There is an array with $n$ places. There is a stream of $n$ unique numbers that arrive at a random order (permutation selected uniformly at random).

Whenever a number arrives, we must put it somewhere in the array, and we are not allowed to move it later. The goal is to have as many numbers as possible (in expectation) in their correct location in the final array. The "correct location" is defined as the location where it would appear if the numbers were sorted.

What is known about this problem? What is the best expected success rate, and what algorithm attains it?


  • A related question is: What is the fastest online sorting algorithm? . It discusses a situation in which items can be moved when new items arrive, and the goal is to minimize the running time.

  • I am mainly interested in maximizing the expected number of the correct positions. However, it is also interesting if there is a way to improve the probability that all numbers are in the correct position, above O(1/n!)

  • 2
    $\begingroup$ 1. Have you tried working through small cases ($n=2$, $n=3$, ...) to see what you can say for small values of $n$? 2. Should we assume the $n$ numbers are unique? If not, I suggest editing the question to define "correct location" more carefully. $\endgroup$
    – D.W.
    Jun 24, 2016 at 16:45
  • 1
    $\begingroup$ Do you have distribution of data? The running time depends on more than just perfectly hitting the proper placement of elements. $\endgroup$
    – Evil
    Jun 24, 2016 at 17:09
  • $\begingroup$ My gut says that nothing can be done if you can not estimate ranks. For the first element you have no chance. Knowing some elements already, better estimates may possible. $\endgroup$
    – Raphael
    Jun 26, 2016 at 11:50
  • $\begingroup$ Maybe minimizing the number of inversions is easier than maximizing the number of correct positions. $\endgroup$
    – adrianN
    Jun 27, 2016 at 9:26
  • $\begingroup$ Do your original numbers (before the permutation) come from some random process, e.g., some particular distribution? Or are they arbitrary (i.e., they could be chosen by an adversary, and we want to know about worst-case performance for the worst possible set of numbers)? This will lead to different answers. $\endgroup$
    – D.W.
    Jun 27, 2016 at 18:03

2 Answers 2


Define a configuration to be a state of the array $A[]$ after receiving some of the numbers. Define a configuration to be obviously wrong if there exists indices $i<j$ such that $A[i],A[j]$ have already been filled in with numbers such that $A[i]>A[j]$.

A randomized algorithm

Here is a natural algorithm: guess where to put each number uniformly at random, conditioned on the one constraint that you never place a number in an obviously wrong position.

The analysis of this approach seems pretty tricky. Based on some back-of-the-envelope estimates I conjecture that this attains a $\sim e^{-\Theta(n)}$ probability of placing all numbers in the correct position, which is better than the naive $1/n!$ algorithm. However I have no proof, so this could be wrong.

A deterministic algorithm

Here's what I can show. Let's consider a variation on the above algorithm: when you receive each number, find the range of array indices that wouldn't be obviously wrong, and then place that number in the exact middle (median) of that range. This algorithm is now deterministic.

For this algorithm, I can show that the probability of placing all numbers in the correct position is asymptotically larger than $1/n!$. In particular, it is at least something like $1/2^{\Theta(n)}$. Here's a hand-wavy analysis. For simplicity, let's assume $n$ is one less than a power of two. Place all $n$ numbers in a complete binary search tree. Now consider all sequences where you enumerate all the numbers at depth $1$ in some order, then all the numbers at depth $2$ in some order, then all the numbers at depth $3$ in some order, and so on. A hand-wavy estimate suggests that there are something in the vicinity of $n!/2^{\Theta(n)}$ such sequences, i.e., a random permutation has very roughly a $1/2^{\Theta(n)}$ probability of having this form. Moreover, the second algorithm above always succeeds in placing every number in the correct position. Therefore, the second algorithm above achieves a success rate of something in the vicinity of $1/2^{\Theta(n)}$. In particular, the success rate is significantly larger than $1/n!$.

For values of $n = 2^i-1$, we can characterize the exact probability of being correct: letting $P(n)$ denote the probability that the second algorithm succeeds in placing every number in the correct position, we find

$$P(n) = {1 \over n} \times P(\lfloor n/2 \rfloor)^2,$$

because the first element needs to be the median, and then you have two problems of half the size (the subsequence of numbers smaller than the median has to be a sequence with the same property, and the same for the subsequence of numbers larger than the median). We have the base cases $P(1)=1$ and $P(3)=1/3$. This recurrence relation grows like $P(n) \sim 2^{-e/2 \cdot (n+1)}$. Here $e/2 \approx 1.359\ldots$, so $P(n)$ grows roughly like $2^{-1.36 n}$, and in particular, much faster than $1/n!$.

In particular, here's one way to analyze this recurrence relation. Let $Q(n) = -\lg P(n)$. Then we find

$$Q(n) = 2 Q(\lfloor n/2 \rfloor) + \lg n,$$

where $Q(1)=0$ and $Q(3) = \lg 3$. Letting $R(i) = Q(2^i-1)$, we find

$$R(i) = 2 R(i-1) + \lg(2^i - 1),$$

with base case $R(1)=0$. Expanding, we find

$$R(i) = 2^{i-1} \times \left(\lg(1) + {\lg 3 \over 2} + {\lg 7 \over 4} + {\lg 15 \over 8} + {\lg 31 \over 16} + \dots + {\lg (2^i-1) \over 2^{i-1}}\right).$$

Summing the series, we find that to an excellent approximation

$$R(i) \approx 2^{i-1} \times e.$$

Therefore, $Q(n) \approx e/2 \cdot (n+1)$ and $P(n) \approx 2^{-e/2 \cdot (n+1)}$.

Credits: My thanks to @Algorithms with Attitude for the recurrence relation and the idea of analyzing the probability in this way.


maybe the title of the question should be a little different, like "Online sorting with the lowest number of modifications of the position of a number".

If you do not have the distribution of the numbers, than you should try to estimate and update the number distribution according to the numbers that come. There are algorithms to create a distribution function according to some data set (the numbers received until now) https://www.mathworks.com/matlabcentral/answers/33917-how-do-i-determine-the-probability-distribution-of-data

Run the probability calculation algorithm after every number and then insert the new number accordingly in your array.

I have no idea where you should put the first number, probably in the middle.

  • $\begingroup$ @D.W. If you have a set of numbers and then those numbers forma distribution. The more numbers you have from the dataset the more accurate you can estimate the distributition. There is no such thing as D1 distribution and D2 distribution and so on. There is distribution D. And the purpose is to get closer and closer to that D from the numbers you have. For this purpose you calculate the pdf (probability distribution function) at every step. It does not increase the complexity of the algorithm and it helps you to insert the numbers more accurate in the array. $\endgroup$
    – yoyo_fun
    Jun 27, 2016 at 16:58
  • $\begingroup$ On the long run past inputs are EXTREMELY good predictors for the future. Even at the stock market. There are extremely complex algorithms to estimate the way the stock market will evolve. Maybe for just one data set you can get errors but on enough datasets you will see that you can estimate future inputs from previous ones more and more accurate. That is the reason statistics exist and that is why it is studied at every computer science university program. I suggest reading a probability and statistics course to understand better how statistics work. $\endgroup$
    – yoyo_fun
    Jun 27, 2016 at 17:06
  • $\begingroup$ For example when you do an online survey previous answers are extremely good indicators for future answers. There are even specific distribution tests to see if the answers are real and not auto-completed randomly by the person who should have give the surveys to people. This works exactly because previous inputs influences future results. If it does not that is called (random uniform distribution) but if you get inputs from users this is NEVER the case. Only when you have a random number generator you can get a random uniform distribution. $\endgroup$
    – yoyo_fun
    Jun 27, 2016 at 17:12
  • $\begingroup$ OK. I see what you mean. I think my question should have been this: What is the expected success rate of your proposal? I agree that your approach seems like it should be better than blind guessing, so it seems like a promising approach. That said... the question asks what expected success rate is attainable; until we have an analysis that indicates what success rate is attained by this approach, this doesn't seem like a complete answer. $\endgroup$
    – D.W.
    Jun 27, 2016 at 17:45
  • $\begingroup$ @D.W. I understand what you mean. I do not know what the success rate might be. It depends on the type of the initial distribution (variance and expected value) the size of the data set and other parameters. The only way to find out is to make the program and run some tests. Unfortunately in this moment I do not have the time to do this. Maybe in a few weeks. $\endgroup$
    – yoyo_fun
    Jun 27, 2016 at 18:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.