# 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.

## 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/… Dec 12 '17 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. Dec 12 '17 at 12:51
• Could a case be made for locking this challenge and posting a new, less restrictive one? Jun 25 '18 at 12:59
• @Shaggy Seems like a question for meta. Jun 25 '18 at 13:44
• Yeah, that's what I was thinking. I'll let you do the honours, seeing as it's your challenge. Jun 25 '18 at 13:45

1
7 8 9 10
11

Probably not the shortest program:

 #
W
I
P
v
>~1=
~
0
=


Another possible program is, 30 bytes:

#WIP>v~0=
~
1
=


Information:

• # = Starting point, pointer direction is to the right
• W = User input, directly adds this to the stack
• I = If-statement, or is equal to
• P = If the stack is prime
• v = If false, assign the pointer direction to the right
• > = If true, assign the pointer direction to the bottom
• ~ = Reset the stack
• 0 / 1 = Assign the stack to 0 / 1
• = = Print the result of the stack

The following paths are executed in this program:

• If W = prime, path = WIP~1=
• If W != prime, path = WIP~0=

This is a 2D esolang I created, which looks a bit like a magic square. This is a programming language in it's early stage, so a lot of functions aren't implemented yet. Also, a lot of bugs might occur, if you find a bug, please report it.

## DStack, 66 bytes

025SSd01kKCccscS0kT0cK1kAsd34SSd1ccd0sd1ddsScsdk0cD0cS0kTcdsKkdtcK


# Gogh, 1 byte

e


Run using:

$./gogh oi "e" <input>  The operator e pushes the primality of the TOS. Implicit input and output. # Pylongolf2, 4 bytes cnz~  Explanations: cnz~ c read input n convert to integer z primality test (beta8+) ~ print it.  # Racket, 31 bytes (require math)(λ(n)(prime? n))  A standard library has a function for it. If we want to keep the import but write our own, we could do this for 51 bytes: (require math)(λ(n)(equal?(list 1 n)(divisors n)))  On the other hand, if builtins were disallowed, for 64 55 bytes, thanks to @MatthewButterick, we can write our own: (λ(n)(=(count(λ(i)(=(modulo n(+ i 1))0))(range n))2))  Old one, for 64: (λ(n)(equal?'(1)(filter(λ(i)(zero?(modulo n i)))(range 1 n))))  • 55 bytes: (λ(n)(=(count(λ(i)(=(modulo n(+ i 1))0))(range n))2)) Apr 11 '16 at 5:49 • @MatthewButterick Nice! I'll add that. – cat Apr 11 '16 at 10:47 • @MatthewButterick from the docunentation, it seens like range needs two args, but I guess I was very tired when I was reading it. – cat Apr 11 '16 at 10:57 • Also, (λ(n)(prime? n)) is the same as prime?, so if (require math)(λ(n)(prime? n)) is an acceptable 31-byte answer, then (require math)prime? ought to be an acceptable 20-byte answer Apr 11 '16 at 17:29 • @MatthewButterick I was very tired when I edited that in last night and I swore I read your name correctly... sorry about that :P – cat Apr 11 '16 at 19:20 ## Sage, 22 bytes lambda N:N in Primes()  Try it online • (Super old but...) Submissions have to be full programs. Sep 2 '17 at 2:17 # UGL, 26 bytes id?u:cuul^^/_c=?_=ocd$:_u:


Try it online!

## How it works:

id?u:cuul^^/_c=?_=ocd$:_u: id?u: int i = input(); if i==1: i=0; cuu int test = 2; l _u: while test != -1: ^^/_c=? if i % test == 0: _=o print(i==test); cd$              test = -1;


# reticular, 6 bytes

in@pp;


A four-byte function:

[@p]


## Explanation

in@pp;
i       take input
n      convert to number
@p    check for primality
p   print
;  terminate

• (non-competing) language is newer than the challenge. Sep 22 '16 at 19:39
• @mbomb007 is that really necessary? This is a catalog. Sep 22 '16 at 20:24
• Oh, idk. I guess not then Sep 22 '16 at 20:29

# R, 19 Bytes

 ->n;sum(!n%%1:n)==2


Starts with a right assign to n. i.e. precede the code with the desired n. then it merely checks primeness of n. Prints TRUE or FALSE implicitly

If you don't like right assign as an input method then for 24 bytes you get:

sum(!(n=scan())%%1:n)==2
or
n=scan();sum(!n%%1:n)==2


Which assigns within the operator.

If you want explicit answer printing, then for 28 bytes:

 p=function(n)sum(!n%%1:n)==2


# RProgN, 102 Bytes

 'i' asoc true i 1 - ] 1 > while [ 'a' asoc i a / i a // = if [ [ false else [ end a 1 - ] 1 > end [ [


# Explanation

 'i' asoc           # Associate the implicit input with i
true i              # Push true to the stack, push i to the stack
1 -                 # Subtact 1 from the top of the stack
] 1 >               # Duplicate the top value of the stack, compare that it's larger than 1
while               # While the top of the stack contains a truthy value
[               # Pop the top of the stack (The conditional in this case)
'a' asoc        # Associate the top of the stack with a
i a /           # Push i divided by a to the top of the stack
i a // =        # Push i integeral divided by a to the top of the stack, compare the top and underneith the top for equality
if              # if i/a = floor(i/a), essentially
[ [ false   # Pop the conditional, and the 'true' we slipped in earlier, push false in in place of it
else            #
[           # Pop the conditional anyway
end             #
a 1 - ] 1 >     # Subtract 1 from a, duplicate it, and compare that it's larger than 1
end                 #
[ [                 # Pop the conditional, the value of a, which leaves only the bottom value we previously inserted, which is implicitly printed


RProgN doesn't have any method of input as of yet, as such, Any input needs to be written in pure form at the very start of the code, such that 23 'i' asoc .... An Extra byte is added to the score, because a space is required in front of any command.

# Pyret, 58 bytes

i=1
(i == 1) or any({(x):num-modulo(i,x) == 0},range(2,i))


You can try this online by copying the code into the online Pyret editor!

Pyret programs are designed to be run in the online editor, so there's no way to read from STDIN. Here, the variable i represents the input. The program returns true if i is not prime and false if i is prime.

• If there's no input from STDIN, maybe its best to do a function instead
– Jo King
Jun 20 '18 at 0:14
• @JoKing This challenge doesn't allow functions. Jun 20 '18 at 1:46

# Yabasic, 43 bytes

Input""n
For d=2To n
p=p+!Mod(n,d)Next
?p=1


Try it online!

# Pepe, 53 bytes

Uses trial-and-error modulus, so gets slower in larger integers.

REeEREErEEEEErREEeEeEReeREEEEeEeEReRererrEEreEErEeree


Try it online! Outputs 0 for truthy and none for falsy.

## strict, 122 bytes

num t
num i 2
num r
get t
jen t 1 n
jen t 2
mrk m
set r t
mod r i
jen t i
jgn r 0 m
mrk n
out: not
mrk
out: prime


Reads input from stdin. Because the language doesn't have truthy/falsy values, it outputs either "prime" or "not prime".

Ungolfed:

num target
num i 2
num remainder
get target

# exceptions
jen target 1 not
jen target 2 prime

# loop from 2 to target
mrk main
set remainder target
mod remainder i
jen target i prime
jgn remainder 0 main

mrk not
out: not\s

mrk prime
out: prime\n

• Welcome to Code Golf Stack Exchange! I'm not sure how best to test this; do you have a link to some sort of online interpreter? I can't access the link thanks to my stupid corporate network restrictions. Jun 28 '19 at 15:50
• @Giuseppe Unfortunately, there is no online interpreter. How much are you restricted? Would a direct mega.nz download link work? Jun 28 '19 at 16:12

## Spice, 93 bytes

;i;m;t;c;r@REA i;ADD 0 1 c;SUB i 1 m;ADD c 1 c;MOD i c t;SWI t 1 9;SWI c m 3;ADD 0 1 r;OUT r;


This outputs [1] for a prime, and [] for a non-prime.

## Un-golfed explanation:

;i;m;t;c;r@ - Declare vars (@ marks end of declarations)
(line 0) REA i;      - Read input into i
(line 1) ADD 0 1 c;  - Set counter c = 1
(line 2) SUB i 1 m;  - Set store input-1 as m
(line 3) ADD c 1 c;  - Start of loop, c+=1
(line 4) MOD i c t;  - Store mod of  t=i%c
(line 5) SWI t 1 8;  - if t < 1 (if non-prime, leave loop) to line 8
(line 6) SWI c m 3;  - if c < m continue loop (go to line 3)
(line 7) ADD 0 1 r;  - set result as truthy value - r = 1
(line 8) OUT r;      - output result, r


## tq, 10 bytes

?-1!xp*p%?


## Explanation

Uses Wilson's theorem to do the prime-checking.

?-1!        # Generate all numbers from 2 to input - 1
x,      # Preserve this value, prevent it from being printed
p*p   # Generate all possible combinations of those numbers
%? # Does the product of any combination equal to the input?


## C# (Visual C# Interactive Compiler), 62 bytes

int n=int.Parse(ReadLine()),i=2;while(n%i>0&i++<n);Write(n<i);


Try it online!

# LCGFuckinputNumeric=true, 138 bytes

(Will turn this into a GitHub repository soon)

1 1im1-1
1 1om2-1
1 1om1 1
<{>>>o+1 1m0<<om
<}>>1-1om0>+om
<<<++1 1 1 32
1 4 12 69 4
1 1 7 78 2
{>>>+++++c+c+++++c+++<<c<{+}}>>>c++c<c+c+c


Try it online through this TIO JS interpreter

A brainfuck variant invented by me but uses a list of linear congruential generators (LCGs) instead of cells.

The $$\N=1\$$ case took too much more bytes to deal with.

### Ungolfed and Explanation

1 1 im 1 -1                    # Storage for the input, since the input can only be retrieved once
1 1 om 2 -1                    # Used to prevent wrapping. Necessary for handling N = 1 case
1 1 om 1 1                     # Counts up from 2 to N
<{                             # While N % counter == 0 (If N == 1 the loop is skipped)
>>>o+                         #  Gets and increments the counter value
1 1 m 0 <<om                  #  Use a new LCG to check N % counter
<}                             # Loop
>>
1 -1 om 0 >+om                 # The output of this LCG will be 2 if N is prime
<<<++                          # Move two states forward (2 - 2 = 0 if prime)
1 1 1 32                       # LCG for space
1 4 12 69 4                    # LCG for E, I, M
1 1 7 78 2                     # LCG for N, O, P, R, T
{>>>+++++c+c+++++c+++<<c<{+}}  # Output "NOT " if this LCG is not at state 0
>>>c++c<c+c+c                  # Output "PRIME"


### Technical Details

(Copied from my obsolete challenge proposal on an LCGFuck interpreter)

An LCG is defined with 5 numbers, like 12345 678 90 -123 45. The last two numbers are optional and default to 0.

LCGFuck has 3 storage variables, the LCG list, the number list and the number memory. The functions are as follows:

• LCG list: Cyclic list that stores all LCGs created. Every entry is in turn a list of 5 numbers in the order [a, b, c, d, e] as notated in the introduction. It has a pointer that controls which LCG is chosen at the moment, and moves in a cycle through the list.
• Number list: Ordinary list that stores the numbers for LCG definition. It can store at most 5 numbers.
• Number memory: A single variable that can be read and written with an input or an LCG output.

LCGFuck has 14 operators, namely:

• \n (newline): Creates an LCG with the numbers in the order of [a, b, c(, d(, e))] and clears the list if the number list contains 3 or more numbers. No-op if the number list contains no more than 2 numbers.
• <: Shifts to the previous LCG circularly (back to the last when moving from the first).
• >: Shifts to the next LCG circularly (back to the first when moving from the last).
• +: Moves the current LCG to the next state (calculates x' = (a * x + b) mod c)
• c: Prints the output of the current LCG as a character (with the output as the codepoint).
• n: Prints the output of the current LCG as a number (with a trailing space)
• o: Writes the output of the current LCG to number memory
• i: Reads a value from the input and writes it to number memory. There is two input mode, one reading a character one time (inputNumeric=false), and one reading a number one time (inputNumeric=true)
• s: Reads the value from number memory and seeds the current LCG with it
• m: Reads the value from number memory and pushes it to the number list
• []: Output loop. Executes the loop if the current LCG is non-zero
• {}: State loop. Executes the loop if the state of the current LCG is non-zero

An integer, optionally with a negative sign at the front, pushes the number to the number list. Any characters other than the newline \n and any of -0123456789[]{}<>+cimnos are no-ops.

The code runs linearly from the first characters, except when loop ends are encountered. Loops can be nested but must be paired accordingly from the innermost loop to the outermost loop.

# Help, WarDoq!, 1 byte

p


Outputs 1 if the input is prime (positive or negative, differs from the other solution), 0 otherwise.

I initially naively wrote this as a Perl answer in 2018-12-08, and then deleted it when I saw there was already a Perl answer. In retrospect, I think there is value in this as a pure-regex answer, exploring its history and various alternative expressions.

First, the shortest form:

^(?!(xx+)\1+$)  There does not exist any $$\q \ge 2\$$ that divides $$\n\$$ with a quotient of $$\2\$$ or greater. Try it online! This is the standard, well-known primality test for numbers expressed in unary. I first became aware of it when arriving upon this 14 byte form independently in the course of solving one of the levels of Regex Golf on 2014-02-12 (which kick-started my interest in regex code golf). But in there, the set of test cases only started at 2 (xx), thus glossing over the fact that this regex matches the non-primes 0 and 1. I later learned that this regex dates back to at least as early as a 1997-10-22 Perl post by Abigail. # Regex (ECMAScript), 16 bytes ^(?!(xx+)\1+$)xx


There does not exist any $$\q \ge 2\$$ that divides $$\n\$$ with a quotient of $$\2\$$ or greater, and $$\n \ge 2\$$.

Try it online!

## Regex (ECMAScript), 18 bytes

(?=(x(x*))\1+$)\2^  The largest $$\q \ge 1\$$ that divides $$\n\$$ with a quotient of $$\2\$$ or greater exists, and $$\q-1=0\$$. Try it online! There may be circumstances in which this could be useful in the context of a larger regex. For example, if the very first thing a regex needs to do is assert that the input is prime, without consuming any characters, this would be 2 bytes shorter than ^(?=(?!(xx+)\1+$)xx).

## Regex (PCRE), 18 bytes

^(?>(x(x*))\1+$)\2  The largest $$\q \ge 1\$$ that divides $$\n\$$ with a quotient of $$\2\$$ or greater exists, and $$\q-1=0\$$. Try it online! It works in any engine supporting atomic grouping, which includes Perl, PCRE, .NET, Java, Boost, Python, Ruby, and more. It could conceivably be useful within the context of a larger regex. It consumes the entire prime number, like ^(?=(xx+?)\1*$)\1$, but its captures seem less likely to be useful; they are just \1=1 and \2=0. However, there may be niche circumstances where this could be coupled with NPCG (non-participating capture group) behavior, such that \1 or \2 matches if the path including the primality expression was crossed, but doesn't match otherwise. This would be equal in length to ^(?!(xx+)\1+$)xx() but capture two potentially useful optionally-unset groups instead of just one.

# 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
`
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