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Isomorphism of Weighted Trees and Stanley’s Isomorphism Conjecture for Caterpillars
Martin Loebl, Jean-Sébastien Sereni
To cite this version:
Martin Loebl, Jean-Sébastien Sereni. Isomorphism of Weighted Trees and Stanley’s Isomorphism
Conjecture for Caterpillars. Annales de l’Institut Henri Poincaré (D) Combinatorics, Physics and
their Interactions, European Mathematical Society, In press. �hal-00992104v4�
CONJECTURE FOR CATERPILLARS
MARTIN LOEBL AND JEAN-SÉBASTIEN SERENI
Abstract. This paper contributes to a programme initiated by the first author: ‘How much information about a graph is revealed in its Potts partition function?’. We show that the W-polynomial distinguishes non-isomorphic weighted trees of agoodfamily. The framework developed to do so also allows us to show that theW-polynomial distinguishes non-isomorphic caterpillars. This establishes Stanley’s isomorphism conjecture for caterpillars, an extensively studied problem.
1. Introduction
Consider the following data setD(T)associated with a treeT: for every integernand every partitionP ofn, we are given the number of subsetsX of edges ofT such thatP is equal to the multiset formed by the orders of the components ofT−X. Note that this number is0ifnis not the number of vertices of T. Note also that ifP is composed oftintegers, the corresponding subsetsX, if any, all have cardinalityt−1. For instance, one can determine the number of vertices ofT by checking, for each positive integern, whether the trivial partition {n} returns a non-zero value (which, necessarily, will be1). Once the numbernof vertices ofT is known, the number of leaves ofT is precisely the number returned by the partition{n−1,1}, which corresponds to the number of edgesesuch thatT−ehas one component of order 1. The problem is to know whether this information distinguishes non-isomorphic trees. In other words, ifT andT0 are two trees such thatD(T) =D(T0), is it true that necessarilyT andT0 are isomorphic? That such a reconstruction is always possible was suggested by different authors. We note that there could be non-constructive proofs of the statement. Thus it is a different (harder) problem to be able to effectively recover the treeT from the knowledge of D(T). We explain in subsections 2.1, 2.2 and 2.3 why studying the strength of the information contained in D(T)for an arbitrary treeT helps to understand the strength of the partition function of the Potts model in a magnetic field, for general graphs.
1.1. State of the Art. Extensive efforts were dedicated (personal communication with Noble) to proving thatD(T)distinguishes non-isomorphic caterpillars — acaterpillar is a tree where all edges not incident with a leaf form a path, and aleaf is a vertex of degree one. Part of the Ph.D. thesis of Zamora [18] (under the supervision of M. L.) is dedicated to this problem. In addition, Aliste-Prieto and Zamora [1], established the statement restricted to the class of proper caterpillars: a caterpillar isproper if every vertex is a leaf or adjacent to a leaf. Prior to that, partial results had been obtained by Martin, Morin and Wagner [9] who had established the
Date: June 19, 2018.
2010Mathematics Subject Classification. 05C31, 05C60.
Key words and phrases. W-polynomial, tree, graph reconstruction, graph isomorphism,U-polynomial, Stanley’s isomorphism conjecture, Potts partition function.
This work was done within the scope of the International Associated Laboratory STRUCO.
The authors were partially supported by the Czech Science Foundation under the contract number P202-13- 21988S (M. L.) and by P.H.C. Barrande 31231PF of the French M.A.E. (J.-S. S.).
1
statement for a subclass of proper caterpillars (where no two non-leaf vertices are adjacent to the same number of leaves) and also to the class of spiders, which is composed of all trees with a unique vertex of degree greater than two. Other related results were obtained by Orellana and Scott [12], Smith, Smith and Tian [15] or can be found in the undergraduate thesis by Fougere [5]
and the MSc thesis by Morin [10]. Finally, Hell and Ji [6] have verified by computer that Stanley’s isomorphism conjecture [16], which we present in Subsection 2.2, is true for trees with at most29 vertices. Previously, Russel has verified by computer that Stanley’s isomorphism conjecture is true for trees with at most25vertices (the code is available at https://github.com/keeler/csf) and it was reported (see [9, p. 238]) that Tan verified it for trees with at most23vertices.
1.2. Main Contribution. We solve affirmatively Stanley’s isomorphism conjecture restricted to the class of caterpillars. We also investigate a weighted version of the problem, bearing in mind its connections with graph polynomials, graph colouring and the Potts model. First we summarise the background and motivations.
2. Motivation
In this section we summarise the background (the Noble and Welsh conjecture and the Stanley conjecture) and describe our motivation.
2.1. The Noble and Welsh Conjecture. Motivated by the combinatorial aspects of the rela- tionship between chord diagrams and Vassiliev invariants of knots, Noble and Welsh [11] introduced a polynomial of weighted graphs, theW-polynomial, which includes several specialisations in combinatorics, such as the Tutte polynomial, the matching polynomial (of ordinary graphs) and the polymatroid polynomial of Oxley and Whittle [13]. We need to introduce some terminology to defineW.
A weighted graph is a graphG= (V, E)together with a function w:V →Z+. Theweight of a subset V0 of vertices is w(V0) := P
v∈V0w(v). If A ⊆ E, we let cV(A) be the number of components of the graph(V, A), where we may omit the subscript when there is no risk of confusion. Further, letn1, . . . , nc(A) be the weights of the vertex sets of these components, listed in decreasing order: n1>· · ·>nc(A). We writex(A)to meanQc(A)
i=1 xni. Let WG(z, x1, x2, . . .) := X
A⊆E
x(A)(z−1)|A|−|V|+c(A).
In particular,WG depends onzif and only ifGcontains a cycle [11, Proposition 5.1]. Unlike the Tutte polynomial, the W-polynomial is#P-hard to compute even for trees [11, Theorems 7.3 and 7.12] and for complete graphs [11, Theorems 7.11 and 7.14].
In the case of unweighted graphs, which corresponds here to the weight function w being identically1, Noble and Welsh refer to theW-polynomial as theU-polynomial. While computingW is hard for complete graphs, Annan [2] proved that UKn(z, x1, x2, . . .) can be computed in polynomial time, which is also the case for the Tutte polynomial. However, U also exhibits differences with the Tutte polynomial: while finding two non-isomorphic graphs with the same Tutte polynomial is easy, the same problem is harder for U. Brylawski [4] found two non- isomorphic graphs with the same polychromate, and Sarmiento [14] proved that theU−polynomial is equivalent to Brylawski’s polychromate. But the question remains open for trees: does the U-polynomial distinguishes non-isomorphic trees? That this is the case became known as theNoble and Welsh conjecture. This is clearly equivalent to our initial question: ‘DoesD(T)distinguish non-isomorphic trees?’
Noble and Welsh demonstrated theU-polynomial to be equivalent to thesymmetric function generalisation of the chromatic polynomial, a function introduced by Stanley [16].
2.2. Stanley’s isomorphism Conjecture. To introduce Stanley’s isomorphism conjecture let us first define graph colouring. Acolouring of a graphG= (V, E)is a mappings:V →N+. We defineb(s)to be the number ofmonochromatic edges in s, that is, the number of edgesuvsuch thats(u) =s(v). The mappingsis ak-colouring ifs(V)⊆ {1, . . . , k} andsisproper ifb(s) = 0, that is,s(u)6=s(v)wheneveruandv are two adjacent vertices ofG. We letCol(G;k)be the set of properk-colourings ofGandCol(G)be the set of all proper colourings ofG.
In the mid 1990s, Stanley [16] introduced thesymmetric function generalization of the chromatic polynomial, defined to be
XG(x1, x2, . . .) := X
s∈Col(G)
Y
v∈V
xs(v).
This is a homogeneous symmetric function in(x1, x2, . . .)of degree|V|. As one might expect, XG does not distinguish non-isomorphic graphs: there exist two non-isomorphic graphs on 5 vertices with the same function X. However, Stanley [16] asked whether the polynomial XG distinguishes non-isomorphic trees. The assertion that it does became known asStanley’s isomorphism conjecture.
Further, Stanley [17] later initiated the study of a common generalisation ofX and the Tutte polynomial, namely thesymmetric function generalisation of the bad colouring polynomial, defined for every graphG= (V, E)by
XG(t, x1, x2, . . .) := X
s:V→N+
(1 +t)b(s) Y
v∈V
xs(v).
Note that the sum runs over all colourings of G, not only the proper ones. Noble and Welsh [11, Theorem 6.2] provedXG(t, x1, x2, . . .)to be equivalent to theU-polynomial ofG.
2.3. Loebl’s Conjectures. Loebl [8] introduced the q-chromatic functions. Letk ∈N. The q-chromatic function of a graphG= (V, E)is
(2.1) MG(k, q) := X
s∈Col(G;k)
qPv∈Vs(v).
It is known [8] that
MG(k, q) = X
A⊂E
(−1)|A| Y
C∈C(A)
(k)q|C|,
where thequantum integer (k)ris rk−1+· · ·+r+ 1andC(A)is the set of components of the spanning subgraph (V, A) while|C| is the number of vertices in the component C. Moreover Loebl also introduced theq-dichromate, defined as
BG(x, y, q) := X
A⊂E
x|A| Y
C∈C(A)
(y)q|C|.
Loebl [8] conjectured the following.
• Theq-dichromate is equivalent to theU-polynomial.
• TheU-polynomial distinguishes non-isomorphic chordal graphs.
There could be a close link between the latter conjecture and that of Stanley: chordal graphs have a very distinguished tree structure. Indeed, a folklore theorem [3] states that the class of chordal graphs is precisely the class of intersection graphs of subtrees of a tree, that is, for each chordal graphG, there exists a treeT and a mappingf that assigns to each vertex ofGa subtree T such that: two vertices uandv ofGare adjacent if and only iff(u)∩f(v)6=∅.
The motivation for Loebl’s conjectures is formula (2.2) below, which connects thek-state Potts model partition function and theq-dichromate.
The Potts model. We consider a standard model where magnetic materials are represented as lattices: vertices are atoms and weighted edges are nearest-neighbourhood interactions. We assume that each atom has one out ofkpossible magnetic moments, for a fixed positive integerk. Thus we setS :={0, . . . , k−1}. Every element ofS is called aspin. Astate of a graphG= (V, E)is then an assignment of a single spin to each vertex ofG, that is, a functions: V →S. We assume that all the coupling constants (nearest-neighbourhood interactions) are equal to a constantJ. For each states, thePotts model energy of the state sis defined to beE(Pk)(s) :=P
uv∈EJ δ(s(u), s(v)) where, as is customary, δ is the Kronecker delta function defined by δ(a, b) := 1 if a=b and δ(a, b) := 0otherwise. Thek-state Potts model partition function is
X
s:V→S
M(s, J)eE(Pk)(s)
whereM(s, J)is a function describing the magnetic field contribution.
Loebl proved that for each realJ,
(2.2) BG(eJ−1, k, q) = X
s:V→S
qPv∈Vs(v)eE(Pk)(s).
This means that theq-dichromate specializes to thek-state Potts model partition function with a certain magnetic field contribution.
Recently a variant of the q-dichromate, Br,G(x, k, q), was proposed by Klazar, Loebl and Moffatt [7]:
Br,G(x, k, q) := X
A⊆E
x|A| Y
C∈C(A) k−1
X
i=0
r|C|qi.
They established that if(k, r)∈N2withr >1andx:= eβJ−1, then
(2.3) Br,G(x, k, q) = X
σ:V→S
eβPuv∈E(G)J δ(σ(u),σ(v))
rPv∈Vqσ(v).
HenceBr,G(x, k, q)is thek-state Potts model partition function with magnetic field contribution rPv∈Vqσ(v). They also provedBr,G to be equivalent to UG, which can be seen as a first step towards Loebl’s programme:
The polynomialUG is equivalent to the Potts partition function of Gwith a magnetic field contribution.
A well-known fact is that the isomorphism problem for general graphs is equivalent to the isomorphism problem restricted to chordal graphs: given a graphG= (V, E), consider the chordal graph G0 = (V0, E0)so that V0 := V ∪E and E0 = V2
∪ {{u, e},{v, e} : {u, v}=e∈E}. It clearly holds thatGandH are isomorphic if and only ifG0 andH0 are isomorphic. It thus seems particularly interesting to determine whether theU-polynomial does distinguish non-isomorphic chordal graphs, as conjectured by Loebl. If true, we would obtain a surprising conclusion:
The Potts partition function with a magnetic field contribution contains essentially (modulo a simple preprocessing) all the information about the underlying graph.
In that respect, it seems natural to study weighted trees. The tree mentioned in the char- acterisation of the class of chordal graphs can be chosen to be aclique-tree, where the vertices of the tree are the maximal cliques of the graph. Now, ifv is a vertex of a weighted tree with weightw(v), one can think ofv as a clique of orderw(v), thus obtaining an unweighted chordal graph. This is what motivates working in the (seemingly harder) setting of weighted trees.
2.4. Main Results. Two weighted graphs are isomorphic if there is an isomorphism of the graphs that preserves the vertex weights. We also consider weighted trees rooted at a vertex:
an isomorphism between rooted weighted trees preserves the weights by definition, but may not preserve the roots. If it does preserve the roots, then it is anr-isomorphism. (In particular, two rooted weighted trees that are r-ismomorphic are isomorphic but the converse is not necessarily true.)
The first purpose of this work is to prove that theW-polynomial distinguishes non-isomorphic weighted trees when restricting to collections of weighted trees satisfying some properties made precise later. We call any such collection a good family. We consider this result as a first observation towards understanding Stanley’s isomorphism conjecture for the class of chordal graphs; even though we do not know natural examples of good families of weighted trees which were studied before. We remark that theW-polynomial does not distinguish general weighted trees; a simple example consists of two paths with weight sequences1,2,1,3,2 and1,3,2,1,2.
Let(T, w)be a weighted tree. We writeV(T)andE(T)for the vertex set and the edge set ofT, respectively. We defineEx(T)to be the multi-set composed of all the vertex weights (with multiplicities) ofT. Ife∈E(T), thenT−eis the disjoint union of two trees, which we consider to be weighted and rooted at the endvertex ofethat they contain. A rooted weighted tree(S, wS) is a shape of(T, w)if26|V(S)|6|V(T)| −2 and there exists an edgee∈E(T)such thatS is one of the two components ofT −e; moreoverwS is the restriction ofw to the vertex set ofS.
We consider S to be rooted at the end-vertex ofe. We usually shorten the notation and writeS for the shape(S, wS). In a tree, a vertex of degree one is called aleaf.
Definition 2.1. A setT of weighted trees(T, w)isgood if it satisfies the following properties.
(1) If a vertex ofT is adjacent to a leaf, then all its neighbours but possibly one are leaves.
(2) Ifv is a leaf or has a neighbour that is a leaf, thenw(v) = 1.
(3) Let(T, w),(T0, w0)∈ T and letS be a shape ofT and such thatw(S)6w(T)/2. LetS0 be a shape ofT0 such thatEx(S0) = Ex(S). ThenS0 and S are r-isomorphic.
Theorem 1. The W-polynomial distinguishes non-isomorphic weighted trees in any good set.
Our proof of Theorem 1 is not constructive in the sense that we are not able to reconstruct the weighted tree(T, w)fromW(T ,w). The difficulty in proving the theorem is that while the main defining property of a good family is about shapes, theW-polynomial does not “see” shapes.
However, shapes turn out to be a useful and rather powerful notion: it allowed us to unlock the case of general caterpillars, thereby confirming Stanley’s isomorphism conjecture for the class of (general) caterpillars.
Theorem 2. Each caterpillar can be reconstructed from its U-polynomial.
Note that Theorem 2, contrary to Theorem 1, allows for a full reconstruction of the tree.
3. The Structure of the Proofs
We write down a procedure and with its help prove both theorems. The rest of the paper then describes our realisation of the procedure. We fix a good set of weighted trees and, from now on, we say that a weighted tree isgood if it belongs to this set.
A j-form is an r-isomorphism class of rooted weighted trees with total weight j. Thus a j-formF is a collection of r-isomorphic rooted weighted trees and, viewing a shape of a treeT as a rooted weighted tree, a shape can belong to aj-form. Note in particular that two shapesS and S0 of a weighted tree belong to the same j-form for some j if and only if S andS0 are r-isomorphic. We start with two observations.
Observation 3.1. Let T1 and T2 be two shapes of a tree T such that w(T1) +w(T2)6w(T).
Then either T1⊆T2 orT2⊆T1 or T1∩T2=∅.
Proof. Fork∈ {1,2}letekbe the edge ofT associated toTk, that is,Tk is a component ofT−ek. Ife1 =e2, then eitherT1 =T2 orT1∩T2 =∅. Assume that e1 6=e2. Then eithere2 ∈E(T1) or e2 ∈ E(T −T1). If e2 ∈ E(T −T1), then either T1 ⊆ T2 or T2 ⊆ T −T1 in which case T1∩T2 =∅. If e2∈ E(T1), thenT2 ⊆T1: otherwise, T1∩T2 6=∅andT −T1 ⊆T2, so that
w(T1) +w(T2)> w(T), contrary to the assumption.
Observation 3.2. Let (T, w)be a weighted tree such that every leaf has weight 1. Assume that we know the total weightw(T)of T and that, for eachj6w(T)/2 and eachj-formF, we know the number of shapes of(T, w)that belong to F. Then we know T.
Proof. We use Observation 3.1. We order the shapes of (T, w) of weight at most w(T)/2 decreasingly according to their weights. Letmbe the maximum weight of such a shape ofT and letS1, . . . , Sa be the shapes with weightm. Note that we know precisely theseatrees. In addition, either the shapesS1, . . . , Sa are joined inT to the same vertex, ora= 2andm=w(T)/2. In the latter case (m=w(T)/2) we know thatT consists of the two weighted rooted treesS1 andS2
(each of weightm) with an edge between their roots: this ends the proof for this case. Assume thatm < w(T)/2. We letrbe the additional vertex to which we link each ofS1, . . . , Sa.
We show by descending induction onj∈ {2, . . . , m}that we know the subtree ofT induced by all shapes ofT with weight in{j, . . . ,bW(T)/2c}. The induction has thus been initialized above, so assume thatj6m−1. LetS1, . . . , Stbe the shapes ofT with weight in{j+ 1, . . . ,bW(T)/2c}.
Note that we know, in particular, each of thesettrees. The shapes ofT of weight equal toj, if any, are either shapes ofS1, . . . , St or joined torby an edge from their root. Fix aj-form F.
Since we do know the total number of shapes belonging toF and contained in each ofS1, . . . , St
(because we know precisely those subtrees), we can deduce the number of shapes that belong toF and are attached tor. As this argument applies to allj-formsF, we infer that we know the subtree ofT formed by all shapes with weight contained in{j, . . . ,bw(T)/2c}. The reconstruction ofT is almost finished: lettingw0 be the total weight of the tree we built so far, it only remains to addw(T)−w0new leaves, each joined to the vertexr. This concludes the proof.
Let(T, w)be a weighted tree. Letα(T) = (α1, . . . , αn)be the weights of the shapes ofT, with α1<· · ·< αn. The definition of a shape implies thatα1>2.
We shall considerconnected partitions of the treeT,i.e., partitions of the vertex set ofT into connected subsets. Later in the paper we refer to connected partitions ofT simply as partitions ofT. We shall also consider the partitions of the integerw(T). To distinguish between them clearly, partitions of an integer are referred to asexpressions. For each partition P of T, the weights of the parts ofT form an expression ofw(T), which we call thecharacteristic ofP.
• Aj-expression of an integermis a partition ofmwhere one of the parts is equal tom−j.
• Fori∈ {1, . . . , `}, letmi be an integer andEi an expression ofmi. We define[E1, . . . , E`] to be the expression ofP`
i=1mi equal to the concatenation ofE1, . . . , E`. In particular, ifS is a shape of T with weightαj, then[Ex(S), w(T)−αj]is an αj-expression ofw(T).
• Aj-partition ofT is a partition ofT whose characteristic is aj-expression ofw(T). In other words, one of the components of the partition has weightw(T)−j.
• Aj-partition(T0, . . . , Tk)ofT withw(T0) =w(T)−j isshaped if there exists an edgee ofT such thatT0is one of the components of T−e. Any such edgeeis thenassociated to(T0, . . . , Tk).
• IfS is a shape ofT with weightαj and vertex setV(S) ={v1, . . . , vs}, we defineP(S) to be(V(T)\V(S),{v1}, . . . ,{vs}), which is a shapedαj-partition ofT.
For an expressionE of a positive integer, we letθ(T, w, E)be the number of partitions of (T, w) with characteristicE. Note that this number is0ifE is not an expression ofw(T). We note that there is a bijection between connected partitions and edge subsets given by taking all edges ofT joining two vertices in different blocks of the connected partition and thenceθ(T, w, E)turns out to be the coefficient ofxE in the W-polynomial of(T, w).
We note that among the partitions ofT corresponding to a given expression, some are shaped and others are not. If all the vertex weights are equal to one, we abbreviateθ(T, w, E)asθ(T, E).
The proofs of both theorems rely on the following procedure.
Procedure 1.
input: The polynomialW(T ,w); an integerj∈ {α2, . . . , α`}, where`is the least integerisuch that αi> w(T)/2; aj-expressionE ofw(T)and, for eachj0< j and eachj0-formF, the number of shapesSofT that are isomorphic to a member ofF (hence, according to the notation introduced above, possibly but not necessarily r-isomorphic, and hence not necessarily a member of F).
output: The number of shaped j-partitions ofT with characteristic[w(T)−j, E].
Let us see how this procedure allows us to establish Theorem 1.
Proof of Theorem 1. Fix two good weighted trees (T, w) and (T0, w0) with W(T ,w) =W(T0,w0). By Observation 3.2,(T, w)and(T0, w0)are isomorphic ifw(T) =w0(T0)and for eachj-formF wherej6w(T)/2, the numbers of shapes ofT and ofT0 that belong toF are equal. To establish this, first note that the vector α(T) = (α1, . . . , αn) can be computed fromW(T ,w), since the coordinates correspond to the partitions ofT into two subtrees (each with at least two vertices).
Thusα(T0) =α(T).
We prove by induction on j ∈ {α1, . . . ,bw(T)/2c} that for everyj-form F, the numbers of shapes ofT and ofT0 that belong toF are the same. So suppose first, as the base case of the induction, that j=α1. Recall thatα1>2. Furthermore, a shape S of T orT0 belongs to an α1-form if and only ifS is the star onα1vertices rooted at its centre. This is because the leaves and their neighbours have weight1. It follows that the number of shapes ofT of weightα1 can be calculated fromW(T ,w)and thus this number is the same for(T0, w0).
Now we establish the induction step. For convenience, ifF is aj-form, letnT(F)be the number of shapes ofT that belong toF; we use a similar notation forT0. Letj∈ {α1+ 1, . . . ,bw(T)/2c}.
The induction hypothesis is thatnT(F0) =nT0(F0) for everyj0-formF0 and everyj0 < j. We want to establish that
(3.1) for every j-formF , nT(F) =nT0(F).
This will prove Theorem 1 by Observation 3.2.
We first set a partial order on thej-forms, which allows us to link tree partitions withj-forms.
Given a j-formF, we defineEx(F)to beEx(f)for an arbitrary representative f of F. (This definition is valid, since all representatives of aj-form are r-isomorphic rooted weighted trees.) A j-formF0issmaller thanaj-formFifEx(F0)is a proper refinement ofEx(F). IfP = (T0, . . . , Tk) is a shapedj-partition ofT wherew(T0) =w(T)−j, we defineS(P)to be the shape ofT formed by the union of all parts ofT different fromT0, that is,S(P) :=∪ki=1Ti=T−T0, rooted at the end-vertex of the edge associated toP.
A key observation is that ifP is a shapedj-partition ofT with characteristic[Ex(F), w(T)−j]
for somej-form F, then Ex(S(P))is a refinement ofEx(F), possibly equal toEx(F).
We prove (3.1) by induction on the j-formF considered (with respect to the partial order defined above).
We first deal with the case whereT has no shape that belongs to aj-formF0 such thatEx(F0) is a proper refinement ofEx(F). We demonstrate the following assertion.
Assertion 3.3. If T has no j-formF0 such thatEx(F0)is a proper refinement ofEx(F), then the number of shapedj-partitions ofT with characteristic[Ex(F), w(T)−j] is equal to nT(F).
This assertion implies that nT(F) = nT0(F) since by Procedure 1 and by the induction hypothesis, the number of shapedj-partitions ofT with characteristic[Ex(F), w(T)−j]is equal to the number of shapedj-partitions of T0 with characteristic[Ex(F), w(T)−j].
To establish Assertion 3.3, we first note that each shape of T that belongs to F provides exactly one shapedj-partition ofT with characteristic[Ex(F), w(T)−j]. On the other hand, if P is a shaped j-partition of T with characteristic [Ex(F), w(T)−j], then Ex(S(P)) is a refinement ofEx(F), which by our hypothesis onF must be equal toEx(F). So S(P)gives rise to precisely one shapedj-partition of T with characteristic [Ex(F), w(T)−j], namely P. As Ex(F) = Ex(S(P)), it follows from Definition 2.1(3) thatS(P)belongs to F, which ends the proof of Assertion 3.3.
In the induction step we assume thatnT(F0) =nT0(F0)for everyj-formF0 such thatEx(F0)is a proper refinement ofEx(F). Observe that for eachj-formF0 withF0 < F, each shape ofT that belongs toF0 gives rise to a certain number of shapedj-partitions ofT with characteristicEx(F), and this number depends only onF0. Thus the numbern0T(F)of shaped j-partitions ofT with characteristic[Ex(F), w(T)−j]such thatEx(S(P))is a proper refinement ofEx(F)depends only on the multi-set{nT(F0) : F0< F}. As{nT(F0) : F0 < F}={nT0(F0) : F0< F}, we deduce thatn0T(F) =n0T0(F). We demonstrate the following assertion.
Assertion 3.4. The number of shapedj-partitions of T with characteristic[Ex(F), w(T)−j]is equal to n0T(F) +nT(F).
This assertion follows analogously as Assertion 3.3. Moreover, we established in the para- graph above that n0(T, F) = n0(T0, F). Since the number of shaped partitions of T with characteristic[Ex(F), w(T)−j]is equal to the number of shaped partitions ofT0 with character- istic[Ex(F), w(T)−j]by Procedure 1, we deduce thatnT(F) =nT0(F)by Assertion 3.4. This establishes (3.1), and hence finishes the proof of Theorem 1.
As we see next, the notion of a shape and Procedure 1 turn out to be essential tools to study Stanley’s isomorphism conjecture restricted to caterpillars.
4. Caterpillars
We first observe that Theorem 2 is true for all caterpillars with at most two vertices. Hence we will assume that a caterpillar has at least three vertices in this section, and we only consider weights to be1; since there is then no risk of confusion, we abbreviate|V(T)| to|T| for every treeT. LetT be a caterpillar (with at least three vertices). Thespine ofT is the unique pathP ofT such that every leaf ofT is at distance exactly one from a vertex ofP.
Before proving Theorem 2, we formalize a simple but crucial observation, which is used repeatedly and implicitly in the proof of Theorem 2.
Observation 4.1. Every shape of a caterpillarT is rooted at a vertex of the spine ofT. It follows from Observation 4.1 that for every integerj, the number of shapes of T withj vertices belongs to{0,1,2}.
IfT is a caterpillar, and E is an expression ofj so that no part of E is equal to |T| −j, then we defineθs(T, E)to be the number of shapedj-partitions ofT with characteristic[|T| −j, E].
LetSk be the star onk vertices — thusS1is a single vertex. We always consider a star to be rooted at its center. IfT is a rooted tree then we define Sk →T to be the tree rooted at the center of Sk and obtained by joining the root ofT to that of Sk by an edge. Hence ifT is a rooted caterpillar, thenSk →T is also a rooted caterpillar.
LetAbe the collection of rooted caterpillarsAsuch that
• Ais a single vertex; or
• Ais a rooted edge; or
• |A|>3and the root of A is either an end-vertex of the spine or a leaf attached to an end-vertex of the spine.
IfA∈ Athen thereverseA˜ofAis defined as follows. IfAis a single vertex thenA˜:=A. IfAis a rooted edge then A˜is the same edge rooted at the other end-vertex. If Ahas at least three vertices and the root is an end-vertex of the spine thenA˜is obtained from Aby resetting the root at the other end-vertex of the spine. IfAhas at least three vertices and the root is a leaf attached to an end-vertex of the spine thenA˜ is obtained fromA by resetting the root at an arbitrary leaf attached to the other end-vertex of the spine. (We note that such a leaf always exists by the definition of the spine.)
Observation 4.2. Let A, B∈ Asuch thatA andB are isomorphic but not r-isomorphic. Let o, o1 ando2 be positive integers.
(1) The caterpillars So→AandSo→B are not isomorphic; and (2) neither are the caterpillarsSo2 →So1 →A andSo2→So1 →B.
Proof. The statements are vacuously true if |A| 62, so we assume that A has at least three vertices — and thus so hasB. Given an elementC∈ Awith|C|>3, we letrC be the root ofC and we define the degree sequencesC ofC as follows. Letw1. . . wtbe the spine ofC, wherew1is closest torC. The degree sequence ofCissC:= (deg(w1), . . . ,deg(wt)). ThereverseofsCis then the sequence(deg(wt), . . . ,deg(w1)). We observe that two elementsCandC0 ofA(with at least three vertices) are isomorphic if and only ifsC =sC0 orsC0 is the reverse ofsC. Furthermore,C andC0 are r-isomorphic if and only if sC =sC0 anddeg(rC) = deg(rC0) (that is, either both roots have degree one, or both roots have degree greater than one).
Let us make another preliminary remark. If degA(rA) = 16= degB(rB), then in each of (1) and (2) the caterpillars obtained fromAand fromB have spines of different lengths, so they are not isomorphic. We can thus assume that either both ofrA andrB have degree one, or both have degree greater than one. This implies thatsA6=sB and t >1, as otherwiseAandB would be r-isomorphic. Consequently,sB is the reverse ofsA. Let us writesA= (a1, . . . , at).
(1). For convenience, setA0:=So→AandB0:=So→B. We know thatsB = (at, . . . , a1)6=
sA. Suppose first that degA(rA) = 1 = degB(rB). Then sA0 = (o,2, a1, . . . , at)if o >1while sA0 = (2, a1, . . . , at)ifo= 1. Similarly,sB0= (o,2, at, . . . , a1)ifo >1while sB0 = (2, at, . . . , a1) if o = 1. In either case, we see that sA0 6= sB0 as sA 6= sB. So suppose for a contradiction thatsB0 is the reverse ofsA0. In the former case,i.e.,o >1, this means that(o,2, a1, . . . , at) = (a1, . . . , at,2, o). Thenaj=oforj odd andaj= 2 forj even. In addition,at=oandat−1= 2, showing that t must be odd unless o = 2. However, either way this yields that sA = sB, a contradiction. In the latter case,i.e.,o= 1, we have(2, a1, . . . , at) = (a1, . . . , at,2), soai= 2for eachi∈ {1, . . . , t}which again contradicts thatsA6=sB.
It remains to deal with the case where degA(rA) 6= 1 6= degB(rB). If o > 1, then sA0 = (o,1 +a1, a2, . . . , at)andsB0 = (o,1 +at, at−1, . . . , a1). Ifo= 1, thensA0 = (1 +a1, a2, . . . , at) andsB0 = (1 +at, at−1, . . . , a1). In either case, note thatsA0 6=sB0 becausesA6=sB. Further, ifsB0is the reverse ofsA0, then it implies thato >1,at=o=a1andai=o+1fori∈ {2, . . . , t−1}, leading tosA=sB, a contradiction. This ends the proof of (1).
(2). For convenience, setA0:=So2 →So1 →Aand B0:=So2 →So1 →B. Assume first that degA(rA) = 1 = degB(rB). Then we infer as before that
sA0 =
(2,2, a1, . . . , at) ifo1= 1ando2= 1, (o2,2,2, a1, . . . , at) ifo1= 1ando2>1, (1 +o1,2, a1, . . . , at) ifo1>1 ando2= 1, (o2,1 +o1,2, a1, . . . , at) ifo1>1 ando2>1.
and
sB0 =
(2,2, at, . . . , a1) ifo1= 1ando2= 1, (o2,2,2, at, . . . , a1) ifo1= 1ando2>1, (1 +o1,2, at, . . . , a1) ifo1>1ando2= 1, (o2,1 +o1,2, at, . . . , a1) ifo1>1ando2>1.
We see that in each of the four possible casessA0 6=sB0 assA 6= sB. In addition, in none of these fours cases cansB0 be the reverse ofsA0, showing thatA0 andB0 are not isomorphic. For instance, in the second case it would imply thatt is 1modulo3andai =o2 ifiis equal to 1 modulo3, whileai= 2otherwise; however this would yield thatsA=sB, a contradiction. To check the fourth case, it is useful to consider the value oft modulo3.
It remains to deal with the case wheredegA(rA) 6= 1 6= degB(rB). We infer the following expressions.
sA0 =
(2,1 +a1, a2, . . . , at) ifo1= 1ando2= 1, (o2,2,1 +a1, a2, . . . , at) ifo1= 1ando2>1, (1 +o1,1 +a1, a2, . . . , at) ifo1>1ando2= 1, (o2,1 +o1,1 +a1, a2, . . . , at) ifo1>1ando2>1.
and
sB0 =
(2,1 +at, at−1, . . . , a1) ifo1= 1and o2= 1, (o2,2,1 +at, at−1, . . . , a1) ifo1= 1and o2>1, (1 +o1,1 +at, at−1, . . . , a1) ifo1>1and o2= 1, (o2,1 +o1,1 +at, at−1, . . . , a1) ifo1>1and o2>1.
It follows that in none of the four cases the sequencesB0 ie equal tosA0 or to the reverse ofsA0,
again relying on the fact thatsA6=sB.
We are now ready to proceed with the proof of Theorem 2.
Proof of Theorem 2. LetT be a caterpillar. We proceed by induction on the number of vertices ofT, the theorem being true if |T|<4. We now deal with the inductive step. As before, we note that the vectorα(T) = (α1, . . . , αn)can be computed fromUT, since the coordinates correspond to the partitions ofT into two subtrees (each with at least two vertices). We prove by induction onj ∈ {α1, . . . ,b|T|/2c}that for everyj-formF, we can deduce fromUT the number of shapes ofT that belong toF. Observation 3.2 ensures then that we can reconstruct T. Analogously as in a previous proof the number of shapes ofT of sizeα1can be calculated fromUT. This number is one or two sinceT is a caterpillar.
We proceed inductively and, at each step of the inductive process, we update our knowledge of the two ends ofT, by increasing the size of our knowledge of (at least) one end of T. It is important to note that to know the number of shapes ofT that belong to a givenj-formF for somej>2, it is enough to know both ends ofT of orderj. At any given step, we letR1andR2 be the currently known forms of the two ends of T. Hence after the first step R1 =Sα1 and
R2=∅orR2=R1, depending on whetherθ(T,[|T| −α1, α1])equals1or2. (As reported earlier, this number can be deduced from the U-polynomial of T.)
Letj∈ {α1+ 1, . . . ,b|T|/2c}. We assume that for eachj0∈ {α1, . . . , j−1}and eachj0-formF we know the number of shapes ofT that belong toF. Let us establish this last statement forj0=j.
Ifj /∈ {α2, . . . , αn}, then we know that the sought number is 0, by the definition of(α1, . . . , αn).
So we suppose now thatj=αk for some integerk∈ {2, . . . , n}. We setm:=αk−αk−1. (Recall that this number can be deduced from the U-polynomial.) Let αk−1 = |R1|> |R2|, withR2
possibly empty. Set p:=αk− |R2|, letR01:=Sm→R1andR02:=Sp→R2.
IfR1 and R2 are r-isomorphic andαk = 1then we setR1:=R01and leaveR2 unchanged. If R1 andR2 are r-isomorphic andαk= 2then we setR1:=R01 andR2:=R02. Hence from now on we assume thatR1 andR2 are not r-isomorphic. We distinguish three cases.
[(1)] Let T have two αk-shapes.
Then we update bothR1 andR2, that is, we setR1:=R01andR2:=R02.
[(2)] LetT have exactly oneαk-shape,i.e., eitherR01orR02. Moreover letR01andR20 be not isomorphic.
We recall that αk 6|T|/2. As |R0i|<|T|, we know by induction thatUR01 6=UR02. Hence there is an expressionE0 of|R10|=αk such thatr1:=θ(R01, E0)6=r2:=θ(R02, E0).
Now comes an important observation that will be used repeatedly in this proof: we know there is only oneαk-shape inT, and thus all shapedαk-partitions ofT have to come from partitions where one removes the edge associated to this shape and any subset of edges inside this shape.
Therefore, there is a unique i∈ {1,2} such thatθs(T, E0) =ri and we can determine it by Procedure 1. We setRi:=Ri0 and leaveR3−i unchanged.
[(3)] LetT have exactly oneαk-shape,i.e., eitherR01orR02. Moreover letR01andR20 be isomorphic but not r-isomorphic.
In this case we explicitly know the unique isomorphism class for theαk-shapes ofT. Therefore we know, for eachαk-formF, the number of shapes of T that are isomorphic (but not necessarily r-isomorphic) to a member ofF. We observe thatk < n. We setq:=αk+1−αk.
By Procedure 1, we know for eachαk+1-expressionE the number of shapedαk+1-partitions ofT with characteristicE.
There are four candidates for anαk+1-shape ofT, namelyS1,1:=Sq→Sm→R1=Sq →R10, S2,1:=Sq+m→R1, S1,2:=Sq+p→R2andS2,2:=Sq →Sp→R2=Sq →R02.
We now introduce some labels for the vertices of the starsSq,Sq+m andSq+p. The vertices ofSq are labelledv1, . . . , vq, wherevq is the centre ofSq. By extension, the corresponding vertices ofS1,1andS2,2 inherit those labels. Fori∈ {p, m}, the vertices ofSq+iare labelled v1, . . . , vq+i
where, this time, the labelsv1, . . . , vq are assigned to leaves only. Similarly, the corresponding vertices ofS1,2 and ofS2,1 inherit those labels. Thus, for example, the vertexvq ofS1,1 is the centre of the starSq and hence the root ofS1,1, while the vertexvq ofS1,2is one of the leaves ofSq+p and hence is adjacent to the root of S1,2.
[(3.1)] Let T have two αk+1-shapes.
There are two possibilities for the twoαk+1-shapes ofT: either S1,1, S1,2 orS2,1, S2,2. We note that this implies that αk+16|T|/2. Fori∈ {1,2}, letTi be any caterpillar with|Ti|=|T| whoseαk+1-shapes are exactlySi,1 andSi,2.
Observation 4.3. Ifq >1then Si,j and Si0,j0 are not isomorphic (as unrooted trees) when- everi, i0, j, j0∈ {1,2} with i6=i0.
Proof. Comparing the lengths of the spines, the only possible pairs of isomorphic trees are:
S1,1 withS2,2, andS1,2 withS2,1. However, the fact thatR01 andR02 are isomorphic but not
r-isomorphic prevents each of these pairs to consist of isomorphic trees, using Observation 4.2(1)
for the former one.
LetE be an expression of αk+1. We note that each Ti has exactly two vertices labelled by vq, namely the root ofSi,i and a leaf of Si,3−i attached to the root ofSi,3−i. We classify the shapedαk+1-partitions of Ti, for eachi∈ {1,2}, into four classesC(E, i,1),C(E, i,2),C(E, i,3) and C(E, i,4). To this end, let Ei be the collections of all shaped αk+1-partitions of Ti of characteristic[|T| −αk+1, E]. An elementP ofEipartitionsa subtreeGofTiif a subset (possibly of order one) of the parts ofP forms a partition ofG.
(1) We letC(E, i,1)be the collection of all elements ofEi such that a subset of parts ofP is a partition of the uniqueαk-shape ofTi.
(2) We let C(E, i,2)be the collection of all elements P ∈ Ei\C(E, i,1) such that {vq} ⊂ V(Si,i)∪V(Si,3−i)is not a part ofP.
(3) We letC(E, i,3) be the collection of all elements P ∈ Ei\C(E, i,1) such that {vi} ⊆ V(Si,3−i)is a part ofP for each i6q.
(4) We let C(E, i,4)be the collection of all elements P ∈ Ei\C(E, i,1) such that {vq} ⊆ V(Si,3−i)is a part ofP and there exists`∈ {1, . . . , q−1} such that {v`} ⊆V(Si,3−i) is not a part ofP.
Observation 4.4. Let i∈ {1,2} and letE be an expression of αk+1.
(1) The partitions in C(E, i,1)partition the shapeSi,iofTi. Moreover, there is a bijectionF from C(E,1,1) to C(E,2,1) so that for each P, there is a bijection between the sets of components of P and F(P) that identifies the class of P containing the root of the (αk+1)-shape with the class ofF(P)containing the root of the (αk+1)-shape.
(2) There is a bijectionF fromC(E,1,2)to C(E,2,2)so that if P partitions the shape Si,j
ofTi, thenF(P)partitions the shape S3−i,j ofT3−i and there is a bijection between the sets of components ofP andF(P)that identifies the class ofP containing the root of the (αk+1)-shape with the class ofF(P)containing the root of the (αk+1)-shape.
(3) The partitions inC(E, i,3)partition the shapeSi,3−iofTi. Moreover, there is a bijectionF from C(E,1,3) to C(E,2,3) so that for each P, there is a bijection between the sets of components of P and F(P) that identifies the class of P containing the root of the (αk+1)-shape with the class ofF(P)containing the root of the (αk+1)-shape.
(4) The partitions in C(E, i,4)partition the shape Si,3−i ofTi.
Proof. Items (2) and (4) follow directly from the structure of the shapesSi,j. Items (1) and (3) follow from the assumption thatR01andR02are isomorphic.
LetS2:=Sq+m−1→R1 andS1:=Sq+p−1→R2. We observe that ifq >1thenS1 andS2 are not isomorphic since none of the pairs(R01, R02)and(R1, R2)consists of r-isomorphic trees and, in addition,R01and R02 are isomorphic.
Observation 4.5. Suppose that q > 1 and let E be an expression of αk+1−1 such that r1:=θ(S1, E)6=θ(S2, E) =:r2. Such an expressionE exists by the induction assumption since αk+1−1<|T|. Leti∈ {1,2} such that ri> r3−i. Then θs(Ti,[E,1])> θs(T3−i,[E,1]).
Proof. LetE0 := [E,1]. By Observation 4.4 it suffices to show that|C(E0, i,4)|>|C(E0,3−i,4)|, which can be argued as follows.
We first observe that|C(E0, j,4)|=rj−|C(E0, j,3)|for eachj ∈ {1,2}. Further,|C(E0,1,3)|=
|C(E0,2,3)|by Observation 4.4. As we assumed thatri> r3−i, the observation holds.
[(3.1.1)] Let q > 1. Let E be the expression from Observation 4.5. We recall that by Procedure 1, we know for eachαk+1-expressionEthe number of shapedαk+1-partitions ofT with
characteristicE. Hence we knowθs(T,[E,1])and alsoθs(T,[E,1])∈ {θs(T1,[E,1]), θs(T2,[E,1])}, which contains two values. Hence this case is solved by Observation 4.5.
[(3.1.2)] Letq= 1. ThenSi,i is isomorphic but not r-isomorphic toS3−i,ifor each i∈ {1,2}, andS1,1is not isomorphic toS2,2sinceR01andR02are not r-isomorphic. We observe thatk+1< n sinceαk+16|T|/2 and not allαk+1-shapes ofT are stars.
We now know all the input data of Procedure 1 forT andj =αk+2since for eachj0 6αk+1
and for eachj0−formF the number of shapesS ofT1that are isomorphic to a member ofF is equal to the number of shapesS ofT2that are isomorphic to a member ofF.
Letq0:=αk+2−αk+1. There are four candidates for anαk+2-shape ofT, namelyS0i,j=Sq0 → Si,j for(i, j)∈ {1,2}2.
Observation 4.6. The trees Si,j0 , for (i, j)∈ {1,2}2, are mutually non-isomorphic.
Proof. For S1,10 andS2,20 , this follows from Observation 4.2. Moreover, for each i∈ {1,2}, we know that Si,i0 is isomorphic to neither ofSi,3−10 andS3−i,i0 because the lengths of the spines are different. Finally we consider S1,20 andS02,1. We know that the rooted caterpillarR01 is the reverse ofR02. Recall the degree sequences of caterpillars, defined in the proof of Observation 4.2 on page 9. Let the degree sequencesR0
1 of R01 be(a1, . . . , an)— we know that a1 =m. Then sS2,10 is the sequences2:= (a1, . . . , an+ 1, q0)andsS01,2 is the sequences1:= (an, . . . , a1+ 1, q0).
We observe that ifs1=s2 or ifs1 is the reverse of s2, then (a1, . . . , an)is equal to its reverse, which contradicts the assumption thatR01andR02are not r-isomorphic.
If T has a uniqueαk+2-shape then we can determine which one of the four mutually non- isomorphic candidates it is using the induction assumption (αk+2<|T|) and Procedure 1, which implies that we know for eachαk+2-expressionEthe number of shapedαk+2-partitions ofT with characteristicE. Hence, we assume thatT has twoαk+2-shapes.
There are two possibilities for the twoαk+2-shapes ofT: eitherS1,10 , S1,20 orS2,10 , S2,20 . Fori∈ {1,2}, letTi0 be any caterpillar with|Ti0|=|T|whoseαk+2-shapes are exactlySi,10 andSi,20 .
Next we proceed analogously as in case (3.1.1). Similarly as before, let us label the vertices of the shapeSq0 ofSi,j0 byu1, . . . , uq0 for each (i, j)∈ {1,2}2, whereuq0 is the centre ofSq0.
Let i∈ {1,2} and let E be an expression of αk+2. We classify the shaped αk+2-partitions ofTi0into four classesC0(E, i,1),C0(E, i,2),C0(E, i,3)andC0(E, i,4). To this end, letEi0 be the collection of all shapedαk+2-partitions ofTiof characteristic[|T| −αk+2, E]. An elementP ofEi0 partitions a subtreeGofTi0 if a subset (possibly of order one) of the parts ofP forms a partition ofG.
(1) We letC0(E, i,1)be the collection of all elements ofEi0 such that a subset of parts ofP is a partition the uniqueαk-shape ofTi0.
(2) We let C0(E, i,2)be the collection of all elements P ∈ Ei0\C0(E, i,1)such that{vq} ⊆ V(Si,3−i)is not a part ofP.
(3) We let C0(E, i,3)be the collection of all elements P ∈ Ei0\C0(E, i,1)such that{vq} ⊆ V(Si,3−i)is a part ofP anduq0 ∈V(Si,3−i0 )does not belong to the same part ofP as the root ofSi,3−i ⊂Si,3−i0 .
(4) We let C0(E, i,4)be the collection of all elements P ∈ Ei0\C0(E, i,1)such that{vq} ⊆ V(Si,3−i)is a part ofP anduq0 ∈V(Si,3−i0 )belongs to the same part ofP as the root ofSi,3−i ⊂ Si,3−i0 .
Observation 4.7. Let i∈ {1,2}and letE be an expression ofαk+2.
(1) The partitions inC0(E, i,1)partition the shape Si,i0 . Moreover, there is a bijection F from C0(E,1,1) toC0(E,2,1)so that for each P, there is a bijection between the sets
