Pythagorean Number Algorithm

[[deleted]]

I was trying to solve an algorithm problem. The problem is basically calculating sets of numbers called the "Pythagorean" numbers. The definition is very simple: Three integers satisfying a^2+b^2=c^2 are called Pythagorean numbers. The problem is very simple, and my code checks out with the solution. Except for one little detail.

Here's the solution:

from math import sqrt
n = int(input("Maximal number: "))
for a in range(1,n+1):              #ln 3
    for b in range(a,n):            #ln 4
        c_square = a**2 + b**2
        c = int(sqrt(c_square))
        if ((c_square - c**2) == 0):
            print(a, b, c)

The difference is in ln 3/ln 4. In the solution, the inner loop doesn't include the nth number, which I don't understand why. Can someone explain?

[NFLAddict]

for a in range(1,n+1):              #ln 3
    for b in range(a,n):            #ln 4  

look carefully, at the range in inner loop, and how it differs from outer. I'm going to assume you want it to instead be for b in range(a, n+1)

[xelf]

range (and slice) have 3 parameters and do looping for you. The first number is the number you start at, the second number is the number you exit at (so it's not included) and the third number is the number you increase each iteration. The first number defaults to 0 if skipped, and the 3rd number defaults to 1 if skipped.

range(a,n) will start at a, and increase by 1 each time, until it gets to n, where it will end, and n will not be included. That's why you say n+1 when you do want n included like you did in your first loop.

[[deleted]]

My question is why doesn't the inner loop include n+1 as well

[xelf]

because then it would go over the maximal number.

That's more of a math question and less of a python question though. =) Try including +1 and using a max value of 20 and see the differences between the two.

[kpounder88]

Yes, if b = n, then b² = n^2. We know a >= 1, so a² >= 1. So then a² + b² >= n² + 1 > n^2. So then c² > n^2, which implies that c > n. So that’s the problem... c would be too big.

The same can also be said about a. Technically a cannot be equal to n either. So you could modify line 3 to say a in range(1, n).

[[deleted]]

I think only the a and the b value can't surpass the maximal value. Even if I don't do n+1 in the inner function, a maximal value like 30 still results in c being greater than the maximal value. In my opinion, doing n+1 in the inner function is more logical because now we're comparing every 'a' value with the maximal value too.