CODEWITHSUNDEEP
python-3.6comparison-operators
MarkS
MarkS

Reputation: 1539

Multiple != versus == operators

The following program works exactly as expected:

data = input("Input the length of three sides of a triangle: ").strip()

if data.count(',') != 2:
    print("You did not enter the lengths of three sides: ")

a, b, c = data.split(', ')


def triangle_type(s1, s2, s3):
    if s1 == s2 == s3:
        return print("The values entered represent an equilateral triangle.")
    elif s1 != s2 and s1 != s3 and s2 != s3:
        return print("The values entered represent a scalene triangle.")
    else:
        if s1 == s2 or s1 == s3 or s2 == s3:
            return print("The values entered represent an isosceles triangle.")


triangle_type(int(a), int(b), int(c))

However, if I change the second condition to the following:

elif s1 != s2 != s3:

it does NOT work as expected.

If I input 3, 6, 3 I get:

Input the length of three sides of a triangle: 3, 6, 3
The values entered represent a scalene triangle.

This is incorrect. This should clearly match condition #3 (isosceles).

It works correctly with the first version of the code.

I am confused on two things:

1) Why does the second version work improperly?

2) If the second version is invalid, then why does the first condition work correctly?

Given the above behavior with the != operator, I would expect to have to treat a == b == c like a == b and a == c and b == c, but I don't have to.

Upvotes: 1

Views: 95

Answers (2)

Bill the Lizard
Bill the Lizard

Reputation: 406125

When you chain comparisons in Python, it ands them together, and it only does as many comparisons as you have comparison operators. So

if s1 == s2 == s3:

is equivalent to 

if s1 == s2 and s2 == s3:

That works because if both of those conditions are True, then s1 must also be equal to s3 (because they're both equal to s2).

However

if s1 != s2 != s3:

is equivalent to 

if s1 != s2 and s2 != s3:

As you've discovered, for the values 3, 6, and 3, the above two conditions are True, but the third assumed condition (s1 != s3) is not True. This doesn't work as you expect because Python isn't making that third comparison.

Upvotes: 1

chepner
chepner

Reputation: 532418

The chained version is equivalent to s1 != s2 and s2 != s3; it does not require that s1 != s3, because inequality is not transitive.

Upvotes: 1

Related Questions