Data Modelling & AI Data Structure & Algorithm

Check whether two numbers are in golden ratio

23 June 2025

1

Given two numbers A and B, the task is to check that A and B are in the golden ratio.
Golden Ratio: Two numbers are said to be in the golden ratio if their ratio is the same as the ratio of the sum of the two numbers to the larger number. Here a > b > 0, Below is the geometric representation of the Golden ratio:

$\frac{A+B}{A} = \frac{A}{B} = \varphi = \frac{1+ \sqrt{5}}{2} = 1.618$

Examples:

Input: A = 1, B = 0.618
Output: Yes
Explanation:
These two numbers together forms Golden ratio

$\frac{A}{B} = \frac{A + B}{A} = \frac{1.618}{1} = 1.618$

Input: A = 61.77, B = 38.22

Output Yes

Explanation:

These two numbers together forms Golden ratio

$\frac{A}{B} = \frac{A + B}{A} = \frac{99.99}{61.77} = 1.618$

Approach: The idea is to find two ratios and check that this ratio is equal to the Golden ratio. That is 1.618.

// Here A denotes the larger number

$\frac{A}{B} = \frac{A + B}{A} = 1.618$

Below is the implementation of the above approach:

C++

// C++ implementation to check  
// whether two numbers are in  
// golden ratio with each other 
#include <bits/stdc++.h>
using namespace std;
 
// Function to check that two  
// numbers are in golden ratio 
bool checkGoldenRatio(float a, 
                      float b)
{
  // Swapping the numbers such  
  // that A contains the maximum 
  // number between these numbers 
  if(a <= b)
  {
    float temp = a;
    a = b;
    b = temp;
  }
 
  // First Ratio
  std::stringstream ratio1;
  ratio1 << std :: fixed << 
            std :: setprecision(3) << 
            (a / b);
 
  // Second Ratio 
  std::stringstream ratio2;
  ratio2 << std :: fixed << 
            std :: setprecision(3) << 
            (a + b) / a;
 
  // Condition to check that two 
  // numbers are in golden ratio 
  if((ratio1.str() == ratio2.str()) && 
      ratio1.str() == "1.618")
  {
    cout << "Yes" << endl;
    return true;
  }
  else
  {
    cout << "No" << endl;
    return false;
  }
}
  
// Driver code
int main()
{
  float a = 0.618;
  float b = 1;
 
  // Function Call 
  checkGoldenRatio(a, b); 
 
  return 0;
}
 
// This code is contributed by divyeshrabadiya07

Java

// Java implementation to check  
// whether two numbers are in  
// golden ratio with each other 
class GFG{
     
// Function to check that two  
// numbers are in golden ratio 
public static Boolean checkGoldenRatio(float a,
                                       float b)
{
     
    // Swapping the numbers such  
    // that A contains the maximum 
    // number between these numbers 
    if (a <= b)
    {
        float temp = a;
        a = b;
        b = temp;
    }
     
    // First Ratio
    String ratio1 = String.format("%.3f", a / b);
     
    // Second Ratio 
    String ratio2 = String.format("%.3f", (a + b) / a);
     
    // Condition to check that two 
    // numbers are in golden ratio
    if (ratio1.equals(ratio2) && 
        ratio1.equals("1.618"))
    {
        System.out.println("Yes");
        return true;
    }
    else
    {
        System.out.println("No");   
        return false;
    }
}
 
// Driver code 
public static void main(String []args)
{
    float a = (float)0.618;
    float b = 1;
     
    // Function Call 
    checkGoldenRatio(a, b);
}
}
 
// This code is contributed by rag2127

Python3

# Python3 implementation to check 
# whether two numbers are in 
# golden ratio with each other
 
# Function to check that two 
# numbers are in golden ratio
def checkGoldenRatio(a, b):
     
    # Swapping the numbers such 
    # that A contains the maximum
    # number between these numbers
    a, b = max(a, b), min(a, b)
     
    # First Ratio
    ratio1 = round(a/b, 3)
     
    # Second Ratio
    ratio2 = round((a+b)/a, 3)
    # Condition to check that two
    # numbers are in golden ratio
    if ratio1 == ratio2 and\
       ratio1 == 1.618:
        print("Yes")
        return True
    else:
        print("No")
        return False
         
# Driver Code
if __name__ == "__main__":
    a = 0.618
    b = 1
     
    # Function Call
    checkGoldenRatio(a, b)

C#

// C# implementation to check  
// whether two numbers are in  
// golden ratio with each other 
using System;
using System.Collections.Generic;
class GFG {
     
    // Function to check that two  
    // numbers are in golden ratio 
    static bool checkGoldenRatio(float a, 
                          float b)
    {
      // Swapping the numbers such  
      // that A contains the maximum 
      // number between these numbers 
      if(a <= b)
      {
        float temp = a;
        a = b;
        b = temp;
      }
      
      // First Ratio
      string ratio1 = String.Format("{0:0.000}", a / b);
      
      // Second Ratio 
      string ratio2 = String.Format("{0:0.000}", (a + b) / a);
 
      // Condition to check that two 
      // numbers are in golden ratio 
      if(ratio1 == ratio2 && ratio1 == "1.618")
      {
        Console.WriteLine("Yes");
        return true;
      }
      else
      {
        Console.WriteLine("No");
        return false;
      }
    } 
   
  // Driver code  
  static void Main() {
      float a = (float)0.618;
      float b = 1;
      
      // Function Call 
      checkGoldenRatio(a, b); 
  }
}
 
// This code is contributed by divyesh072019

Javascript

<script>
 
// Javascript implementation to check 
// whether two numbers are in 
// golden ratio with each other
 
// Function to check that two 
// numbers are in golden ratio
function checkGoldenRatio(a, b)
{
      
    // Swapping the numbers such 
    // that A contains the maximum
    // number between these numbers
    if (a <= b)
    {
        let temp = a;
        a = b;
        b = temp;
    }
      
    // First Ratio
    let ratio1 = (a / b).toFixed(3);
      
    // Second Ratio
    let ratio2 = ((a + b) / a).toFixed(3);
      
    // Condition to check that two
    // numbers are in golden ratio
    if ((ratio1 == ratio2) &&
        ratio1 == "1.618")
    {
        document.write("Yes");
        return true;
    }
    else
    {
        document.write("No");  
        return false;
    }
}
 
// Driver Code
     
    let a = 0.618;
    let b = 1;
      
    // Function Call
    checkGoldenRatio(a, b);
 
</script>

Output:

Yes

Time Complexity: O(1)
Auxiliary Space: O(1)

References: https://en.wikipedia.org/wiki/Golden_ratio

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Check whether two numbers are in golden ratio

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