Kembalikan Nomor Perdana Terdekat



Ini sederhana: Diberikan bilangan bulat positif hingga 1.000.000, kembalikan bilangan prima terdekat.

Jika nomor itu sendiri prima, maka Anda harus mengembalikan nomor itu; jika ada dua bilangan prima yang sama dekat dengan nomor yang disediakan, kembalikan yang lebih rendah dari keduanya.

Input dalam bentuk integer tunggal, dan output juga harus dalam bentuk integer.

Saya tidak peduli bagaimana Anda menerima input (fungsi, STDIN, dll.) Atau menampilkan output (fungsi, STDOUT, dll.), Asalkan berfungsi.

Ini adalah kode golf, jadi aturan standar berlaku — program dengan byte terkecil menang!

Uji Kasus

Input  =>  Output
------    -------
80     =>      79
100    =>     101
5      =>       5
9      =>       7
532    =>     523
1      =>       2

Hai dan selamat datang di PPCG !. Untuk menghindari pemungutan suara karena kurangnya kualitas, saya sarankan Anda untuk mengirimnya ke kotak pasir terlebih dahulu dan setelah beberapa hari mempostingnya di sini
Luis felipe De jesus Munoz

Ini adalah salah satu output yang diminta dalam tantangan ini .

Sangat terkait erat tetapi tidak cukup identik.

@Arnauld saya melihat yang itu, tapi saya pikir mereka cukup berbeda untuk mendapatkan pertanyaan baru.
Nathan Dimmer

See also OEIS A051697.
Eric Towers



Gaia, 3 bytes


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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


JavaScript (ES6), 53 bytes


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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)


Octave, 40 bytes


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

How it works


@(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


Japt, 5 bytes


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


Wolfram Language (Mathematica), 31 bytes


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                              & (*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).

Nearest[Prime@Range@#,#,1]& for 27


Brachylog, 7 5 bytes


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Saved 2 bytes thanks to @DLosc.


;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

@DLosc Mostly because I am stupid. Thanks.

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. :)


Pyth, 10 bytes


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


Jelly, 9 7 bytes


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

Hey, that's clever! Save two by substituting _A¥ with (absolute difference). Oh, and can really be .
Erik the Outgolfer

@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.
Nick Kennedy

Hm, that's a concern.
Erik the Outgolfer


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

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A recursive function that uses the Wilson's Theorem prime generator. The product p tracks (k1)!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.


Tidy, 43 bytes


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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).


APL (Dyalog Extended), 20 15 bytesSBCS

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


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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)
  | the magnitude (absolute value)
  - the differences

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


Perl 6, 35 bytes

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

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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.

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
Jo King

@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


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!

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
Embodiment of Ignorance

@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 :)
Lince Assassino

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:…
Embodiment of Ignorance


APL(NARS), 38 chars, 76 bytes


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

  f¨80 100 5 9 532 1
79 101 5 7 523 2 


Perl 5, 59 bytes


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

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


MathGolf, 10 bytes


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            # 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)

@JoKing Thanks! I knew Max thought about changing it, but didn't knew he actually did. The docs still state the old one.
Kevin Cruijssen

Ah, I use the mathgolf.txt file as reference, since it seems to be more up to date
Jo King

@JoKing Yeah, he told me yesterday about that file as well. Will use it from now on. :)
Kevin Cruijssen


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;}

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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,...)
        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


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.


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


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!

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.
Kevin Cruijssen

@KevinCruijssen Oops, rollbacked!
Olivier Grégoire

Sorry about that (and thanks for the -1 byte). I like your version as well, though. Already upvoted before I saw NaturalNumberGuy's answer.
Kevin Cruijssen


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

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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)

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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 (!)

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.

Anonymous functions are generally allowed, so the initial f= does not need to be counted. Also the parenthesis enclosing (-1+n) can be dropped.

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.

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.


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:

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})


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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