Turing machine

The Turing machine is an abstract model of computer execution and storage introduced in 1936 by Alan Turing to give a mathematically precise definition of algorithm or 'mechanical procedure'. As such it is still widely used in theoretical computer science, especially in complexity theory and the theory of computation. The thesis that states that Turing machines indeed capture the informal notion of effective or mechanical method in logic and mathematics is known as the Church-Turing thesis.

Turing machines shouldn't be confused with the Turing test, Turing's attempt to capture the notion of artificial intelligence.

A Turing machine that is able to simulate any other Turing machine is called a universal Turing machine.

Table of contents 1 Definition 1.1 Example: 2 The Universal Turing Machine 3 A Physical Turing Machine 4 A mathematical version 4.1 Executions 4.2 Termination, running time 4.3 Computability 4.4 Equivalences 4.5 Polynomial time 1 References and external links: 2 See also:

Definition

A Turing machine consists of:

1. A tape which is divided into cells, one next to the other. Each cell contains a symbol from some finite alphabet. The alphabet contains a special blank symbol (here written as '0') and one or more other symbols. The tape is assumed to be arbitrarily extendible to the left and to the right, i.e., the Turing machine is always supplied with as much tape as it needs for its computation. Cells that have not been written to before are assumed to be filled with the empty symbol.
2. A head that can read and write symbols on the tape and move left and right.
3. A state register that stores the state of the Turing machine. The number of different states is always finite and there is one special start state with which the state register is initialized.
4. An action table that tells the machine what symbol to write, how to move the head ('L' for one step left, and 'R' for one step right) and what its new state will be, given the symbol it has just read on the tape and the state it is currently in. If there is no entry in the table for the current combination of symbol and state then the machine will halt.

Note that every part of the machine is finite, but it is the potentially unlimited amount of tape that gives it an unbounded amount of storage space.

Example:

The following Turing machine has an alphabet {'0', '1'}, with 0 being the blank symbol. It expects a series of 1's on the tape, with the head initially on the leftmost 1, and doubles the 1's with a 0 in between, i.e., "111" becomes "1110111". The set of states is {s1, s2, s3, s4, s5} and the start state is s1. The action table is as follows.

  Old Read   Wr.      New     Old Read   Wr.      New
  St. Sym.   Sym. Mv. St.     St. Sym.   Sym. Mv. St.
  - - - - - - - - - - - -     - - - - - - - - - - - -
   s1  1  ->  0    R   s2      s4  1  ->  1    L   s4
   s2  1  ->  1    R   s2      s4  0  ->  0    L   s5
   s2  0  ->  0    R   s3      s5  1  ->  1    L   s5
   s3  0  ->  1    L   s4      s5  0  ->  1    R   s1
   s3  1  ->  1    R   s3

A computation of this Turing machine might for example be: (the position of the head is indicated by displaying the cell in bold face)

  Step State Tape   Step State Tape
  - - - - - - - -   - - - - - - - - -
     1  s1   11        9  s2   1001 
     2  s2   01       10  s3   1001
     3  s2   010      11  s3   10010
     4  s3   0100     12  s4   10011
     5  s4   0101     13  s4   10011
     6  s5   0101     14  s5   10011
     7  s5   0101     15  s1   11011
     8  s1   1101       -- halt --

The behavior of this machine can be described as a loop: it starts out in s1, replaces the first 1 with a 0, then uses s2 to move to the right, skipping over 1's and the first 0 encountered. S3 then skips over the next sequence of 1's (initially there are none) and replaces the first 0 it finds with a 1. S4 moves back to the left, skipping over 1's until it finds a 0 and switches to s5. s5 then moves to the left, skipping over 1's until it finds the 0 that was originally written by s1. It replaces that 0 with a 1, moves one position to the right and enters s1 again for another round of the loop. This continues until s1 finds a 0 (this is the 0 right in the middle between the two strings of 1's) at which time the machine halts.

The Universal Turing Machine

Every Turing machine computes a certain fixed partial function from the input strings over its alphabet. In that sense it behaves like a computer with a fixed program. However, as Alan Turing already described, we can encode the action table of any Turing machine in a string. Thus we might try to construct a Turing machine that expects on its tape a string describing an action table followed by a string describing the input tape, and then computes the tape that the encoded Turing machine would have computed. As Turing showed, such a Turing machine is indeed possible and since it is able to simulate any other Turing machine it is called a universal Turing machine.

With this encoding of action tables as strings, it becomes in principle possible for Turing machines to answer questions about the behavior of other Turing machines. Most of these questions however are undecidable, see for instance the Halting problem, which was already shown to be undecidable in Turing's original paper, and Rice's theorem.

If we broaden the definition to include any Turing machine that simulates some Turing-complete computational model, not just Turing machines that directly simulate other Turing machines, a universal Turing machine can be fairly simple, using just a few states and a few symbols. For example, only 2 states are needed, since a 2×18 (meaning 2 states, 18 symbols) universal Turing machine is known. A complete list of the smallest known universal Turing machines is: 2×18, 3×10, 4×6, 5×5, 7×4, 10×3, 22×2. These simulate a computational model called tag-systems.

A universal Turing machine is Turing-complete. It can calculate any recursive function, decide any recursive language, and accept any recursively enumerable language. According to the Church-Turing thesis, the problems solvable by a universal Turing machine are exactly those problems solvable by an algorithm or an effective method of computation, for any reasonable definition of those terms.

A Physical Turing Machine

It is not difficult to simulate a Turing Machine on a modern computer (except for the limited memory of actually existing computers).

It is also possible to build a Turing Machine on a purely mechanical basis. The mathematician Karl Scherer[?] has indeed built such a machine in 1986 using metal and plastic construction sets, and some wood. The 1.5 meter high machine uses the pulling of strings to read, move and write the data (which is represented using ball bearing balls).

The machine is now exhibited in the entrance of the Department of Computer Science of the University of Heidelberg, Germany.

A mathematical version

Let A and S be finite, nonempty sets (the "alphabet" and the "set of states," respectively) and Z be the set of integers. One of the elements of S is singled out as the start state and we will denote it by *. Let T be the set of doubly-infinite sequences of elements of A (functions from Z to A, "set of tapes.") Let X=S×Z×T be the set of all triplets (s,n,t) where s is in S, n is in Z and t is in T (the set of "complete machine data.") Let r be a total binary relation from S×A to {-1,0,1}×S×A (the "transition rule".) (Note that there are finitely many such relations.) The set X₀ is the subset of X so that, for each (s,n,t) in X₀, s=* and n=0 (the set of "admissible initial conditions.") For any x=(a,b,c) in X, we will denote its tape by t(x) (=c).

Such a set-up (the data (A,S,r), or more implicitly, r) is called a Turing machine.

We define a new relation R from X to X as follows. (a,b,c)×(d,e,f) is in R if, and only if:

1. The sequences (or tapes) c and f are identical, except perhaps that c_b may be different from f_b.
2. e-b is either -1, 0 or +1.
3. The relation r contains (a,c_b)×(e-b,d,f_b)

If the relation r is a function, then so is R and the Turing machine is said to be deterministic, otherwise it is said to be nondeterministic.

Executions

If x₀,x₁,x₂,... is a sequence in X so that, for every k:

1. x_k×x_k+1 is in R and
2. x₀ is in X₀ (i.e., it is an admissible initial condition)

the sequence x₀,x₁,x₂,... is said to be a run or execution of the Turing machine r. If the Turing machine is deterministic, then such a sequence is always completely determined by x₀ since x_k+1=R(x_k) for everk k. On the other hand, if the Turing machine is nondeterministic, there will likely be two (or more) runs of the Turing machine, say x₀,x₁,x₂,... and y₀,y₁,y₂,..., which are different but such that x₀=y₀.

Termination, running time

If x is an element of X such that x×x is in R, we say that x is a stopping state.

If x₀,x₁,x₂,... is a run of the Turing machine and if, for some j, x_j is a stopping state, we say that the execution has terminated at time j. The running time of the execution is the smallest j which is a termination time for x. If x does not terminate, we say that its running time is ∞. If x is an execution, we will denote its running time by |x|

In the case of deterministic Turing machines, observe that the running time of an execution depends only on x₀ and not on all of the data in x₁,x₂,... On the other hand, if the Turing machine is nondeterministic, x₀ is insufficient to compute the running time of the entire sequence. However, we wish to define the running time given only x₀ and so a supplemental step is taken.

We will define g(z), a function of X₀, to be the set of all executions x₀,x₁,x₂,... with the shortest running time and with x₀=z. If the Turing machine is deterministic, then there is only one possible execution. If the Turing machine is nondeterministic, many executions will satisfy x₀=z and g(z) is the set of those satisfactory executions with the least running time.

Then, the running time of the Turing machine given the initial situation z in X₀ is the running time of any execution in g(z), whether the machine is deterministic or not. In the deterministic case, g(z) also gives the only possible execution, but in the nondeterministic case, g(z) lists all possible executions. We shall now write |z| for |g(z)|, the running time of z for any z in X₀.

Computability

Let f be any function from U to T where U is some subset of T. We say that f is computable if there exists a Turing machine such that, for all a in U:

1. There is a run x=x₀,x₁,x₂,... so that
2. t(x₀)=a, the tape is initially a.
3. x terminates, with running time n.
4. t(x_n)=f(a), the tape is f(a) at termination.

An arbitrary function from U to T is usually thought of as a "problem". For instance, A might contain the digits from 0 to 9 and the decimal point, and U might be the set of all real numbers, and f might compute the square root of any number written on U. This function f would be the "square root problem." This problem is clearly not computable because, for instance, the square root of two is irrational, and hence no Turing machine could ever write it out completely on the tape in finite time.

The prototypical non-computable problem is the halting problem.

Equivalences

x₀,x₁,x₂,... and y₀,y₁,y₂,... are executions of two possibly different Turing machines with the same alphabet A, we say that these executions are equivalent if

1. t(x₀)=t(y₀)
2. x and y both terminate
3. t(x_m)=t(y_n) where m is the running time of x and n is the running time of y

If r and s are Turing machines with the same alphabet, we say that s generalizes or subsumes r when, for any execution x of r, there is an execution y of s so that x and y are equivalent.

If two Turing machines subsume each other, we say that they are equivalent. The reader may check that if r is any nondeterministic Turing machine, it is equivalent to some deterministic Turing machine s. However, there is a more interesting notion of equivalence.

Polynomial time

Let @ be some special symbol in A. If t is a tape in T, then by size of the input (or size of t, or |t|) we shall mean the smallest n so that, for any |k|>n, t_k is @. The size of t may be ∞.

We say that a Turing machine r runs in polynomial time if for any z in X₀, the running time |z| is bounded above by some polynomial in the length of the tape |t(z)|. The set of deterministic Turing machines which run in polynomial time is denoted by P, while the set of nondeterministic Turing machines that run in polynomial time is denoted NP. It is fairly clear that any polynomial time deterministic Turing machine is equivalent to some polynomial time nondeterministic Turing machine (the only thing to do is to add some nondeterminism without changing the shortest executions.)

Some well known problems (satisfiability, the traveling salesman problem, etc...) are known to be solvable by machines in NP, but so far we have been unable to find machines in P to solve them. One of the most fundamental open questions in theory of computation is:

For every r in NP, is there an s in P which is equivalent to r?

This is known as the "P=NP?" problem. See Complexity classes P and NP.

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References and external links:

• Turing, A., On Computable Numbers, With an Application to the Entscheidungsproblem, Proceedings of the London Mathematical Society, Series 2, Volume 42, 1936; reprinted in M. David (ed.), The Undecidable, Hewlett, NY: Raven Press, 1965; online: http://www.abelard.org/turpap2/tp2-ie.asp
• Boolos, G. and Jeffrey, R., Computability and Logic, 2nd ed., Cambridge: Cambridge University Press, 1980.
• Rogozhin, Yurii, "A Universal Turing Machine with 22 States and 2 Symbols", Romanian Journal Of Information Science and Technology, 1(3), 259-265, 1998. http://www.imt.ro/Romjist/Volum1/Vol1_3/turing.htm (surveys known results about small universal Turing machines)

See also:

• Langton's ant, a simple two-dimensional Turing machine
• Church's Conjecture, effective Turing machine can perform any computation in any language