5

This is supposedly a thought-provoking interview question asked, and I though I have an idea of a possible solution, I can't prove it.

The question is the following:

You have $n$ cars that are all traveling the same direction on an infinitely long one-lane highway. Unfortunately, they are all going different speeds, and cannot pass each other. Eventually the cars will clump up in one or more traffic jams. In terms of $n$, what is the expected number of clumps of cars?

How would you go about solving a question like this mathematically?

Neil
  • 446
  • You can't answer this question without knowing anything about the distribution of the speeds among the cars. – Dominik Sep 07 '15 at 15:08
  • @Dominik True, so there is a statistics component here. Though I imagine like the sum of throwing two die, a pattern tends to emerge. How about this, what would the average number of clumps of cars be, given $n$? – Neil Sep 07 '15 at 15:12
  • 1
    @Dominik As long as the speeds of the cars are iid and the distribution is continuous, it doesn't matter. It's really asking how many cars are slower than all cars in front of it, which means that if we assume all cars have distinct speeds it's more of a combinatorics question. – Arthur Sep 07 '15 at 15:15
  • I retract my comment; As long as the speeds are all pairwise different the solution doesn't depend on their distribution. You basically ask for how many $i \in {1, \ldots, n - 1}$ you have $\sigma(i) > \sigma(i - 1)$, where $\sigma$ is a random permutation on $n$ elements. – Dominik Sep 07 '15 at 15:16
  • How can the solution not depend on the distribution? What if the speed of car $1$ is $n$ with probability $1$, and the speed of car $2$ is $n-1$ and so on? – 5xum Sep 07 '15 at 15:21
  • @5xum: I assumed the cars would just have any (pairwise different) speeds and then get "mixed" by applying a random permutation. – Dominik Sep 07 '15 at 15:23
  • @5xum You are right that that would change the question, and we would certainly need more information, but we can still answer it as asked if we assume that the speeds of all the cars are drawn from the same distribution. – Plutoro Sep 07 '15 at 15:23
  • Oh yeah. I agree with both of you. – 5xum Sep 07 '15 at 15:24
  • Does a single car count as a 'clump'? For instance if the fastest car is in the front, it will stay by itself in front without any clumping. – paw88789 Sep 07 '15 at 16:37
  • Possible related question: http://math.stackexchange.com/questions/1040479/shooting-bullets – Hetebrij Sep 07 '15 at 16:37
  • @paw88789 Yes, it is considered to be a clump. A clump meaning a group of 1 or more cars. – Neil May 26 '16 at 07:20

4 Answers4

3

Here's my thoughts. This isn't really my area, so please critique me if necessary. There will be a slowest car, and it will lie in the back as often as the front, so its expected position is the very center, forming a clump of $n/2$ cars behind it and $n/2$ cars driving in front of it. The cars in front will also have a slowest, which will, on average, split the group into $n/4$ behind it and $n/4$ in front. Continuing the process, we expect about $\log_2(n)$ clumps.

Plutoro
  • 23,378
3

It is easy to see that the $i^\mbox{th}$ car from front is at the head of a mini-jam if and only if it is slower than all cars before it.

Assuming a continuous distribution of speeds, the probability of this is simply $1/i$ as the cars are sampled from the same distribution and hence each one of them is equally likely to be the slowest.

Using linearity of expectation, expected number of clumps, \begin{align*} E_N &= \sum_{i=1}^N 1/i \\ &= \text{Harmonic number } H_N \end{align*}

CutestPenguin
  • 46
2

Let $c(n)$ denote the expected number of clusters. As written in the comments, I will assume that the cars have pairwise different speeds and then a random permutation is applied to rearrange the cars.

So wlog the cars are all ordered in the initial distribution [before we apply our permutation]. This means car $1$ is the slowest, while car $n$ is the fastest. Let $\sigma$ be our random permutation. If we interpret a single car as a cluster, then there will be a cluster ending in car $i < n$ iff $\sigma(i) < \sigma(i + 1)$. Additionally, there will be a cluster at position $n$ iff $\sigma(n) = n$.

So if $\sigma(n) = n$, then we will have $1$ cluster, plus the number of clusters in the first $n - 1$ cars. If $\sigma(n) \ne n$, then we will have the same number of clusters as in the first $n - 1$ cars, because the car number $n$ will always be the fastest and therefore end up in one of the clusters of the first $n - 1$ cars.

This means: $$\begin{align*}c(n) &= P(\sigma(n) = n)(c(n - 1) + 1) + P(\sigma(n) \ne n) c(n)\\ &= c(n - 1) + P(\sigma(n) = n) = c(n - 1) + \frac{1}{n}\end{align*}$$

Together with $c(1) = 1$ this yields $c(n) = H_n = \sum \limits_{i = 1}^n \frac{1}{i} \approx \ln(n) + \gamma$.

Dominik
  • 20,241
  • Why would it be a natural log and not a log base 2? A car can either be a part of the cluster in front or a part of the cluster behind, so it would seem to follow a trend of base 2, wouldn't it? – Neil Sep 08 '15 at 06:20
  • I don't have an intuitive explanation, but the considerations with the base 2 logarithm are too simple. – Dominik Sep 08 '15 at 06:53
-1

The logical answer for this question is that if we suppose that there are 4 cars 1,2,3 and 4 and they are going in an order like 

1

   2

      3

         4

1 is the slowest car 2 is the second in matters of speed 3 is third in the matters of speed
4 is the fastest one when 2 aligns with 1 sod o 3 and 4 because they come closer to one faster than 2 and then they are going parallel They will eventually form a clump lie 1 2 3 4

Neil
  • 446
Tanmay N
  • 1