186
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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>

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12
  • 1
    \$\begingroup\$ Nothing can be printed to STDERR. Is this true only when running, or also when compiling (assuming that is a separate step?) \$\endgroup\$
    – AShelly
    Sep 24, 2015 at 20:47
  • 1
    \$\begingroup\$ @AShelly Only when running \$\endgroup\$
    – Beta Decay
    Sep 24, 2015 at 20:48
  • 1
    \$\begingroup\$ 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. \$\endgroup\$
    – Timwi
    Sep 24, 2015 at 23:28
  • 9
    \$\begingroup\$ @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. \$\endgroup\$
    – Geobits
    Sep 25, 2015 at 0:50
  • 72
    \$\begingroup\$ A "vanilla fizzbuzz" sounds delicious. \$\endgroup\$ Sep 25, 2015 at 15:12

395 Answers 395

1 2
3
4 5
14
7
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MediaWiki markup + WP templates, 142 bytes

{{for nowiki|<br>|<nowiki>{{If then show|{{#ifeq:{{mod|{{{1}}}|3}}|0|Fizz|}}{{#ifeq:{{mod|{{{1}}}|5}}|0|Buzz|}}|{{{1}}}}}</nowiki>|count=100}}

Explanation

This will only work on the English Wikipedia.

{{for nowiki <!-- Simple loop -->
|<br> <!-- seperator for each line -->
|<nowiki>{{If then show| <!-- main looping code, wrapped in <nowiki> tags to prevent processing except by {{for nowiki}} -->
{{#ifeq:{{mod|{{{1}}}|3}}|0|Fizz|}}{{#ifeq:{{mod|{{{1}}}|5}}|0|Buzz|}}|{{{1}}}}}</nowiki> <!-- {{{1}}} is substituted with the index -->
|count=100}}

{{For nowiki}}'s source

{{If then show}}'s source

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3
6
\$\begingroup\$

GolfScript, 37 bytes

100,{)..3%!'Fizz'*\5%!'Buzz'*+\or n}/
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6
\$\begingroup\$

Snowman 1.0.2, 97 chars

)1vn101nR:du*_/3NmO0eQ)(#5NmO0eQ}~(~%@or(%nO?_/)#%@{%@tS?)aRsP@@"Fizz"_aRsP\"Buzz"aRsP?)10wRsP;aE

How does it work, you ask? ... I have no idea. I might edit in a full explanation at some point if I ever decide to try to understand this again.

(Pulled directly from Snowman's examples directory.)

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6
+500
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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.

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1
6
\$\begingroup\$

aussie++, 223 190 187 bytes

G'DAY MATE!
I RECKON x IS A WALKABOUT FROM[1TO 100]<YA RECKON x %15<1?<GIMME "FizzBuzz";>WHATABOUT x %5<1?<GIMME "Buzz";>WHATABOUT x %3<1?<GIMME "Fizz";>WHATABOUT ?<GIMME x;>>CHEERS C***!

Fair dinkum, this was hard yakka cause there's no way to golf the whitespace because the parser seems to be strict. I coulda sworn it never worked when I tried to remove whitespace.

-33 thanks to @Bubbler

-3 thanks to @JoKing

Wanna hear what it actually sounds like? Wonder no more: https://youtu.be/ewu1CmjZcmQ

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2
  • 3
    \$\begingroup\$ Idea: Aussie++ on Code Review \$\endgroup\$
    – emanresu A
    Nov 5, 2021 at 23:29
  • \$\begingroup\$ Just a note that Aussie++ has an online playground so everyone can try out the code easily. (No permalinks though) \$\endgroup\$
    – Bubbler
    Nov 11, 2021 at 3:36
6
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HTML & CSS, 502 bytes

body{counter-reset:n}
p::after{counter-increment:n;content:counter(n)}
p:nth-child(3n)::after{content:"Fizz"}
p:nth-child(5n)::after{content:"Buzz"}
p:nth-child(15n)::after{content:"FizzBuzz"}
<p><p><p><p><p><p><p><p><p><p>
<p><p><p><p><p><p><p><p><p><p>
<p><p><p><p><p><p><p><p><p><p>
<p><p><p><p><p><p><p><p><p><p>
<p><p><p><p><p><p><p><p><p><p>
<p><p><p><p><p><p><p><p><p><p>
<p><p><p><p><p><p><p><p><p><p>
<p><p><p><p><p><p><p><p><p><p>
<p><p><p><p><p><p><p><p><p><p>
<p><p><p><p><p><p><p><p><p><p>

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2
  • 1
    \$\begingroup\$ Nice answer! You don't need the <div> elements and you can skip the closing </p> tags as well, it should still render fine. \$\endgroup\$
    – emanresu A
    Nov 22, 2021 at 2:53
  • \$\begingroup\$ Thanks @emanresuA. I still needed an element to apply the counter-reset to but it could just have been the body. Removing the closing </p> helped shave off a lots of bytes. \$\endgroup\$ Nov 22, 2021 at 9:02
6
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rSNBATWPL, 57 bytes

for{i=1;i<101}$print{(("Fizz"*1>i%3)+"Buzz"*1>i%5)or++$i}

RSNBATWPL, or Radvylf Should Not Be Allowed To Write Programming Languages, is a "practical" programming language I've designed over the last week or so. It aims to be a sort of "super-JS/PHP/C++": a language that seems practical to the uninitiated and allows writing nice code, but deep down, is horribly cursed.

This is a significantly golfed version of my more readable FizzBuzz program:

n = cast.int{input{"Input: "}};

for {i = 1; i <= n; ++$i} {
    cond {(i % 3) and (i % 5)} {
        print{i};
    } {
        s = "";

        if {not{i % 3}} {
            s += "Fizz";
        };

        if {not{i % 5}} {
            s += "Buzz";
        };

        print{s};
    };
};

It uses a few interesting golfs, like replacing function calls (with {}) with the $ operator (which calls a function with an argument). It also relies on rSN's weird parsing order, with things like +"Buzz"*1>i%5 only being possible because rSN is always parsed right-to-left. In this case, we also make use of multiplying a string by a bool, a type hack similar to what you'd see in JS.

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6
\$\begingroup\$

Cubestack, 178 162 bytes

M R2 f' M' S S' y S S' y2 M R' M' R E M r M' E2 M r2 M' E' U M R M' D E S R' b' R2 b r U2 r U2 S' E2 S R' B r l r U2 r U2 S' E' L S S' u S S' y2 M R' M' R F2 b y'

Try it Online!

Final cube state:

enter image description here

Another fine addition to the collection of FizzBuzz I've written.

Explained

M R2 f' M' S S' y      # For each item in the range [0, 100)
S S' y2 M R' M' R      #   Push that item + 1 to the stack
E M r M' E2 M r2 M' E' #   Push the list [3, 5] to the stack
U                      #   Calculate [(item + 1) % 3, (item + 1) % 5]
M R M' D               #   and check if each of those results is equal to 0
E ... E2               #   Push the list ["Fizz", "Buzz"] to the stack
L S S' u               #   and multiply the strings with the modulo equality list. 
                       #   This implements the ("Fizz" * (item % 3 == 0)) + ("Buzz" * (item % 5 == 0)) 
                       #   I use in my other fizzbuzzes to concisely do the multiplicity check
S S' y2 M R' M' R      #   Push the item + 1 to the stack again. Needed because there's no DUP yet
F2 b                   #   Logically or that with the FizzBuzz string and print with a newline. 
                       #   If the item isn't divisible by 3 or 5, then the fizzbuzz string will be empty
                       #   meaning the logical or here will return the number. Otherwise it'll return whatever word is needed.
y'                     # End the loop
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6
\$\begingroup\$

D, 90 bytes

import std;void main(){foreach(i;1..101)i.text.max("FizzBuzz"[i%3?4:0..i%5?4:8]).writeln;}

Try it online!

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1
  • 1
    \$\begingroup\$ Welcome to Code Golf, and nice answer! \$\endgroup\$
    – Steffan
    Aug 19 at 15:45
5
\$\begingroup\$

Befunge-93, 82 81 bytes

I'm sure this could be golfed, but I think this is a good start.

1+:::3%:#v_"zziF"v>*|>25*,:"!"3*`#@_
v _v#:%5\<   ,,,,<^ >^
<  >\"zzuB",,,,   ^ .#

Try it in this online interpreter.


Attempts that didn't work

1+:::3%: #v_"zziF" v>*#v_>25*,:"!"3*`#@_
v _v#:%5\ <>#,,,,,#<^  >.^
<  >\"zzuB"^        ^

Tries to combine the printing of Fizz and Buzz. Ends up at 88 bytes.

Vertical rendition of the above

Forgot about newlines. 122 bytes. Ick. Without newlines it would be 122-41=81 bytes. Welp.

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5
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SQL (PostgreSQL flavour), 107 bytes

SELECT(array[n||'','Fizz','Buzz','FizzBuzz'])[1+(n%3=0)::int+(n%5=0)::int*2]FROM generate_series(1,100)a(n)

Same sort of logic as my R answer

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5
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bc, 83 bytes

Undeclared variables are zero by default, so i=0 can be omitted. The three line breaks are required.

for(;++i<101;){if(!i%15)"FizzBuzz
"else if(!i%5)"Buzz
"else if(!i%3)"Fizz
"else i;}
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5
\$\begingroup\$

JavaScript, 79 bytes

After a long time I tried to use JS again...

Thanks to @ShadowCat we made it to 79 bytes:

for(i=0,s="";i++<100;s+=(i%5?i%3?i:'Fizz':i%3?'Buzz':'FizzBuzz')+"\n");alert(s)

Old solution, 87 bytes:

for(i=2,s="1";i<101;s+="\n"+["FizzBuzz","Buzz","Fizz",i][(i%3>0)+2*(i++%5>0)]);alert(s)
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6
  • \$\begingroup\$ I recommend dropping the semi-colon at the end. Also I just used conditionals i%5?i%3?i:'Fizz':i%3?'Buzz':'FizzBuzz' and I got 79 bytes, so I recommend trying that. I would also avoid s="1" and just start i=1,s="" to save an extra byte \$\endgroup\$
    – ShadowCat7
    Sep 24, 2015 at 20:44
  • 1
    \$\begingroup\$ I was working on a ?: solution but I was not able to get it to work before you commented=) Thank you very much! (Honestly I am still proud of my array indexing solution=P) \$\endgroup\$
    – flawr
    Sep 24, 2015 at 20:50
  • \$\begingroup\$ A trailing newline is acceptable, but a leading newline is not. I don't like this rule, but it's a rule nevertheless \$\endgroup\$
    – edc65
    Sep 24, 2015 at 21:16
  • \$\begingroup\$ 77: change starting part i=s="". 76 using ES6 and template string for newline \$\endgroup\$
    – edc65
    Sep 24, 2015 at 21:42
  • \$\begingroup\$ 75: i recomment this for(i=0,s="";i++<100;s+=((i%3?'':'Fizz')+(i%5?'':'Buzz')||i)+"\n");alert(s), because it saves 6 \$\endgroup\$ Sep 29, 2015 at 12:00
5
\$\begingroup\$

Haskell, 105 97 bytes

main=mapM(putStrLn.f)[1..100]
a%b=a`rem`b<1
f n|n%15="FizzBuzz"|n%3="Fizz"|n%5="Buzz"
f n=show n

I'm kinda new to Haskell, so any advice would be appreciated!

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4
  • \$\begingroup\$ first line could be main=mapM(print . f)[1..100] \$\endgroup\$
    – wlad
    Sep 25, 2015 at 11:34
  • \$\begingroup\$ You can write all the guards of f in a single line: f n|n%15="FizzBuzz"|n%3="Fizz"|n%5="Buzz". \$\endgroup\$
    – nimi
    Sep 25, 2015 at 15:21
  • \$\begingroup\$ I count 96 bytes. If you write f 3++f 5 instead of "FizzBuzz", you can save two more. Also, henkma on anarchy golf has 82 somehow. \$\endgroup\$
    – Lynn
    Sep 26, 2015 at 22:12
  • 1
    \$\begingroup\$ I posted an 85 byte answer. \$\endgroup\$
    – Lynn
    Sep 27, 2015 at 14:45
5
\$\begingroup\$

Fortran, 213 bytes

character(len=8)::o
do i=1,100
if(mod(i,15)==0)then;write(*,*)'FizzBuzz'
elseif(mod(i,3)==0)then;write(*,*)'Fizz'
elseif(mod(i,5)==0)then;write(*,*)'Buzz'
else;write(o,'(i8)')i;write(*,*)adjustl(o)
endif;enddo;end

Not as graceful as the golf languages. I could save bytes using print instead of write, but print indents 1 space without a format specifier which would increase the byte count instead. Likewise I lose bytes printing the number because Fortran doesn't like left-justified output for numbers. I didn't bother sticking it all on one line as newlines and semicolons are both 1 byte -- no savings.

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1
  • \$\begingroup\$ Could possibly save bytes by setting a variable and writing it out once at the end, instead of multiple write statements. \$\endgroup\$
    – roblogic
    Jun 30, 2021 at 23:59
5
\$\begingroup\$

MATLAB, 94 bytes

for i=1:100 t=mod(i,3);f=mod(i,5);s=[(t&f)*num2str(i) ~t*'Fizz' ~f*'Buzz' ''];disp(s(s>0));end

So this new code is a slight improvement on the one below. Rather than using arrayfun() which is quite costly in characters as it requires 'UniformOutput','false' to get it to work, I have simply made it a for loop - because the range of numbers is hard coded, there is no need to use a function as I had done in my last edit. Removing it from the function saves another 10 characters.

This does basically the same thing, but rather than making all the strings first, in makes them one by one in a for loop and displays them. This actually also means char() only has to be used once (in the other one it was used a second time to display everything). Having the loop means I can use variables to store the results of mod(i,3) and mod(i,5) so they don't need calculating twice. The nonzeros() function has also now been removed, instead opting for storing to a variable then only printing anything which is not equal to zero. This solution when you run it also doesn't print ans= before the first line.

Thanks to @flawr for the tips, saved 4 bytes.


Old code:

MATLAB, 118 bytes

char(arrayfun(@(x) char(nonzeros([(mod(x,3)&&mod(x,5))*num2str(x) ~mod(x,3)*'Fizz' ~mod(x,5)*'Buzz'])'),1:100,'Un',0))

A bit of fun with multiplying strings with scalars. Basically the output of ~mod(x,5) and ~mod(x,3) are multiplied by 'Fizz' and 'Buzz' respectively which produces either zeros (blanks) or one or both of the words. (mod(x,3)&&mod(x,5))) is basically when the number is neither a multiple of 3 nor 5 which is multiplied by the number as a string to get either zeros or the number.

These are then concatenated into an array which then has all of the zeros removed using nonzeros() and then resulting array transposed to be in the right direction for conversion to a character string.

Finally once all numbers have been processed by arrayfun(), the resulting cell array of arrays is passed to char() which converts it to a cell array of strings. Because there is no ; at the end of the string, the output is dumped to the console.


It might be possible to make it smaller, I'm looking ;)

\$\endgroup\$
3
  • \$\begingroup\$ I like your solution! Just some hints, you can use [s,''] instead of char(s) (I'd add that in the disp.) Then if you have arguments like UniformOutput it usually suffices to just write U or Un (just as many letters needed for discerning the different possible arguments) and instead of true,false you can use 1,0 (or for true any finite value bigger but zero). This also means instead of s~=0 you can use ~s or if this does not work s>0 or s<0 depending on the application. \$\endgroup\$
    – flawr
    Sep 25, 2015 at 22:50
  • \$\begingroup\$ @flawr Thanks for the tips. I hadn't realised that appending '' converted to a char string. I've saved 4 more bytes. Added the '' to the s=[ ] bit as it does the same as adding it in disp() but without the cost of the extra []. Also the s>0 was a bit of a duh moment on my part! \$\endgroup\$ Sep 26, 2015 at 4:29
  • \$\begingroup\$ There is a Tipps for golfing in Matlab (and linked there one for golfing in Octave) might be worth reading (and expanding=). Happy golfing=) \$\endgroup\$
    – flawr
    Sep 26, 2015 at 10:04
5
\$\begingroup\$

TI-BASIC, 59 bytes

For(X,1,ᴇ2
int(ln(gcd(X,15→J              ;[3 divides X] + [5 divides X]
X
If J
sub("FizzBuzz",7-2gcd(X,3),4J
Disp Ans
End

Or at the same length:

For(X,1,ᴇ2
gcd(X,15→J
X
If ln(J
sub("FizzBuzz",5^(J=5),4int(ln(J
Disp Ans
End

Both programs use the fact that ⌊ln(3)⌋ = ⌊ln(5)⌋ = 1 and ⌊ln(15)⌋ = 2.

There could be another byte to golf off somewhere, but I can't find it. By comparison, here's the naïve approach at 67 bytes:

For(X,1,ᴇ2
"Fizz
If fPart(X/3:X
If not(fPart(X/5:"Buzz
If not(fPart(X/15:"FizzBuzz
Disp Ans
End

TI-BASIC's quirks lengthen the program in two ways:

  • TI-BASIC needs two bytes to encode every lowercase letter other than i (which represents the imaginary unit).
  • Empty strings are not supported: sub("FizzBuzz",5,0 and ""+"Buzz" both throw errors.
\$\endgroup\$
4
  • \$\begingroup\$ Wouldn't moving X to the 2nd to last line shave a byte? (Disp X?) (For the seond one) \$\endgroup\$ Sep 29, 2015 at 0:08
  • \$\begingroup\$ Here are my suggested revisions (I don't have my calculator handy, so I'm not sure if it remains the same.) \$\endgroup\$ Sep 29, 2015 at 0:49
  • \$\begingroup\$ Ohhhh I see. The Ans functionality sometimes confuses me. Sorry if I sounded pretentious ;) \$\endgroup\$ Sep 29, 2015 at 0:53
  • \$\begingroup\$ I found a byte on the "naive" solution by storing "Buzz" to Str2 then "Fizz" to Str1 at the beginning of the loop, leaving "Fizz in Ans so lines 5 and 6 can use Str2 and Str1+Str2. Also by my byte counts following token size here: tibasicdev.wikidot.com/one-byte-tokens The naive solution is 70 bytes to the above 69. Between the optimal solutions, the one that does not precompute int(ln( on gcd(X,15 should be the smallest at 60 bytes to 61 for the other. \$\endgroup\$
    – TiKevin83
    Nov 22, 2019 at 16:34
5
\$\begingroup\$

Groovy, 69 Bytes

(1..100).each{i->println i%15?(i%5?(i%3?i:'Fizz'):'Buzz'):'FizzBuzz'}
\$\endgroup\$
1
  • 3
    \$\begingroup\$ Can save 2 bytes using 1.upto(100){} and no parenthesis are needed in your answer, resulting in: 1.upto(100){i->println i%15?i%5?i%3?i:'Fizz':'Buzz':'FizzBuzz'}​ +1 though :). \$\endgroup\$ Oct 13, 2016 at 17:49
5
\$\begingroup\$

Fortran, 188 bytes

do i=1,100
if(mod(i,3)==0.and.mod(i,5)==0)then;print'(a)','fizzbuzz'
elseif(mod(i,3)==0)then;print'(a)','fizz'
elseif(mod(i,5)==0)then;print'(a)','buzz'
else;print'(i0)',i
end if;enddo;end
\$\endgroup\$
3
  • 3
    \$\begingroup\$ Welcome to PPCG! \$\endgroup\$ Aug 3, 2017 at 15:32
  • 2
    \$\begingroup\$ ...change print'(a)' etc. to print* to save 16 bytes \$\endgroup\$
    – roblogic
    Apr 5, 2019 at 14:49
  • 2
    \$\begingroup\$ I never used print* as it prints the characters with a space or two before hand, thus it doesn't match the required format. \$\endgroup\$
    – lewisfish
    Apr 15, 2019 at 14:09
5
\$\begingroup\$

Transact-SQL, 163 143 124 110 bytes

This requires SQL Server 2012+

(thanks MickyT for the unnamed variable and the IIF suggestions, changed to muqo's GOTO loop instead of WHILE)

declare @ int=1a:print iif(@%3*@%5>0,ltrim(@),iif(@%3=0,'Fizz','')+iif(@%5=0,'Buzz',''))set @+=1IF @<101GOTO a

Formatted and explained:

declare @ int=1                      --@ is a valid variable name
a:                                   --shorter than WHILE
    print iif(@%3*@%5>0, ltrim(@),   --ltrim is shorter than explicit cast
          iif(@%3=0,'Fizz','')       --nest the IIFs
        + iif(@%5=0,'Buzz',''))
    set @+=1
IF @<101 GOTO a
\$\endgroup\$
12
  • \$\begingroup\$ Hello, and welcome to PPCG! Great answer! Can you please add an explanation? \$\endgroup\$ May 6, 2016 at 2:06
  • \$\begingroup\$ Sure thing I'll include the ungolfed version, it's just a while loop \$\endgroup\$
    – Phrancis
    May 6, 2016 at 2:06
  • \$\begingroup\$ I feel a bit privileged that in many languages you can't use expressions in case statements, but also a bit sad that the Oracle SQL person got fewer bytes... \$\endgroup\$
    – Phrancis
    May 6, 2016 at 6:15
  • 1
    \$\begingroup\$ Good work, but this needs more golfing. Use GOTO, not WHILE. Skip CONCAT because you need to force only @ to be a string, so just enclose it in LTRIM or something. Use + for the Fizz and Buzz IIFs, and nest that in the IIF for the other numbers. Remove as many spaces as possible. I can get your entry down to 110 doing all that. \$\endgroup\$
    – Muqo
    Jul 21, 2016 at 13:59
  • 1
    \$\begingroup\$ Code and explanation didn't match. Took the opportunity to edit in @Muqo's suggestions. \$\endgroup\$
    – BradC
    Apr 19, 2018 at 17:32
5
\$\begingroup\$

MathGolf, 24 22 bytes

♀{î╕Σ╠δ╕┌╠δ`+Γî35α÷ä§p

Try it online!

Explanation

♀                         Push 100
 {                        Start block
  î                       Push loop counter (1-indexed)
   ╕Σ╠δ                   Decompress "Σ╠" and capitalize to get "Fizz"
       ╕┌╠δ               Decompress "┌╠" and capitalize to get "Buzz"
           `              Duplicate top 2 elements of stack
            +             Add (creating "fizzbuzz")
             Γ            Wrap top 4 elements of stack in array
              î           Push loop counter (1-indexed)
               3          Push 3
                5         Push 5
                 α        Wrap last 2 elements in array
                  ÷       Check divisibility (implicit mapping)
                   ä      Convert from binary to int
                    §     Get array item
                     p    print
                          End block on code end, for loop implicit (100 iterations)
\$\endgroup\$
2
  • \$\begingroup\$ @dzaima Darn, I forgot about that completely. I have however had a capitalization operator in mind for a while. That should add one byte to the solution, but it is not yet implemented. \$\endgroup\$
    – maxb
    Sep 12, 2018 at 19:33
  • \$\begingroup\$ I fixed the capitalization, at the cost of two bytes. \$\endgroup\$
    – maxb
    Sep 13, 2018 at 6:58
5
\$\begingroup\$

brainfuck, 628 499 bytes

+[>>+>>+++<<[>+>->+<[>]>[<+>-]<<[<]>-]>>>[>>]<[>-[------->+<]>---.[--<+++>]<.-[---<+>]<-..<<<<[-]>>>>>>]<[-]<[-]<[-<+>]>+++++<<[>+>->+<[>]>[<+>-]<<[<]>-]>>>[>>]<[-[++++>---<]>-.++[-----<+>]<+.+++++..<<<<[-]>>>>>>]<[-]<[-]<[-<+>]<<<[[-]>>[-<+>>+<]>>>++++++++++<<[->+>-[>+>>]>[+[-<+>]>+>>]<<<<<<]>>>>>++++++++++<[->-[>+>>]>[+[-<+>]>+>>]<<<<<]>[-]>>[>++++++[-<++++++++>]<.<<+>+>]<[<[->-<]++++++[->++++++++<]>.[-]]<<++++++[-<++++++++>]<.[-]<[-]<[-<+>]<[-]<<[->+<]<]++++++++++.>>[-<+>>+<]-[<-->-----]<++]

This took me WAY too long. It's an extremely naive implementation with two divmods and a printing bit for Fizz and Buzz, as well as an equality check for 100. All in all not really golfed. At all. Not even a little bit. But it was fun, as this was my first ever real brainfuck program.

I started this by purposely not looking at any of the other brainfuck answers, partly so I wouldn't get discouraged, and partly so I wouldn't even subconsciously use any other ideas.

Any feedback or shrinkings are appreciated!

Badly commented source code

Try it online!

\$\endgroup\$
4
  • 1
    \$\begingroup\$ Some quick golfing; There's some leftover whitespace in the code, and all the [-]s can be replaced with ,s (if you're okay with EOF being 0). At least one of the [-]s isn't necessary. The not of the modulo will probably be shorter with forking the movement instead ([>>]<), making sure the tape ends up the same afterwards \$\endgroup\$
    – Jo King
    Feb 28, 2019 at 2:26
  • 1
    \$\begingroup\$ I don't think any of my implementations support EOF as 0, but I am interested in the unnecessary [-] clear cell and the [>>]<. Is there a chance you can suggest an edit? \$\endgroup\$ Feb 28, 2019 at 5:35
  • \$\begingroup\$ 596 bytes without whitespace and redundant [-]s. I think you'll find that most implementations use 0 as the EOF, and even the ones that don't usually use -1/255, which is still one byte shorter. I guess there's also no change, but that's rare. I'll work on the [>>]< part \$\endgroup\$
    – Jo King
    Feb 28, 2019 at 6:11
  • \$\begingroup\$ 499 bytes. I've changed it to the [>>]< method I mentioned above, and golfed some other parts. I didn't really touch the print as a number part, so you can probably golf something there. \$\endgroup\$
    – Jo King
    Feb 28, 2019 at 8:02
5
\$\begingroup\$

Seed, 6015 bytes

186 83944644497775792185807323999861330742900673481712359255839929374667216476568023851764637813248013688693896717275687232066153870107749175160194707761486575005885313952906442913325967953944535811512056079183251514390746052490451307246213606447865255212286265116943188690839445165359981774554198082946825007384980162551436390435260155582466874315352220541140434954811560506556477075852586817070090764184077321583489797358147025926992986354865415277367374043398355163643621296927936732212800075642321038031920620549375319366961667358107695950468557182519902148699776620624078332688843597417318685229781090303723160175745852091928346724348001020777912570918121340035523698691520422590527296248455779012094897281946771732844646467975422651655880202725095981882371807518829844157636552230152953535332201239998966830698645115246907049972204010095894629001206397276344975636564553238199728981112245810583644580269593382612869323054013019350314231784293448249910440692956496060253971880373279161390518237118507378691330341470412624728172421405109963475552981299919837091765136501358887135317778357279769289706821629750840213676581513947891276551415821187603189444230553630205086966993211879390093045277089486589385876312541940492767981701353769007246001400130719416118275492304807198011860072050303044225722194337717896831470037434122578382321765362062455130027125953869723625862069785410531287781468150359623495258126317807476517184801674197008747267394632928708383370640309530427431150403028006421518064783676469959724647961395869176187323826585310379142853088186864023708457159764123364029018721951815248751431170413579315400528983481702298849209797951129827235189965388114007169146563257209800975334610910397591190300324319540160364733177122449656422045835924029442449258975425381755648345550014417207471617919736685787448764321614059138375253980386335770138273578332418379763049794151811098188219754733324389355566317376598492318359989945589480418455750455469859956801589082779968271578522034926104268403877968397758570110853038536860556853388613997493905742068279137736678857073233174739380697685907170886092925173057794582407478105690770633669349947638020604803238951194381273510993928897230197983256465537488284258216944902276278712201715008765995404107536372438956009631805114079400698265046614085687812688769068889878215788959245601084883372435344000194364054441952170474833494604130017222942459218768152451353838900515646024764453476453552863938005162079649698291854043301771100462916706520075039044805956181705047266329114921035274711980897862579454154525144505597823262540025190878917613039796282312555295659294071407630514652646787462515667390800389392114746442068094785615537685660298361617697365653764669207754605147347324074541222276275532258194690926246437097055706104108319469003781845674958634242078791475826910418111428349167152839025683587469305219626924263962611094845308744968793376752461012553325189830937237777556407663277494722334379175044299145527685100424059993166500276116324651034389342857099326382267792125696063602436504279969923412378117936327869017861642388185619118723658181098871790872093500520584752433465443515989483851486224923341213990357424279098636563794743080022540640677640107176034203326304825856510242291526956869439836743210321616337131260522898066320086593786007891762080406650341513081903510025534886914361255147039463158189995500627220823997958074917925862170348467342212390436963261050291735696261617106196786273505443408966191019872244618002281427523777003325474230066462489531195157755555344637353155356279454281010364750875012491070816465270498678197434494849001514802132418781410405222031814823702582955264273367316636558580379660009282902598817700470535664545017234816209988594892835104965686956867851308392004044845547855039571721575449154495443479356960829759209226823917197487855397504141411230632224369201022681688301238065753737278463677847356448662974398206270856674118500884955014781944009536761710654880983265123955150729759406311220705702903728217724237231225228243114081826682040349685205121446193821261139703572900811243143912445467975694459999888915923116875430564637390061881889161512304819029656014338113772447267072899546161543780769850725712979853265124144813517511575434597175140160445249538753447856073028845070404104820699013418827593400549228758486817648053770622330987136941236768786640613654490823959024174575959620155957372041756651311988259843620347948677131103848960623371212015137053539005334518693964703094514744873758157905983139622925587291886231049371800498250086745691716656644176053612172113439907334397406577774498757112425437339205922869531575921611548158798375762386782843394828401126984185115349224303420458491251207091698634206841378497320259972637725826224808114267917428717816623368840721796784587349258491132128347823644310226457190477267965710821892576687103122134712656694901146343938508484308662925632874468476850854977933467327214644382364732022885458172449643330918925157997124690072012401561398600599725096512808715148150880033416494352767432615596760722206416684175988533830899565642405659319720681197534141300323187308210290375660950200938389996580887564670049743784962037885099514826194024014058093062057706139814777110125812527054724707373942053107785831307633110641398607263726549612409340127167852023607787965086360296071952107945138419803900924977980375619565210182772121849111987355981226968062629667279278470647719759267716675652568099185449478479471691254158308189631416224208590609150412906273490206303890853713104183863050119383440553687111132467944761189779633340481533657859382354192428154204487886313458547651002779223927726608436086877353829832057341247094095114657693381028772025327489391112902949559367893876211447144802835381133403893955836438518252858893637260806405255643702579004661210305919494653119109150790218729956868835489222681506976514342204634574313468575534020458319909578144280137077342941741913678288702984697189445809519253013406175446129538246339672293879388869701830984873958965248

Try it online!

\$\endgroup\$
2
  • \$\begingroup\$ Oh, my generator must have glitched. I'll look on this \$\endgroup\$ Sep 8, 2019 at 11:46
  • \$\begingroup\$ @JoKing Well, I should have tested my stuff before. Now it works \$\endgroup\$ Sep 8, 2019 at 12:25
5
\$\begingroup\$

Arn -hm, 27 22 18 bytes

─=█BƒHXåõÝü»ÝÑÕ=!&

Try it!

Explained

Unpacked: "Fizz"^!%3|`#&`^!%5||

Set STDIN to the range [1, 100]

      "Fizz" Literal string
    ^ Repeat
      ! NOT
          _ Implied variable
        % Modulo
          3
  | Concatenated with
      `#&` Compressed string "Buzz"
    ^
      !
          _ Implied
        %
          5
|| OR
  _ Implied

Mapped over STDIN with key`_

I wasn't going to update this, but might as well.

\$\endgroup\$
5
+400
\$\begingroup\$

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.

\$\endgroup\$
5
\$\begingroup\$

Squire, 92 bytes

n=N whilst C>n{t=""n=n+I if!(n%3){t="Fizz"}if!(n%V){t=t+"Buzz"}proclaim((t||arabic(n))+"
")}

Hear ye, hear ye! Unbeknownst to some, Sampersand hath also created Squire, a companion language to ye lowly language Knight. Squire hath many possessions, but alas, it doth not golfeth as well as its brethren. Thou shalt check it out as well, though.

I cannot provideth thee with a demo link yet as the interpreter is a very lengthy incantation.

Ungolfeth:

# Createth a variable known by most as "n",
# He shalt begin his journey as a mere
# peasant with naught but the value zero.
# (Behold! Squire permits thee to write numbers
# in Roman numerals)
n = N
# Loop thine program unto n approacheth 100
whilst C > n {
    # Lo! I spotteth another variable,
    # who goes by the name of "t".
    # It shall start off as a lonely empty string.
    t = ""
    # Increase n's wealth by bestowing upon it the number one
    n = n + I
    # If n divideth evenly into 3...
    if !(n % 3) {
        # Replaceth t with the string "Fizz"
        t = "Fizz"
    }
    # If n also divideth evenly into 5...
    if !(n % V) {
        # Accompany t with the string "Buzz"
        t = t + "Buzz"
    }
    # Proclaim to ye standard output...
    proclaim(
        # Either...
        (
           #  t if it is not a lonely empty string...
           t
           # or n, converteth to ye foreign Arabic
           # number system, used in lands afar.
           || arabic(n)
        )
        # Appendeth upon it a newline to start
        # another journey.
        + "\n"
    )
}

A more favorable program is as follows:

n=N whilst C>n{t=""n=n+I if!(n%III){t="Fizz"}if!(n%V){t=t="Buzz"}proclaim((t||string(n))+"
")}

This program useth Roman numerals, but nay, thine great quest beseecheth that one must outputeth thy program using this unusual number system.

\$\endgroup\$
5
\$\begingroup\$

Pip, 32 31 30 bytes

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

Attempt this online!

Here's the 31-byte equivalent in Pip Classic: Try it online!

Explanation

LhP J["Fizz""Buzz"]X!*Ui%^35|i
                                Preinitialized variables: h=100, i=0
Lh                              Loop 100 times:
                         ^35     Split 35 into a list of digits: [3 5]
                      Ui         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
\$\endgroup\$
0
5
\$\begingroup\$

Vyxal, Hj, 25 19 12 bytes

-6 bytes thanks to Aaron
-7 bytes thanks to lyxal

Flagless 15 bytes
Longer than lyxal's answer, but uses a different technique

ƛ₃kf*n₅kb*+∴               stack is preset to a 100 because of the `H` flag

ƛ                          lambda map with variable n
 ₃                         push 1 if n%3 == 0 (we'll call the return a)
  kf                       push constant Fizz to the stack
    *                      push a*kF
     n                     push n
      ₅                    push 1 if n%5 == 0 (we'll call it b)
       kb                  push Buzz to the stack
         *                 push b*kB
          +                add last two elements of the stack ""/Fizz/Buzz
           ∴               push the maximum of n and ""/Fizz/Buzz/FizzBuzz

Vyxal prints out the last element of the stack by default (the `j` flag joins the list with new lines)

Try it Online!

\$\endgroup\$
6
  • 1
    \$\begingroup\$ You can use the register instead of variables, and the H flag, for 19 bytes \$\endgroup\$ Jun 27, 2021 at 17:30
  • \$\begingroup\$ Try it Online! for 16 bytes using the register, H flag and string multiplication instead of if statements \$\endgroup\$
    – lyxal
    Jun 27, 2021 at 23:01
  • \$\begingroup\$ Try it Online! for 12 \$\endgroup\$
    – lyxal
    Jun 28, 2021 at 1:34
  • \$\begingroup\$ @lyxal and Aaron Thanks! \$\endgroup\$
    – mathcat
    Jun 28, 2021 at 14:35
  • \$\begingroup\$ @lyxal would you please check the explanation? I didn't get the last part so i guessed \$\endgroup\$
    – mathcat
    Jun 28, 2021 at 15:34
5
\$\begingroup\$

Jellyfish, 124 105 98 97 92 bytes

P
L,L,  v"Buzz
1 1v"Fizz
   E $X$,,S
 LvLjXX   *4
 1 0*rE101b6
  $R  * E4
  ,1N1b19252
  100

Try it online!
Try the 97B version online!
Try the 98B version online!
Try the 105B version online!
Try the 124B version online!

explanation

The main idea is to concatenate each string from 3 string-arrays, numstr_array, fizz_array, buzz_array, to create 1 string-array, and matrix-print that with "P" function.

while a code-block like \$\left|\begin{array}{ll}L&,~~~~s\_ary2\\1&s\_ary1\end{array}\right|\$ create a string-array like
[ s_ary1[i]+s_ary2[i] for i in range(min(len(s_ary1),len(s_ary2))) ]
in Python,
\$\left|\begin{array}{ll}L&,~~L~~~,~~buzz\_array\\1&\downarrow~1~~fizz\_array\\~&numstr\_array\end{array}\right|\$ creates such a "concatenated" array.

The 3 string-arrays above should have these contents;

  • numstr_array: [ "1" "2" "" "4" "" "" "7" "8" "" "" ... ]
  • fizz_array: [ "" "" "Fizz" "" "" "Fizz" "" "" "Fizz" ""... ]
  • buzz_array: [ "" "" "" "" "Buzz" "" "" "" "" "Buzz" ... ]

and the last 2 arrays are simply created from drop function "v".

2-parameter version of "v" takes num-array and a string, and create a string-array whose elements are "dropped" from the original string, like [ string[i:] for i in num_array ] in Python.
For example, \$\left|\begin{array}{}v&"&F&i&z&z\\E&~&$&X\\~&~&X&X&~&~&~&*&4\\~&~&~&E&1&0&1&b&6\end{array}\right|\$ code block creates the fizz_array. The "num-array" in this block is a 101 elements of [ 4 4 0 4 4 0 4 4 0 ...].

The first array ( numstr_array ) is a little more complicated from the others. because the original strings ( before dropped ) are not the same unlike fizz_array or buzz_array are created from one original string for each.
To solve this problem, use \$\left|\begin{array}{ll} L & v ~~ str\_array \\ 1 & 2d\_num\_array\end{array}\right|\$ to "drop" each string independently.
str_array is created from a 0-array and a 1..100 sequence combined with \$\left|\begin{array}{ll} L & j ~~ sequence \\ 0 & 0\_array \end{array}\right|\$.
2d_num_array is a 100x1 2d-array like [ [0] [0] [4] [0] [4] [4] [0] ... ], which is created as a repetition of a 15-length pattern [0 0 4 0 4 4 0 ...].

\$\endgroup\$
5
  • \$\begingroup\$ Heyo, welcome to code golf! This is quite a neat first answer, so keep up the good work! \$\endgroup\$
    – lyxal
    Nov 16 at 10:41
  • \$\begingroup\$ Thx! I will add some code explanation after I prepare it. \$\endgroup\$
    – angel_p_57
    Nov 16 at 11:13
  • 1
    \$\begingroup\$ i'm afraid 92 byte version is not working on TIO because of handling null input \$\endgroup\$ Nov 26 at 10:00
  • 1
    \$\begingroup\$ Thank you for letting me know about the old sample on TIO was not working. I fixed it and update this post. \$\endgroup\$
    – angel_p_57
    Nov 26 at 13:31
  • 1
    \$\begingroup\$ I misunderstood the "header" area on TIO as for title or some comment, and added a string on it. That's why the code was not working. \$\endgroup\$
    – angel_p_57
    Nov 26 at 13:37
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\$\begingroup\$

Perl 6, 109 93 bytes

for$i(1..100){if($i%3<1){say"Fizz"};if($i%5<1){say"Buzz"};if($i%3>0&&$i%5>0){say$i};say"\n";}

There might be some more golfing potential here.
Takes advantage the fact that 0 is the only way x % y can be less than 1 (thanks to Alex. A for shaving off 4 bytes with this) and Perl 6's say keyword.

\$\endgroup\$
1
  • \$\begingroup\$ I didn't spot this somehow... Sorry, If you'd like to adopt some of the ideas in my answer you can save more bytes! say is available in Perl 5 too for free (meta.codegolf.stackexchange.com/q/273#answer-274) and it includes a newline (which would further shorten mine by 5 bytes). Happy to remove mine (since it's newer than yours). \$\endgroup\$ Sep 24, 2015 at 20:26
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