Goldbach's conjecture

Question

Write a program that prompts the user for an even integer greater than 2.

Given Goldbach’s conjecture that every even integer greater than 2 can be expressed as the sum of two primes, print out two prime numbers which, when added together, provide the requested even number. Edit: the program only has to print A PAIR of primes, not all. For example:

4 : 2 + 2

6: 3 + 3

8: 3 + 5

10: 5 + 5 or 3 + 7

Answer 1

APL, 34 or 44 bytes

The first version is 34 symbol long and is restricted to characters from the original single-byte APL charsets, such as the one still supported in Dyalog APL:

↑c/⍨n=+/¨c←,∘.,⍨v/⍨~v∊v∘.×v←1↓⍳n←⎕

Explanation:

                               n←⎕   ⍝ ask for a number, store as n
                          v←1↓⍳n     ⍝ generate all integers from 2 to n
                      v∘.×v          ⍝ compute the product table of any two such integers
                v/⍨~v∊               ⍝ select those that don't appear in the product table 
         c←,∘.,⍨                     ⍝ generate all possible pairs of these primes
    n=+/¨c                           ⍝ check which pairs have a sum equal to n
↑c/⍨                                 ⍝ take the first that does

---

The second version is only 22 symbol long, because it exploits the π function to check for prime numbers, but that's only available in NARS2000 which uses Unicode, so the byte count is 44 in UCS-2:

2⍴(⌿⍨{∧/0π⍵})(⍪,⌽)⍳⎕-1

Explanation:

                   ⎕    ⍝ ask for a number N
                  ⍳ -1  ⍝ generate all naturals from 1 to N-1
             (⍪,⌽)      ⍝ arrange it into a table of all pairs of naturals with sum N
     {∧/0π⍵}            ⍝ check which pairs are made of all primes
2⍴(⌿⍨       )           ⍝ return the first pair that does

Examples

(⎕: is the prompt asking for a number)

      2⍴(⌿⍨{∧/0π⍵})(⍪,⌽)⍳⎕-1
⎕:
      4
2 2
      2⍴(⌿⍨{∧/0π⍵})(⍪,⌽)⍳⎕-1
⎕:
      6
3 3
      2⍴(⌿⍨{∧/0π⍵})(⍪,⌽)⍳⎕-1
⎕:
      8
3 5
      2⍴(⌿⍨{∧/0π⍵})(⍪,⌽)⍳⎕-1
⎕:
      124
11 113

Answer 2

Python 2, 75 71 bytes

n=input();k=m=1;p={0}
while{n-k,k}-p:m*=k*k;k+=1;p|={m%k*k}
print n-k,k

Test it on Ideone.

How it works

We use a corollary of Wilson's theorem:

At all times, the variable m is equal to the square of the factorial of k - 1; k starts at value 1 and m at value 0!² = 1. The set p will consist of 0 and all prime numbers up to the current value of k.

In each iteration, we first check if both n - k and k belong to p, which is true if and only if the set difference of {n-k, k} and p is empty. If it is, the condition is falsy and the loop continues.

Note that k > 0, and {n - k, k} will satisfy the condition for some positive value of n - k (assuming that Goldbach's conjecture is true), so the 0 in p won't lead to false positives.

In the loop, we update k and m. The new value of m is m × k² = (k - 1)!² × k² = k!², and the new value of k is k + 1, so m = (k - 1)!² still holds before and after the update.

Then, we perform set union to add the value of m % k × k to p. By the corollary of Wilson's theorem, this will add 1 × k = k if k is prime and 0 × k = 0 if not.

When the loop ends, we print the last values of n - k and k, which will be primes with sum n.

Answer 3

Ruby 2.0 (65)

require'prime'
n=gets.to_i
Prime.find{|i|p [i,n-i]if(n-i).prime?}

Answer 4

PHP - 73 bytes

<?for(;@($n%--$$n?:$o=&$argv[1]>$$n=++$n)||${++$b}^${--$o};);echo"$b+$o";

Input is taken as a command line argument.

Sample usage:

$ php goldbach.php 7098
19+7079

Answer 5

GolfScript 41 33 32

~(,2>.-1%]zip{{.,2>\{\%!}+,},!}?

Accepts command line argument e.g.

echo "14" | ruby golfscript.rb goldbach.gs
-> [2 12]

Finds all relevant partitions of the input number with:

(,2>.-1%]zip  #If only zip were a one-character command!  It is so often useful.

and then finds the first partition where no numbers are NOT prime with:

{np,!}? #For each partition, filter down to elements that are not prime, and only accept if there are no such results (since [] is falsey).

where the composite-checking block np is:

{.,2>\{\%!}+,}

this block filters down to all numbers that evenly divide a given number. If there are no such numbers (so the number is prime), the result is [], which is falsey in GolfScript.

Answer 6

R, 170 112 83 characters

a=scan();b=2:a;p=b[rowSums(!outer(b,b,`%%`))<2];q=p[(a-p)%in%p][1];cat(a,":",q,a-q)

Indented:

a=scan() #Take user input as a numeric
b=2:a
p=b[rowSums(!outer(b,b,`%%`))<2] #Find all primes from 2 to user input
q=p[(a-p)%in%p][1] #Check which a-p also belong to p and takes the first one
cat(a,":",q,a-q)

Usage:

> a=scan();b=2:a;p=b[rowSums(!outer(b,b,`%%`))<2];q=p[(a-p)%in%p][1];cat(a,":",q,a-q)
1: 72
2: 
Read 1 item
72 : 5 67 

Old solution at 112 characters, for posterity

a=scan();b=2:a;p=b[rowSums(!outer(b,b,`%%`))<2];w=which(outer(p,p,`+`)==a,T);cat(a,":",p[w[1,1]],p[w[1,2]])

Indented:

a=scan()
b=2:a
p=b[rowSums(!outer(b,b,`%%`))<2]
w=which(outer(p,p,`+`)==a,T) #Find the index of valid combinations
cat(a,":",p[w[1,1]],p[w[1,2]]) #Prints the first valid combination

Answer 7

perl 6: 69

$/=get;for grep &is-prime,^$/ {exit say $_,$_-$/ if ($/-$_).is-prime}

Answer 8

JavaScript (ES6) (Regex), 105

a=/^(xx+?)(?!(xx+)\2+$)x*(?=\1$)(?!(xx+)\3+$)/.exec("x".repeat(prompt()));alert(a[1].length+"+"+a[0].length)

Now you have a regex that tests the Goldbach conjecture, which has low requirement on special features (basic back-reference support, positive and negative look-ahead).

This uses String.prototype.repeat(), which is part of EcmaScript 6th edition proposal. Currently, this code only works on Firefox.

I really need a better language that has terse command when working with regex...

Answer 9

Python - 107

Basically an improvement on the second part of nutria's answer (I ran this on 2.7 but I think it should also work for 3.x)

p=lambda x:all(x%i!=0 for i in range(2,x))
n=input()
for i in range(2,n-1):
    if p(i)&p(n-i): print i,n-i

Answer 10

Scala, 286 192 172 148 chars

Not the fastest but it works. Call g(10) to obtain the list of goldbach pairs for 10.

def g(n:Int)={def p(n:Int,f:Int=2):Boolean=f>n/2||n%f!=0&&p(n,f+1)
s"$n : "+(for(i<-2 to n/2;j=n-i if p(i)&&p(j))yield s"$i + $j").mkString(" or ")}

Conversion to C++ is straightforward.

Answer 11

newLISP - 169 148 chars

(define(p n)(=(length(factor n))1))
(define(g n)(when(even? n)(for(i 3 n 2)
(and(p i)(p(- n i))(println n {: } i { }(- n i))))))
(g(int(read-line)))

includes the code to run it. The results are over-generous...

72: 5 67
72: 11 61
72: 13 59
72: 19 53
72: 29 43
72: 31 41
72: 41 31
72: 43 29
72: 53 19
72: 59 13
72: 61 11
72: 67 5

Answer 12

C - 139 129 characters

a,b;i(x,y){return x>y?x%y?i(x,y+1):0:x>1;}main(){scanf("%i",&a);for(b=a/2;b-->1;)i(b,2)&&i(a-
b,2)&&printf("%i:%i+%i\n",a,b,a-b);}

Answer 13

Sage, 65 62

n=input()
i=0
p=is_prime
while p(i)*p(n-i)==0:i+=1
print i,n-i

With the above in file goldbach.sage, execute it with Sage running in a terminal:

sage: %runfile goldbach.sage 

Thanks to @boothby for the p=is_prime idea.

Answer 14

Mathematica 56

This returns all of the solutions for the input integer.

Select[Tuples[Prime@Range@PrimePi[n = Input[]], 2], Tr@# == n &]

For example, when 1298 is input…

{{7, 1291}, {19, 1279}, {61, 1237}, {67, 1231}, {97, 1201}, {127, 1171}, {181, 1117}, {211, 1087}, {229, 1069}, {277, 1021}, {307, 991}, {331, 967}, {379, 919}, {421, 877}, {439, 859}, {487, 811}, {541, 757}, {547, 751}, {571, 727}, {607, 691}, {691, 607}, {727, 571}, {751, 547}, {757, 541}, {811, 487}, {859, 439}, {877, 421}, {919, 379}, {967, 331}, {991, 307}, {1021, 277}, {1069, 229}, {1087, 211}, {1117, 181}, {1171, 127}, {1201, 97}, {1231, 67}, {1237, 61}, {1279, 19}, {1291, 7}}

As written, it returns each solution twice.

Union[Sort/@ %]

{{7, 1291}, {19, 1279}, {61, 1237}, {67, 1231}, {97, 1201}, {127, 1171}, {181, 1117}, {211, 1087}, {229, 1069}, {277, 1021}, {307, 991}, {331, 967}, {379, 919}, {421, 877}, {439, 859}, {487, 811}, {541, 757}, {547, 751}, {571, 727}, {607, 691}}

Answer 15

Sage, 60

Similar in score and feel to r.e.s.'s solution, but I think it's different enough to post.

i=n=input()
while not{i,n-i}<set(primes(n)):i-=1
print i,n-i

Answer 16

Haskell, 97C

g n=head[(a,b)|let q=p n,a<-q,b<-q,a+b==n]
p n=filter c[2..n]
c p=null[x|x<-[2..p-1],p`mod`x==0]

Explanation:

• g is the "goldbach" function. Calling g n gives you the pair of primes that add up to n.
• p is a function that generates a list of primes less than n.
• c is the prime checker function used to define p.

Example runs:

*Main> g 4
(2,2)
*Main> g 6
(3,3)
*Main> g 8
(3,5)
*Main> g 10
(3,7)
*Main> g 12
(5,7)
*Main> map g [4,6..100]
[(2,2),(3,3),(3,5),(3,7),(5,7),(3,11),(3,13),(5,13),(3,17),(3,19),(5,19),(3,23),(5,23),(7,23),(3,29),(3,31),(5,31),(7,31),(3,37),(5,37),(3,41),(3,43),(5,43),(3,47),(5,47),(7,47),(3,53),(5,53),(7,53),(3,59),(3,61),(5,61),(7,61),(3,67),(5,67),(3,71),(3,73),(5,73),(7,73),(3,79),(5,79),(3,83),(5,83),(7,83),(3,89),(5,89),(7,89),(19,79),(3,97)]

Answer 17

Python 3 - 150 143 characters

Old version (150 characters):

p=lambda n:0 in[n % i for i in range(2,n)]
n=int(input())
[print('%d+%d'%(a, b))for b in range(2,n)for a in range(2,n)if not(a+b!=n or p(a) or p(b))]

New version (thanks to ProgramFOX):

p=lambda n:0 in[n%i for i in range(2,n)]
n=int(input())
[print('%d+%d'%(a,b))for b in range(2,n)for a in range(2,n)if not((a+b!=n)|p(a)|p(b))]

It prints every combination, for example:
4 2+2
10 7+3 5+5 3+7

Answer 18

JavaScript, 139 137 136

a=prompt();function b(n){for(i=2;i<n;i++)if(n%i<1)return;return 1}for(c=2,r=1;c<a&&r;c++)if(b(c)&&b(a-c))alert(a+": "+c+" + "+(a-c)),r=0

Answer 19

Julia, 62 Chars (85 with prompt)

julia> g(n)=collect(filter((x)->sum(x)==n,combinations(primes(n),2)))
g (generic function with 1 method)

julia> g(88)
4-element Array{Array{Int64,1},1}:
 [5,83] 
 [17,71]
 [29,59]
 [41,47]

Answer 20

GTB, 31

For your TI-84 Calculator

`A:.5A→B@%;A,4)4$~B+1,B-1#~B,B&

No prime built-ins.

Example runs

?4
               2
               2
?6
               3
               3
?8
               3
               5
?10
               5
               5

Answer 21

q [116 chars]

y where all each{{2=count where 0=(x mod)each til x+1}each x}each y:{a where x=sum each a:a cross a:til x}"I"$read0 0

No inbuilt function to find prime number.

Input

72

Output

5  67
11 61
13 59
19 53
29 43
31 41
41 31
43 29
53 19
59 13
61 11
67 5

Answer 22

J - 35 32 char

"Prompt the user" is the bane of every J golfer. There go all my hard-earned characters!

p:(n,n)#:i.&n,+/~p:i.n=:".1!:1]1

Explained:

• ".1!:1]1 - Read in a string (1!:1) from input (file handle 1) and convert it to a number (".).
• p:i.n=: - Assign this number to the variable n, and then take the first n primes.
• +/~ - Make an addition table, n wide and n high.
• i.&n, - Turn the table into a single list, and then find the index of the first occurrence of n, which exists if Goldbach's conjecture is true.
• p:(n,n)#: - Retrieve the row and column from the index, and take the corresponding primes.

Usage:

   p:(n,n)#:i.&n,+/~p:i.n=:".1!:1]1
666
5 661
   p:(n,n)#:i.&n,+/~p:i.n=:".1!:1]1
1024
3 1021

Had the prompt not been a requirement, here's a 25 character verb:

(,~p:@#:]i.~&,+/~@:p:@i.)

Answer 23

Python - 206

A little late to the party but I am practicing my golfing skills.

I actually coded this before I found this question! So mine doesn't include the beautiful lambda that the other Python solutions use.

import math
def p(n):
    if n%2==0&n>2:return False
    for i in range(3,n):
        if n%i==0:return False
    return True 
X=int(input())
for i in range(2,X):
    if p(i)&p(X-i):print i,X-i;break

Answer 24

Jelly, 8 bytes (non-competing)

_ÆRfÆR.ị

Try it online! or verify all test cases.

How it works

_ÆRfÆR.ị  Main link. Argument: n (integer)

 ÆR       Prime range; yield all primes in [1, ..., n].
_         Subtract all primes from n.
   fÆR    Filter; intersect the list of differences with the prime range.
      .ị  At-index 0.5; yield the last and first element.

Answer 25

Julia, 50 49 bytes

~=primes;n=ARGS[]|>int
(n-~n)∩~n|>extrema|>show

Try it online!

If a function were acceptable, the code could be shortened to 32 bytes:

~=primes
!n=(n-~n)∩~n|>extrema

How it works

~=primes creates an alias for the built-in primes function which returns a list of all prime numbers up to its argument. n=ARGS[]|>int parses the first command-line argument as saves it in n.

To find a suitable pair of primes, we first compute the aforementioned prime range with ~n. Then, n-~n yields all differences of these primes and n.

By intersecting (∩) the result with the prime range itself, we make sure that the remaining primes p are such that n - p is also a prime.

Finally, extrema takes the lowest and highest prime in the intersection, so their sum must be n.

Answer 26

Batch - 266

@echo off&setLocal enableDelayedExpansion&for /L %%a in (2,1,%1)do (set/aa=%%a-1&set c=&for /L %%b in (2,1,!a!)do set/ab=%%a%%%%b&if !b!==0 set c=1
if !c! NEQ 1 set l=!l!%%a,)&for %%c in (!l!)do for %%d in (!l!)do set/ad=%%c+%%d&if !d!==%1 set o=%%c + %%d
echo !o!

Set out neatly -

@echo off
setLocal enableDelayedExpansion
for /L %%a in (2,1,%1) do (
    set /a a=%%a-1
    set c=
    for /L %%b in (2,1,!a!) do (
        set /a b=%%a%%%%b
        if !b!==0 set c=1
    )
    if !c! NEQ 1 set l=!l!%%a,
)
for %%c in (!l!) do for %%d in (!l!) do (
    set /a d=%%c+%%d
    if !d!==%1 set o=%%c + %%d
)
echo !o!

Answer 27

SQL, 295 284

In postgresql:

create function c(c int) returns table (n int, m int) as $$ 
with recursive i(n) as
(select 2 union all select n+1 from i where n<c), 
p as (select * from i a where not exists 
(select * from i b where a.n!=b.n and mod(a.n,b.n)=0))
select * from p a, p b where a.n+b.n=c 
$$ language sql;

Should be able to do it in half the space though, if it weren't for things like "no left outer join in recursion", "no subquery in recursion"...

Here's the output:

postgres=# select c(10);
   c   
-------
 (3,7)
 (5,5)
 (7,3)
(3 rows)

postgres=# select c(88);
   c    
---------
 (5,83)
 (17,71)
 (29,59)
 (41,47)
 (47,41)
 (59,29)
 (71,17)
 (83,5)
(8 rows)

Answer 28

Perl 5, 58 bytes

57, plus 1 for -nE

/^(11+?)(?!(11+)\2+$)1*(?=\1$)(?!(11+)\3+$)/;say for$1,$&

Input and output are in unary. Example:

$ perl -nE'/^(11+?)(?!(11+)\2+$)1*(?=\1$)(?!(11+)\3+$)/;say for$1,$&'
1111111111
111
1111111

Hat-tip.

Answer 29

Oracle SQL 11.2, 202 bytes

WITH v(x,y,s)AS(SELECT LEVEL,LEVEL,0 FROM DUAL CONNECT BY LEVEL<=:1 UNION ALL SELECT x,y-1,s+SIGN(MOD(x,y))FROM v WHERE y>1),p AS(SELECT x FROM v WHERE x-s=2)SELECT a.x,b.x FROM p a,p b WHERE:1=a.x+b.x;   

Un-golfed

WITH v(x,y,s) AS
(
  SELECT LEVEL,LEVEL,0 FROM DUAL CONNECT BY LEVEL<=:1 
  UNION ALL 
  SELECT x,y-1,s+SIGN(MOD(x,y))FROM v WHERE y>1
)
,p AS (SELECT x FROM v WHERE x-s=2)
SELECT a.x,b.x 
FROM p a,p b 
WHERE :1=a.x+b.x;   

Answer 30

Python 3, 107 bytes

b=lambda x:all(x%y for y in range(2,x))
g=lambda x,i=2:((i,x-i)if b(i)&b(x-i)else g(x,i+1))if(i<x)else 0

b(x) is a primality test for x, and g(x) recurses through numbers to find two primes that fit.