J.W. Moon. Counting Labelled Trees. Canadian Mathematical Congress, Montreal, 1970.  Home/Search   Document Not in Database   Summary   Related Articles   Check

....that the exponential generating function satisfies the functional equation which implies that U is symmetric in ff and fi. We first discuss what is already known about the polynomials u n (ff; fi) Since there are (n 1) forests of rooted trees with vertex set [n] see, e.g. Moon [8] for many proofs) we have u n (1; 1) n 1) It is also known that the number of forests of rooted trees on [n] with i leaves is (n =i )S(n; n Gamma i 1) where S(n; k) is the Stirling number of the second kind. See Moon [8, p. 20, Theorem 3.5] or Knuth [7; exercise 19, p. 397; ....

J. W. Moon, Counting Labelled Trees, Canadian Mathematical Congress, 1970.

....) 1 the electronic journal of combinatorics 3 (2) 1996) #R8 2 which implies that U is symmetric in ff and fi. We first discuss what is already known about the polynomials un (ff; fi) Since there are (n 1) n Gamma1 forests of rooted trees with vertex set [n] see, e.g. Moon [8] for many proofs) we have un (1; 1) n 1) n Gamma1 . It is also known that the number of forests of rooted trees on [n] with i leaves is (n =i )S(n; n Gamma i 1) where S(n; k) is the Stirling number of the second kind. See Moon [8, p. 20, Theorem 3.5] or Knuth [7; exercise 19, p. 397; ....

J. W. Moon, Counting Labelled Trees, Canadian Mathematical Congress, 1970.

....trees and random trees 8.1 Kirchho , Cayley, Kasteleyn and Tutte Cayley proved that the number of trees on n labeled vertices is n n 2 . There are many beautiful proofs which demonstrate various principal techniques in enumeration theory and, amazingly, new proofs are still being found. See [162, 84] [Stanley] The matrix tree theorem, asserting that the number of spanning trees for a graph G is (essentially) the determinant of the Laplacian of G, is even earlier and is attributed to Kirchho . Of the vast knowledge on tree enumeration, let me mention two additional results. Kasteleyn (see ....

....of the symmetric group S n on H n . Haiman s conjecture turned out to be related to exciting issues in algebraic geometry and representation theory. Many tried to solve it, but very recently Haiman himself proved his conjecture [Haiman] 8. 7 Some links and references Tree enumeration [162, 84, 156]; trees and probability [161, 106] random spanning trees and forests [146, 160, 163, 106] enumerative combinatorics [83, 84, 147] Zeilberger ] Stanley] Schramm s processes, Brownian motion [164, 159, 167] Haiman s conjectures [150] Haiman] MACAULAY [158] Bayer] approximate enumeration ....

J. W. Moon, Counting labelled trees, Canadian Mathematical Congress, Montreal, Que. 1970.

....#U)e x(1 # # ##U) 1 ### # ## ####### ####### ## ############# # ### ####### ### 2 which implies that U is symmetric in # and #. We first discuss what is already known about the polynomials u n (#, #) Since there are (n 1) n 1 forests of rooted trees with vertex set [n] see, e.g. Moon [8] for many proofs) we have un (1, 1) n 1) n 1 . It is also known that the number of forests of rooted trees on [n]withi leaves is (n i )S(n, n i 1) whereS(n, k) is the Stirling number of the second kind. See Moon [8, p. 20, Theorem 3.5] or Knuth [7; exercise 19, p. 397; solution, p. ....

J. W. Moon, Counting Label led Trees, Canadian Mathematical Congress, 1970.

....(x; z)g: 18) Even though the equation appears to be extremely hard to solve in a closed form, it is ideally suited to determine (asymptotically) the mean and variance of Y n . For example we will first use it to derive a differential equation (23) for E(x) in terms of T (x) the tree function ([11]) defined for jxj e Gamma1 , which is easily solved (24) Using Cauchy s formula, we can now write E n as a contour integral which we can then estimate. Now for the details. Let E n = E(Y n ) and E(x) 1 X n=1 E n n n Gamma1 n x n : Since E n = f 0 n (1) we have E(x) F z ....

.... p n= log n (a; b) a; b) The degree sequence of a random tree d 1 ; d 2 ; d is such that d 1 Gamma 1; d 2 Gamma 1; d Gamma 1 have the joint distribution of the occupancy numbers in the uniform allocation of Gamma 2 distinct balls in distinct boxes (see e.g. Moon [11]) So in particular d 1 has distribution 1 BINOMIAL( Gamma 2; 1= Thus, Pr(d 1 d = q n= log n ) Gamma 2 d Gamma 1 1 d Gamma1 = expf Gamma Omega Gamma q n log n)g: But for u = O(1= p n) a; b) e O( p n) and so 2 = e O( p n) Pr(d or ffi ....

J. W. Moon, Counting labelled trees, Canadian Mathematical Congress, Montreal, 1970.

.... Gamma Gamma1 ) is the number of ways of choosing a forest in F [ and then choosing a mapping from its roots to its vertices. The middle factor in Upsilon is the number of forests on [ with rooted trees, each of which contains one of a prescribed set of vertices as its root, see Moon [8], for instance) Furthermore Theta = Gamma Gamma1 # Theta X t=0 t Gamma Gamma t Gamma t Gamma ( Gamma ) Gamma ) # = Gamma Gamma1 # Theta Gamma 1 Gamma 1 ( Gamma ) Gamma ....

J.W.Moon, Counting labelled trees, Canadian Mathematical Congress, Montreal, 1970.

....However, the main point of this paragraph is to indicate that at least the basics of the Polya enumeration theory are fairly well understood. Beautiful and useful as the Polya theory is, it only applies to a restricted class of tree enumeration problems, and there is a wide literature (cf. [4,16,26]) on other methods. Here we will be concerned with some recent developments that involve nonlinear iteration. As a simple example, we consider the problem of height distribution among binary trees. Binary trees are rooted trees in which every node has either 0 or 2 sons, and left and right sons ....

J. W. Moon, Counting Labelled Trees, Canadian Mathematical Congress, 1970.

....but is not very appealing. One simple feature is P (D = r 1) 1 1 r) 2ar 2a 1 r) 1 (2ar) 2 r . 12 Note that the distribution depends only on (r, a) and not on the size n of the graph. The asymptotics are simpler: see Proposition 10. For the complete graph it is known ([17] eq. 7.3) that D has Binomial(n 2, 1 n) distribution: here Proposition 9 applies with r = n 1 and a = n (n 1) For the cube graph Q d one can calculate a = 1 2 1 d) 1 1 d ) 9) and deduce the distribution of D from the Proposition. Finally, recall that a priori ED = 1 ....

J.W. Moon. Counting Labelled Trees. Canadian Mathematical Congress, Montreal, 1970.

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J. W. Moon. Counting Labelled Trees. Canadian Mathematical Congress, Montreal, 1970.

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