I came across this problem:

The following iterative sequence is defined for the set of positive integers:

    n → n/2 (n is even)
    n → 3n + 1 (n is odd)

Using the rule above and starting with 13, we generate the following sequence:

  13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1

It can be seen that this sequence (starting at 13 and finishing at 1) contains 10 terms. Although it has not been proved yet (Collatz Problem), it is thought that all starting numbers finish at 1.

Which starting number, under one million, produces the longest chain?

and to this I gave this solution:

def number(n):
    ctr = 1
    if n % 2 == 0:
       k = n / 2
       ctr += 1
       k = 3 * n + 1
       ctr += 1
    while k > 1:
       if k % 2 == 0:
           k /= 2
           ctr += 1
           k = k * 3 + 1
           ctr += 1
    return ctr

def k(upto):
    for num in range(1,upto):
         if number(num)>last_ctr:
    return largest_number

print k(999999)

I am new to Python and don't think this is a good solution to this problem and maybe its complexity is \$\mathcal{O}(n^2)\$. So what is the best code for this problem if the number is really huge? How do I reduce the complexity?

And yes I read other questions related to this topic but they were not satisfactory answers to this in Python.

  • 2
    \$\begingroup\$ Note: This is Project Euler's 14th problem \$\endgroup\$
    – tobias_k
    Commented Mar 3, 2016 at 20:45
  • \$\begingroup\$ Your current solution is a faithful representation of the "brute force" approach. While there's room for improvement programming-wise, any radical change can only be achieved by doing the maths to find a better approach, I'm afraid. \$\endgroup\$
    – ivan_pozdeev
    Commented Mar 3, 2016 at 20:45
  • 3
    \$\begingroup\$ If you don't mind using lots of memory, you could go for a sort of dynamic programming approach, and store the values for each number as you find them, so once you reach the start of a chain you've already calculated, you don't have to do the work again. \$\endgroup\$
    – StephenTG
    Commented Mar 3, 2016 at 20:45
  • \$\begingroup\$ related: stackoverflow.com/q/13219320/674039 \$\endgroup\$
    – wim
    Commented Mar 3, 2016 at 21:09

3 Answers 3


The key to solving this problem (and many others) is using memoization: The numbers of the collatz sequence converge in a kind of tree-like structure. Once you have calculated the value of, say collatz(100) on one path, you do not need to calculate it again when you encounter it again on another path.

One way to do this would be to have (a) a cache and (b) a list of all the numbers on the current path. As soon as you encounter a number that is already in the cache, go your path backwards and add all the numbers to the cache.

cache = {1: 1}
def collatz(n):
    path = [n]
    while n not in cache:
        if n % 2:
            n = 3 * n + 1
            n = n / 2
    for i, m in enumerate(reversed(path)):
        cache[m] = cache[n] + i
    return cache[path[0]]

Another, maybe more elegant way would be using recursion and a decorator function. You can define a @memo decorator and use it again and again for all sorts of function you want to apply memoization to. This decorator automatically replaces your function with a version that looks in the cache, first.

def memo(f):
    f.cache = {}
    def _f(*args):
        if args not in f.cache:
            f.cache[args] = f(*args)
        return f.cache[args]
    return _f

def collatz(n):
    if n == 1:
        return 1
    if n % 2 == 0:
        return 1 + collatz(n / 2)
    if n % 2 == 1:
        return 1 + collatz(3 * n + 1)

In both cases, call like this and after a few seconds you should get your result.

print(max(range(1, 10**6), key=collatz))

Like some of the comments said, you can use dynamic approach:

def k(upto):
    #save the reults you already have
    results=[0 for i in range(upto+1)]

    #for every other number
    for num in range(2,upto+1):

        #if we don't already know how long the sequence is

            #one round of collatz
            if col%2==0:
                col=col >> 1
            #if our number is small enough to be in the list
            if col<upto:
                #try to take a shortcut

    return results[1:]

print max(k(1000000))

the idea is that if you start with 16 for example

You get to 8 after one iteration

But we have already calculated the sequence length for 8, so why do it again?

So we save all lengths and return the steps it took to get to 8 + the length of the sequence from 8 on.

This should run in around O(n) but it is hard to say with Collatz

(If Collatz is wrong it doesn't ever finish, but I don't think you have enough memory to reach those numbers)


Faster thanks to the help of dynamic programming, at the cost of some extra memory usage. We save the lengths of each chain we calculate, and reuse them so we only need to calculate them once. This implementation only saves lengths for values below the number we're going up to, but since the collatz function will go above this, we could raise the range of values we save if we don't mind increasing space used

def k(upto):
    def collatz(n):
        if n < upto and lst[n] > 0:
            return lst[n]
        if n % 2 == 0:
            val = collatz(n/2) + 1
            val = collatz((3*n + 1)/2) + 2
        if n < upto:
            lst[n] = val
        return val

    lst = [0]*upto
    lst[1] = 1
    lst[0] = 1
    for i in range(upto):
    return max(lst)
  • 1
    \$\begingroup\$ Wow that's beautifully short. Might I suggest using return collatz((3*n+1)/2) +2 instead of return collatz(3*n + 1) + 1, as 3n+1 is always divisible by 2? @StephenTG \$\endgroup\$
    – JeD
    Commented Mar 3, 2016 at 21:28
  • \$\begingroup\$ @JeD I suppose. Since if 3n+1 is in our list, so is 3n+1 / 2, right? \$\endgroup\$
    – StephenTG
    Commented Mar 3, 2016 at 21:35
  • 1
    \$\begingroup\$ Yep, It's even more probable that 3n+1/2 is in the list since it is smaller. @StephenTG \$\endgroup\$
    – JeD
    Commented Mar 3, 2016 at 21:36

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