Computer Science Engineering (CSE) Exam > Computer Science Engineering (CSE) Questions > Which of the following is TRUE?a)Every subset...
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Which of the following is TRUE?
- a)Every subset of a regular set is regular.
- b)Every finite subset of a non-regular set is regular.
- c)The union of two non-regular sets is not regular.
- d)Infinite union of finite sets is regular.
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
Which of the following is TRUE?a)Every subset of a regular set is regu...
Some points for Regular Sets:
- A set is always regular if it is finite.
- A set is always regular if a DFA/NFA can be drawn for it.
Option A: Every subset of a regular set is regular is False. For input alphabets a and b, a*b* is regular. A DFA can be drawn for a*b* but a n b n for n≥0 which is a subset of a*b* is not regular as we cannot define a DFA for it.
Option B: Every finite subset of a non-regular set is regular is True. Each and every set which is finite can have a well-defined DFA for it so whether it is a subset of a regular set or non-regular set it is always regular.
Option C: The union of two non-regular sets is not regular is False. For input alphabets a and b, an bn for all n≥0 is non-regular as well as an bm for n≠m is also non- regular but their union is a*b* which is regular.
Option D: TInfinite union of finite sets is regular is False. For input alphabets a and b sets {ab}, {aabb}, {aaabbb}…….. are regular but their union {ab} U {aabb} U {aaabbb} U …………………….. gives {a n b n for n>0} which is not regular.
Option B: Every finite subset of a non-regular set is regular is True. Each and every set which is finite can have a well-defined DFA for it so whether it is a subset of a regular set or non-regular set it is always regular.
Option C: The union of two non-regular sets is not regular is False. For input alphabets a and b, an bn for all n≥0 is non-regular as well as an bm for n≠m is also non- regular but their union is a*b* which is regular.
Option D: TInfinite union of finite sets is regular is False. For input alphabets a and b sets {ab}, {aabb}, {aaabbb}…….. are regular but their union {ab} U {aabb} U {aaabbb} U …………………….. gives {a n b n for n>0} which is not regular.
Most Upvoted Answer
Which of the following is TRUE?a)Every subset of a regular set is regu...
Explanation:
Regular Set: A set that can be recognized by a finite automaton is called a regular set.
Non-Regular Set: A set that cannot be recognized by a finite automaton is called a non-regular set.
a) Every subset of a regular set is regular: This statement is true. Suppose we have a regular set R and a subset of R called S. Since R is regular, it can be recognized by a finite automaton. Since S is a subset of R, any string that is in S must also be in R. Therefore, S can also be recognized by the same finite automaton that recognizes R, and hence S is also regular.
b) Every finite subset of a non-regular set is regular: This statement is true. Suppose we have a non-regular set N and a finite subset of N called F. Since N is non-regular, it cannot be recognized by any finite automaton. However, F is a finite set, and therefore, we can list all the strings in F. We can then construct a finite automaton that recognizes all the strings in F, which means that F is regular.
c) The union of two non-regular sets is not regular: This statement is false. There exist non-regular sets whose union is regular. For example, let A be the set of all strings of the form 0^n1^n (where n is a positive integer), and let B be the set of all strings of the form 1^n0^n. Both A and B are non-regular sets, but their union (which is the set of all strings of the form 0^n1^n or 1^n0^n) is regular.
d) Infinite union of finite sets is regular: This statement is false. There exist infinite unions of finite sets that are non-regular. For example, let A_n be the set of all strings of length n over the alphabet {0, 1}. Each A_n is a finite set, but their infinite union (which is the set of all binary strings) is non-regular. This can be proved using the pumping lemma for regular sets.
Regular Set: A set that can be recognized by a finite automaton is called a regular set.
Non-Regular Set: A set that cannot be recognized by a finite automaton is called a non-regular set.
a) Every subset of a regular set is regular: This statement is true. Suppose we have a regular set R and a subset of R called S. Since R is regular, it can be recognized by a finite automaton. Since S is a subset of R, any string that is in S must also be in R. Therefore, S can also be recognized by the same finite automaton that recognizes R, and hence S is also regular.
b) Every finite subset of a non-regular set is regular: This statement is true. Suppose we have a non-regular set N and a finite subset of N called F. Since N is non-regular, it cannot be recognized by any finite automaton. However, F is a finite set, and therefore, we can list all the strings in F. We can then construct a finite automaton that recognizes all the strings in F, which means that F is regular.
c) The union of two non-regular sets is not regular: This statement is false. There exist non-regular sets whose union is regular. For example, let A be the set of all strings of the form 0^n1^n (where n is a positive integer), and let B be the set of all strings of the form 1^n0^n. Both A and B are non-regular sets, but their union (which is the set of all strings of the form 0^n1^n or 1^n0^n) is regular.
d) Infinite union of finite sets is regular: This statement is false. There exist infinite unions of finite sets that are non-regular. For example, let A_n be the set of all strings of length n over the alphabet {0, 1}. Each A_n is a finite set, but their infinite union (which is the set of all binary strings) is non-regular. This can be proved using the pumping lemma for regular sets.
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Which of the following is TRUE?a)Every subset of a regular set is regular.b)Every finite subset of a non-regular set is regular.c)The union of two non-regular sets is not regular.d)Infinite union of finite sets is regular.Correct answer is option 'B'. Can you explain this answer?
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