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I'm not pruning any board symmetries. So, yes, the 4x4 position with only one Black stone adjacent to a corner would be counted 8 times (with the Black stone at A2 or A3 or B1 or B4 or C1 or C4 or D2 or D3). The only symmetry that the algorithm takes advantage of is color symmetry; but even there nothing is pruned; the count for a color-normalized state is simply doubled.

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I apologize. I had skimmed the paper, but not read it in depth, and it was clear that you had considered symmetries (such as Figure 4 in the paper); I assumed, then, that your algorithm would prune symmetries and simply account for them by multiplying the results in each class by the number of symmetries. I hadn't yet finished reading the paper, however.

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