3 approaches to solving FizzBuzz

FizzBuzz.rb

require 'benchmark'

# using simple conditional statements
def conditional(n)
  if (n % 3 == 0) && (n % 5 != 0)
    'Fizz'
  elsif (n % 3 != 0) && (n % 5 == 0)
    'Buzz'
  elsif (n % 3 == 0) && (n % 5 == 0)
    'FizzBuzz'
  else n
  end
end

# no conditionals, just boolean logic
def declarative(n)
  h = {
    'Fizz' => (n % 3 == 0) && (n % 5 != 0),
    'Buzz' => (n % 3 != 0) && (n % 5 == 0),
    'FizzBuzz' => (n % 3 == 0) && (n % 5 == 0),
    n => (n % 3 != 0) && (n % 5 != 0)
  }
  h.key(true)
end

# conditionals used first time to build up the hash,
# after that it's just a hash look-up
memoization ||= Hash.new do |hash,key|
  hash[key] = case
  when (key % 3 == 0) && (key % 5 != 0)
    'Fizz'
  when (key % 3 != 0) && (key % 5 == 0)
    'Buzz'
  when (key % 3 == 0) && (key % 5 == 0)
    'FizzBuzz'
  else key
  end
end


r = (1..1_000_000)


Benchmark.bm do |bm|
  bm.report(:conditional) { r.each  { |n| conditional(n) }}
  bm.report(:conditional) { r.each  { |n| conditional(n) }}
  bm.report(:declarative) { r.each  { |n| declarative(n) }}
  bm.report(:declarative) { r.each  { |n| declarative(n) }}
  bm.report(:memoization) { r.each  { |n| memoization[n] }}
  bm.report(:memoization) { r.each  { |n| memoization[n] }}
end