1 Turing Machines and Equivalent Models Section 13.1 Turing Machines.

1

Turing Machines and Equivalent Models

Section 13.1 Turing Machines

2

Section 13.1 Turing MachinesA Turing machine (TM) is a simple computer

that has an infinite amount of storage in the form of cells on an infinite tape. There is a control unit that contains a finite set of state transition instructions.

3

Operations

• An instruction reads the symbol in the current tape cell, writes a symbol to the same cell,

• and then moves the tape head one cell left or right or remains at the same cell, after which the machine enters a new state.

4

Assumptions

• The symbol Λ denotes a blank cell.• The tape head is initially at the left end of

a nonempty input string (unless specifiedotherwise)

• There is a designated start state.• There is one “halt” state.• The moves are denoted by L, S, R for

move left, stay, and move right.

5

Example

• Instructions are represented in graphical form or as 5-tuples, as shown in the example.

• This instruction executes if the current state of the TM is i and the current input symbol is a. The TM writes b into that cell, moves right one cell and enters state j.

6

Acceptance• An input string on the tape is accepted by

the TM if the machine enters the halt state.• The language of a TM is the set of all

strings accepted by the machine.

7

Example/Quiz• Given the following simple TM.

(0, a, a, R, halt).What is the language of the TM?

• Answer: If an input string begins with the letter a, then the TM executes the instruction andhalts. So the string is accepted. Therefore, the language of the TM is the set of all strings thatbegin with the letter a.

8

Example• Construct a TM to accept the language {abn | n ∊ N}.

• Solution. Let the start state be state 0. The idea is to see whether a is in the current cell.If so, move right to scan b’s until Λ is found.(0, a, a, R, 1)(1, Λ, Λ, S, halt)(1, b, b, R, 1).

9

Quiz• Construct a TM to accept the language {abna | n ∊ N}.

• Solution. Let the start state be state 0. The idea is to see whether a is in the current cell.If so, move right to scan b’s until a is found. Then make sure there is Λ to the right.(0, a, a, R, 1)(1, a, a, R, 2)(1, b, b, R, 1)(2, Λ, Λ, S, halt).

10

Quiz• Construct a TM to accept the language of the

regular expression a(a + b)*.• Solution. Let the start state be state 0. The idea

is to see whether a is in the current cell.If so, move right to scan any a’s or b’s until Λ is found.(0, a, a, R, 1)(1, a, a, R, 1)(1, b, b, R, 1)(1, Λ, Λ, S, halt).

11

Example/Quiz• Find a TM to accept {anbn | n ∊ N}.• Solution. Let the start state be state 0. The idea is to see whether a is in the current

cell.If so, write X and scan right looking for a b to replace by Y.(0, Λ, Λ, S, halt) accept Λ(0, a, X, R, 1) mark a with X(0, Y, Y, R, 3) no more a’sScan right looking for b to pair with a:(1, a, a, R, 1)(1, Y, Y, R, 1)(1, b, Y, L, 2) mark b with YScan back left looking for X:(2, a, a, L, 2)(2, Y, Y, L, 2)(2, X, X, R, 0)Scan right looking for Λ and halt:(3, Y, Y, R, 3)(3, Λ, Λ, S, halt)

12

TMs are very powerful

For example, the preceding example can be generalized to a TM that accepts the non-context free language {anbncn | n ∊ N}. (See Example 13.2 in the Text.) So a TM can handle two stacks. In fact, a TM can handle any number of stacks.

13

Turing Machines with Output• Specify the form of the output on the tape when the machine halts.• Example/Quiz. Find a TM to add 4 to a natural number represented in binary. Start

with the tape head at the right end of the input string and halt with the tape head at the left end of the output string.

• Solution. Let the start state be 0.Move two cells left:(0, 0, 0, L, 1)(0, 1, 1, L, 1)(1, 0, 0, L, 2)(1, 1, 1, L, 2)(1, Λ, 0, L, 2)Add 1:(2, 0, 1, L, 3) Move left(2, 1, 0, L, 2) Carry(2, Λ, 1, S, halt) DoneFind left end of the string:(3, 0, 0, L, 3)(3, 1, 1, L, 3)(3, Λ, Λ, R, halt) Done

14

Quiz• How would you construct a TM to add 5 to a binary number?• One Solution: Add 1, move back to right end, and then use the

preceding solution.For this purpose, let the start state be 4.Add 1:(4, 0, 1, R, 5) Move right(4, 1, 0, L, 4) Carry(4, Λ, 1, R, 5) Move rightFind right end of the string:(5, 0, 0, R, 5)(5, 1, 1, R, 5)(5, Λ, Λ, L, 0) Go add 4

15

Example/Quiz• Find a TM to move any string over {a, b} to the left one cell position.

Assume the tape head ends at the left end of any nonempty output string.• Solution. Let the start state be 0.

Find a or b to move:(0, a, Λ, L, 1) Go write a(0, b, Λ, L, 2) Go write b(0, Λ, Λ, L, 4) DoneWrite a or b:(1, Λ, a, R, 3) Write a(2, Λ, b, R, 3) Write b(3, Λ, Λ, R, 0) Skip ΛMove to left end of output:(4, Λ, Λ, L, 5)(5, a, a, L, 5)(5, b, b, L, 5)(5, Λ, Λ, R, halt).

16

A trace for input aba (tape head is underlined):

Λ a b a Λ (0, a, Λ, L, 1) Λ Λ b a Λ (1, Λ, a, R, 3) a Λ b a Λ (3, Λ, Λ, R, 0) a Λ b a Λ (0, b, Λ, L, 2) a Λ Λ a Λ (2, Λ, b, R, 3) a b Λ a Λ . (3, Λ, Λ, R, 0) a b Λ a Λ (0, a, Λ, L, 1) a b Λ Λ Λ (1, Λ, a, R, 3) a b a Λ Λ (3, Λ, Λ, R, 0) a b a Λ Λ (0, Λ, Λ, L, 4) a b a Λ Λ (4, Λ, Λ, L, 5) a b a Λ Λ (5, a, a, L, 5) a b a Λ Λ (5, b, b, L, 5) a b a Λ Λ (5, a, a, L, 5)Λ a b a Λ Λ (5, Λ, Λ, R, halt)Λ a b a Λ Λ Output.

17

Quiz• (Exercise 2). Find a TM to search for the symbol # on an otherwise

empty tape, where the tape head starts at a random cell.• A Solution: Use a new symbol, say x, to mark cells that have been

searched as the machine alternately looks left and right.A Sample Trace:# Λ Λ Λ Λ# Λ x Λ Λ# x x Λ Λ# x x Λ Λ# x x x Λ# x x x Λ# x x x Λ.

18

Alternative Definitions of Turing Machine

19

Quiz

• Why are they all equal in power to the original definition of a Turing machine?

• An n-head or n-tape instruction contains n-tuples for the cell contents and move operations.

• Example. A typical instruction for a 2-head or a 2-tape TM is(0, (a, b), (c, d), (L, R), 1).

20

Example/QuizImplement some sample

PDA instructions as TM instructions for a 2-tape TM. Assume one tape holds the input string and the other tape holds the stack. Assume the input is scanned left to right and the stack grows from right to left.

21

Sample Instructions

PDA Instruction:1. (i, a, X, nop, j)2. (i, a, X, pop, j)3. (i, a, X, push(Y), j)4. (i, Λ, X, pop, j).

TM Instruction:1. (i, (a, X), (a, X), (R, S), j)2. (i, (a, X), (a, Λ), (R, R), j)3. (i, (a, X), (a, X), (S, L), k)

and (k, (a, Λ), (a, Y), (R, S), j)

4. For each letter a: (i, (a, X), (a, Λ), (S, R), j).

22

Nondeterministic Turing Machines

• Example. Generate an arbitrary string over {a, b} of length n.

• Solution: Assume the input is any string over {a, b} of length n. The tape head is at the right end of the output string.(0, a, a, R, 0)(0, a, b, R, 0)(0, b, a, R, 0)(0, b, b, R, 0)(0, Λ, Λ, L, halt).

23

Nondeterministic TMs have the same power as deterministic TMs

The idea: Any nondeterministic TM can be simulated by a deterministic TM that simulates all 1-step computations, then all 2-step computations, and so on. The process stops if some computation enters the halt state.

24

The End of Section 13-1