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For the quite similar puzzle Atomix, it also seems like the branching factor would be much higher for backward search because upper bounds are weaker, but you can show that on average the branching factor is actually the same [1]. I wonder if the same argument would work here.

[1] http://hueffner.de/falk/hueffner-studienarbeit-atomix.pdf Section 5.5

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I wonder if it would make sense to do backward search, even if the forward and backward branching factors are very different. For example if the branching factor for forward search is 10 vs. 100 for backwards search, wouldn’t it make sense to do one step of backward search for every two steps of forward search? Or more generally log(b)/log(f) backward search steps for every forward search step, where the forward branching factor is f and backward branching factor is b?

This is all based on spontaneous intuitive ideas of mine and very superficial reasoning (and probably not even new).

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