It's worth thinking about what happens if we try the same thing that we've done with our previous machines and make some modifications to Turing machines. For instance, if we remember the finite automaton, we have the deterministic and nondeterministic variants. As it turned out, the power of the machines were exactly the same, in that they both recognized exactly the same class of languages. However, there was a tradeoff in the number of states between the two models. Similarly, we recall that this same variation yields vastly different results for the pushdown automaton. The deterministic pushdown automata could recognize fewer languages that could be recognized by the nondeterministic machine.
We now explore the same kinds of ideas with Turing machines. The goal is to show that the Turing machine is actually fairly powerful and to give us some tools when working with them.
As with PDAs, we need to define a notion of the current state of the machine, or the configuration.
A configuration is a string $uqv$, where $q \in Q$ is the current state, and the current tape contents is $uv$, where $u,v \in \Gamma^*$, and the tape head is located on top of the tape cell corresponding to the first symbol of $v$. For example, the initial configuration with input word $w$ is the word $q_0 w$. An accepting configuration is a word $u q_A v$ with $u,v \in \Gamma^*$.
Unlike with PDAs, we don't need a separate notion of the input to be consumed because that information is on the tape and can be changed during the course of computation.
One example of such a modification is what if instead of using only a single tape, we allowed the Turing machine to have multiple tapes with tape heads that move independently of each other? That is, for $k$ tapes, we modify the definition so we have the transition function $\delta : Q \times \Gamma^k \to Q \times \Gamma^k \times \{L,R\}^k$.
Every language recognized by a multi-tape Turing machine can be recognized by a single tape Turing machine.
The idea that we use to show this is one that we will be using a lot in the rest of this course: simulation. We show how to simulate a multi-tape Turing machine on a single-tape Turing machine. The idea is to store all the contents of each tape on the single tape and separate each tape as required. We also keep track of the tape head positions on each of the virtual tapes.
What does this machine look like? We describe a Turing machine which simulates a $k$-tape TM $M$ as follows:
Our current Turing machine is defined to be deterministic. However, we can show that we can introduce nondeterminism without increasing or decreasing the power of the model.
In a nondeterministic Turing machine, the transition function is a function $\delta: Q \times \Gamma \to 2^{Q \times \Gamma \times \{L,R\}}$. We can view the computation of a nondeterministic Turing machine as a tree, where each branch represents a possible guess. A nondeterministic Turing machine accepts a word if at least one branch of the computation tree enters an accepting state. It's important to note that we only require that at least one computation accepts, since some branches may lead to infinitely long computation paths.
Every nondeterministic Turing machine has an equivalent deterministic Turing machine.
The idea behind this proof is to construct a Turing machine that performs a breadth-first search of the computation tree. If the word is accepted by the nondeterministic machine, then a search will eventually find the branch where the computation is accepted. Otherwise, the machine may run forever.
We use a 3-tape deterministic machine to simulate the nondeterministic machine $N$. We can do this because we have just conveniently showed that a $k$-tape machine is equivalent to a single tape machine. The tapes are used as follows:
The third tape is what will allow us to navigate through the tree. For each node, we assign a value from $1$ to $k$, where $k$ is the size of the largest set of choices in the transition function of $N$. Then by considering a string over the alphabet $\{1,\dots,k\}$, we follow a particular branch at each set of choices, starting from the root of the tree. However, not every string over $\{1,\dots,k\}$ will correspond to a valid address, since not every nondeterministic set of choices is guaranteed to have as many as $k$ choices. This is fine, since that address simply will not correspond to a node. Then by enumerating through every string, we are guaranteed to reach every node of the computation tree.
Now, the machine operates as follows:
This gives us the following corollary.
Corollary 20.4. A language is recognizable if and only if some nondeterministic Turing machine recognizes it.
However, if we know that a nondeterministic Turing machine will always halt on every branch of its computation, then the corresponding deterministic machine will be also always halt. This leads to the following result.
Corollary 20.5. A language is decidable if and only if some nondeterministic Turing machine decides it.
Something you might notice or object to is that simulating a nondeterministic Turing machine with a deterministic machine is going to take a very long time. Since we're performing a breadth-first search, this means that we're looking at a computation that's exponential in the length of the shortest accepting path in the nondeterministic computation tree. But that's okay, because we only care whether our deterministic machine is able to recognize or decide any language that a nondeterministic machine is capable of recognizing.
But for argument's sake, let's suppose that we do care about the time that it takes. We can measure this: how many steps is the computation on our machine? We can apply our notions of efficiency from other CS courses and consider polynomially many steps to be good or efficient. So suppose that I have a nondeterministic TM that has an accepting computation branch that's polynomially long in the size of the input. Is it possible to simulate or come up with a way to solve the same problem deterministically, but still end up with only a polynomially long computation?
The answer is: we don't know! Because this question is exactly the P vs. NP problem.