£ CS 373: Theory of Computation £ Madhusudan Parthasarathy Lecture 4: The product construction: Closure un- der intersection and union 28 January 2010 This lecture nishes section 1.1 of Sipser and also covers the start of 1.3. 1 Product Construction 1.1 Product Construction: Example Let Σ = fa; bg and L is the set of strings in Σ∗ that have the form a∗b∗ and have even ∗ ∗ ∗ length. L is the intersection of two regular languages L1 = a b and L2 = (ΣΣ) . We can show they are regular by exhibiting DFAs that recognize them. r0 a b a,b a,b a,b b a q0 q1 drain r1 L1 L2 We can run these two DFAs together, by creating states that remember the states of both machines. (q0,r0) (q1,r0) (drain, r0) b a a a a b b b a,b a,b (q0,r1) (q1,r1) (drain, r1) Notice that the nal states of the new DFA are the states (q; r) where q is nal in the rst DFA and r is nal in the second DFA. To recognize the union of the two languages, rather than the intersection, we mark all the states (q; r) such that either q or r are accepting states in the their respective DFAs. State of a DFA after reading a word w. In the following, given a DFA M = (Q; Σ; δ; q0;F ), we will be interested in the state the DFA M is in, after reading the a string w. Let us denote 1 by δ∗ : Q × Σ∗ ! Q the function such that δ∗(q; w) is the state the DFA will land in, if started from state q and fed the word w. Formally, we dene δ∗ : Q × Σ∗ ! Q using the following inductive denition: • δ∗(q; ) = q for every q 2 Q, • δ∗(q; wa) = δ(δ∗(q; w); a) for each q 2 Q; w 2 Σ∗; a 2 Σ. ∗ ∗ Note that by the above denition of δ , we have that w 2 L(M) i δ (q0; w) 2 F . 2 Product Construction: Formal construction We are given two DFAs and 0 0 0 0 0 both working over M = (Q; Σ; δ; q0;F ) M = (Q ; Σ; δ ; q0;F ) the same alphabet Σ.A product automaton of M and M 0 is an automaton 0 N = Q; Σ; δN ; (q0; q0) ;FN ; 0 0 0 where Q = Q × Q , and δN : Q × Σ !Q. Also, for any q 2 Q, q 2 Q and c 2 Σ, we require 0 0 0 δN ((q; q ) ; c ) = δ(q; c); δ (q ; c) : (1) | {z } state of N The set FN ⊆ Q of accepting states is free to be whatever we need it to be, depending on what we want N to recognize. For example, if we would like N to accept the intersection 0 0 L(M) \ L(M ) then we will set FN = F × F . If we want N to recognize the union language 0 0 0 L(M) [ L(M ) then FN =(F × Q ) [[(Q × F ). Lemma 2.1 For any input word w 2 Σ∗, the product automata N of the DFAs M = and 0 0 0 0 0 , is in state 0 after reading , if and only (Q; Σ; δ; q0;F ) M = (Q ; Σ; δ ; q0;F ) (q; q ) w if (i) M in the state q after reading w, and (ii) M 0 is in the state q0 after reading w. In other words, ∗ 0 ∗ 0∗ 0 . δN ((q0; q0); w) = (δ (q0; w); δ (q0; w)) Proof: The proof is by induction on the length of the word w. If is the empty word, then is initially in the state 0 by construction, where w = N (q0; q0) (resp. 0 ) is the initial state of (resp. 0). As such, the claim holds in this case. q0 q0 M M Formally, ∗ 0 0 ∗ 0∗ 0 (by denition of ∗ , ∗ and 0∗). δN ((q0; q0); ) = (q0; q0) = (δ (q0; ); δ (q0; )) δN δ δ ∗ Otherwise, jwj > 0, and let us hence assume w = w1a (w1 2 Σ ; a 2 Σ), and assume the induction hypothesis that the claim is true for all input words of length strictly smaller than jwj (in particular jw1j). Let 0 be the state that is in after reading the string . By the induction (qk−1; qk−1) N w1 hypothesis, as , we know that is in the state after reading , and 0 is jw1j = k − 1 M qk−1 wb M in the state 0 after reading . qk−1 wb ∗ Let qk = δ(qk−1; a) = δ(δ (q0; w1) ; a) = δ(q0; w) and 0 0 0 0 0 0 0 0 qk = δ (qk−1; a) = δ (δ (q0; w1) ; a) = δ (q0; w) : 2 As such, by denition, (resp. 0) would in the state (resp. 0 ) after reading . M M qk qk w Also, by the denition of its transition function, after reading w the DFA N would be in the state 0 ∗ 0 0 δN ((q0; q0); w) = δN (δN ((q0; q0); w1) ; a) = δN (qk−1; qk−1); a 0 0 = δ(qk−1; a); δ(qk−1; a) =(qk; qk) ; (see Eq. (1)). This establishes the claim. Lemma 2.2 Let and 0 0 0 0 0 be two given DFAs. Let M = (Q; Σ; δ; q0;F ) M = (Q ; Σ; δ ; q0;F ) N be a product automaton with set of accepting states is F × F 0. Then L(N) = L(M) \ L(M 0). Proof: i ∗ 0 0 w 2 L(N) δN ((q0; q0); w) 2 F × F i ∗ 0∗ 0 0 (by Lemma 2.1) (δ (q0; w); δ (q0; w)) 2 F × F i ∗ and 0∗ 0 0 δ (q0; w) 2 F δ (q0; w) 2 F i w 2 L(M) and w 2 L(M 0). More verbosely: If 0 , then let ∗ and 0 0∗ 0 0. By w 2 L(M) \ L(M ) qw = δ (q0; w) 2 F qw = δ (q0; w) 2 F Lemma 2.1, this implies that ∗ 0 0 0. Namely, accepts the δN ((q0; q0); w) = (qw; qw) 2 F × F N word w, implying that w 2 L(N), and as such L(M) \ L(M 0) ⊆ L(N). Similarly, if , then 0 ∗ 0 must be an accepting state of w 2 L(N) (pw; pw) = δN ((q0; q0); w) . But the set of accepting states of is 0. That is 0 0, implying N N F × F (pw; pw) 2 F × F that and 0 0. Now, by Lemma 2.1, we know that ∗ and pw 2 F pw 2 F δ (q0; w) = pw 2 F 0∗ 0 0 0. Thus, and 0 both accept , which implies that and δ (q0; w) = pw 2 F M M w w 2 L(M) w 2 L(M 0). Namely, w 2 L(M) \ L(M 0), implying that L(N) ⊆ L(M) \ L(M 0). Putting the above together proves the lemma. 3.