LING 388 Language and Computers Lecture 17 10/28/03 Sandiway FONG

Administrivia Next Time … Next Time …  Computer Laboratory  SBS 224  October 30th and November 4th

Last Time … We saw how to implement the filter: We saw how to implement the filter:  All variables must be bound by an operator ( x)  Blocks the example :  *John hits(np(john),vp(v(hit),np(x))) The implementation was : The implementation was :  filter(X) :- \+ ( variable(X,F), var(F) ).  where:  variable(X,F) is an np(x) finder (tree-walker, disjunctive form) and F reports on whether there is a dominating lambda in Xis an np(x) finder (tree-walker, disjunctive form) and F reports on whether there is a dominating lambda in X

Today’s Lecture Explore how to implement the second filter defined in Lecture 14 … Explore how to implement the second filter defined in Lecture 14 …  Operator ( x) can only bind one variable np(x) Example: Example:  *the cat that saw  *the cat that [ NP e i ] saw [ NP e i ]  (the cat)( x.x saw x)  i.e. the cat that saw itself Pseudo-Logical Form: Pseudo-Logical Form:  np(np(det(the),n(cat)),lambda(x,s(np(x),vp(v(saw),np(x))))) Note: the first filter (below) fails to rule this example out Note: the first filter (below) fails to rule this example out  All variables must be bound by an operator

Today’s Lecture Another Example: Another Example:  *the cat that John saw Mary  (the cat)( x.John saw Mary) Pseudo-Logical Form: Pseudo-Logical Form:  np(np(det(the),n(cat)),lambda(x,s(np(john),vp(v(saw),np(mary)))))  vacuous x The first filter fails to rule this out The first filter fails to rule this out  because there are no np(x) variables

Filter Implementation Filter 2: Filter 2:  Operator ( x) can only bind one variable Basic Implementation Idea: Basic Implementation Idea:  Look for structure lambda(x,Y)  Check Y to see that there is at most one np(x) inside

Filter Implementation Question: Question:  How to look for lambda(x,Y) Answer: Answer: from Lectures 15 and 16, we have that code available already  Use the tree-walker (disjunctive form) operator/1  operator(X) holds if there is a lambda expression in X

Filter Implementation Code: Code:  operator(lambda(x,_)).  operator(X) :- X =.. [F,A1,A2],X =.. [F,A1,A2], operator(A1).operator(A1).  operator(X) :- X =.. [F,A1,A2],X =.. [F,A1,A2], operator(A2).operator(A2).  operator(X) :- X =.. [F,A],X =.. [F,A], operator(A).operator(A).  Predicate operator(X) holds if X contains lambda(x,_) inside Notes: 1.All clauses except for the 1st recursively call operator/1 on a subtree 2.Successful stopping condition is when we match with lambda(x,_)

Filter Implementation Step 1: Modify operator/1 to add in a sub-call to check for a variable Step 1: Modify operator/1 to add in a sub-call to check for a variable  operator(lambda(x,Y)).  operator(X) :-  X =.. [F,A1,A2],  operator(A1).  operator(X) :-  X =.. [F,A1,A2],  operator(A2).  operator(X) :-  X =.. [F,A],  operator(A).  operator(lambda(x,Y)):-  variable(Y).  operator(X) :-  X =.. [F,A1,A2],  operator(A1).  operator(X) :-  X =.. [F,A1,A2],  operator(A2).  operator(X) :-  X =.. [F,A],  operator(A).

Filter Implementation Re-using the definition of variable/1 from last time … Re-using the definition of variable/1 from last time …  operator/1 holds if lambda(x, … np(x) …) exists  operator(lambda(x,Y)):-  variable(Y).  operator(X) :-  X =.. [F,A1,A2],  operator(A1).  operator(X) :-  X =.. [F,A1,A2],  operator(A2).  operator(X) :-  X =.. [F,A],  operator(A).  variable(np(x)).  variable(X) :-  X =.. [F,A1,A2],  variable(A1).  variable(X) :-  X =.. [F,A1,A2],  variable(A2).  variable(X) :-  X =.. [F,A],  variable(A).

Filter Implementation Now, operator/1 admits both: Now, operator/1 admits both:  lambda(x, … np(x) …)  *lambda(x, … np(x) … np(x) … ) So: So:  we need to modify variable/1 to count and fail when it finds two (or more) np(x)s Basic Idea: Basic Idea:  Use a conjunctive tree-walker  We need to count, so we need make sure we visit every node  Disjunctive tree-walking doesn’t work because it has a choice of which branch to take when faced with binary branching

Filter Implementation Question: Question:  How to implement a counter? Answer: Answer:  Use the extra argument device to pass around a counter which gets incremented when needed Simpler Answer (for counting up to two only): Simpler Answer (for counting up to two only):  Use the feature setting mechanism we used to detect lambda in Lecture 16  Instantiate the extra argument to be a constant, e.g. one, when we find out first np(x)  When we see an np(x), check to see if the extra argument is still a variable or one - fail in the latter case [For yet another solution suggested in class: see appendix] [For yet another solution suggested in class: see appendix]

Filter Implementation Basic tree-walker (conjunctive form): Basic tree-walker (conjunctive form):  visit(X) :-  X =.. [F,A1,A2],  visit(A1),  visit(A2).  visit(X) :-  X =.. [F,A],  visit(A).  visit(X) :- atom(X). s npvp vnp detn theman saw john

Filter Implementation Step 1: Add a clause for np(x) Step 1: Add a clause for np(x)  visit(np(x)).  visit(X) :-  X =.. [F,A1,A2],  visit(A1),  visit(A2).  visit(X) :-  X =.. [F,A],  visit(A).  visit(X) :- atom(X).  visit(X) :-  X =.. [F,A1,A2],  visit(A1),  visit(A2).  visit(X) :-  X =.. [F,A],  visit(A).  visit(X) :- atom(X).

Filter Implementation Step 2: Add an extra argument Step 2: Add an extra argument  visit(np(x)).  visit(X) :-  X =.. [F,A1,A2],  visit(A1),  visit(A2).  visit(X) :-  X =.. [F,A],  visit(A).  visit(X) :- atom(X).  visit(np(x),F).  visit(X,F) :-  X =.. [_,A1,A2],  visit(A1,F),  visit(A2,F).  visit(X,F) :-  X =.. [_,A],  visit(A,F).  visit(X,F) :- atom(X).

Filter Implementation Step 3: Modify clause for np(x) to test and set the extra argument Step 3: Modify clause for np(x) to test and set the extra argument  visit(np(x),F).  visit(X,F) :-  X =.. [_,A1,A2],  visit(A1,F),  visit(A2,F).  visit(X,F) :-  X =.. [_,A],  visit(A,F).  visit(X,F) :- atom(X).  visit(np(x),F) :-  var(F), F = one.  visit(X,F) :-  X =.. [_,A1,A2],  visit(A1,F),  visit(A2,F).  visit(X,F) :-  X =.. [_,A],  visit(A,F).  visit(X,F) :- atom(X).

Filter Implementation Problem: Problem:  What happens if we see a 2nd np(x)?  1st clause will correctly block, but np(x) also matches the 3rd clause  visit(np(x),F) :-  var(F), F = one.  visit(X,F) :-  X =.. [_,A1,A2],  visit(A1,F),  visit(A2,F).  visit(X,F) :-  X =.. [_,A],  visit(A,F).  visit(X,F) :- atom(X). Succeeds for the 2nd np(x) But we want to prevent this!

Filter Implementation Step 4: Add a pre-condition to clause 3 to block np(x) from matching X Step 4: Add a pre-condition to clause 3 to block np(x) from matching X  visit(np(x),F) :-  var(F), F = one.  visit(X,F) :-  X =.. [_,A1,A2],  visit(A1,F),  visit(A2,F).  visit(X,F) :-  \+ X = np(x),  X =.. [_,A],  visit(A,F).  visit(X,F) :- atom(X). Pre-condition prevents X = np(x) from succeeding with this clause Note: = and == can be used interchangeably here because the test is in the scope of the negation operator \+. X \== np(x) is possible as well

Filter Implementation Step 5: Rename visit/2 as lt2vars/2 (less than 2 variables) to reflect the revised semantics Step 5: Rename visit/2 as lt2vars/2 (less than 2 variables) to reflect the revised semantics  lt2vars(np(x),F) :-  var(F), F = one.  lt2vars(X,F) :-  X =.. [_,A1,A2],  lt2vars(A1,F),  lt2vars(A2,F).  lt2vars(X,F) :-  \+ X = np(x),  X =.. [_,A],  lt2vars(A,F).  lt2vars(X,F) :- atom(X).  operator(lambda(x,Y)):-  lt2vars(Y,_).  operator(X) :-  X =.. [F,A1,A2],  operator(A1).  operator(X) :-  X =.. [F,A1,A2],  operator(A2).  operator(X) :-  X =.. [F,A],  operator(A).

Filter Implementation Filter 2: Filter 2:  Operator ( x) can only bind one variable Implementation: Implementation:  filter2(X) :- operator(X).  almost but not quite correct, do you see when it fails to work?  see Computer Laboratory to come … Use: Use:  ?- s(X,Sentence,[]), filter2(X).  X is the phrase structure returned by the DCG  Sentence is the input sentence encoded as a list  filter/2 is a condition on representation

Filter Implementation: Summary Summary: Start with a lambda(x,Y) finder that calls visit(Y) Make visit/1 a conjunctive tree-walker 1. We’ll be interested in making np(x) a special case  Add a clause for np(x) to visit/1 2. We’ll need to count during the search  Add an extra argument to visit/1 => visit/2 3. We’ll count when we match the special case np(x)  Modify clause for np(x) to test and set the extra argument 4. Prevent np(x) from matching any clause other than the special case  Modify the unary branching clause, i.e. when X is F(A), to block it from matching np(x) 5. We have new semantics for the conjunctive tree-walker  Rename it!

Bijection Principle Implementation Bijection Principle (Koopman): Bijection Principle (Koopman):  Operators and variables must be in one-to-one correspondence Filters: Filters:  (filter1) All variables must be bound by an operator ( x)  (filter2) Operator ( x) can only bind one variable Implementation: Implementation:  bijectionPrinciple(X) :- filter(X), filter2(X). Use: Use:  ?- s(X,Sentence,[]), bijectionPrinciple(X).  X is the phrase structure returned by the DCG  Sentence is the input sentence encoded as a list

Tree-Walker Summary Question: Question:  Which tree-walker to use? Answers: Answers:  Need to find some element in the tree?  Use the disjunctive form E.g. variable(X) holds if there exists some np(x) in XE.g. variable(X) holds if there exists some np(x) in X  Need to count the number of occurrences of some element in the tree?  Use the conjunctive form E.g. lt2vars(X,_) holds there are zero or one occurrences of np(x) in XE.g. lt2vars(X,_) holds there are zero or one occurrences of np(x) in X

Appendix

Filter Implementation Revisited Question: Question:  How to implement a counter (to count up to two)? Answer (in preceding slides): Answer (in preceding slides):  Use an extra argument in a conjunctive tree-walker; test and set the argument 1st time around Another Solution: Another Solution:  Use negation  Advantage: no need for extra argument

Filter Implementation Revisited Add a test to confirm that if variable/1 holds for one branch, it cannot hold for the other branch: Add a test to confirm that if variable/1 holds for one branch, it cannot hold for the other branch:  variable(np(x)).  variable(X) :-  X =.. [F,A1,A2],  variable(A1) -> \+ variable(A2) ; variable(A2).  variable(X) :-  X =.. [F,A],  variable(A). Truth Table: Truth Table: variable(X)variable(A1)variable(A2) F (False) FF F T (True) T TFT TTF Prolog’s “if then else” construct: -> ;

Exercise 1: Relative Clauses Example: Pseudo-Logical Form for *the cat that saw: Example: Pseudo-Logical Form for *the cat that saw: np det the n cat lambda xs np vp vx sawx TF TT F F variable(X) fails for s