decidability

Question

A = { <M> | M is a DFA that doesn't accept any string containing an odd number of 1s }. Then A will be -

a) Recursive enumerable

b) Turing co-recognizable

c) decidable

d) none of these

Comments

1.as M is a DFA accepting all string excepting odd no. of 1's ....(regular)
and
2. A is the complement of M  (accepting string having odd no. of 1's)...(regular)

checking of decidability ...

1. is any given string being accepted by the language A ???
here
for above question we can say yes or no...

therefor A is decidable.....

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@ hs_yadav , thanks for proper explanation

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@hs_yadav, may you please provide any online ref to read about regularity problem of all the languages.

Answer 1

The language is Regular/DFA can be constructed. Hence Decidable.