On the intersection of submonoids of the free monoid

1. On the intersection of submonoids of the free monoid L. Giambruno A. Restivo

2. Content • Introduction • Some useful definitions • Correspondence submonoids-monoidal automata • Product of two flower automata • Prefix case • Conclusions

3. Introduction • Characterization of the intersection of two free submonoids of rank two • Upper bound for the rank of the intersection of two finitely generated submonoids if the intersection is finitely generated. The purpose of our research is to study the intersection of two finitely generated submonoids of the free monoid on a finite alphabet. In particular:

4. Motivation • The study of the intersection of two free submonoids of fixed rank is not trivial. In fact by a result of M.Latteux and J.Leguy every regular language is obtained as omomorphic image of the intersection of two finitely generared monoids: Theorem Let A be an alphabet and R a language of A*. R is a regular language if and only if there exist two finite languages F1 , F2 and a morphism g such that • Karhumaki’s characterization on the intersection of two free submonoids of rank two. Theorem (Prefix case) Let H,K free submonoids of A*generated by prefix sets of two words. If then is generated by at most two elements. If then .

5. The ‘Hanna Neumann conjecture’ for subgroups of the free group: ‘If H,K are finitely generated subgroups of a free group F then , where for every T rk’(T)=max(rk(T)-1, 0) with rk(T) the rank of T.’ • Meakin and Weil in 2002 proved the Hanna Neumann conjecture for subgroups positively generated of a free group. • The ‘Hanna Neumann conjecture for submonoids’ : ‘Let H,K are finitely generated submonoids of A* with A finite alphabet. If is finitely generated then ,where for every T rk’(T)=max(rk(T)-1, 0) with rk(T) the rank of T.’ • We can conjecture also the Hanna Neumann conjecture in the free case. In this case by the analogies with the conjecture in free groups we think that it is more probably true. • Our approach is with automata.

6. b c Example 1:X={c,ab,acb} i 2 a b a AX 4 c 3 Some useful definition The flower automaton Let A be a finite alphabet. By Berstel and Perrin, given a finite language X in A* we can associate an automaton AX that recognizes the submonoid X* of A* called theflower automaton. Such automaton has the properties that: 1) It has a unique initial and final state (i) 2) All the cycles visit (i) 3) All the cycles intersect themselves only in (i) 4)The cycles in (i) without (i) as intermediate vertex have as labels the words of X. We can construct AX creating a fixed vertex (i), creating for every xiin X a cycle in (i) with label xi and letting (i) be the unique initial and final vertex.

7. b Example 2: c 2 i a c b 3 Correspondence submonoids- monoidal automata Definition: Let A=(A,Q,i,F,t) be a non deterministic trim automaton. A is a semiflower automatonif A is a monoidal automaton such that every cycle in A visit i. Definition: Let A=(A,Q,i,F,t) be a non deterministic trim automaton. A is a monoidal automatonif F={i}. If A is a monoidal automaton then H=L(A) is a submonoid. If H is a finitely generated submonoid then there exists a monoidal automaton recognizing H. Classes closed under this correspondence Let Abe a monoidal automaton recognizing a submonoid H, then: • A is a semiflower automaton if and only ifH is a finitely generated submonoid • Ais anunambiguos automaton if and only ifH is a free submonoid • If A is a deterministic monoidal automaton thenH is a submonoid generated by a prefix set.

8. b c A Example 3: L(A)={c,ab,acb}* rk(L(A))=3 e=6, v=4, rk(L(A))=6-4+1=3 i 2 a b a 3 4 c Theorem: If A is a semi-flower automaton with v vertices and e edges and H=L(A) then rk(H) e-v+1. Moreover, if A is unambiguos then rk(H) = e-v+1. We remark that a similar result holds for free groups. • Definition: Given a graph G, we say that a vertex v is a branch point (bp in short) if the number of edges going out from v is greater than two. • In Example 3 the vertex (i) is a bp.

9. b c c c {2,3} 4 i 2 i a a b Example 4:X={c,ab,acb} a b 4 c 3 AX AXD b Theorem: Let A2={a,b}. If Ais a deterministic semi-flower automaton with language non empty and with v vertices and e edges on A2 then e - v = # bp. Remark: Trivially if L(A) is empty then e – v=-1 Given a submonoid H=X* with X finite set, let AXD be the automaton obtained from AXby applying the subset construction and then considering only the set of states accessible and coaccessible.

10. Example 5: X={a,aab} a AXD AX a a a {1,2} {1,2,3} i 1 2 a a b b 3 • If X is not a prefix set then AXD is not necessarly a semi-flower automaton, as in the following example Theorem: LetH=X*with X finite prefix set then AXD is a deterministic semi-flower automaton. Given a free finitely generated submonoid T, let rk’(T)=max(rk(T)-1, 0). Unifying the previous results we get: Theorem: LetH=X*with X finite prefix set of A2*, then AXD is a deterministic semi-flower automaton and rk’(H)=# bp. In Example 4: rk’(H)=2=# bp

11. Product of two flower automata Let A1 and A2 be two automata and let A1XA2 be the product automaton of A1 and A2.. • L(A1 X A2)= L(A1) L(A2) • The product of two automata with a unique initial and final state is still an automaton with a unique initial and final state • The product of two deterministic automata is still a deterministic automaton Remark1: The product of two semi-flower automata is not necessarly a semi-flower automaton. Remark2: The product of two trim automata is not necessarly a trim automaton. • Given an automaton A,let AT denote the set of accessible and coaccessible states of A.

12. 3 a b a b a 3’ 1’ 2’ 2 1 a a a b b a 5’ 4’ A1 X A2 4 a b a (1,1’) (2,1’) (3,1’) (1,2’) a (2,3’) a a b a (4,2’) (4,5’) (1,5’) (3,4’) b a b (2,4’) (1,3’) a A2 A1 Example 6:

13. 4’ a b a a b 3’ 1’ 2’ b a 6’ 5’ a A1X A2 a (2,6’) (1,5’) (A1X A2 )T b b b a a b (1,1’) (2,2’) (1,3’) (2,4’) (1,1’) A2 b A1 Example 7: a 1 2 b

14. We let bp(A1 X A2) be the set of vertices (u,v) in A1XA2 bp for A1XA2. • We let bp(A1 ) X bp(A2) be the set of vertices (u,v) in A1XA2 such that u is a bp of A1 and v is a bp of A2. Lemma 1: If A1 and A2 are deterministic automata then bp(A1 X A2)= bp(A1) X bp(A2). • As remarked before the product of two trim automata is not necessarly trim. Lemma 2: If A1 and A2 are deterministic automata then bp(A1 X A2)T bp(A1) X bp(A2).

15. Prefix case Let H and K be submonoids finitely generated by prefix sets of A2* . • All the results proved in the binary alphabet A2 are also proved in a finite arbitrary alphabet A. Let AH and AK be the associated flower automata and AHDand AKD as before defined. • In Example 6 A1= AH with H={aab,aba}* and A2= AK K={a,baaba}* In Example 7 A1= AH with H={b,ab}* and A2= AK K={b,abbaa,abaa}* Applying Lemma 2 to (AHDX AKD)T we get the following theorem: Theorem Let H,K are submonoids finitely generated by prefix sets of A2* with A2 ={a,b}.If is finitely generated then .

16. B D E A H AHD KAKD 1’ v 1 u C F b a (1,1’) (u,v) d (AHDXAKD)T b1 a1 a2 (1,1’) (x,y) (u,v) d Prefix case: rank two Theorem (Karhumaki,Prefix case) Let H,K are free submonoids of A*generated by prefix sets of two words. If then is generated by at most two elements. If then . where a=a1a2 , b=a2b1

17. AYD AXD 3 a b a b a 3’ 1’ 2’ 2 1 a a a b b a AXD X AYD 5’ 4’ 4 a b a (1,1’) (2,1’) (3,1’) (1,2’) a (2,3’) a a b a (4,2’) (4,5’) (1,5’) (3,4’) b a b AXD X AXD = (AXD X AYD)T L((AXD X AYD)T )= (2,4’) (1,3’) a Example 4: X={aab,aba}* , Y={a,baaba}*

18. Conclusions Not prefix case: • We use the same techniques • Non-deterministic approach • Partial result

19. Thanks!