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# Fizzbuzz Without Ifs in 90 Characters -- I'll Buy You a Beer if You Can Beat It

09.20.2012
| 6251 views |

Yesterday I got nerd-sniped into solving the FizzBuzz problem without using if sentences.

Unfortunately, a more careful inspection of the problem specification reveals this solution was incorrect.

Write a program that prints the numbers from 1 to 100. 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”.

Luckily, that’s a very simple fix:

```# codegolf version (90 characters):
[{0:"fizzbuzz"}.get(x%(3*5),{0:"fizz"}.get(x%3,{0:"buzz"}.get(x%5,x)))for x in range(101)]

[{0: "fizzbuzz"}.get(x%(3*5),
{0: "fizz"}.get(x%3,
{0: "buzz"}.get(x%5, x)))
for x in range(101)]```

There you have it, a solution to the fizzbuzz problem that relies on python’s interactive command line interpreter thingy to do the printing, shaving off 4 characters for print.

The “trick”, if it can be called that, is relying on the native get method to return the correct string or fall back to the default value, which checks for the next string in the if-then-else chain.

I wish never to see code like that in something I have to maintain. Ever.

## The challenge

Your challenge, should you choose to accept it, is to produce a shorter fizzbuzz solution without using if sentences. I will buy a bottle of beer for the winner.

The best Scala solution so far is 106 characters, by @HairyFotr:

`1 to 100 map(a=>println((a%3,a%5)match{case(0,0)=>"fizzbuzz"case(0,_)=>"fizz"case(_,0)=>"buzz"case _=>a})) /cc`

No really, I will send a bottle of beer anywhere in the world if you write a shorter solution. Of course, if feasible, I would prefer to take you out for a beer than use snail mail

Published at DZone with permission of Swizec Teller, author and DZone MVB. (source)

(Note: Opinions expressed in this article and its replies are the opinions of their respective authors and not those of DZone, Inc.)

### Tomasz Nurkiewicz replied on Sun, 2012/09/23 - 1:01pm

Squeezed by 6 characters (in Scala): println(1 to 100 map(_%15 match{case 0=>"FizzBuzz"case 3|6|9|12=>"Fizz"case 5|10=>"buzz"case x=>x}))

### Łukasz Wolski replied on Mon, 2012/09/24 - 3:52am

89 chars:

[{0:"fizzbuzz"}.get(x%(15),{0:"fizz"}.get(x%3,{0:"buzz"}.get(x%5,x)))for x in range(101)]

### Will Fi replied on Mon, 2012/09/24 - 8:56am

Groovy: 67 chars

(1..100).each{println((!(it%3)?'Fizz':'')+(!(it%5)?'Buzz':'')?:it)}

### Gabe Schenz replied on Tue, 2012/09/25 - 10:44pm

Shamelessly stealing from your example... but the output does look prettier. (73 characters)

`for i in range(101):print i,{0:'Fizz'}.get(i%3,'')+{0:'Buzz'}.get(i%5,'')`

Or working in bash or ksh (65 characters):

`a=Fizz;b=Buzz;for i in {1..100};do echo \$i\${a[i%3]}\${b[i%5]};done `

### Garcia Mario replied on Tue, 2012/09/25 - 4:57pm in response to: Will Fi

Almost Will !!!

But you're missing 'FizzBuzz'. Just replace the "it" at the end with 'FizzBuzz'. I have three different versions:

1. 64 chars: (1..100).each{n->print"\${!(n%3)?'Fi':!(n%5)?'Bu':'FizzBu'}zz\n"}
2. 64 chars: (1..100).each{n->println"\${n%3==0?'Fi':n%5==0?'Bu':'FizzBu'}zz"}
3. 59 chars: 1..100>>{n->print"\${n%3==0?'Fi':n%5==0?'Bu':'FizzBu'}zz\n"}
All right I may have cheated a little bit in the third, to make it work you just have to add first the following line:
• Range.metaClass.rightShift = {closure-> delegate.each closure}

### Will Fi replied on Wed, 2012/09/26 - 8:59am in response to: Garcia Mario

No, @Garcia Mario, mine is correct. It prints 'FizzBuzz' when the result is divisible by both, thus getting printed 6 times. Yours is printing either 'Fizz', 'FizzBuzz' or 'Buzz' everytime, and the number needs to be printed when it's not divisible by neither of the three.

http://groovyconsole.appspot.com/script/763001

It's getting the same result as in this one in rubyforge: http://fizzbuzz.rubyforge.org/

### Garcia Mario replied on Wed, 2012/09/26 - 3:08pm in response to: Will Fi

@Will Completely right!!! I didn't read it carefully, my apologies :)

This time I've just done it using collections no ifs (not even the ternary operator). This time 103 chars of pure Groovy.

(1..100).each{n->print"\${[([0:'fizzbuzz'][n%(15)]),([0:'fizz'][n%3]),([0:'buzz'][n%5]),n].find{it}}\n"}

### Will Fi replied on Thu, 2012/09/27 - 10:25am in response to: Garcia Mario

@Garcia Mario

Ah! searching for an index in little maps!

Without ternary operators, the  best i could is Groovy: 96 :

(1..100).each{a->println([(15):'FizzBuzz',(3):'Fizz',(5):'Buzz',(a):a].find{k,v->!(a%k)}.value)}

It searches the map for an entry whose key is divisible by the number from 1 to 100. I was trying to work on a solution to ignore the (15):'FizzBuzz' using findAll, but no luck so far :-)

### Willie Wheeler replied on Fri, 2012/12/21 - 4:03pm

Late response, but Groovy in 61 characters:

(1..100).each{println((it%3?'':'Fizz')+(it%5?'':'Buzz')?:it)}

This is obviously just a slightly modified version of Will's solution.