Models of Computation

1
Chapter 10

• Models of Computation

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What is a model?

• Captures the essence the important properties
  of the real thing
• Probably differs in scale from the real thing
• Suppresses details of the real thing
• Lacks the full functionality of the real thing

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Examples

• Physical models
• Model cars
• Dolls
• Velcro-covered balls stuck together in a certain
  way to represent the molecular structure.
• Mathematical models
• drxt
• Weather system

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Models

• Models
• Can be used to predict behavior of existing
  phenomena
• can provide environments for learning and
  practicing interactions with various phenomena.
• Can be used as design tools.
• But, the information gained is only as good as
  the model used.

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A Model of Computing Agent

• We want to construct a model for the computing
  agent that will capture its fundamental
  properties and enable us to explore the
  capabilities and limitations of computation in
  the most general sense.

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Properties of A Computing Agent

• We shall require that any computing agent
• Can accept input
• Can store information in and retrieve it from
  memory
• Can take actions according to algorithm
  instructions, and that the choice of what action
  to take may depend on the present state of the
  computing agent as well as on the input item
  presently being processed
• Can produce output

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The Turing Machine Historical Prospective

• Interest in the theoretical nature of computation
  far predated the advent of modern computers.
• Mathematicians are interested in formalizing the
  nature of proof, with two goals in mind
• Guarantee the correctness of a proof
• Allow for mechanical theorem proving
• Leads to the study of the nature of computation
  procedure itself.
• We will look at the model proposed by Alan Turing.

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The Turing Machine

• A Turing machine includes a (conceptual) tape
  that extends infinitely in both directions.
• The tape is divided into cells, each of which
  contains one symbol.
• The symbols must come from a finite set of
  symbols called the alphabet.
• The alphabet always contains a special symbol b
  (for blank), usually both of the symbols 0 and
  1, and others.
• At any point in time, only a finite number of
  cells contain nonblank symbols.

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The Turing Machine (contd)

• Example
• The tape will be used to hold input to the Turing
  machine.
• The Turing machine will write its output on the
  tape.
• The tape will also serve as memory.

.
b
b
0
1
1
b
b
.
.
.
.
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The Turing Machine (contd)

• The rest of the Turing machine consists of a unit
  that reads one cell of the tape at a time and
  writes a symbol in that cell.
• There is a finite number k of states of the
  machine, labeled 1,2,,k, and at any moment, the
  unit is in one of these states.
• A Turing machine configuration

.
b
b
0
1
1
b
b
.
.
.
.
1 (current state of the machine)
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The Turing Machine (contd)

• The Turing machine is designed to carry out only
  one type of primitive operation.
• Each time such an operation is done, three
  actions take place
• Write a symbol in the cell
• Go into a new state
• Move one cell left or right
• The details of the actions depend on the current
  state of the machine and on the contents of the
  tape cell currently being read (the input).

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Primitive Operation (Instruction)

• If you are in state Iand you are reading
  symbol jthen write symbol k onto the tape go
  into state s move in direction d

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Shorthand Notation for Turing Machine Instructions

• Five components
• Current state
• Current symbol
• Next symbol
• Next state
• Direction of move
• (current state, current symbol, next symbol, next
  state, direction of move)

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Example

• (1,0,1,2,R)

.
b
b
0
1
1
b
b
.
.
.
.
1 (current state of the machine)
.
b
b
1
1
1
b
b
.
.
.
.
2 (current state of the machine)
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More on Turing Machine Instructions

• What if we have (1,0,1,2,R) and (1,0,0,3,L)
• Avoid ambiguity by requiring that a set of
  instructions for a Turing machine can never
  contain two different instructions of the form
  (i,j,-,-,-) and (i,j,-,-,-)
• If there is no instruction that applies to the
  current state-current symbol for the machine,
  then the machine halts.
• Conventions about initial configuration
• The start-up will always be state 1
• The machine will be reading the leftmost nonblank
  cell on the tape.

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Example

1. (1,0,1,2,R)
2. (1,1,1,2,R)
3. (2,0,1,2,R)
4. (2,1,0,2,R)
5. (2,b,b,3,L)

.
b
b
0
1
1
b
b
.
.
.
.
1
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Example (contd)
.
b
b
1
1
1
b
b
.
.
.
.
2
.
b
b
1
0
1
b
b
.
.
.
.
2
.
b
b
1
0
0
b
b
.
.
.
.
2
.
b
b
1
0
0
b
b
.
.
.
.
3
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Turing Machine as a Computing Agent

• The Turing machine can
• Read symbols on its tape.
• Write symbols on its tape and, by moving around
  over the tape, can go back and read those symbols
  at a later time.
• The present state and symbol determine the
  appropriate instruction
• If the Turing machine halts, what is written on
  the tape at that time can be considered output.

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Summary

• A Turing machine is a general computing machine.
• It has no limit on the amount of memory.

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A Model of An Algorithm

• An algorithm is a collection of instructions
  intended for a computing agent to follow.
• Requirements
• Be a well-ordered collection
• Consists of unambiguous and effectively
  computable operations
• Halts in a finite amount of time
• Produce a result
• Does an arbitrary collection of Turing machine
  instructions satisfy all the requirements?

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Turing Machine Examples

• Bit converter
• (1,0,1,1,R) and (1,1,0,1,R)
• A state diagram is a visual representation of a
  Turing machine algorithm where circles represent
  states, and arrows represent transitions from one
  state to another.
• Along each transition arrow, we show three
  things
• The input symbol that causes the transition
• The corresponding output
• The direction of move

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A Parity Bit Machine

• Odd parity bit will be set so that the number of
  1s in the whole string of bits, including the
  parity bit, is odd.
• (1,1,1,2,R) even parity state reading 1, change
  state
• (1,0,0,1,R)even parity state reading 0, dont
  change state
• (2,1,1,1,R) odd parity state reading 1, change
  state
• (2,0,0,2,R)odd parity state reading 0, dont
  change state
• (1,b,1,3,R) end of string in even parity state,
  write 1 and go to state 3
• (2,b,0,3,R) end of string in even parity state,
  write 1 and go to state 3

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Unary Incrementing

• Unary representation of numbersNumber Turing
  machine tape representation0 11 112 111
• Incrementing (algorithm 1)
• (1,1,1,1,R) pass to the right over 1s
• (1,b,1,2,R) Add a single one at the right hand
  side of the string

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b
b
1
1
1
b
b
.
.
.
.
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Incrementing Algorithm 2

• (1,1,1,1,L) pass to the left over 1s
• (1,b,1,2,L) Add a single 1 at the left-hand end
  of the string
• 2 steps vs. n2 steps

.
b
b
1
1
1
b
b
.
.
.
.
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Unary Addition Machine

• (1,1,b,2,R) erase the leftmost 1 and move right
• (2,1,b,3,R) erase the second 1 and move right
• (3,1,1,3,R)pass over any 1s until a blank is
  found
• (3,b,1,4,R)write a 1 over the blank and halt.

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The Church-Turing Thesis

• If there is an algorithm to do a symbol
  manipulation task, then there is a Turing machine
  to do that task.
• If we find a symbol manipulation task that no
  Turing machine can perform, then there is no
  algorithm for this task.
• Turing machines define the limits of
  Computability.

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Unsolvable Problems

• Example(1,0,1,2,R)(1,1,0,2,R)(2,0,0,2,R)(2,b,
  b,2,L)
• b11bbb 2
• b1bbb 1

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The Halting Problem

• Formal statement Decide, given any collection of
  Turing machine instructions together with any
  initial tape contents, whether that Turing
  machine will ever halt if started on that tape.
• The halting problem is unsolvable, and we will
  prove it by contradiction.

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Proof

• Assume that P is a Turing machine that solves the
  halting problem.
• On the initial for P we will have to put a
  description of a collection T of Turing machine
  instructions, as well as the initial tape content
  t on which those instructions run.
• Tbtbbb (T symbolize the binary form of T)

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Proof (contd)

• P will give us an answer
• Yes, halts (1)
• No, never halts (0)
• When begun on a tape containing T and t,
• P halts with 1 on its tape exactly when T
  eventually halts when begun on t
• P halts with 0 on its tape exactly when T never
  halts when begun on t
• Refer to Figure 10.10

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Proof (contd)

• Lets imagine adding more instructions to P to
  create a new machine Q that behaves just like P
  except that when it reaches this sme
  configuration, it moves forever to the right on
  the tape instead of halting. (Refer to Figure
  10.11)
• Example Pick some state not in P, say 52, and
  add two new instructions to P
• (9,1,1,52,R)
• (52,b,b,52,R)

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Proof (contd)

• Finally, create a new machine S which first makes
  a copy of what appears on its input.
• After S is finished with its copying job, it uses
  the same instruction as machine Q.
• What happens when machine S is run on a tape that
  contains S, the binary representation of S?
• S first makes a copy of S and then turn the
  computation over to Q, which is now running on a
  tape containing S and S.

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Proof (contd)
S
Input
SbS
S
Never halts exactly when S eventually halts on S
Halts with 0 on tape exactly when S never halts
on S
output
Contradiction!
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Practical Unsolvable Problems

• No C program can be written to decide whether
  any given C program always stops eventually, no
  matter what input.
• No C program can be written to decide whether
  any two C programs are equivalent.
• No C program can be written to decide whether
  any given C program on any given input will
  ever produce some specific output.