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Trying to figure out how to make a deterministic single-tape Turing machine for the language of words comprising as many $a$'s as $b$'s.

I am very confused, because I can't figure out how I would keep track of how many $a$'s and $b$'s there would be. The method for finding $a^nb^n$ goes right and back left to do this but this language would include things like $abaaaababab$.

How can I keep track of number of $a$'s and $b$'s?

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  • $\begingroup$ Sorry if I was unclear, it means the number of a's cannot equal the number of b's $\endgroup$ – John Mar 4 '18 at 15:49
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    $\begingroup$ When clarifying the question, you should not use a comment, but edit your question $\endgroup$ – chi Mar 4 '18 at 21:07
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You dont need to keep track of them. At each iteration you have three possibilities. Either you first see an A, or a B or a nothing. If nothing accept. If an A first, replace it with some special character and search for a B. If you see one replace that one as well with some special character go back to beginning but if you dont just reject. Do the same for B first. After you cross our 2 characters loop back to beginning. It is actually very similar to $a^nb^n$.

The TM design should be straightforward actually. Similar to $a^nb^n$ but considers $b$ first option as well at each iteration.

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  • $\begingroup$ I don't think this approach will work because you'd be introducing blank characters in your input string. When you're trying to go back to the beginning and you get to a blank, you won't be able to tell whether this is a blank you wrote to the tape (and there are more characters to the left of that blank) or whether you're actually at the beginning. Instead of replacing with a blank, you could cross off with some other character like an X. $\endgroup$ – roctothorpe Mar 5 '18 at 20:31
  • $\begingroup$ Yeah you are right it's a careless mistake but the algorithm TM design remains same. Thanks. $\endgroup$ – sunnytheit Mar 5 '18 at 20:36

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