Obfuscated FizzBuzz Golf

50

15

Create the shortest possible obfuscated FizzBuzz implementation.

To be considered obfuscated, it should satisfy at least one of the following:

  1. Does not contain any of the words "Fizz", "Buzz", or "FizzBuzz"
  2. Does not contain the numbers 3, 5, or 15.
  3. Use any of the above in a misleading way.

Remember: The goal is to be short and hard to follow.

The code sample which inspired this question follows:

public class Default
{
        enum FizzBuzz
        {
            Buzz = 1,
            Fizz,
            FizzBuzz
        }
        public static void Main(string[] args)
        {
            byte[] foo = 
              Convert.FromBase64String("IAmGMEiCIQySYAiDJBjCIAmGMEiCIQySYA==");
            MemoryStream ms = new MemoryStream(foo);
            byte[] myByte = new byte[1];
            do
            {
                FizzBuzz fb;
                ms.Read(myByte, 0, 1);
                for (int i = 0; i < 4; i++)
                {
                    fb = (FizzBuzz)(myByte[0] >> (2 * i) 
                         & (int)FizzBuzz.FizzBuzz);
                    Console.Out.WriteLine( (((int)fb > 0) ? "" + fb : "" 
                         + ((ms.Position - 1) * 4 + i + 1)));
                }
            } while (ms.Position < ms.Length);
        }
}

mootinator

Posted 2011-01-28T01:55:09.600

Reputation: 1 171

Question was closed 2016-09-20T21:49:25.003

How do you know the bound ? In your solution you have ms.Length but in some solutions there is no such bound... – Labo – 2015-11-10T09:20:29.683

I'm voting to close this question as off-topic because it does not have an objective validity criterion, making it impossible to indisputably decide whether a submission follows the rules. – Dennis – 2016-09-20T21:49:25.003

Answers

25

GolfScript, 75 69 65 60 59 chars

100,{)6,{.(&},{1$1$%{;}{4*35+6875*25base{90\-}%}if}%\or}%n*

So, you'd think GolfScript by itself is already obfuscated, right? Well, just to follow the spec, I decided to have the program not contain "fizz", "buzz", nor the numbers 3, 5, nor 15. :-)

Yes, there are some numbers with multiples of 5, like 25, 35, 90, 100, and 6875. Are they red herrings? You decide. ;-)

Chris Jester-Young

Posted 2011-01-28T01:55:09.600

Reputation: 4 464

The numbers 3 and 5 appear in your code, so it is not correct !!! – Labo – 2015-11-10T19:12:52.220

@Labo Only one of the criteria needs to be satisfied, not all three. Read the question again. :-) – Chris Jester-Young – 2015-11-10T20:39:28.540

Is it a JOKE ? I spend several hours on that ! Though I still managed to have a 58 chars long Python code :p http://codegolf.stackexchange.com/a/63543/47040

– Labo – 2015-11-10T20:57:47.610

Without this constraint, I achieve 54 chars :D – Labo – 2015-11-11T00:35:33.330

@Labo Okay, I upvoted your answer. I'm not the OP so I can't change the accepted answer. Lol. – Chris Jester-Young – 2015-11-11T00:59:53.223

Lol no problem, thanks :) – Labo – 2015-11-11T01:01:14.793

4@Labo: I can see the digits 3 and 5 but not the numbers 3 and 5. – David Ongaro – 2016-01-27T00:30:13.283

2

Although I've written commentary for all my other GolfScript submissions, none will be forthcoming for this one. Rationale: http://chat.stackexchange.com/transcript/message/436819#436819 :-D

– Chris Jester-Young – 2011-01-31T04:34:00.493

65

Javascript 97 chars - no numbers at all

Numbers ? Who needs number when you have Javascript !

a=b=!![]+![],a--,c=b+b;while(++a)e=!(a%(c+c+b)),alert(!(a%(c+b))?e?"FizzBuzz":"Fizz":e?"Buzz":a);

Note: There is an infinite loop that will alert you the sequence.

Bonus (666 chars)

  • No number
  • No letter (only zfor has been use in the whole script)

.

_=$=+!![];$__=((_$={})+'')[_+$+_+$+_];__$=((![])+'')[$];_$_=((_$={})+'')
[_+$+_+$+_+$];____=[][$__+((_$={})+'')[$]+(($)/(![])+'')[$]+$__+__$+_$_];$__$=(!![]+"")
[$+$+$]+([][(![]+"")[$+$+$]+(+[]+{})[$+$]+(!![]+"")[$]+(!![]+"")[+[]]]+"")[($+$)+""+
($+$+$)]+(![]+"")[$]+(![]+"")[$+$];$_$_=____()[$-$][$__$]("\"\\"+($)+($+$+$+$+$+$+$)+
($+$)+"\"");_$=(![]+'')[$-$]+([][[]]+[])[$+$+$+$+$]+$_$_+$_$_;$_=(_+{})[$+$+$]+(!![]+'')
[_+$]+$_$_+$_$_;_--,$$=$+$;____()[$-$][$__$]((![]+"")[+[]]+(+[]+{})[$+$]+(!![]+"")[$]+
"(;++_;)$$$=!(_%("+($$+$$+$)+")),____()[+[]][__$+((![])+'')["+($+$)+"]+((!![])+'')["+
($+$+$)+"]+((!![])+'')[+!![]]+_$_](!(_%("+($$+$)+"))?$$$?_$+$_:_$:$$$?$_:_);");

HoLyVieR

Posted 2011-01-28T01:55:09.600

Reputation: 839

I really want to understand how the second one works. – Steve Robbins – 2012-08-07T20:39:25.400

3@stevether It's mostly about abusing type conversion (ex.: +!![] is the same as 1 and ({}+"")[5] is the same as c) and abusing array notation to access method (ex.: window['eval']( is the same eval( ). – HoLyVieR – 2012-08-19T05:06:21.810

1Appropriate character count if I've ever seen one. – captncraig – 2013-01-03T20:31:10.293

1congratulations for having the 100th post on codegolf.SE! (http://codegolf.stackexchange.com/q/100) – Doorknob – 2013-09-23T21:54:54.050

@Nabb I took the eval trick even further now. – HoLyVieR – 2011-03-10T19:50:54.533

18Real programmers code just like the second one. – None – 2011-01-29T11:57:50.943

1The second one doesn't run on latest version of Firefox (Type error, can't convert undefined to object). Would you please check it? – n̴̖̋h̷͉̃a̷̭̿h̸̡̅ẗ̵̨́d̷̰̀ĥ̷̳ – 2014-02-15T18:08:49.600

1@nhahtdh Looks like they changed the behavior of "Array.prototype.concat" and the new behavior is actually how it's suppose to be implemented. The code used a common misbehavior of browser, where "Array.prototype.concat()[0]" returns "window". – HoLyVieR – 2014-02-15T22:51:38.033

9@M28: Yep. That's one way to build job security...'cause finding someone who can maintain this code wouldn't be the easiest thing. – Andy – 2011-02-02T05:07:07.703

I think the first version violates rule number 1? – flawr – 2014-06-17T07:57:48.593

1

You can use window"eval" for z.

– Nabb – 2011-02-03T02:25:27.117

1looks like BrainF*ck on Java(Script) – Malachi – 2014-07-18T13:18:00.297

26

Python - 78 chars

i=0
while 1:i+=1;print"".join("BzuzzizF"[::2*j]for j in(-1,1)if 1>i%(4+j))or i

gnibbler

Posted 2011-01-28T01:55:09.600

Reputation: 14 170

1Took me 10 minutes to understand what you did there. Nice and twisted. – Trufa – 2011-06-08T15:04:26.617

23

PostScript, 96 bytes

So obfuscated it looks like random garbage.

1<~0o0@eOuP7\C+tf6HS7j&H?t`<0f>,/0TnSG01KZ%H9ub#H@9L>I=%,:23M].P!+.F6?RU#I;*;AP#XYnP"5~>cvx exec

Usage: $ gs -q -dNODISPLAY -dNOPROMPT file.ps

KirarinSnow

Posted 2011-01-28T01:55:09.600

Reputation: 801

5

I bet that passes diehard.

– kaoD – 2013-02-06T04:27:35.607

23

C++: 886 chars

I've tried to hide the 'fizz' and the 'buzz'. Can you spot them?

#include <iostream>
#define d(a,b) a b
#define _(a,b) d(#b,#a)
#define b(b) _(b,b)
#define y _(i,f)c
#define x _(u,b)c
#define c b(z)
#define i int
#define p main
#define s char
#define q 810092048
#define h for
#define m 48
#define a ++
#define e ==
#define g 58
#define n 49
#define l <<
#define oe std::cout<<
#define v '\n'

int  p   (i,  s*t     ){i   j  =   q;h   (*(
i    *     )    t     =  m  ;  2     [     t
]?   0    :    1      ??(   t  ]    ?     a
1    [   t    ]       e  g  ?  1   [     t
]    =   48,  ++0     ??(    t]e   g?0   ??(

t]=  n   ,1[  t]=
2    [     t    ]
=m   :    1    :
1    :   a    0
[    t   ??)  ==g

?0[   t  ]   =49   ,1[
t  ]  =  m     :     1
;j=   (  j    /     4
)  |  (  (   j     &
3)l    28)   )oe   (j&

3?j  &   1?j  &2?
y    x     :    y
:x   :    t    )
l    v   ;    }
i    f   =m&  ~g;

Skizz

Posted 2011-01-28T01:55:09.600

Reputation: 2 225

5That confuzzlesing my brane. – Mateen Ulhaq – 2011-03-11T03:17:21.877

2I think you meant membrane – Korvin Szanto – 2011-12-14T22:34:44.473

16

DC (256 255 bytes)

Here it is, I tried (rather successfully, if I may say so myself) to hide anything except for letters, and +-[];:= (which are vital and impossible to obfuscate). It does segfault after getting to about 8482 or so on my machine, but that is to do with stack issues related to the way the recursion is implemented. The solution itself is correct. 255 bytes if you remove the whitespace (included for ease of reading) Enjoy:

Izzzdsa+dsbrsc+dsdd+sozdsezzsm+s
nloddd++splbz++ddaso+dln-dstsqlm
d+-sr[PPPP]ss[IP]su[lpdlqlrlsxlu
x]lm:f[lpdltdI+lm+rlblm+-lsxlux]
ln:f[[]sulm;fxln;f[IP]sux]la:f[;
fsk]sg[lmlgx]sh[lnlgx]si[lalgx]s
j[lc[lcp]sklerldlolclerlblolcler
lalox=hx=ix=jlkxclcz+scllx]dslx

Hiato

Posted 2011-01-28T01:55:09.600

Reputation: 731

+1 for dc. Even unobfuscated, of course, it's not especially readable.

– Jesse Millikan – 2011-03-14T21:29:09.880

12

This was a bit tricky to embed using the indentation so a gist:

Ruby, 4312 chars

https://gist.github.com/dzucconi/1f88a6dffa2f145f370f

eval("                                                 

















































                                                                                                                             ".split(/\n/).map(&:size).pack("C*"))

dzuc

Posted 2011-01-28T01:55:09.600

Reputation: 221

How does this even work? – addison – 2016-06-22T14:57:17.550

._. How this thing work... – TuxCrafting – 2016-06-26T10:15:18.147

"string of your program".unpack("C*").map { |x| " " * x }.join("\n") – dzuc – 2016-06-26T14:14:59.243

11

Brainfuck - 626 656

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

Goes from 1 to 255

captncraig

Posted 2011-01-28T01:55:09.600

Reputation: 4 373

9"waste" is a strong word ... – Claudiu – 2014-08-22T23:00:42.570

1Turns out this actually does BuzzFizz. It gets FizzBuzz correct for %15, but it swaps %3 and %5. I may try to fix it, but for now my brain is officially F'ed – captncraig – 2011-12-15T23:32:36.930

2Fixed at cost of 30. Could have golfed it more with effort, but I have already wasted enough time on this. – captncraig – 2011-12-15T23:36:43.540

10

Brainfuck, 708 characters

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

Description of how it works is available in my Code Review question

Simon Forsberg

Posted 2011-01-28T01:55:09.600

Reputation: 198

9

Haskell - 147 142 138 characters

fi=zz.bu
bu=zz.(:).(++"zz")
[]#zz=zz;zz#__=zz
zZ%zz=zZ zz$zZ%zz
zz=(([[],[]]++).)
z=zipWith3(((#).).(++))(bu%"Fi")(fi%"Bu")$map show[1..]

The code is 19 characters longer than it needs to be, but I thought the aesthetics were worth it! I believe all three "objectives" are satisfied.

> take 20 z
["1","2","Fizz","4","Buzz","Fizz","7","8","Fizz","Buzz","11","Fizz","13","14",
"FizzBuzz","16","17","Fizz","19","Buzz"]

MtnViewMark

Posted 2011-01-28T01:55:09.600

Reputation: 4 779

Hi, I try to get to understand your code, but I am unable to run it! The function zZ' is applied to six arguments, but its type(a0 -> b0 -> c0) -> [a0] -> [b0] -> [c0]' has only three – RobAu – 2013-01-31T20:07:32.470

And I, while being able to run it, only get ["1","2","3","4","5","6"...]. – Artyom – 2013-02-01T18:14:53.023

Fixed - The right version was still on my disk... must have mis-pasted the text long ago! – MtnViewMark – 2013-02-05T04:52:03.477

please specify how can it be shortened by 19 characters, or show the code. i am curious and i have no idea – proud haskeller – 2014-07-18T15:31:00.080

There are 19 occurrances of 2 letter names: bu, fi, zz, and zZ. These could be shortened to one letter names. – MtnViewMark – 2014-07-19T01:17:55.453

8

, 33 chars / 92 bytes (noncompetitive)

ѨŃ(1,ṥ)ć⇀ᵖɘƃ႖סР깜 #ē($%3⅋4,$%5?4:8)⋎$⸩

Try it here (Firefox only).

This language is way too OP for restricted source challenges.

Mama Fun Roll

Posted 2011-01-28T01:55:09.600

Reputation: 7 234

6

Javascript, 469 bytes

This was probably the most fun I've ever had.

z=0;_=(function(){b=0;window[0xA95ED.toString(36)]((function(){yay="&F bottles of beer on the wall, &F bottles of beer. Take one down, pass it around, &z Bottles of beer on the wall.";return atob("eisrOyAg") + "console.log(((function(y){if((y%0xf)==0){return [1,72,84,84,86,78,84,84]}else if(y%0b11==0){return [1,72,84,84]}else if(y%0b101==0){return [86,78,84,84]}else{b=1;return [y]}})(z).map(function(x){return b==0?yay[x]:x}) ).join(''))"})())});setInterval(_,1000);

Try it here

anOKsquirrel

Posted 2011-01-28T01:55:09.600

Reputation: 361

Dang, I just realized that the objective was to be short and hard to follow... Sorry :P – anOKsquirrel – 2015-11-02T23:55:12.510

+1 Might have missed on the shortness, but at least you didn't have fizz buzz in there – MickyT – 2015-11-03T00:00:46.863

4

Perl 6 (52 bytes)

say "Fizz"x$_%%(2+1)~"Buzz"x$_%%(4+1)||$_ for 1..100

Let me put an explanation here. It's the worst rule abuse I've done in such task. I know what you are saying - there is obvious Fizz and Buzz here. But let's take a look at the rules.

To be considered obfuscated, it should satisfy at least one of the following:

This avoids 3, 5 and 15. Therefore, it's valid and really short solution.

Konrad Borowski

Posted 2011-01-28T01:55:09.600

Reputation: 11 185

4

Ruby - 165 characters

(1..100).each{|i|i%0xF==0? puts(["46697A7A42757A7A"].pack("H*")):i%(0xD-0xA)==0? puts(["46697A7A"].pack("H*")):i%(0xF-0xA)==0? puts(["42757A7A"].pack("H*")):puts(i)}

This was my first attempt at code golf. I had a lot of fun. =)

Splendor

Posted 2011-01-28T01:55:09.600

Reputation: 141

3

C (237 209 characters)

#include<stdlib.h>
#define e printf  
a=50358598,b=83916098,c=1862302330;_(m,n){return(m%((c&n)>>24))
||!(e(&n)|e(&c));}main(_);(*__[])(_)={main,exit};main(i){_(i,a)
&_(i,b)&&e("%i",i);e("\n");__[i>=100](++i);}

Though I'm not sure this conforms to the C standard :)
It works, though. On Linux using GCC, that is.

marinus

Posted 2011-01-28T01:55:09.600

Reputation: 30 224

3

Scala, 295 characters

object F extends Application{var(f,i,z)=("",('z'/'z'),"FBiuzzzz");while(i<(-'b'+'u'+'z'/'z')*('¥'/'!')){if(i%(-'f'+'i'/('z'/'z'))==0)f+=z.sliding(1,2).mkString;if(i%((-'b'+'u'+'z'/'z')/('f'/'f'+'i'/'i'+'z'/'z'+'z'/'z'))==0)f+=z.drop(1).sliding(1,2).mkString;if(f=="")f+=i;println(f);i+=1;f="";}}

Gareth

Posted 2011-01-28T01:55:09.600

Reputation: 11 678

3

Python 3 - 338

import sys
def fibu():
        (F,I,B,U),i,u,z=sys._getframe(0).f_code.co_name,0xf,0xb,lambda x,y:x%((i//u)+(i^u))==u>>i if y>u else x%(((u<<(u>>2))&i)>>(u>>2))==i>>u
        A,RP = "",chr(ord(U)+((i//u)+(i^u)))*2
        for x in range(100):print(x if not (z(x,u)or z(x,i))else A.join((F+I+RP if z(x,u)else A,B+U+RP if z(x,i)else A)))
fibu()

This is my first golf. Not the shortest, but it's pretty ugly! None of the forbidden numbers or string literals. Firp, Burp!

LexyStardust

Posted 2011-01-28T01:55:09.600

Reputation: 131

3

Python - 157

from itertools import cycle as r
c=str.replace
[c(c(c(z+y,'x','fix'),'y','bux'),'x','zz').strip() or x for z,y,x in zip(r('  y'),r('    x'),range(1,101))]

Not quite the shortest, but I hope the reader will appreciate the pure functional style and extensibility to arbitrarily long counts.

Marcin

Posted 2011-01-28T01:55:09.600

Reputation: 131

3

Python 2 - 54 chars

i=0
while 1:i+=1;print'FizzBuzz'[i%~2&4:12&8+i%~4]or i

Python 3 - 56 chars

i=0
while 1:i+=1;print('FizzBuzz'[i%~2&4:12&8+i%~4]or i)

If you do not want 'FizzBuzz' to appear :

Python 2 - 58 chars

i=0
while 1:i+=1;print' zzuBzziF'[12&8+i%~2:i%~4&4:-1]or i

Python 3 - 60 chars

i=0
while 1:i+=1;print(' zzuBzziF'[12&8+i%~2:i%~4&4:-1]or i)

Or how to beat GolfScript with Python ;)

Labo

Posted 2011-01-28T01:55:09.600

Reputation: 229

The first two seem to do nothing, since i=0 means the while loop is never entered. – xnor – 2015-11-10T21:45:03.777

Lol I used my test version, in which the condition is i<20. – Labo – 2015-11-10T21:53:44.197

But now it works :) – Labo – 2015-11-10T22:39:49.797

Shouldn't it stop at 100 according to the original FizzBuzz problem? – David Ongaro – 2016-01-27T00:33:52.163

3

K, 155

{m:{x-y*x div y};s:{"c"$(10-!#x)+"i"$x};$[&/0=m[x]'(2+"I"$"c"$49;4+"I"$"c"$49);s"<`rs<pvw";0=m[x;2+"I"$"c"$49];s"<`rs";0=m[x;4+"I"$"c"$49];s"8lrs";x]}'!100

I could golf it quite a bit but I'd rather it be more obfuscated.

tmartin

Posted 2011-01-28T01:55:09.600

Reputation: 3 917

2

C# - 218 characters

using System;class D{static void Main(){int l,i,O=1;l++;string c="zz",a="fi",b="bu";l++;l++;i=l;i++;i++;for(;O<101;O++)Console.WriteLine(((O%l)>0&&1>(O%i))?a+c:(1>(O%l)&&(O%i)>0)?b+c:(1>(O%l)&&1>(O%i))?a+c+b+c:O+"");}}

Could be shortened if I introduced other numbers like so: (210 characters total)

using System;class D{static void Main(){int l=1,i,O=1;string c="zz",a="fi",b="bu";l+=2;i=l;i+=2;for(;O<101;O++)Console.WriteLine(((O%l)>0&&1>(O%i))?a+c:(1>(O%l)&&(O%i)>0)?b+c:(1>(O%l)&&1>(O%i))?a+c+b+c:O+"");}}

Decided to remove the obvious word fizz and buzz and go for slightly more obfuscation. Second one is shorter than the first one but is slightly more direct on what's occurring in the addition.

jcolebrand

Posted 2011-01-28T01:55:09.600

Reputation: 1 683

2

Python, 1 line, 376 characters

pep8-E501 ignored. Only works in python3.

print(*((lambda x=x: ''.join(chr(c) for c in (102, 105)) + (2 * chr(122)) + ''.join(chr(c) for c in (98, 117)) + (2 * chr(122)) + '\n' if x % (30 >> 1) == 0 else ''.join(chr(c) for c in (102, 105)) + (2 * chr(122)) + '\n' if x % (6 >> 1) == 0 else ''.join(chr(c) for c in (98, 117)) + (2 * chr(122)) + '\n' if x % (10 >> 1) == 0 else str(x) + '\n')() for x in range(1, 101)))

David Watson

Posted 2011-01-28T01:55:09.600

Reputation: 121

2

Alternative Ruby (126 characters)

(1..100).map{|i|(x="\xF\3\5\1Rml6ekJ1eno=".unpack('C4m'))[-1]=~/(.*)(B.*)/
[*$~,i].zip(x).map{|o,d|i%d>0||(break $><<o<<?\n)}}

Short and obscure, just how we like it. The 3 and the 5 are actually in there but not as integer literals so I think that still counts.

Note that the this is the shortest Ruby version without literal 'Fizz', 'Buzz', 'FizzBuzz' on here.

Arne Brasseur

Posted 2011-01-28T01:55:09.600

Reputation: 131

2

This isn't exactly golfed, its about 120 lines.

I thought I'd do something that took advantage of all the fun potential for undefined behavior with C++ memory management.

#include <iostream>
#include <string>

using namespace std;

class Weh;
class HelloWorld;

class Weh
{
public:

    string value1;
    string value2;
    void (*method)(void * obj);

    Weh();

    string getV1();

    static void doNothing(void * obj);
};

class HelloWorld
{
public:
    static const int FOO = 1;
    static const int BAR = 2;
    static const int BAZ = 4;
    static const int WUG = 8;

    string hello;
    string world;
    void (*doHello)(HelloWorld * obj);

    HelloWorld();

    void * operator new(size_t size);

    void tower(int i);
    const char * doTower(int i, int j, int k);

    static void doHe1lo(HelloWorld * obj);
};

Weh::Weh()
{
    method = &doNothing;
}

void Weh::doNothing(void * obj)
{
    string s = ((Weh *) obj)->getV1();
    ((HelloWorld *) obj)->tower(1);
}

string Weh::getV1()
{
    value1[0] += 'h' - 'j' - 32;
    value1[1] += 'k' - 'g';
    value1[2] += 'u' - 'g';
    value1[3] = value1[2];
    value2 = value1 = value1.substr(0, 4);

    value2[0] += 'd' - 'h';
    value2[1] += 'w' - 'k';
    value2[2] = value1[2];
    value2[3] = value1[3];

    return "hello";
}

void * HelloWorld::operator new(size_t size)
{
    return (void *) new Weh;
}

HelloWorld::HelloWorld()
{
    hello = "hello";
    world = "world";
}

void HelloWorld::doHe1lo(HelloWorld * obj)
{
    cout << obj->hello << " " << obj->world << "!" << endl;
}

void HelloWorld::tower(int i)
{
    doTower(0, 0, i);
    tower(i + (FOO | BAR | BAZ | WUG));
}

const char * HelloWorld::doTower(int i, int j, int k)
{
    static const char * NOTHING = "";
    int hello = BAR;
    int world = BAZ;
    int helloworld = FOO | BAR | BAZ | WUG;

    if ((hello & i) && (world & j))
        cout << this->hello << this->world << endl;
    else if (hello & i)
    {
        cout << this->hello << endl;
        cout << doTower(0, j + 1, k + 1);
    }
    else if (world & j)
    {
        cout << this->world << endl;
        cout << doTower(i + 1, 0, k + 1);
    }
    else
    {
        cout << k << endl;
        cout << doTower(i + 1, j + 1, k + 1);
    }

    return NOTHING;
}

int main()
{
    HelloWorld * h = new HelloWorld;
    h->doHello(h);
}

Wug

Posted 2011-01-28T01:55:09.600

Reputation: 1 607

2

Ruby - 89 chars

puts (0..99).map{|i|srand(1781773465)if(i%15==0);[i+1,"Fizz","Buzz","FizzBuzz"][rand(4)]}

I can't take credit for this piece of brilliance, but I couldn't leave this question without my favorite obfuscated implementation :)

The implementation above was written by David Brady and is from the fizzbuzz ruby gem. Here is the explanation from the source code:

Uses the fact that seed 1781773465 in Ruby's rand will generate the 15-digit sequence that repeats in the FizzBuzz progression. The premise here is that we want to cleverly trick rand into delivering a predictable sequence. (It is interesting to note that we don't actually gain a reduction in information size. The 15-digit sequence can be encoded as bit pairs and stored in a 30-bit number. Since 1781773465 requires 31 bits of storage, our cleverness has actually cost us a bit of storage efficiency. BUT THAT'S NOT THE POINT!

Ruby - 87 chars

puts (0..99).map{|i|srand(46308667)if(i%15==0);["FizzBuzz","Buzz",i+1,"Fizz"][rand(4)]}

Here's a different version which uses a shorter seed but the lookup table is in a different order. Here is the explanation from the source code:

The first implementation (89 chars) adheres to the specific ordering of 0=int, 1=Fizz, 2=Buzz, 3=FizzBuzz. It may be possible to find a smaller key if the ordering is changed. There are 24 possible permutations. If we assume that the permutations are evenly distributed throughout 2*31 space, and about a 50% probability that this one is "about halfway through", then we can assume with a decent confidence (say 20-50%) that there is a key somewhere around 1.4e+9 (below 2*28). It's not much gain but it DOES demonstrate leveraging rand's predefined sequence to "hide" 30 bits of information in less that 30 bits of space.

Result: The permutation [3,2,0,1] appears at seed 46308667, which can be stored in 26 bits.

Jacob Stanley

Posted 2011-01-28T01:55:09.600

Reputation: 129

2very sweet, but does contain a literal "Fizz", "Buzz" etc so not valid according to the rules – Arne Brasseur – 2014-06-28T20:05:59.853

2

JavaScript 111 chars - no key numbers

a=b=c=0;while(a++<99)document.write((b>1?(b=0,"Fizz"):(b++,""))+(c==4?(c=0,"Buzz"):(c++,""))+(b*c?a:"")+"<br>")

aaaaaaaaaaaa

Posted 2011-01-28T01:55:09.600

Reputation: 4 365

1

Squeak (4.4) Smalltalk 206 bytes

|f i zz b u z|z:=''.b:=28r1J8D0LK. 1to:100do:[:o|0<(f:=(i:=(zz:=b\\4)//2*4)+(u:=zz\\2*4))or:[z:=z,o].b:=zz<<28+(b//4).z:=z,((z first:f)replaceFrom:1to:f with:28r1A041FHQIC7EJI>>(4-i*u*2)startingAt:1),'
'].z

Or same algorithm with less explicit messages, same number of characters

|l f i zz b u z|z:=#[].b:=36rDEB30W. 1to:100do:[:o|0<(f:=(i:=(zz:=b\\4)//2)+(u:=zz\\2)*4)or:[z:=z,('',o)].b:=zz<<28+(b//4).l:=36r2JUQE92ONA>>(1-u*i*24).1to:f do:[:k|z:=z,{l-((l:=l>>6)-1<<6)}].z:=z,'
'].'',z

My apologizes to Alan Kay for what I did to Smalltalk.
Some of these hacks are portable across Smalltalk dialects, some would require a Squeak compatibility layer...

Note that if you execute in a Workspace, you can omit declarations |f i zz b u z| and gain 14 characters.

If we can afford 357 characters (315 with single letter vars), then it's better to avoid trivial #to:do: loop:

|fizz buzz if f fi zz b u bu z|f:=fizz:=buzz:=0.z:=#[].b:=814090528.if:=[:i|i=0or:[fi:=28.zz:=27<<7+i.u:=26.(fizz:=[zz=0or:[z:=z,{(u:=u//2)\\2+1+(zz+((fi:=fi//2)\\2+2-(zz:=zz//8)*8)*4)}.fizz value]])value]].(buzz:=[(f:=f+1)>100or:[(fi:=(zz:=b\\4)//2*17)+(bu:=zz\\2*40)>0or:[z:=z,('',f)].b:=zz<<28+(b//4).if value:fi;value:bu.z:=z,'
'.buzz value]])value.'',z

aka.nice

Posted 2011-01-28T01:55:09.600

Reputation: 411

1

Haskell 226 bytes, including the whitespace for layout ;)

z=[fI$ (++)            \ 
(fi zz 1 "Fi" )        \  
(fi zz 2 "Bu" )        \ 
:[show zz]  | zz<-[1..]]
fI (zZ:zz)  | zZ==[]   \
= concat zz | 1==1=zZ  
fi zZ bu zz | zZ%bu=   \
(zz++"zz")  | 1==1=[] 
bu%zz=mod bu (zz*2+1)==0

The 'real' code is 160 bytes and can be compressed, but loses fizz-buzz-ness then.

Run it (for nice output):

putStrLn (unwords (take 20 z ))

Output:

1 2 Fizz 4 Buzz Fizz 7 8 Fizz Buzz 11 Fizz 13 14 FizzBuzz 16 17 Fizz 19 Buzz 

RobAu

Posted 2011-01-28T01:55:09.600

Reputation: 641

0

Python, 180 chars

from random import*;seed(0);_,y=''.join(sample('zzzzuiFB',8)),5**2<<2
print chr(int(`y`[::2])).join(filter(bool,map(lambda z:(not z%(y>>5))*_[:1<<2:]+(1>z%5)*_[1<<2::],range(1,y))))

Daniel Lubarov

Posted 2011-01-28T01:55:09.600

Reputation: 301

0

C Solution

    #include  <stdio.h>
    #define a b b b b b
    #define b c c c c c
    #define c d d d d d
   #define  h(k,s)k?s:""
  #define x !(n%('0'>>4))
  #define y !(n%('P'>>4))
  #define d if(100>=++n)\
printf("%s%s%s",h(x,"Fizz"\
),h(y,"Buzz"),h(x|y,"\n"));
int main(){static int n; a}

Daniel Lubarov

Posted 2011-01-28T01:55:09.600

Reputation: 301

0

Perl

use MIME::Base64;print map{map{(++$i,'Fizz','Buzz','FizzBuzz')[$_]."\n"}(3&ord,3&ord>>2,3&ord>>4,3&ord>>6)}split//,decode_base64"EAZJMIRBEgxhkARDGCTBEAZJMIRBEgxhkA"

One I made in 2009. It's pretty easy to figure out, though.

Edit: Darn, it uses "Fizz" and "Buzz!" :( I thought I changed that. Nevermind then.

Ry-

Posted 2011-01-28T01:55:09.600

Reputation: 5 283

0

K

-1',/'$(+`fizz`buzz,,1+!100)@'&:'+(a,,~|/[a:100#'(001b;00001b)])

mollmerx

Posted 2011-01-28T01:55:09.600

Reputation: 229

0

Here's a fun one in C. It can be shorter, but then it won't be as much fun!

#define i(bu,zz,fi) printf("?"[1]|""[0]?"fi":"%"#zz"\n",bu)
#define x(a,i,p) {-0xf##i&(3+5-1),{z(z,a,p,),""}}
#define z(b,f,z,l) #z#f""#b""#l#b
main(b){struct{char z;char*u[2];}u[]={x(i,d,f),x(u,b,b),{0}};for(b=1;b<'e';b++)
{int z=0;char f[3+5+1]={0};for(b;u[z].z;z++)strcat(f,u[z].u[!!(b%u[z].z)]);*f&&
i(f,s,3)||i(b,d,5);}}

cmchick

Posted 2011-01-28T01:55:09.600

Reputation: 1

0

JavaScript, 80 characters

for(i=0;i<100;a=++i%3,b=i%5,alert(i+"FizzBuzz".slice(t=a&&!b*4,t?8:!a*4+!b*4)));

Frederik.L

Posted 2011-01-28T01:55:09.600

Reputation: 101

Your code uses "FizzBuzz" which is disallowed by the rules. – Fatalize – 2015-08-27T11:16:06.273

Fatalize's comment isn't the full story: you could either remove the FizzBuzz, or the 3 and5`, or you could use any of those "in a misleading way" (whatever that means). – Martin Ender – 2015-08-27T12:17:34.817

Ok, I'll see what I can come up to in that sense. Thanks for exposing the rules, didn't paid attention to those on the first run. – Frederik.L – 2015-08-27T20:06:32.867

0

C, 283 bytes

#include "stdio.h"
#include "string.h"
#define z 122
int main(){char f[]={70,105,z,z,0},b[]={66,117,z,z,0},i=1,B[z],g;for(i=1;i<=100;i++){g=0;if(i%3<1)strcpy(B,f),g=1;if(i%5<1){if(g>0)strcat(B,b);else if(g<1)strcpy(B,b),g=1;}if(g<1)printf("%d\n",i);else printf("%s\n",B);}return 0;}

Obscure? Maybe. Golfed? A little more so than it is obscure. Ugly? YES.

James Murphy

Posted 2011-01-28T01:55:09.600

Reputation: 267

0

C 216 bytes

#define t(b) putchar(p+=b);
main(p,v,c){p=70;for(v=c=1;v<=p*2-40&&!(c=0);++v){if(!(v%(p/23))){t(0)t(35)t(17)t(0)++c;}if(!(v%(p/(14+c*9)))){t(-56+!c*52)t(51)t(5)t(0);++c;}if(c){t(-112)p+=60;}else printf("%i\n",v);}}

Kaslai

Posted 2011-01-28T01:55:09.600

Reputation: 641

0

PHP - 157 154 95 chars

<?$x="zz";for(;++$i<=100;){?> <?$j=$i%(4-1)^0?"":"Fi$x";$j.=$i%(4+1)^0?"":"Bu$x";echo$j?$j:$i;}

And no whitespace! Only one space!

TwoScoopsofPig

Posted 2011-01-28T01:55:09.600

Reputation: 131

0

C 213 Bytes

int c=1,m,n,O,z='z';void main(){O=0;m=c%5-c%3;if(c>2){O=!m&&n?printf("Fiz%cBuz%c",z,z):O;O=m<0&&n+1?printf("Buz%c",z):O;O=m!=n&&m+1&&!O?printf("Fiz%c",z):O;}n=m;O?0:printf("%i",c);putchar(10);c++;if(c<101)main();}

Everything is based on counter % 5 - counter % 3.

Élektra

Posted 2011-01-28T01:55:09.600

Reputation: 284

0

C++ (478 Bytes)

It beats the previous C++ code easily.

#include<iostream>
#include<string>
using namespace std;int main() {string K = "FBiuzzzz";for (int i = 1; i <= 100; i++) {if (i % ((35 / 10)*(35 % 10))==0) {for (int j = 0; j < 8; j += 2)cout << K.at(j);for (int j = 1; j < 8; j += 2)cout << K.at(j);cout << endl;}else if (i % (35 / 10) == 0) {for (int j = 0; j < 8; j += 2)cout << K.at(j);cout << endl;}else if (i % (35 % 10) == 0) {for (int j = 1; j < 8; j += 2)cout << K.at(j);cout << endl;} else cout << i << endl;}return 0;}

user54200

Posted 2011-01-28T01:55:09.600

Reputation:

Lol, I have made FBiuzzzz instead of FizzBuzz (Scrambling trick intensifies) – None – 2016-06-26T02:54:44.800

0

gs2, 1 byte

f

prints fizzbuzz, gs2 has a very simple function

Shaun Wild

Posted 2011-01-28T01:55:09.600

Reputation: 2 329

0

JavaScript - 166 chars

Here's another JavaScript solution which uses an encoded for loop body to hide 3, 5, Fizz, and Buzz. It's relatively terse, but you can still kind of see what's going on.

for(i=0;++i<101;)console.log(eval(String.fromCharCode(...'286925333f27273a2746697a7a27292b286925353f27273a2742757a7a27297c7c69'.match(/../g).map(n=>parseInt(n,16)))))

For something a bit more interesting, we can encode all the programming logic in zero-width whitespace characters and have something that looks concise but is actually not at all. This is similar to dzuc's Ruby solution.

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Personally, I like playing with fire.

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