Reducibility
1
Problem
A
is reduced to problem
If we can solve problem
we can solve problem A
B
B then
B
A
2
Problem
If
If
A
A
is reduced to problem
B
B is decidable then A is decidable
is undecidable then
B is undecidable
3
Example:
the halting problem
is reduced to
the state-entry problem
4
The state-entry problem
Inputs:
•Turing Machine
•State
q
•String
w
M
Question: Does
M enter state q
on input w ?
5
Theorem:
The state-entry problem is undecidable
Proof:
Reduce the halting problem to
the state-entry problem
6
Suppose we have an algorithm (Turing Machine)
that solves the state-entry problem
We will construct an algorithm
that solves the halting problem
7
Assume we have the state-entry algorithm:
M
w
q
Algorithm for
state-entry
problem
YES
M enters q
NO
doesn’t
enter
M
q
8
We want to design the halting algorithm:
M
w
Algorithm for
Halting problem
YES
M halts on w
NO
doesn’t
halt on
M
w
9
Modify input machine
•Add new state
M :
q
•From any halting state add transitions to
q
M
M
halting states
q
Single
halt state
10
M halts
if and
only if
M  halts on state q
11
Algorithm for halting problem:
Inputs:
machine
1. Construct machine
M and string w
M  with state q
2. Run algorithm for state-entry problem
with inputs: M  , q , w
12
Halting problem algorithm
M
Generate
M
M
q State-entry
w
algorithm
YES
YES
NO
NO
w
13
We reduced the halting problem
to the state-entry problem
Since the halting problem is undecidable,
it must be that the state-entry problem
is also undecidable
END OF PROOF
14
Another example:
the halting problem
is reduced to
the blank-tape halting problem
15
The blank-tape halting problem
Input:
Turing Machine
Question: Does
M
M halt when started with
a blank tape?
16
Theorem:
The blank-tape halting problem is undecidable
Proof: Reduce the halting problem to the
blank-tape halting problem
17
Suppose we have an algorithm
for the blank-tape halting problem
We will construct an algorithm
for the halting problem
18
Assume we have the
blank-tape halting algorithm:
YES
M
Algorithm for
blank-tape
halting problem NO
M halts on
blanks tape
M doesn’t halt
on blank tape
19
We want to design the halting algorithm:
M
w
Algorithm for
halting problem
YES
M halts on w
NO
doesn’t
halt on
M
w
20
Construct a new machine
Mw
• On blank tape writes
w
• Then continues execution like
M
Mw
step 1
if blank tape
step2
execute M
then write
with input
w
w
21
M halts on input string w
if and
only if
M w halts when started with blank tape
22
Algorithm for halting problem:
Inputs:
machine
1. Construct
Mw
M and string w
2. Run algorithm for
,
blank-tape halting, problem
with input M w
23
M
w
Halting problem algorithm
Generate
Mw
Mw
blank-tape
halting
algorithm
YES
YES
NO
NO
24
We reduced the halting problem
to the blank-tape halting problem
Since the halting problem is undecidable,
the blank-tape halting problem is
also undecidable
END OF PROOF
25
Summary of Undecidable Problems
Halting Problem:
Does machine
M halt on input w ?
Membership problem:
Does machine M accept string
In other words: Is a string
w
w?
member of a
recursively enumerable language
L?
26
Blank-tape halting problem:
Does machine M halt when starting
on blank tape?
State-entry Problem:
Does machine
on input w ?
M enter state q
27
Uncomputable Functions
28
Uncomputable Functions
Domain
f
Values
region
A function is uncomputable if it cannot
be computed for all of its domain
29
An uncomputable function:
f (n) 
maximum number of moves until
any Turing machine with n states
halts when started with the blank tape
30
Theorem: Function
Proof:
f (n) is uncomputable
Assume for contradiction that
f (n) is computable
Then the blank-tape halting problem
is decidable
31
Algorithm for blank-tape halting problem:
Input: machine
M
1. Count states of
2. Compute
M: m
f (m)
3. Simulate M for f (m) steps
starting with empty tape
If
M halts then return YES
otherwise return NO
32
Therefore, the blank-tape halting
problem is decidable
However, the blank-tape halting
problem is undecidable
Contradiction!!!
33
Therefore, function
f (n) in uncomputable
END OF PROOF
34
Rice’s Theorem
35
Definition:
Non-trivial properties of
recursively enumerable languages:
any property possessed by some (not all)
recursively enumerable languages
36
Some non-trivial properties of
recursively enumerable languages:
• L is empty
• L is finite
• L contains two different strings
of the same length
37
Rice’s Theorem:
Any non-trivial property of
a recursively enumerable language
is undecidable
38
We will prove some non-trivial properties
without using Rice’s theorem
39
Theorem:
For any recursively enumerable language
it is undecidable to determine whether
L is empty
L
Proof:
We will reduce the membership problem
to this problem
40
Let
M be the machine that accepts L
L( M )  L
Assume we have the empty language algorithm:
M
Algorithm for
empty language
problem
YES
L(M ) empty
NO
L(M ) not empty
41
We will design the membership algorithm:
M
w
Algorithm for
membership
problem
YES
NO
M accepts w
M rejects w
42
First construct machine
M w:
When M enters a final state,
compare original input string with
w
Accept if original input is
the same with w
43
w L
if and
only if
L( M w ) is not empty
L( M w )  {w}
44
Algorithm for membership problem:
Inputs: machine
1. Construct
M and string w
Mw
2. Determine if
L( M w ) is empty
YES: then w L(M )
NO: then w L(M )
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Membership algorithm
M
w
construct
Mw
Mw
Check if
YES
NO
NO
YES
L( M w )
is empty
END OF PROOF
46