Parsing Regular Expressions and Regular Grammars
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1 Regular Expressions and Regular Grammars Laura Heinrich-Heine-Universität Düsseldorf Sommersemester 2011 Regular Expressions (1) Let Σ be an alphabet The set of regular expressions over Σ is recursively defined: is a regular expression denoting the set is a regular expression denoting the set {} For each a Σ: a is a regular expression denoting {a} If r and s are regular expressions denoting R and S, then (r + s), (rs) and (r ) are also regular expressions denoting R S, RS and R respectively Regular Grammars April 2011 Regular Grammars April 2011 Overview 1 Regular expressions 2 NFA with empty transitions 3 FSA and regular expressions 4 Regular Grammars 5 Regular Grammars and NFAs Regular Expressions (2) Assume that has a higher priority than concatenation which in turn has a higher priority than + a + is an abbreviation for aa For a regular expression x, we define L(x) as the set denoted by x Regular Grammars April 2011 Regular Grammars April 2011
2 Regular Expressions (3) The following holds: For each language L: there is a regular expression x with L = L(x) iff there is a FSA M with L = L(M) In order to show this, we define NFAs with empty transitions and show their equivalence to NFAs; show how to construct an equivalent NFA with empty transitions for a given regular expression; NFA with empty transitions (2) Let Q, Σ, δ, q 0, F be an NFA with empty transitions Definition of ˆδ: -closure(q) := {q there is a path from q to q containing only -transitions} {q} ˆδ(q, ) := -closure(q) ˆδ(q, wa) := p P -closure(p) where P = {p there is an r ˆδ(q, w) such that p δ(r, a)} show how to construct an equivalent regular expression for a given DFA Regular Grammars April 2011 Regular Grammars April 2011 NFA with empty transitions (1) A NFA with empty transitions is defined like an NFA except that δ allows for ε as second argument: A NFA with empty transitions M is a quintuple Q, Σ, δ, q 0, F such that Q, Σ, q 0 and F are like in a NFA δ : Q (Σ {ε}) P(Q) is the transition function NFA with empty transitions (3) Construction of an equivalent NFA Q, Σ, δ, q 0, F (without empty transitions) for a given NFA Q, Σ, δ, q 0, F with empty transitions: F := F {q 0 } if -closure(q 0 ) F Otherwise, F := F δ (q, a) := ˆδ(q, a) DFAs, NFAs and NFAs with empty transitions are equivalent Regular Grammars April 2011 Regular Grammars April 2011
3 FSA and regular expressions (1) To show: 1 For a given regular expression r there exists an NFA with empty transitions that accepts L(r) The set of languages denoted by regular expressions is a subset of the set of languages accepted by FSAs 2 For a given DFA, there exists a regular expression r such that L(r) is exactly the language acctepted by the DFA The set of languages accepted by FSAs is a subset of the set of languages denoted by regular expressions FSA and regular expressions (3) r = r 1 + r 2 : 0 r = r 1 r 2 : f1 r = r 1 : q 0 qf1 qf2 2 1 q f1 qf2 qf0 Regular Grammars April 2011 Regular Grammars April 2011 FSA and regular expressions (2) Construction of an equivalent NFA with empty transitions for a given regular expression r: r = : q0 r = : 0 qf r = a, a Σ: a 0 qf FSA and regular expressions (4) To show: For each DFA, there is an equivalent regular expression Assume that {q 1,, q n } it the set of states Define Ri,j k as the set of sequences of input symbols that can be obtained from following all paths from q i to q j that do not pass through states q l with l > k Question: What do the Ri,j k look like and what are the corresponding regular expressions ri,j k? k = 0: Only paths of length 0 or 1 can be considered, since, between q i and q j, no other states q l are possible Each r 0 i,j has the form a 1 + a a m or (if i = j) a 1 + a a m + or (if i j and there is no edge from q i to q j, ) Regular Grammars April 2011 Regular Grammars April 2011
4 FSA and regular expressions (5) In general, R k ij contains all elements from R k 1 ij, all concatenation of a) an element from R k 1 ik, b) any number (including 0) of elements from R k 1 kk and c) an element from R k 1 Consequently, r k i,j = rk 1 ij + r k 1 ik (rk 1 kk ) r k 1 Finally, the regular expression for the language accepted by the automaton is r n 1f1 + rn 1f2 + + rn 1fm if F = {q f1,, q fm } FSA and regular expressions (7) The JFLAP conversion uses a different definition: Rij k is the set obtained by using no states q l with l < k Then ri,j k = rk+1 ij + r k+1 r 1 i,j ik (rk+1 kk ) r k+1 is the regular expression for the language accepted by the DFA With this, we obtain (ab c) ab for the sample automaton Regular Grammars April 2011 Regular Grammars April 2011 FSA and regular expressions (6) Regular Grammars (1) Example: abcsjff A type 2 grammar is called r 2 12 = r r 1 12(r 1 22) r 1 22 right-linear iff for all productions A β: β T or β = β X with β T, X N r12 1 = r r0 11 (r0 11 ) r12 0 = a + ( )a = a left-linear iff for all productions A β: β T or β = Xβ with β T, X N r 1 22 = r r0 21 (r0 11 ) r 0 12 regular if it is left-linear or right-linear = (b + ) + c() a = + b + ca For each left-linear grammar there is an equivalent right-linear grammar and vice versa r 2 12 = a + a( + b + ca)+ = a( + b + ca) = a(b + ca) Regular Grammars April 2011 Regular Grammars April 2011
5 Regular Grammars (2) The following holds: L = L(G) for a regular grammar G iff L is a regular set (ie, can be described by a regular expression) To show this, we show that for each right-linear grammar, there exists an equivalent NFA; and for each DFA, there is an equivalent right-linear grammar Regular Grammars (4) Construction of a right-linear grammar for given DFA: The states are the nonterminals The start state is the start symbol For each transition a Q Q add a production Q aq and, if Q F, a production Q a If we construct the right-linear grammar for L R (L reversed) and we mirror all right-hand sides of productions, we obtain a left-linear grammar for L Regular Grammars April 2011 Regular Grammars April 2011 Regular Grammars (3) Construction of equivalent NFA (with empty transitions) for given right-linear G: Start state [S] For all productions A β 1 β 2 there is a state [β 2 ] Transitions simulate top-down left-to-right traversal of derivation tree: For every A β: [A] [β] For every state [aβ] with a T: a [aβ] [β] Final state is [] Regular Grammars April 2011
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