Regular Expressions

CS 130: Theory of Computation

HMU textbook, Chapter 3

Machines versus expressions Finite automata are machine-like

descriptions of languages Alternative: declarative description

Specifying a language using expressions and operations

Example: 01* + 10* defines the language containing strings such as 01111, 100, 0, 1000000; * and + are operators in this “algebra”

Regular expressions defined The simplest regular expressions are

The empty set Φ The empty string A symbol a from an alphabet

Given that R1 and R2 are regular expressions, regular expressions are built from the following operations

Union: R1+R2 Concatenation: R1R2 Closure: R1* Parentheses (to enforce precedence): (R1)

Nothing else is a regular expression unless it is built from the above rules

Examples 1 + (ab)* + (ba)* (0+1+2+3+4+5+6+7+8+9)*(0+5) (x+y)*x(x+y)* (01)* 0(1*) 01* equivalent to 0(1*)

Equivalence between regular expressions and finite automataStrategy: Convert regular expression to an -

NFA Convert a DFA to a regular

expression

Regular Expression to -NFA Recursive construction Base cases follow base case definitions of

regular expressions : -NFA that accepts the empty string –

a single state that is the start and end state a: -NFA that accepts {a} – two-state

machine (start and final state) with an a-transition

Note: technically, we also need an -NFA for the empty language {} – easy

Regular Expression to -NFA Recursive step: build -NFA from smaller -

NFAs that correspond to the operand regular expressions

To simplify construction, we may ensure the following characteristics for the automata we build Only one final state, with no outgoing transitions No transitions into the start state Note: the base cases satisfy these

characteristics

Regular Expression to -NFA Suppose -NFA1 and -NFA2 are the automata for

R1 and R2 Three operations to worry about: union

R1 + R2, concatenation R1R2), closure R1* With -transitions, construction is straightforward

Union: create a new start state, with -transitions into the start states of -NFA1 and -NFA2; create a new final state, with -transitions from the two final states of -NFA1 and -NFA2

Concatenation: -transition from final state of -NFA1 to the start state of -NFA2

Closure: closure can be supported by an -transition from final to start state; need a few more -transitions (why?)

DFA to Regular Expression More difficult construction Build the regular expression “bottom

up” starting with simpler strings that are acceptable using a subset of states in the DFA

Define Rki,j as the expression for strings

that have an admissible state sequence from state i to state j with no intermediate states greater than k Assume no states are numbered 0, but k

can be 0

R0i,j

Observe that R0i,j describes strings of length

1 or 0: {a1, a2, a3, … }, where, for each ax, (i,ax) = j Add to the set if i = j R0

i,j is the union (+) of applicable symbols in Σ (and perhaps )

The 0 in R0i,j means no intermediate states

are allowed, so either no transition is made (just stay in state i to accept if i = j) or make a single transition from state i to state j

These are the base cases in our construction

Rki,j

Recursive step: for each k, we can build Rk

i,j as follows:Rk

i,j = Rk-1i,j + Rk-1

i,k (Rk-1k,k)* Rk-1

k,j Intuition: since the accepting sequence

contains one or more visits to state k, break the path into pieces that first goes from i to its first k-visit (Rk-1

i,k) followed by zero or more revisits to k (Rk-1

k,k) followed by a path from k to j (Rk-1

k,j )

And finally… We get the regular expression(s) that

represent all strings with admissible sequences that start with the initial state (state 1) and end with a final state

Resulting regular expression built from the DFA: the union of all Rn

1,f where f is a final state Note: n is the number of states in the DFA

meaning there are no more restrictions for intermediate states in the accepting sequence

Example

31 2

a a a,b

b b

Describes strings with at least 2 b’s

31 2

a a a,b

b b

i \ j 1 2 31 + a b Φ2 Φ + a b3 Φ Φ + a +

b

R0i,j

31 2

a a a,b

b b

i \ j 1 2 31 a* a*b Φ2 Φ + a b3 Φ Φ + a +

b

R1i,j

31 2

a a a,b

b b

i \ j 1 2 31 a* a*ba* a*ba*b2 Φ a* a*b3 Φ Φ + a +

b

R2i,j

31 2

a a a,b

b b

i \ j 1 2 31 a* a*ba* a*ba*b(a+b)

*2 Φ a* a*b(a+b)*3 Φ Φ (a+b)*

R3i,j

31 2

a a a,b

b b

i \ j 1 2 31 a* a*ba* a*ba*b(a+b)

*2 Φ a* a*b(a+b)*3 Φ Φ (a+b)*

Final answer: R31,3