# Elixir FizzBuzz Golf

## Instructions

Write an elixir program that will print the numbers from 1 to 100, replacing numbers divisible by 3 with Fizz, numbers divisible by 5 with Buzz and numbers divisible by both 3 and 5 with FizzBuzz. All other numbers should be returned as normal.

For example:

iex> --enter your code here--
1
2
Fizz
4
Buzz
Fizz
7
8
Fizz
Buzz
11
Fizz
13
14
FizzBuzz
..
..
97
98
Fizz
Buzz

## Solutions

### Winner

The winning solution was submitted by @andrei_here:

char count: 89

iex(1)> for x<-1..?d,z=(rem(x,3)<1&&"Fizz"||"")<>(rem(x,5)<1&&"Buzz"||""),do: IO.puts z==""&&x||z
1
2
..
..
97
98
Fizz
Buzz

This solution uses a comprehension for x<-1.. to iterate over the range of numbers. The ?d is a clever trick that saves one character, this is because the character 'd' is 100 in ASCII and by using the ? we get the integer value of the character.

Comprehensions can consist of between 2 and 3 components: a generator, a filter (optional) and a body. In this solution Andrei abuses the filter, where he builds up the string "FizzBuzz" and binds it to the variable z. The "Fizz" string is determined when the range passes a value x into the rem function. If x / 3 then the remainder is 0. Andrei manages to save one character by using <1 instead of ==0. If x is divisible by 3 then the && returns a "Fizz" string otherwise the || returns an empty string.

The <> concatenates the "Fizz" string with a similar "Buzz" string, at which point the body of the comprehension is called. All this does is check to see whether the z is an empty string if so it returns the value of the variable x otherwise it returns the appropriate "Fizz", "Buzz", "FizzBuzz".

This all weighs in at 89 characters! Well done.

To see some of Andrei’s other attempts checkout his Gist: https://gist.github.com/zyro/e60e75ef3a3c3f08e5ef

### Honourable mentions

Submitted by @henrik:

char count: 99

for n<-1..?d,r=&(rem(n,&1)==0&&IO.write&2),do: (f=r.(3,"Fizz");b=r.(5,"Buzz");f||b||r.(1,n);:io.nl)
for n<-1..?d,r=&(rem(n,&1)==0&&IO.puts&2),do: r.(15,"FizzBuzz")||r.(3,"Fizz")||r.(5,"Buzz")||r.(1,n)
for n<-1..?d,do: Enum.find [{15,"FizzBuzz"},{3,"Fizz"},{5,"Buzz"},{1,n}],fn({d,t})->rem(n,d)==0&&IO.puts(t)end

Checkout Henrik’s Gist and see further explainations: https://gist.github.com/henrik/6ca434c0c5cb66ed29d7

Submitted by @emson:

char count: 129

s=&(&1==1&& &2||String.slice("FizzBuzz",&1==10&&4||0,&1==6&&4||8));Enum.each 1..?d,&IO.puts s.(rem(trunc(:math.pow &1,4),-15),&1)

This attempt is influenced by the HackerRank article https://blog.hackerrank.com/fizzbuzz/. Here we create an anonymous function that slices the "FizzBuzz" string in different ways according to the numbers passed to it. The numbers are generated using Fermat’s little theorem, which proved to be an interesting way of calculating whether the numbers cleanly divide the values from 1 to 100.