# Is this number a prime?

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.

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.

• 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>

• Is there a reason for the full program requirement, rather than allowing the full range of default input types? E.g. answering with a function that takes its input as an argument, is currently disallowed? codegolf.meta.stackexchange.com/questions/2447/… Commented Dec 12, 2017 at 6:21
• @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. Commented Dec 12, 2017 at 12:51
• Could a case be made for locking this challenge and posting a new, less restrictive one? Commented Jun 25, 2018 at 12:59
• @Shaggy Seems like a question for meta. Commented Jun 25, 2018 at 13:44
• Yeah, that's what I was thinking. I'll let you do the honours, seeing as it's your challenge. Commented Jun 25, 2018 at 13:45

# F#, 79 84 bytes

Fixed as per Dennis's suggestion.

stdin.ReadLine()|>int|>fun n->not(Seq.exists(fun i->n%i=0)[2..n-1]||n=1)|>printf"%b"


Now works for input = 1. Can be run same as before.

Original submission:

I didn't see an F# answer so...

stdin.ReadLine()|>int|>fun n->not(Seq.exists(fun i->n%i=0)[2..n-1])|>printf"%b"


Can be compiled with fsc or run directly in fsi. Takes a newline-terminated string from stdin and spits out a bool to stdout when run.

• Unfortunately, that doesn't work for input 1. Try it online! Commented Dec 27, 2018 at 20:59
• Remember, lambdas are allowed, so... 44? Commented Jan 31, 2019 at 8:38

C#, 116 bytes

Non fancy full program entry (cherry-picking some stuff from java):

class P{static void Main(string[] a){int i=2,n=int.Parse(a[0]);for(;i<n;)n=n%i++<1?0:n;System.Console.Write(n>1);}}


# C++ (gcc), 85 bytes

#include<iostream>
int main(){int n,d=1;for(std::cin>>n;n%++d%n;);std::cout<<(n==d);}


Try it online!

• -2 bytes thanks to @ceilingcat

# Ungolfed

#include <iostream>
int main() {
int n,d=1;
for (std::cin >> n; n % ++d      // Trial division. If n dividible by d, end.
% n        /* No effect when n > 1.
But required to make sure the loop ever ends if n == 1
*/
);
) { /* Empty loop body */ }  /* d is now the lowest number greater than 1 which
divides n (or 2 if n == 1)
*/
std::cout << (n==d);             // If that number is n, n must be prime.

}


Port of my Ink answer. Competitive in multiple other languages. I wasn't going to post this, but I couldn't find a C++ compiler that would compile the existing C++ answer.

# Ink, 68 51 bytes

-(l)
+(n)[+]
->l
*(r)[{n}]
{n%(r+1)%n:->r|{n==(r+1)}}


Ink was made for writing interactive fiction. It was never meant to be a general-purpose programming language. So it only has one way to recieve input.
Choice prompts.
So the way it works is, if option 2 is labelled with the number you want to check, you pick option 2. Otherwise you pick option 1 to increment option 2's label. So, if you want to check if 15 is prime, you'd input

1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
2


TIO doesn't handle this, unfortunately.

-17 bytes by using read counts of named options instead of creating variables. This brings the overhead of taking input this way down to something like 24 characters (or even 18 if we don't care about minimising the text from the prompts), compared to the 10 for defining a parametrised stitch and returning from it.

Since full programs can use global variables and read counts, but stitches that have to be reusable can't, using full programs might actually pay off in places where I thought it wouldn't. Neat.

# T-SQL, 116 bytes

WITH t AS(SELECT 2n UNION ALL SELECT n+1FROM t WHERE n<5e4)
SELECT*FROM z,t WHERE p%n=0AND n<p OPTION(MAXRECURSION 0)


There is an Oracle SQL answer here, but couldn't find a Microsoft T-SQL version for this classic question.

Notes:

• Line break is for readability only.
• Returns no rows if prime, returns 1 or more rows otherwise. This is allowed per this output rule.
• Input is taken via pre-existing table $$\z\$$ with INT field $$\p\$$, per this input rule.
• Handles values of $$\p\$$ up to the max value of the INT type, $$\2^{31}-1\$$

Explanation:

Generates an in-memory number table $$\t\$$ from 2 to 50k (which exceeds the square root of the max int), then joins to the input table $$\z\$$ and uses the modulo operator % to test for divisibility. Any values that divide evenly are returned, so an empty result means the input is prime.

OPTION(MAXRECURSION 0) is needed to recursively generate a table with more than 100 rows, unless you've altered this configuration setting on your SQL server.

Optimized for bytes, not speed; checks more values than is strictly necessary.

# Wren, 48 43 bytes

I feel surprised that I think better when using W instead of Wren. Also, the 1-primalty problem is fixed.

Fn.new{|i|(1..i).where{|x|i%x==0}.count==2}


Try it online!

# Symbolic Raku, 17 bytes

$_=?($_%%[^] ^$_)  Try it online! Suprisingly, this is shorter than the my previous Raku answer, though one byte longer than using the built-in ### Explanation: $_=                # Set the output to
?(           )  # The boolean of
$_ # If the input is %% # Divisible by [^] # Only one of ^$_   # The range from 0 to input-1


## W d, 5 bytes

(Possibly the only <=5-byter that does not use a prime or number factorization built-in?) (The two spaces here represent a tab)

àTéBΦ


repr'd:'‡T\tB⑸'

Unpacked:

m!Wk2=


## Explanation

This is really difficult to read, so an explanation will be nice.

    W    % (implicit) Foreach the inclusive range from 1 to the input:
bam      % The "bam" magic! Just kidding. Evaluates b % a
% Since b is out-of-bounds, this takes the upper-level input
% as the operand.
!     % Remove those that aren't valid multiples.
k   % Is the length of the array ...
2= % ... equal to 2?
% i.e. primes have exactly 2 factors: 1 and itself

• Crax! That sounds like a good golfing language name. Commented Jan 1, 2020 at 6:16
• Looks like there is a bug in the interpreter and I had to change the compression...
– user85052
Commented Jan 20, 2020 at 0:14

# Keg, 43366 4 bytes (SBCS)

@MPƒ


-2 bytes using a BFL function

:;¡²$%  Try it online! -30 bytes thanks to @JoKing No TIO as it needs to be updated due to a bug found with the ² operator. Transpiles to: from KegLib import * from Stackd import Stack stack = Stack() printed = False duplicate(stack) decrement(stack) factorial(stack) square(stack) swap(stack) maths(stack, '%') if not printed: printing = "" for item in stack: if type(item) is Stack: printing += str(item) elif type(item) is str: printing += custom_format(item) elif type(item) == Coherse.char: printing += item.v elif item < 10 or item > 256: printing += str(item) else: printing += chr(item) print(printing, end="")  ### 36 bytes (SBCS)  ¿®n21®t{:©n≠©t0≠*|:©n$%[1|0]®t1+}©t.


Try it online!

-7 bytes by not using a variable to keep track of the count.

Explained

¿®n21®t{:©n≠©t0≠*|:©n$%[1|0]®t1+}©t. ¿®n #Take the number to check and store it in variable n 2 #This will act as the divisor counter 1®t #This will act as the return value, indicating the primality {:©n≠©t0≠*| #While the counter isn't equal to n and there isn't a divisor yet :©n$%             #Take the counter and number, then swap for modulus
[1|0]®t      #Depending upon the result set t to 1 if it doesn't divide otherwise 0
1+}   #Increment the counter


### 43 bytes (SBCS)

¿®n2®c1®t{©c©n≠©t0≠*|©n©c%[1|0]®t©n1+®c}©t.


It's a real ugly mess, but it works. I'll explain it all later. Also, doesn't work on TIO due to a bug that I found. It should work soon however.

Use the latest github interpreter

# cQuents, 3 bytes

?pZ


Try it online!

## Explanation

Usually, I'd put the commands here, but I feel like that wouldn't be enough.

? outputs if the input is in the sequence. Item 1: Z returns 0, as Z returns the previous item, and Item 0 is always 0. p finds the next prime number (2). Item 2: Z returns 2. p finds the next prime number, which I thought would be 2, but no, it returns 3.

# Haskell + Parsec, 181 bytes, no math

import Text.Parsec
a=anyChar
g s=a>>= \c->try(many1$string$c:s)*>eof<|>g(c:s)
main=readLn>>= \x->print$0<case parse(try(a*>eof)<|>(pure<$>a>>=g))""$replicate x '0'of{Left _->1;_->0}  Try it online! ### Ungolfed and explanation: import Data.Either (isLeft) import Text.Parsec (eof, (<|>), try, anyChar, many1, string, parse) import Text.Parsec.String (Parser) -- Successfully parses any string of nonprime length, and fails on all others. -- Exactly one character is not prime. If more than one character exists, we recurse with go. notPrime :: Parser () notPrime = try (anyChar *> eof) <|> (pure <$> anyChar >>= go)
-- go cs reads a character c and determines whether the remaining input is c:cs repeated.
-- If so, the parse succeeds. If not, we call go (c:cs) to read another character and test again.
where go :: String -> Parser ()
go cs = do
c <- anyChar
try (many1 $string$ c:cs) *> eof <|> go (c:cs)

-- Checks whether the parser notPrime fails on a uniform string of the specified length
isPrime :: Int -> Bool
isPrime n = isLeft $parse notPrime "" (replicate n '\0')  Try it online! # qalc/Qalculate, 8 bytes If reading from any variable is allowed: p-1)!²%p  Otherwise 12 bytes: ans−1)!²%ans  The great thing about "programming" in a calculator is that you can basically just enter formulas directly. Example usage in Bash for the "ans" case: echo "5" | qalc -t -f - "ans−1)! ²%ans"  Note that I added a space after the exclamation mark only because Bash otherwise tries to interpret "!²" as an "event". It's not actually required by qalc, which can be seen by for example creating this file: 5 (ans−1)!²%ans  …and then executing it with qalc -t -f test.txt. Other consoles or other ways of calling qalc than Bash might not have this problem. The calculation can also be entered in "interactive mode" by simply running qalc -t, entering the number and then entering the expression on the next input. This also works in the GUI program "Qalculate". ### Meta explanation: The output is 0 for non-prime numbers and a positive integer for prime numbers. I let this run in a loop for all numbers up to 10000 and it was always 1 for prime numbers, but I haven't checked whether that is guaranteed yet. This solution is based on this Labyrinth answer. I essentially just found something (qalc) that has all the parts of the formula as builtins instead of having to program all the parts myself on a lower level. The -t argument hides the interpreted calculation, so that the output of the example command looks like this: 5 1  …instead of like this: 5 = 5 rem(factorial(ans - 1)^2, ans) = 1  Reading the number in in qalc is difficult. It has a system of variables, but I haven't figured out how to set those in the CLI yet. In the GUI it's pretty easy. That's one of the main reasons why I also provided a solution with "ans". Inserting the number into the formula two times would mean modifying the "program", which is surely not allowed. There seems to be no way to execute multiple expressions non-interactively with qalc, except by reading in a file. In this case, I used - as a "file", which is StdIn, and used that only to store the input in ans, which is implicitly done by qalc. You could also use an actual file and use the number in that file as input. Or you could even say that ans is already the input of the "program", in which case the program is simply the 12 bytes seen above, or even any variable, then the 8 byte solution. ### Algorithm explanation: The Wikipedia article linked in the Labyrinth answer looks pretty confusing, but the principle is easy: (6-1)!²%6 = 1·1·2·2·3·3·4·4·5·5%6 = 1·1· 6 · 6 ·4·4·5·5%6 = 6 · something %6 = 0  This works because every non-prime number is the product of at least two factors that are smaller than that number. Prime numbers aren't, which leads to something like this: (5-1)!²%5 = 1·1·2·2·3·3·4·4%5 = 1·1·2·2·3·3·4·4%5 (no multiplication of those factors results in anything divisible by 5) = 1  The squaring is necessary in case there is only one prime factor and that occurs twice: (9-1)!²%9 = 1·1·2·2·3·3·4·4·5·5·6·6·7·7·8·8%9 = 1·1·2·2· 9 ·4·4·5·5·6·6·7·7·8·8%9 = 9 · something %9 = 0  I originally wrote this answer for TI-Basic: fPart((Ans-1)!²/Ans  That's just 10 bytes, but it fails on the input 13 already, due to floating point imprecisions in those large numbers (12!²=229442532802560000). So the only difference to an even/odd checker within the working range is 9 and it does not fulfil the criterion to work up to 255. For inputs above 42 it even crashed from an overflow (42!²>10¹⁰⁰). # Java (JDK), 105 bytes interface P{static void main(String[]a){long i=2,n=new Long(a[0]);while(n%i++>0);System.out.print(i>n);}}  Try it online! # Duocentehexaquinquagesimal, 23 bytes 7 ÇÇW∞DćƒΣZ;ndž“~3¦€н¸k  Try it online! # PPL 1.0.5, 85 bytes fnp(n){ declareb=1 declaref=4<5 loop n - 2{ b=b+1 ifn%b<1{ f=3<2 } } unsetb returnf }  It is a "good practice" (defined by me, of course) to unset all unneeded variables before returning. The unset statement is like delete in JavaScript. The fn statement defines a function and returns the flag. Still many bugs to iron out (remember, this is still being developed) but seeing as this is a catalogue of primality tests, thought I might post one. You have to unset n and f each time you recall the function, though. # SmallBasic, 163 bytes n=TextWindow.ReadNumber() t=1 If n<2 Then t=0 Goto e EndIf For i=2 to Math.Power(n,.5) If Math.Remainder(n,i)=0 Then t=0 Goto e EndIf EndFor e: TextWindow.Write(t)  This is so self-explanatory. But some key points:- • A simple algorithm, Set the prime check value to 1, iterate through 2 to square root of input. If there are any divisors set prime check value to 0, else continue. At last output it. • What a shame, its Square root builtin is shorter than power by 1/2. # Pxem, Content: 0 bytes + Filename: 54 bytes. • Filename (escaped): ._.c\002.y.d.a.c\002\003.t.x.c.m.%\001.y.d.a.c.m\001.+.c.t.c.!\001.-.a.n • Content is empty. ## Usage • Input is given from stdin, as a string of decimal integer. • Outputs the input number for truthy; no output for falsey. ## How it works ._XX.z # let input be n # get out if <2 .a.c\002.y.d.aXX.z # let i be 2 # while i*i<=n (or i*i-1<n); do .a.c\002\003.t.xXX.z # if n%i<1; then exit; fi .a.c.m.%\001.y.d.aXX.z # i++ .a.c.m\001.+.c.t.c.!\001.-XX.z # done; print n .a.a.n  Try it online! # MMIX, 14 bytes (56 instrs) (jxd) 00000000: 31ff0001 7a00ff00 31ff0002 6300ff03 1”¡¢z¡”¡1”¡£c¡”¤ 00000010: e3010002 f7010000 1e020001 feff0006 ẉ¢¡£ẋ¢¡¡œ£¡¢“”¡© 00000020: 7a00ff00 32ff0201 e7010001 59fffffb z¡”¡2”£¢ḃ¢¡¢Y””» 00000030: 7b000001 f8010000 {¡¡¢ẏ¢¡¡  Disassembled: prime CMP$255,$0,1 ZSNZ$0,$255,$0     // if(n == 1) n = 0
CMP  $255,$0,2
CSZ  $0,$255,3      // if(n == 2) n = 3
SETL $1,2 // i = 2 PUT rD,0 0H DIVU$2,$0,$1       // loop: q = n / i
GET  $255,rR // r = n % i ZSNZ$0,$255,$0     // if(!r) n = 0
CMPU $255,$2,$1 INCL$1,1
PBNN $255,0B // if(q >= i++) goto loop ZSNZ$0,$0,1 POP 1,0 // return (bool)n  # Knight, 35 bytes ;=y;=x+0P1;W&=z>x=y+1y%x y 0O&-xT!z  Try it online! ### Edit: Outgolfed by sampersand Rough. Prints true for primes, false or 0 for non-primes. # prompt for a number, coerce to integer ; = x + 0 PROMPT # loop counter ; = i 1 # Loop while x > y and x % y != 0 ; WHILE & (= cmp > x y) (% x y) : NOP # Output result. The - x 1 returns false for 1. : OUTPUT & (- x 1) (!cmp)  # ErrLess, 47 bytes q@1=1-[{3+}2xY@;>1-[{@;%0=1-[{0#.};1+xy}2]y1_#.  Could probably get golfed a lot more... Prints 0 if the input is not prime, -1 if it is, no trailing newline. It uses -1 as a truthy value, since that is used for if-constructs, and what you get if a comparison is true. ## Explanation q { Input a number: (p) } @1= 1-[{ { If p == 1 } 3+ { 1 -> 4, so that it correctly identifies 1 as non-prime } } 2x { Push a stack containing the number, and two: ((p 2)) } Y @;> 1-[{ { While p > i } @;%0= 1-[{ { If p%i == 0 } 0#. { Print 0 and halt } } ;1+x { Increment i: ((p i++)) } y}2]y 1_#. { Print -1 (ErrLess's truthy value) and halt }  ErrLess is a stack-based language I made for fun over the last few months. You can read the docs here, and I also started a tutorial. Try it online! # MathGolf, 1 byte ¶   -- for each line in input ¶ -- 1 if it's prime or 0 if not (works with lists too) -- implicit print  Try it online! # Excel, 34 bytes =SUM((MOD(A1,SEQUENCE(A1))=0)*1)=2  Link to Spreadsheet Limited to $$\n <= 2^{20}\$$ because of SEQUENCE(). A 40-byte alternative that works for $$\n<(2^{20}+1)^2\$$ is below. =SUM((MOD(A2,SEQUENCE(SQRT(A2)))=0)*1)=1  • I think that =AND(MOD(A1,SEQUENCE(A1-2,,2))>0) should work for 33 bytes - basically, invert the logic and search only the range from 0...n-1 for divisors Commented Feb 3, 2022 at 22:35 • I like the idea but SEQUENCE(A1-2,,2) breaks for A1 < 3. Commented Feb 3, 2022 at 23:41 • Oh that's a good catch there - in that case the only improvement I can think of is pretty boring - -SUM(-(MOD(A1,SEQUENCE(A1))<1))=2 for 33 bytes Commented Feb 4, 2022 at 0:02 # Vyxal, 1 byte æ  Try it Online! # Vyxal, 7 bytes ‹¡›$%¬


No shady built-ins, just pure, beautiful maths.
Basically Wilson's theorem

$$\\large{(n-1)!+1 } \equiv 0 \left(\mod n\right)\$$

Try it Online!

Explanation:

‹¡›$%¬ ‹¡ push n-1 factorial › push n+1$%    modulo top two items swapped
¬   apply logical not


# Python3, 53 bytes

(lambda n:all(n%i for i in range(2,n)))(int(input()))


# Thunno, 1 byte

(Technically it's $$\1 \log_{256}(96) \approx\$$ 0.82 bytes, but that doesn't show up on the leaderboard)

N


Attempt This Online!

# Scratch, 81 bytes

Try it on Scratch!

Scratchblocks syntax:

define(n
set[i v]to((n)-(1
repeat until<(0)=((n)mod(i
change[i v]by(-1
say<(i)=(1


This works best if the procedure is checked as "Run without screen refresh." If the option is unchecked it will still work, but will take n frames to output the correct result (i.e. the result will be incorrect for n frames).

# hyperscript, 82 69 65 bytes

def f(n)set x to n-1 repeat while n mod x decrement(x)end log x<2


This is shorter than a recursive solution because hyperscript doesn’t have a great way to do optional parameters.

<script src="https://unpkg.com/[email protected]"></script>
<script type="text/hyperscript">
def f(n)set x to n-1 repeat while n mod x decrement(x)end log x<2
</script>

<label style="font-family:monospace">n: <input type="number"style="font-size:1.5rem;width:4.5ch"_="on change call f(my value)" /></label>

Ungolfed:

def f(n)
set x to n - 1
repeat while n mod x
decrement(x)
end
return x is 1
end


# Python, 73 Bytes

import math
y=lambda b:b>1and math.factorial(b-1)%b==b-1
print(y(input()))


-37 from The Thonnu :O
+20 to fulfil challenge requirements -6 by using math.factorial.

This code is based on Wilson's Theorem.

• 59 by using an anonymous function Commented Apr 7, 2023 at 9:14
• @TheThonnu The rules say it needs to be a full program, so this 59 byte version is invalid. But it can be 82 bytes by using an anonymous function in that full program: Try it online!, and 75 bytes in Python 2, because its input() is the same as Python 3+'s eval(input()): Try it online! And incidentally, your 59 byter is not a true anonymous function, but it can be in Python 3.8+: Try it online! Commented Apr 7, 2023 at 19:00
• Also, just in case you didn't know about it, there is a way to search for other Python answers to this question. Commented Apr 7, 2023 at 19:10
• @Deadcode I did, in fact, search through all the Python answers to see how it compared with the others. Also, thanks for telling me about the full program thing, it's really annoying that old challenges have that requirement. Commented Apr 7, 2023 at 21:18
• Yours also only works in Python 2 since you aren't converting the input() to int. But anyway, 66 bytes, works in Python 2 and Python 3. Commented May 13, 2023 at 15:54

# Vyxal 3.0.0-beta.2, 8 bytes

1?R%0C1=


Try it online! (link is to literate version)

alternatively

1 input a->b mod 0 count 1 equal?


The fun of pre-releases

## Explained

1?R%0C1=­⁡​‎‎⁪⁡⁪⁠⁪⁡⁪‏⁠‎⁪⁡⁪⁠⁪⁢⁪‏⁠‎⁪⁡⁪⁠⁪⁣⁪‏‏​⁡⁠⁡‌⁢​‎‎⁪⁡⁪⁠⁪⁤⁪‏‏​⁡⁠⁡‌⁣​‎‎⁪⁡⁪⁠⁪⁢⁡⁪‏⁠‎⁪⁡⁪⁠⁪⁢⁢⁪‏‏​⁡⁠⁡‌⁤​‎‎⁪⁡⁪⁠⁪⁢⁣⁪‏⁠‎⁪⁡⁪⁠⁪⁢⁤⁪‏‏​⁡⁠⁡‌­
1?R       # ‎⁡Range [1, input)
%      # ‎⁢input % each in ^
0C    # ‎⁣Count the number of 0s. This will return 0 for input = 1, because range [1, 1) is []
1=  # ‎⁤Does that equal 1?
💎
`

Created with the help of Luminespire.

• Why not just use the built-in? Has it not been implemented yet? Commented Aug 1, 2023 at 14:00
• @TheEmptyStringPhotographer that's right. It hasn't been implemented yet Commented Aug 1, 2023 at 14:17