192
\$\begingroup\$

Believe it or not, we do not yet have a code golf challenge for a simple primality test. While it may not be the most interesting challenge, particularly for "usual" languages, it can be nontrivial in many languages.

Rosetta code features lists by language of idiomatic approaches to primality testing, one using the Miller-Rabin test specifically and another using trial division. However, "most idiomatic" often does not coincide with "shortest." In an effort to make Programming Puzzles and Code Golf the go-to site for code golf, this challenge seeks to compile a catalog of the shortest approach in every language, similar to "Hello, World!" and Golf you a quine for great good!.

Furthermore, the capability of implementing a primality test is part of our definition of programming language, so this challenge will also serve as a directory of proven programming languages.

Task

Write a full program that, given a strictly positive integer n as input, determines whether n is prime and prints a truthy or falsy value accordingly.

For the purpose of this challenge, an integer is prime if it has exactly two strictly positive divisors. Note that this excludes 1, who is its only strictly positive divisor.

Your algorithm must be deterministic (i.e., produce the correct output with probability 1) and should, in theory, work for arbitrarily large integers. In practice, you may assume that the input can be stored in your data type, as long as the program works for integers from 1 to 255.

Input

  • If your language is able to read from STDIN, accept command-line arguments or any other alternative form of user input, you can read the integer as its decimal representation, unary representation (using a character of your choice), byte array (big or little endian) or single byte (if this is your languages largest data type).

  • If (and only if) your language is unable to accept any kind of user input, you may hardcode the input in your program.

    In this case, the hardcoded integer must be easily exchangeable. In particular, it may appear only in a single place in the entire program.

    For scoring purposes, submit the program that corresponds to the input 1.

Output

Output has to be written to STDOUT or closest alternative.

If possible, output should consist solely of a truthy or falsy value (or a string representation thereof), optionally followed by a single newline.

The only exception to this rule is constant output of your language's interpreter that cannot be suppressed, such as a greeting, ANSI color codes or indentation.

Additional rules

  • This is not about finding the language with the shortest approach for prime testing, this is about finding the shortest approach in every language. Therefore, no answer will be marked as accepted.

  • Submissions in most languages will be scored in bytes in an appropriate preexisting encoding, usually (but not necessarily) UTF-8.

    The language Piet, for example, will be scored in codels, which is the natural choice for this language.

    Some languages, like Folders, are a bit tricky to score. If in doubt, please ask on Meta.

  • Unlike our usual rules, feel free to use a language (or language version) even if it's newer than this challenge. If anyone wants to abuse this by creating a language where the empty program performs a primality test, then congrats for paving the way for a very boring answer.

    Note that there must be an interpreter so the submission can be tested. It is allowed (and even encouraged) to write this interpreter yourself for a previously unimplemented language.

  • If your language of choice is a trivial variant of another (potentially more popular) language which already has an answer (think BASIC or SQL dialects, Unix shells or trivial Brainfuck derivatives like Headsecks or Unary), consider adding a note to the existing answer that the same or a very similar solution is also the shortest in the other language.

  • Built-in functions for testing primality are allowed. This challenge is meant to catalog the shortest possible solution in each language, so if it's shorter to use a built-in in your language, go for it.

  • Unless they have been overruled earlier, all standard rules apply, including the http://meta.codegolf.stackexchange.com/q/1061.

As a side note, please don't downvote boring (but valid) answers in languages where there is not much to golf; these are still useful to this question as it tries to compile a catalog as complete as possible. However, do primarily upvote answers in languages where the author actually had to put effort into golfing the code.

Catalog

The Stack Snippet at the bottom of this post generates the catalog from the answers a) as a list of shortest solution per language and b) as an overall leaderboard.

To make sure that your answer shows up, please start your answer with a headline, using the following Markdown template:

## Language Name, N bytes

where N is the size of your submission. If you improve your score, you can keep old scores in the headline, by striking them through. For instance:

## Ruby, <s>104</s> <s>101</s> 96 bytes

If there you want to include multiple numbers in your header (e.g. because your score is the sum of two files or you want to list interpreter flag penalties separately), make sure that the actual score is the last number in the header:

## Perl, 43 + 2 (-p flag) = 45 bytes

You can also make the language name a link which will then show up in the snippet:

## [><>](http://esolangs.org/wiki/Fish), 121 bytes

<style>body { text-align: left !important} #answer-list { padding: 10px; width: 290px; float: left; } #language-list { padding: 10px; width: 290px; float: left; } table thead { font-weight: bold; } table td { padding: 5px; }</style><script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script> <link rel="stylesheet" type="text/css" href="//cdn.sstatic.net/codegolf/all.css?v=83c949450c8b"> <div id="language-list"> <h2>Shortest Solution by Language</h2> <table class="language-list"> <thead> <tr><td>Language</td><td>User</td><td>Score</td></tr> </thead> <tbody id="languages"> </tbody> </table> </div> <div id="answer-list"> <h2>Leaderboard</h2> <table class="answer-list"> <thead> <tr><td></td><td>Author</td><td>Language</td><td>Size</td></tr> </thead> <tbody id="answers"> </tbody> </table> </div> <table style="display: none"> <tbody id="answer-template"> <tr><td>{{PLACE}}</td><td>{{NAME}}</td><td>{{LANGUAGE}}</td><td>{{SIZE}}</td><td><a href="{{LINK}}">Link</a></td></tr> </tbody> </table> <table style="display: none"> <tbody id="language-template"> <tr><td>{{LANGUAGE}}</td><td>{{NAME}}</td><td>{{SIZE}}</td><td><a href="{{LINK}}">Link</a></td></tr> </tbody> </table><script>var QUESTION_ID = 57617; var ANSWER_FILTER = "!t)IWYnsLAZle2tQ3KqrVveCRJfxcRLe"; var COMMENT_FILTER = "!)Q2B_A2kjfAiU78X(md6BoYk"; var OVERRIDE_USER = 12012; var answers = [], answers_hash, answer_ids, answer_page = 1, more_answers = true, comment_page; function answersUrl(index) { return "https://api.stackexchange.com/2.2/questions/" + QUESTION_ID + "/answers?page=" + index + "&pagesize=100&order=desc&sort=creation&site=codegolf&filter=" + ANSWER_FILTER; } function commentUrl(index, answers) { return "https://api.stackexchange.com/2.2/answers/" + answers.join(';') + "/comments?page=" + index + "&pagesize=100&order=desc&sort=creation&site=codegolf&filter=" + COMMENT_FILTER; } function getAnswers() { jQuery.ajax({ url: answersUrl(answer_page++), method: "get", dataType: "jsonp", crossDomain: true, success: function (data) { answers.push.apply(answers, data.items); answers_hash = []; answer_ids = []; data.items.forEach(function(a) { a.comments = []; var id = +a.share_link.match(/\d+/); answer_ids.push(id); answers_hash[id] = a; }); if (!data.has_more) more_answers = false; comment_page = 1; getComments(); } }); } function getComments() { jQuery.ajax({ url: commentUrl(comment_page++, answer_ids), method: "get", dataType: "jsonp", crossDomain: true, success: function (data) { data.items.forEach(function(c) { if (c.owner.user_id === OVERRIDE_USER) answers_hash[c.post_id].comments.push(c); }); if (data.has_more) getComments(); else if (more_answers) getAnswers(); else process(); } }); } getAnswers(); var SCORE_REG = /<h\d>\s*([^\n,<]*(?:<(?:[^\n>]*>[^\n<]*<\/[^\n>]*>)[^\n,<]*)*),.*?(\d+)(?=[^\n\d<>]*(?:<(?:s>[^\n<>]*<\/s>|[^\n<>]+>)[^\n\d<>]*)*<\/h\d>)/; var OVERRIDE_REG = /^Override\s*header:\s*/i; function getAuthorName(a) { return a.owner.display_name; } function process() { var valid = []; answers.forEach(function(a) { var body = a.body; a.comments.forEach(function(c) { if(OVERRIDE_REG.test(c.body)) body = '<h1>' + c.body.replace(OVERRIDE_REG, '') + '</h1>'; }); var match = body.match(SCORE_REG); if (match) valid.push({ user: getAuthorName(a), size: +match[2], language: match[1], link: a.share_link, }); else console.log(body); }); valid.sort(function (a, b) { var aB = a.size, bB = b.size; return aB - bB }); var languages = {}; var place = 1; var lastSize = null; var lastPlace = 1; valid.forEach(function (a) { if (a.size != lastSize) lastPlace = place; lastSize = a.size; ++place; var answer = jQuery("#answer-template").html(); answer = answer.replace("{{PLACE}}", lastPlace + ".") .replace("{{NAME}}", a.user) .replace("{{LANGUAGE}}", a.language) .replace("{{SIZE}}", a.size) .replace("{{LINK}}", a.link); answer = jQuery(answer); jQuery("#answers").append(answer); var lang = a.language; lang = jQuery('<a>'+lang+'</a>').text(); languages[lang] = languages[lang] || {lang: a.language, lang_raw: lang.toLowerCase(), user: a.user, size: a.size, link: a.link}; }); var langs = []; for (var lang in languages) if (languages.hasOwnProperty(lang)) langs.push(languages[lang]); langs.sort(function (a, b) { if (a.lang_raw > b.lang_raw) return 1; if (a.lang_raw < b.lang_raw) return -1; return 0; }); for (var i = 0; i < langs.length; ++i) { var language = jQuery("#language-template").html(); var lang = langs[i]; language = language.replace("{{LANGUAGE}}", lang.lang) .replace("{{NAME}}", lang.user) .replace("{{SIZE}}", lang.size) .replace("{{LINK}}", lang.link); language = jQuery(language); jQuery("#languages").append(language); } }</script>

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  • \$\begingroup\$ Can I take inputs as negative numbers, where abs(input) would be the number I am testing? \$\endgroup\$ – Stan Strum Sep 6 '17 at 3:40
  • \$\begingroup\$ No, the input is a strictly positive integer. \$\endgroup\$ – Dennis Sep 6 '17 at 3:44
  • 1
    \$\begingroup\$ @LyndonWhite This was intended as a catalog (like “Hello, World!”) of primality tests, so a unified submission format seemed preferable. It's one of two decisions about this challenge that I regret, the other being only allowing deterministic primality tests. \$\endgroup\$ – Dennis Dec 12 '17 at 12:51
  • 1
    \$\begingroup\$ @Shaggy Seems like a question for meta. \$\endgroup\$ – Dennis Jun 25 '18 at 13:44
  • 1
    \$\begingroup\$ Yeah, that's what I was thinking. I'll let you do the honours, seeing as it's your challenge. \$\endgroup\$ – Shaggy Jun 25 '18 at 13:45

287 Answers 287

1
\$\begingroup\$

Molecule, 3 bytes

Inp

Explanation:

Inp
I   read input
 n  convert to number
  p primality test
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1
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LiveCode 8, 708 bytes

on mouseUp
    ask ""
    put p(it) into field "a"
end mouseUp
function p n
    if n is 1 then return false
    repeat with i=2 to n-1
        if n mod i is  0 then return false
    end repeat
    return true
end p

This code can be placed inside any button. It will print true or false to a field named a. It should work for small-ish integers, but will probably freeze/crash on anything too large. Byte count is size of saved LiveCode stack with one button and one field and with this code in the button.

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1
\$\begingroup\$

Unipants' Golfing Language, 53 47 bytes

i$cu^d^-l_u^^/%cu%?%d%:_?coc$d$$:__^d^-:_u?cuo:

Try it online!

(It has been implemented as a standard example in the interpreter.)

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1
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C, 54 bytes

i;main(j){for(scanf("%d",&i);i%++j;);putchar(49^j<i);}
  • Must be run without parameters, so j is initialized to 1.
  • Reads from standard input.
  • Prints 1 for primes, 0 for composites (no newline).
  • Could save 2 bytes with unprintable output - I'm not sure if \x00 and \x01 qualify as falsy/truthy.
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  • \$\begingroup\$ '\0' and '\1' are truthy and falsy according to our definition, so printing NUL or SOH should be fine. \$\endgroup\$ – Dennis Jun 16 '16 at 22:58
1
\$\begingroup\$

brainfuck, 208 202 bytes

This is the code I wrote for my Sesos answer. See that answer for a detailed explanation.

This code assumes input fits in a single cell. Attempting to take input from an empty file must set the cell to zero. It works on an unbounded tape with no wrapping, including one that can go negative. It works for all positive numbers. Output will be a byte with a value of 0 or 1.

[->+>+<<]++>[-<->]+<[>-<,]>[->+<]>[[->>>+<<<]>>>[->>>+<<<<<<+>>>]>>>-]<<<<<<+[<<<<<<]>>>>>>[-<+>]++[<[->>>+<<<]>>>[->>>+<<<<<<+>>>]>>>>][[->-[>+>>]>[+[-<+>]>+>>]<<<<<]<<<<<<]>>>>>>>>[>]>[<+>,]+<[>-<-]>.
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1
\$\begingroup\$

Idris, 78 bytes

main:IO()
main=let S n=length!getLine in print$0<(pow(product[1..n])2`mod`S n)

This takes input as a unary string (it uses the string length as input).

Compile this with idris -O3 or you’ll spend forever in Nat-computey hell.

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1
\$\begingroup\$

RPN, 16 bytes

lambda x isprime

This define a function checking the primality of it's argument.

Call like this:

<number> lambda x isprime eval
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1
\$\begingroup\$

Python 3.5, 58 52 58 bytes

import math;n=int(input());print(math.factorial(n-1)**2%n)

I unassumed. Thanks Dennis.

I assumed something about the way the input would be given. Thanks daHugLenny.

import math;lambda n:print(math.factorial(n-1)**2%n)

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  • \$\begingroup\$ You can save 13 bytes by using a lambda: import math;lambda n:math.factorial(n-1)**2%n \$\endgroup\$ – acrolith Aug 7 '16 at 2:21
  • \$\begingroup\$ In this particular challenge, lambdas are not allowed. The spec says Write a full program that... (CC @daHugLenny) \$\endgroup\$ – Dennis Aug 10 '16 at 2:57
1
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Nim, 70 56 bytes

import os,math
let x=1.paramStr.len
echo fac(x-1)^2mod x

Uses Wilson's theorem; that is, x is prime if (x - 1)!² mod x is 1.

Takes input in unary (any character) via the first command-line argument. Outputs 1 if the input is prime, and 0 otherwise. To test:

$ nim c prime.nim
$ ./prime 11111111
0
$ ./prime 1111111
1
$ ./prime 1
0

Note that according to the Nim os docs, this solution will not work on POSIX as paramStr isn't available for some reason.

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1
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PHP, 70 65 bytes

for($i=2;$i<$n=$argv[1];$i++)if(is_int($n/$i)?1:0){echo 0;break;}

Empty output if the number is prime, print 0 if the number is not prime.

Not very original or best answer, but this is it...

Test online

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1
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Ruby, 33 bytes

A little longer than other solutions, but doesn't require flags:

require'prime';p$*[0].to_i.prime?

Can be run putting in a file, say prime.rb:

ruby prime.rb 97

Or directly on the command line:

ruby -e "require'prime';p$*[0].to_i.prime?" 97

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1
\$\begingroup\$

Logy, 82 bytes

f[X]->X<2&1|X*f[X-1];p[&1]->0<0;p[X]->f[X-1]%X==X-1;main[A]->print[p[atoi[A(1)]]];

I have no idea what I am doing.

Use Wilson's theorem.

Ungolfed code:

factorial[X] -> X < 2 & 1 | X*factorial[X - 1];
prime[&1] -> FALSE;
prime[X] -> factorial[X - 1]%X == X - 1;
main[Args] -> print[prime[atoi[Args(1)]]];
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1
\$\begingroup\$

Batch, 121 bytes

set i=2
if %1%==2 exit /b
if %1%==1 echo 0
:b
set /a a=%1%%%i%
set /a i=%i%+1
if %i%==%1% exit /b
if %a%==0 echo 0
goto b

Outputs 0 if it's no prime. This works only for inputs input > 0.

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1
\$\begingroup\$

Dip, 1 byte

P

Pretty self-explanatory...

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1
\$\begingroup\$

Unary, 165192711826249169871716905363504385377479408598339493836 bytes

Well, basically 165192711826249169871716905363504385377479408598339493836 zeros. Here is a Java program that prints the actual program:

public class Main{
  public static void main(String[]args){
    for(BigInteger i = new BigInteger("165192711826249169871716905363504385377479408598339493836"),BigInteger j = BigInteger.ZERO;j.compareTo(i)!=0;j = j.add(BigInteger.ONE)){
      System.out.println("0");
    }
  }
}

This a port of this Brainfuck answer.

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1
\$\begingroup\$

memes, 2 bytes

I decided to hop on the language train. Code:

}"

Explanation:

} //Takes next input.
" //Returns `True` or `False`, representing if the input is prime.
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1
\$\begingroup\$

Ceylon, 107 112

That was my first try:

shared void p(){if(exists a=process.readLine(),exists c=parseInteger(a)){print(!(2:c-2).any((d)=>c%d<1));}}

This is a simple trial division of all numbers between 2 and c - 1, using the range operator : and the any method for iterating.

But this will also note 1 as prime. So here the corrected version:

shared void p(){if(exists a=process.readLine(),exists c=parseInteger(a)){print(1<c&&!(2:c-2).any((d)=>c%d<1));}}

That works for numbers up to 2^63-1 ... and takes quite long for larger primes (or composite numbers with large factors) – I think it took about half a minute to confirm 2147483647 (=2^31-1) as prime.

The formatted version:

shared void p() {
    if (exists a = process.readLine(),
        exists c = parseInteger(a)) {
        print(1 < c && !(2 : c - 2).any((d) => c % d < 1));
    }
}

I originally wrote the check as d.divides(c), but the modulo operator is so much shorter.

As a bonus, here is a variant for arbitrary size integers, though there is not really a point in using this due to the slowness.

import ceylon.math.whole {
    w=parseWhole,
    t=two,
    o=one
}
shared void q() {
    if (exists a = process.readLine(),
        exists c = w(a)) {
        print(c == t || !(t .. c - t).any((d) => c % d < o));
    }
}

The : range operator needs an Integer (the number of elements) as a second argument, so we use the .. operator, which takes the same type as the first argument, i.e. here a Whole, as second argument (the upper limit). 2..c-1 is the same as 2:c-2 for large c, but this becomes an empty list for two (and a non-empty list for one), so we need a different exception here than in the Integer version. It becomes mainly longer due to the imports.

The space-reduced version has length 153:

import ceylon.math.whole{w=parseWhole,t=two,o=one}shared void q(){if(exists a=process.readLine(),exists c=w(a)){print(c==t||!(t..c-o).any((d)=>c%d<o));}}
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1
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Maverick, 25 bytes

(1:(<>@-1)//$*)%<>@=<>@-1

Try it online!

This is a pretty fun language IMO. Infix and esoteric.

(1:(<>@-1)//$*)%<>@=<>@-1
 1:                         range from 1 to
   (<>@                     first command line arg (<> called with no args)
       -1)                  minus 1
          //$*              folded over multiplication
(             )%            modulus
                <>@         the input
                   =        does the above equal
                    <>@-1   the input minus 1? If so, yields prime
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1
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Valyrio, 14 bytes

s∫main [ipo]

Outputs 1 if the input is prime. Otherwise outputs 0.

Explanation

s∫ tells the interpreter to enter stack mode

main [ starts the main code block

i takes an input and evaluates it

p pushes 1 or 0 to the stack depending on primality

o outputs the top item on the stack

] ends the main code block

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1
\$\begingroup\$

Racket 75 bytes

(let p((c 2))(cond[(> c(sqrt n))#t][(= 0(modulo n c))#f][else(p(+ 1 c))]))

Ungolfed:

(define (f n)                 ; COMMENTS: 
  (let loop ((c 2))
    (cond
      [(> c (sqrt n)) #t]     ; if no divisor found till sq root of number, it must be a prime
      [(= 0 (modulo n c)) #f] ; if divisor found, it is not a prime
      [else (loop (add1 c))]  ; try with next number
      )))

Following version is longer but more efficient for checking larger numbers since it keeps & uses previously found prime numbers:

(λ(N)(define(p n(o'(2))(c 2))(cond[(> c n)o][(ormap(λ(x)(= 0(modulo c x)))o)
(p n o(+ 1 c))][else(p n(cons c o)(+ 1 c))]))(= N(car(p N))))

Ungolfed version:

(p=subfunction to build prime number list till n; o= list of prime numbers found; c=current number being checked)

(define f
  (λ (N) 
    (define (p n (o '(2)) (c 2))
      (cond
        [(> c n) o]
        [(ormap (lambda(x) (= 0 (modulo c x))) o)
                (p n o (add1 c)) ]
        [else (p n (cons c o) (add1 c))]))
    (= N (first (p N)))))

Testing:

(f 109)
(f 10)
(f 11)
(f 12)
(f 13)
(f 49)
(f 43)
(f 57)
(f 47)

Output:

#t
#f
#t
#f
#t
#f
#t
#f
#t
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  • \$\begingroup\$ If I understand correctly, the solution given is a snippet, not a full program, because it inputs from the variable n. \$\endgroup\$ – Esolanging Fruit Apr 24 '17 at 6:43
1
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PHP, 43 39 54 44 43 bytes

improved my answer from stackoverflow:

for($i=$n=$argv[1];--$i&&$n%$i;);echo$i==1;

Run with -r. prints 1 if argument is prime, empty string else.

loops $i down from $n-1 until it finds a divisor of $n; $n is prime if that divisor is 1.

6 bytes extra for the 1 case. almost happy.

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1
\$\begingroup\$

√ å ı ¥ ® Ï Ø ¿ , 3 bytes

Ipo

p is the primality test.

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1
\$\begingroup\$

NO!, 34 bytes

This was taken from the NO! GitHub page. It isn't mine

NOOOOOOO?NOOOOOOOOOOO
NOOOOOOOO?no
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  • \$\begingroup\$ This should be marked as community wiki if you didn't write it. \$\endgroup\$ – Esolanging Fruit May 11 '17 at 14:09
1
\$\begingroup\$

Ruby 16+15 bytes

require 'prime';p->e{e.prime?}
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1
\$\begingroup\$

QBIC, 21 20 3 bytes (nc)

Since the purpose of the challenge is to build a catalog, I feel it's important to keep the QBIC post up-to-date with the current state of the language:

­?µ:

Explanation:

?    PRINT
 µ   QBIC's prime test, returns -1 for primes, 0 otherwise
  :  Read a number from the command line, insert that here
     The function is closed automatically because of EOF.

Previously, at 20 bytes:

:[a|~a%b|\p=p+1}?p<3

Simple trial divider. Every time b (the loop counter) cleanly divides a (our prime-candidate), p is increased. Primes will end with p at 2 (or p=1 for a=1). This then prints -1 for primes, and 0 for non-primes, which are QBasic default values for true and false resp.


Original entry, 21 bytes. This prints 1 for primes, and 0 for others. This feels more 'natural' to me than QBasic's default -1/0. Also, this only does slightly less than a/2 divisions for primes (and quits when it detects a non-prime), instead of doing a divisions regardlessly.

:[2,a/2|~a%b=0|_Xp}?q

Explanation:

:       Get the input number, 'a'
[2,a/2| FOR(b=2, b<=(a/2), b++)
~a%b=0  IF a MOD b == 0     --> Clean division == no prime
|_Xp    THEN exit program, printing 'p' (which never gets set and is 0 by default)
}       Close all language constructs: IF/END IF, FOR/NEXT
?q      We've made it through the FOR loop without division, N is prime.
        In QBIC, 'q' is auto-initialised to 1, '?' prints it.
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1
\$\begingroup\$

Add++, 14 17 bytes

D,f,@,P
+?
$f,x
O

Try it online!

10 bytes are boilerplate for the full program requirement.

Function, 7 bytes

D,f,@,P

Fairly basic, just defines a function that performs a primality test

Built-in, 1 byte

P

As functions and main code use both different memory models and different commands, this only works in function mode and so would be invalid.

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1
\$\begingroup\$

cQuents, 3 bytes

?pz

Try it online!

Explanation

?     Mode ? (query)
 p    Builtin: next prime after parameter
  z   Previous item in sequence
   )  Implicit closing parenthesis

Query mode returns true if the input is in the sequence, and false if it is not.

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  • \$\begingroup\$ Note current version uses Z instead of z \$\endgroup\$ – Stephen Feb 1 at 4:53
1
\$\begingroup\$

TI-BASIC, 37 bytes

Prompt N
N=1
For(A,2,√(N
Ans+not(fPart(N/A
End
not(Ans

Prompts for N and displays 1 if N is prime or 0 otherwise.

Also in TI-BASIC:

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  • 1
    \$\begingroup\$ I rewrote it. This would have done better as a comment as well. \$\endgroup\$ – Zenohm Sep 13 '15 at 20:19
  • \$\begingroup\$ Nice. I modified my answer. Incidentally, I would have commented on your post but I created this account just to contribute this and didn't have enough reputation to comment. \$\endgroup\$ – Jakob Sep 13 '15 at 21:40
1
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TI-BASIC (nspire), 15 bytes

f(n):=isPrime(n

Uses the CAS isPrime() builtin. The environment by default adds an extra parenthesis on the end of the input, making it f(n):=isPrime(n) when the function is defined in the interpreter.

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  • \$\begingroup\$ I'm assuming nspire doesn't allow unclosed parentheses? \$\endgroup\$ – Zacharý Aug 24 '17 at 0:24
  • \$\begingroup\$ Yeah, it autocloses them. \$\endgroup\$ – user50198 Aug 24 '17 at 0:36
  • \$\begingroup\$ I don't think that the TI-Nspire runs TI-BASIC. I was under the impression that it used a variant of Lua... \$\endgroup\$ – Scott Milner Aug 24 '17 at 0:38
  • \$\begingroup\$ It also uses a form of lua for code in documents, but it does use basic for on-calculator programming (without an interpreter page) \$\endgroup\$ – user50198 Aug 24 '17 at 0:39
  • \$\begingroup\$ If it autocloses them, then isPrime(n should work. \$\endgroup\$ – Zacharý Aug 24 '17 at 0:44
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Pyth, 3 2 bytes

I can't see a simpler way of doing this. My ignorance continues to amaze me.

P_

Explanation (if it needs one):

P   Prime function: factorization if positive, isPrime if negative
 _  Negate the implicit input

Test suite

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  • \$\begingroup\$ No need to subtract from zero, as _ performs negation. \$\endgroup\$ – Dennis Sep 6 '17 at 3:53
  • \$\begingroup\$ @Dennis I tried that earlier, it didn't seem to work... \$\endgroup\$ – Stan Strum Sep 6 '17 at 3:54

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