A k-bridge hypergraph is an h-uniform linear hypergraph consisting of klinear paths having common ends. In this note it is shown that every two chromat- ically equivalent k-bridge hypergraphs are isomorphic ifk ≥3. This solves in affirmative an open question raised by Bokhary et al. [2], where a supplementary condition on the multiplicities of path lengths was imposed.
AMS 2010 Subject Classification: Primary 05C15.
Key words: chromatic polynomial, linear path,k-bridge hypergraph, chromati- cally equivalent hypergraphs.
1.NOTATION AND PRELIMINARY RESULTS
An h-uniform hypergraph (h ≥ 2) H = (V,E) of order n = |V| and size m = |E|, consists of a vertex set V(H) = V and edge set E(H) = E, where E ⊂ V and |E| = h for each edge E in E. H is said to be linear if 0≤ |E∩F| ≤1 for any two distinct edgesE, F ∈ E(H) [1].
Let Pph,1 denote the linear path consisting of p ≥ 1 edges E1, . . . , Ep such that |E1| = . . . = |Ep| = h, |Ek ∩El| = 1 if {k, l} = {i, i+ 1} for any 1≤i≤p−1 and 0 otherwise.
Each vertex from E1\E2 or from Ep\Ep−1 will be called an end vertex of Pph,1.
For any positive integersa1, . . . , ak ∈Nandh≥2 we denote byθ(h;a1, . . . , ak) theh-uniform linear hypergraph consisting ofklinear pathsPah,11 , Pah,12 , . . . , Pah,1k of lengths a1, a2, . . . , ak respectively, having in common only two fixed ends. It is a parallel hypergraph [4] of order (h−1)(a1 +. . .+ak)−k+ 2.
For example, θ(3; 4,3,5) is depicted in Fig. 1 with x and y common ends of the paths. This linear hypergraph will be called a k-bridge hypergraph;
θ(2;a1, . . . , ak) is a notation for ak-bridge graph (see [3]).
A λ-coloring of a hypergraphH is a functionf :V(H)→ {1, . . . , λ}such that each edgeE ∈ E(H) contains two verticesx andy having different colors f(x)6=f(y). The number ofλ-colorings ofHis given by a polynomialP(H, λ)
MATH. REPORTS15(65),3(2013), 281–285
Fig. 1 – 3-bridge hypergraphθ(3; 4,3,5).
in λ, called the chromatic polynomial of H, of degree equal to |V(H)|. Two hypergraphsH andGare said to be chromatically equivalent if they have the same chromatic polynomial,i.e.,P(H, λ) =P(G, λ).
The chromatic polynomial of k-bridge hypergraphs was deduced in [2]:
Lemma 1.1. For every h≥2 and k, a1, . . . , ak ≥1 we have P(θ(h;a1, . . . , ak), λ) = 1
λk−1
k
Y
i=1
((λh−1−1)ai+ (−1)ai(λ−1))
+ λ−1 λk−1
k
Y
i=1
((λh−1−1)ai−(−1)ai).
In [2] it was proved that if k, h≥3; 2≤a1 ≤. . .≤ak; 2≤b1 ≤. . .≤bk and any number in the multisets{a1, . . . , ak}and{b1, . . . , bk}has a multiplic- ity less than 2h−1 −1, then chromatic equivalence between θ(h;a1, . . . , ak) and θ(h;b1, . . . , bk) implies ai = bi for all i = 1, . . . , k. This means that θ(h;a1, . . . , ak) andθ(h;b1, . . . , bk) are isomorphic hypergraphs.
An open question raised in [2] was to decide whether the condition on the multiplicities of path lengths can be removed, since this can be done at least forh= 3 and a similar property also holds for k-bridge graphs (h= 2) [3].
In the next section, we shall prove that this can be done for all k, h≥3.
2. MAIN RESULT
First, we need a lemma concerning a non-divisibility property in a poly- nomial ring.
Lemma 2.1. Let m, n ≥ 2 be natural numbers. The polynomial λn−1 does not divide (λ−1)m+ (−1)m(λ−1).
Proof. Suppose thatλn−1 divides (λ−1)m+(−1)m(λ−1). It follows that m≥nand P(λ) =λn−1+λn−2+. . .+ 1 dividesQ(λ) = (λ−1)m−1+ (−1)m. This means that all roots of P(λ) are also roots of Q(λ). The roots of P(λ) are all roots of ordern of unity which are different from 1,i.e., complex
1+cosβ+isinβ = cosα+isinα. We deduce cosα= 1+cosβand sinα= sinβ, hence (1 + cosβ)2+ sin2β= 1, which implies cosβ =−12. We getβ1 = 2π3 and β2 = 4π3 and corresponding values α1= π3 and α2= 5π3 .
Consequently, only two roots of P(λ) can possibly be roots of Q(λ). It follows thatP(λ) does not divideQ(λ) for n≥4.
It remains to verify this property forn= 2 andn= 3. Forn= 2 we have P(λ) =λ+ 1 and Q(−1)6= 0. If n= 3 the roots ofP(λ) are εand ε2, where ε= −1+i
√ 3
2 is a cubic root of unity. Suppose thatQ(ε) = 0. This would imply
|ε−1|= 1. But|ε−1|=√
3, a contradiction.
Theorem 2.2. Let k, h ≥ 3. If k-bridge hypergraphs H1 and H2 are chromatically equivalent, then they are isomorphic hypergraphs.
Proof. Suppose that H1 =θ(h;a1, . . . , ak) andH2 =θ(h;b1, . . . , bk) with 1≤a1≤. . .≤ak and 1≤b1 ≤. . .≤bk.
By hypothesis we haveP(H1, λ) =P(H2, λ). It is necessary to prove that ai=bi for all i= 1, . . . , k.
By Lemma 1.1 we get after reductions (1) λk−1P(H1, λ) =λ(λh−1−1)a1+...+ak+
k
X
i=2
X
K⊂{1,...,k}
|K|=i
((λ−1)i+ (−1)i(λ−1))×
(−1)aK(λh−1−1)aK,
where we have denoted K = {1, . . . , k} \K, ak = Σ
j∈Kaj and a∅ = 0. Since P(H1, λ) = P(H2, λ) it follows that Σk
i=1ai = Σk
i=1bi because the terms of the highest degree in both polynomials must be equal.
We denote S1=
k−1
X
i=2
X
K⊂{1,...,k}
|K|=i
((λ−1)i+ (−1)i(λ−1))(−1)aK(λh−1−1)aK
and let S2 be the corresponding sum for b1, . . . , bk.
By equating λk−1P(H1, λ) and λk−1P(H2, λ), from (1) we get S1 = S2 since the free terms in (1) coincide for H1 and H2.
Suppose thata16=b1. Without loss of generality we can considera1< b1. Let p denote the multiplicity of a1(1 ≤ p ≤ k). If p = k it follows that a1=. . .=ak. We deduce that Σk
i=1ai=ka1 < kb1 ≤ Σk
i=1bi, a contradiction.
Consequently, p ≤ k−1 hence, a1 = . . . = ap < ap+1 ≤ . . . ≤ ak. It can be seen that the term of S1 containing the smallest power of λh−1 −1 corresponds to i=k−1 and choices K ={1},{2}, . . . ,{p} and is equal to
R(λ) =p((λ−1)k−1+ (−1)k−1(λ−1))(−1)
k i=1Σ ai−a1
(λh−1−1)a1. Since b1 > a1, or b1 ≥a1+ 1, the expression S2 is divisible by (λh−1− 1)a1+1. Also,S1−R(λ) is divisible by (λh−1−1)a1+1sinceap+1, . . . , ak≥a1+1.
It follows thatR(λ) =S2−(S1−R(λ)) is also divisible by (λh−1−1)a1+1, which means that (λ−1)k−1+ (−1)k−1(λ−1) is divisible byλh−1−1. This is impossible by Lemma 2.1 sinceh, k ≥3. It follows that a1=b1.
Let mbe such that 2≤m≤k−1. Suppose we have proved that ai =bi fori= 1, . . . , m−1 andam< bm holds.
By canceling all equal terms from both sides of the equation S1 =S2 we get another equation, denoted byS11 =S21.
In S11 and S21 the second sum is over all subsets K ⊂ {1, . . . , k} with
|K| = i such that K * {1, . . . , m−1}, since the corresponding terms in S1 and S2 have been canceled because Σk
i=1
ai = Σk
i=1
bi. In this case, the minimum of aK is reached only for i = k−1 and K = {m},{m + 1}, . . . , or {m+ q−1}, and the corresponding term of S11 equals q((λ−1)k−1+ (−1)k−1(λ− 1))(−1)
k i=1Σ ai−am
(λh−1−1)am if the multiplicity of am isq ≥1.
A similar argument as above shows that am=bm. If m = k, then ak = Σk
i=1ai −k−1Σ
i=1ai = Σk
i=1bi −k−1Σ
i=1bi = bk. Therefore, ai=bi fori= 1, . . . , k, which concludes the proof.
Note that for k = 2 the property is not true for a1+a2 ≥ 4 since all 2-bridge hypergraphs θ(h;a1, a2) having a1 +a2 = m≥ 4 represent the same linear cycle withm edges.
Acknowledgment. This research was partially supported by Higher Education Com- mission of Pakistan.
REFERENCES [1] C. Berge,Hypergraphes. Bordas, Paris, 1987.
[2] S.A. Bokhary, I. Tomescu and A.A. Bhatti, On the chromaticity of multi-bridge hyper- graphs.Graphs Combin. 25(2009), 145–152.
“Abdus Salam” School of Mathematical Sciences, Lahore-Pakistan
unique.sana@hotmail.com University of Bucharest,
Faculty of Mathematics and Computer Science, 010014 Bucharest, Romania
ioan@fmi.unibuc.ro
