Notes on Paths, Trees and Lagrange Inversion

Transcription

1 Notes on Paths, Trees and Lagrange Inversion Today we are going to start with a problem that may seem somewhat unmotivated, and solve it in two ways. From there, we will proceed to discuss applications to counting trees and computing power series defined by functional equations. This subject has the unfortunate aspect that it makes the most logical sense to present the theory first, and then the motivation. My presentation is heavily influenced by Richard Stanley s Enumerative Combinatorics II, and by the discussion at A problem about counting sequences Consider sequences of nonnegative integers (d, d 2,..., d n ). Define a 0 = 0, a = a 0 d +, a 2 = a d 2 +,..., a n = a n d n +. (We can only climb one step at a time, but we can fall down arbitrarily fast.) Consider those sequences such that. a i, for i 2. a n =. These are called Lukasiewicz words.

2 How many such sequences are there? More precisely, build a generating function where we record the monomial w d w d2 w dn for each such sequence. So, for example, if n = 2m+, then the coefficient of w0 m+ w2 m is the Catalan number ( ) 2m+ m /(2m + ). Let Pn (w 0, w,...) be this generating function. We will take two of our proofs of the Catalan formula a slick one and a generating function based one, and repurpose them to address this more general question. For the slick proof, take any sequence of nonnegative integers (d, d 2,..., d n ) with d k = n (so a n = ). Extend the d s to a periodic infinite sequence d, d 2,..., d n, d, d 2,..., d n, d,.... Keep defining the a s by a m = a m + d m. So a m+n = a m +. Exactly one of the cyclic rotations of the sequence will have a, a 2,..., a n. So, out of all the ways to choose d s such that d k = n, precisely /n of them obey a i for i n. We can write this in terms of generating functions: P n = n [coefficient of x ] ( w 0 x + w + w 2 x + w 3 x 2 + )n. On the other hand, we can build a recurrence for P as we did with Catalan numbers. Let P (x) = P n x n. If d n = 0, we are not Lukasiewicz, unless the sequence is just (0). That contributes w 0. If d n =, then (d,..., d n ) is Lukasiewicz if and only if (d,..., d n ) is. The number of ways for this to happen is w P n. If d n = 2, then (d,..., d n ) if and only if (d,..., d n ) can be separated into the concatenation of two Lukasiewicz words, and there is precisely one such separation. So the number of ways for this to happen is w 2 n 2 k= P kp n k. Continuing in this way, we see that P (x) = xw 0 + xw P + xw 2 P 2 + xw 3 P 3 +. Writing W (y) = w i t i, we have P (x) = xw (P (x)). In summary: 2

3 If P (x) = xw (P (x)) then n [coefficient of x ] (W (x)/x) n = [coefficient of x n ]P (x). We will move in two directions from here: First we will clean up this formula and generalize it, to give a formula known as the Lagrange inversion formula. Then we will explain why this result is useful, in applications to trees and other subjects. These topics are largely independent of each other. 2 Lagrange Inversion Let s study a more general collection of sequences. We will now have two parameters, m and n, and we will be studying sequences of nonnegative integers (d, d 2,..., d n ) such that. a i, for i 2. a n = m. So our old problem was m =. We ll temporarily introduce the symbol L(w 0, w,...) for the generating function for this problem. Again, we take two approaches. before, we obtain: Using the same periodicity argument as L = m n [coefficient of x m ] ( w 0 x + w + w 2 x + w 3 x 2 + )n = m n [coefficient of x m ] (W (x)/x) n. But also, every such sequence is uniquely a concatenation of m Lukasiewicz sequences. So L = [Coefficient of x n ]P (x) m. Combining these, we get: 3

4 Lagrange Inversion: If P (x) = xw (P (x)) then m n [coefficient of x m ] (W (x)/x) n = [coefficient of x n ]P (x) m. This can be restated in many ways. For example, your textbook rewrites [coefficient of x m ] (W (x)/x) n as [coefficient of x n m ] (W (x)) n. The way I find prettiest is to set Q(x) = x/w (x). So Q(P (x)) = x or, in other words, Q is the inverse function of P. Lagrange Inversion, prettiest form: Let P (x) and Q(x) be inverse functions, with P (0) = Q(0) = 0. Then m[coefficient of x m ]Q(x) n = n[coefficient of x n ]P (x) m. This appears to have a symmetry under switching P and Q, and sending (m, n) to ( n, m). But be careful! As yet, we have only proved the result for m and n positive integers, because those are the only values for which our combinatorial argument applies. We now explain how to remedy this. When m > n, both sides are 0. This is easy to see directly in both cases, and also makes sense in terms of our sequence interpretation: If you can only climb up one step every time, then you can t get above height n in n steps. So let s restrict our attention to the case where n = m + k. Write P (x) = px P (x) and Q(x) = p x Q(x), with P (0) = Q(0) =. We want to show that m[coefficient of x m ]x n p n Q(x) n = n[coefficient of x n ]x m p m P (x) m or, in other words, that mp k [coefficient of x k ] Q(x) m+k = (m + k)[coefficient of x k ] P (x) m But P (x) looks like + ax + bx 2 + cx 3 +, so P (x) k looks like + akx + (a 2( k 2) + bk)x 2 +. For fixed k, the coefficient of x k in P (x) m is a polynomial in m of degree k. And the left hand side is also a polynomial in m of degree k. We have already seen that these polynomials are equal for every positive k, so they are equal for all m. Reversing every step, we have no proved Lagrange inversion for all integers (m, n). In particular, the previous symmetry claims now make perfect sense. 4

5 3 Trees Let s look back at the recursion P (x) = xw 0 + xw P + xw 2 P 2 + xw 3 P 3 +. If we were given this recursion, what sort of structure might we think P was a generating function for? Well, that sum means that we start out by choosing an integer k say 4. We record the weight w 4, and now we are supposed find 4 smaller structures whose sizes add up to n. Maybe for the first of those structures, we take k = 2, so now we must find two smaller structures which together make up a. And it keeps going, in a branching tree-like structure. W4 W2 W W2 In fact, we see that P n (w 0, w, w 2,...) is the generating function for rooted planar trees with n-vertices and the exponent of w i is the number of vertices with i children. In particular, P n (,,,...) is the n-th Catalan number. (Computer Science interpretation: View this tree as an expression to be computed, with some k-ary function at each vertex with k-children. Compute this function in a depth-first traversal, storing partial results on a stack. Then a i is the height of the stack at the i-th step. So a n =, because you get a unique answer, and a i, because the stack is never completely empty.) 5

6 Let s see some examples: Suppose we wanted to count trees where every vertex has either 0 or 2 children. Then P (x) = x( + P 2 ). Or trees where every node has either 0 or 5 descendents? No problem Q(x) = xq(x) + xq(x) 5. Let s turn this around. We have just proven that a root of the above quintic equation is Q(x) = ( ) 5m + x 5m+. 5m + m This is an equation which is not solvable by radicals. In fact, the solution of a general quintic can be expressed in terms of this power series! Or suppose we wanted to count trees where every vertex has 0, or 2 children, with the children labeled LEFT and RIGHT, so that there are two ways to have single child. We get R(x) = x( + 2R(x) + R(x) 2 ). By the way, P (x) = x + xr(x 2 ). Can you give a combinatorial proof? Let s generalize the above example. Suppose we have r labels, instead of the two labels LEFT and RIGHT. We want to look at trees where each vertex has r children, and they are labeled by a subset of [r]. As above, we get F (x) = x ( + ( r ) F + ( r 2 ) )F F r = x( + F ) r. We deduce that the coefficient of x n in F is n [coefficient of x ] ( x + x r ) ( ) n rn =. n n 6

7 Let s fix n = 3, let r grow, and consider the above example. Every such tree looks like one of the two trees below: There are r 2 ways to label the first tree and ( r 2) ways to label the second. So our computation says that r 2 + This is true, as you can easily check. ( ) r = 2 3 ( ) 3r. 2 We will now continue analyzing this situation, and wind up proving a famous result of Cayley. Consider a rooted tree T, with n vertices, and with the children of each vertex coming in order. Let d j be the number of children of the j-th vertex. The number of ways to label such a tree is ( )( ) ( ) r r r d d 2 d n = ( r d i + lower powers of r (d i )! ) = r d i d!d 2! d n! +. Since every vertex except the root is the child of a unique parent, we have di = n. So the contribution from trees of type T is r n /(d!d 2! d n!). We know that, summing over all trees, we are supposed to get (/(rn)) ( ) rn n. As a polynomial in r, we have ( ) rn = nn r n rn n n! + lower powers of r. So, T d!d 2! d n! = nn n! 7

8 or T n!!d!d 2! d n! = nn That! is there for elegance, you ll see why soon. Using our n = 3 example of above, this identity says that = 9. For n = 4, there are 5 terms in the sum, and we have = 64. (n)! So, what is going on here? The multinomial coefficient!(d )!(d 2 )! (d n)! counts the number of ways to take [n] and divide it up into a set of size, a set of size d, a set of size d 2 and so forth. Take that first set and use its single element to label the root. Take the next set and use its elements to label the d children of vertex, placing the labels in order. Take the next set and use its elements to label the d 2 children of vertex 2. So the sum counts all ways to take the elements of [n] and use them to label an n vertex rooted planar tree, with the children of each vertex labeled in order. But this is just the number of rooted trees with [n] labeling their vertices! There is always exactly one way to take such a rooted tree and order the children of each vertex into order. So we have proven: Cayley s Theorem There are n n rooted trees on the vertex set [n]. Since a rooted tree is just a tree with one vertex selected, this is often stated as: Cayley s Theorem There are n n 2 trees on the vertex set [n]. We will see many more proofs of this result throughout the term. 8