Project Euler #23 – Python Non Abundant Sums

I have done a couple of changes to the code mentioned above and i had the correct answer within 13 seconds. My CPU is intel core i5. Here is my code:

from array import array
import math


def find_abundant(number):
    sum_factor = 1
    value = math.ceil(math.sqrt(number))
    if number == 1:
        return False
    for i in range(2, value):
        if number % i == 0:
            sum_factor += (i+(number//i))
    if value**2 == number:
        sum_factor += value
    if sum_factor > number:
        return True
    else:
        return False


numbers = [0]*28123
abundant_numbers = array("i", [])
for abundant in range(1, 28124):
    if find_abundant(abundant):
        abundant_numbers.append(abundant)

for x in abundant_numbers:
    for y in abundant_numbers[0:abundant_numbers.index(x)+1]:
        z = x+y
        if z < 28124 and numbers[z-1] == 0:
            numbers[z-1] = z


_sum = 0
for vary in range(1, len(numbers)+1):
    if numbers[vary-1] == 0:
        _sum += vary
print(_sum)}

This code works 6 seconds on my computer:

from math import sqrt
import itertools
import functools
import operator
#simplicity test
def fuc(a):
    d = 1
    z = 0
    while d<= sqrt(a):
        d = d + 1
        if a == 2:
            z = a
        elif (a%d) == 0:
            z = False
            break

        else:
            z = a
    return z
#prime number divisors
def func(value):
    v = []
    d = 1
    value1= value# for optimization
    while d <= sqrt(value1):
        d+=1
        if fuc(d)!= False and value1 % d == 0:
            v.append(d)
            value1 = value1/d
            d = 1
    if value1 != value and value1 != 1:
        v.append(value1)
    return v
# all number divisors without 1 and source number
def calculate_devisors(n):
    prime_multiples_list = func(n)
    unique_combinations = set()
    for i in range(1, len(prime_multiples_list)):
        unique_combinations.update(set(itertools.combinations(prime_multiples_list,i)))
        combinations_product = list(functools.reduce(operator.mul,i) for i in unique_combinations)
        combinations_product.sort()
    try:
        return combinations_product
    except:
        return []

abundentNums = []

for n in range(1,28123):
    if sum(calculate_devisors(n))+1>n:
        abundentNums.append(n)

sums = [0]*28124
for x in range (0, len(abundentNums)):
    for y in range (x, len(abundentNums)):
        sumOf2AbundantNums = abundentNums[x]+abundentNums[y]
        if (sumOf2AbundantNums<= 28123):
            if (sums[sumOf2AbundantNums] == 0):
                sums[sumOf2AbundantNums] = sumOf2AbundantNums
ans = 0
for i in range(1,len(sums)):
    if sums[i]==0:
        ans+=i
print(ans)

I found these small tweaks might help reducing the overall time taken. I got the in about 4 seconds on my computer.

from math import sqrt
from itertools import compress
start = time.process_time()

abundant_nos = []
for i in range(12,28123):
    factor = 0
    for j in range(2,int(sqrt(i))+1):
        if i%j == 0:
            factor+= j + i//j
            if j == sqrt(i):
                factor -= j
    if factor>i:
        abundant_nos.append(i)
num_list = [True]*28123
k = 0
for i in abundant_nos:
    for j in abundant_nos[k:]:
        if(i+j>28123): break
        num_list[i+j-1] = False
    k+=1
answer = sum(compress(range(1,28124),num_list))

print(answer)

A possible solution with numpy that avoids the nested loop, and generally times less than 3 seconds on my laptop, is the following.

Description

Take your list of abundant Numbers and transform to numpy array A. Make an additional array containing the numbers from 1 to the limit posed in the problem B. Use an outer sum to create matrix that contains all possible sum-combinations of abundant numbers below the limit. And delete duplicate entries. And finally the difference between the sum of all numbers to limit in B and sum of all numbers that CAN be expressed as sum of two abundant numbers in C is the quantity looked for.

Code

A = np.array(abundentNums)
B = np.linspace(1,28123,28123)
C = np.unique(np.add.outer(A, A)[(np.add.outer(A, A) < 28124)])
ans = int(np.sum(B) - np.sum(C))

So therefore the entire code including the corrected getDivisors function by Miriam Farber is

import math
import numpy as np

def getDivisors(num):
    if num==1:
        return 1
    n = math.ceil(math.sqrt(num))
    total = 1
    divisor = 2
    while (divisor < n):
        if (num%divisor == 0):
            total += divisor
            total += num//divisor
        divisor+=1
    if n**2==num:
        total+=n
    return total

def isAbundant(num):
    if (getDivisors(num) > num):
        return True
    else:
        return False

abundentNums = []
for x in range (0,28124):
    if (isAbundant(x)):
        abundentNums.append(x)
del abundentNums[0]


A = np.array(abundentNums)
B = np.linspace(1,28123,28123)
C = np.unique(np.add.outer(A, A)[(np.add.outer(A, A) < 28124)])
ans = int(np.sum(B) - np.sum(C))

A further speed tweak would be list comprehension to make the list of abundant numbers.