An Easier Sufficient Condition

Though we have now seen a complete answer to the decidability problem, putting this method into practical use may, in some cases, be more work than is actually required. We can in fact derive a much easier-to-check set of conditions by recalling the unavoidable substructures result of Chapter2:

Theorem2.17. Every sufficiently long simple permutation contains an alternation of lengthk or an oscillation of lengthk.

Thus a permutation class without arbitrarily long alternations or arbitrarily long oscil- lations necessarily contains only finitely many simple permutations. First note that these strong conditions are not necessary; for example, the juxtaposition

Av(21) Av(12) contains arbitrarily long (wedge) alternations, yet the only simple permutations in this class are 1, 12, and 21. The work of Albert, Linton, and Ruˇskuc [5] also attests to the strength of these conditions; they prove that classes without long alternations have ratio- nal generating functions.

As we have already shown how to decide ifAv(B)contains arbitrarily long alterna- tions, to convert Theorem2.17from a theorem about unavoidable substructures to an eas- ily checked sufficient condition for containing only finitely many simple permutations we need to decide if Av(B) contains arbitrarily long oscillations. As with the parallel and wedge alternations from Section7.2, the increasing oscillations nearly form a chain in the pattern-containment order, so we need only compute the class of permutations that are contained in some increasing oscillation, or equivalently, that are order isomorphic to a subset of the increasing oscillating sequence. This computation is given without proof in Murphy’s thesis [97]. Here we provide the proof.

Proposition 7.8. The class of all permutations contained in all but finitely many increasing oscil- lations isAv(321,2341,3412,4123).

Proof. It is straightforward to see that every oscillation avoids321, 2341, 3412, and4123, so it suffices to show that every permutation avoiding this quartet is contained in the in-

7.6 OTHERCONTEXTS. 151

creasing oscillation sequence. We use therank encoding2_{for this. The rank encoding of the}

permutationπof lengthnis the wordd(π) =d1· · ·dnwhere

di=|{j :j > iandπ(j)< π(i)}|,

i.e., di is the number of points below and to the right of π(i). It is easy to verify that a

permutation can be reconstructed from its rank encoding. Now consider the rank encoding for someπ _∈Av(321,2341,3412,4123). Routinely, one may check:

• d(π)_{∈ {}0,1,2_}∗,

• d(π)does not end in1,2, or20,

• d(π)does not contain21,22,111,112,2011, or2012factors.

We now describe how to embed a permutation with rank encoding satisfying these rules into the increasing oscillating sequence. Suppose that we have embeddedπ(1), . . . , π(i−1). If di ≥ 1 then we embedπ(i) as the next even entry in the sequence. If di = 0then we

embed π(i) as the next odd entry if it ends a 20, 110, or 2010 factor, and as the second next odd entry otherwise. See Figure 7.7for an example. It remains to show that this is indeed an embedding ofπ; to do this it suffices to verify that the number of points of this

embedding below and to the right of our embedding of π(i)isdi. This follows from the

rules above.

7.6 Other Contexts.

To the best of our knowledge, no analogue of Theorem 7.1 is known for other relational structures. If we were to follow the pattern laid down in this thesis, our approach would be to decompose the simple structures and then establish an algorithmic method to avoid these structures. We discussed in Section 2.6some possibilities to generalise the decom- position methods of Chapter 2, and saw in particular the problems encountered in the

. ..

Figure 7.7:The filled points show the embedding of2153647, with rank encoding1020100, given by the proof of Proposition7.8.

S F C ε,L ε,A L, L Ł , A I, L E, A ε, L ε, A I, L E, A L, L A, A ε, L ε, A

Figure 7.8:The prototype transducer for graphs.

graph case. On the assumption that these difficulties may be overcome (particularly in the graph case, but perhaps more generally) it seems likely that decidability would most likely follow. Our approach, therefore, remains furtive.

Determining the Language of Pins in Graphs. Assuming the existing definition from Section2.6for pin sequences in the graph case is nearly correct, it will actually turn out to be somewhat easier to construct an analogue to Lemma7.7. To begin with, recall that the al- phabet for the language of pins in graphs consists of only four letters, namely{L, A, I, E},

7.7 THEPINCLASS 153

Figure 7.9:The basis elements of length6for the pin class (up to symmetry).

an independent point (i.e. connected to nothing) andE a point connected to everything.

The transducer producing all strict pin words for graphs is thus much smaller than the permutation case of Figure 7.6, and a prototype is given in Figure7.8. Note that since we do not have the issue of quadrants in graphs, there is only one fabrication stateF and one

copy stateC.