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Swizec

I recently challenged people to codegolf fizzbuzz (http://swizec.com/blog/fizzbuzz-without-ifs-in-90-char-i-wil...)

The Haskell solution was really cool:

[max(show x)(concat[n|(f,n)<-[(3,"Fizz"),(5,"Buzz")],mod x f==0])|x<-[1..100]]

This is much simpler and it looks easier to extend as well.

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I wrote a JS one-liner solution the other day, though JS isn't quite as concise as Haskell:

function fizzbuzz (n) { return new Array(n + 1).join().split(',').map(function (j, i) { return (i % 3 ? '' : ' fizz') + (i % 5 ? '' : ' buzz') || ' ' + i; }).slice(1).join().slice(1); }

185 characters

Edit: If I move from a general solution to only 1-100 and remove spaces and semicolons:

new Array(101).join().split(',').map(function(j, i){return (i%3?'':' fizz')+(i%5?'':' buzz')||' '+i}).slice(1).join().slice(1)

Down to 126, but a lot of that is related to precision with the whitespace and handling a JS quirk with map on undefined values; without the extra joins/split/slices, it goes down to 83 characters (but also doesn't work).

That max looks suspicious. It (ab)uses the fact that the strings "Fizz", "Buzz" compare higher than any printed integer. If you try to extend it to (7, " Bazz") it fails, because " " comes before digits.

Python: under 90 characters and no explicits if statements

    for i in range(100):print ''.join([s*(i%m==0)for m,s in[(3,"Fizz"),(5,"Buzz")]])or i

66 characters:

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

Unfortunately not below 90 characters, but without loops or ifs:

    l=range(101)
    l[3::3]=100/3*['Fizz']
    l[5::5]=100/5*['Buzz']
    l[15::15]=100/15*['FizzBuzz']
    print l[1:]

This one is 86 characters by using constants:

    l=range(101)
    l[::3]=34*['Fizz']
    l[::5]=21*['Buzz']
    l[::15]=7*['FizzBuzz']
    print l[1:]

Perl:

    # 97 characters
    for my $i (1..100) { say join("", map { {3=>'fizz',5=>'buzz'}->{$_} unless $i % $_ } 3,5) || $i }

    # 81, if those pesky spaces are removed
    for my$i(1..100){say join("",map{{3=>'fizz',5=>'buzz'}->{$_}unless$i%$_}3,5)||$i}

My goto ruby solution looks something like:

(1..100).map{|i|(f=[["Fizz"][i%3],["Buzz"][i%5]].join).empty? ? i:f}

60 character solution in LiveScript[0], without using 'if's, also easily extensible!

[1 to 100]map(->[s for s,n of{Fizz:3,Buzz:5}|it%n<1]*''||it)

[0]http://gkz.github.com/LiveScript/

In JS:

i=0;while(i++<100)console.log(((i%3?'':'Fizz')+(i%5?'':'Buzz'))||i)

Pardon the global.

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