Solution verification request - "Two robbers steal a necklace of black and white pearls" question

Question

I have the following problem.Below is the description of the problem, my attempt and question related to my solution:

Description

Two robbers steal a necklace of black and white pearls. There are equal numbers of white and black pearls in the necklace. The pearls are evenly spaced. However, the colors are not necessarily alternating or symmetric.

Prove that no matter what the necklace looks like, it is always possible to cut it into two pieces each having half the black and half the white pearls.

Attempt

To prove that it is always possible to cut the necklace into two pieces, each having half the black and half the white pearls, we can construct the following strategy:

Let's choose an arbitrary pearl on the necklace and mark it. We use this marked pearl as a reference point.

Next, we move along the necklace, counting the number of white and black pearls encountered, starting from the marked pearl. While moving along, we keep track of the difference between the number of white and black pearls counted so far. Let's call this difference $d$.

(e.g. suppose we start at the marked pearl and move along the necklace in a clockwise direction. Let's say we encounter 3 black pearls and 5 white pearls before we return to the marked pearl. Then we counted 3 black pearls and 5 white pearls. So here $d = 5 - 3 = 2$.)

We can repeat this process for all possible starting points on the necklace.

Since there are an even number of pearls, there is an even number of possible starting points.

Therefore, there are only two possible cases:

1. There exists a starting point where $d = 0$.

Then, we have already found a cut that divides the necklace into two pieces, each with half the black and half the white pearls. We can cut the necklace at the marked pearl and at the starting point where $d = 0$. We have two pieces with the desired properties.

2. There does not exist a starting point where $d = 0$.

Then, we know that $d$ must be even for all starting points, since there is an even number of pearls and each pearl contributes to $d$ exactly once. Wlog $d = 2k$ for some $k \in \mathbb{N}$.

Then we can cut the necklace at the marked pearl and at any starting point where $d = k$. This gives two pieces, each with $k$ black pearls and $k$ white pearls, which is exactly half of each color.

Therefore, it is always possible to cut the necklace into two pieces, each having half the black and half the white pearls, no matter how the pearls are arranged.

Question

Is my above reasoning/proof correct? I would appreciate feedbacks from MathStack Exchange community!

Hope my solution posted is going to enrich content of MathStack Exchange platform since I did not find any similar problem to the one described above.

Answer 1

A much simpler solution is to pick any point in between two pearls and draw a line through the diameter

Suppose one half has $(b-x)$ black pearls, then this half has $x$ white pearls, while the other half has $x$ b)ack pearls

Answer 2

Your proof is not valid. Here is an example to illustrate with eight pearls, numbered $1$ to $8$. Bead number $4$ is marked.

1 2 3 4 5 6 7 8 
W W B W B B B W
      *

This necklace falls into the first case, because there exists a starting place where $d=0$. Specifically, if you start at pearl number $3$, and then proceed until pearl number $4$, then you encounter exactly one white bead and one black pearl, so $d=1-1=0$. You then say that you should cut the necklace at the starting pearl and the marked pearl. When you do this, the first piece has $1$ black pearl and $1$ white pearl, while the remaining piece has $3$ of each color. This is not a fair division.

There are several other places where your proof is incorrect, but I will not list them all, because this counterexample alone is sufficient to show that you need an entirely new strategy.