Theory of Computation  Lectures 4 and 5

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Transcript of Theory of Computation  Lectures 4 and 5
Formal Definition of Computation
Let M = (Q, ∑ , δ, q0, F) be a finite automaton
and let w = w1w
2.....w
n be a string where each
wi is a member of the alphabet ∑. Then M accepts w if a sequence of states r
0, r
1, ….,r
n In Q
exists with three conditions:1 r
0= q
0,
2 δ(ri, w
i+1)=r
i+1, for i=0,...,n1, and
3 rn
F∈
Cont. Condition 1 says that the machine starts in
the start state. Condition 2 says that the machine goes
from state to another according to the transition function
Condition 3 says that the machine accepts its input if it ends up in an accept state. We say that M recognizes language A if
A = {w  M accepts w}
Regular Language
Definition:
A language is called a regular language if some finite automaton recognizes it.
See example 1.17 p. 41
Designing Finite Automata Just like artwork, design is a creative
process... Suppose you are given some language,
and you want to design a FA that recognizes it...pretend that you are the machine!
Take symbols, one by one, and and after each symbol, you should decide whether the string so far is in the language or not.
Designing Finite Automata –Cont. What if the string is from earth to moon
long? And your memory (as a FA) is like a sheet of paper? Worryless, you will have to memorize the crucial info only, like: “What's my status now?”
Example: suppose: alphabet = {0,1}, and that the language consists of all strings with an odd number of 1s. You want to construct a FA E
1 to recognize this language
Designing Finite Automata –Cont.
By pretending that you're the FA, get the symbols (1 by 1), do you need to remember all the string? No! Just remembering the number of 1s so far is even or not is enough, so that if next is 0, don't change, if 1, change!
Therefore, you have 2 states, say: qodd
, q
even.
Designing Finite Automata –Cont.
Next, assign transitions. Assign a start state (q
even) in our example,
since no symbol has been entered so far (0) and 0 is even.
What about Final state (accept)?
See example 1.21  P.43
Regular Operations
We use them to study properties of regular languages, those properties will help us develop a toolbox that will include ways of proving certain other languages are nonregular (i.e. beyond the capability of FA)
Regular Operations If A and B are languages, then: Union: A B = {xx ∪ A or x B∈ ∈ } Concatenation: A ○ B = {xy  x A and y ∈ ∈
B} Star: A* = {x
1x
2....x
k  k>=0 and each x
iA}∈
The star operation is unary operation (on 1 language) while the union and conc. are binary operations
Regular Operations – cont.
Let ∑ = {a,b,....,z}, if A = {good, bad}, and B = {boy, girl} the:
A B = {good, bad, boy, girl}∪
A ○ B = {goodboy, goodgirl, badboy, badgirl}
A*= {ε, good, bad, goodgood, badbad, badgood, goodgoodgood, goodgoodbad, etc.....}
Regular Operations Cont.
Closed under multiplication:
example, if x and y are two numbers from N, then x*y results a number also from N.
But N is not closed under division, because x/y may result a number which isn't member of N, ex. X=1, y=2, x/y=1/2 which isn't member of N
Please read theorems 1.25, 1.26 p.45, 46
Nondeterminism DFA vs. NFA:
DFA has exactly one exiting transition arrow for each symbol in the alphabet, NFA violets that.
DFA uses symbols from alphabet only for arrow labels, NFA may use others, such as ε.
What will happen if a state has two arrows for same input? Or if input is ε?
NFA
q1 q2 q3 q41 0,ε 1
0,1 0,1
q4
Examples:
1.30, 1.33, 1.35 P: 5153
Formal definition of NFA
Similar to DFA, the NFA has the 5tuple, yet the difference is in δ (The Transition Function)
In DFA, δ takes a state and an input symbol and produces the next state.
In NFA, δ takes a state and an an input symbol OR the empty string and produces the set of possible next states.
Cont.
Therefore, point 3 of the tuple would be as:
δ : Q X ∑ε → P(Q), where
∑ε is the result of ∑ ε∪
Example
The formal definition of N1 (slide 13):
1 Q = {q1,q
2,q
3,q
4}
2 ∑ = {0,1}
3 δ is given as:
4 q1 =start state
5 F ={q4}
0 1 εq
1{q
1} {q
1,q
2} Ф
q2
{q3} Ф {q
3}
q3
Ф {q4} Ф
q4
{q4} {q
4} Ф
Reading
Every NFA has an equivalent DFA...
To prove, read P.54 P.58 Based on your previous reading (P.45),
read Closure under the regular operations (P.5863). Which is similar to, yet easier than example P.45.
Regular Expressions In math, we use + and x to build up
expressions like: (5+3)x4 (Gives 32) Similarly, we use and * to build up ∪
expressions describing languages, called Regular Expressions (RegEx):
(0 1)0*∪ … (Gives a language)
What does (0 1)0* mean?∪ It's a language consisting of all strings starting
with a 0 or 1 followed by any number of 0s.
RegEx – Cont.(0 1)0* is:∪
 (0 1) = ({0} {1}) = {0,1}∪ ∪
 0* means {0}*, means a language consisting of of any number of 0s.
The concatenation between (0 1) and 0* is a ∪shorthand for for (0 1) ○ 0*∪
Applications of RegEx: many, Grep (textsearch)in Unix and Linux, Perl, C++, text editors etc....
In C++, you can use regex r("(\\+)?[[:digit:]]+");, but why?
Regex Example(0 1)* = the language consisting of all possible ∪
strings of 0s and 1s.
If ∑ ={0,1}, then ∑ is a shorthand for (0 1).∪ Generally, if ∑ is any alphabet, the regex ∑
describes the language consisting of all strings of length 1 over the alphabet, and ∑* describes the language consisting of all strings over the alphabet.
∑*1 is the language that contains all strings that end with 1
Regex Examplecont.
The language (0∑*) (∪ ∑*1) consists of all strings that either start with a 0 or end with a 1.
* similar to mathematical precedence ( x before + for example), in regex, * is done first ,then concatenation, finally the union. Unless parentheses are used.
Formal Definition of regex
We say that R is a regex if R:
1. a for some a in the alphabet
2. ε,
3. ᶲ
4. (R1 R∪
2)
5. (R1 ○ R
2)
6. (R1*), where R
1 is a regex