We can think of this as a game of "omega-nim;" to more precise since the game you are describing is impartial, operating under the normal play convention, and finite we have that the Sprague-Grundy Theorem applies. In other words, to every "hydra-ordinal" there is an "omega-nimber."

Already this suggests thinking of the plus signs in the hydra as being something roughly analogous to the heaps in Nim. Let me make this precise.

**Definition** The set of **hydras** (or hydrae?), $\mathcal{H}$, is defined recursively as

- $0 \in \mathcal{H}$
- $ n \in \mathbb{N} \implies \omega^n \in \mathcal{H}$
- $ \kappa_0,...,\kappa_{n-1} \in \mathcal{H} \implies \sum_{i}\omega^{\kappa_i} \in \mathcal{H}$

We define the "winner function" as:

**Definition** We denote by $w(\kappa, n, i )$ the "**winner** of the game $\kappa$ at the $n^\text{th}$ step if it is currently $i^\text{th}$ player's ($i \in \{0,1\}$) turn to move," *i.e.* $n $ is the number of hydra heads that will grow out if player $i$ cuts a head off of $\kappa$ and $w(\kappa, n, i )$ is the winner under optimal play. Since $w(\kappa, n, i )= 1- w(\kappa, n, 1- i )$ we will sometimes just consider the case $w(\kappa, n) \overset{\text{def}}{=}w(\kappa, n, 0) $ for simplicity.

The answer to your question is to compute $w(\kappa,1,i)$; but of course, in order to answer it we are going to need to define the following

**Definition** A **strategy** is a $\sigma: (\mathcal{H} \setminus \{0\}) \times \mathbb{N} \longrightarrow \mathcal{H} $ such that $\sigma(\kappa,n)$ is a legal position at step $n+1$ that succeeds a legal position at $n$. Equivalently we could have allowed for a "virtual position/move" $-1$ and defined $\sigma': \mathcal{H} \times \mathbb{N} \longrightarrow \mathcal{H} \cup \{-1\}$ as $\sigma'(0,n) = -1$ and $\sigma'(\kappa,n) = \sigma(\kappa,n)$ otherwise. We let $\mathcal{S}$ be the set of all such strategies.

Let us try to define the winning $\sigma$ by taking cases on the different possible "heaps."

**The heap has size 0 or 1**

Since $w(0,n,i) = 1-i$ the strategy $\sigma(0,n)$ is meaningless (that's we defined $\sigma $ on $(\mathcal{H} \setminus \{0\})$); likewise we see that $w(1,n,i) = i$ so that $\sigma(1,n)$ is forced to be $0$. This easily generalizes to the following case

**The hydra is a natural number**

It is straightforward to prove by induction that $w(k,n,i) = 1- ((k+i) \% 2)$ where $\%$ is remainder after division since by the rules of the game no new hydras grow after cutting a head off a hydra of the form $\omega^0 + \omega^0 +... + \omega^0 = \omega \cdot k $. Likewise $\sigma(1,n)$

**The hydra is of the form $\kappa + 1$**

If $\kappa ' = \left( \sum_{i}\omega^{\kappa_i}\right) + 1 = \kappa + 1$ then $w(\kappa' ,n,i) = 1 - w(\kappa' - 1 ,n,i) = 1-w(\kappa ,n,i) $. This can proven by a "sum of games" style proof. The idea is the following: if it is $i$'s turn then either:

- making a cut on $\kappa$ wins the game,
- making a cut on $\omega^0 = 1$ wins the game,
- or none of the above

but these are respectively true if and only if

- The cut $\sigma(\kappa,n)$ loses the game (for the game $\kappa$ not $\kappa+1$) for all $\sigma$
- proof by induction using the following two observations: 1) $\omega^0=1$ (or "parity") is a "loop-invariant" of the game and 2) $\sigma(\kappa,n) < \kappa$ (see the proof of thm 2 of [Kirby; Paris] for the proof of the inequality) 3) eventually we will hit $\kappa \in \omega $ for any strategy $\sigma$ ( once again see [Kirby; Paris]) which is the previous case

- the position $\kappa$ is a losing position,
*but this is true iff the first case is true*
- or none of the above,
*but this is true iff the first case is false*

Therefore the game is lost or won regardless of what move is made; it only depends on the "parity."

**The hydra is of the form $\kappa + \lambda $**

By a proof by induction and using the fact that $\sigma(\kappa,n) < \kappa$ and $\sigma(\lambda,n) < \lambda $ (see thm 2 of [Kirby; Paris]) we have that

\begin{equation}
w(\kappa + \lambda , n ) = w(\kappa , n ) \oplus w( \lambda , n ) \oplus 1 .
\end{equation}
Since we have already proven it for $\lambda =1$ and we can assume $\kappa > 2$ we can take the following cases in the induction

- Player 1 cuts $\kappa$ then Player 2 cuts $\kappa$
- Player 1 cuts $\kappa$ then Player 2 cuts $\lambda$
- Player 1 cuts $\lambda$ then Player 2 cuts $\kappa$
- Player 1 cuts $\lambda$ then Player 2 cuts $\lambda$

Which all lead to smaller cases to which we can apply the induction hypothesis.

Lets verify for simple examples: since $\sigma(\omega,n) = n $ we have that $w(\omega,n) = n \% 2 $ and also $w(0,n) =1$ which agrees with $w(\omega+ 0 ,n) = n \% 2 = w(\omega , n ) \oplus w( 0 , n ) \oplus 1$. Similarly $w(\omega+ 1 ,n) = n \oplus 1 = w(\omega , n ) \oplus w( 1 , n ) \oplus 1$ and more generally we have that $w(\omega+ k ,n) = n \oplus k = w(\omega , n ) \oplus w( k , n ) \oplus 1$. Likewise by a second induction we have that
\begin{equation}
w\left(\sum_i \kappa_i , n \right) = 1 \oplus \bigoplus_i w( \kappa_i , n ) .
\end{equation}

**The hydra is of the form $\omega ^ \kappa $**

The point here is that is $\omega ^ \kappa \neq \omega ^ {\lambda +1}$ for all $\lambda \in \mathcal{H}$ then all of the cuts are made inside of the $\kappa$. By induction, we also see that if $\omega ^ \kappa = \omega ^ {\lambda +1}$ then $w(\omega^\kappa,n)$ only depends on the parity of $n$ since if $\sigma$ is the "subtract 1" cut then $\sigma(\omega^\kappa,n) = \omega^\lambda \cdot n$ and $w(\omega^\lambda \cdot n, n+1, 1 ) = \bigoplus_i w( \omega^\lambda , n+1 ) $ by the previous section, so that $w(\omega^\kappa, n ) = \min\{1 \oplus \bigoplus_i w( \omega^\lambda , n + 1) ,1 \oplus w(\kappa, n )\}$ by a similar proof to the last section. The second term corresponds to the following: if player 0 can lose/win the game $\kappa$ then $\kappa$ corresponds to an odd/even game, but when the time finally comes to split $\omega^1$ player 1 will be left with an even/odd game.

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