User blog:Wythagoras/Upper bound for Hydra(4)

Here I'll present a upper bound on Hydra(4) using King's method.

Let n be the steps that are already done.

The hydra inside root pair just adds a step, corresponding to f 0 (n).

The () inside root pair expands to n 's, corresponding to f 1 (n).

The following table makes more comparisons like that. To make the right bound, we must add a number to get a correct bound.

Look for example to the evaluation of hydra(3).

((()))

(())

(()()())

This would result in an upper bound of f 1 (f 1 (f 1 (2))) = 16, which isn't correct, as hydra(3) = 37.

If we add 2 to every step we do, it will be right:

((()))

(())

(()()())

(()()), there are four not two.

(()())

(()())

(()())

(()())

(()()), adding 1 here would be fine, but adding two is somewhat more secure. We must make sure that it is an upper bound.

We have this modified fast-growing hierarchy:

\(f'_0(n) = n+1\)

\(f'_{\alpha+1}(n) = f'_{\alpha}^{n+2}(n+2)\)

\(f'_{\alpha}(n) = f'_{\alpha[n+2]}(n+2)\)

For ordinals \(\alpha≥1\), the inequality \(f_{\alpha}(n+2) < f'_{\alpha}(n) < f_{\alpha}(n+3)\) holds.

(((())))

((()))

((()()()))

((()()))

\(\text{Hydra}(4) < f'_{\omega2+4}(4) < f_{\omega2+4}(7)\)