# Winner: Aditsu's CJam answer! A whopping 25 bytes! Nice!

You may continue to submit your answer, however you can no longer win. Original post kept for posterity:

The "Bzzt" game is a game where you must count to a number (in this case 500). However, if the number has a 3 in it, or is divisible by 3, you don't say the number. Instead, you say "Bzzt".

Rules:

• You can't hard-code the numbers.
• The number only has to satisfy at least 1 of the following requirements
• Divisible by 3
• Number contains a 3
• Some type of separator is mandatory (12bzzt14 doesn't count)
• Score is measured in bytes.
• You must count exactly to 500, starting at 1 or 0 (you chose).
• The numbers must be output, but it doesn't matter how (e.g., stdout, writing to a text file, etc.).
• 0 can be divisible by 3 or not divisible. You can choose.
• You can output the numbers one at a time (ex, output 1, then 2, then bzzt, then 4, etc) or all at once (e.g., output 1 2 bzzt 4 5).
• You must replace the letter 3 with the word "bzzt". This is not case sensitive (bZzt, Bzzt, bzzt are all okay).

• This is a challenge, so the shortest code wins.
• This contest ends June 30th 2014 (7 days from posting).
• For purposes of this question, is 0 divisible by 3? Jun 23, 2014 at 7:07
• Is it "buzz" or "bzzt"? You wrote "buzz" twice so far. Jun 23, 2014 at 8:40
• Please clarify. Do I have to output buzz or bzzt if both of the requirements apply? Do I have to output 12bzzt4 or bzzt for 1234? Jun 23, 2014 at 12:41
• I'd say bzzt for 1234. It's a 'common' drinking game here (only we often do it with 7) Jun 23, 2014 at 13:38
• "0 can be divisible by 3 or not divisible. You can choose." I really don't think you can choose. 0 mod 3 is 0, that's not really a matter of opinion. Jun 23, 2014 at 15:54

# Scala, 70

## Code

1 to 500 map(x=>if(x%3<1||(""+x).contains(51))"bzzt"else x)map println


## Usage

scala -e '1 to 500 map(x=>if(x%3<1||(""+x).contains(51))"bzzt"else x)map println'


## Demo

http://ideone.com/KfcO1M

## Explanations

• 1 to 500: produces values from 0 to 500
• map: function which takes iterable argument at the left, and applies the function given at the right
• x=>...: lambda expression
• if(condition) value_if_condition_is_true else value_if_condition_is_false: returns a value (functional behavior) which can be either a java.lang.String or a scala.Int. The expression type is scala.Any in this case, because this is the lower type the two types can coerce to (see the Scala Class Hierarchy)
• map println: takes each object (java.lang.String or scala.Int) and prints it with line feeds

### Swift - 74

for i in 0..501 {println(i%3<1||"\(i)".rangeOfString("3") ?"bzzt":"\(i)")}


# 05AB1E, 2623 21 bytes (non-competing)

501FND3ÖN3å~i"bzzt"},


Try it online!

# JavaScript - 57

Another JavaScript example:

for(i=0;i<500;)alert((++i+'').indexOf(3)<0&&i%3?i:'bzzt')


## PROLOG (144)

b:-b(R,1),print([0|R]).
b([500],500).
b([bzzt|R],N):-((0is N mod 3);(name(N,C),member(51,C))),N1 is N+1,b(R,N1).
b([N|R],N):-N1 is N+1,b(R,N1).


Usage:

b.


## Microsoft Small Basic (115 chars, 127 bytes)

For i = 1 To 500
j=i
If(Math.Remainder(i,3)=0 Or Text.IsSubText(i,"3"))then
j="bzzt"
EndIf
TextWindow.Write(j+" ")
EndFor


# Haskell 78

main=mapM(putStrLn.f)[0..500];f x|xmod3<1||'3'elemshow x="bzzt";f x=show x


Python - 53

for i in range(1,501):print[i,'Bzzt']['3'[:i%3]ini]


Can't add comments to above answer so posting down here. Starting range at 0 produces Bzzt at 0, need to start at 1.

# Python2 (59)

Simple case of list comprehension

print[n if n%3or '3'in str(n)else"bzzt"for n in range(501)]

• With some tips you can get to 52 bytes at least. Nov 4, 2016 at 3:09

# MathGolf, 19 18 bytes

♪½{î3╧î3÷+î╕▄┴_ß§p


Try it online!

## Explanation

♪                     push 1000
½                    divide by 2, making 500
{                   start for loop
î                  index of current loop (1-based)
3                 push 3
╧                pop a, b, a.contains(b)
î               index of current loop (1-based)
3              push 3
÷             is divisible
+            pop a, b : push(a+b)
î           index of current loop (1-based)
╕▄┴        decompress "bzzt"
_       duplicate TOS
ß      wrap last three elements in array (makes [<index>, "bzzt", "bzzt"])
§     get from array
p    print with newline

• How does the string compression work, and is there some kind of generator program I could use for other strings? Just posted an answer with two strings, but not sure how to do string compression in MathGolf. Jun 3, 2019 at 12:27
• The way I do it is that I check the reference page and Code page 437, and I match the column and the row of the character within the code page with the indices of the string I want. So, ╕ means indexing in the string etaoinsrdluczbfp. In that string, b has index 13, z has index 12, and t has index 1. Thus, I look at row 13 and column 12 for bz, which is character ▄. Then I look at row 12 and column 1 for zt, which is character ┴.
– maxb
Jun 3, 2019 at 12:39
• Be wary though that it is not strictly Code page 437, since I swapped two characters (the null byte and the non-breaking space). Both of these characters should be replaced with their new counterparts (× and Þ). It should be fairly straight-forward to make a generator program for short strings, I just haven't made one. The compression is nothing fancy, but in cases like this challenge it can save you 2 bytes over the standard ÿbzzt.
– maxb
Jun 3, 2019 at 12:42
• Ah ok. And compressed strings of mixed characters? So partially from etaoinsrdluczbfp and partially from gwymvkxjqh ?*#.,? Or that isn't possible (yet)? To be more concrete, what would the compressed strings for ÿduck and "goose be? (Here a link to my current answer.) Jun 3, 2019 at 12:45
• I've been able to figure out how to compress "goose to 'g╕3 . As for ÿduck, I guess I could use a 3-char compression for "duc" and a loose 'k, but how does that work when they are in pairs of row-column? EDIT: Never mind, for 3-char it doesn't matter what the column is, but now I have the problem that the strings aren't joined to print. Maybe a different question: how to do a ranged loop without pushing the indexes of the loop? I now have this. Jun 3, 2019 at 12:51

# bash – 74

I know it won't win anything, not even within bash, but thought I'd submit one where 500 or 501 is not in the solution.

seq 0 166|awk '{print"bzzt\n"($1)*3+1"\n"($1)*3+2;}'|sed 's/.*3.*/bzzt/g'


# F# - 86

for i in 0..500 do if(string i).Contains"3"||i%3=0 then printfn"Bzzt"else printfn"%i"i


# C# (110)

Enumerable.Range(1,500).All(x =>{Console.Write((x%3==0)||(x + "").Contains("3")?"Bzzt ":x+" ");return true;});

• You can get a lower byte count by removing unnecessary whitespace. Actually, you don't need any whitespace here. Jun 23, 2014 at 16:31
• What version of C# has an IEnumerable<T>.ForEach method? Jun 23, 2014 at 16:38
• I get an error on Mono's C# compiler: Unexpected symbol: Enumerable' Jun 23, 2014 at 16:54
• (x+"").Contains("3") should work instead of the ToString() to shorten it. Jun 23, 2014 at 19:15
• Made the changes, removed ForEach which was a extension method from a 3rd party library. Jun 23, 2014 at 19:39

## Ruby, 57

Not a competitive Ruby answer, but I like it because it's weird.

i=-1;([b=:bzzt,p,p]*167).map{|a|puts"#{i+=1}"[?3]?b:a||i}


PHP, no seperator - 53

while(++$i<501)echo$i%3<1||strpos(-$i,'3')?'bzzt':$i;


It's a improved version of @Martijn's code. I would like to add comment to his answer, but I haven't enough reputation to do that...

# seq and awk 48 38

seq 500|awk '!($1%3)||/3/{$1="bzzt"}1'

• The awk code part could be $1%3&&!/3/||$1="bzzt". Jul 25, 2015 at 15:30

# Python2 (63)

Corrected version of @idiot.py's code (I don't have enough reputation to comment on his post).

print[n if n%3and'3'not in str(n)else"bzzt"for n in range(501)]


or, equivalently,

print["bzzt"if not n%3or'3'in str(n)else n for n in range(501)]


You have to negate the second condition, otherwise it gives incorrect output.

output:

['bzzt', 1, 2, 'bzzt', 4, 5, 'bzzt', 7, 8, 'bzzt', 10, 11, 'bzzt', 'bzzt', 14, 'bzzt', 16, 17, 'bzzt', 19, 20, 'bzzt', 22, 'bzzt', 'bzzt', 25, 26, 'bzzt', 28, 29, 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 40, 41, 'bzzt', 'bzzt', 44, 'bzzt', 46, 47, 'bzzt', 49, 50, 'bzzt', 52, 'bzzt', 'bzzt', 55, 56, 'bzzt', 58, 59, 'bzzt', 61, 62, 'bzzt', 64, 65, 'bzzt', 67, 68, 'bzzt', 70, 71, 'bzzt', 'bzzt', 74, 'bzzt', 76, 77, 'bzzt', 79, 80, 'bzzt', 82, 'bzzt', 'bzzt', 85, 86, 'bzzt', 88, 89, 'bzzt', 91, 92, 'bzzt', 94, 95, 'bzzt', 97, 98, 'bzzt', 100, 101, 'bzzt', 'bzzt', 104, 'bzzt', 106, 107, 'bzzt', 109, 110, 'bzzt', 112, 'bzzt', 'bzzt', 115, 116, 'bzzt', 118, 119, 'bzzt', 121, 122, 'bzzt', 124, 125, 'bzzt', 127, 128, 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 140, 'bzzt', 142, 'bzzt', 'bzzt', 145, 146, 'bzzt', 148, 149, 'bzzt', 151, 152, 'bzzt', 154, 155, 'bzzt', 157, 158, 'bzzt', 160, 161, 'bzzt', 'bzzt', 164, 'bzzt', 166, 167, 'bzzt', 169, 170, 'bzzt', 172, 'bzzt', 'bzzt', 175, 176, 'bzzt', 178, 179, 'bzzt', 181, 182, 'bzzt', 184, 185, 'bzzt', 187, 188, 'bzzt', 190, 191, 'bzzt', 'bzzt', 194, 'bzzt', 196, 197, 'bzzt', 199, 200, 'bzzt', 202, 'bzzt', 'bzzt', 205, 206, 'bzzt', 208, 209, 'bzzt', 211, 212, 'bzzt', 214, 215, 'bzzt', 217, 218, 'bzzt', 220, 221, 'bzzt', 'bzzt', 224, 'bzzt', 226, 227, 'bzzt', 229, 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 241, 242, 'bzzt', 244, 245, 'bzzt', 247, 248, 'bzzt', 250, 251, 'bzzt', 'bzzt', 254, 'bzzt', 256, 257, 'bzzt', 259, 260, 'bzzt', 262, 'bzzt', 'bzzt', 265, 266, 'bzzt', 268, 269, 'bzzt', 271, 272, 'bzzt', 274, 275, 'bzzt', 277, 278, 'bzzt', 280, 281, 'bzzt', 'bzzt', 284, 'bzzt', 286, 287, 'bzzt', 289, 290, 'bzzt', 292, 'bzzt', 'bzzt', 295, 296, 'bzzt', 298, 299, 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 400, 401, 'bzzt', 'bzzt', 404, 'bzzt', 406, 407, 'bzzt', 409, 410, 'bzzt', 412, 'bzzt', 'bzzt', 415, 416, 'bzzt', 418, 419, 'bzzt', 421, 422, 'bzzt', 424, 425, 'bzzt', 427, 428, 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 'bzzt', 440, 'bzzt', 442, 'bzzt', 'bzzt', 445, 446, 'bzzt', 448, 449, 'bzzt', 451, 452, 'bzzt', 454, 455, 'bzzt', 457, 458, 'bzzt', 460, 461, 'bzzt', 'bzzt', 464, 'bzzt', 466, 467, 'bzzt', 469, 470, 'bzzt', 472, 'bzzt', 'bzzt', 475, 476, 'bzzt', 478, 479, 'bzzt', 481, 482, 'bzzt', 484, 485, 'bzzt', 487, 488, 'bzzt', 490, 491, 'bzzt', 'bzzt', 494, 'bzzt', 496, 497, 'bzzt', 499, 500]


# Tcl 90

set i 0;while {$i < 501} {if {[regexp 3$i]||($i %3)<1} {puts "bzzt"} {puts "$i"};incr i}


I'm sure other tclers could make this more terse, but the interpreter is funny about whitespaces.

# C, 9086 81

i,s;main(){while(++i<501)sprintf(&s,"%d",i),puts(i%3&&!strchr(&s,51)?&s:"bzzt");}


Not a serious contender. This is essentially bacchusbeale's answer golfed better, as I suggested in comments to his answer. As well as those changes, I used strchr as a shorter test than a for loop to see if the string version of the integer contains the character 3 (character 51 in ASCII).

Edit: down to 81 characters incorporating edc65's sick trick from the comments (assumes int is at least 32 bits). Thanks!

• 81: i,s;main(){while(++i<501)sprintf(&s,"%d",i),puts(i%3&&!strchr(&s,51)?&s:"bzzt"); (need a 32 bits int or longer) Jun 25, 2014 at 21:43

## C, 79

i;main(s){while(++i<498+sprintf(&s,"%d",i))puts(i%3*!strchr(&s,51)?&s:"bzzt");}


Improved version of squeamish ossifrage's solution. Now it's the shortest C solution.

# JavaScript, 70

First entry into a code golf

i=0;while(i<501)console.log(((""+i).match(/3/g)||i%3==0)?'bzzt':i),i++


Can probably be made shorter, but this one doesn't do alert and make you want to kill the developer.

• Count up to 500, for better than while as you use i=0;i++;,alert(once!) better than console.log, test better than match,invert condition to avoid ==0, then you have [codegolf.stackexchange.com/questions/32267/… Jun 26, 2014 at 15:54

# C#

Code can be executed directly in LinQPad.

## Number is divisible by 3 (49)

var i=0;while(i++<501)(i%3<1?"bzzt":i+"").Dump();


This one should count as the question stated:

The number only has to satisfy at least 1 of the following requirements.

## Number contains 3 (64)

var i=0;while(i++<501)((i+"").Contains("3")?"bzzt":i+"").Dump();


## Both conditions apply (70)

var i=0;while(i++<501)(i%3<1|(i+"").Contains("3")?"buzz":i+"").Dump();


## Delphi - 119

repeat ShowMessage(IfThen((Tag mod 3=0) or (pos('3',IntToStr(Tag))>0),'bzzt',IntToStr(Tag))); Tag:=Tag+1 until Tag=500


Uses the fact that "self" is assumed so the form's Tag property can be accessed directly without actually declaring a variable. I had critiqued another Delphi answer here a few minutes ago so I thought it was only fair I posted one too.

# Delpi - 128

procedure x;var I: integer;begin for I:=0to 500do writeln(iif((IntToStr(i).Contains('3')or(i mod 3=0)),'bzzt',inttostr(i)));end;

• That won't compile. You never declared i. Also there's no iif in Delphi. You must use IfThen().
– Paul
Jun 27, 2014 at 21:37
• I don't understand why I didnt declare I, maybe I made it a global variable which shouldnt work either.. I'll correct it. But there is an iif in delphi, its in the unit IdGlobal. Jun 30, 2014 at 7:16

# Java, 127124 95 bytes

class B{static{for(int i=0;i++<500;System.out.println(i%3<1||(i+"").contains("3")?"Bzzt":i));}}


It throws an exception at the beginning but it works, at least for me (Java version 1.7.0_60-b19 on Mac OS X 10.8.5).

• That trick doesn't work in Java 7 or later. It looks for a main method before executing static initializers. Jun 23, 2014 at 21:48
• @tbodt I have Java 7 and it works Jun 23, 2014 at 21:51
• Is there any way you can get a modern toolset to compile and run this, please add it to the answer? (I use 1.7.0_55 on linux) If you need a specific version you need to mention it in your answer. Jun 23, 2014 at 21:52
• I tested it on Java 6 (works then throws exception) and Java 8 (doesn't work). Jun 23, 2014 at 22:09

# LiveScript (54)

for n to 500=>alert if n%3==0||/3/test n=>\bzzt else n


# JAGL V1.0 - 35

1 501r{ddg@3%nSg51e|"bzzt"SnI32+p}/


Probably could have done better. May try to revise it.

Explaination:

1 501r                                     Make a range of 1 to 500 inclusive
{ddg                                 Duplicate twice and convert top to string
@3%n                             Rotate top 3 on stack, and push not(top mod 3)
Sg                           Swap top two values, and convert top to string
51e                        Check if 51 (3) is in the string rep
|"bzzt"                 Take the logical "or", and push "bzzt"
SnI              Swap, reverse, and if not divisible, drop "bzzt"
32+p          Add a space, and print
}/        End block, and map over range


## Java, 139132128 127 bytes

class A{public static void main(String[] a){for(int i=0;i++<500;)System.out.print(i%3==0|(i+"").contains("3")?"bzzt ":i+" ");}}

• You can shave off a couple of characters by removing the {} braces on your for loop and doing a i++<501;) check. Jul 2, 2015 at 14:14
• Sure @NathanMerrill....shortened 'args' variable name as well Jul 2, 2015 at 14:28
• Too many parenthesis: both % and == have higher precedence than || and all of them have higher precendece than ?:: System.out.print(i%3==0||(i+"").contains("3")?"bzzt ":i+" "). Jul 2, 2015 at 14:50
• well said @manatwork I never cared/thought about using precedence before. Will keep in mind for code-golf Jul 2, 2015 at 15:13
• As now it became too similar to Jop V.'s Java answer, here is another trick: use bitwise | instead of logical ||. In this case gives the same result. Jul 2, 2015 at 15:21

# Mouse-2002, 34 bytes

I'm proud that lil' ol' Mouse, first implemented in 1979 and the most recent iteration updated Aug '07 (excluding the one by me), can do so well! :')

(x.501<^x.d\[x.!J!'|"bzzt!"]x.1+x:)


Explained:

(             ~ while true
x. 501 < ^  ~ push x; push 1 if less than 501 and break if 0
x. d \      ~ push x mod 3
[           ~ if 0; then
x. ! J!'  ~ push x and print, push the ascii code 9 and print
|           ~ else
"bzzt!"   ~ print this string
]           ~ endif
x. b+ x:    ~ push x, increment by 1
)             ~ endwhile
$~ (implicit for scripts) end prog  Sample run: $ mouse bzzt.mou | head
bzzt
1       2       bzzt
4       5       bzzt
7       8       bzzt
10      11      bzzt
13      14      bzzt
16      17      bzzt
19      20      bzzt
22      23      bzzt
25      26      bzzt


Tab chars because I say so.

My reboot of this language's legacy can do this exact thing in 27 bytes:

(x501<^xd\[x.|"bzzt".]1!x+)


And I'm not even aiming for it to be a golfing lang, just a small syntax! No interpreter yet.

# jq: 59 bytes

(56 characters code + 3 characters command line option.)

range(501)|select(.%3<1or("\(.)"|test("3"))|not)//"bzzt"


Sample run:

bash-4.3\$ jq -n -r 'range(501)|select(.%3<1or("\(.)"|test("3"))|not)//"bzzt"' | head
bzzt
1
2
bzzt
4
5
bzzt
7
8
bzzt


On-line test (Passing -r through URL is not supported – check Raw Output yourself.)