[Ruby] FizzBuzz problem

Overview

As a study of Ruby, I solved the ** FizzBuzz problem ** using ** while ** and ** each **. There are other ways to solve it, but this time I tried to solve the problem with ** while **, which I personally use less frequently, and ** each **, which I often use.

** (Added on 2020.1.13) I also practiced the method using the upto method and arguments. ** **

table of contents

--Practice --Problem --Answer (while) --Answer (each) --Answer (up to)

• bonus

--Supplement

--Summary

--References

Practice

problem

Output the numbers 1 ~ 100 to the terminal.

【conditions】

--Output as " Fizz " when the value is a multiple of 3 --Output as " Buzz " when the value is a multiple of 5. --Output as " FizzBuzz " when the values ​​are multiples of 3 and 5.

Answer (while)

def fizz_buzz
  num = 0

  while (num <= 100) do    #Conditions to repeat up to 100
    num += 1               #Add 1 each time you repeat

    if (num % 15) == 0     #When it is a multiple of 15
      p 'FizzBuzz'
    elsif (num % 3) == 0   #When it is a multiple of 3
      p 'Fizz'
    elsif (num % 5) == 0   #When it is a multiple of 5
      p 'Buzz'
    else                   #At other times
      p num
    end
  end
end

fizz_buzz

Answer (each)

def fizz_buzz

  (1..100).each do |num|     # 1~Up to 100
    if (num % 15) == 0       #When it is a multiple of 15
      p 'FizzBuzz'
    elsif (num % 3) == 0     #When it is a multiple of 3
      p 'Fizz'
    elsif (num % 5) == 0     #When it is a multiple of 5
      p 'Buzz'
    else                     #At other times
      p num
    end
  end
end

fizz_buzz

Answer (up to)

def fizz_buzz
  1.upto(100) do |num|  #From 1 to 100
    if num % 15 == 0
      p 'FizzBuzz'
    elsif num % 3 == 0
      p 'Fizz'
    elsif num % 5 == 0
      p 'Buzz'
    else
      p num
    end
  end
end

fizz_buzz

bonus

I wrote a program hoping to create a pattern of 1 to any number instead of 1 to 100.

The method doesn't change much, it just iterates ** up to the value specified by the argument **.

def fizz_buzz(max_num)
  1.upto(max_num) do |num| #1 to the number specified by the argument (arbitrary number)
    if num % 15 == 0
      p 'FizzBuzz'
    elsif num % 3 == 0
      p 'Fizz'
    elsif num % 5 == 0
      p 'Buzz'
    else
      p num
    end
  end
end

p 'How many do you count?'
num = gets.to_i
fizz_buzz(num)

Supplement

Outputs " FizzBuzz " when the values ​​are multiples of 3 and 5.

num % 3 == 0 && num % 5 == 0

The answer says (num% 15) == 0, but you can replace it as above.

** Reasons to write the condition of multiples of 3 or multiples of 5 (multiples of 15) ** first

As you can see in the problem statement, ** "multiples of 3 and 5" is FizzBuzz **, so it must be a multiple of ** 3 and a multiple of 5 (a multiple of 15) **. If you write this condition at the end, the conditions of ** "multiple of 3" ** and ** "multiple of 5" ** will be evaluated before that, so add ** multiple of 15 ** at the beginning. doing.

Summary

--There are other ways to solve the FizzBuzz problem, and you can solve other than the ones mentioned above. --When there is a condition like this time, it will not be evaluated unless ** a multiple of 15 ** is added first.

References

-IT Glossary (FizzBuzz Problem) that makes you feel like "I understand" but "I don't understand"

-Ruby 3.0.0 Reference Manual (Integer # upto)