33
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Challenge

This is a simple one: Given a positive integer up to 1,000,000, return the closest prime number.

If the number itself is prime, then you should return that number; if there are two primes equally close to the provided number, return the lower of the two.

Input is in the form of a single integer, and output should be in the form of an integer as well.

I don't care how you take in the input (function, STDIN, etc.) or display the output (function, STDOUT, etc.), as long as it works.

This is code golf, so standard rules apply—the program with the least bytes wins!

Test Cases

Input  =>  Output
------    -------
80     =>      79
100    =>     101
5      =>       5
9      =>       7
532    =>     523
1      =>       2
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  • 5
    \$\begingroup\$ Hi and welcome to PPCG!. To avoid down voting due to lack of quality I suggest you to post it to the sandbox first and after a couple of days post it here \$\endgroup\$ – Luis felipe De jesus Munoz Mar 27 at 15:17
  • \$\begingroup\$ This is one of the outputs requested in this challenge. \$\endgroup\$ – Arnauld Mar 27 at 15:24
  • \$\begingroup\$ Very closely related but not quite identical. \$\endgroup\$ – Giuseppe Mar 27 at 15:35
  • \$\begingroup\$ @Arnauld I saw that one, but I thought that they were different enough to warrant a new question. \$\endgroup\$ – Nathan Dimmer Mar 27 at 15:37
  • 2
    \$\begingroup\$ See also OEIS A051697. \$\endgroup\$ – Eric Towers Mar 28 at 8:52

37 Answers 37

8
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Gaia, 3 bytes

ṅD⌡

Try it online!

Rather slow for large inputs, but works given enough memory/time.

I'm not sure why D⌡ implicitly pushes z again, but it makes this a remarkably short answer!

ṅ	| implicit input z: push first z prime numbers, call it P
 D⌡	| take the absolute difference between P and (implicit) z,
	| returning the smallest value in P with the minimum absolute difference
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12
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JavaScript (ES6), 53 bytes

n=>(g=(o,d=N=n+o)=>N%--d?g(o,d):d-1?g(o<0?-o:~o):N)``

Try it online!

Commented

n => (            // n = input
  g = (           // g = recursive function taking:
    o,            //   o = offset
    d =           //   d = current divisor, initialized to N
    N = n + o     //   N = input + offset
  ) =>            //
    N % --d ?     // decrement d; if d is not a divisor of N:
      g(o, d)     //   do recursive calls until it is
    :             // else:
      d - 1 ?     //   if d is not equal to 1 (either N is composite or N = 1):
        g(        //     do a recursive call with the next offset:
          o < 0 ? //       if o is negative:
            -o    //         make it positive (e.g. -1 -> +1)
          :       //       else:
            ~o    //         use -(o + 1) (e.g. +1 -> -2)
        )         //     end of recursive call
      :           //   else (N is prime):
        N         //     stop recursion and return N
)``               // initial call to g with o = [''] (zero-ish)
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10
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05AB1E, 5 bytes

Åps.x

Try it online! or as a Test Suite

Inefficient for big numbers

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  • 3
    \$\begingroup\$ Too bad Ån is "In case of a tie, the higher prime is pushed" Didn't even knew we had this builtin, tbh. \$\endgroup\$ – Kevin Cruijssen Mar 28 at 7:51
  • \$\begingroup\$ @KevinCruijssen: Neither did I until now :) \$\endgroup\$ – Emigna Mar 28 at 9:15
7
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Octave, 40 bytes

@(n)p([~,k]=min(abs(n-(p=primes(2*n)))))

Try it online!

This uses the fact that there is always a prime between n and 2*n (Bertrand–Chebyshev theorem).

How it works

@(n)p([~,k]=min(abs(n-(p=primes(2*n)))))

@(n)                                      % Define anonymous function with input n
                       p=primes(2*n)      % Vector of primes up to 2*n. Assign to p
                abs(n-(             ))    % Absolute difference between n and each prime
      [~,k]=min(                      )   % Index of first minimum (assign to k; not used)
    p(                                 )  % Apply that index to p
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6
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Japt, 5 bytes

_j}cU

Try it or run all test cases

_j}cU     :Implicit input of integer U
_         :Function taking an integer as an argument
 j        :  Test if integer is prime
  }       :End function
   cU     :Return the first integer in [U,U-1,U+1,U-2,...] that returns true
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5
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05AB1E, 4 bytes

z-Ån

Try it online!

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5
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Wolfram Language (Mathematica), 31 bytes

Nearest[Prime~Array~78499,#,1]&

Try it online!

                              & (*pure function*)
        Prime~Array~78499       (*among the (ascending) first 78499 primes*)
                            1   (*select one*)
Nearest[                 ,#, ]  (*which is nearest to the argument*)

1000003 is the 78499th prime. Nearest prioritizes values which appear earlier in the list (which are lower).

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  • 4
    \$\begingroup\$ Nearest[Prime@Range@#,#,1]& for 27 \$\endgroup\$ – Ben Mar 29 at 16:10
5
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Brachylog, 7 5 bytes

;I≜-ṗ

Try it online!

Saved 2 bytes thanks to @DLosc.

Explanation

;I≜      Label an unknown integer I (tries 0, then 1, then -1, then 2, etc.)
   -     Subtract I from the input
    ṗ    The result must be prime
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  • \$\begingroup\$ @DLosc Mostly because I am stupid. Thanks. \$\endgroup\$ – Fatalize Mar 29 at 8:26
  • \$\begingroup\$ I think we just approached it from different directions. You were thinking about from the start, I assume, whereas I was thinking about pairing and subtracting and only later realized I'd need to make it work. :) \$\endgroup\$ – DLosc Mar 29 at 21:27
4
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Pyth, 10 bytes

haDQfP_TSy

Try it online here, or verify all the test cases at once here.

haDQfP_TSyQ   Implicit: Q=eval(input())
              Trailing Q inferred
         yQ   2 * Q
        S     Range from 1 to the above
    f         Filter keep the elements of the above, as T, where:
     P_T        Is T prime?
  D           Order the above by...
 a Q          ... absolute difference between each element and Q
                This is a stable sort, so smaller primes will be sorted before larger ones if difference is the same
h             Take the first element of the above, implicit print
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4
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Jelly, 9 7 bytes

ḤÆRạÞµḢ

Try it online!

Slow for larger input, but works ok for the requested range. Thanks to @EriktheOutgolfer for saving 2 bytes!

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  • \$\begingroup\$ Hey, that's clever! Save two by substituting _A¥ with (absolute difference). Oh, and can really be . \$\endgroup\$ – Erik the Outgolfer Mar 27 at 19:08
  • \$\begingroup\$ @EriktheOutgolfer thanks. Surely using won’t always work? It means that only primes up to n+1 will be found, while the closest might be n+2. \$\endgroup\$ – Nick Kennedy Mar 27 at 22:39
  • \$\begingroup\$ Hm, that's a concern. \$\endgroup\$ – Erik the Outgolfer Mar 27 at 22:40
4
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C (gcc), 87 76 74 72 bytes

Optimization of innat3's C# (Visual C# Interactive Compiler), 100 bytes

f(n,i,t,r,m){for(t=0,m=n;r-2;t++)for(r=i=1,n+=n<m?t:-t;i<n;n%++i||r++);}

Try it online!

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  • \$\begingroup\$ Hello and welcome to PPCG. A few tips: r!=2 is equivalent to r-2, n%++i?0:r++ can most likely be n%++i||r++. \$\endgroup\$ – Jonathan Frech Mar 29 at 13:11
  • \$\begingroup\$ I didn't immediately see that. Thanks. \$\endgroup\$ – Natural Number Guy Mar 29 at 15:13
3
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Tidy, 43 bytes

{x:(prime↦splice(]x,-1,-∞],[x,∞]))@0}

Try it online!

Explanation

This is a lambda with parameter x. This works by creating the following sequence:

[x - 1, x, x - 2, x + 1, x - 3, x + 2, x - 4, x + 3, ...]

This is splicing together the two sequences ]x, -1, -∞] (left-closed, right-open) and [x, ∞] (both open).

For x = 80, this looks like:

[79, 80, 78, 81, 77, 82, 76, 83, 75, 84, 74, 85, ...]

Then, we use f↦s to select all elements from s satisfying f. In this case, we filter out all composite numbers, leaving only the prime ones. For the same x, this becomes:

[79, 83, 73, 71, 89, 67, 97, 61, 59, 101, 103, 53, ...]

Then, we use (...)@0 to select the first member of this sequence. Since the lower of the two needs to be selected, the sequence which starts with x - 1 is spliced in first.

Note: Only one of x and x - 1 can be prime, so it is okay that the spliced sequence starts with x - 1. Though the sequence could be open on both sides ([x,-1,-∞]), this would needlessly include x twice in the sequence. So, for sake of "efficiency", I chose the left-closed version (also because I like to show off Tidy).

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3
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Python 2, 71 bytes

f=lambda n,k=1,p=1:k<n*3and min(k+n-p%k*2*n,f(n,k+1,p*k*k)-n,key=abs)+n

Try it online!

A recursive function that uses the Wilson's Theorem prime generator. The product p tracks \$(k-1)!^2\$, and p%k is 1 for primes and 0 for non-primes. To make it easy to compare abs(k-n) for different primes k, we store k-n and compare via abs, adding back n to get the result k.

The expression k+n-p%k*2*n is designed to give k-n on primes (where p%k=1), and otherwise a "bad" value of k+n that's always bigger in absolute value and so doesn't affect the minimum, so that non-primes are passed over.

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3
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APL (Dyalog Extended), 20 15 bytesSBCS

Tacit prefix function inspired by Galen Ivanov's J answer.

⊢(⊃⍋⍤|⍤-⊇⊢)¯2⍭⍳

Try it online!

ɩndices one through the argument.

¯2⍭ nth primes of that

⊢() apply the following tacit function to that, with the original argument as left argument:

 the primes

 indexed by:

   the ascending grade (indices which would sort ascending)
   of
  | the magnitude (absolute value)
   of
  - the differences

 pick the first one (i.e. the one with smallest difference)

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3
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Perl 6, 35 bytes

{$_+=($*=-1)*$++until .is-prime;$_}

Try it online!

This uses Veitcel's technique for generating the list of 0, -1, 2, -3 but simplifies it greatly to ($*=-1)*$++ using the anonymous state variables available in P6 (I originally had -1 ** $++ * $++, but when golfed the negative loses precedence). There's a built in prime checker but unfortunately the until prevents the automagically returned value so there's an extra $_ hanging around.

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  • \$\begingroup\$ I'd usually use a sequence operator approach to something like this, but comes out to one byte longer, so nice work finding a shorter method \$\endgroup\$ – Jo King Mar 29 at 0:39
  • \$\begingroup\$ @JoKing good catch. The things that happen when I golf too quickly after getting a working solution. I had a similar one but the damned lack of [-1] haha \$\endgroup\$ – guifa Mar 29 at 0:49
3
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C, 122 121 104 bytes

p(a,i){for(i=1;++i<a;)if(a%i<1)return 0;return a>1;}c(a,b){for(b=a;;b++)if(p(--a)|p(b))return p(b)?b:a;}

Use it calling function c() and passing as argument the number; it should return the closest prime.

Thanks to Embodiment of Ignorance for 1 byte saved a big improvement.

Try it online!

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  • \$\begingroup\$ But c() receives two parameters... Also, you can probably shorten the while(1) to for(;;) (untested, since I don't get how to run your code \$\endgroup\$ – Embodiment of Ignorance Mar 28 at 5:03
  • \$\begingroup\$ @EmbodimentofIgnorance I wrote it and tested it all on an online c compiler, I could call c() passing only the first parameter. And you are right, for(;;) saves me a byte, only 117 left to get first place :) \$\endgroup\$ – Lince Assassino Mar 28 at 11:30
  • \$\begingroup\$ 110 bytes: #define r return p(a,i){i=1;while(++i<a)if(a%i<1)r 0;r a>1;}c(a,b){b=a;for(;;b++){if(p(--a))r a;if(p(b))r b;}}. Here is a TIO link: tio.run/… \$\endgroup\$ – Embodiment of Ignorance Mar 28 at 21:06
  • \$\begingroup\$ 106: tio.run/… \$\endgroup\$ – Embodiment of Ignorance Mar 28 at 21:12
  • \$\begingroup\$ 101: tio.run/… \$\endgroup\$ – Embodiment of Ignorance Mar 28 at 21:17
3
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Wolfram Language (Mathematica), 52 bytes

If[PrimeQ[s=#],s,#&@@Nearest[s~NextPrime~{-1,1},s]]&

Try it online!

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  • \$\begingroup\$ You have an extra space that can be removed to save a byte. \$\endgroup\$ – Ben Mar 29 at 16:03
  • \$\begingroup\$ @Ben you are right. thanx \$\endgroup\$ – J42161217 Mar 29 at 23:08
2
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APL(NARS), 38 chars, 76 bytes

{⍵≤1:2⋄0π⍵:⍵⋄d←1π⍵⋄(d-⍵)≥⍵-s←¯1π⍵:s⋄d}

0π is the test for prime, ¯1π the prev prime, 1π is the next prime; test:

  f←{⍵≤1:2⋄0π⍵:⍵⋄d←1π⍵⋄(d-⍵)≥⍵-s←¯1π⍵:s⋄d}
  f¨80 100 5 9 532 1
79 101 5 7 523 2 
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2
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J, 19 15 bytes

(0{]/:|@-)p:@i.

Try it online!

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2
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Perl 5, 59 bytes

$a=0;while((1x$_)=~/^.?$|^(..+?)\1+$/){$_+=(-1)**$a*($a++)}

Try it online!

/^.?$|^(..+?)\1+$/ is tricky regexp to check prime

(-1)**$a*($a++) generate sequence 0,-1, 2,-3 ...

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2
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MathGolf, 10 bytes

∞╒g¶áÅ-±├Þ

Try it online.

Explanation:

∞            # Double the (implicit) input-integer
 ╒           # Create a list in the range [1, 2*n]
  g¶         # Filter so only the prime numbers remain
    áÅ       # Sort this list using the next two character:
      -±     #  The absolute difference with the (implicit) input-integer
        ├    # Push the first item of the list
             # (unfortunately without popping the list itself, so:)
         Þ   # Discard everything from the stack except for the top
             # (which is output implicitly as result)
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  • \$\begingroup\$ @JoKing Thanks! I knew Max thought about changing it, but didn't knew he actually did. The docs still state the old one. \$\endgroup\$ – Kevin Cruijssen Mar 28 at 8:53
  • \$\begingroup\$ Ah, I use the mathgolf.txt file as reference, since it seems to be more up to date \$\endgroup\$ – Jo King Mar 28 at 23:21
  • \$\begingroup\$ @JoKing Yeah, he told me yesterday about that file as well. Will use it from now on. :) \$\endgroup\$ – Kevin Cruijssen Mar 29 at 7:23
2
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Factor, 91 bytes

: p ( x -- x ) [ nprimes ] keep dupd [ - abs ] curry map swap zip natural-sort first last ;

Try it online!

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2
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Python 2 (Cython), 96 bytes

l=lambda p:min(filter(lambda p:all(p%n for n in range(2,p)),range(2,p*3)),key=lambda x:abs(x-p))

Try it online!

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  • \$\begingroup\$ Might be able to save a couple bytes via r=range;... \$\endgroup\$ – Skyler Mar 28 at 20:33
  • 1
    \$\begingroup\$ @Arnauld, it now works for x=1 \$\endgroup\$ – Snaddyvitch Dispenser Mar 29 at 8:20
2
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C# (Visual C# Interactive Compiler), 104 100 bytes

n=>{int r=0,t=0,m=n;while(r!=2){n+=(n<m)?t:-t;t++;r=0;for(int i=1;i<=n;i++)if(n%i==0)r++;}return n;}

Try it online!

Explanation:

int f(int n)
{
    int r = 0; //stores the amount of factors of "n"
    int t = 0; //increment used to cover all the integers surrounding "n"
    int m = n; //placeholder to toggle between adding or substracting "t" to "n"

    while (r != 2) //while the amount of factors found for "n" is different to 2 ("1" + itself)
    {
        n += (n < m) ? t : -t; //increment/decrement "n" by "t" (-0, -1, +2, -3, +4, -5,...)
        t++;
        r = 0;
        for (int i = 1; i <= n; i++) //foreach number between "1" and "n" increment "r" if the remainder of its division with "n" is 0 (thus being a factor)
            if (n % i == 0) r++; 
    }
    return n;
}

Console.WriteLine(f(80)); //79
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2
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Java 8, 88 87 bytes

n->{for(int c=0,s=0,d,N=n;c!=2;s++)for(c=d=1,n+=n<N?s:-s;d<n;)if(n%++d<1)c++;return n;}

Port of @NaturalNumberGuy's (first) C answer, so make sure to upvote him!!
-1 byte thanks to @OlivierGrégoire.

Try it online.

Explanation:

n->{               // Method with integer as both parameter and return-type
  for(int c=0,     //  Counter-integer, starting at 0
          s=0,     //  Step-integer, starting at 0 as well
          d,       //  Divisor-integer, uninitialized
          N=n;     //  Copy of the input-integer
      c!=2;        //  Loop as long as the counter is not exactly 2 yet:
      s++)         //    After every iteration: increase the step-integer by 1
    for(c=d=1,     //   (Re)set both the counter and divisor to 1
        n+=n<N?    //   If the input is smaller than the input-copy:
            s      //    Increase the input by the step-integer
           :       //   Else:
            -s;    //    Decrease the input by the step-integer
        d<n;)      //   Inner loop as long as the divisor is smaller than the input
      if(n%++d     //    Increase the divisor by 1 first with `++d`
              <1)  //    And if the input is evenly divisible by the divisor:
        c++;       //     Increase the counter-integer by 1
  return n;}       //  Return the now modified input-integer as result
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2
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Java (JDK), 103 bytes

n->{int p=0,x=0,z=n,d;for(;p<1;p=p>0?z:0,z=z==n+x?n-++x:z+1)for(p=z/2,d=1;++d<z;)p=z%d<1?0:p;return p;}

Try it online!

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  • \$\begingroup\$ Umm.. I had already create a port of his answer.. ;) Although yours is 1 byte shorter, so something is different. EDIT: Ah, I have a result-integer outside the loop, and you modify the input inside the loop, hence the -1 byte for ;. :) Do you want me to delete my answer?.. Feel free to copy the explanation. \$\endgroup\$ – Kevin Cruijssen Mar 29 at 12:50
  • \$\begingroup\$ @KevinCruijssen Oops, rollbacked! \$\endgroup\$ – Olivier Grégoire Mar 29 at 12:59
  • \$\begingroup\$ Sorry about that (and thanks for the -1 byte). I like your version as well, though. Already upvoted before I saw NaturalNumberGuy's answer. \$\endgroup\$ – Kevin Cruijssen Mar 29 at 12:59
2
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Haskell, 79 74 bytes (thanks to Laikoni)

72 bytes as annonymus function (the initial "f=" could be removed in this case).

f=(!)(-1);n!x|x>1,all((>0).mod x)[2..x-1]=x|y<-x+n=last(-n+1:[-n-1|n>0])!y

Try it online!


original code:

f=(!)(-1);n!x|x>1&&all((>0).mod x)[2..x-1]=x|1>0=(last$(-n+1):[-n-1|n>0])!(x+n)

Try it online!

Explanation:

f x = (-1)!x

isPrime x = x > 1 && all (\k -> x `mod` k /= 0)[2..x-1]
n!x | isPrime x = x            -- return the first prime found
    | n>0       = (-n-1)!(x+n) -- x is no prime, continue with x+n where n takes the 
    | otherwise = (-n+1)!(x+n) -- values -1,2,-3,4 .. in subsequent calls of (!)
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  • 1
    \$\begingroup\$ Inside a guard you can use , instead of &&. (last$ ...) can be last(...), and the second guard 1>0 can be used for a binding to save parenthesis, e.g. y<-x+n. \$\endgroup\$ – Laikoni Mar 30 at 10:57
  • \$\begingroup\$ Anonymous functions are generally allowed, so the initial f= does not need to be counted. Also the parenthesis enclosing (-1+n) can be dropped. \$\endgroup\$ – Laikoni Mar 30 at 11:22
  • \$\begingroup\$ Thanks for the suggestions. I didn't know "," and bindings are allowed in function guards! But i don't really like the idea of an annonymous function as an answer. It doesn't feel right in my opinion. \$\endgroup\$ – Sachera Mar 31 at 2:58
  • \$\begingroup\$ You can find more tips in our collection of tips for golfing in Haskell. There is also a Guide to Golfing Rules in Haskell and dedicated chat room: Of Monads and Men. \$\endgroup\$ – Laikoni Mar 31 at 22:13
2
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VDM-SL, 161 bytes

f(i)==(lambda p:set of nat1&let z in set p be st forall m in set p&abs(m-i)>=abs(z-i)in z)({x|x in set{1,...,9**7}&forall y in set{2,...,1003}&y<>x=>x mod y<>0})

A full program to run might look like this - it's worth noting that the bounds of the set of primes used should probably be changed if you actually want to run this, since it will take a long time to run for 1 million:

functions
f:nat1+>nat1
f(i)==(lambda p:set of nat1&let z in set p be st forall m in set p&abs(m-i)>=abs(z-i)in z)({x|x in set{1,...,9**7}&forall y in set{2,...,1003}&y<>x=>x mod y<>0})

Explanation:

f(i)==                                        /* f is a function which takes a nat1 (natural number not including 0)*/
(lambda p:set of nat1                         /* define a lambda which takes a set of nat1*/
&let z in set p be st                         /* which has an element z in the set such that */
forall m in set p                             /* for every element in the set*/
&abs(m-i)                                     /* the difference between the element m and the input*/
>=abs(z-i)                                    /* is greater than or equal to the difference between the element z and the input */
in z)                                         /* and return z from the lambda */
(                                             /* apply this lambda to... */
{                                             /* a set defined by comprehension as.. */
x|                                            /* all elements x such that.. */ 
x in set{1,...,9**7}                          /* x is between 1 and 9^7 */
&forall y in set{2,...,1003}                  /* and for all values between 2 and 1003*/
&y<>x=>x mod y<>0                             /* y is not x implies x is not divisible by y*/
} 
)
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1
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C# (Visual C# Interactive Compiler), 112 bytes

g=>Enumerable.Range(2,2<<20).Where(x=>Enumerable.Range(1,x).Count(y=>x%y<1)<3).OrderBy(x=>Math.Abs(x-g)).First()

Try it online!

Left shifts by 20 in submission but 10 in TIO so that TIO terminates for test cases.

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1
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Swift, 186 bytes

func p(a:Int){let b=q(a:a,b:-1),c=q(a:a,b:1);print(a-b<=c-a ? b:c)}
func q(a:Int,b:Int)->Int{var k=max(a,2),c=2;while k>c && c != a/2{if k%c==0{k+=b;c=2}else{c=c==2 ? c+1:c+2}};return k}

Try it online!

\$\endgroup\$

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