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.
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.
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?
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.
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.