LectureNotes18 - ECS 120 Lesson 18 Decidable Problems the...

ECS 120 Lesson 18 – Decidable Problems, the
Halting Problem
Oliver Kreylos
Friday, May 11th, 2001
In the last lecture, we had a look at a problem that we claimed was not
solvable by an algorithm – the problem whether a given C program prints
the message “Hello World” and then exits. Today we are going to look at a
very similar problem and prove formally that there exists no Turing Machine
that can decide it.
1 Problems and Languages
Let us start by defining some terminology for problems. A problem P is a
question that can be asked about entities E of some type (Turing Machines,
integers, graphs, etc.) and answers either “yes” or “no.” For example, the
“Hello World” problem can be applied to C programs and asks whether an
entity E (a certain C program) prints “Hello World” and then exits. We
denote the answer that problem P gives about entity E as P ( E ). If P ( E ) is
“yes,” we call E an instance of P .
At first glance, it seems that the theory about languages we have investi-
gated so thoroughly cannot be applied to reason about problems. A language
is merely a set of words over some alphabet, whereas a problem is a question
about some entity, that can either be answered “yes” or “no.” As it turns out,
this difference is not fundamental, but just a matter of notation. We have al-
ready seen that any entity of some type we can reason about mathematically
can also be converted into words over some alphabet by the encoding mech-
anism discussed earlier; this means that any entity E can be converted to a
word E ∈ Σ * for some alphabet Σ. We can now define a language L ( P ) for
problem P as follows: L ( P ) := O ∈ Σ
*
E is an instance of P , where
1

Σ is the alphabet needed to encode entity O . In other words, the language for
a problem P consists of exactly those words that are encodings of instances
of P , i. e., that are encodings of entities for which P ’s answer is “yes.” Being
a set of words, L ( P ) is a language in the usual sense.
We can now directly apply language theory to solving problems: If and
only if P is a problem that can be solved by an algorithm, i. e., there exists
an algorithm that always terminates and either answers “yes” or “no,” then
the language L ( P ) must be Turing-decidable: Given any word w ∈ Σ
*
, there
is a Turing Machine D ( P ) that can decode w = E back to an entity E ,
and can then apply the algorithm to answer P to that entity. An algorithm
comes up with an answer after a finite number of steps; therefore, D ( P ) can
accept or reject the word after a finite number of steps. This means that
D ( P ) is a decider for L ( P ).
2 The Halting Problem Revisited
We defined the halting problem as the question whether a given Turing
Machine M halts on input word w . Using the conversion strategy from
above, we convert the halting problem to a language H
TM
:= M, w
TM M halts on input w . The words in H
TM
are encodings of Turing Ma-
chines and the input words they halt on. Let us first look at a slightly modi-
fied version of the halting problem: Let A
TM
:= M, w TM M accepts w