Formulas counting spanning trees in line graphs and their extensions
aa r X i v : . [ m a t h . C O ] J u l Formulas counting spanning trees in line graphs andtheir extensions
Fengming Dong ∗ National Institute of EducationNanyang Technological University, Singapore
Abstract
For any connected multigraph G = ( V, E ) and any M ⊆ E , if M induces anacyclic subgraph of G and removing all edges in M yields a subgraph of G whosecomponents are complete graphs, a formula for τ G ( M ) is obtained, where τ G ( M )is the number of spanning trees in G which contain all edges in M . Applying thisresult, we can easily obtain a formula for the number of spanning trees in the linegraph or the middle graph of an arbitrary graph. Applying this result, we also showthat for any connected graph G with a clique U which is a cut-set of G , the numberof spanning trees in G has a factorization which is analogous to a property of thechromatic polynomial of G . MSC : 05A15, 05C05, 05C30, 05C76
Keywords : Graph, clique, Spanning tree, Cayley’s formula
The graphs considered in this article are multigraphs without loops. For any graph G , let V ( G ) and E ( G ) be the vertex set and the edge set of G respectively. For any non-empty V ′ ⊆ V ( G ), let G [ V ′ ] denote the subgraph of G induced by V ′ , and when V ′ = V ( G ), let G − V ′ be the subgraph G [ V ( G ) − V ′ ] (i.e., the subgraph of G obtained by deleting allvertices in V ′ ). Let N G ( V ′ ) = S v ∈ V ′ N G ( v ), where N G ( v ) is the set of neighbours of v in G , and N G [ V ′ ] = V ′ ∪ N G ( V ′ ). For any E ′ ⊆ E ( G ), let G h E ′ i be the spanning subgraph ∗ Corresponding author. Email: [email protected] G with edge set E ′ , let G [ E ′ ] be the subgraph of G induced by E ′ when E ′ = ∅ (i.e.,the graph obtained from G h E ′ i by removing all isolated vertices), let G/E ′ be the graphobtained from G by contracting all edges in E ′ and G − E ′ be the subgraph G h E ( G ) − E ′ i (i.e., the graph obtained from G by removing all edges in E ′ ).For any graph G , let ST G be the set of spanning trees of G and let τ G = |ST G | . Clearly, τ G = 0 if and only if G is disconnected. It is well-known that τ ( K n ) = n n − , dueto Cayley [1], where K n is the complete graph of order n . This beautiful formula wasextended by Moon [10, 11, 12] (also see Lov´asz [7, Problem 4 in page 34]) for countingthe number of spanning trees T ∈ ST K n which contain all edges of a given forest in K n .For any M ⊆ E ( G ), let ST G ( M ) be the set of those members T ∈ ST G with M ⊆ E ( T )and let τ G ( M ) = |ST G ( M ) | . Thus ST G ( M ) ⊆ ST G , where ST G ( M ) = ST G holdswhenever M consists of bridges of G . Clearly, τ G ( M ) = 0 if and only if either G isdisconnected or G h M i contains cycles. Theorem 1.1 (Lov´asz [7] and Moon [10, 11, 12])
For any M ⊆ E ( K n ) , if K n h M i is a forest with c components whose orders are n , n , · · · , n c , then τ K n ( M ) = n c − c Y i =1 n i . (1.1)It is natural to consider a suitable extension of Theorem 1.1. In this article, we assume that G = ( V, E ) is a connected graph, where V can be partitioned into subsets V , V , · · · , V k and V i is a clique of G (i.e., G [ V i ] is a complete graph) for all i = 1 , , · · · , k . Thus G [ V i ]has no parallel edges for all 1 ≤ i ≤ n , although G may have parallel edges. Note that V may be an empty set and G [ V ] may be not complete and may have parallel edges also.For any U , U ⊆ V , let E G ( U , U ) denote the set of those edges in G with one end in U and another end in U , and let E G ( U ) = E G ( U , V − U ). In the case that V = ∅ and M = S ≤ i Let H be a connected and loopless graph with vertices v , v , · · · , v k .Then τ L ( H ) = k Y i =1 | d ( v i ) | | d ( v i ) |− X T ∈ST H Y e ∈ E ( H ) − E ( T ) (cid:0) | d ( v a ( e ) ) | − + | d ( v b ( e ) ) | − (cid:1) . (1.3) where d ( v i ) is the degree of v i in H and v a ( e ) and v b ( e ) are the two ends of e in H . For any connected graph H , the middle graph M ( H ) of H is the one obtained from H bysubdividing each edge in H exactly once and adding a new edge joining each pair of newvertices u , u which subdivide a pair of adjacent edges in H (see [3]).Observe that if V = ∅ , M is a matching of G and exactly one vertex in each V i isnot incident with M , where 1 ≤ i ≤ k , then G/M is actually the middle graph of G ∗ , and thus, by the equality τ G ( M ) = τ G/M , a formula for τ M ( H ) follows directly fromTheorem 1.2. Remark : The study of a relation between τ L ( H ) and τ H for a connected graph H wasstarted in 1966 when Vahovskii [15] first established such a relation for a r -regular graph H : τ L ( H ) = 2 m − n +1 r m − n − τ H , (1.4)where n = | V ( H ) | and m = | E ( H ) | . When H is a graph in which each vertex is ofdegree 1 or r , where r is a constant, a similar relation between τ L ( H ) and τ H was foundby Yan [17] in 2013. When H is an ( a, b )-semiregular bipartite graph, such a relation wasfound by Cvetkovi´c (see [8, see Theorem 3.9], [9, § τ L ( H ) and τ H for an arbitrary connected graph H , whichimplies all these known results.In this article, we will further extend Theorem 1.2. Recall that V , V , · · · , V k is a partitionof V , where V i is a clique of G for all i = 1 , , · · · , k . Let M = M ∪ S ≤ i ≤ k E G ( V , V i ),where M = S ≤ i G/e ( E ′ − { e } ). (v) follows from (iv)directly. ✷ For any W ⊆ E , let G ⋆ W denote the graph obtained from G by adding a new vertex w i and new edges joining w i to all vertices in W i for all i = 1 , , · · · , r , where W , · · · , W r are the components of G [ W ]. An example of G ⋆ W is shown in Figure 1, where G [ W ]has three components. Thus, V ( G ⋆ W ) = V ( G ) ∪ { w , w , · · · , w r } and E ( G ⋆ W ) = E ( G ) ∪ S ≤ i ≤ r E G⋆W ( w i ). Also note that { w , w , · · · , w r } is an independent set in G ⋆ W . (cid:0)(cid:0) (cid:0)(cid:0) ✈ ✈✈ ✈ ✈✈ ✈✈ ✈✈ ✈✈ ✈✈ ✈✈ ✈✈ ✈✈ ✈✈ ✈✈ ✈ w w w ✁✁✁ ✁✁✁ ✬✫ ✩✪ ✬✫ ✩✪ W W W W W W (a) G (b) G ⋆ W Figure 1: Graph G ⋆ W , where G [ W ] has 3 components Lemma 2.2 Let W ⊆ E such that G [ W ] is a forest. For any W ⊆ W , τ G ( W ) = τ G⋆W ( W ′ ) = τ G⋆W − W ( W ′ ) , (2.1) where W ′ = E ( G ⋆ W ) − E ( G ) . roof . Observe that τ G⋆W ( W ′ ) = τ G⋆W − W ( W ′ ) follows from Lemma 2.1 (vi) directly,while τ G ( W ) = τ G⋆W ( W ′ ) follows from Lemma 2.1 (v) and the fact that the two graphsobtained respectively from G/W and ( G ⋆ W ) /W ′ by removing their loops are isomorphic.Thus the result follows. ✷ Recall that V , V , · · · , V k is a partition of V such that each V i is a clique of G for all i = 1 , , · · · , k . Let M = S ≤ i Let G ′ denote the graph G ⋆ W − M , V ′ = V ∪ ( V ( G ⋆ W ) − V ( G )) and W ′ = E ( G ⋆ W ) − E ( G ) . The following properties hold: (i) τ G ( W ) = τ G ′ ( W ′ ) ; (ii) V ′ , V , · · · , V k is a partition of V ( G ′ ) , where V i is a clique of G ′ for all i = 1 , , · · · , k ; (iii) E G ′ ( V i , V j ) = ∅ for all i, j with ≤ i < j ≤ k ; (iv) each component of G ′ [ W ′ ] is a star with a center in V ( G ′ ) − V ( G ) ⊆ V ′ . Lemma 2.3 (i) follows from Lemma 2.2 while Lemma 2.3 (ii)-(iv) follow directly from thedefinitions of G ′ and W ′ . By Lemma 2.3, the study of τ G ( W ) can be restricted to thespecial case that V ( G ) has a partition V , V , · · · , V k satisfying the following conditions:(i) V i is a clique for all i = 1 , , · · · , k and E G ( V i , V j ) = ∅ holds for each pair of i, j with1 ≤ i < j ≤ k ;(ii) for M = S ≤ i ≤ k E G ( V , V i ), each component of G [ M ] is a star with a center in V (i.e., d G ( u ) ≤ | V i | holds for each u ∈ V ( G ) − V );(iii) W = M ∪ N for some N ⊆ E ( G [ V ]).When the above three conditions holds, G has its structure as shown in Figure 2.6 ✫ ✩✪ · · · V V V k V ✈✈✉✈ Figure 2: E G ( V i , V j ) = ∅ for all 1 ≤ i < j ≤ k and each component of G [ M ] is a star witha center in V U Let U be a clique of a connected graph G = ( V, E ). In Subsection 3.1, we will deduce aformula for τ G ( W ) in the case that G [ W ] is a forest, where W = E − E ( G [ U ]). Let G • U denote the graph G/G [ U ]. In Subsection 3.2, we will give a relation between τ G ( M ∪ N )and τ G • U ( N ), where M = E G ( U ) and N ⊆ E ( G − U ), under the condition that eachcomponent of G [ M ] is a star with a center in V − U . G − E ( G [ U ]) is a forest Note that G − E ( G [ U ]) is a forest if and only if G [ W ] is a forest, where W = E − E ( G [ U ]).When G [ W ] is a forest, applying Theorem 1.1, we get a formula for τ G ( W ) below. Proposition 3.1 Let U be a clique of G with U = V and W = E − E ( G [ U ]) . If F = G [ W ] is a forest with components F , F , · · · , F t , then τ G ( W ) = | U | | U |− t − n − n −···− n t t Y i =1 n i , (3.1) where t is the number of components of F and n i = | V ( F i ) ∩ U | for i = 1 , , · · · , t .Proof . Note that for any i = 1 , , · · · , t , | E ( F i ) | ≥ n i , and | E ( F i ) | = n i if and only if F i is a star with a center in V − U and E ( F i ) ⊆ E G ( U ).We shall prove this result by the following claims.7 laim 1 : (3.1) holds when each F i is a star with a center at V − U and E ( F i ) ⊆ E G ( U ).Assume that each F i is a star with a center at V − U and E ( F i ) ⊆ E G ( U ). Then G − U is an empty graph, V = U ∪ N G ( U ) and F is the bipartite graph G [ E G ( U )].Let E ′ = { e , e , · · · , e t } , where e i is an edge in F i . Applying Lemma 2.1 (iv) and (iii)repeatedly, we have τ G ( W ) = τ G/E ′ ( E ( F/E ′ )) = τ G [ U ] ( E ( F/E ′ )) , (3.2)where F/E ′ is considered as a subforest of G [ U ] whose components’s vertex sets are U ∩ V ( F i ) for i = 1 , , · · · , t . Note that F = G h E ( F/E ′ ) i is a spanning forest of G [ U ]with | U | − n − n − · · · − n t + t components with the following orders: n , · · · , n t , , , · · · , | {z } | U |− t numbers . (3.3)By Theorem 1.1, we have τ G [ U ] ( F ) = | U | | U |− t − n −···− n t t Y i =1 n i . (3.4)Hence Claim 1 holds. Claim 2 : (3.1) holds when each F i is a star with a center at V − U .Assume that each F i is a star with a center at V − U , as shown in Figure 3. If G − U is an ✇ ✇✇ ✁✁✁✁✁ · · · · · ·· · · U · · · ✬✫ ✩✪ F F F t · · · · · · · · · Figure 3: Each F i is a star with a center in V − U independent set of G , then each F i is a star with a center in V − U and E ( F i ) ⊆ E G ( U ),implying that the claim holds by Claim 1.Now assume that V − U is not independent in G , i.e., E = E ( G − U ) = ∅ . Since eachcomponent F i is a star with a center in V − U , each edge e ∈ E is incident with two8ertices in V − U one of which is an end-vertex. Thus U is still a clique of G/E and F/E is a forest with t components F ′ , F ′ , · · · , F ′ t each of which is a star with a centerin V ( G/E ) − U and each edge in E ( F/E ) is incident with some vertex in U , where F ′ i = F i / ( E ∩ E ( F i )). By Claim 1, the result holds for G/E , i.e., τ G/E ( E ( F/E )) = | U | | U |− t − n ′ − n ′ −···− n ′ t t Y i =1 n ′ i , (3.5)where n ′ i = | U ∩ V ( F ′ i ) | . Clearly n ′ i = | U ∩ V ( F ′ i ) | = | U ∩ V ( F i ) | = n i . ApplyingLemma 2.1 (iv) repeatedly, we have τ G ( E ( F )) = τ G/E ( E ( F/E )). Thus Claim 2 holds. Claim 3 : (3.1) holds whenever F is a forest.Let W = E ( F ) = E ( G − G [ U ]) and G ′ = G ⋆ W − W . Observe that U is a clique of G ′ and G ′ [ E ( G ′ ) − E ( G ′ [ U ])] is a forest with t components F ′ , · · · , F ′ t each of which is a starwith a center in V ( G ′ ) − U such that V ( F ′ i ) ∩ U = V ( F i ) ∩ U holds for all i = 1 , , · · · , t .By Claim 2, the result holds for G ′ , i.e., τ G ′ ( W ′ ) = | U | | U |− t − n − n −···− n t t Y i =1 n i , (3.6)where W ′ = E ( G ⋆ W ) − W . By Lemma 2.2, we have τ G ( W ) = τ G⋆W − W ( W ′ ).Thus Claim 3 holds and the result is proved. ✷ Remark : For any forest M in K n with components M , · · · , M t , let G = K n ⋆ M . ByLemma 2.2, τ K n ( M ) = τ G ( W ), where W = E ( G ) − E ( K n ). Note that each component F i of G [ W ] is a star with a center in V ( G ) − U and E ( F i ) ⊆ E G ( U ), where U = V ( K n ) and V ( F i ) ∩ U = V ( M i ). Theorem 1.1 corresponds to Proposition 3.1 for the case that eachcomponent F i of F is a star with a center in V − U and E ( F i ) ⊆ E G ( U ). τ G ( M ∪ N ) and τ G • U ( N ) In this subsection, we assume that U is a clique of G and each component of G [ E G ( U )]is a star with a center in V − U . Note that each component of G [ E G ( U )] is a star witha center in V − U if and only if each vertex in U is incident with at most one edge in E G ( U ).Let u be the new vertex in G • U created after contracting all edges in E ( G [ U ]). So thevertex set of G • U is ( V − U ) ∪ { u } . Note that G • U may have parallel edges incident9ith vertex u , as all edges in E G ( U ) = E G ( U, V − U ) are the edges in G • U incident with u . For each v ∈ V − U , the number of parallel edges in G • U joining u and v is equal to | N G ( v ) ∩ U | .Now we are going to establish the main result in this section. Theorem 3.1 Let M = E G ( U ) and N ⊆ E ( G − U ) . If each component of G [ M ] is astar with a center in V − U , then τ G ( M ∪ N ) = | U | | U |− −| M | X T ∈ST G • U ( N ) | U | | E T ( u ) | . (3.7) Proof . By the given condition on M , G [ M ∪ N ] contains cycles if and only if G • U [ N ]contains cycles, implying that (3.7) holds whenever G [ M ∪ N ] contains cycles. Thus, itsuffices to consider the case that G [ M ∪ N ] is a forest. We will prove (3.7) by completingthe following claims. Claim 1 : (3.7) holds if G − E ( G [ U ]) is a forest and N = E ( G − U ). ✤✣ ✜✢ ✤✣ ✜✢✇ ✇✇ ✇ ✁✁✁✁✁ ✁✁✁✁✁ · · · · · ·· · · · · ·· · · · · · U · · · ✬✫ ✩✪ F F t Figure 4: G [ M ∪ N ] is forest with t components F , · · · , F t Assume that N = E ( G − U ) and G [ M ∪ N ] is a forest with components F , F , · · · , F t , asshown in Figure 4. By Proposition 3.1, τ G ( M ∪ N ) = | U | | U |− t − n −···− n t t Y i =1 n i , (3.8)where n i = | V ( F i ) ∩ U | for i = 1 , , · · · , t , implying that (3.7) holds if and only if thefollowing equality holds: | U | t − n −···− n t t Y i =1 n i = U −| M | X T ∈ST G • U ( N ) | U | | E T ( u ) | . (3.9)10y the given condition, each vertex in U is incident with at most one edge in M . Since n + n + · · · + n t is the number of vertices in U which are incident with edges in M , we have n + n + · · · + n t = | M | . For any T ∈ ST G • U ( N ), we have | E ( T ) ∩ E G ( U, V ( F i ) − U ) | = 1for all i = 1 , , · · · , t , implying that | E T ( u ) | = t .It remains to show that τ G • U ( N ) = Q ti =1 n i . Let T ∈ ST G • U ( N ). Observe that T − u isactually the graph G − U , which consists of t components F i − V ( F i ) ∩ U for i = 1 , , · · · , t .Also note that T contains exactly t edges e , e , · · · , e t , where each e i with u and somevertex in F i − V ( F i ) ∩ U for i = 1 , , · · · , t . Observe that each e i can be any one of theedges in the set M ∩ E ( F i ) whose size is exactly | V ( F i ) ∩ M | = n i . Hence τ G • U ( N ) = t Y i =1 | V ( F i ) ∩ U | = t Y i =1 n i . (3.10)Thus (3.9) holds and Claim 1 follows. Claim 2 : (3.7) holds for any N ⊆ E ( G − U ) such that G [ M ∪ N ] is a forest.For any T ∈ ST G ( M ∪ N ), T − U is a forest with N ⊆ E ( T − U ). Let E = E ( G ( U ))and let N be the family of those subsets N ′ of E ( G − U ) with N ⊆ N ′ such that G h N ′ i is a forest and G h M ∪ N ′ ∪ E i is connected.Clearly, ST G h M ∪ N ∪ E i ( M ∪ N ) and ST G h M ∪ N ∪ E i ( M ∪ N ) are disjoint for any pair ofdistinct members N , N ∈ N , and ST G ( M ∪ N ) = [ N ′ ∈N ST G h M ∪ N ′ ∪ E i ( M ∪ N ′ ) . (3.11)Similarly, for any pair of distinct members N , N ∈ N , ST G • U h M ∪ N i ( N ) and ST G • U h M ∪ N i ( N )are disjoint, and ST G • U ( N ) = [ N ′ ∈N ST G • U h M ∪ N ′ i ( N ′ ) . (3.12)By Claim 1, the following identity holds for any N ′ ∈ N : τ G h M ∪ N ′ ∪ E i ( M ∪ N ′ ) = | U | | U |− −| M | X T ∈ST G • U h M ∪ N ′i ( N ′ ) | U | | E T ( u ) | . (3.13)Thus Claim 2 follows from (3.11), (3.12) and (3.13). ✷ V , V , · · · , V k are disjoint cliques of G In this section, we always assume that G = ( V, E ) is a connected and loopless multigraph,where V is partitioned into non-empty subsets V , V , · · · , V k satisfying the following con-11itions:(i) V i is a clique for all i = 1 , , · · · , k ;(ii) E G ( V i , V j ) = ∅ for each pair i, j with 1 ≤ i < j ≤ k ;(iii) each component of G [ M ] is a star with a center in V , where M = S ≤ i ≤ k E G ( V , V i ).The structure of G under conditions (i), (ii) and (iii) above is as shown in Figure 5(a).Note that condition (iii) above is equivalent to that each vertex in V i is incident with atmost one edge in M for all i = 1 , , · · · , k . All parallel edges of G must be in the subgraph G [ V ]. ✤✣ ✜✢ ✤✣ ✜✢ · · · V V V k V V t tt tt t t ✉t · · · v v v k · · · · · · · · ·· · · · · ·· · · · · · · · · (a) G (b) G • U Figure 5: G − V consists of k cliques and G [ M ] consists of stars with centers in V Let M i = E G ( V i , V ) for all i = 1 , , · · · , k . Then M = E G ( V ) = S ≤ i ≤ k M i . In this section,our main purpose is to apply Theorem 3.1 to find an expression for τ G ( M ∪ N ) for any N ⊆ E ( G [ V ]). Applying this result, we are able to get an expression of τ G ( R ∪ N ) forany R ⊆ M . τ G ( M ∪ N ) for N ⊆ E ( G [ V ]) Let U = V ∪ · · · ∪ V k . Recall that G • U is defined to be the graph G/E ( G [ U ]). As G [ U ]has k components G [ V ] , G [ V ] , · · · , G [ V k ], G • U can be obtained from G by removingall edges in G [ U ] and identifying all vertices in each V i as one vertex, denoted by v i , for i = 1 , , · · · , k , as shown in Figure 5 (b). Thus V ( G • U ) = V ∪ { v , v , · · · , v k } and E ( G • U ) = E ( G ) − S ≤ i ≤ k E ( G [ V i ]) = M ∪ E ( G [ V ]).12 heorem 4.1 For any N ⊆ E ( G [ V ]) , τ G ( M ∪ N ) = k Y i =1 | V i | | V i |− −| M i | X T ∈ST G • U ( N ) k Y i =1 | V i | | E T ( v i ) | , (4.1) where M i = E G ( V i , V ) for i = 1 , , · · · , k and M = M ∪ · · · ∪ M k .Proof . If k = 1, the result follows directly from Theorem 3.1. Assume that the resultholds for k < n , where n ≥ 2. Now consider the case that k = n .By Theorem 3.1, τ G ( M ∪ N ) | V k | | V k |− −| M k | = X T ∈ST G • Vk (( M − M k ) ∪ N ) | V k | | E T ( v k ) | = X B k ⊆ M k X T ∈ST G • Vk (( M − Mk ) ∪ N ) ET ( vk )= Mk − Bk | V k | | M k − B k | = X B k ⊆ M k | V k | | M k − B k | τ ( G − B k ) • V k (( M − B k ) ∪ N ) , (4.2)where the last equality follows from the fact that T ∈ ST G • V k (( M − M k ) ∪ N ) with E T ( v k ) = M k − B k if and only if T ∈ ST ( G − B k ) • V k (( M − B k ) ∪ N ), as M − B k =( M − M k ) ∪ ( M k − B k ).For any B k ⊆ M k , we have M − B k = M ∪ · · · ∪ M k − ∪ ( M k − B k ). By the inductiveassumption, τ ( G − B k ) • V k (( M − B k ) ∪ N ) Q k − i =1 | V i | | V i |− −| M i | = X T ∈ST ( G − Bk ) • U ( N ∪ ( M k − B k )) k − Y i =1 | V i | | E T ( v i ) | . (4.3)By (4.2) and (4.3), we have τ G ( M ∪ N ) Q ki =1 | V i | | V i |− −| M i | = X B k ⊆ M k X T ∈ST ( G − Bk ) • U ( N ∪ ( M k − B k )) | V k | | M k − B k | k − Y i =1 | V i | | E T ( v i ) | = X B k ⊆ M k X T ∈ST ( G − Bk ) • U ( N ) ET ( vk )= Mk − Bk | V k | | E T ( v k ) | k − Y i =1 | V i | | E T ( v i ) | = X T ∈ST G • U ( N ) k Y i =1 | V i | | E T ( v i ) | . (4.4)13 Now we give a proof of Theorem 1.2 by applying Theorem 4.1 directly. Proof of Theorem 1.2. Assume that G = ( V, E ) is a graph satisfying the conditionsassumed in the beginning of this section, V is an independent set of G and each componentof G [ M ] is a star of size 2.Let V = { w , w , · · · , w r } . Then G [ M ] consists of exactly r components S , S , · · · , S r which are stars of size 2 with centers w , w , · · · , w r respectively. Let e i, and e i, denotethe two edges in S i .Clearly, { e i, , e i, } ∩ E ( T ) = ∅ holds for any 1 ≤ i ≤ r and any T ∈ ST G • U . For any I ⊆ { , , · · · , r } , let ST IG • U be the set of members T ∈ ST G • U such that { ≤ i ≤ r : { e i, , e i, } ⊆ E ( T ) } = I. Observe that ST IG • U = ∅ if and only if the edge set { e i, , e i, : i ∈ I } induces a spanningtree of the subgraph ( G • U ) − { w j : 1 ≤ j ≤ r, j / ∈ I } . When ST IG • U = ∅ , there areexactly 2 r −| I | members in ST IG • U and for each T ∈ ST IG • U , | E ( T ) ∩ { e i, , e i, }| = 1 holdsfor all i ∈ { , , · · · , r } − I .Let G ′ be the graph obtained from G by contracting exactly one edge in each S i for all i = 1 , , · · · , r , and let M ′ = M ∩ E ( G ′ ). By Lemma 2.1 (iv), τ G ′ ( M ′ ) = τ G ( M ).For any e ∈ M , let l ( e ) = i such that e is incident with a vertex in V i . By Theorem 4.1, τ G ( M ) = k Y i =1 | V i | | V i |− −| M i | X T ∈ST G • U k Y i =1 | V i | | E T ( v i ) | = k Y i =1 | V i | | V i |− X I ⊆{ , , ··· ,r } X T ∈ST IG • U k Y i =1 | V i | −| M i − E T ( v i ) | = k Y i =1 | V i | | V i |− X I ⊆{ , , ··· ,r } X T ∈ST IG • U Y e ∈ M − E ( T ) | V l ( e ) | − = k Y i =1 | V i | | V i |− X I ⊆{ , , ··· ,r }ST IG • U = ∅ Y i ∈{ , , ··· ,r }− I ( | V l ( e i, ) | − + | V l ( e i, ) | − )= k Y i =1 | V i | | V i |− X T ′ ∈ST G ∗ Y e ∈ M ′ − E ( T ′ ) ( | V a ( e ) | − + | V b ( e ) | − ) , (4.5)where G ∗ is the graph obtained from G ′ by identifying all vertices in each V i as a vertex,14enoted by v i , and removing all loops, and a ( e ) and b ( e ) are numbers in { , , · · · , k } suchthat v a ( e ) and v b ( e ) are the two ends of e in G ∗ which correspond to V a ( e ) and V b ( e ) .Since |ST G ′ ( M ′ ) | = τ G ( M ), Theorem 1.2 is proven by (4.5). ✷ τ G ( R ∩ N ) for R ⊆ M and N ⊆ E ( G − U ) In this subsection, we will find an expression for τ G ( R ∪ N ) for any R ⊆ E G ( U ) and N ⊆ E ( G − U ). Theorem 4.2 For any R ⊆ M and N ⊆ E ( G − U ) , τ G ( R ∪ N ) = k Y i =1 | V i | | V i |− X T ∈ST G • U ( N ) k Y i =1 | V i | −| M i − E ( T ) | (1 + | V i | ) | ( M i − R ) − E ( T ) | . (4.6) Proof . Let R be a fixed subset of M . Note that T ∈ ST G ( R ∪ N ) if and only if T ∈ ST G − B (( M − B ) ∪ N ) for some B with B ⊆ M − R . Obviously, for distinct subsets B , B of M − R , ST G − B (( M − B ) ∪ N ) and ST G − B (( M − B ) ∪ N ) are disjoint. Thus, τ G ( R ∪ N ) = X B ⊆ M − R τ G − B (( M − B ) ∪ N ) . (4.7)Then, by Theorem 4.1, τ G ( R ∪ N ) = X B ⊆ M − R X T ∈ST ( G • U ) − B ( N ) k Y i =1 | V i | | V i |− −| M i − B | + | ( M i − B ) ∩ E T ( v i ) | = k Y i =1 | V i | | V i |− X B ⊆ M − R X T ∈ST ( G • U ) − B ( N ) k Y i =1 | V i | −| M i − B | + | ( M i − B ) ∩ E T ( v i ) | = k Y i =1 | V i | | V i |− X B ⊆ M − R X R ′ ⊆ M − B X T ∈ST ( G • U ) − B ( N ) E ( T ) ∩ M = R ′ k Y i =1 | V i | −| M i − B | + | R ′ ∩ M i | = k Y i =1 | V i | | V i |− X R ′ ⊆ M k Y i =1 | V i | | R ′ ∩ M i | Φ( R ′ ) (4.8)where Φ( R ′ ) = X B ⊆ M − ( R ′ ∪ R ) X T ∈ST ( G • U ) − B ( N ) E ( T ) ∩ M = R ′ k Y i =1 | V i | −| M i − B | k Y i =1 | V i | −| R ∩ M i | X B ⊆ M − ( R ′ ∪ R ) X T ∈ST ( G • U ) − B ( N ) E ( T ) ∩ M = R ′ k Y i =1 | V i | −| ( M i − B ) − R | = k Y i =1 | V i | −| R ∩ M i | X T ∈ST G • U ( N ) E ( T ) ∩ M = R ′ X B ⊆ M − ( R ′ ∪ R ) k Y i =1 | V i | −| ( M i − R ) − B | = k Y i =1 | V i | −| ( R ∪ R ′ ) ∩ M i | X T ∈ST G • U ( N ) E ( T ) ∩ M = R ′ X B ⊆ M − ( R ′ ∪ R ) k Y i =1 | V i | −| M i − ( R ∪ R ′ ) − B | = k Y i =1 | V i | −| ( R ∪ R ′ ) ∩ M i | X T ∈ST G • U ( N ) E ( T ) ∩ M = R ′ k Y i =1 (1 + | V i | − ) | M i − ( R ∪ R ′ ) | . (4.9)By (4.9) and (4.8), we have τ G ( R ∪ N ) Q ki =1 | V i | | V i |− = X R ′ ⊆ M X T ∈ST G • U ( N ) E ( T ) ∩ M = R ′ k Y i =1 | V i | −| ( R − R ′ ) ∩ M i | (1 + | V i | − ) | M i − ( R ∪ R ′ ) | = X T ∈ST G • U ( N ) k Y i =1 | V i | −| ( M i ∩ R ) − E ( T ) | (1 + | V i | − ) | ( M i − R ) − E ( T )) | . (4.10)Thus we can verify that the result holds. ✷ When R = ∅ , a direct application of Theorem 4.2 gives an expression for τ G ( N ). Corollary 4.1 For any N ⊆ E ( G − U ) , τ G ( N ) = k Y i =1 | V i | | V i |− X T ∈ST G • U ( N ) k Y i =1 (1 + 1 / | V i | ) | M i − E ( T ) | . (4.11) Note that (4.6) can be changed to τ G ( R ∪ N ) = k Y i =1 | V i | | V i |− −| M i | (1 + | V i | ) | M i − R | × X T ∈ST G • U ( N ) k Y i =1 | V i | | M i ∩ R ∩ E ( T ) | (cid:18) | V i | | V i | (cid:19) | ( M i − R ) ∩ E ( T ) | . (4.12)16or any R ⊆ M , let ω be the mapping from E ( G • U ) (i.e., M ∪ E ( G − U )) to N = { , , , · · ·} defined below: ω R ( e ) = | V i | , e ∈ M i ∩ R ; | V i | / (1 + | V i | ) , e ∈ M i − R ;1 , otherwise . (4.13)Then (4.6) can be expressed as τ G ( R ∪ N ) = k Y i =1 | V i | | V i |− −| M i | (1 + | V i | ) | M i − R | X T ∈ST G • U ( N ) Y e ∈ E ( T ) ω R ( e ) . (4.14)Let w , w , · · · , w r be the vertices in the set N G ( U ) − U . These vertices are actually centersof the components of G [ M ], as each component of G [ M ] is a star.As there may be more than one edge in E G • U ( v i , w j ) ∩ R or E G • U ( v i , w j ) − R for 1 ≤ i ≤ k and 1 ≤ j ≤ r , (4.14) can be further simplified.Given R ⊆ M , let G ◦ R U denote the graph obtained from G • U by removing | E G • U ( v i , w j ) ∩ R |− E G • U ( v i , w j ) ∩ R whenever | E G • U ( v i , w j ) ∩ R | ≥ | E G • U ( v i , w j ) − R | − E G • U ( v i , w j ) − R whenever | E G • U ( v i , w j ) − R | ≥ i, j : 1 ≤ i ≤ k and 1 ≤ j ≤ r . Thus, in the graph G ◦ R U , there are at most two paralleledges joining each pair of vertices v i and w j . If this case happens, then exactly one of thetwo edges joining v i and w j is contained in R .Let ω ′ R be the mapping from E ( G ◦ R U ) to N = { , , , · · ·} defined below: ω ′ R ( e ) = | V i | · | E G • U ( v i , w j ) ∩ R | , e ∈ R and e joins v i and w j ; | V i | | V i | · | E G • U ( v i , w j ) − R | , e / ∈ R and e joins v i and w j ;1 , otherwise , (4.15)where 1 ≤ i ≤ k and 1 ≤ j ≤ r . Then (4.14) can be replaced by the following expression: τ G ( R ∪ N ) = k Y i =1 | V i | | V i |− −| M i | (1 + | V i | ) | M i − R | X T ∈ST G ◦ RU ( N ) Y e ∈ E ( T ) ω ′ R ( e ) . (4.16)17 When E , E , · · · , E k is a partition of E such thateach G [ E i ] is a complete graph τ G = |S T G ⋄S | holds for a graph G ⋄ S Let v be any vertex in G and E ⊆ E G ( v ). Let G v⊳E denote the graph obtained from G − ( E G ( v ) − E ) by adding a new vertex v ′ and a new edge joining v and v ′ and finallychanging the end v of all edges in E G ( v ) − E to v ′ , as shown in Figure 6. Clearly G ⊳ E /e ′ ∼ = G , where e ′ = vv ′ is the only edge in E ( G ⊳ E ) − E ( G ). By Lemma 2.1 (iv), τ G ( W ) = |ST G v⊳E ( W ∪ e ′ ) | holds. ✈ ✈✈ v v ′ e e e e e s e s e ′ e ′ e ′ e ′ e ′ t e ′ t ... ...... ... v e ′ (a) G (b) G v⊳E Figure 6: Graphs G and G v⊳E , where E = { e , e , · · · , e s } For any subgraph G of G , let G ⋄ G be the graph below: G ⋄ G = ( · · · (( G v ⊳E ) v ⊳E ) · · · ) v r ⊳E r , where v , v , · · · , v r are those vertices in G with E G ( v i ) = E G ( v i ) and E i = E G ( v i ).Clearly, G is the subgraph of G ⋄ G induced by V ( G ) and the edges in E ( G ⋄ G ) − E ( G ) form a matching of G ⋄ G . An example is shown in Figure 7, where G = G [ { v , v , v , v , v , v } ] − { v v , v v } and the new edges in G ⋄ G are expressed by lines.If G ∼ = K and G is a 5-cycle, then G ⋄ G is the Petersen graph.Note that E ( G ⋄ G ) − E ( G ) is a matching of G ⋄ G . Since G is actually the graph obtainedfrom G ⋄ G by contracting all edges in E ( G ⋄ G ) − E ( G ), applying Lemma 2.1 (iv)repeatedly on all edges in E ( G ⋄ G ) − E ( G ) implies the following result. Lemma 5.1 Let G be a subgraph of G and M = E ( G ⋄ G ) − E ( G ) . Then, for any N ⊆ E ( G ) , τ G ( N ) = τ G ⋄ G ( M ∪ N ) . (5.1)18 ✈✈ ✈✈ ✈✈ ✈✈ ✈✈ ✈✈ ✈ ✈✈✈✈ ✈✈✈ ✈ ✈✈ ✈✈✈ ✈ ✈✈ v v v v v v v v v v v v v ′ v ′ v ′ v ′ v ′ · · · · ······· ······ ················ v ′ (a) G is a subgraph of G (b) G ⋄ G Figure 7: Graphs G and G ⋄ G For a family S = { E , E , · · · , E k } of pairwise disjoint subsets of E ( G ), let G ⋄S denote thefollowing graph obtained by a sequence of ⋄ -operations on subgraphs G [ E ] , G [ E ] , · · · , G [ E k ]:( · · · (( G ⋄ G [ E ]) ⋄ G [ E ]) · · · ) ⋄ G [ E k ] . (5.2)Note that G ⋄ S is irrelevant to the order of E , E , · · · , E k and ( G ⋄ S ) /W ∼ = G , where W = E ( G ⋄ S ) − E ( G ). An example of G ⋄ S is shown in Figure 8, where S = { E , E , E , E } , E = E ( G [ v , v , v ]), E = E ( G [ v , v , v , v ]), E = E ( G [ v , v , v ])and E = E ( G [ v , v , v , v ]). ✈ ✈✈ ✈✈ ✈✈ ✈✈ ✈ ✈✈ ✈ ✈✈ ✈ ✈✈ ✈✈✈ ✈✈ v v v v v v v v v t ttt t v , v , v , v , v , v , v , v , v , v , v , v , v , v , w w w w w (a) G (b) G ⋄ S Figure 8: Graphs G and G ⋄ S Assume that V ( G ) = { v , v , · · · , v n } . By the above definition, if S = { E , E , · · · , E k } is19 partition of E ( G ), then G ⋄ S is actually the graph with vertex set: { w i : v i ∈ V ′ } ∪ k [ j =1 V j , (5.3)where V ′ = V ( G ) − { v i ∈ V ( G ) : ∃ j, N G ( v i ) = E j } and V j = { v i,j : v i ∈ V ( G [ E j ]) } , andedge set: k [ j =1 E ′ j ∪ [ v i ∈ V ′ { w i v i,j : E G ( v i ) ∩ E j = ∅ , ≤ j ≤ k } , (5.4)where each E ′ j is a copy of E j by changing the ends v s and v t of each edge e in E j to v s,j and v t,j . As each edge in G has exactly one copy in G ⋄ S , E ( G ⋄ S ) is also considered asthe union of E ( G ) and S v i ∈ V ′ { w i v i,j : E G ( v i ) ∩ E j = ∅ , ≤ j ≤ k } . An example for thelabels of vertices and edges in G ⋄ S is given in Figure 8.Some basic facts on G ⋄ S follow directly. Lemma 5.2 Let M = E ( G ⋄ S ) − E ( G ) . Then (i) | V ( G ⋄ S ) − V ( G ) | = | V ′ | ; (ii) the component number of ( G ⋄ S )[ M ] is equal to | V ′ | ; (iii) each component of ( G ⋄ S )[ M ] is a star S j with its center w j ∈ V ( G ⋄ S ) − V ( G ) and its size equal to the number of different sets E i with E G ( v j ) ∩ E i = ∅ ; (iv) { w i : v i ∈ V ′ } is an independent set in G ⋄ S and its removal from G ⋄ S results in k components isomorphic to G [ E ] , G [ E ] , · · · , G [ E k ] respectively; (v) G ⋄ S [ V j ] ∼ = G [ E j ] for each j ; (vi) E G ⋄S ( V j , V j ) = ∅ for all ≤ j < j ≤ k . Applying Lemma 5.1, we get the following conclusion. Lemma 5.3 For any partition S = { E , E , · · · , E k } of E ( G ) and any N ⊆ E ( G ) , wehave τ G ( N ) = τ G ⋄S ( M ∪ N ) , (5.5) where M = E ( G ⋄ S ) − E ( G ) . G [ E i ] is a complete graph in G for all i = 1 , , · · · , k , applying Lemma 5.3 and Theo-rem 4.1 gets the following expression on τ G . Theorem 5.1 Assume that S = { E , E , · · · , E k } is a partition of E ( G ) such that G [ E i ] is a complete graph for all i = 1 , , · · · , k . Then τ G = k Y i =1 n n i − −| n ′ i | i X T ∈ST ( G ⋄S ) /E ( G ) k Y i =1 n | E T ( v ′ i ) | i , (5.6) where n i = | V ( G [ E i ]) | , n ′ i is the size of the set { v s ∈ V ( G [ E i ]) : E G ( v s ) E i } and v ′ i isthe new vertex in ( G ⋄ S ) /E ( G ) produced by contracting V i in G ⋄ S . Note that ( G ⋄ S ) /E ( G ) is the bipartite graph with a bipartition { w i : v i ∈ V ′ } and { v ′ j : 1 ≤ j ≤ k } and edge set { w i v ′ j : E G ( v i ) ∩ E j = ∅} . Obviously Theorem 5.1 can be applied to the graph in Figure 8 (a). Actually this graphis the middle graph of K − e (i.e., the graph with one edge removed from K ). For agraph H with vertex set { u , u , · · · , u k } , the middle graph of H , denoted by M ( H ), isthe graph obtained from its line graph L ( H ) and the empty graph H − E ( H ) by addingedges joining each vertex u i in H − E ( H ) to all those vertices in L ( H ) which correspondto edges in the set E H ( u i ).Applying generalized Wye-Delta transform and Delta-Wye transform, Yan [16] gave arelation between ST M ( H ) and ST S ( H ) , where S ( H ) is the graph obtained from H by sub-dividing each edge in H exactly once. Such a relation actually follows from Theorem 5.1directly. Observe that the edge set of M ( H ) has a partition S = { E , E , · · · , E k } , whereeach E i is the set of edges in the subgraph of M ( H ) induced by { u i } ∪ N M ( H ) ( u i ). Clearlyeach M ( H )[ E i ] is a complete graph of order d H ( u i ) + 1 and contains exactly d H ( u i ) ver-tices u such E M ( H ) ( u ) M ( H )[ E i ]. Also note that M ( H ) ⋄ S /E ( M ( H )) is actually thegraph S ( H ). Thus, applying Theorem 5.1, we have τ M ( H ) = X T ∈ST S ( H ) k Y i =1 ( d H ( u i ) + 1) | E T ( u i ) |− . (5.7)Similarly, a relation between ST L ( H ) and ST S ( H ) can be obtained: τ L ( H ) = X T ∈ST S ( H ) k Y i =1 d H ( u i ) | E T ( u i ) | . (5.8)21t is not difficult to verify that (1.3) can be obtained from (5.8). τ G ( W ) If a simple graph G = ( V, E ) contains a clique U and a partition S , S of V − U with S ∩ N G ( S ) = ∅ , then the chromatic polynomial χ ( G, λ ) has the following factorizationdue to Zykov [18] (see [4, 13] also): χ ( G, λ ) = χ ( G [ U ∪ S ] , λ ) χ ( G [ U ∪ S ] , λ ) χ ( G [ U ] , λ ) . (6.1)In this section, we find a similar expression for τ G ( W ) for any W ⊆ E ( G ) − E ( G [ U ]) byapplying results in Sections 2, 3 and 4. Theorem 6.1 Let G = ( V, E ) be a connected and loopless multigraph and U be a cliqueof G . If S and S form a partition of V − U with N G [ S ] ∩ N G [ S ] = ∅ , as shown inFigure 9, then, for any W ⊆ E − E ( G [ U ]) , τ G ( W ) = τ G [ U ∪ S ] ( W ) · τ G [ U ∪ S ] ( W ) | U | | U |− , (6.2) where W i = W ∩ E ( G [ U ∪ S i ]) . US S ssssssssssss ... ... ❤❤❤❤❤❤❤ Figure 9: A clique U and a partition S , S of V − U with N G [ S ] ∩ N G [ S ] = ∅ Proof . Let M = E G ( U ), R = W ∩ M , G i = G [ U ∪ S i ], M i = M ∩ E ( G i ), R i = R ∩ E ( G i )and N i = ( W − R ) ∩ E ( G i ) for i = 1 , 2. Thus W i is the disjoint union of R i and N i . Wewill prove this result by the following claims. Claim 1 : (6.2) holds if each component of G [ M ] is a star with a center in S ∪ S .Let u i represent the vertex in G i • U after identifying all vertices in U as one vertex.For T i ∈ ST G i • U ( N i ) for i = 1 , 2, let T · T denote the tree obtained from T and T by22dentifying u and u as one vertex. By the definition of G • U and the given condition,we have ST G • U ( N ∪ N ) = { T · T : T i ∈ ST G i • U ( N i ) , i = 1 , } . Thus, for i = 1 , 2, by Theorem 4.2 with k = 1, τ G i ( W i ) = | U | | U |− X T i ∈ST Gi • U ( N i ) | U | −| M i − E ( T i ) | (1 + | U | ) | ( M i − R i ) − E ( T i ) | (6.3)and τ G ( W ) = | U | | U |− X T ∈ST G • U ( N ∪ N ) | U | −| M − E ( T ) | (1 + | U | ) | ( M − ( R ∪ R )) − E ( T ) | = | U | | U |− Y i =1 X T i ∈ST Gi • U ( N i ) | U | −| M i − E ( T i ) | (1 + | U | ) | ( M i − R i ) − E ( T i ) | = τ G ( W ) τ G ( W ) | U | | U |− . (6.4)Thus Claim 1 holds. Claim 2 : (6.2) holds if R = M and G [ M ] is a forest.Let G ′ = G ⋆ W − M and W ′ = E ( G ⋆ W ) − E ( G ). For i = 1 , 2, let W i = W ∩ E ( G i ), G ′ i = G i ⋆ W i − R i and W ′ i = E ( G ⋆ W i ) − E ( G i ). By Lemma 2.3 (i), τ G ( W ) = τ G ′ ( W ′ ) | . (6.5)and τ G i ( W i ) = |ST G ′ i ( W ′ i ) | , i = 1 , . (6.6)Note that U is a clique of G ′ and G ′ [ E G ′ ( U )] is a star. By Claim 1, τ G ′ ( W ′ ) | = τ G ′ ( W ′ ) τ G ′ ( W ′ ) | U | | U |− . (6.7)Thus, Claim 2 follows from (6.5), (6.6) and (6.7). Claim 3 : (6.2) holds.For any M ′ ⊆ M , G [ M ′ ] is not a forest if and only if G i [ M ′ i ] is not a forest for some i = 1 , 2, where M ′ i = M ′ ∩ E G ( U ). Thus τ G ( M ′ ∩ N ) = τ G ( M ′ ∩ N ) τ G ( M ′ ∩ N ) | U | | U |− = 0 . (6.8)23et R be a fixed subset of M such that G [ R ] is a forest. For i = 1 , 2, let R i = { R ′ i : R i ⊆ R ′ i ⊆ M ∩ E ( G i ) , G [ R ′ i ] is a forest } , where R i = R ∩ E ( G i ). Note that ST G ( R ∪ N ) = [ R ′ i ∈R ii =1 , ST G − ( M − ( R ′ ∪ R ′ )) ( R ′ ∪ R ′ ∪ N ∪ N ) , (6.9)where for distinct order pairs ( R ′ , R ′ ) and ( R ′′ , R ′′ ) in the above union, the two corre-sponding sets ST G − ( M − ( R ′ ∪ R ′ )) ( R ′ ∪ R ′ ∪ N ∪ N ) and ST G − ( M − ( R ′′ ∪ R ′′ )) ( R ′′ ∪ R ′′ ∪ N ∪ N )are disjoint. Similarly, ST G i ( R i ∪ N i ) = [ R ′ i ∈R i ST G i − ( M ∩ E ( G i ) − R ′ i ) ( R ′ i ∪ N i ) , i = 1 , , (6.10)where the above union is disjoint union for both i = 1 , 2. By Claim 2, for any R ′ i ∈ R i for i = 1 , 2, we have τ G − ( M − ( R ′ ∪ R ′ )) ( R ′ ∪ R ′ ∪ N ∪ N ) = Q ≤ i ≤ τ G i − ( M ∩ E ( G i ) − R ′ i ) ( R ′ i ∪ N i ) | U | | U |− . (6.11)Thus, Claim 3 follows from (6.9), (6.10) and (6.11), and the result is proven. ✷ Corollary 6.1 Let G = ( V, E ) be any connected multigraph and U be a clique of G . If w is a vertex in V − U with N G [ w ] ∩ N G [ V − ( U ∪ { w } )] = ∅ , then τ G = τ G − w (cid:0) d ( w )(1 + 1 / | U | ) d ( w ) − (cid:1) . (6.12) Proof . Let S = { w } and S = V − { w } . As N G [ w ] ∩ N G [ V − ( U ∪ { w } )] = ∅ , byapplying Theorem 6.1, τ G = τ G · τ G | U | | U |− , (6.13)where G = G [ U ∪ { w } ] and G = G [ U ∪ S ] = G − w . By Theorem 4.2, τ G = d ( w ) · | U | | U |− · (1 + 1 / | U | ) d ( w ) − . (6.14)The result then follows from (6.13) and (6.14). ✷ Remarks : (a) The condition for (6.1) is weaker than the one for (6.2), as (6.2) holdswith an extra condition N G [ S ] ∩ N G [ N ] = ∅ .(b) Note that when W = ∅ , (6.2) is equivalent to the following equality T G ( x, y ) T K | U | ( x, y ) = T G [ U ∪ S ] ( x, y ) T G [ U ∪ S ] ( x, y ) (6.15)24hen ( x, y ) = (1 , T G ( x, y ) is the Tutte polynomial of G .(c) Under the condition that S ∩ N G ( S ) = ∅ , (6.1) implies that (6.15) holds when y = 0,as T G (1 − x, 0) = x − c ( G ) ( − | V |− c ( G ) χ ( G, x ) holds for any simple graph G (see [2, 6]),where c ( G ) is the number of components of G . Furthermore, (6.15) also holds for graph G satisfying condition N G [ S ] ∩ N G [ N ] = ∅ when ( x, y ) = (2 , T G (2 , 2) = 2 | E ( G ) | holds for any graph G . We have verified (6.15) for some graphs G satisfying the samecondition when ( x, y ) = (0 , − G satisfyingthis condition. Problem 6.1 Let U be a clique of G = ( V, E ) . 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