written 7.5 years ago by | modified 2.7 years ago by |

**Mumbai University > Informatica Technology > Sem 4 > Automata Theory**

**Marks: 05**

**Year:May 16**

**1 Answer**

0

7.1kviews

Short note on Unsolvable Problems

written 7.5 years ago by | modified 2.7 years ago by |

**Mumbai University > Informatica Technology > Sem 4 > Automata Theory**

**Marks: 05**

**Year:May 16**

ADD COMMENT
EDIT

0

69views

written 7.5 years ago by |

One of well known unsolvable problems is the halting problem. It asks the following question: Given an arbitrary Turing machine M over alphabet = { a , b } , and an arbitrary string w over , does Mhalt when it is given w as an input ? It can be shown that the halting problem is not decidable, hence unsolvable.

**Theorem 1** : The halting problem is undecidable.

**Proof (by M. L. Minsky)**: This is going to be proven by "proof by contradiction".

Suppose that the halting problem is decidable. Then there is a Turing machine T that solves the halting problem. That is, given a description of a Turing machine M (over the alphabet ) and a string w, T writes "yes" if M halts on w and "no" if M does not halt on w, and then T halts.

We are now going to construct the following new Turing machine Tc.

First we construct a Turing machine Tm by modifying T so that if T accepts a string and halts, then Tm goes into an infinite loop (Tm halts if the original T rejects a string and halts).

Next using Tm we are going to construct another Turing machine Tc as follows:
Tc takes as input a description of a Turing machine M, denoted by d(M), copies it to obtain the string d(M)*d(M), where * is a symbol that separates the two copies of d(M) and then supplies d(M)*d(M) to the Turing machine Tm .

Let us now see what Tc does when a string describing Tc itself is given to it.

When Tc gets the input d(Tc) , it makes a copy, constructs the string d(Tc)*d(Tc) and gives it to the modified T. Thus the modified T is given a description of Turing machine Tc and the string d(Tc).

The way T was modified the modified T is going to go into an infinite loop if Tc halts on d(Tc) and halts if Tc does not halt on d(Tc).

Thus Tc goes into an infinite loop if Tc halts on d(Tc) and it halts if Tc does not halt on d(Tc). This is a contradiction.

This contradiction has been deduced from our assumption that there is a Turing machine that solves the halting problem. Hence that assumption must be wrong. Hence there is no Turing machine that solves the halting problem.

**Program correctness and Halting Problem**

Note that for any computer program a Turing machine can be constructed that performs the task of the program.

Thus the question of whether or not a program halts for a given input is nothing but the halting problem.

Thus one implication of the halting problem is that there can be no computer programs (Turing machines) that check whether or not any arbitrary computer program stops for a given input.

ADD COMMENT
EDIT

Please log in to add an answer.