Your SlideShare is downloading. ×
Introduction to Computer theory (Automata Theory) 2nd Edition By Denial I.A. COHEN. Chapter 2 Problems
Upcoming SlideShare
Loading in...5
×

Thanks for flagging this SlideShare!

Oops! An error has occurred.

×
Saving this for later? Get the SlideShare app to save on your phone or tablet. Read anywhere, anytime – even offline.
Text the download link to your phone
Standard text messaging rates apply

Introduction to Computer theory (Automata Theory) 2nd Edition By Denial I.A. COHEN. Chapter 2 Problems

4,257
views

Published on

solution to some problems of Automata

solution to some problems of Automata

Published in: Data & Analytics

0 Comments
4 Likes
Statistics
Notes
  • Be the first to comment

No Downloads
Views
Total Views
4,257
On Slideshare
0
From Embeds
0
Number of Embeds
6
Actions
Shares
0
Downloads
174
Comments
0
Likes
4
Embeds 0
No embeds

Report content
Flagged as inappropriate Flag as inappropriate
Flag as inappropriate

Select your reason for flagging this presentation as inappropriate.

Cancel
No notes for slide

Transcript

  • 1. By: F.A 4/1/2014 AUTOMATA CHAPTER 2: LANGUAGES (PROBLEMS)
  • 2. Chapter 2: LANGUAGES Problems: 1. Consider the language S*, where S = {a, b}. How many words does this language have of length 2? Of length 3? Of length n? Solution: S = {a, b} S* = {, a, b, aa, ab, bb, ba, aaa, aab, aba, abb, bbb, bba, bab, baa,…. }  So this language have words of length 2 = 4 Of length 3 = 8 Of length n = 2n  We can solve this question in the following way too Let number of words = nm Then words of length 2 = 22 = 4 Of length 3 = 23 = 8 Of length n = 2n 2. Consider the language S*, where S = {aa, b}. how many words does this language have of length 4? Of length 5? Of length 6? What can be said in general? Solution: S = {aa, b} S* = {, aa, b, aaaa, aab, bb, baa, aaaaaa, aaaab, aabb, aabaa, baaaa, baab, bbb, bbaa, aaaaaaaa, aaaaaab, aaaabb, aaaabaa, aabaaaa, aabaab, aabbb, aabbaa,
  • 3. bbbb, bbbaa, bbaaaa, bbaab, baabb, baabaa, baaaaaa, baaaab, aaaaaaaaaa, …….}  So words of length 4 = 24 = 16 Of length 5 = 25 = 32 Of length 5 = 26 = 64 In general: In general we can say that 3. Consider the language S*, where S = {ab, ba}. Write out all the words in S* that have seven or fewer letters. Can any word in this language contain the substrings aaa or bbb? What is the smallest word that is not in this language? Solution: S = {ab, ba} S* = {, ab, ba, abab, abba, baba, baab, ababab, ababba, abbaba, abbaab, baabab, baabba, bababa, babaab, abababab, abababba, ababbaba, ababbaab, abbaabab, abbaabba, abbababa, abbabaab, baababab, baababba, baabbaba, baabbaab, babaabab, babaabba, babababa, bababaab, ababababab, ababababba, abababbaba, abababbaab, ababbaabab, ababbaabba, ababbababa, ababbabaab, abbaababab, abbaababba, abbaabbaba, abbaabbaab, abbabaabab, abbabaabba, abbabababa, abbababaab, baabababab, baabababba, baababbaba, baababbaab, baabbaabab, baabbaabba, baabbababa, baabbabaab, babaababab, babaababba, babaabbaba, babaabbaab, bababaabab, bababaabba, bababababa, babababaab, abababababab, abababababba, ababababbaba, ababababbaab, abababbaabab, abababbaabba, abababbababa, abababbabaab, ababbaababab, ababbaababba, ababbaabbaba, ababbaabbaab, ababbabaabab, ababbabaabba, ababbabababa, ababbababaab, abbaabababab, abbaabababba, abbaababbaba, abbaababbaab, abbaabbaabab, abbaabbaabba, abbaabbababa, abbaabbabaab, abbabaababab, abbabaababba, abbabaabbaba, abbabaabbaab, abbababaabab, abbababaabba, abbababababa, abbabababaab, ababababababab, ababababababba, abababababbaba, abababababbaab, ababababbaabab, ababababbaabba, ababababbababa, ababababbabaab, abababbaababab, abababbaababba, ababbaabbaba, ababbaabbaab, ababbabaabab,
  • 4. abababbabaabba, abababbabababa, abababbababaab, ababbaabababab, ababbaabababba, ababbaababbaba, ababbaababbaab, ababbaabbaabab, ababbaabbaabba, ababbaabbababa, ababbaabbabaab, ababbabaababab, ababbabaababba, ababbabaabbaba, ababbabaabbaab, ababbababaabab, ababbababaabba, ababbababababa, ababbabababaab, baabababababab, baabababababba, baababababbaba, baababababbaab, baabababbaabab, baabababbaabba, baabababbababa, baabababbabaab, baababbaababab, baababbaababba, baababbaabbaba, baababbaabbaab, baababbabaabab, baababbabaabba, baababbabababa, baababbababaab, baabbaabababab, baabbaabababba, baabbaababbaba, baabbaababbaab, baabbaabbaabab, baabbaabbaabba, baabbaabbababa, baabbaabbabaab, baabbabaababab, baabbabaababba, baabbabaabbaba, baabbabaabbaab, baabbababaabab, baabbababaabba, baabbababababa, baabbabababaab,…}  All words in S* that have seven or fewer letters: S* = { abababababab, abababababba, ababababbaba, ababababbaab, abababbaabab, abababbaabba, abababbababa, abababbabaab, ababbaababab, ababbaababba, ababbaabbaba, ababbaabbaab, ababbabaabab, ababbabaabba, ababbabababa, ababbababaab, abbaabababab, abbaabababba, abbaababbaba, abbaababbaab, abbaabbaabab, abbaabbaabba, abbaabbababa, abbaabbabaab, abbabaababab, abbabaababba, abbabaabbaba, abbabaabbaab, abbababaabab, abbababaabba, abbababababa, abbabababaab, ababababababab, ababababababba, abababababbaba, abababababbaab, ababababbaabab, ababababbaabba, ababababbababa, ababababbabaab, abababbaababab, abababbaababba, ababbaabbaba, ababbaabbaab, ababbabaabab, abababbabaabba, abababbabababa, abababbababaab, ababbaabababab, ababbaabababba, ababbaababbaba, ababbaababbaab, ababbaabbaabab, ababbaabbaabba, ababbaabbababa, ababbaabbabaab, ababbabaababab, ababbabaababba, ababbabaabbaba, ababbabaabbaab, ababbababaabab, ababbababaabba, ababbababababa, ababbabababaab, baabababababab, baabababababba, baababababbaba, baababababbaab, baabababbaabab, baabababbaabba, baabababbababa, baabababbabaab, baababbaababab, baababbaababba, baababbaabbaba, baababbaabbaab, baababbabaabab, baababbabaabba, baababbabababa, baababbababaab, baabbaabababab, baabbaabababba, baabbaababbaba, baabbaababbaab, baabbaabbaabab, baabbaabbaabba, baabbaabbababa, baabbaabbabaab, baabbabaababab, baabbabaababba, baabbabaabbaba, baabbabaabbaab, baabbababaabab, baabbababaabba, baabbababababa, baabbabababaab }
  • 5.  No, No words can contain aaa or bbb because the first a in string ab and the a in ba never allow to make aaa or bbb.  The smallest word is of length zero (0) that is  (capital lambda) and it is present in the language so other than it there is no smallest word that is not in the language. 4. Consider the language S*, where S = {a, ab, ba}. Is the string (abbba) a word in this language? Write out all the words in this language with six or fewer letters. What is another way in which to describe the words in this language? Be careful this is not simply the language of all words without bbb. Solution:  No, the string abbba a word is not present in the language, because b individually does not exist in the given string S, so (ab b ba) can’t exist in this language.  All the words with six or fewer letters in this language are written below: S* = {, a, ab, ba, aa, aab, aba, aba, abab, abba, baa, baab, baba, aaa, aaab, aaba, aaba, aabab, aabba, abaa, abaab, ababa, abaa, abaab, ababa, ababa, ababab, ababba, abbaa, abbaab, abbaba, baaa, baaab, baaba, baaba, baabab, baabba, babaa, babaab, bababa, aaaa, aaaab, aaaba, aaaba, aaabab, aaabba, aabaa, aabaab, aababa, aabaa, aabaab, aababa, aababa, aababab, aababba, aabbaa, aabbaab, aabbaba, abaaa, abaaab, abaaba, abaaba, abaabab, abaabba, ababaa, ababaab, abababa, abaaa, abaaab, abaaba, abaaba, abaabab, abaabba, ababaa, ababaab, abababa, ababaa, ababaab, abababa, abababa, abababab, abababba, ababbaa, ababbaab, ababbaba, abbaaa, abbaaab, abbaaba, abbaaba, abbaabab, abbaabba, abbabaa, abbabaab, abbababa, baaaa, baaaab, baaaba, baaaba, baaabab, baaabba, baabaa, baabaab, baababa, baabaa, baabaab, baababa, baababa, baababab, baababba, baabbaa, baabbaab, baabbaba, babaaa, babaaab, babaaba, babaaba, babaabab, babaabba, bababaa, bababaab, babababa, aaaaa, aaaaab, aaaaba, aaaaba, aaaabab, aaaabba, aaabaa, aaabaab, aaababa, aaabaa, aaabaab, aaababa, aaababa, aaababab, aaababba, aaabbaa, aaabbaab, aaabbaba, aabaaa, aabaaab, aabaaba, aabaaba, aabaabab, aabaabba, aababaa, aababaab, aabababa, aabaaa, aabaaab, aabaaba, aabaaba, aabaabab, aabaabba, aababaa, aababaab, aabababa, aababaa, aababaab, aabababa, aabababa, aabababab, aabababba, aababbaa, aababbaab, aababbaba, aabbaaa, aabbaaab, aabbaaba, aabbaaba, aabbaabab, aabbaabba, aabbabaa, aabbabaab,
  • 6. aabbababa, abaaaa, abaaaab, abaaaba, abaaaba, abaaabab, abaaabba, abaabaa, abaabaab, abaababa, abaabaa, abaabaab, abaababa, abaababa, abaababab, abaababba, abaabbaa, abaabbaab, abaabbaba, ababaaa, ababaaab, ababaaba, ababaaba, ababaabab, ababaabba, abababaa, abababaab, ababababa, abaaaa, abaaaab, abaaaba, abaaaba, abaaabab, abaaabba, abaabaa, abaabaab, abaababa, abaabaa, abaabaab, abaababa, abaababa, abaababab, abaababba, abaabbaa, abaabbaab, abaabbaba, ababaaa, ababaaab, ababaaba, ababaaba, ababaabab, ababaabba, abababaa, abababaab, ababababa, ababaaa, ababaaab, ababaaba, ababaaba, ababaabab, ababaabba, abababaa, abababaab, ababababa, abababaa, abababaab, ababababa, ababababa, ababababab, ababababba, abababbaa, abababbaab, abababbaba, ababbaaa, ababbaaab, ababbaaba, ababbaaba, ababbaabab, ababbaabba, ababbabaa, ababbabaab, ababbababa, abbaaaa, abbaaaab, abbaaaba, abbaaaba, abbaaabab, abbaaabba, abbaabaa, abbaabaab, abbaababa, abbaabaa, abbaabaab, abbaababa, abbaababa, abbaababab, abbaababba, abbaabbaa, abbaabbaab, abbaabbaba, abbabaaa, abbabaaab, abbabaaba, abbabaaba, abbabaabab, abbabaabba, abbababaa, abbababaab, abbabababa,…} [upto six letters]  Another way…. 5. Consider the language S*, where S = {xx, xxx}. In how many ways can x 19 be written as the product of words in S? This means: How many different factorizations are there of x 19 into xx and xxx? Solution: (xx) (xx) (xx) (xx) (xx) (xx) (xx) (xx) + (xxx) = x16 + x3 = x19 x19 can consist of 8 double xx combinations and 1 triple xxx combination i.e = 8 * 2 + 1 * 3 = 19 x19 can consist of 5 double xx combinations and 3 triples xxx combinations = 5 * 2 + 3 * 3 = 19 x19 can consist of 2 double xx combinations and 5 triples xxx combinations = 2 * 2 + 5 * 3 = 19 3 double xx combinations can be replaced by 2 triple xxx combinations like (xx)(xx)(xx) = (xxx)(xxx)

×