How to solve this 8×8 logic puzzle

8×8 · severe · 32 steps

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This 8×8 board rates as severe (difficulty score 1617). Solving it from scratch takes 41 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 G2, 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 B6, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
Then, 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.
After that, marking A1 would force F2 (the last open cell in region 3), and then region 2 would be left with no open cell for its marker. So A1 can't be the marker there.
Now, if B1 were the marker, it would force F2 (the last open cell in region 3), then C4 (the last open cell in region 1), and region 5 would end up with no open cell for its marker, which can't happen. So it can't go there.
From there, every remaining candidate in this region touches D2, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
Following that, every remaining candidate in this region touches B3, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
First, that one can't be the marker: at C1 it would force F2 (the last open cell in region 3), and region 7 would be left with no open cell for its marker.
Next, if E1 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.
Then, that one can't be the marker: at F1 it would force E3 (the last open cell in region 2), then D5 (the last open cell in region 5), then A6 (the last open cell in region 6), and region 7 would be left with no open cell for its marker.
After that, if H1 were the marker, it would force F2 (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.
Now, that one can't be the marker: at A2 it would force G1 (the last open cell in region 3), then E3 (the last open cell in region 2), then D5 (the last open cell in region 5), and region 6 would be left with no open cell for its marker.
From there, marking B2 would force G1 (the last open cell in region 3), then E3 (the last open cell in region 2), then D5 (the last open cell in region 5), and the chain continues, and then region 7 would be left with no open cell for its marker. So B2 can't be the marker there.
Following that, if C2 were the marker, it would force E3 (the last open cell in region 2), then G1 (the last open cell in region 3), then D5 (the last open cell in region 5), and the chain continues, and column B would end up with no open cell for its marker, which can't happen. So it can't go there.
First, marking H2 would immediately leave region 3 with no open cell for its marker, so it can't be the marker there.
Next, marking C3 would force A4 (the last open cell in region 1), and then region 6 would be left with no open cell for its marker. So C3 can't be the marker there.
Then, if D3 were the marker, region 2 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: place it at E3, and row 2 would be left with no open cell for its marker.
Now, marking F3 would immediately leave row 2 with no open cell for its marker, so it can't be the marker there.
From there, every remaining candidate in this region touches E5, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
Following that, if G3 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.
First, that one can't be the marker: at H3 it would force D5 (the last open cell in region 5), then E2 (the last open cell in region 2), then G1 (the last open cell in region 3), and the chain continues, and region 7 would be left with no open cell for its marker.
Next, place a marker at A3. Row 3 has exactly one open cell left. Placing here clears the rest of row 3, column A, and its region.
Then, place a marker at B7. Region has exactly one open cell left. Placing here clears the rest of row 7, column B, and its region. Placing here also clears its 2 touching neighbours.
After that, 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. Placing here also clears its 2 touching neighbours.
Now, 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.
From there, every open cell left in this region sits in row 6, so its marker has to land there; that clears every other open cell in the row.
Following that, place a marker at H8. Column 8 has exactly one open cell left. Placing here clears the rest of row 8, column H, and its region.
First, place a marker at D1. Column 4 has exactly one open cell left. Placing here clears the rest of row 1, column D, and its region.
Next, place a marker at F2. Region has exactly one open cell left. Placing here clears the rest of row 2, column F, and its region.
Then, place a marker at E4. Region has exactly one open cell left. Placing here clears the rest of row 4, column E, and its region.
Finally, place a marker at G6. Region has exactly one open cell left.

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