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Asked: 2026-05-26T10:04:18+00:00

I’m creating an iterative algorithm (Monte Carlo method). The algorithm returns a value on

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I’m creating an iterative algorithm (Monte Carlo method).
The algorithm returns a value on every iteration, creating a stream of values.

I need to analyze these values and stop the algorithm when say 1000 returned values are withing some epsilon.

I decided to implement it calculation the max and min values of the last 1000 values, and then calculate the error using this formula (max-min)/min and compare it to epsilon: error<=epsilon. And if this condition is reached, stop the iterations and return the result.

  1. The first hare-brained idea was to use a list and append new values to it, calculating the max and min values for the last 1000 values of it after each appending.

  2. Then I decided there is no use of keeping more that 1000 last values. So I remembered of deque. It was a very good idea since the complexity on adding and deleting on both ends of deque object is O(1). But it didn’t solve the problem of needing to go through all the last 1000 values on each iteration to calculate min and max.

  3. Then I remembered there is the heapq module. It keeps the data in such a way as to efficiently return the smallest one at every moment. But I need both the smallest and the largest ones. Furthermore I need to preserve the order of the elements so that I can keep 1000 last returned elements of the algorithm, and I don’t see how I can achieve it with heapq.

Having all those thoughts in mind I decided to ask here:

How can I solve this task the most efficiently?

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1 Answer

  1. Editorial Team

    If you are free / willing to change your definition of error, you might want to consider using the variance instead of (max-min)/min.

    You can compute the variance incrementally.
    True, using this method, you are not deleting any values from your stream — the variance will depend on all the values. But so what? With enough values, the first few won’t matter a great deal to the variance, and the variance of the average, variance/n, will become small when enough values cluster around some fixed value.

    So, you can choose to halt when the variance/n < epsilon.

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