How to solve this 8×8 logic puzzle

8×8 · severe · 20 steps

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This 8×8 board rates as severe (difficulty score 1181). Solving it from scratch takes 30 logical steps, using 7 techniques: Single cell, Row/column exclusion, Adjacency clear, Region confinement, Paired regions, Shape squeeze, Forced chain. Every step below is forced: nothing here is a guess.

First, place a marker at B8. Region has exactly one open cell left. Placing here clears the rest of row 8, column B, and its region. Placing here also clears its 2 touching neighbours.
Next, 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.
Then, 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.
After that, that one can't be the marker: at C1 it would force H2 (the last open cell in row 2), and row 3 would be left with no open cell for its marker.
Now, marking D1 would force H2 (the last open cell in row 2), and then row 3 would be left with no open cell for its marker. So D1 can't be the marker there.
From there, that one can't be the marker: at F1 it would force C2 (the last open cell in row 2), then E3 (the last open cell in region 1), and region 4 would be left with no open cell for its marker.
Following that, 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.
First, every remaining candidate in this region touches E6 and E7, so whichever candidate ends up marked will clear them, and they can be ruled out now regardless of which one wins.
Next, if C2 were the marker, it would force D7 (the last open cell in column D), then E1 (the last open cell in region 1), then G5 (the last open cell in column G), and row 4 would end up with no open cell for its marker, which can't happen. So it can't go there.
Then, marking G1 would force E2 (the last open cell in row 2), and then row 3 would be left with no open cell for its marker. So G1 can't be the marker there.
After that, 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.
Now, marking E2 would force H1 (the last open cell in row 1), and then row 3 would be left with no open cell for its marker. So E2 can't be the marker there.
From there, if H1 were the marker, it would force F2 (the last open cell in row 2), and column E would end up with no open cell for its marker, which can't happen. So it can't go there.
Following that, place a marker at E1. Row 1 has exactly one open cell left. Placing here clears the rest of row 1, column E, and its region. Placing here also clears its 1 touching neighbour.
First, place a marker at H2. Row 2 has exactly one open cell left. Placing here clears the rest of row 2, column H, and its region.
Next, place a marker at F3. Row 3 has exactly one open cell left. Placing here clears the rest of row 3, column F, and its region. Placing here also clears its 1 touching neighbour.
Then, 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.
After that, place a marker at G5. Region has exactly one open cell left. Placing here clears the rest of row 5, column G, and its region.
Now, place a marker at C4. Region has exactly one open cell left.
Finally, place a marker at A6. Region has exactly one open cell left.

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