4 ms·
It very much is. sqrt(x) is just x^(1/2) which is x^(2^-1). Dirac's solution is using iterated square root of 2, effectively generating a sequence similar to wh
by karparov 2y ago
It very much is. sqrt(x) is just x^(1/2) which is x^(2^-1). Dirac's solution is using iterated square root of 2, effectively generating a sequence similar to what's used in this post.
- yorwba 2y agoOkay, but iterating square roots like √√2 = (2^(2^-1))^(2^-1) recurses into the base, whereas the equivalent iterated log is 2^(2^-1 × 2^-1) = 2^(2^-2) = +2^(+2^(-2^(+2^0))) with the bit representation [1 1 0 0 1 0 0 ...], i.e. it recurses into the exponent.
- DannyBee 2y agoSo uh, how is that not a variation, exactly? Dirac's solution was also arbitrarily restricted by the problem definition. Do you believe that if you asked someone familiar with the solution to come up with a bit efficient variant they would not have trivially come up with the encoding in this post and called it a variation? I don't, for a second, believe that.