CODEWITHSUNDEEP
python
Jugo de mora
Jugo de mora

Reputation: 35

Is there a faster way to know if it's a real triangle?

The following code is intended to return 0 if it's not possible to form a triangle with three given sides and return 1 if it is possible. Taking into account that it's not possible for a triangle to have an area of zero or less, first I calculated the semiperimeter of this hypothetical triangle and then return 1 if the result is greater than the greatest value in the given list. I did the previous because if the semiperimeter is less than the greatest value in the list, then when calculating the area with the Heron formula it would be an undefined root.

def isTriangle(aList):
    semiperimeter = (aList[0] + aList[1] + aList[2]) / 2
    if semiperimeter > max(aList):
        return 1
    else:
        return 0

So I want to know if there is a faster way in Python to perform this (or at least with fewer lines of code).

Upvotes: 0

Views: 508

Answers (3)

Mad Physicist
Mad Physicist

Reputation: 114468

You can check if the sum of the two smaller sides is at least as large as the longest side:

def check(aList)
    return sum(aList) - max(aList) > max(aList)

With python 3.8+, you can avoid recomputing the maximum twice using a walrus operator:

def check(aList):
    return (m := max(aList)) < sum(aList) - m

Both solutions rely on numerical comparisons returning a big boolean. In python bool is a subclass of int that has only two possible singleton values: True == 1 and False == 0. It's standard to return a bool rather than 0 or 1 for something like this. In fact, your original conditional could be replaced by

return semiperimeter > max(aList)

Or, absorbing the previous line into the return:

return sum(aList) / 2 > max(aList)

You could also write

return sum(aList) > 2 * max(aList)

This version avoids any type conversions, regardless if whether you pass in floats or integers (/ 2 always results in a float).

Upvotes: 4

holydragon
holydragon

Reputation: 6728

To determine if given 3 sides can form a valid triangle, I would suggest using Triangle Inequality Theorem. It roughly states that the sum of any two sides must always be equal or greater than the other side to be able to form a valid triangle.

Therefore, I would instead write something along this line. [Edited: initialization according to Avinash]

def isTriangle(aList):
    a,b,c = aList
    return (a + b >= c and a + c >= b and b + c >= a)

In case you really want to make it return 0 or 1 instead of boolean then you can use this return instead.

return 1 if a + b >= c and a + c >= b and b + c >= a else 0

Upvotes: -1

ex4
ex4

Reputation: 2438

You asked for shorter code. That one is possible to do as a oneliner.

def isTriangle(aList):
    return ((aList[0] + aList[1] + aList[2]) / 2) > max(aList) and 1 or 0

Exatly the same code, but just written shorter

Upvotes: -1

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