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232021316499

The treewidth of 2-section of hypergraphs

Ke Liu [email protected]    Mei Lu [email protected] Department of Mathematical Sciences, Tsinghua University, Beijing 100084, China.
(2020-05-24; 2021-06-16; 2021-06-30)
Abstract

Let H=(V,F)H=(V,F) be a simple hypergraph without loops. HH is called linear if |fg|1|f\cap g|\leq 1 for any f,gFf,g\in F with fgf\not=g. The 22-section of HH, denoted by [H]2[H]_{2}, is a graph with V([H]2)=VV([H]_{2})=V and for any u,vV([H]2)u,v\in V([H]_{2}), uvE([H]2)uv\in E([H]_{2}) if and only if there is fFf\in F such that u,vfu,v\in f. The treewidth of a graph is an important invariant in structural and algorithmic graph theory. In this paper, we consider the treewidth of the 22-section of a linear hypergraph. We will use the minimum degree, maximum degree, anti-rank and average rank of a linear hypergraph to determine the upper and lower bounds of the treewidth of its 22-section. Since for any graph GG, there is a linear hypergraph HH such that [H]2G[H]_{2}\cong G, we provide a method to estimate the bound of treewidth of graph by the parameters of the hypergraph.

keywords:
Linear hypergraph, treewidth, 22-section, supertree width

1 Introduction

The treewidth of a graph is an important invariant in structural and algorithmic graph theory. The concept of treewidth was originally introduced by Bertelé and Brioschi [2] under the name of dimension. It was later rediscovered by Halin [7] in 1976 and by Robertson and Seymour [13] in 1984. Now it has been studied by many other authors (see for example [5]-[12]). Treewidth is commonly used as a parameter in the parameterized complexity analysis of graph algorithms, since many NP-complete problems can be solved in polynomial time on graphs of bounded treewidth [3]. The relation between the treewidth and other graph parameters has been explored in a number of papers (see [10] for a recent survey). In [11], Harvey and Wood studied the treewidth of line graphs. They proved sharp lower bounds of the treewidth of the line graph of a graph GG in terms of both the minimum degree and the average degree of GG. Motivated by their work, in this paper, we study the treewidth of 22-section of linear hypergraphs.

A hypergraph is a pair H=(V,F)H=(V,F), where VV is a finite set of vertices and FF is a family of subsets of VV such that for any fFf\in F, ff\neq\emptyset and V=fFfV=\cup_{f\in F}f. The size of HH is the cardinality of FF. A simple hypergraph is a hypergraph HH such that if fgf\subseteq g, then f=gf=g, where f,gFf,g\in F. If |f|=1|f|=1, we call ff a loop. In this paper, we just consider simple and no loop hypergraphs. The rank and anti-rank of HH is defined as r(H)=maxfF|f|r(H)=\max\limits_{f\in F}|f| and s(H)=minfF|f|s(H)=\min\limits_{f\in F}|f|, respectively. If r(H)=s(H)=2r(H)=s(H)=2, then HH is a graph. For any vVv\in V, denote F(v)={fF|vf}F(v)=\{f\in F|v\in f\}. Then the degree of vv, denoted by deg(v)deg(v), is |F(v)||F(v)|. The maximum and minimum degree of HH will be denoted by Δ=maxvVdeg(v)\Delta=\max_{v\in V}deg(v) and δ=minvVdeg(v)\delta=\min_{v\in V}deg(v), respectively. If δ=Δ=k\delta=\Delta=k, then we call the hypergraph kk-regular. The average rank of HH is defined as l(H)=fF|f|/|F|l(H)=\sum_{f\in F}|f|/|F|. A hypergraph HH is called linear if |fg|1|f\cap g|\leq 1 for any f,gFf,g\in F with fgf\not=g.

Let H=(V,F)H=(V,F) be a hypergraph (or graph), where V={v1,,vn}V=\{v_{1},\ldots,v_{n}\} and F={f1,f2,,fm}F=\{f_{1},f_{2},\ldots,f_{m}\}. The dual of HH, denoted by H=(V,F)H^{*}=(V^{*},F^{*}), is a hypergraph whose vertices u1,u2,,umu_{1},u_{2},\ldots,u_{m} correspond to the edges of HH and with edges gi={uj|vifj}g_{i}=\{u_{j}|v_{i}\in f_{j}\}, 1in1\leq i\leq n. The line graph of a (hyper)graph HH, denoted by L(H)L(H) is a graph whose vertices w1,w2,,wmw_{1},w_{2},\ldots,w_{m} correspond to the edges of HH and with edges wiwjw_{i}w_{j} if fifjf_{i}\cap f_{j}\neq\emptyset. Terminology and notation concerning hypergraph not defined here can be found in [1]. Now we give the definitions of treewidth. Let TT be a tree. We will use TT to denote the vertex set of TT for short.

Definition 1.1 A tree decomposition of a graph G=(V,E)G=(V,E) is a pair (T,(Bt)tT)(T,(B_{t})_{t\in T}), where TT is a tree and (Bt)tT(B_{t})_{t\in T} (BtB_{t} is called a bag) is a family of subsets of VV such that:

(T1) for every vVv\in V, the set B1(v)={tT|vBt}B^{-1}(v)=\{t\in T|v\in B_{t}\} is nonempty and connected in TT;

(T2) for every edge uwE(G)uw\in E(G), there is tTt\in T such that u,wBtu,w\in B_{t}.

The width of the decomposition (T,(Bt)tT)(T,(B_{t})_{t\in T}) is the number

max{|Bt||tT}1.\max\{\left|B_{t}\right||t\in T\}-1.

The treewidth tw(G)tw(G) of GG is the minimum of the widths of the tree decompositions of GG.

In order to cite the definition of a generalized hypertree decomposition of a hypergraph which was given in [4], we introduce the definition of the 2-section of a hypergraph.

Definition 1.2 The 2-section of a hypergraph H=(V,F)H=(V,F), denoted by [H]2[H]_{2}, is a graph with V([H]2)=VV([H]_{2})=V and for any u,vV([H]2)u,v\in V([H]_{2}), uvE([H]2)uv\in E([H]_{2}) if and only if there is fFf\in F such that u,vfu,v\in f.

Definition 1.3 [4] A generalized hypertree decomposition of a hypergraph H=(V,F)H=(V,F) is a 3-tuple (T,(Bt)tT,(λt)tT)(T,(B_{t})_{t\in T},(\lambda_{t})_{t\in T}), where TT is a tree, (Bt)tT(B_{t})_{t\in T} is a family of subsets of VV and (λt)tT(\lambda_{t})_{t\in T} is a family of subsets of FF such that:

(TI) (T,(Bt)tT)(T,(B_{t})_{t\in T}) is a tree decomposition of [H]2[H]_{2};

(TII) for every tTt\in T, BtfλtfB_{t}\subseteq\cup_{f\in\lambda_{t}}f.

Now we give a new definition of decomposition of a hypergraph which is called a supertree decomposition of a hypergraph.

Definition 1.4 A supertree decomposition of a hypergraph H=(V,F)H=(V,F) is a 3-tuple (T,(Bt)tT,(λt)tT)(T,(B_{t})_{t\in T},(\lambda_{t})_{t\in T}), where (T,(Bt)tT,(λt)tT)(T,(B_{t})_{t\in T},(\lambda_{t})_{t\in T}) is a generalized hypertree decomposition of HH such that:

(TIII) for every fFf\in F, the set λ1(f)={tT|fλt}\lambda^{-1}(f)=\{t\in T|f\in\lambda_{t}\} is nonempty and connected in TT;

(TIV) for every f1,f2Ff_{1},f_{2}\in F with f1f2f_{1}\cap f_{2}\neq\emptyset, there is a tTt\in T such that f1,f2λtf_{1},f_{2}\in\lambda_{t}.

The width of the decomposition (T,(Bt)tT,(λt)tT)(T,(B_{t})_{t\in T},(\lambda_{t})_{t\in T}) of HH is the number

max{|λt||tT}.\max\{\left|\lambda_{t}\right||t\in T\}.

The supertree width stw(H)stw(H) of HH is the minimum of the widths of the supertree decompositions of HH.

By Definition 1.4, if (T,(Bt)tT,(λt)tT)(T,(B_{t})_{t\in T},(\lambda_{t})_{t\in T}) is a supertree decomposition of HH, then (T,(λt)tT)(T,(\lambda_{t})_{t\in T}) is a tree decomposition of L(H)L(H). When (T,(λt)tT)(T,(\lambda_{t})_{t\in T}) is a tree decomposition of L(H)L(H), we let Bt=fλtfB_{t}=\cup_{f\in\lambda_{t}}f. Then (T,(Bt)tT,(λt)tT)(T,(B_{t})_{t\in T},(\lambda_{t})_{t\in T}) is a supertree decomposition of HH. So we have stw(H)=tw(L(H))+1stw(H)=tw(L(H))+1.

From the Definition 1.2, we can get the following lemma.

Lemma 1.1 Let HH be a 22-regular linear hypergraph. Then [H]2L(H)[H]_{2}\cong L(H^{*}).

Proof 1.1.

Let H=(V,F)H=(V,F) be a 22-regular linear hypergraph. Then HH^{*} is a simple graph. By the definition of dual hypergraph, there is a bijection σ\sigma between the edge set F(H)F(H) (resp. the vertex set V(H)V(H)) and the vertex set V(H)V(H^{*}) (resp. the edge set of E(H)E(H^{*})) such that for any f,gF(H)f,g\in F(H) (resp. for any u,vV(H)u,v\in V(H)), σ(f)σ(g)E(H)\sigma(f)\sigma(g)\in E(H^{*}) if and only if fgf\cap g\not=\emptyset (resp. σ(u)σ(v)\sigma(u)\cap\sigma(v)\not=\emptyset if and only if there is fF(H)f\in F(H) with u,vfu,v\in f). By the definition of line graph, there is a bijection ϕ:E(H)V(L(H))\phi:E(H^{*})\rightarrow V(L(H^{*})) such that for any e1,e2E(H)e_{1},e_{2}\in E(H^{*}), ϕ(e1)ϕ(e2)E(L(H))\phi(e_{1})\phi(e_{2})\in E(L(H^{*})) if and only if e1e2e_{1}\cap e_{2}\not=\emptyset.

We will show that [H]2L(H)[H]_{2}\cong L(H^{*}). Let ϕσ:V([H]2)V(L(H))\phi\sigma:V([H]_{2})\rightarrow V(L(H^{*})). Then ϕσ\phi\sigma is a bijection. For any u,vV([H]2)u,v\in V([H]_{2}), uvE([H]2)uv\in E([H]_{2}) if and only if there is fF(H)f\in F(H) such that u,vfu,v\in f if and only if σ(u)σ(v)\sigma(u)\cap\sigma(v)\not=\emptyset if only if ϕ(σ(u))ϕ(σ(v))E(L(H))\phi(\sigma(u))\phi(\sigma(v))\in E(L(H^{*})). Thus [H]2L(H)[H]_{2}\cong L(H^{*}).

By the definitions of line graph and the dual, we can easily get the following lemma.

Lemma 1.2 Let HH be a 22-regular linear hypergraph. Then HL(H)H^{*}\cong L(H).

Let HH be a hypergraph. There are two elementary lower bounds on treewidth of [H]2[H]_{2}. First,

tw([H]2)r(H)1\begin{split}&tw([H]_{2})\geq r(H)-1\end{split} (1)

since the vertices in a hyperedge form a clique in [H]2[H]_{2}. Second, given a minimum width tree decomposition of [H]2[H]_{2}, replace each vertex with all the hyperedges containing the vertex to obtain a supertree decomposition of HH. It follows that

tw([H]2)1Δstw(H)1.\begin{split}&tw([H]_{2})\geq\frac{1}{\Delta}stw(H)-1.\end{split} (2)

We prove the following lower bound on tw([H]2)tw([H]_{2}) in terms of the minimum degree δ\delta, maximum degree Δ\Delta and average rank l(H)l(H) of a linear hypergraph HH.

Theorem 1.1 Let HH be a linear hypergraph with minimum degree δ\delta, maximum degree Δ\Delta and average rank l(H)l(H). Let Δδ2\Delta\geq\delta\geq 2. Suppose Δ2δ22δ\Delta\leq 2\delta^{2}-2\delta. Then

tw([H]2)>{(2δ22δΔ)l(H)2+(2Δ+4δ24δ)l(H)4Δδ(δ1)1if δ2δΔ2Δ/l(H),(2δ22δΔ)l(H)2+6Δl(H)8Δ4Δδ(δ1)1otherwise.tw([H]_{2})>\left\{\begin{array}[]{ll}\frac{(2\delta^{2}-2\delta-\Delta)l(H)^{2}+(2\Delta+4\delta^{2}-4\delta)l(H)}{4\Delta\delta(\delta-1)}-1&\mbox{if $\delta^{2}-\delta\leq\Delta-2\Delta/l(H)$,}\\ \frac{(2\delta^{2}-2\delta-\Delta)l(H)^{2}+6\Delta l(H)-8\Delta}{4\Delta\delta(\delta-1)}-1&\mbox{otherwise.}\end{array}\right.

In [11], Harvey and Wood showed that for every graph GG, tw(L(G))>18d(G)2+34d(G)2tw(L(G))>\frac{1}{8}d(G)^{2}+\frac{3}{4}d(G)-2, where d(G)d(G) is the average degree of GG. Let HH be a 2-regular linear hypergraph of order nn and size mm. By Lemma 1.1, [H]2L(H)[H]_{2}\cong L(H^{*}). Note that d(H)=2nm=l(H)d(H^{*})=\frac{2n}{m}=l(H). By Theorem 1.1, tw(L(H))=tw([H]2)>18d(H)2+34d(H)2tw(L(H^{*}))=tw([H]_{2})>\frac{1}{8}d(H^{*})^{2}+\frac{3}{4}d(H^{*})-2, just as the result in [11].

We also prove two lower bounds on tw([H]2)tw([H]_{2}) in terms of the anti-rank s(H)s(H) based on different condition of minimum degree of the given hypergraph HH.

Theorem 1.2 For every linear hypergraph HH with anti-rank s(H)s(H) and minimum degree δ3\delta\geq 3, we have

tw([H]2){38s(H)2+34s(H)1whens(H) is even,38s(H)2+12s(H)78whens(H) is odd.tw([H]_{2})\geq\left\{\begin{array}[]{ll}\frac{3}{8}s(H)^{2}+\frac{3}{4}s(H)-1&\mbox{when}\ s(H)\mbox{ is even,}\\ \frac{3}{8}s(H)^{2}+\frac{1}{2}s(H)-\frac{7}{8}&\mbox{when}\ s(H)\mbox{ is odd}.\end{array}\right.

Theorem 1.3 For every linear hypergraph HH with anti-rank s(H)s(H) and minimum degree δ=2\delta=2, we have

tw([H]2){14s(H)2+s(H)1whens(H) is even,14s(H)2+s(H)54whens(H) is odd.tw([H]_{2})\geq\left\{\begin{array}[]{ll}\frac{1}{4}s(H)^{2}+s(H)-1&\mbox{when}\ s(H)\mbox{ is even,}\\ \frac{1}{4}s(H)^{2}+s(H)-\frac{5}{4}&\mbox{when}\ s(H)\mbox{ is odd}.\end{array}\right.

In [11], Harvey and Wood showed that for every graph GG with minimum degree δ(G)\delta(G),

tw(L(G)){14δ(G)2+δ(G)1whenδ(G) is even,14δ(G)2+δ(G)54whenδ(G) is odd.tw(L(G))\geq\left\{\begin{array}[]{ll}\frac{1}{4}\delta(G)^{2}+\delta(G)-1&\mbox{when}\ \delta(G)\mbox{ is even,}\\ \frac{1}{4}\delta(G)^{2}+\delta(G)-\frac{5}{4}&\mbox{when}\ \delta(G)\mbox{ is odd}.\end{array}\right.

Let HH be a 2-regular linear hypergraph. By Lemma 1.1, [H]2L(H)[H]_{2}\cong L(H^{*}). Then tw(L(H))tw(L(H^{*})) =tw([H]2)=tw([H]_{2}). Note that δ(H)=s(H)\delta(H^{*})=s(H). Thus we can obtain the same result as that in [11] by Theorem 1.3. Now we consider upper bounds on tw([H]2)tw([H]_{2}). It is easy to show that

tw([H]2)r(H)stw(H)1.\begin{split}&tw([H]_{2})\leq r(H)stw(H)-1.\end{split} (3)

To see this, we consider a minimum width supertree decomposition of HH, and replace each bag λt\lambda_{t} by the vertices that are incident to an hyperedge of λt\lambda_{t}. This creates a tree decomposition of [H]2[H]_{2}, where each bag contains at most r(H)stw(H)r(H)stw(H) vertices. In Section 5, we improve this bound as follows.

Theorem 1.4 For every linear hypergraph HH, we have

tw([H]2)23stw(H)r(H)+13(stw(H)1)2+13r(H)1.tw([H]_{2})\leq\frac{2}{3}stw(H)r(H)+\frac{1}{3}(stw(H)-1)^{2}+\frac{1}{3}r(H)-1.

Theorem 1.4 is of primary interest when r(H)stw(H)r(H)\gg stw(H), in which case the upper bound is (23stw(H)+13)r(H)(\frac{2}{3}stw(H)+\frac{1}{3})r(H). When r(H)<stw(H)1r(H)<stw(H)-1, the bound in (3) is better than that in Theorem 1.4.

In [11], Harvey and Wood showed that for every graph GG, tw(L(G))23(tw(G)+1)Δ(G)+13tw(G)2+13Δ(G)1tw(L(G))\leq\frac{2}{3}(tw(G)+1)\Delta(G)+\frac{1}{3}tw(G)^{2}+\frac{1}{3}\Delta(G)-1. Let HH be a 2-regular linear hypergraph. Recall that stw(H)=tw(L(H))+1stw(H)=tw(L(H))+1. By Lemmas 1.1 and 1.2, [H]2L(H)[H]_{2}\cong L(H^{*}) and HL(H)H^{*}\cong L(H). Then tw(H)=stw(H)1tw(H^{*})=stw(H)-1. Note that Δ(H)=r(H)\Delta(H^{*})=r(H). By Theorem 1.4, tw(L(H))=tw([H]2)23(tw(H)+1)Δ(H)+13tw(H)2+13Δ(H)1tw(L(H^{*}))=tw([H]_{2})\leq\frac{2}{3}(tw(H^{*})+1)\Delta(H^{*})+\frac{1}{3}tw(H^{*})^{2}+\frac{1}{3}\Delta(H^{*})-1, just the same as that in [11].

The rest of this paper is organized as follows. In Section 2, some properties of tree decompositions of 2-section of hypergraphs are given. The proof of Theorem 1.1 is given in Section 3. In Section 4, we will prove Theorems 1.2 and 1.3. In Section 5, we prove Theorem 1.4. Section 6 concludes this paper.

2 Tree decomposition of 2-section of hypergraphs

In this section, we first give some properties of tree decompositions of 2-section of hypergraphs which will be used in next sections.

Let H=(V,F)H=(V,F) be a hypergraph. For any vVv\in V, recall that F(v)={fF|vf}F(v)=\{f\in F|v\in f\}. Let (T,(Bt)tT)(T,(B_{t})_{t\in T}) be a tree decomposition of [H]2[H]_{2}. For u,vTu,v\in T, we use Path(u,v)Path(u,v) to denote the path in TT connecting uu and vv.

Lemma 2.1 For every hypergraph H=(V,F)H=(V,F), there exists a minimum width tree decomposition (T,(Bt)tT)(T,(B_{t})_{t\in T}) of [H]2[H]_{2} together with an assignment b:FTb:F\rightarrow T such that for each vertex vVv\in V, B1(v)=V(STv)B^{-1}(v)=V(ST_{v}), where STvST_{v} is the subtree of TT induced by fi,fjF(v)Path(b(fi),b(fj))\cup_{f_{i},f_{j}\in F(v)}Path(b(f_{i}),b(f_{j})).

Proof 2.1.

Let (T,(Bt)tT)(T,(B_{t})_{t\in T}) be a minimum width tree decomposition of [H]2[H]_{2} such that vV|B1(v)|\sum_{v\in V}\left|B^{-1}(v)\right| is minimized. For each fFf\in F, the vertices in ff form a clique in [H]2[H]_{2}. Thus there exists a bag BtB_{t} containing all the vertices in ff, where tTt\in T. Hence for each fFf\in F choose one such node and declare it b(f)b(f).

Let vVv\in V. For any fF(v)f\in F(v), we have vBb(f)v\in B_{b(f)}. It follows that V(STv)B1(v)V(ST_{v})\subseteq B^{-1}(v). If |V(STv)|<|B1(v)||V(ST_{v})|<|B^{-1}(v)|, then we remove vv from all bags BtB_{t} for tB1(v)V(STv)t\in B^{-1}(v)\setminus V(ST_{v}). Since each vertex incident to vv appears in fF(v)Bb(f)\cup_{f\in F(v)}B_{b(f)}, the removal of the aforementioned vertices yields another tree decomposition of [H]2[H]_{2}. However, the existence of such a tree decomposition would contradict our choice of (T,(Bt)tT)(T,(B_{t})_{t\in T}). Hence V(STv)=B1(v)V(ST_{v})=B^{-1}(v), as required.

We call b(f)b(f) the base node of ff. From the proof of Lemma 2.1, we can construct a tree decomposition of [H]2[H]_{2} so that we can assign a base node for each fFf\in F and all the vertices in ff are placed in the corresponding bag Bb(f)B_{b(f)}. In fact, we can obtain a slightly stronger result that will be used to prove our theorems.

Given (T,(Bt)tT)(T,(B_{t})_{t\in T}) and bb guaranteed by Lemma 2.1, we can also ensure that each base node is a leaf and that bb is a bijection between edges of HH and leaves of TT. If b(f)b(f) is not a leaf, then we add a leaf adjacent to b(f)b(f) and let b(f)b(f) be this leaf instead. If some leaf tt is the base node for several edges of HH, then we add a leaf adjacent to tt for each edge assigned to tt. Finally, if tt is a leaf that is not a base node, then delete tt; this maintains the desired properties since a leaf is never an internal node of a subtree.

We can improve this further. Given a tree TT, we can root it at a node and orient all edges away from the root. In such a tree, a leaf is a node with outdegree 0. Say a rooted tree is binary if every non-leaf node has outdegree 2. Given a tree decomposition, by the same way described in [11], we can root it and then modify the underlying tree so that each non-leaf node has outdegree 2. The above results give the following lemma.

Lemma 2.2 For every hypergraph H=(V,F)H=(V,F) there exists a minimum width tree decomposition (T,(Bt)tT)(T,(B_{t})_{t\in T}) of [H]2[H]_{2} together with an assignment b:FTb:F\rightarrow T such that TT is a binary tree, bb is a bijection onto the leaves of TT and for each vVv\in V, B1(v)=V(STv)B^{-1}(v)=V(ST_{v}).

By Lemma 2.2, we have the following lower bound on tw([H]2)tw([H]_{2}) that is slightly stronger than (2).

Lemma 2.3 Let H=(V,F)H=(V,F) be a hypergraph with Δ2\Delta\geq 2. Then we have tw([H]2)1Δ1(stw(H)1)1tw([H]_{2})\geq\frac{1}{\Delta-1}(stw(H)-1)-1.

Proof 2.2.

Let k=tw([H]2)+1k=tw([H]_{2})+1 and (T,(Bt)tT)(T,(B_{t})_{t\in T}) be a tree decomposition of [H]2[H]_{2} of width k1k-1, together with an assignment bb as ensured by Lemma 2.2. For each vVv\in V with deg(v)2deg(v)\geq 2, we can assume that |B1(v)|2|B^{-1}(v)|\geq 2; otherwise, we can simply add a leaf tt^{\prime} adjacent to tt and let Bt=BtB_{t^{\prime}}=B_{t}, where tB1(v)t\in B^{-1}(v). We now partially construct a supertree decomposition (T,(Bt)tT,(λt)tT)(T^{\prime},(B^{\prime}_{t})_{t\in T^{\prime}},(\lambda_{t})_{t\in T^{\prime}}) of HH as follows: first let (T,(Bt)tT)=(T,(Bt)tT)(T^{\prime},(B^{\prime}_{t})_{t\in T^{\prime}})=(T,(B_{t})_{t\in T}). Let vVv\in V. If deg(v)=1deg(v)=1, then we just put the hyperedge adjacent to vv in λt\lambda_{t}, where tB1(v)t\in B^{-1}(v). Assume that deg(v)2deg(v)\geq 2. We arbitrarily choose two hyperedges in F(v)F(v), say fv1f_{v}^{1} and fv2f_{v}^{2}. We put F(v){fv2}F(v)\setminus\{f_{v}^{2}\} in λt\lambda_{t} for all tB1(v){b(fv2)}t\in B^{-1}(v)\setminus\{b(f_{v}^{2})\} and put F(v){fv1}F(v)\setminus\{f_{v}^{1}\} in λb(fv2)\lambda_{b(f_{v}^{2})}. Then for all tt, |λt|k(Δ1)|\lambda_{t}|\leq k(\Delta-1) since each vertex contributes at most Δ1\Delta-1 hyperedges to a given bag. An edge tttt^{\prime} in TT is called the edge corresponding to a 3-tuple (v,fv1,fv2)(v,f_{v}^{1},f_{v}^{2}) if fv1λtλtf_{v}^{1}\in\lambda_{t}\setminus\lambda_{t^{\prime}} and fv2λtλtf_{v}^{2}\in\lambda_{t^{\prime}}\setminus\lambda_{t}. If tttt^{\prime} is an edge corresponding to (v,fv1,fv2)(v,f_{v}^{1},f_{v}^{2}) and (v,fv1,fv2)(v^{\prime},f_{v^{\prime}}^{1},f_{v^{\prime}}^{2}) simultaneously, say fv1,fv1λtf_{v}^{1},f_{v^{\prime}}^{1}\in\lambda_{t}, then we subdivide tttt^{\prime} by adding a new node t′′t^{\prime\prime} and let Bt′′=BtBt,λt′′=(λt{fv1}){fv2}B^{\prime}_{t^{\prime\prime}}=B^{\prime}_{t}\cap B^{\prime}_{t^{\prime}},\lambda_{t^{\prime\prime}}=(\lambda_{t}\setminus\{f_{v}^{1}\})\cup\{f_{v}^{2}\}. Repeat this process so that every edge in TT^{\prime} corresponds to at most one 3-tuple. We also use TT^{\prime} to denote the tree after all of these subdivisions. Note that (T,(Bt)tT)(T^{\prime},(B^{\prime}_{t})_{t\in T^{\prime}}) is still a tree decomposition of [H]2[H]_{2}. Now if there is an edge tttt^{\prime} corresponding to (v,fv1,fv2)(v,f_{v}^{1},f_{v}^{2}), then we add fv1f_{v}^{1} into λt\lambda_{t^{\prime}}, which increases the size of λt\lambda^{\prime}_{t} at most 1.

We are going to show that (T,(Bt)tT,(λt)tT)(T^{\prime},(B^{\prime}_{t})_{t\in T^{\prime}},(\lambda_{t})_{t\in T^{\prime}}) is a supertree decomposition of HH. By the construction of (T,(Bt)tT,(λt)tT)(T^{\prime},(B^{\prime}_{t})_{t\in T^{\prime}},(\lambda_{t})_{t\in T^{\prime}}), (TI) and (TIV) in Definitions 2.3 and 2.4 hold. So we just need to show (TII) and (TIII).

First, we prove that for all tTt\in T^{\prime}, BtfλtfB^{\prime}_{t}\subseteq\cup_{f\in\lambda_{t}}f. If tTt\in T, then the conclusion holds because for each vBtv\in B_{t}, we put at least one edge incident to vv in λt\lambda_{t}. If tt is the subdivided node, then the result holds by the construction.

Now we show that for any fFf\in F, λ1(f)\lambda^{-1}(f) is connected in TT^{\prime}. Suppose there is fFf\in F such that λ1(f)\lambda^{-1}(f) is nonconnected in TT^{\prime}. Then there must be t,tTt,t^{\prime}\in T^{\prime} such that for all t′′Path(t,t){t,t}t^{\prime\prime}\in Path(t,t^{\prime})\setminus\{t,t^{\prime}\}, f(λtλt)λt′′f\in(\lambda_{t}\cap\lambda_{t^{\prime}})\setminus\lambda_{t^{\prime\prime}}. If there is a vertex vfv\in f such that Path(t,t)B1(v)Path(t,t^{\prime})\subseteq B^{-1}(v), then we should have fλt′′f\in\lambda_{t^{\prime\prime}} for all t′′Path(t,t)t^{\prime\prime}\in Path(t,t^{\prime}) even if t′′t^{\prime\prime} is a subdivided node by the construction, a contradiction. Suppose there exist two vertices v,vfv,v^{\prime}\in f such that tB1(v)B1(v),tB1(v)B1(v)t\in B^{-1}(v)\setminus B^{-1}(v^{\prime}),t^{\prime}\in B^{-1}(v^{\prime})\setminus B^{-1}(v), then there is t′′Path(t,t){t,t}t^{\prime\prime}\in Path(t,t^{\prime})\setminus\{t,t^{\prime}\} such that t′′B1(v)B1(v)t^{\prime\prime}\in B^{-1}(v)\cap B^{-1}(v^{\prime}) by vvE([H]2)vv^{\prime}\in E([H]_{2}) which implies that fλt′′f\in\lambda_{t^{\prime\prime}}, a contradiction.

Thus we have stw(H)(Δ1)k+1=(Δ1)(tw([H]2)+1)+1stw(H)\leq(\Delta-1)k+1=(\Delta-1)(tw([H]_{2})+1)+1, as required.

3 Lower bound in terms of average rank

In this section, we prove Theorem 1.1. Let H=(V,F)H=(V,F) be a hypergraph with size mm. Let Vi={vV(H)|deg(v)=i}V_{i}=\{v\in V(H)|deg(v)=i\} and ni=|Vi|n_{i}=|V_{i}|, where i=δ,,Δi=\delta,\ldots,\Delta. By the definition of average rank, l(H)=(δnδ++ΔnΔ)/ml(H)=(\delta n_{\delta}+\ldots+\Delta n_{\Delta})/m. Given two sets X,YFX,Y\subseteq F. Let σij(X)=#{vVi|degX(v)=j}\sigma^{j}_{i}(X)=\#\{v\in V_{i}|deg_{X}(v)=j\} and let σij,l(X,Y)=#{vVi|degX(v)=j and degY(v)=l}\sigma_{i}^{j,l}(X,Y)=\#\{v\in V_{i}|deg_{X}(v)=j\mbox{~{}and~{}}deg_{Y}(v)=l\}, where degX(v)=#{f|fF(v)X}deg_{X}(v)=\#\{f|f\in F(v)\cap X\}. We say a hypergraph HH is minimal if l(HS)<l(H)l(H_{S})<l(H) for every nonempty proper subset SS of FF, where

l(HS)=[δ(nδj=1δσδj(S))++Δ(nΔj=1ΔσΔj(S))]/(m|S|).l(H_{S})=\left[\delta\left(n_{\delta}-\sum\limits_{j=1}^{\delta}\sigma^{j}_{\delta}(S)\right)+\ldots+\Delta\left(n_{\Delta}-\sum\limits_{j=1}^{\Delta}\sigma^{j}_{\Delta}(S)\right)\right]/(m-|S|).

Firstly, we need some lemmas before we start bounding the tree width of [H]2[H]_{2}.

Lemma 3.1 If HH is a minimal hypergraph and SS is a nonempty proper subset of F(H)F(H), then

1Δl(H)<1|S|(fS|f|i=δΔj=1iσij(S)(jiΔ)).\frac{1}{\Delta}l(H)<\frac{1}{|S|}\left(\sum\limits_{f\in S}|f|-\sum\limits_{i=\delta}^{\Delta}\sum\limits_{j=1}^{i}\sigma^{j}_{i}(S)\left(j-\frac{i}{\Delta}\right)\right).
Proof 3.1.

Since HH is minimal, we have l(H)>l(HS)l(H)>l(H_{S}) which implies

i=δΔinim>i=δΔinii=δΔij=1iσij(S)m|S|.\frac{\sum\limits_{i=\delta}^{\Delta}in_{i}}{m}>\frac{\sum\limits_{i=\delta}^{\Delta}in_{i}-\sum\limits_{i=\delta}^{\Delta}i\sum\limits_{j=1}^{i}\sigma^{j}_{i}(S)}{m-|S|}.

Hence l(H)=(i=δΔini)/m<(i=δΔj=1iiσij(S))/|S|.l(H)=\left(\sum\limits_{i=\delta}^{\Delta}in_{i}\right)/m<\left(\sum\limits_{i=\delta}^{\Delta}\sum\limits_{j=1}^{i}i\sigma^{j}_{i}(S)\right)/|S|. We easily get fS|f|=i=δΔj=1ijσij(S)\sum\limits_{f\in S}\left|f\right|=\sum\limits_{i=\delta}^{\Delta}\sum\limits_{j=1}^{i}j\sigma^{j}_{i}(S). Thus

1Δl(H)<1|S|i=δΔj=1iiΔσij(S)=1|S|(fS|f|i=δΔj=1iσij(S)(jiΔ)),\frac{1}{\Delta}l(H)<\frac{1}{|S|}\sum\limits_{i=\delta}^{\Delta}\sum\limits_{j=1}^{i}\frac{i}{\Delta}\sigma^{j}_{i}(S)=\frac{1}{|S|}\left(\sum\limits_{f\in S}\left|f\right|-\sum\limits_{i=\delta}^{\Delta}\sum\limits_{j=1}^{i}\sigma^{j}_{i}(S)\left(j-\frac{i}{\Delta}\right)\right),

and we have the conclusion.

Let (T,(Bt)tT)(T,(B_{t})_{t\in T}) be a tree decomposition of [H]2[H]_{2} as guaranteed by Lemma 2.2. For each node tt of TT, let TtT_{t} denote the subtree of TT rooted at tt containing exactly tt and the descendants of tt. And let z(Tt)z(T_{t}) be the set of edges of HH with the base nodes in TtT_{t}. (Recall all base nodes are leaves.)

Lemma 3.2 Let H=(V,F)H=(V,F) be a minimal hypergraph and (T,(Bt)tT)(T,(B_{t})_{t\in T}) be a tree decomposition of [H]2[H]_{2} as guaranteed by Lemma 2.2. If tTt\in T is a non-leaf, non-root node and a,ba,b are the children of tt, then

|Bt|>1Δ(|z(Ta)|+|z(Tb)|)l(H)i=δΔσii(z(Ta))i=δΔσii(z(Tb)).|B_{t}|>\frac{1}{\Delta}(|z(T_{a})|+|z(T_{b})|)l(H)-\sum\limits_{i=\delta}^{\Delta}\sigma_{i}^{i}(z(T_{a}))-\sum\limits_{i=\delta}^{\Delta}\sigma_{i}^{i}(z(T_{b})).
Proof 3.2.

Denote

g(z(Ta),z(Tb))=fz(Ta)|f|+fz(Tb)|f|i=δΔj=1iu+w=j,u,w0σiu,w(z(Ta),z(Tb))(jiΔ).g(z(T_{a}),z(T_{b}))=\sum\limits_{f\in z(T_{a})}|f|+\sum\limits_{f\in z(T_{b})}|f|-\sum\limits_{i=\delta}^{\Delta}\sum\limits_{j=1}^{i}\sum\limits_{u+w=j,u,w\geq 0}\sigma^{u,w}_{i}(z(T_{a}),z(T_{b}))\left(j-\frac{i}{\Delta}\right).

We first show that g(z(Ta),z(Tb))>1Δ(|z(Ta)|+|z(Tb)|)l(H)g(z(T_{a}),z(T_{b}))>\frac{1}{\Delta}(|z(T_{a})|+|z(T_{b})|)l(H).

Since tTt\in T is a non-leaf, non-root node, z(Ta),z(Tb)z(T_{a}),z(T_{b})\neq\emptyset, z(Ta)z(Tb)=z(T_{a})\cap z(T_{b})=\emptyset and z(Tt)=z(Ta)z(Tb)Fz(T_{t})=z(T_{a})\cup z(T_{b})\subsetneq F. By Lemma 3.1, we have

1Δl(H)<1|z(Ta)z(Tb)|(fz(Ta)z(Tb)|f|i=δΔj=1iσij(z(Ta)z(Tb))(jiΔ)).\frac{1}{\Delta}l(H)<\frac{1}{|z(T_{a})\cup z(T_{b})|}\left(\sum\limits_{f\in z(T_{a})\cup z(T_{b})}|f|-\sum\limits_{i=\delta}^{\Delta}\sum\limits_{j=1}^{i}\sigma^{j}_{i}(z(T_{a})\cup z(T_{b}))\left(j-\frac{i}{\Delta}\right)\right).

So g(z(Ta),z(Tb))>1Δ(|z(Ta)|+|z(Tb)|)l(H)g(z(T_{a}),z(T_{b}))>\frac{1}{\Delta}(|z(T_{a})|+|z(T_{b})|)l(H). We consider BtB_{t}, the bag of the target node tt, which consists of vertices covered by edges in z(Ta)z(T_{a}) (resp. z(Tb)z(T_{b})) and Fz(Ta)F\setminus z(T_{a}) (resp. Fz(Tb)F\setminus z(T_{b})) at the same time. Thus,

|Bt|\displaystyle|B_{t}| =\displaystyle= i=δΔj=1iσij(z(Ta))+i=δΔj=1iσij(z(Tb))i=δΔj=1iu+w=j,u,w>0σiu,w(z(Ta),z(Tb))\displaystyle\sum\limits_{i=\delta}^{\Delta}\sum\limits_{j=1}^{i}\sigma^{j}_{i}(z(T_{a}))+\sum\limits_{i=\delta}^{\Delta}\sum\limits_{j=1}^{i}\sigma^{j}_{i}(z(T_{b}))-\sum\limits_{i=\delta}^{\Delta}\sum\limits_{j=1}^{i}\sum\limits_{u+w=j,u,w>0}\sigma^{u,w}_{i}(z(T_{a}),z(T_{b}))
i=δΔσii(z(Ta))i=δΔσii(z(Tb))\displaystyle-\sum\limits_{i=\delta}^{\Delta}\sigma_{i}^{i}(z(T_{a}))-\sum\limits_{i=\delta}^{\Delta}\sigma_{i}^{i}(z(T_{b}))
=\displaystyle= fz(Ta)|f|i=δΔj=1iσij(z(Ta))(j1)+fz(Tb)|f|i=δΔj=1iσij(z(Tb))(j1)\displaystyle\sum\limits_{f\in z(T_{a})}|f|-\sum\limits_{i=\delta}^{\Delta}\sum\limits_{j=1}^{i}\sigma^{j}_{i}(z(T_{a}))(j-1)+\sum\limits_{f\in z(T_{b})}|f|-\sum\limits_{i=\delta}^{\Delta}\sum\limits_{j=1}^{i}\sigma^{j}_{i}(z(T_{b}))(j-1)
i=δΔj=1iu+w=j,u,w>0σiu,w(z(Ta),z(Tb))i=δΔσii(z(Ta))i=δΔσii(z(Tb)).\displaystyle-\sum\limits_{i=\delta}^{\Delta}\sum\limits_{j=1}^{i}\sum\limits_{u+w=j,u,w>0}\sigma^{u,w}_{i}(z(T_{a}),z(T_{b}))-\sum\limits_{i=\delta}^{\Delta}\sigma_{i}^{i}(z(T_{a}))-\sum\limits_{i=\delta}^{\Delta}\sigma_{i}^{i}(z(T_{b})).

Notice that σij(z(Ta))=u=0ijσij,u(z(Ta),z(Tb))\sigma_{i}^{j}(z(T_{a}))=\sum\limits_{u=0}^{i-j}\sigma_{i}^{j,u}(z(T_{a}),z(T_{b})) and σij(z(Tb))=u=0ijσij,u(z(Tb),z(Ta))\sigma_{i}^{j}(z(T_{b}))=\sum\limits_{u=0}^{i-j}\sigma_{i}^{j,u}(z(T_{b}),z(T_{a})). We have i=δΔj=1iσij(z(Tx))=i=δΔj=1iu=0ijσij,u(z(Tx),z(Ty))=i=δΔj=1iu+w=j,u>0,w0σiu,w(z(Tx),z(Ty))\sum\limits_{i=\delta}^{\Delta}\sum\limits_{j=1}^{i}\sigma_{i}^{j}(z(T_{x}))=\sum\limits_{i=\delta}^{\Delta}\sum\limits_{j=1}^{i}\sum\limits_{u=0}^{i-j}\sigma_{i}^{j,u}(z(T_{x}),z(T_{y}))=\sum\limits_{i=\delta}^{\Delta}\sum\limits_{j=1}^{i}\sum\limits_{u+w=j,u>0,w\geq 0}\sigma^{u,w}_{i}(z(T_{x}),z(T_{y})) holds for x=a,y=bx=a,y=b or x=b,y=ax=b,y=a.

So the above formula can be written as

|Bt|\displaystyle|B_{t}| =\displaystyle= fz(Ta)z(Tb)|f|i=δΔj=1iu+w=j,u,w>0σiu,w(z(Ta),z(Tb))(1+u1+w1)\displaystyle\sum\limits_{f\in z(T_{a})\cup z(T_{b})}|f|-\sum\limits_{i=\delta}^{\Delta}\sum\limits_{j=1}^{i}\sum\limits_{u+w=j,u,w>0}\sigma^{u,w}_{i}(z(T_{a}),z(T_{b}))(1+u-1+w-1)
\displaystyle- i=δΔσii(z(Ta))i=δΔσii(z(Tb))i=δΔj=1iu+w=j,uw=0σiu,w(z(Ta),z(Tb))(j1)\displaystyle\sum\limits_{i=\delta}^{\Delta}\sigma_{i}^{i}(z(T_{a}))-\sum\limits_{i=\delta}^{\Delta}\sigma_{i}^{i}(z(T_{b}))-\sum\limits_{i=\delta}^{\Delta}\sum\limits_{j=1}^{i}\sum\limits_{u+w=j,u\cdot w=0}\sigma^{u,w}_{i}(z(T_{a}),z(T_{b}))(j-1)
=\displaystyle= fz(Ta)z(Tb)|f|i=δΔj=1iu+w=j,u,w0σiu,w(z(Ta),z(Tb))(u1+w1+1)\displaystyle\sum\limits_{f\in z(T_{a})\cup z(T_{b})}|f|-\sum\limits_{i=\delta}^{\Delta}\sum\limits_{j=1}^{i}\sum\limits_{u+w=j,u,w\geq 0}\sigma^{u,w}_{i}(z(T_{a}),z(T_{b}))(u-1+w-1+1)
\displaystyle- i=δΔσii(z(Ta))i=δΔσii(z(Tb))\displaystyle\sum\limits_{i=\delta}^{\Delta}\sigma_{i}^{i}(z(T_{a}))-\sum\limits_{i=\delta}^{\Delta}\sigma_{i}^{i}(z(T_{b}))
=\displaystyle= fz(Ta)z(Tb)|f|i=δΔj=1iu+w=j,u,w0σiu,w(z(Ta),z(Tb))(j1)i=δΔσii(z(Ta))\displaystyle\sum\limits_{f\in z(T_{a})\cup z(T_{b})}|f|-\sum\limits_{i=\delta}^{\Delta}\sum\limits_{j=1}^{i}\sum\limits_{u+w=j,u,w\geq 0}\sigma^{u,w}_{i}(z(T_{a}),z(T_{b}))(j-1)-\sum\limits_{i=\delta}^{\Delta}\sigma_{i}^{i}(z(T_{a}))
\displaystyle- i=δΔσii(z(Tb))\displaystyle\sum\limits_{i=\delta}^{\Delta}\sigma_{i}^{i}(z(T_{b}))

Furthermore, we have

|Bt|\displaystyle|B_{t}| \displaystyle\geq fz(Ta)z(Tb)|f|i=δΔj=1iu+w=j,u,w0σiu,w(z(Ta),z(Tb))(jiΔ)i=δΔσii(z(Ta))\displaystyle\sum\limits_{f\in z(T_{a})\cup z(T_{b})}|f|-\sum\limits_{i=\delta}^{\Delta}\sum\limits_{j=1}^{i}\sum\limits_{u+w=j,u,w\geq 0}\sigma^{u,w}_{i}(z(T_{a}),z(T_{b}))\left(j-\frac{i}{\Delta}\right)-\sum\limits_{i=\delta}^{\Delta}\sigma_{i}^{i}(z(T_{a}))
\displaystyle- i=δΔσii(z(Tb))\displaystyle\sum\limits_{i=\delta}^{\Delta}\sigma_{i}^{i}(z(T_{b}))
=\displaystyle= g(z(Ta),z(Tb))i=δΔσii(z(Ta))i=δΔσii(z(Tb))\displaystyle g(z(T_{a}),z(T_{b}))-\sum\limits_{i=\delta}^{\Delta}\sigma_{i}^{i}(z(T_{a}))-\sum\limits_{i=\delta}^{\Delta}\sigma_{i}^{i}(z(T_{b}))
>\displaystyle> 1Δ(|z(Ta)|+|z(Tb)|)l(H)i=δΔσii(z(Ta))i=δΔσii(z(Tb)),\displaystyle\frac{1}{\Delta}(|z(T_{a})|+|z(T_{b})|)l(H)-\sum\limits_{i=\delta}^{\Delta}\sigma_{i}^{i}(z(T_{a}))-\sum\limits_{i=\delta}^{\Delta}\sigma_{i}^{i}(z(T_{b})),

thus we have the conclusion.

Theorem 1.1 follows from the following lemma since every hypergraph HH with δ2\delta\geq 2 contains a minimal subgraph HH^{\prime} with l(H)l(H)l(H^{\prime})\geq l(H), in which case [H]2[H]2[H^{\prime}]_{2}\subseteq[H]_{2} and tw([H]2)tw([H]2)tw([H]_{2})\geq tw([H^{\prime}]_{2}). Lemma 3.3 Let HH be a minimal linear hypergraph with minimum degree δ\delta, maximum degree Δ\Delta and average rank l(H)l(H). Let Δδ2\Delta\geq\delta\geq 2. Suppose Δ2δ22δ\Delta\leq 2\delta^{2}-2\delta. Then

tw([H]2)>{(2δ22δΔ)l(H)2+(2Δ+4δ24δ)l(H)4Δδ(δ1)1if δ2δΔ2Δ/l(H),(2δ22δΔ)l(H)2+6Δl(H)8Δ4Δδ(δ1)1otherwise.tw([H]_{2})>\left\{\begin{array}[]{ll}\frac{(2\delta^{2}-2\delta-\Delta)l(H)^{2}+(2\Delta+4\delta^{2}-4\delta)l(H)}{4\Delta\delta(\delta-1)}-1&\mbox{if $\delta^{2}-\delta\leq\Delta-2\Delta/l(H)$,}\\ \frac{(2\delta^{2}-2\delta-\Delta)l(H)^{2}+6\Delta l(H)-8\Delta}{4\Delta\delta(\delta-1)}-1&\mbox{otherwise.}\end{array}\right.
Proof 3.3.

If 0l(H)<20\leq l(H)<2, we have δ2δ>0Δ2Δ/l(H)\delta^{2}-\delta>0\geq\Delta-2\Delta/l(H). Then tw([H]2)02Δ1=22(2δ22δΔ)+62Δ8Δ4Δδ(δ1)1>(2δ22δΔ)l(H)2+6Δl(H)8Δ4Δδ(δ1)1tw([H]_{2})\geq 0\geq\frac{2}{\Delta}-1=\frac{2^{2}(2\delta^{2}-2\delta-\Delta)+6\cdot 2\cdot\Delta-8\Delta}{4\Delta\delta(\delta-1)}-1>\frac{(2\delta^{2}-2\delta-\Delta)l(H)^{2}+6\Delta l(H)-8\Delta}{4\Delta\delta(\delta-1)}-1, as required. So we assume that l(H)2l(H)\geq 2. Since δ2\delta\geq 2 and HH is a linear hypergraph, we have l(H)<|F|l(H)<|F|.

Let (T,(Bt)tT)(T,(B_{t})_{t\in T}) be a tree decomposition of [H]2[H]_{2} as guaranteed by Lemma 2.2. Call a node tt of TT significant if |z(Tt)|>l(H)2|z(T_{t})|>\frac{l(H)}{2} but |z(Tt)|l(H)2|z(T_{t^{\prime}})|\leq\frac{l(H)}{2} for each child tt^{\prime} of tt.

Claim 1. There exists a non-root, non-leaf significant node tt.

Proof 3.4 (of Claim 1).

Starting at the root of TT, begin traversing down the tree by the following rule: if some child tt of the current node has |z(Tt)|>l(H)2|z(T_{t})|>\frac{l(H)}{2}, then traverse to tt; otherwise stop. Clearly this algorithm halts.

Since |z(Tt)|=1l(H)2|z(T_{t})|=1\leq\frac{l(H)}{2} for any leaf tt, the algorithm will stop at a non-leaf.

Let tt be the node where the above algorithm stops. Suppose that tt is the root. Then |z(Tt)|=|F||z(T_{t})|=|F|. Assume t1t_{1} and t2t_{2} are the children of tt. Then |z(Tt1)|,|z(Tt2)|l(H)2|z(T_{t_{1}})|,|z(T_{t_{2}})|\leq\frac{l(H)}{2}. Thus |z(Tt)|=|z(Tt1)|+|z(Tt2)|l(H)<|F||z(T_{t})|=|z(T_{t_{1}})|+|z(T_{t_{2}})|\leq l(H)<|F|, a contradiction. Hence the algorithm does not stop at the root.

Let tt be a non-root, non-leaf significant node and a,ba,b be the children of tt. Set A=z(Ta)A=z(T_{a}) and B=z(Tb)B=z(T_{b}). Then |A|,|B|l(H)2|A|,|B|\leq\frac{l(H)}{2} but |z(Tt)|=|AB|>l(H)21|z(T_{t})|=|A\cup B|>\frac{l(H)}{2}\geq 1. By Lemma 3.2, we can get

|Bt|>1Δ(|A|+|B|)l(H)i=δΔσii(A)i=δΔσii(B).|B_{t}|>\frac{1}{\Delta}(|A|+|B|)l(H)-\sum\limits_{i=\delta}^{\Delta}\sigma_{i}^{i}(A)-\sum\limits_{i=\delta}^{\Delta}\sigma_{i}^{i}(B).

We consider the following linear programming.

𝐦𝐚𝐱i=δΔσii(A)𝐬.𝐭i=δΔi(i1)2σii(A)|A|(|A|1)2,σii(A)0,δiΔ,\begin{split}{\bf max}\quad&\sum\limits_{i=\delta}^{\Delta}\sigma_{i}^{i}(A)\\ {\bf s.t}\quad&\sum\limits_{i=\delta}^{\Delta}\frac{i(i-1)}{2}\sigma_{i}^{i}(A)\leq\frac{|A|(|A|-1)}{2},\\ &\sigma_{i}^{i}(A)\geq 0,\quad\delta\leq i\leq\Delta,\end{split}

where the constraint condition is based on HH being a linear hypergraph, that is, each pair (fif_{i}, fjf_{j}) can only be calculated at most one in σii(A)\sigma_{i}^{i}(A). From the above linear programming, we can easily know that i=δΔσii(A)|A|(|A|1)δ(δ1)\sum\limits_{i=\delta}^{\Delta}\sigma_{i}^{i}(A)\leq\frac{|A|(|A|-1)}{\delta(\delta-1)}. Thus

|Bt|>1Δ(|A|+|B|)l(H)|A|(|A|1)δ(δ1)|B|(|B|1)δ(δ1).|B_{t}|>\frac{1}{\Delta}(|A|+|B|)l(H)-\frac{|A|(|A|-1)}{\delta(\delta-1)}-\frac{|B|(|B|-1)}{\delta(\delta-1)}.

Define α,β,s\alpha,\beta,s such that |A|=αl(H)|A|=\alpha l(H), |B|=βl(H)|B|=\beta l(H) and s=1l(H)s=\frac{1}{l(H)}. Recall |A|,|B|12l(H)|A|,|B|\leq\frac{1}{2}l(H) and |A|+|B|>12l(H)|A|+|B|>\frac{1}{2}l(H). Hence |A|,|B|1|A|,|B|\geq 1. Thus sα,β12s\leq\alpha,\beta\leq\frac{1}{2} and α+β>12\alpha+\beta>\frac{1}{2}. Now we have

|Bt|>1Δ(αl(H)+βl(H))l(H)αl(H)(αl(H)1)δ(δ1)βl(H)(βl(H)1)δ(δ1)=l(H)2(αΔ+βΔ+1δ(δ1)(α2+sαβ2+sβ)).\begin{split}|B_{t}|&>\frac{1}{\Delta}(\alpha l(H)+\beta l(H))l(H)-\frac{\alpha l(H)(\alpha l(H)-1)}{\delta(\delta-1)}-\frac{\beta l(H)(\beta l(H)-1)}{\delta(\delta-1)}\\ &=l(H)^{2}\left(\frac{\alpha}{\Delta}+\frac{\beta}{\Delta}+\frac{1}{\delta(\delta-1)}(-\alpha^{2}+s\alpha-\beta^{2}+s\beta)\right).\\ \end{split}

In Appendix A, we prove that f(α,β)=αΔ+βΔ+1δ(δ1)(α2+sαβ2+sβ)min{f(12,s),f(12s,s)}f(\alpha,\beta)=\frac{\alpha}{\Delta}+\frac{\beta}{\Delta}+\frac{1}{\delta(\delta-1)}(-\alpha^{2}+s\alpha-\beta^{2}+s\beta)\geq\min\left\{f\left(\frac{1}{2},s\right),f\left(\frac{1}{2}-s,s\right)\right\}. Since f(12,s)>f(12s,s)f(\frac{1}{2},s)>f(\frac{1}{2}-s,s) if and only if δ2δ>Δ2Δs\delta^{2}-\delta>\Delta-2\Delta s and tw([H]2)+1|Bt|tw([H]_{2})+1\geq|B_{t}|, the result holds immediately.

By Theorem 1.1, we can get the following corollary.

Corollary 3.4 Let H=(V,F)H=(V,F) be a hh-regular linear hypergraph with average rank l(H)l(H). Suppose h2h\geq 2. Then

tw([H]2)>(2h3)l(H)2+6l(H)84h(h1)1.tw([H]_{2})>\frac{(2h-3)l(H)^{2}+6l(H)-8}{4h(h-1)}-1.

Let k2k\geq 2 be a positive integer. We construct a hypergraph H=(V,F)H=(V,F) where F={f1,f2,,fn}F=\{f_{1},f_{2},\ldots,f_{n}\}. The vertex set of HH is determined as the following rule: for any positive integers i,ji,j with iji\not=j, if |ij|k|i-j|\leq k, then we put the vertex vi,jv_{i,j} into both fif_{i} and fjf_{j}, where we let vi,j=vj,iv_{i,j}=v_{j,i}. Then V=i=1nfiV=\cup_{i=1}^{n}f_{i}. We can easily know that HH is a 22-regular linear hypergraph with l(H)=2kγl(H)=2k-\gamma, where γ0\gamma\to 0 when nn\to\infty. By Corollary 3.4, tw([H]2)>12k2+32k2γ(12k+3418γ)tw([H]_{2})>\frac{1}{2}k^{2}+\frac{3}{2}k-2-\gamma(\frac{1}{2}k+\frac{3}{4}-\frac{1}{8}\gamma). Since 12k2+32p2\frac{1}{2}k^{2}+\frac{3}{2}p-2 is an integer, tw([H]2)12k2+32k2tw([H]_{2})\geq\frac{1}{2}k^{2}+\frac{3}{2}k-2. Note that [H]2L(Pnk)[H]_{2}\cong L(P_{n}^{k}), where PnkP_{n}^{k} is the kthk^{th}-power of an nn-vertex path with vertex set {v1,v2,,vn}\{v_{1},v_{2},\ldots,v_{n}\} and edge set {(vi,vj)||ij|k,1i<jn}\{(v_{i},v_{j})~{}|~{}|i-j|\leq k,1\leq i<j\leq n\}. In [11], Harvey and Wood showed that tw(L(Pnk))12k2+32k1tw(L(P_{n}^{k}))\leq\frac{1}{2}k^{2}+\frac{3}{2}k-1. Thus when h=2h=2, Corollary 3.4 is almost precisely sharp for treewidth.

4 Lower bounds in terms of anti-rank

We use similar techniques to those in Section 3 to prove a lower bound on tw([H]2)tw([H]_{2}) in terms of s(H)s(H) instead of l(H)l(H).

Proof 4.1 (of Theorems 1.2 and 1.3 ).

If s(H)<2s(H)<2, then Theorems 1.2 and 1.3 are trivial, since tw([H]2)0tw([H]_{2})\geq 0 whenever [H]2[H]_{2} contains at least one vertex. Now we assume that s(H)2s(H)\geq 2. Since HH is a linear hypergraph and δ2\delta\geq 2, s(H)<|F|s(H)<|F|.

Let (T,(Bt)tT)(T,(B_{t})_{t\in T}) be a tree decomposition for [H]2[H]_{2} as guaranteed by Lemma 2.2. For each node tt of TT, let TtT_{t} denote the subtree of TT rooted at tt containing exactly tt and the descendants of tt. Let z(Tt)z(T_{t}) be the set of edges of HH with the base nodes in TtT_{t}. (Recall all base nodes are leaves.)

Call a node tt of TT significant if |z(Tt)|>s(H)2|z(T_{t})|>\frac{s(H)}{2} but |z(Tt)|s(H)2|z(T_{t^{\prime}})|\leq\frac{s(H)}{2} for each child tt^{\prime} of t. By a argument similar to Claim 1, there exists a non-root, non-leaf significant node tt. Let a,ba,b be the children of tt, and let A=z(Ta)A=z(T_{a}) and B=z(Tb)B=z(T_{b}). Then |A|,|B|1|A|,|B|\geq 1, |A|,|B|s(H)2|A|,|B|\leq\frac{s(H)}{2} but |AB|>s(H)2|A\cup B|>\frac{s(H)}{2}. Since |A|,|B||A|,|B| are integers, if s(H)s(H) is odd (resp. even) then |A|+|B|s(H)2+12|A|+|B|\geq\frac{s(H)}{2}+\frac{1}{2} (resp. |A|+|B|s(H)2+1|A|+|B|\geq\frac{s(H)}{2}+1). Define α,β,s\alpha,\beta,s such that |A|=αs(H),|B|=βs(H)|A|=\alpha s(H),|B|=\beta s(H) and s=1s(H)s=\frac{1}{s(H)}. Then sα,β12s\leq\alpha,\beta\leq\frac{1}{2} and

α+β{12+swhens(H) is even,12+s2whens(H) is odd.\alpha+\beta\geq\left\{\begin{array}[]{ll}\frac{1}{2}+s&\mbox{when}\ s(H)\mbox{ is even,}\\ \frac{1}{2}+\frac{s}{2}&\mbox{when}\ s(H)\mbox{ is odd}.\end{array}\right.

We first give the proof of Theorem 1.2 where δ3\delta\geq 3. As in Lemma 3.2,

|Bt|=fAB|f|i=δΔj=1iu+w=j,u,w0σiu,w(A,B)(j1)i=δΔσii(A)i=δΔσii(B)=fAB|f|i=δΔj=1iu+w=j,u,w0σiu,w(A,B)(j1)i=δΔu+w=i,uw=0σiu,w(A,B)\begin{split}|B_{t}|&=\sum\limits_{f\in A\cup B}|f|-\sum\limits_{i=\delta}^{\Delta}\sum\limits_{j=1}^{i}\sum\limits_{u+w=j,u,w\geq 0}\sigma^{u,w}_{i}(A,B)(j-1)-\sum\limits_{i=\delta}^{\Delta}\sigma_{i}^{i}(A)-\sum\limits_{i=\delta}^{\Delta}\sigma_{i}^{i}(B)\\ &=\sum\limits_{f\in A\cup B}|f|-\sum\limits_{i=\delta}^{\Delta}\sum\limits_{j=1}^{i}\sum\limits_{u+w=j,u,w\geq 0}\sigma^{u,w}_{i}(A,B)(j-1)-\sum\limits_{i=\delta}^{\Delta}\sum_{u+w=i,uw=0}\sigma_{i}^{u,w}(A,B)\end{split}

Furthermore, we have

|Bt|fAB|f|i=δΔj=1iu+w=j,u,w0σiu,w(A,B)(j1)i=δΔu+w=i,u,w0σiu,w(A,B)=fAB|f|i=δΔj=1i1u+w=j,u,w0σiu,w(A,B)(j1)i=δΔu+w=i,u,w0iσiu,w(A,B)=fAB|f|i=δΔj=1i1σij(AB)(j1)i=δΔiσii(AB)=fAB|f|j=2Δi=j+1Δσij(AB)(j1)i=δΔiσii(AB),\begin{split}|B_{t}|&\geq\sum\limits_{f\in A\cup B}|f|-\sum\limits_{i=\delta}^{\Delta}\sum\limits_{j=1}^{i}\sum\limits_{u+w=j,u,w\geq 0}\sigma^{u,w}_{i}(A,B)(j-1)-\sum\limits_{i=\delta}^{\Delta}\sum_{u+w=i,u,w\geq 0}\sigma_{i}^{u,w}(A,B)\\ &=\sum\limits_{f\in A\cup B}|f|-\sum\limits_{i=\delta}^{\Delta}\sum\limits_{j=1}^{i-1}\sum\limits_{u+w=j,u,w\geq 0}\sigma^{u,w}_{i}(A,B)(j-1)-\sum\limits_{i=\delta}^{\Delta}\sum_{u+w=i,u,w\geq 0}i\sigma_{i}^{u,w}(A,B)\\ &=\sum\limits_{f\in A\cup B}|f|-\sum\limits_{i=\delta}^{\Delta}\sum\limits_{j=1}^{i-1}\sigma^{j}_{i}(A\cup B)(j-1)-\sum\limits_{i=\delta}^{\Delta}i\sigma_{i}^{i}(A\cup B)\\ &=\sum\limits_{f\in A\cup B}|f|-\sum\limits_{j=2}^{\Delta}\sum\limits_{i=j+1}^{\Delta}\sigma^{j}_{i}(A\cup B)(j-1)-\sum\limits_{i=\delta}^{\Delta}i\sigma_{i}^{i}(A\cup B),\end{split}

where σij(AB)=0\sigma^{j}_{i}(A\cup B)=0 if i<δi<\delta or i>Δi>\Delta. In Appendix B, we show that

j=2Δi=j+1Δσij(AB)(j1)i=δΔiσii(AB)|AB|(|AB|1)2.-\sum\limits_{j=2}^{\Delta}\sum\limits_{i=j+1}^{\Delta}\sigma^{j}_{i}(A\cup B)(j-1)-\sum\limits_{i=\delta}^{\Delta}i\sigma_{i}^{i}(A\cup B)\geq-\frac{|A\cup B|(|A\cup B|-1)}{2}.

Thus we have

|Bt|fAB|f||AB|(|AB|1)2(|A|+|B|)s(H)(|A|+|B|)(|A|+|B|1)2=((1+12s)α12α2+(1+12s)β12β2αβ)s(H)2.\begin{split}|B_{t}|&\geq\sum\limits_{f\in A\cup B}|f|-\frac{|A\cup B|(|A\cup B|-1)}{2}\\ &\geq(|A|+|B|)s(H)-\frac{(|A|+|B|)(|A|+|B|-1)}{2}\\ &=\left(\left(1+\frac{1}{2}s\right)\alpha-\frac{1}{2}\alpha^{2}+\left(1+\frac{1}{2}s\right)\beta-\frac{1}{2}\beta^{2}-\alpha\beta\right)s(H)^{2}.\end{split}

Now we calculate the minimum value of f(α,β)=(1+12s)α12α2+(1+12s)β12β2αβf(\alpha,\beta)=(1+\frac{1}{2}s)\alpha-\frac{1}{2}\alpha^{2}+(1+\frac{1}{2}s)\beta-\frac{1}{2}\beta^{2}-\alpha\beta under the condition 0sα,β120\leq s\leq\alpha,\beta\leq\frac{1}{2} and

α+β{12+swhens(H) is even,12+s2whens(H) is odd.\alpha+\beta\geq\left\{\begin{array}[]{ll}\frac{1}{2}+s&\mbox{when}\ s(H)\mbox{ is even,}\\ \frac{1}{2}+\frac{s}{2}&\mbox{when}\ s(H)\mbox{ is odd}.\end{array}\right.

Consider the partial derivative of f(α,β)f(\alpha,\beta) with respect to α\alpha and β\beta, we have

f(α,β)α=s2αβ+1>0,f(α,β)β=s2αβ+1>0.\frac{\partial f(\alpha,\beta)}{\partial\alpha}=\frac{s}{2}-\alpha-\beta+1>0,~{}~{}\frac{\partial f(\alpha,\beta)}{\partial\beta}=\frac{s}{2}-\alpha-\beta+1>0.

Since f(α,β)=f(β,α)f(\alpha,\beta)=f(\beta,\alpha), we have f(α,β)min{f(12,12),f(12,s)}f(\alpha,\beta)\geq\min\left\{f\left(\frac{1}{2},\frac{1}{2}\right),f\left(\frac{1}{2},s\right)\right\} when s(H)s(H) is even and f(α,β)min{f(12,12),f(12,s),f(s,1212s)}f(\alpha,\beta)\geq\min\left\{f\left(\frac{1}{2},\frac{1}{2}\right),f\left(\frac{1}{2},s\right),f\left(s,\frac{1}{2}-\frac{1}{2}s\right)\right\} when s(H)s(H) is odd. Since f(12,12)=s2+12f\left(\frac{1}{2},\frac{1}{2}\right)=\frac{s}{2}+\frac{1}{2}, f(12,s)=3s4+38f\left(\frac{1}{2},s\right)=\frac{3s}{4}+\frac{3}{8} and f(1212s,s)=s28+s2+38,f\left(\frac{1}{2}-\frac{1}{2}s,s\right)=\frac{s^{2}}{8}+\frac{s}{2}+\frac{3}{8}, we have

f(α,β){3s4+38whens(H) is even,s28+s2+38whens(H) is odd.f(\alpha,\beta)\geq\left\{\begin{array}[]{ll}\frac{3s}{4}+\frac{3}{8}&\mbox{when}\ s(H)\mbox{ is even,}\\ \frac{s^{2}}{8}+\frac{s}{2}+\frac{3}{8}&\mbox{when}\ s(H)\mbox{ is odd}.\end{array}\right.

Thus

tw([H]2)+1|Bt|{38s(H)2+34s(H)whens(H) is even,38s(H)2+12s(H)+18whens(H) is odd.tw([H]_{2})+1\geq|B_{t}|\geq\left\{\begin{array}[]{ll}\frac{3}{8}s(H)^{2}+\frac{3}{4}s(H)&\mbox{when}\ s(H)\mbox{ is even,}\\ \frac{3}{8}s(H)^{2}+\frac{1}{2}s(H)+\frac{1}{8}&\mbox{when}\ s(H)\mbox{ is odd}.\end{array}\right.

Now we give the Proof of Theorem 1.3 where δ=2\delta=2. We have

|Bt|=fAB|f|i=2Δj=1iu+w=j,u,w0σiu,w(A,B)(j1)i=2Δσii(A)i=2Δσii(B)=fAB|f|i=3Δj=1iu+w=j,u,w0σiu,w(A,B)(j1)i=3Δσii(A)i=3Δσii(B)j=12u+w=j,u,w0σiu,w(A,B)(j1)σ22(A)σ22(B)=fAB|f|j=2Δi=j+1Δ(j1)σij(AB)i=3Δiσii(AB)σ22(AB)σ22(A)σ22(B).\begin{split}|B_{t}|&=\sum\limits_{f\in A\cup B}|f|-\sum\limits_{i=2}^{\Delta}\sum\limits_{j=1}^{i}\sum\limits_{u+w=j,u,w\geq 0}\sigma^{u,w}_{i}(A,B)(j-1)-\sum\limits_{i=2}^{\Delta}\sigma_{i}^{i}(A)-\sum\limits_{i=2}^{\Delta}\sigma_{i}^{i}(B)\\ &=\sum\limits_{f\in A\cup B}|f|-\sum\limits_{i=3}^{\Delta}\sum\limits_{j=1}^{i}\sum\limits_{u+w=j,u,w\geq 0}\sigma^{u,w}_{i}(A,B)(j-1)-\sum\limits_{i=3}^{\Delta}\sigma_{i}^{i}(A)-\sum\limits_{i=3}^{\Delta}\sigma_{i}^{i}(B)\\ &\quad-\sum\limits_{j=1}^{2}\sum\limits_{u+w=j,u,w\geq 0}\sigma^{u,w}_{i}(A,B)(j-1)-\sigma^{2}_{2}(A)-\sigma^{2}_{2}(B)\\ &=\sum\limits_{f\in A\cup B}|f|-\sum\limits_{j=2}^{\Delta}\sum\limits_{i=j+1}^{\Delta}(j-1)\sigma^{j}_{i}(A\cup B)-\sum\limits_{i=3}^{\Delta}i\sigma_{i}^{i}(A\cup B)-\sigma_{2}^{2}(A\cup B)-\sigma^{2}_{2}(A)\\ &-\sigma^{2}_{2}(B).\end{split}

In Appendix C, we show that

j=2Δi=j+1Δ(j1)σij(AB)i=3Δiσii(AB)σ22(AB)σ22(A)σ22(B)|AB|(|AB|1)2|A|(|A|1)2|B|(|B|1)2.\begin{split}&-\sum\limits_{j=2}^{\Delta}\sum\limits_{i=j+1}^{\Delta}(j-1)\sigma^{j}_{i}(A\cup B)-\sum\limits_{i=3}^{\Delta}i\sigma_{i}^{i}(A\cup B)-\sigma_{2}^{2}(A\cup B)-\sigma^{2}_{2}(A)-\sigma^{2}_{2}(B)\\ &\geq-\frac{|A\cup B|(|A\cup B|-1)}{2}-\frac{|A|(|A|-1)}{2}-\frac{|B|(|B|-1)}{2}.\end{split}

Thus

|Bt|fAB|f||AB|(|AB|1)2|A|(|A|1)2|B|(|B|1)2(|A|+|B|)s(H)(|A|+|B|)(|A|+|B|1)2|A|(|A|1)2|B|(|B|1)2=((1+s)αα2+(1+s)ββ2αβ)s(H)2.\begin{split}|B_{t}|&\geq\sum\limits_{f\in A\cup B}|f|-\frac{|A\cup B|(|A\cup B|-1)}{2}-\frac{|A|(|A|-1)}{2}-\frac{|B|(|B|-1)}{2}\\ &\geq(|A|+|B|)s(H)-\frac{(|A|+|B|)(|A|+|B|-1)}{2}-\frac{|A|(|A|-1)}{2}-\frac{|B|(|B|-1)}{2}\\ &=((1+s)\alpha-\alpha^{2}+(1+s)\beta-\beta^{2}-\alpha\beta)s(H)^{2}.\end{split}

Let f(α,β)=(1+s)αα2+(1+s)ββ2αβf(\alpha,\beta)=(1+s)\alpha-\alpha^{2}+(1+s)\beta-\beta^{2}-\alpha\beta. By the same argument as above, we have

f(α,β){s+14whens(H) is even,s24+s+14whens(H) is odd.f(\alpha,\beta)\geq\left\{\begin{array}[]{ll}s+\frac{1}{4}&\mbox{when}\ s(H)\mbox{ is even,}\\ \frac{-s^{2}}{4}+s+\frac{1}{4}&\mbox{when}\ s(H)\mbox{ is odd}.\end{array}\right.

Thus

tw([H]2){14s(H)2+s(H)1whens(H) is even,14s(H)2+s(H)54whens(H) is odd,tw([H]_{2})\geq\left\{\begin{array}[]{ll}\frac{1}{4}s(H)^{2}+s(H)-1&\mbox{when}\ s(H)\mbox{ is even,}\\ \frac{1}{4}s(H)^{2}+s(H)-\frac{5}{4}&\mbox{when}\ s(H)\mbox{ is odd},\end{array}\right.

and we have the conclusion.

When s(H)s(H) is even, let H=(Cnk)H=(C_{n}^{k})^{*}, where CnkC_{n}^{k} is the kthk^{th}-power of an nn-vertex cycle with vertex set {v1,v2,,vn}\{v_{1},v_{2},\ldots,v_{n}\} and edge set {(vi,vj)|min{|ij|,i+nj}k,1i<jn}\{(v_{i},v_{j})~{}|~{}\min\{|i-j|,i+n-j\}\leq k,1\leq i<j\leq n\}. We can see that in this case HH is a 2-regular linear hypergraph and s(H)=δ(Cnk)=2ks(H)=\delta(C_{n}^{k})=2k. By Theorem 1.3, tw([H]2)14s(H)2+s(H)1=k2+2k1tw([H]_{2})\geq\frac{1}{4}s(H)^{2}+s(H)-1=k^{2}+2k-1. In [11], Harvey and Wood showed that tw(L(Cnk))k2+2k1tw(L(C_{n}^{k}))\leq k^{2}+2k-1. Hence Theorem 1.3 is precisely sharp when s(H)s(H) is even.

When s(H)s(H) is odd, choose two matchings X1={1(nk+1),2(nk+2),,kn}X_{1}=\{1(n-k+1),2(n-k+2),\ldots,kn\} and X2={(k+1)(k+2),(k+3)(k+4),,(nk1)(nk)}X_{2}=\{(k+1)(k+2),(k+3)(k+4),\ldots,(n-k-1)(n-k)\} in CnkC_{n}^{k}. If nn is odd (resp. even), let H=(CnkX1)H=(C_{n}^{k}\setminus X_{1})^{*} (resp. H=(Cnk(X1X2))H=(C_{n}^{k}\setminus(X_{1}\cup X_{2}))^{*}). Then we have s(H)=2k1s(H)=2k-1. By Theorem 1.3, tw([H]2)k2+k1tw([H]_{2})\geq k^{2}+k-1. In [11], Harvey and Wood showed that

tw([H]2)=tw(L(H)){k2+k1whenn is even,k2+k2whenn is odd.tw([H]_{2})=tw(L(H^{*}))\leq\left\{\begin{array}[]{ll}k^{2}+k-1&\mbox{when}\ n\mbox{ is even,}\\ k^{2}+k-2&\mbox{when}\ n\mbox{ is odd}.\end{array}\right.

Thus when s(H)s(H) is odd, Theorem 1.3 is precisely sharp when nn is even; and within ‘+1’ when nn is odd.

5 Upper bound

Proof 5.1 (of Theorem 1.4).

Let (T,(Bt)tT,(λt)tT)(T,(B_{t})_{t\in T},(\lambda_{t})_{t\in T}) be a supertree decomposition of a linear hypergraph HH with width kk such that TT has maximum degree at most 3. By the discussion in Section 1, we may assume that r(H)k1r(H)\geq k-1. (The existence of such a supertree decomposition (T,(Bt)tT,(λt)tT)(T,(B_{t})_{t\in T},(\lambda_{t})_{t\in T}) is well known and follows by a similar argument to Lemma 2.2.)

Say a hyperedge fFf\in F is small if |f|k1|f|\leq k-1 and large otherwise. For each fFf\in F, we use T(f)T(f) to denote the subtree of TT induced by λ1(f)\lambda^{-1}(f). For each edge ee in TT, let A(e),B(e)A(e),B(e) denote the two subtrees of TeT-e. If ee is also an edge of T(f)T(f) for some fF(H)f\in F(H), then let A(e,f),B(e,f)A(e,f),B(e,f) denote two subtrees of T(f)eT(f)-e, where A(e,f)A(e)A(e,f)\subseteq A(e) and B(e,f)B(e)B(e,f)\subseteq B(e). For tλ1(f)t\in\lambda^{-1}(f), let γt(f)={vV(H)|vfg,gλt{f}}\gamma_{t}(f)=\{v\in V(H)|v\in f\cap g,g\in\lambda_{t}\setminus\{f\}\}. Since HH is a linear hypergraph, we have |γt(f)|k1|\gamma_{t}(f)|\leq k-1. Denote α(e,f)=tA(e,f)γt(f)\alpha(e,f)=\cup_{t\in A(e,f)}\gamma_{t}(f) and β(e,f)=tB(e,f)γt(f)\beta(e,f)=\cup_{t\in B(e,f)}\gamma_{t}(f). We have the following claim. Claim 2. For every large fFf\in F there is an edge ee in T(f)T(f) such that |α(e,f)|,|β(e,f)|23|f|+13(k1)|\alpha(e,f)|,|\beta(e,f)|\leq\frac{2}{3}|f|+\frac{1}{3}(k-1).

Proof 5.2 (of Claim 2).

Assume for the sake of a contradiction that no such ee exists. Hence for all ee in T(f)T(f), either |α(e,f)||\alpha(e,f)| or |β(e,f)||\beta(e,f)| is too “large”. Direct the edge ee towards A(e,f)A(e,f) or B(e,f)B(e,f) respectively. (If both |α(e,f)||\alpha(e,f)|, |β(e,f)||\beta(e,f)| are too large, then direct ee arbitrarily.) Given this orientation of T(f)T(f), there must be a sink (all the edges incident to it direct to it), which we label t0t_{0}.

Let e1,,ede_{1},\ldots,e_{d} be the edges in T(f)T(f) incident to t0t_{0}, where d{1,2,3}d\in\{1,2,3\}. Without loss of generality say that eie_{i} was directed towards B(ei,f)B(e_{i},f) for all eie_{i}. Hence |β(ei,f)|>23|f|+13(k1)|\beta(e_{i},f)|>\frac{2}{3}|f|+\frac{1}{3}(k-1) for all ii. Let α(ei,t0,f)=α(ei,f)γt0(f)\alpha^{\prime}(e_{i},t_{0},f)=\alpha(e_{i},f)\setminus\gamma_{t_{0}}(f). Then α(ei,t0,f)α(ej,t0,f)=\alpha^{\prime}(e_{i},t_{0},f)\cap\alpha^{\prime}(e_{j},t_{0},f)=\emptyset when iji\neq j.

If d=3d=3, then i=13|β(ei,f)|>2|f|+k1\sum_{i=1}^{3}|\beta(e_{i},f)|>2|f|+k-1. But i=13|β(ei,f)|=i=13ji(|γt0(f)|+|α(ej,t0,f)|)=2(|γt0(f)|+j=13|α(ej,t0,f)|)+|γt0(f)|2|f|+k1\sum_{i=1}^{3}|\beta(e_{i},f)|=\sum_{i=1}^{3}\sum_{j\neq i}(|\gamma_{t_{0}}(f)|+|\alpha^{\prime}(e_{j},t_{0},f)|)=2(|\gamma_{t_{0}}(f)|+\sum_{j=1}^{3}|\alpha^{\prime}(e_{j},t_{0},f)|)+|\gamma_{t_{0}}(f)|\leq 2|f|+k-1, a contradiction. If d=2d=2, then i=12|β(ei,f)|>43|f|+23(k1)\sum_{i=1}^{2}|\beta(e_{i},f)|>\frac{4}{3}|f|+\frac{2}{3}(k-1). However, i=12|β(ei,f)|=i=12(ji(|α(ej,t0,f)|)+|γt0(f)|)=|γt0(f)|+j=12|α(ej,t0,f)|+|γt0(f)||f|+k1\sum_{i=1}^{2}|\beta(e_{i},f)|=\sum_{i=1}^{2}(\sum_{j\neq i}(|\alpha^{\prime}(e_{j},t_{0},f)|)+|\gamma_{t_{0}}(f)|)=|\gamma_{t_{0}}(f)|+\sum_{j=1}^{2}|\alpha^{\prime}(e_{j},t_{0},f)|+|\gamma_{t_{0}}(f)|\leq|f|+k-1, a contradiction with |f|>k1|f|>k-1. If d=1d=1, then |β(e1,f)|>23|f|+13(k1)|\beta(e_{1},f)|>\frac{2}{3}|f|+\frac{1}{3}(k-1). However, |β(e1,f)|=|γt0(f)|k1|\beta(e_{1},f)|=|\gamma_{t_{0}}(f)|\leq k-1, a contradiction with |f|>k1|f|>k-1.

For each small hyperedge ff of HH, arbitrarily select a base node in λ1(f)\lambda^{-1}(f). For each large hyperedge ff of HH, select an edge ee in the subtree T(f)T(f) as guaranteed by Claim 2. Subdivide ee and declare the new node to be b(f)b(f), the base node of ff. If ee is selected for several different hyperedges, then subdivide it multiple times and assign a different base node for each hyperedge of HH that selected ee. Denote the tree after all of these subdivisions as TT^{\prime}. Together, this underlying tree TT^{\prime} and the assignment bb gives a tree decomposition of [H]2[H]_{2} in the same form as Lemma 2.1. Label the set of bags for this tree decomposition by BB^{\prime}. So the tree decomposition of [H]2[H]_{2} is (T,(Bt)tT)(T^{\prime},(B^{\prime}_{t^{\prime}})_{t^{\prime}\in T^{\prime}}) and for each vertex vVv\in V, B1(v)=V(STv)B^{\prime-1}(v)=V(ST_{v}), where STvST_{v} is the subtree of TT^{\prime} induced by fi,fjF(v)Path(b(fi),b(fj))\cup_{f_{i},f_{j}\in F(v)}Path(b(f_{i}),b(f_{j})). It remains to bound the width of this tree decomposition of [H]2[H]_{2}.

For each bag BtB^{\prime}_{t^{\prime}} of TT^{\prime}, define a corresponding bag in TT as follows. If tTt^{\prime}\in T^{\prime} is also in TT, then the corresponding bag of BtB^{\prime}_{t^{\prime}} is simply λt\lambda_{t}. If tTt^{\prime}\in T^{\prime} is a subdivision node created by subdividing the edge t1t2t_{1}t_{2}, then the corresponding bag of BtB^{\prime}_{t^{\prime}} is λt1\lambda_{t_{1}} or λt2\lambda_{t_{2}}, chosen arbitrarily.

The following two claims give enough information to bound the width of (T,(Bt)tT)(T^{\prime},(B^{\prime}_{t^{\prime}})_{t^{\prime}\in T^{\prime}}).

Claim 3. If BtB^{\prime}_{t^{\prime}} is a bag of TT^{\prime} with corresponding bag λt\lambda_{t} (tTt\in T) and vBtv\in B^{\prime}_{t^{\prime}}, then there is an edge fF(v)f\in F(v) such that fλtf\in\lambda_{t}.

Proof 5.3 (of Claim 3).

Suppose that fλtf\notin\lambda_{t} for all fF(v)f\in F(v). Then tfF(v)λ1(f)t\notin\cup_{f\in F(v)}\lambda^{-1}(f). If there are fi,fjF(v)f_{i},f_{j}\in F(v) such that T(fi)T(f_{i}) and T(fj)T(f_{j}) are contained in different components of TtT-t, then λ1(fi)λ1(fj)=\lambda^{-1}(f_{i})\cap\lambda^{-1}(f_{j})=\emptyset a contradiction with vfifjv\in f_{i}\cap f_{j}. Thus for all fF(v)f\in F(v), T(f)T(f) are all contained in the same component of TtT-t. Note that b(f)b(f) is assigned inside of λ1(f)\lambda^{-1}(f) (perhaps after some edges are subdivided, but this doesn’t alter their positions relative to λt\lambda_{t}). Hence the subtree STvST_{v} in TT^{\prime} doesn’t include tt^{\prime} which implies vBtv\notin B^{\prime}_{t^{\prime}}, a contradiction.

Claim 4. Suppose ff is a large hyperedge and tTt^{\prime}\in T^{\prime} is not b(f)b(f). If λt\lambda_{t} (tTt\in T) is the corresponding bag of BtB^{\prime}_{t^{\prime}}, then we have |γt(f)|23|f|+13(k1)|\gamma_{t}(f)|\leq\frac{2}{3}|f|+\frac{1}{3}(k-1).

Proof 5.4 (of Claim 4).

Since tt^{\prime} is not b(f)b(f), there exists a component of Tb(f)T^{\prime}-b(f), say T′′T^{\prime\prime}, containing tt^{\prime}. Let vBtfv\in B^{\prime}_{t^{\prime}}\cap f. Then BtB^{\prime}_{t^{\prime}} is a bag in the subtree STvST_{v} in TT^{\prime}. Hence there is fF(v){f}f^{\prime}\in F(v)\setminus\{f\} such that b(f)V(T′′)b(f^{\prime})\in V(T^{\prime\prime}) since vBtv\in B^{\prime}_{t^{\prime}} .

Since ff is a large hyperedge, b(f)b(f) is a subdivision node. Let ee be the edge in T(f)T(f) that was subdivided to create b(f)b(f). (The edge ee is also guaranteed by Claim 2.) Hence V(T′′)V(T^{\prime\prime}) has non-empty intersection with exactly one of V(A(e))V(A(e)) and V(B(e))V(B(e)), say V(T′′)V(A(e))V(T^{\prime\prime})\cap V(A(e))\neq\emptyset. Since b(f)V(T′′)b(f^{\prime})\in V(T^{\prime\prime}), there must be vα(e,f)v\in\alpha(e,f) by vffv\in f\cap f^{\prime}. Then |α(e,f)|23|f|+13(k1)|\alpha(e,f)|\leq\frac{2}{3}|f|+\frac{1}{3}(k-1) by Claim 2. Hence BtB^{\prime}_{t^{\prime}} contains at most 23|f|+13(k1)\frac{2}{3}|f|+\frac{1}{3}(k-1) vertices in ff, which means γt(f)23|f|+13(k1)\gamma_{t}(f)\leq\frac{2}{3}|f|+\frac{1}{3}(k-1).

We now determine an upper bound on the size of a bag Bt,tTB^{\prime}_{t^{\prime}},t^{\prime}\in T^{\prime}. We count the vertices of BtB^{\prime}_{t^{\prime}} by considering the number of vertices in a given hyperedge ff of HH contributes to BtB^{\prime}_{t^{\prime}}. By Claim 3, at most kk hyperedges of the corresponding bag λt\lambda_{t} contribute to BtB^{\prime}_{t^{\prime}}.

If ff is small, it contributes at most k1k-1 vertices to BtB^{\prime}_{t^{\prime}}. If ff is large and tb(f)t^{\prime}\neq b(f), by Claim 4, ff contributes at most 23r(H)+13(k1)k1\frac{2}{3}r(H)+\frac{1}{3}(k-1)\geq k-1 vertices to BtB^{\prime}_{t^{\prime}}. If ff is large and t=b(f)t^{\prime}=b(f), then ff contributes at most r(H)23r(H)+13(k1)r(H)\geq\frac{2}{3}r(H)+\frac{1}{3}(k-1) vertices. Therefore, we conclude the highest possible contribution of a large ff with t=b(f)t^{\prime}=b(f) is greater than that of a large ff with tb(f)t^{\prime}\neq b(f) which is greater than that of a small ff. Note that t=b(f)t^{\prime}=b(f) for at most one fFf\subseteq F. Hence

|Bt|(k1)(23r(H)+13(k1))+r(H)=23kr(H)+13(k1)2+13r(H),|B^{\prime}_{t^{\prime}}|\leq(k-1)\left(\frac{2}{3}r(H)+\frac{1}{3}(k-1)\right)+r(H)=\frac{2}{3}kr(H)+\frac{1}{3}(k-1)^{2}+\frac{1}{3}r(H),

where there exists k1k-1

If we set (T,(Bt)tT,(λt)tT)(T,(B_{t})_{t\in T},(\lambda_{t})_{t\in T}) to be a minimum width hypertree decomposition of HH, then k=stw(H)k=stw(H) and so

tw([H]2)23stw(H)r(H)+13(stw(H)1)2+13r(H)1.tw([H]_{2})\leq\frac{2}{3}stw(H)r(H)+\frac{1}{3}(stw(H)-1)^{2}+\frac{1}{3}r(H)-1.

Remark If tw(L(H))ctw(L(H))\leq c for some constant cc, we can get the exact value of stw(H)stw(H) in polynomial time by stw(H)=tw(L(H))+1stw(H)=tw(L(H))+1. Then Theorem 1.4 can give a useful upper bound for tw([H]2)tw([H]_{2}).

6 Conclusion

Treewidth is an important graph parameter in structural graph theory and in algorithmic graph theory. This paper studies the treewidth of corresponding graphs of linear hypergraphs. Let G=(V,E)G=(V,E) be a graph. There is a linear hypergraph H=(V,F)H=(V,F) such that [H]2G[H]_{2}\cong G. For example, we first find cliques C1,C2,,CsC_{1},C_{2},\ldots,C_{s} in GG such that i=1sV(Ci)=V\cup_{i=1}^{s}V(C_{i})=V and |V(Ci)V(Cj)|1|V(C_{i})\cap V(C_{j})|\leq 1 for iji\not=j and 1i,js1\leq i,j\leq s. Now we construct a hypergraph H=(V,F)H=(V,F) with V(H)=V(G)V(H)=V(G) and F={C1,,Cs}{{x,y}|x,yV,xyE(G)i=1sE(Ci)}F=\{C_{1},\ldots,C_{s}\}\cup\{\{x,y\}|x,y\in V,xy\in E(G)\setminus\cup_{i=1}^{s}E(C_{i})\}. Obviously, HH is a linear hypergraph and [H]2G[H]_{2}\cong G. Thus we provide a method to estimate the bound of treewidth of graph by the parameters of the hypergraph.

Acknowledgements.
This work is partially supported by the National Natural Science Foundation of China (Grant 11771247 & 11971158) and Tsinghua University Initiative Scientific Research Program. Many thanks to the anonymous referees for their many helpful comments and suggestions, which have considerably improved the presentation of the paper.

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Appendix A

We calculate the minimum value of f(α,β)=αΔ+βΔ+1δ(δ1)(α2+sαβ2+sβ)f(\alpha,\beta)=\frac{\alpha}{\Delta}+\frac{\beta}{\Delta}+\frac{1}{\delta(\delta-1)}(-\alpha^{2}+s\alpha-\beta^{2}+s\beta), under the condition sα,β12s\leq\alpha,\beta\leq\frac{1}{2}, α+β>12\alpha+\beta>\frac{1}{2}, 0<s120<s\leq\frac{1}{2} and Δδ2\Delta\geq\delta\geq 2.

Note that f(α,β)=f(β,α)f(\alpha,\beta)=f(\beta,\alpha). Consider the partial derivative of f(α,β)f(\alpha,\beta) with respect to α\alpha and β\beta, we have

f(α,β)α=1Δ2αδ(δ1)+sδ(δ1),f(α,β)β=1Δ2βδ(δ1)+sδ(δ1).\frac{\partial f(\alpha,\beta)}{\partial\alpha}=\frac{1}{\Delta}-\frac{2\alpha}{\delta(\delta-1)}+\frac{s}{\delta(\delta-1)},~{}~{}\frac{\partial f(\alpha,\beta)}{\partial\beta}=\frac{1}{\Delta}-\frac{2\beta}{\delta(\delta-1)}+\frac{s}{\delta(\delta-1)}.

In the boundary α+β=12\alpha+\beta=\frac{1}{2}, we have

f(α,12α)=8Δα2+4Δα+2δ22δΔ+2Δs4Δδ(δ1)f\left(\alpha,\frac{1}{2}-\alpha\right)=\frac{-8\Delta\alpha^{2}+4\Delta\alpha+2\delta^{2}-2\delta-\Delta+2\Delta s}{4\Delta\delta(\delta-1)}

and

f(α,12α)α=14αδ(δ1).\frac{\partial f\left(\alpha,\frac{1}{2}-\alpha\right)}{\partial\alpha}=\frac{1-4\alpha}{\delta(\delta-1)}.

Thus we have f(α,β)min{f(12,12),f(12,s),f(12s,s)}f(\alpha,\beta)\geq\min\left\{f\left(\frac{1}{2},\frac{1}{2}\right),f\left(\frac{1}{2},s\right),f\left(\frac{1}{2}-s,s\right)\right\}. Note that

f(12,12)\displaystyle f\left(\frac{1}{2},\frac{1}{2}\right) =\displaystyle= 2Δs+2δ22δΔ2Δ(δ1),\displaystyle\frac{2\Delta s+2\delta^{2}-2\delta-\Delta}{2\Delta(\delta-1)},
f(12,s)\displaystyle f\left(\frac{1}{2},s\right) =\displaystyle= 2Δs2δΔ4δs+4δ2s+2δ24Δ(δ1),\displaystyle\frac{2\Delta s-2\delta-\Delta-4\delta s+4\delta^{2}s+2\delta^{2}}{4\Delta(\delta-1)},
f(12s,s)\displaystyle f\left(\frac{1}{2}-s,s\right) =\displaystyle= 8Δs2+6Δs+2δ22δΔ4Δ(δ1).\displaystyle\frac{-8\Delta s^{2}+6\Delta s+2\delta^{2}-2\delta-\Delta}{4\Delta(\delta-1)}.

Since Δ2δ22δ\Delta\leq 2\delta^{2}-2\delta, f(12,12)f(12,s)f(\frac{1}{2},\frac{1}{2})\geq f(\frac{1}{2},s). Thus f(α,β)min{f(12,s),f(12s,s)}f(\alpha,\beta)\geq\min\left\{f\left(\frac{1}{2},s\right),f\left(\frac{1}{2}-s,s\right)\right\}.

Appendix B

Here we show that j=2Δi=j+1Δσij(AB)(j1)i=δΔiσii(AB)|AB|(|AB|1)2-\sum\limits_{j=2}^{\Delta}\sum\limits_{i=j+1}^{\Delta}\sigma^{j}_{i}(A\cup B)(j-1)-\sum\limits_{i=\delta}^{\Delta}i\sigma_{i}^{i}(A\cup B)\geq-\frac{|A\cup B|(|A\cup B|-1)}{2}.

We consider the following linear programming

𝐦𝐢𝐧j=2Δi=j+1Δ(j1)σij(AB)i=δΔiσii(AB)𝐬.𝐭j=2Δi=jΔj(j1)2σij(AB)n(n1)2,σij(AB)0,2jiΔ,\begin{split}{\bf min}\quad&-\sum\limits_{j=2}^{\Delta}\sum\limits_{i=j+1}^{\Delta}(j-1)\sigma_{i}^{j}(A\cup B)-\sum\limits_{i=\delta}^{\Delta}i\sigma_{i}^{i}(A\cup B)\\ {\bf s.t}\quad&\sum\limits_{j=2}^{\Delta}\sum\limits_{i=j}^{\Delta}\frac{j(j-1)}{2}\sigma_{i}^{j}(A\cup B)\leq\frac{n^{\prime}(n^{\prime}-1)}{2},\\ &-\sigma_{i}^{j}(A\cup B)\leq 0,\quad 2\leq j\leq i\leq\Delta,\end{split}

where n=|AB|n^{\prime}=\left|A\cup B\right|. The KKTKKT condition of this linear programming is

1j+uj(j1)2ui,j=0,2j<iΔ,i+ui(i1)2ui,i=0,i=δ,,Δ,u,ui,j0,2jiΔ,u(j=2Δi=jΔj(j1)2σij(AB)n(n1)2)=0,ui,jσij(AB)=0,2jiΔ.\begin{split}&1-j+u\frac{j(j-1)}{2}-u_{i,j}=0,\quad 2\leq j<i\leq\Delta,\\ &-i+u\frac{i(i-1)}{2}-u_{i,i}=0,\quad i=\delta,\ldots,\Delta,\\ &u,u_{i,j}\geq 0,\quad 2\leq j\leq i\leq\Delta,\\ &u\left(\sum\limits_{j=2}^{\Delta}\sum\limits_{i=j}^{\Delta}\frac{j(j-1)}{2}\sigma_{i}^{j}(A\cup B)-\frac{n^{\prime}(n^{\prime}-1)}{2}\right)=0,\\ &u_{i,j}\sigma_{i}^{j}(A\cup B)=0,\quad 2\leq j\leq i\leq\Delta.\end{split}

We can easily get u=1u=1, ui0,2=0u_{i_{0},2}=0 and σi02(AB)=n(n1)2\sigma_{i_{0}}^{2}(A\cup B)=\frac{n^{\prime}(n^{\prime}-1)}{2} for some δi0Δ\delta\leq i_{0}\leq\Delta, and any other ui,j=σij(AB)=0u_{i,j}=\sigma_{i}^{j}(A\cup B)=0. Thus j=2Δi=j+1Δσij(AB)(j1)i=δΔiσii(AB)|AB|(|AB|1)2-\sum\limits_{j=2}^{\Delta}\sum\limits_{i=j+1}^{\Delta}\sigma^{j}_{i}(A\cup B)(j-1)-\sum\limits_{i=\delta}^{\Delta}i\sigma_{i}^{i}(A\cup B)\geq-\frac{|A\cup B|(|A\cup B|-1)}{2}.

Appendix C

Here we show the lower bound of

j=2Δi=j+1Δ(j1)σij(AB)i=3Δiσii(AB)σ22(AB)σ22(A)σ22(B).-\sum\limits_{j=2}^{\Delta}\sum\limits_{i=j+1}^{\Delta}(j-1)\sigma^{j}_{i}(A\cup B)-\sum\limits_{i=3}^{\Delta}i\sigma_{i}^{i}(A\cup B)-\sigma_{2}^{2}(A\cup B)-\sigma^{2}_{2}(A)-\sigma^{2}_{2}(B).

We consider the following linear programming

𝐦𝐢𝐧j=2Δi=j+1Δ(j1)σij(AB)i=δΔiσii(AB)σ22(AB)σ22(A)σ22(B)𝐬.𝐭j=2Δi=jΔj(j1)2σij(AB)n(n1)2,σ22(A)+σ22(B)σ22(AB),0σ22(A)n1(n11)2,0σ22(B)n2(n21)2,σij(AB)0,2jiΔ,\begin{split}{\bf min}\quad&-\sum\limits_{j=2}^{\Delta}\sum\limits_{i=j+1}^{\Delta}(j-1)\sigma_{i}^{j}(A\cup B)-\sum\limits_{i=\delta}^{\Delta}i\sigma_{i}^{i}(A\cup B)-\sigma_{2}^{2}(A\cup B)-\sigma^{2}_{2}(A)-\sigma^{2}_{2}(B)\\ {\bf s.t}\quad&\sum\limits_{j=2}^{\Delta}\sum\limits_{i=j}^{\Delta}\frac{j(j-1)}{2}\sigma_{i}^{j}(A\cup B)\leq\frac{n^{\prime}(n^{\prime}-1)}{2},\\ &\sigma^{2}_{2}(A)+\sigma^{2}_{2}(B)\leq\sigma_{2}^{2}(A\cup B),\\ &0\leq\sigma^{2}_{2}(A)\leq\frac{n^{\prime}_{1}(n^{\prime}_{1}-1)}{2},\\ &0\leq\sigma^{2}_{2}(B)\leq\frac{n^{\prime}_{2}(n^{\prime}_{2}-1)}{2},\\ &-\sigma_{i}^{j}(A\cup B)\leq 0,\quad 2\leq j\leq i\leq\Delta,\end{split}

where n=|AB|n^{\prime}=\left|A\cup B\right|, n1=|A|n^{\prime}_{1}=|A| and n2=|B|n^{\prime}_{2}=|B|. The KKTKKT condition of this linear programming is

1j+u1j(j1)2ui,j=0,2j<iΔ,i+u1i(i1)2ui,i=0,i=3,,Δ,1+u1u2u2,2=0,1+u2+u3u5=0,1+u2+u4u6=0,ui,j0,2jiΔ,ul0,l=1,,5,u1(j=2Δi=jΔj(j1)2σij(AB)n(n1)2)=0,u2(σ22(A)+σ22(B)σ22(AB))=0,u3(σ22(A)n1(n11)2)=0,u4(σ22(B)n2(n21)2)=0,u5σ22(A)=0,u6σ22(B)=0,ui,jσij(AB)=0,2jiΔ.\begin{split}&1-j+u_{1}\frac{j(j-1)}{2}-u_{i,j}=0,\quad 2\leq j<i\leq\Delta,\\ &-i+u_{1}\frac{i(i-1)}{2}-u_{i,i}=0,\quad i=3,\ldots,\Delta,\\ &-1+u_{1}-u_{2}-u_{2,2}=0,\\ &-1+u_{2}+u_{3}-u_{5}=0,\\ &-1+u_{2}+u_{4}-u_{6}=0,\\ &u_{i,j}\geq 0,\quad 2\leq j\leq i\leq\Delta,\\ &u_{l}\geq 0,\quad l=1,\ldots,5,\\ &u_{1}(\sum\limits_{j=2}^{\Delta}\sum\limits_{i=j}^{\Delta}\frac{j(j-1)}{2}\sigma_{i}^{j}(A\cup B)-\frac{n^{\prime}(n^{\prime}-1)}{2})=0,\\ &u_{2}(\sigma_{2}^{2}(A)+\sigma_{2}^{2}(B)-\sigma_{2}^{2}(A\cup B))=0,~{}u_{3}(\sigma_{2}^{2}(A)-\frac{n^{\prime}_{1}(n^{\prime}_{1}-1)}{2})=0,\\ &u_{4}(\sigma_{2}^{2}(B)-\frac{n^{\prime}_{2}(n^{\prime}_{2}-1)}{2})=0,u_{5}\sigma_{2}^{2}(A)=0,u_{6}\sigma_{2}^{2}(B)=0,\\ &u_{i,j}\sigma_{i}^{j}(A\cup B)=0,\quad 2\leq j\leq i\leq\Delta.\end{split}

We can get σ22(AB)=n(n1)2\sigma_{2}^{2}(A\cup B)=\frac{n^{\prime}(n^{\prime}-1)}{2}, σ22(A)=n1(n11)2\sigma_{2}^{2}(A)=\frac{n^{\prime}_{1}(n^{\prime}_{1}-1)}{2}, σ22(B)=n2(n21)2\sigma_{2}^{2}(B)=\frac{n^{\prime}_{2}(n^{\prime}_{2}-1)}{2} and all other σij(AB)=0\sigma_{i}^{j}(A\cup B)=0. Thus

j=2Δi=j+1Δ(j1)σij(AB)i=3Δiσii(AB)σ22(AB)σ22(A)σ22(B)|AB|(|AB|1)2|A|(|A|1)2|B|(|B|1)2.\begin{split}&-\sum\limits_{j=2}^{\Delta}\sum\limits_{i=j+1}^{\Delta}(j-1)\sigma^{j}_{i}(A\cup B)-\sum\limits_{i=3}^{\Delta}i\sigma_{i}^{i}(A\cup B)-\sigma_{2}^{2}(A\cup B)-\sigma^{2}_{2}(A)-\sigma^{2}_{2}(B)\\ &\geq-\frac{|A\cup B|(|A\cup B|-1)}{2}-\frac{|A|(|A|-1)}{2}-\frac{|B|(|B|-1)}{2}.\end{split}