Ponder This Challenge - July 2025 - Swallows on a Wire

This puzzle was suggested by Latchezar Christov - thanks!

A wire with length $L$ is stretched between two poles. Swallows come, one by one, and land on the wire. All swallows have a width equal to 1 and cannot overlap on the wire, and each new swallow selects where to land at random with uniform probabilities (taken over all the perimissible options). This repeats until no swallow can land anymore.

In this moment, there are $S$ swallows on the wire. The number $S$ is a discrete random variable whose expectancy $N=\mathbf{E}[S]$ depends on $L$. We will call "flock density" (FD) the ratio $\mathbf{FD}(L) = \frac{N(L)}{L}$. For example

etc.

Your goal: Find the lengths $L_1$ and $L_2$ ($L_1,L_2>2$) for which the flock density is accordingly $\mathbf{FD}(L_1)=\frac{36}{51}$ and $\mathbf{FD}(L_2)=\frac{38}{51}$. Provide eight decimal digits of accuracy.

A bonus "*" will be given for finding the median of the distribution of the distance between two consecutive landing points for $L\to\infty$, where a "landing point" of a swallow is in the middle of the space it occupies. Provide four decimal digits after the decimal point.

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Solution

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  July 2025 Solution:

  The numerical solutions are:

  L1=6.0507220
  L2=100.96564
  

  Bonus:

  m=1.26278296766
  

  This problem is better known as "Rényi's Parking Problem" which deals with cars parking randomally along the road. When the cars are of unit length $1$ and we're interested in the mean number of cars that will be able to park on an interval of length $x$, we have the following recursive formula:

  $M\left(x\right)=\begin{cases} 0 & 0\le x<1\\ 1 & x=1\\ 1+\frac{2}{x-1}\int_{0}^{x-1}M\left(t\right)dt & x>1 \end{cases}$ This formula can be obtained in the following (rather informal) manner: Consider an interval of length $x+1$, and put a car on $t$ ($0\le t\le x$), with the effect of splitting the interval into two smaller intervals of lengths $t$ and $x-t$. The expected number of cars on the interval $(0,x+1)$ in this case is $M\left(t\right)+1+M\left(x-t\right)$ (since we also count the car already placed) And now, using The symmetry of $M(t)$ and $M(x-t)$ we obtain M\left(x+1\right)=\frac{1}{x}\int_{0}^{x}\left[M\left(t\right)+1+M\left(x-t\right)\right]dt$$=1+\frac{2}{x}\int_{0}^{x}M\left(t\right)dt

  There are several methods to numerically compute this integral. One nice approach described in "Numerical Solution of Some Classical Differential-Difference Equations" computes a power series representation for $M$ in the interval $[N,N+1]$ based on a power series representation for the previous integral. Implementing this approach yields an extremely fast and accurate algorithm for directly computing $FD$ and a simple binary search algorithm (based on $FD$'s monotonicity in the relevant domain) yields

  A common answer to the bonus was $1.337617\ldots$, which is $\frac{1}{C_r}$ where $C_r$ is Rényi's Parking Constant, $C\_r=\lim\_{x\to\infty}\frac{M(x)} {x}$. This is due to an interpretation of the bonus question as requesting the **mean** distance between two swallows. To compute it, first note the the distance between two consecutive swallows is equal to half the first swallow + half the second one + the empty space between them. So if we compute the mean empty space + 1, we obtain the requested result. The mean total empty space for a wire of length $x$ is $x-M(x)$, and dividing by the number of swallows yields the result $1+\frac{x-M(x)}{M(x)}=\frac{x}{M(x)}$ which, taken to the limit, is indeed $\frac{1}{C_r}$.

  The intended meaning of the bonus was to find the median, not the mean. This can be done via a Monte Carlo simulation.

Solvers

• *Bertram Felgenhauer (1/7/2025 4:27 PM IDT)
• *Bert Dobbelaere (1/7/2025 11:03 PM IDT)
• *Lazar Ilic (2/7/2025 4:39 AM IDT)
• *Alper Halbutogullari (2/7/2025 9:52 AM IDT)
• Jiri Spitalsky (2/7/2025 8:20 PM IDT)
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• *Daniel Bitin (2/7/2025 11:45 PM IDT)
• *Guangxi Liu (3/7/2025 6:34 AM IDT)
• *Ankit Chakraborty (3/7/2025 9:57 AM IDT)
• *King Pig (3/7/2025 4:54 PM IDT)
• *Juergen Koehl (4/7/2025 9:54 PM IDT)
• *Prashant Wankhede (5/7/2025 9:09 PM IDT)
• *Hugo Pfoertner (5/7/2025 9:18 PM IDT)
• Kang Jin Cho (7/7/2025 8:51 PM IDT)
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• *Daniel Chong Jyh Tar (9/7/2025 8:31 PM IDT)
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• *Shirish Chinchalkar (13/7/2025 11:16 PM IDT)
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• Gary M. Gerken (15/7/2025 12:50 PM IDT)
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• *Jordan Payette (17/7/2025 6:58 PM IDT)
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• *Camden Chappelle (18/7/2025 3:20 PM IDT)
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• *Hubert Puszklewicz (30/7/2025 8:50 PM IDT)
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• *Motty Porat (1/8/2025 1:53 AM IDT)
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