20 comments

  • macintux 6 hours ago ago

    One of the few lessons I distinctly remember from college was finite automata in my PL class. I really enjoyed exploring the concepts and writing a grep tool; we were supposed to write either a NFA or DFA processing application, but I decided to write both.

    20 years later I got to apply some of the same ideas to a language processing application, and it was such a pleasure to actually use something conceptual like that. Made me briefly regret landing in more hybrid infrastructure/automation roles instead of pure software development.

    Somewhere I may still have my copy of Preperata and Yeh that my professor recommended at the time for further reading. Like most of my books, it was never actually read, just sat around for years.

  • praptak an hour ago ago

    "But on a day-to-day basis, if asked to recognize balanced parentheses?"

    On day-to-day basis you will never encounter this problem in pure form. As the consequence the solutions are not good for the day-to-day stuff.

    Even if you only are only writting a verifier (which is already a bit unrealistic), you'll need to say something more than "not balanced". Probably rather something along the lines of "closing brace without a matching opening at [position]" or "[n] unclosed parentheses at <end of stream>" which rules out the simple recursive regex approach (counter still works).

    • jibal 23 minutes ago ago

      To report the location of an unclosed opener you need a stack.

      • vidarh 20 minutes ago ago

        Depends. You want a stack, as it's certainly more efficient, but if you can rewind the position pointer you don't need one (you can count backwards).

        EDIT: It gets complicated if you need to count multiple different types of openers. In that case I think you need the stack, at least unless there are constraints on which openers can occur within others - you at the very least need to know which closer you're looking for right now, but if you can't deduce what is outside, you obviously then need to keep track of it.

        In practice, of course, we'll generally use a stack because it's just pointless to make life harder by not using one for this.

  • userbinator 3 hours ago ago

    we’ll ask, “What’s the simplest possible computing machine that can recognize balanced parentheses?”

    A counter. That's the difference between theory and practice. Because in practice, everything is finite.

    • jibal 31 minutes ago ago

      The counter is simply the stack depth without bothering with the actual stack. If the stack is empty when you encounter a closer then it's unbalanced. If the stack isn't empty when you reach the end of the input then the items in the stack are unbalanced.

      If you have multiple kinds of brackets then you need the same number of counters. Each counter corresponds to the number of openers of that type currently on the stack.

      If you're writing a parser and you want to report the location of an unclosed opening bracket then you need the actual stack.

      • vidarh 12 minutes ago ago

        You need the actual stack, I think, in the case of multiple types of openers without additional constraints, because if you just have raw counters you'd get tripped up by ([)] or similar.

        So to generalise your point you need a counter for each transition to a different type of opener.

        So (([])) needs only 2 counters, not 3.

        You could constrain it further if certain types of openers are only valid in certain cases so you could exclude certain types of transitions.

      • gpderetta 13 minutes ago ago

        Wouldn't two counters report "([)]" as being properly balanced?

    • nmadden 2 hours ago ago

      > Because in practice, everything is finite.

      Indeed! https://neilmadden.blog/2019/02/24/why-you-really-can-parse-...

    • testaccount28 3 hours ago ago

      you don't need a full counter. increment, decrement, and check_if_zero are enough. no need for get_value.

      • stellalo 2 hours ago ago

        you also need check_if_negative to detect close-before-open

        • jibal an hour ago ago

          The counter is at 0, which indicates an error ... that plus the counter being non-zero when reaching the end of input is the entire point.

    • pfortuny 2 hours ago ago

      Yes. Actually, a more interesting example which does not complicate the statement (not the problem) too much is to check for nested parenthesis and brackets:

      (([[()])) -> ok ((([](])) -> not ok

      Hope OP gets this message.

      • jibal an hour ago ago

        You just need as many counters as there are types of brackets. If any counter is non-zero at EOT or when a different type of right bracket is encountered then there's an unclosed left bracket of that type. And if a right bracket is encountered when its counter is zero then it's missing a left bracket of that type.

        • immibis 10 minutes ago ago

          Your solution incorrectly fails ({}). You need the stack.

  • firechickenbird an hour ago ago

    The proof of non-regularity is a bit convoluted. You can easily apply the pumping lemma there

  • senorqa 3 hours ago ago

    The pictures of Brutalist architecture are awesome!

  • Antibabelic 2 hours ago ago

    What is some further reading y'all could recommend on formal languages?