# Robots on a Plane

A proof that, should you ever be attacked by a giant robotic bishop while on a chessboard, you can survive by standing in the right spot.

$P(n):$ at any time $n\in\mathbb{N},x+y$ is even.

# Basis

$P(1): (1,1),(1,-1),(-1,1),(-1,-1)$

$x+y = 2,0,0,-2$.

Thus, $P(1)$ is true.

# Inductive Step

Assume $1\leq k$ and $P(k)$ is true, we show that $P(k+1)$ is true.

The coordinates $(a,b)$ resemble the location of the robot at time $k$.

Because $P(k)$ is true, $a+b$ is even.

Thus, $\exists l\in\mathbb{Z}\ni a+b=2l$ because $P(k)$ is true.

At time $k+1$ there are three cases:

## Case 1: The robot moves up and to the right from $(a,b)$

Therefore, $P(k+1)$ coordinates are $(a+1,b+1)$

thus, $x+y=(a+1)+(b+1)$

$\Rightarrow x+y=a+b+2$

$\Rightarrow x+y=2l+2$ by substitution

$\Rightarrow x+y=2(l+1)$ where $l+1\in\mathbb{Z}$

Therefore, $x+y$ is even.

## Case 2: The robot moves either left and up or right and down from $(a,b)$

Therefore, $P(k+1)$ coordinates are either $(a+1,b-1)$ or $(a-1,b+1)$.

Thus, $x+y=(a+1)+(b-1)$ or $x+y=(a-1)+(b+1)$

$\Rightarrow x+y=2l$

Therefore, $x+y$ is even.

## Case 3: The robot moves left and down from $(a,b)$

Therefore, $P(k+1)$ coordinates are $(a-1,b-1)$

Thus, $x+y=(a-1)+(b-1)$

$x+y=a+b-2$

$x+y=2l-2$

$x+y=2(l-1)$ where $(l-1)\in\mathbb{Z}$

Therefore, $x+y$ is even.

# Conclusion

Therefore, $P(k)\Rightarrow P(k+1)$.

Thus, no matter which way the robot moves, $x+y$ will always be even.

And thus, no matter how much time the robot has to move around, it will never be able to reach the point $(1,0)$, since $1+0=1$ is odd.