Count number of right triangles possible with a given perimeter

Given a perimeter P, the task is to find the number of right triangles possible with perimeter equal to p.


Input: P = 12
Output: number of right triangles = 1 
The only right angle possible is with sides 
hypotenuse = 5, perpendicular = 4 and base = 3. 

Input: p = 840
Output: number of right triangles = 8

So the aim is to find the number of solutions which satisfy equations a + b + c = p and a2 + b2 = c2.

A naive approach is to run two loops for a(1 to p/2) and b(a+1 to p/3) then make c=p-a-b and increase count by one if . This will take time.

An efficient approach can be found by little algebraic manipulation :

Since a + c > b or, p – b > b or, b < p/2. Thus iterating b from 1 to p/2, calculating a and storing only the whole number a would give all solutions for a given p. There are no right triangles are possible for odd p as right angle triangles follow the Pythagoras theorem. Use a list of pairs to store the values of a and band return the count at the end.

Below is the implementation of the above approach.





# python program to find the number of
# right triangles with given perimeter
# Function to return the count 
def countTriangles(p):
    # making a list to store (a, b) pairs
    store =[]
    # no triangle if p is odd
    if p % 2 != 0 : return 0
    else :
        count = 0
        for b in range(1, p // 2):
            a = p / 2 * ((p - 2 * b) / (p - b))
            inta = int(a)
            if (a == inta ):
                # make (a, b) pair in sorted order 
                ab = tuple(sorted((inta, b)))
                # check to avoid duplicates
                if ab not in store :
                    count += 1
                    # store the new pair 
        return count
# Driver Code
p = 840
print("number of right triangles = "+str(countTriangles(p)))

number of right triangles = 8

Time complexity: O(P)

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Improved By : siddhanthapliyal

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