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I think this works perfectly if your list has 2^n elements. Otherwise, you have to resort to multiplying by imprecise fractions.
by umdiff 8y ago
I think this works perfectly if your list has 2^n elements. Otherwise, you have to resort to multiplying by imprecise fractions.
- _bxg1 8y agoYou would only need one special case: if a list (top level or not) has 2n+1 numbers, weight the last one differently.