You are a student, assigned to work in the cafeteria today, and it is your duty to divide the available food between all students. The food today is a sausage of 1m length, and you need to cut it into as many pieces as students come for lunch, including yourself.
The problem is, the knife is operated by the rotating door through which the students enter, so every time a student comes in, the knife comes down and you place the cut. There is no way for you to know if more students will come or not, so after each cut, the sausage should be cut into pieces of approximately equal length.
So here the question – is it possible to place the cuts in a manner to ensure the ratio of the largest and the smallest piece is always below 2?
And if so, what is the smallest possible ratio?
Example 1 (unit is cm):
- 1st cut: 50 : 50 ratio: 1
- 2nd cut: 50 : 25 : 25 ratio: 2 – bad
Example 2
- 1st cut: 40 : 60 ratio: 1.5
- 2nd cut: 40 : 30 : 30 ratio: 1.33
- 3rd cut: 20 : 20 : 30 : 30 ratio: 1.5
- 4th cut: 20 : 20 : 30 : 15 : 15 ratio: 2 – bad
Sorry for the awful analogy, I think this is a math problem but I have no real idea how to formulate this in a proper mathematical way.
Best Answer
TLDR: $a_n=\log_2(1+1/n)$ works, and is the only smooth solution.
This problem hints at a deeper mathematical question, as follows. As has been observed by Pongrácz, there is a great deal of possible variation in solutions to this problem. I would like to find a "best" solution, where the sequence of pieces is somehow as evenly distributed as possible, given the constraints.
Let us fix the following strategy: at stage $n$ there are $n$ pieces, of lengths $a_n,\dots,a_{2n-1}$, ordered in decreasing length. You cut $a_n$ into two pieces, forming $a_{2n}$ and $a_{2n+1}$. We have the following constraints:
$$a_1=1\qquad a_n=a_{2n}+a_{2n+1}\qquad a_n\ge a_{n+1}\qquad a_n<2a_{2n-1}$$
I would like to find a nice function $f(x)$ that interpolates all these $a_n$s (and possibly generalizes the relation $a_n=a_{2n}+a_{2n+1}$ as well).
First, it is clear that the only degree of freedom is in the choice of cut, which is to say if we take any sequence $b_n\in (1/2,1)$ then we can define $a_{2n}=a_nb_n$ and $a_{2n+1}=a_n(1-b_n)$, and this will completely define the sequence $a_n$.
Now we should expect that $a_n$ is asymptotic to $1/n$, since it drops by a factor of $2$ every time $n$ doubles. Thus one regularity condition we can impose is that $na_n$ converges. If we consider the "baseline solution" where every cut is at $1/2$, producing the sequence
$$1,\frac12,\frac12,\frac14,\frac14,\frac14,\frac14,\frac18,\frac18,\frac18,\frac18,\frac18,\frac18,\frac18,\frac18,\dots$$ (which is not technically a solution because of the strict inequality, but is on the boundary of solutions), then we see that $na_n$ in fact does not tend to a limit - it varies between $1$ and $2$.
If we average this exponentially, by considering the function $g(x)=2^xa_{\lfloor 2^x\rfloor}$, then we get a function which gets closer and closer to being periodic with period $1$. That is, there is a function $h(x):[0,1]\to\Bbb R$ such that $g(x+n)\to h(x)$, and we need this function to be constant if we want $g(x)$ itself to have a limit.
There is a very direct relation between $h(x)$ and the $b_n$s. If we increase $b_1$ while leaving everything else the same, then $h(x)$ will be scaled up on $[0,\log_2 (3/2)]$ and scaled down on $[\log_2 (3/2),1]$. None of the other $b_i$'s control this left-right balance - they make $h(x)$ larger in some subregion of one or the other of these intervals only, but preserving $\int_0^{\log_2(3/2)}h(x)\,dx$ and $\int_{\log_2(3/2)}^1h(x)\,dx$.
Thus, to keep these balanced we should let $b_1=\log_2(3/2)$. More generally, each $b_n$ controls the balance of $h$ on the intervals $[\log_2(2n),\log_2(2n+1)]$ and $[\log_2(2n+1),\log_2(2n+2)]$ (reduced$\bmod 1$), so we must set them to $$b_n=\frac{\log_2(2n+1)-\log_2(2n)}{\log_2(2n+2)-\log_2(2n)}=\frac{\log(1+1/2n)}{\log(1+1/n)}.$$
When we do this, a miracle occurs, and $a_n=\log_2(1+1/n)$ becomes analytically solvable: \begin{align} a_1&=\log_2(1+1/1)=1\\ a_{2n}+a_{2n+1}&=\log_2\Big(1+\frac1{2n}\Big)+\log_2\Big(1+\frac1{2n+1}\Big)\\ &=\log_2\left[\Big(1+\frac1{2n}\Big)\Big(1+\frac1{2n+1}\Big)\right]\\ &=\log_2\left[1+\frac{2n+(2n+1)+1}{2n(2n+1)}\right]\\ &=\log_2\left[1+\frac1n\right]=a_n. \end{align}
As a bonus, we obviously have that the $a_n$ sequence is decreasing, and if $m<2n$, then \begin{align} 2a_m&=2\log_2\Big(1+\frac1m\Big)=\log_2\Big(1+\frac1m\Big)^2=\log_2\Big(1+\frac2m+\frac1{m^2}\Big)\\ &\ge\log_2\Big(1+\frac2m\Big)>\log_2\Big(1+\frac2{2n}\Big)=a_n, \end{align}
so this is indeed a proper solution, and we have also attained our smoothness goal — $na_n$ converges, to $\frac 1{\log 2}=\log_2e$. It is also worth noting that the difference between the largest and smallest piece has limit exactly $2$, which validates Henning Makholm's observation that you can't do better than $2$ in the limit.
It looks like this (rounded to the nearest hundred, so the numbers may not add to 100 exactly):
If you are working with a sequence of points treated$\bmod 1$, where the intervals between the points are the "sausages", then this sequence of segments is generated by $p_n=\log_2(2n+1)\bmod 1$. The result is beautifully uniform but with a noticeable sweep edge:
A more concrete optimality condition that picks this solution uniquely is the following: we require that for any fraction $0\le x\le 1$, the sausage at the $x$ position (give or take a sausage) in the list, sorted in decreasing order, should be at most $c(x)$ times smaller than the largest at all times. This solution achieves $c(x)=x+1$ for all $0\le x\le 1$, and no solution can do better than that (in the limit) for any $x$.