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 2298). Solving it from scratch takes 50 logical steps, using 6 techniques: Single cell, Row/column exclusion, Adjacency clear, Region confinement, Shape squeeze, Forced chain. Every step below is forced: nothing here is a guess.

First, every remaining candidate in this region touches J5, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
Next, marking A1 would force I2 (the last open cell in region 3), then K7, K8, K9, and 7 more cleared (together, two regions have open cells only in columns K and J), then H11 (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 A1 can't be the marker there.
Then, if B1 were the marker, it would force I2 (the last open cell in region 3), then K7, K8, K9, and 7 more cleared (together, two regions have open cells only in columns K and J), then H11 (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.
After that, that one can't be the marker: at C1 it would force I2 (the last open cell in region 3), then K7, K8, K9, and 7 more cleared (together, two regions have open cells only in columns K and J), then H11 (the last open cell in region 10), and the chain continues, and region 9 would be left with no open cell for its marker.
Now, marking D1 would force I2 (the last open cell in region 3), then K7, K8, K9, and 7 more cleared (together, two regions have open cells only in columns K and J), then H11 (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 D1 can't be the marker there.
From there, if E1 were the marker, it would force I2 (the last open cell in region 3), then K7, K8, K9, and 7 more cleared (together, two regions have open cells only in columns K and J), then H11 (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.
Following that, marking G1 would force I2 (the last open cell in region 3), then A3, B3, C3, and 2 more cleared (every open cell left in this region sits in row 3, so its marker has to land there), then J4 (the last open cell in region 4), 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.
First, that one can't be the marker: at I1 it would force K7, K8, K9, and 7 more cleared (together, two regions have open cells only in columns K and J), then H11 (the last open cell in region 10), then G3, G4, G6, 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 column K would be left with no open cell for its marker.
Next, marking J1 would force I4, I7, I8, and 8 more cleared (together, two regions have open cells only in columns I and K), then H11 (the last open cell in region 10), then G3, G4, G6, 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 then region 5 would be left with no open cell for its marker. So J1 can't be the marker there.
Then, if K1 were the marker, it would force 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), then I4, I7, I8, and 8 more cleared (together, two regions have open cells only in columns I and J), then H11 (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.
After that, every remaining candidate in this region touches H2, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
Now, that one can't be the marker: place it at G2, and row 1 would be left with no open cell for its marker.
From there, that one can't be the marker: at J2 it would force H1 (the last open cell in region 3), then I4, I7 and I8 cleared (together, two regions have open cells only in columns K and I), then G5 (the last open cell in region 2), and region 9 would be left with no open cell for its marker.
Following that, marking K2 would force H1 (the last open cell in region 3), 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), then I4, I7, I8, and 2 more cleared (together, two regions have open cells only in columns I and J), and the chain continues, and then region 9 would be left with no open cell for its marker. So K2 can't be the marker there.
First, marking F3 would force H1 (the last open cell in row 1), and then row 2 would be left with no open cell for its marker. So F3 can't be the marker there.
Next, if G3 were the marker, it would force I2 (the last open cell in row 2), then J4 (the last open cell in region 4), then K6 (the last open cell in region 6), 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.
Then, that one can't be the marker: place it at H3, and row 1 would be left with no open cell for its marker.
After that, marking I3 would force H1 (the last open cell in region 3), then G5 (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), and the chain continues, and then region 10 would be left with no open cell for its marker. So I3 can't be the marker there.
Now, if J3 were the marker, it would force H1 (the last open cell in region 3), then G5 (the last open cell in region 2), then A6, B6, C6, and 2 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 region 10 would end up with no open cell for its marker, which can't happen. So it can't go there.
From there, marking D4 would force K3 (the last open cell in row 3), then I2 cleared (every open cell left in this region sits in row 2, so its marker has to land there), then H1 (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 D4 can't be the marker there.
Following that, if E4 were the marker, it would force K3 (the last open cell in row 3), then I2 cleared (every open cell left in this region sits in row 2, so its marker has to land there), then H1 (the last open cell in region 3), 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.
First, that one can't be the marker: at F4 it would force H1 (the last open cell in row 1), and region 2 would be left with no open cell for its marker.
Next, marking G4 would force I2 (the last open cell in row 2), then K3 (the last open cell in region 4), then J6 (the last open cell in region 6), 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.
Then, if A3 were the marker, it would force I2 cleared (every open cell left in this region sits in row 2, so its marker has to land there), then H1 (the last open cell in region 3), then K4, D5 and K5 cleared (together, two regions have open cells only in rows 4 and 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, if C2 were the marker, it would force H1 (the last open cell in region 3), then K3 (the last open cell in row 3), then A6, B6, D6, and 3 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 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 B3 it would force I2 cleared (every open cell left in this region sits in row 2, so its marker has to land there), then H1 (the last open cell in region 3), then K4, A5, D5, and 1 more cleared (together, two regions have open cells only in rows 4 and 5), and the chain continues, and region 10 would be left with no open cell for its marker.
From there, that one can't be the marker: at D2 it would force H1 (the last open cell in region 3), then K3 (the last open cell in row 3), then A6, B6, C6, and 3 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 region 9 would be left with no open cell for its marker.
Following that, marking C3 would force I2 cleared (every open cell left in this region sits in row 2, so its marker has to land there), then H1 (the last open cell in region 3), then K4, A5, D5, and 1 more cleared (together, two regions have open cells only in rows 4 and 5), 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.
First, marking E2 would force H1 (the last open cell in region 3), then K3 (the last open cell in row 3), then A6, B6, C6, and 3 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 9 would be left with no open cell for its marker. So E2 can't be the marker there.
Next, if D3 were the marker, it would force I2 cleared (every open cell left in this region sits in row 2, so its marker has to land there), then H1 (the last open cell in region 3), then K4, A5 and K5 cleared (together, two regions have open cells only in rows 4 and 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.
Then, if H1 were the marker, it would force E3 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 K3 (the last open cell in row 3), 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 region 9 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 I2. Region has exactly one open cell left. Placing here clears the rest of row 2, column I, and its region.
Now, place a marker at E3. Region has exactly one open cell left. Placing here clears the rest of row 3, column E, and its region.
From there, 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.
Following that, place a marker at K6. Region has exactly one open cell left. Placing here clears the rest of row 6, column K, and its region.
First, place a marker at H11. Region has exactly one open cell left. Placing here clears the rest of row 11, column H, and its region. Placing here also clears its 1 touching neighbour.
Next, place a marker at F1. Row 1 has exactly one open cell left. Placing here clears the rest of row 1, column F, and its region.
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 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.
Now, every remaining candidate in this region touches B8, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
From there, marking A7 would force G9 (the last open cell in region 9), then D10 (the last open cell in region 11), and then row 8 would be left with no open cell for its marker. So A7 can't be the marker there.
Following that, that one can't be the marker: place it at C7, and row 8 would be left with no open cell for its marker.
First, marking D7 would force A8 (the last open cell in region 7), and then row 10 would be left with no open cell for its marker. So D7 can't be the marker there.
Next, marking B5 would force G7 (the last open cell in row 7), then D8 cleared (every open cell left in this region sits in row 8, so its marker has to land there), then D10 cleared (every remaining candidate in this region touches D10), and the chain continues, and then column C would be left with no open cell for its marker. So B5 can't be the marker there.
Then, place a marker at C5. Region has exactly one open cell left. Placing here clears the rest of row 5, column C, and its region.
After that, marking G7 would force A8 (the last open cell in region 7), then D10 (the last open cell in region 11), and then row 9 would be left with no open cell for its marker. So G7 can't be the marker there.
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 B7. Row 7 has exactly one open cell left. Placing here clears the rest of row 7, column B, and its region.
Following that, place a marker at D8. Region has exactly one open cell left. Placing here clears the rest of row 8, column D, and its region.
Finally, place a marker at A10. Region has exactly one open cell left.

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