Jumping tools in a lattice

Question

a 7 * 7 lattice. In the inner area of the frame of the lattice is a structure of 3 * 3 size of 9 tools placed on the lattice points. When there are three adjacent lattice points, either horizontally or vertically, two adjacent tools , and the third is free (without tool), one tool can be jumped over the other, linearly, into the free point, and the other tool is removed. it is impossible to leave one tool in the board , how we can prove it?

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Answer 1

It's a fairly easy programming exercise to enumerate all configurations that can occur after a certain number of moves. There are none that end with one tool after $8$ moves.

Answer 2

The move you want is impossible.

On each lattice point, assign it to an integral coordinate. On any specific time, we can define the following sets:

$$ \begin{aligned} T &:= \{(x,y) \ | \ (x,y) \text{ has a tool on it}\}\\ S_0 &:= \{(x, y) \in T \ | \ x + y \equiv 0 \pmod 3\}\\ S_1 &:= \{(x, y) \in T \ | \ x + y \equiv 1 \pmod 3\}\\ S_2 &:= \{(x, y) \in T \ | \ x + y \equiv 2 \pmod 3\}. \end{aligned} $$

and let $(n_0, n_1, n_2) := (|S_0|\pmod 2, |S_1|\pmod 2, |S_2|\pmod 2)$. At the beginning, $(n_0, n_1, n_2) := (1,1,1)$. After each valid step, two of $S_i$ got one element removed, while the third got one element inserted, so after each step $(n_0, n_1, n_2)$ is either $(0,0,0)$ or $(1,1,1)$.

So there is no chance that exactly one tool remains.