Remaining Proofs of Section 4

Lemma 10. For two NFAs AandB, the following conditions are equivalent.

1. AutomataA andBare synchronizable.

2. There exist synchronized languagesK^A⊆L(A)andK^B⊆L(B).

Proof. The implication from left to right is immediate. To prove the opposite impli-cation, letK^A=D^A₁ . . . D_k^AandK^B=D^B₁ . . . D^B_k, where D^A_i andD^B_i are synchro-nized in one step, for all 1≤i≤k. Define the n-th canonical word of a singleton language as its unique word, and of a cycle language v_pref(v_mid)^∗v_suff as the word v_pref(v_mid)ⁿv_suff. LetN be the maximum number of states of automataAand B, and let w^A_i and w^B_i be the N-th canonical words of languages D_i^A and D^B_i, re-spectively, for 1≤i ≤k. Let w^A =w^A₁ · · ·w^A_k and w^B =w^B₁ · · ·w_k^B. Notice that w^A∈K^A⊆L(A) and w^B∈K^B ⊆L(B). Consider some of the accepting runs of Aonw^Aand ofBonw^B, respectively,

q^A₁ ^w

A

−−→1 q₂^{A w}

A

−−→2 . . . ^w

A

−−−→k−1 q_k^{A w}

A

−−→k q_k+1^A

and

By definition of run, statesq₁^Aandq₁^Bare initial respectively inAandB, and states q^A_k+1 and q_k+1^B are accepting in A and B, respectively. Thus, to show that A and Bare synchronizable, it is sufficient to show that pairs (q^A_i , q^B_i) and (q_i+1^A , q_i+1^B ) are synchronizable, for all 1≤ i ≤k. Fix some 1 ≤i ≤ k, then there are two cases.

Either bothD^A_i andD_i^B are singletons, or they are cycle languages. Consider first the situation when they are singletons. Then we have w_i^A=w^B_i ∈D_i^A=D_i^B and pairs (q_i^A, q_i^B) and (q_i+1^A , q_i+1^B ) are clearly synchronizable.

Focus now on the situation whereD^A_i andD_i^Bare cycle languages. In this case, D_i^A=v_pref^A (v_mid^A )^∗v^A_suff and D_i^B=v^B_pref(v^B_mid)^∗v_suff^B ,

which shows that statesq_i^Aand q_i+1^A areAlph(v_mid^A )-connected inA, since we have Alph(v_pref^A )∪Alph(v^A_suff)⊆ Alph(v_mid^A ) by definition of languages synchronized in one step. Similarly we can show that q_i^B and q_i+1^B are Alph(v_mid^B )-connected in B.

However, by definition of synchronization in one step, the cycle alphabets ofD_i^Aand D_i^B are the same, so Alph(v_mid^A ) = Alph(v_mid^A ). This shows that the pairs (q^A_i , q^B_i) and (q_i+1^A , q^B_i+1) are synchronizable and completes the proof.

Theorem 11. Let AandBbe two NFAs. Then the languages L(A)andL(B)are separable by piecewise testable languages if and only if the automata A and B are not synchronizable.

Proof. This theorem follows from the previous results. Namely, by Theorem 3, the languagesL(A) andL(B) are separable by piecewise testable languages if and only if there is no infinite zigzag between them. Lemma 9 shows that the existence of a zigzag is equivalent to the existence of two synchronized sublanguages K^A ⊆ L(A) and K^B ⊆ L(B). Finally, by Lemma 10, the existence of two synchronized sublanguages is equivalent to the fact that the automataAandBare synchronizable, which concludes the proof.

Theorem 12. Given two NFAsA andB, it is possible to test in polynomial time whetherL(A) andL(B)can be separated by a piecewise testable language.

Proof. By Theorem 11 it is enough to check whetherAandB are synchronizable.

Let A = (Q^A,Σ, δ^A, q₀^A, F^A) and B = (Q^B,Σ, δ^B, q^B₀, F^B). We will consider the graphSYNCHfor which the vertices are pairs of states ofQ^A×Q^Band the edges cor-respond to pairs of vertices synchronizable in one step. Specifically, there is an edge (p^A, p^B)−→(q^A, q^B) inSYNCHif and only if (p^A, p^B) and (q^A, q^B) are synchronizable in one step. Thus,AandBare synchronizable if and only if a vertex consisting of accepting states is reachable inSYNCHfrom the pair of initial states (q^A₀, q^B₀). Since reachability is testable in PTIME, it is thus sufficient to show how we compute the edges ofSYNCH.

The definition of synchronizability in one step (page 8) consists of two cases.

We refer to the first case assymbol synchronization and to the second case ascycle synchronization.

For two symbol-synchronizable pairs of states (p^A, p^B), (q^A, q^B), there should be an edge (p^A, p^B)→(q^A, q^B) inSYNCH if there exists an a∈Σ such that p^A−→^a q^A andp^B−→^a q^B. Since it is easy to find all these pairs in polynomial time, these edges inSYNCH can be easily constructed.

We now show how to construct the edges for cycle-synchronized states. For two pairs (p^A, p^B) and (q^A, q^B) to be cycle-synchronizable, we require thatp^Aand q^A are Σ0-connected in A, and p^B and q^B are Σ0-connected in B, for (the same) Σ0⊆Σ. We now rephrase this definition using other notions that will be useful in the algorithm.

A pair of states (p^A, p^B)∈Q^A×Q^B has a saturatedΣ0-cycle if there exist two wordsv^A, v^B satisfying

1. Alph(v^A) =Alph(v^B) = Σ₀; 2. p^{A v}

A

−−→p^Ain A; and 3. p^{B v}

B

−−→p^B inB.

We say that there is a Σ0-route from (p^A, p^B)∈Q^A×Q^B to (q^A, q^B)∈Q^A×Q^B if there exist words v^A and v^B in Σ^∗₀ such thatp^{A v}

A

−−→ q^A andp^{B v}

B

−−→q^B. So, in contrast to saturated Σ0-cycles, here we do not require that the alphabetsAlph(v^A) andAlph(v^B) are equal to Σ0.

Note that if a pairV = (q^A, q^B) has a saturated Σ0-cycle and a saturated Σ1 -cycle, then it also has a saturated (Σ0∪Σ1)-cycle (obtained by the concatenation of the two cycles). Thus, for every pair V = (q^A, q^B), there exists a unique maximal alphabet Σ₀ ⊆ Σ such that it has a saturated Σ₀-cycle. (This unique maximal alphabet can be empty if no such saturated cycle exists.) We call this alphabet the saturated cycle alphabet ofV and denote it by Σ^V₀. This means thatV_p= (p^A, p^B) andV_q= (q^A, q^B) are cycle synchronizable if and only if there is aV such that there are Σ^V₀-routes fromV_p toV and from V toV_q.

To find all the cycle synchronizable pairs we can first compute, for every V = (p^A, p^B), the saturated cycle alphabet Σ^V₀. This can be done in polynomial time in the following manner. LetC₀^AandC₀^B be the strongly connected components of Aand Bcontainingp^A andp^B, respectively. For a strongly connected component C, let Alph(C) be the union of all symbols a of Σ that label transitions of the form p −→^a q, where both p and q belong to C. If Alph(C₀^A) = Alph(C₀^B), then

Σ^V₀ equals Alph(C₀^A). Otherwise, set Σ1 = Alph(C₀^A) ∩ Alph(C₀^B) and consider automata A1 and B1 obtained from A and B by removing all transitions labeled by symbols from Σ\Σ₁. Consider the strongly connected componentsC₁^AandC₁^B ofA₁ and B₁ containingp^A and p^B, respectively, and proceed in the same way as before. Continuing this procedure we obtain a sequence of decreasing alphabets Σ₁)Σ₂). . ., hence we perform at most |Σ|iterations. If we arrive at the empty alphabet then we say Σ^V₀ =∅.

We argue that we compute Σ^V₀ correctly. Clearly, if the algorithm returns a set Σ⁰, then Σ⁰⊆Σ^V₀. Conversely, we have that Σ^V₀ ⊆Σ⁰ because, at each point in the algorithm, the alphabet under consideration contains Σ^V₀. (In the first iteration, Σ^V₀ ⊆Alph(C₀^A) and Σ^V₀ ⊆ Alph(C₀^B). Furthermore, at each iteration i, if Σ^V₀ ⊆ Alph(C_i^A) and Σ^V₀ ⊆Alph(C_i^B), then Σ^V₀ ⊆(Alph(C_i^A)∩Alph(C_i^B)).)

Once we know, for each pair V = (q^A, q^B), its saturated cycle alphabet Σ^V₀, we can find all verticesVp such that there is a Σ^V₀-route from Vpto V and all vertices Vq such that there is a Σ^V₀-route from V to Vq, and add edges Vp → Vq to the graphSYNCH. This concludes the construction ofSYNCHand the presentation of the algorithm. We note that our algorithm is clearly not yet time-optimal.