Is there an algorithmically better way of doing this?

import numpy as np
def gcd(a, b):

    while b:      
        a, b = b, a % b
    return a
def lcm(a, b):

    return a * b // gcd(a, b)

while z<10:
    """count how many times we have crossed (-1-ed) a number from p"""
    while p[k]==-1 and k<9 :
        """find the first non -1 element in p (it's here that I feel that the code is especially clumsy) """

    """k is the first element in the cycle """
    while l[i]!=k:
        """go from i to l[i] until we get back to k, closing the cycle """
        """g is just used to find the lcm """


3 Answers 3


Use the built in gcd

In Python 3, you can just do:

from math import gcd

Consider keeping with the standard library

Instead of using np.random.permutation, consider just using random.shuffle to generate a random permutation. Removes the numpy dependency altogether.

Just use # instead of """

Comments such as:

"""count how many times we have crossed (-1-ed) a number from p"""

Are usually used for docstrings, you are using them to elaborate on certain steps in an algorithm, I would use # for this.

Place the cycle decomposition into a function

The while loop introduces a lot of global variables and makes it hard to use, I would put the contents into a function. Also explain that this function displays a permutation as disjoint cycles.

Please use better names.

I'm still not sure what every variable does. Like why is p=list(range(10))? I would think p is short for permutation, but l seems to be what the actual permutation is. This program is really hard to understand simply because of your name choice.


There's room for improvement. Apart from what @Dair already suggested, I'd also recommend a few other changes.

Magic numbers

You're using 10 in many places, so why not defining it at the top of your program? (keep in mind that constants should be upper-cased)

Code style


z = z + 1

Can be replaced by this:

z += 1

Since you're generating a shuffled list already, why not sort that to get the p ? So:

p = list(range(10))

Can become:

p = sorted(l)

While I understand the reason @Dair suggested shuffle instead of numpy, I'd suggest you keep it the way you did, as it's the fastest way you can get a shuffled list.

  • 2
    \$\begingroup\$ One thing I would like to note: When I say "consider" I usually mean that there are reasons to disregard the advice I gave but you should, well, consider my point and decide whether it is right for situation. (This is regarding your last point and numpy) \$\endgroup\$
    – Dair
    Commented Apr 25, 2017 at 16:26

In addition to what @Dair said (especially about variable names):

  • You have a lot of hardcoded numbers in there (going from 10 to some other upper limit means replacing quite a few other numbers). Better keeping those in a constant like UPPER_LIMIT = 10 and using that.
  • You don't need that while p[k]==-1 and k<9 : loop. You know which element is the first -1 because you're the one setting it, just keep track of the index and you're good to go (BTW, what if you had a million elements, would you go through all of them every time?).

I didn't test it thoroughly, but this should give you the same result without that loop:

k = 0
while z < 10:
    i = k
    p[k] = -1
    z = z+1
    j = 1
    old_k = k
    k_set = False
    while l[i] != old_k:
        p[l[i]] = -1
        k += 1
        z += 1
        i = l[i]
        j += 1
    prod = lcm(prod,j)

  • \$\begingroup\$ thanks, but you are wrong in the second point (the first element, after the one that is set to -1 at the end of the inner while loop, may be a -1 already) \$\endgroup\$
    – Nesa
    Commented Apr 25, 2017 at 13:18
  • \$\begingroup\$ I see. In that case you should still consider what I said: keep track of the minimum index of the element that is not -1. As I said, you're the one setting it, so you can do that. \$\endgroup\$
    – ChatterOne
    Commented Apr 25, 2017 at 14:37
  • \$\begingroup\$ I don't think that's possible, consider a permutation where by my method we -1 out the following elements during the first cycle: 0,8,7,6,5,4 , so the minimal element that is not -1is 1, then after we cross out 2 and 3 how do we know if 9 is still there? I don't know if my approach would run faster if we did it with a set instead of 1-ing in a list like here: gist.github.com/begriffs/2211881 \$\endgroup\$
    – Nesa
    Commented Apr 25, 2017 at 16:53

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