def test_number_nodes(self, d):
"""if the root is an interior node, it must be numbered
two less than the number of symbols"""
# a complete tree has one fewer interior nodes than
# it has leaves, and we are numbering from 0
# NB: this also tests huffman_tree indirectly
t = huffman_tree(d)
assume(not t.is_leaf())
count = len(d)
number_nodes(t)
self.assertEqual(count, t.number + 2)
def test_num_nodes_to_bytes(self, b):
"""num_nodes_to_bytes returns a bytes object that
has length 1 (since the number of internal nodes cannot
exceed 256)"""
# NB: also indirectly tests make_freq_dict and huffman_tree
d = make_freq_dict(b)
assume(len(d) > 1)
t = huffman_tree(d)
number_nodes(t)
n = num_nodes_to_bytes(t)
self.assertTrue(isinstance(n, bytes))
self.assertEqual(len(n), 1)
def test_tree_to_bytes(self, b):
"""tree_to_bytes generates a bytes representation of
a post-order traversal of a trees internal nodes"""
# Since each internal node requires 4 bytes to represent,
# and there are 1 fewer internal node than distinct symbols,
# the length of the bytes produced should be 4 times the
# length of the frequency dictionary, minus 4"""
# NB: also indirectly tests make_freq_dict, huffman_tree, and
# number_nodes
d = make_freq_dict(b)
assume(len(d) > 1)
t = huffman_tree(d)
number_nodes(t)
output_bytes = tree_to_bytes(t)
dictionary_length = len(d)
leaf_count = dictionary_length
self.assertEqual(4 * (leaf_count - 1), len(output_bytes))
