*nod* Yours is the second pointer to "Mastering Regular Expressions" I've received, and theorbtwo said substantially the same thing (doable if the regexps are deterministic) in the Chatterbox earlier.

*ponder* of course, any NFA can be mechanistically converted to a DFA (by creating a state for every reachable combination of NFA states), so really the answers should be equivalent. But since NFA/DFA's aren't actually Turing-complete (barring a Turing-machine-style memory tape), we don't run into the halting problem. Huh. Sounds like it should be solvable.

But it's sounding like I may have to write the actual code myself; at least, nobody has spoken up and said "Oh, yes, that's been solved, and here's a link" yet. Which is, honestly, what I was hoping for. :-)

Of course, even the fact that two deterministic regexps can be proven to be identical doesn't actually answer the original question. abc and ab[c|d] will have different DFA representations, but all strings that abc matches will also be matched by ab[c|d]. But it does suggest some approaches for resolving the issue.

Anyone got more advice?

Thanks 10e6,

Mickey.

In reply to Re^2: Comparative satisfiability of regexps. by Meowse
in thread Comparative satisfiability of regexps. by Meowse

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