A Blossoming Algorithm for Tree Volumes of Composite Digraphs

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1 A Blossoming Algorithm for Tree Volumes of Composite Digraphs Victor J. W. Guo Center for Combinatorics, LPMC, Nankai University, Tianjin 30007, People s Republic of China jwguo@eyou.com Abstract. A weighted composite digraph is obtained from some weighted digraph by replacing each vertex with a weighted digraph. In this paper, we give a beautiful combinatorial proof of the formula for forest volumes of composite digraphs obtained by Kelmans-Pak-Postnikov [7]. Moreover, a generalization of this formula is present. Keywords: composite digraph, tree volume, oriented tree volume, bijection Introduction The composite graph G(H,..., H n ) is obtained from the graph G on [n] by replacing each vertex i by a graph H i. It is a generalization of multipartite graph, whose complexity was studied by Knuth [8] and Kelmans [6]. The number of spanning trees of G(H,..., H n ) was obtained by Pak-Postnikov [9]. These papers generalized the encoding of Prüfer [0]. Recently, Kelmans-Pak-Postnikov [7] investigated the tree and forest volumes of weighted digraphs with algebraic methods, and deduced a nice formula for forest volumes of composite weighted digraphs. To give a combinatorial interpretation of Kelmans-Pak-Postnikov s formula, we notice that the composite digraph can always be obtained from a series of digraph-substitutions. This fact discloses the essence of a composite digraph. Based on it, we construct a bijection between two sets related to oriented trees. The bijection uses the ideas of Joyal [5] and Goulden-Jackson [3], who independently constructed an elegant encoding for bi-rooted trees. The Joyal encoding was also used by Stanley [2], Eǧecioǧlu-Remmel [, 2] and Guo [4] in their approachs to bipartite or multipartite trees. Although a bijective proof of Kelmans-Pak-Postnikov s formula may be given by using the Prüfer code, the proof applying the Joyal code is much more simple and straightforward. We introduce the concept of the oriented tree volume of a weighted digraph, and obtain a formula for oriented tree volumes of weighted composite digraphs, which is equivalent to Kelmans-Pak-Postnikov s formula. Moreover, a generalized form of this formula is present. 2 Notation and terminology A directed graph or simply a digraph G is a graph G with each edge uv endowed with a direction from u to v or v to u. The edge set and vertex set of G are denoted by V (G)

2 and E(G), respectively. The edge (or called arc) with direction from u to v is denoted by (u, v), and we call u the initial vertex and v the final vertex. The outdegree of a vertex v, denoted deg + G (v), is the number of edges of G with initial vertex v. Similarly, the indegree of v, denoted deg G (v), is the number of edges with final vertex v. An oriented tree (or in-tree) with root v is a digraph T with v as one of its vertices, such that there is a unique directed path from any vertex u to v. The set of spanning oriented trees of G is denoted by Sp(G). A weighted digraph G is a digraph G, such that each edge e of G is associated with an indeterminate t e (or an element of a commutative ring). For each vertex v of a weighted digraph G, we associate an indeterminate x v. Let T be a spanning oriented tree of G with root r. Define the weight ω(t ) and oriented weight ω(t ) of T by ω(t ) := x deg T v (v) t e, (2.) ω(t ) := v V (T ) v V (T ) x deg T (v) v e E(T ) e E(T ) t e, (2.2) where t e is the weight of the edge e. Now define the tree volume f G (x, t) and oriented tree volume f G (x, t) of G to be the polynomials in the variables (x v ) v V (G) and (t e ) e E(G) by f G (x, t) := f G (x, t) := T Sp(G) T Sp(G) ω(t ), (2.3) ω(t ). (2.4) Let G = (V, E) be a weighted digraph, G = {G v : v V } and H = {H v : v V } be two families of disjoint weighted digraphs such that H v is a subgraph of G v for every v V. We construct a new digraph Γ = G(G, H) as follows: (i) V (Γ) = v V V (G v ); (ii) E(Γ) = v V E(G v ) {(x, y): x V (H u ), y V (H v ), and (u, v) E(G)}; (iii) For x V (H u ), y V (H v ), and (u, v) E(G), the edge (x, y) is endowed with the weight t (x,y) = t (u,v). We call Γ the composition of G through (G, H). Denote by G(H) = G(H, H). If G u = G, H u = H, and G v = H v = {v} for v u, then G(G, H) is denoted by G[G, H, u], and we call G[G, H, u] a substitution of (G, H ) into G in place of u. When V (H ) = V (G ), G[G, H, u] is denoted by G[G, u]. Let G be a weighted digraph, we construct a weighted digraph G as follows: (i) V (G ) = V (G), where / V (G); 2

3 (ii) E(G ) = E(G) {(v, ): v V (G)}; (iii) Each edge (v, ) is weighted by t (v, ) =. We call G the cone of G. For convenience, we also write G for the cone of G with the new vertex. Now define the forest volume of G(H) to be the tree volume of the digraph G(H). Theorem 2. (Kelmans-Pak-Postnikov [7, Theorem.2]) We have f G(H) (x, t) = f G (x, t) xv=hv(x) f H v (x, t), (2.5) x =x +g v(x,t) where h v (x) = u V (H v) x u, and g v (x, t) = (v,a) E(G) h a(x)t (v,a). v V We will prove the above theorem in the last section. 3 The complete bipartite digraph Let R = {, 2,..., r}, S = {r +, r + 2,..., r + s}, and let K p,q r,s be the complete bipartite digraph with vertex set partitioned into R S, and for any i R and j S, the arcs (i, j) and (j, i) are weighted by t (i,j) = p i and t (j,i) = q i, respectively. Lemma 3. We have f K p,q r,s (x, t) = p p r (q x + + q r x r ) s (x r+ + + x r+s ) r (x q /p + + x r q r /p r + x r+ + + x r+s ). (3.) We will give a bijective proof of Lemma 3.. The bijection established here is a little different from Eǧecioǧlu-Remmel [] and the second solution of Stanley [2, pp ]. The following definitions and lemma play an important role. Definition 3.2 An element a i is called a left-to-right maximum of a permutation a a 2 a n, if a i > a j for every j < i. Let S n denote the set of all permutations on [n] := {, 2,..., n}. It is well known that a permutation can be written as a product of its distinct cycles. Definition 3.3 A standard representation of a permutation is a product of its distinct cycles satisfying that (a) each cycle is written with its largest element first, and (b) the cycles are written in increasing order of their largest element. Definition 3.4 For a permutation π, we define φ(π) to be the permutation obtained from π by writing it in standard form and erasing the parentheses. 3

4 Lemma 3.5 (cf. Stanley [, Proposition.3.]) The map φ: S n S n defined above is a bijection. If π S n has k cycles, then φ(π) has k left-to-right maxima. Example 3.6 For π = with standard form π = (42)(6)(73)(85), we have φ(π) = with left-to-right maxima 4, 6, 7, 8. Note that φ (or φ) is well-defined for any permutation π = a a 2 a k over [n] for k n. For instance, if π = 57249, then φ (π) = (5)(724)(9). Proof of Lemma 3.. Recall that R = {, 2,..., r} and S = {, 2,..., s } = {r +, r + 2,..., r + s}. We linearly order R and S by < 2 < < r and < 2 < < s, respectively, and we would not compare the elements between R and S. Suppose T Sp(Kr,s p,q ) with root v. There is a unique directed path π = a b a 2 b 2 a m b m from v 0 to v, where v 0 = if v R, and v 0 = if v S. By Lemma 3.5, we obtain φ (π ) = C C l from π = b b 2 b m, where φ is defined in Definition 3.4 and C,..., C l are cycles. Then define C k to be the cycle obtained from C k by replacing each b i with b i a i+. Specifically, the cycle C k = (b ib i+ b j ) corresponds to C k = (b i a i+ b i+ a i+2 b j a j+ ). Let D π be the disjoint union of the directed cycles C,..., C l. When we remove all the edges of the path π, we obtain a disjoint union of oriented trees. Merge these oriented trees and D π by identifying vertices with the same label. Then we obtain a weighted digraph θ v (T ) by endowing each arc e with initial (respectively, final) vertex i R with weight t e = p i (respectively, t e = q i ). It is clear that θ v : T θ v (T ) is a bijection. If v R, then we define the weight of θ v (T ) by ω(θ v (T )) = q v x v e E(θ v(t )) t e while if v S, then we define the weight of θ v (T ) by ω(θ v (T )) = p x v e E(θ v(t )) t e i R S i R S x deg (i) i. x deg (i) i, It is easy to see ω(t ) = ω(θ v (T )), and ω(θ v (T )) is a term in the expansion of p p r (q x + +q r x r ) s (x r+ + +x r+s ) r x v q v /p v if v R, or a term in p p r (q x + + q r x r ) s (x r+ + + x r+s ) r x v if v S. The lemma follows by summing over all v. Examples for r = 5, s = 8 are shown in Figure. 4 The bijections for substituted digraphs The substituted digraph plays an important part in composite digraphs. We modify the bijection in the argument of Lemma 3. to count oriented tree volumes of weighted substituted digraphs. The key idea is to visualize each small tree as a vertex, and a 4

5 Figure : The two cases for Lemma 3.. spanning oriented tree of a weighted substituted digraph is then something like a bipartite oriented tree. Let H be a subgraph of G. Define Sp(G, H) to be the set of spanning oriented trees of G with root in H, and denote by f G, H (x, t) := ω(t ). T Sp(G,H) Lemma 4. Let G, G be disjoint weighted digraphs, H V (G ), and u V (G). We have f G[G,H,u], G[H,u](x, t) = f G (x, t) xu=h (x) f G H (x, t) x =g u(x,t), (4.) where h (x) = v H x v, and g u (x, t) = (u,v) E(G) x vt (u,v). Proof. Let G 2 = G \ {u} and H 2 be the the set of vertices v such that (u, v) E(G), and let Ω be the set of four-tuples (T, T 2, w (), w (2) ) such that (i) T Sp(G H ), and T 2 Sp(G); (ii) w () and w (2) are words on H and H 2, respectively; (iii) w () = deg T 2 (u), and w (2) = deg T ( ). We want to construct a bijection between Sp(G[G, H, u], G[H, u]) and Ω. First linearly order H and V (G 2 ), respectively. Suppose T is a spanning oriented tree of G[G, H, u] with root r in G[H, u]. Deleting the edges of T between H and G 2, we obtain weighted oriented forests F and F 2, which are contained in G and G 2, respectively. The oriented tree in F or F 2 with root v is denoted by R v, and the root sets of F and F 2 are denoted by M and M 2, respectively. If r M, then we define v 0 to be min M 2, while if r M 2, then we define v 0 to be min M. Assume that r is the first vertex on the path from v 0 to r in T such that r V (R r ). We may obtain an oriented tree T from F by adding the vertex, and edges (v, ) (v M ), each weighted by. To obtain the oriented tree T 2 from F 2, we add the vertex 5

6 u. If r M, then we add edges (v, u) (v M 2 ) with weights in G, while if r M 2, then we add edges (v, u) (v M 2 \ {r}) and (u, r ) with weights in G. It remains to find out the words w () and w (2). Identifying each R v with v in T, we obtain a weighted oriented tree T rooted at r. Assume that the directed path from v 0 to r in T is π = p p 2 p 2m, and p i < p i2 < < p il are all the left-to-right maxima of p 2 p 4 p 2m 2. Let a k = p ik ( k l), and a l+ = p 2m. For any vertex v v 0, r of T, let D(v) denote the vertex z such that { (ak, z) E(T ), if v = a k+ ( k l), (v, z) E(T ), otherwise. Write M \{v 0, r} = {u, u 2,..., u q } and M 2 \{v 0, r} = {v, v 2,..., v s } in increasing order. Put { D(v )D(v w () 2 ) D(v s )r, if r M, = D(v )D(v 2 ) D(v s ), if r M 2, w (2) = D(u )D(u 2 ) D(u q ). It is clear that w () = deg T 2 (u) and w (2) = deg T ( ). The procedure from Ω to Sp(G[G, H, u], G[H, u]) is as follows: Given T = (T, T 2, w (), w (2) ) Ω, deleting the vertices and u of T and T 2, we get oriented forests F and F 2, respectively. The oriented tree in F or F 2 with root v is denoted by R v, and the root sets of F and F 2 are denoted by M and M 2, respectively. Suppose the root of T 2 is r. If r = u, then we define v 0 to be min M 2, r the entry in the last position of w (), and r the root of the oriented tree in F that contains r. Otherwise, (u, r ) E(T 2 ) for some vertex r, we denote r = r, and define v 0 to be min M. Write M \ {v 0, r} = {u, u 2,..., u q } and M 2 \ {v 0, r} = {v, v 2,..., v s } in increasing order. For the words w () and w (2), if we identify each letter (vertex) with the root of the oriented tree in F or F 2 that contains it, then we obtain words w () and w (2) on M and M 2, respectively. By the definition of Ω, we have w () = s + or s, according to r = u or not, and w (2) = q. Regard the following function ( ) v,..., v s, u,..., u q D 0 = w (),..., w () s, w (2),..., w (2) q as a digraph on M M 2, such that deg + (v 0 ) = deg + (r) = 0. We may recover the oriented tree T = θr (D 0 ) with root r by Lemma 3.. Let ( ) v,..., v s, u,..., u q D =. w (),..., w s (), w (2),..., w q (2) 6

7 Assume that the path from v 0 to r in T is π = p p 2 p 2m, and p i < p i2 < < p il are all the left-to-right maxima of p 2 p 4 p 2m 2. Let a k = p ik ( k l), and a l+ = p 2m. We now connect all the oriented trees in F or F 2 by drawing the following edges: (a k, D(a k+ )) ( k l), and (p 2m, r ); (v, D(v)), for v V (T ) \ {a, a 2,..., a l+, r}. Finally, endow the above edges with weights in G[G, H, u]. Thus we obtain a spanning oriented tree T of G[G, H, u] with root r (in G[H, u]). It is not difficult to see that the above two procedures are inverse to each other, therefore we obtain a bijection between Sp(G[G, H, u], G[H, u]) and Ω. We now define the weight of T = (T, T 2, w (), w (2) ) Ω by ω( T ) = ω(t )ω(t 2 )x deg T ( ) x deg T (u) 2 u k w () x k v w (2) x v t (u,v). It is straightforward to see that the above bijection T T is weight-preserving, that is, ω(t ) = ω( T ). Clearly, ω(t ) is a term of f G[G,H,u], G[H,u](x), while ω( T ) is a term in the expansion of fg (x, t) xu= P v H x v f G H (x, t) x = P v H 2 x vt (u,v). This completes the proof of (4.). Examples for Lemma 4. are given in Figures 2, 3, 4, and 5, where H = {, 2,..., 4 } = {2, 3,..., 25} and N = {, 2,..., } is the set of vertices adjacent to u in G. We leave out labels of those vertices in V (G ) \ H or V (G 2 ) \ N. But the root is labeled by r if necessary. 7

8 T u T T 2 w () = 6, 20, 2, 3, w (2) =, 6, 3, 5, 2, 8 Figure 2: Example for Lemma 4. in the case r H T w () = 6, 20, 2, 3, w (2) =, 6, 3, 5, 2, 8 T, T 2 ( ) D = ( θ 4 (T ) Figure 3: Explanation of w () and w (2) in Figure 2. 8 )

9 T 0 r r u T T 2 w () = 6, 20, 3, w (2) =, 3, 5, 7, 2, 3 5 Figure 4: Example for Lemma 4. in the case r V (G 2 ). 4 6 r T 4 6 r θ r (T ) 7 w () = 6, 20, 3, w (2) =, 3, 5, 7, 2, 3 T, T 2 ( ) D = ( r 6 4 r Figure 5: Explanation of w () and w (2) in Figure 4. 9 )

10 Note that, in the proof of Lemma 4., the bijection T (T, T 2, w (), w (2) ) transforms the root of T into the root of T 2 when it belongs to V (G 2 ), and vice versa. Hence we actually prove the following assertion: Lemma 4.2 Let G, G be disjoint weighted digraphs, G 0 a subgraph of G, u V (G), and H V (G ). If u V (G 0 ), then we have f G[G,H,u], G 0 [H,u](x, t) = f G, G0 (x, t) xu=h (x) f G H (x, t) x =g u(x,t), (4.2) while if u / V (G 0 ), we have f G[G,H,u], G 0 (x, t) = f G, G0 (x, t) xu=h (x) f G H (x, t) x =g u(x,t), (4.3) where h (x) = v H x v, and g u (x, t) = (u,v) E(G) x vt (u,v). 5 The blossoming theorem In the previous section, we obtain a formula for counting the oriented tree volume of a substituted digraph. By using Lemmas 4., 4.2, and a series of substitutions of digraphs and variables, we can deduce the main theorem of this paper. Theorem 5. (The Blossoming Theorem) Let G = (V, E) be a weighted digraph, and let G = {G v : v V }, H = {H v : v V } be two families of disjoint weighted digraphs such that H v is a subgraph of G v for every v V. We have f G(G,H), G(H) (x, t) = f G (h, t) hv=h v(x) v V f Gv H v (x, t) x =g v(x,t), (5.) where h v (x) = u V (H v) x u and g v (x, t) = (v,a) E(G) h a(x)t (v,a). In particular, f G(H) (x, t) = f G (h, t) hv=h v(x) v V f H v (x, t) x =g v(x,t). (5.2) Proof. Let V = {v, v 2,..., v n }. We introduce the following notation: G k := {G vi : i k} {({v j }, ): k + j n}, H k := {H vi : i k} {({v j }, ): k + j n}, { h (k) hv (x) = u V (H v (x) := x v) u, if v {v,..., v k }, x v, otherwise, g v (k) (x, t) := h a (k) (x)t (v,a), v V (G). (v,a) E(G) 0

11 By the definition of composite digraphs, it is easy to see G(G k, H k ) = G(G k, H k )[G vk, H vk, v k ], G(H k ) = G(H k )[H vk, v k ]. As we call (G vk, H vk ) a flower, the process from G = G(G 0, H 0 ) to G(G, H) = G(G n, H n ) is something like blossoming, and any linear ordering of V (G) leads to the same result G = G(G, H). Hence, it is proper to call the following proof the blossoming algorithm. By Lemma 4.2, for k m, we have f G(Gk,H k ), G(H k ) = f G(Gk,H k ), G(H k )(x, t) xvk =h vk (x) It follows that f G(Gk,H k ), G(H k )(x, t) xvi =h vi (x), k+ i n f Gvk H v k (x, t) x =g (k) v k (x,t). = f G(Gk,H k ), G(H k )(x, t) xvi =h vi (x), k i n f G vk H v k (x, t) x =g vk (x,t). By iteration of the above equation, we have f G(G,H), G(H) (x, t) = f G(Gn,H n), G(H n)(x, t) = f G(Gn,H n ), G(H n )(x, t) xvn =h vn (x) f G vn H vn (x, t) x =g vn (x,t) = f G(Gn 2,H n 2 ), G(H n 2 )(x, t) xvj =h vj (x, t) (x), n j n f Gvn H v n (x, t) x =g vn (x,t) f G vn H vn (x, t) x =g vn (x,t) = = f G(G0,H 0 ), G(H 0 )(x, t) xvi =h vi (x), i n = f G (h, t) hu=h u(x) v V j n f Gv H v (x, t) x =g v(x,t). f Gvj H v j (x, t) x =g vj (x,t) This completes the proof of (5.). An illustration of the blossoming process is shown in Figure 6, where the directed connection of H i and H j (or j) is meant that every vertex of H i is connected to every vertex of H j (or the vertex j) with the same direction and weight as the edge ij. Proof of Theorem 2.. The proof follows from (5.2) by replacing G with G and H with H { }, and the follwing obvious relations: f G(H) (x, t) = f G(H) (x, t) x, f G (x, t) xv=h v(x) = f G (x, t) xv=h v(x) x.

12 G 2 G G G G 4 G 2 G 3 G G G 2 G H G 2 5 G 2 H G G 4 Figure 6: The blossoming procedure. It is less obvious to see that (2.5) also leads to (5.2) by comparing the terms independent of x on both sides of (2.5). Thus they are in fact equivalent to each other. For a (weighted) graph G, G is understood to be an undirected graph, and the tree volume f G (x, t) is defined as (2.3), where Sp(G) denotes the set of spanning trees of G. Eq. (5.2) has many conclusions, we would mention the following: Corollary 5.2 Let G = (V, E) be a weighted graph, and let H = {H v : v V } be a family of disjoint weighted graphs. Then f G(H) (x, t) = f G (h, t) hv=h v(x) v V f H v (x, t) x =g v(x,t), where h v (x) = u V (H v) x u and g v (x, t) = (v,a) E(G) h a(x)t (v,a). Corollary 5.3 (Kelmans [6, Theorem ]) Let H v be the empty graph on n v vertices. Then f G(H) (x) = f G (h) hu=hu(x) (g v (x)) nv, where f G (x) = f G (x, t) te=. v V (G) Corollary 5.4 (Pak-Postnikov [9]) The number of spanning trees of G(H,..., H n ) is equal to H v H v deg T (v) f i (H v )g(v) i, T Sp(G) v V 2 v V i=

13 where f i (H v ) is the number of forests in H v with i roots and g(v) = References (v,a) E H a. [] O. Eǧecioǧlu and J. B. Remmel, Bijections for Cayley trees, spanning trees, and their q-analogues, J. Combin. Theory Ser. A 42 (986) [2] O. Eǧecioǧlu and J. B. Remmel, A bijection for spanning trees of complete multipartite graphs, Congr. Numer. 00 (994) [3] I. P. Goulden and D. M. Jackson, Combinatorial Enumeration, John Wiely, New York, 983. [4] V. J. W. Guo, Spanning trees of complete multipartite graphs and combinatorial identities, preprint. [5] A. Joyal, Une théorie combinatoire des séries formelles, Adv. Math. 42 (98) 82. [6] A. K. Kelmans, Spanning trees of extended graphs, Combinatorica 2 () (992) [7] A. Kelmans, I. Pak, and A. Postnikov, Tree and forest volumes of graphs, available as DIMACS Technical Report , January [8] D. E. Knuth, Another enumeration of trees, Canad. J. Math. 20 (968) [9] I. M. Pak and A. E. Postnikov, Enumeration of spanning trees of certain graphs, Russian Math. Surveys 45 (3) (990) [0] H. Prüfer, Neuer Beweis eines Satzes über Permutationen, Arch. Math. Phys. 27 (98) [] R. P. Stanley, Enumerative Combinatorics, Vol., Wadsworth & Brooks/Cole, Monterey, California, 986. [2] R. P. Stanley, Enumerative Combinatorics, Vol. 2, Cambridge University Press, Cambridge,