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I have the following code for Project Euler Problem 12. However, it takes a very long time to execute. Does anyone have any suggestions for speeding it up?

n = input("Enter number: ")
def genfact(n):
    t = []
    for i in xrange(1, n+1):
        if n%i == 0:
            t.append(i)
    return t

print "Numbers of divisors: ", len(genfact(n))
print

m = input("Enter the number of triangle numbers to check: ")
print
for i in xrange (2, m+2):
    a = sum(xrange(i))
    b = len(genfact(a))
    if b > 500:
        print a

For n, I enter an arbitrary number such as 6 just to check whether it indeed returns the length of the list of the number of factors. For m, I enter entered 80 000 000

It works relatively quickly for small numbers. If I enter b > 50 ; it returns 28 for a, which is correct.

TopGun
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5 Answers5

5

My answer here isn't pretty or elegant, it is still brute force. But, it simplifies the problem space a little and terminates successfully in less than 10 seconds.

Getting factors of n: Like @usethedeathstar mentioned, it is possible to test for factors only up to n/2. However, we can do better by testing only up to the square root of n:

let n = 36
=> factors(n) : (1x36, 2x18, 3x12, 4x9, 6x6, 9x4, 12x3, 18x2, 36x1)

As you can see, it loops around after 6 (the square root of 36). We also don't need to explicitly return the factors, just find out how many there are... so just count them off with a generator inside of sum():

import math

def get_factors(n):
    return sum(2 for i in range(1, round(math.sqrt(n)+1)) if not n % i)

Testing the triangular numbers

I have used a generator function to yield the triangular numbers:

def generate_triangles(limit):
    l = 1
    while l <= limit:
        yield sum(range(l + 1))
        l += 1

And finally, start testing:

def test_triangles():
    triangles = generate_triangles(100000)
    for i in triangles:
        if get_factors(i) > 499:
            return i

Running this with the profiler, it completes in less than 10 seconds:

$ python3 -m cProfile euler12.py 

361986 function calls in 8.006 seconds

The BIGGEST time saving here is get_factors(n) testing only up to the square root of n - this makes it heeeaps quicker and you save heaps of memory overhead by not generating a list of factors.

As I said, it still isn't pretty - I am sure there are more elegant solutions. But, it fits the bill of being faster :)

Nick Burns
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  • wow - when you complete this you get to read Project Euler's overview on the problem. They have some very cool approaches :) – Nick Burns Jul 19 '13 at 11:55
  • Hi, thanks for the reply. def get_factors(n): return sum(2 for i in range(1, round(math.sqrt(n)+1)) if not n % i) You said you use a generator within the sum. Could you explain this please – TopGun Jul 19 '13 at 12:18
  • My solution already counted to sqrt(n) instead of n/2?, and the fact of not writing them to memory but just counting doesnt improve much? – usethedeathstar Jul 19 '13 at 13:52
  • Yeah this definitely helps. I just wanted to understand how the 'generator' works 'within the sum'. – TopGun Jul 19 '13 at 14:01
  • @TopGun, hey, the `sum(...)` expression works like this: it tests every value `i` from `1 -> sqrt(n) + 1` and if `i` divides `n` evenly (`not n% i`), then it adds `2` to the sum. Putting it inside the sum(), means that it works in a similar way to `xrange` in Python2x, it evaluates one value at a time without needing to evaluate the entire sequence at once – Nick Burns Jul 19 '13 at 20:16
  • @NickBurns, I know the comments section says not to write 'Thanks' comments, but thank you for the detailed clarification. :D – TopGun Jul 31 '13 at 10:14
  • @NickBurns this solution will over count perfect squares. 36 properly has 9 unique factors not 10. otherwise very nice. – David Chan Dec 18 '14 at 15:47
4

I got my answer to run in 1.8 seconds with Python.

import time
from math import sqrt


def count_divisors(n):
    d = {}
    count = 1

    while n % 2 == 0:
        n = n / 2
        try:
            d[2] += 1
        except KeyError:
            d[2] = 1

    for i in range(3, int(sqrt(n+1)), 2):
        while n % i == 0 and i != n:
            n = n / i
            try:
                d[i] += 1
            except KeyError:
                d[i] = 1

    d[n] = 1

    for _,v in d.items():
        count = count * (v + 1)

    return count

def tri_number(num):
  next = 1 + int(sqrt(1+(8 * num)))
  return num + (next/2)


def main():
    i = 1
    while count_divisors(i) < 500:
        i = tri_number(i)
    return i


start = time.time()
answer = main()
elapsed = (time.time() - start)

print("result %s returned in %s seconds." % (answer, elapsed))

Here is the output showing the timedelta and correct answer:

$ python ./project012.py
result 76576500 returned in 1.82238006592 seconds.

Factoring

For counting the divisors, I start by initializing an empty dictionary and a counter. For each factor found, I create key of d[factor] with value of 1 if it does not exist, otherwise, I increment the value d[factor].

For example, if we counted the factors 100, we would see d = {25: 1, 2: 2}

The first while loop, I factor out all 2's, dividing n by 2 each time. Next, I begin factoring at 3, skipping two each time (since we factored all even numbers already), and stopping once I get to the square root of n+1.

We stop at the square_root of n because if there's a pair of factors with one of the numbers bigger than square_root of n, the other of the pair has to be less than 10. If the smaller one doesn't exist, there is no matching larger factor. https://math.stackexchange.com/questions/1343171/why-only-square-root-approach-to-check-number-is-prime

while n % 2 == 0:
        n = n / 2
        try:
            d[2] += 1
        except KeyError:
            d[2] = 1

    for i in range(3, int(sqrt(n+1)), 2):
        while n % i == 0 and i != n:
            n = n / i
            try:
                d[i] += 1
            except KeyError:
                d[i] = 1
d[n] = 1

Now that I have gotten each factor, and added it to the dictionary, we have to add the last factor (which is just n).

Counting Divisors

Now that the dictionary is complete, we loop through each of the items, and apply the following formula: d(n)=(a+1)(b+1)(c+1)... https://www.wikihow.com/Determine-the-Number-of-Divisors-of-an-Integer

All this formula means is taking all of the counts of each factor, adding 1, then multiplying them together. Take 100 for example, which has factors 25, 2, and 2. We would calculate d(n)=(a+1)(b+1) = (1+1)(2+1) = (2)(3) = 6 total divisors

for _,v in d.items():
    count = count * (v + 1)

return count

Calculate Triangle Numbers

Now, taking a look at tri_number(), you can see that I opted to calculate the next triangle number in a sequence without manually adding each whole number together (saving me millions of operations). Instead I used T(n) = n (n+1) / 2 http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/runsums/triNbProof.html

We are providing a whole number to the function as an argument, so we need to solve for n, which is going to be the whole number to add next. Once we have the next number (n), we simply add that single number to num and return

S=n(n+1)2

S=n2+n2

2S=n2+n

n2+n−2S=0

At this point, we use the quadratic formula for : ax2+bx+c=0.

n=−b±√b2−4ac / 2a

n=−1±√1−4(1)(−2S) / 2

n=−1±√1+8S / 2

https://socratic.org/questions/how-do-you-solve-for-n-in-s-n-n-1-2

So all tri_number() does is evaluate n=1+√1+8S / 2 (we ignore the negative equation here). The answer that is returned is the next triangle number in the sequence.

def tri_number(num):
  next = 1 + int(sqrt(1+(8 * num)))
  return num + (next/2)

Main Loop

Finally, we can look at main(). We start at whole number 1. We count the divisor of 1. If it is less than 500, we get the next triangle number, then try again and again until we get a number with > 500 divisors.

def main():
    i = 1
    while count_divisors(i) < 500:
        i = tri_number(i)
    return i

I am sure there are additional ways to optimize but I am not smart enough to understand those ways. If you find any better ways to optimize python, let me know! I originally solved project 12 in Golang, and that run in 25 milliseconds!

$ go run project012.go
76576500
2018/07/12 01:56:31 TIME: main() took 23.581558ms
Brandon
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one of the hints i can give is 

def genfact(n):
    t = []
    for i in xrange(1, n+1):
        if n%i == 0:
            t.append(i)
    return t

change that to 

def genfact(n):
    t=[]
    for i in xrange(1,numpy.sqrt(n)+1):
        if(n%i==0):
           t.append(i)
           t.apend(n/i)

since if a is a divisor than so is b=n/a, since a*b=a*n/b=n, That should help a part already (not sure if in your case a square is possible, but if so, add another case to exclude adding the same number twice)

You could devise a recursive thing too, (like if it is something like for 28, you get 1,28,2,14 and at the moment you are at knowing 14, you put in something to actually remember the divisors of 14 (memoize), than check if they are alraedy in the list, and if not, add them to the list, together with 28/d for each of the divisors of 14, and at the end just take out the duplicates

If you think my first answer is still not fast enough, ask for more, and i will check how it would be done to solve it faster with some more tricks (could probably make use of erastothenes sieve or so too, and some other tricks could be thought up as well if you would wish to really blow up the problem to huge proportions, like to check the first one with over 10k divisors or so)

usethedeathstar
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  • Thanks for the reply. I understand the logic behind your modification. However, as you suspected, it's still too slow. :) – TopGun Jul 19 '13 at 12:27
0
while True:
    c=0
    n=1
    m=1
    for i in range(1,n+1):
        if n%i==0:
            c=c+1
            m=m+1    
            n=m*(m+1)/2
            if c>500:
                break
print n
Tunaki
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Chanaka Hasun
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0

this is not my code but it is so optimized. source: http://code.jasonbhill.com/sage/project-euler-problem-12/

import time


def num_divisors(n):
    if n % 2 == 0: n = n / 2
    divisors = 1
    count = 0
    while n % 2 == 0:
        count += 1
        n = n / 2
    divisors = divisors * (count + 1)
    p = 3
    while n != 1:
        count = 0
        while n % p == 0:
            count += 1
            n = n / p
        divisors = divisors * (count + 1)
        p += 2
    return divisors


def find_triangular_index(factor_limit):
    n = 1
    lnum, rnum = num_divisors(n), num_divisors(n + 1)
    while lnum * rnum < 500:
        n += 1
        lnum, rnum = rnum, num_divisors(n + 1)
    return n


start = time.time()
index = find_triangular_index(500)
triangle = (index * (index + 1)) / 2
elapsed = (time.time() - start)

print("result %s returned in %s seconds." % (triangle, elapsed))
jackson jones
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