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 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: 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.

4. 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 F 1, F 2 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. Motivation

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. 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 A X that recognizes the submonoid X * of A * called the flower 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 A X creating a fixed vertex (i), creating for every x i in X a cycle in (i) with label x i and letting (i) be the unique initial and final vertex. Example 1: X={c,ab,acb} b i 4 3 2 c c b a a AXAX

7. Definition: Let A =(A,Q,i,F,t) be a non deterministic trim automaton. A is a monoidal automaton if F={i}. Correspondence submonoids- monoidal automata 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 A be a monoidal automaton recognizing a submonoid H, then: A is a semiflower automaton if and only if H is a finitely generated submonoid A is an unambiguos automaton if and only if H is a free submonoid If A is a deterministic monoidal automaton then H is a submonoid generated by a prefix set. Definition: Let A =(A,Q,i,F,t) be a non deterministic trim automaton. A is a semiflower automaton if A is a monoidal automaton such that every cycle in A visit i. b c i 2 c a 3 b Example 2:

8. 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. b i 2 4 3 a c b a c Example 3: L( A )={c,ab,acb} * rk(L( A ))=3 e=6, v=4, rk(L( A ))=6-4+1=3 A

9. Theorem: Let A 2 ={a,b}. If A is a deterministic semi-flower automaton with language non empty and with v vertices and e edges on A 2 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 A X D be the automaton obtained from A X by applying the subset construction and then considering only the set of states accessible and coaccessible. Example 4: X={c,ab,acb} AXAX i 4 3 2 c c b a a b i 4 c c b a b {2,3} AXDAXD

10. Theorem: Let H=X* with X finite prefix set then A X D 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: Let H=X* with X finite prefix set of A 2 *, then A X D is a deterministic semi-flower automaton and rk’(H)=# bp. In Example 4: rk’(H)=2=# bp Example 5: X={a,aab} i a b a {1,2} a 1 3 2 a a b {1,2,3} a AXAX AXDAXD If X is not a prefix set then A X D is not necessarly a semi-flower automaton, as in the following example

11. Product of two flower automata Let A 1 and A 2 be two automata and let A 1 X A 2 be the product automaton of A 1 and A 2.. L( A 1 X A 2 )= L( A 1 ) L( A 2 ) 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 A T denote the set of accessible and coaccessible states of A.

12. Example 6: A1A1 A2A2 1 4 b a a 2 3 b a 1’ 4’ 3’ 2’ a a b ab a 5’ A 1 X A 2 (1,1’) a (2,1’) (4,2’) (1,3’)(2,4’) (4,5’) (2,3’) (1,2’)(3,1’) (1,5’)(3,4’) a a b a a b b a a b a

13. Example 7: A1A1 A2A2 1’ 5’ 3’ 2’ a b ba b 6’ a 4’ a a A 1 X A 2 a (2,2’) (2,6’)(1,5’) (2,4’)(1,3’) b b a a (1,1’) b b ( A 1 X A 2 ) T b 1 2 b a

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

15. Prefix case Let H and K be submonoids finitely generated by prefix sets of A 2 *. All the results proved in the binary alphabet A 2 are also proved in a finite arbitrary alphabet A. Let A H and A K be the associated flower automata and A H D and A K D as before defined. In Example 6 A 1 = A H with H={aab,aba} * and A 2 = A K K={a,baaba} * In Example 7 A 1 = A H with H={b,ab} * and A 2 = A K K={b,abbaa,abaa} * Applying Lemma 2 to ( A H D X A K D ) T we get the following theorem: Theorem Let H,K are submonoids finitely generated by prefix sets of A 2 * with A 2 ={a,b}. If is finitely generated then.

16. 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. (AHDXAKD)T(AHDXAKD)T    ( 1,1’ ) (u,v)   ( 1,1’ )  (x,y)(u,v)   1 u 1’ v H A H D K A K D C  E D F where        

17. Example 4: X={aab,aba} *, Y={a,baaba} * AXDAXD AYDAYD 1 4 b a a 2 3 b a 1’ 4’ 3’ 2’ a a b ab a 5’ A X D X A Y D (1,1’) a (2,1’) (4,2’) (1,3’)(2,4’) (4,5’) (2,3’) (1,2’)(3,1’) (1,5’)(3,4’) a a b a a b b a a b a A X D X A X D = (A X D X A Y D ) T L((A X D X A Y D ) T )=

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

19. Thanks!