4 ms·

Everett is an extremely controversial figure within linguistics, to put it very mildly. Recursive embedding is one of the foundational marks of human language

Everett is an extremely controversial figure within linguistics, to put it very mildly. Recursive embedding is one of the foundational marks of human language (though not the only), and finding a natural language without it would cast long shadows across the existing literature while making that language's "discoverer" very famous. His data have been challenged by a number of other linguists, despite Everett's attempts to keep others from access to first hand sources. Note too that the existence of alternate constructions such as "John has a brother. This brother has a house." do not preclude the language's ability to accommodate embedding. Anyway, I'm no orthodox Chomskyist (in part because I'm not well enough educated to be:) but I think all of Everett's claims should be taken with as many grains of salt as we can find in our immediate vicinity.

derriz 6y ago
Traces of recursion have been found in animal "language" also - https://www.nature.com/scitable/blog/cognoculture/the_less_human_humans_recursion/ https://www.nature.com/scitable/blog/cognoculture/the_less_h... describes recursive elements to certain bird song. More recently here's a recent experiment - https://cosmosmagazine.com/people/behaviour/complex-linguistic-feature-not-unique-to-humans/ https://cosmosmagazine.com/people/behaviour/complex-linguist... - which seems to show that some primates have similar "recursive" reasoning skills as young (under 4) children.
- foldr 6y ago
  The studies on song birds have for the most part been very poorly conceived and based on a simple misunderstanding. (The fact that a grammar can be defined using recursive rules doesn't entail that its string language can't be recognized by non-recursive procedures.) For example, the common A^nB^n string language (which can be defined using a recursive CFG) can obviously also be recognized by counting.
  - YeGoblynQueenne 6y ago
    But the a^nb^n language is context free so there is no regular grammar that can represent it. If a couting mechanism can be used to represent a^nb^n the result will still have to be a context-free grammar, otherwise a^nb^n would not be context-free.
    - foldr 6y ago
      No, that's exactly the mistake. You can check that the counts are equal without parsing the string at all. So the ability to recognize the string language is not evidence that the string is being parsed according to a particular CFG (or any other grammar). In other words, it is not evidence that the birds are pairing up each A with the corresponding B. Most of the literature on birds stems from a confusion of the distinction between regular string languages and context free string languages with the distinction between grammars with and without recursive rules. The two distinctions are largely orthogonal. It is certainly possible, for example, to define certain regular string languages using recursive grammars, and to define certain context free string languages without using recursive grammars. When you say that "the result must be a context free grammar" I think what you mean to say is that the string language defined must be a context free (and non-regular) string language. But that does not in any way entail that the only way to recognize the string language is by means of a particular context free grammar.
      - deleted 6y ago
        [deleted]
      - YeGoblynQueenne 6y ago
        Ah, I see - my mistake. You're saying that one can count n "a"s followed by n "b"s without the need for a context-free grammar (or any grammar!). I agree. What I meant by "the result will still have to be a context free grammar" was a context-free grammar that incorporates a counting mechanism, as part of its definition. Something like this, perhaps (in Definite Clause Grammars notation so you can actually run it as a Prolog program): 'S' --> 'A'(N), 'B'(N). 'A'([1]) --> 'A'. 'A'([1|As]) --> 'A', 'A'(As). 'B'([1]) --> 'B'. 'B'([1|Bs]) --> 'B', 'B'(Bs). 'A' --> [a]. 'B' --> [b]. Though that one is recursive (or at least the counting mechanism is) and I'm not sure how you'd do the same thing without recursion to be honest (or without going up a couple of levels to Turing-completeness so you can call arbitrary functions). So, never mind- I just misunderstood what you meant.
jfengel 6y ago
I've never understood why it was supposed to be so earth-shaking. Even if true, there's no reason that you have to use embedding. And human beings don't really parse via stacks: we can't parse many deeply center-embedded sentences and other garden path sentences even if they're "valid" under some context-free grammar. That CFGs model so many languages so well is fascinating and puts constraints on the language faculty, but if some language happens to not use the full power of it, that shouldn't come as quite such a surprise. Semantics are not context-free and we know that the underlying brain mechanism is far more complicated than finite-state-machine-plus-stack. A full study of Piraha would likely be illuminating if it really is as different from other languages as Everett says -- and it's practically useless to have him say only "It doesn't have feature X". Unless he can also show that a Piraha child can't learn another language -- that would really mean something. The fact that linguists make a controversy out of it means I must be missing something, but I can't figure out what.
- WaxProlix 6y ago
  My understanding is that Everett has done his best to make a controversy of it, with most linguists preferring to try to ignore him. Unfortunately that's hard when he's doing the Pinker jig, engaging in pop linguistics, etc. So people are forced to engage with it. I generally agree; finding a language which opts not to use recursion would be a weird phenomenon, and might make certain claims about UG weaker. But it wouldn't invalidate all of traditional linguistic research for the past however many years or totally upend our current theoretical frameworks.

derriz 6y ago

Traces of recursion have been found in animal "language" also - https://www.nature.com/scitable/blog/cognoculture/the_less_human_humans_recursion/ https://www.nature.com/scitable/blog/cognoculture/the_less_h... describes recursive elements to certain bird song. More recently here's a recent experiment - https://cosmosmagazine.com/people/behaviour/complex-linguistic-feature-not-unique-to-humans/ https://cosmosmagazine.com/people/behaviour/complex-linguist... - which seems to show that some primates have similar "recursive" reasoning skills as young (under 4) children.

foldr 6y ago
The studies on song birds have for the most part been very poorly conceived and based on a simple misunderstanding. (The fact that a grammar can be defined using recursive rules doesn't entail that its string language can't be recognized by non-recursive procedures.) For example, the common A^nB^n string language (which can be defined using a recursive CFG) can obviously also be recognized by counting.
- YeGoblynQueenne 6y ago
  But the a^nb^n language is context free so there is no regular grammar that can represent it. If a couting mechanism can be used to represent a^nb^n the result will still have to be a context-free grammar, otherwise a^nb^n would not be context-free.
  - foldr 6y ago
    No, that's exactly the mistake. You can check that the counts are equal without parsing the string at all. So the ability to recognize the string language is not evidence that the string is being parsed according to a particular CFG (or any other grammar). In other words, it is not evidence that the birds are pairing up each A with the corresponding B. Most of the literature on birds stems from a confusion of the distinction between regular string languages and context free string languages with the distinction between grammars with and without recursive rules. The two distinctions are largely orthogonal. It is certainly possible, for example, to define certain regular string languages using recursive grammars, and to define certain context free string languages without using recursive grammars. When you say that "the result must be a context free grammar" I think what you mean to say is that the string language defined must be a context free (and non-regular) string language. But that does not in any way entail that the only way to recognize the string language is by means of a particular context free grammar.
    - deleted 6y ago
      [deleted]
    - YeGoblynQueenne 6y ago
      Ah, I see - my mistake. You're saying that one can count n "a"s followed by n "b"s without the need for a context-free grammar (or any grammar!). I agree. What I meant by "the result will still have to be a context free grammar" was a context-free grammar that incorporates a counting mechanism, as part of its definition. Something like this, perhaps (in Definite Clause Grammars notation so you can actually run it as a Prolog program): 'S' --> 'A'(N), 'B'(N). 'A'([1]) --> 'A'. 'A'([1|As]) --> 'A', 'A'(As). 'B'([1]) --> 'B'. 'B'([1|Bs]) --> 'B', 'B'(Bs). 'A' --> [a]. 'B' --> [b]. Though that one is recursive (or at least the counting mechanism is) and I'm not sure how you'd do the same thing without recursion to be honest (or without going up a couple of levels to Turing-completeness so you can call arbitrary functions). So, never mind- I just misunderstood what you meant.

jfengel 6y ago

I've never understood why it was supposed to be so earth-shaking. Even if true, there's no reason that you have to use embedding. And human beings don't really parse via stacks: we can't parse many deeply center-embedded sentences and other garden path sentences even if they're "valid" under some context-free grammar. That CFGs model so many languages so well is fascinating and puts constraints on the language faculty, but if some language happens to not use the full power of it, that shouldn't come as quite such a surprise. Semantics are not context-free and we know that the underlying brain mechanism is far more complicated than finite-state-machine-plus-stack. A full study of Piraha would likely be illuminating if it really is as different from other languages as Everett says -- and it's practically useless to have him say only "It doesn't have feature X". Unless he can also show that a Piraha child can't learn another language -- that would really mean something. The fact that linguists make a controversy out of it means I must be missing something, but I can't figure out what.

WaxProlix 6y ago
My understanding is that Everett has done his best to make a controversy of it, with most linguists preferring to try to ignore him. Unfortunately that's hard when he's doing the Pinker jig, engaging in pop linguistics, etc. So people are forced to engage with it. I generally agree; finding a language which opts not to use recursion would be a weird phenomenon, and might make certain claims about UG weaker. But it wouldn't invalidate all of traditional linguistic research for the past however many years or totally upend our current theoretical frameworks.