I have the following code:

n - number of 1-d points (real numbers)
numbers[n] - array of numbers n
list, se_list - list variables to store cluster starting points


find_cluster(start, end, ci): \\ ci - cluster index
  best_se = +infinity \\ best squared eror
  list = {start}
  for i from start to end−ci+1:
    if ci > 1:
      (se, se_list) = find_cluster(i+1, end, ci-1)
      se = get_se(i+1, end) 
      se_list += {i+1} 
  new_se = get_se(start, i)
  if new_se + se < best_se:
    best_se = new_se + se
    list = list + se_list
  return (best_se, list)

\\ this computes variance of points from start to end
get_se(start, end):
  sum = 0
  for i from start to end:
    sum = sum + numbers[i]
  mean = sum/(end-start+1)
  se = 0
  for i from start_to_end:
    se = se + (numbers[i]−mean)*(numbers[i]−mean)
  return se

As far a my analysis goes, this algorithm take time O(n^(2k)), where k = ci, but I am not sure if I am correct. I first proved it to myself as

T(n) = n*(n + T(n-1)) = n^2 + n*(n-1)*((n-1) + T(n-2)) + ...

so it tends to n^k, but there are smaller arguments which are for sure smaller than n^k, so overall time is O(n^(2k)). But I am not very experienced so what is the running time of it?

Details: the original question for each I made this algorithm is "there are n points in 1 dimension (real numbers) which you need to cluster as k-means in polynomial time". K (i called it ci in this algorithm) is the number of clusters.

Details 2: By smaller arguments i meant that when we start opening parenthesis in T(n) = n^2 + n*(n-1)*((n-1) + T(n-2)) -> there will appear 1 argument n^k and a lot of smaller arguments. Though n^k is the largest power in this equation, if I take O(n^k) < O(n^2k), then this will also account for smaller arguments (just because), but i dont think that "just because" is a valid argument especially if a constant should do. But still the question holds O(n^2k).

NB. I need to check the links in comments, but if I understood this well enough there wouldn't be a need to ask this question right? So I do not think that referencing me to other materials unless they site exactly the same algorithm is right. So I am asking someone with good understanding of algorithms either verify my claim (better with small proof) or just say i'm plain wrong and possibly provide an explanation why.

  • $\begingroup$ Your analysis is very terse. Can you edit the question to clarify where you got those equations from? You want the running time of invoking what function, with what arguments? What is T(n) intended to represent? How did you get that recurrence relation? Where did you get that solution to the recurrence? I don't think you've solved the recurrence accurately. What do you mean by "there are smaller arguments which are for sure smaller than n^k"? Have you read our reference questions? cs.stackexchange.com/q/2789/755, cs.stackexchange.com/q/23593/755 $\endgroup$ – D.W. Dec 4 '16 at 16:33
  • $\begingroup$ T(n) is abstract running time, complexity, etc. described as a function of n - number of points on the real line. I need particularly O(n) complexity, which I am trying to get from T(n). I got this solution from recurrence just by substituting T(n-1) with T(n-2), etc. I do not need a super rigorous proof of this - I either confirmation of O(n^(2k)) or refusal. Better with a small proof. $\endgroup$ – Tom Dec 4 '16 at 16:52
  • $\begingroup$ If you just want to know what the running time of k-means clustering is, that's already answered on Wikipedia: en.wikipedia.org/wiki/K-means_clustering#Complexity. I haven't looked at your code carefully but I'm skeptical that it is actually computing a k-means clustering. $\endgroup$ – D.W. Dec 4 '16 at 17:05
  • $\begingroup$ No, I know the k-means algorithm that is np-hard. In this case it is in 1 dimension, and this algorithm is not k-means. The problem can be solved in the best time of k*n^2 with dynamic programming. But i didnt use this technique. Instead mine is a brute-force algorithm that trims visited combinations of points due to sorting them. And that sure should provide polynomial result in 1 dimension. But i am not sure if i didnt make a blunder in my reasoning somewhere $\endgroup$ – Tom Dec 4 '16 at 17:23

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