Given an input N as the size of a centrifuge, I need to find out the number of balanced configurations. The full description is here.

Centrifuge is a piece of equipment that puts multiple test tubes in rotation at very high speeds. It consists of a cylindrical rotor with holes situated evenly along the circumference of the rotor. Because the device is operating at high speeds, the center of mass of all tubes must coincide with the center of the rotor. We assume all tubes have the same mass.

For the sake of simplicity, we also assume that tubes are point masses, fixed firmly to the rotor. All at the same distance from the center. You may safely assume the (x,y) coordinates of tube k are \$R\sin a, R\cos a\$, where \$a = 2\pi\frac{k}{n}\$

The problem is: Given a centrifuge with N holes and K tubes, is it possible to balance it?

example 1: given N = 6 possible values for K are 2,3,4,6.

example 2: given N = 12, it is possible to balance 5 tubes: put 3 evenly dispersed (every 4th hole), then find a pair of opposite unoccupied holes and insert remaining 2 tubes there.

example 3: given N = 10, it is impossible to balance 7 tubes: put 5 evenly dispersed, and there's no opposite unoccupied holes left!


Line 1: An integer N for the capacity of the centrifuge.


Line 1: An integer M for the number of different possible values of K.

This is my code:

import numpy as np

angle = 2*np.pi/N
z = complex(np.cos(angle), np.sin(angle))

import itertools
combs = []
lst = range(1,N)
for i in range(2, N-1):
    els = [list(x) for x in itertools.combinations(lst, i)]

lis = []
while count<=len(combs):
    for a in range(len(combs[count])):
        for b in range(len(combs[count][a])):
        if abs(s)<1e-10:

lengths = []
for i in lis:
    if len(i) not in lengths:


The code works fine, but slows down after input size of above 20, and is practically unusable after the input grows to 50 or above, due to the for loops. How do I optimize it?

  • \$\begingroup\$ Welcome to Code Review. The current question title, which states your concerns about the code, obfuscates the original intend of your code. The site standard is for the title to simply state the task accomplished by the code. Please see How to Ask for examples, and revise the title accordingly. \$\endgroup\$
    – Zeta
    Commented Dec 20, 2018 at 7:58
  • \$\begingroup\$ @Zeta Is it better now? \$\endgroup\$ Commented Dec 20, 2018 at 8:06
  • 2
    \$\begingroup\$ Now your title only states the concerns and wishes you have for your code. A better title would be something similar to "Balancing centrifuges" or "Balanced centrifuge configurations". \$\endgroup\$
    – Zeta
    Commented Dec 20, 2018 at 8:10
  • \$\begingroup\$ Ok, got it. Replacing it now with your suggested title. \$\endgroup\$ Commented Dec 20, 2018 at 8:11
  • 1
    \$\begingroup\$ Please do not update the code in your question to incorporate feedback from answers, doing so goes against the Question + Answer style of Code Review. This is not a forum where you should keep the most updated version in your question. Please see what you may and may not do after receiving answers. \$\endgroup\$
    – Mast
    Commented Jul 23, 2019 at 7:38

3 Answers 3


This is not a review, but an extended comment.

The code cannot be meaningfully optimized. Few notes though.

  • First, thou shalt not bruteforce. It is almost inevitably wrong.

  • Next, you must realize that it is a number-theoretical problem. Even the names of the test cases suggest so. Among other things it means that testing the sum against 1e-10 is not right.

  • Finally, do watch the video referenced in the problem. It is a big spoiler, and it doesn't provide a proof of the claim. The proof is far from trivial; if you are interested in the proof, follow up with this is a fascinating reading (trigger warning: heavy-duty math).

  • \$\begingroup\$ Well, about the 1e-10, I used it because of the floating point error. So, I thought 1e-10 should be a sufficiently small difference between 0 and s to establish their equality (i.e., s=0). \$\endgroup\$ Commented Dec 20, 2018 at 9:18
  • \$\begingroup\$ @Kristada673 Sufficiently small for which Ns? \$\endgroup\$
    – vnp
    Commented Dec 20, 2018 at 9:20
  • 1
    \$\begingroup\$ "First, thou shalt not bruteforce. It is almost inevitably wrong." In what context? This is definitely not true in real life (wiki.c2.com/?PrematureOptimization), and often not true in coding challenges either. I was in a coding competition in college. Our team was doing practice problems, and I was working on an efficient way to solve the next one. My teammate beat me handily by using a much simpler, brute-force approach. It all depends on the size and complexity of the data and the algorithm. In this case, yes, brute-force is the wrong approach. \$\endgroup\$
    – LarsH
    Commented Dec 20, 2018 at 19:50
  • 1
    \$\begingroup\$ The link now 404s. \$\endgroup\$ Commented Jul 23, 2019 at 7:21
  • \$\begingroup\$ Was it one of the references mentioned in oeis.org/A322366 ? \$\endgroup\$ Commented Jul 23, 2019 at 7:57

There are two main problems:

  1. Enumerating all combinations, even those that are obviously unbalanced

  2. The use of floating point calculus in a combinatorics problem

Re-read the task carefully, especially examples 2 and 3, which hint a better algorithm:

  1. Factorize N. The factors of N=12 are [2,3], which tells us that all tubes are members of either a balanced pair or triplet.

  2. Find all combinations of factors that sum to K. For N=12, K=10, there are two solutions: 2+2+2+2+2=10 and 2+2+3+3=10.

  3. Find out if the solutions are valid. For N=10, K=7 the only solution is 2+5=7, which is not valid (see example 3).

For large values of K, balance holes instead of tubes. So N=12, K=10 is the same as N=12, K=2.


Revisiting this problem exactly 1 year after asking it, the following is a dynamic-programming based solution, thanks to commenter Niako in the original post (linked in the question):

N = int(input())

def primeFactorization(n):
    """ Return the prime factors of n """
    p_factors = []
    lastresult = n
    while 1:
        if lastresult == 1:
        c = 2
        while 1:
            if lastresult % c == 0:
            c += 1
        if c not in p_factors:
        lastresult /= c
    return p_factors

F = primeFactorization(N)
R = [False]*(N+1)
R[0] = True
for p in F:
    for n in range(p,N+1):
        R[n] = R[n] or R[n-p]

sols = []
for i in range(2,int(N/2)+1):
    if R[i]==True:

This works, except for 2 test cases, cases 7 and 8 (i.e., inputs 33 and 35, where the expected outputs are 13 and 11 respectively, but this program outputs 15 for both). It would be great if someone can point out why.


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.