So yesterday I was looking around searching for proof of cut-elimination theorem, and I've accidentally came across this paper. This gives exact values of Hydra function, in terms of Hardy hierarchy. I'll replicate all the details here, with all details filled.

Let's fix some hydra H. Let the number \(n\) appearent in hydra game be called *index*. We will, to every node of H, assign an ordinal *value* to it: all leaf nodes get 0 associated. If a node N which isn't leaf has nodes with ordinals \(\alpha_1,\alpha_2,...,\alpha_n\) assigned to them, in that order, then N will get \(\omega^{\alpha_1}+\omega^{\alpha_2}+...+\omega^{\alpha_n}\). The value of hydra H is value of its root.

We also call a node *nice* if it's either a leaf, or all of nodes above it are nice and their values are arranged in decreasing order, from left to right. We call H nice if it's root (equivalently, all its nodes) are nice. It's easily seen that hydra is nice if it's value, when computed using algorithm above and without using any ordinal arithmetic, is already in CNF. We will only care about trees which are nice, because of the following:

**Lemma 1** If H is nice, and we cut it's rightmost head, we get a nice hydra again.

- Suppose we cut a leaf sticking from the node with value \(\alpha+1\), which in turn sticks out of node with value \(\beta+\omega^{\alpha+1}\). If we cut this head, we will get a subhydra with value \(\alpha\), which, from CNF interpretation, is easily seen to be nice, and we get \(n\) copies of above subhydra. So the other node now changes it's value from \(\beta+\omega^{\alpha+1}\) to \(\beta+\omega^\alpha n\), which, again, via CNF, is nice.

You might have noticed, by the above, that cutting the hydra looks a lot like taking fundamental sequence of the ordinal, by using standard fundamental sequences defined as follows: \((\beta+\omega^{\gamma+1})[n]=\beta+\omega^\gamma n\) and, for \(\gamma\) limit, \((\beta+\omega^\gamma)[n]=\beta+\omega^{\gamma[n]}\).

This, maybe unsurprisingly, is exactly the case!

**Lemma 2** Cutting the rightmost head of hydra with limit value \(\alpha\) leaves us with hydra with value \(\alpha[n]\).

- Note that \(\alpha\) being limit means exactly that there are no leaves connected directly to the root. The proof of lemma 1 actually shows that this works in the case that leaf is in the distance 2 of the root. But if we, for a moment, ignore everything in H except for a subhydra rooted at the node 2 below the leaf we are cutting, we can see that the value of this subhydra actually follows the rule. So we can proceed by noticing that we can express the ordinal as \(\beta_1+\omega^{\beta_2+\omega^{...^{\beta_k+\omega^{\gamma+1}}}}\), and \((\beta_1+\omega^{\beta_2+\omega^{...^{\beta_k+\omega^{\gamma+1}}}})[n]=\beta_1+\omega^{(\beta_2+\omega^{...^{\beta_k+\omega^{\gamma+1}}})[n]}=...=\beta_1+\omega^{\beta_2+\omega^{...^{(\beta_k+\omega^{\gamma+1})[n]}}}=\beta_1+\omega^{\beta_2+\omega^{...^{\beta_k+\omega^\gamma n}}}\), which is exactly the ordinal we get from cutting the hydra.

Thanks to this, we can almost emulate the Hardy hierarchy using hydras. There is a slight problem with boundary conditions though. Either way, we can define a following hierarchy: \(L_0(n)=0, L_{\alpha+1}(n)=L_\alpha(n+1)+1, L_\alpha(n)=L_{\alpha[n]}(n+1)+1\), where last rule applies ofr limit ordinals.

**Lemma 3** \(L_\alpha(n)\) is the number of steps to kill the hydra with value \(\alpha\) when starting with index equal to \(n\).

- If \(\alpha=0\), then game already ended and \(0=L_0(n)\) steps were made. Both other rules mean the same - ordinal changes according to what is the next hydra we will get (in the first case it's cutting leaf next to root, in the other some higher node). We then get \(n+1\) in the parentheses because index has increased, and there is \(+1\) at the end because additional step was made.

Now that we can concisely express how many steps we need to kill hydra, we can relate this hierarchy to Hardy hierarchy, which, in my opinion, is the most beautiful result shown in the paper (this equality is due to Kirby and Paris though, I believe). Recall Hardy hierarchy: \(H_0(n)=n,H_{\alpha+1}(n)=H_\alpha(n+1), H_\alpha(n)=H_{\alpha[n]}(n+1)\) (this actually is a slightly stronger version of HH, but bigger numbers do no harm).

**Lemma 4** \(H_\alpha(n)=H_{L_\alpha(n)}(n)=n+L_\alpha(n)\)

- First equality is proven by transfinite recursion: for 0 we have \(0=L_0(n)\), so it's clear that \(H_0(n)=H_{L_0(n)}(n)\). For \(\alpha+1\) we take \(H_{\alpha+1}(n)=H_\alpha(n+1)=H_{L_\alpha(n+1)}(n+1)=H_{L_\alpha(n+1)+1}(n)=H_{L_{\alpha+1}(n)}(n)\) - first equality is from definition of H, second from induction hypothesis, third from the definition of H for successors and fourth from definition of L. For \(\alpha\) limit we have \(H_\alpha(n)=H_{\alpha[n]}(n+1)=H_{L_{\alpha[n]}(n+1)}(n+1)=H_{L_{\alpha[n]}(n+1)+1}(n)=H_{L_\alpha(n)}(n)\), where equalities are the same as above. For second equality, this is trivial from the fact that for finite \(m\) we have \(H_m(n)=n+m\), which can be shown with simple finite induction, and that L_\alpha(n) is finite.

So there we go, we have expressed hydra game in terms of (slightly modified) HH. Now we can make some bounds. Let \(\alpha=\omega^{...^\omega}\) with k-2 \(\omega\)s. With induction we could show \(H_\alpha(2)>k+1\), but this would get quite messy, so I'll avoid it here. Now we can have \(1+L_{\omega^{\omega^\alpha}}(1)=H_{\omega^{\omega^alpha}}(1)=H_{\omega^\alpha}(2)\) (one \(\omega\) on top becomes 1, and it cancels, next inequality could also be proven with induction) \(>H_{\alpha+\alpha}(2)=H_\alpha(H_\alpha(2))>H_\alpha(k+1)\geq H_\alpha(k)+1\), so we get \(L_{\omega^{\omega^alpha}}(1)>H_\alpha(k)\). But term on the left is exactly the time we need to destroy the hydra which is composed of a path with k+1 edges, so we get an actual bound \(Hydra(k)>H_{\omega\uparrow\uparrow k-3}(k-1)\).

We can refine this bound in many ways, but this isn't a thing I'm going to bother myself with.