Halting problem

In computability theory the halting problem is a decision problem which can be informally stated as follows:


 * Given a description of a program and a finite input, decide whether the program finishes running or will run forever, given that input.

Alan Turing proved in 1936 that a general algorithm to solve the halting problem for all possible program-input pairs cannot exist. We say that the halting problem is undecidable over Turing machines. (See with respect to attribution of the halting problem to Turing.)

Formal statement
A decision problem is a set of natural numbers; the "problem" is to determine whether a particular number is in the set.

Given a Gödel numbering $$\varphi$$ of the computable functions (such as Turing's description numbers) and a pairing function $$\langle i, x \rangle$$,  the halting problem (also called the halting set) is the decision problem for the set
 * $$K_{\varphi}^{0} := \{ \langle i, x \rangle | \varphi_i(x) \ \mathrm{exists} \}$$

with $$\varphi_i$$ the i-th computable function under the Gödel numbering $$\varphi$$.

Although the set K is recursively enumerable, the halting problem is not solvable by a computable function.

There are many equivalent formulations of the Halting problem; any set whose Turing degree is the same as that of the Halting problem can be thought of as such a formulation. Examples of such sets include:
 * $$\{ i \mid \varphi_i(0) \ \mathrm{halts} \}$$
 * $$\{ i \mid \exists n ( \varphi_i(n) \ \mathrm{halts})\}$$

Importance and consequences
The historical importance of the halting problem lies in the fact that it was one of the first problems to be proved undecidable. (Turing's proof went to press in May 1936, whereas Church's proof of the undecidability of a problem in the lambda calculus had already been published in April 1936.) Subsequently, many other such problems have been described; the typical method of proving a problem to be undecidable is with the technique of reduction. To do this, the computer scientist shows that if a solution to the new problem was found, it could be used to decide an undecidable problem (by transforming instances of the undecidable problem into instances of the new problem). Since we already know that no method can decide the old problem, no method can decide the new problem either.

One such consequence of the halting problem's undecidability is that there cannot be a general algorithm that decides whether a given statement about natural numbers is true or not. The reason for this is that the proposition that states that a certain algorithm will halt given a certain input can be converted into an equivalent statement about natural numbers. If we had an algorithm that could solve every statement about natural numbers, it could certainly solve this one; but that would determine whether the original program halts, which is impossible, since the halting problem is undecidable.

Yet another consequence of the undecidability of the halting problem is Rice's theorem which states that the truth of any non-trivial statement about the function that is defined by an algorithm is undecidable. So, for example, the decision problem "will this algorithm halt for the input 0" is already undecidable. Note that this theorem holds for the function defined by the algorithm and not the algorithm itself. It is, for example, quite possible to decide if an algorithm will halt within 100 steps, but this is not a statement about the function that is defined by the algorithm.

Gregory Chaitin has defined a halting probability, represented by the symbol Ω, a type of real number that informally is said to represent the probability that a randomly produced program halts. These numbers have the same Turing degree as the halting problem. It is a normal and transcendental number which can be defined but cannot be completely computed. This means one can prove that there is no algorithm which produces the digits of Ω, although its first few digits can be calculated in simple cases.

While Turing's proof shows that there can be no general method or algorithm to determine whether algorithms halt, individual instances of that problem may very well be susceptible to attack. Given a specific algorithm, one can often show that it must halt for any input, and in fact computer scientists often do just that as part of a correctness proof. But each proof has to be developed specifically for the algorithm at hand; there is no mechanical, general way to determine whether algorithms on a Turing machine halt. However, there are some heuristics that can be used in an automated fashion to attempt to construct a proof, which succeed frequently on typical programs. This field of research is known as automated termination analysis.

Turing's introduction of the machine model that has become known as the Turing machine, introduced in the paper, has proven a convenient model for much theoretical computer science since.

Sketch of proof
The method of proof is reductio ad absurdum. You are given a programming language, a scheme that associates every program that is realizable on a Turing machine with a string that describes that program. Suppose that someone claims to have found a program halt(p, i) that returns true if the string p describes a program that halts when given as input the string i, and returns false if p describes a program that does not halt on i. The remainder of the proof shows how to construct a program on which halt cannot behave as claimed.

If two programs are realizable on a Turing machine, then executing both of them in sequence is also realizable, and so is executing one depending on a condition. This means that you can construct a program called trouble, taking one input, that does the following:


 * 1) Duplicate the input.
 * 2) Apply halt to the two copies of the input.
 * 3) If halt returned true, then loop forever.

Since all programs have string descriptions, there is a string t that represents the program trouble. Does trouble</tt> halt when its input is t?

Consider both cases:
 * If trouble(t)</tt> halts, it must be because halt(t, t)</tt> returned false</tt>, but that would mean that trouble(t)</tt> should not have halted.
 * If trouble(t)</tt> runs forever, it is either because halt</tt> itself runs forever, or because it returned true</tt>. This would mean either that halt</tt> does not give an answer for every program and input, or that trouble(t)</tt> should have halted.

Either case concludes that halt</tt> did not give a correct answer, contrary to the original claim. Since the same reasoning applies to any program that someone might offer as a solution to the halting problem, there can be no solution.

This classic proof is typically referred to as the diagonalization proof, so called because if one imagines a grid containing all the values of halt(p, i)</tt>, with every possible p value given its own row, and every possible i value given its own column, then the values of halt(s, s)</tt> are arranged along the main diagonal of this grid. The proof can be framed in the form of the question: what row of the grid corresponds to the string t? The answer is that the trouble</tt> function is devised such that halt(t, i)</tt> differs from every row in the grid in at least one position: namely, the main diagonal, where t=i. This contradicts the requirement that the grid contains a row for every possible p value, and therefore constitutes a proof by contradiction that the halting problem is undecidable.

Common pitfalls
Many students, upon analyzing the above proof, ask whether there might be an algorithm that can return a third option for some programs, such as "undecidable" or "would lead to a contradiction." This reflects a misunderstanding of decidability. It is easy to construct one algorithm that always answers "halts" and another that always answers "doesn't halt." For any specific program and input, one of these two algorithms answers correctly, even though nobody may know which one. The difficulty of the halting problem lies not in particular programs, but in the requirement that a solution must work for all programs.

There are programs (interpreters) that simulate the execution of whatever source code they are given. Such programs can demonstrate that a program does halt if this is already known to be the case. However, an interpreter will not halt if its input program does not halt, so this approach cannot solve the halting problem as stated.

The halting problem is, in theory if not in practice, decidable for deterministic machines with finite memory. A machine with finite memory has a finite number of states, and thus any deterministic program on it must eventually either halt or repeat a previous state:
 * "...any finite-state machine, if left completely to itself, will fall eventually into a perfectly periodic repetitive pattern. The duration of this repeating pattern cannot exceed the number of internal states of the machine..."(italics in original, Minsky 1967, p. 24)

Minsky warns us, however, that machines such as computers with e.g. a million small parts, each with two states, will have on the order of 2^1,000,000 possible states:
 * "This is a 1 followed by about three hundred thousand zeroes ... Even if such a machine were to operate at the frequencies of cosmic rays, the aeons of galactic evolution would be as nothing compared to the time of a journey through such a cycle" (Minsky p. 25)

Minsky exhorts the reader to be suspicious -- although a machine may be finite, and finite automata "have a number of theoretical limitations":
 * "...the magnitudes involved should lead one to suspect that theorems and arguments based chiefly on the mere finiteness [of] the state diagram may not carry a great deal of significance" (ibid).

For more on this issue of "intractability" see the article Busy beaver.

It can also be decided automatically whether a nondeterministic machine with finite memory halts on no, some, or all possible sequences of nondeterministic decisions, by enumerating states after each possible decision.

Formalization of the halting problem
In his original proof Turing formalized the concept of algorithm by introducing Turing machines. However, the result is in no way specific to them; it applies equally to any other model of computation that is equivalent in its computational power to Turing machines, such as Markov algorithms, Lambda calculus, Post systems or register machines.

What is important is that the formalization allows a straightforward mapping of algorithms to some data type that the algorithm can operate upon. For example, if the formalism lets algorithms define functions over strings (such as Turing machines) then there should be a mapping of these algorithms to strings, and if the formalism lets algorithms define functions over natural numbers (such as computable functions) then there should be a mapping of algorithms to natural numbers. The mapping to strings is usually the most straightforward, but strings over an alphabet with n characters can also be mapped to numbers by interpreting them as numbers in an n-ary numeral system.

Relationship with Gödel's incompleteness theorem
The concepts raised by Gödel's incompleteness theorems are very similar to those raised by the halting problem, and the proofs are quite similar. In fact, a weaker form of the First Incompleteness Theorem is an easy consequence of the undecidability of the halting problem. This weaker form differs from the standard statement of the incompleteness theorem by asserting that a complete, consistent and sound axiomatization of all statements about natural numbers is unachievable. The "sound" part is the weakening: it means that we require the axiomatic system in question to prove only true statements about natural numbers (it's very important to observe that the statement of the standard form of Gödel's First Incompleteness Theorem is completely unconcerned with the question of truth, but only concerns the issue of whether it can be proven).

The weaker form of the theorem can be proved from the undecidability of the halting problem as follows. Assume that we have a consistent and complete axiomatization of all true first-order logic statements about natural numbers. Then we can build an algorithm that enumerates all these statements. This means that there is an algorithm N(n) that, given a natural number n, computes a true first-order logic statement about natural numbers such that, for all the true statements, there is at least one n such that N(n) yields that statement. Now suppose we want to decide if the algorithm with representation a halts on input i. We know that this statement can be expressed with a first-order logic statement, say H(a, i). Since the axiomatization is complete it follows that either there is an n such that N(n) = H(a, i) or there is an n&#39; such that N(n&#39;) = ¬ H(a, i). So if we iterate over all n until we either find H(a, i) or its negation, we will always halt. This means that this gives us an algorithm to decide the halting problem. Since we know that there cannot be such an algorithm, it follows that the assumption that there is a consistent and complete axiomatization of all true first-order logic statements about natural numbers must be false.

Can humans solve the halting problem?
It might seem like humans could solve the halting problem. After all, a programmer can often look at a program and tell whether it will halt. It is useful to understand why this cannot be true. For simplicity, we will consider the halting problem for programs with no input, which is also undecidable.

To "solve" the halting problem means to be able to look at any program and tell whether it halts. It is not enough to be able to look at some programs and decide. Humans may also not be able to solve the halting problem, due to the sheer size of the input (a program with millions of lines of code). Even for short programs, it isn't clear that humans can always tell whether they halt. For example, we might ask if this pseudocode function, which corresponds to a particular Turing machine, ever halts:

function searchForOddPerfectNumber var int n = 1    // arbitrary-precision integer loop { var int sumOfFactors = 0 for factor from 1 to n - 1 { if factor is a factor of n then sumOfFactors = sumOfFactors + factor }         if sumOfFactors = n then exit loop n = n + 2 }     return

This program searches until it finds an odd perfect number, then halts. It halts if and only if such a number exists, which is a major open question in mathematics. So, after centuries of work, mathematicians have yet to discover whether a simple, ten-line program halts. This makes it difficult to see how humans could solve the halting problem.

More generally, it's usually easy to see how to write a simple brute-force search program that looks for counterexamples to any particular conjecture in number theory; if the program finds a counterexample, it stops and prints out the counterexample, and otherwise it keeps searching forever. For example, consider the famous (and still unsolved) twin prime conjecture. This asks whether there are arbitrarily large prime numbers p and q with p+2 = q. Now consider the following program, which accepts an input N:

function findTwinPrimeAbove(int N)     int p = N      loop if p is prime and p + 2 is prime then return else p = p + 1

This program searches for twin primes p and p+2 both at least as large as N. If there are arbitrarily large twin primes, it will halt for all possible inputs. But if there is a pair of twin primes P and P+2 larger than all other twin primes, then the program will never halt if it is given an input N larger than P. Thus if we could answer the question of whether this program halts on all inputs, we would have the long-sought answer to the twin prime conjecture. It's similarly straightforward to write programs which halt depending on the truth or falsehood for many other conjectures of number theory.

Because of this, one might say that the unsolvability of the halting problem is unsurprising. If there were a mechanical way to decide whether arbitrary programs would halt, then many apparently difficult mathematical problems would succumb to it. A counterargument to this, however, is that even if the halting problem were decidable over Turing machines, as it is over physical computers and other LBAs, it might still be infeasible in practice because it takes too much time or memory to execute. For example, there are some very large upper bounds on numbers with certain properties in number theory (the predominant example of this being Graham's number), but it's not feasible to check all values below this bound in a naïve way with a computer &mdash; they can't even hold some of these numbers in memory.

Recognizing partial solutions
There are many programs that either return a correct answer to the halting problem or do not return an answer at all. If it were possible to decide whether a program gives only correct answers, one might hope to collect a large number of such programs and run them in parallel, in the hope of being able to determine whether many programs halt. However, recognizing such partial halting solvers (PHS) is just as hard as the halting problem itself.

Suppose someone claims that program PHSR is a partial halting solver recognizer. Construct a program H: input a program P X := "input Q. if Q = P output 'halts' else loop forever" run PHSR with X as input

If PHSR recognizes the constructed program X as a partial halting solver, that means that P, the only input for which X produces a result, halts. If PHSR fails to recognize X, then it must be because P does not halt. Therefore H can decide whether an arbitrary program P halts; it solves the halting problem. Since this is impossible, the program PHSR could not have been a partial halting solver recognizer as claimed. Therefore no program can be a partial halting solver recognizer.

Another example, HT, of a Turing machine which gives correct answers only for some instances of the halting problem can be described by the requirements that, if HT is started scanning a field which carries the first of a finite string of a consecutive "1"s, followed by one field with symbol "0" (i. e. a blank field), and followed in turn by a finite string of i consecutive "1"s, on an otherwise blank tape, then While its existence has not been refuted (essentially: because there's no Turing machine which would halt only if started on a blank tape), such a Turing machine HT would solve the halting problem only partially either (because it doesn't necessarily scan the symbol "1" in the final state, if the Turing machine represented by a does halt when given the starting state and input represented by i, as explicit statements of the halting problem for Turing machines may require).
 * HT halts for any such starting state, i. e. for any input of finite positive integers a and i;
 * HT halts on a completely blank tape if and only if the Turing machine represented by a does not halt when given the starting state and input represented by i; and
 * HT halts on a nonblank tape, scanning an appropriate field (which however does not necessarily carry the symbol "1") if and only if the Turing machine represented by a does halt when given the starting state and input represented by i. In this case, the final state in which HT halted (contents of the tape, and field being scanned) shall be equal to some particular intermediate state which the Turing machine represented by a attains when given the starting state and input represented by i; or, if all those intermediate states (including the starting state represented by i) leave the tape blank, then the final state in which HT halted shall be scanning a "1" on an otherwise blank tape.

History of the halting problem
In the following: U refers to the source "Undecidable".


 * For more see History.

1900 -- Hilbert poses his "23 questions" cf Hilbert problems at the Second International Congress of Mathematicians in Paris, "Of these, the second was that of proving the consistency of the 'Peano axioms' on which, as he had shown, the rigour of mathematics depended" (Hodges p. 83, Davis' commentary in U p. 108).

1928 -- Hilbert recasts his 'Second Problem' at the Bologna International Congress (cf Reid pp. 188-189). Hodges claims he posed three questions: i.e. #1: Was mathematics complete? #2: Was mathematics consistent? #3: Was mathematics decidable? (Hodges p. 91). The third question is known as the Entscheidungsproblem (Decision Problem) (Hodges p. 91, Penrose p. 34)

1930 -- Gödel announces a proof as an answer to the first two of Hilbert's 1928 questions [cf Reid p. 198]. "At first he [Hilbert] was only angry and frustrated, but then he began to try to deal constructively with the problem... Gödel himself felt -- and expressed the thought in his paper -- that his work did not contradict Hilbert's formalistic point of view" (Reid p. 199).

1931 -- The paper of Kurt Gödel appears: "On Formally Undecidable Propositions of Principia Mathematica and Related Systems I", (reprinted in U p. 5ff)

19 April 1935 -- Paper of Alonzo Church "An Unsolvable Problem of Elementary Number Theory" identifies what it means for a function to effective calculable. Such a function will have an algorithm, and "...the fact that the algorithm has terminated becomes effectively known ..." (italics added, U p. 100).

1936 -- Alonzo Church publishes the first proof that the Entscheidungsproblem is unsolvable [A Note on the Entsheidungsproblem, reprinted in U p. 110].

7 October 1936 -- Emil Post's paper "Finite Combinatory Processes. Formulation I" is received. Post adds to his "process" an instruction "(C) Stop". He called such a process "type 1 ... if the process it determines terminates for each specific problem." (U. p.289ff)

1937-- Alan Turing's paper On Computable Numbers With an Application to the Entscheidungsproblem reaches print in January 1937 (reprinted in U, p. 115). Turing's proof departs from calculation by recursive functions and introduces the notion of computation by machine. Stephen Kleene (1952) refers to this as one of the "first examples of decision problems proved unsolvable".

1939 -- J.B. Rosser observes the essential equivalence of "effective method" defined by Gödel, Church, and Turing (Rosser in U p. 273, "Informal Exposition of Proofs of Gödel's Theorem and Church's Theorem").

1943 -- In his 1943 paper Stephen Kleene states that "In setting up a complete algorithmic theory, what we do is describe a procedure ... which procedure necessarily terminates and in such manner that from the outcome we can read a definite answer, "Yes" or "No," to the question, "Is the predicate value true?"."

1952 -- Stephen Kleene (1952) Chapter XIII ("Computable Functions") includes a discussion of the unsolvability of the halting problem for Turing machines and reformulates it in terms of machines that "eventually stop", i.e. halt: "... there is no algorithm for deciding whether any given machine, when started from any given situation, eventually stops." (Kleene (1952) p.382)

1952 -- "Davis [ Martin Davis ] thinks it likely that he first used the term 'halting problem' in a series of lectures that he gave at the Control Systems Laboratory at the University of Illinois in 1952 (letter from Davis to Copeland, 12 Dec. 2001.)" (Footnote 61 in Copeland (2004) pp.40ff)

Humor
Larry Wall once remarked, "It's easy to solve the halting problem with a shotgun." (Larry Wall in <199801151836.KAA14656@wall.org>)