The Jumping Peg Problem

1. Problem Statement

Ref: link

Imagine the \(\mathbb{Z}^2\) lattice. You have 3 pegs at (0,0), (0,1), and (1,0). At each time step you are allowed to reflect any one peg w.r.t any other peg. For example here (0,1) reflected in (0,0) would go to (0,-1), reflecting again in (1,0) will put it at (2,1). Can a peg ever reach (1,1)?

2. Solution

With some thought this ones an easy solve. Consider the looser problem - (think of pegs as coloring the latttice point) whenever you reflect you color both the original position and the reflected position. In our base problem we only color the lattice corresponding to the reflected point and decolor the original.

A trivial claim is: If you can color (1,1) in former problem, you can color (1,1) in later problem (Just follow those reflection, extra points to reflect from doesn’t hinder you!). We will show (1,1) isn’t colored in the simpler problem.

You can easily reach the following state with minimal effort (there might be stuff outside of it, but we ignore those for a moment) -

Tile for our lattice

Now reflecting this (instead of reflecting in points, you can reflect in line to get faster) we can tile the whole \(\mathbb{Z}^2\) as follows -

Tile the lattice

The question now becomes is this stable under reflections or there is some clever reflection that will send the colored point to a hole? 3 things before proceeding -

1. Any arbitary reflection can be thought of composition of vertical and horizontal reflection. If our diagram is stable in both horizontal and vertical reflection we are okay.
2. Coloring any hole implies coloring all holes (easy to check just keep reflecting in vertical and horizontal lines). So coloring any hole implies coloring (1,1).
3. Sending a colored point to hole is same as sending the hole to colored point. We will use this viewpoint.

Fix a hole (WLOG as the diagram is symmeteric), pick any point to reflect from. (If it helps rotate the plane so that the point is on y-axis). It is easy to see reflection through the point sends hole to hole or our diagram is fixed in any horizontal & vertical reflection. Proving our claim.