Reputation: 47
How to make a Turing machine graph which recognizes all words with an even number of a's
but which is not a decider
and nothing else. Also, how to make a decider
Turing machine graph for the same language.
Upvotes: 1
Views: 334
Reputation: 28312
The language of all words with an even number of a's is a regular language, so a simple two-state DFA will accept the language. The same machine is a two-state decider TM that happens not to use the tape for anything except reading input.
To turn such a decider into a recognizer-non-decider (since all deciders are recognizers, too) we must make the TM explicitly recognize strings with an odd number of a's and, rather than simply halt-reject, go into an infinite loop. This can easily be done with a single state.
Upvotes: 2