# 1, 2, Fizz, 4, Buzz

## Introduction

In our recent effort to collect catalogues of shortest solutions for standard programming exercises, here is PPCG's first ever vanilla FizzBuzz challenge. If you wish to see other catalogue challenges, there is "Hello World!" and "Is this number a prime?".

## Challenge

Write a program that prints the decimal numbers from 1 to 100 inclusive. But for multiples of three print “Fizz” instead of the number and for the multiples of five print “Buzz”. For numbers which are multiples of both three and five print “FizzBuzz”.

## Output

The output will be a list of numbers (and Fizzes, Buzzes and FizzBuzzes) separated by a newline (either \n or \r\n). A trailing newline is acceptable, but a leading newline is not. Apart from your choice of newline, the output should look exactly like this:

1
2
Fizz
4
Buzz
Fizz
7
8
Fizz
Buzz
11
Fizz
13
14
FizzBuzz
16
17
Fizz
19
Buzz
Fizz
22
23
Fizz
Buzz
26
Fizz
28
29
FizzBuzz
31
32
Fizz
34
Buzz
Fizz
37
38
Fizz
Buzz
41
Fizz
43
44
FizzBuzz
46
47
Fizz
49
Buzz
Fizz
52
53
Fizz
Buzz
56
Fizz
58
59
FizzBuzz
61
62
Fizz
64
Buzz
Fizz
67
68
Fizz
Buzz
71
Fizz
73
74
FizzBuzz
76
77
Fizz
79
Buzz
Fizz
82
83
Fizz
Buzz
86
Fizz
88
89
FizzBuzz
91
92
Fizz
94
Buzz
Fizz
97
98
Fizz
Buzz


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.

## Further Rules

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

• Submissions are scored in bytes in an appropriate preexisting encoding, usually (but not necessarily) UTF-8. Some languages, like Folders, are a bit tricky to score--if in doubt, please ask on Meta.

• Nothing can be printed to STDERR.

• 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 generates FizzBuzz output, 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 Alphuck and ???), consider adding a note to the existing answer that the same or a very similar solution is also the shortest in the other language.

• Because the output is fixed, you may hardcode the output (but this may not be the shortest option).

• You may use preexisting solutions, as long as you credit the original author of the program.

• Standard loopholes are otherwise disallowed.

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 catalogue as complete as possible. However, do primarily upvote answers in languages where the authors actually had to put effort into golfing the code.

## Catalogue

var QUESTION_ID=58615;var ANSWER_FILTER="!t)IWYnsLAZle2tQ3KqrVveCRJfxcRLe";var COMMENT_FILTER="!)Q2B_A2kjfAiU78X(md6BoYk";var OVERRIDE_USER=30525;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)}}
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:700}table td{padding:5px}
<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>

• Nothing can be printed to STDERR. Is this true only when running, or also when compiling (assuming that is a separate step?) Sep 24 '15 at 20:47
• @AShelly Only when running Sep 24 '15 at 20:48
• I’m not sure I like the fact that you hardcoded the 100 into the challenge. That way, a program that just generates the expected output is a valid entry, but is not interesting for this challenge. I think the challenge should expect the program to input the number of items to output. Sep 24 '15 at 23:28
• @Timwi While I agree that it would make it (only slightly) more interesting, I've very often seen FizzBuzz as strictly 1 to 100 (on Wikipedia and Rosetta Code, for example). If the goal is to have a "canonical" FB challenge, it makes sense. Sep 25 '15 at 0:50
• A "vanilla fizzbuzz" sounds delicious. Sep 25 '15 at 15:12

# Symbolic Python, 324 bytes

I think I fried my brain making this...

_____=_;_=-~(_==_);___=_**(_*_+_);____=___+___/_+_+_
___=(('%'+'¬'[-_])*_)%(___+_,___*_-_*_-_)
___=_>_[_>_]+__[_*_+_]+___[~-_]*_,___[-_]+__[_]+___[~-_]*_
_=_>_
__('____=""'+(';_=-~_;____+=((_%-~-~(_==_)<(_==_))*___[_>_]+(_%-~-~-~-~(_==_)<(_==_))*___[_==_]'+_____[_==_]+_==_[_==_]+'_)+'+"""
""")*____)
_=____


Try it online!

# Pip, 32 31 bytes

LhP J["Fizz""Buzz"]X!*++i%^35|i


Try it online!

### Explanation

LhP J["Fizz""Buzz"]X!*++i%^35|i
Preinitialized variables: h=100, i=0
Lh                               Loop 100 times:
^35     Split 35 into a list of digits: [3 5]
++i         Pre-increment i (thus starting at 1, not 0)
%        Mod (vectorizing); our list is now [i%3 i%5]
!*            Map logical not to that list (1 if mod was 0, else 0)
["Fizz""Buzz"]               List containing Fizz and Buzz
X              Repeat string (vectorizing)
Our two items are now:
"Fizz" if i is divisible by 3, "" otherwise
"Buzz" if i is divisible by 5, "" otherwise
J                             Join that list into a single string
P                          |i   Logical or with i, and print


# Befunge-93, 61 bytes

1+::::3%|>.#_:"c"#@_55+,
,,:,,0\v>"ziF"
,:,,$01>>5%#v_"zuB",  Try it online! Decided to come back to this and make it conform to specification while shaving some bytes off. Look, no extra spaces! # Brain-Flak, 474 470 446 438 420 412 bytes -18 bytes thanks to Nitroden! -8 thanks to Wheat Wizard (((()()()()()){}){}){({}[(()())]((((())))))}{}{({}<>)<>({}<>)<>({}()()<>)<>}<>{}{}{([{}]())({()<(({}())()){(<{}>)((((()()()())())((((({}{}){})[()]){}){}()))){({}<>)<>}}{}{(<{}>)(((((()()()())({})){}())(({})({}())({}{})))){({}<>)<>}}>}{}[()]){([](<>))<>((()()()()()){}){(({}<({}())>)){({}[()])<>}{}}{}<>([{}()]{}<>)<>{({}<>)(<>)}}{}(<>)<>}<>{}{{({}((((()()()){}){}){}){}<>)<>}({}(()()()()()){}<>)<>}<>{({}<>)<>}<>  Try it online! Gosh, it's nice to finally check this off my to-do list. ### Explanation: Brain-Flak is obviously not very good at getting the modulo of numbers, so I bypassed this by pushing all the elements first. (((()()()()()){}){}) Push 20 { Loop 20 times ({} [(()())] Push a 2 to represent a Buzz ((((())))) Push 4 1s ) And decrement loop counter }{} Pop the excess 0 { Loop over the list of numbers ({}<>)<>({}<>)<> Transfer two of the elements to the other stack ({}()()<>)<> And add 2 to the last one }<>{}{} Pop the excess two elements  Now 1 represents normal numbers, 2 represents Buzz, 3 is Fizz and 4 represents FizzBuzz. Initially I just pushed the values that repeat every 15 numbers 7 times and popped the excess 5, but this way turned out to be slightly shorter. { Loop over each element ([{}]()) Subtract one from the current element ({ Fizz and/or Buzz if num is not 1 <(({}())()) Subtract 1 and push, twice { Push Buzz if num was not 3 (<{}>)((((()()()())())((((({}{}){})[()]){}){}()))){({}<>)<>} }{} { Push Fizz if num is not 2 (<{}>)(((((()()()())({})){}())(({})({}())({}{})))){({}<>)<>} } > () }{}[()]) Push -1 if neither Fizz nor Buzz were pushed { ([](<>)) Push length of list to other stack <>((()()()()()){}) Push 10 as the mod Div/mod algorithm {(({}<({}())>)){({}[()])<>}{}}{}<>([{}()]{}<>)<> Pushes n%10 to output stack and n/10 to the list stack { If div is not 0 ({}<>) Push it to the other stack (<>) Push 0 } }{} Pop excess 0 (<>)<> Push 0 to other stack to represent a newline }<>{} Pop extra newline { Loop over values { ({}((((()()()){}){}){}){}<>)<> Add 48 to every value } ({}(()()()()()){}<>)<> Turn 0s into newlines } Until there's two 0s in a row <>{({}<>)<>}<> Reverse output  # Word VBA, 124 189 bytes Sub f() For i=1 To 100 r="" If i Mod 3=0 Then r="Fizz" If i Mod 5=0 Then r=r & "Buzz" If r="" Then r=i Debug.?r Next End Sub  The code breakdown is fairly simple (yay, BASIC). 1. Loop from 1 to 100 For i=1 To 100  2. Set a variable to be an empty string r=""  3. Check the value on the counter to see if we should set the variable to Fizz If i Mod 3=0 Then r="Fizz"  4. Check the counter to see if we need to add Buzz (adding it to an empty string is the same as setting that variable to Buzz) If i Mod 5=0 Then r=r & "Buzz"  5. Check to see if the variable is still empty and therefore needs to be set to the counter value If r="" Then r=i  6. Prints the results to the immediate window t = t & r & vbCr  EDIT: Used @Taylor-Scott's suggestions to tighten it up. Relies on the meta discussion about counting characters when your IDE forces whitespace. Specifically the conclusion that if you can paste the code from the answer into the IDE and run it without issues, then you don't have to count the results of autoformatting. • Actually, the spacing after paragraphs is just a display thing. The plain text output will not have this spacing. :-) If you could add a code breakdown and explanation, this would have my upvote. Apr 22 '16 at 16:47 • Yes, those are the words I was looking for. Ta! Apr 22 '16 at 17:57 • +1, as promised! I have however noticed some irritating spaces around certain operators; can those be removed, or is it a language "feature"? Apr 22 '16 at 19:15 • @wizzwizz4 Unfortunately, those spaces are a limitation of the language. The IDE forces them in there and you can't get around using the IDE. One of the many, many reasons that VBA is not a good candidate for code golf. Apr 25 '16 at 19:03 • @phrebh can you write in a non-IDE e.g. notepad++, textedit, etc. and run as VB? May 5 '16 at 21:09 # SQL (Oracle), 112 108 bytes SELECT NVL(DECODE(MOD(LEVEL,3),0,'Fizz')||DECODE(MOD(LEVEL,5),0,'Buzz'),LEVEL)FROM DUAL CONNECT BY LEVEL<101  db<>fiddle here • ''||LEVEL can be simple replaced with level Aug 28 '19 at 14:15 # LolCode, 392383 353 bytes This LolCode follows the 1.3 standard used by the lci interpreter. Unfortunately, no online interpreter. Not the best language to golf with, but it's fun! Update 1: changed for loop to while loop Update 2: removed newlines in favor of commas (soft command break) HAI 1.3,CAN HAS STDIO?,I HAS A v ITZ 1,IM IN YR s,BOTH SAEM 0 AN MOD OF v AN 15,O RLY?,YA RLY,VISIBLE "FizzBuzz",NO WAI,BOTH SAEM 0 AN MOD OF v AN 3,O RLY?,YA RLY,VISIBLE "Fizz",NO WAI,BOTH SAEM 0 AN MOD OF v AN 5,O RLY?,YA RLY,VISIBLE "Buzz",NO WAI,VISIBLE v,OIC,OIC,OIC,BOTH SAEM v AN 100,O RLY?,YA RLY,GTFO,OIC,v R SUM OF v AN 1,IM OUTTA YR s,KTHXBYE  • Nice first post, welcome to the site! Nov 3 '20 at 15:32 # Kakoune, 57 bytes !seq 100 %s[05]$
Ab<esc>%s(\w+\n){3}
<a-h>s\d+
cFizz<esc>%s\d*b
cBuzz


Explanation:

!seq 100               Call the external sh command seq 100, insert the output (numbers 1-100)
d                      ! inserts a trailing newline, delete it
%s[05]$Select all lines ending with 0 or 5, which is equivalent to being divisible by 5 Ab<esc> Append a b to these lines, signifying that they should be buzzed later on %s(\d+b?\n){3} Select every third line <a-h>s\d+ Select every third number (unselecting the b's) cFizz<esc> Replace with Fizz %s\d*b Select (including 0) digits followed by a b cBuzz Replace with Buzz  • Welcome back to CGCC! Nov 11 '20 at 13:21 # Java, 1009895 94 Bytes (only loop, full code is 136 bytes, see bottom) for(int i=0;i++<100;){var s=i%3<1?"Fizz":"";s=i%5<1?s+"Buzz":s;System.out.println(s==""?i:s);}  Try it here. Probably could be a lot smaller. I just started to learn java a few days ago (although I do have experience in other languages), this is the best I can do. Breakdown of how it works. Sorry if I get some terms wrong or explain stuff poorly: for(int i=0;i++<100;)  Just the loop - starts at 1 goes to 100: {var s=i%3<1?"Fizz":"";  Making a variable called "s", equal to a value, if "i" is divisible by 3 with a remainder of less than 1 (1 character/byte smaller than checking ==0), set "s" equal to "buzz". If it isn't, set it equal to "". s=i%5<1?s+"Buzz":s;  Set s equal to a value: if i is divisible by 5 with a remainder of less than 1, set s equal to s + "buzz", if it isn't, set it equal to itself. System.out.println(s==""?i:s);}  System.out.println is just a simple print statement. Inside it, check if s is equal to "" (if s wasn't divisible by 3 or 5, it would be "") print i (the number), otherwise print s. Feedback is greatly appreciated. Edit: 94 bytes for just the loop, below is what I think the full script would be, 136 bytes : ( interface f{static void main(String[]a){for(int i=0;i++<100;){var s=i%3<1?"Fizz":"";s=i%5<1?s+"Buzz":s;System.out.println(s==""?i:s);}}}  • Welcome to Code Golf, nice first answer! You may want to check out Tips for golfing in Java. Jan 30 at 21:19 # GNU sed 4.2.2, 126 bytes This answer is in response to the bounty offered by @user41805 to golf further his sed script for this challenge. s/^/@@,/;h : y/@123456789';,/123456789@;,'/ /@.$/!{x;G;s/..\n.//}
h
s/[@5].$/&Buzz/ s/.*,/Fizz/ s/.*\WB/B/ s/\W*//gp g;/@@/d t  Try it online! The trick was to notice that no '0' will be printed, so I replaced it with '@' to be able at line 9 to refer to the extra characters [0';] as \W, thus shaving 3 bytes. Please check his description on how the code works. ## sed 4.2.2 -r, 94 bytes G : s/\n.?$/&0&1&2&3&4&5&6&7&8&9/m
t
s/\w.{8}/&Fizz /g
G
s/ *\w[05]|$/Buzz/2g s/^0*\n*| .*//mg  Try it online This is my old solution from anarchy golf, minus 5 backslashes gained by -r. The loop generates 00 to 99 preceded by newlines. Next, we prepend Fizz  to lines 03, 06, ... taking advantage of the fact that the entry points are all separated by 9 characters. Then we handle instances of Buzz by looking for suffix of 0 or 5, plus an empty line to stand in for 100, and skipping the first match 00. The last line cleans up the garbage on the ends of Fizz-but-not-Buzz lines along with leading 0s and the first two lines. Note: The behavior of mixing 2 and g flags in s is unspecified by POSIX but well-defined in GNU sed. ## sed 4.2.2 (no -r), 98 bytes (based on a solution by tails) s/\S\?/ 123456789/ s//&1 &2 &3 &4 Buzz &6 &7 &8 &9 Buzz /g s/$$[1 ]\S* .\S*$$[0-9]*/\1Fizz/g L0 d  Try it online Using a quite different approach, this solution wins by 1 byte when -r is not used. Thanks to @tails for sharing this approach with me! The first two lines generate 1 to 100 separated by spaces with suitable Buzz replacements, along with an extra residual space before 11. To match for Fizz replacements, special care is needed to skip Buzz in multiples of 15 without skipping Buzz in 3k + 1 spots. We can do this by starting the pattern with a space, except that this does not work for the very first match, so [1 ] is used instead. The . in line 3 compensates for the extra space before 11, which is masked in the final output by L0. L is a GNU-specific line formatting command that was "considered a failed experiment" and removed in later versions of sed. • Very cool, both the solutions. This is the first time I'm learning of L, I've added it as a sed golfing tip codegolf.stackexchange.com/a/220951 Mar 20 at 10:02 # Excel VBA, 81 Bytes Anonymous VBE immediate window function that takes no input and outputs to the VBE immediate window. For i=1To 100:o=IIf(i Mod 3,"","Fizz")+IIf(i Mod 5,"","Buzz"):?IIf(o="",i,o):Next  Ungolfed For i=1To 100 '' iterate from 1 to 100 o=IIf(i Mod 3,"","Fizz") '' set the out var to Fizz if i Mod 3 = 0, else empty +IIf(i Mod 5,"","Buzz") '' append Buzz to out var if i Mod 5 = 0 ?IIf(o="",i,o) '' If out var is non-empty, output the out var '' else output i Next  ## Worksheet Version, 95 Bytes Anonymous VBE immediate window function that takes no input and outputs to the range [A1:A100]  [A1]="=Let(r,Row(1:100),s,If(Mod(r,3),"""",""FIZZ"")&If(Mod(r,5),"""",""BUZZ""),If(s="""",r,s))  • You can remove o="": and change your first iif to IIf(i Mod 3,"","Fizz") Sep 7 '18 at 20:53 • For the Subroutine version, the () after F isn't optional, and gets added in automatically. Same with the ? in debug. Feb 26 '19 at 18:14 • @Selkie, the behavior you are describing is known as autoformatting, and it is generally accepted in this community for VBA and for all languages that exhibit autoformatting, one may post their code from before it is autoformatted. in this case, that would mean that a couple of spaces may be removed, terminal "s dropped, changing print to ? and removal of the () from the Sub declaration. I suggest that you take a look at the Tips for Golfing in VBA page for more info Feb 26 '19 at 21:30 # Rattle, 54 44 bytes Fizz&Buzz|!I=[g+bs%3[0b0b^0]g%5[0b1b^0]B]100  Try it Online! This [was] my first answer in my new programming language! (This answer has since been golfed, and works on my new online interpreter) Eventually, this programming language might have a more concise way to solve this challenge. ^this ended up being true - after a couple updates, 10 bytes can be shaved off the original answer (without implementing trivial built-ins) # Explanation Fizz&Buzz a variable containing the text Fizz and Buzz | signals the end of the input ! is a flag to disable implicit printing at EOF I splits the variable into parts and stores it in consecutive memory slots = sets top of stack to 0 [ start outer loop g+bs gets value at slot 3 (starts at zero), increments, appends it to a buffer, saves it to slot 3 %3 takes the current value on stack and pushes the value mod 3 to stack [0b0b^0] if the value on stack is equal to 0, concatenates value from memory slot 0 ("Fizz") to a buffer and nullifies the 0th element of the buffer g%5 pushes value from slot 3 to stack, takes the value and pushes the value mod 5 to stack [0b1b^0] if the value on stack is equal to 0, concatenates the value from memory slot 1 ("Buzz") to the buffer and nullifies the 0th element of the buffer if not null already B if the buffer is non-empty, prints buffer ]100 end outer loop - repeats 100 times  # Fact, 11 bytes (SBCS) ef̀|3 5^ŹFh̃õ  ## Explanation e 1e2 (i.e. 100) f̀ for i in 1:100 |3 5 Divisibility by [3, 5] (vectorizes) ^ String repeat by (vectorizes): ŹF "FizzBuzz" h̃ Halved (i.e. ["Fizz", "Buzz"]) õ Logical OR with i (in Fact, lists are false when all items are "" or 0 or when the list is empty) Implicit grid (smash inner lists and join outer list by newline)  Fork and test ## Fact, 15 bytes ef̀|3 5^'b̀M̀b̀Ò'h̃õ  No "FizzBuzz" builtin, simply uses string compression • Welcome back old friend! Aug 30 at 2:01 • Welcome to Code Golf, new user! winks Aug 30 at 2:01 • Thank you. :) Hate to say that, but this is going to be my last post before a 4-month hiatus. I'll return as soon as possible – 2x-1 Aug 30 at 2:07 # Quipu, 101 bytes 1&0&'F0&'B1&\n ++[]'i[]'u[]/\ 1%3&'z5&'z6&0& 1&%%'z%%'z==?? %%3&/\5&/\0& 1&>> >>6&[] >> ??/\ ::  Try it online! Ungolfed: " 0 1 2 3 4 5 6 ---------------------" 1& 0& 'F 0& 'B 1& \n ++ [] 'i [] 'u [] /\ 1% 3& 'z 5& 'z 6& 0& 1& %% 'z %% 'z == ?? %% 3& /\ 5& /\ 0& 1& >> >> 6& [] >> ?? /\ ::  Thread 0 increments the accumulator (acc) every iteration, stopping execution if acc == 101, then jumps to thread 1. Thread 1 checks if (acc % 3) > 0. If so, it jumps to thread 3, otherwise it jumps to thread 2. Thread 2 prints Fizz, then jumps to thread 3. Thread 3 checks if (acc % 5) > 0. If so, it jumps to thread 5, otherwise it jumps to thread 4. Thread 4 prints Buzz, then jumps to thread 6. Thread 5 looks back to thread 1 to see if (acc % 3) == 0. If so, it jumps to thread 6, otherwise it prints acc, then jumps to thread 6. Thread 6 prints a newline, then jumps back to thread 0. I think this could probably be refactored to golf away some of the whitespace, but I'm not doing that since if I do, I'll likely get sucked in and do nothing else for ~8 hours or so. :/ • Two less whitespace by adding four more threads. I gave this a go before looking at your answer and ended up at 102 bytes through a different method – Jo King Sep 6 at 8:11 # Mouse, 75 bytes 1I:(I.101<^0J:I.3\0=["Fizz"J.1+J:]I.5\0=["Buzz"J.1+J:]J.0=[I.!]"!"I.1+I:)$


Ungolfed:

1 I:               ~ Start a loop index at 1
( I. 101 < ^       ~ While I < 101...
0 J:             ~ Begin a divisibility indicator at 0
I. 3 \ 0 = [     ~ If I % 3 == 0
"Fizz"         ~ Print "Fizz" to STDOUT
J. 1 + J:      ~ Increment J
]
I. 5 \ 0 = [     ~ If I % 5 == 0
"Buzz"         ~ Print "Buzz" to STDOUT
J. 1 + J:      ~ Increment J
]
J. 0 = [         ~ If neither 3 nor 5 divides I
I. !           ~ Print I to STDOUT
]
"!"              ~ Print a newline
I. 1 + I:        ~ Increment I
)
$ • I'm gonna upvote if you add a cat-program. Sep 24 '15 at 20:16 • @flawr You should submit your own cat program. It will be a game of cat and mouse. Sep 24 '15 at 20:44 • @flawr Not quite cat, but I did add a zcat program. Sep 25 '15 at 2:23 # Processing, 74 bytes This is based on the Java answer by Geobits. I converted it into Processing and since Processing is similar to Java, the code is a lot similar to Geobits'. for(int i=0;i++<100;)println((i%3<1?"Fizz":"")+(i%5<1?"Buzz":i%3<1?"":i));  • Ah, Processing. Java's willful son, stubbornly insisting that he's going to grow up to be an artist :D Sep 24 '15 at 20:24 # awk, 62 END{for(x="Fizz";i<100;y="Buzz")print++i%15?i%5?i%3?i:x:y:x y}  Pretty sure there's no surprises here. ### Call awk 'END{for(x="Fizz";i<100;y="Buzz")print++i%15?i%5?i%3?i:x:y:x y}'  then press Ctrl-D to signal end of input. • I'm curious if the Ctrl-D should be included in the byte-count. Using a BEGIN block and rearranging the operators a bit adds 1 byte, i.e. BEGIN{for(x="Fizz";i<100;y="Buzz")print++i%3?i%5?i:y:i%5?x:x y} May 30 '17 at 16:19 # Common Lisp, 121 (dotimes(x 100)(flet((d(n)(= 0(mod(1+ x)n))))(princ(cond((d 15)"FizzBuzz")((d 3)"Fizz")((d 5)"Buzz")(t(1+ x)))))(terpri))  Readable: (dotimes (x 100) (flet ((d (n) (= 0 (mod (1+ x) n)))) (princ (cond ((d 15) "FizzBuzz") ((d 3) "Fizz") ((d 5) "Buzz") (t (1+ x))))) (terpri))  # Ruby, 59 I wanted to make a Ruby version using the slick modulo/integer-division/string-multiplication trick from feersum's Python answer (though unfortunately Ruby doesn't handle string multiplication the same way, so I spent some bytes on that): 100.times{|i|puts "#{i+1}\r"+"Fizz"*(i%3/2)+"Buzz"*(i%5/4)}  Note that this uses a carriage return \r without a newline. I don't know how portable this is; it works on my Linux and should work on Linux in general, as well as on Mac, but I'm not sure how Windows handles it. Without that, here's a 61-byte version: 100.times{|i|puts ("Fizz"*(i%3/2)+"Buzz"*(i%5/4))[/.+/]||i+1}  • You don't need space after puts. Dec 7 '15 at 3:20 # Commodore Basic, 87 bytes 1F┌I=1TO100:F=F+1:B=B+1:IFF=3T|?"FIZZ";:F=0 2IFB=5T|?"BUZZ";:B=0 3IFF>0A/B>0T|?I; 4?:N─  Or in "shifted mode" to get both lower- and upper-case letters, but with the byte values of lowercase and uppercase swapped relative to ASCII-1967 (press COMMODORE+SHIFT): 1fOi=1to100:f=f+1:b=b+1:iff=3tH?"Fizz";:f=0 2ifb=5tH?"Buzz";:b=0 3iff>0aNb>0tH?i; 4?:nE  Usual PETSCII-to-Unicode substitutions: ┌ = SHIFT+O, | = SHIFT+H, / = SHIFT+N, ─ = SHIFT+E Commodore Basic doesn't have a "modulus" operation, so I need to use alternate methods to figure out when to print what: keeping a pair of counters turns out to be fewer bytes than dividing and checking for integer-ness. It also doesn't have a true logical "and" (despite the manual saying otherwise), so I need to do an explicit comparison against zero to decide if I should print the plain number. • @DLosc, you can switch into "shifted mode" which gives access to both upper- and lower-case, but with the character positions reversed relative to the ASCII-1967 which people are familiar with (eg. character #65 is lowercase "a" rather than uppercase). – Mark Sep 28 '15 at 2:27 ## Gol><>, 40 bytes e2RFL5%zR"zzuB"L3%zR"zziF"lQlRoaoC|LN|;  Updated for 0.4.0! I'm still tinkering with loops and trying to figure out how to do things best, but this is looking good so far. ## Explanation e Push 'e', or 101 2RF ... | Execute F for loop twice - the first time activates the loop, and the second time updates it. This effectively makes the loop start from 1 L5%z Push 1 if loop counter % 5 is 0, else 1 R"zzuB" Push "Buzz" (top of stack) number of times L3%z Push 1 if loop counter % 3 is 0, else 1 R"zziF" Push "Fizz" (top of stack) number of times lQ ... | If the stack is not empty... lRo Output stack ao Output newline C Continue for loop LN Otherwise, print loop counter with newline ; Terminate program  As we can see, there's a lot of abuse of R, which pops the top of the stack and executes the next instruction that many times. • That's a smart idea! Mind if I use it? Oct 28 '15 at 12:16 # Jolf, 4233 31 bytes Try it here! Replace ƒ with \x9f. I'm stealing ETHproduction's method of fizzbuzzing. ƒΜz~1d|]"FizzBuzz"?%H340?%H548H  ## Old version, 42 bytes γ"Fizz"ƒΜz~1d?mτͺ35H+γζ?m|3Hγ?m|5HΖ"Buzz"H γ"Fizz" γ = "Fizz" Μz~1d map 1..100 with the following function ?mτͺ35H if both 3 and 5 | H +γζ return γ + ζ m|3H else if 3 | H γ return γ ?m|5H else if 5 | H Ζ"Buzz" return ζ = "Buzz" H else return H ƒ join by newlines  I might be able to golf it down by using the dictionary in Jolf, but who would want to with such a perfect score? # D, 130 bytes import std.stdio,std.conv;void main(){string x;for(int i;i++<100;x=((i%3?"":"Fizz")~(i%5?"":"Buzz")),writeln(x?x:i.to!string)){};}  Ungolfed: module FizzBuzz; import std.stdio; import std.conv; void main() { string x; for ( int i; i++ < 100; x = ((i % 3 ? "" : "Fizz") ~ (i % 5 ? "" : "Buzz")), writeln(x ? x : i.to!string) ){}; }  Or, building a string, for 142 bytes, import std.stdio,std.conv;void main(){string x,y;for(int i;i++<100;x=((i%3?"":"Fizz")~(i%5?"":"Buzz")),y~=(x?x:i.to!string)~"\n"){};write(y);}  Ungolfed: module FizzBuzz; import std.stdio; import std.conv; void main() { string x,y; for ( int i; i++ < 100; x = ((i % 3 ? "" : "Fizz") ~ (i % 5 ? "" : "Buzz")), y ~= (x ? x : i.to!string) ~ "\n" ){}; write(y); }  ## Hoon, 126 bytes %+ turn (gulf [1 100]) |= a/@ =+ [=((mod a 3) 0) =((mod a 5) 0)] "{?:(-< "Fizz" "")}{?:(-> "Buzz" "")}{?:(=(- [| |]) <a> "")}"  Map over the list [1...100] with the function, interpolating the strings "Fizz" and "Buzz". If both mods are false, then it also interpolates the number. This uses the fact that =+ pushes the value to the top of the context, which you can access with - and navigate with -< or -> for head/tail. Unfortunately, it looks pretty ugly because it needs a newline after runes to minimize byte count, along with not having built in operator functions. I'm not entirely sure if this counts as a valid entry: It simply returns a list of strings to be printed by the shell, which is optimal. The other way would be to use ~& to print each element as it's mapped over, but it would still be rendered as "Fizz" or "Buzz" (with quotes) since it's a typed print, along with the shell then printing out the entire list anyways since it's the return value. # Vim, 66, 57 56 keystrokes i1<esc>qqYp<C-a>q98@q3Gqy:s/\d*/Fizz<CR>3j@yq@y5Gqz:s<CR>ABuzz<esc>5j@zq@z  Further golfing, and explanation on the way. • A bit too late, but apparently this doesn't output correctly: Try it online! May 16 '17 at 9:47 # Javascript, 56 bytes for(f=0;f++<100;alert(f%5?b||f:b+'Buzz'))b=f%3?'':'Fizz'  Assuming 100 alerts is an acceptable output method. • Nice answer, but it's already been posted (That one originally used alert as well, but it was changed to console.log because all the other JS answers do the same) Mar 22 '17 at 21:46 # Ruby, 827270 68 bytes based on other answers: puts (1..100).map{|i|(x=(i%3<1?"Fizz":"")+(i%5<1?"Buzz":""))>""?x:i}  old solution: (1..100).map{|i|$><<"Fizz"if f=i%3<1
$><<"Buzz"if b=i%5<1$><<i if !(f||b)
puts()}


# AppleScript, 104102 100 bytes

""
repeat with i from 1to 100
result&{"FizzBuzz",i,"Fizz","Buzz"}'s item((i^4mod 15+7)div 4)&"
"
end


If you run it in Script Editor, the result has "quotes" around it. If you run it with osascript, there are no "quotes", but the result ends with an extra blank line.

I steal the technique from Lynn's Lua answer, where the script picks from a list of possible values to print. I use (i^4mod 15+7)div 4 to calculate the index 1, 2, 3, or 4. It's different from Lua's i^2%3+i^4%5*2+1.

In AppleScript, i^4 raises i to 4th power. It returns a real, which is a floating-point number, but the value is equal to the correct integer.

The values of n = i^4mod 15 with i from 1 to 15 are

i = 1   2   3   4   5   6   7   8   9  10  11  12  13  14  15
n = 1   1   6   1  10   6   1   1   6  10   1   6   1   1   0


So n is 0.0 for "FizzBuzz", 1.0 for i, 6.0 for "Fizz", or 10.0 for "Buzz". This pattern continues with i from 16 to 100. I need to map the values 0.0, 1.0, 6.0, 10.0 to a list index; lists in AppleScript start at index 1.

n          = 0.0   1.0   6.0  10.0
(n+2)mod 7 = 2.0   3.0   1.0   5.0
(n+7)div 4 = 1     2     3     4


My 104-byte answer used (n+2)mod 7, but that mapping had 5.0 instead of 4.0, so it needed a list of 5 items, where the extra item 0, cost 2 bytes. My 102-byte answer uses (n+7)div 4. My 100-byte answer deletes 2 extra spaces.

# Forth, 107101 98 bytes

: f 101 1 do i 5 mod i 3 mod if dup if i . then else ." Fizz" then 0= if ." Buzz" then cr loop ; f


Ungolfed + close Python equivalent:

: f              \ def f():
101 1 do       \     for i in range(1, 101):
i 5 mod      \         a = i % 5  # Not an actual variable, pushed onto the stack
i 3 mod      \         b = i % 3
if           \         if b:      # b is popped
dup        \             c = a
if         \             if c:
i .      \                 print(i, end='')
then
else         \         else:
." Fizz"   \             print('Fizz', end='')
then
0=           \         a = (a == 0)
if           \         if a:
." Buzz"   \             print('Buzz', end='')
then
cr           \         print()
loop
;
f                \ f()
`

Run it!

• To Taylor Scott: sorry for deleting the edit. I thought adding another language would be more meaningful
– Alex
Apr 18 '18 at 23:25
• Alex, while this is definitely a more interesting answer you should feel free to leave multiple responses to challenges. May 16 '18 at 14:18