Make Project Euler 27 solution idiomatic Ruby

I learned to program in Java and C# and in my free time I am using Ruby because it is fun. Unfortunately I have the feeling that I write Ruby code like I am making Java or C# code. I have learned to use regex instead of strings for comparison, using each instead of for-loops, keeping method names lowercase, and how to use code blocks (which are quite similar to lambda expressions in C#).

I would love to improve the Rubiness of my code and I hope you would be willing to give me one or more pointers on a piece of code I made. It answers Project Euler problem 27.

class Integer
  def prime?
    return false if self  max_primes then
        max_primes = primes
        max_product = a*b
      end
    end
  end
  max_product
end

start = Time.now
answer = get_product_of_coefficients_that_produce_maximum_number_of_primes_for_consecutive_values()

puts "The answer is #{answer} and it took #{Time.now-start} seconds."

I think I can improve on the if-then statement and write it more concise, and also the two "upto-loops" where I first declare the variables max_primes and max_product can be written in a more Ruby-way I am sure.

I would be very grateful if you could let me know how to write more like Ruby!

Links that ask similar questions which I am reading in the meantime:

• Advice on making ruby code more ruby-like

• Ruby method return values - how to be more Ruby-like?

• How to make my script nicer / more Ruby?

Succinctness

As rubyists, we love being succinct, and we love playing with enumerations.

You will see very few literal false and true in ruby code, as well as very few explicit return calls.

For example:

Instead of writing return false if self = 1 && ... which will do the same thing, but we "save" return false.

The power of Enumeration

Ruby has a very powerful Enumerable, and is used widely, often more than once in a line (using method chaining).

For example:

2.upto(Math.sqrt(self)) do |i|
  return false if self % i == 0
end

Here you check if any of the numbers in the range are a divisor for self, and break if there is any. A more ruby way of doing it will be:

return false if 2.upto(Math.sqrt(self)).any? { |i| self % i == 0 }

We'll also prefer to more succinct range syntax (2..Math.sqrt(self)), which is simply shorter...

So now, our def prime? method could be reduced to a one-liner:

class Integer
  def prime?
    self > 1 && !(2..Math.sqrt(self)).any? { |i| self % i == 0 }
  end
end

Mapping

Anywhere in the code I see the following pattern:

result = []
some_loop do
  result << something
end

A red flag is raised, and I look for a way to use map to do the same thing:

result = some_loop.map { something }

Your code goes over all the non-negative integers, and takes counts how many of them result in a prime, until the first non-prime.

"All the non-negative integers" can be expressed in ruby as (0..Float::INFINITY), so we can write:

(0..Float::INFINITY).map { |n| n**2 + a*n + b }.take_while { |result| result.prime? }.count

This code takes each integer, maps it into the result of n**2 + a*n + b, takes all the results until they are no longer prime, and counts how many are there.

Cool! Right? The only problem with the code above, is that it will take infinity to complete it, as it takes all the numbers and maps them, and then checks for how many to take.

To solve this problem ruby now has...

Lazy Enumerables

As of ruby 2.0, lazy enumerables allows you to calculate values in an infinite stream only as needed.

To solve the problem above, all we need to do now is to add the lazy operator on the range:

(0..Float::INFINITY).lazy.map { |n| n**2 + a*n + b }.take_while(&:prime?).count

And we have another one-liner!

Everything is an enumerable

So you want to save on your "upto-loops"? Let's do it!

You want to enumerate over each pair of numbers from -999 to 1000, so what you actually want is to have a long matrix of those pairs:

[[-999, -999], [-999, -998],...,[1000, 1000]].do_something_smart

To do that, you can use product:

(-999..1000).to_a.product((-999..1000).to_a)

But since both a and b have the same range, we can even DRY this up, and use repeated_permutation:

(-999..1000).to_a.repeated_permutation(2)

Both of these solutions will give you the needed matrix, so we can move on the see what we should do with it...

We want to get the coeffiecients that produce the number of primes, so let's do just that:

a, b = (-999..1000).to_a.repeated_permutation(2).max_by { |a, b| get_amount_of_primes_from_quadratic_formula(a,b) }

Now all we need to do is multiply them with each other!

Method naming

Your names are very verbose, which is a good thing, but ruby idiom frowns upon get_ prefixes. Also, prefer using verbs already in the language (count) over those which are not in the language (amount_of)

So now the code will look like:

class Integer
  def prime?
    self > 1 && !(2..Math.sqrt(self)).any? { |i| self % i == 0 }
  end
end

def count_quadratic_formula_primes(a,b)
  (0..Float::INFINITY).lazy.map { |n| n**2 + a*n + b }.take_while(&:prime?).count
end

def product_of_coefficients_that_produce_maximum_number_of_primes_for_consecutive_values()
  a, b = (-999..1000).to_a.repeated_permutation(2).max_by { |a, b| count_quadratic_formula_primes(a,b) }
  a * b
end

start = Time.now
answer = product_of_coefficients_that_produce_maximum_number_of_primes_for_consecutive_values

puts "The answer is #{answer} and it took #{Time.now-start} seconds."

15 lines of hard-core ruby-style code!

Enjoy!

Update

It seems that lazy adds considerable overhead to the performance of the code. So it is not advisable to use it.

Fortunately this works:

(0..Float::INFINITY).take_while { |n| (n**2 + a*n + b).prime? }.count

My code still runs ~2 times slower than the original (ends in 18 seconds), but it is more reasonable than with lazy...

StackExchange Code Review Q#47656, answer score: 6