1. LING/C SC/PSYC 438/538
Lecture 21
Sandiway Fong

2. Today's Topics Homework 8 Review
Optional Homework 9 (make up on Homework 7)

3. Homework 8 Review
Question1: write a Prolog regular grammar for the following language:
Σ = {0, 1}
L = strings from Σ* with an odd number of 0's and an even number of 1's
assume 0 is an even number
Hint: start with a FSA
How many grammar rules do you have?
submit your program and sample runs
Examples:
*[] (empty string)
*1 *10 *01 *11

4. Homework 8 Review # 0's can be odd or even only
Four possible states (abbreviated):
e0e1: even 0 even 1
o0e1: odd 0 even 1
e0o1: even 0 odd 1
o0o1: odd 0 odd 1
Start state: e0e1
empty string: zero 0's and 1's
Accept state: o0e1
e0e1 --> [0], o0e1.
e0e1 --> [1], e0o1.
o0e1 --> [].
o0e1 --> [0], e0e1.
o0e1 --> [1], o0o1.
o0o1 --> [0], e0o1.
o0o1 --> [1], o0e1.
e0o1 --> [0], o0o1.
e0o1 --> [1], e0e1.

5. Homework 8 Review
?- [hw8]. true. ?- e0e1([0],[]). true ; false. ?- e0e1([0,0,0],[]). ?- e0e1([0,0],[]). ?- e0e1([0,0,0,0],[]). ?- e0e1([],[]). ?- e0e1([1],[]). ?- e0e1([1,1],[]). ?- e0e1([1,1,1],[]). ?- e0e1([0,1,1],[]). ?- e0e1([1,0,1],[]). ?- e0e1([1,1,0],[]). ?- e0e1([1,0],[]). ?- e0e1([0,1],[]).

6. Homework 8 Review
Recall Σ* Prolog code?

7. Homework 8 Review e0e1 --> [0]. e0e1 --> [0], o0e1.
o0e1 --> [0], e0e1.
o0e1 --> [1], o0o1.
o0o1 --> [0], e0o1.
o0o1 --> [1].
o0o1 --> [1], o0e1.
e0o1 --> [0], o0o1.
e0o1 --> [1], e0e1.
e0e1 --> [0], o0e1.
e0e1 --> [1], e0o1.
o0e1 --> [].
o0e1 --> [0], e0e1.
o0e1 --> [1], o0o1.
o0o1 --> [0], e0o1.
o0o1 --> [1], o0e1.
e0o1 --> [0], o0o1.
e0o1 --> [1], e0e1.
right recursive canonical format for regular grammars

8. Homework 8 Review (D)FSA:
Optional Homework 9 (make-up opportunity for Homework 7)
Give a regex for this DFSA
Implement it in Perl or Python and show it works correctly
Due next Monday

9. Homework 8 Review
What happens when you run the following Prolog query on your grammar?
s(String,[]).

10. Homework 8 Review
?- sigmas(L), e0e1(L,[]). L = [0] ; L = [0, 0, 0] ; L = [1, 1, 0] ; L = [1, 0, 1] ; L = [0, 1, 1] ; L = [0, 0, 0, 0, 0] ; L = [1, 1, 0, 0, 0] ; L = [1, 0, 1, 0, 0] ; L = [0, 1, 1, 0, 0] ; L = [1, 0, 0, 1, 0] ; L = [0, 1, 0, 1, 0] ; L = [0, 0, 1, 1, 0] ; L = [1, 1, 1, 1, 0] ; L = [1, 0, 0, 0, 1] ; L = [0, 1, 0, 0, 1] ; L = [0, 0, 1, 0, 1] ; L = [1, 1, 1, 0, 1] ; L = [0, 0, 0, 1, 1] ; L = [1, 1, 0, 1, 1] ; L = [1, 0, 1, 1, 1] ; L = [0, 1, 1, 1, 1] ; L = [0, 0, 0, 0, 0, 0, 0] ; L = [1, 1, 0, 0, 0, 0, 0] ; L = [1, 0, 1, 0, 0, 0, 0]

11. Homework 8 Review
Use iterative deepening (material beyond this introductory course):
?- [id_meta, hw8].
String = [0, 0, 1, 1, 0] ;
String = [1, 1, 1, 1, 0] ;
true.
String = [0, 1, 0, 0, 1] ;
String = [0, 0, 0, 0, 0, 0, 0] ;
String = [0, 1, 0, 1, 0] ;
String = [0, 0, 0, 0, 0, 1, 1] ;
?- id(e0e1(String,[])).
String = [0, 1, 1, 0, 0] ;
String = [0, 0, 0, 0, 1, 0, 1] ;
String = [0] ;
String = [0, 1, 1, 1, 1] ;
String = [0, 0, 0, 0, 1, 1, 0] ;
String = [0, 0, 0] ;
String = [1, 0, 0, 0, 1] ;
String = [0, 0, 0, 1, 0, 0, 1] ;
String = [0, 1, 1] ;
String = [1, 0, 0, 1, 0] ;
String = [0, 0, 0, 1, 0, 1, 0] ;
String = [1, 0, 1] ;
String = [1, 0, 1, 0, 0] ;
String = [0, 0, 0, 1, 1, 0, 0] 
String = [1, 1, 0] ;
String = [1, 0, 1, 1, 1] ;
String = [0, 0, 0, 0, 0] ;
String = [1, 1, 0, 0, 0] ;
String = [0, 0, 0, 1, 1] ;
String = [1, 1, 0, 1, 1] ;
String = [0, 0, 1, 0, 1] ;
String = [1, 1, 1, 0, 1] ;

12. Left Recursion and Set Enumeration
Question
What is the language of this grammar?
Example:
s --> a, [!].
a --> ba, [a].
a --> a, [a].
ba --> b, [a].
b --> [b].
Grammar is:
a regular grammar
left recursive (nonterminal on left)
Answer: Sheeptalk!
ba..a! (# a's > 1)
Sentential forms: s a!
baa!
Underscoring used
here to indicate a nonterminal

13. Left Recursion and Set Enumeration
Example:
s --> a, [!].
a --> ba, [a].
a --> a, [a].
ba --> b, [a].
b --> [b].
Prolog query:
?- s([b,a,a,!],[]).
true
?- s([b,a,a,a,a,!],[]).
But it doesn’t halt when faced with a string not in the language
?- s([b,a,!],[]).
ERROR: Out of local stack

14. Left Recursion and Set Enumeration
In fact…
Example:
s --> a, [!].
a --> ba, [a].
a --> a, [a].
ba --> b, [a].
b --> [b].

15. Left Recursion and Set Enumeration
Example:
s --> a, [!].
a --> ba, [a].
a --> a, [a].
ba --> b, [a].
b --> [b].
?- s([b,a,!],[]).
ERROR: Out of local stack
Why?

16. Left Recursion and Set Enumeration
left recursive regular grammar:
s --> a, [!].
a --> ba, [a].
a --> a, [a].
ba --> b, [a].
b --> [b].
Behavior
halts when presented with a string that is in the language
doesn’t halt when faced with a string not in the language
unable to decide the language membership question
Surprisingly, the query:
?- s(L,[]).
enumerates the strings in the language just fine.
L = [b, a, a, !] ;
L = [b, a, a, a, !] ;
L = [b, a, a, a, a, !] ;
L = [b, a, a, a, a, a, !] ;
L = [b, a, a, a, a, a, a, !] ;
L = [b, a, a, a, a, a, a, a, !] ;
L = [b, a, a, a, a, a, a, a, a|...] w
L = [b, a, a, a, a, a, a, a, a, !]

17. Left Recursion and Set Enumeration
derivation tree for
?- s(L,[]).
[Powerpoint
animation]
left recursive regular grammar:
s --> a, [!].
a --> ba, [a].
a --> a, [a].
ba --> b, [a].
b --> [b].
Behavior
halts when presented with a string that is in the language
doesn’t halt when faced with a string not in the language
L = [b,a,a,!]
L = [b,a,a|..]
L = [b,a|..]
L =
L = [b|..]
Choice point s s a ! a ! ba a a a
and so on b a ba a b a b b

18. Left Recursion and Set Enumeration
[Powerpoint
animation]
However, this slightly re-ordered left recursive regular grammar:
s --> a, [!].
a --> a, [a].
a --> ba, [a].
ba --> b, [a].
b --> [b].
(rules 2 and 3 swapped) won’t halt when enumerating
Why? s a a a a
...
descends
infinitely
using rule #2

19. So far … We've talked about:
1. Declarative (logical) reading of grammar rules
2. Prolog query: s(String,[]).
Case 1. String is known: Is String ∈ L(G)?
Case 2. String is unknown: enumerate L(G)
3. Different search strategies
Prolog's (left-to-right) depth-first search
Iterative deepening

20. Beyond Regular Languages
anbn = {ab, aabb, aaabbb, aaaabbbb, ... } n≥1
is not a regular language
That means no FSA, RE or RG can be built for this language
We only have a finite number of states to play with …
We’re only allowed simple free iteration (looping)
Pumping Lemma proof
Perl regex:

21. Beyond Regular Languages
Example:
Language
anbn = {ab, aabb, aaabbb, aaaabbbb, ... } n>=1
A regular grammar extended to allow both left and right recursive rules can accept/generate it:
anbn.pl
a --> [a], b.
b --> [b].
b --> a, [b].
Set
membership
Set
enumeration

22. Beyond Regular Languages
anbn = {ab, aabb, aaabbb, aaaabbbb, ... } n>=1
A regular grammar extended to allow both left and right recursive rules can accept/generate it:
a --> [a], b.
b --> [b].
b --> a, [b].
Intuition:
grammar implements the stacking of partial trees balanced for a’s and b’s: B A b a

23. Beyond Regular Languages
anbn = {ab, aabb, aaabbb, aaaabbbb, ... } n>=1
A regular grammar extended to allow both left and right recursive rules can accept/generate it:
a --> [a], b.
b --> [b].
b --> a, [b].
A type-2 or context-free grammar (CFG) has no restrictions on what can go on the RHS of a grammar rule
Note:
CFGs still have a single nonterminal limit for the LHS of a rule
Example:
s --> [a], [b].
s --> [a], s, [b].

24. Extra Argument: Parse Tree
Recovering a parse tree
when want Prolog to return more than just true/false answers
in case of true, we can compute a syntax tree representation of the parse
by adding an extra argument to nonterminals
applies to all grammar rules (not just regular grammars)
Example
sheeptalk again
DCG (non-regular, context-free):
s --> [b], [a], a, [!].
a --> [a]. (base case)
a --> [a], a. (recursive case) s a ! b

25. Extra Argument: Parse Tree! b
Prolog term data structure:
hierarchical
allows sequencing of arguments
functor(arg1,..,argn)
each argi could be another term or simple atom
Tree:
s(b,a,a(a,a(a)),!)

26. Extra Arguments: Parse Tree
DCG
s --> [b],[a], a, [!].
a --> [a]. (base case)
a --> [a], a. (right recursive case) s a ! b
base case
a --> [a].
a(subtree) --> [a].
a(a(a)) --> [a].
s(b,a,a(a,a(a)),!)
recursive case
a --> [a], a.
a(subtree) --> [a], a(subtree).
a(a(a,A)) --> [a], a(A).
Idea: for each nonterminal, add an argument to store its subtree

27. Extra Arguments: Parse Tree
Prolog grammar
s --> [b], [a], a, [!].
a --> [a]. (base case)
a --> [a], a. (right recursive case) s a ! b
base and recursive cases
a(a(a)) --> [a].
a(a(a,A)) --> [a], a(A).
start symbol case
s --> [b], [a], a, [!].
s(tree) --> [b], [a], a(subtree), [!].
s(s(b,a,A,!) ) --> [b], [a], a(A), [!].
s(b,a,a(a,a(a)),!)

28. Extra Arguments: Parse Tree
Prolog grammar
s --> [b], [a], a, [!].
a --> [a]. (base case)
a --> [a], a. (right recursive case)
Equivalent Prolog grammar computing a parse
s(s(b,a,A,!)) --> [b], [a], a(A), [!].
a(a(a)) --> [a].
a(a(a,A)) --> [a], a(A).

29. Quick Homework 10
Modify this grammar to build parse trees for strings in the language anbn, n≥1.
anbn.pl
a --> [a], b.
b --> [b].
b --> a, [b].
Show your code and give example runs.
Due next Monday night