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by pfdietz 1048 days ago
Addition can be performed with circuits of depth O(log n), where n is the number of bits, and with a little more work also of size O(n).
1 comments

I'm confused. Since O(log n) is faster than O(n) why would O(n) require a little more work? Did I misread something or did you miswrite?
I'm talking about constructing boolean circuits. The usual parallel prefix algorithm yields a circuit of depth O(log n) with O(n log n) gates. Using some trickery, however, this can be reduced to O(n) gates, with a constant factor increase in depth.

https://en.wikipedia.org/wiki/Prefix_sum https://en.wikipedia.org/wiki/Adder_(electronics)

Constant depth requires unbounded fan-in, which may not be available in practice
Yes, I was talking about using gates of bounded fan-in.