Post on 25-Jul-2020
Prof. Busch - LSU 1
Busch Complexity Lectures:
Turing Machines
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The Language Hierarchy
*aRegular Languages
Context-Free Languages nnba Rww
nnn cba ww?
**ba
?
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*aRegular Languages
Context-Free Languages nnba Rww
nnn cba ww
**ba
Languages accepted by Turing 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
ak
3. 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 position of 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 caTime 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 transition for input symbol c
Allowed:
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Halting
The machine halts in a state if there is no 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 transition from 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 the symbols in the string
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Turing Machine Example
Accepts the language: *a
0q
Raa ,→
L,◊→◊1q
Input alphabet },{ ba=Σ
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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 }{a=ΣA 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
0qa
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,◊→◊
1≥n
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Match a’s with b’s: Repeat: replace leftmost a with x find leftmost b and replace it with y Until 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 Definitions for
Turing Machines
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Transition Function
1q 2qRba ,→
),,(),( 21 Rbqaq =δ
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1q 2qLdc ,→
),,(),( 21 Ldqcq =δ
Transition Function
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Turing Machine:
),,,,,,( 0 FqQM ◊ΓΣ= δ
States
Input alphabet
Tape alphabet
Transition function
Initial state
blank
Accept states
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Configuration
◊ ◊ ◊ba
1q
a
Instantaneous description:
c◊
baqca 1
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◊ ◊yx
2q
a bTime 4
◊ ◊yx
0q
a bTime 5
◊ ◊
A Move: aybqxxaybq 02 ≻
(yields in one mode)
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◊ ◊yx
2q
a bTime 4
◊ ◊yx
0q
a bTime 5
◊ ◊
bqxxyybqxxaybqxxaybq 1102 ≻≻≻
◊ ◊yx
1q
x bTime 6
◊ ◊yx
1q
x bTime 7
◊ ◊
A computation
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bqxxyybqxxaybqxxaybq 1102 ≻≻≻
bqxxyxaybq 12∗≻Equivalent notation:
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Initial configuration: wq0
◊ ◊ba
0q
a b ◊
wInput 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 machine then we say that is:
• Turing Acceptable • Recursively Enumerable
ML
L
• Turing Recognizable
Other names used: