Is there an equation for this?

Question

So I want to have a grouping where 50% of the right-most length of a line (segment) contains 25% of the 'units' being grouped.

I then want the remaining 75% of the units grouped in to the remaining 50% of the line.

Next, I want to repeat the process about 3 more times (but could be arbitrary), with that 75% of the units, and the left segment of the line - placing 25% of them in to 50% of the right-most length of the line segment, and so on.

This sounds like something that could be reduced to an equation. I, however, don't know of any such equation, so I look to the brilliant minds on stack exchange.

Answer 1

Let's call $a_k$ ($resp$. $b_k$) the proportion in the group generated in the left($resp$. right)-most end at the $k^{th}$ application of the process.

$b_n=0.75b_{n-1}$ : The right-most end at step n is 75% of the right-most end at the previous $a_n=0.25b_{n-1}$ : The left-most end at step n is 75% of the right-most end at the previous

From the previous you can derive :

$\left\{ \begin{array}{ll} a_n=0.25 \times 0.75^{n-1} \\ b_n=0.75^n \end{array} \right.$

After applying exactly $n$ times the process the $n+1$ intervals you end up with a partition in the following proportions: $$(a_1,\dots,a_{n},b{n})$$ with $a_1+\dots+a_{n}+b{n}=1$

Answer 2

So essentially you are dividing into smaller portions of $3/4$ and $1/4$ four times. Since $4^4=256$ you can imagine that the initial quantity is divided into $256$ parts apportioned as follows.

Steps:

1. $192\in[0,8]$, $64\in[8,16]$
2. $144\in[0,4]$, $48\in[4,8]$
3. $108\in[0,2]$, $36\in[2,4]$
4. $81\in[0,1]$, $27\in[1,2]$

So the final result is

1. $81\in[0,1]$
2. $27\in[1,2]$
3. $36\in[2,4]$
4. $48\in[4,8]$
5. $64\in[8,16]$

As a graph, the distribution looks like this

Notice that as you move from the right-most interval towards the left, the intervals contain the following ratios of the total number of initial amounts:

$$ \frac{1}{4}, \frac{3}{16}, \frac{9}{64}, \frac{27}{256} $$

This is a geometric series with initial value $a=\frac{1}{4}$ and common ratio $r=\frac{3}{4}$. Notice that the first interval is left with $\frac{81}{256}$ of the initial amount, three times the amount of the second interval.

You can continue the process as many times as you wish. The first $n$ intervals counting from the right will be the $n$ terms of the geometric series above and the remaining interval will be three times the amount of the interval to its right.

For example, if you continued the process one more time the ratios of the initial amounts beginning at the right and moving to the intervals to the left would be

$$ \frac{1}{4},\frac{3}{16},\frac{9}{64},\frac{27}{256},\frac{81}{1024},\frac{243}{1024} $$

Notice that the last fraction, representing the amount remaining in the leftmost interval has the same denominator but three times the numerator of the previous interval.

Answer 3

I'll suppose the length of your line segment is $L$ and you want to place $N$ "units" in $k$ groups.

Numbering the groups from $1$ (the leftmost) to $k$ (the rightmost), and measuring distances from the leftmost end of the segment, group $1$ occupies the portion of the segment from distance $0$ to distance $\left(\frac12\right)^{k-1} L$ and contains $\left(\frac34\right)^{k-1} N$ units. For $i > 1,$ group $i$ occupies the portion of the segment from distance $\left(\frac12\right)^{k-i+1} L$ to distance $\left(\frac12\right)^{k-i} L$ and contains $\frac14\left(\frac34\right)^{k-i} N$ units.

Of course that works exactly only if $N$ is a multiple of a large enough power of $4$ to multiply it by a factor of $\frac34$ that many times while still getting an integer result. Otherwise you have to do some kind of rounding off in a consistent fashion.