Given a number ‘n’, how to check if n is a Fibonacci number. First few Fibonacci numbers are 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, ..
Examples :
Input : 8
Output : Yes
Input : 34
Output : Yes
Input : 41
Output : No
Approach 1:
A simple way is to generate Fibonacci numbers until the generated number is greater than or equal to ‘n’. Following is an interesting property about Fibonacci numbers that can also be used to check if a given number is Fibonacci or not.
A number is Fibonacci if and only if one or both of (5*n2 + 4) or (5*n2 – 4) is a perfect square (Source: Wiki). Following is a simple program based on this concept.
C++
// C++ program to check if x is a perfect square#include <bits/stdc++.h>using namespace std;// A utility function that returns true if x is perfect// squarebool isPerfectSquare(int x){ int s = sqrt(x); return (s * s == x);}// Returns true if n is a Fibonacci Number, else falsebool isFibonacci(int n){ // n is Fibonacci if one of 5*n*n + 4 or 5*n*n - 4 or // both is a perfect square return isPerfectSquare(5 * n * n + 4) || isPerfectSquare(5 * n * n - 4);}// A utility function to test above functionsint main(){ for (int i = 1; i <= 10; i++) isFibonacci(i) ? cout << i << " is a Fibonacci Number \n" : cout << i << " is a not Fibonacci Number \n"; return 0;}// This code is contributed by Sania Kumari Gupta (kriSania804) |
C
// C program to check if x is a perfect square#include <math.h>#include <stdbool.h>#include <stdio.h>// A utility function that returns true if x is perfect// squarebool isPerfectSquare(int x){ int s = sqrt(x); return (s * s == x);}// Returns true if n is a Fibonacci Number, else falsebool isFibonacci(int n){ // n is Fibonacci if one of 5*n*n + 4 or 5*n*n - 4 or // both is a perfect square return isPerfectSquare(5 * n * n + 4) || isPerfectSquare(5 * n * n - 4);}// A utility function to test above functionsint main(){ for (int i = 1; i <= 10; i++) { if (isFibonacci(i)) printf("%d is a Fibonacci Number \n", i); else printf("%d is a not Fibonacci Number \n", i); } return 0;}// This code is contributed by Sania Kumari Gupta (kriSania804) |
Java
// Java program to check if x is a perfect squareclass GFG{ // A utility method that returns true if x is perfect square static boolean isPerfectSquare(int x) { int s = (int) Math.sqrt(x); return (s*s == x); } // Returns true if n is a Fibonacci Number, else false static boolean isFibonacci(int n) { // n is Fibonacci if one of 5*n*n + 4 or 5*n*n - 4 or both // is a perfect square return isPerfectSquare(5*n*n + 4) || isPerfectSquare(5*n*n - 4); } // Driver method public static void main(String[] args) { for (int i = 1; i <= 10; i++) System.out.println(isFibonacci(i) ? i + " is a Fibonacci Number" : i + " is a not Fibonacci Number"); }}//This code is contributed by Nikita Tiwari |
Python
# python program to check if x is a perfect squareimport math# A utility function that returns true if x is perfect squaredef isPerfectSquare(x): s = int(math.sqrt(x)) return s*s == x# Returns true if n is a Fibonacci Number, else falsedef isFibonacci(n): # n is Fibonacci if one of 5*n*n + 4 or 5*n*n - 4 or both # is a perfect square return isPerfectSquare(5*n*n + 4) or isPerfectSquare(5*n*n - 4) # A utility function to test above functionsfor i in range(1,11): if (isFibonacci(i) == True): print i,"is a Fibonacci Number" else: print i,"is a not Fibonacci Number " |
C#
// C# program to check if // x is a perfect squareusing System;class GFG { // A utility function that returns // true if x is perfect square static bool isPerfectSquare(int x) { int s = (int)Math.Sqrt(x); return (s * s == x); } // Returns true if n is a // Fibonacci Number, else false static bool isFibonacci(int n) { // n is Fibonacci if one of // 5*n*n + 4 or 5*n*n - 4 or // both are a perfect square return isPerfectSquare(5 * n * n + 4) || isPerfectSquare(5 * n * n - 4); } // Driver method public static void Main() { for (int i = 1; i <= 10; i++) Console.WriteLine(isFibonacci(i) ? i + " is a Fibonacci Number" : i + " is a not Fibonacci Number"); }}// This code is contributed by Sam007 |
Javascript
<script>// javascript program to check if x is a perfect square// A utility function that returns true if x is perfect squarefunction isPerfectSquare( x){ let s = parseInt(Math.sqrt(x)); return (s * s == x);}// Returns true if n is a Fibonacci Number, else falsefunction isFibonacci( n){ // n is Fibonacci if one of 5*n*n + 4 or 5*n*n - 4 or both // is a perfect square return isPerfectSquare(5 * n * n + 4) || isPerfectSquare(5 * n * n - 4);}// A utility function to test above functions for (let i = 1; i <= 10; i++) isFibonacci(i)? document.write( i + " is a Fibonacci Number <br/>"): document.write(i + " is a not Fibonacci Number <br/>") ; // This code is contributed by Rajput-Ji </script> |
PHP
<?php// PHP program to check if// x is a perfect square// A utility function that// returns true if x is // perfect squarefunction isPerfectSquare($x){ $s = (int)(sqrt($x)); return ($s * $s == $x);}// Returns true if n is a// Fibonacci Number, else falsefunction isFibonacci($n){ // n is Fibonacci if one of // 5*n*n + 4 or 5*n*n - 4 or // both is a perfect square return isPerfectSquare(5 * $n * $n + 4) || isPerfectSquare(5 * $n * $n - 4);}// Driver Codefor ($i = 1; $i <= 10; $i++)if(isFibonacci($i))echo "$i is a Fibonacci Number \n";elseecho "$i is a not Fibonacci Number \n" ;// This code is contributed by mits?> |
Output
1 is a Fibonacci Number 2 is a Fibonacci Number 3 is a Fibonacci Number 4 is a not Fibonacci Number 5 is a Fibonacci Number 6 is a not Fibonacci Number 7 is a not Fibonacci Number 8 is a Fibonacci Number 9 is a not Fibonacci Number 10 is a not Fibonacci Number
Time Complexity: O(log N), where N is is the number that we square-root.
Auxiliary Space: O(1)
Approach 2:
In this approach, we first handle the special case where the input number is 0 (which is a Fibonacci number). Then, we use a while loop to generate Fibonacci numbers until we find a Fibonacci number greater than or equal to the input number. If the generated Fibonacci number is equal to the input number, we return true. Otherwise, we check if either (5 * n * n + 4) or (5 * n * n – 4) is a perfect square, as per the formula mentioned in the original code.
This approach may be more efficient than the original code in some cases, especially for larger input values, as it generates Fibonacci numbers on-the-fly and stops as soon as it finds a Fibonacci number greater than or equal to the input number.
C++
#include <bits/stdc++.h>using namespace std;bool isPerfectSquare(int n) { int root = sqrt(n); return (root * root == n);}bool isFibonacci(int n) { if (n == 0) { return true; } int a = 0, b = 1, c = 1; while (c < n) { a = b; b = c; c = a + b; } return (c == n || isPerfectSquare(5 * n * n + 4) || isPerfectSquare(5 * n * n - 4));}int main() { for (int i = 1; i <= 10; i++) { if (isFibonacci(i)) { cout << i << " is a Fibonacci number.\n"; } else { cout << i << " is not a Fibonacci number.\n"; } } return 0;} |
Java
import java.util.*;public class Main { public static boolean isPerfectSquare(int n) { int root = (int) Math.sqrt(n); return (root * root == n); } public static boolean isFibonacci(int n) { if (n == 0) { return true; } int a = 0, b = 1, c = 1; while (c < n) { a = b; b = c; c = a + b; } return (c == n || isPerfectSquare(5 * n * n + 4) || isPerfectSquare(5 * n * n - 4)); } public static void main(String[] args) { for (int i = 1; i <= 10; i++) { if (isFibonacci(i)) { System.out.println(i + " is a Fibonacci number."); } else { System.out.println(i + " is not a Fibonacci number."); } } }} |
Python3
import mathdef is_perfect_square(n): root = int(math.sqrt(n)) return (root * root == n)def is_fibonacci(n): if n == 0: return True a, b, c = 0, 1, 1 while c < n: a = b b = c c = a + b return c == n or is_perfect_square(5 * n * n + 4) or is_perfect_square(5 * n * n - 4)for i in range(1, 11): if is_fibonacci(i): print(i, "is a Fibonacci number.") else: print(i, "is not a Fibonacci number.") |
C#
// C# program for the above approachusing System;public class Program { static bool IsPerfectSquare(int n) { int root = (int)Math.Sqrt(n); return (root * root == n); } static bool IsFibonacci(int n) { if (n == 0) { return true; } int a = 0, b = 1, c = 1; while (c < n) { a = b; b = c; c = a + b; } return (c == n || IsPerfectSquare(5 * n * n + 4) || IsPerfectSquare(5 * n * n - 4)); } public static void Main() { for (int i = 1; i <= 10; i++) { if (IsFibonacci(i)) { Console.WriteLine(i + " is a Fibonacci number."); } else { Console.WriteLine(i + " is not a Fibonacci number."); } } }}// This code is contributed by adityasha4x71 |
Javascript
function is_perfect_square(n) { let root = Math.floor(Math.sqrt(n)); return (root * root === n);}function is_fibonacci(n) { if (n === 0) { return true; } let a = 0, b = 1, c = 1; while (c < n) { [a, b] = [b, c]; c = a + b; } return c === n || is_perfect_square(5 * n * n + 4) || is_perfect_square(5 * n * n - 4);}for (let i = 1; i <= 10; i++) { if (is_fibonacci(i)) { console.log(i + " is a Fibonacci number."); } else { console.log(i + " is not a Fibonacci number."); }}// Contributed by adityasha4x71 |
Output
1 is a Fibonacci number. 2 is a Fibonacci number. 3 is a Fibonacci number. 4 is not a Fibonacci number. 5 is a Fibonacci number. 6 is not a Fibonacci number. 7 is not a Fibonacci number. 8 is a Fibonacci number. 9 is not a Fibonacci number. 10 is not a Fibonacci number.
Time Complexity: O(log N), where N is is the number that we square-root.
Auxiliary Space: O(1)
Approach 3:
This is another approach to check if a given number is Fibonacci number or not.
Steps:
To check if a given number is Fibonacci number or not, we do the following steps:
- First check if the number is 0 or 1, then return true.
- Then till the number comes do while loop.
- In each iteration:
- First calculate Fibonacci of that iteration.
- Then check if it matches with given number or not.
- If matches, return true.
- If the value goes beyond, given number then return false.
- Otherwise continue.
Below is the implementation of the above approach:
C++
// C++ program to check if a given number is// Fibonacci number or not#include <iostream>using namespace std;// Function to check Fibonacci numberbool isFibonacci(int N){ if (N == 0 || N == 1) return true; int a = 0, b = 1, c; while (true) { c = a + b; a = b; b = c; if (c == N) return true; else if (c >= N) { return false; } }}int main(){ for (int i = 1; i <= 10; i++) { if (isFibonacci(i)) { cout << i << " is a Fibonacci number.\n"; } else { cout << i << " is not a Fibonacci number.\n"; } } return 0;}// This code is contributed by Susobhan Akhuli |
Java
public class GFG { // Function to check if a given number is a Fibonacci // number static boolean isFibonacci(int N) { // Fibonacci numbers start with 0 and 1, so they are // already Fibonacci if (N == 0 || N == 1) return true; // Initialize two variables to track Fibonacci // numbers int a = 0, b = 1, c; // Generate Fibonacci numbers until we reach N or a // number greater than N while (true) { // Calculate the next Fibonacci number in the // sequence c = a + b; a = b; b = c; // If the current Fibonacci number is equal to // N, it is a Fibonacci number if (c == N) return true; // If the current Fibonacci number is greater // than N, it is not a Fibonacci number else if (c >= N) { return false; } } } public static void main(String[] args) { // Loop from 1 to 10 to check if each number is a // Fibonacci number for (int i = 1; i <= 10; i++) { // Call the isFibonacci function to check if the // number is a Fibonacci number if (isFibonacci(i)) { System.out.println( i + " is a Fibonacci number."); } else { System.out.println( i + " is not a Fibonacci number."); } } }}// This code is contributed by shivamgupta310570 |
Python3
# Python program to check if a given number is# Fibonacci number or not# Function to check Fibonacci numberdef isFibonacci(N): if N == 0 or N == 1: return True a, b = 0, 1 while True: c = a + b a = b b = c if c == N: return True elif c >= N: return False# Driver Codeif __name__ == '__main__': for i in range(1, 11): if isFibonacci(i): print(i, "is a Fibonacci number.") else: print(i, "is not a Fibonacci number.")# This code is contributed by Aaysi Mishra |
C#
// C# program to check if a given number is// Fibonacci number or notusing System;public class GFG { // Function to check if a given number is a Fibonacci // number static bool IsFibonacci(int N) { // Fibonacci numbers start with 0 and 1, so they are // already Fibonacci if (N == 0 || N == 1) return true; // Initialize two variables to track Fibonacci // numbers int a = 0, b = 1, c; // Generate Fibonacci numbers until we reach N or a // number greater than N while (true) { // Calculate the next Fibonacci number in the // sequence c = a + b; a = b; b = c; // If the current Fibonacci number is equal to // N, it is a Fibonacci number if (c == N) return true; // If the current Fibonacci number is greater // than N, it is not a Fibonacci number else if (c >= N) { return false; } } } static void Main(string[] args) { // Loop from 1 to 10 to check if each number is a // Fibonacci number for (int i = 1; i <= 10; i++) { // Call the IsFibonacci function to check if the // number is a Fibonacci number if (IsFibonacci(i)) { Console.WriteLine( $"{i} is a Fibonacci number."); } else { Console.WriteLine( $"{i} is not a Fibonacci number."); } } }}// This code is contributed by Susobhan Akhuli |
Output
1 is a Fibonacci number. 2 is a Fibonacci number. 3 is a Fibonacci number. 4 is not a Fibonacci number. 5 is a Fibonacci number. 6 is not a Fibonacci number. 7 is not a Fibonacci number. 8 is a Fibonacci number. 9 is not a Fibonacci number. 10 is not a Fibonacci number.
Time Complexity: O(N), for iteration.
Auxiliary Space: O(1)
This article is contributed by Abhay Rathi. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
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