Ordering problem in intervals

[jefftrull]

Intervals implement the special comparison methods, but in some corner cases they behave in surprising ways:

import portion as P

a = P.openclosed(0, 1)
b = P.closedopen(1, 2)

print(b > a)
print(b == a)
print(b >= a)

prints:

False
False
True

Similarly,

print(a < b)
print(a == b)
print(a <= b)

prints

False
False
True

This violates the expectation that (x >= y) == ((x > y) or (x == y)).

In my view the right fix is to make each of the first expressions above - that is, b > a and a < b - return true for these cases.

[AlexandreDecan]

Hello,

While it may seem incorrect, this is the expected behaviour of comparison operators. As stated in the documentation (see https://github.com/AlexandreDecan/portion#comparison-operators):

Moreover, intervals are comparable using >, >=, < or <=. These comparison operators have a different behaviour than the usual ones. For instance, a < b holds if a is entirely on the left of the lower bound of b and a > b holds if a is entirely on the right of the upper bound of b. [...] Similarly, a <= b holds if a is entirely on the left of the upper bound of b, and a >= b holds if a is entirely on the right of the lower bound of b. [...] Note that all these semantics differ from classical comparison operators. As a consequence, some intervals are never comparable in the classical sense.

For example:

>>> P.closed(0, 4) <= P.closed(1, 2) or P.closed(0, 4) >= P.closed(1, 2)
False
>>> P.closed(0, 4) < P.closed(1, 2) or P.closed(0, 4) > P.closed(1, 2)
False
>>> P.empty() < P.empty()
True
>>> P.empty() < P.closed(0, 1) and P.empty() > P.closed(0, 1)
True

[jefftrull]

Wow! OK, thanks. I missed that part in the docs.

[AlexandreDecan]

No problem ;-)