Lecture 13 The Halting Problem

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CSci 4011
INHERENT LIMITATIONS OF COMPUTER PROGRAMS
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QUIZ 3
A Turing Machine decides the language L if
It accepts all w ? L and rejects all w ? L.
The language L is Turing-recognizable if
There is a TM that accepts all and only the
strings in L.
w M accepts w
If M is a TM then L(M) is the set
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SET THEORY 101
A function f A ! B is
1-1 (or injective) if
f(x)f(y) ? xy
onto (or surjective) if
?y ?x y f(x)
bijective if it is 1-1 and onto.
f can help us count. If f is
1-1 then A B

onto then A B

bijective then B A.

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Let S be any set and P(S) be the power set of S
Theorem There is no onto map from S to P(S)
Proof
Assume, for a contradiction, that there is an
onto map f S ? P(S)
We construct an D ? S that cannot be the
output, f(y), for any y ?S.
Let D x ? S x ? f(x)
If D f(y) then
y ? D if and only if y ? D
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Y
Y
N
Y
(yi 2 D) (yi ? f(yi))
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The process of constructing a counterexample
by contradicting the diagonal is called
DIAGONALIZATION
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No matter what, P(S) always has more elements
than S
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THE CHURCH-TURING THESIS
L is recognized by a program for some
computer , L is recognized by a TM
The computer must be reasonable
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There are languages over 0,1 that are not
decidable
If we believe the Church-Turing Thesis, this
means there are problems that computers
inherently cannot solve.
We proved this using a simple counting
argument
There are more Languages than Turing Machines.
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Languages over 0,1
Turing Machines
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Not all languages over 0,1 are decidable
Turing Machines
Languages over 0,1
Sets of strings of 0s and 1s
Strings of 0s and 1s
S
P(S)
OK, but what does an undecidable language look
like?
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THE DIAGONAL LANGUAGE
DTM ltMgt M does not accept ltMgt is
undecidable.
Y
Y
N
Y
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Programs can accept themselves as input
grep grep lt/usr/bin/grep Binary file (standard
input) matches
cat gt /tmp/example.pl ltltEOP !/usr/bin/perl
n print Yippee!\n if // EOP
/tmp/example.pl lt/tmp/example.pl Yippee! Yippee!
Programs can reject themselves as input
/usr/bin/python lt/usr/bin/python SyntaxError
invalid syntax
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THE ACCEPTANCE PROBLEM
ATM ltM,wgt M is a TM that accepts string w
Theorem ATM is Turing-recognizable but NOT
decidable
ATM is Turing-recognizable
Define TM U as follows On input ltM,wgt, U runs
M on w. If M ever accepts accept. If M ever
rejects reject.
If M(w) loops forever U(ltM,wgt) loops forever.
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ATM ltM,wgt M is a TM that accepts string w
ATM is undecidable
(proof by contradiction)
Assume TM acceptsInput decides ATM
acceptsInput(M,w)
Construct a new TM LLPF as follows on input ltMgt
run acceptsInput(M,ltMgt) and output the opposite
LLPF
LLPF
LLPF( M )
LLPF
LLPF
LLPF
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OUTPUT OF acceptsInput(M,N)

M1
M2
M3
M4
LLPF
M1
accept
accept
M2
reject
accept
M3
accept
reject
M4
accept
reject

?
LLPF
reject
accept
reject
accept
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Theorem ATM is Turing-recognizable (r.e.) but
NOT decidable
Theorem ?ATM is not even Turing-recognizable!
Proof
Suppose ATM is recognized by TM V.
We know ATM is recognized by U.
Build Z to decide ATM run V(M,w) and U(M,w) in
parallel. If V accepts, reject if U accepts,
accept.
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A language is called Turing-recognizable,
semi-decidable, or recursively enumerable if some
TM recognizes it
A language is called decidable or recursive if
some TM decides it
all languages
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THE HALTING PROBLEM
HALTTM ltM,wgt M is a TM that halts on string
w
Theorem HALTTM is undecidable
Proof
Assume, for a contradiction, that TM
haltsOnInput decides HALTTM
haltsOnInput(M,w)
liarLiar(ltliarLiargt) halts iff liarLiar(ltliarLiargt
) loops!
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We can also prove that ATM is undecidable
by reusing the proof that DTM is undecidable.
Theorem. If ATM is decidable, then so is DTM.
Proof Suppose we have a program,
acceptsInput, that decides ATM, e.g.
acceptsInput(M,w) returns true if ltM,wgt ?ATM and
returns false otherwise.
We can build a TM notDiagonal to decide DTM

notDiagonal(ltMgt) return acceptsInput(M,ltMgt)
Theorem. If HALTTM is decidable, then so is ATM.
Proof acceptsInput(M,w) let M M with
qreject replaced by qloop. return
haltsOnInput(M,w)