Theory of computing, part 1

Von NeumannTuring machineFinite state machinesNP complete problems-maximum clique-travelling salesman problem-colour graph

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

Old computers

Abacus

Abacus

Born December 26, 1791 in Teignmouth, Devonshire UK,

Died 1871, London; Known to some as the Father of

Computing for his contributions to the basic design of

the computer through his Analytical machine..

The difference engine (1832)

The ENIAC machine occupied a room 30x50 feet. The

controls are at the left, and a small part of the

output device is seen at the right.

ENIAC (1946)

The IBM 360 was a revolutionary advance in

computer system architecture, enabling a family

of computers covering a wide range of price and

performance.

ENIAC (1946)

Books

Books

Introduction

Finite state machines (automata)

Pattern recognition

Simple circuits (e.g. elevators, sliding doors)

Automata with stack memory (pushdown autom.)

Parsing computer languages

Automata with limited tape memory

Automata with infinite tape memory

Called `Turing machines’,

Most powerful model possible

Capable of solving anything that is solvable

Models of computing

Regular grammars

Context free grammars

Context sensitive grammars

Unrestricted grammars

Chomsky hierarchy of grammars

Computers can be made to recognize, or accept, the

strings of a language.

There is a correspondence between the power of the

computing model and the complexity of languages

that it can recognize!

Finite automata only accept regular grammars.

Push down automata can also accept context free

grammars.

Turing machines can accept all grammars.

Computers can recognise languages

Languages

A set is a collection of things called its elements. If

x is an element of set S, we can write this:

x S A set can be represented by naming all its elements,

for example:

S = {x, y, z}

There is no particular order or arrangement of the

elements, and it doesn’t matter if some appear more

than once. These are all the same set:

{x, y, z}={y, x, z}={y, y, x, z, z}

Sets

A set with no elements is called the empty set, or

a null set. It is denoted by ={}. If every element of set A is also an element of

set B, then A is called a subset of B :

A B The union of two sets is the set with all elements

which appear in either set:

C = A B The intersection of two sets is the set with all

the elements which appear in both sets:

C = A B

Combining sets

A string a sequence of symbols that are placed

next to each other in juxtaposition

The set of symbols which make up a string are

taken from a finite set called an alphabet

E.g. {a, b, c} is the alphabet for the string

abbacb.

A string with no elements is called an empty

string and is denoted . If Σ is an alphabet, the infinite set of all

strings made up from Σ is denoted Σ*.

E.g., if Σ ={a}, then Σ *={, a, aa, aaa, …}

Alphabet and strings

A language is a set of strings.

If Σ is an alphabet, then a language over Σ

is a collection of strings whose components

come from Σ.

So Σ* is the biggest possible language over

Σ, and every other language over Σ is a

subset of Σ*.

Languages

Four simple examples of languages over an

alphabet Σ are the sets , {}, Σ, and Σ *.

For example, if Σ={a} then these four simple

languages over Σ are

 , {}, {a}, and {, a, aa, aaa, …}.

Recall {} is the empty string while is the empty set. Σ* is an infinite set.

Examples of languages

The alphabet is Σ = {a,b,c,d,e…x,y,z}

The English language is made of strings

formed from Σ: e.g. fun, excitement.

We could define the English Language as the

set of strings over Σ which appear in the

Oxford English dictionary (but it is clearly

not a unique definition).

Example: English

The natural operation of concatenation

of strings places two strings in

juxtaposition.

For example, if then the concatenation

of the two strings aab and ba is the

string aabba.

Use the name cat to denote this

operation. cat(aab, ba) = aabba.

Concatenation

Languages are sets of strings, so they

can be combined by the usual set

operations of union, intersection,

difference, and complement.

Also we can combine two languages L

and M by forming the set of all

concatenations of strings in L with

strings in M.

Combining languages

This new language is called the product of L

and M and is denoted by L M.

A formal definition can be given as follows:

L M = {cat(s, t) | s L and t M}

For example, if L = {ab, ac} and M = {a, bc,

abc}, then the product L•M is the language

 L M = {aba, abbc, ababc, aca, acbc, acabc}

Products of languages

The following simple properties hold for any

language L:

L {} = {} L = L

L = L =

The product is not commutative. In other

words, we can find two languages L and M such

that

L M M L

The product is associative. In other words,

if L, M, and N are languages, then

L (M N) = (L M) N

 

Properties of products

If L is a language, then the product

L L is denoted by L2.

The language product Ln for every n {0, 1, 2, …} is as follows:

L0 = {}

Ln = L Ln-1 if n > 0

Powers of languages

For example, if L = { a, bb} then the first

few powers of L are

L0 = {}

L1 = L = {a, bb}

L2 = L L = {aa, abb, bba, bbbb}

L3 = L L2 = {aaa, aabb, abba, abbbb,

bbaa, bbabb, bbbba, bbbbbb}

Example

If L is a language over Σ (i.e. L A*) then the closure of L is the language

denoted by L* and is defined as follows:

L* = L0 L1 L2 …

The positive closure of L is the language

denoted by L+ and defined as follows: 

L+ = L1 L2 L3 …

Closure of a language

It follows that L* = L+ {}. But it’s not necessarily true that

L+ = L* - {}

For example, if we let our alphabet

be Σ = {a} and our language be L =

{, a}, then

L+ = L*

Can you find a condition on a

language L such that

L+ = L* - {}?

L* vs. L+

The closure of Σ coincides with our

definition of Σ* as the set of all

strings over Σ. In other words, we have

a nice representation of Σ* as follows:

A* = A0 A1 A2 ….

 

Closure of an alphabet

Let L and M be languages over the

alphabet Σ. Then: 

a) {}* = * = {}

b) L* = L* L* = (L*)*

c) L if and only if L+ = L*

d) (L* M*)* = (L* M*)* = (L M)*

e) L (M L)* = (L M)* L

Properties of closure

Grammars

A grammar is a set of rules used to

define the structure of the strings in a

language.

If L is a language over an alphabet Σ,

then a grammar for L consists of a set of

grammar rules of the following form: 

where and denote strings of symbols taken from Σ and from a set of grammar

symbols (non-terminals) that is disjoint

from Σ

Grammars

A grammar rule is often called a production, and it can be read in any of

several ways as follows:

replace by

produces

rewrites to

reduces to

Productions

Every grammar has a special grammar symbol

called a start symbol, and there must be at

least one production with left side

consisting of only the start symbol.

For example, if S is the start symbol for a

grammar, then there must be at least one

production of the form

S

Where to begin ……

1. An alphabet N of grammar symbols called

non-terminals. (Usually upper case

letters.)

2. An alphabet T of symbols called

terminals. (Identical to the alphabet of

the resulting language.)

3. A specific non-terminal called the start

symbol. (Usually S. )

4. A finite set of productions of the form , where and are strings over the alphabet N T

The 4 parts of a grammar

Let Σ = {a, b, c}. Then a grammar for the

language Σ* can be described by the

following four productions:

S

S aS

S bS

S cS

Or in shorthand:

S | aS | bS | cS

S can be replaced by either , or aS, or bS, or cS.

Example

A string made up of grammar symbols

and language strings is called a

sentential form.

If x and y are sentential forms and is a production, then the

replacement of by in xy is

called a derivation, and we denote it

by writing

xy xy

String

S | aS | bS | cS,

S aS

S aS aaS.

S aS aaS aacS aacbS..

S aS aaS aacS aacbS aacb

= aacb

A short hand way of showing a derivation

exists:

S * aacb

Sample derivation

The following three symbols with their

associated meanings are used quite often in

discussing derivations:

  derives in one step,

 + derives in one or more steps,

 * derives in zero or more steps.

Other shorthand

S AB

A | aA

B | bB

 

We can deduce that the grammar non-terminal

symbols are S, A, and B, the start symbol is

S, and the language alphabet includes , a, and b.

A more complex grammar

Let's consider the string aab.

The statement S + aab means that there

exists a derivation of aab that takes one or

more steps.

For example, we have

S AB aAB aaAB aaB aabB aab

Another derivation

If G is a grammar, then the language of G

is the set of language strings derived

from the start symbol of G. The language

of G is denoted by:

L(G)

Grammar specifies the language

If G is a grammar with start symbol S and

set of language strings T, then the

language of G is the following set:

 

L(G) = {s | s T* and S + s}

Grammar specifies the language

If the language is finite, then a grammar can

consist of all productions of the form

S w

for each string w in the language.

For example, the language {a, ba} can be

described by the grammar S a | ab.

Finite languages

If the language is infinite, then some

production or sequence of productions must be

used repeatedly to construct the derivations.

Notice that there is no bound on the length of

strings in an infinite language.

Therefore there is no bound on the number of

derivation steps used to derive the strings.

If the grammar has n productions, then any

derivation consisting of n + 1 steps must use

some production twice

Infinite languages

For example, the infinite language {anb | n

0 } can be described by the grammar,

S b | aS

To derive the string anb, use the production

S aS

repeatedly --n times to be exact-- and then

stop the derivation by using the production

S b

The production S aS allows us to say

If S derives w, then it also derives aw

Finite languages

A production is called recursive if its

left side occurs on its right side.

For example, the production S aS is

recursive.

A production S is indirectly

recursive if S derives (in two or more

steps) a sentential form that contains

S.

Recursion

For example, suppose we have the

following grammar:

S b | aA

A c | bS

The productions S aA and A bS are

both indirectly recursive  

S aA abS 

A bS baA

Indirect recursion

A grammar is recursive if it contains

either a recursive production or an

indirectly recursive production.

A grammar for an infinite language must

be recursive!

However, a given language can have many

grammars which could produce it.

Recursion

Suppose M and N are languages. We can

describe them with grammars have disjoint

sets of nonterminals.

Assign the start symbols for the grammars of

M and N to be A and B, respectively:

M : A N : B

… …

Then we have the following rules for creating

new languages and grammars:

Combining grammars

The union of the two languages, M N, starts with the two productions

S A | B

followed by the grammars of M and N

A

…

B

…

Combining grammars, union rule

Similarly, the language M N starts with the production

S AB

followed by, as above,

A

…

B

…

Combining grammars, product rule

Finally, the grammar for the closure of a

language, M*, starts with the production

S AS |

followed by the grammar of M

A

…

Combining grammars, closure rule

Suppose we want to write a grammar for the

following language: 

L = {, a, b, aa, bb, ..., an, bn, ..}

L is the union of the two languages

M = { an | n N}

N = { bn | n N}

Example, union

Thus we can write a grammar for L as follows:

S A | B union rule,

A | aA grammar for M,

B | bB grammar for N.

Example, union

Suppose we want to write a grammar for the

following language: 

L = { ambn | m,n N}

L is the product of the two languages

M = {am | m N}

N = {bn | n N}

Example, product

Thus we can write a grammar for L

as follows:

  S AB

product rule,

A | aA grammar for M

B | bB grammar for N

Example, product

Suppose we want to construct the

language L of all possible strings made

up from zero or more occurrences of aa

or bb.

L = {aa, bb}* = M*

M = {aa, bb}

Example, closure

So we can write a grammar for L as follows: 

S AS | closure rule,A aa | bb grammar for {aa, bb}.

Example, closure

We can simplify this grammar:

Replace the occurrence of A in S

AS by the right side of A aa to

obtain the production S aaS.

Replace A in S AS by the right

side of A bb to obtain the

production S bbS.

This allows us to write the the

grammar in simplified form as:

  S aaS | bbS |

An equivalent grammar

Language Grammar

{a, ab, abb, abbb} S a | ab | abb | abbb

{, a, aa, aaa, …} S aS |

{b, bbb, bbbbb, … b2n+1} S bbS | b

{b, abc, aabcc, …, anbcn} S aSc | b

{ac, abc, abbc, …, abnc} S aBc

B bB |

Some simple grammars

Regular languages

There are many possible ways of describing the

regular languages:

Languages that are accepted by some finite

automaton

Languages that are inductively formed from

combining very simple languages

Those described by a regular expression

Any language produced by a grammar with a

special, very restricted form

What is a regular language?

We start with a very simple basis of

languages and build more complex ones by

combining them in particular ways:

 

Basis: , {} and {a} are regular

languages for all a Σ.

 

Induction: If L and M are regular

languages, then the following languages

are also regular:

L M, L M and L*

Building a regular language

For example, the basis of the definition

gives us the following four regular

languages over the alphabet Σ = {a,b}:

, {}, {a}, {b}

Sample building blocks

Regular languages over {a, b}.

 

Language {, b}

We can write it as the union of the two

regular languages {} and {b}:

{,b} = {} {b}

Example 1

Language {a, ab }

We can write it as the product of the

two regular languages {a} and {, b}:

{a, ab} = {a} {, b}

Example 2

Language {, b, bb, …, bn,…}

It's just the closure of the regular

language {b}:

{b}* = {, b, bb, ..., bn ,...}

Example 3

{a, ab, abb, ..., abn, ...}

= {a} {, b, bb, ..., bn, ... }

= {a} {b}*

Example 4

{, a, b, aa, bb, ..., an, …, bm, ...}

= {a}* {b}*

Example 5

A regular expression is basically a

shorthand way of showing how a regular

language is built from the basis

The symbols are nearly identical to

those used to construct the languages,

and any given expression has a language

closely associated with it

For each regular expression E there is

a regular language L(E)

Regular expressions

The symbols of the regular expressions are

distinct from those of the languages

Regular expression Language 

L () =

L () = {}

a L {a} = {a}

Basis of regular Basis of regular

expressions languages

Regular expressions versus languages

There are two binary operations on regular expressions

(+ and ) and one unary operator (*)

Regular expression Language 

R + S L (R + S ) = L (R ) L (S )

R S , R S L (R S ) = L (R ) L (S )  

R* L (R* ) = L (R )*

These are closely associated with the union, product and

closure operations on the corresponding languages

Operators on regular expressions

Like the languages they represent, regular

expressions can be manipulated inductively to

form new regular expressions

 

Basis: , and a are regular expressions for all a Σ.

Induction: If R and S are regular expressions,

then the following expressions are also regular:

(R), R + S, R S and R*

Building regular expressions

For example, here are a few of the infinitely

many regular expressions over the alphabet Σ

= {a, b }:

 

, , a, b

+ b, b*, a + (b a), (a + b) a,

a b*, a* + b*

Regular expressions

To avoid using too many parentheses, we

assume that the operations have the

following hierarchy:

* highest (do it first)

+ lowest (do it last)

For example, the regular expression

a + b a* 

can be written in fully parenthesized form

as 

(a + (b (a*)))

Order of operations

Use juxtaposition instead of whenever no confusion arises. For example, we can

write the preceding expression as

a + ba*

This expression is basically shorthand

for the regular language

{a} ({b} ({a}*)) 

So you can see why it is useful to write

an expression instead!

Implicit products

Find the language of the regular expression

a + bc*

L(a + bc*) = L(a) L(bc*) 

= L(a) (L(b) L(c*)) 

= L(a) (L(b) L(c)*) 

= {a} ({b} {c}*) 

= {a} ({b) {, c, c2, ., cn,…}) 

= {a} {b, bc, bc2, .. .... bcn,…}

= {a, b, bc, bc2, ..., bcn, ...}.

Example

Many infinite languages are easily seen to

be regular. For example, the language

{a, aa, aaa, ..., an, ... }

is regular because it can be written as the

regular language {a} {a}*, which is

represented by the regular expression aa*.

Regular language

The slightly more complicated language

{, a, b, ab, abb, abbb, ..., abn, ...}

is also regular because it can be represented

by the regular expression

+ b + ab*

 

However, not all infinite languages are

regular!

Regular language

Distinct regular expressions do not always

represent distinct languages.

For example, the regular expressions a + b

and b + a are different, but they both

represent the same language, 

L(a + b) = L(b + a) = {a, b} 

Regular language

We say that regular expressions R and

S are equal if

L(R) = L(S)

and we denote this equality by

writing the following familiar

relation:

R = S

Regular language

For example, we know that

L (a + b) = {a, b} = {b, a} = L (b + a)

Therefore we can write

a + b = b + a

We also have the equality 

(a + b) + (a + b) = a + b

Regular language

Properties of regular language

Additive (+) properties:

R + T = T + R

R + = + R= R

R + R = R

(R +S) +T = R + (S+ T)

These follow simply from the

properties of the union of sets

Properties of regular language

Product (•) properties

R = R =

R = R = R

(RS)T =R(ST)

Distributive properties

R(S + T) = RS + RT

(S + T)R = SR +TR

* = * =

R* = R*R* = (R*)* = R+R*

RR* = R*R

R(SR)* = (RS)* R

(R+S)* = (R*S*)* = (R* + S*)* = R*(SR*)*

Closure properties

Show that ( + a + b)* = a*(ba*)*  

( + a + b)* = (a + b)* (+ property)

= a*(ba*)* (closure property)

Example

Regular grammars

A regular grammar is one where each

production takes one of the following

forms:

S

S w

S T

S wT

where the capital letters are non-

terminals and w is a non-empty string

of terminals

Regular grammar

Only one nonterminal can appear on the

right side of a production. It must

appear at the right end of the right

side.

Therefore the productions

A aBc and S TU

are not part of a regular grammar, but

the production A abcA is.

Regular grammar