1, 2, Fizz, 4, Buzz

Question

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

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Answer 1

Minkolang, 50 bytes

"d"[i1+d5%6&"Buzz"0c3%6&"Fizz"I1-3&N6@0gx(O)25*O].

Try it here.

Explanation

"d"[...].        For loop that loops from 0 to 99, then stops
i1+              Loop counter + 1 (so it's 1 to 100)
d5%6&"Buzz"      Divisibility test by 5, skips "Buzz" if not divisible
0c3%6&"Fizz"     Divisibility test by 3, skips "Fizz" if not divisible
I1-              Length of stack minus 1 (0 if there's no Fizz or Buzz)
3&N6@            Output as integer if ^ is 0, skip character output otherwise
0gx(O)           Dump the loop counter and output "Fizz"/"Buzz"/"FizzBuzz"
25*O             Print newline

Answer 2

Swift, 75 bytes

for n in 1...100{print(n%3*n%5>0 ?n:(n%3>0 ?"":"Fizz")+(n%5>0 ?"":"Buzz"))}

Answer 3

scg, 51 bytes

1á01°r{d[[d"Buzz"]["Fizz"d"Buzz"+]]\3%!@\5%!@"
"}m

So, does this mean that scg is a real language now? Explanation:

1                         .- adds 1 to the stack
 á01                      .- adds 101 to the stack
    °r                    .- range, adds array with 1-100 on the stack
      {                   .- start function for use in map
       d                  .- duplicates number
       [                  .- array
        [
         d                .- duplicate number again, ends up in array
          "Buzz"          .- wonder what this does
                ]         .- end array
        [
         "Fizz"
               d          .- duplicate fizz
         "Buzz"+          .- ends up with "FizzBuzz"
                ]
                 ]        .- end array. Ends up with a 2D array
        \                 .- gets number to calculate to the top
        3%                .- mod 3
          !               .- not, so any above 0 int turns to 0 and 0 turns to1
           @              .- get array value. Now you have two choices for output
            \5%!@         .- same as above but for 5.
                          .- now we have the correct fizzbuzz value
                 "\n"     .- pushes newline. I do not have variables yet so no shortcuts
                     }m   .- end function, map. output is implicit

Answer 4

𝔼𝕊𝕄𝕚𝕟, 32 chars / 47 bytes

⩥Ṥⓜᵖ`FizzBuzz`ė⧺_%3⅋4,_%5?4:8)⋎_

Try it here (Firefox only).

Answer 5

C#, 174 bytes

void A(){for(int x=1;x<101;x++){if(x%15<1)Console.Write("FizzBuzz\n");else if(x%3<1)Console.Write("Fizz\n");else if(x%5<1)Console.Write("Buzz\n");else Console.WriteLine(x);}}

Ungolfed:

void A(){
    for (int x = 1; x < 101; x++) {
        if (x % 15 < 1) Console.Write("FizzBuzz\n");
        else if (x % 3 < 1) Console.Write("Fizz\n");
        else if (x % 5 < 1) Console.Write("Buzz\n");
        else Console.WriteLine(x);
    }
}

Answer 6

Mathematica, 103 86 73 bytes

With 30 bytes saved thanks to @A Simmons!

f="Fizz";b="Buzz";Range@100/.{x_/;15∣x->f<>b,x_/;3∣x->f,x_/;5∣x->b}

Answer 7

XQuery 3, 172 bytes

declare option output:method "text";string-join(for$x in 1 to 100 return if($x mod 15=0)then"FizzBuzz"else if($x mod 3=0)then"Fizz"else if($x mod 5=0)then"Buzz"else $x,"
")

There were only 7 XQuery answers on the whole site, I thought it could at least have its FizzBuzz ! Granted it's not very golfy, in particular when you need to add a 36 bytes preface so that it does not output an XML header.

I tested it with Saxon-HE's command-line XQuery tool (java net.sf.saxon.Query fizzbuzz.xq), with which I had to replace the w3-defined option declaration with declare option saxon:output "method=text";.

Answer 8

Javascript, 191 bytes

for(var i=1; i<=100; i++){
  var r = "";
  if( i%15 == 0 ? r = "FizzBuzz" : (i%5 == 0 ? r = "Buzz" : (i%3 == 0 ? r = "Fizz" : r = i)) ){
    console.log(r);
  }
}

Answer 9

Tcl, 136 bytes

set f {Fizz 3 Buzz 5}
while {[incr n]<=100} {set s ""
foreach {m d} $f {if {$n%$d==0} {append s $m}}
if {$s eq ""} {puts $n} {puts $s}}

This solution, incidentally, is easily extensible to any combination of multiples. See The Smart Person's Mirage golf, where gnibbler posted the same idea (but in Python).

set iterations 100

set fizzies {
  Fizz 3
  Jazz 4
  Buzz 5
}

while {[incr n] <= $iterations} {
  set s ""
  foreach {name divisor} $fizzies {
    if {$n % $divisor == 0} {append s $name}
  }
  if {$s eq ""} {puts $n} {puts $s}
}

Answer 10

Oration, 98 bytes

literally, for i in range(1,100):x=""if i%3 else'Fizz';x+=""if i%5 else "Buzz";print x if x else i

Answer 11

ROOP, 187 bytes

1
V!         !<
(102)      1|
 e#r3##r5# a|
#H  #   # Y-<
   N   N  !
"Fizz""Buzz"

   mX  mX
### V-->  !
   A V---->
    #
  ' C
 V'e  "\n"
 |# M  #
 <V# #v
 C  A  C
  #X ##
    w
    O#

I will try to explain each section of code:

V!         !<  Add 1 to each number that goes to the left of the a
           1|  and sends it to the bottom of the V
           a|
           -<

---

(102)          The 102 falls to the left of the e and each number
 e             that passes over is compared to 102.
#H             If a number is equal then the H runs and ends the program

---

  #r3#         With each number that goes above the r
    #          the remainder of dividing by 3 is obtained.
   N           The N returns 1 if the number is 0, and 0 otherwise.
"Fizz"         The string "Fizz" falls and moves to the left of the m.
               The number is multiplied by the string
   mX          ("" or "Fizz" if it is 0 or 1 respectively)
###            The X removes the number when it moves to the right

               The same is done with 5 and "Buzz"

---

   mX  mX      
    V-->      Both strings are concatenated with the A
   A          getting "", "Fizz", "Buzz" or "FizzBuzz"
    #

---

  ' C         The C changes the direction of advance of string, to the left.
  'e          At the same time the "e" compares the string with the empty string.
  #           The single quotes are a vertical literal string.

---

 V            The V and pipes redirects the string to the right of the C
 |            that changes the direction again in order that comes to the left of the A
 <V
 C  A

---

          Y   The original number is converted to a string with Y.
          !



          !   Pipes and teleporters (!) Redirects the string to below the V
     V---->

---

    M         The string falls to the right of the M and multiplies
   # #        with the number previously obtained by the e
    A         The result is above the A

---

      "\n"   The string "\n" falls on the v which makes a copy
    M  #     whenever there is a space below.
      v      The C changes the direction of the string so it goes to the left.
    A  C     It waits to the right of the A

---

    A        The A concatenate all 3 strings, the result is on the w that sends it
  #X ##      to the O representing the output. At the same time the X deletes the string.
    w
    O#

I hope it is comprehensible, English is not my main language.

Answer 12

Nim, 100 76 73 bytes

for i in 1..100:echo max(["Fizz","",""][i%%3]&["Buzz",""][ord i%%5>0],$i)

Hm... still trying to learn Nim, and I'm thinking there's got to be a better way...

Answer 13

Befunge, 65 bytes

_1+::3%^>55+,:"c"`#@
>"zuB"vv.#,,:,,<
|!:%5\ _:!"ziF"^
<,,:,,<:|*

Try it online!

Answer 14

Valyrio, 13 bytes

s∫main [CF]

This is a fairly basic (and slightly unimaginative) answer.

Explanation

C pushes 100 to the stack, which means that ...

F is the FizzBuzz builtin. This was mainly added in as a basic stack based program but got left in as a command and I never got rid of it.

Answer 15

Go, 162 158 145 143 142 139 bytes

package main;import."fmt";func main(){for i,p:=1,Println;i<101;i++{s:="";if i%3<1{s+="Fizz"};if i%5<1{s+="Buzz"};if s!=""{p(s)}else{p(i)}}}

Go Playground Link

Answer 16

Cardinal 217 bytes

%x>+ v  >+++++~M! 8# "buzz"
0    #~0#+++~>M! # V  "fizz"
+ ^jM<  ~        V>#
+       V      > #xV
= 0              V
t ~             >#x
t t       v      V
* =   > #}#}     /
> ^              >}/
                   .

Try it Online

Explanation

%x
0
+
+
= 0
t ~
t t
* =
> ^

Initializes a pointer with an inactive value of 100 and active value of 0

>+ v
   #
^jM<

Loops around, incrementing the active value of the pointer and sending out a duplicate until the active value is equal to the inactive value (100)

  >+++++~M!
~0#+++~>M!

Checks if the value mod 3 or mod 5 is equal to 0

                   # "buzz"
        #        # V  "fizz"
        ~        V>#
        V      > #xV
                 V
                >#x
          v      V
      > #}#}     /
                 >}/
                   .

Uses reflectors and splitters in order to print only numbers that are not divisible by 3 or 5. Printing fizz when divisible by 3 and buzz when divisible by 5.

Answer 17

Java, 153 145 bytes

enum f{;static{for(int i=1;i<101;i++)System.out.println(i%3>0&i%5>0?i:(i%3<1&i%5<1?"FizzBuzz":(i%3<1?"Fizz":(i%5<1?"Buzz":0))));System.exit(0);}}

Not as long as I expected it to be, maybe because I ommited main function and used an enum. Quite overcomplicated print statement.

Ungolfed version(Modified to be readable) with comments:

enum f {;                                       // Required because it is an enum
    static {                                    // static block
        for (int i = 1; i < 101; i++) {         // loop from 1 to 100 inclusive
            boolean by3 = i % 3 == 0;
            boolean by5 = i % 5 == 0;

            if (!by3 && !by5)                   // Not divisible by 3 and 5
                System.out.println(i);          // Just show the number
            else if (by3 && by5)                // Divisible by both
                System.out.println("FizzBuzz"); // Show FizzBuzz
            else if (by3)                       // Divisible only by 3
                System.out.println("Fizz");     // Show Fizz
            else                                // Divisible only by 5
                System.out.println("Buzz");     // Show Buzz
        }
        System.exit(0);                         // Required because there can't be output to stderr and it would crash trying to run main function
    }
}

Ungolfed not modified:

enum f {;
    static {
        for (int i = 1; i < 101; i++)
            System.out.println(i % 3 > 0 & i % 5 > 0 ? i : (i % 3 < 1 & i % 5 < 1 ? "FizzBuzz" : (i % 3 < 1 ? "Fizz" : (i % 5 < 1 ? "Buzz" : 0))));
        System.exit(0);
    }
}

EDIT: Saved 8 bytes by replacing ==0 with <1, !=0 with >0 and && with &

Answer 18

Julia, 70 bytes

for i=1:100 println("$(i%3<1?"Fizz":"")$(i%5<1?"Buzz":i%3>0?i:"")")end

Julia has perl-like string formatting. Very nice for challenges like these.

Answer 19

REXX 81 Bytes

f.=""
f.0="buzz"
t.=""
t.0="fizz"
do i=1 to 100
  f=i//5
  t=i//3
  say overlay(t.t||f.f,i)
end

Answer 20

tcl, 72

time {puts [expr [incr i]%3?$i%5?$i:"":"Fizz"][expr $i%5?"":"Buzz"]} 100

I think it is more golfable, to avoid the repetition of i%5

demo

Answer 21

Swift, 97 bytes

for i in 1...100{i%15==0 ?print("FizzBuzz"):i%3==0 ?print("Fizz"):i%5==0 ?print("Buzz"):print(i)}

Answer 22

q/kdb+, 58 56 49 bytes

Solution:

0{$[sum i:0=y mod 3 5;`Fizz`Buzz(&)i;y]}'1+(!)100

Example:

q)0{$[sum i:0=y mod 3 5;`Fizz`Buzz(&)i;y]}'1+(!)100
1
2
,`Fizz
4
,`Buzz
,`Fizz
7
8
,`Fizz
,`Buzz
11
,`Fizz
13
14
`Fizz`Buzz
16
...etc

Explanation:

0{$[sum i:0=y mod 3 5;`Fizz`Buzz where i;y]}'1+til 100  / ungolfed
 {                                         }'           / anonymous function that takes each-left and each-right
0                                                       / this would be parameter 'x' but we dont use it
                                               til 100  / til generates a list of 0..99
                                             1+         / adds 1 to every item in the list, thus 1..100
  $[                 ;                  ; ]             / switch, $[condition;true;false]
            y mod 3 5                                   / modulo operation on input for 3 and 5, mod[1;3 5] = 1 1
          0=                                            / is 0 equal to this result (basically a 'not' operation)
        i:                                              / save in i for later
    sum                                                 / add these, will get 0, 1 or 2. 0 is interpretted as false
                                 where i                / where gives indices where i is true
                      `Fizz`Buzz                        / 2 item list which gets indexed into (and implicitly returned)
                                     y                  / return the input if the condition was false

Notes:

This ^^ is pretty much a q version of the k solution, so I've written in a different way.. unfortunately it's about 50% slower :(

0{(`Fizz;`Buzz;y)(&)(0=a),all a:mod[y;3 5]}'1+(!)100

Here we are indexing into a list of Fizz, Buzz, based on the result of the modulo operation... The k solution style is better.

Answer 23

MATLAB, 172 bytes

s=char(arrayfun(@(n){num2str(n)},[1:100 1e7])');s(3:3:end,1:4)=repmat('Fizz',33,1);s(5:5:100,1:4)=repmat('Buzz',20,1);s(15:15:100,:)=repmat('FizzBuzz',6,1);disp(s(1:100,:))

Answer 24

groovy, 55 bytes

Borrows idea from answer by @feersum https://codegolf.stackexchange.com/a/58623

100.times{println'Fizz'*(it%3/2)+'Buzz'*(it%5/4)?:it+1}

Answer 25

Python 2, 57 bytes

for i in range(100):print i%3/2*"Fizz"+i%5/4*"Buzz"or i+1

Not the most beautiful code but it works at least with

python -c

Answer 26

tinylisp repl, 130 bytes

(d f(q((n # %)(i(e n 100)(q Buzz)(i(disp(i #(i % n(q Buzz))(i %(q Fizz)(q FizzBuzz))))0(f(a n 1)(i #(s # 1)2)(i %(s % 1)4
(f 1 2 4

Try it online! (Note that the repl auto-completes parentheses at the ends of lines; these have been filled in on the TIO version.)

Explanation

It took me a while to realize I could do FizzBuzz in tinylisp, since the language doesn't have strings. However, it does have the Name type, which can be any run of non-whitespace, non-parenthesis characters; and it also has the disp command, which outputs a value followed by a newline. Together with some creative abuse of recursion, we can get the desired output.

We define a function f with three arguments:

• n is the counter. We will start at 1 and recurse until 100.
• # is the number of iterations till the next divisible-by-3 number. (Mnemonic: Shift-3)
• % is the number of iterations till the next divisible-by-5 number. (Mnemonic: Shift-5)

The first thing f does is check whether n is 100. If so, we return Buzz. This value gets passed all the way up the call stack and printed at the end.

If not, we need to use disp to print either the number, Fizz, Buzz, or FizzBuzz, depending on the values of # and %:

• If # and % are both nonzero (truthy), the number is not divisible by 3 or 5. Display the number, n.
• If # is nonzero but % is zero, the number is divisible by 5 but not 3. Display Buzz.
• If # is zero but % is nonzero, the number is divisible by 3 but not 5. Display Fizz.
• If # and % are both zero, the number is divisible by 3 and 5. Display FizzBuzz.

Since tinylisp does not have an equivalent to Common Lisp's progn, disp'ing something and then returning a value requires a little trickery. Since disp always returns (), which is falsey, we can use the disp call as the condition of an if (i). The true branch will never be evaluated (0 is a convenient placeholder), and we can put the actual return value in the false branch.

This return value is a recursive call. We add 1 to n, and we subtract 1 from # and % unless they are zero, in which case we reset them to (3-1) or (5-1), respectively.

Ungolfed, using the standard library for long names of builtins (TIO):

(load library)

(def fizz-buzz
 (lambda (counter steps-till-fizz steps-till-buzz)
  (if (equal? counter 100)
   (q Buzz)
   (if
    (disp
     (if steps-till-fizz
      (if steps-till-buzz counter (q Buzz))
      (if steps-till-buzz (q Fizz) (q FizzBuzz))))
    (comment Return val of disp is always falsey, so the true branch is never executed)
    (fizz-buzz
     (add2 counter 1)
     (if steps-till-fizz (sub2 steps-till-fizz 1) 2)
     (if steps-till-buzz (sub2 steps-till-buzz 1) 4))))))

(fizz-buzz 1 2 4)

Answer 27

Java (OpenJDK 8), 127 bytes

interface J{static void main(String[]a){for(int i=0;i++<100;)System.out.println((i%3<1?"Fizz":"")+(i%5<1?"Buzz":i%3<1?"":i));}}

Try it online!

ungolfed:

interface J {
    static void main(String[] a) {
        for (int i = 0; i++ < 100;)
            System.out.println((i % 3 < 1 ? "Fizz" : "") + (i % 5 < 1 ? "Buzz" : i % 3 < 1 ? "" : i));
    }
}

Answer 28

SmileBASIC, 70 bytes

Printing "Fizz" and "Buzz" is easy, the slightly more difficult part is to only print the number when required. There are basically 2 ways to do this (and they end up being the same length)

1: Print the number when I isn't divisible by 3 or 5

FOR I=0TO 100A=I MOD 3B=I MOD 5?"Fizz"*!A;"Buzz"*!B;STR$(I)*(A&&B)NEXT

2: Print the number if the cursor is at column 0:

FOR I=1TO 100?"Fizz"*!(I MOD 3);"Buzz"*!(I MOD 3);
?STR$(I)*!CSRX
NEXT

In a previous version of SB, % was used for MOD, making the program shorter:

Petit Computer BASIC, 62 bytes

FOR I=0TO 100?"Fizz"*!(I%3);"Buzz"*!(I%5);
?STR$(I)*!CSRX
NEXT

Answer 29

KoopaScript, 250 characters

def i 1 if \%va is \%vu set a 1;setath b \%va %% 3;setath c \%va %% 5;setath e \%va %% 15;if \%vb is 0 if \%vc not 0 print Fizz;if \%vb not 0 if \%vc is 0 print Buzz;if \%ve is 0 print FizzBuzz;if \%vb not 0 if \%vc not 0 print \%va;setath a \%va + 1

Guide to reading: Don't. Either learn KoopaScript more or less entirely by looking at the code (my documentation isn't that great) or just take my word for it that it works. By the way, KoopaScript is an interpreted language I made (not specifically for this) that runs inside ActionScript 2, and doesn't actually have else statements, or... most of the stuff that makes other examples short. All functions are one line, so this was pretty easy. Here's the GitHub repo.

Answer 30

Husk, 35 bytes

m§Ysλṁ!⁰₇ḣ3)ḣ100
¶Σfm%⁰NC2¨¶¶¶⌈iΩZu

Try it online!

I feel like I'm missing some opportunities for cleverness here. Like, I'm not really pleased with that lambda sitting there on the first line and having to be manually closed. But anyway, here we are.

Explanation

I'll explain the second line first. It's a function TNum -> [[TChar]] that maps

1 -> []
2 -> ["","","Fizz"]
3 -> ["","","","","Buzz"]

It works like this:

¶Σfm%⁰NC2¨¶¶¶⌈iΩZu
         ¨¶¶¶⌈iΩZu --The compressed string "\n\n\n\nFiBuzz"
       C2¨¶¶¶⌈iΩZu --cut; the list ["\n\n", "\n\n", "Fi", "Bu", "zz] 
   m%⁰N            --nats mod input, i.e, with input 3, [1,2,0,1,2,...]
  fm%⁰NC2¨¶¶¶⌈iΩZu --filter;so grab (1) nothing, (2) pairs 135, (3) pairs 1245
 Σfm%⁰NC2¨¶¶¶⌈iΩZu --concat;now at (1) [] (2) ["\n\nFizz"] (3) ["\n\n\n\nBuzz"]
¶Σfm%⁰NC2¨¶¶¶⌈iΩZu --split by newlines to return claimed result

Next, the lambda in the first line. It has type TNum -> [TChar]; given an integer, it returns "", "Fizz", "Buzz", or "FizzBuzz" as appropriate. In detail:

λṁ!⁰₇ḣ3)
     ḣ3  --heads; the list [1,2,3]
    ₇ḣ3  --call 2nd line with map overflow: [[],["","","Fizz"],["","","","","Buzz"]]
 ṁ!⁰₇ḣ3  --index the lambda argument to each list (modularly) and concat the results

And pulling it together:

m§Ysλṁ!⁰₇ḣ3)ḣ100 
    λṁ!⁰₇ḣ3)      --"","Fizz","Buzz",or"FizzBuzz"
   s              --string representation of the number
 §Ysλṁ!⁰₇ḣ3)      --apply both and take max; "" < numeric strings < alphabet
m§Ysλṁ!⁰₇ḣ3)ḣ100  --map across [1,..,100]

Then Husk's default output style for an object of type [[TChar]] turns out to be just what we want.