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Re-enter Chomsky

Re-enter Chomsky. More about grammars. Consider L = { a m b n | m , n > 0 }. (one/more a ’s followed by one/more b ’s). A. B. S  A B  a A B  aa A B  aaa B  aaa bB  aaa bb. S  A B  A bB  A bb  aA bb  aaA bb  aaa bb. a. A. b. B.

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Re-enter Chomsky

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  1. Re-enter Chomsky More about grammars

  2. Consider L = {am bn| m, n > 0 } (one/more a’s followed by one/more b’s) A B S A B  a A B  aa A B  aaa B  aaa bB  aaa bb S A B  A bB  A bb  aA bb  aaA bb  aaa bb a A b B a A Parse trees S  A B A  aA | a B  bB | b A Parse Tree (for the same string) This is a “common” representation where the order of derivation is not explicit—there is no such thing as “left parse tree” or “right parse tree”! Consider the string “aaabb” which is a valid string in L and can be derived from the grammar. S Left-most derivation Right-most derivation b a The leaves of the tree form the input string.

  3. a b b a a b b a L b L L b L L L L L L a L a a L b L b L a L ε ε ε ε Parse trees Consider L = { wє {a,b}* | w contains a ‘b’ } (any combination of a’s & b’s that contains at least one b somewhere) S  L b L L  aL | bL |ε Consider the string “abba” which is a valid string in L and can be derived/generated from/by the grammar. Parse Tree 1 Parse Tree 2 S S The grammar can generate the input string in two different ways. In other words, there are two different parse trees for the string. Since it’s unclear as to how exactly the grammar should generate the string, the grammar is said to be ambiguous *. Note that the grammar on the previous slide is not ambiguous. * This example is based on an observation by Mr. Hui Zhang, a COMPSCI 220 student.

  4. a b b a X b Y X Y a X b Y a Y ε ε Parse trees An unambiguous grammar for L = { wє {a,b}* | w contains a ‘b’ } has only zero/more a’s first occurrence of b zero/more a’s and b’s S  X b Y X  a X |ε Y  a Y | b Y | ε S Consider the string “abba” which is a valid string in L and can be derived/generated from/by the grammar. There is only one way in which you can “group” the input string this time!

  5. Ambiguous grammars A grammar is said to be ambiguous if it generates some string wєΣ* in more than one way, i.e. if the string has more than one parse tree.

  6. What is wrong with ambiguity? Ambiguous grammars can be undesirable, for instance, in Compiler Design*, where the code generated by the compiler might depend on the particular way in which the input string (a statement in a programming language) is generated. This will be demonstrated in the examples that follow. * Grammars are used to describe the syntax of statements in a programming language.

  7. Grammar for IF-ELSE statement Is condition true? yes no Is condition true? yes no (i) IF (condition) Statement 1; Statement 2; Statement 1 Statement 2 (ii) IF (condition) Statement 1; ELSE Statement 2; Statement 3; Statement 1 Statement 2 Statement 3 Statement  IF_statement | … IF_statement if (Cond) Statement IF_statement if (Cond) Statement else Statement

  8. Grammar for IF-ELSE statement if ( Cond ) Statement else Statement if ( Cond ) Statement else Statement Statement  IF_statement | … IF_statement if (Cond) Statement IF_statement if (Cond) Statement else Statement Parse tree for the statement Statement Consider the statement: IF_Statement IF ( C1) S1; ELSE IF ( C2 ) S2; ELSE S3; IF_Statement C1 S1 C2 S2 S3

  9. ?  if ( Cond ) Statement if ( Cond ) Statement Statement else S3 if ( Cond ) Statement else Statement if ( Cond ) Statement Grammar for IF-ELSE statement Consider the statement: IF ( C1) IF ( C2 ) S2; ELSE S3; IF ( C1) IF ( C2 ) S2; ELSE S3; Parse tree 1 Parse tree 2 AMBIGUITY ! (the same expression can be generated in a different way) Statement Statement IF_Statement IF_Statement IF_Statement IF_Statement C1 C1 C2 S2 S3 C2 S2

  10. Grammar for arithmetic expressions VERSION I Consider arithmetic expressions with only one or two variables (that use +, -, * only). e.g. a, a + b, a – b, a * b E Var E Var + Var E Var – Var E Var * Var

  11. Var * E Var + E Wrong grouping! (wrong order of precedence) 1 a * b + c 2 Grammar for arithmetic expressions VERSION II Consider arithmetic expressions with any number of variables (more realistic!). e.g. a + b – c * d Try generating the expression: a * b + c E Var E Var + E E Var – E E Var * E E a b Var c

  12. E * E E + E AMBIGUITY ! (the same expression can be generated in a different way) E + E E * E looks OK! Grammar for arithmetic expressions VERSION III Try to generate arithmetic expressions—preserving the order of precedence. E Var E E + E E E – E E E * E Try generating the same expression again: a * b + c E E Var Var a c Var Var Var Var b c a b Note: Each parse tree conveys a different “meaning”; each of them corresponds to a different code (therefore possibly different results) generated by the compiler.

  13. E + Term Var Term * Grammar for arithmetic expressions VERSION IV Try to generate arithmetic expressions—preserving the order of precedence and also avoiding ambiguity. Try generating the same expression again: a * b + c E E + Term E E – Term E Term Term Term * Var Term Var E Term Var c Var b a

  14. What Context Free Grammars (CFGs) can’t express Examples of languages that can’t be generated by CFGs: { an bn cn| n, m > 0} { an bm cn dm | n, m > 0} {w c w| w єΣ* }

  15. Four classes of grammar Type 3 (Regular grammar) Right side: (i) a single terminal symbol OR (ii) a single terminal followed by a single non-terminal Left side: a single non-terminal symbol A  a A  a B Type 2 (Context-free grammar) Right side: no restriction (any string of terminals and non-terminals). Left side: a single non-terminal symbol A α Type 1 (Context-sensitive grammar) Right, left sides: no restriction except that length( α ) <= length ( β ) αβ Type 0 (Phrase-structure grammar) A language is said to be type i(i = 0, 1, 2, 3) if it can be specified by type i grammar and cannot specified by type (i +1) grammar. Right, left sides: no restriction at all! αβ

  16. Converting FSA into equivalent grammar b a i a j b L = {strings of a’s and b’s—with at least one ‘a’} Any given FSA can be mechanically converted into grammar rules that generate the exactly the same language recognized by the FSA. Rules: (i) For an a-transition from state i to state j, generate the production rule: A i a A j (i) For the final state f, generate the production rule: A fε Grammar rules that generate L A i a A j A i b A i A j a A j A j b A j A jε

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