Theory of computing, part 3

1 Introduction

2 Theoretical background Biochemistry/molecular biology

3 Theoretical background computer science

4 History of the field

5 Splicing systems

6 P systems

7 Hairpins

8 Detection techniques

9 Micro technology introduction

10 Microchips and fluidics

11 Self assembly

12 Regulatory networks

13 Molecular motors

14 DNA nanowires

15 Protein computers

16 DNA computing - summery

17 Presentation of essay and discussion

Course outline

Finite automata

Introduction

Deterministic finite automata (DFA’s)

Non-deterministic finite automata (NFA’s)

NFA’s to DFA’s

Simplifying DFA’s

Regular expressions finite automata

Outline

Consider the control system for a

one-way swinging door:

There are two states: Open and

Closed

It has two inputs, person detected

at position A and person detected

at position B

If the door is closed, it should

open only if a person is detected

at A but not B

Door should close only if no one is

detected

A B

Automatic one way door

Open Closed

A, no B

No A or B

A and BA, no BB, no A

A and BB, no ANo A or B

Control schematic

A finite automaton is usually represented

like this as a directed graph

Two parts of a directed graph:

The states (also called nodes or

vertices)

The edges with arrows which

represent the allowed transitions

One state is usually picked out as the

starting point

For so-called ‘accepting automata,’ some

states are chosen to be final states

Finite automaton

The input data are represented by a string

over some alphabet and it determines how the

machine progresses from state to state.

Beginning in the start state, the characters

of the input string cause the machine to

change from one state to another.

Accepting automata give only yes or no

answers, depending on whether they end up in

a ‘final state.’ Strings which end in a

final state are accepted by the automaton.

Strings and automata

The labeled graph in the figure above represents a FA over the alphabet Σ = {a, b} with start state 0 and final state 3.

Final states are denoted by a double circle.

Example

The previous graph was an example of a

deterministic finite automaton – every node

had two edges (a and b) coming out

A DFA over a finite alphabet Σ is a finite

directed graph with the property that each

node emits one labeled edge for each distinct

element of Σ.

Deterministic finite automata (DFA’s)

A DFA accepts a string w in Σ* if there is a

path from the start state to some final state

such that w is the concatenation of the labels

on the edges of the path.

Otherwise, the DFA rejects w .

The set of all strings accepted by a DFA M is

called the language of M and is denoted by

L(M)

More formally

Construct a DFA to recognize the regular

languages represented by the regular

expression (a + b)* over alphabet Σ = {a, b}.

This is the set {a, b}* of all strings over

{a, b}. This can be recognised by

Example: (a+b)*

Find a DFA to recognize the language

represented by the regular expression

a(a + b)* over the alphabet Σ = {a, b}.

This is the set of all strings in Σ*

which begin with a. One possible DFA

is:

Example: a(a+b)*

Build a DFA to recognize the regular language

represented by the regular expression

(a + b)*abb over the alphabet Σ = {a, b}.

The language is the set of strings that begin

with anything, but must end with the string

abb.

Effectively, we’re looking for strings which

have a particular pattern to them

Example: pattern recognition

The diagram below shows a DFA to recognize this language.

If in state 1: the last character was a

If in state 2 : the last two symbols were ab

If in state 3: the last three were abb

Solution: (a+b)*abb

State transition function

We can also represent a DFA by a state transition

function, which we'll denote by T, where any state

transition of the form

is represented by: T(i,a) = j

To describe a full DFA we need to know:

what states there are,

which are the start and final ones,

the set of transitions between them.

State transition function

The class of regular languages is

exactly the same as the class of

languages accepted by DFAs!

Kleene (1956)

For any regular language, we can find

a DFA which recognizes it!

Regular languages

DFA’s are very often used for pattern matching,

e.g. searching for words/structures in strings

This is used often in UNIX, particularly by the

grep command, which searches for combinations of

strings and wildcards (*, ?)

grep stands for Global (search for) Regular

Expressions Parser

DFA’s are also used to design and check simple

circuits, verifying protocols, etc.

They are of use whenever significant memory is

not required

Applications of DFA’s

DFA’s are called deterministic because

following any input string, we know exactly

which state its in and the path it took to

get there

For NFA’s, sometimes there is more than one

direction we can go with the same input

character

Non-determinism can occur, because following

a particular string, one could be in many

possible states, or taken different paths to

end at the same state!

Non-deterministic finite automata

A non-deterministic finite automaton (NFA)

over an alphabet Σ is a finite directed graph

with each node having zero or more edges,

Each edge is labelled either with a letter

from Σ or with .

Multiple edges may be emitted from the same

node with the same label.

Some letters may not have an edge associated

with them. Strings following such paths are

not recognised.

NFA’s

If an edge is labelled with the empty string , then we can travel the edge without consuming an

input letter. Effectively we could be in either

state, and so the possible paths could branch.

If there are two edges with the same label, we

can take either path.

NFA’s recognise a string if any one of its many

possible states following it is a final state

Otherwise, it rejects it.

Non-determinism

DFA for a*a :

Why is the top an NFA while the bottom is a DFA?

NFA for a*a :

NFA’s versus DFA’s

Draw two NFAs to recognize the language of the

regular expression ab + a*a.

This NFA has a edge, which allows us to travel to state 2 without consuming an input letter.

The upper path corresponds to ab and the lower one

to a*a

Example

This NFA also recognizes the same language. Perhaps

it's easier to see this by considering the equality

ab + a*a = ab + aa*

An equivalent NFA

Since there may be non-determinism, we'll let the

values of this function be sets of states.

For example, if there are no edges from state k

labelled with a, we'll write

T(k, a) =

If there are three edges from state k, all labelled

with a, going to states i, j and k, we'll write 

T(k, a) = {i, j, k}

NFA transition functions

All digital computers are deterministic; quantum

computers may be another story!

The usual mechanism for deterministic computers is

to try one particular path and to backtrack to the

last decision point if that path proves poor.

Parallel computers make non-determinism almost

realizable. We can let each process make a random

choice at each branch point, thereby exploring

many possible trees.

Comments on non-determinism

The class of regular languages is exactly the

same as the class of languages accepted by NFAs!

Rabin and Scott (1959)

Just like for DFA’s!

Every NFA has an equivalent DFA which recognises

the same language.

Some facts

We prove the equivalence of NFA’s and DFA’s by showing

how, for any NFA, to construct a DFA which recognises

the same language

Generally the DFA will have more possible states than

the NFA. If the NFA has n states, then the DFA could

have as many as 2n states!

Example: NFA has three states {A}, {B}, {C}

the DFA could have eight: {}, {A}, {B}, {C},

{A, B}, {A, C}, {B, C}, {A, B, C}

These correspond to the possible states the NFA could

be in after any string

From NFA’s to DFA’s

Begin in the NFA start state, which could be a

multiple state if its connected to any by

Determine the set of possible NFA states you could be

in after receiving each character. Each set is a new

DFA state, and is connected to the start by that

character.

Repeat for each new DFA state, exploring the possible

results for each character until the system is closed

DFA final states are any that contain a NFA final

state

DFA construction

The start state is A, but following an a you could be in A

or B; following a b you could only be in state A

A B Ca b

a,b

A A,Ba b

b

A,C

ab

a

NFA

DFA

Example (a+b)*ab

Regular expressions represent the regular languages.

DFA’s recognize the regular languages.

NFA’s also recognize the regular languages.

Summary

So far, we’ve introduced two kinds of automata:

deterministic and non-deterministic.

We’ve shown that we can find a DFA to recognise

anything language that a given NFA recognises.

We’ve asserted that both DFA’s and NFA’s recognise

the regular languages, which themselves are

represented by regular expressions.

We prove this by construction, by showing how any

regular expression can be made into a NFA and vice

versa.

Finite automata

Given a regular expression, we can find an automata

which recognises its language.

Start the algorithm with a machine that has a start

state, a single final state, and an edge labelled

with the given regular expression as follows:

Regular expressions finite automata

1. If an edge is labelled with , then erase the edge.

2. Transform any diagram like

into the diagram

Four step algoritm

3. Transform any diagram like

into the diagram

Four step algoritm

4. Transform any diagram like

into the diagram

Four step algoritm

Construct a NFA for the regular expression, a* + ab

Start with

Apply rule 2

a* + ab

ab

a*

Example a*+ab

ab

a

a

a b

Apply rule 4 to a*

Apply rule 3 to ab

Example a*+ab

1 Create a new start state s, and draw a

new edge labelled with from s to the original start state.

2 Create a new final state f, and draw new

edges labelled with from all the

original final states to f

Finite automata regular expressions

3 For each pair of states i and j that have

more than one edge from i to j, replace all

the edges from i to j by a single edge

labelled with the regular expression formed

by the sum of the labels on each of the

edges from i to j.

4 Construct a sequence of new machines by

eliminating one state at a time until the

only states remaining are s and the f.

Finite automata regular expressions

As each state is eliminated, a new machine is

constructed from the previous machine as follows:

Let old(i,j) denote the label on edge i,j of the current machine. If no edge exists, label

it .

Assume that we wish to eliminate state k. For

each pair of edges i,k (incoming edge) and

k,j (outgoing edge) we create a new edge label new(i, j)

Eliminating states

The label of this new edge is given by:

new(i,j) = old(i,j) + old(i, k) old(k, k)* old(k,j)

All other edges, not involving state k, remain the

same:

new(i, j) = old(i, j)

After eliminating all states except s and f, we wind up

with a two-state machine with the single edge s, f labelled with the desired regular expression new(s, f)

Eliminate state k

Initial DFA

Steps 1 and 2

Add start and final states

Example

Eliminate state 2

(No path to f)

Eliminate state 0

Eliminate state 1

Final regular expression

Example

Sometimes our constructions lead to more

complicated automata than we need, having more

states than are really necessary

Next, we look for ways of making DFA’s with a

minimum number of states

Myhill-Nerode theorem:

‘Every regular expression has a unique* minimum

state DFA’

* up to a simple renaming of the states

Finding simpler automata

Two steps to minimizing DFA:

1 Discover which, if any, pairs of states are

indistinguishable. Two states, s and t, are

equivalent if for all possible strings w,

T(s,w) and T(t,w) are both either final or

non-final.

2 Combine all equivalent states into a single

state, modifying the transition functions

appropriately.

Finding minimum state DFA

States 1 and 2 are indistinguishable! Starting in either,

b* is rejected and anything with a in it is accepted.

a ab

b a

b

1

2

a,b

aa,b a,b

b

Consider the DFA

1. Remove all inaccessible states, where no path

exists to them from start.

2. Construct a grid of pairs of states.

3. Begin by marking those pairs which are clearly

distinguishable, where one is final and the other

non-final.

4. Next eliminate all pairs, which on the same

input, lead to a distinguishable pair of states.

Repeat until you have considered all pairs.

5. The remaining pairs are indistinguishable.

Part 1, finding indistinguishable pairs

1. Construct a new DFA where any pairs of

indistinguishable states form a single state

in the new DFA.

2. The start state will be the state containing

the original start state.

3. The final states will be those which contain

original final states.

4. The transitions will be the full set of

transitions from the original states (these

should all be consistent.)

Part 2, construct minimum DFA

a

a

a,b

a

a

b b b

0

3

2

1

4

What are the distinguishable pairs of states?

Clearly, {0, 4} {1, 4} {2, 4} {3, 4} are all

distinguishable because 4 is final but none of

the others are.

b

Example

We eliminate these as possible

indistinguishable pairs.

Next consider {0, 1}. With input

a, this becomes {3, 4} which is

distinguishable, so {0, 1} is as

well.

Similarly, we can show {0, 2} and

{0, 3} are also distinguishable,

leading to the modified grid…

1234

? ? ? ? ? ?x x x x

0 1 2 3

1234

xx ? x ? ?x x x x

0 1 2 3

Grid of pairs of state

We are left with

{1, 2} given a {4, 4}

given b {2, 1}

{2, 3} given a {4, 4}

given b {1, 2}

{1, 3} given a {4, 4}

given b {2, 2}

These do not lead to pairs we know to

be distinguishable, and are therefore

indistinguishable!

Remaining pairs

States 1, 2, and 3 are all indistinguishable, thus

the minimal DFA will have three states:

{0} {1, 2, 3} {4}

Since originally T(0, a) = 3 and T(0, b) = 1, the

new transitions are T(0, a) = T(0, b) = {1,2,3}

Similarly,

T({1,2,3}, a) = 4 and T({1,2,3}, b) = {1,2,3}

Finally, as before,

T(4, a) = 4 and T(4, b) = 4

Construct minimal DFA

The resulting DFA is much simpler:

aa,b a,b

b

0 1, 2, 3 4

This recognises regular expressions of the form, (a +

b) b* a (a + b)*

This is the simplest DFA which will recognise this

language!

Resulting minimal DFA

We now have many equivalent ways of representing

regular languages: DFA’s, NFA’s, regular expressions

and regular grammars.

We can also now simply(?!) move between these

various representations.

We’ll see next lecture that the automata

representation leads to a simple way of recognising

some languages which are not regular.

We’ll also begin to consider more powerful language

types and correspondingly more powerful computing

models!

conclusions

M = (Q, Σ, δ, q0, F)

Q = states a finite setΣ = alphabet a finite setδ = transition function a total function in Q Σ

Qq0 = initial/starting state q0 Q

F = final states F Q

Formal definition