3 ms·

Isn't this just a variant of Dirac's solution for representing any number using sqrt, log, and the number 2?

by DannyBee 2y ago

yorwba 2y ago
It is not. (There are no square roots, for instance.)
- 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 ago
    Okay, 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 ago
      So 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.
snarkconjecture 2y ago
Not really. Dirac's trick works entirely at a depth of two logs, using sqrt like unary to increment the number. It requires O(n) symbols to represent the number n, i.e. O(2^n) symbols to represent n bits of precision. This thing has arbitrary nesting depth of logs (or exps), and can represent a number to n bits of precision in O(n) symbols.
- DannyBee 2y ago
  Sure, but that was because that was an arbitrary restriction on the problem dirac solved. Still seems a fairly simple variation once you remove the arbitrary restriction. To the point: I don't believe for a second that anyone familiar with his solution, asked to make it more bit efficient, would not have come up with this. Nor do I believe they would call it anything other than a variation. That doesn't make it less cool but I don't think it's like amazingly novel.

Isn't this just a variant of Dirac's solution for representing any number using sqrt, log, and the number 2?

yorwba 2y ago

It is not. (There are no square roots, for instance.)

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 ago
  Okay, 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 ago
    So 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.

snarkconjecture 2y ago

Not really. Dirac's trick works entirely at a depth of two logs, using sqrt like unary to increment the number. It requires O(n) symbols to represent the number n, i.e. O(2^n) symbols to represent n bits of precision. This thing has arbitrary nesting depth of logs (or exps), and can represent a number to n bits of precision in O(n) symbols.

DannyBee 2y ago
Sure, but that was because that was an arbitrary restriction on the problem dirac solved. Still seems a fairly simple variation once you remove the arbitrary restriction. To the point: I don't believe for a second that anyone familiar with his solution, asked to make it more bit efficient, would not have come up with this. Nor do I believe they would call it anything other than a variation. That doesn't make it less cool but I don't think it's like amazingly novel.