Combinatorial interpretations for TG(1, −1)

CombinatorialInterpretations forTG(1,−1)

Jun Ma1,2 and Yeong-Nan Yeh2

1DEPARTMENT OF MATHEMATICS, SHANGHAIJIAO TONG UNIVERSITY, SHANGHAI 200240

PEOPLE’S REPUBLIC OF CHINAE-mail: [email protected],

[email protected]

2INSTITUTE OF MATHEMATICS, ACADEMIASINICA, TAIPEI, TAIWAN

E-mail: [email protected]

Received November 26, 2009; Revised November 16, 2010

Published online 9 February 2011 in Wiley Online Library (wileyonlinelibrary.com).DOI 10.1002/jgt.20588

Abstract: Let G be a connected and simple graph with vertex set{1,2, . . . ,n+1} and TG(x,y ) the Tutte polynomial of G. In this paper, wegive combinatorial interpretations for TG(1,−1). In particular, we givethe definitions of even spanning tree and left spanning tree. We proveTG(1,−1) is the number of even-left spanning trees of G. We associatea permutation with a spanning forest of G and give the definition of oddG-permutations. We show TG(1,−1) is the number of odd G-permutations.We give a bijection from the set of odd Kn+1-permutations to the set of

Contract grant sponsor: NSC; Contract grant number: 98-2115-M-001-010-MY3(to Y.-N. Y.)Journal of Graph Theory� 2011 Wiley Periodicals, Inc.

341

342 JOURNAL OF GRAPH THEORY

alternating permutations on the set {1,2, . . . ,n}. � 2011 Wiley Periodicals, Inc. J

Graph Theory 69: 341–348, 2012

Keywords: alternating permutation; spanning tree; Tutte polynomial

1. INTRODUCTION

Let G be a connected graph with vertex V(G) and edge set E(G). Suppose we are givena total ordering of edges in G. Fix a spanning tree T of G. Given an edge e∈E(G)\E(T),we call the edge e an externally active edge of T if it is the smallest edge in the uniquecycle contained in T ∪e. Define EAG(T) as the set of externally active edges for T andlet eaG(T)=|EAG(T)|. Given an edge e∈E(T), define UT (e) as the set of edges e inE(G) such that (T −e)∪ e is a spanning tree. An edge e in E(T) is internally active ifit is the smallest edge in UT (e). Define IAG(T) as the set of internally active edges inT and let iaG(T)=|IAG(T)|. In [9], the Tutte polynomial was defined as follows:

TG(x,y)= ∑T∈TG

xiaG(T)yeaG(T),

where TG denotes the set of spanning trees of G.In general, the Tutte polynomial encodes information about subgraphs of G.

Following [1], we list some special values of the Tutte polynomial in Table I.Let Sn denote the set of permutations on the set {1,2, . . . ,n}. A permutation �=

(�(1) . . .�(n))∈Sn is said to be alternating if �(1)>�(2)<�(3)>�(4)<�(5)>�(6)< · · ·�(n). Let An be the set of alternating permutations in Sn. We put an =|An|. Thesequence {an}n≥0 satisfies the following recurrence relation:

an =� n−1

2 �∑i=0

(n−1

2i

)a2ian−2i−1

with initial condition a0 =1, see [6, 8].Let Kn+1 be the complete graph on n+1 vertices. In [6, 8], it is proved that

TKn+1 (1,−1) is equal to the number of alternating permutations on the set {1,2, . . . ,n},follows from the fact that satisfies the same recurrence relation as the sequence {an}n≥0.We are interested in combinatorial interpretations of TKn+1 (1,−1). A tree on vertex set{1,2, . . . ,n} and rooted at 1 is said to be increasing if its vertices increase along thepaths away from the root. A increasing tree is even if every non-root vertex has aneven number of children. Kuznetsov et al. [8] constructed an involution that is fixedon even increasing trees and changes the parity of the number of inversions of the

TABLE I. Some special values of the Tutte polynomial.

TG(1,1) is the number of spanning trees of G.TG(2,1) is the number of spanning forests of G.TG(1,2) is the number of connected spanning

subgraphs of G.TG(2,2) is the number of spanning subgraphs of G.

Journal of Graph Theory DOI 10.1002/jgt

COMBINATORIAL INTERPRETATIONS FOR TG(1, −1) 343

remaining trees and proved TKn+1 (1,−1) is equal to the number of even increasing treesin Kn+1. Moreover, A 0-1-2 increasing tree is an increasing tree where all the verticesincluding the root have at most 2 children. An important fact is TKn+1 (1,−1) is equalto the number of 0-1-2 increasing trees in Kn. This was first formulated by Foata [4]and was proved by Foata and Strehl [5]. Donaghey [3] gave a simple bijective prooffor this fact.

Let G be a connected and simple graph with vertex set {1,2, . . . ,n+1}. In this paper,we give combinatorial interpretations for TG(1,−1). In particular, for any spanning treeT of G, let �T be a sequence on vertices of G formed by the vertex-adding order, whichis the third example of a proper tree order given in Chebikin and Pylyavskyy [2], anddescribed in Algorithm A below. With respect to the sequence �T , we define NT as theset of T-redundant edges of G, in the sense of Kostic and Yan [7, Section 4.1] and asdefined below after Algorithm A. In [7, p 91] it is shown that TG(1,y)=∑T∈TG

y|NT |.We then proceed to define even spanning trees and left spanning trees. Let IG be alleven-left spanning trees of G and TG =TG \TG. By establishing an involution on theset TG, we show TG(1,−1) is the number of even-left spanning trees of G.

We associate a permutation in Sn with a spanning forest of the graph G and givethe definition of odd G-permutations. Let SG,odd be the set of odd G-permutations.We give a bijection from TG to SG,odd. Hence TG(1,−1) is also the number of oddG-permutations inSn. We consider the case in which the graph G is the complete graphKn+1. We give a bijection from the set SKn+1,odd to the set An, where An denotes theset of alternating permutations of Sn.

We organize this paper as follows. In Section 2, we prove that TG(1,−1) is thenumber of even-left spanning trees of G. In Section 3, we give the definition of oddG-permutations and prove TG(1,−1) is the number of odd G-permutations in Sn. LetG=Kn+1. We give a bijection from the set SKn+1,odd to the set An.

2. EVEN-LEFT SPANNING TREES OF G

Let [n] :={1,2, . . . ,n} and G be a simple and connected graph with vertex set [n+1]and edge set E(G). In this section, we give the definition of even-left spanning trees ofG and prove TG(1,−1) is the number of even-left spanning trees of G.

A subtree of G is a connected subgraph of G without cycles. A spanning tree of Gis a subtree of G containing all the vertices of G. Let TG be the set of the spanningtrees of G. Given disjoint subsets U and W of the vertex set of G, we write EG(U,W)for the set of U-W edges, that is, for the set of edges joining a vertex in U to a vertexin W.

Given a spanning tree T of G, we associate T with a sequence

�T = (�T (0),�T (1), . . . ,�T (n))

on vertices of G by the following algorithm:

Algorithm A.Step 1. Let �T (0)=n+1.Step 2. Assume that �T (0),�T (1), . . . ,�T (i) are determined. Let Ui ={�T (j) | j=0,1, . . . , i}. Let v=min{w∈V(G)\Ui |ET (Ui,{w}) �=∅} and set �T (i+1)=v.



The sequence �T obtained by Algorithm A is the same as the vertex-adding orderon vertices of G rooted at n+1 defined by Chebikin and Pylyavskyy [2]. Note that �Tcan be viewed as a bijection from the set {0,1, . . . ,n} to [n+1]=V(G). Furthermore,for any vertex v∈V(G)\{n+1}, there exists a unique vertex u such that {u,v}∈E(T)and �−1

T (u)<�−1T (v). We say u is the predecessor of v and write u=pT (v). Fixing a

vertex v, suppose u is its predecessor. Define NT ,v as the set of vertices w such that{w,v}∈E(G)\E(T) and �−1

T (u)<�−1T (w)<�−1

T (v). For any w∈NT ,v, the edge {w,v} iscalled a T-redundant edge in [7]. Let NT be the set of T-redundant edges in G, i.e.NT =⋃v∈[n]{{w,v} |w∈NT ,v}.

Lemma 2.1 (Kostic and Yan [7]). TG(1,y)=∑T∈TGy|NT |.

Fix a spanning tree T . Let �T be the sequence of vertices of G obtained byAlgorithm A. We say a vertex v in [n] is T-even if the number of T-redundant edgesincident with v in G is even, i.e., |NT ,v|≡0(mod 2). If every vertex in [n] is T-even,then T is called an even spanning tree.

Given a vertex v∈ [n], for any w∈ [n+1], we say (w,v) is a inversion of the sequence�T if w>v and �−1

T (w)<�−1T (v). Since �T (0)=n+1, there exists a unique vertex u such

that (u,v) is a inversion of �T and �T (j)<v for any �−1T (u)<j<�−1

T (v). We say the vertexu is a (T;v)-rightmost inversion vertex. Let w=pT (v). It is easy to see �−1

T (u)≤�−1T (w)

by Algorithm A. Define MT ,v as the set of vertices s such that {s,v}∈E(G)\E(T) and�−1

T (u)≤�−1T (s)<�−1

T (w). If MT ,v =∅, then we say the vertex v is T-left. If every vertexin [n] is T-left, then we say T is a left spanning tree.

Example 2.2. Let us consider the complete graph Kn+1. Let T and �T be a spanningtree and the sequence of vertices of Kn+1 obtained by Algorithm A, respectively. Forany vertex v∈ [n], let w and u be the predecessor of v and the (T;v)-rightmost inversionvertex, respectively. Then v is T-even if and only if the number of vertices betweenw and v in the sequence �T , not including w and v, is even; v is T-left if and only ifu=w.

Example 2.3. We consider the graph G in Figure 1. We give all the spanning treesof the graph G in Figure 2. See Table II. For the spanning tree T1 in Figure 2,by Algorithm A, we have �T1 = (4,1,2,3). It is easy to see NT1,1 =∅, NT1,2 =∅, andNT1,3 ={1,2}. Hence, T1 is an even spanning tree in G. Note that MT1,v =∅ for allv=1,2,3. So T1 is a left spanning tree in G.

We say T is even-left if T is a spanning tree of G that is both even and left. DefineTG as the set of even-left spanning trees of G. Let TG =TG \TG.

4

1 2

3

FIGURE 1. A graph G.



4

1 2

3

4

1 2

3

4

1 2

3

4

1 2

3

4

1 2

3

4

1 2

3

4

1 2

3

4

1 2

3

T1 T2 T3 T4

T5 T6 T7 T8

FIGURE 2. All the spanning trees of G.

TABLE II. Even and left spanning trees of G.

�Ti NTi ,1 NTi ,2 NTi ,3 MTi ,1 MTi ,2 MTi ,3

T1 (4,1,2,3) ∅ ∅ {1,2} Even ∅ ∅ ∅ LeftT2 (4,1,2,3) ∅ ∅ {2} ∅ ∅ {4}T3 (4,1,2,3) ∅ ∅ ∅ Even ∅ ∅ {1,4}T4 (4,1,3,2) ∅ ∅ {1} ∅ ∅ ∅ LeftT5 (4,1,3,2) ∅ ∅ ∅ Even ∅ ∅ {4}T6 (4,3,1,2) ∅ {1} ∅ ∅ ∅ ∅ LeftT7 (4,3,1,2) ∅ ∅ ∅ Even ∅ {3} ∅T8 (4,3,2,1) ∅ ∅ ∅ Even ∅ ∅ ∅ Left

Lemma 2.4. There is an involution �=�G from the set TG to itself such that||NT |−|N�(T)||=1 for any T ∈TG.

Proof. For any T ∈TG, let �T be the sequence of vertices of G obtained byAlgorithm A. Since T is not an even-left spanning tree, there are some vertices w suchthat either |NT ,w|≡1(mod 2) or MT ,w �=∅. Let v be the first such vertex in the sequence�T . We discuss the following two cases.

Case 1. |NT ,v|≡1(mod 2).Clearly, NT ,v �=∅. Let u be the vertex in NT ,v such that �−1

T (u)≤�−1T (w) for any

w∈NT ,v. We construct a new spanning tree of G, denoted �(T), by adding the edge{u,v} and deleting the edge {pT (v),v}. It is easy to see �T =��(T), |N�(T),v|≡0 (mod 2),M�(T),v �=∅, and |NT |−|N�(T)|=1.

Case 2. |NT ,v|≡0 (mod 2) and MT ,v �=∅.



Let u be the vertex in MT ,v such that �−1T (u)≥�−1

T (w) for any w∈MT ,v since MT ,v �=∅.Adding the edge {u,v} and deleting the edge {pT (v),v}, we denote the obtained spanningtree by �(T). Then �T =��(T), |N�(T),v|≡1 (mod 2), and |NT |−|N�(T)|=−1. �

Example 2.5. Let G be the graph in Figure 1. See Table II. We have TG ={T1,T8}and TG ={Ti |2≤ i≤7}. We easily obtain �(T2)=T3, �(T4)=T5, and �(T6)=T7.

Theorem 2.6. The number of even-left spanning trees of G is TG(1,−1), i.e.,TG(1,−1)=|TG|.

Proof. Note that |NT |≡0 (mod 2) for any T ∈TG. By Lemma 2.4, we have

TG(1,−1) = ∑T∈TG

(−1)|NT |

= ∑T∈TG

(−1)|NT |+ ∑T∈TG

(−1)|NT |

= ∑T∈TG

1

= |TG|. �

3. ODD G-PERMUTATIONS

In this section, we give the definition of odd G-permutations and prove TG(1,−1) isthe number of odd G-permutations in Sn. We conclude with a bijection from the setof odd G-permutations to the set of alternating permutations on the set [n].

Fix a simple graph G with vertex set [n+1]. Given a permutation �= (�(1) . . .�(n))∈Sn, we associate � with a spanning forest of G as follows.

Algorithm B.Step 1. Set �(0)=n+1. Let E0 =∅.Step 2. At time i≥1, let j=max{k |0≤k<i,�(k)>�(i)}. Define N�,i as the set ofindexes k such that j≤k<i and {�(k),�(i)}∈E(G). If N�,i =∅, then let Ei =Ei−1.Otherwise let Ei =Ei−1 ∪{�(m),�(i)}, where m=min N�,i.

Iterating Step 2 until i=n, we obtain a spanning forest of G with edge set En. Denotethis spanning forest by F�. If F� is a spanning tree, then we say � is a G-permutation.Clearly, a permutation � is a G-permutation if and only if N�,i �=∅ for all i∈ [n]. LetSG be the set of all G-permutations. Furthermore, we say a G-permutation � is odd if|N�,i|≡1 (mod 2) for each i∈ [n]. Let SG,odd be the set of odd G-permutations.

Example 3.1. Let G be the graph in Figure 1. We consider all the permutations inS3 and give Table III.

It is easy to see SG ={�1,�2,�5,�6} and SG,odd ={�1,�6}.Lemma 3.2. There is a bijection from TG to SG,odd.



TABLE III. All the permutations in S3.

� N�,1 N�,2 N�,3

�1 (1,2,3) {0} {1} {0,1,2}�2 (1,3,2) {0} {0,1} {2}�3 (2,1,3) ∅ {1} {0,1,2}�4 (2,3,1) ∅ {0,1} {2}�5 (3,1,2) {0} {1} {1,2}�6 (3,2,1) {0} {1} {2}

Proof. For any T ∈TG, let �T be the sequence of vertices of G obtained byAlgorithm A. Consider a permutation � such that �(i)=�T (i) for any i∈ [n]. It is easyto see �∈SG,odd.

Conversely, for any �= (�(1) . . .�(n))∈SG,odd, let T be the spanning tree of Gobtained by Algorithm B. Let �T be the sequence of vertices of G obtained byAlgorithm A. To prove T ∈TG, it is sufficient to show �T (i)=�(i) for all i∈{0,1, . . . ,n}.Note that �T (0)=�(0)=n+1. By Algorithm A, for the spanning tree T , we assume�T (0), . . . ,�T (i) are determined and �T (j)=�(j) for all 0≤ j≤ i. Let Ui ={�T (j) | j=0, . . . , i} and W ={w |w /∈Ui,ET (Ui,{w}) �=∅}. By Algorithm B, we have �(i+1)∈Wsince |N�,i+1|≡1 (mod 2). We claim �(i+1)=minW. Otherwise, there exists a vertexu=�(k) for some k≥ i+2 such that u<�(i+1) and pT (u)∈Vi. u=�(k)<�(i+1) impliesN�,k ⊆{�(s) | i+1≤s≤k−1}. Hence, by Algorithm B, pT (u) /∈Vi, a contradiction. So�T (i)=�(i) for all 0≤ i≤n. By Algorithm B, we have |NT ,�(i)|=|N�,i|−1 and MT ,�(i) =∅for all i∈ [n]. Hence, T ∈TG. �

Combining Theorem 2.6 and Lemma 3.2, we obtain the following corollary.

Corollary 3.3. The number of odd G-permutations is TG(1,−1), i.e., TG(1,−1)=|SG,odd|.

We now consider the case when G is the complete graph Kn+1. Let opn be the numberof odd Kn+1-permutations, i.e., opn =|SKn+1,odd|. By Corollary 3.3, we immediatelyobtain the following corollary.

Corollary 3.4. TKn+1 (1,−1)=opn.

Recall An is the set of alternating permutations in Sn and an=|An|. By Corol-lary 3.4, we have opn =an since TKn+1 (1,−1)=an. It is easy to see the sequence {opn}n≥0satisfies the same recurrence relation and the initial condition as the sequence {an}n≥0. Inparticular, for any �= (�(1) . . .�(n))∈Sn, let �= (�(i1)�(i2) . . .�(ik)) be a subsequenceof � and “red” the increasing bijection of {�(i1),�(i2), . . . ,�(ik)} onto [k]. Define thereduction of the subsequence �, denoted by red(�), as (red(�(i1))red(�(i2)) . . .red(�(ik))).For any �= (�(1) . . .�(n))∈Sn, suppose �(2i+1)=n for some 0≤ i≤�(n−1) /2�. Weobtain two subsequences �1 = (�(1) . . .�(2i)) and �2 = (�(2i+2) . . .�(n)) of �. Clearly,red(�1)∈SK2i+1,odd and red(�2)∈SKn−2i,odd. There are

(n−12i

)ways to form the set

{�(1), . . . ,�(2i)}. Hence, we have opn =∑�(n−1)/2�i=0

(n−12i

)op2iopn−2i−1.



For any �∈Sn, suppose �=�(i1)�(i2) . . .�(ik) be a subsequence of �. Let � denotethe sequence (n+1−�(i1))(n+1−�(i2)) . . . (n+1−�(ik)).

Lemma 3.5. There is a bijection �n from SKn+1,odd to An.

Proof. By induction. Clearly, �1(1)=1. Now, we assume that �k is the bijectionfrom SKk+1,odd to Ak for all k≤n. For any �= (�(1) . . .�(n+1))∈Sn+1, suppose�(2k+1)=n+1, �1 = (�(1) . . .�(2k)) and �2 = (�(2k+2) . . .�(n+1)) for some 0≤k≤�n /2�. Note that red(�1)∈SK2k+1,odd and red(�2)∈SKn−2k+1,odd. It is easy to see that�2k(red(�1))∈A2k and �n−2k(red(�2))∈An−2k. Let �m =red−1 ◦�m ◦red and �m =red−1 ◦(�m ◦red) for short. Let �n(�)=�2k(�1)(n+1)�n−2k(�2). Then �n(�)∈An. �

Example 3.6. Given �= (42531)∈SK6,odd, we have �1 =42 and �2 =31. So,�2(42)=red−1 ◦�2 ◦red(42)=red−1 ◦�2(21)=red−1(21)=42 and �2(31)=red−1 ◦(�2 ◦red(31))=red−1 ◦(�2(21))=red−1 ◦21=red−1(12)=13. Then �5(�)=42513.

REFERENCES

[1] B. Bollobas, Modern Graph Theory, Springer, New York, 1998.[2] D. Chebikin and P. Pylyavskyy, A family of bijections between G-parking

functions and spanning trees, J Combin Theory Ser A 110 (2005), 31–41.[3] R. Donaghey, Alternating permutations and binary increasing trees, J Combin

Theory Ser A 18 (1975), 141–148.[4] D. Foata, Groupes de reearrangements et nombres d’Euler, C R Acad Sci Paris

Sr A-B 275 (1972), A1147–A1150.[5] D. Foata and V. Strehl, Rearrangements of the symmetric group and

enumerative properties of the tangent and secant numbers, Math Z 137 (1974),257–264.

[6] I. P. Goulden and D. M. Jackson, Combinatorial Enumeration, Wiley,Chichester, 1983.

[7] D. Kostic and C. H. Yan, Multiparking functions, graph searching, and theTutte polynomial, Adv Appl Math 40 (2008), 73–97.

[8] A. G. Kuznetsov, I. M. Pak, and A. E. Postnikov, Increasing trees andalternating permutations, Russian Math Surveys 49(6) (1994), 79–114.

[9] W. T. Tutte, A contribution to the theory of chromatic polynomials, Canad JMath 6 (1953), 80–91.


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Combinatorial interpretations for TG(1, −1)