Day 10 of 30 Ruby Coding Challenge: Recursive Factorial Numbers

Hey!

Today, we’re going to solve the previous Factorial problem using recursion in Ruby. This is the written version of the Youtube video :).

Just a reminder, it doesn’t matter if you’re using Ruby, Python, JavaScript, or your favorite language — what really matters is the concepts and logic used to solve problems.

If you haven’t watched the previous video, I just want to remember the definition of a Factorial number

A Factorial number N is the product of all positive integers less than or equal to the number N

We indicate a Factorial number as N!

Real example:

xxxxxxxxxx

5! = 5*4*3*2*1

# result in 120

A possible (and not so beautiful) solution to this problem is the solution in the first video:

xxxxxxxxxx

def factorial(number)

  result = number

  while number > 1

    result = result * (number - 1)

    number = number - 1

  end

  return result

end

puts factorial(5) # 120

Recursion

Again, just recalling what recursion is:

Recursion is when you solve a problem by breaking the problem down into smaller versions of the same problem

And when it comes to coding, the method or function calls on itself :)

Before going to code, let’s see the first example:

xxxxxxxxxx

5! = 5*4*3*2*1

But you can actually calculate this way:

xxxxxxxxxx

5! = 5*4!

Because 4! = 432*1, which is a small part of the 5! problem

You can go further and say:

xxxxxxxxxx

5! = 5*4*3!

A little bit more

xxxxxxxxxx

5! = 5*4*3*2!

And finally

xxxxxxxxxx

5! = 5*4*3*2*1

Notice that we’ve broken that down into smaller Factorial calculations.

The recursive Ruby method will be like the following:

xxxxxxxxxx

def factorial(number)

  return 1 if number == 0

  return number * factorial(number - 1)

end

Notice that the function:

• Returns 1 if the number is equal to 0. That’s the criteria to stop the method execution.
• Returns number * factorial(number - 1) which, as we saw, is a method calling on itself.

A leaner version of this method in Ruby would be:

xxxxxxxxxx

def factorial(number)

  number == 0 ? 1 : number * factorial(number - 1)

end

Also notice that:

• We don’t need to explicitly use return in Ruby.
• We’re using a ternary condition, which in our case makes the readability (in my opinion) much better.

That’s it!

Hope you liked it and see you tomorrow in the next challenge :).

Don’t forget to come by and say hi!

Twitter Youtube Instagram Linkedin GitHub