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Your challenge, should you chose to accept it, is to code-golf a function that returns true or false (or some similar meaningful representation of yes and no) if a number meets the following criteria:

  1. The integer itself is a prime number OR
  2. Either of its neighbor integers are prime

For example:
An input of 7 would return True.
An input of 8 would also return True.
An input of 15 would return False. (Neither 14, 15, or 16 are prime)

The input must be able to return correctly for numbers between 2^0 and 2^20 inclusive, so there's no need to worry about sign issues or integer overflows.

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32-bit number overflows, not buffer overflows, I guess. –  user unknown Jan 14 '12 at 1:29
    
Whoops, meant "integer overflow". Brain went on autopilot. –  GigaWatt Jan 16 '12 at 14:57

17 Answers 17

up vote 10 down vote accepted

J, 17

*/<:$&q:(<:,],>:)

Returns booleans encoded as process return codes: zero for true, nonzero for false. Sample use:

   */<:$&q:(<:,],>:) 7
0
   */<:$&q:(<:,],>:) 8
0
   */<:$&q:(<:,],>:) 15
3
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*/0 p:<:,],>: is shorter and a proper (lambda) function is ([:*/0 p:<:,],>:) –  randomra Jul 9 '13 at 10:40

Haskell, 47 characters

f n=any(\k->all((>0).mod k)[2..k-1])[n-1..n+1]
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Python 85 80

def f(n):g=lambda n:all(n%i!=0for i in range(2,n));return g(n)or g(n-1)or g(n+1)

First time on Code Golf so there's probably some tricks I'm missing.

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You can remove the []. all will be more than happy to work with a generator expression. If you don't mind your code being ugly, you can also remove the spaces between 0 and for, and ) and or. –  stranac Jan 14 '12 at 0:14
    
@stranac Awesome. Thank you very much. –  Kris Harper Jan 14 '12 at 1:12
3  
Made a few straightforward changes, hopefully it still works: f=lambda n:any(all(m%i for i in range(2,m))for m in[n,n-1,n+1]) –  Nabb Jan 14 '12 at 3:35
    
@Nabb Very nice. Well done. –  Kris Harper Jan 14 '12 at 4:30

Not a real contender in code shortness by any means, but still submitting since determining primeness by regular expression is twisted in many ways!

Python (2.x), 85 characters

import re
f=lambda n:any(not re.match(r"^1?$|^(11+?)\1+$","1"*x)for x in[n,n-1,n+1])
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You can remove the for loop and build it into the regexp by testing "1"*(n+1) but starting with ^1?1? instead. –  Howard Jan 20 '12 at 15:11

Ruby (55, or 50 as lambda)

def f q;(q-1..q+1).any?{|n|(2..n-1).all?{|d|n%d>0}};end

or as lambda (use g[23] to call it)

g=->q{(q-1..q+1).any?{|n|(2..n-1).all?{|d|n%d>0}}}

Coffeescript (53)

p=(q)->[q-1..q+1].some (n)->[2..n-1].every (d)->n%d>0
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<pedantic> It should be "proc" not "lambda"</pedantic> ;-) –  Doorknob Dec 27 '13 at 14:47

The boring Mathematica, 35 solution!

PrimeQ[n-1]||PrimeQ[n]||PrimeQ[n+1]
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13  
At least you may golf it into Or@@PrimeQ/@{n-1,n,n+1}. –  Howard Jan 14 '12 at 10:04

C, 112 82 72 characters

Following Ilmari Karonen's comment, saved 30 chars by removing main, now P returns true/false. Also replaced loop with recursion, and some more tweaks.

p(n,q){return++q==n||n%q&&p(n,q);}P(n){return p(-~n,1)|p(n,1)|p(~-n,1);}

Original version:

p(n,q,r){for(r=0,q=2;q<n;)r|=!(n%q++);return!r;}
main(int n,int**m){putchar(48|p(n=atoi(*++m))|p(n-1)|p(n+1));}
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You could save 2 chars with main(n,m)int**m;. –  Ilmari Karonen Jan 17 '12 at 18:02
    
...and besides, the challenge says "code-golf a function". –  Ilmari Karonen Jan 17 '12 at 18:04

JavaScript (71 73 80)

n=prompt(r=0);for(j=n-2;p=j++<=n;r|=p)for(i=1;++i<j;)p=j%i?p:0;alert(r)

Demo: http://jsfiddle.net/ydsxJ/3/

Edit 1: Change for(i=2;i<j;i++) to for(i=1;++i<j;) (thanks @minitech). Convert if statement to ternary. Moved r|=p and p=1 into outer for to eliminate inner braces. Saved 7 characters.

Edit 2: Combine p=1 and j++<=n to p=j++<=n, save 2 chars (thanks @ugoren).

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You can use for(i=1;++i<j;) instead of for(i=2;i<j;i++) to save 1 more character. –  minitech Jan 14 '12 at 1:18
1  
@minitech: !j%i won't work because of precedence. A working alternative is j%i<1. –  Nabb Jan 14 '12 at 3:37
    
@Nabb: Wow, you're right. That's silly. –  minitech Jan 14 '12 at 3:40
    
How about p=j++<=n? If Javascript is like C here, it should work. –  ugoren Jan 18 '12 at 11:34
    
@ugoren: Looks like it worked, thanks! –  mellamokb Jan 18 '12 at 16:07

GolfScript: 26

)0\{.:i,{i\%!},,2=@|\(}3*;

Explanation: The innermost block {.:i,{i\%!},,2=@|\(} determines if the top of the stack is prime by checking if there are exactly 2 factors less than the top of the stack. It then disjuncts this with the second item on the stack, which holds the state of whether a prime has been seen yet. Finally, it decrements the number on the top of the stack.

Start by incrementing the input, initializing the prime-seen state, and repeat the block 3 times. Since this will decrement twice, but we started by incrementing, this will cover n+1 and n-1.

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C#, 87 97 chars

bool p(int q){return new[]{q-1,q,q+1}.Any(x=>Enumerable.Range(2,Math.Abs(x-2)).All(y=>x%y!=0));}
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I don't think this works with 1 or 2 as input –  Ben Reich Dec 29 '13 at 4:30
    
@BenReich It didn't. I had to add ten chars to fix it :( –  Steve Clanton Dec 31 '13 at 16:34

R, 68 chars

f=function(n){library(gmp);i=isprime;ifelse(i(n-1)|i(n)|i(n+1),1,0)}

Usage (1 for TRUE, 0 for FALSE):

f(7)
[1] 1
f(8)
[1] 1
f(15)
[1] 0
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1  
I don’t really know how R works, but could you just do i(n-1)|i(n)|i(n+1) instead of ifelse(i(n-1)|i(n)|i(n+1),1,0)? –  minitech Jul 9 '13 at 2:07
    
You are right: g=function(n){library(gmp);i=isprime;i(n-1)|i(n)|i(n+1)} - down to 56 characters! ;-) –  Paolo Sep 4 '13 at 12:50

C++

k=3;cin>>i;i--;
while(k)
{l[k]=0;
  for(j=2;j<i;j++)
   if(!(i%j))
     l[k]++;
  k--;
  i++;
}
if(!l[1]|!l[2]|!l[3])
     cout<<"1";
else cout<<"0";
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Welcome to CodeGold.SE. If you look at the other answers you'll notice a common format used for answers to [code-golf] questions. You might want to apply it to your answers as well. –  dmckee Jan 31 '12 at 20:48

Q, 43 chars 36

{any min each a mod 2_'til each a:x+-1 0 1}
{any(min')a mod 2_'(til')a:x+-1 0 1}
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J, 16 chars

   (_2&<@-4 p:]-2:)

   (_2&<@-4 p:]-2:) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19
1 1 1 1 1 1 1 1 0 1 1 1 1 1 0 1 1 1 1
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C#, 96

It returns -1,0,1 for true, anything else is false.

Any suggestions to make it shorter would be wonderful!

int p(int q){var r=q-1;for(var i=2;i<r&r<q+2;i++){if(i==r-1)break;if(r%i==0)r+=i=1;}return r-q;}

Expanded form:

int p(int q){
    var r=q-1;
    for(var i=2;i<r&r<q+2;i++){
        if(i==r-1)break;
        if(r%i==0)r+=i=1;
    }
    return r-q;     
}
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Python, 69 67 chars

any(all(i%j for j in range(2,i))for i in range(input()-1,8**7)[:3])

8**7 > 2**20 while being slightly shorter to write

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Ruby, 47 chars but very readable

require'Prime'
f=->x{[x-1,x,x+1].any? &:prime?}
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