Chess Board and Knights I can't solve this question. Anyone knows how I must solve?
You have a 6x3 chess board. How many forms exist to put a Knight in a square and with valid moviments pass in all squares but only one time in each one?
And with in a LxC chess board?
 A: For a $6 \times 3$ board, it's impossible. Consider the board below, and the squares, marked with an $\times$.
$$\begin{matrix} 
\cdot & \cdot & * & \cdot & \cdot & \cdot \\ 
\times & \cdot & \cdot & \cdot & \times & \cdot \\ 
\cdot & \cdot & * & \cdot & \cdot & \cdot \\ \end{matrix}$$
The only way to get to these squares is through the squares marked with an $\ast$. But if you use both to get to one of them and to get out again, then you cannot reach the other marked square anymore. The only way to solve this is to start at one of these marked squares. By symmetry, we have two other such squares:
$$\begin{matrix} 
\cdot & \cdot & \cdot & \ast & \cdot & \cdot \\ 
\cdot & \times & \cdot & \cdot & \cdot & \times \\ 
\cdot & \cdot & \cdot & \ast & \cdot & \cdot \\ \end{matrix}$$
So we have to finish at another of these squares as well. So the first four and final four moves are determined, as, one of the following two scenarios:
$$\begin{matrix} 
\cdot & \cdot & 2 & 15 & \cdot & \cdot \\ 
1 & 16 & \cdot & \cdot & 3 & 18 \\ 
\cdot & \cdot & 4 & 17 & \cdot & \cdot \\ \end{matrix} \qquad \text{or} \qquad \begin{matrix} 
\cdot & \cdot & 2 & 17 & \cdot & \cdot \\ 
1 & 16 & \cdot & \cdot & 3 & 18 \\ 
\cdot & \cdot & 4 & 15 & \cdot & \cdot \\ \end{matrix}$$
In both cases, we need the remaining $10$ squares to connect the two ends of the chain to form a knight's path. This means that each of these $10$ points is connected to two other of these points (except for the ones next to $2$ and $17$). By starting with the corners, we get two chains of length $5$:
$$\begin{matrix} 
d & a & \cdot & \cdot & A & D \\ 
\cdot & \cdot & c & C & \cdot & \cdot \\ 
b & e & \cdot & \cdot & E & B \end{matrix}$$
There's just no way to connect $a/e$ to $A/E$, so there's no chain of length $10$ connecting these squares. So there is no knight's path on this board.
A: In the book "Across the Board" (Watkins, 2004) there's a nice explanation why no knight's tour (i.e. starting and ending at the same square) exists on the $6 \times 3$ chessboard:

Suppose there is a knight's tour. One can connect the squares formed by the tour to form a necklace. If you remove one piece of a necklace, the necklace breaks, but is still in one piece. Taking another piece from the necklace then breaks the necklace into at most two parts. But removing the two squares marked $\ast$ above disconnects the two square marked $\times$ from the rest of the board by knight moves. So we get three components, which is a contradiction. So no knight's tour exists on a $6 \times 3$ chessboard.

This could be extended to show that no knight's path exists. If we have a chain (i.e. an open necklace), and we remove $6$ pieces of the chain, we should be left with at most $7$ components. But if we remove the six squares marked below:
$$\begin{matrix} 
\cdot & \cdot & \times & \times & \cdot & \cdot \\ 
\cdot & \cdot & \times & \times & \cdot & \cdot \\ 
\cdot & \cdot & \times & \times & \cdot & \cdot \\ \end{matrix}$$
then we are left with $8$ components:
$$\begin{matrix} 
1 & 2 & \times & \times & 5 & 6 \\ 
3 & 4 & \times & \times & 7 & 8 \\ 
2 & 1 & \times & \times & 6 & 5 \\ \end{matrix}$$
So there is no knight's path on a $6 \times 3$ chessboard either.
