 1 touching neighbour.
Following that, place a marker at J2. Region has exactly one open cell left. Placing here clears the rest of row 2, column J, and its region. Placing here also clears its 1 touching neighbour.
First, 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.
Next, place a marker at B3. Region has exactly one open cell left. Placing here clears the rest of row 3, column B, and its region.
Then, place a marker at H6. Region has exactly one open cell left. Placing here clears the rest of row 6, column H, and its region.
After that, marking D1 would force C11 (the last open cell in region 10), then F5 cleared (every open cell left in this region sits in column F, so its marker has to land there), then E5 (the last open cell in region 5), and the chain continues, and then column F would be left with no open cell for its marker. So D1 can't be the marker there.
Now, if E1 were the marker, it would force F5 (the last open cell in region 5), then G11 (the last open cell in region 11), then D10 (the last open cell in region 10), 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, that one can't be the marker: at F1 it would force E5 (the last open cell in region 5), then G11 (the last open cell in region 11), then D10 (the last open cell in region 10), and row 9 would be left with no open cell for its marker.
Following that, place a marker at C1. Region has exactly one open cell left. Placing here clears the rest of row 1, column C, and its region.
First, 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.
Next, every remaining candidate in this region touches F9 and F10, so whichever candidate ends up marked will clear them, and they can be ruled out now regardless of which one wins.
Then, every open cell left in this region sits in row 11, so its marker has to land there; that clears every other open cell in the row.
After that, 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. Placing here also clears its 2 touching neighbours.
Now, place a marker at G9. Region has exactly one open cell left. Placing here clears the rest of row 9, column G, and its region.
From there, place a marker at F11. Region has exactly one open cell left. Placing here clears the rest of row 11, column F, and its region.
Finally, place a marker at E5. Region has exactly one open cell left.

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