Kabul etmeyi seçmeniz durumunda, bir sayı aşağıdaki ölçütleri karşılıyorsa, doğru veya yanlış (veya evet ve hayır gibi benzer anlamlı bir gösterim) döndüren bir işlevi kodla golf etmektir:

1. Tamsayının kendisi asal bir sayı VEYA
2. Her iki komşu tamsayısı asal

Örneğin:
girişi 7True döndürür.
Girişi 8de True döndürür. Girdisi False
değerini 15döndürür. (Ne 14, 15, ne de 16 asal)

Giriş, 2 ^ 0 ile 2 ^ 20 dahil sayılar için doğru şekilde dönebilmelidir, bu nedenle işaret sorunları veya tamsayı taşmaları hakkında endişelenmenize gerek yoktur.

J, 17

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

Süreç dönüş kodları olarak kodlanmış booleans döndürür: true için sıfır, false için sıfır olmayan. Örnek kullanım:

   */<:$&q:(<:,],>:) 7
0
   */<:$&q:(<:,],>:) 8
0
   */<:$&q:(<:,],>:) 15
3

Haskell, 47 karakter

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

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)

Code Golf ilk kez bu yüzden muhtemelen bazı hileler eksik.

Herhangi bir yolla kod kısalığı açısından gerçek bir yarışmacı değil, ama düzenli ifadeyle önceliği belirlemeye başladığından beri hala gönderme , birçok yönden bükülmüştür!

Python (2.x), 85 karakter

import re
f=lambda n:any(not re.match(r"^1?$|^(11+?)\1+$","1"*x)for x in[n,n-1,n+1])

Yakut (lambda olarak 55 veya 50)

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

veya lambda olarak ( g[23]aramak için kullanın )

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

Kahve (53)

p=(q)->[q-1..q+1].some (n)->[2..n-1].every (d)->n%d>0

Sıkıcı Mathematica, 35 çözüm!

PrimeQ[n-1]||PrimeQ[n]||PrimeQ[n+1]

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));}

Mathematica, 24 bytes

Don't know why this old post showed up in my list today, but I realized Mathematica is competitive here.

Or@@PrimeQ/@{#-1,#,#+1}&

Unnamed function taking an integer argument and returning True or False. Direct implementation.

Stax, 6 bytes

Ç▀<╝ºΩ

Run and debug it

Explanation (unpacked):

;v:Pv<! Full program, implicit input  Example: 7
;v      Copy input and decrement               7 6
  :P    Next prime                             7 7
    v   Decrement                              7 6
     <! Check if input is >= result            1

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).

Regex (ECMAScript), 20 bytes

^x?x?(?!(x+)(x\1)+$)

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The above version does not correctly handle zero, but that only takes 1 extra byte:

^x?x?(?!(x+)(x\1)+$)x

As an added bonus, here's a version that gives a return match of 1 for one less than a prime, 2 for prime, and 3 for one more than a prime:

^x?x??(?!(x+)(x\1)+$)x

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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;     
}

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.

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));}

CJam, 12 bytes

CJam is much younger than this challenge, so this answer is not eligible for the green checkmark (which should be updated to randomra's answer anyway). However, golfing this was actually quite fun - I started at 17 bytes and then changed my approach completely three times, saving one or two bytes each time.

{(3,f+:mp:|}

This is a block, the closest equivalent to a function in CJam, which expects the input on the stack, and leaves a 1 (truthy) or 0 (falsy) on the stack.

Test it here.

Here is how it works:

(3,f+:mp:|
(          "Decrement the input N.";
 3,        "Push an array [0 1 2].";
   f+      "Add each of those to N-1, to get [N-1 N N+1].";
     :mp   "Test each each element for primality, yielding 0 or 1.";
        :| "Fold bitwise OR onto the list, which gives 1 if any of them was 1.";

F#, 68 bytes (non-competing)

let p n=Seq.forall(fun x->n%x>0){2..n-1}
let m n=p(n-1)||p n||p(n+1)

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This is why I love code golf. I'm still very green with F# but I learn so much about how the language works and what it can do from these kinds of challenges.

APL (Dyalog Classic), 20 bytes

0∊¯1 0 1∘+∊(∘.×⍨1↓⍳)

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Retina, 22 bytes

$
1
^1?1?(?!(11+)\1+$)

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Takes unary as input

Java 8, 83 bytes

n->n==1|p(n-1)+p(n)+p(n+1)>0int p(int n){for(int i=2;i<n;n=n%i++<1?0:n);return--n;}

Returns true / false as truthy/falsey values.

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Explanation:"

n->                    // Method with integer parameter and boolean return-type
  n==1                 //  Return whether the input is 1 (edge-case)
  |p(n-1)+p(n)+p(n+1)>0//  Or if the sum of `n-1`, `n`, and `n+1` in method `p(n)` is not 0

int p(int n){          // Separated method with integer as both parameter and return-type
  for(int i=2;i<n;     //  Loop `i` in the range [2, `n`)
    n=n%i++<1?         //   If `n` is divisible by `i`
       0               //    Change `n` to 0
      :                //   Else:
       n);             //    Leave `n` as is
                       //  (After the loop `n` is either 0, 1, or unchanged,
                       //   if it's unchanged it's a prime, otherwise not)
  return--n;}          //  Return `n` minus 1

So int p(int n) will result in -1 for n=0 and non-primes, and will result in n-1 for n=1 or primes. Since p(0)+p(1)+p(2) will become -1+0+1 = 0 and would return false (even though 2 is a prime), the n=1 is an edge case using this approach.

A single loop without separated method would be 85 bytes:

n->{int f=0,j=2,i,t;for(;j-->-1;f=t>1?1:f)for(t=n+j,i=2;i<t;t=t%i++<1?0:t);return f;}

Returns 1 / 0 as truthy/falsey values.

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Explanation:

n->{              // Method with integer as both parameter and return-type
  int f=0,        //  Result-integer, starting at 0 (false)
      j=2,i,      //  Index integers
      t;          //  Temp integer
  for(;j-->-1;    //  Loop `j` downwards in range (2, -1]
      f=          //    After every iteration: Change `f` to:
        t>1?      //     If `t` is larger than 1 (`t` is a prime):
         1        //      Change `f` to 1 (true)
        :         //     Else:
         f)       //      Leave `f` the same
    for(t=n+j,    //   Set `t` to `n+j`
        i=2;i<t;  //   Inner loop `i` in the range [2, t)
      t=t%i++<1?  //    If `t` is divisible by `i`:
         0        //     Change `t` to 0
        :         //    Else:
         t);      //     Leave `t` the same
                  //   (If `t` is still the same after this inner loop, it's a prime;
                  //   if it's 0 or 1 instead, it's not a prime)
  return f;}      //  Return the result-integer (either 1/0 for true/false respectively)

Japt, 7 bytes

´Uô2 dj
´Uô2    // Given the range [U-1..U+1],
     dj // sing "Daddy DJ", uh, I mean, check if any are a prime.

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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

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";

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}

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

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

Ruby, 47 chars but very readable

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

C++ 97

ugoren seems to have beat me to the clever solution. So he's a shortish version on the loop three times approach:

P(int k){int j=1;for(int i=2;i<k;){j=k%i++&&j;}return j;}
a(int b){return P(b)|P(b+1)|P(b-1);}

Forth (gforth), 104 bytes

: p dup 1 > if 1 over 2 ?do over i mod 0> * loop else 0 then nip ;
: f dup 1- p over 1+ p rot p + + 0< ;

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Explanation

Prime check (p)

dup 1 > if          \ if number to check if greater than 1
   1 over 2 ?do     \ place a 1 on the stack to act as a boolean and loop from 2 to n
      over i  mod   \ take the modulo of n and i
      0> *          \ check if greater than 0 (not a divisor) and multiply result by boolean
   loop             \ end the loop, result will be -1 if no divisor was found (prime)
else                \ if n is less than 2
   0                \ put 0 on the stack (not prime)
then                \ end the loop
nip                 \ drop n from the stack

Main Function (f)

dup 1- p             \ get n-1 and check if prime
over 1+ p            \ get n+1 and check if prime
rot p                \ rotate stack to put n on top and check if prime
+ + 0<               \ add the three results and check if less than 0 (at least 1 was prime)

Julia 0.4, 23 bytes

n->any(isprime,n-1:n+1)

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Jelly, 5 bytes

‘r’ẒẸ

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How it works

‘r’ẒẸ    Monadic main link. Input: integer n
‘r’      Generate range n-1..n+1
   ẒẸ    Is any of them a prime?