aa r X i v : . [ m a t h . C O ] F e b A recursive algorithm for trees and forests
Song Guo and Victor J. W. Guo School of Mathematical Sciences, Huaiyin Normal University, Huai’an,Jiangsu 223300, People’s Republic of China [email protected], [email protected]
Abstract.
Trees or rooted trees have been generously studied in the litera-ture. A forest is a set of trees or rooted trees. Here we give recurrence relationsbetween the number of some kind of rooted forest with k roots and that with k + 1 roots on { , , . . . , n } . Classical formulas for counting various trees suchas rooted trees, bipartite trees, tripartite trees, plane trees, k -ary plane trees, k -edge colored trees follow immediately from our recursive relations. Keywords : forests; rooted trees; bipartite trees; tripartite trees; plane trees; k -edge colored trees. MR Subject Classifications : 05C05, 05A15, 05A19.
The famous Cayley’s formula for counting trees states that the number oflabeled trees on [ n ] := { , . . . , n } is n n − . Clarke [7] first gives a refinedversion for Cayley’s formula by setting up a recurrence relation. Erd´elyi andEtherington [11] gives a bijection between semilabeled trees and partitions.This bijection was also discovered by Haiman and Schmitt [14]. A generalbijective algorithm was given by Chen [4]. Aigner and Ziegler’s book [1]collected four different proofs of Cayley’s formula. We refer the reader to[5, 6, 8, 13, 15, 20] for several recent results on the enumeration of trees. Thegoal of this paper is to establish simple linear recurrences between certainforests with roots 1 , . . . , k and forests with roots 1 , . . . , k + 1, from which onecan deduce several classical results on counting trees.The set of forests of k rooted trees on [ n ] with roots 1 , . . . , k is denotedby F kn . Suppose that F ∈ F kn and x is a vertex of F , the subtree rooted at x is denoted by F x . We say that a vertex y of F is a descendant of x , if y is avertex of F x , i.e., x is on the path from the root of T to the vertex y , and isdenoted by y ≺ x . Corresponding author. e = ( x, y ) of a tree T in a forest F , if y is a vertex of T x , wecall x the father vertex of e , y the child vertex of e , x the father of y , and y a child of x , sometimes we also say e is out of x . The degree of a vertex x ina rooted tree T is the number of children of x , and is denoted by deg T ( x ), ordeg F ( x ). As usual, a vertex with degree zero is called a leaf.An unrooted labeled tree will be treated as a rooted tree in which thesmallest vertex is chosen as the root. Moreover, if A is a set of trees, then wewill use A [ P ] to denote the subset of all elements of A satisfying the condition P . One of our main results is as follows.
Theorem 2.1
For k n − , we have the following recurrence relation: |F k − n [ n ≺ | = n |F kn [ n ≺ | . (2.1) Proof.
Suppose that F ∈ F k − n [ n ≺ F k from F and add it to be a new tree in the forest. Second if n is not a descendant of 1in the new forest, n must be a descendant of k , exchange labels of the vertices1 and k . Thus, we obtain a forest F ′ ∈ F kn [ n ≺ F ′ ∈ F kn [ n ≺ F ′ k to any vertex ofthe other trees in F as a subtree, or attach F ′ to any vertex of F ′ k as a subtreeand exchange labels of the vertices 1 and k . The proof then follows from thefact that F ′ has n vertices altogether. (cid:3) It is clear that |F n − n [ n ≺ | = 1 . (2.2)We have the following corollaries. Corollary 2.2 (Cayley [3])
The number of labeled trees on n vertices is n n − . Corollary 2.3 (Cayley [3], Clarke [7])
The number of rooted trees on n +1 vertices with a specific root and root degree k is (cid:0) n − k − (cid:1) n n − k .Proof. It follows from (2.1) and (2.2) that |F kn [ n ≺ | = n n − k − . ✁✁ ❆❆❆ s s s s s s s s s s s s s s s s s s s s ✁✁✁ ❆❆❆ s s s s s s s s s s n = 5 and k = 3.Exchanging the labels of the vertices j and 1 for 1 < j k < n , weestablish a bijection between F kn [ n ≺
1] and F kn [ n ≺ j ]. Therefore, |F kn | = kn n − k − , from which one can see that the number of forests with n vertices and k treesis (cid:18) nk (cid:19) |F kn | = (cid:18) nk (cid:19) kn n − k − = (cid:18) n − k − (cid:19) n n − k . (cid:3) Remark.
It is worth mentioning the third and fourth proofs of Cayley’s formulain Aigner and Ziegler’s book [1, Chapter 30]. The third proof in [1, Chapter30], essentially due to Riordan [19] and R´enyi [18], is as follows: Let T n,k denote the number of forests on [ n ] consisting k trees where the vertices of[ k ] appear in different trees. Consider such a forest F and suppose that 1 isadjacent to i vertices. Removing the vertex 1, we obtain a forest of k − i trees. As we can reconstruct F by first fixing i , then selecting the i neighborsof 1, and then the forest F \
1, this gives T n,k = n − k X i =0 (cid:18) n − ki (cid:19) T n − ,k − i , from which we can prove Cayley’s formula by induction on n .The fourth proof in [1, Chapter 30], due to Pitman [17], is as follows:Let F n,k denote the set of all forests that consist of k rooted trees on [ n ]. Asequence F , . . . , F k of forests is called a refining sequence if F i ∈ F n,i and F i +1 is obtained from F i by removing one edge for all i . Fix F k ∈ F n,k anddenote by 3 N ( F k ) the number of rooted trees containing F k , • N ∗ ( F k ) the number of refining sequences ending in F k Then it is easy to see that N ∗ ( F k ) = N ( F k )( k − N ∗ ( F k ) = n k − ( k − N ∗ ( F k ) in two ways, first by starting a tree and secondly bystarting at F k . Hence, N ( F k ) = n k − for any F k ∈ F n,k .Since F n contains n isolated nodes, we get F n, = n n − , and thus Cayley’sformula.Therefore, both our proof of Cayley’s formula and the third proof in [1,Chapter 30] are recursive. But the latter requires more algebraic computa-tions. Our proof is very similar to Pitman’s proof. The difference is thatPitman considers all forests with k rooted trees while we only consider forestswith roots 1 , . . . , k . Pitman’s proof uses the idea of double counting and ourproof is a little more straightforward and combinatorial. A complete k -partite graph K n ,...,n k is a graph G = ( V, E ) with the vertexset V partitioned into k parts V , . . . , V k , where | V i | = n i , i k , any twovertices x, y ∈ V are adjacent if and only if x and y do not belong to the samepart V i . A spanning tree of K n ,...,n k is called a k -partite tree. Corollary 3.1 (Fiedler and Sedl ´aˇc ek [12])
The number c ( K r,s ) of span-ning trees of the complete bipartite graph K r,s is r s − s r − .Proof. Let V = { , . . . , r } and V = { r + 1 , . . . , r + s } . Denote by F kr,s the setof forests of k rooted bipartite trees on V = V ∪ V with roots 1 , . . . , k .Note there is no edge connecting vertices x and y if x and y belong to thesame set V i . In a similar way to Theorem 2.1, we have |F k − r,s [ r + 1 ≺ | = s |F kr,s [ r + 1 ≺ | , k r. (3.3)It is easy to see that |F rr,s [ r + 1 ≺ | = r s − . (3.4)We complete the proof by (3.3) and (3.4). (cid:3) orollary 3.2 For r > , s > , we have r − X i =1 s X j =1 (cid:18) r − i − (cid:19)(cid:18) s − j − (cid:19) i j − j i − ( r − i ) s − j − ( s − j ) r − i − = r s − s r − . Here we set = 1 .Proof. It follows from (3.3) and (3.4) that |F r,s [ r + 1 ≺ | = r s − s r − . On the other hand, both of the trees in a forest F of F r,s [ r + 1 ≺
1] arebipartite trees. Applying Corollary 3.1, we complete the proof. (cid:3)
Corollary 3.3 (Austin [2])
The number c ( K r,s,t ) of spanning trees of thecomplete tripartite graph K r,s,t is ( r + s + t )( r + s ) t − ( s + t ) r − ( t + r ) s − .Proof. Let V = { , . . . , r } , V = { r + 1 , . . . , r + s } , and V = { r + s +1 , . . . , r + s + t } . Denote by F kr,s,t (respectively, ˇ F kr,s,t ) the set of forests of k rooted tripartite trees on V = V ∪ V ∪ V with roots 1 , . . . , k (respectively,2 , . . . , k + 1).In exactly the same way as in the proof of Corollary 3.1, we have |F k − r,s,t [ r + 1 ≺ | = ( s + t ) |F kr,s,t [ r + 1 ≺ | , k r. It follows that c ( K r,s,t ) = |F r,s,t [ r + 1 ≺ | = ( s + t ) r − |F rr,s,t [ r + 1 ≺ | . (3.5)We have the following important fact: |F rr,s,t [ r + 1 ≺ | = | ˇ F rr,s,t [1 ≺ r + 1] | , (3.6)because 1 and r + 1 are in the same tree in each forest, what is different isthat we choose 1 as the root of the tree on the left-hand side, while we choose r + 1 on the right-hand side. Again, similarly to the proof of Corollary 3.1,we get | ˇ F k − r,s,t [1 ≺ r + 1] | = ( r + t ) | ˇ F kr,s,t [1 ≺ r + 1] | , r + 1 k r + s − . (3.7)On the other hand, we have | ˇ F r + s − r,s,t [1 ≺ r + 1] | = ( r + s + t )( r + s ) t − . (3.8)5n fact, there are ( r + s ) t forests such that 1 is a child of r + 1, and t ( r + s ) t − forests such that the father of 1 is a child of r + 1, which comes from V .The proof then follows from combining (3.5)–(3.8). (cid:3) Remark.
A bijective proof of Corollary 3.1 was found by Stanley [21, pp. 125-126]. Austin [2] proved that the number of spanning trees of the complete k -partite graph K n ,...,n k is n k − ( n − n ) n − · · · ( n − n k ) n k − , where n = n + · · · + n p . Eˇgecioˇglu and Remmel [9] found a bijective proof of the enumerationof bipartite trees and tripartite trees and later they [10] found a bijection forthe k -partite trees. The reader is encouraged to find a simple proof of Austin’sformula by modifying our recursive algorithm for trees and forests. A labeled plane tree is a rooted labeled tree for which the children of anyvertex are linearly ordered. Denote by P kn the set of forests of k labeled planetrees on n vertices with roots 1 , . . . , k . Corollary 4.1
For k n − , the following holds |P k − n [ n ≺ | = (2 n − k ) |P kn [ n ≺ | . (4.9) Proof.
The first step is the same as Theorem 2.1. For the reversal step, wesuppose that F ′ ∈ P kn [ n ≺ x ) + 1 waysfor us to attach F ′ k to the vertex x of the other trees in F as a subtree. Thesecond case is the same. The proof then follows from the fact X x ∈ V ( F ′ ) (deg F ′ ( x ) + 1) = n − k + n = 2 n − k. Corollary 4.2
The number of labeled plane trees on n +1 vertices is (2 n )! /n ! ,i.e., the number of unlabeled plane trees on n + 1 vertices is the well-knownCatalan number n n = 1 n + 1 (cid:18) nn (cid:19) . Proof.
It follows from (4.9) and |P nn +1 [ n + 1 ≺ | = 1 that the number oflabeled plane trees on [ n +1] with root 1 is (2 n )! / ( n +1)!, but the total numberof plane trees is n + 1 times that. (cid:3) We say that a plane tree is non-leaf vertex labeled if all the leaves are notlabeled, while all of the other vertices are labeled. Denote by P rn,p the set offorests of k non-leaf vertex labeled plane trees on n vertices with p leaves andwith roots 1 , . . . , r . 6 heorem 4.3 For r < n − p , the following holds |P r − n,p [ n − p ≺ | = ( p + 1) |P rn +1 ,p +1 [ n − p ≺ | . (4.10) Proof.
The proof is analogous to that of Theorem 2.1. But this time when asubtree is removed, its root is left and is considered as an unlabeled vertex.Thus, the number of the leaves (unlabeled vertices) of the forest is increasedby one. The reverse is clear and straightforward. (cid:3)
For n = 12 , p = 7 , r = 3, an example is given in Figure 2. ✁✁✁✁✁✁✁ ❆❆❆❆❆❆❆❆❆ r r r r r rr r r r r rr ✁✁✁✁✁ ❆❆❆❆❆❆❆❆✁✁✁ r rr r r r r r r r rr ✁✁✁✁✁ ❙❙❙❆❆❆❅❅❅ r r r r rr r r r r rr ✁✁✁✁✁ ❆❆❆❆❆✁✁✁ r r r r r r r r r r rr ✁✁✁✁✁ ❆❆❆❆❆❆ ✁✁✁ ❆❆❆ r r r r r r r r r r rr ✁✁✁✁✁✁✁ ❆❆❆❆❆❆❆❆❆ r rr r r r r r r r rr ✁✁✁ ✁✁✁✁✁ ❙❙❙❆❆❆❆❆❆ r rr r r rr r r r rr ✁✁✁ ❆❆❆❆❆✁✁✁✁✁ ❆❆❆ r r r r r r r r r r rr ✁✁✁✁✁ ❆❆❆❆❆❆ ✁✁✁ ❆❆❆ r r r r r r r r r r r rr Figure 2: An example for Theorem 4.3.A plane tree is called a k -ary plane tree if every non-leaf vertex has degree k . Corollary 4.4
The number of forests of r non-leaf vertex labeled k -ary planetrees on kn + r vertices with roots , . . . , r equals rn (cid:18) knn − r (cid:19) ( n − r )! . roof. The recurrence (4.10) also holds for k -ary plane trees. But in this case,( n, p ) is replaced by ( kn + 1 , ( k − n + 1). Hence, we have |P rkn +1 , ( k − n + r [ n ≺ | = (( k − n + r + 1)(( k − n + r + 2) · · · ( kn − · k. Here the factor k occurs because there are k possibilities for n to be a child of 1.The proof then follows from the fact that |P rkn +1 , ( k − n +1 | = r |P rkn +1 , ( k − n +1 [ n ≺ | . (cid:3) Corollary 4.5
The number of unlabeled k -ary plane trees on kn + 1 verticesequals n (cid:18) knn − (cid:19) = 1 kn + 1 (cid:18) kn + 1 n (cid:19) . Proof.
For any k -ary plane tree on kn + 1 vertices, it has n non-leaf vertices.We have ( n − kn + 1vertices with root 1 by ( n − k -ary plane trees. (cid:3) Corollary 4.6
For p, q > , n > p + q , we have n − q X i = p pqi ( n − i ) (cid:18) kii − p (cid:19)(cid:18) k ( n − i ) n − i − q (cid:19) = p + qn (cid:18) knn − ( p + q ) (cid:19) . Proof.
Let r = p + q . The forest of p + q non-leaf labeled trees may be treatedas the union of two small forests, namely, the forest of p trees with roots1 , . . . , p , and the forest of q trees with roots p + 1 , . . . , p + q . It follows fromCorollary 4.4 that n − q X i = p (cid:18) n − ( p + q ) i − p (cid:19) pi (cid:18) kii − p (cid:19) ( i − p )! qn − i (cid:18) k ( n − i ) n − i − q (cid:19) ( n − i − q )!= p + qn (cid:18) knn − ( p + q ) (cid:19) ( n − ( p + q ))! . Dividing both sides of the identity by ( n − ( p + q ))!, we complete the proof. (cid:3) Corollary 4.7 (Narayana [16])
The number of unlabeled plane trees on n +1 vertices with p leaves equals n + 1 (cid:18) n + 1 p (cid:19)(cid:18) n − n − p (cid:19) . roof. It follows from (4.10) that |P n +1 ,p [ n + 1 − p ≺ | = ( p + 1)( p + 2) · · · ( n − |P n − p n − p,n − [ n + 1 − p ≺ | = ( n − p ! |P n − p n − p,n − [ n + 1 − p ≺ | . (4.11)For any F ∈ P n − p n − p,n − [ n + 1 − p ≺ n + 1 − p is a child of1 and so the number of forests in P n − p n − p,n − [ n + 1 − p ≺
1] with the vertex i having degree x i is equal to x (there are x possibilities for n + 1 − p to be achild of 1). It follows that |P n − p n − p,n − [ n + 1 − p ≺ | = X x + ··· + x n − p +1 = n x . (4.12)It is well known that the number of solutions in positive integers to theequation x + · · · + x n − p +1 = n, is (cid:0) n − n − p (cid:1) . So the right-hand side of (4.12) is equal to nn − p + 1 (cid:18) n − n − p (cid:19) . Together with (4.11), we obtain |P n +1 ,p | = ( n − p )! n + 1 (cid:18) n + 1 p (cid:19)(cid:18) n − n − p (cid:19) . Dividing this number by ( n − p )!, we get the desired formula for unlabeledplane trees. (cid:3) Corollary 4.8
For any nonnegative integer sequence ( d , d , . . . , d n ) satisfy-ing d + d + · · · + d n = n − , there are ( n − plane trees on [ n ] with thedegree of i being d i .Proof. Without loss of generality, suppose that d i > i n − p . Just like (4.11), we see that the number of non-leaf vertex labeledplane trees with root i and with the degree of j being d j for 1 j n − p is d i ( n − /p !, so the total number is X d i > d i ( n − p ! = ( n − p ! . But there are p ! ways to label the leaves of each plane tree with { n − p +1 , . . . , n } . This completes the proof. (cid:3) orollary 4.9 For any nonnegative integer sequence ( d , d , . . . , d n ) satisfy-ing d + d + · · · + d n = n − , the number of rooted trees on [ n ] with the degreeof i being d i is (cid:18) n − d , d , . . . , d n (cid:19) . Proof.
For any rooted tree on [ n ] with the degree of i being d i , there are d ! d ! . . . d n ! plane trees with the same degree sequence. The proof then followsfrom Corollary 4.8. (cid:3) Corollary 4.10 (Erd ´e lyi-Etherington [11]) Let n n · · · m n m be a parti-tion of n − , i.e., n +2 n + · · · + mn m = n − , and n = n − ( n + n + · · · + n m ) .Then the number of unlabeled plane trees having n i vertices with degree i isequal to n (cid:18) nn , n , . . . , n m (cid:19) . Proof.
It follows from the proof of Corollary 4.7 (replace n by n −
1) that thenumber of such non-leaf vertex labeled plane trees is equal to the product of( n − /n ! and P x , where x , x , . . . , x n − n ranges over all the permutationsof the multiset { n , n , . . . , m n m } . It is not hard to see that X x = m X j =1 ( n + · · · + n m − n ! · · · n m ! /n j j = ( n + · · · + n m − n ! · · · n m ! ( n − . That is, the multiplication equals( n + · · · + n m − n ! n ! · · · n m ! ( n − . Dividing it by ( n − − n )!, we obtain the desired number for unlabeled planetrees. (cid:3) k -Edge colored trees A k -edge colored tree is a tree whose edges are colored from a set of k colorssuch that any two edges with a common vertex have different colors. Forconvenience, the set of k colors is denoted by { , , . . . , k } .A special k -edge colored tree is a rooted k -edge colored tree such thatall the edges out of the root are colored with the first k − { , , . . . , k − } . Note that for such a tree, the degree of the root is lessthan or equal to k −
1. Denote by E rn,k the set of forests of r special k -edgetrees with roots 1 , . . . , r . 10 heorem 5.1 For r n − , we have the following recurrence relations: |E r − n,k [ n ≺ | = ( kn − n + r ) |E rn,k [ n ≺ | . (5.13) Proof.
For F ∈ E r − n,k [ n ≺ r has thecolor x , then none of the edges with father vertex r could have color x . • First, remove the subtree F r from F and add it to be a new tree in theforest. • Second, if one edge with father vertex r is colored by k , we recolor it by x . • Third, if n is not a descendant of 1 in the new forest, n must be adescendant of r , exchange the labels of vertices 1 and r .Thus we obtain a forest F ′ ∈ E rn,k [ n ≺ F ′ ∈ E rn,k [ n ≺ F : • First, attach F ′ r back to any vertex of the other trees in F as a subtree,or attach F ′ to any vertex of F ′ r as a subtree and exchange the labels ofthe vertices 1 and r . Note that there is an uncolored edge. i.e., the edge e with child vertex r . Denote by v the father vertex of e , and C ( v ) theset of colors of the other edges incident with v . • Second, if v is one of the roots in the forest, i.e., 1 v r , we can color e with any color in { , , . . . , k − } \ C ( v ), whenever it is not empty.Otherwise, we color e with any color in { , , . . . , k } \ C ( v ), wheneverit is not empty. • Third, if the color of e , say x , is the same as some edge f incident withvertex r , then recolor f by k .Note that if 1 v r , we have C ( v ) ⊆ { , , . . . , k − } , and | C ( v ) | =deg F ′ ( v ). Otherwise, we have | C ( v ) | = deg F ′ ( v ) + 1. Therefore, the numberof ways to recover F ′ is given by r X v =1 ( k − − deg F ′ ( v )) + n X v = r +1 ( k − (1 + deg F ′ ( v )))= ( k − n − n X v =1 deg F ′ ( v ) = kn − n + r. (cid:3) Figure 3 gives an example for the case n = 6 , r = 3 , k = 3. In order tomake the figure clear, we replace the colors set { , , } by { x , y , z } . ✁✁✁ ❆❆❆ s s s s s s x yyx ✁✁✁ ❆❆❆ s s s s s s x yzx ✁✁✁ ❆❆❆ s s s s s s x yxz ✁✁✁ ❆❆❆ s s s s s s x yzx ✁✁✁ ❆❆❆❆❆✁✁✁ s s s s s s x yx z ✁✁✁ ❆❆❆ s s s s s s xzx y ✁✁✁ ❆❆❆ s s s s s s xyx z ✁✁✁ ❆❆❆ s s s s s s x y yx ✁✁✁ ❆❆❆ s s s s s s x y xz ✁✁✁ ❆❆❆ s s s s s s x y x Figure 3: An example for the recurrence relation (5.13).
Corollary 5.2
The number of forests of r special k -edge colored trees on [ n ] with roots , . . . , r is r ( k − kn − n − r + 1)( kn − n − r + 2) · · · ( kn − n − r ( k − n − r − (cid:18) kn − n − n − r − (cid:19) . Proof.
It follows immediately from (5.13) and the fact |E n − n,k [ n ≺ | = k − (cid:3) The following result is due to Gessel (see [21, pp. 124]), and can also bededuced in the same way as before. 12 orollary 5.3
The number of k -edge colored trees on [ n ] is k ( n − (cid:18) kn − nn − (cid:19) . Proof.
The case r = 1 of Corollary 5.2 implies that the number of special k -colored trees with root 1 is( k − n − (cid:18) kn − n − n − (cid:19) . Note that a k -colored tree is not special if and only if one of the edges out ofthe root is colored by k , and we can delete this edge to obtain a forest of twospecial k -colored trees such that one of the roots is 1. But Corollary 5.2 saysthat the number of forests of two special k -edge colored trees with roots 1 , k − n − (cid:18) kn − n − n − (cid:19) , so is the number of forests with roots 1 , x , where x = 1. Hence, the numberof k -colored trees on [ n ] (with root 1) is given by( k − n − (cid:18) kn − n − n − (cid:19) + 2( n − k − n − (cid:18) kn − n − n − (cid:19) = k ( kn − n )( kn − n − · · · ( kn − n + 3) . This completes the proof. (cid:3)
Corollary 5.4
The number of k -edge colored trees on [ n ] with a specific rootand root degree r is k ( n − (cid:18) k − r − (cid:19)(cid:18) ( k − n − n − r − (cid:19) . Proof.
Suppose that the root of the tree is 1. Then there are (cid:0) n − k (cid:1) ways tochoose the children of 1 from the set { , . . . , n } , and (cid:0) kr (cid:1) r ! ways to color theedges out of the root. Obviously, each of subtrees of 1 may be regarded asa special k -edge colored tree, because the edges out of its root are coloredfrom a definite set of k − r ( k − n − r − (cid:0) k ( n − − nn − r − (cid:1) ways to construct the forest of r special k -edgetrees on n − r specific roots. Therefore, the number of therequired trees is (cid:18) n − r (cid:19)(cid:18) kr (cid:19) r ! r ( k − n − r − (cid:18) k ( n − − nn − r − (cid:19) ,
13s desired. (cid:3)
Acknowledgments.
The authors would like to thank the anonymous refereesfor helpful comments on a previous version of this paper. The first authorwas partially supported by the Natural Science Research Project of OrdinaryUniversities in Jiangsu Province of China (grant no. 13KJB110001).
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