Another way of saying this is that G4, G5, H5, and H6 only have 3 options between them. The reason this is now broken is because the 6 in A4, the 1 in G1, the 3 in F5, and the 4 in box 2’s second column together imply 1,3,4, and 6 to be confined to H4, J4, and J6, which is only 3 squares. I say the computer should be able to see this because none of these implications depend on each other, so I believe so far this would be classified as chaining, and not bifurcation. That’s because it implies F2 is a 1, which implies F5 is a 3 as well as that E1 is a 6, which implies B2 is a 6, which implies A4 is a 6, which implies there is a 4 in box 2’s second column. After letting the solver run its course, one deduction I think the computer should be able to see is that G1 cannot be a 1.
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