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Unify the two branches in the consistency proof check #149
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As I'm re-implementing this logic, I found a panic in the original version of my code when
n = 0due to an underflow. The fact that[start, end)is a subtree of the tree of sizenimplies that0 <= start < end <= nso it must be the case thatend > 0andn > 0, but maybe we should add an explicit note about that?There was a problem hiding this comment.
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Hmm... how about we just make the function check the preconditions? It's fallible anyway, and makes it consistent with the inclusion proof version. Arguably it's redundant with the "type signature" but people will be sloppy.