Draw flowchart in java to accept three numbers and check whether they are Pythagorean triplet or not
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Answer:
GEEKSFORGEEKS
Generate Pythagorean Triplets
A Pythagorean triplet is a set of three positive integers a, b and c such that a2 + b2 = c2. Given a limit, generate all Pythagorean Triples with values smaller than given limit.
Input : limit = 20
Output : 3 4 5
8 6 10
5 12 13
15 8 17
12 16 20
A Simple Solution is to generate these triplets smaller than given limit using three nested loop. For every triplet, check if Pythagorean condition is true, if true, then print the triplet. Time complexity of this solution is O(limit3) where ‘limit’ is given limit.
An Efficient Solution can print all triplets in O(k) time where k is number of triplets printed. The idea is to use square sum relation of Pythagorean triplet, i.e., addition of squares of a and b is equal to square of c, we can write these number in terms of m and n such that,
a = m2 - n2
b = 2 * m * n
c = m2 + n2
because,
a2 = m4 + n4 – 2 * m2 * n2
b2 = 4 * m2 * n2
c2 = m4 + n4 + 2* m2 * n2
We can see that a2 + b2 = c2, so instead of iterating for a, b and c we can iterate for m and n and can generate these triplets.
Below is the implementation of above idea :
// C++ program to generate pythagorean
// triplets smaller than a given limit
#include <bits/stdc++.h>
// Function to generate pythagorean
// triplets smaller than limit
void pythagoreanTriplets(int limit)
{
// triplet: a^2 + b^2 = c^2
int a, b, c = 0;
// loop from 2 to max_limitit
int m = 2;
// Limiting c would limit
// all a, b and c
while (c < limit) {
// now loop on j from 1 to i-1
for (int n = 1; n < m; ++n) {
// Evaluate and print triplets using
// the relation between a, b and c
a = m * m - n * n;
b = 2 * m * n;
c = m * m + n * n;
if (c > limit)
break;
printf("%d %d %d\n", a, b, c);
}
m++;
}
}
// Driver Code
int main()
{
int limit = 20;
pythagoreanTriplets(limit);
return 0;
}
Output :
3 4 5
8 6 10
5 12 13
15 8 17
12 16 20
Time complexity of this approach is O(k) where k is number of triplets printed for a given limit (We iterate for m and n only and every iteration prints a triplet)
Note: The above method doesn’t generate all triplets smaller than a given limit. For example “9 12 15” which is a valid triplet is not printed by above method. Thanks to Sid Agrawal for pointing this out.