Let $\alpha \geq 1$ be a real number. Hephaestus and Poseidon play a turn-based game on an infinite grid of unit squares. Before the game starts, Poseidon chooses a finite number of cells to be [i]flooded[/i]. Hephaestus is building a [i]levee[/i], which is a subset of unit edges of the grid (called [i]walls[/i]) forming a connected, non-self-intersecting path or loop*.

The game then begins with Hephaestus moving first. On each of Hephaestus’s turns, he adds one or more walls to the levee, as long as the total length of the levee is at most $\alpha n$ after his $n$th turn. On each of Poseidon’s turns, every cell which is adjacent to an already flooded cell and with no wall between them becomes flooded as well. Hephaestus wins if the levee forms a closed loop such that all flooded cells are contained in the interior of the loop — hence stopping the flood and saving the world. For which $\alpha$ can Hephaestus guarantee victory in a finite number of turns no matter how Poseidon chooses the initial cells to flood?
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[size=75]*More formally, there must exist lattice points $\mbox{\footnotesize \(A_0, A_1, \dotsc, A_k\)}$, pairwise distinct except possibly $\mbox{\footnotesize \(A_0 = A_k\)}$, such that the set of walls is exactly $\mbox{\footnotesize \(\{A_0A_1, A_1A_2, \dotsc , A_{k-1}A_k\}\)}$. Once a wall is built it cannot be destroyed; in particular, if the levee is a closed loop (i.e. $\mbox{\footnotesize \(A_0 = A_k\)}$) then Hephaestus cannot add more walls. Since each wall has length $\mbox{\footnotesize \(1\)}$, the length of the levee is $\mbox{\footnotesize \(k\)}$.[/size]

[i]Nikolai Beluhov[/i]
We claim that the answer is all $\boxed{\alpha > 2}$. Call Poseidon "Po" and call Hephaestus "You, the engineer" (yes, you).

1. **Winning Strategy for $\alpha > 2$:**
   - Define a round to be two turns, one by you and the one after by Po.
   - Label the squares with coordinates and suppose that if Po flooded squares $F= \{(x,y)\}$, then $n$ is a positive integer such that $n \geq |x|+|y|$ for all $(x,y)\in F$.
   - Let $t_0\in \mathbb{N}$ such that 
     \[
     2t_0 + 4n + 3 \leq \alpha t_0.
     \]
   - We claim that for the first $t_0-1$ turns, you can just extend the levee so that it blocks the squares $(-n-1,y)$ from $(-n,y)$ for all $|y| < t$ (after the $t$-th turn).
   - After $t$ rounds, the flood has spread to exactly $(x,y)$ with $|x|+|y| \leq n+t$ and $x \geq -n$.
   - On the $t_0$-th turn, Po will extend the levee so that there is a wall to the left of $(-n,y)$ for $|y| \leq n+t_0$ and one above $(x,n+t_0)$ and one below $(x,-n-t_0)$ for each of $-n \leq x \leq 0$.
   - On the following $i$ turns, you just build a wall above $(i, n+t_0)$ and one below $(i, -n-t_0)$, extending the levee.
   - After the $t_0+i$-th round, the flooded points $(x,y)$ are those satisfying $|y| \leq t_0+n$, $x \geq -n$, and $|x|+|y| \leq n+t_0+i$.
   - Continue this way until 
     \[
     2(3t_0 + 4n + 2 + I) \leq \alpha(t_0 + I),
     \]
     at which point on the $t_0+I$-th turn we bound the entire rectangle $R_I$ and win.

2. **Poseidon's Winning Strategy for $\alpha \leq 2$:**
   - We call a **snorlax** a simply connected [animal](http://mathworld.wolfram.com/LatticeAnimal.html).
   - Let $R$ be a snorlax with perimeter $P$ and adjacency graph $G$.
   - Since the lattice points can be two-colored and that walking around the perimeter with $P$ unit steps starts and ends at the same color, $P$ is even, say $P = 2k$.

   **Lemma 1:** If $R$ is a snorlax and $G$ does not have a bridge, and $AB$ are adjacent cells in $R$, and $C,D$ are next to $B$ perpendicular to $AB$, then one of $C,D$ is in $R$:
   \[
   \begin{array}{c c c}
   & A & \\
   C & B & D \\
   & E & \\
   \end{array}
   \]

   *Proof:* Suppose by contradiction that neither $C$ nor $D$ is in $R$. Since $AB$ is not a bridge, $B$ must have some other neighbor in $R$ on the other side from $A$, say $E$. Then, we must have some other path from $A$ to $E$ that is not through $B$. In that case, the path either goes around $C$ or around $D$. Whichever one it goes around must be in $R$ because $R$ is simply connected. $\blacksquare$

   **Lemma 2:** $\operatorname{diam} G \leq k-2$.

   *Proof:* We strong induce on $k$. If $k = 2$, then $R$ is a unit square and so $G$ is a single vertex.

   If $G$ is a bridge, then $R$ can be split into $R_1, R_2$ with perimeters $P_1, P_2$ and adjacency graphs $G_1, G_2$. Then,
   \[
   \operatorname{diam} G \leq \operatorname{diam} G_1 + \operatorname{diam} G_2 + 1 \leq \frac{P_1}{2} - 2 + \frac{P_2}{2} - 2 + 1 = \frac{P+2}{2} - 3 = k-2
   \]
   indeed.

   Else, note that when we walk around the perimeter of $R$, we will have two consecutive 90 degree turns. Therefore, after suitable rotation, we have squares $S_1, \ldots, S_m$ so that 
   \[
   \begin{array}{c c c c c}
   & S_1 & S_2 & \cdots & S_m \\
   & T_1 & T_2 & \cdots & T_m \\
   \end{array}
   \]
   where the red segments are part of the boundary of $R$. We claim now that the grid squares $T_1, \ldots, T_m$ are all in $R$. For $1 \leq i \leq m-1$, $T_i \in R$ follows from Lemma 1 since $S_i, S_{i+1}$ are both in $R$ but the square above $S_i$ is not. For $i = m$, note that $S_{m-1}, S_m$ are in $R$ but the square above $S_m$ is not. Thus, if $R'$ is $R$ with the $S_i$'s removed, then $R'$ is still a snorlax. Furthermore, $R'$'s perimeter is $2(k-1)$. Finally, if $R'$ has adjacency graph $G'$, then
   \[
   \begin{aligned}
   \operatorname{diam} G &= \max\left(\operatorname{diam} G', m-1, \max_{\substack{v \in G' \\ 1 \leq i \leq m}} d(v, S_i)\right) \\
   &\leq \max\left(\operatorname{diam} G', m-1, \max_{\substack{v \in G' \\ 1 \leq i \leq m}} d(v, T_i) + 1\right) \\
   &\leq \max\left(\operatorname{diam} G', m-1, \operatorname{diam} G' + 1\right) \\
   &= \operatorname{diam} G' + 1
   \end{aligned}
   \]
   where we note that $\operatorname{diam} G' \geq m-1$ since $d(T_1, T_m) = m-1$. Thus,
   \[
   \operatorname{diam} G \leq \operatorname{diam} G' + 1 \leq k-3 + 1 = k-2,
   \]
   and the induction is complete. $\blacksquare$

   Now, suppose $\alpha = 2$ and Po just floods $O = (0,0)$ initially. Let $S_t$ be the set of squares flooded on the $t$-th turn. Then, $d(O, A) = t$ (as vertices of $G$) for any $A \in S_t$. In particular, suppose you win on your $n$-th turn. Then, the enclosed region $R$ has perimeter at most $2n$. Furthermore,
   \[
   R \supset S_0 \cup S_1 \cup \cdots \cup S_{n-1},
   \]
   and $S_{n-1}$ is nonempty. So, $\operatorname{diam} G \geq n-1$ since $d(O, A) = n-1$ for $A \in S_{n-1}$, while $R$'s perimeter is at most $2n$, contradicting Lemma 2. Thus, you cannot win in finitely many turns. $\blacksquare$