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Lecture # 8 Chapter # 4: Syntax Analysis. Practice Context Free Grammars a) CFG generating alternating sequence of 0’s and 1’s b) CFG in which no consecutive.

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Presentation on theme: "Lecture # 8 Chapter # 4: Syntax Analysis. Practice Context Free Grammars a) CFG generating alternating sequence of 0’s and 1’s b) CFG in which no consecutive."— Presentation transcript:

1 Lecture # 8 Chapter # 4: Syntax Analysis

2 Practice Context Free Grammars a) CFG generating alternating sequence of 0’s and 1’s b) CFG in which no consecutive b’s can occur but consecutive a’s can occur c) CFG for the following language: L(G)= {a n b 2n | n>=0}

3 Practice Answers a) S  0A | 1B c) S  aSbb | є A  1B | 0 B  0A | 1 b)S  aS | bT |a |b T  aS | a

4 Example Design a CFG for the language L(G)= {0 n 1 m | n <> m} There are two cases: – For n>m – For n

5 Example For n>m S1  AB B  0A1 | Combining both: A  0A | 0 S  S1 | S2 For n

6 Practice CFG Design a CFG for the language L(G)= {0 i 1 j 2 k | i=j or j=k} There are two cases: – For i=j – For j=k – Write two separate set of rules and combine them

7 Solution of Practice For i=j S1  AB A  0A1 | ε Combining both: B  2B | ε S  S1 | S2 For j=k S2  XY X  0X | ε Y  1Y2 | ε

8 Derivations The one-step derivation is defined by  A      where A   is a production in the grammar In addition, we define –  is leftmost  lm if  does not contain a nonterminal –  is rightmost  rm if  does not contain a nonterminal – Transitive closure  * (zero or more steps) – Positive closure  + (one or more steps) The language generated by G is defined by L(G) = {w  T* | S  + w}

9 9 Derivation (Example) Grammar G = ({E}, {+,*,(,),-,id}, P, E) with productions P =E  E + E E  E * E E  ( E ) E  - E E  id E  - E  - id E * EE * E E  + id * id + id E  rm E + E  rm E + id  rm id + id Example derivations: E  * id + id

10 Example For the given grammar derive the string using Left most derivation Then derive the same string using Right most derivation

11 11 Example Grammar list  list + digit list  list - digit list  digit digit  0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 G = with productions P = Context-free grammar for simple expressions:

12 12 Derivation for the Example Grammar list  list + digit  list - digit + digit  digit - digit + digit  9 - digit + digit  digit  This is an example leftmost derivation, because we replaced the leftmost nonterminal (underlined) in each step. Likewise, a rightmost derivation replaces the rightmost nonterminal in each step

13 13 Chomsky Hierarchy: Language Classification A grammar G is said to be – Regular if it is right linear where each production is of the form A  w Bor A  w or left linear where each production is of the form A  B wor A  w – Context free if each production is of the form A   where A  N and   (N  T)* – Context sensitive if each production is of the form  A      where A  N, , ,   (N  T)*, |  | > 0 – Unrestricted

14 14 Chomsky Hierarchy L (regular)  L (context free)  L (context sensitive)  L (unrestricted) Where L (T) = { L(G) | G is of type T } That is: the set of all languages generated by grammars G of type T L 1 = { a n b n | n  1 } is context free L 2 = { a n b n c n | n  1 } is context sensitive Every finite language is regular! (construct a FSA for strings in L(G)) Examples:

15 15 Parse Trees The root of the tree is labeled by the start symbol Each leaf of the tree is labeled by a terminal (=token) or  Each interior node is labeled by a nonterminal If A  X 1 X 2 … X n is a production, then node A has immediate children X 1, X 2, …, X n where X i is a (non)terminal or  (  denotes the empty string)

16 16 Parse Tree for the Example Grammar Parse tree of the string using grammar G list digit list digit The sequence of leafs is called the yield of the parse tree

17 Example of Parse Tree Suppose we have the following grammar E → E + E E → E * E E → ( E ) E → - E E → id Perform Left most derivation, right most derivation and construct a parse tree for the string id+id*id

18 Two possible Parse Trees using Leftmost derivation

19 Parse Tree via Right most derivation

20 Ambiguity Grammar is ambiguous if more than one parse tree is possible for some string as shown in the previous example. If there are more than one left most derivations or more than one right most derivations. Ambiguity is not acceptable – Unfortunately, it’s undecidable to check whether a given CFG is ambiguous – Some CFLs are inherently ambiguous (do not have an unambiguous CFG)

21 21 Ambiguity (cont’d) string  string + string | string - string | 0 | 1 | … | 9 G = with production P = Consider the following context-free grammar: This grammar is ambiguous, because more than one parse tree represents the string 9-5+2

22 22 Two Parse Trees for the same string string

23 Practice Show that the following grammar is ambiguous: (Find out strings and two parse trees) 1) S  AB | aaB 2) S  a | abSb |aAb A  a | Aa A  bS | aAAb B  b 3) S  aSb | SS | ε


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