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Definition of a language L with alphabet {a} is as follows:

L= { a^{nk} | k > 0,n >0}.

There were four options to this question

k+1

n+1

2^(n+1)

2^(k+1)

k > 0, and n is a positive integer constant

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Akshat Sharma
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1 Answers1

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I got the answer

n is a constant and k is any positive integer.

For example, if n is given as 3, then the DFA must be able to accept 3a, 6a, 9a, 12a, ..

To build such a DFA, we need 4 states. ans=(n+1)

Yassin Hajaj
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user19940105
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    But maybe `k` is a constant and `n` is any positive integer? – anatolyg Nov 01 '15 at 09:12
  • according to me k>0 and n is a positive integer – user19940105 Nov 01 '15 at 09:17
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    The language as specified is just L = {a^m | m > 0} since any m can be represented as a product of n and k. Nowhere is mentioned that n is a constant. – Henry Nov 01 '15 at 09:20
  • @user19940105 Maybe you should explain this part in your answer - this will make it clearer and more useful. Maybe you have access to the original version of OP's question, and it specifies that `n` is a constant? Maybe you have access to OP's book, and it has this convention of `k` being a parameter? Maybe it's a trick question, and all 4 possible answers are indeed incorrect (question author was in a bad mood)? Maybe you are the author of the question, and this was your intent? – anatolyg Nov 01 '15 at 09:24
  • I completely concur. Please share how you arrived at (n+1). It just doesn't make any sense from the problem statement given above. – Dinesh Nov 01 '15 at 09:26
  • k > 0, and n is a positive integer constant sorry for the misprint – Akshat Sharma Nov 01 '15 at 09:27
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    I solved this question because this question is similar to a question which came in GATE paper few years ago and i remembered the methodology of solving these questions when i practiced it earlier... That is why i assumed the values as k>0 and n is a positive integer...that's all!!! – user19940105 Nov 01 '15 at 12:10