Zero-dependency TypeScript library for regex utilities that go beyond string matching. These are surprisingly hard to come by for any programming language. β¨
- π Documentation
- π RegExp Equivalence Checker
- π Set-like operations:
- .and(...) - Compute intersection of two regex.
- .not() - Compute the complement of a regex.
- .without(...) - Compute the difference of two regex.
 
- β
 Set-like predicates:
- .isEquivalent(...) - Check whether two regex match the same strings.
- .isSubsetOf(...)
- .isSupersetOf(...)
- .isDisjointFrom(...)
 
- π Generate strings:
- .sample(...) - Generate random strings matching a regex.
- .enumerate() - Exhaustively enumerate strings matching a regex.
 
- π§ Miscellaneous:
- .size() - Count the number of strings that a regex matches.
- .derivative(...) - Compute a Brzozowski derivative of a regex.
 
npm install @gruhn/regex-utilsimport { RB } from '@gruhn/regex-utils'Generate 5 random email addresses:
const email = RB(/^[a-z]+@[a-z]+\.[a-z]{2,3}$/)
for (const str of email.sample().take(5)) {
  console.log(str)
}ky@e.no
cc@gg.gaj
z@if.ojk
vr@y.ehl
e@zx.hzq
Generate 5 random email addresses, which have exactly 20 characters:
const emailLength20 = email.and(/^.{20}$/)
for (const str of emailLength20.sample().take(5)) {
  console.log(str)
}kahragjijttzyze@i.mv
gnpbjzll@cwoktvw.hhd
knqmyotxxblh@yip.ccc
kopfpstjlnbq@lal.nmi
vrskllsvblqb@gemi.wc
Say we identified a regex in the code base that is prone to catastrophic backtracking and came up with a new version:
const oldRegex = /^(?:[a-zA-Z]\:\\|\\\\)([^\\\/\:\*\?\<\>\"\|]+(\\){0,1})+$/
const newRegex = /^(?:[a-zA-Z]:\\|\\\\)([^\\\/\:*?<>"|]+\\?)+$/Using .isEquivalent we can verify that the refactored version matches exactly the same strings as the old version.
That is, whether oldRegex.test(str) === newRegex.test(str) for every possible input string:
RB(oldRegex).isEquivalent(newRegex) // trueThere is also a web interface for checking regex equivalence which also shows counterexample strings if the two regular expressions are not equivalent. The source code is a single HTML file: ./equiv-checker.html.
 
How do you write a regex that matches HTML comments like:
<!-- This is a comment -->
A straight forward attempt would be:
<!--.*-->The problem is that .* also matches the end marker -->,
so this is also a match:
<!-- This is a comment --> and this shouldn't be part of it -->We need to specify that the inner part can be any string that does not contain -->.
With .not() (aka. regex complement) this is easy:
import { RB } from '@gruhn/regex-utils'
const strContainingCommentEnd = RB(/.*-->.*/)
const commentStart = RB('<!--')
const commentInner = RB(/^.*-->.*$/).not()
const commentEnd = RB('-->')
const comment = commentStart.concat(commentInner).concat(commentEnd)With .toRegExp() we can convert back to a native JavaScript regex:
comment.toRegExp()/^<!--(---*[^->]|-?[^-])*---*>$/
It's difficult to write a single regex for multiple independent constraints. For example, to specify a valid password. But with regex intersections it's very natural:
import { RB } from '@gruhn/regex-utils'
const passwordRegex = RB(/^[a-zA-Z0-9]{12,32}$/) // 12-32 alphanumeric characters
  .and(/[0-9]/) // at least one number
  .and(/[A-Z]/) // at least one upper case letter   
  .and(/[a-z]/) // at least one lower case letterWe can convert this back to a native JavaScript RegExp with:
passwordRegex.toRegExp()Note
The output RegExp can be very large.
We can also use other utilities like .size() to determine how many potential passwords match this regex:
console.log(passwordRegex.size())2301586451429392354821768871006991487961066695735482449920n
With .sample() we can generate some of these matches:
for (const str of passwordRegex.sample().take(10)) {
  console.log(str)
}NEWJIAXQISWT0Wwm
lxoegadrzeynezkmtfcIBzzQ9e
ypzvhvtwpWk4u6
MSZXXKIKEKWKXLQ8HQ7Ds
BCBSFBSMNOLKlgQN5L
8950244600709IW1pg
UOTQBLVOTZQWFSAJYBXZNQBEeom0l
520302447164378435bv4dp4ysC
71073970686490eY2Jt4
afgpnxqwUK5B
In the coding puzzle Advent Of Code 2023 - Day 12
you are given pairs of string patterns.
An example pair is .??..??...?##. and 1,1,3.
Both patterns describe a class of strings and the task is to count the number of strings that match both patterns.
In the first pattern, . and # stand for the literal characters "dot" and "hash".
The ? stands for either . or #.
This can be written as a regular expression:
- for #we simply write#
- for .we writeo(since.has a special meaning in regular expressions)
- for ?we write(o|#)
So the pattern .??..??...?##. would be written as:
const firstRegex = /^o(o|#)(o|#)oo(o|#)(o|#)ooo(o|#)##o$/In the second pattern, each digit stands for a sequence of # separated by at least one o.
This can also be written as a regular expression:
- For a digit like 3we write#{3}.
- Between digits we write o+.
- Additionally, arbitrary many oare allowed at the start and end, so we addo*at the start and end.
Thus, 1,1,3 would be written as:
const secondRegex = /^o*#{1}o+#{1}o+#{3}o*$/To solve the task and find the number of strings that match both regex,
we can use .and(...) and .size() from regex-utils.
.and(...) computes the intersection of two regular expressions.
That is, it creates a new regex which exactly matches the strings matched by both input regex.
const intersection = RB(firstRegex).and(secondRegex)With .size() we can then determine the number of matched strings:
console.log(intersection.size())4n
While at it, we can also try .enumerate() to list all these matches:
for (const str of intersection.enumerate()) {
  console.log(str)
}oo#ooo#ooo###o
o#oooo#ooo###o
oo#oo#oooo###o
o#ooo#oooo###o
For a full solution checkout: ./benchmark/aoc2023-day12.ts.
Heavily informed by these papers: