Simplify a complex regular expression

Question

I am looking for a way to simplify a regular expression which consists of values (e.g. 12345), relation signs (<,>,<=,>=) and junctors (&,!). E.g. the expression:

>= 12345 & <=99999 & !55555 

should be matched. I have this regular expression:

(^<=|^<= | ^>= | ^>= |^<|^>|^< |^> |^)((!|)([0-9]{1,5}))( & > | & < |& >=|&>=|&<=||&<=|&>=|&<|&>|&| &| & |$))*

I am especially unhappy with the repetition of <=, >=, <, > at the beginning and end of the expression. I would be glad to get a hint how to make it simpler e.g. look ahead, look back.

Answer 1

Starting from your regex, you can do this simplification steps:

 (^<=|^<= | ^>= | ^>= |^<|^>|^< |^> |^)((!|)([0-9]{1,5}))( & > | & < |& >=|&>=|&<=||&<=|&>=|&<|&>|&| &| & |$))*

1. Move the anchor out of the alternation

   ^(<=|<= |>= |>= |<|>|< |> |)((!|)([0-9]{1,5}))( & > | & < |& >=|&>=|&<=||&<=|&>=|&<|&>|&| &| & |$))*
   

   Why has there been whitespace before the anchor? (removed that)

2. Move the following whitespace outside and make it optional

   ^(<=|<=|>=|>=|<|>|<|>|) ?((!|)([0-9]{1,5}))( & > | & < |& >=|&>=|&<=||&<=|&>=|&<|&>|&| &| & |$))*
   

3. Remove the duplicates in the alternations

   ^(<=|>=|<|>|) ?((!|)([0-9]{1,5}))( & > | & < |& >=|&>=|&<=||&<=|&>=|&<|&>|&| &| & |$))*
   

4. The empty alternative at the end would match the empty string ==> this alternation is optional

   ^((<=|>=|<|>)? ?)?((!|)([0-9]{1,5}))( & > | & < |& >=|&>=|&<=||&<=|&>=|&<|&>|&| &| & |$))*
   

5. Make the equal sign optional and remove the duplicates

   ^((<|>)=? ?)?((!|)([0-9]{1,5}))( & > | & < |& >=|&>=|&<=||&<=|&>=|&<|&>|&| &| & |$))*
   

6. The alternation with single characters can be replaced with a character class

   ^([<>]=? ?)?((!|)([0-9]{1,5}))( & > | & < |& >=|&>=|&<=||&<=|&>=|&<|&>|&| &| & |$))*
   

7. Do similar things with the alternation at the end and you end up with something like this:

   ^([<>]=? ?)?((!|)([0-9]{1,5}))( ?(& ?([<>]=?)?)?|$)
   

This is untested, I did not change the semantic (I think so), but I did this just here in the editor.