How to solve this 8×8 logic puzzle

8×8 · severe · 30 steps

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This 8×8 board rates as severe (difficulty score 1464). Solving it from scratch takes 38 logical steps, using 7 techniques: Single cell, Row/column exclusion, Adjacency clear, Region confinement, Line confinement, Shape squeeze, Forced chain. Every step below is forced: nothing here is a guess.

First, every open cell left in this region sits in row 8, so its marker has to land there; that clears every other open cell in the row.
Next, every remaining candidate in this region touches D6, 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 C7 and D7, so whichever candidate ends up marked will clear them, and they can be ruled out now regardless of which one wins.
After that, 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.
Now, every remaining candidate in this region touches C5, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
From there, every remaining candidate in this region touches F6 and F7, so whichever candidate ends up marked will clear them, and they can be ruled out now regardless of which one wins.
Following that, marking A1 would force H4 (the last open cell in region 4), then G2 (the last open cell in region 2), then F5 (the last open cell in region 5), and the chain continues, and then row 8 would be left with no open cell for its marker. So A1 can't be the marker there.
First, if B1 were the marker, it would force H4 (the last open cell in region 4), and row 3 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 C1 it would force D8 (the last open cell in region 8), then B6 (the last open cell in region 6), and column E would be left with no open cell for its marker.
Then, 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.
After that, marking D1 would force C8 (the last open cell in region 8), then B6 (the last open cell in region 6), then A4 (the last open cell in region 3), and the chain continues, and then column H 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 E7 (the last open cell in region 7), then C8 (the last open cell in region 8), then D5 (the last open cell in region 6), 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.
From there, every remaining candidate in this region touches F2, so whichever candidate ends up marked will clear it, and it can be ruled out now regardless of which one wins.
Following that, marking B2 would immediately leave column A with no open cell for its marker, so it can't be the marker there.
First, if C2 were the marker, it would force D8 (the last open cell in region 8), then B6 (the last open cell in region 6), and column E 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 F1 (the last open cell in region 2), then G7 (the last open cell in column G), then E6 (the last open cell in region 7), and then region 6 would be left with no open cell for its marker. So H2 can't be the marker there.
Then, 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.
After that, that one can't be the marker: place it at B3, and column A would be left with no open cell for its marker.
Now, place a marker at H3. Row 3 has exactly one open cell left. Placing here clears the rest of row 3, column H, and its region.
From there, that one can't be the marker: at F1 it would force G7 (the last open cell in column G), then E6 (the last open cell in region 7), and region 6 would be left with no open cell for its marker.
Following that, place a marker at G1. Region has exactly one open cell left. Placing here clears the rest of row 1, column G, and its region.
First, every open cell in column F belongs to the same region, so that region's marker has to be somewhere in this column; that clears every other open cell in the region.
Next, if B4 were the marker, it would force F5 (the last open cell in region 5), then C6 (the last open cell in region 6), then E7 (the last open cell in region 7), and row 8 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 A2 it would force F4 (the last open cell in row 4), then D5 (the last open cell in region 6), and row 6 would be left with no open cell for its marker.
After that, place a marker at D2. Region has exactly one open cell left. Placing here clears the rest of row 2, column D, and its region.
Now, place a marker at C8. Region has exactly one open cell left. Placing here clears the rest of row 8, column C, and its region. Placing here also clears its 1 touching neighbour.
From there, place a marker at B6. Region has exactly one open cell left. Placing here clears the rest of row 6, column B, and its region. Placing here also clears its 2 touching neighbours.
Following that, place a marker at A4. Region has exactly one open cell left. Placing here clears the rest of row 4, column A, and its region.
First, place a marker at F5. Region has exactly one open cell left.
Finally, place a marker at E7. Region has exactly one open cell left.

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