Compact way of writing (a + b == c or a + c == b or b + c == a)

Question

If we look at the Zen of Python, emphasis mine:

The Zen of Python, by Tim Peters

Beautiful is better than ugly.
Explicit is better than implicit.
Simple is better than complex.
Complex is better than complicated.
Flat is better than nested.
Sparse is better than dense.
Readability counts.
Special cases aren't special enough to break the rules.
Although practicality beats purity.
Errors should never pass silently.
Unless explicitly silenced.
In the face of ambiguity, refuse the temptation to guess.
There should be one-- and preferably only one --obvious way to do it.
Although that way may not be obvious at first unless you're Dutch.
Now is better than never.
Although never is often better than right now.
If the implementation is hard to explain, it's a bad idea.
If the implementation is easy to explain, it may be a good idea.
Namespaces are one honking great idea -- let's do more of those!

The most Pythonic solution is the one that is clearest, simplest, and easiest to explain:

a + b == c or a + c == b or b + c == a

Even better, you don't even need to know Python to understand this code! It's that easy. This is, without reservation, the best solution. Anything else is intellectual masturbation.

Furthermore, this is likely the best performing solution as well, as it is the only one out of all the proposals that short circuits. If a + b == c, only a single addition and comparison is done.

Solving the three equalities for a:

a in (b+c, b-c, c-b)

Python has an any function that does an or on all the elements of a sequence. Here I've converted your statement into a 3-element tuple.

any((a + b == c, a + c == b, b + c == a))

Note that or is short circuiting, so if calculating the individual conditions is expensive it might be better to keep your original construct.

If you know you're only dealing with positive numbers, this will work, and is pretty clean:

a, b, c = sorted((a, b, c))
if a + b == c:
    do_stuff()

As I said, this only works for positive numbers; but if you know they're going to be positive, this is a very readable solution IMO, even directly in the code as opposed to in a function.

You could do this, which might do a bit of repeated computation; but you didn't specify performance as your goal:

from itertools import permutations

if any(x + y == z for x, y, z in permutations((a, b, c), 3)):
    do_stuff()

Or without permutations() and the possibility of repeated computations:

if any(x + y == z for x, y, z in [(a, b, c), (a, c, b), (b, c, a)]:
    do_stuff()

I would probably put this, or any other solution, into a function. Then you can just cleanly call the function in your code.

Personally, unless I needed more flexibility from the code, I would just use the first method in your question. It's simple and efficient. I still might put it into a function:

def two_add_to_third(a, b, c):
    return a + b == c or a + c == b or b + c == a

if two_add_to_third(a, b, c):
    do_stuff()

That's pretty Pythonic, and it's quite possibly the most efficient way to do it (the extra function call aside); although you shouldn't worry too much about performance anyway, unless it's actually causing an issue.

If you will only be using three variables then your initial method:

a + b == c or a + c == b or b + c == a

Is already very pythonic.

If you plan on using more variables then your method of reasoning with:

a + b + c in (2*a, 2*b, 2*c)

Is very smart but lets think about why. Why does this work?
Well through some simple arithmetic we see that:

a + b = c
c = c
a + b + c == c + c == 2*c
a + b + c == 2*c

And this will have to hold true for either a,b, or c, meaning that yes it will equal 2*a, 2*b, or 2*c. This will be true for any number of variables.

So a good way to write this quickly would be to simply have a list of your variables and check their sum against a list of the doubled values.

values = [a,b,c,d,e,...]
any(sum(values) in [2*x for x in values])

This way, to add more variables into the equation all you have to do is edit your values list by 'n' new variables, not write 'n' equations