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Questions

Page history last edited by PBworks 16 years, 7 months ago

Questions & Answers

 

This is a space for starting up some discussion and exchange of ideas. Enjoy it!

 

 


I was able to run the Prolog parser using SWI Prolog. I highly recommend the experience. For example:

$ swipl

?- lg.

?- cbn_lex(molly*(thinks*(someone*(teases*leopold))), s).

 

 

With LaTeX installed, this generates a PDF file "display.pdf" with two readings for the sentence, de re and de dicto.

 

However, I couldn't seem to get scope ambiguity in "someone teases everyone". The following query generates a PDF file "display.pdf" that contains two derivations along with the same reading:

?- cbn_lex(someone*(teases*everyone), s).

 

 

What did I do wrong? -Ken Shan

 

Michael responding:

Thanks, Ken -- just to reassure you, here is the ambiguity you want to see.

I'll check whether I made a cut-and-paste error putting together the LG-light version I made for the class.

And I've put the version that seems to behave well on the Software page.

 

 


(Thanks! It works now!)

 

 

May I recommend the following books for people who don't mind gaining intuition about continuations from programming:

You need absolutely no programming experience to read these books. Special ESSLLI offer: I will buy them for anyone who gives me their shipping address.

 

-Ken Shan

 

 


Today in class we saw the denotation for "someone", whose type is "(s minus s) from np" before the CBN CPS transformation. Another type for "someone" was mentioned, for quantifiers that take scope only locally: "(s minus np) from s", if I recall correctly. What would a quantificational denotation be for that latter type? -Ken Shan

 

Michael: Here's a possibility -- don't take us to seriously propose this, but it is a solution to the typing constraints imposed by the cps-cbn image of the type.

 

lex(cbn,someone,oslash(s,obslash(np,s)),lambda(Q,appl(Q,pair(lambda(Q,N,appl('$exists$',N)),lambda(C,true))))).

 

 


Q. Sorry, but I understand CBN-> and CBV->, but I don't understand CBN<- and CBV<-, why (slide 5.1) have we: CBV<- : KB --> KA and CBN<- : KB --> (CA --> R)?

 

- Alain

 

A. Here are two answers, I post both, choose the one you like better.

 

Take A |-- B in CBV-> is V_A ---> C_B.

 

1.

* By currying it: (V_A x K_B) ---> R

* By swapping the pair: (K_B x V_A) ---> R.

* By uncurrying it: K_B --> K_A

 

these three steps give the CBV <- reading of A |-- B.

 

2. A -- B in CBV -> iff not B -- not A STILL in CBV -> .

 

Read: not B as B --> FALSE I.e. K_B and similarly not A=K_A.

 

Hence, in CBV -> not B |-- not A is K_B --> K_A

 

This gives the CBV <- reading of A |- B.

 

Raffa

 

 

Thanks! it is the kind of small thing that helps a lot!

 

Alain


Hi! Michael and Raffa!

I should really be happy if I could understand intuitively the terms and co-terms as they are built by the rules.

I try to compare with Herbelin's proposal. Do you agree if, for instance I say that what you name M |x K corresponds

to a directional version of its v @ e, meaning that this is the new environment created by applying v and inserting

the result into the environment e?

 

Alain

 

Michael responding:

That is exactly right. CH have v.e for the introduction of A->B on the left: the intuitive interpretation

is what you say. Because we want to distinguish AB from B/A, we write vltimes e (for AB) versus ertimes v

(for B/A): in that way, we keep track of what the context is, and what the value to put into this context.

 

Alain continuing:

this is my interpretation of M >- K:

the term which allows you to build the result of putting in any context the term you obtain by

putting M into the context K, right?

 

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