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Re^4: Perl is not Dynamically Parseable

by blokhead (Monsignor)
on Oct 13, 2009 at 18:15 UTC ( #800954=note: print w/ replies, xml ) Need Help??


in reply to Re^3: Perl is not Dynamically Parseable
in thread Perl is not Dynamically Parseable

I don't know what the halting problem has to do with non-deterministic or randomized programs. The perl parser is deterministic, and it does not toss random coins. You can certainly insert randomness into a BEGIN block, but that's not what's going on here and it doesn't mean that parsing itself is randomized. Jeffrey's examples & undecidability proofs all involve deterministic BEGIN phases. If you run the parser a billion times on a deterministic program, you will get the same result every time. Different instances of the parser do not give different parses. The only way to get different parses is by having different behavior within the BEGIN block.

blokhead


Comment on Re^4: Perl is not Dynamically Parseable
Re^5: Perl is not Dynamically Parseable
by ikegami (Pope) on Oct 13, 2009 at 19:43 UTC

    The perl parser is deterministic,

    Parsing is the activity of assigning meaning to code. Since it's impossible to predict the meaning Perl will assign to code in some circumstances, the parser is non-determinisitc.

      it's impossible to predict the meaning Perl will assign to code in some circumstances
      That's what I thought you must have meant. I guess we have different working definitions of what "deterministic" means.

      In this case, one could even argue that it's not really a matter of determining what meaning Perl assigns to some code, but a matter of determining whether Perl will get around to assigning meaning to some code.

      blokhead

        I guess we have different working definitions of what "deterministic" means.

        Wikipedia says:

        A deterministic algorithm is an algorithm which, in informal terms, behaves predictably. Given a particular input, it will always produce the same output, and the underlying machine will always pass through the same sequence of states.

        Perl's parser, due to it's pluggable nature, does not adhere to that definition. The output can vary for a given input, and it varies because the underlying states visited vary.

        I've looked at more format definitions elsewhere and reached the same conclusion.

        In this case, one could even argue that it's [...] a matter of determining whether Perl will get around to assigning meaning to some code.

        When one discuses whether Perl can be parsed or not, one is not referring to the time it takes or even if it's finite. One is discussing whether the output is stable enough to study or store for future use.

        Let's take C++ for example. One would consider C++ to be parseable. One can perform static analysis on C++ code. C++ code can be compiled. Yet, it's possible to introduce an infinite loop in the parser because C++'s template system is turing complete.

        The difference between C++ and Perl is that C++'s parser is confined to making decision based on its input, whereas Perl's parser can make decision based on external data.

Re^5: Perl is not Dynamically Parseable
by ikegami (Pope) on Oct 13, 2009 at 23:19 UTC

    Something I missed:

    Different instances of the parser do not give different parses.

    They do for some program. Did you (incorrectly) think I said it would happen for all programs? I said that once one finds a program where this happens, one can prove the non-parsability of Perl.

      Did you (incorrectly) think I said it would happen for all programs?
      Nope.
      I said that once one finds a program where this happens, one can prove the non-parsability of Perl.
      If this example (from this previous node of yours) proves that perl is unparseable (because it sometimes defines foo, it sometimes doesn't):
      BEGIN { eval "sub foo {}" if rand() < 0.5; } foo();
      then this next example proves that the halting problem is undecidable (because sometimes it halts, sometimes it doesn't):
      if (rand() < 0.5) { 1 while 1; }
      But neither is a proof of the undecidability of anything satisfying. They assume an interpretation/model of programs that employ randomness (and by extension, any external calls) which completely trivializes the halting problem into a meaningless statement. I think this is my main objection with your examples.

      Turing machines are deterministic, so if you want to model things like random choices and externally-obtained data, you model them (without loss of generality) as being provided as input on separate tapes. Indeed, this is exactly what is done in the theory of computation. For this reason, I have a gut reaction against random choices and external data being used to "prove undecidability." Those issues are red herrings and don't speak to undecidability in any meaningful way. From the perspective of undecidability, they should be considered as simply additional, different kinds of input to a program.

      So in the first example above, you can confidently say that whenever the random tape contains such-and-such, the program is syntactic, and otherwise it is not. In the second example, you can confidently say that if the random tape contains such-and-such, then the program halts, and otherwise it does not. You can efficiently solve the parseability problem and the halting problem for classes of programs of this simple form, for the way of modeling a program's random choices that doesn't already trivialize the halting problem. So they are not satisfying demonstrations of undecidability.

      On the other hand, it is impossible to decide the parseability problem (however you want to define it) for the class of programs of the following form:

      BEGIN { my $x = "... arbitrary perl code ..."; eval "sub foo {}" if eval $x; } foo();
      And this does not rely on things like external system calls, or random choices, since the halting problem is still quite meaningful when those things are not used.

      blokhead

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