# Express a number as sum of 2 primes

Problem: Given an even number (greater than 2), return two prime numbers whose sum will be equal to given number.

NOTE 1: A solution will always exist. (Goldbach's conjecture)

Note 2: If there are more than one solutions possible, return the lexicographically smaller solution.

This solution becomes slower for higher order numbers. How can I optimise this code?

import math
# Function to create a list of all prime numbers uptil given number
def list_prime(A):
primes = []  # initialise an empty list of primes
bool_a = [True for i in range(2,A+3)]  # creates boolean list of index
for  i in range(2, int(math.sqrt(A))+1):
if bool_a[i]:
for j in range( i * i, A +1 , i):  # eliminate all multiples of i*i+i, i*i+2i and so on uptil A+1
if bool_a[j]:
bool_a[j] = False;

for i in range(2,A):  # the index of reamaining bools are prime numbers
if bool_a[i]:
primes.append(i)

return primes

#Function that returns two prime numbers whose sum is equal to the given number
def prime_sum(A):
solution_set = []

for i in (list_prime(A)):
for j in (list_prime(A)[::-1]):
if i + j == A:
solution_set.append((i,j))
break

return min(solution_set)

• You calculate all the primes on each iteration. Caching this result will help tremendously – Maarten Fabré May 12 '18 at 17:00
• (Goldbach's conjecture is yet unproven). – Daniel May 12 '18 at 17:29
• @Jamal: How does changing the wording from "This is my solution: [solution] It becomes slower for […]" to "This solution becomes slower for […]: [solution]" make the question better? I agree that most of your changes increased the readability of the question or otherwise improved it, but the change mentioned above seems more like an adaptation to your personal preference than an objective improvement in readability or otherwise. I find that this slightly goes against the policy to always respect the author's intent, even if, in this case, it applies more to the language style than the content. – Stingy May 13 '18 at 12:07
• @Stingy: Because "this is my solution" is a bit unnecessary as it's already obvious that a solution is given. The real concern is then moved up so that it stands out more. – Jamal May 13 '18 at 19:24

## 2 Answers

Using the Sieve of Eratosthenes seems like a reasonable way to generate a large number of primes.

One thing that you could do is directly return your bitset representation of the primes, rather than copying it to a condensed list. This would save a bit of time in the short term, and be helpful for the next step.

for i in (list_prime(A)):
for j in (list_prime(A)[::-1]):


A few comments.

First, doing something twice is slower than doing it once! Since list_prime(A) is repeated in both lines, it will have to re-run the Sieve of Eratosthenes: even though you know that the list of prime numbers isn't going to change, Python doesn't! Even worse, because the second line is in a loop, you're running list_prime again for every value of i. So you can speed it up a bit by storing that list in a variable before you get to the loops.

Second, when you have nested loops like this, you make your algorithm quadratic in the size of list_prime(A). The issue is that you're doing the imprecise test for j before the precise test, which is normally the slow way around. That is, you're considering all the values of j which meet the rule "j is a prime number" and then testing whether "j + i == A". Instead consider all the values of j for which "j + i == A" first. There is exactly one of those for each i, which is found by simple subtraction. Then check whether that j is in your list of prime numbers. If so, you have a match. If not, you can immediately check the next i.

(The reason that I suggested returning the bitset representation of the list of primes is that it makes it much faster to check whether a specific number is prime.)

A few less substantial optimisations

solution_set.append((i,j))


It is good to consider the tie-breaking rules for problems like this, and listing candidates first and checking which wins afterwards is a valid way to do so.

However, it is worth thinking about the order of your algorithm before implementing the tie-breaking check. If, as is the case here, the first satisfying value is guaranteed to be the one that wins the tie break because it has the smallest i value, you might as well just return the first one you get to.

if bool_a[j]:
bool_a[j] = False;


If the end result of these two lines is that bool_a[j] must be False, just set it to false. if statements are surprisingly slow on modern CPUs.

primes.append(i)


Whenever you find yourself appending something to a list one element at a time, consider whether you can rewrite it using list comprehensions. For example something like the following:

primes = [index for (index, is_prime) in enumerate(bool_a) if is_prime]


(However, as above, my actual recommendation is that you completely remove that step and directly return bool_a)

Restating and condensing what @Josiah already said in their answer:

1. Don't do unnecessary calculations. This includes not re-doing the calculations for all prime as well as not going through all pairs of numbers just to find two that sum to the final number.

2. Simplify the sieve of Eratosthenes to be a generator that yields each prime number as it finds it.

With these two changes your code would become:

def prime_sieve(limit):
prime = [True] * limit
prime[0] = prime[1] = False

for i, is_prime in enumerate(prime):
if is_prime:
yield i
for n in range(i * i, limit, i):
prime[n] = False

def sum_of_primes(n):
primes = list(prime_sieve(n))
# for fast in lookups:
primes_set = set(primes)
for a in primes:
b = n - a
if b in primes_set:
# this is the lexicographically smallest by design
return a, b


This outperforms your original code by a huge margin:

In [5]: %timeit prime_sum(1002)
27.1 ms ± 256 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

In [6]: %timeit sum_of_primes(1002)
142 µs ± 699 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)


It also scales a lot better (O(n) instead of O(n^2)):

• Although I am generally in favour of generators, is there much advantage to a generator that is immediately packed into a list? – Josiah May 12 '18 at 21:50
• @Josiah: No, not really. But there is an advantage over the current approach by the OP which goes through all numbers first and then goes through the array again, accumulating all numbers which are prime. – Graipher May 12 '18 at 21:53
• @Josiah: Generator code is usually a couple of lines shorter, and there are more opportunities for re-use (it might not always be the case that it gets immediately turned into a list). – Gareth Rees May 13 '18 at 8:05
• This could be further modularized by having a method two_sum(numbers, target) that finds two numbers in a list that sum up to a given target. The method neither knows nor cares about prime numbers. – Rainer P. May 13 '18 at 13:20