Formal Languages, Automata and Compilers

Transcription

1 Formal Languages, Automata and Compilers Lecure LFAC ( ) Lecture 4 1 / 31

2 Curs 4 1 Grammars of type 3 and finite automata 2 Closure properties for type 3 languages 3 Regular Expressions 4 The automaton equivalent to a regular expression LFAC ( ) Lecture 4 2 / 31

3 Grammars of type 3 and finite automata Lecture 4 1 Grammars of type 3 and finite automata 2 Closure properties for type 3 languages 3 Regular Expressions 4 The automaton equivalent to a regular expression LFAC ( ) Lecture 4 3 / 31

4 Grammars of type 3 and finite automata From regular grammars to finite automata For any regular grammar G (in normal form) there exists a nondeterministic finite automaton A such that L(A) = L(G): In grammar G In automaton A T Σ = T N Q = N {f}, F = {f} S q 0 = S q ap p δ(q, a) q a f δ(q, a) if S ǫ add S to F LFAC ( ) Lecture 4 4 / 31

5 Grammars of type 3 and finite automata From finite automata to regular grammars For any deterministic finite automaton there exists a regular grammar G such that L(A) = L(G): In automaton A Σ Q In grammar G T = Σ N = Q q 0 S = q 0 δ(q, a) = p δ(q, a) F if q 0 F q ap q a add rule q 0 ǫ LFAC ( ) Lecture 4 5 / 31

6 Closure properties for type 3 languages Lecture 4 1 Grammars of type 3 and finite automata 2 Closure properties for type 3 languages 3 Regular Expressions 4 The automaton equivalent to a regular expression LFAC ( ) Lecture 4 6 / 31

7 Closure properties for type 3 languages Closure under complement and set difference operations If L L 3 then L = (Σ \ L) L 3 Let A = (Q,Σ,δ, q 0, F) be a finite automaton with L(A) = L. The automaton A which accepts L = L(A): A = (Q,Σ,δ, q 0, Q \ F) LFAC ( ) Lecture 4 7 / 31

8 Closure properties for type 3 languages Closure under complement and set difference operations If L L 3 then L = (Σ \ L) L 3 Let A = (Q,Σ,δ, q 0, F) be a finite automaton with L(A) = L. The automaton A which accepts L = L(A): A = (Q,Σ,δ, q 0, Q \ F) If L 1, L 2 L 3 then L 1 \ L 2 L 3 : L 1 \ L 2 = L 1 L 2 LFAC ( ) Lecture 4 7 / 31

9 Closure properties for type 3 languages Closure under product Let A 1 = (Q 1,Σ,δ 1, q 01,{f 1 }) and A 2 = (Q 2,Σ,δ 2, q 02,{f 2 }) automata with a single final state such that L 1 = L(A 1 ) and L 2 = L(A 2 ). Automaton A (with ǫ-transitions) which accepts L 1 L 2 : A = (Q 1 Q 2,Σ,δ, q 01,{f 2 }) LFAC ( ) Lecture 4 8 / 31

10 Closure properties for type 3 languages Closure under union Let A 1 = (Q 1,Σ 1,δ 1, q 01,{f 1 }) and A 2 = (Q 2,Σ 2,δ 2, q 02,{f 2 }) automata with a single final state such that L 1 = L(A 1 ) and L 2 = L(A 2 ). Automaton A (with ǫ-transitions) which accepts L 1 L 2 : A = (Q 1 Q 2 {q 0, f},σ 1 Σ 2,δ, q 0,{f}) LFAC ( ) Lecture 4 9 / 31

11 Closure properties for type 3 languages Closure under iteration Let A = (Q,Σ,δ, q 01,{q f }) automaton with a single final state such that L(A) = L. Automaton A (with ǫ-transitions) which accepts L (= L(A) ): A = (Q {q 0, f},σ,δ, q 0,{f}) LFAC ( ) Lecture 4 10 / 31

12 Regular Expressions Lecture 4 1 Grammars of type 3 and finite automata 2 Closure properties for type 3 languages 3 Regular Expressions 4 The automaton equivalent to a regular expression LFAC ( ) Lecture 4 11 / 31

13 Regular Expressions Regular expressions - definition Definition 1 If Σ is an alphabet, then a regular expression over Σ can be defined by induction:, ǫ, a (a Σ) are regular expressions which describe the languages:, {ǫ}, {a}. If E, E 1, E 2 are regular expressions, then: (E 1 E 2 ) is a regular expression which describes the language L(E 1 ) L(E2) (E 1 E 2 ) is a regular expression which describes the language L(E 1 )L(E 2 ) (E ) is a regular expression which describes the language L(E) The priority order for the operators is:,, LFAC ( ) Lecture 4 12 / 31

14 Regular Expressions Examples (a b) (c d) {a, b, c, d} (0 1) (0 1) {00, 01, 10, 11} a b {a n b k n, k 0} (a b) {a, b} ( )( ) describes the set of positive integers (a b c... z)(a b c... z ) describes the set of identifiers Two regular expressions E 1, E 2 are equivalent, written E 1 = E 2, if L(E 1 ) = L(E 2 ) LFAC ( ) Lecture 4 13 / 31

15 Regular Expressions Properties (p q) r = p (q r) (pq)r = p(qr) p q = q p p ǫ = ǫ p = p p = p p = p p = p = p(q r) = pq pr (p q)r = pr qr ǫ pp = p ǫ p p = p LFAC ( ) Lecture 4 14 / 31

16 Regular Expressions From a regular expression to a finite automaton Theorem 1 For any regular expression E over the alphabet Σ, there existis a finite automaton (with ǫ - transitions) A, such that L(A) = L(E). Proof : structural induction if E {,ǫ, a} (a Σ) then the corresponding automata: LFAC ( ) Lecture 4 15 / 31

17 Regular Expressions Proof E = E 1 E 2 E = E 1 E 2 E = E 1 LFAC ( ) Lecture 4 16 / 31

18 Regular Expressions Representing regular expressions as trees Input: the regular expression E = e 0 e 1...e n 1 Precedence of operators: prec( ) = 1, prec( ) = 2, prec( ) = 3 (prec(()= 0). Output: The corresponding tree: t. Method: Two stacks: STACK1 operators stack STACK2 trees stack (contains trees for sub-expressions of the initial regular expression) Method tree(r, tl, tr), whre r is the root node, tl the left sub-tree tr the right sub-tree LFAC ( ) Lecture 4 17 / 31

19 Regular Expressions i = 0; while(i < n) { c = e i ; switch(c) { case ( : { STACK1.push(c); break; } case symbol (from alphabet): { STACK2.push(tree(c,NULL,NULL)); break; } case operator: { while (prec(stack1.top())>=prec(c)) build tree(); STACK1.push(c); break; } case ) : { do { build tree();} while(stack1.top()!= ( ); STACK1.pop(); break; } } i++; } while(stack1.not empty()) build tree(); t = STACK2.pop(); LFAC ( ) Lecture 4 18 / 31

20 Regular Expressions build tree() op = STACK1.pop(); t1 = STACK2.pop(); switch (op) { case : { new tree = tree(op, t1, NULL); STACK2.push(new tree); break; } case, : { t2 = STACK2.pop(); new tree = tree(op, t2, t1); STACK2.push(new tree); break; } } LFAC ( ) Lecture 4 19 / 31

21 Regular Expressions Example a b a (b c) LFAC ( ) Lecture 4 20 / 31

22 The automaton equivalent to a regular expression Lecture 4 1 Grammars of type 3 and finite automata 2 Closure properties for type 3 languages 3 Regular Expressions 4 The automaton equivalent to a regular expression LFAC ( ) Lecture 4 21 / 31

23 The automaton equivalent to a regular expression The automaton equivalent to a regular expression E = E 1 E 2 E = E 1 E 2 E = E 1 LFAC ( ) Lecture 4 22 / 31

24 The automaton equivalent to a regular expression Remarks for any occurrence of a symbol from Σ or ǫ in E, 2 states must be added in the equivalent automaton. for every occurrence of and in a regular expression E, two more states must be added for operator no additional states must be added if n is the number of symbols from E and m is the number of brackets and operators, then the number of states of the automaton equivalent to E is p = 2(n m). LFAC ( ) Lecture 4 23 / 31

25 The automaton equivalent to a regular expression Input: The regular expression E with n symbols, from which m are brackets and product operators; Output: The automaton with ǫ - transitions equivalent to E; Method: 1. Build the tree corresponding to expression E; 2. Traverse the tree in pre-order and attach to visited nodes (except those labeled with the product operator) the numbers 1, 2,...,n m; LFAC ( ) Lecture 4 24 / 31

26 The automaton equivalent to a regular expression Example E = a b c LFAC ( ) Lecture 4 25 / 31

27 The automaton equivalent to a regular expression 3. Traverse the tree in post-order and attach to each node N a pair of numbers (N.i, N.f) which represent the initial and the final state of the automaton equivalent to the expression corresponding to the sub-tree with root N, as follows: If the node has the number k (from step 2) then: N.i = 2k 1, N.f = 2k; If the node is labeled with the product operator, and L is the left child of N, R the right child, then: N.i = L.i and N.f = R.f LFAC ( ) Lecture 4 26 / 31

28 The automaton equivalent to a regular expression Example E = a b c LFAC ( ) Lecture 4 27 / 31

29 The automaton equivalent to a regular expression 4. Traverse again the tree in post-order. If N is the current node and L and R its children, then: LFAC ( ) Lecture 4 28 / 31

30 The automaton equivalent to a regular expression 4. Traverse again the tree in post-order. If N is the current node and L and R its children, then: If N is labelled with a (is a leaf): δ(n.i, a) = N.f LFAC ( ) Lecture 4 28 / 31

31 The automaton equivalent to a regular expression 4. Traverse again the tree in post-order. If N is the current node and L and R its children, then: If N is labelled with a (is a leaf): δ(n.i, a) = N.f If N is labelled with : δ(n.i,ǫ) = {L.i, R.i}, δ(l.f,ǫ) = N.f, δ(r.f,ǫ) = N.f LFAC ( ) Lecture 4 28 / 31

32 The automaton equivalent to a regular expression 4. Traverse again the tree in post-order. If N is the current node and L and R its children, then: If N is labelled with a (is a leaf): δ(n.i, a) = N.f If N is labelled with : If N is labelled with : δ(n.i,ǫ) = {L.i, R.i}, δ(l.f,ǫ) = N.f, δ(r.f,ǫ) = N.f δ(l.f,ǫ) = R.i LFAC ( ) Lecture 4 28 / 31

33 The automaton equivalent to a regular expression 4. Traverse again the tree in post-order. If N is the current node and L and R its children, then: If N is labelled with a (is a leaf): δ(n.i, a) = N.f If N is labelled with : If N is labelled with : δ(n.i,ǫ) = {L.i, R.i}, δ(l.f,ǫ) = N.f, δ(r.f,ǫ) = N.f δ(l.f,ǫ) = R.i If N is labelled with (R does not exist in this case): δ(n.i,ǫ) = {L.i, N.f}, δ(l.f,ǫ) = {L.i, N.f} LFAC ( ) Lecture 4 28 / 31

34 The automaton equivalent to a regular expression 4. Traverse again the tree in post-order. If N is the current node and L and R its children, then: If N is labelled with a (is a leaf): δ(n.i, a) = N.f If N is labelled with : If N is labelled with : δ(n.i,ǫ) = {L.i, R.i}, δ(l.f,ǫ) = N.f, δ(r.f,ǫ) = N.f δ(l.f,ǫ) = R.i If N is labelled with (R does not exist in this case): δ(n.i,ǫ) = {L.i, N.f}, δ(l.f,ǫ) = {L.i, N.f} 5. The initial state of the automaton is N.i, the final state is N.f, where N is the root node of the tree LFAC ( ) Lecture 4 28 / 31

35 The automaton equivalent to a regular expression Example E = a b c LFAC ( ) Lecture 4 29 / 31

36 The automaton equivalent to a regular expression Example δ a b c ǫ 1 {3, 5} {2} 5 {6, 7} 6 {9} {6, 7} {2} LFAC ( ) Lecture 4 30 / 31

37 The automaton equivalent to a regular expression The correctness of the algorithm Theorem 2 The automaton with ǫ - transitions built by the algorithm is equivalent to the expression E. Proof: The pairs of states (i, f ) for each node of the tree correspond to the constructions in Theorem 1. The transitions defined in the step 5 of the algorithm correspond to the constructions in Theorem 1. LFAC ( ) Lecture 4 31 / 31