Given an integer N, the task is to print the first N terms of the Lower Wythoff sequence.
Examples:
Input: N = 5
Output: 1, 3, 4, 6, 8Input: N = 10
Output: 1, 3, 4, 6, 8, 9, 11, 12, 14, 16
Approach: Lower Wythoff sequence is a sequence whose nth term is a(n) = floor(n * phi) where phi = (1 + sqrt(5)) / 2. So, we run a loop and find the first n terms of the sequence.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to print the first n terms// of the lower Wythoff sequencevoid lowerWythoff(int n){ // Calculate value of phi double phi = (1 + sqrt(5)) / 2.0; // Find the numbers for (int i = 1; i <= n; i++) { // a(n) = floor(n * phi) double ans = floor(i * phi); // Print the nth numbers cout << ans; if (i != n) cout << ", "; }}// Driver codeint main(){ int n = 5; lowerWythoff(n); return 0;} |
Java
// Java implementation of the approachclass GFG{// Function to print the first n terms// of the lower Wythoff sequencestatic void lowerWythoff(int n){ // Calculate value of phi double phi = (1 + Math.sqrt(5)) / 2.0; // Find the numbers for (int i = 1; i <= n; i++) { // a(n) = floor(n * phi) double ans = Math.floor(i * phi); // Print the nth numbers System.out.print((int)ans); if (i != n) System.out.print(" , "); }}// Driver codepublic static void main(String[] args) { int n = 5; lowerWythoff(n);}}// This code is contributed by 29AjayKumar |
Python3
# Python3 implementation of the approach # from math import sqrt,floorfrom math import sqrt, floor# Function to print the first n terms # of the lower Wythoff sequence def lowerWythoff(n) : # Calculate value of phi phi = (1 + sqrt(5)) / 2; # Find the numbers for i in range(1, n + 1) : # a(n) = floor(n * phi) ans = floor(i * phi); # Print the nth numbers print(ans,end=""); if (i != n) : print( ", ",end = ""); # Driver code if __name__ == "__main__" : n = 5; lowerWythoff(n);# This code is contributed by AnkitRai01 |
C#
// C# implementation of the approachusing System;class GFG{ // Function to print the first n terms// of the lower Wythoff sequencestatic void lowerWythoff(int n){ // Calculate value of phi double phi = (1 + Math.Sqrt(5)) / 2.0; // Find the numbers for (int i = 1; i <= n; i++) { // a(n) = floor(n * phi) double ans = Math.Floor(i * phi); // Print the nth numbers Console.Write((int)ans); if (i != n) Console.Write(" , "); }}// Driver codestatic public void Main (){ int n = 5; lowerWythoff(n);}}// This code is contributed by ajit. |
Javascript
<script>// Javascript implementation of above approach// Function to print the first n terms// of the lower Wythoff sequencefunction lowerWythoff(n){ // Calculate value of phi var phi = (1 + Math.sqrt(5)) / 2.0; // Find the numbers for(var i = 1; i <= n; i++) { // a(n) = floor(n * phi) var ans = Math.floor(i * phi); // Print the nth numbers document.write( ans); if (i != n) document.write(", "); }}// Driver codevar n = 5;lowerWythoff(n);// This code is contributed by SoumikMondal</script> |
Output:
1, 3, 4, 6, 8
Time Complexity: O(n)
Auxiliary Space: O(1)
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