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lyudmila [28]
3 years ago
8

The set of three integer values for the lengths of the sides of a right triangle is called a Pythagorean triple. These three sid

es must satisfy the relationship that the sum of the two sides is equal to the square of the hypotenuse. Find all integer Pythagorean triples for side1, side2, and the hypotenuse, all no larger than 500. Use a triple-nested for loop that tries all possibilities. This program is an example of brute force computing. You will learn in more advanced computer science courses that there are many interesting problems for which there is no algorithmic approach other than using sheer brute force. This program does not need any input from the user. Write at least one bool function that takes the 3 sides of the triangle and returns true or false based on if it is a right triangle or not.
Computers and Technology
1 answer:
Sloan [31]3 years ago
8 0

Answer:

In Python:

def  Pythagorean_triple(hyp,side1,side2):

   if hyp**2 == side1**2 + side2*2:

       return True

   else:

       return False

               

print("Hypotenuse\tSide 1\t Side 2\t Return Value")

for i in range(1,501):

   for j in range(1,501):

       for k in range(1,501):

           print(str(i)+"\t"+str(j)+"\t"+str(k)+"\t"+str(Pythagorean_triple(i,j,k)))

Explanation:

This defines the function

def  Pythagorean_triple(hyp,side1,side2):

This checks for pythagorean triple

   if hyp**2 == side1**2 + side2*2:

Returns True, if true

       return True

   else:

Returns False, if otherwise

       return False

               

The main method begins

This prints the header

print("Hypotenuse\tSide 1\t Side 2\t Return Value")

The following is a triple-nested loop [Each of the loop is from 1 to 500]

for i in range(1,501): -->The hypotenuse

   for j in range(1,501): -->Side 1

       for k in range(1,501):-->Side 2

This calls the function and prints the required output i.e. the sides of the triangle and True or False

           print(str(i)+"\t"+str(j)+"\t"+str(k)+"\t"+str(Pythagorean_triple(i,j,k)))

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{  

int even_count = 0;    

while (n > 0)  

{  

 int rem = n % 10;  

 if (rem % 2 == 0)  

  even_count++;  

 n = n / 10;  

}  

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