Question

How do __hash__ and __eq__ use in identification in sets?
For example some code that should help to solve some domino puzzle:

class foo(object):
    def __init__(self, one, two):
        self.one = one
        self.two = two
    def __eq__(self,other):
        if (self.one == other.one) and (self.two == other.two): return True
        if (self.two == other.one) and (self.one == other.two): return True
        return False
    def __hash__(self):
        return hash(self.one + self.two)

s = set()

for i in range(7):
    for j in range(7):
        s.add(foo(i,j))
len(s) // returns 28 Why?

If i use only __eq__() len(s) equals 49. Its ok because as i understand objects (1-2 and 2-1 for example) not same, but represent same domino. So I have added hash function.
Now it works the way i want, but i did not understand one thing: hash of 1-3 and 2-2 should be same so they should counted like same object and shouldn’t added to set. But they do! Im stuck.

Answer 1

Equality for dict/set purposes depends on equality as defined by __eq__. However, it is required that objects that compare equal have the same hash value, and that is why you need __hash__. See this question for some similar discussion.

The hash itself does not determine whether two objects count as the same in dictionaries. The hash is like a “shortcut” that only works one way: if two objects have different hashes, they are definitely not equal; but if they have the same hash, they still might not be equal.

In your example, you defined __hash__ and __eq__ to do different things. The hash depends only on the sum of the numbers on the domino, but the equality depends on both individual numbers (in order). This is legal, since it is still the case that equal dominoes have equal hashes. However, like I said above, it doesn’t mean that equal-sum dominoes will be considered equal. Some unequal dominoes will still have equal hashes. But equality is still determined by __eq__, and __eq__ still looks at both numbers, in order, so that’s what determines whether they are equal.

It seems to me that the appropriate thing to do in your case is to define both __hash__ and __eq__ to depend on the ordered pair — that is, first compare the greater of the two numbers, then compare the lesser. This will mean that 2-1 and 1-2 will be considered the same.