# E :: Halting Machine

Time Limit: 7 Seconds Memory Limit: 32768 KB

“In computability theory, the halting problem is the problem of determining, from a description of an arbitrary computer program and an input, whether the program will finish running or continue to run forever.

Alan Turing proved in 1936 that a general algorithm to solve the halting problem for all possible program-input pairs cannot exist. A key part of the proof was a mathematical definition of a computer and program, which became known as a Turing machine; the halting problem is undecidable over Turing machines. It is one of the first examples of a decision problem.

The halting problem is historically important because it was one of the first problems to be proved undecidable. (Turing’s proof went to press in May 1936, whereas Alonzo Church’s proof of the undecidability of a problem in the lambda calculus had already been published in April 1936 (Church, 1936).) Subsequently, many other undecidable problems have been described; the typical method of proving a problem to be undecidable is with the technique of reduction. To do this, it is sufficient to show that if a solution to the new problem were 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. Often the new problem is reduced to solving the halting problem. (Note: the same technique is used to demonstrate that a problem is NP complete, only in this case, rather than demonstrating that there is no solution, it demonstrates there is no polynomial time solution, assuming P ≠ NP).

For example, 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 stating that a certain program will halt given a certain input can be converted into an equivalent statement about natural numbers. If we had an algorithm that could find the truth value of every statement about natural numbers, it couldcertainly find the truth value of this one; but that would determine whether the original program halts, which is impossible, since the halting problem is undecidable.

Rice’s theorem generalizes the theorem that the halting problem is unsolvable. It states that for any non-trivial property, there is no general decision procedure that, for all programs, decides whether the partial function implemented by the input program has that property. (A partial function is a function which may not always produce a result, and so is used to model programs, which can either produce results or fail to halt.) For example, the property ”halt for the input 0” is undecidable. Here, ”non-trivial” means that the set of partial functions that satisfy the property is neither the empty set nor the set of all partial functions. For example, ”halts or fails to halt on input 0” is clearly true of all partial functions, so it is a trivial property, and can be decided by an algorithm that simply reports ”true.” Also, note that this theorem holds only for properties of the partial function implemented by the program; Rice’s Theorem does not apply to properties of the program itself. For example, ”halt on input 0 within 100 steps” is not a property of the partial function that is implemented by the program—it is a property of the program implementing the partial function and is very much decidable.

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.

Since the negative answer to the halting problem shows that there are problems that cannot be solved by a Turing machine, the Church–Turing thesis limits what can be accomplished by any machine that implements effective methods. However, not all machines conceivable to human imagination are subject to the Church–Turing thesis (e.g. oracle machines). It is an open question whether there can be actual deterministic physical processes that, in the long run, eludesimulation by a Turing machine, and in particular whether any such hypothetical process could usefully be harnessed in the form of a calculating machine (a hypercomputer) that could solve the halting problem for a Turing machine amongst other things. It is also an open question whether any such unknown physical processes are involved in the working of the human brain, and whether humans can solve the halting problem (Copeland 2004, p. 15).”

–Wikipedia

However, you can solve the halting problem for a program in language X under certain assumptions.

Each program in language X consists of N goto statements with the following structure (there are no white-spaces in conditions and the spaces are only after ‘:’, ‘,’, and the goto keyword):

goto l1: c1, l2: c2, …, lk: ck;

This means that the program goes to line number li ≥ 0 if condition ci is true, 0 < i ≤ k;

Assuming that the conditions are independent and possible with some input, your task is to find out if the program halts, and if so, in how many steps. A program halts if and only if it reaches line number N.

## Input

The first line of the input contains one integer T(0 ≤ T ≤ 11), the number of test cases.

Each test case is a program:

The first line : integer N(0 ≤ N ≤ 1000) the number of lines in that program.

The next N lines: each contains exactly one goto statement.

## Output

For each test case, output the number of steps in the worst case or ”infinity” (without quotations) if the program does not halt in some case.

## Sample Input

2 2 goto 1: a>b; goto 0: 10<x; 3 goto 2: a<b, 1: b<a; goto 3: a>0; goto 1: b>0;

## Sample Output

infinity 3Submit

Source: 14th Iran Nationwide Internet Contest