GATE CS 2019 Set 1 — Question 44
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Engineering Mathematics → Sets & Combinatorics → Recursive & RE Languages
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Question
Consider the following sets:
S1. Set of all recursively enumerable languages over the alphabet {0,1}
S2. Set of all syntactically valid C programs
S3. Set of all languages over the alphabet {0,1}
S4. Set of all non-regular languages over the alphabet {0,1}Which of the above sets are uncountable?
S1. Set of all recursively enumerable languages over the alphabet {0,1}
S2. Set of all syntactically valid C programs
S3. Set of all languages over the alphabet {0,1}
S4. Set of all non-regular languages over the alphabet {0,1}Which of the above sets are uncountable?
Correct answer
(B) S3 and S4
Solution
S1: The set of recursively enumerable languages is a subset of the set of Turing Machines. Since the set of Turing Machines is countable (each TM can be encoded as a finite string), S1 is countable.
S2: The set of all syntactically valid C programs is a subset of the set of all finite strings over the ASCII alphabet. Since the set of finite strings is countable, S2 is countable.
S3: The set of all languages over {0,1} is the power set of . Since is countably infinite, its power set is uncountable (Cantor's Theorem). Thus, S3 is uncountable.
S4: The set of all languages (uncountable) is the union of regular languages (countable) and non-regular languages. If the set of non-regular languages were countable, the union would be countable, which is a contradiction. Thus, S4 is uncountable.Therefore, S3 and S4 are uncountable.
S2: The set of all syntactically valid C programs is a subset of the set of all finite strings over the ASCII alphabet. Since the set of finite strings is countable, S2 is countable.
S3: The set of all languages over {0,1} is the power set of . Since is countably infinite, its power set is uncountable (Cantor's Theorem). Thus, S3 is uncountable.
S4: The set of all languages (uncountable) is the union of regular languages (countable) and non-regular languages. If the set of non-regular languages were countable, the union would be countable, which is a contradiction. Thus, S4 is uncountable.Therefore, S3 and S4 are uncountable.
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