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I'm trying to create a TM that changes all $a$'s to $b$'s and all $b$'s to $a$'s in a given string and then halts on the first blank space it encounters. What I have is:

$$ \langle q_1,a,b,q_1\rangle \\ \langle q_1,b,R,q_2\rangle \\ \langle q_2,a,L,q_1\rangle \\ \langle q_2,b,a,q_2\rangle $$

I'm new to Turing machines, so I'm wondering if it makes sense to convert $a$'s to $b$'s and then send $b$'s into a different state; is there a sort of temporal sequencing here that will send the $q_1$-manufactured $b$'s (former $a$'s) into $q_2$ as well? If so, how can you get around this problem to construct a suitable machine?

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    this belongs to computer science – 2017-02-14
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    @cloudchamber Are you using the Boolos and Jeffrey book? – 2017-02-14

2 Answers 2

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If you want just to exchange 'a' and 'b' symbols on the input string then you can just design a TM that always go rightward and changes the symbols on the fly:

\begin{align*} (q_1,a)\rightarrow(q_1,b,R)\\ (q_1,b)\rightarrow(q_1,a,R)\\ (q_1,\#)\rightarrow(q_2,\#,R) \end{align*}

Here $q_2$ is final and $\#$ is the blank symbol. $q_1$ is the initial state. $R$ means "move the head to the right".

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    is mine wrong, then, or just inefficient? – 2017-02-14
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    It is incomplete. You didn't specify the halting state nor its rules. And yes, it is inefficient ;-) – 2017-02-14
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I see that you are using the quadruple definition, rather than the more commonly used quintuple definition. But that is fine of course!

And yes, in many Turing-machine definitions you can have the machine halt without defining an explicit halting state; it just halts when there is not transition that applies.

Your machine will not work, however. If the input is $aa$ and the machine starts on the left $a$, then it will change the first $a$ to a $b$, and after that it will forever go back and forth between that $b$ and the $a$ to the right of it.

Try this instead:

$$ < q_1,a,b,q_2> \\ < q_1,b,a,q_2> \\ < q_2,a,R,q_1> \\ < q_2,b,R,q_1> \\ $$