Part II Permutation Classes
5.2 Enumeration
Probably the largest active area in the study of permutation classes is enumeration: given a classC, how many permutations are there of lengthn, and is this sequence well-behaved?
Once these questions are answered, we may be interested in finding out what other com- binatorial structures are enumerated by this sequence, and whether bijections can be es- tablished between them. In the first instance, this may be done by looking at the Online Encyclopaedia of Integer Sequences [110].
For a permutation classC, we denote byCnthe setC ∩Sn, i.e. the permutations inCof
lengthn, and we refer tof(x) = P
|Cn|xnas thegenerating function forC. The generating
functionf isalgebraic if it solves an equation of the formpn(x)fn+pn−1(x)fn−1 +· · ·+
p0(x)f0 = 0 for polynomials pi. Similarly, a rational generating function is one that may
inCis an inflation of a member ofSi(C)so it follows (e.g., inductively) thatC ⊆ W(Si(C)). ThusW(C) ⊆ W(Si(C)), establishing thatSi(C) = Si(W(C)). As wreath closed classes are uniquely determined by their sets
of simple permutations,W(C)is the largest class with this property.
5The analogous question for graphs was raised by Giakoumakis [60] and has received a sizable amount of
be written as a rational function, i.e. a function of the form p(x)
q(x) wherep(x) andq(x)are polynomials inxover the field of rational numbers.
As a trivial first example, consider the classI ={1,12,123, . . .}. There is precisely one
permutation of each length, and so its generating function isf(x) = ∞ X
n=0
xn= 1+x+x2+· · ·, or, in other words,f = 1
1−x, a rational function. Note that here our sum begins atn= 0,
implying that we are including the single permutation of length zero in the class. This is a convention that may or may not always be used – there are cases where including the empty permutation is convenient (particularly when considering recursive structures), while in other cases we may specifically not want it. It will be our convention to include the empty permutation unless required to do otherwise.
Our next example is somewhat more complicated, and the method employed to derive the enumeration is a classic recursive technique relying on knowledge of the structure of a permutation in the specified class. This is, of course, precisely where the rˆole of simple permutations and the substitution decomposition will become invaluable.
Example 5.11(The Stack Sortable Permutations). As seen in Example 5.3, the set of stack sortable permutations is precisely the class Av(231). Within this class, the permutations of lengths1,2,3,4,5. . . are enumerated by the sequence1,2,5,14,42, . . ., which looks en-
couragingly like the sequence of Catalan Numbers (sequence A000108 of [110]), with gen- eral term (2n)!
n!(n+ 1)!.
We prove this fact by considering a permutation π ∈ Av(231) of lengthn. Since π
must avoid 231, every point coming before the value n in π must lie below every point
coming after the valuen, i.e.π =α⊕(1 β)for someαandβ, which also of course must
themselves avoid231(see Figure5.3). Thusαandβ must lie inAv(231), but there are no other restrictions onαandβsave that we must of course have|α|+|β|+ 1 =n. Note also
that this decomposition into αandβis unique, and hence can be used to decompose (or
construct) every permutation inAv(231).
5.2 ENUMERATION 93
β∈Av(132)
α∈Av(132)
Figure 5.3:Generic structure of a231-avoider.
then we can use the above consideration to derive the recursion
f =xf2+ 1.
Note that here we have included the empty permutation, as we must allow α and/orβ
to be empty. Note further that the empty permutation cannot be decomposed as we did above because it has no maximum entry, hence the appearance of the “+1” term. Solving this algebraic equation is then straightforward, and gives
f = 1− √ 1−4x 2x = 1 +x+ 2x 2+ 5x3+ 14x4+. . . as required.
Central to the enumeration problem is the classification of permutation classes with the same enumeration. We say that two permutationsαandβareWilf equivalentif|Av(α)n|=
|Av(β)n| for all n, i.e. the classes Av(α) and Av(β) are enumerated by the same gener-
ating function. We may also say that the permutationsα andβ belong to the sameWilf class. For example, the permutations231 and123 are Wilf equivalent, a fact which may be proved using several different bijections – see Richards [102], Rotem [104], Simion and Schmidt [109] or West [120] for various approaches to this problem. Since enumeration is then preserved under symmetry, this proves that all the permutations of length3belong to the same Wilf class. The computation of the Wilf classes up to length7were completed in 2001 by Stankova and West [112].
This term has since been extended in the natural way to sets of permutations – the permutation sets A andB are Wilf-equivalent if|Av(A)n| = |Av(B)n|. While this may
open up an endless but for the most part uninteresting variety of problems, there are some very surprising results. Notably, B´ona [18] shows that the classAv(1342) has generating
functionf = 32x
−8x2+ 20x+ 1−(1−8x)3/2. This is the same as the class of permutations which may be sorted with two ordered stacks in series, whose basis is infinite:
B={(2,2m−1,4,1,6,3,8,5, . . . ,2m,2m−3)|m= 2,3,4, . . .}.
(This problem was previously discussed at the end of Example5.3.)
Another approach to the problem of enumeration is that of asymptotics – how many permutations of lengthnare there in a given permutation class asnapproaches infinity?
In other words, we want to be able to say something about lim
n→∞|Cn|, or, somewhat more usefully, lim
n→∞
n p
|Cn|. As a first step, we have the “Stanley-Wilf conjecture”, namely that
for a given classCnot containing every permutation, there exists a constantKsuch that
lim sup n→∞
n p
|Cn|=K.
This result was proved in 2004 by Marcus and Tardos [87]. The constantKis known as the upper growth rate of the permutation class. We may similarly define thelower growth rate, lim infn→∞ n
p
|Cn| = K. This naturally begs the question whether the upper and lower
growth rates coincide, in which case limn→∞ n p
|Cn| = K is called the growth rate of C.
It is conjectured that the growth rate always exists, a fact that has been shown in some cases. Arratia [6] proves this for sum or skew complete classes, among which are all of the permutation classes defined by a single basis element.
For example, the growth rate of the stack sortable permutations Av(231) is4, a fact easily seen by recalling that|Av(231)n| =
(2n)!
n!(n+ 1)!, and using Stirling’s approximation
n!≈√2nπn e
n .
5.3 ANTICHAINS, PARTIALWELLORDER ANDATOMICITY 95