From 34b316db74e00e4c2f8802b4163485beca340e3c Mon Sep 17 00:00:00 2001
From: Vitor Oliveira <vbrazo@gmail.com>
Date: Sun, 7 Mar 2021 12:27:36 -0800
Subject: [PATCH] Solve fibonacci with golden ratio formula

---
 maths/fibonacci.rb | 43 +++++++++++++++++++++++++++++++++++++++++++
 1 file changed, 43 insertions(+)
 create mode 100644 maths/fibonacci.rb

diff --git a/maths/fibonacci.rb b/maths/fibonacci.rb
new file mode 100644
index 0000000..19e72dd
--- /dev/null
+++ b/maths/fibonacci.rb
@@ -0,0 +1,43 @@
+# 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 fib(n)
+  golden_ratio = (1 + 5**0.5) / 2
+  ((golden_ratio**n + 1) / 5**0.5).to_i
+end
+
+n = 2
+# Output: 1
+# Explanation: F(2) = F(1) + F(0) = 1 + 0 = 1.
+puts(fib(n))
+
+n = 3
+# Output: 2
+# Explanation: F(3) = F(2) + F(1) = 1 + 1 = 2.
+puts(fib(n))
+
+n = 4
+# Output: 3
+# Explanation: F(4) = F(3) + F(2) = 2 + 1 = 3.
+puts(fib(n))