1 / 14

练习 : 1. 文法 : Z → U0 | V1 U →Z1 | 1 V → Z0 | 0 (1) 请写出全部由此文法描述的只含有四个符号的句子 .

练习 : 1. 文法 : Z → U0 | V1 U →Z1 | 1 V → Z0 | 0 (1) 请写出全部由此文法描述的只含有四个符号的句子 . (2) 该文法是 Chomsky 几型文法 ?. 1. 文法 : Z → U0 | V1 U →Z1 | 1 V → Z0 | 0 请写出全部由此文法描述的只含有四个符号的句子: 0101, 1010, 1001, 0110 . 该文法是 Chomsky 几型文法 ? 3 型文法. 2 、 给定前缀表示的表达式文法G: (1) E → -EE (2) E → -E

alpha
Download Presentation

练习 : 1. 文法 : Z → U0 | V1 U →Z1 | 1 V → Z0 | 0 (1) 请写出全部由此文法描述的只含有四个符号的句子 .

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. 练习: • 1. 文法: Z → U0 | V1 U →Z1 | 1 V → Z0 | 0 (1)请写出全部由此文法描述的只含有四个符号的句子. (2)该文法是Chomsky几型文法?

  2. 1. 文法: Z → U0 | V1 U →Z1 | 1 V → Z0 | 0 请写出全部由此文法描述的只含有四个符号的句子: 0101,1010,1001,0110 . 该文法是Chomsky几型文法? 3型文法

  3. 2、给定前缀表示的表达式文法G: (1)E→-EE (2)E→-E (3)E→a (4)E→b (5)E→c 试问--a-bc是否L(G)的句子?若是,请给出该句子所有可能的分析树;若不是,请说明理由.

  4. 可能的分析树(1): E - E E - E c E a - E b

  5. 可能的分析树(2): E - E E E E - - E b c a

  6. 可能的分析树3): E - E - E E a E - E c b

  7. S ( L ) ( a , ( a , a ) ) L , S • 4.1 P267 S ( L ) 最左推导:S =lm> (L) =lm> (L,S) =lm> (S,S) =lm> (a,S) =lm> (a,(L)) =lm> (a,(L,S)) =lm> (a,(S,S)) =lm> (a,(a,S)) =lm> (a,(a,a)) a L , S S a a 最右推导:S =rm> (L) =rm> (L,S) =rm> (L,(L)) =rm> (L,(L,S)) =rm> (L,(L,a)) =rm> (L,(S,a)) =rm> (L,(a,a)) =rm> (S,(a,a)) =rm> (a,(a,a))

  8. abab S S • 4.2 a S b S a S b S ε b S a S ε a S b S ε ε ε ε 最左推导:S =lm> aSbS =lm> abSaSbS =lm> abaSbS =lm> ababS =lm> abab 最左推导:S =lm> aSbS =lm> abS =lm> abaSbS =lm> ababS =lm> abab

  9. 文法G: P → PaP | PbP | cP | Pe | f 证明文法G是二义文法. ** 通过证明句型 f b f b f 存在两棵分析树.

  10. P P P P b P P b P P f b P P f b f f f f 最左推导:P =lm> PbP =lm> PbPbP =lm> fbPbP =lm> fbfbP =lm> fbfbf 最左推导:P =lm> PbP =lm> fbP =lm> fbPbP =lm> fbfbP =lm> fbfbf

  11. 作业: • 1. 文法: S → (L) | a L → L,S | S 分析句型: ( L , a , S ) 求:短语、直接短语、句柄、素短语、最左素短语

  12. S ( ) L • 句型:( L , a , S ) L S , • 短语: ( L , a , S )、 L , a , S、 L , a、a L S , • 直接短语:a a • 句柄:a • 素短语:a • 最左素短语: a

  13. 2. 文法: S → a│^│( T ) T → T , S│S 分析句型: ( T , ( ^ , a ) ) 求:短语、直接短语、句柄、素短语、最左素短语 ftp://192.168.151.129 pwb pwb

  14. S ( ) T • 句型: ( T , ( ^ , a ) ) T S , • 短语: ( T , ( ^ , a ) )、 ^ , a T , ( ^ , a )、 ( ^ , a )、 ^、a ( ) T • 直接短语: ^ 、a T S , • 句柄: ^ a • 素短语: ^ 、a S • 最左素短语: ^ ^

More Related