The generating function of a weighted set

i∏∈I

fi

◦g = _∏

i∈I

(fi◦g). Proof of Proposition 3.11.32 (sketched). See Section B.2.

3.12. The generating function of a weighted set

So far, we have built a theory of FPSs, but their application to combinatorics (via generating functions) was a trick. I will now explain one way to make this latter trick into a theory as well. This theory isn’t magic – it is just rewriting our arguments in a more uniform way; but it makes some of them easier to manage.

I will briefly survey this theory (as it isn’t a focus of this course), following Fink’s introduction [Fink17, §3.3–§3.4] (but speaking offinite-type weighted sets where Fink speaks of “combinatorial classes”).

3.12.1. The theory

Definition 3.12.1. (a) A weighted set is a set A equipped with a function w : A → N, which is called the weight function of this weighted set. For each a ∈ A, the value w(a) is denoted |a| and is called the weight of a (in our weighted set).

Usually, instead of explicitly specifying the weight function w as a func-tion, we will simply specify the weight |a| for each a ∈ A. The weighted set consisting of the set A and the weight function w will be called (A,w) or simply Awhen the weight function wis clear from the context.

(b) A weighted set A is said to be finite-type if for each n ∈ N, there are only finitely many a∈ Ahaving weight |a| =n.

(c) If A is a finite-type weighted set, then the weight generating function of A is defined to be the FPS

a

∑

∈A

x^|a| =

∑

n∈_N

(# of a∈ Ahaving weight n)·xⁿ ∈ _Z[[x]]. This FPS is denoted by A.

(d) An isomorphismbetween two weighted sets A and B means a bijection ρ : A →B that preserves the weight (i.e., eacha ∈ Asatisfies |ρ(a)|=|a|).

(e) We say that two weighted sets A and B are isomorphic (this is written A ∼=B) if there exists an isomorphism between A and B.

Example 3.12.2. Let B be the weighted set of all binary strings, i.e., finite tuples consisting of 0s and 1s. Thus,

B={(), (0), (1), (0, 0), (0, 1), (1, 0), (1, 1), (0, 0, 0), (0, 0, 1), . . .}. The weight of a k-tuple is defined to be k. This weighted setB is finite-type, since for each k ∈ N, there are only finitely many binary strings of length k (namely, there are 2^k such strings). The weight generating function of Bis

B=

∑

n∈_N

(# of a∈ B having weightn)

| {z }

=(# of binary strings of lengthn)

=2ⁿ

·xⁿ

=

∑

n∈_N

2ⁿxⁿ = ¹ 1−2x.

Proposition 3.12.3. Let Aand Bbe two isomorphic finite-type weighted sets.

Then, A =B.

Proof. This is almost trivial: The weighted sets A and B are isomorphic; thus, there exists an isomorphismρ : A → B. Consider this ρ. Then, ρ is a bijection

and preserves the weight (since ρ is an isomorphism of weighted sets). The latter property says that we have

|ρ(a)| =|a| for each a∈ A. (126) Now, the definition ofByields

B =

∑

b∈B

x^|b| =

∑

a∈A

x^|ρ(a)|

| {z }

=x^|a|

(by (126))

here, we have substitutedρ(a) forb in the sum, sinceρis a bijection

=

∑

a∈A

x^|a|.

Comparing this with A=

∑

a∈A

x^|a| by the definition of A , we obtain A =B. This proves Proposition 3.12.3.

Note that Proposition 3.12.3 has a converse: If A= B, then A∼= B.

Recall that the disjoint union of two sets A and B is “the union of A and B where we pretend that A and B are disjoint even if they aren’t”. Formally, it is defined as the set({0} ×A)∪({1} ×B), but we think of the elements (0,a) of this set as copies of the respective elementsa ∈ A, and we think of the elements (1,b) of this set as copies of the respective elements b ∈ B. If Aand B are two finite sets, then their disjoint union always has size|A|+|B|.

Definition 3.12.4. Let A and B be two finite-type weighted sets. Then, the weighted set A+B is defined to be the disjoint union of A and B, with the weight function inherited from A and B (meaning that each element of A has the same weight that it had in A, and each element of B has the same weight that it had in B). Formally speaking, this means that A+Bis the set ({0} ×A)∪({1} ×B), with the weight function given by

|(0,a)| =|a| for each a∈ A (127) and

|(1,b)| =|b| for eachb ∈ B. (128) Proposition 3.12.5. Let AandBbe two finite-type weighted sets. Then, A+B is finite-type, too, and satisfies A+B= A+B.

Proof. The formal definition of A+Bsays that A+B= ({0} ×A)∪({1} ×B). Thus, the set A+B is the union of the two disjoint sets {0} ×A and {1} ×B (indeed, these two sets are clearly disjoint, since 06=1).

Let us first check that A+Bis finite-type.

For each n ∈ N, there are only finitely many a ∈ A having weight |a| = n (since A is finite-type), and there are only finitely many b ∈ _B having weight

|b| = n (since B is finite-type). Hence, there are only finitely many c ∈ A+B

(here, we have substituted(0,a) forcin the sum, since the

(here, we have substituted(1,b) forcin the sum, since the mapB→{1}×B, b7→(1,b)is a bijection)





here, we have split the sum, since the set A+_B is the union of the two disjoint

Thus, the proof of Proposition 3.12.5 is complete.

We can easily extend Definition 3.12.4 and Proposition 3.12.5 to disjoint unions of any number (even infinite) of weighted sets. (However, in the case of infinite disjoint unions, we need to require the disjoint union to be finite-type in order for Proposition 3.12.5 to make sense.)

Definition 3.12.6. Let A and B be two weighted sets. Then, the weighted set A×B is defined to be the Cartesian product of A and B (that is, the set{(a,b) | a∈ Aand b ∈ B}), with the weight function defined as follows:

For any (a,b)∈ A×B, we set

|(a,b)|=|a|+|b|. (129)

Proposition 3.12.7. Let AandBbe two finite-type weighted sets. Then, A×B is finite-type, too, and satisfies A×B= A·B.

Proof of Proposition 3.12.7 (sketched). The proof that A×Bis finite-type is left to the reader.

The definition ofA×B yields

A×B =

∑

(a,b)∈A×B

x^|(a,b)|

| {z }

=x^|a|+|b|

(by (129))

since all elements of A×B have the form (a,b) for some aand b

=

∑

(a,b)∈A×B

x^|a|+|b|

| {z }

=x^|a|·x^|b|

=

∑

(a,b)∈A×B

x^|a|·x^|b| =

∑

a∈A

x^|a|

!

| {z }

=A

b

∑

∈B

x^|b|

!

| {z }

=B

=A·B.

The proof of Proposition 3.12.7 is thus complete.

We can easily extend Definition 3.12.6 and Proposition 3.12.7 to Cartesian products of k weighted sets. In particular, we obtain Cartesian powers when we multiplykcopies of the same weighted set:

Definition 3.12.8. Let A be a weighted set. Then, A^k (fork ∈ N) means the weighted set A×A× · · · ×A

| {z }

ktimes

.

Proposition 3.12.9. Let Abe a finite-type weighted set. Let k∈ _{N. Then,} A^k is finite-type, too, and satisfies A^k = A^k.

Note that the 0-th Cartesian power A⁰of a weighted set A always consists of a single element – namely, the empty 0-tuple(), which has weight 0.

3.12.2. Examples

Now, let us use this theory to revisit some of the things we have already counted:

• Fixk ∈ _{N, and let}

C_k ={compositions of lengthk}

={(a₁,a₂, . . . ,a_k) | a₁,a₂, . . . ,a_k are positive integers}

=_P^k (whereP={1, 2, 3, . . .}).

This becomes a finite-type weighted set if we set |(a1,a2, . . . ,a_k)| = a1+ a₂+· · ·+a_k for every (a₁,a₂, . . . ,a_k) ∈ C_k. What is its weight generating function C_k ? We can turn P itself into a weighted set, by defining the

weight of a positive integer n by |n| = n. Then, C_k = _P^k not just as sets, but as weighted sets. Hence,

C_k =_P^k =_P^k (by Proposition 3.12.9)

=x¹+x²+x³+· · ·^k sinceP= x¹+x²+x³+· · ·

= x

1−x

k

sincex¹+x²+x³+· · · = ^x 1−x

. This recovers the equality (92).

• Recall the notion of Dyck paths (as defined in Example 2 in Section 3.1), as well as the Catalan numbers c0,c1,c2, . . . (defined in the same place).

Let

D ={Dyck paths from (0, 0) to (2n, 0) for somen∈ _N}. This setDbecomes a finite-type weighted set if we set

|P| =n whenever Pis a Dyck path from (0, 0) to (2n, 0). Thus,

D =

∑

n∈_N

(# of Dyck paths from (0, 0) to (2n, 0))

| {z }

=cn

xⁿ =

∑

n∈_N

cnxⁿ. This is the generating function we called C(x) in Example 2 of Section 3.1.

Recall that there is only one Dyck path from (0, 0) to (0, 0), namely the trivial path. All the other Dyck paths inDare nontrivial. We let

Dtriv ={trivial Dyck paths in D} and Dnon={nontrivial Dyck paths inD}.

These two setsDtriv andDnon are subsets ofD, and thus are weighted sets themselves (we define their weight functions by restricting the one ofD).

The setD_triv consists of a single Dyck path, which has weight 0; thus, its weight generating function is

Dtriv = x⁰ =1.

In Example 2 of Section 3.1, we have seen that any nontrivial Dyck path π has the following structure:⁴¹

41The colors are referring to the following picture:

– a NE-step,

– followed by a (diagonally shifted) Dyck path (drawn in green), – followed by a SE-step,

– followed by another (horizontally shifted) Dyck path (drawn in pur-ple).

If we denote the green Dyck path by α and the purple Dyck path by β, then we thus obtain a bijection

Dnon →D×D, π 7→(α,β).

Alas, this bijection is not an isomorphism of weighted sets, since it fails to preserve the weight. Indeed,|(α,β)| =|α|+|β| =|π| −16=|π|.

Fortunately, we can fix this rather easily. Define a weighted set X :={1} with |1|=1.

This is a one-element set, so the only real difference between the weighted setsX×D×Dand D×D is in the weights. Indeed, the sets D×D and X×D×D are in bijection (any pair (α,β) ∈ D×D corresponds to the triple (1,α,β) ∈ X×D×D), but the weights of corresponding elements differ by 1 (namely,|(1,α,β)| = |1|

|{z}

=1

+|α|+|β|

| {z }

=|(α,β)|

=₁+|(α,β)|).

Thus, by replacing D×D by X×D×D, we can fix the degrees in our above bijection. We thus obtain a bijection

Dnon →X×D×D, π 7→(1,α,β)

that does preserve the weight. This bijection is thus an isomorphism of weighted sets. Hence,Dnon ∼=X×D×D.

Each Dyck path is either trivial or nontrivial. Hence, D∼=Dtriv+ Dnon

| {z }

∼=X×D×D

∼=Dtriv+X×D×D,

so that

D =Dtriv+X×D×D (by Proposition 3.12.3)

=Dtriv

| {z }

=1

+ X

|{z}=x

·D·D

| {z }

=D²

(by Proposition 3.12.5 and Proposition 3.12.7)

=1+x·D².

This is precisely the quadratic equation C(x) = 1+x(C(x))² that we obtained in Section 3.1. This time, we obtained it from a more abstract technique.

3.12.3. Domino tilings

Let us now apply the theory of weight generating functions to something we haven’t already counted. Namely, we shall count the domino tilings of a rect-angle. Informally, these are defined as follows:

• For anyn,m ∈_{N, we let} Rn,m be a rectangle with widthn and heightm.

• Adominomeans a rectangle that is an R1,2 or an R2,1.

• Adomino tiling of a shape S means a tiling of S by dominos (i.e., a set of dominos that coverS and whose interiors don’t intersect).

For example, here is a domino tiling of R_8,8:

(note that the colors are purely ornamental here: we are coloring a domino pink if it lies horizontally and green if it stands vertically, for the sake of con-venience).

This sounds geometric, but actually is a combinatorial object hiding behind geometric language. Our rectangles and dominos all align to a square grid.

Thus, rectangles can be modeled simply as finite sets of grid squares, and dominos are unordered pairs of adjacent grid squares. This allows us to re-define domino tilings combinatorially as follows:

Definition 3.12.10. (a)Ashapemeans a subset ofZ².

We draw each (i,j) ∈ _Z² as a unit square with center at the point (i,j) (in Cartesian coordinates); thus, a shape can be drawn as a cluster of squares.

(b)For anyn,m∈ N, the shape Rn,m (called then×m-rectangle) is defined to be

{1, 2, . . . ,n} × {1, 2, . . . ,m}=ⁿ(i,j) ∈ _Z² | 1≤i ≤nand 1≤j ≤mo . (c)Adominomeans a size-2 shape of the form

{(i,j), (i+1,j)} (a “horizontal domino”) or {(i,j), (i,j+1)} (a “vertical domino”)

for some(i,j) ∈_Z².

(d) Adomino tiling of a shape S is a set partition ofS into dominos (i.e., a set of disjoint dominos whose union isS).

(e)For any n,m∈ _{N, let} dn,m be the # of domino tilings of Rn,m. Can we computedn,m ?

The casem =1 is a bit too simple (do it!), so let us start with the casem =2.

Here are thedn,2 forn ∈ {0, 1, . . . , 4}: n dn,2 domino tilings

0 d0,2 =1

1 d_1,2 =1

2 d2,2 =_{2 ,}

3 d3,2 =3 , ,

4 d_4,2 =_{5 , , ,}

,

.

What do we expectdn,2 to be in general?

Aheight-2 rectangle shall mean a rectangle of the form R_n,2 with n ∈ _{N. Let} us define the weighted set

D:={domino tilings of height-2 rectangles}

={domino tilings ofRn,2 with n ∈_N}.

We define the weight of a tiling T of R_n,2 to be |T| := n (that is, the width of the rectangle tiled by this tiling).

Thus, Dis a finite-type weighted set, with generating function D=

∑

n∈_N

d_n,2xⁿ.

So we want to compute D. Let us define a new weighted set that will help us at that.

Namely, we say that afaultof a domino tiling T is a vertical line `such that

• each domino of T lies either left of ` or right of ` (but does not straddle

`_{), and}

• there is at least one domino of Tthat lies left of `, and at least domino of T that lies right of`.

For example, in the following picture:

,

you see two domino tilings. The tiling on the left has a fault (namely, the vertical line separating the 2nd from the 3rd column), but the tiling on the right has none (a fault must be a vertical line by definition; a horizontal line doesn’t count).

A domino tiling will be calledfaultfree if it is nonempty and has no fault. So the tiling on the right is faultfree.

Observation: Any domino tiling of a height-2 rectangle can be de-composed uniquely into a tuple of faultfree tilings of (usually smaller) height-2 rectangles, by cutting it along the faults. For example:

decomposes as



 , ,



. (Note that if the original tiling was faultfree, then this will be a

1-tuple. If the original tiling was empty, then this will be a 0-1-tuple.) Moreover, the sum of the weights of the faultfree tilings in the tuple is the weight of the original tiling.

Thus, if we define a new weighted set

F:={faultfreedomino tilings ofR_n,2 with n ∈_N} (with the same weights as in D), then

D∼= F⁰+F¹+F²+F³+· · ·

(here, the right hand side is an infinite disjoint union). Thus, D =F⁰+F¹+F²+F³+· · ·= F⁰+F¹+F²+F³+· · ·

(by the infinite analogue of Proposition 3.12.5 and by Proposition 3.12.9)

= ¹

1−F by (5), withF substituted for x . Thus, if we can compute F, then we can compute D.

In order to computeF, let us see how a faultfree domino tiling of a height-2 rectangle looks like. Here are two such tilings:

and .

I claim that these two tilings are theonlyfaultfree tilings of height-2 rectan-gles. Indeed, consider any faultfree tiling of a height-2 rectangle. In this tiling, look at the domino that covers the box(1, 1). If it is a vertical domino, then this vertical domino must constitute the entire tiling, since otherwise there would be a fault to its right. If it is a horizontal domino, then there must be a second horizontal domino stacked atop it, and these two dominos must then constitute the entire tiling, since otherwise there would be a fault to their right. This leads to the two options we just named.

Thus, the weighted set F consists of just the two tilings shown above: one tiling of weight 1 and one tiling of weight 2. Hence, its weight generating function isF =x+x². So

D = ¹

1−F = ¹

1−(x+x²) = ¹

1−x−x² = f1+ f2x+ f3x²+ f4x³+· · · , where (f0, f₁, f2, . . .) is the Fibonacci sequence. Thus, comparing coefficients, we find

d_n,2 = f_n+1 for eachn ∈_N.

There are, of course, more elementary proofs of this (see [19fco, Proposition 1.1.11]).

Remark: There is an alternative argument forD = ¹

1−F that runs as follows:

Any tiling in D is either empty, or can be uniquely split into a pair of a faultfree tiling and an arbitrary tiling (just split it along its leftmost fault, or along the right end if there is no fault). Thus,

D ∼=1+F×D.

Hence,

D =1+F×D=1+F·D.

Solving this forD, we find

D= ¹

1−F.

Now, let us try to solve the analogous problem for height-3 rectangles. Forget about the Dand Fwe defined above. Instead, define a new weighted set

D:={domino tilings of height-3 rectangles}

={domino tilings ofRn,3 with n ∈_N}.

The weight of a tiling T of Rn,3 is defined as before (i.e., it is |T| = n). Thus, D is a finite-type weighted set, with generating function

D=

∑

n∈_N

dn,3xⁿ. We want to compute thisD.

Set

F :={faultfreedomino tilings ofRn,3 withn ∈_N}.

How does a faultfree domino tiling of a height-3 rectangle look like? Again, let us distinguish cases according to the kind of dominos that occupy the first column of the tiling:

• The faultfree domino tilings of a height-3 rectangle that contain a vertical domino in thetoptwo cells of the first column are

· · · (this is an infinite sequence of tilings, each obtained from the previous by inserting two columns in the middle by a fairly self-explanatory proce-dure). The weights of these tilings are 2, 4, 6, . . ., so their total contribution to the weight generating function Fof Fis x²+x⁴+x⁶+· · ·.

• The faultfree domino tilings of a height-3 rectangle that contain a vertical domino in thebottomtwo cells of the first column are

· · · (these are the top-down mirror images of the previously classified tilings).

The weights of these tilings are 2, 4, 6, . . ., so their total contribution to the weight generating function Fof F isx²+x⁴+x⁶+· · ·.

• The faultfree domino tilings of a height-3 rectangle that containnovertical domino in the first column are

(yes, there is only one such tiling). The weight of this tiling is 2, so its total contribution to the weight generating functionF ofF is x².

This classification of faultfree domino tilings entails F =x²+x⁴+x⁶+· · ·+x²+x⁴+x⁶+· · ·+x²

=x²· ¹

1−x² +x²· ¹

1−x² +x²

since x²+x⁴+x⁶+· · ·= x²· ¹ 1−x²

= ^3x

2−x⁴ 1−x² . Thus,

D = ¹

1−F = ¹

1− ^3x

2−x⁴ 1−x²

= ¹−x² 1−4x²+x⁴

=1+3x²+11x⁴+41x⁶+153x⁸+· · · .

One thing you see right away is that only even powers of x appear in this FPS. In other words,

dn,3 =0 whennis odd.

This is not surprising, since ifn is odd, then the rectangle Rn,3 has an odd # of squares, and thus cannot be tiled by dominos.

But we can also compute dn,3 for even n. Indeed, using the same method (partial fractions) that we used for the Fibonacci sequence in Section 3.1, we can expand 1−x²

1−4x²+x⁴ as a sum of geometric series. Thus, after some com-putation, we find

dn,3 = ³+√ 3 6

2+√

3n/2

+³−√ 3 6

2−√

3n/2

for any even n.

Now, what about computingdn,m in general? The above reasoning leading up toD = ¹

1−F can be applied for anym ∈ N, but describing F becomes harder and harder as m grows larger. The generating function D is still a quotient of

two polynomials for anym(see, e.g., [KlaPol79]), but this requires more insight to prove. For m ≥ 6, it appears that there is no formula for dn,m that requires only quadratic irrationalities.

Let me mention a different formula for dn,m, found by Kasteleyn in 1961 (motivated by a theoretical physics model):

Theorem 3.12.11 (Kasteleyn’s formula). Assume that m is even and n ≥ 1.

Then,

See [Loehr11, Theorem 12.85] or [Stucky15] for proofs of this formula. Note that it can indeed be used for exact computation ofdn,m (as there are algorithms for exact manipulation of “cyclotomic integers” such as cos jπ

m+1); for example, it yieldsd_8,8 =12 988 816.

Im Dokument Algebraic Combinatorics (Seite 149-163)