Step by Step solution about How to resolve the algorithm Stern-Brocot sequence step by step in the Ruby programming language

The provided Ruby code defines a method called sb that generates the Fibonacci sequence and includes some operations on the generated sequence.

1. sb Method:

   • This method represents a lazy enumerator for the Fibonacci sequence.
   • It takes an optional block as an argument.
   • If a block is not provided, it returns an enumerator that can be iterated manually.
   • If a block is provided, it yields each Fibonacci number to the block indefinitely.

2. Fibonacci Sequence Generation:

   • The sequence starts with a=[1,1].
   • It uses a step loop to iterate indefinitely, starting from 0.
   • Inside the loop:
     • It yields the Fibonacci number at index i from the a array.
     • It calculates the next two Fibonacci numbers and appends them to the a array.

3. Printing the First 15 Fibonacci Numbers:

   • The code uses the first method to retrieve the first 15 Fibonacci numbers from the sb enumerator.
   • It prints them as a comma-separated string.

4. Finding the First Occurrence of Numbers:

   • It iterates over the numbers from 1 to 10 and 100.
   • For each number, it uses the find_index method on the sb enumerator to find its first occurrence.
   • It prints the result as "N first appears at M." where N is the number and M is the index of its first occurrence.

5. Checking if All GCD's Are 1:

   • It uses the take(1000) method to obtain the first 1000 Fibonacci numbers.
   • It then uses the each_cons(2) method to iterate over pairs of consecutive Fibonacci numbers.
   • For each pair, it calculates the greatest common divisor (GCD) using the gcd method.
   • If all the GCD's are 1, it prints "All GCD's are 1", otherwise it prints "Whoops, not all GCD's are 1!".

def sb
  return enum_for :sb unless block_given?
  a=[1,1]
  0.step do |i|
    yield a[i]
    a << a[i]+a[i+1] << a[i+1]
  end
end

puts "First 15: #{sb.first(15)}"

[*1..10,100].each do |n| 
  puts "#{n} first appears at #{sb.find_index(n)+1}."
end

if sb.take(1000).each_cons(2).all? { |a,b| a.gcd(b) == 1 }
  puts "All GCD's are 1"
else
  puts "Whoops, not all GCD's are 1!"
end