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In python 3.x, are there some a, b such that max(a, b) != max(b, a)?
a and b should be of a built-in python type and do not contain any conversion.
The type of a and b can be:
num, float, bool, list, tuple ...
Example:
a = 1 # Correct
a = [1, 2] # Correct
a = float('nan') # Wrong
This satisfies all criteria:
>>> max(1.0, float("nan"))
1.0
>>> max(float("nan"), 1.0)
nan
This is because any comparisons other than != or is with float("nan") always return false.
More generally, we can have max(a, b) != max(b, a) when the type(s) of a, b do not provide a total order. As Daniel Mesejo points out, sets have this property as well. For sets < means 'strict subset', but with {1, 2} and {3, 4} neither is a strict subset of the other, so:
>>> max({1, 2}, {3, 4})
{1, 2}
>>> max({3, 4}, {1, 2})
{3, 4}
This one technically doesn't have max(a, b) != max(b, a) because 1 == True, but there's still an observable difference:
>>> max(True, 1)
True
>>> max(1, True)
1
According to the documentation, using max on sets produces undefined results:
Accordingly, sets are not appropriate arguments for functions which depend on total ordering (for example, min(), max(), and sorted() produce undefined results given a list of sets as inputs).
Here is an example,
a = {1, 2}
b = {2, 3}
print(max(b, a) != max(a, b))
Output
True