Can regex match intersection between two regular expressions? - regex

Question

Given several regular expressions, can we write a regular expressions which is equal to their intersection?
For example, given two regular expressions c[a-z][a-z] and [a-z][aeiou]t, their intersection contains cat and cut and possibly more. How can we write a regular expression for their intersection?
Thanks.

Answer 1

A logical AND in regex is represented by
(?=...)(?=...)
So,
(?=[a-z][aeiou]t)(?=c[a-z][a-z])

Answer 2

The lookahead examples are easy to use, but technically are no longer regular languages. However it is possible to take the intersection of two regular languages, and that complement is regular.
First note that Regular Expressions can be converted to and from NFAs; they both are ways of expressing regular languages.
Second, by DeMorgan's law,
Thus these are the steps to compute the intersection of two RegExs:
Convert both RegExs to NFAs.
Compute the complement of both NFAs.
Compute the union of the two complements.
Compute the complement of that union.
Convert the resulting NFA to a RegEx.
Some sources:
Union and RegEx to NFA: http://courses.engr.illinois.edu/cs373/sp2009/lectures/lect_06.pdf
NFA to RegEx: http://courses.engr.illinois.edu/cs373/sp2009/lectures/lect_08.pdf
Complement of NFA: https://cs.stackexchange.com/questions/13282/complement-of-non-deterministic-finite-automata

Answer 3

Mathematically speaking, an intersection of two regular languages is regular, so there has to be a regular expression that accepts it.
Building it via corresponding NFAs is probably the easiest. Consider the two NFAs that correspond to the two regexes. The new states Q are pairs (Q1,Q2) from the two NFAs. If there is a transition (P1,x,Q1) in the first NFA and (P2,x,Q2) in the second NFA, then and only then there is a transition ((P1,P2),x,(Q1,Q2)) in the new NFA. A new state (Q1,Q2) is initial/final iff both Q1 and Q2 are initial/final.
If you use NFAs with ε-moves, then also for each transition (P1,ε,Q1) there will be a transition ((P1,P2),ε,(Q1,P2)) for all states P2. Likewise for ε-moves in the second NFA.
Now convert the new NFA to a regular expression with any known algorithm, and that's it.
As for PCRE, they are not, strictly speaking, regular expressions. There is no way to do it in the general case. Sometimes you can use lookaheads, like ^(?=regex1$)(?=regex2$) but this is only good for matching the entire string and is no good for either searching or embedding in other regexps. Without anchoring, the two lookaheads may end up matching strings of different lengths. This is not intersection.

Answer 4

First, let's agree on terms. My syntactical assumption will be that
The intersection of several regexes is one regex that matches strings
that each of the component regexes also match.
The General Option
To check for the intersection of two patterns, the general method is (pseudo-code):
if match(regex1) && match(regex2) { champagne for everyone! }
The Regex Option
In some cases, you can do the same with lookaheads, but for a complex regex there is little benefit of doing so, apart from making your regex more obscure to your enemies. Why little benefit? Because the engine will have to parse the whole string multiple times anyway.
Boolean AND
The general pattern for an AND checking that a string exactly meets regex1 and regex2 would be:
^(?=regex1$)(?=regex2$)
The $ in each lookahead ensures that each string matches the pattern and nothing more.
Matching when AND
Of course, if you don't want to just check the boolean value of the AND but also do some actual matching, after the lookaheads, you can add a dot-star to consume the string:
^(?=regex1$)(?=regex2$).*
Or... After checking the first condition, just match the second:
^(?=regex1$)regex2$
This is a technique used for instance in password validation. For more details on this, see Mastering Lookahead and Lookbehind.
Bonus section: Union of Regexes
Instead of working on an intersection, let's say you are interested in the union of the following regexes, i.e., a regex that matches either of those regexes:
catch
cat1
cat2
cat3
cat5
This is accomplished with the alternation | operator:
catch|cat1|cat2|cat3|cat5
Furthermore, such a regex can often be compressed, as in:
cat(?:ch|[1-35])

Answer 5

For And operation, we have something like this in RegEx
(REGEX)(REGEX)
Taking your example
'Cat'.match(/^([A-Za-z]+)([aeiouAEIOU]+)([A-Za-z]+)$/)
["Cat", "C", "a", "t"]
'Ca'.match(/^([A-Za-z]+)([aeiouAEIOU]+)([A-Za-z]+)$/)
//null
'Cat123'.match(/^([A-Za-z]+)([aeiouAEIOU]+)([A-Za-z]+)$/)
//null
where
([A-Za-z]+) //Match All characters
and
([aeiouAEIOU]+) //Match all vowels
Combine them both will match
([A-Za-z]+)([aeiouAEIOU]+)([A-Za-z]+)
eg:
'Hmmmmmm'.match(/^([A-Za-z]+)([aeiouAEIOU]+)([A-Za-z]+)$/)
//null
'Stckvrflw'.match(/^([A-Za-z]+)([aeiouAEIOU]+)([A-Za-z]+)$/)
null
'StackOverflow'.match(/^([A-Za-z]+)([aeiouAEIOU]+)([A-Za-z]+)$/)
["StackOverflow", "StackOverfl", "o", "w"]