medium | discrete |

On a table you have a square made of 4 coins at the corner at distance 1. So, the square is of size 1×1. In a valid move, you can choose any two coin let’s call them mirror and jumper. Now, you move the jumper in a new position which is its mirror image with respect to mirror. That is, imagine that mirror is a centre of a circle and the jumper is on the periphery. You move the jumper to a diagonally opposite point on that circle. With any number of valid moves, can you form a square of size 2×2? If yes, how? If no, why not?

Invariance

No!

Note, that every valid move is reversible. That is if you make a move in one direction, the other direction is also a possibility. So, if you can scale a square of size 1×1 into a square of size 2×2, then you should also be able to shrink it into arbitrary small square. Say 1/2 x 1/2 or 1/1024 x 1/1024.

So, now we need to show that you cannot shrink the square. If we show that distance between any two coins is always greater than 1 unit, we are done. And here is the simplest part!

Imagine a grid on the 2-d plane. Let the coins be at (0,0) (0,1) (1,0) (1,1) forming a square of size 1×1. Our grid cells are of size 1×1. Note, that no matter how many moves you make, the coins will always have integer co-ordinates ( basically stays on the grid ). Since, the shortest distance on the grid is 1 unit, no two coins can have distance shorter than 1 unit. And we are done!