Edward’s college Peshawar
Department of computer science
Assignment
Submitted To:
Professor : Ali raza
Dept of bs(cs)
Submitted by:
Mohammad ashiq
Roll no#19563
Dept of bs(cs)
Date of submission: 24/7/2020
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Automata theory assignment, Assignments of Theory of Automata
short assignment of automata theory in simple form
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2019/2020
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Edward’s college Peshawar
Department of computer science
Assignment
Submitted To:
Professor : Ali raza
Dept of bs(cs)
Submitted by:
Mohammad ashiq
Roll no#
Dept of bs(cs)
Date of submission: 24/7/
Q. 1 Σ= {B, aB, bab, d}B, aB, bab, d} S = BaBbabBd Find the reverse of a string S Sol: s=BaBbabd Tokenizing (B),(aB),(bab),(B),(d) Rev(s)=dBbabaBB Q:2 Prove that: a) Let S = {B, aB, bab, d}ab, bb} and T={B, aB, bab, d}ab, bb, bbbb} Show that S
= T
s={B, aB, bab, d}ab, bb} t={B, aB, bab, d}ab, bb, bbbb} the Kleene closure of s is as fallow S={B, aB, bab, d}^,ab,bb,abbb,bbab,bbbb,abab…….} the Kleene closure of t is as fallow t={B, aB, bab, d}^,ab,bb,bbbb,abbb,bbbbbbb,bbbbab,abab,bbbbbbbb…….} the language is generated by s* and t* are the same so s=t as s* ⊆ t* and t* ⊆ s* b) Let S = {B, aB, bab, d}ab, bb} and T={B, aB, bab, d}ab, bb, bbb} Show that S
≠ T
But S
T consider the following data s={B, aB, bab, d}ab, bb} t={B, aB, bab, d}ab, bb, bbbb} the Kleene closure of s is as fallow
Q4. Prove that for any set of strings of s i. (S+)=(S)* ii. (S
)
=S
iii. Is (S
)
=(S
)
iv. (S
)
=(S
)
Every factor from s+ is made up of factor from s, excluding if it’s not in the set .Every factor from(s) is made up of factor from S,including. Every factor from S* is made up of factor from S, including. Every factor from S(S) is made up of factor from S* , including. Therefor every word in (S+)* and (S) is made up of factor from S. Therefor every word in(S+)* is also a word in(s) v. (S
)
=S
Lets say I have the below language: S = {B, aB, bab, d}a, b} So if we apply Kleene plus to that language, it is something like: S+ = {B, aB, bab, d}a, b, aa, bb, ... ..} If we apply Kleene plus to above language again, It is something like: (S+)+ = {B, aB, bab, d}a, b, aa, bb, aaa, bbb ... ..}
S+ is all concatenations of words in S excluding the /\ (the empty string) In this case, (S+)+ is all concatenations of words in S+ excluding /
So that means (S+)+ includes the words in S+. vi. Is (S
)
=(S
)
S* = {B, aB, bab, d}Λ and all concatenations of words in S}(S)+ = {B, aB, bab, d}all concatenations of words in S, but not Λ}S* already contains Λ, so (S)+ contains Λ too, since it’s part of the language.No new words are added with the + operator, so (S)+ = S. S+ = {B, aB, bab, d}all concatenations of words in S, without Λ} (S+) = {B, aB, bab, d}Λ and all concatenations of words in S+}= {B, aB, bab, d}Λ and all concatenations of (all concatenations of words in S, not Λ)}= {B, aB, bab, d}Λ and all concatenations of words in S}= SThe external * operator only added Λ to the language.Therefore (S)+ = (S+)* = S*.