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 $K_m+H_n$ and $K_m+K_{n,p}$, where $H_n$ 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 ).
www.ejgta.org Enumeration for spanning trees and forests of join graphs ... | Sung Sik U 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.
2. 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 {x 1 , x 2 , · · · , x m }.
be the given root set of K m . There are m l ways to choose the l roots in V (K m ). Also, let X ′ = X\{x i 1 , x i 2 , · · · , x i l } be 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 x i 1 , x i 2 , · · · , x i l ∈ X and D * (K m ; x i 1 , x i 2 , · · · , x i l ) be the set of the labelled spanning forests of K m with l roots x i 1 , x i 2 , · · · , x i l ∈ X. Now any spanning forest in D(m, |{x i 1 , x i 2 , · · · , x i l }|; m − l, 0) containing G gives rise to a spanning forest in Conversely, any forest in D * (K m ; x i 1 , x i 2 , · · · · · · , x i l ) can be extended to a forest in D(m, |{x i 1 , x i 2 , · · · · · · , x i l }|; m − l, 0) containing G by inserting vertex x ′′ ∈ X ′′ after x ′ ∈ X ′ . 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 |X ′ | = m−l and the number of components not containing choices for the second edge, · · · , and (m − l)(m − l − t) choices for the tth edge.
The order in which the t edges are added to G is immaterial, so it follows that there are Enumeration for spanning trees and forests of join graphs ... | Sung Sik U 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 K m + H n is y n } be the vertex sets of K m , H n , respectively, and y 1 ∈ V (H n ) be the given root of K m + H n . Let D(m, 0; n, |{y 1 }|) be the set of the labelled spanning trees of K m +H n with root y 1 and T (m, n) be the set of the labelled spanning trees of K m + H n . Clearly, |T (m, n)| = |D(m, 0; n, |{y 1 }|)|. 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 Proof Let V (H n ) = {y 1 , y 2 , · · · , y n } 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 V (H n ). Let V (K m ) = {x 1 , x 2 , · · · , x m } be the vertex set of K m and Y ′ = V (H n )\{y i 1 , y i 2 , · · · , y i k } be a subset of V (H n ).
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 y i 1 , y i 2 , · · · , y i k . There are 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 www.ejgta.org Enumeration for spanning trees and forests of join graphs ... | Sung Sik U 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. Thus, this corollary is true. ✷ 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 K m + K n,p . Clearly, K m + K n,p = (K m + H n ) + H p . Let D(m, l; n, k) be the set of the labelled spanning forests of K m + H n with l roots in K m and k roots in H n , i.e., g(m, l; n, k) = |D(m, l; n, k)|. (9) Theorem 3.1. The number g(m, n, p) of the spanning trees of K m + K n,p is equal to g(m, n, p) = (m + n) p−1 (m + p) n−1 (m + n + p) m .

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Enumeration for spanning trees and forests of join graphs ... | Sung Sik U 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. Therefore, we get the required result. ✷ Theorem 3.2. The number S(m, n, p) of all spanning forests of the join graph K m + K n,p is equal to S(m, n, p) = (m + n + p + 1) m+1 (m + n + 1) p−1 (m + p + 1) n−1 .
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 (K m + H n ) + H p with r roots in V (H p ) as follows. Let z i 1 , z i 2 , · · · , z ir ∈ V (H p ) be root vertices. The number of ways to select r roots in V (H p ) is equal to p r . Let Z ′ = V (H p )\{z i 1 , z i 2 , · · · , z ir }. 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 www.ejgta.org Enumeration for spanning trees and forests of join graphs ... | Sung Sik U not containing z in the graph already constructed, we repeat this procedure t times. There are [(p − r)(l + k − 1)][(p − r)(l + k − 2)] · · · [(p − r)(l + k − t)] t! = l + k − 1 t (p − r) t 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 l+k−1 t=0 l + k − 1 t (p − r) t (r + 1) l+k−t = (r + 1)(p + 1) l+k−1 forests in B(p, r) that contain G. Hence, by (9) and Theorem 2. p r (m + n) p−r (r + 1)(p + 1) l+k−1 = (m + n + p + 1) m+1 (m + n + 1) p−1 (m + p + 1) n−1 .
Thus, this theorem is true. ✷