# A curious incident in the flea circus

The ringmaster of a flea circus puts three fleas $A$, $B$, $C$ on three different numbers on the real number line, so that flea $B$ sits exactly in the middle between $A$ and $C$.

• Whenever the ringmaster shouts "Hop!", one of the three fleas jumps over one of the other fleas to the mirror point on the other side. (In other words, a flea sitting in point $x-y$ may jump over a flea sitting in point $x$ to the new point $x+y$.)
• While the fleas are jumping around, sometimes two of them may be sitting simultaneously on the same real number. (This is fine, as these fleas are infinitesimally small.)

After some time, the ringmaster notices that the three fleas again occupy the three starting points, but they are now sitting in a different order.

Question: Is it possible that flea $A$ is now sitting on the starting point of flea $B$?

No.

I can describe their co-ordinates at the start as -1, 0, 1 on some scale where 0 is defined to be where B is sitting and the unit 1 is half the distance between A and C. There is a simple transformation from this scale to the real numbers where they were placed.

Switching their positions means that A has to reach 0 and B has to reach -1. But A will be jumping in multiples of 2. For example if A is one unit from a flea (as it is at the start) it will jump 2 units and will still be one unit from that flea. It starts on an odd number and will therefore stay on an odd number (on my transformed scale.) Similarly B starts on an even number and will stay on an even number.

By jumping past each other they can move arbitrarily up and down the numberscale but they cannot change the odd/even state of their position using my transformed scale, nor move to a point that is not an integer on my scale.

Let us assume the fleas to be sitting on consecutive terms of an infinite AP. No flea can leave the AP. This means that a flea on an odd number term can never reach an even number term (and vice versa)'

Hence it is not possible.

No, as others have said

For me the easiest way to see it is that if they are all starting on integers (say 0,1 and 2), then jumping leaves their polarity unchanged. So an odd number jumping over another number lands on an odd number, and an even number jumping over an even number stays on an even number.

This is simply because a jump moves them a distance of $2d$, where $d$ is the distance between the jumper and the jumpee.

Hence not only can they not swap, but they can't even reach each other's spaces. A and C are doomed to stay on the even numbers, and B on the odd.

It is not possible

$A$ and $B$ have a distance of $d$

The same distance $d$ is between $B$ and $C$

The distance between $A$ and $C$ is therefore $2d$

The following rules hold true:

• If before a jump the distance between all pairs of fleas is a multiple of any number $N$ then any flea can only jump a distance equal to a multiple of $2N$
• if before a jump the distance between all pairs of fleas is a multiple of any number $N$ and any flea jumps a distance equal to a multiple of $N$ then the distance between all pairs of fleas after the jump is still a multiple of $N$

Using both rules we can see, that if rule 1 holds true before any jump $J$ then it also holds true for jump $J+1$

Since rule 1 holds true for the first jump with $N=d$ we can see that it holds true for all jumps.

We see now that because $A$ is exactly a distance of $k$ from the starting point of $B$ and has to jump multiples of $2k$ with every jump it can never have a distance of $0$ from the starting point of $B$ and can therefore never be in the starting point of $B$