Finding Not-Quite-Prime Numbers

Question

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.

Answer 1

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

Answer 2

Haskell, 47 characters

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

Answer 3

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.

Answer 4

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])

Answer 5

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

Answer 6

The boring Mathematica, 35 solution!

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

Answer 7

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));}

Answer 8

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.

Answer 9

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

Answer 10

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).

Answer 11

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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Answer 12

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;     
}

Answer 13

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.

Answer 14

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));}

Answer 15

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.";

Answer 16

APL (Dyalog Classic), 20 bytes

0∊¯1 0 1∘+∊(∘.×⍨1↓⍳)

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Answer 17

Retina, 22 bytes

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

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Takes unary as input

Answer 18

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.

Try it online.

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)

Answer 19

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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Answer 20

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?

Answer 21

F#, 68 bytes

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.

Answer 22

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

Answer 23

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";

Answer 24

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}

Answer 25

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

Answer 26

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

Answer 27

Ruby, 47 chars but very readable

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

Answer 28

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);}

Answer 29

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)

Answer 30

Julia 0.4, 23 bytes

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

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