1 / 18

Chapter 3 - 2

Chapter 3 - 2. Construction Techniques. Section 3.3 Grammars. A grammar is a finite set of rules, called productions, that are used to describe the strings of a language.

topper
Download Presentation

Chapter 3 - 2

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Chapter 3 - 2 Construction Techniques

  2. Section 3.3 Grammars • A grammar is a finite set of rules, called productions, that are used to describe the strings of a language. • Notational Example. The productions take the form a→ ß, where a and ß are strings over an alphabet of terminals and nonterminals. Read a→ ß as, “a produces ß,” “a derives ß,” or “a is replaced by ß.” The following four expressions are productions for a grammar. S → aSB Alternative Short form S → Λ S → aSB | Λ B → bB B → bB | b B → b.

  3. Terms • Terminals: {a, b}, the alphabet of the language. • Nonterminals: {S, B}, the grammar symbols (uppercase letters), disjoint from terminals. • Start symbol: S, a specified nonterminal alone on the left side of some production. • Sentential form: any string of terminals and/or nonterminals.

  4. Derivation • a transformation of sentential forms by means of productions as follows: If xay is a sentential form and a→ß is a production, then the replacement of a by ß in xay to obtain xßy is a derivation step, which we denote by xay → xßy. • Example Derivation: S ➯ aSB ➯ aaSBB ➯ aaBB ➯ aabBB ➯ aabbB ➯ aabbb. This is a leftmost derivation, where each step replaces the leftmost nonterminal. • The symbol ➯+ means one or more steps and ➯* means zero or more steps. So we could write S ➯+ aabbb or S ➯* aabbb or aSB ➯* aSB, and so on.

  5. The Language of a Grammar • The language of a grammar is the set of terminal strings derived from the start symbol. • Example. Can we find the language of the grammar S → aSB | Λ and B → bB | b? • Solution: Examine some derivations to see if a pattern emerges. S ➯ Λ S ➯ aSB ➯ aB ➯ ab S ➯ aSB ➯ aB ➯ abB ➯ abbB ➯ abbb S ➯ aSB ➯ aaSBB ➯ aaBB ➯ aabB ➯ aabb S ➯ aSB ➯ aaSBB ➯ aaBB ➯ aabBB ➯ aabbBB ➯ aabbbB ➯ aabbbb. So we have a pretty good idea that the language of the grammar is {anbn+k | n, k ∊ N}. • Quiz (1 minute). Describe the language of the grammar S → a | bcS. • Solution: {(bc)na | n ∊ N}.

  6. Construction of Grammars • Example. Find a grammar for {anb | n ∊ N}. • Solution: We need to derive any string of a’s followed by b. The production S → aS can be used to derive strings of a’s. The production S → b will stop the derivation and produce the desired string ending with b. So a grammar for the language is S → aS | b.

  7. Quizzes • Quiz (1 minute). Find a grammar for {ban | n ∊ N}. • Solution: S → Sa | b. • Quiz (1 minute). Find a grammar for {(ab)n | n ∊ N}. • Solution: S → Sab | Λ or S → abS | Λ.

  8. Rules for Combining Grammars • Let L and M be two languages with grammars that have start symbols A and B, respectively, and with disjoint sets of nonterminals. Then the following rules apply. • L ∪ M has a grammar starting with S → A | B. • LM has a grammar starting with S → AB. • L* has a grammar starting with S → AS | Λ.

  9. Example • Find a grammar for {ambmcn | m, n ∊ N}. • Solution: The language is the product LM, where L = {ambm | m ∊ N} and M = {cn | n ∊ N}. So a grammar for LM can be written in terms of grammars for L and M as follows. S → AB A → aAb | Λ B → cB | Λ.

  10. Example • Find a grammar for the set Odd, of odd decimal numerals with no leading zeros, where, for example, 305 ∊ Odd, but 0305 ∉ Odd. • Solution: Notice that Odd can be written in the form Odd = (PD*)*O, where O = {1, 3, 5, 7, 9}, P = {1, 2, 3, 4, 5, 6, 7, 8, 9}, and D = {0} ∪ P. Grammars for O, P, and D can be written with start symbols A, B, and C as: A → 1 | 3 | 5 | 7 | 9, B → A | 2 | 4 | 6 | 8, and C → B | 0. Grammars for D* and PD* and (PD*)* can be written with start symbols E, F, and G as: E → CE | Λ, F → BE, and G → FG | Λ. So a grammar for Odd with start symbol S is S → GA.

  11. Example • Find a grammar for the language L defined inductively by, • Basis: a, b, c ∊ L. • Induction: If x, y ∊ L then ƒ(x), g(x, y) ∊ L. • Solution: We can get some idea about L by listing some of its strings. a, b, c, ƒ(a), ƒ(b), …, g(a, a), …, g(ƒ(a), ƒ(a)), …, ƒ(g(b, c)), …, g(ƒ(a), g(b, ƒ(c))), … So L is the set of all algebraic expressions made up from the letters a, b, c, and the function symbols ƒ and g of arities 1 and 2, respectively. A grammar for L can be written as S → a | b | c | ƒ(S) | g(S, S). For example, a leftmost derivation of g(ƒ(a), g(b, ƒ(c))) can be written as S ➯ g(S, S) ➯ g(ƒ(S), S) ➯ g(ƒ(a), S) ➯ g(ƒ(a), g(S, S)) ➯ g(ƒ(a), g(b, S)) ➯ g(ƒ(a), g(b, ƒ(S))) ➯ g(ƒ(a), g(b, ƒ(c))).

  12. Parse Tree S • A Parse Tree is a tree that represents a derivation. The root is the start symbol and the children of a nonterminal node are the symbols(terminals, nonterminals, or Λ) on the right side of the production used in the derivation step that replaces that node. • Example. The tree shown in the picture is the parse tree for the following derivation: S ➯ g(S, S) ➯ g(ƒ(S), S) ➯ g(ƒ(a), S) ➯ g(ƒ(a), b). g ( , S S ) ( S ) b f a

  13. Ambiguous Grammar • Means there is at least one string with two distinct parse trees, or equivalently, two distinct leftmost derivations or two distinct rightmost derivations. • Example. Is the grammar S → SaS | b ambiguous? • Solution: Yes. For example, the string babab has two distinct leftmost derivations: S ⇒ SaS ⇒ SaSaS ⇒ baSaS ⇒ babaS ⇒ babab. S ⇒ SaS ⇒ baS ⇒ baSaS ⇒ babaS ⇒ babab. The parse trees for the derivations are pictured in the next slide.

  14. parse trees S S S a S S a S S a S b b S a S b b b b

  15. Quiz (2 minutes) • Show that the grammar S → abS | Sab | c is ambiguous. • Solution: The string abcab has two distinct leftmost derivations: S ⇒ abS ⇒ abSab ⇒ abcab S ⇒ Sab ⇒ abSab ⇒ abcab.

  16. Unambiguous Grammar • Sometimes one can find a grammar that is not ambiguous for the language of an ambiguous grammar. • Example. The previous example showed S → SaS | b is ambiguous. The language of the grammar is {b, bab, babab, …}. Another grammar for the language is S → baS | b. It is unambiguous because S produces either baS or b, which can’t derive the same string.

  17. Example • The previous quiz showed S → c | abS | Sab is ambiguous. Its language is {(ab)mc(ab)n | m, n ∈ N}. Another grammar for the language is S → abS | cT and T → abT | ٨. • It is unambiguous because S produces either abS or cT, which can’t derive the same string. • Similarly, T produces either abT or ٨, which can’t derive the same string.

  18. The End of Chapter 3 - 2

More Related