Goldbach’s Conjecture

March 2, 2010

Christian Goldbach (1690-1764) was a Prussian mathematician and contemporary of Euler. One of the most famous unproven conjectures in number theory is known as Goldbach’s Conjecture, which states that every even number greater than two is the sum of two prime numbers; for example, 28 = 5 + 23.

Your task is to write a function that finds the two primes that add to a given even number greater than two. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below.

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10 Responses to “Goldbach’s Conjecture”

  1. [...] Praxis – Goldbach’s Conjecture By Remco Niemeijer In today’s Programming Praxis problem we have to test Golbach’s conjecture that every even number [...]

  2. Remco Niemeijer said

    My Haskell solution (see http://bonsaicode.wordpress.com/2010/03/02/programming-praxis-goldbach%E2%80%99s-conjecture/ for a version with comments):

    import Data.Numbers.Primes
    
    goldbach :: Integer -> (Integer, Integer)
    goldbach n = head [(p, n - p) | p <- takeWhile (< n) primes, isPrime (n - p)]
    
  3. Jason said

    Here’s my quick and naive C implementation.

    #include <stdio.h>
    #include <stdlib.h>
    
    void show_usage();
    unsigned long *create_sieve_to_number(unsigned long number);
    
    int main (int argc, const char * argv[]) {
    
        if (argc != 2) {
    		show_usage();
    		return -1;
    	}
    	
    	
    	unsigned long number = atol(argv[1]);
    	
    	if ((number % 2) != 0) {
    		show_usage();
    		return -1;
    	}
    	
    	unsigned long *sieve = create_sieve_to_number(number);
    	
    	for (unsigned long i = 2; i < number; i++) {
    		if (sieve[i] == 1) {
    			for (unsigned long j = i; j < number; j++) {
    				if (sieve[j] == 1) {
    					if (i + j == number) {
    						printf("Solution found: %d + %d\n", i, j);
    						return 0;
    					}
    				}
    			}
    		}
    	}
    	
    	printf("no solution found!  pick up your Fields Medal!\n");
    	
    		
    	return 0;
    	
    }
    
    void show_usage(void) {
    	printf("usage: goldbach [even number]\n");
    }
    
    unsigned long *create_sieve_to_number(unsigned long number) {
    	unsigned long *sieve;
    	
    	sieve = (unsigned long *)malloc(sizeof(unsigned long) * (number + 1));
    	
    	for (int i = 0; i < number; i++) {
    		sieve[i] = 1;
    	}
    	
    	for (unsigned long i = 2; i < number; i++) {
    		if (sieve[i] == 1) {
    			for (unsigned long j = i * i; j < number; j = j + i) {
    				sieve[j] = 0;
    			}
    		}
    	}
    	
    	return sieve;
    }
    
  4. [...] Today’s Praxis is on the Goldbach Conjecture which states that any even number greater than 2 can be expressed as the sum of two primes. The challenge is to write a program that will take in an even number, and spit out the two primes that can be added together to make it. [...]

  5. Jason said

    D’oh! The prime flags of the sieve should obviously be bool, and not ulong. Way to waste memory there, Jason. :)

  6. Dave said

    Here’s a managed code solution. No sieve is used, because it seemed like that would be a waste in most cases. A more optimized IsPrime() function could be swapped in.

    public static class Goldbach
    {
    private static List primes = new List() { 2, 3 };

    public static void GetGoldbachPrimes(int value, out int prime1, out int prime2)
    {
    Debug.Assert(value > 2 && value % 2 == 0, “value > 2 && value % 2 == 0″);
    if (value <= 2 || value % 2 != 0)
    {
    throw new ArgumentException("value must be even number greater than 2", "value");
    }

    foreach (var prime in GetPrimes())
    {
    int difference = value – prime;
    if (IsPrime(difference))
    {
    prime1 = prime;
    prime2 = difference;
    return;
    }
    }

    throw new InvalidOperationException("Primes could not be found");
    }

    private static IEnumerable GetPrimes()
    {
    for (int i = 0; i < primes.Count; i++)
    {
    yield return primes[i];
    }

    for (int i = primes[primes.Count - 1] + 2; i < int.MaxValue; i += 2)
    {
    if (IsPrime(i))
    {
    primes.Add(i);
    yield return i;
    }
    }
    }

    private static bool IsPrime(int value)
    {
    if (value = 0)
    {
    return true;
    }

    int sqrt = Convert.ToInt32(Math.Ceiling(Math.Sqrt(value)));

    for (int i = 3; i <= sqrt; i += 2)
    {
    if (value % i == 0)
    {
    return false;
    }
    }

    return true;
    }
    }

  7. I just came across this site and had done some related work with GC, in terms of symmetric prime pair solutions
    eg for 24, we have (11/13), (7/17),(5/19) as solutions, where each pair is symmetric about N/2=12
    it uses Pari/GP built-in function is ispseudoprime

    mirrors(N)=
    {
    \\ N is the number of interest to find symmetric prime pairs
    \\ written as pari/GP script
    i=1;cnt=0;Q1=999;

    if(N%2!=0|N5! “);return(2) );

    print(“selected N = “,N);

    while(Q1>5,

    Q1=N-i;Q2=N+i;

    if(ispseudoprime(Q1)&&ispseudoprime(Q2),
    \\## print(” mirror pair at “,Q1,” / “,Q2); \\disable for speed or if N> BB1
    cnt++ ); \\ end IF

    i=i+2; \\ skip multiples if 5! for speed
    if(i%5==0,i=i+2) \\ BB1

    ); \\ end WHILE

    print(“# pairs found = “,cnt);
    print(“other candidate N: 6,12,30,60,180,210,360,420,1260,2310,2520,4620,”);

    return(0);
    }

  8. Mike said

    Two python versions.

    For many values of n, it is faster to generate all the primes less than n than it is to generate each prime (p) up to n/2 and test whether n-p is prime. So the second routine runs faster than the first.

    Reuses is_prime and primes_to from earlier problems.

    from primes import primes_to, is_prime
    
    def prime_pairs1( x ):
        return ((p,x-p) for p in primes_to(x/2) if is_prime( x-p ))
    
    
    def prime_pairs2( x ):
        prime = list( primes_to( x ) )
        
        lo, hi = 0, len(prime)-1
        while lo <= hi:
    	n = prime[hi] + prime[lo]
    	if n < x:
    		lo += 1
    	elif n == x:
    		yield prime[lo], prime[hi]
    		lo += 1
    	else:
    		hi -= 1
    
  9. Not as elegant but I think this will do for now.

    def goldbach(n):
    	prime = getPrime(n)
    	try:
    		return [(prime[i], prime[j]) for i in range(len(prime)) for j in range(i, len(prime)) if prime[i] + prime[j] == n][0]
    	except IndexError:
    		pass
    	
    #get prime less than n
    def getPrime(n):
    	#list odd numbers
    	odd = []
    	a = 3
    	while a <= n: 
    		odd.append(a)
    		a = a + 2
    
    	#list prime numbers via sieve of eratosthenes
    	for p in odd:
    		q = 2*p 
    		while p <= odd[-1]:
    			p = p + q
    			if p in odd: 
    				odd.remove(p)
    
    	odd.insert(0,2)
    	return odd
    

    Sample output:

    4 (2, 2)
    6 (3, 3)
    8 (3, 5)
    10 (3, 7)
    12 (5, 7)
    14 (3, 11)
    16 (3, 13)
    18 (5, 13)
    20 (3, 17)
    22 (3, 19)
    24 (5, 19)
    26 (3, 23)
    28 (5, 23)
    30 (7, 23)
    32 (3, 29)
    34 (3, 31)
    36 (5, 31)
    38 (7, 31)
    40 (3, 37)
    42 (5, 37)
    44 (3, 41)
    46 (3, 43)
    48 (5, 43)
    50 (3, 47)
    52 (5, 47)
    54 (7, 47)
    56 (3, 53)
    58 (5, 53)
    60 (7, 53)
    62 (3, 59)
    64 (3, 61)
    66 (5, 61)
    68 (7, 61)
    70 (3, 67)
    72 (5, 67)
    74 (3, 71)
    76 (3, 73)
    78 (5, 73)
    80 (7, 73)
    82 (3, 79)
    84 (5, 79)
    86 (3, 83)
    88 (5, 83)
    90 (7, 83)
    92 (3, 89)
    94 (5, 89)
    96 (7, 89)
    98 (19, 79)
    

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