Enumeration for spanning trees and forests of join graphs based on the combinatorial decomposition

This paper discusses the enumeration for rooted spanning trees and forests of the labelled join graphs Km +Hn and Km +Kn,p, where Hn is a graph with n isolated vertices.


Introduction
In this paper we consider the enumeration problem of rooted spanning trees and forests of two labelled join graphs.In [2], the number of spanning forests of the labelled complete bipartite graph K m,n on m and n vertices has been enumerated by combinatorial method.In [1] and [3], it has been given the enumeration of spanning trees of the complete tripartite graph K m,n,p on m, n and p vertices and the complete multipartite graph, respectively.In [4], by using the multivariate Lagrange inverse, the number of spanning forests of the labelled complete multipartite graph was derived.And, in [5], it has been found the asymptotic number of labeled spanning forests of the complete bipartite graph K m,n as m → ∞ when m ≤ n and n = o(m 6/5 ).
Given two graphs G 1 = (V 1 , E 1 ) and G 2 = (V 2 , E 2 ) with disjoint vertex sets, we let G 1 + G 2 denote the join of G 1 and G 2 , that is, the graph Enumeration for spanning trees and forests of join graphs ...
Clearly, by the definition of a join graph, the complete bipartite graph K m,n is a join graph H m + H n and the complete tripartite graph K m,n,p is a join graph H m + H n + H p , where H m , H n and H p are graphs with m isolated vertices, n isolated vertices and p isolated vertices, respectively.
The goal of this paper first is to give a combinatorial proof of the enumeration for the spanning trees and forests of a labelled join graph K m + H n , where K m is the complete graph on m vertices and H n is the graph with n isolated vertices.Second, this paper also gives a combinatorial proof of the enumeration for the spanning trees and all forests of another labelled join graph K m + K n,p , where K n,p is the complete bipartite graph on n vertices and p vertices.

Enumeration for spanning trees and forests of a join graph K m + H n
Let V (G) denote the vertex set of graph G. Throughout this paper, we will consider only the labelled graphs.In this section, we consider a join graph K m + H n where K m is the complete graph on the vertex set There are m l ways to choose the l roots in V (K m ).Also, let a subset of X, and X be another copy of X and let x ∈ X denote copy of x ∈ X .Take the complete bipartite graph K m,m−l with the partition (X, X ) of its vertex set.Consider the subgraph G of K m,m−l that contains only the directed edges of the form (x , x ), x ∈ X , x ∈ X .The number of the components of G is equal to m − l and G is a forest of K m−l,m−l = (X , X ).Let D(m, |{x i 1 , x i 2 , • • • , x i l }|; m − l, 0) be the set of the labelled spanning forests of K m,m−l = (X, X ) with l roots be the set of the labelled spanning forests of K m with l roots Therefore, from G, we will construct the rooted spanning forests of K m,m−l with l roots in X as follows.
For any fixed integer t ∈ [0, m − l − 1], add t edges consecutively to G as follows.At each step we add an edge of the form (v, x ) between x ∈ X and a (unique)vertex v ∈ X of outdegree zero in any component not containing x in the graph already constructed.The number of components decreases by one each time such an edge is added.
Since The order in which the t edges are added to G is immaterial, so it follows that there are ways.
Every graph we obtained will have m − l − t (weakly) connected components each of which has a unique vertex in X of out-degree zero.Link edges from m − l − t vertices of out-degree zero in these components to l given roots Let D(m, l) be the set of the labelled spanning forests of K m with l roots, i.e., Theorem 2.1.The number g(m, n) of the labelled spanning trees of From every graph F ∈ D(m, l), we will construct the rooted spanning trees of K m + H n as follows.Link an edge (y, x) between every y ∈ V (H n )\{y 1 } and some x ∈ V (F ).There are m n−1 ways.Notice that the obtained graph G has l (weakly) connected components each of which has a unique vertex in V (K m ) of out-degree zero.Now, for any fixed integer t, let G denote a graph obtained by adding t edges consecutively to G as follows.At each step we add an edge of the form (x, y) where y is any vertex of y ∈ V (H n )\{y 1 } and x ∈ V (K m ) is a vertex of out-degree zero in any component not containing y in the graph already constructed.The number of components decreases by one each time such an edge is added.
Since |V (H n )\{y 1 }| = n − 1 and the number of components not containing y in the graph G already constructed is l − 1, there are (n − 1)(l − 1) choices for the first such edge.Similarly, there are (n − 1)(l − 2) choices for the second edge, • • • , and (n − 1)(l − t) choices for the tth edge, where, 0 ≤ t ≤ l − 1, because the number of components in the graph G is l.The graph G thus constructed has l − t components each of which has a unique vertex in V (K m ) of out-degree zero and the remaining vertices all have out-degree one; if we add edges from these vertices of out-degree zero to y 1 , we obtain a tree T in D(m, 0; n, |{y 1 }|) that contains G and in which the in-degree of y 1 equals to l − t .The order in which the t edges are added to G to form G is immaterial, so it follows that there are rooted spanning trees T for fixed integer t.This implies that there are spanning trees T in D(m, 0; n, |{y 1 }|) that contain G. Hence, by (2) and Lemma 2.1, we have Theorem 2.2.The number g(m, l; n, k) of the labelled spanning forests of K m + H n with l roots in K m and k roots in From every graph F ∈ D(m, s)(s ≥ l), we will construct the rooted spanning forests of K m + H n with l roots in K m and k roots in H n as follows.Link an edge (y, v) between every y ∈ Y and some v ∈ V (F ).There are m n−k ways.Notice that the obtained graph G has s (weakly) connected components each of which has a unique vertex in V (K m ) of out-degree zero and the remaining vertices all have out-degree one.
As in the proof of former theorem, link an edge (v, y) between y ∈ Y and a vertex v ∈ V (K m ) of out-degree zero in any component not containing y in the graph already constructed, we repeat this procedure i times, where, 0 ≤ i ≤ s − l, because the required forests have l roots in V (K m ).
There are ways.
Every graph G we obtained will have s − i components each of which has a unique vertex in V (K m ) of out-degree zero.Now, choose the s − i − l vertices of out-degree zero in these s − i components and link edges from these s − i − l vertices to k roots ways.Therefore, by ( 5) and ( 6), the number of the rooted spanning forests of K m + H n which are obtained from F is equal to Hence, by ( 2), (7) and Lemma 2.1, the number g(m, l; n, k) of the labelled spanning forests of K m + H n with l roots in K m and k roots in H n is as follows.
We get the required result. 2 Corollary 2.1.The number S(m, n) of all spanning forests of the join graph K m + H n is equal to Proof By Theorem 2.
Thus, this corollary is true. 2 3. Enumeration for spanning trees and forests of a join graph K m + K n,p In this section, we consider another join graph K m + K n,p where K m is the complete graph and K n,p is the complete bipartite graph.We will show how to count the number of the spanning trees of a join graph Theorem 3.1.The number g(m, n, p) of the spanning trees of We shall obtain the spanning trees in D(m, 0; n, 0; p, |{z 1 }|) from every graph F ∈ D(m, l; n, k).As in the proof of former theorem, link an edge (z, v) between every z ∈ Z and some v ∈ V (F ).There are (m + n) p−1 ways.Notice that the obtained graph G has l + k (weakly) connected components each of which has a unique vertex in V (K m ) ∪ V (H n ) of out-degree zero and the remaining vertices all have out-degree one.
For any fixed integer t such that 0 ≤ t ≤ l + k − 1, link an edge (v, z) between z ∈ Z and a vertex v ∈ V (K m ) ∪ V (H n ) of out-degree zero in any component not containing z in the graph already constructed, we repeat this procedure t times.
There are ways.Therefore, the number of the spanning trees which are obtained from F is equal to Hence, by (9) and Theorem 2.2, Therefore, we get the required result. 2 Theorem 3.2.The number S(m, n, p) of all spanning forests of the join graph Proof Let B(p, r) denote the set of spanning forests of the join graph K m +K n,p = (K m +H n )+H p which r roots are in V (H p ) and remaining roots are in V (K m ) or V (H n ).
From every graph F ∈ D(m, l; n, k), we will construct the rooted spanning forests of Link an edge (z, v) between every v ∈ Z and some v ∈ V (F ).There are (m+n) p−r ways.Notice that the obtained graph G has l + k (weakly) connected components each of which has a unique vertex in V (K m ) ∪ V (H n ) of out-degree zero and the remaining vertices all have out-degree one.
As in the proof of former theorem, for any fixed integer t such that 0 ≤ t ≤ l + k − 1, link an edge (v, z) between z ∈ Z and a vertex v ∈ V (K m )∪V (H n ) of out-degree zero in any component not containing z in the graph already constructed, we repeat this procedure t times.There are ways.
The graph G thus constructed has l + k − t components each of which has a unique vertex in V (K m ) ∪ V (H n ) of out-degree zero and the remaining vertices all have out-degree one; if we add edges from some vertices of these vertices of out-degree zero to z i 1 , z i 2 , • • • , z ir ∈ Z, we obtain a forest in B(p, r) that contains G.There are (r + 1) l+k−t ways.Therefore, this implies that there are Thus, this theorem is true. 2 |X | = m−l and the number of components not containing x in the graph G is m−l −1, there are (m − l)(m − l − 1) choices for the first such edge.Similarly, there are (m − l)(m − l − 2) choices for the second edge, • • • , and (m − l)(m − l − t) choices for the tth edge.
be the vertex set of H n and {y i 1 , y i 2 , • • • , y i k } be the given root set of H n .There are n k ways to choose the k roots in y 2 , ..., y n } be the vertex set of K m + H n and V (H p ) = {z 1 , z 2 , ..., z p } be the vertex set of H p .Let z 1 ∈ V (H p ) be the given roots of K m + K n,p and Z = V (H p )\{z 1 }, D(m, 0; n, 0; p, |{z 1 }|) be the set of the labelled spanning trees of K m + K n,p with root z 1 .Clearly, g(m, n, p) = |D(m, 0; n, 0; p, |{z 1 }|)|.