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bpo-41972: Use the two-way algorithm for string searching #22904

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Feb 28, 2021
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39 changes: 39 additions & 0 deletions Lib/test/string_tests.py
Original file line number Diff line number Diff line change
Expand Up @@ -5,6 +5,7 @@
import unittest, string, sys, struct
from test import support
from collections import UserList
import random

class Sequence:
def __init__(self, seq='wxyz'): self.seq = seq
Expand Down Expand Up @@ -317,6 +318,44 @@ def test_rindex(self):
else:
self.checkraises(TypeError, 'hello', 'rindex', 42)

def test_find_periodic_pattern(self):
"""Cover the special path for periodic patterns."""
def reference_find(p, s):
for i in range(len(s)):
if s.startswith(p, i):
return i
return -1

rr = random.randrange
choices = random.choices
for _ in range(1000):
p0 = ''.join(choices('abcde', k=rr(10))) * rr(10, 20)
p = p0[:len(p0) - rr(10)] # pop off some characters
left = ''.join(choices('abcdef', k=rr(2000)))
right = ''.join(choices('abcdef', k=rr(2000)))
text = left + p + right
with self.subTest(p=p, text=text):
self.checkequal(reference_find(p, text),
text, 'find', p)

def test_find_shift_table_overflow(self):
"""When the table of 8-bit shifts overflows."""
N = 2**8 + 100

# first check the periodic case
# here, the shift for 'b' is N + 1.
pattern1 = 'a' * N + 'b' + 'a' * N
text1 = 'babbaa' * N + pattern1
self.checkequal(len(text1)-len(pattern1),
text1, 'find', pattern1)

# now check the non-periodic case
# here, the shift for 'd' is 3*(N+1)+1
pattern2 = 'ddd' + 'abc' * N + "eee"
text2 = pattern2[:-1] + "ddeede" * 2 * N + pattern2 + "de" * N
self.checkequal(len(text2) - N*len("de") - len(pattern2),
text2, 'find', pattern2)

def test_lower(self):
self.checkequal('hello', 'HeLLo', 'lower')
self.checkequal('hello', 'hello', 'lower')
Expand Down
Original file line number Diff line number Diff line change
@@ -0,0 +1 @@
Substring search functions such as ``str1 in str2`` and ``str2.find(str1)`` now sometimes use the "Two-Way" string comparison algorithm to avoid quadratic behavior on long strings.
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