def test_ambig_highlight(self):
p = self.p
p.add(
ParseRule(
"top", "top",
[Terminal("a"), NonTerminal("a"),
Terminal("a")]))
p.add(ParseRule("1", "a", [Terminal("a")]))
p.add(ParseRule("2", "a", [Terminal("a")]))
self.ambig("a a a", start_index=1, end_index=2)
def test_complexity(self):
# Correctly written an Earley parse should be O(n) for several sorts of grammar
# and O(n^3) worse case
p = self.p
p.add(ParseRule("1", "top", [NonTerminal("a", star=True)]))
p.add(ParseRule("2", "a", [Terminal("a")]))
p.add(ParseRule("3", "a", [Terminal("a")]))
# Assuming the EarleyParser is correct, this should only take linear time/space to compute, despite 2^n possible
# parses. Note that you can set this value well beyond Python's recursion depth
n = 1000
forest = parse(self.p, "top", lex("a " * n))
self.assertEqual(forest.count(), 2**n)
self.assertEqual(forest.internal_node_count, 4 + 5 * n)
