Reducibility
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Problem
A
is reduced to problem
If we can solve problem
we can solve problem A
B
B then
B
A
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Problem
If
If
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A
A
is reduced to problem
B
B is decidable then A is decidable
is undecidable then
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B is undecidable
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Example:
the halting problem
is reduced to
the state-entry problem
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The state-entry problem
Inputs:
•Turing Machine
•State
q
•String
w
M
Question: Does
M enter state q
on input w ?
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Theorem:
The state-entry problem is undecidable
Proof:
Reduce the halting problem to
the state-entry problem
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Suppose we have a Decider
for the state-entry algorithm:
M
w
q
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state-entry
problem
decider
YES
M enters q
NO
doesn’t
enter
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M
q
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We want to build a decider
for the halting problem:
M
w
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Halting problem
decider
YES
M halts on w
NO
doesn’t
halt on
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M
w
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We want to reduce the halting problem to
the state-entry problem:
Halting problem decider
M
w
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M
q
w
YES
YES
State-entry
problem
NO NO
decider
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We need to convert one problem instance
to the other problem instance
Halting problem decider
M
w
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Convert
Inputs
?
M
q
w
YES
YES
State-entry
problem
NO NO
decider
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M to M  :
q
•Add new state
Convert
•From any halting state of
M
add transitions to
q
M
q
M
halting states
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Single
halt state
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M halts on input w
if and
only if
M  halts on state q on input w
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Halting problem decider
M
Generate
M
M
q
w
YES
YES
State-entry
problem
NO NO
decider
w
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We reduced the halting problem
to the state-entry problem
Since the halting problem is undecidable,
the state-entry problem is undecidable
END OF PROOF
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Another example:
the halting problem
is reduced to
the blank-tape halting problem
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The blank-tape halting problem
Input:
Turing Machine
Question: Does
M
M halt when started with
a blank tape?
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Theorem:
The blank-tape halting problem is undecidable
Proof: Reduce the halting problem to the
blank-tape halting problem
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Suppose we have a decider for the
blank-tape halting problem:
YES
M
blank-tape
halting problem
NO
decider
M halts on
blank tape
M doesn’t halt
on blank tape
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We want to build a decider
for the halting problem:
M
w
Fall 2004
halting problem
decider
YES
M halts on w
NO
doesn’t
halt on
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M
w
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We want to reduce the halting problem to
the blank-tape halting problem:
Halting problem decider
M
YES
M
w
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Blank-tape
w halting problem
NO
decider
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YES
NO
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We need to convert one problem instance
to the other problem instance
Halting problem decider
M
w
Fall 2004
Convert
Inputs
?
YES
M
Blank-tape
w halting problem
NO
decider
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YES
NO
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Construct a new machine
Mw
• When started on blank tape, writes
• Then continues execution like
w
M
Mw
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step 1
if blank tape
step2
execute M
then write
with input
w
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w
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M halts on input string w
if and
only if
M w halts when started with blank tape
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Halting problem decider
M
w
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Generate
Mw
M
YES YES
blank-tape
w halting problem
NO
NO
decider
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We reduced the halting problem
to the blank-tape halting problem
Since the halting problem is undecidable,
the blank-tape halting problem is undecidable
END OF PROOF
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Summary of Undecidable Problems
Halting Problem:
Does machine
M halt on input w ?
Membership problem:
Does machine M accept string
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w?
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Blank-tape halting problem:
Does machine M halt when starting
on blank tape?
State-entry Problem:
Does machine
on input w ?
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M enter state q
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Uncomputable Functions
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Uncomputable Functions
Domain
f
Values
region
A function is uncomputable if it cannot
be computed for all of its domain
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An uncomputable function:
f (n) 
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maximum number of moves until
any Turing machine with n states
halts when started with the blank tape
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Theorem: Function
Proof:
f (n) is uncomputable
Assume for contradiction that
f (n) is computable
Then the blank-tape halting problem
is decidable
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Decider 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
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Therefore, the blank-tape halting
problem is decidable
However, the blank-tape halting
problem is undecidable
Contradiction!!!
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Therefore, function
f (n) in uncomputable
END OF PROOF
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Undecidable Problems
for
Recursively Enumerable Languages
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Take a recursively enumerable language L
Decision problems:
• L is empty?
• L is finite?
• L contains two different strings
of the same length?
All these problems are undecidable
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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
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Let
M be the TM with L( M )  L
Suppose we have a decider for the
empty language problem:
M
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empty language
problem
decider
YES
L(M ) empty
NO
L(M ) not empty
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We will build the decider for the
membership problem:
M
w
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membership
problem
decider
YES
NO
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M accepts w
M rejects w
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We want to reduce the membership problem to
the empty language problem:
Membership problem decider
M
w
Fall 2004
Mw
empty language
problem
decider
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YES
NO
NO
YES
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We need to convert one problem instance
to the other problem instance
Membership problem decider
M
w
Fall 2004
Convert
inputs
?
Mw
empty language
problem
decider
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YES
NO
NO
YES
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Construct machine
Mw :
On arbitrary input string
s
M w executes the same as with M
When M enters a final state,
compare s with w
Accept only if
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sw
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w L
if and
only if
L( M w ) is not empty
L( M w )  {w}
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Membership problem decider
M
w
construct M
w
Mw
empty language
problem
decider
YES
NO
NO
YES
END OF PROOF
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