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Turing Machine

This document provides an overview of Turing machines including their components, operation, and how they are used to recognize formal languages. Key points covered include: Turing machines have a tape, read/write head, and set of states, and transition between states based on the symbol being read on the tape. They can move left/right on the tape and write new symbols. Examples are provided to demonstrate how Turing machines can accept or reject strings and recognize formal languages like {anbn} through transitions that match symbols. Infinite loops cause strings to be rejected. Formal definitions of the transition function and components of a Turing machine are also given.

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Introduction to Turing Machines and language hierarchy including Regular and Context-Free Languages.

Overview of Turing machine components including tape, read-write head, and input string structure.

Explanation of states and transitions in Turing machines, detailing how reads, writes, and moves function.

Characteristics of deterministic Turing machines, including the absence of lambda transitions and halting conditions.

Definition of accepting states and conditions for accepting input strings and the process of acceptance.

Example of Turing machine acceptance of a specific language and how it operates over input.

Examples of rejection scenarios for Turing machines and the conditions under which they halt.

Discussion of infinite loops in Turing machines, consequences of not reaching an accepting state.

Computation of functions using Turing machines, defining what makes a function computable.

Detailed examples of Turing machines that compute functions including addition and comparison.

Examining the combined operations of Turing machines, including block diagrams and function comparisons.

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Fall 2006 Costas Busch - RPI 1 Turing Machines
Fall 2006 Costas Busch - RPI 2 The Language Hierarchy *a Regular Languages Context-Free Languages nn ba R ww nnn cba ww? **ba ?
Fall 2006 Costas Busch - RPI 3 *a Regular Languages Context-Free Languages nn ba R ww nnn cba ww **ba Languages accepted by Turing Machines
Fall 2006 Costas Busch - RPI 4 A Turing Machine ............ Tape Read-Write head Control Unit
Fall 2006 Costas Busch - RPI 5 The Tape ............ Read-Write head No boundaries -- infinite length The head moves Left or Right
Fall 2006 Costas Busch - RPI 6 ............ Read-Write head The head at each transition (time step): 1. Reads a symbol 2. Writes a symbol 3. Moves Left or Right
Fall 2006 Costas Busch - RPI 7 ............ Example: Time 0 ............ Time 1 1. Reads 2. Writes a a cb a b k c a k 3. Moves Left
Fall 2006 Costas Busch - RPI 8 ............ Time 1 a b k c ............ Time 2 a k cf 1. Reads 2. Writes b f 3. Moves Right
Fall 2006 Costas Busch - RPI 9 The Input String ............ ◊ ◊ ◊ ◊ Blank symbol head ◊a b ca Head starts at the leftmost position of the input string Input string
Fall 2006 Costas Busch - RPI 10 States & Transitions 1q 2qLba ,→ Read Write Move Left 1q 2qRba ,→ Move Right
Fall 2006 Costas Busch - RPI 11 Example: 1q 2qRba ,→ ............ ◊ ◊ ◊ ◊◊a b ca Time 1 1q current state
Fall 2006 Costas Busch - RPI 12 ............ ◊ ◊ ◊ ◊◊a b ca Time 1 1q 2qRba ,→ ............ ◊ ◊ ◊ ◊◊a b cb Time 2 1q 2q
Fall 2006 Costas Busch - RPI 13 ............ ◊ ◊ ◊ ◊◊a b ca Time 1 1q 2qLba ,→ ............ ◊ ◊ ◊ ◊◊a b cb Time 2 1q 2q Example:
Fall 2006 Costas Busch - RPI 14 ............ ◊ ◊ ◊ ◊◊a b ca Time 1 1q 2qRg,→◊ ............ ◊ ◊ ◊ ◊ga b cb Time 2 1q 2q Example:
Fall 2006 Costas Busch - RPI 15 Determinism 1q 2qRba ,→ Allowed Not Allowed 3qLdb ,→ 1q 2qRba ,→ 3qLda ,→ No lambda transitions allowed Turing Machines are deterministic
Fall 2006 Costas Busch - RPI 16 Partial Transition Function 1q 2qRba ,→ 3qLdb ,→ ............ ◊ ◊ ◊ ◊◊a b ca 1q Example: No transition for input symbol c Allowed:
Fall 2006 Costas Busch - RPI 17 Halting The machine halts in a state if there is no transition to follow
Fall 2006 Costas Busch - RPI 18 Halting Example 1: ............ ◊ ◊ ◊ ◊◊a b ca 1q 1q No transition from HALT!!! 1q
Fall 2006 Costas Busch - RPI 19 Halting Example 2: ............ ◊ ◊ ◊ ◊◊a b ca 1q 1q 2qRba ,→ 3qLdb ,→ No possible transition from and symbol HALT!!! 1q c
Fall 2006 Costas Busch - RPI 20 Accepting States 1q 2q Allowed 1q 2q Not Allowed •Accepting states have no outgoing transitions •The machine halts and accepts
Fall 2006 Costas Busch - RPI 21 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
Fall 2006 Costas Busch - RPI 22 Observation: In order to accept an input string, it is not necessary to scan all the symbols in the string
Fall 2006 Costas Busch - RPI 23 Turing Machine Example Accepts the language: *a 0q Raa ,→ L,◊→◊ 1q Input alphabet },{ ba=Σ
Fall 2006 Costas Busch - RPI 24 ◊ ◊ ◊ ◊aaTime 0 0q a 0q Raa ,→ L,◊→◊ 1q
Fall 2006 Costas Busch - RPI 25 ◊ ◊ ◊ ◊aaTime 1 0q a 0q Raa ,→ L,◊→◊ 1q
Fall 2006 Costas Busch - RPI 26 ◊ ◊ ◊ ◊aaTime 2 0q a 0q Raa ,→ L,◊→◊ 1q
Fall 2006 Costas Busch - RPI 27 ◊ ◊ ◊ ◊aaTime 3 0q a 0q Raa ,→ L,◊→◊ 1q
Fall 2006 Costas Busch - RPI 28 ◊ ◊ ◊ ◊aaTime 4 1q a 0q Raa ,→ L,◊→◊ 1q Halt & Accept
Fall 2006 Costas Busch - RPI 29 Rejection Example 0q Raa ,→ L,◊→◊ 1q ◊ ◊ ◊ ◊baTime 0 0q a
Fall 2006 Costas Busch - RPI 30 0q Raa ,→ L,◊→◊ 1q ◊ ◊ ◊ ◊baTime 1 0q a No possible Transition Halt & Reject
Fall 2006 Costas Busch - RPI 31 Accepts the language: *a 0q but for input alphabet }{a=Σ A simpler machine for same language
Fall 2006 Costas Busch - RPI 32 ◊ ◊ ◊ ◊aaTime 0 0q a 0q Halt & Accept Not necessary to scan input
Fall 2006 Costas Busch - RPI 33 Infinite Loop Example 0q Raa ,→ L,◊→◊ 1q Lbb ,→ A Turing machine for language *)(* baba ++
Fall 2006 Costas Busch - RPI 34 ◊ ◊ ◊ ◊baTime 0 0q a 0q Raa ,→ L,◊→◊ 1q Lbb ,→
Fall 2006 Costas Busch - RPI 35 ◊ ◊ ◊ ◊baTime 1 0q a 0q Raa ,→ L,◊→◊ 1q Lbb ,→
Fall 2006 Costas Busch - RPI 36 ◊ ◊ ◊ ◊baTime 2 0q a 0q Raa ,→ L,◊→◊ 1q Lbb ,→
Fall 2006 Costas Busch - RPI 37 ◊ ◊ ◊ ◊baTime 2 0q a ◊ ◊ ◊ ◊baTime 3 0q a ◊ ◊ ◊ ◊baTime 4 0q a ◊ ◊ ◊ ◊baTime 5 0q a Infiniteloop
Fall 2006 Costas Busch - RPI 38 Because of the infinite loop: •The accepting state cannot be reached •The machine never halts •The input string is rejected
Fall 2006 Costas Busch - RPI 39 Another Turing Machine Example Turing machine for the language }{ nn ba 0q 1q 2q3q Rxa ,→ Raa ,→ Ryy ,→ Lyb ,→ Laa ,→ Lyy ,→ Rxx ,→ Ryy ,→ Ryy ,→ 4q L,◊→◊ 1≥n
Fall 2006 Costas Busch - RPI 40 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:
Fall 2006 Costas Busch - RPI 41 0q 1q 2q3q Rxa ,→ Raa ,→ Ryy ,→ Lyb ,→ Laa ,→ Lyy ,→ Rxx ,→ Ryy ,→ Ryy ,→ 4q L,◊→◊ ◊ ◊ba 0q a bTime 0 ◊
Fall 2006 Costas Busch - RPI 42 0q 1q 2q3q Rxa ,→ Raa ,→ Ryy ,→ Lyb ,→ Laa ,→ Lyy ,→ Rxx ,→ Ryy ,→ Ryy ,→ 4q L,◊→◊ ◊ ◊bx 1q a b ◊Time 1
Fall 2006 Costas Busch - RPI 43 0q 1q 2q3q Rxa ,→ Raa ,→ Ryy ,→ Lyb ,→ Laa ,→ Lyy ,→ Rxx ,→ Ryy ,→ Ryy ,→ 4q L,◊→◊ ◊ ◊bx 1q a b ◊Time 2
Fall 2006 Costas Busch - RPI 44 0q 1q 2q3q Rxa ,→ Raa ,→ Ryy ,→ Lyb ,→ Laa ,→ Lyy ,→ Rxx ,→ Ryy ,→ Ryy ,→ 4q L,◊→◊ ◊ ◊yx 2q a b ◊Time 3
Fall 2006 Costas Busch - RPI 45 0q 1q 2q3q Rxa ,→ Raa ,→ Ryy ,→ Lyb ,→ Laa ,→ Lyy ,→ Rxx ,→ Ryy ,→ Ryy ,→ 4q L,◊→◊ ◊ ◊yx 2q a b ◊Time 4
Fall 2006 Costas Busch - RPI 46 0q 1q 2q3q Rxa ,→ Raa ,→ Ryy ,→ Lyb ,→ Laa ,→ Lyy ,→ Rxx ,→ Ryy ,→ Ryy ,→ 4q L,◊→◊ ◊ ◊yx 0q a b ◊Time 5
Fall 2006 Costas Busch - RPI 47 0q 1q 2q3q Rxa ,→ Raa ,→ Ryy ,→ Lyb ,→ Laa ,→ Lyy ,→ Rxx ,→ Ryy ,→ Ryy ,→ 4q L,◊→◊ ◊ ◊yx 1q x b ◊Time 6
Fall 2006 Costas Busch - RPI 48 0q 1q 2q3q Rxa ,→ Raa ,→ Ryy ,→ Lyb ,→ Laa ,→ Lyy ,→ Rxx ,→ Ryy ,→ Ryy ,→ 4q L,◊→◊ ◊ ◊yx 1q x b ◊Time 7
Fall 2006 Costas Busch - RPI 49 0q 1q 2q3q Rxa ,→ Raa ,→ Ryy ,→ Lyb ,→ Laa ,→ Lyy ,→ Rxx ,→ Ryy ,→ Ryy ,→ 4q L,◊→◊ ◊ ◊yx x y 2q ◊Time 8
Fall 2006 Costas Busch - RPI 50 0q 1q 2q3q Rxa ,→ Raa ,→ Ryy ,→ Lyb ,→ Laa ,→ Lyy ,→ Rxx ,→ Ryy ,→ Ryy ,→ 4q L,◊→◊ ◊ ◊yx x y 2q ◊Time 9
Fall 2006 Costas Busch - RPI 51 0q 1q 2q3q Rxa ,→ Raa ,→ Ryy ,→ Lyb ,→ Laa ,→ Lyy ,→ Rxx ,→ Ryy ,→ Ryy ,→ 4q L,◊→◊ ◊ ◊yx 0q x y ◊Time 10
Fall 2006 Costas Busch - RPI 52 0q 1q 2q3q Rxa ,→ Raa ,→ Ryy ,→ Lyb ,→ Laa ,→ Lyy ,→ Rxx ,→ Ryy ,→ Ryy ,→ 4q L,◊→◊ ◊ ◊yx 3q x y ◊Time 11
Fall 2006 Costas Busch - RPI 53 0q 1q 2q3q Rxa ,→ Raa ,→ Ryy ,→ Lyb ,→ Laa ,→ Lyy ,→ Rxx ,→ Ryy ,→ Ryy ,→ 4q L,◊→◊ ◊ ◊yx 3q x y ◊Time 12
Fall 2006 Costas Busch - RPI 54 0q 1q 2q3q Rxa ,→ Raa ,→ Ryy ,→ Lyb ,→ Laa ,→ Lyy ,→ Rxx ,→ Ryy ,→ Ryy ,→ 4q L,◊→◊ ◊ ◊yx 4q x y ◊ Halt & Accept Time 13
Fall 2006 Costas Busch - RPI 55 If we modify the machine for the language }{ nn ba we can easily construct a machine for the language }{ nnn cba Observation:
Fall 2006 Costas Busch - RPI 56 Formal Definitions for Turing Machines
Fall 2006 Costas Busch - RPI 57 Transition Function 1q 2qRba ,→ ),,(),( 21 Rbqaq =δ
Fall 2006 Costas Busch - RPI 58 1q 2qLdc ,→ ),,(),( 21 Ldqcq =δ Transition Function
Fall 2006 Costas Busch - RPI 59 Turing Machine: ),,,,,,( 0 FqQM ◊ΓΣ= δ States Input alphabet Tape alphabet Transition function Initial state blank Accept states
Fall 2006 Costas Busch - RPI 60 Configuration ◊ ◊ ◊ba 1q a Instantaneous description: c◊ baqca 1
Fall 2006 Costas Busch - RPI 61 ◊ ◊yx 2q a b Time 4 ◊ ◊yx 0q a b Time 5 ◊ ◊ A Move: aybqxxaybq 02  (yields in one mode)
Fall 2006 Costas Busch - RPI 62 ◊ ◊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
Fall 2006 Costas Busch - RPI 63 bqxxyybqxxaybqxxaybq 1102  bqxxyxaybq 12 ∗ Equivalent notation:
Fall 2006 Costas Busch - RPI 64 Initial configuration: wq0 ◊ ◊ba 0q a b ◊ w Input string
Fall 2006 Costas Busch - RPI 65 The Accepted Language For any Turing Machine M }:{)( 210 xqxwqwML f ∗ =  Initial state Accept state
Fall 2006 Costas Busch - RPI 66 If a language is accepted by a Turing machine then we say that is: •Turing Acceptable •Recursively Enumerable M L L •Turing Recognizable Other names used:
Fall 2006 Costas Busch - RPI 67 Computing Functions with Turing Machines
Fall 2006 Costas Busch - RPI 68 A function )(wf Domain: Result Region: has: D Dw∈ S Swf ∈)( )(wf
Fall 2006 Costas Busch - RPI 69 A function may have many parameters: yxyxf +=),( Example: Addition function
Fall 2006 Costas Busch - RPI 70 Integer Domain Unary: Binary: Decimal: 11111 101 5 We prefer unary representation: easier to manipulate with Turing machines
Fall 2006 Costas Busch - RPI 71 Definition: A function is computable if there is a Turing Machine such that: f M Initial configuration Final configuration Dw∈ Domain ◊ 0q ◊w ◊ fq ◊)(wf accept stateinitial state For all
Fall 2006 Costas Busch - RPI 72 )(0 wfqwq f ∗  Initial Configuration Final Configuration A function is computable if there is a Turing Machine such that: f M In other words: Dw∈ DomainFor all
Fall 2006 Costas Busch - RPI 73 Example The function yxyxf +=),( is computable Turing Machine: Input string: yx0 unary Output string: 0xy unary yx, are integers
Fall 2006 Costas Busch - RPI 74 ◊ 0 0q 1 1 ◊1 1 x y 1 Start initial state The 0 is the delimiter that separates the two numbers
Fall 2006 Costas Busch - RPI 75 ◊ 0 0q 1 1 ◊1 1 x y 1  ◊ 0 fq 1 ◊1 yx +  11 Start Finish final state initial state
Fall 2006 Costas Busch - RPI 76 ◊ 0 fq 1 ◊1 yx +  11Finish final state The 0 here helps when we use the result for other operations
Fall 2006 Costas Busch - RPI 77 0q Turing machine for function 1q 2q 3qL,◊→◊ L,01→ L,11→ R,◊→◊ R,10 → R,11→ 4q R,11→ yxyxf +=),(
Fall 2006 Costas Busch - RPI 78 Execution Example: 11=x 11=y ◊ 0 0q 1 1 ◊1 1 Time 0 x y Final Result ◊ 0 4q 1 1 ◊1 1 yx + (=2) (=2)
Fall 2006 Costas Busch - RPI 79 ◊ 0 0q 1 1Time 0 ◊ 0q 1q 2q 3qL,◊→◊ L,01→ L,11→ R,◊→◊ R,10 → R,11→ 4q R,11→ 1 1
Fall 2006 Costas Busch - RPI 80 ◊ 0q ◊ 0q 1q 2q 3qL,◊→◊ L,01→ L,11→ R,◊→◊ R,10 → R,11→ 4q R,11→ 01 11 1Time 1
Fall 2006 Costas Busch - RPI 81 0q 1q 2q 3qL,◊→◊ L,01→ L,11→ R,◊→◊ R,10 → R,11→ 4q R,11→ ◊ 0 0q 1 1 ◊1 1Time 2
Fall 2006 Costas Busch - RPI 82 0q 1q 2q 3qL,◊→◊ L,01→ L,11→ R,◊→◊ R,10 → R,11→ 4q R,11→ ◊ 1q ◊1 11 11Time 3
Fall 2006 Costas Busch - RPI 83 0q 1q 2q 3qL,◊→◊ L,01→ L,11→ R,◊→◊ R,10 → R,11→ 4q R,11→ ◊ 1q 1 1 ◊1 11Time 4
Fall 2006 Costas Busch - RPI 84 0q 1q 2q 3qL,◊→◊ L,01→ L,11→ R,◊→◊ R,10 → R,11→ 4q R,11→ ◊ 1q ◊1 11 11Time 5
Fall 2006 Costas Busch - RPI 85 0q 1q 2q 3qL,◊→◊ L,01→ L,11→ R,◊→◊ R,10 → R,11→ 4q R,11→ ◊ 2q 1 1 ◊1 11Time 6
Fall 2006 Costas Busch - RPI 86 0q 1q 2q 3qL,◊→◊ L,01→ L,11→ R,◊→◊ R,10 → R,11→ 4q R,11→ ◊ 3q ◊1 11 01Time 7
Fall 2006 Costas Busch - RPI 87 0q 1q 2q 3qL,◊→◊ L,01→ L,11→ R,◊→◊ R,10 → R,11→ 4q R,11→ ◊ 3q 1 1 ◊1 01Time 8
Fall 2006 Costas Busch - RPI 88 0q 1q 2q 3qL,◊→◊ L,01→ L,11→ R,◊→◊ R,10 → R,11→ 4q R,11→ ◊ 3q ◊1 11 01Time 9
Fall 2006 Costas Busch - RPI 89 0q 1q 2q 3qL,◊→◊ L,01→ L,11→ R,◊→◊ R,10 → R,11→ 4q R,11→ ◊ 3q 1 1 ◊1 01Time 10
Fall 2006 Costas Busch - RPI 90 0q 1q 2q 3qL,◊→◊ L,01→ L,11→ R,◊→◊ R,10 → R,11→ 4q R,11→ ◊ 3q ◊1 11 01Time 11
Fall 2006 Costas Busch - RPI 91 0q 1q 2q 3qL,◊→◊ L,01→ L,11→ R,◊→◊ R,10 → R,11→ 4q R,11→ ◊ 4q 1 1 ◊1 01 HALT & accept Time 12
Fall 2006 Costas Busch - RPI 92 Another Example The function xxf 2)( = is computable Turing Machine: Input string: x unary Output string: xx unary x is integer
Fall 2006 Costas Busch - RPI 93 ◊ 0q 1 ◊1 x 1 ◊ 1 fq 1 ◊1 x2  11 Start Finish accept state initial state
Fall 2006 Costas Busch - RPI 94 Turing Machine Pseudocode for xxf 2)( = • Replace every 1 with $ • Repeat: • Find rightmost $, replace it with 1 • Go to right end, insert 1 Until no more $ remain
Fall 2006 Costas Busch - RPI 95 0q 1q 2q 3q R,1$ → L,1→◊ L,◊→◊ R$,1→ L,11→ R,11→ R,◊→◊ Turing Machine for xxf 2)( =
Fall 2006 Costas Busch - RPI 96 0q 1q 2q 3q R,1$ → L,1→◊ L,◊→◊ R$,1→ L,11→ R,11→ R,◊→◊ Example ◊ 0q 1 ◊1 ◊ 3q ◊1 11 1 Start Finish
Fall 2006 Costas Busch - RPI 97 Another Example The function =),( yxf is computable 0 1 yx > yx ≤ if if Input: yx0 Output: 1 0or
Fall 2006 Costas Busch - RPI 98 Turing Machine Pseudocode: Match a 1 from with a 1 fromx y • Repeat Until all of or is matchedx y • If a 1 from is not matched erase tape, write 1 else erase tape, write 0 x )( yx > )( yx ≤
Fall 2006 Costas Busch - RPI 99 Combining Turing Machines
Fall 2006 Costas Busch - RPI 100 Block Diagram Turing Machine input output
Fall 2006 Costas Busch - RPI 101 Example: =),( yxf 0 yx + yx > yx ≤ if if Comparator Adder Eraser yx, yx, yx > yx ≤ yx + 0

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Turing Machine

  • 1.
    Fall 2006 CostasBusch - RPI 1 Turing Machines
  • 2.
    Fall 2006 CostasBusch - RPI 2 The Language Hierarchy *a Regular Languages Context-Free Languages nn ba R ww nnn cba ww? **ba ?
  • 3.
    Fall 2006 CostasBusch - RPI 3 *a Regular Languages Context-Free Languages nn ba R ww nnn cba ww **ba Languages accepted by Turing Machines
  • 4.
    Fall 2006 CostasBusch - RPI 4 A Turing Machine ............ Tape Read-Write head Control Unit
  • 5.
    Fall 2006 CostasBusch - RPI 5 The Tape ............ Read-Write head No boundaries -- infinite length The head moves Left or Right
  • 6.
    Fall 2006 CostasBusch - RPI 6 ............ Read-Write head The head at each transition (time step): 1. Reads a symbol 2. Writes a symbol 3. Moves Left or Right
  • 7.
    Fall 2006 CostasBusch - RPI 7 ............ Example: Time 0 ............ Time 1 1. Reads 2. Writes a a cb a b k c a k 3. Moves Left
  • 8.
    Fall 2006 CostasBusch - RPI 8 ............ Time 1 a b k c ............ Time 2 a k cf 1. Reads 2. Writes b f 3. Moves Right
  • 9.
    Fall 2006 CostasBusch - RPI 9 The Input String ............ ◊ ◊ ◊ ◊ Blank symbol head ◊a b ca Head starts at the leftmost position of the input string Input string
  • 10.
    Fall 2006 CostasBusch - RPI 10 States & Transitions 1q 2qLba ,→ Read Write Move Left 1q 2qRba ,→ Move Right
  • 11.
    Fall 2006 CostasBusch - RPI 11 Example: 1q 2qRba ,→ ............ ◊ ◊ ◊ ◊◊a b ca Time 1 1q current state
  • 12.
    Fall 2006 CostasBusch - RPI 12 ............ ◊ ◊ ◊ ◊◊a b ca Time 1 1q 2qRba ,→ ............ ◊ ◊ ◊ ◊◊a b cb Time 2 1q 2q
  • 13.
    Fall 2006 CostasBusch - RPI 13 ............ ◊ ◊ ◊ ◊◊a b ca Time 1 1q 2qLba ,→ ............ ◊ ◊ ◊ ◊◊a b cb Time 2 1q 2q Example:
  • 14.
    Fall 2006 CostasBusch - RPI 14 ............ ◊ ◊ ◊ ◊◊a b ca Time 1 1q 2qRg,→◊ ............ ◊ ◊ ◊ ◊ga b cb Time 2 1q 2q Example:
  • 15.
    Fall 2006 CostasBusch - RPI 15 Determinism 1q 2qRba ,→ Allowed Not Allowed 3qLdb ,→ 1q 2qRba ,→ 3qLda ,→ No lambda transitions allowed Turing Machines are deterministic
  • 16.
    Fall 2006 CostasBusch - RPI 16 Partial Transition Function 1q 2qRba ,→ 3qLdb ,→ ............ ◊ ◊ ◊ ◊◊a b ca 1q Example: No transition for input symbol c Allowed:
  • 17.
    Fall 2006 CostasBusch - RPI 17 Halting The machine halts in a state if there is no transition to follow
  • 18.
    Fall 2006 CostasBusch - RPI 18 Halting Example 1: ............ ◊ ◊ ◊ ◊◊a b ca 1q 1q No transition from HALT!!! 1q
  • 19.
    Fall 2006 CostasBusch - RPI 19 Halting Example 2: ............ ◊ ◊ ◊ ◊◊a b ca 1q 1q 2qRba ,→ 3qLdb ,→ No possible transition from and symbol HALT!!! 1q c
  • 20.
    Fall 2006 CostasBusch - RPI 20 Accepting States 1q 2q Allowed 1q 2q Not Allowed •Accepting states have no outgoing transitions •The machine halts and accepts
  • 21.
    Fall 2006 CostasBusch - RPI 21 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
  • 22.
    Fall 2006 CostasBusch - RPI 22 Observation: In order to accept an input string, it is not necessary to scan all the symbols in the string
  • 23.
    Fall 2006 CostasBusch - RPI 23 Turing Machine Example Accepts the language: *a 0q Raa ,→ L,◊→◊ 1q Input alphabet },{ ba=Σ
  • 24.
    Fall 2006 CostasBusch - RPI 24 ◊ ◊ ◊ ◊aaTime 0 0q a 0q Raa ,→ L,◊→◊ 1q
  • 25.
    Fall 2006 CostasBusch - RPI 25 ◊ ◊ ◊ ◊aaTime 1 0q a 0q Raa ,→ L,◊→◊ 1q
  • 26.
    Fall 2006 CostasBusch - RPI 26 ◊ ◊ ◊ ◊aaTime 2 0q a 0q Raa ,→ L,◊→◊ 1q
  • 27.
    Fall 2006 CostasBusch - RPI 27 ◊ ◊ ◊ ◊aaTime 3 0q a 0q Raa ,→ L,◊→◊ 1q
  • 28.
    Fall 2006 CostasBusch - RPI 28 ◊ ◊ ◊ ◊aaTime 4 1q a 0q Raa ,→ L,◊→◊ 1q Halt & Accept
  • 29.
    Fall 2006 CostasBusch - RPI 29 Rejection Example 0q Raa ,→ L,◊→◊ 1q ◊ ◊ ◊ ◊baTime 0 0q a
  • 30.
    Fall 2006 CostasBusch - RPI 30 0q Raa ,→ L,◊→◊ 1q ◊ ◊ ◊ ◊baTime 1 0q a No possible Transition Halt & Reject
  • 31.
    Fall 2006 CostasBusch - RPI 31 Accepts the language: *a 0q but for input alphabet }{a=Σ A simpler machine for same language
  • 32.
    Fall 2006 CostasBusch - RPI 32 ◊ ◊ ◊ ◊aaTime 0 0q a 0q Halt & Accept Not necessary to scan input
  • 33.
    Fall 2006 CostasBusch - RPI 33 Infinite Loop Example 0q Raa ,→ L,◊→◊ 1q Lbb ,→ A Turing machine for language *)(* baba ++
  • 34.
    Fall 2006 CostasBusch - RPI 34 ◊ ◊ ◊ ◊baTime 0 0q a 0q Raa ,→ L,◊→◊ 1q Lbb ,→
  • 35.
    Fall 2006 CostasBusch - RPI 35 ◊ ◊ ◊ ◊baTime 1 0q a 0q Raa ,→ L,◊→◊ 1q Lbb ,→
  • 36.
    Fall 2006 CostasBusch - RPI 36 ◊ ◊ ◊ ◊baTime 2 0q a 0q Raa ,→ L,◊→◊ 1q Lbb ,→
  • 37.
    Fall 2006 CostasBusch - RPI 37 ◊ ◊ ◊ ◊baTime 2 0q a ◊ ◊ ◊ ◊baTime 3 0q a ◊ ◊ ◊ ◊baTime 4 0q a ◊ ◊ ◊ ◊baTime 5 0q a Infiniteloop
  • 38.
    Fall 2006 CostasBusch - RPI 38 Because of the infinite loop: •The accepting state cannot be reached •The machine never halts •The input string is rejected
  • 39.
    Fall 2006 CostasBusch - RPI 39 Another Turing Machine Example Turing machine for the language }{ nn ba 0q 1q 2q3q Rxa ,→ Raa ,→ Ryy ,→ Lyb ,→ Laa ,→ Lyy ,→ Rxx ,→ Ryy ,→ Ryy ,→ 4q L,◊→◊ 1≥n
  • 40.
    Fall 2006 CostasBusch - RPI 40 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:
  • 41.
    Fall 2006 CostasBusch - RPI 41 0q 1q 2q3q Rxa ,→ Raa ,→ Ryy ,→ Lyb ,→ Laa ,→ Lyy ,→ Rxx ,→ Ryy ,→ Ryy ,→ 4q L,◊→◊ ◊ ◊ba 0q a bTime 0 ◊
  • 42.
    Fall 2006 CostasBusch - RPI 42 0q 1q 2q3q Rxa ,→ Raa ,→ Ryy ,→ Lyb ,→ Laa ,→ Lyy ,→ Rxx ,→ Ryy ,→ Ryy ,→ 4q L,◊→◊ ◊ ◊bx 1q a b ◊Time 1
  • 43.
    Fall 2006 CostasBusch - RPI 43 0q 1q 2q3q Rxa ,→ Raa ,→ Ryy ,→ Lyb ,→ Laa ,→ Lyy ,→ Rxx ,→ Ryy ,→ Ryy ,→ 4q L,◊→◊ ◊ ◊bx 1q a b ◊Time 2
  • 44.
    Fall 2006 CostasBusch - RPI 44 0q 1q 2q3q Rxa ,→ Raa ,→ Ryy ,→ Lyb ,→ Laa ,→ Lyy ,→ Rxx ,→ Ryy ,→ Ryy ,→ 4q L,◊→◊ ◊ ◊yx 2q a b ◊Time 3
  • 45.
    Fall 2006 CostasBusch - RPI 45 0q 1q 2q3q Rxa ,→ Raa ,→ Ryy ,→ Lyb ,→ Laa ,→ Lyy ,→ Rxx ,→ Ryy ,→ Ryy ,→ 4q L,◊→◊ ◊ ◊yx 2q a b ◊Time 4
  • 46.
    Fall 2006 CostasBusch - RPI 46 0q 1q 2q3q Rxa ,→ Raa ,→ Ryy ,→ Lyb ,→ Laa ,→ Lyy ,→ Rxx ,→ Ryy ,→ Ryy ,→ 4q L,◊→◊ ◊ ◊yx 0q a b ◊Time 5
  • 47.
    Fall 2006 CostasBusch - RPI 47 0q 1q 2q3q Rxa ,→ Raa ,→ Ryy ,→ Lyb ,→ Laa ,→ Lyy ,→ Rxx ,→ Ryy ,→ Ryy ,→ 4q L,◊→◊ ◊ ◊yx 1q x b ◊Time 6
  • 48.
    Fall 2006 CostasBusch - RPI 48 0q 1q 2q3q Rxa ,→ Raa ,→ Ryy ,→ Lyb ,→ Laa ,→ Lyy ,→ Rxx ,→ Ryy ,→ Ryy ,→ 4q L,◊→◊ ◊ ◊yx 1q x b ◊Time 7
  • 49.
    Fall 2006 CostasBusch - RPI 49 0q 1q 2q3q Rxa ,→ Raa ,→ Ryy ,→ Lyb ,→ Laa ,→ Lyy ,→ Rxx ,→ Ryy ,→ Ryy ,→ 4q L,◊→◊ ◊ ◊yx x y 2q ◊Time 8
  • 50.
    Fall 2006 CostasBusch - RPI 50 0q 1q 2q3q Rxa ,→ Raa ,→ Ryy ,→ Lyb ,→ Laa ,→ Lyy ,→ Rxx ,→ Ryy ,→ Ryy ,→ 4q L,◊→◊ ◊ ◊yx x y 2q ◊Time 9
  • 51.
    Fall 2006 CostasBusch - RPI 51 0q 1q 2q3q Rxa ,→ Raa ,→ Ryy ,→ Lyb ,→ Laa ,→ Lyy ,→ Rxx ,→ Ryy ,→ Ryy ,→ 4q L,◊→◊ ◊ ◊yx 0q x y ◊Time 10
  • 52.
    Fall 2006 CostasBusch - RPI 52 0q 1q 2q3q Rxa ,→ Raa ,→ Ryy ,→ Lyb ,→ Laa ,→ Lyy ,→ Rxx ,→ Ryy ,→ Ryy ,→ 4q L,◊→◊ ◊ ◊yx 3q x y ◊Time 11
  • 53.
    Fall 2006 CostasBusch - RPI 53 0q 1q 2q3q Rxa ,→ Raa ,→ Ryy ,→ Lyb ,→ Laa ,→ Lyy ,→ Rxx ,→ Ryy ,→ Ryy ,→ 4q L,◊→◊ ◊ ◊yx 3q x y ◊Time 12
  • 54.
    Fall 2006 CostasBusch - RPI 54 0q 1q 2q3q Rxa ,→ Raa ,→ Ryy ,→ Lyb ,→ Laa ,→ Lyy ,→ Rxx ,→ Ryy ,→ Ryy ,→ 4q L,◊→◊ ◊ ◊yx 4q x y ◊ Halt & Accept Time 13
  • 55.
    Fall 2006 CostasBusch - RPI 55 If we modify the machine for the language }{ nn ba we can easily construct a machine for the language }{ nnn cba Observation:
  • 56.
    Fall 2006 CostasBusch - RPI 56 Formal Definitions for Turing Machines
  • 57.
    Fall 2006 CostasBusch - RPI 57 Transition Function 1q 2qRba ,→ ),,(),( 21 Rbqaq =δ
  • 58.
    Fall 2006 CostasBusch - RPI 58 1q 2qLdc ,→ ),,(),( 21 Ldqcq =δ Transition Function
  • 59.
    Fall 2006 CostasBusch - RPI 59 Turing Machine: ),,,,,,( 0 FqQM ◊ΓΣ= δ States Input alphabet Tape alphabet Transition function Initial state blank Accept states
  • 60.
    Fall 2006 CostasBusch - RPI 60 Configuration ◊ ◊ ◊ba 1q a Instantaneous description: c◊ baqca 1
  • 61.
    Fall 2006 CostasBusch - RPI 61 ◊ ◊yx 2q a b Time 4 ◊ ◊yx 0q a b Time 5 ◊ ◊ A Move: aybqxxaybq 02  (yields in one mode)
  • 62.
    Fall 2006 CostasBusch - RPI 62 ◊ ◊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
  • 63.
    Fall 2006 CostasBusch - RPI 63 bqxxyybqxxaybqxxaybq 1102  bqxxyxaybq 12 ∗ Equivalent notation:
  • 64.
    Fall 2006 CostasBusch - RPI 64 Initial configuration: wq0 ◊ ◊ba 0q a b ◊ w Input string
  • 65.
    Fall 2006 CostasBusch - RPI 65 The Accepted Language For any Turing Machine M }:{)( 210 xqxwqwML f ∗ =  Initial state Accept state
  • 66.
    Fall 2006 CostasBusch - RPI 66 If a language is accepted by a Turing machine then we say that is: •Turing Acceptable •Recursively Enumerable M L L •Turing Recognizable Other names used:
  • 67.
    Fall 2006 CostasBusch - RPI 67 Computing Functions with Turing Machines
  • 68.
    Fall 2006 CostasBusch - RPI 68 A function )(wf Domain: Result Region: has: D Dw∈ S Swf ∈)( )(wf
  • 69.
    Fall 2006 CostasBusch - RPI 69 A function may have many parameters: yxyxf +=),( Example: Addition function
  • 70.
    Fall 2006 CostasBusch - RPI 70 Integer Domain Unary: Binary: Decimal: 11111 101 5 We prefer unary representation: easier to manipulate with Turing machines
  • 71.
    Fall 2006 CostasBusch - RPI 71 Definition: A function is computable if there is a Turing Machine such that: f M Initial configuration Final configuration Dw∈ Domain ◊ 0q ◊w ◊ fq ◊)(wf accept stateinitial state For all
  • 72.
    Fall 2006 CostasBusch - RPI 72 )(0 wfqwq f ∗  Initial Configuration Final Configuration A function is computable if there is a Turing Machine such that: f M In other words: Dw∈ DomainFor all
  • 73.
    Fall 2006 CostasBusch - RPI 73 Example The function yxyxf +=),( is computable Turing Machine: Input string: yx0 unary Output string: 0xy unary yx, are integers
  • 74.
    Fall 2006 CostasBusch - RPI 74 ◊ 0 0q 1 1 ◊1 1 x y 1 Start initial state The 0 is the delimiter that separates the two numbers
  • 75.
    Fall 2006 CostasBusch - RPI 75 ◊ 0 0q 1 1 ◊1 1 x y 1  ◊ 0 fq 1 ◊1 yx +  11 Start Finish final state initial state
  • 76.
    Fall 2006 CostasBusch - RPI 76 ◊ 0 fq 1 ◊1 yx +  11Finish final state The 0 here helps when we use the result for other operations
  • 77.
    Fall 2006 CostasBusch - RPI 77 0q Turing machine for function 1q 2q 3qL,◊→◊ L,01→ L,11→ R,◊→◊ R,10 → R,11→ 4q R,11→ yxyxf +=),(
  • 78.
    Fall 2006 CostasBusch - RPI 78 Execution Example: 11=x 11=y ◊ 0 0q 1 1 ◊1 1 Time 0 x y Final Result ◊ 0 4q 1 1 ◊1 1 yx + (=2) (=2)
  • 79.
    Fall 2006 CostasBusch - RPI 79 ◊ 0 0q 1 1Time 0 ◊ 0q 1q 2q 3qL,◊→◊ L,01→ L,11→ R,◊→◊ R,10 → R,11→ 4q R,11→ 1 1
  • 80.
    Fall 2006 CostasBusch - RPI 80 ◊ 0q ◊ 0q 1q 2q 3qL,◊→◊ L,01→ L,11→ R,◊→◊ R,10 → R,11→ 4q R,11→ 01 11 1Time 1
  • 81.
    Fall 2006 CostasBusch - RPI 81 0q 1q 2q 3qL,◊→◊ L,01→ L,11→ R,◊→◊ R,10 → R,11→ 4q R,11→ ◊ 0 0q 1 1 ◊1 1Time 2
  • 82.
    Fall 2006 CostasBusch - RPI 82 0q 1q 2q 3qL,◊→◊ L,01→ L,11→ R,◊→◊ R,10 → R,11→ 4q R,11→ ◊ 1q ◊1 11 11Time 3
  • 83.
    Fall 2006 CostasBusch - RPI 83 0q 1q 2q 3qL,◊→◊ L,01→ L,11→ R,◊→◊ R,10 → R,11→ 4q R,11→ ◊ 1q 1 1 ◊1 11Time 4
  • 84.
    Fall 2006 CostasBusch - RPI 84 0q 1q 2q 3qL,◊→◊ L,01→ L,11→ R,◊→◊ R,10 → R,11→ 4q R,11→ ◊ 1q ◊1 11 11Time 5
  • 85.
    Fall 2006 CostasBusch - RPI 85 0q 1q 2q 3qL,◊→◊ L,01→ L,11→ R,◊→◊ R,10 → R,11→ 4q R,11→ ◊ 2q 1 1 ◊1 11Time 6
  • 86.
    Fall 2006 CostasBusch - RPI 86 0q 1q 2q 3qL,◊→◊ L,01→ L,11→ R,◊→◊ R,10 → R,11→ 4q R,11→ ◊ 3q ◊1 11 01Time 7
  • 87.
    Fall 2006 CostasBusch - RPI 87 0q 1q 2q 3qL,◊→◊ L,01→ L,11→ R,◊→◊ R,10 → R,11→ 4q R,11→ ◊ 3q 1 1 ◊1 01Time 8
  • 88.
    Fall 2006 CostasBusch - RPI 88 0q 1q 2q 3qL,◊→◊ L,01→ L,11→ R,◊→◊ R,10 → R,11→ 4q R,11→ ◊ 3q ◊1 11 01Time 9
  • 89.
    Fall 2006 CostasBusch - RPI 89 0q 1q 2q 3qL,◊→◊ L,01→ L,11→ R,◊→◊ R,10 → R,11→ 4q R,11→ ◊ 3q 1 1 ◊1 01Time 10
  • 90.
    Fall 2006 CostasBusch - RPI 90 0q 1q 2q 3qL,◊→◊ L,01→ L,11→ R,◊→◊ R,10 → R,11→ 4q R,11→ ◊ 3q ◊1 11 01Time 11
  • 91.
    Fall 2006 CostasBusch - RPI 91 0q 1q 2q 3qL,◊→◊ L,01→ L,11→ R,◊→◊ R,10 → R,11→ 4q R,11→ ◊ 4q 1 1 ◊1 01 HALT & accept Time 12
  • 92.
    Fall 2006 CostasBusch - RPI 92 Another Example The function xxf 2)( = is computable Turing Machine: Input string: x unary Output string: xx unary x is integer
  • 93.
    Fall 2006 CostasBusch - RPI 93 ◊ 0q 1 ◊1 x 1 ◊ 1 fq 1 ◊1 x2  11 Start Finish accept state initial state
  • 94.
    Fall 2006 CostasBusch - RPI 94 Turing Machine Pseudocode for xxf 2)( = • Replace every 1 with $ • Repeat: • Find rightmost $, replace it with 1 • Go to right end, insert 1 Until no more $ remain
  • 95.
    Fall 2006 CostasBusch - RPI 95 0q 1q 2q 3q R,1$ → L,1→◊ L,◊→◊ R$,1→ L,11→ R,11→ R,◊→◊ Turing Machine for xxf 2)( =
  • 96.
    Fall 2006 CostasBusch - RPI 96 0q 1q 2q 3q R,1$ → L,1→◊ L,◊→◊ R$,1→ L,11→ R,11→ R,◊→◊ Example ◊ 0q 1 ◊1 ◊ 3q ◊1 11 1 Start Finish
  • 97.
    Fall 2006 CostasBusch - RPI 97 Another Example The function =),( yxf is computable 0 1 yx > yx ≤ if if Input: yx0 Output: 1 0or
  • 98.
    Fall 2006 CostasBusch - RPI 98 Turing Machine Pseudocode: Match a 1 from with a 1 fromx y • Repeat Until all of or is matchedx y • If a 1 from is not matched erase tape, write 1 else erase tape, write 0 x )( yx > )( yx ≤
  • 99.
    Fall 2006 CostasBusch - RPI 99 Combining Turing Machines
  • 100.
    Fall 2006 CostasBusch - RPI 100 Block Diagram Turing Machine input output
  • 101.
    Fall 2006 CostasBusch - RPI 101 Example: =),( yxf 0 yx + yx > yx ≤ if if Comparator Adder Eraser yx, yx, yx > yx ≤ yx + 0