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](https://twitter.com/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 Andreiâs other attempts checkout his Gist: [https://gist.github.com/zyro/e60e75ef3a3c3f08e5ef](https://gist.github.com/zyro/e60e75ef3a3c3f08e5ef) ### Honourable mentions Submitted by [@henrik](https://twitter.com/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 Henrikâs Gist and see further explainations: [https://gist.github.com/henrik/6ca434c0c5cb66ed29d7](https://gist.github.com/henrik/6ca434c0c5cb66ed29d7) * * * Submitted by [@emson](https://twitter.com/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/](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 [Fermatâs little theorem](https://en.wikipedia.org/wiki/Fermat's_little_theorem), which proved to be an interesting way of calculating whether the numbers cleanly divide the values from 1 to 100. * * * Resources --------- * Elixir docs `Enum.at`: [http://elixir-lang.org/docs/stable/elixir/Enum.html#at/3](http://elixir-lang.org/docs/stable/elixir/Enum.html#at/3) * Comprehension docs `for`: [http://elixir-lang.org/getting-started/comprehensions.html](http://elixir-lang.org/getting-started/comprehensions.html) * More comprehension docs: [http://elixir-lang.org/docs/stable/elixir/Kernel.SpecialForms.html#for/1](http://elixir-lang.org/docs/stable/elixir/Kernel.SpecialForms.html#for/1) * HackerRank FizzBuzz: [https://blog.hackerrank.com/fizzbuzz/](https://blog.hackerrank.com/fizzbuzz/) * Fermatâs little theorem: [https://en.wikipedia.org/wiki/Fermat's\_little\_theorem](https://en.wikipedia.org/wiki/Fermat's_little_theorem) [comments powered by](https://disqus.com/) ||||I|||| \* Home \* About \* Puzzles \* Search elixirgolf.com Solve elixir code problems, using the shortest number of characters. Don't miss a cunning puzzle, signup to our weekly newsletter. SAVE 30%: www.udemy.com/elixir-for-beginners/?couponCode=Save30%25 Elixir FizzBuzz Golf By elixirgolf November 17, 2015 Comment Tweet Like +1 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 Andreiâs 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 Henrikâs 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 Fermatâs little theorem, which proved to be an interesting way of calculating whether the numbers cleanly divide the values from 1 to 100. Resources \* Elixir docs Enum.at: http://elixir-lang.org/docs/stable/elixir/Enum.html#at/3 \* Comprehension docs for: http://elixir-lang.org/getting-started/comprehensions.html \* More comprehension docs: http://elixir-lang.org/docs/stable/elixir/Kernel.SpecialForms.html#for/1 \* HackerRank FizzBuzz: https://blog.hackerrank.com/fizzbuzz/ \* Fermatâs little theorem: https://en.wikipedia.org/wiki/Fermat's\_little\_theorem Please enable JavaScript to view the comments powered by Disqus. comments powered by Disqus Previous Next © 2016 elixirgolf. Powered by Jekyll using the So Simple Theme .

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