An Expert 10×10 Puzzle Solved with Forced Chain in 41 Steps

10×10 · expert · 41 steps

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This 10×10 board rates as expert (difficulty score 2011). Solving it from scratch takes 41 logical steps, using 6 techniques: Single cell, Row/column exclusion, Adjacency clear, Region confinement, Paired regions, Forced chain. Every step below is forced: nothing here is a guess.

First, marking A1 would force B9, D9, E9, and 7 more cleared (together, two regions have open cells only in rows 9 and 10), then B8, C8, D8, and 3 more cleared (every open cell left in this region sits in row 8, so its marker has to land there), then B7, C7, H7, and 2 more cleared (every open cell left in this region sits in row 7, so its marker has to land there), and the chain continues, and then column D would be left with no open cell for its marker. So A1 can't be the marker there.
Next, marking G1 would force H5, H6, H7, and 6 more cleared (together, two regions have open cells only in columns H and F), then E2, E3, E4, and 4 more cleared (every open cell left in this region sits in column E, so its marker has to land there), then A9, B9, C9, and 9 more cleared (together, two regions have open cells only in rows 9 and 10), and the chain continues, and then row 3 would be left with no open cell for its marker. So G1 can't be the marker there.
Then, if H1 were the marker, it would force G5, G6, G7, and 6 more cleared (together, two regions have open cells only in columns G and F), then E2, E3, E4, and 4 more cleared (every open cell left in this region sits in column E, so its marker has to land there), then A9, B9, C9, and 9 more cleared (together, two regions have open cells only in rows 9 and 10), 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.
After that, if F2 were the marker, it would force I3 cleared (every open cell in row 1 belongs to the same region, so that region's marker has to be somewhere in this row), then H5, H6, H7, and 6 more cleared (together, two regions have open cells only in columns H and G), then E6, E7 and E8 cleared (every open cell left in this region sits in column E, 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.
Now, marking H2 would force B3, C3, D3, and 6 more cleared (every open cell in row 1 belongs to the same region, so that region's marker has to be somewhere in this row), then F7, F8, F9, and 4 more cleared (together, two regions have open cells only in columns F and G), then E1, E6, E7, and 1 more cleared (every open cell left in this region sits in column E, so its marker has to land there), and the chain continues, and then row 3 would be left with no open cell for its marker. So H2 can't be the marker there.
From there, marking F3 would force H4 (the last open cell in region 4), then G6, G7 and G8 cleared (every open cell left in this region sits in column G, so its marker has to land there), then E6, E7 and E8 cleared (every open cell left in this region sits in column E, so its marker has to land there), and the chain continues, and then row 6 would be left with no open cell for its marker. So F3 can't be the marker there.
Following that, marking A4 would force B9, D9, E9, and 7 more cleared (together, two regions have open cells only in rows 9 and 10), then B8, C8, D8, and 3 more cleared (every open cell left in this region sits in row 8, so its marker has to land there), then H7, I7 and J7 cleared (every open cell left in this region sits in row 7, so its marker has to land there), 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.
First, that one can't be the marker: at A5 it would force B9, D9, E9, and 7 more cleared (together, two regions have open cells only in rows 9 and 10), then B8, C8, D8, and 3 more cleared (every open cell left in this region sits in row 8, so its marker has to land there), then H7, I7 and J7 cleared (every open cell left in this region sits in row 7, so its marker has to land there), and the chain continues, and row 8 would be left with no open cell for its marker.
Next, that one can't be the marker: at G5 it would force H3 (the last open cell in region 4), then F10 (the last open cell in region 10), and region 9 would be left with no open cell for its marker.
Then, marking H5 would force G7, G8, F1, and 3 more cleared (together, two regions have open cells only in columns G and F), then E1, E2, E3, and 4 more cleared (every open cell left in this region sits in column E, so its marker has to land there), then A9, B9, C9, and 9 more cleared (together, two regions have open cells only in rows 9 and 10), and the chain continues, and then row 9 would be left with no open cell for its marker. So H5 can't be the marker there.
After that, that one can't be the marker: at A2 it would force B9, D9, E9, and 7 more cleared (together, two regions have open cells only in rows 9 and 10), then B8, C8, D8, and 3 more cleared (every open cell left in this region sits in row 8, so its marker has to land there), then H7, I7 and J7 cleared (every open cell left in this region sits in row 7, so its marker has to land there), and the chain continues, and row 8 would be left with no open cell for its marker.
Now, if A3 were the marker, it would force B9, D9, E9, and 7 more cleared (together, two regions have open cells only in rows 9 and 10), then B8, C8, D8, and 3 more cleared (every open cell left in this region sits in row 8, so its marker has to land there), then H7, I7 and J7 cleared (every open cell left in this region sits in row 7, so its marker has to land there), 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, if I2 were the marker, it would force B3, C3, D3, and 5 more cleared (every open cell in row 1 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), then H8 and H9 cleared (together, two regions have open cells only in columns J and H), 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 C3 would force J2 cleared (every open cell in row 1 belongs to the same region, so that region's marker has to be somewhere in this row), then G2 (the last open cell in row 2), then I5, J5, H6, and 1 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), and the chain continues, and then row 4 would be left with no open cell for its marker. So C3 can't be the marker there.
First, if B4 were the marker, it would force J2 and I3 cleared (every open cell in row 1 belongs to the same region, so that region's marker has to be somewhere in this row), then G2 (the last open cell in row 2), then J3 (the last open cell in row 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.
Next, if A6 were the marker, it would force B9, D9, E9, and 7 more cleared (together, two regions have open cells only in rows 9 and 10), then B8, C8, D8, and 3 more cleared (every open cell left in this region sits in row 8, so its marker has to land there), then D7, E7, F7, and 1 more cleared (every open cell left in this region sits in row 7, so its marker has to land there), and region 7 would end up with no open cell for its marker, which can't happen. So it can't go there.
Then, marking F6 would force I7, I8, I9, and 5 more cleared (together, two regions have open cells only in columns I and J), then H8 and H9 cleared (every open cell left in this region sits in column H, so its marker has to land there), then G8 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 7 would be left with no open cell for its marker. So F6 can't be the marker there.
After that, if G6 were the marker, it would force H8, H10, F1, and 2 more cleared (together, two regions have open cells only in columns H and F), then E1, E2, E3, and 4 more cleared (every open cell left in this region sits in column E, so its marker has to land there), then B2, D4, A7, and 2 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 region 6 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 I1 it would force H3, H4, H8, and 1 more cleared (together, two regions have open cells only in columns J and H), then G7, G8, F7, and 2 more cleared (together, two regions have open cells only in columns G and F), then E2, E3, E4, and 4 more cleared (every open cell left in this region sits in column E, so its marker has to land there), and the chain continues, and column H would be left with no open cell for its marker.
From there, marking J1 would force H3, H4, H8, and 1 more cleared (together, two regions have open cells only in columns I and H), then G7, G8, F7, and 2 more cleared (together, two regions have open cells only in columns G and F), then E2, E3, E4, and 4 more cleared (every open cell left in this region sits in column E, so its marker has to land there), and the chain continues, and then region 6 would be left with no open cell for its marker. So J1 can't be the marker there.
Following that, that one can't be the marker: at D2 it would force F1 (the last open cell in row 1), then G7, G8, H6, and 3 more cleared (together, two regions have open cells only in columns G and H), then E6, E7 and E8 cleared (every open cell left in this region sits in column E, so its marker has to land there), and the chain continues, and region 7 would be left with no open cell for its marker.
First, marking E2 would immediately leave row 1 with no open cell for its marker, so it can't be the marker there.
Next, that one can't be the marker: at G2 it would force I3 (the last open cell in region 2), then C4, E4 and E5 cleared (every open cell in row 1 belongs to the same region, so that region's marker has to be somewhere in this row), then D4 (the last open cell in row 4), and the chain continues, and row 6 would be left with no open cell for its marker.
Then, if B1 were the marker, it would force J2 (the last open cell in row 2), then F4, G4, H4, and 1 more cleared (every open cell in row 3 belongs to the same region, so that region's marker has to be somewhere in this row), then H3, H8 and H9 cleared (together, two regions have open cells only in columns I and H), 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 J2 (the last open cell in row 2), then F4, G4, H4, and 1 more cleared (every open cell in row 3 belongs to the same region, so that region's marker has to be somewhere in this row), then H3, H8 and H9 cleared (together, two regions have open cells only in columns I and H), 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 J2 it would force B3, D3, E3, and 3 more cleared (every open cell in row 1 belongs to the same region, so that region's marker has to be somewhere in this row), then F4, G4, H4, and 1 more cleared (every open cell in row 3 belongs to the same region, so that region's marker has to be somewhere in this row), then H3, H8 and H9 cleared (together, two regions have open cells only in columns I and H), and the chain continues, and region 9 would be left with no open cell for its marker.
From there, marking D1 would force I3 (the last open cell in region 2), then B2 (the last open cell in row 2), then F5 cleared (every open cell in row 4 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 D1 can't be the marker there.
Following that, if E1 were the marker, it would force I3 (the last open cell in region 2), then B2 (the last open cell in row 2), then F5 cleared (every open cell in row 4 belongs to the same region, so that region's marker has to be somewhere in this row), 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.
First, 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.
Next, together, two regions have open cells only in columns G and H; 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.
Then, every open cell left in this region sits in column E, 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 9 and 10; 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 A8. Region has exactly one open cell left. Placing here clears the rest of row 8, column A, and its region. Placing here also clears its 1 touching neighbour.
From there, place a marker at D7. Region has exactly one open cell left. Placing here clears the rest of row 7, column D, and its region. Placing here also clears its 1 touching neighbour.
Following that, place a marker at J6. Region 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.
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 G3. Region has exactly one open cell left. Placing here clears the rest of row 3, column G, and its region.
Then, place a marker at C2. Region has exactly one open cell left. Placing here clears the rest of row 2, column C, and its region.
After that, place a marker at B5. Region has exactly one open cell left.
Now, 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.
Finally, place a marker at E10. Region has exactly one open cell left.

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