# Bowser and Mario in a Pinch! [duplicate]

Our little hero starts off in a round (or square) pit with very high walls, an escape does not seem possible. However also in the pit is Bowser! How did that happen? Both of them are now facing off each other, which means, Mario is on the run and Bowser chases him. Both of them run at the same constant speed. The question now is: Can Mario outrun Bowser and will never be caught or is he sooner or later doomed to be caught by it? Do both of them have an optimal strategy to run? (No attack moves, jumps, power ups etc. allowed! Just running!)

If this pit goes through the entire planet in one long line, then Bowser will not be able to catch Mario.

But if the pit is round, then Mario has no chance. No matter where he runs, Bowser will run towards him. When Mario reaches the wall, he has to go left or right. Depending where he goes, Bowser will slightly move too that side too making the distance smaller.

Mario will probably keep on running in circles, but Bowser will make some corrections and shorten the distance with every turn of Mario.

So, Mario is doomed.

If the pit is a square, Mario is doomed.

Suppose the pit is a square 1 m x 1 m. Mario ($M$) has to maximize the distance from Bowser ($B$) and $B$ minimize it. The strategy for $B$ is to stand in the middle of the square. The most distant point for $M$ to stand is at one of the corners which is at ${\sqrt2 \over 2}m$ which is greater than ${1\over2}m$, the distance from the center to one of the sides.

If $M$ starts to run following one of the sides, $B$ has only to run, at the same speed, towards that side. They will met at $1\over2$ of the side itself, and $M$ is captured.

$B$ has to run towards $M$ following the diagonal of the square, decreasing the dimension of the square until $M$ is captured.

When running towards the edge of the pit, $B$ has to run in a straight orthogonal line. If $M$ runs back $B$ runs on a $45°$ line orthogonal to the diagonal of the square. In this way $B$ runs a shorter path since he now runs on the cathetus instead of the hypotenuse of the triangle made with the diagonal of the pit, assures that $M$ doesn't escape and decreases the distance between them when getting back to the diagonal.

If $M$ at some point decide to run diagonally, $B$ only has to decrease the run and wait $M$ intersect his path.