Triangular Numbers Up To 100
A number is termed as triangular number if we tin can represent information technology in the form of triangular grid of points such that the points grade an equilateral triangle and each row contains equally many points equally the row number, i.e., the first row has 1 signal, 2nd row has two points, third row has 3 points and then on. The starting triangular numbers are one, 3 (1+2), 6 (1+2+three), x (1+2+3+4).
How to check if a number is Triangular?
The idea is based on the fact that due north'th triangular number can be written every bit sum of due north natural numbers, that is n*(n+1)/2. The reason for this is simple, base line of triangular filigree has n dots, line above base has (n-1) dots so on.
Method 1 (Simple)
We start with one and check if the number is equal to 1. If it is not, we add two to make it 3 and recheck with the number. We repeat this process until the sum remains less than or equal to the number that is to be checked for being triangular.
Following is the implementations to check if a number is triangular number.
C++
#include <iostream>
using namespace std;
bool isTriangular( int num)
{
if (num < 0)
return imitation ;
int sum = 0;
for ( int northward=1; sum<=num; n++)
{
sum = sum + due north;
if (sum==num)
return truthful ;
}
return false ;
}
int main()
{
int n = 55;
if (isTriangular(n))
cout << "The number is a triangular number" ;
else
cout << "The number is NOT a triangular number" ;
return 0;
}
Java
course GFG
{
static boolean isTriangular( int num)
{
if (num < 0 )
return false ;
int sum = 0 ;
for ( int north = 1 ; sum <= num; n++)
{
sum = sum + n;
if (sum == num)
return true ;
}
return false ;
}
public static void main (String[] args)
{
int n = 55 ;
if (isTriangular(n))
Organization.out.print( "The number "
+ "is a triangular number" );
else
System.out.print( "The number"
+ " is NOT a triangular number" );
}
}
Python3
def isTriangular(num):
if (num < 0 ):
render False
sum , n = 0 , 1
while ( sum < = num):
sum = sum + due north
if ( sum = = num):
render Truthful
north + = 1
render Imitation
due north = 55
if (isTriangular(north)):
print ( "The number is a triangular number" )
else :
print ( "The number is NOT a triangular number" )
C#
using System;
class GFG {
static bool isTriangular( int num)
{
if (num < 0)
return false ;
int sum = 0;
for ( int north = i; sum <= num; n++)
{
sum = sum + northward;
if (sum == num)
return true ;
}
return imitation ;
}
public static void Main ()
{
int north = 55;
if (isTriangular(n))
Console.WriteLine( "The number "
+ "is a triangular number" );
else
Console.WriteLine( "The number"
+ " is NOT a triangular number" );
}
}
PHP
<?php
function isTriangular( $num )
{
if ( $num < 0)
return false;
$sum = 0;
for ( $n = one; $sum <= $num ; $n ++)
{
$sum = $sum + $n ;
if ( $sum == $num )
render truthful;
}
return false;
}
$n = 55;
if (isTriangular( $n ))
echo "The number is a triangular number" ;
else
repeat "The number is Non a triangular number" ;
?>
Javascript
<script>
function isTriangular(num)
{
if (num < 0)
return faux ;
let sum = 0;
for (let n = 1; sum <= num; n++)
{
sum = sum + north;
if (sum == num)
return true ;
}
return simulated ;
}
allow n = 55;
if (isTriangular(n))
document.write( "The number is a triangular number" );
else
document.write( "The number is NOT a triangular number" );
</script>
Output:
The number is a triangular number
Fourth dimension Complexity: O(√n)
Auxiliary Space: O(1)
Method 2 (Using Quadratic Equation Root Formula)
We form a quadratic equation by equating the number to the formula of sum of first 'n' natural numbers, and if we get atleast i value of 'n' that is a natural number, we say that the number is a triangular number.
Permit the input number be 'num'. We consider, n*(n+ane) = num as, ntwo + n + (-two * num) = 0
Beneath is the implementation of higher up idea.
C++
#include <bits/stdc++.h>
using namespace std;
bool isTriangular( int num)
{
if (num < 0)
return fake ;
int c = (-ii * num);
int b = 1, a = 1;
int d = (b * b) - (four * a * c);
if (d < 0)
return false ;
float root1 = ( -b + sqrt (d)) / (2 * a);
float root2 = ( -b - sqrt (d)) / (2 * a);
if (root1 > 0 && floor (root1) == root1)
return true ;
if (root2 > 0 && flooring (root2) == root2)
return true ;
return false ;
}
int main()
{
int num = 55;
if (isTriangular(num))
cout << "The number is a triangular number" ;
else
cout << "The number is NOT a triangular number" ;
return 0;
}
Coffee
import java.io.*;
class GFG {
static boolean isTriangular( int num)
{
if (num < 0 )
return fake ;
int c = (- two * num);
int b = 1 , a = i ;
int d = (b * b) - ( four * a * c);
if (d < 0 )
render fake ;
float root1 = ( -b +
( bladder )Math.sqrt(d)) / ( 2 * a);
float root2 = ( -b -
( bladder )Math.sqrt(d)) / ( 2 * a);
if (root1 > 0 && Math.floor(root1)
== root1)
render true ;
if (root2 > 0 && Math.floor(root2)
== root2)
return true ;
render false ;
}
public static void main (Cord[] args) {
int num = 55 ;
if (isTriangular(num))
Organization.out.println( "The number is"
+ " a triangular number" );
else
System.out.println ( "The number "
+ "is NOT a triangular number" );
}
}
Python3
import math
def isTriangular(num):
if (num < 0 ):
render Fake
c = ( - 2 * num)
b, a = 1 , ane
d = (b * b) - ( 4 * a * c)
if (d < 0 ):
return False
root1 = ( - b + math.sqrt(d)) / ( two * a)
root2 = ( - b - math.sqrt(d)) / ( 2 * a)
if (root1 > 0 and math.floor(root1) = = root1):
return Truthful
if (root2 > 0 and math.flooring(root2) = = root2):
return Truthful
return False
n = 55
if (isTriangular(due north)):
print ( "The number is a triangular number" )
else :
print ( "The number is NOT a triangular number" )
C#
using System;
grade GFG {
static bool isTriangular( int num)
{
if (num < 0)
return false ;
int c = (-2 * num);
int b = 1, a = one;
int d = (b * b) - (4 * a * c);
if (d < 0)
return faux ;
float root1 = ( -b + ( float )Math.Sqrt(d))
/ (two * a);
float root2 = ( -b - ( float )Math.Sqrt(d))
/ (2 * a);
if (root1 > 0 && Math.Floor(root1) == root1)
render true ;
if (root2 > 0 && Math.Flooring(root2) == root2)
render truthful ;
return imitation ;
}
public static void Main () {
int num = 55;
if (isTriangular(num))
Console.WriteLine( "The number is a "
+ "triangular number" );
else
Panel.WriteLine ( "The number is Non "
+ "a triangular number" );
}
}
PHP
<?php
office isTriangular( $num )
{
if ( $num < 0)
return false;
$c = (-2 * $num );
$b = 1; $a = 1;
$d = ( $b * $b ) - (4 * $a * $c );
if ( $d < 0)
return faux;
$root1 = (- $b + (float)sqrt( $d )) / (2 * $a );
$root2 = (- $b - (float)sqrt( $d )) / (two * $a );
if ( $root1 > 0 && flooring ( $root1 ) == $root1 )
render truthful;
if ( $root2 > 0 && floor ( $root2 ) == $root2 )
return true;
return faux;
}
$num = 55;
if (isTriangular( $num ))
echo ( "The number is" .
" a triangular number" );
else
echo ( "The number " .
"is Not a triangular number" );
?>
Javascript
<script>
function isTriangular(num)
{
if (num < 0)
return false ;
var c = (-two * num);
var b = one, a = 1;
var d = (b * b) - (four * a * c);
if (d < 0)
return imitation ;
var root1 = (-b + Math.sqrt(d)) / (ii * a);
var root2 = (-b - Math.sqrt(d)) / (2 * a);
if (root1 > 0 && Math.floor(root1) == root1)
render true ;
if (root2 > 0 && Math.floor(root2) == root2)
render truthful ;
render false ;
}
var num = 55;
if (isTriangular(num))
certificate.write("The number is " + " a triangular number ");
else
document.write(" The number " + " is NOT a triangular number");
</script>
Output:
The number is a triangular number
Fourth dimension Complexity: O(logn)
Auxiliary Space: O(ane), since no extra space has been taken.
Please write comments if you discover anything incorrect, or yous want to share more information about the topic discussed in a higher place
Triangular Numbers Up To 100,
Source: https://www.geeksforgeeks.org/triangular-numbers/
Posted by: graydowits.blogspot.com
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