Algorithms & Models of Computation What stack got to do with it? CS/ECE 374, Fall 2017 What’s a stack but a second hand memory? DFA / NFA /Regular expressions. Context Free Languages and 1 ≡ constant memory computation. Grammars NFA + stack 2 ≡ context free grammars ( CFG ). Lecture 7 Turing machines DFA / NFA + unbounded memory. 3 ≡ a standard computer/program. Tuesday, September 19, 2017 ≡ NFA with two stacks. Sariel Har-Peled (UIUC) CS374 1 Fall 2017 1 / 36 Sariel Har-Peled (UIUC) CS374 2 Fall 2017 2 / 36 Context Free Languages and Grammars Programming Languages Programming Language Specification Parsing Natural language understanding Generative model giving structure . . . Sariel Har-Peled (UIUC) CS374 3 Fall 2017 3 / 36 Sariel Har-Peled (UIUC) CS374 4 Fall 2017 4 / 36
Natural Language Processing Models of Growth L -systems http://www.kevs3d.co.uk/dev/lsystems/ Sariel Har-Peled (UIUC) CS374 5 Fall 2017 5 / 36 Sariel Har-Peled (UIUC) CS374 6 Fall 2017 6 / 36 Kolam drawing generated by grammar Context Free Grammar ( CFG ) Definition Definition A CFG is a quadruple G = ( V , T , P , S ) V is a finite set of non-terminal symbols T is a finite set of terminal symbols (alphabet) P is a finite set of productions, each of the form A → α where A ∈ V and α is a string in ( V ∪ T ) ∗ . Formally, P ⊂ V × ( V ∪ T ) ∗ . S ∈ V is a start symbol � � G = Variables , Terminals , Productions , Start var Sariel Har-Peled (UIUC) CS374 7 Fall 2017 7 / 36 Sariel Har-Peled (UIUC) CS374 8 Fall 2017 8 / 36
Example Example formally... V = { S } V = { S } T = { a , b } T = { a , b } P = { S → ǫ | a | b | aSa | bSb } P = { S → ǫ | a | b | aSa | bSb } (abbrev. for S → ǫ, S → a , S → b , S → aSa , S → bSb ) (abbrev. for S → ǫ, S → a , S → b , S → aSa , S → bSb ) S → ǫ, S � aSa � abSba � abbSbba � abb b bba S → a , G = { S } , { a , b } , S → b S S → aSa What strings can S generate like this? S → bSb Sariel Har-Peled (UIUC) CS374 9 Fall 2017 9 / 36 Sariel Har-Peled (UIUC) CS374 10 Fall 2017 10 / 36 Palindromes Examples L = { 0 n 1 n | n ≥ 0 } Madam in Eden I’m Adam Dog doo? Good God! S → ǫ | 0 S 1 Dogma: I am God. A man, a plan, a canal, Panama Are we not drawn onward, we few, drawn onward to new era? Doc, note: I dissent. A fast never prevents a fatness. I diet on cod. http://www.palindromelist.net Sariel Har-Peled (UIUC) CS374 11 Fall 2017 11 / 36 Sariel Har-Peled (UIUC) CS374 12 Fall 2017 12 / 36
Notation and Convention “Derives” relation Let G = ( V , T , P , S ) then Formalism for how strings are derived/generated a , b , c , d , . . . , in T (terminals) Definition A , B , C , D , . . . , in V (non-terminals) Let G = ( V , T , P , S ) be a CFG . For strings α 1 , α 2 ∈ ( V ∪ T ) ∗ u , v , w , x , y , . . . in T ∗ for strings of terminals we say α 1 derives α 2 denoted by α 1 � G α 2 if there exist strings β, γ, δ in ( V ∪ T ) ∗ such that α, β, γ, . . . in ( V ∪ T ) ∗ X , Y , X in V ∪ T α 1 = β A δ α 2 = βγδ A → γ is in P . Examples: S � ǫ , S � 0 S 1 , 0 S 1 � 00 S 11 , 0 S 1 � 01 . Sariel Har-Peled (UIUC) CS374 13 Fall 2017 13 / 36 Sariel Har-Peled (UIUC) CS374 14 Fall 2017 14 / 36 “Derives” relation continued Context Free Languages Definition Definition For integer k ≥ 0 , α 1 � k α 2 inductive defined: The language generated by CFG G = ( V , T , P , S ) is denoted by L ( G ) where L ( G ) = { w ∈ T ∗ | S � ∗ w } . α 1 � 0 α 2 if α 1 = α 2 α 1 � k α 2 if α 1 � β 1 and β 1 � k − 1 α 2 . Definition Alternative definition: α 1 � k α 2 if α 1 � k − 1 β 1 and β 1 � α 2 A language L is context free ( CFL ) if it is generated by a context free grammar. That is, there is a CFG G such that L = L ( G ) . ∗ is the reflexive and transitive closure of � . � ∗ α 2 if α 1 � k α 2 for some k . α 1 � ∗ 0000011111 . ∗ ǫ , 0 S 1 � Examples: S � Sariel Har-Peled (UIUC) CS374 15 Fall 2017 15 / 36 Sariel Har-Peled (UIUC) CS374 16 Fall 2017 16 / 36
Example Closure Properties of CFLs L = { 0 n 1 n | n ≥ 0 } G 1 = ( V 1 , T , P 1 , S 1 ) and G 2 = ( V 2 , T , P 2 , S 2 ) Assumption: V 1 ∩ V 2 = ∅ , that is, non-terminals are not shared S → ǫ | 0 S 1 Theorem L = { 0 n 1 m | m > n } CFLs are closed under union. L 1 , L 2 CFLs implies L 1 ∪ L 2 is a CFL . � � ∗ � � � L = w ∈ ( , ) � w is properly nested string of parenthesis � Theorem CFLs are closed under concatenation. L 1 , L 2 CFLs implies L 1 · L 2 is a CFL . Theorem CFLs are closed under Kleene star. ⇒ L ∗ is a CFL . If L is a CFL = Sariel Har-Peled (UIUC) CS374 17 Fall 2017 17 / 36 Sariel Har-Peled (UIUC) CS374 18 Fall 2017 18 / 36 Closure Properties of CFLs Closure Properties of CFLs Union Concatenation G 1 = ( V 1 , T , P 1 , S 1 ) and G 2 = ( V 2 , T , P 2 , S 2 ) Theorem Assumption: V 1 ∩ V 2 = ∅ , that is, non-terminals are not shared. CFLs are closed under concatenation. L 1 , L 2 CFLs implies L 1 · L 2 Theorem is a CFL . CFLs are closed under union. L 1 , L 2 CFLs implies L 1 ∪ L 2 is a CFL . Sariel Har-Peled (UIUC) CS374 19 Fall 2017 19 / 36 Sariel Har-Peled (UIUC) CS374 20 Fall 2017 20 / 36
Closure Properties of CFLs Exercise Stardom (i.e, Kleene star) Prove that every regular language is context-free using previous Theorem closure properties. CFLs are closed under Kleene star. Prove the set of regular expressions over an alphabet Σ forms a ⇒ L ∗ is a CFL . If L is a CFL = non-regular language which is context-free. Sariel Har-Peled (UIUC) CS374 21 Fall 2017 21 / 36 Sariel Har-Peled (UIUC) CS374 22 Fall 2017 22 / 36 Closure Properties of CFLs continued Canonical non- CFL Theorem Theorem L = { a n b n c n | n ≥ 0 } is not context-free. CFLs are not closed under complement or intersection. Proof based on pumping lemma for CFLs . Technical and outside Theorem the scope of this class. If L 1 is a CFL and L 2 is regular then L 1 ∩ L 2 is a CFL . Sariel Har-Peled (UIUC) CS374 23 Fall 2017 23 / 36 Sariel Har-Peled (UIUC) CS374 24 Fall 2017 24 / 36
Parse Trees or Derivation Trees Example ∗ w . A tree to represent the derivation S � S A derivation tree for abbaab Rooted tree with root labeled S (also called “parse tree”) Non-terminals at each internal node of tree a S b Terminals at leaves Children of internal node indicate how non-terminal was b S a S à aSb | bSa | SS | ab | ba | ε expanded using a production rule A picture is worth a thousand words S S A corresponding derivation of abbaab b a ε S è aSb è abSab è abSSab è abbaSab è abbaab Sariel Har-Peled (UIUC) CS374 25 Fall 2017 25 / 36 Sariel Har-Peled (UIUC) CS374 26 Fall 2017 26 / 36 Ambiguity in CFLs Ambiguity in CFLs Definition Original grammar: S → S − S | 1 | 2 | 3 Unambiguous grammar: A CFG G is ambiguous if there is a string w ∈ L ( G ) with two S → S − C | 1 | 2 | 3 different parse trees. If there is no such string then G is C → 1 | 2 | 3 unambiguous. Example: S → S − S | 1 | 2 | 3 S S S – S C – The grammar forces a parse S S – S S corresponding to left-to-right S – C 1 evaluation. S – S S – S 3 1 3 2 1 3 2 2 (3–2)–1 3–(2–1) (3–2)–1 Sariel Har-Peled (UIUC) CS374 27 Fall 2017 27 / 36 Sariel Har-Peled (UIUC) CS374 28 Fall 2017 28 / 36
Inherently ambiguous languages Inductive proofs for CFGs Question: How do we formally prove that a CFG L ( G ) = L ? Definition A CFL L is inherently ambiguous if there is no unambiguous CFG Example: S → ǫ | a | b | aSa | bSb G such that L = L ( G ) . Theorem There exist inherently ambiguous CFLs . Example: L = { a n b m c k | n = m or m = k } L ( G ) = { palindromes } = { w | w = w R } Given a grammar G it is undecidable to check whether L ( G ) is Two directions: inherently ambiguous. No algorithm! ∗ w then w = w R L ( G ) ⊆ L , that is, S � L ⊆ L ( G ) , that is, w = w R then S � ∗ w Sariel Har-Peled (UIUC) CS374 29 Fall 2017 29 / 36 Sariel Har-Peled (UIUC) CS374 30 Fall 2017 30 / 36 L(G) ⊆ L L ⊆ L(G) ∗ w then w = w R Show that if w = w R then S � ∗ w . Show that if S � By induction on length of derivation, meaning By induction on | w | ∗ k w implies w = w R . That is, for all k ≥ 0 , | w | = k and w = w R implies S � ∗ w . For all k ≥ 1 , S � If S � 1 w then w = ǫ or w = a or w = b . Each case Exercise: Fill in proof. w = w R . Assume that for all k < n , that if S → k w then w = w R Let S � n w (with n > 1 ). Wlog w begin with a . ◮ Then S → aSa � k − 1 aua where w = aua . ◮ And S � n − 1 u and hence IH, u = u R . ◮ Therefore w r = ( aua ) R = ( ua ) R a = au R a = aua = w . Sariel Har-Peled (UIUC) CS374 31 Fall 2017 31 / 36 Sariel Har-Peled (UIUC) CS374 32 Fall 2017 32 / 36
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