Computability
Decidability
1
Unsolvable problems
Computer scientists usually deal with problems which are
solvable.
We are supposed to find an algorithm to solve a particular
problem.
Why do we need to know about problems which are
unsolvable ?
we will not waste time in finding an non-existed algorithm.
we will realize that the problem needs to be simplified or altered
before we can find an algorithmic solution.
unsolvable problems can stimulate our imagination and help us
gain an important perspective on computation.
2
Decidable Languages
We have seen in the last chapter an undecidable
problem, the 10th problem of Hilbert.
Now we will show some other problems and
consider whether they are decidable by algorithm
or not.
3
Decidable Problems Concerning Regular Languages
Acceptance problem
Whether a finite automata accepts a string ?
ADFA = {<B,w>| B is a DFA that accepts input string w}
Whether a nondeterministic finite automata accepts
a string ?
ANFA = {<B,w>| B is a NFA that accepts input string w}
Whether a regular expression generates a given string.
AREX = {<R,w>| R is a regular expression that generates string w}
4
Decidable Problems concerning Regular Lang.
Empty Language problem
Whether a language of a finite automata is empty ?
EDFA = {<A>| A is a DFA and L(A) = }
Equivalence problem
Whether two finite automata are equivalent ?
EQDFA = {<A,B>| A and B are DFAs and L(A) = L(B)}
5
Theorem 1
ADFA is a decidable language
ADFA = {<B,w>| B is a DFA that accepts input string w}
Show that the problem of testing whether a given finite
automata accepts a given string is decidable
Proof idea: Present a TM M that decides ADFA.
6
Proof
M = “on input <B,w>, where B is a DFA and w is a string:
Simulate B on input w.
If the simulation ends in an accept state, accept. If it ends in a nonaccepting state, reject.”
Notes:
input of M is an encoded string <B,w>
B can be represented reasonably as a list of its five components,
Q, , , q0 and F.
before M starts its computation, M first checks on whether it
properly represents a DFA B and a string w. If not, M rejects.
7
Proof:
Input Tape
w
M keeps track of B’s
current state and B’s
current position in the
input w by writing this
information down on its
tape.
B
reject
accept
Machine M
8
Theorem 2
ANFA is a decidable language
ANFA = {<B,w>| B is a NFA that accepts input string w}
Proof idea:
Present a TM N that decides ANFA.
N has M as a subroutine.
Because M is designed to work with DFAs, N first converts
the NFA B it receives as input to a DFA before passing it
to M.
9
Proof
N = “on input <B,w>, where B is a NFA and w is a
string:
Convert NFA B to an equivalent DFA C
Run TM M from Theorem 1 on input <C,w>.
If M accepts, accept; otherwise reject.”
10
Proof:
Input Tape
w
Machine N
C
reject
accept
Machine M
11
Theorem 3
AREX is a decidable language
AREX = {<R,w>| R is a regular expression that generates string w}
Proof idea:
Present a TM P that decides AREX.
12
Proof
P = “on input <R,w>, where R is a regular
expression and w is a string:
Convert regular expression R to an equivalent DFA A.
Run TM M on input <A,w>.
If M accepts, accept; if M rejects , reject.”
13
Theorem 4
EDFA is a decidable language
EDFA = {<A>| A is a DFA and L(A) = }
Show that the problem of testing whether a given finite
automata accepts any strings at all is decidable.
Proof idea:
A DFA accepts some string if and only if reaching an accept state
from the start state by traveling alone the arrows of the DFA is
possible.
Present a TM T that uses a marking algorithm.
14
Proof
T = “on input <A>, where A is a DFA:
Mark the start state of A.
Repeat until no new states get marked:
Mark any state that has a transition
coming into it from any state that is
already marked.
If no accept state is marked, accept; otherwise reject.”
A is empty
15
Theorem 5
EQDFA is a decidable language
EQDFA = {<A,B>| A and B are DFAs and L(A) = L(B)}
Show that the problem of testing whether 2 DFAs recognize
the same language is decidable.
Proof idea:
Construct a new DFA C from A and B, where C accepts only those
strings that are accepted by either A or B but not by both.
If A and B recognize the same language, C will accept nothing.
Test whether L(C) is empty by the theorem 4.4
16
L(C)
Symmetric difference of L(A) and L(B)
L(C) = (L(A)  L(B))  (L(A)  L(B))
L(A)
L(B)
C is also a DFA as the class of regular language is closed under
complementation, union and intersection
17
Proof
F = “on input <A,B>, where A and B are DFAs:
Construct DFA C as described.
Run TM T from Theorem 4 on input <C>
If T accepts, accept. If T rejects, reject.”
18
Proof
Machine F
Machine T
C
reject
reject
accept
accept
19
Decidable Problems concerning to Context-Free
Language
Acceptance problem
Whether a CFG accepts a string ?
ACFG = {<G,w>| G is a CFG that accepts input string w}
Emptiness problem
Whether a CFG generates any strings at all ?
ECFG = {<G>| G is a CFG and L(G) = }
Equivalent problem
Whether two CFGs generate the same language ?
EQCFG = {<G,H>| G and H are CFGs and L(G) = L(H)}
20
Theorem 6
ACFG is a decidable language
ACFG = {<G,w>| G is a CFG that generates string w}
Proof idea:
Naïve idea :
Go through every derivations to determine whether any is a derivation of w.
But this algorithm will not halt if w is not a member of G.
Better approach:
We need to ensure that the algorithm tries only finitely many derivations, so
use CNF form.
any derivation of w has 2n-1 steps, where n is the length of w.
21
Proof
S = “on input <G,w>, where G is a CFG and w is a
string:
Convert G to an equivalent grammar in Chomsky
Normal Form.
List all derivations with 2n-1 steps, where n is the length
of w.
If any of these derivations generate w, accept; if not,
reject.”
22
CFLs and PDAs
As we can convert back and forth between CFLs and
PDAs.
Hence, everything we say about the decidability of
problems concerning CFLs applies well to PDAs.
23
Theorem 7
ECFG is a decidable language
ECFG = {<G>| G is a CFG and L(G) = }
Proof idea:
Naïve idea : to determine whether L(G) = 
try going through all possible w’s, one by one.
But there are infinitely many w’s to try, so this algorithm could end up running
forever.
Better approach:
We need to test whether the start variable can generate a string of terminals.
Determine for each variable whether that variable is capable of generating a
string of terminals, the algorithm keeps track of this information by placing a
mark on that variable.
24
Proof
R = “on input <G>, where G is a CFG :
Mark all terminal symbols in G.
Repeat until no new variables get marked
Mark any variable A where G has a rule
AU1 U2… Uk and each symbol U1 , …, Uk
has already been marked.
If the start symbol is not marked, accept; otherwise
reject. ”
 G is empty
25
EQCFG
Consider the problem of testing whether
2 CFGs generate the same language.
ECFG = {<G,w>| G is a CFG that generates string w}
Proof idea:
Naïve idea :
use the same method as theorem 4.5
But CFL is not closed under complementation or intersections.
In fact, ECFG is not decidable. We’ll prove it in the next
chapter.
26
Theorem 8
“Every context-free language is decidable.”
We show that every CFL is decidable.
Proof idea:
Naïve (bad) idea :
try to convert a PDA for A directly to a TM.
the PDA for A may be nondeterministic, so we get an NTM which we can
convert into a deterministic TM.
however, PDA can have some branches go on forever, reading and writing
the stack without coming to a halt. therefore our TM will not be a
decider.
Better approach:
We prove this theorem with the TM S in Theorem 6 to decide ACFG.
27
Proof
Let G be a CFG for A and design a TM MG that
decides A.
We build a copy of G into MG.
It works as follow.
MG = “on input w:
Run TM S on input <G,w>
If this machine accepts, accept; if it rejects, reject.”
28
Proof
Machine MG
Machine S
1. Convert G to CNF form.
2. List all derivations with 2n-1 steps, where
n is the length of w.
3. If any of these derivations generate w,
accept; if not, reject.
accept
reject
accept
reject
29
Relationship among classes
of Language
Turing-recognizable
Regular
Context-free
Decidable
30
Non Recursively Enumerator language
We want to find a language that is not Recursively
Enumerable
This language is not accepted by any Turing Machine
It can’t be implemented on (contemporary) computer.
Example:
Verify that the program we wrote performs as specified (that it is
correct).
Halting problem : Testing whether a TM accepts a given input
string (ATM)
31
Theorem 9
ATM = {<M,w>| M is a TM and M accepts w}.
ATM is undecidable.
Proof: observe that ATM is Turing-recognizable.
We try to find a TM (U) that recognizes ATM.
32
Halting Problem
U = “On input <M,w>, where M is a TM and w is a
string:
Simulate M on input w.
If M ever enters its accept state, accept; if M ever enters
its reject state, reject.”
Note: U loops on input <M,w> if M loops on w,
that’s why U doesn’t decide ATM.
33
Universal Turing Machine
proposed by Turing.
a machine is called universal if it is capable to
simulate any other Turing machine from the
description of that machine.
played an important early role in stimulating the
development of stored-program computers.
34
The Diagonalization method
discovered by Georg Cantor in 1873.
is the proof of the undecidability of the halting
problem.
using countable and uncountable set to prove.
Idea is : we want to prove that
set of Turing-recognizable language is uncountable
while set of Turing machine is countable.
35
Cantor’s problem
Concerned with the problem of measuring the sizes
of infinite sets.
If we have 2 infinite sets, how can we tell whether
one is larger than the other or whether they are of
the same size ?
Naïve: try to count the elements of the infinite set,
we will never finish!
36
Cantor’s problem
Better approach:
Observe that 2 finite sets are of the same size if the
elements of the one set can be paired with the
element of the other set.
We can extend this idea to infinite sets.
37
Definition: Correspondence
Consider 2 sets, A and B, and a function f from A
to B.
Say that f is one-to-one if it never maps 2 different
elements to the same place that is, f(a)  f(b)
whenever a  b.
Say that A and B are the same size if there is a
one-to-one, onto function
f: AB.
one-to-one and onto is called Correspondence
38
Example
Let
N be the set of Natural numbers {1,2,3,…}
 be the set of even natural numbers {2,4,6,…}
Using cantor’s method, we can see that N and 
have the same size.
n
f(n)
1
2
2
4
3
...
6
...
39
Definition : Countable
A set A is countable if either it
is finite or it has the same size
as N.
40
Example
Let
N be the set of Natural numbers {1,2,3,…}
Q be the set of positive rational numbers
Q = {m/n |m,n N }
Q seems the be bigger than N but actually they have the
same size.
Using cantor’s method, we list all the elements of Q and
then pair each element with the element of N.
We have to ensure that every member of Q appears only
once on the list.
41
Correspondence
of N and Q
1
1
1
2
1
3
1
4
1
5
2
1
2
2
2
3
2
4
2
5
3
1
3
2
3
3
3
4
3
5
4
1
4
2
4
3
4
4
4
5
5
1
5
2
...
.
.
.
42
Trick
We pair
the first element of Q on the list with the number 1 from N.
the second element of Q on the list with the number 2 from N.
…
After seeing the corresponding of N and Q, we can see
that as each curve Q has finite number of elements
which can be mapped with elements of N, thus both
sets have same size.
43
Theorem 4.14
R is uncountable
R is the set of real number.
Proof by the diagonalization method.
44
Proof
by contradiction : suppose that a correspondence f
existed between N and R.  means that we must
pair all the members of N with all the members of
R.
show that f fails to work as it should.
How to show ?  finding an x in R that is not
paired with anything in N.
45
Proof
Finding x by constructing it.
choose each digit of x to make x different from one
of the real numbers that is paired with an element
of N
Sure that x is different from any real number that is
paired.
46
Illustration of the idea
suppose we have a corresponding f:
we construct x by
let the first digit of x be anything different
from the first fractional digit of f(1).
let the second digit of x be anything
different from the second fractional digit
of f(2).
…
thus we can construct x which is
different from f(n) for any n.
n
f(n)
1
0.123456
2
0.500000
3
...
0.759685
...
n
f(n)
1
0.423456
2
0.530000
3
...
0.756685
...
47
Corollary 4.15
Some languages are not Turing-recognizable.
Proof idea:
set of Turing-recognizable language is uncountable while set of
Turing machine is countable.
Because each Turing machine can recognize a single language
and there are more languages than TMs, some languages are
not recognized by any TM.
Thus, there exists some languages which are not Turingrecognizable.
48
set of all TMs is countable.
Proof idea:
first of all
we have to observe that the set of all * is countable, for
any alphabet .
with only finitely many strings of each length, we may
form a list of * by written down all strings of length
0, length 1, length 2 and so on.
49
set of all TMs is countable.
Secondly
We can encode each TM into a string <M>
as the set of all strings is countable therefore the set of all
TMs is countable.
50
set of all languages is uncountable
Proof idea:
first of all
we have to observe that the set of all infinite binary
sequence is uncountable
An infinite binary sequence is an unending sequence of
0s and 1s.
we can prove by the diagonalization method.
(imagine as the same way as R)
51
set of all languages is uncountable
Secondly
Let B be the set of all infinite binary sequences.
Let L be the set of all languages over alphabet .
Show that L is uncountable by giving correspondence
with B.
the idea is that “as B is uncountable thus L is also
uncountable.”
52
set of all languages is uncountable
Correspondence of L and B
Show that both sets have the same size.
Let * = {s1,s2,…}.
Each language A L has a unique sequence of B. (oneto-one and onto)
Characteristic sequence of A, (A)
The ith bit of that sequence is a 1 if si  A and is a 0 if si  A.
53
set of all languages is uncountable
Example : Characteristic sequence
*
= {  , 0 , 1 , 00 , 01 , 10 ,
A
= {
(A) = { 0
0 ,
1
1
1
000 , 001 , … } ;
00 , 01
0
1
1
, 000 , 001 , … } ;
0
0
1
1
… } ;
The function f: L  B, where f(A) equals the Characteristic
sequence of A, is one-to-one and onto and hence a
correspondence. Therefore, as B is uncountable, L is
uncountable as well.
54
The Halting Problem
We have proved that there are some languages which
are not Turing-recognizable.
Now, we want to show that such a TM which can
recognize the non Turing-recognizable language
doesn’t exist.
55
Proof of Theorem 9
ATM is undecidable.
Proof:
Consider  = {a}.
The set of all strings is
*
= {,a, aa, aaa, aaaa…}
= {,a1, a2, a3, a4…}
the set of all TMs is countable, thus we can order them as M1, M2,
M3,…
to indicate which language is accepted by what machine, we can
again use the characteristic sequence of that language.
56
Proof of Theorem 9
Example:
assume a machine Mi accepts a language L(Mi)
L(Mi) = {a2, a4, a6}
we can write
*
L(Mi)
= { a1 , a2 , a3 ,
a4
= { 0
1
1
0
, a5 , a6 , a7 ,
0
1
0
a8
0
,
,
… } ;
,
… } ;
57
Proof of Theorem 9
For each machine, we can write its characteristic
sequence
a1
a2
a3
a4

 L(M )
1
0
1
0
1

 L(M )
2
1
0
0
1

 L(M )
3
0
1
1
1

 L(M )
4
0
0
0
1

58
Proof of Theorem 9
Consider the language
L  {a : a  L( M i )}
i
i
L consists from the 1’s in the diagonal
59
1
2
3
4
a
a
 L(M )
1
0
1
0
1

 L(M )
2
1
0
0
1

 L(M )
3
0
1
1
1

 L(M )
4
0
0
0
1



a
L  {a , a ,}
3
4
a

 
60
Proof of Theorem 9
Since L is a properly defined language,
we can consider its complement
Consider the language
i
i
L
L  {a : a  L( M i )}
L  {a : a  L( M i )}
i
i
L consists from the 0’s in the diagonal
61
1
2
3
4

a
a
 L(M )
1
0
1
0
1

 L(M )
2
1
0
0
1

 L(M )
3
0
1
1
1

 L(M )
4
0
0
0
1



a
L  {a , a ,}
1
2
a
 
62
Proof of Theorem 4.9
Claim 1:
Language
L is not recursively enumerable
If we can prove that it is not recursively
enumerable, we can now find a language which
is not Turing recognizable.
63
Proof: Claim 1
Assume for contradiction that
is recursively enumerable
There must exist some machine
that accepts
L
L
Mk
L( M k )  L
64
1
2
3
4

a
a
 L(M )
1
0
1
0
1

 L(M )
2
1
0
0
1

 L(M )
3
0
1
1
1

 L(M )
4
0
0
0
1


a
a
 

Question:
M k  M1 ?
65
1
2
3
4

a
a
 L(M )
1
0
1
0
1

 L(M )
2
1
0
0
1

 L(M )
3
0
1
1
1

 L(M )
4
0
0
0
1


Answer:

M k  M1
a
a
a  L( M k ) 
1
1
a  L ( M1 )
66
1
2
3
4

a
a
 L(M )
1
0
1
0
1

 L(M )
2
1
0
0
1

 L(M )
3
0
1
1
1

 L(M )
4
0
0
0
1



Question: M k  M 2 ?
Answer: M k  M 2
a
a

a  L( M k )
2
2
a  L( M 2 )
67
1
2
3
4

a
a
 L(M )
1
0
1
0
1

 L(M )
2
1
0
0
1

 L(M )
3
0
1
1
1

 L(M )
4
0
0
0
1



Question: M k  M 3 ?
Answer: M k  M 3
a
a

a  L( M k )
3
3
a  L( M 3 )
68
Proof: Claim 1
Similarly:
Mk  Mi
for any
i
Because either:
a  L( M k )
i
a  L( M i )
i
a  L( M k )
i
or
a  L( M i )
i
69
Proof: Claim 1
Therefore, the machine
M k cannot exist
Therefore, the language L
is not recursively enumerable
End of Proof
70
Observation:
There is no algorithm that describes
L
(otherwise L would be accepted by
some Turing Machine)
71
Non Recursively Enumerable
L
Recursively Enumerable
Recursive
72
Testing whether TM accepts a given string :
ATM
Try to construct a TM (H) to decide whether
another TM (M) accepts a string (w) or not.
H is decider on (M,w)
accept
if M accepts w
H(M,w)
=
reject
if M rejects w
73
Testing whether TM accepts a given string :
ATM
If w = the encoded string <M>
H is decider on (M,<M>)
H(M,<M>)
accept
if M accepts <M>
reject
if M rejects <M>
=
74
Testing whether TM accepts a given string :
ATM
Now we construct a new TM (D) with H as a
subroutine.
D does opposite to what its subroutine does.
D = “On input <M>, where M is a TM,
Run H on input <M,<M>)
Output the opposite to what H outputs; that is, if H
accepts, reject and if H rejects, accept.”
75
Testing whether TM accepts a given string :
ATM
accept
if H rejects <M>
reject
if H accepts <M>
accept
if M rejects <M>
reject
if M accepts <M>
D(<M>) =
D(<M>) =
76
Testing whether TM accepts a given string :
ATM
D(<M>)
H (M,<M>)
M (<M>)
if M accepts <M> then H accepts
if M rejects <M> then H rejects
accept
reject
reject
accept
77
Testing whether TM accepts a given string :
ATM
running M on <M>
 M 1   M 2   M 3  M 4  
M1
1
1
M2
1
M3
1
1
M4
1
1





1
1

 
78
Testing whether TM accepts a given string :
ATM
output of H
 M 1   M 2   M 3  M 4  
M1
1
0
0
1

M2
0
0
1
0

M3
1
1
0
0

M4
1
1
1
1



 
79
Testing whether TM accepts a given string :
ATM
Output of D
 M 1   M 2   M 3  M 4   D 
M1
1
0
0
1
M2
0
0
1
M3
1
1
M4
1
1
D


0
 1
 1
0
0
 0

1
1
 1

0


0
1


1
 
?


80
A Language which is Recursively
Enumerable and not Recursive
81
We want to find a language which
Is recursively
enumerable
There is a
Turing Machine
that accepts
the language
But not
recursive
The machine
doesn’t halt
on some input
82
Proof
We will prove that the language
L  {a : a  L( M i )}
i
i
Is recursively enumerable
but not recursive
83
1
2
3
4

a
a
 L(M )
1
0
1
0
1

 L(M )
2
1
0
0
1

 L(M )
3
0
1
1
1

 L(M )
4
0
0
0
1

a
L  {a , a ,}
3
a
4
84
Theorem:
The language
L  {a : a  L( M i )}
i
i
is recursively enumerable
Proof:
We will give a Turing Machine that accepts
L
85
Turing Machine that accepts
For any input string
L
w
• Compute i for which
• Find Turing machine
wa
i
Mi
(using the enumeration procedure for Turing Machines)
i
on
input
Mi
a
• If M i accepts, then accept
• Simulate
End of Proof
w
86
Observation:
Recursively enumerable
i
i
L  {a : a  L( M i )}
Not recursively enumerable
i
i
L  {a : a  L( M i )}
(Thus, also not recursive)
87
Theorem:
The language
L  {a : a  L( M i )}
i
i
is not recursive
Proof:
Assume for contradiction that
L is recursive
88
Proof
Then
L is recursive
But we know:
L is not recursively enumerable
thus, not recursive
CONTRADICTION!
Therefore, L is not recursive
End of Proof
89
Non Recursively Enumerable
L
Recursively Enumerable
L
Recursive
90
Theorem
A language is decidable (recursive) if and only
if it is both Turing-recognizable and CoTuring-recognizable.
Co-Turing-recognizable language is the the
complement of a Turing-recognizable
language.
91
Corollary
ATM is not Turing-recognizable.
Proof:
ATM is Turing-recognizable.
if ATM is Turing-recognizable,
ATM would be decidable.
But ATM is not decidable, thus ATM
must not be Turing-recognizable.
92
Recursively Enumerable
and
Recursive
Languages
93
Definition:
A language is recursively enumerable
(rekursivt uppräknelig)
if some Turing machine accepts it
94
Let
L be a recursively enumerable language
and
M the Turing Machine that accepts it
For string
w:
if
w L
then
M halts in a final state
if
w L
then
M halts in a non-final state
or loops forever
95
Definition:
A language is recursive
if some Turing machine accepts it
and halts on any input string
In other words:
A language is recursive if there is
a membership algorithm for it
96
Let
L be a recursive language
and
M the Turing Machine that accepts it
For string
w:
if
w L
then
if
w L
then M halts in a non-final state
M halts in a final state
97
We will prove:
1. There is a specific language
which is not recursively enumerable
(not accepted by any Turing Machine)
2. There is a specific language
which is recursively enumerable
but not recursive
98
Non Recursively Enumerable
Recursively Enumerable
Recursive
99
We will first prove:
• If a language is recursive then
there is an enumeration procedure for it
• A language is recursively enumerable
if and only if
there is an enumeration procedure for it
100
Theorem:
if a language L is recursive then
there is an enumeration procedure for it
101
Proof:
Enumeration Machine
~
M
Enumerates all
strings of input alphabet
M
Accepts
L
102
If the alphabet is {a, b} then
~
M can enumerate strings as follows:
a
b
aa
ab
ba
bb
aaa
aab
......
103
Enumeration procedure
Repeat:
~
M
generates a string
w
M checks if w L
w to output
NO: ignore w
YES: print
End of Proof
104
Example:
~
M
a
b
aa
ab
ba
bb
aaa
aab
......
L  {b, ab, bb, aaa,....}
L(M )
Enumeration
Output
b
b
ab
ab
bb
aaa
bb
aaa
......
......
105
Theorem:
if language L is recursively enumerable then
there is an enumeration procedure for it
106
Proof:
Enumeration Machine
~
M
Enumerates all
strings of input alphabet
M
Accepts
L
107
If the alphabet is {a, b} then
~
M can enumerate strings as follows:
a
b
aa
ab
ba
bb
aaa
aab
108
NAIVE APPROACH
Enumeration procedure
Repeat:
~
M
generates a string
w
M checks if w L
w to output
NO: ignore w
YES: print
Problem:
If w L
machine M may loop forever
109
BETTER APPROACH
~
M
Generates first string
w1
M executes first step on w1
~
M
Generates second string
w2
M executes first step on w2
second step on
w1
110
~
M
Generates third string
M
w3
executes first step on
w3
second step on
w2
third step on w1
And so on............
111
Step
in
string

w1
w2
w3
w4
1
1
1
1
2
2
2
2
3
3
3
3

112
If for any string wi
machine M halts in a final state
then it prints wi on the output
End of Proof
113
Theorem:
If for language L
there is an enumeration procedure
then L is recursively enumerable
114
Proof:
Input Tape
w
Machine that
accepts L
Enumerator
for L
Compare
115
Turing machine that accepts L
For input string
w
Repeat:
• Using the enumerator,
generate the next string of
L
• Compare generated string with
w
If same, accept and exit loop
End of Proof
116
We have proven:
A language is recursively enumerable
if and only if
there is an enumeration procedure for it
117
A Language which
is not
Recursively Enumerable
118
We want to find a language that
is not Recursively Enumerable
This language is not accepted by any
Turing Machine
119
{a}
Consider alphabet
Strings:
a, aa, aaa, aaaa, 
1
a a
2
a
3
a
4

120
Consider Turing Machines
that accept languages over alphabet
{a}
They are countable:
M1, M 2 , M 3 , M 4 , 
121
Mi
Example language accepted by
L( M i )  {aa, aaaa, aaaaaa}
L( M i )  {a , a , a }
2
4
6
Alternative representation
1
a
a
L( M i ) 0
1
2
a
3
0
5
4
a
1
0
a
a
6
1
a
7
0


122
3
1
a
a
2
a
L( M1)
0
1
0
1

L( M 2 )
1
0
0
1

L( M 3 )
0
1
1
1

L( M 4 )
0
0
0
1

a
4

123
Consider the language
L  {a : a  L( M i )}
i
i
L consists from the 1’s in the diagonal
124
1
2
3
4

a
a
L( M1)
0
1
0
1

L( M 2 )
1
0
0
1

L( M 3 )
0
1
1
1

L( M 4 )
0
0
0
1

a
L  {a , a ,}
3
a
4
125
L
Consider the language
L  {a : a  L( M i )}
i
i
L  {a : a  L( M i )}
i
i
L consists from the 0’s in the diagonal
126
1
a
a
2
a
L( M1)
0
1
0
1

L( M 2 )
1
0
0
1

L( M 3 )
0
1
1
1

L( M 4 )
0
0
0
1

3
L  {a , a ,}
1
a
4

2
127
Theorem:
Language
L is not recursively enumerable
128
Proof:
Assume for contradiction that
L is recursively enumerable
There must exist some machine
that accepts L
Mk
L( M k )  L
129
1
a
a
2
a
L( M1)
0
1
0
1

L( M 2 )
1
0
0
1

L( M 3 )
0
1
1
1

L( M 4 )
0
0
0
1

Question:
3
a
4

M k  M1 ?
130
1
a
a
2
a
L( M1)
0
1
0
1

L( M 2 )
1
0
0
1

L( M 3 )
0
1
1
1

L( M 4 )
0
0
0
1

3
a
4

1
Answer:
M k  M1
a  L( M k )
1
a  L ( M1 )
131
1
a
a
2
a
L( M1)
0
1
0
1

L( M 2 )
1
0
0
1

L( M 3 )
0
1
1
1

L( M 4 )
0
0
0
1

Question:
3
a
4

Mk  M2 ?
132
1
a
a
2
a
L( M1)
0
1
0
1

L( M 2 )
1
0
0
1

L( M 3 )
0
1
1
1

L( M 4 )
0
0
0
1

3
a
4

2
Answer:
Mk  M2
a  L( M k )
2
a  L( M 2 )
133
1
a
a
2
a
L( M1)
0
1
0
1

L( M 2 )
1
0
0
1

L( M 3 )
0
1
1
1

L( M 4 )
0
0
0
1

Question:
3
a
4

M k M 3 ?
134
1
a
a
2
a
L( M1)
0
1
0
1

L( M 2 )
1
0
0
1

L( M 3 )
0
1
1
1

L( M 4 )
0
0
0
1

3
a
4

3
Answer:
M k  M3
a  L( M k )
3
a  L( M 3 )
135
Similarly:
Mk  Mi
for any
i
Because either:
a  L( M k )
i
a  L( M i )
i
a  L( M k )
i
or
a  L( M i )
i
136
Thus the machine
M k cannot exist
Therefore, the language L
is not recursively enumerable
End of Proof
137
Observation:
There is no algorithm that describes
L
(otherwise L would be accepted by
some Turing Machine)
138
Non Recursively Enumerable
L
Recursively Enumerable
Recursive
139
A Language which is
Recursively Enumerable
and not Recursive
140
We want to find a language which
Is recursively
enumerable
There is a
Turing Machine
that accepts
the language
But not
recursive
The machine
doesn’t halt
on some input
141
We will prove that the language
L  {a : a  L( M i )}
i
i
Is recursively enumerable
but not recursive
142
1
2
3
4

a
a
L( M1)
0
1
0
1

L( M 2 )
1
0
0
1

L( M 3 )
0
1
1
1

L( M 4 )
0
0
0
1

a
L  {a , a ,}
3
a
4
143
Theorem:
The language
L  {a : a  L( M i )}
i
i
is recursively enumerable
144
Proof:
We will give a Turing Machine that
accepts L
145
Turing Machine that accepts
For any input string
L
w
• Compute i , for which
• Find Turing machine
wa
i
Mi
(using the enumeration procedure for Turing Machines)
• Simulate
• If
M i on input a
i
M i accepts, then accept w
End of Proof
146
Observation:
Recursively enumerable
i
i
L  {a : a  L( M i )}
Not recursively enumerable
i
i
L  {a : a  L( M i )}
(Thus, also not recursive)
147
Theorem:
The language
L  {a : a  L( M i )}
i
i
is not recursive
148
Proof:
Assume for contradiction that
L is recursive
Take the Turing Machine M that accepts
L
M halts on any input:
If
If
Then
M accepts then reject
M rejects then accept
L is recursive
149
L is recursive
But we know:
L is not recursively enumerable
thus, not recursive
CONTRADICTION!
Therefore,
L is not recursive
End of Proof
150
Non Recursively Enumerable
L
Recursively Enumerable
L
Recursive
151