The Leibniz harmonic triangle is a triangular arrangement of unit fractions in which the outermost diagonals consist of the reciprocals of the row numbers and each inner cell is the cell diagonally above and to the left minus the cell to the left. To put it algebraically, L(r, 1) = 1/r, where r is the number of the row, starting from 1, and c is the number, never more than r and L(r, c) = L(r – 1, c – 1) – L(r, c – 1).
Relation with pascal’s triangle
Whereas each entry in Pascal’s triangle is the sum of the two entries in the above row, each entry in the Leibniz triangle is the sum of the two entries in the row below it. For example, in the 5th row, the entry (1/30) is the sum of the two (1/60)s in the 6th row.
Just as Pascal’s triangle can be computed by using binomial coefficients, so can Leibniz’s:
Properties
If one takes the denominators of the nth row and adds them, then the result will equal n.2n-1. For example, for the 3rd row, we have 3 + 6 + 3 = 12 = 3 × 22.
Given a positive integer n. The task is to print Leibniz harmonic triangle of height n.
Examples:
Input : n = 4 Output : 1 1/2 1/2 1/3 1/6 1/3 1/4 1/12 1/12 1/4 Input : n = 3 Output : 1 1/2 1/2 1/3 1/6 1/3
Below is the implementation of printing Leibniz harmonic triangle of height n based on above relation with Pascal triangle.
C++
// CPP Program to print Leibniz Harmonic Triangle #include <bits/stdc++.h> using namespace std; // Print Leibniz Harmonic Triangle void LeibnizHarmonicTriangle( int n) { int C[n + 1][n + 1]; // Calculate value of Binomial Coefficient in // bottom up manner for ( int i = 0; i <= n; i++) { for ( int j = 0; j <= min(i, n); j++) { // Base Cases if (j == 0 || j == i) C[i][j] = 1; // Calculate value using previously // stored values else C[i][j] = C[i - 1][j - 1] + C[i - 1][j]; } } // printing Leibniz Harmonic Triangle for ( int i = 1; i <= n; i++) { for ( int j = 1; j <= i; j++) cout << "1/" << i * C[i - 1][j - 1] << " " ; cout << endl; } } // Driven Program int main() { int n = 4; LeibnizHarmonicTriangle(n); return 0; } |
Java
// Java Program to print // Leibniz Harmonic Triangle import java.io.*; import java.math.*; class GFG { // Print Leibniz Harmonic Triangle static void LeibnizHarmonicTriangle( int n) { int C[][] = new int [n + 1 ][n + 1 ]; // Calculate value of Binomial // Coefficient in bottom up manner for ( int i = 0 ; i <= n; i++) { for ( int j = 0 ; j <= Math.min(i, n); j++) { // Base Cases if (j == 0 || j == i) C[i][j] = 1 ; // Calculate value using // previously stored values else C[i][j] = C[i - 1 ][j - 1 ] + C[i - 1 ][j]; } } // printing Leibniz Harmonic Triangle for ( int i = 1 ; i <= n; i++) { for ( int j = 1 ; j <= i; j++) System.out.print( "1/" + i * C[i - 1 ][j - 1 ] + " " ); System.out.println(); } } // Driven Program public static void main(String args[]) { int n = 4 ; LeibnizHarmonicTriangle(n); } } // This code is contributed by Nikita Tiwari |
Python3
# Python3 Program to print # Leibniz Harmonic Triangle # Print Leibniz Harmonic # Triangle def LeibnizHarmonicTriangle(n): C = [[ 0 for x in range (n + 1 )] for y in range (n + 1 )]; # Calculate value of Binomial # Coefficient in bottom up manner for i in range ( 0 , n + 1 ): for j in range ( 0 , min (i, n) + 1 ): # Base Cases if (j = = 0 or j = = i): C[i][j] = 1 ; # Calculate value using # previously stored values else : C[i][j] = (C[i - 1 ][j - 1 ] + C[i - 1 ][j]); # printing Leibniz # Harmonic Triangle for i in range ( 1 , n + 1 ): for j in range ( 1 , i + 1 ): print ( "1/" , end = ""); print (i * C[i - 1 ][j - 1 ], end = " " ); print (); # Driver Code LeibnizHarmonicTriangle( 4 ); # This code is contributed # by mits. |
C#
// C# Program to print Leibniz Harmonic Triangle using System; class GFG { // Print Leibniz Harmonic Triangle static void LeibnizHarmonicTriangle( int n) { int [,]C = new int [n + 1,n + 1]; // Calculate value of Binomial // Coefficient in bottom up manner for ( int i = 0; i <= n; i++) { for ( int j = 0; j <= Math.Min(i, n); j++) { // Base Cases if (j == 0 || j == i) C[i,j] = 1; // Calculate value using // previously stored values else C[i,j] = C[i - 1,j - 1] + C[i - 1,j]; } } // printing Leibniz Harmonic Triangle for ( int i = 1; i <= n; i++) { for ( int j = 1; j <= i; j++) Console.Write( "1/" + i * C[i - 1,j - 1] + " " ); Console.WriteLine(); } } // Driven Program public static void Main() { int n = 4; LeibnizHarmonicTriangle(n); } } // This code is contributed by vt_m. |
PHP
<?php // PHP Program to print // Leibniz Harmonic Triangle // Print Leibniz Harmonic Triangle function LeibnizHarmonicTriangle( $n ) { // Calculate value of // Binomial Coefficient in // bottom up manner for ( $i = 0; $i <= $n ; $i ++) { for ( $j = 0; $j <= min( $i , $n ); $j ++) { // Base Cases if ( $j == 0 || $j == $i ) $C [ $i ][ $j ] = 1; // Calculate value // using previously // stored values else $C [ $i ][ $j ] = $C [ $i - 1][ $j - 1] + $C [ $i - 1][ $j ]; } } // printing Leibniz // Harmonic Triangle for ( $i = 1; $i <= $n ; $i ++) { for ( $j = 1; $j <= $i ; $j ++) echo "1/" , $i * $C [ $i - 1][ $j - 1], " " ; echo "\n" ; } } // Driver Code $n = 4; LeibnizHarmonicTriangle( $n ); // This code is contributed by aj_36 ?> |
Javascript
<script> // JavaScript Program to print // Leibniz Harmonic Triangle // Print Leibniz Harmonic Triangle function LeibnizHarmonicTriangle(n) { let C = new Array(n + 1); // Loop to create 2D array using 1D array for (let i = 0; i < C.length; i++) { C[i] = new Array(2); } // Calculate value of Binomial // Coefficient in bottom up manner for (let i = 0; i <= n; i++) { for (let j = 0; j <= Math.min(i, n); j++) { // Base Cases if (j == 0 || j == i) C[i][j] = 1; // Calculate value using // previously stored values else C[i][j] = C[i - 1][j - 1] + C[i - 1][j]; } } // printing Leibniz Harmonic Triangle for (let i = 1; i <= n; i++) { for (let j = 1; j <= i; j++) document.write( "1/" + i * C[i - 1][j - 1] + " " ); document.write( "<br/>" ); } } // Driver Code let n = 4; LeibnizHarmonicTriangle(n); // This code is contributed by avijitmondal1998. </script> |
Output:
1/1 1/2 1/2 1/3 1/6 1/3 1/4 1/12 1/12 1/4
Time complexity: O(n2) for given n
Auxiliary Space: O(n2)
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