الانتقال إلى المحتوى الرئيسي
eLearner.app
الوحدة 2 · الدرس 4 من 48/32 في الدورة~12 min
دروس الوحدة (4/4)

التراجع: نظرة عامة

When a greedy quantifier "overshoots", the engine performs backtracking: it steps back, tries another configuration, retries. On simple patterns it is invisible; on poorly written patterns it can become exponential and freeze the browser for seconds. It is the root of ReDoS (Regular expression Denial of Service).

Code
Pattern: .*\d
Sample:  abc123def
Steps:
  1.  .*  consumes everything: "abc123def"
  2.  \d  fails (end of string).
  3.  Backtrack: .* drops 'f' -> "abc123de", \d fails on 'f'.
  4.  Backtrack: .* drops 'ef' -> "abc123d", \d fails on 'e'.
  5.  ... and so on until \d finds a digit: match = "abc123" + "?"... actually
       the pattern matches "abc123" because .* leaves "def" and \d looks at '3'? No: the
       backtrack continues and .* ends up being "abc12", \d=3.  Final match: "abc123".

The engine always gets there, but if a greedy quantifier contains another greedy quantifier (e.g. (a+)+), the number of configurations to try can explode.

Catastrophic patterns

Code
Pattern: (a+)+b
Sample:  aaaaaaaaaaaaaaaaaaaaa

No trailing b → the pattern fails, but to get there the engine tries every possible partition of the as. Exponential growth: with 30 as you can wait for hours.

Perils of backtracking and ReDoS

When a pattern contains overlapping or nested quantifiers (e.g. (a+)+), the engine evaluates an exponential number of paths before failing. Replacing generic wildcards with excluding character classes dramatically reduces backtracking paths.

Try it

تمرين#regex.m2.l4.e1
المحاولات: 0جارٍ التحميل…

Extract the content between square brackets for every occurrence in the text. Use the lazy version of the quantifier to avoid jumping to later closings.

جارٍ تحميل المحرر…
إظهار التلميح

\\[.*?\\] stops at the first closing ]: no heavy backtracking.

الحل متاح بعد 3 من المحاولات

Review exercise

تمرين#regex.m2.l4.e2
المحاولات: 0جارٍ التحميل…

Find every `http://...` or `https://...` URL in the text, using the `?` quantifier for the optional `s` and `\\S+` for the non-space characters of the path.

جارٍ تحميل المحرر…
إظهار التلميح

https? means 'http' followed by an optional 's'. Then //, then \\S+ (non-spaces).

الحل متاح بعد 3 من المحاولات

Additional challenge

تمرين#regex.m2.l4.e3
المحاولات: 0جارٍ التحميل…

Match strings enclosed in double quotes, avoiding excessive backtracking by using the negated class `[^"]*` instead of `.*`.

جارٍ تحميل المحرر…
إظهار التلميح

Replace the dot with the negated class [^"] to restrict the engine's backtracking paths.

الحل متاح بعد 3 من المحاولات