I want to find out if there could ever be conflicts between two known regular expressions, in order to allow the user to construct a list of mutually exclusive regular expressions.
For example, we know that the regular expressions below are quite different but they both match xy50
:
'^xy1\d' '[^\d]\d2$'
Is it possible to determine, using a computer algorithm, if two regular expressions can have such a conflict? How?
We say that two regular expressions R and S are equivalent if they describe the same language. In other words, if L(R) = L(S) for two regular expressions R and S then R = S. Examples. Are R and S equivalent?
To match a character having special meaning in regex, you need to use a escape sequence prefix with a backslash ( \ ). E.g., \. matches "." ; regex \+ matches "+" ; and regex \( matches "(" . You also need to use regex \\ to match "\" (back-slash).
Literal Characters and Sequences For instance, you might need to search for a dollar sign ("$") as part of a price list, or in a computer program as part of a variable name. Since the dollar sign is a metacharacter which means "end of line" in regex, you must escape it with a backslash to use it literally.
There's no halting problem involved here. All you need is to compute if the intersection of ^xy1\d
and [^\d]\d2$
in non-empty.
I can't give you an algorithm here, but here are two discussions of a method to generate the intersection without resorting the construction of a DFA:
And then there's RAGEL
which can compute the intersection of regular expressions too.
UPDATE: I just tried out Ragel with OP's regexp. Ragel can generate a "dot" file for graphviz from the resulting state machine, which is terrific. The intersection of the OP's regexp looks like this in Ragel syntax:
('xy1' digit any*) & (any* ^digit digit '2')
and has the following state machine:
While the empty intersection:
('xy1' digit any*) & ('q' any* ^digit digit '2')
looks like this:
So if all else fails, then you can still have Ragel compute the intersection and check if it outputs the empty state machine, by comparing the generated "dot" file.
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