Merge pull request #92 from TheAlgorithms/solve-fibonacci-with-math

Fibonacci: math approach

DIRECTORY.md

@@ -40,6 +40,7 @@
   * [Ceil](https://github.com/TheAlgorithms/Ruby/blob/master/maths/ceil.rb)
   * [Ceil Test](https://github.com/TheAlgorithms/Ruby/blob/master/maths/ceil_test.rb)
   * [Decimal To Binary](https://github.com/TheAlgorithms/Ruby/blob/master/maths/decimal_to_binary.rb)
+  * [Fibonacci](https://github.com/TheAlgorithms/Ruby/blob/master/maths/fibonacci.rb)
   * [Number Of Digits](https://github.com/TheAlgorithms/Ruby/blob/master/maths/number_of_digits.rb)
   * [Square Root](https://github.com/TheAlgorithms/Ruby/blob/master/maths/square_root.rb)
   * [Square Root Test](https://github.com/TheAlgorithms/Ruby/blob/master/maths/square_root_test.rb)

maths/fibonacci.rb

@@ -0,0 +1,45 @@
+# The Fibonacci numbers, commonly denoted F(n) form a sequence,
+# called the Fibonacci sequence, such that each number is the sum
+# of the two preceding ones, starting from 0 and 1. That is,
+#
+# F(0) = 0, F(1) = 1
+# F(n) = F(n - 1) + F(n - 2), for n > 1.
+# 
+# Given n, calculate F(n).
+
+#
+# Approach: Math
+#
+
+# Intuition: Using the golden ratio, a.k.a Binet's formula
+
+# Algorithm: Use the golden ratio formula to calculate the Nth Fibonacci number.
+# https://demonstrations.wolfram.com/GeneralizedFibonacciSequenceAndTheGoldenRatio/
+
+# Complexity Analysis
+
+# Time complexity: O(1). Constant time complexity since we are using no loops or recursion
+# and the time is based on the result of performing the calculation using Binet's formula.
+# 
+# Space complexity: O(1). The space used is the space needed to create the variable
+# to store the golden ratio formula.
+
+def fibonacci(n)
+  golden_ratio = (1 + 5**0.5) / 2
+  ((golden_ratio**n + 1) / 5**0.5).to_i
+end
+
+n = 2
+puts(fibonacci(n))
+# Output: 1
+# Explanation: F(2) = F(1) + F(0) = 1 + 0 = 1.
+
+n = 3
+puts(fibonacci(n))
+# Output: 2
+# Explanation: F(3) = F(2) + F(1) = 1 + 1 = 2.
+
+n = 4
+puts(fibonacci(n))
+# Output: 3
+# Explanation: F(4) = F(3) + F(2) = 2 + 1 = 3.