Day 2 of 30 - Ruby Coding Challenge - Ugly Prime Algorithm

Hey!

Day 2 of 30 - Ruby Coding Challenges in 30 Days. This is the post version of the Youtube video.

Continuing to stretch our arms and legs to the long run, I’m going to solve a simple and common algorithm:

How Many Prime Numbers Are in a Given Array

How does my brain work when it comes to solving algorithms? Just get the job done!

• Forget about good practices
• Forget about code design
• Especially forget performance

I really love to respect my brain because, well, it can be against me.

Today I’ll solve this problem with poor design and you’ll get the feeling that we should not be friends anymore.

But I promise that I’m going to write AT LEAST four versions of this Algorithm and maybe you might reconsider our friendship.

Let’s start this conversation with refreshing our brain.

What is a Prime Number

Short answer:

A natural number greater than 1 that is evenly divided only by 1 and itself.

Wikipedia has a really sexy answer.

Going through some examples:

• 1 is prime.
• 2 is prime.
• 3 is prime.
• 4 is not prime. It’s divisible by 2.
• 5 is prime.
• 6 is not prime. It’s divisible by 2 and 3.
• 7 is prime.
• 8 is not prime. It’s divisible by 2 and 4.
• 9 is not prime. It’s divisible by 3.
• 10 is not prime. It’s divisible by 2 and 5.

Yep, you got it.

Prime Number Algorithm in Ruby

Are you familiar with Ruby? If so, then probably you’re already aware of the Prime library. This is an unfair advantage.

I’m going to talk about that later but for now, let’s stick with the old and good manual process.

It’s time to create the code skeleton. A method waiting for an array:

def count_prime_number_version_1(array)

    # magic will happen here

end

array = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

puts count_prime_number_version_1(array)

Just for clarity, I’m going to create a variable, which indicates the total count:

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def count_prime_number_version_1(array)

  prime_count = 0

    # a loop expression here

  return prime_count

end

array = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

puts count_prime_number_version_1(array)

It’s time to go through the array and ask for each item if they are prime or not

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def count_prime_number_version_1(array)

  prime_count = 0

  for item in array

  end

  return prime_count

end

We know that 1 is not prime, so why not just ignore it?

In Ruby, we use next to say: “Hey Ruby, forget this item and jump to the next item”

Are you coming from Java, Python or Javascript? That’s the same with continue

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def count_prime_number_version_1(array)

  prime_count = 0

  for item in array

        next if item == 1

  end

  return prime_count

end

How can we validate in Ruby if a number is divided by another number, resulting in a remainder zero?

With percent %:

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# checking if the remainder value is zero when dividing an item by count

item % count == 0

Ok, now we can write a while loop that

• given an item
• check if there’s a number that divides the item, remaining in zero

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is_prime = true

number = item - 1

while number > 1

  if item % number != 0

    number = number - 1

  else

    is_prime = false

    break

  end

end

Notice that:

• we’re using is_prime to indicate when an item is, well, prime
• we’re using break to stop the while loop once the item is divided by a number, which already means that it’s not prime

It’s time to risk our virtual friendship. The complete code:

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def count_prime_number_version_1(array)

  prime_count = 0

  for item in array

    next if item == 1

    is_prime = true

    number = item - 1

    while number > 1

      if item % number == 0

        is_prime = false

        break

      else

        number = number - 1

      end

    end

    if is_prime

      prime_count = prime_count + 1

    end

  end

  return prime_count

end

array = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

puts count_prime_number_version_1(array)

What do I want to prove?

• I write terrible code at first and I’m not even sorry
• I don’t worry about design, performance, etc on the first version. Just get the job done (and lose some friends on the way)
• For beginners, it’s probably a good way to go through this process

As I said, this algorithm will become beautiful (hopefully), or at least better in the upcoming video and we’ll be friends again.

See you!

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