On a flat plane in Camelot, King Arthur builds a labyrinth $\mathfrak{L}$ consisting of $n$ walls, each of which is an infinite straight line. No two walls are parallel, and no three walls have a common point. Merlin then paints one side of each wall entirely red and the other side entirely blue.

At the intersection of two walls there are four corners: two diagonally opposite corners where a red side and a blue side meet, one corner where two red sides meet, and one corner where two blue sides meet. At each such intersection, there is a two-way door connecting the two diagonally opposite corners at which sides of different colours meet.

After Merlin paints the walls, Morgana then places some knights in the labyrinth. The knights can walk through doors, but cannot walk through walls.

Let $k(\mathfrak{L})$ be the largest number $k$ such that, no matter how Merlin paints the labyrinth $\mathfrak{L},$ Morgana can always place at least $k$ knights such that no two of them can ever meet. For each $n,$ what are all possible values for $k(\mathfrak{L}),$ where $\mathfrak{L}$ is a labyrinth with $n$ walls?
We claim that the number \( k(\mathfrak{L}) \) (or the image of \( \mathfrak{L} \) under the function \( k \)) is always equal to \( \boxed{n+1} \). This means that:
1. For every possible labyrinth \( \mathfrak{L} \) King Arthur makes,
2. No matter how Merlin colors it, Morgana can always place \( n+1 \) knights, and
3. Merlin can color it in such a way so that Morgana is forced to place exactly \( n+1 \) knights and no more.

To prove this, we will proceed with the following steps:

1. **Graph Representation**:
   - Construct a graph \( G(\mathfrak{L}) \) where the vertices represent the regions formed by the walls, and the edges represent the doors connecting these regions.
   - Since no two walls are parallel and no three walls intersect at a single point, the number of regions formed by \( n \) walls is given by:
     \[
     R = \frac{n(n+1)}{2} + 1
     \]
   - The number of edges (doors) is given by the number of intersections, which is:
     \[
     E = \binom{n}{2} = \frac{n(n-1)}{2}
     \]

2. **Euler's Formula for Planar Graphs**:
   - For a connected planar graph, Euler's formula states:
     \[
     V - E + F = 2
     \]
     where \( V \) is the number of vertices, \( E \) is the number of edges, and \( F \) is the number of faces (regions).
   - In our case, \( V = R \) and \( E \) is as calculated above. Substituting these into Euler's formula, we get:
     \[
     \left( \frac{n(n+1)}{2} + 1 \right) - \frac{n(n-1)}{2} + F = 2
     \]
     Simplifying, we find:
     \[
     F = 1
     \]
     This indicates that the graph \( G(\mathfrak{L}) \) is connected and has exactly one face (the outer region).

3. **Maximizing the Number of Knights**:
   - The maximum number of knights Morgana can place is equal to the number of connected components in \( G(\mathfrak{L}) \). This is because one knight can access all regions connected to its starting point.
   - To ensure that no two knights can meet, each knight must be placed in a separate connected component.

4. **Merlin's Coloring Strategy**:
   - Merlin can color the walls in such a way that the graph \( G(\mathfrak{L}) \) has no cycles. This can be achieved by ensuring that each connected component of \( G(\mathfrak{L}) \) is a tree.
   - In a tree, the number of vertices \( V \) and edges \( E \) satisfy \( V = E + 1 \). Therefore, the number of connected components \( C \) is given by:
     \[
     C = V - E = \left( \frac{n(n+1)}{2} + 1 \right) - \frac{n(n-1)}{2} = n + 1
     \]

5. **Conclusion**:
   - Since Merlin can always color the walls such that \( G(\mathfrak{L}) \) has \( n+1 \) connected components, Morgana can always place \( n+1 \) knights.
   - Therefore, \( k(\mathfrak{L}) = n+1 \).

\(\blacksquare\)