Decidability

We are now in a position to prove our main result. We wish to decide whether the finitely based class Av(B) contains only finitely many simple permutations. Propositions 7.2– 7.4 show how to decide ifAv(B) contains arbitrarily long parallel alternations or wedge simple permutations, so by Theorem2.14(repeated in this chapter on page138) it suffices to decide whetherAv(B)contains arbitrarily long proper pin sequences.

We first recall two lemmas concerning pin words that we will require. The first shows that we may convert every proper pin sequence to a strict pin word. The proof is given on Page37.

Lemma2.15. Every proper pin sequence corresponds to a strict pin word.

The other lemma we must recall shows us how to relate subsequences of proper pin sequences with pin words, and vice versa. The proof may be found on Page38.

Lemma 2.16. If the pin wordwcorresponds to the permutationπ andσ _≤π then there is a pin worducorresponding toσwithuw. Conversely, ifu wthen the permutation corresponding touis contained in the permutation corresponding tow.

Now, consider a permutationπ that is order isomorphic to a proper pin sequence and

thus, by Lemma2.15, corresponds to at least one strict pin word, sayw. Ifπ _6∈Av(B)then

π ≥βfor someβ ∈B. By Lemma2.16,βcorresponds to a pin worduw. Conversely, if w ufor someucorresponding toβ _∈ B, then Lemma2.16shows thatπ _≥β. Therefore

the set

{strict pin wordsw:wufor someucorresponding to aβ ∈B}

consists of all strict pin words which represent permutations not inAv(B), so by removing this set from the regular language of all strict pin words we obtain the language of all strict pin words corresponding to permutations in Av(B). In the upcoming lemma, we prove that for any pin wordu, the set_{strict pin wordsw: wu_}forms a regular language, and

thus the language of strict pin words inAv(B)is regular. It remains only to check if this language is finite or infinite, which can be determined by Proposition7.5.

Lemma 7.7. For any pin wordu, the set{strict pin wordsw: wu_}forms a regular language, and a finite accepting automaton for this language can be effectively constructed.

Proof. LetT denote the transducer in Figure7.6. We claim that a strict pin wordwlies in T(u)if and only ifw u. The lemma then follows by intersectingT(u)with the regular language of all strict pin words.

We begin by noting several prominent features ofT:

(T1) Every transition writes a symbol.

(T2) Other than the start stateS, the automaton is divided into two parts, the “fabrication”

statesFiand the “copy” statesCi.

(T3) Every transition to a fabrication state hasεinput.

(T4) Every transition from a fabrication state to a copy state reads a numeral and writes a direction, and except for the transitions fromS, these are the only transitions that

7.4 DECIDABILITY 147 S F₁ F₂ F₃ F₄ C₁ C₂ C₃ C₄ ε,R ε,U ε, L ε, U ε, L ε, D ε,R ε,D R, R U,_U L,L U,U L, L D , D R,R D, D ε, L ε, D ε, R ε, U ε, D ε, L ε, U ε, R L, L D ,D R, R U ,U D ,D L, L U ,U R, R 1, R 1, U 2_{, L} 2_{, U} 3, L 3, D 4 , R 4 , D ε,R ε,U ε, L ε, U ε,L ε,D _ε, R ε, D 2, L ε, R 4,D _ε,U 1_{, R} ε, L 3 ,D ε,U 2,U ε,D 4, R ε, L 1 ,U ε, D 3_{, L} ε, R 1_, 1 2,2 4,4 ε,4 ε,₁ ε,2 3_, 3 ε, 3

(T5) All transitions between copy states read a direction and write the same direction, these are the only transitions that read a direction, and there is such a transition for every copy state and every direction.

(T6) From every fabrication and copy state, each direction can be output via a transition to a fabrication state with inputε.

(T7) The subscripts of the fabrication and copy states indicate quadrants: if the strict pin wordw1· · ·wn, corresponding to the pin sequencep1, . . . , pn, has just been written

by the transducer and the transducer is currently in stateCk or Fk, thenpn lies in

quadrantk. Moreover, if the pin wordu1· · ·um, corresponding to the pin sequence q1, . . . , qm, has been read and the transducer currently lies in the copy stateCk, then qmlies in quadrantk.

(T8) From any state, any copy state can be reached by two transitions, the first being a transition to a fabrication state; for example:C2

ε,D

−→F3 4,R

−→C4.

First we prove thatw ufor every strict pin wordwproduced from inputu by this

transducer. We prove this by induction on the number of strong numeral-led factors inu.

The base case is whenuconsists of precisely one strong numeral-led factor. Suppose that

the output right before the first letter ofuis read isv(1). There are two cases. If_v(1)is empty, then the transducer is currently in stateS, and must both read and write the first letter ofu,

moving the transducer into stateCu1. At this point, (T5) shows that the transducer could

continue to transition between copy states, outputting a wordw = uv(2) u. The only

other option available to the transducer (again, by (T5)) is to transition to a fabrication state, but then (T4) shows that the transducer can never again reach a copy state (because

u has only one numeral), and thus by (T3), it can never finish reading u. In the other

case, where v(1) is nonempty, the transducer lies in a fabrication state by (T4). The next

transition must then by (T4) be into a copy state, and (T7) guarantees that the letter written corresponds to a point in quadrantu1. The same argument as in the previous case shows that the transducer is now confined to copy states until the rest of u has been read, and

7.4 DECIDABILITY 149

thus the transducer will outputv(1)w(1)v(2) u.

Now suppose thatudecomposes intoj _≥ 2strong numeral-led factors asu(1)_{· · ·u}(j). By induction, at the point whereu(j−1) has just been read, the transducer has output a

wordv(1)_w(1)_{· · ·v}(j−1)_w(j−1) and lies in a copy state. Since the first letter of_u(j)is a nu- meral, the transducer is forced by (T4) to transition to a fabrication state, and this transition will write but not read by (T3). The transducer can then transition freely between fabrica- tion states. Let us suppose thatv(1)w(1)· · ·v(j−1)w(j−1)v(j)has been output at the moment

just before the transducer begins readingu(j). As in our second base case above, the trans- ducer must at this point transition to a copy state by (T4), which it will do by reading the numeral that beginsu(j) and writing a letter that — by (T7) — corresponds to a point in this quadrant. The situation is then analogous to the base case, and the transducer will outputv(1)w(1)_{· · ·}v(j−1)w(j−1)v(j)w(j)v(j+1)u.

Now we need to verify that the transducer produces every strict pin wordwwithw

u. Breakuinto its strong numeral-led factorsu(1)_{· · ·u}(j)and suppose that the factorisation

w = v(1)w(1)· · ·v(j−1)w(j−1)v(j)w(j)v(j+1)satisfies (O1) and (O2). Ifv(1) is nonempty then

it can be output immediately by a sequence of transitions to fabrication states by (T6); by (O2) and (T7), the first letter ofw(1) (which must be a direction becausew is a strict pin

word) can then be output by transitioning to a copy state, from which (T5) shows that the rest ofu(1)can be read and the rest ofw(1) can be written. Ifv(1)is empty thenu(1) =w(1)

by (O1). The transducer can, by (T5), readu(1) and writew(1) by transitioning fromS to

a copy state and then transitioning between copy states. Becausew is a strict pin word,

(O2) shows that v(2) must be nonempty, and (T6) shows that_v(2) can be output without reading any more letters of u. We then must output w(2) whilst reading u(2). The only

possible obstacle would be reaching the correct copy state, but (T8) guarantees that this can be done. The rest ofucan be read, and the rest ofwwritten, in the same fashion.

The proof of Theorem 7.1 now follows from the discussion at the beginning of the section.