Lecture 24
Decidability and Undecidability
COT 4420
Theory of Computation
Section 12.1
Slides from Prof. Busch
Definition:
A language L is decidable if there is a Turing
machine (decider) M which accepts L and halts
on every input string
Also known as recursive languages
For any string
w:
w∈ L
M halts in an accept state
w∉ L
M halts in a non-accept state
Every decidable language is Recursively Enumerable
Sometimes, it is convenient to have Turing
machines with single accept and reject states
qaccept
qreject
These are the only halting states
We can convert any Turing machine to
have single accept and reject states
Old machine
New machine
qaccept
x → x ,R
x → x ,R
x → x ,L
x → x ,R
Multiple
accept states
For each tape symbol
One accept state
x
Do the following for each possible
halting state:
New machine
Old machine
For each
qi
Multiple
reject states
qi x → x, R
qreject
For all tape symbols x
not used for read in the
other transitions of qi
One reject state
For a decidable language L :
Decider for L
qaccept
Input
string
qreject
Decision
On Halt:
Accept
Reject
For each input string, the computation
halts in the accept or reject state
Transform accept state to reject
and vice-versa
M
M′
qaccept
′
qreject
qreject
qr
′
qaccept
qr
qa
qa
Turing Machine
M′
On each input string w do:
1. Let M be the decider for L
2. Run M with input string w
If M accepts then reject
If M rejects then accept
Accepts L and halts on every input string
Problems
• Informally, a “problem” is a yes/no
question about an infinite set of possible
instances.
• Example: “Does graph G have a Hamilton
cycle (cycle that touches each node exactly
once)?
– Each undirected graph is an instance of the
“Hamilton-cycle problem.”
Problems
• Formally, a problem is a language.
• Each string encodes some instance.
• The string is in the language if and only if the
answer to this instance of the problem is
“yes.”
Decidable Problems
• A problem is decidable if there is an algorithm
to answer it.
– Recall: An “algorithm,” formally, is a TM that halts
on all inputs, accepted or not.
– Put another way, “decidable problem” =
“recursive language.”
• Otherwise, the problem is undecidable.
Decidable Problems
• A problem is decidable if there is an algorithm
to answer it.
– Recall: An “algorithm,” formally, is a TM that halts
on all inputs, accepted or not.
– Put another way, “decidable problem” =
“recursive language.”
• Otherwise, the problem is undecidable.
Undecidable Problem
• There is no Turing machine that accepts the
language and makes a decision (halts) for
every input string. (for every input instance)
• Machine may make decision for some input
strings. (for some input instances)
Examples: Undecidable Problems
• Can a particular line of code in a program ever
be executed?
• Is a given context-free grammar ambiguous?
• Do two given CFG’s generate the same
language?
Halting Problem
Given the description of a Turing machine M and
an input w, does M, when started in the initial
configuration q0w, perform a computation that
eventually halts?
Halting Problem
Input:
•Turing Machine M
•String w
Question:
Does M halt while
processing input string w?
Corresponding language:
HALTTM = { M ,w : M is a Turing machine that
halts on input string w }
Theorem:
HALTTM is undecidable
(The halting problem is unsolvable)
Proof:
Basic idea:
Assume for contradiction that
the halting problem is decidable;
we will obtain a contradiction
Suppose that HALTTM is decidable
There exists Turing machine H that solves the
halting problem
M
w
H
YES
M halts on w
NO
doesn’t
halt on
M
w
Construction of H
Input
Initial tape content
wM w
H
q0
YES
M halts on w?
NO
Encoding of M
String w
Construction of H’
Construct machine H’:
• If H returns YES then loop forever
• If H returns NO then halt
Construct machine H ′ :
H′
H
wM w
Loop forever
YES
qa
q0 M halts on w?
NO
If M halts on input w
Then Loop Forever
Else Halt
qb
Construction of
Construct machine
:
Input: wM (machine M)
• If M halts on input wM then loop forever
• Else halt
Construct machine
wM
Copy wM
on tape
wM wM
:
H′
If M halts on input wM
Then loop forever
Else halt
Run
with input itself
Copy
on tape
If
halts on input
Then
Else
loops forever on input
halts on input
CONTRADICTION!!!
H′
Another proof of the halting problem
Basic idea:
If the halting problem was decidable then
every Recursively enumerable language
would be recursive/decidable
Suppose that HALTTM is decidable
There exists Turing machine H that solves the
halting problem
Input
M
YES
M
halts on
input
w
M
doesn’t halt
on input
w
H
w
NO
Let L be a recursively enumerable language
Let ML be the Turing Machine that accepts L
We will prove that L is also decidable:
We will describe a Turing machine that
accepts L and halts on any input
Decider for
L
Decider for
ML
w
HALTTM
ML halts on w?
NO
YES
ML halts
Run ML
with input w
and accepts
ML halts
and rejects
Reject w
and halt
Accept w
and halt
Reject w
and halt
Therefore, L is decidable/recursive
Since L is chosen arbitrarily, every Recursively
enumerable language is also recursive
But we have shown before that there are
recursively enumerable languages which are not
recursive
Contradiction!!!!
Therefore, the halting problem is
undecidable!
We can show:
Recursively enumerable
HALTTM
Recursive/Decidable
HALTTM is Recursively enumerable
Turing machine that accepts HALTTM :
1. Run M on input w
M ,w
2. If M halts on w
then accept M ,w
The Membership Problem
Membership Problem
Input:
•Turing Machine
•String w
Question:
M
Does M accept w ?
w∈ L(M ) ?
Corresponding language:
ATM = { M ,w : M is a Turing machine that
accepts string w }
Theorem:
ATM
is undecidable
There are M and w for which we cannot
decide whether w is in L
Suppose that ATM is decidable. Thus, there
exists a Turing machine H that solves the
membership problem
Decider
for ATM
M
w
H
YES
M accepts w
NO
M rejects
w
Let L be a Recursively enumerable language
Let ML be the Turing Machine that accepts L
We will prove that L is also recursive:
we will build a decider for L that halts on
every input
Decider for L
ML
w
Input
string
Decider for ATM
H
ML accepts w?
YES
Accept w
(and halt)
NO
Reject w
(and halt)
Therefore, L is decidable
Since L is chosen arbitrarily, every recursively
enumerable language is also
decidable/recursive
But there is a Recursively enumerable language
which is not recursive
Contradiction!!!!