prof. busch - lsu1 turing machines. prof. busch - lsu2 the language hierarchy regular languages...
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Prof. Busch - LSU 1
Turing Machines
Prof. Busch - LSU 2
The Language Hierarchy
*aRegular Languages
Context-Free Languagesnnba Rww
nnn cba ww?
**ba
?
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*aRegular Languages
Context-Free Languagesnnba Rww
nnn cba ww
**ba
Languages accepted byTuring Machines
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A Turing Machine
............Tape
Read-Write head
Control Unit
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The Tape
............
Read-Write head
No boundaries -- infinite length
The head moves Left or Right
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............
Read-Write head
The head at each transition (time step):
1. Reads a symbol 2. Writes a symbol 3. Moves Left or Right
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............
Example:Time 0
............Time 1
1. Reads
2. Writes
a a cb
a b k c
a
k3. Moves Left
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............Time 1
a b k c
............Time 2
a k cf
1. Reads
2. Writes
bf
3. Moves Right
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The Input String
............
Blank symbol
head
a b ca
Head starts at the leftmost positionof the input string
Input string
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States & Transitions
1q 2qLba ,
Read Write Move Left
1q 2qRba ,
Move Right
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Example:
1q 2qRba ,
............ a b ca
Time 1
1qcurrent state
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............ a b caTime 1
1q 2qRba ,
............ a b cbTime 2
1q
2q
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............ a b caTime 1
1q 2qLba ,
............ a b cbTime 2
1q
2q
Example:
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............ a b caTime 1
1q 2qRg,
............ ga b cbTime 2
1q
2q
Example:
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Determinism
1q
2qRba ,
Allowed Not Allowed
3qLdb ,
1q
2qRba ,
3qLda ,
No lambda transitions allowed
Turing Machines are deterministic
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Partial Transition Function
1q
2qRba ,
3qLdb ,
............ a b ca
1q
Example:
No transitionfor input symbol c
Allowed:
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Halting
The machine halts in a state if there isno transition to follow
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Halting Example 1:
............ a b ca
1q
1q No transition from
HALT!!!
1q
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Halting Example 2:
............ a b ca
1q
1q
2qRba ,
3qLdb ,
No possible transitionfrom and symbol
HALT!!!
1q c
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Accepting States
1q 2q Allowed
1q 2q Not Allowed
•Accepting states have no outgoing transitions•The machine halts and accepts
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Acceptance
Accept Input If machine halts in an accept state
Reject Input
If machine halts in a non-accept state or If machine enters an infinite loop
string
string
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Observation:
In order to accept an input string,it is not necessary to scan all thesymbols in the string
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Turing Machine Example
Accepts the language: *a
0q
Raa ,
L,1q
Input alphabet },{ ba
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aaTime 0
0q
a
0q
Raa ,
L,1q
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aaTime 1
0q
a
0q
Raa ,
L,1q
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aaTime 2
0q
a
0q
Raa ,
L,1q
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aaTime 3
0q
a
0q
Raa ,
L,1q
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aaTime 4
1q
a
0q
Raa ,
L,1q
Halt & Accept
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Rejection Example
0q
Raa ,
L,1q
baTime 0
0q
a
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0q
Raa ,
L,1q
baTime 1
0q
a
No possible Transition
Halt & Reject
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Accepts the language: *a
0q
but for input alphabet }{aA simpler machine for same language
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aaTime 0
0q
a
0q
Halt & Accept
Not necessary to scan input
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Infinite Loop Example
0q
Raa ,
L,1q
Lbb ,
A Turing machine for language *)(* baba
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baTime 0
0q
a
0q
Raa ,
L,1q
Lbb ,
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baTime 1
0q
a
0q
Raa ,
L,1q
Lbb ,
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baTime 2
0q
a
0q
Raa ,
L,1q
Lbb ,
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baTime 2
0q
a
baTime 3
0q
a
baTime 4
0q
a
baTime 5
0q
a
Infinite
loop
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Because of the infinite loop:
•The accepting state cannot be reached
•The machine never halts
•The input string is rejected
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Another Turing Machine Example
Turing machine for the language }{ nnba
0q 1q 2q3qRxa ,
Raa ,Ryy ,
Lyb ,
Laa ,Lyy ,
Rxx ,
Ryy ,
Ryy ,4q
L,
1n
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Match a’s with b’s:Repeat: replace leftmost a with x find leftmost b and replace it with yUntil there are no more a’s or b’s
If there is a remaining a or b reject
Basic Idea:
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0q 1q 2q3qRxa ,
Raa ,Ryy ,
Lyb ,
Laa ,Lyy ,
Rxx ,
Ryy ,
Ryy ,4q
L,
ba
0q
a bTime 0
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0q 1q 2q3qRxa ,
Raa ,Ryy ,
Lyb ,
Laa ,Lyy ,
Rxx ,
Ryy ,
Ryy ,4q
L,
bx
1q
a b Time 1
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0q 1q 2q3qRxa ,
Raa ,Ryy ,
Lyb ,
Laa ,Lyy ,
Rxx ,
Ryy ,
Ryy ,4q
L,
bx
1q
a b Time 2
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0q 1q 2q3qRxa ,
Raa ,Ryy ,
Lyb ,
Laa ,Lyy ,
Rxx ,
Ryy ,
Ryy ,4q
L,
yx
2q
a b Time 3
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0q 1q 2q3qRxa ,
Raa ,Ryy ,
Lyb ,
Laa ,Lyy ,
Rxx ,
Ryy ,
Ryy ,4q
L,
yx
2q
a b Time 4
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0q 1q 2q3qRxa ,
Raa ,Ryy ,
Lyb ,
Laa ,Lyy ,
Rxx ,
Ryy ,
Ryy ,4q
L,
yx
0q
a b Time 5
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0q 1q 2q3qRxa ,
Raa ,Ryy ,
Lyb ,
Laa ,Lyy ,
Rxx ,
Ryy ,
Ryy ,4q
L,
yx
1q
x b Time 6
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0q 1q 2q3qRxa ,
Raa ,Ryy ,
Lyb ,
Laa ,Lyy ,
Rxx ,
Ryy ,
Ryy ,4q
L,
yx
1q
x b Time 7
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0q 1q 2q3qRxa ,
Raa ,Ryy ,
Lyb ,
Laa ,Lyy ,
Rxx ,
Ryy ,
Ryy ,4q
L,
yx x y
2q
Time 8
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0q 1q 2q3qRxa ,
Raa ,Ryy ,
Lyb ,
Laa ,Lyy ,
Rxx ,
Ryy ,
Ryy ,4q
L,
yx x y
2q
Time 9
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0q 1q 2q3qRxa ,
Raa ,Ryy ,
Lyb ,
Laa ,Lyy ,
Rxx ,
Ryy ,
Ryy ,4q
L,
yx
0q
x y Time 10
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0q 1q 2q3qRxa ,
Raa ,Ryy ,
Lyb ,
Laa ,Lyy ,
Rxx ,
Ryy ,
Ryy ,4q
L,
yx
3q
x y Time 11
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0q 1q 2q3qRxa ,
Raa ,Ryy ,
Lyb ,
Laa ,Lyy ,
Rxx ,
Ryy ,
Ryy ,4q
L,
yx
3q
x y Time 12
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0q 1q 2q3qRxa ,
Raa ,Ryy ,
Lyb ,
Laa ,Lyy ,
Rxx ,
Ryy ,
Ryy ,4q
L,
yx
4q
x y
Halt & Accept
Time 13
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If we modify the machine for the language }{ nnba
we can easily construct a machine for the language }{ nnn cba
Observation:
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Formal Definitionsfor
Turing Machines
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Transition Function
1q 2qRba ,
),,(),( 21 Rbqaq
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1q 2qLdc ,
),,(),( 21 Ldqcq
Transition Function
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Turing Machine:
),,,,,,( 0 FqQM
States
Inputalphabet
Tapealphabet
Transitionfunction
Initialstate
blank
Acceptstates
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Configuration
ba
1q
a
Instantaneous description:
c
baqca 1
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yx
2q
a b
Time 4
yx
0q
a b
Time 5
A Move: aybqxxaybq 02
(yields in one mode)
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yx
2q
a b
Time 4
yx
0q
a b
Time 5
bqxxyybqxxaybqxxaybq 1102
yx
1q
x b
Time 6
yx
1q
x b
Time 7
A computation
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bqxxyybqxxaybqxxaybq 1102
bqxxyxaybq 12
Equivalent notation:
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Initial configuration: wq0
ba
0q
a b
w
Input string
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The Accepted Language
For any Turing Machine M
}:{)( 210 xqxwqwML f
Initial state Accept state
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If a language is accepted by a Turing machinethen we say that is:
•Turing Acceptable•Recursively Enumerable
ML
L
•Turing Recognizable
Other names used:
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