Upload
barton
View
37
Download
0
Tags:
Embed Size (px)
DESCRIPTION
Introduction to Computability Theory. Lecture14: The Halting Problem Prof. Amos Israeli. The Halting Problem. In this lecture we present an undecidable language. - PowerPoint PPT Presentation
Citation preview
1
Introduction to Computability Theory
Lecture14: The Halting Problem
Prof. Amos Israeli
In this lecture we present an undecidable language.
The language that we prove to be undecidable is a very natural language namely the language consisting of pairs of the form where Mis a TM accepting string w:
The Halting Problem
2
wMwMATM acceptingTM a is ,
wM ,
Since this language requires to decide whether the computation of TM M halts on input w, it is often called The Halting Problem.
Theorem
The halting problem is Turing Recognizable.
The Halting Problem
3
Consider a TM U that gets a pair as input and simulates the run of M on input w. If M accepts or rejects so does U. Otherwise, U loops.
Note: U recognizes ATM ,since it accepts any pair
, that is: any pair in which M accepts input w.
Proof
4
wM ,
LwM ,
On the previous lecture, we detailed the simulation of a DFA by a TM.
Simulating one TM by another, using the encoding of the first TM is a very similar process. In the next slide we review the main characteristics of TM N simulating TM M, using M’s encoding <M>.
Simulating an Input TM
5
TM N works as follows:
1. Mark M’s initial state and w’s initial symbol as the “current state” and “current head location”.
2. Look for M’s next transition on the description of its transition function.
3. Execute M’s transition.
Simulating an Input TM
6
4. Move M’s “current state” and “current head location” to their new places.
5. If M’s new state is a deciding state decide according to the state, otherwise – repeat stages 2-5.
Simulating an Input TM
7
So far we proved the existence of a language which is not Turing recognizable. Now we continue our quest to prove:
Theorem
The language
is undecidable.
The Language ___ Is Undecidable
8
TMA
wMwMATM acceptingTM a is ,
Before we start the proof let us consider two ancient questions:
Question1:
Can god create a boulder so heavy such that he (god) cannot lift?
The Language ___ Is Undecidable
9
TMA
Question2:
In the small town of L.J. there is a single barber:
Over the barber’s chair there is a note saying:“I will shave you on one condition: Thaw shall never shave thyself!!!”
Who Shaves the Barber?
The Language ___ Is Undecidable
10
TMA
Assume, by way of contradiction, that is decidable and let H be a TM deciding .
That is
Define now another TM, D, that uses H as a subroutine as follows:
Proof
11
TMA
wMrejectwMaccept
wMHaccept not does if
accepts if ,
TMA
Define now another TM new D that uses H as a subroutine as follows:
D=“On input where M is a TM:
1. Run H on input .
2. Output the opposite of H’s output namely: If H accepts reject, otherwise accept.“
Proof
12
MMM ,
Note: What we do here is taking advantage of the two facts:
Fact1: TM M should be able to compute with any string as input.
Fact2: The encoding of M, , is a string.
Proof
13
M
Running a machine on its encoding is analogous to using a compiler for the computer language Pascal, to compile itself (the compiler is written in Pascal).
As we recall from the two questions self-reference is sometimes means trouble (god forbid…)
Proof
14
What we got now is:
Consider now the result of running D with input . What we get is:
Proof
15
M accepts if M ejects if
MrejectMaccept
MDr
D
D accepts if D ejects if
DrejectDaccept
DDr
So if D accepts, it rejects and if it rejects it accepts. Double Trouble.
And it all caused by our assumption that TM H exists!!!
Proof
16
D accepts if D ejects if
DrejectDaccept
DDr
1. Define .
2. Assume that id decidable and let H be a TM deciding it.
3. Use H to build TM D that gets a string and behaves exactly opposite to H’s behavior, namely:
Proof Review
17
wMwMATM acceptingTM a is ,
TMA
M accepts if M ejects if
MrejectMaccept
MDr
4. Run TM D on its encoding and conclude:
Contradiction.
Proof Review
18
D accepts if D ejects if
DrejectDaccept
DDr
D
The following table describes the behavior of each machine on some machine encodings:
So Where is the Diagonalization?
19
acceptacceptMM
acceptacceptacceptacceptMacceptacceptM
MMMM
4
3
2
1
4321
This table describes the behavior of TM H. Note: TM H rejects where loops.
So Where is the Diagonalization?
20
rejectrejectacceptacceptMrejectrejectrejectrejectMacceptacceptacceptacceptMrejectacceptrejectacceptMMMMM
4
3
2
1
4321
iM
Now TM D is added to the table…
Proof Review
21
???
4
3
2
1
4321
acceptacceptrejectrejectD
acceptrejectrejectacceptacceptMrejectrejectrejectrejectrejectMacceptacceptacceptacceptacceptMacceptrejectacceptrejectacceptM
DMMMM
So far we proved the existence of a language which is not Turing recognizable. Now we will prove a theorem that will enable us to identify a specific language, namely , as a language that is not Turing recognizable.
__ Is Not Turing Reecognizable
22
TMA
TMA
(Reminder) A Language L is Turing recognizable if there exists a TM recognizing L. That is, a TM that accepts any input .
A Language L is co-Turing recognizable if the complement of L, , is Turing recognizable.
__ Is Not Turing Reecognizable
23
TMA
Lw
L
Theorem
A language L Is decidable, if and only if it is Turing recognizable and co-Turing recognizable.
Proof ->
If L is decidable so is its complement, . Any decidable language is Turing recognizable.
__ Is Not Turing Reecognizable
24
TMA
L
Proof ->
If L is decidable so is its complement, . A TM to decide , is obtained by running a TM deciding L and deciding the opposite. Any decidable language is Turing recognizable, hence is Turing recognizable and L is co-Turing recognizable.
__ Is Not Turing Reecognizable
25
TMA
L
L
L
Proof <-
If L and are both Turing recognizable, there exist two TM-s, , and , that recognize and , respectively. A TM M to decide L can be obtained by running both , and in parallel and halt when one of them accepts.
__ Is Not Turing Recognizable
26
TMA
L
1M 2M
L L
1M 2M
Corollary
The language is not Turing recognizable.
Proof
We already know that is Turing recognizable. Assume is Turing recognizable. Then, is co-Turing recognizable. By the previous theorem is decidable, a contradiction.
__ Is Not Turing Recognizable
27
TMA
TMA
TMATMA
TMA
TMA
This is the diagram we had after we studied CFL-s:
CFL-s Ex: 0| nba nn
RL-s Ex: 0| nan
Discussion
28
Non CFL-s Ex: 0| ncba nnn
???
Now, we can add some more details:
CFL-s Ex: 0| nba nn
RL-s Ex: 0| nan
Discussion
29
Decidable Ex: 0| ncba nnn
Turing recognizableEx: TMA
Non Turing recognizableEx: TMA