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Check whether parens are matched in a string
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import re | |
from hypothesis import strategies as st, given, assume, note, example | |
from typing import List, Optional | |
def is_balanced(mystr): | |
if len(mystr) % 2 != 0: | |
return False | |
stack = [] | |
for i in mystr: | |
if i == '(': | |
stack.append(')') | |
elif i == '[': | |
stack.append(']') | |
elif i == '{': | |
stack.append('}') | |
else: | |
if len(stack) == 0 or stack[-1] != i: | |
return False | |
else: | |
stack.pop() | |
return len(stack) == 0 | |
def is_balanced_brute_force_no_regex(text: str): | |
while (new_text := text.replace( | |
'()', '').replace('[]', '').replace('{}', '')) != text: | |
text = new_text | |
return text == '' | |
def is_balanced_brute_force(text: str): | |
while (new_text := re.sub(r'\(\)|{}|\[\]', '', text)) != text: | |
text = new_text | |
return text == '' | |
PAIRS = {'(': ')', '[': ']', '{': '}'} | |
def is_balanced_matthias(mystr: str): | |
stack: List[Optional[str]] = [None] | |
for c in mystr: | |
if c == stack[-1]: | |
stack.pop() | |
else: | |
stack.append(PAIRS.get(c)) | |
return len(stack) == 1 | |
def extend(part): | |
"""This meta-strategy extends a partial strategy to one that can: | |
- repeat the partial strategy | |
- or put various parens around it | |
""" | |
return (st.lists(part, min_size=2).map(''.join) | |
| part.map(lambda s: f"({s})") | |
| part.map(lambda s: f"[{s}]") | |
| part.map(lambda s: f"{{{s}}}") | |
) | |
# This strategy only produced balanced strings. | |
# Balanced strings are basically either the empty string, | |
# or an already balanced string that we are extending. | |
# | |
# Try in the REPL: | |
# >>> create_balanced.example() | |
# '{(({}{}(){}))}' | |
# >>> create_balanced.example() | |
# '({{{({})({})}}})' | |
create_balanced = st.recursive(st.just(''), extend) | |
@given(s=create_balanced) | |
def test_balanced(s): | |
assert is_balanced(s) | |
assert is_balanced_matthias(s) | |
assert is_balanced_brute_force(s) | |
# Here we either create random strings, | |
# or we deliberately create a balanced string. | |
parens = st.text(alphabet="()[]{}x") | create_balanced | |
@ given(s=parens) | |
def test_against_reference_implementation(s): | |
assert is_balanced_brute_force(s) == is_balanced(s) | |
def reverse_parens(s): | |
"""Turns eg '(hello {w]orld)' into '(dlro[w} olleh)' | |
Ie we reverse the string, and flip all the parens. | |
The input doesn't have to be balanced.""" | |
parens = {'(': ')', ')': '(', '[': ']', ']': '[', '{': '}', '}': '{'} | |
return ''.join(parens.get(c, c) for c in reversed(s)) | |
@ given(text=parens) | |
def test_reverse(text): | |
"""'Reversing' a string shouldn't change whether it's balanced or not.""" | |
reverse_text = reverse_parens(text) | |
note(f"reverse_text: {reverse_text}") | |
assert is_balanced(text) == is_balanced(reverse_text) | |
if __name__ == '__main__': | |
# print(create_balanced.example()) | |
test_against_reference_implementation() | |
test_balanced() | |
test_reverse() |
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