Test script for bidict.frozenbidict::

    >>> from bidict import frozenbidict, frozenorderedbidict
    >>> f1 = frozenbidict(one=1)
    >>> f2 = frozenorderedbidict([('one', 1), ('two', 2)])
    >>> f3 = frozenorderedbidict([('two', 2), ('one', 1)])
    >>> fs = (f1, f2, f3)
    >>> all(hash(f) is not 'an error' for f in fs)
    True

Cached hash value reused (for the sake of the coverage report)::

    >>> all(hash(f) for f in fs)  # should not have to call _compute_hash() again
    True

Insertable into sets and dicts::

    >>> dict.fromkeys(fs) and set(fs) is not 'an error'
    True

Can toggle _USE_ITEMSVIEW_HASH (affects space complexity but should produce the
same result)::

    >>> frozenbidict._USE_ITEMSVIEW_HASH = not frozenbidict._USE_ITEMSVIEW_HASH
    >>> hash(f1.copy()) == f1._hashval  # matches value computed + cached on f1 before toggling
    True

Unlikely that different types of collections comprising the same items hash to
the same value::

    >>> hash(f1) == hash(frozenset(f1.items()))  # very unlikely
    False
    >>> from itertools import chain  # frozenorderedbidict._compute_hash flattens items
    >>> hash(f2) == hash(tuple(chain.from_iterable(f2.items())))  # very unlikely
    False

To reduce hash collisions with unequal frozenorderedbidicts with the same items,
frozenorderedbidict.__hash__ is order-sensitive::

    >>> f1 != f2
    True
    >>> hash(f1) == hash(f2)  # very unlikely
    False

Can customize frozenorderedbidict._compute_hash to limit the number of items
considered::

    >>> frozenorderedbidict._HASH_NITEMS_MAX = 1
    >>> hash(frozenorderedbidict([(1, 1), (2, 2)])) == \
    ... hash(frozenorderedbidict([(1, 1), (3, 3)]))  # second items ignored
    True