Chapter 12 How do you get a derivation from a Phrase Structured - PDF document

Chapter 12 � How do you get a derivation from a Phrase Structured Grammar �G�V,T,S,P� ICS 6B Suppose a phrase‐structured grammar has Boolean Algebra & Logic productions: S �ABa, A �BB, B �ab, AB �b �1� �2� �3� �4� Find the derivation of BBaba: Lecture Notes for Summer Quarter, 2008 ba �4� �1� �3�, �3�, �3� S � ABa � BBBa �2� � abababa Michele Rousseau �3� Aaba �BBaba �3�, �3� �2� Answer: ABa � Aaba � Bbaba Final Review (Some slides inspired and adapted from Alessandra Pantano) ABa � BBBa � BBaba or 2 Different types of Grammars Construct the Language of a Grammar G ��V, T, S, P� with Type AKA Productions Allowed ● V��S, A, a, b� 0 Any production – no restrictions ● T��a, b� Change 1 (non-terminal symbol) between 2 strings AND ● S� the start element S � λ lAr � lwr Context – ● P��S�aA, S �b, A�aa� 1 sensitive A is a N (non-terminal) symbol lr is a string of T & N symbols (may be λ ) l i t i f T & N b l ( b λ ) Find L�G� w is a string of T & N symbols (can’t be empty) Change 1 non-terminal symbol S � aA � aaa A �lr � 2 Context-free A is a single N terminal symbol lr is a string of T & N symbols (may be λ ) b L(G) = {b, aaa} Change 1 non-terminal symbol � very restricted how it can be changed A �aB; A�a & S�λ 3 regular A & B are non terminal, a is terminal 3 4 Example S � AB �CaBa �cbaba � V��S,A,B,a,b� Productions Type AKA Allowed ● Given P: � T��a,b� 0 no restrictions S�AB Context – S � λ S � P� 1 1. A �Ca sensitive lAr � lwr 2. B �Ba �1�S�Aba, AB � a, Context- A �lr 2 A B free 3. B �Cb �2�S �λ, aAb �aab A �aB; A�a & 3 3 regular regular 4. B �b 4 B b S�λ C a a B �3�S�λ, A�aa 5. C �cb �4�S�λ, A �b 6. C �b c b b G1� type 0, but not type 1 G4� type 3 � � Construct a derivation tree for cbaba ● AB is 2 N symbols We see that cbaba can be derived from: G2� type 1, but not type 2 � S � AB � CaBa � cbaba ● Not a single N symbol �has some T symbols too� Terminal elements are at the G3� type 2, but not type 3 end of each branch so we are done � ● Goes to two T & N strings 5 6 1

Remember Vending Machine Vending Maching I/O and States Table Inputs Outputs Transition – f(s,i) Output – g(s,i) Possible inputs Possible outputs �5�Nickel �N� Nada States 5 O A 5 O A Nickel �5� nothing �N� �O� O‐btn �5� Nickel 0 s 0 s 1 s 0 s 0 N N N �A�‐A‐Btn �OJ� O‐button �O� Nickel �5� 5c s 1 s 2 s 1 s 1 N N N …. �AJ� A‐button �A� OJ 10c s 2 s 3 s 2 s 2 N N N 15c s 3 s 3 s 0 s 0 5 OJ AJ A Directed Graph Possible states It is important to �A, AJ� O know how to go from one initial state s 0 �no money in the machine� to the other �O, N� �O, N� state s 1 �5c in the machine� �O, N� Start �5, N� �5, N� �5, N� s 0 s 1 s 2 s 3 �5, 5� state s 2 �10c in the machine� �A, N� �A, N� �A, N� state s 3 �15 in the machine� �O, OJ� 7 8 What strings are recognized by Going from Graph to Table the following machine s 1 � Only 1 final state s 3 s 1 � Think of all possible paths that O originate at s 0 and end at s 3 Transition– f(s,i) Output – g(s,i) O �1,0� start s 0 s 3 States 0 1 0 1 s 3 s 0 s 0 s 1 s 0 1 0 � We can loop as many times � We can loop as many times �0 0� �0,0� s 1 s 3 s 0 1 1 O as we want, s 2 s 1 s 2 0 1 0 ; 1 s 2 s 2 and then we must follow s 3 s 2 s 1 0 0 �1,1� This says the strings recognized s 1 � Also, be able to determine output by the machine are all of the form Also be able to do this O n 11, with n�0 Example with automaton What is the output of 1110? start s 0 s 3 0001 9 10 Given a automaton determine How is a Non-Deterministic L(M) automaton different? 1 O � M: s 1 Start � Final states: s 2, s 3 0 1 0,1 s 0 s 1 s 2 s 3 0,1 Inputs O O,1 O States 0 1 start s 0 1 s 3 s 0 s 0, s 2 s 3 � There’s only 1 final state �s 2 �. So we look y � 2 � s 0 s 1 , s 3 s 1 s for paths from s 0 to s s 0 , s 2 s 2 1 O s 3 s 0 , s 2, s 2 s 1 s 2 0 1 1 OR 1 s 0 s 2 s 0 s 1 s 2 Notice that the transition function with state s 2 and input 0 produces the empty set of states: f�s 2 , 0� does not give any next state � �� L�M� ��0,01� 11 2

Chapter 1: Chapter 1 � Determine whether this proposition is a � In the questions below write the statement in the form “If … , tautology: ��p � �q� � q� � � p then … .” “The team wins if the quarterback can pass.” p q � p � q p �� q � q � � p If the quarterback can pass, then the team wins. T T F F F F T T F F T T F T Write the contrapositive, converse, and inverse of the following: “You sleep late if it is Saturday” following: You sleep late if it is Saturday T T T F T T F Contrapositive: �q � �p T T T F T F F If you do not sleep late, then it is not Saturday. � Also know how to prove equivalences Converse: q � p and how to show using identities If you sleep late, then it is Saturday. �� know the identities Inverse: �p � �q If it is not Saturday, then you do not sleep late. 13 14 CHAPTER 2 English to Logical Expressions F�x�: x is a fleegle Remember the Cartesian Product? S�x�: x is a snurd Let A��1,2� , B� �a,b,c�, C� �y,z� T�x�: x is a thingamabob What is A x B x C U��fleegles, snurds, thingamabobs� � A x B x C� ��1,a,y�, �1,a,z�, �1,b,y�, �1,b,z�, �1,c,y�, �1,c,z�, �2,a,y�, �2,a,z�, �2,b,y�, �2,b,z�, �2,c,y�, �2,c,z�� “ Everything is a fleegle” ∀ x F�x� � � ∃ x � F�x� � � � � “ Nothing is snurd” ∀ x � S�x� � � ∃ x S�x� “ All fleegles are snurds” ∀ x� F�x� �S�x�� � ∀ x�� F�x� � S�x�� � ∀ x �� F�x� � � S�x�� Thus A x B x C is all possible ordered tuples (a,b,c) where a � A , b � B , and c � C � � ∃ x�F�x� � � S�x�� 15 16 Operations on Sets Inverse Functions (Example) U � �0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10� f ‐1 f A B A B A� �1, 2, 3, 4, 5� a B � �4, 5, 6, 7, 8� a V V b b W A � B� �1, 2, 3, 4, 5, 6, 7, 8� W c c X X A �B � A �B �4, 5� �4 5� d d d d Y Y A c � �0, 6, 7, 8, 9, 10� B c � �0, 1, 2, 3, 9, 10� A ‐ B � �1, 2, 3� A bijection is called invertible because you can define B ‐ A � �6, 7, 8� the inverse function. To be invertible it must be a bijection. A�B � �1, 2, 3, 6, 7, 8� 17 18 3