The Bipartition Polynomial of a Graph: Reconstruction, Decomposition, and Applications
aa r X i v : . [ m a t h . C O ] F e b The Bipartition Polynomial of a Graph:Reconstruction, Decomposition, and Applications
Seongmin OkKorea Institute for Advanced Study, SeoulPeter TittmannUniversity of Applied Sciences MittweidaApril 6, 2018
Abstract
The bipartition polynomial of a graph, introduced in [Dod+15], is ageneralization of many other graph polynomials, including the domina-tion, Ising, matching, independence, cut, and Euler polynomial. We showin this paper that it is also a powerful tool for proving graph properties.In addition, we can show that the bipartition polynomial is polynomiallyreconstructible, which means that we can recover it from the multiset ofbipartition polynomials of one-edge-deleted subgraphs.
The bipartition polynomial B ( G ; x, y, z ) of a simple graph G has been intro-duced in [Dod+15]. The bipartition polynomial is related to the set of bipartitesubgraphs of G ; it generalizes the Ising polynomial [AM09], the matching poly-nomial [Far79], the independence polynomial (in case of regular graphs) [LM05;GH83], the domination polynomial [AL00], the Eulerian subgraph polynomial[Aig07], and the cut polynomial of a graph. In this paper, we consider the nat-ural generalization of the bipartition polynomial to graphs with parallel edges.Let G = ( V, E ) be a simple undirected graph with vertex set V and edge set E . The open neighborhood of a vertex v of G is denoted by N ( v ) or N G ( v ). Itis the set of all vertices of G that are adjacent to v . The closed neighborhoodof v is defined by N G [ v ] = N G ( v ) ∪ { v } . The neighborhood of a vertex subset W ⊆ V is: N G ( W ) = [ w ∈ W N G ( w ) \ W,N G [ W ] = N G ( W ) ∪ W. The edge boundary ∂W of a vertex subset W of G is ∂W = {{ u, v } | u ∈ W and v ∈ V \ W } , i.e., the set of all edges of G with exactly one end vertex in W . Throughoutthis paper, we denote by n the order, n = | V | , by m the size, m = | E | , and by k ( G ) the number of components of G . 1he bipartition polynomial of a graph G is defined by B ( G ; x, y, z ) = X W ⊆ V x | W | X F ⊆ ∂W y | N ( V,F ) ( W ) | z | F | . (1)Note that the definitions of neighborhood, edge boundary and Equation (1) canbe easily extended to graphs with parallel edges. From now on, unless otherwisestated, we allow graphs to have parallel edges. Note that adding loops does notchange the bipartition polynomial. FW Y
Figure 1: Illustration of the definition of the bipartition polynomialFigure 1 provides an illustration of the definition given in Equation (1).First we select a vertex subset W , which is located within the left gray-shadedbubble in Figure 1. The cardinality of the set W is counted in the exponent ofthe variable x . The edge boundary ∂W consists of all edges that stick out fromthat bubble. Assume we select the edges shown in bold as subset F ⊆ ∂W .The end vertices of these edges outside W are presented within the next bubblethat is labeled with Y . The cardinality of the set Y is counted in the exponentof variable y of the bipartition polynomial. The third variable, z , counts theedges in F . We see that by the definition of F always a bipartite subgraph of G is defined, which is the reason for the naming ‘bipartition polynomial’. If H = ( S ∪ T, F ) is a connected bipartite graph, then the partition sets S and T are uniquely defined (up to order).Equation (1) implies that we can derive the order and size of a graph fromits bipartition polynomial: n = deg B ( G ; x, ,
1) (2) m = 12 [ xyz ] B ( G ; x, y, z ) , (3)where [ xyz ] B ( G ; x, y, z ) denotes the coefficient of xyz in B . Proposition 1
A loopless graph G is bipartite if and only if
12 [ xyz ] B ( G ; x, y, z ) = deg B ( G ; 1 , , z ) . Proof.
The left-hand side is, according to Equation (3), the number of edgesof G . A graph G = ( V, E ) is bipartite if and only if there is a vertex subset W ⊆ V with ∂W = E . Equation (1) shows that in this case the degree of z in B ( G ; x, y, z ) is equal to m . 2n the remaining part of this paper, we present different representation anddecompositions of the bipartition polynomial (Section 2), derive relations toother graph polynomials (Section 3), prove its polynomial reconstructibility(Section 4), and provide some applications for proving graph properties (Section5). In this section, we provide some different representations of the bipartition poly-nomial and decomposition formulae with respect to vertex and edge deletions.
Theorem 2 (product representation, [Dod+15])
The bipartition polyno-mial of a graph G can be represented as B ( G ; x, y, z ) = X W ⊆ V x | W | Y v ∈ N G ( W ) h y h (1 + z ) | ∂v ∩ ∂W | − i + 1 i . (4) The bipartition polynomial of a simple graph G = ( V, E ) satisfies B ( G ; x, y, z ) = X W ⊆ V x | W | Y v ∈ N G ( W ) h y h (1 + z ) | N G ( v ) ∩ W | − i + 1 i . (5) Corollary 3
The number of components of a graph G is log B ( G ; 1 , , − . Proof.
From Equation (4), we obtain B ( G ; 1 , , −
1) = X W ⊆ V Y v ∈ N G ( W ) | ∂v ∩ ∂W | . The product vanishes for all W ⊆ V with N G ( W ) = ∅ , since ∂v ∩ ∂W = ∅ forall v ∈ N G ( W ). The product equals 1 if N G ( W ) is empty. We have N G ( W ) = ∅ if and only if W is the (possibly empty) union of vertex sets of components of G . For a graph with k components, there are 2 k ways to form a union of thevertex sets of the components. Hence we obtain B ( G ; 1 , , −
1) = 2 k .The proof of the last proposition also yields the following statement. Corollary 4 If G consists of k components G , ..., G k such that the order of G i is k i , then B ( G ; x, , −
1) = k Y i =1 (1 + x k i ) . Consequently, we can derive the order of all components of G by the followingsimple procedure. The order of the first component is the smallest positivepower, say k , of x in B ( G ; x, , − B ( G ; x, , −
1) by (1 + x k )and proceed step by step with the resulting polynomial in the same manneruntil you obtain the constant polynomial 1.A connected bipartite graph with at least one edge is called proper . For anygiven graph G , we denote by Comp( G ) the set of proper components of G . Asan abbreviation, we use Comp( V, E ) instead of Comp((
V, E )). The number ofisolated vertices of a graph G = ( V, E ) is denoted by iso( G ) or by iso( V, E ).3 heorem 5 (bipartite representation, [Dod+15])
The bipartition polyno-mial of a graph G = ( V, E ) satisfies B ( G ; x, y, z ) = X F ⊆ E ( V,F ) is bipartite z | F | (1 + x ) iso( V,F ) Y ( S ∪ T,A ) ∈ Comp ( V,F ) ( x | S | y | T | + x | T | y | S | ) . (6)For another representation of the bipartition polynomial using so-called ac-tivity , we assume that the edge set E = { e , . . . , e m } of the graph G = ( V, E )is linearly ordered, that is e < e < · · · < e m . Let H be a spanning forest of G , which is a forest H = ( V, F ) with the same vertex set as G and F ⊆ E .An edge e ∈ E \ F is externally active with respect to the forest H if it is thelargest edge in a cycle of even length of H + e . We denote by ext( H ) the numberof externally active edges of H . Note that our definition of external activity islittle different from that of Tutte [Tut54]. Theorem 6 (forest representation, [Dod+15])
The bipartition polynomialof a graph G = ( V, E ) satisfies B ( G ; x, y, z ) = X H is spanningforest of G (1 + x ) iso( H ) z n − k ( H ) (1 + z ) ext( H ) × Y ( S ∪ T,A ) ∈ Comp( H ) ( x | S | y | T | + x | T | y | S | ) . (7) Remark 7
In [Dod+15], the Theorems 2,3, and 4 are proven for simple graphsonly. However, the generalization to non-simple graphs is straightforward.
We need also the following result, which is proven in [Dod+15] too.
Theorem 8
Let G be a graph consisting of k components G , . . . , G k . Then B ( G ; x, y, z ) = k Y i =1 B ( G i ; x, y, z ) . First we consider decompositions for the bipartition polynomial of a graph withrespect to local vertex and edge operations.
Theorem 9
The bipartition polynomial of a graph G = ( V, E ) satisfies for eachvertex v ∈ V the relation B ( G ; x, y, z ) = (1 + x ) B ( G − v ; x, y, z )+ X ( S ∪ T,F ) conn. bip. v ∈ S ∪ T z | F | ( x | S | y | T | + x | T | y | S | ) B ( G − ( S ∪ T ); x, y, z )= (1 + x ) B ( G − v ; x, y, z ) + X ( S ∪ T,F ) tree of Gv ∈ S ∪ T z | F |− (1 + z ) ext( S ∪ T,F ) × ( x | S | y | T | + x | T | y | S | ) B ( G − ( S ∪ T ); x, y, z ) , where the first sum is taken over all proper subgraphs, and the second sum isover all nontrivial trees (having at least one edge) of G that contain the vertex v . roof. We show the proof for the first equality and the second one can beshown similarly. From Theorem 5, we obtain B ( G ; x, y, z ) = X F ⊆ E ( V,F ) is bipartite z | F | (1 + x ) iso( V,F ) Y ( S ∪ T,A ) ∈ Comp(
V,F ) ( x | S | y | T | + x | T | y | S | )= X F ⊆ E \ ∂v ( V,F ) is bipartite z | F | (1 + x ) iso( V,F ) Y ( S ∪ T,A ) ∈ Comp(
V,F ) ( x | S | y | T | + x | T | y | S | )+ X F ⊆ E, F ∩ ∂v = ∅ ( V,F ) is bipartite z | F | (1 + x ) iso( V,F ) Y ( S ∪ T,A ) ∈ Comp(
V,F ) ( x | S | y | T | + x | T | y | S | )= (1 + x ) B ( G − v ; x, y, z )+ X ( S ∪ T,F ) conn. bip. v ∈ S ∪ T z | F | ( x | S | y | T | + x | T | y | S | ) B ( G − ( S ∪ T ); x, y, z )The last equality results from factoring out the term of the product that corre-sponds to the component containing v and applying Theorem 8.The proof of the next statement can be performed in the same way. Theorem 10
Let G = ( V, E ) be a graph and e ∈ E ; then B ( G ; x, y, z ) = B ( G − e ; x, y, z )+ X ( S ∪ T,F ) conn. bip. e ∈ F z | F | ( x | S | y | T | + x | T | y | S | ) B ( G − ( S ∪ T ); x, y, z ) . Several well-known graph polynomials can be obtained by substitution of thevariables of the bipartition polynomial and (in some case) by multiplicationwith a certain factor that can easily be obtained from graph parameters likeorder, size, and the number of components. First we recall some results from[Dod+15].
Domination polynomial
The domination polynomial of a graph G = ( V, E ),introduced in [AL00], is the ordinary generating function for the number ofdominating sets of G . Let d k ( G ) be the number of dominating sets of size k of G . We define the domination polynomial of G by D ( G, x ) = n X k =0 d k ( G ) x k . The domination polynomial satisfies, [Dod+15], D ( G, x ) = (1 + x ) n B (cid:18) G ; −
11 + x , x x , − (cid:19) . (8)5 generalized domination polynomial is given by B ( G ; x, − y, −
1) = X W ⊆ V x | W | y | N G ( W ) | , which follows directly from Theorem 2. The variable y counts here the numberof vertices that are dominated by a given set W . Consequently, the coefficientof x i y j in B ( G ; x, − y, −
1) gives the number of vertex subsets of cardinality i that dominate a vertex set of size j . Ising polynomial
The
Ising polynomial of a graph G is defined by Z ( G ; x, y ) = x n y m X W ⊆ V x −| W | y −| ∂W | . The Ising polynomial has been differently introduced in [AM09] by˜ Z ( G ; x, y ) = X σ ∈ Ω x ǫ ( σ ) y M ( σ ) , where σ : V → {− , } is the state of G = ( V, E ), σ ( v ) the magnetization of v ∈ V , Ω the set of all states of G . The sum M ( σ ) = X v ∈ V σ ( v )is called magnetization of G with respect to σ . The parameter ǫ ( σ, e ) definesthe energy of the edge e ∈ E . The energy of G with respect to σ is ǫ ( σ ) = X e ∈ E ǫ ( σ, e ) . The here given notions result from the interpretation of the Ising polynomial(Ising model) in statistical physics. The relation between the two above givenrepresentations of the Ising polynomial is˜ Z ( G ; x, y ) = x − n y − m Z ( G ; x , y ) . Generalizations and modifications of the Ising polynomial and their efficientcomputation in graphs of bounded clique-width are considered in [KM15].The Ising polynomial can be obtained from the bipartition polynomial by,[Dod+15], Z ( G ; x, y ) = x n y m B (cid:18) G ; 1 x , , y − (cid:19) . (9) Cut polynomial
The cut polynomial of a graph G = ( V, E ) is the ordinarygenerating function for the number of cuts of G , C ( G, z ) = 12 k ( G ) X W ∈ V z | ∂W | . The relation between cut polynomial and bipartition polynomial is given by, see[Dod+15], C ( G, z ) = 12 k ( G ) B ( G ; 1 , , z − . (10)6 orollary 11 A graph G of order n with k components is a forest if and onlyif C ( G, z ) = (1 + z ) n − k . Proof.
The statement follows from the fact that a forest is the only graph forwhich any edge subset is a cut.The polynomial B ( G ; x, , z −
1) = X W ⊆ V x | W | z | ∂W | (11)can be considered as a generalized cut polynomial; it is equivalent to the Isingpolynomial. Equation (11) implies also1 ∂x B ( G ; x, , t − (cid:12)(cid:12)(cid:12)(cid:12) x =0 = X v ∈ V t deg v , (12)which is the degree generating function of G . Euler polynomial An Eulerian subgraph of a graph G = ( V, E ) is a spanningsubgraph of G in which all vertices have even degree. The Euler polynomial of G is defined by E ( G, z ) = X F ⊆ E ( V,F ) is Eulerian z | F | . In [Aig07] it is shown that the Euler polynomial is related to the Tutte polyno-mial via E ( G, z ) = (1 − z ) m − n + k ( G ) z n − k ( G ) T (cid:18) G ; 1 z , z − z (cid:19) . There is also a nice direct relation between cut polynomial and Euler polynomial,which is also shown in [Aig07], C ( G, z ) = (1 + z ) | E | | E |−| V | + k ( G ) E (cid:18) G, − z z (cid:19) (13)Solving Equation (13) for E ( G, z ) and substituting C according to Equation(10) yields E ( G, z ) = (1 + z ) | E | | V | B (cid:18) G ; 1 , , − z z (cid:19) . (14)Let G be a plane graph (a planar graph with a given embedding in the plane)and G ∗ its geometric dual. The set of cycles of G is in one-to-one correspondencewith the set of cuts of G ∗ , which yields E ( G, z ) = C ( G ∗ , z )or, corresponding to Equations (10) and (14),(1 + z ) m B (cid:18) G ; 1 , , − z z (cid:19) = 2 n − B ( G ∗ ; 1 , , z − . (15)7 an der Waerden polynomial The definition of this polynomial is pre-sented in [AM09] and based on an idea given in [Wae41]. Let G = ( V, E ) be agraph of order n and size m . Let w ij ( G ) be the number of subgraphs of G withexactly j edges and i vertices of odd degree. The van der Waerden polynomial of G is defined by W ( G ; x, y ) = n X i =0 m X j =0 w ij ( G ) x i y j . From [AM09] (Theorem 2.9), we obtain easily W ( G ; x, y ) = (cid:18) − x (cid:19) n (1 − y ) m Z (cid:18) G ; 1 + x − y , y − y (cid:19) , where Z is the Ising polynomial. The van der Waerden polynomial can bederived from the bipartition polynomial by W ( G ; x, y ) = (cid:18) x (cid:19) n (1 + y ) m B (cid:18) G ; 1 − x x , , − y y (cid:19) , (16)where we use Equation (9). Matching polynomial
The matching polynomial , see [Far79], of G is definedby M ( G, x ) = X F ⊆ EF matching in G x | F | . Notice that the definition that is given here corresponds to matching generatingpolynomial from [LP09]. A subgraph of G with exactly k edges and exactly 2 k vertices of odd degree is a matching in G , which yields M ( G, t ) = lim y → W ( G ; ty − , y ) . Substituting Equation (16) for W , we obtain M ( G, t ) = lim y → (cid:18) √ y + t √ y (cid:19) n (1 + y ) m B (cid:18) G ; √ y − t √ y + t , , − y y (cid:19) . (17) Independence polynomial
The independence polynomial of a graph G =( V, E ) is the ordinary generating function for the number of independent set of G , I ( G, x ) = X W ⊆ VW independent in G x | W | . If G is a simple r -regular graph, then I ( G, t ) = lim x → B (cid:18) G ; tx r , , x − (cid:19) . The proof of this relation is given in [Dod+15]. We can easily rewrite the lastequation in order to avoid the limit: I ( G, t ) = 12 π Z π B ( G ; te irx , , e − ix − dx. x e ix transforms each power of x into a periodic functionwhose period divides 2 π such that the integration over [0 , π ] yields the constantterm (with respect to x ) multiplied by 2 π . One of the important questions about graph polynomials is their distinguishingpower, which can be stated as follows:Let C be a graph class and let P be a polynomial-valued isomorphism in-variant defined on C . Are there nonismorphic graphs G and H in C such that P ( G ) = P ( H )?Although we know that the bipartition polynomial cannot distinguish allgraphs up to isomorphism, see [Dod+15], we do not know yet whether thereare two nonisomorphic trees with the same bipartition polynomial. Instead, astrees are well-known to be ‘reconstructible’ in various senses, we show that thebipartition polynomial of a graph is edge-reconstructible from its polynomial-deck, which shall be defined precisely below.For a graph G , its polynomial-deck is the multiset { B ( G − e ) } e ∈ E ( G ) . Weshow that the bipartition polynomial is ‘edge-reconstructible’ in most cases inthe following sense:A graph G is bp-reconstructible if whenever a graph H has the same poly-nomial-deck as G we have B ( H ) = B ( G ).Unfortunately, there are some graphs with few edges that are not bp-recon-structible. To describe such examples, let P s and C s denote respectively thepath and cycle on s vertices. We denote by C s + tP the disjoint union of C s and t isolated vertices, and the graphs P s + tP , sP + tP etc. are definedsimilarly. The following graphs in each line have the same polynomial-deck buthave different bipartition polynomial. • C + ( t + 2) P , P + ( t + 1) P and 2 P + tP for t ≥ • C + ( t + 1) P and K , + tP for t ≥ t not only have the same poly-nomial-deck but also have the same collection of one-edge-deleted subgraphs.We prove the following in this section. Theorem 12
A graph G is bp-reconstructible unless G is one of the exceptionsabove. In particular, all graphs with at least four edges are bp-reconstructible. We shall use the following information on graphs that are deducible from thebipartition polynomial. The statement combines the results given in Equations(2), (3), (12), Proposition 1, and Corollaries 4, 3, 11.
Theorem 13
Let G be a graph. The bipartition polynomial of G yields | V ( G ) | , | E ( G ) | , k ( G ) , iso ( G ) , the degree sequence, and the multiset of orders of allcomponents of G . We can also decide from B ( G ) whether G is bipartite, aforest, a path, or connected. (The last two properties follow from the otherones.)
9e begin the proof of Theorem 12 with the case when two graphs G and H have different number of isolated vertices but the same polynomial-deck. Notethat from Theorem 13, we know that two graphs with a different number ofisolated vertices have a different bipartition polynomial. Lemma 14
Let G and H be two graphs having different number of isolatedvertices. If G and H have the same polynomial-deck, then there exists t ≥ such that either • { G, H } ⊂ { C + ( t + 2) P , P + ( t + 1) P , P + tP } or • { G, H } = { C + ( t + 1) P , K , + tP } . Proof.
Suppose G and H have the same polynomial-deck and iso( G ) = t while iso( H ) > t . Since iso( G − e ) ≤ t + 2 for every edge e ∈ E ( G ), wehave iso( H ) = t + 1 or t + 2. As iso( H − f ) > t for all f ∈ E ( H ), we haveiso( G − e ) > iso( G ) for all e ∈ E ( G ) implying that every edge of G is incidentwith a vertex of degree 1. That is, the components of G are stars and isolatedvertices. By Theorem 13, we deduce that H − f is a forest for every f ∈ E ( H ).Hence either H itself is a forest or H = C s + t ′ P for some s and t +1 ≤ t ′ ≤ t +2.If iso( H ) = t + 2 then every edge removal from G produces two new isolatedvertices, so that G = s ′ P + tP for some s ′ . Moreover, no edge of H is incidentwith a vertex of degree 1, that is, H = C s + ( t + 2) P . Since G and H havethe same order and an equal number of edges, we conclude G = 2 P + tP and H = C + ( t + 2) P .Now we assume iso( H ) = t + 1. If H = C s + ( t + 1) P for some s , then forall f ∈ E ( H ), iso( H − f ) = t + 1 and H − f has maximum degree at most two.As G and H have the same polynomial-deck, the same holds for G − e for all e ∈ E ( G ). Since G is a disjoint union of stars with t isolated vertices, the onlypossibility for this case is G = K , + tP and H = C + ( t + 1) P .Now we also assume that H is a forest. Theorem 13 states that we candecide the orders of the components from the bipartition polynomial. If G − e for some e ∈ E ( G ) has three P -components, then H − f has it too for some f ∈ E ( H ) and H has a P -component. Removing its edge produces a subgraphwith t + 3 isolated vertices, which cannot be obtained from G by removing onlyone edge. Thus for all e ∈ E ( G ), G − e can have at most two P -componentsand G may have at most three P -components. If G has three P -components,then they are the only nontrivial components of G .On the other hand, as H is a forest, each nontrivial component of H hasat least two leaves which leave t + 2 isolated vertices each when removed. Thenumber of leaves of H must be equal to the number of P -components of G , sothat either G = 3 P + tP or H = P s + ( t + 1) P for some s . It is easy to checkthat for this case the only possibility of non-isomorphic pair G and H with samepolynomial-deck is G = 2 P + tP and H = P + ( t + 1) P . Because of Lemma 14 we only need to compare those graphs without isolatedvertices. The remaining part of our proof of Theorem 12 is presented in thefollowing order.1. Every non-bipartite graph except C + tP for t ≥ C + tP for t ≥ P + ( t + 1) P , 2 P + tP , K , + tP for t ≥ K with bipartition ( U , U ), let m ( K ) =min( | U | , | U | ) and M ( K ) = max( | U | , | U | ). Theorem 5 states B ( G ; x, y, z ) = X F ⊆ E ( V,F ) bipartite z | F | (1 + x ) iso( V,F ) Y K ∈ Comp(
V,F ) [ x M ( K ) y m ( K ) + x m ( K ) y M ( K ) ] . (18)Let us define for each F ⊆ E , a polynomial φ G ( F ; x, y ) or simply φ G ( F ) as φ G ( F ) := ( (1 + x ) iso( V,F ) Q K ∈ Comp(
V,F ) [ x M ( K ) y m ( K ) + x m ( K ) y M ( K ) ] if ( V, F ) is bipartite,0 otherwise.We also write B ( G ) instead of B ( G ; x, y, z ) for convenience. With this definition,Equation (18) simplifies to B ( G ) = X F ⊆ E z | F | φ G ( F ) . (19)We consider the sum B ′ ( G ) = P e ∈ E B ( G − e ) for a given multiset { B ( G − e ) } e ∈ E .For each F ( E , the term z | F | φ G ( F ) appears precisely | E | − | F | times on theright-hand-side so that B ′ ( G ) = X F ( E (cid:0) | E | − | F | (cid:1) z | F | φ G ( F ) . Thus for each k = 0 , , . . . , | E | −
1, the coefficient of z k in B ( G ) is the coefficientof z k in B ′ ( G ) divided by | E | − k . The only remaining term to decide B ( G )is z | E | φ G ( E ). Therefore, to show G is bp-reconstructible it is enough to showthat if H = ( V ′ , E ′ ) is another graph with the same polynomial-deck as G , then φ H ( E ′ ) = φ G ( E ).Now we show that nonbipartite graphs are bp-reconstructible except C + tP for t ≥ Lemma 15
Every nonbipartite graph except C + tP for t ≥ is bp-recon-structible. Proof.
By Lemma 14, it is enough to show that if G is not bipartite, iso( G ) =iso( H ) and H has the same polynomial-deck as G then B ( G ) = B ( H ). We mayadditionally assume that G and H have no isolated vertices.Suppose G = ( V, E ) is not bipartite, iso( G ) = 0 and let D = { B ( G − e ) } e ∈ E .Let H = ( V ′ , E ′ ) be a graph with iso( H ) = 0 whose polynomial-deck is equalto D as a multiset. If G − e is not bipartite for some e ∈ E then from thecorresponding bipartition polynomial, we infer that H − e ′ is nonbipartite for11ome e ′ ∈ E ′ and φ H ( E ′ ) = 0, that is, B ( G ) = B ( H ). If there is no such e , then G is an odd cycle. Applying Theorem 13 to D , we deduce that everyone-edge-deleted subgraph of H is a path consisting of odd number of verticesand the only graph with such property is an odd cycle, and hence φ H ( E ′ ) = 0.We now suppose that G = ( V, E ) is bipartite. Note that the degree of φ G ( F )is precisely | V | . Since x M ( K ) y m ( K ) + x m ( K ) y M ( K ) = ( xy ) m ( K ) (cid:16) x M ( K ) − m ( K ) + y M ( K ) − m ( K ) (cid:17) , to decide φ G ( E ) we only need M ( K ) − m ( K ) for each nontrivial component K in G . If G has k nontrivial components with bipartitions ( U i , V i ) for i = 1 , , . . . , k and t isolated vertices, then we say G has type ( a , a , . . . , a k , ∗ , ∗ , . . . , ∗ )where a i = (cid:12)(cid:12) | U i | − | V i | (cid:12)(cid:12) and the number of ∗ ’s are t . We will ignore the order ofentries in types. From now on we consider the type instead of φ G ( E ). Lemma 16
Let G be a bipartite graph. If G has a cycle then G is bp-recon-structible unless G = C + tP for some t ≥ . Proof.
By Lemmas 14 and 15, it is enough to show that if H = ( V ′ , E ′ )is another bipartite graph with the same polynomial-deck as G and iso( G ) =iso( H ) = 0, then the type of H is uniquely determined. Note that the exceptions C + tP are automatically excluded since t ≥ G is connected, then G − e is connected for some e , since G has a cycle.Let e ′ ∈ E ′ be an edge such that B ( H − e ′ ) = B ( G − e ). We know that H isbipartite and, by Theorem 13, H − e ′ is connected. The coefficient of z | E ′ |− in B ( H − e ′ ) = B ( G − e ) tells us the type of H ′ , which must be the same as thetype of H and hence φ H ( E ′ ) = φ G ( E ).Suppose G is not connected. Then G − e contains a cycle for some e ∈ E , andby Theorem 13, H also has an edge e ′ such that H − e ′ contains a cycle. Thus H has a cycle, and we choose e ′′ ∈ E ′ such that H − e ′′ has minimum numberof components among the one-edge-deleted subgraphs of H . The componentsof H are vertex-wise same as the components of H − e ′′ and have precisely thesame bipartitions. That is, the type of H is the type of H − e ′′ which is againequal to the type of G . Hence φ H ( E ′ ) = φ G ( E ) and G is bp-reconstructible. In this section we prove the following lemma, thereby completing the proof ofTheorem 12.
Lemma 17
Every forest except P + tP , P + ( t + 1) P and K , + tP for t ≥ is bp-reconstructible. To prove Lemma 17 for forests with at least four edges we show the following:
Lemma 18
Let F be a forest. The type of F is uniquely determined fromthe degree sequence of F and the multiset consisting of types of F − e for all e ∈ E ( F ) .
12n [DFR02], it was shown that if G has at least four edges, then the degreesequence of G is completely determined from the degree sequences of one-edge-deleted subgraphs. Theorem 13 states that the degree sequence is obtainablefrom the bipartition polynomial, so that Lemma 17 follows for forests with atleast four edges. The missing cases for Lemma 17 without isolated vertices, P ,3 P , P + P and P as simple graphs and also the non-simple ones are easy tocheck.We shall use some lemmas about trees. For the definition of the type of abipartite graph see the discussion preceding Lemma 16. In a tree, a vertex ofdegree 1 is a leaf and an edge incident with a leaf is a leaf-edge . An edge is internal if it is not a leaf-edge. Lemma 19
Let T be a tree with at least one edge. Let ( U, V ) be the bipartitionof T .(i) If U has all the leaves, then | U | > | V | .(ii) If V has only one leaf, then | U | ≥ | V | .(iii) If T has type ( a ) for a ≥ , then T has two edges e , e such that both T − e and T − e have type ( a − , ∗ ) .(iv) Suppose T has type (0) . If T − e has type (1 , for every internal edge e ,then the degrees of vertices of T are all odd.(v) Suppose T has type (2) . If the types of T − e with ∗ are all (1 , ∗ ) , theneither T is K , or T − f has type (0 , for some edge f . Proof.
Let u be a vertex in U . Consider u as a root and direct every edge of T away from u . Then | U | = 1 + X v ∈ V ( T ) d + ( v ) = 1 + X v ∈ V (cid:0) d ( v ) − (cid:1) = 1 + X v ∈ V (cid:0) d ( v ) − (cid:1) + | V | , so that | U | − | V | = 1 + X v ∈ V (cid:0) d ( v ) − (cid:1) (*)and (i), (ii) follows immediately.Suppose T has type ( a ) for some a ≥
1. We may assume | U | − | V | = a . By(i) and (ii), U contains at least two leaves and removing their incident edgesproduce forests, each of type ( a − , ∗ ). Thus (iii) holds.Now we consider (iv). Suppose T has type (0) and T − e has type (1 ,
1) forevery internal edge e of T . If T consists of only one edge then the conclusionholds. Thus we assume that T has at least one internal edge. Suppose that T has a vertex v of even degree, for contradiction. Since T has type (0) v isincident to at least one internal edge. Let us say v is adjacent to s leaves and isincident to t internal edges e , e , . . . , e t where e i = vu i for i = 1 , , . . . , t . Letus denote by ( U i , V i ) the bipartition of the component of T − e i containing u i such that u i ∈ U i . By the assumptions on T , we know | U i | − | V i | is odd for all i . We consider the component of T − e containing v . Its type is given by (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) t X i =2 | U i | + s − t X i =2 | V i | − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) , T − e has type(1 , T be a tree with bipartition ( U, V ) such that | U | −| V | = 2. Suppose that for every leaf-edge e of T , the forest T − e has type (1 , ∗ ).That is, U contains all the leaves. From Equation ( ∗ ), we deduce that V has aunique vertex of degree 3 and all other vertices in V have degree 2. If V has novertex of degree 2 then | V | = 1 and T is K , . Suppose V has a vertex, say v ,of degree 2. Let e, f be the edges incident with v . If e is a leaf-edge then T − f has type (0 , e and f are internal edges. The graph T − e has two components, say T = ( W , E ) and T = ( W , E ) where f ∈ E .Note that all the leaves of T are in U and all but one leaves of T are in U . By(i) and (ii), we have | U ∩ W | > | V ∩ W | and | U ∩ W | ≥ | V ∩ W | . Since 2 = | U | − | V | = ( | U ∩ W | − | V ∩ W | ) + ( | U ∩ W | − | V ∩ W | ), we haveeither | U ∩ W | − | V ∩ W | = 2 and | U ∩ W | − | V ∩ W | = 0or | U ∩ W | − | V ∩ W | = 1 and | U ∩ W | − | V ∩ W | = 1 . For the former T − e has type (0,2). For the latter T − f has type (0,2). Thus(v) holds. Proof of Lemma 18.
We shall divide the proof of Lemma 18 into threecases, depending on whether the forest F has one component, two componentsor more than two components. In each case we show how to retrieve the typeof F . We call the multiset of the types of F − e for all edges e as the type-deck of F . The sub-multiset consisting of those types with ∗ is called ∗ -deck . We canassume the following in all three cases. The reasoning is given below. • The types in the ∗ -deck contains precisely one ∗ . • If F has more than one component, then at least one type in the ∗ -deckhas a zero as entry. • No type in the ∗ -deck has more than one zero.As the degree sequence of F is given by the assumption, we may assume that F has no isolated vertices. If the the ∗ -deck of F contains a type with two ∗ ’s,then the two isolated vertices came from deleting an edge in a P -component of F , so that we can recover the type of F by replacing the two ∗ ’s with a zero.Thus we may assume that the types in the ∗ -deck contains precisely one ∗ .Let m be the minimum of integral entries in the types in the ∗ -deck. If m >
0, then F cannot induce a component of type ( m ) by Lemma 19 (iii). If m = 1, the entry m is obtained from a component of type ( m + 1), and thetype of F is obtained by replacing ( m, ∗ ) with a m + 1. If m = 1 and F hasmore than one component, then F cannot have a component of type (0) so thatagain, by replacing ( m, ∗ ) with m + 1 we retrieve the type of F . Hence we mayassume that some types in the ∗ -deck have a zero.Suppose that a type in the ∗ -deck has at least two zeros. Then F has acomponent of type (0). Among the types in the ∗ -deck, we choose one with aminimum number of zeros, denote this type by X . The zeros in X came directly14rom F and there must be a 1 and a ∗ which was obtained by removing a leaf-edge of a component with type (0). Thus we replace (1 , ∗ ) by (0) to retrieve thetype of F . Now we may assume that no type in the ∗ -deck has more than onezero.Now we prove Lemma 18. Case 1. F is a tree.If the the ∗ -deck has ( a, ∗ ) and ( a + 2 , ∗ ) for some a then the only possibletype of F is ( a + 1). Suppose ( a, ∗ ) is the unique type in the ∗ -deck. If a = 1then by Lemma 19 (iii) the type of F is ( a + 1). Suppose (1 , ∗ ) is the only typein the ∗ -deck. F has two possibilities: (0) and (2). If the type-deck has (0 , F is (3,1,1,1) then F is K , . If the type-deck does not have (0 ,
2) and F is not K , then by Lemma19 (v) the type of F is (0). This completes Case 1. Case 2. F has precisely two components. By the assumptions the ∗ -deckhas (0 , a, ∗ ) for some a ≥
1. If the ∗ -deck has (0 , a + 2 , ∗ ) also then F has type(0 , a + 1). Thus we assume (0 , a, ∗ ) is the unique type in the ∗ -deck with a 0.First we assume a ≥ a = 1.Suppose a ≥
2. If F had type (0 , a − ∗ -deckmust have (0 , a − , ∗ ) which is a contradiction. Thus F has type (0 , a + 1) or(1 , a ). If the ∗ -deck has (1 , a − , ∗ ) then the type of F cannot be (0 , a + 1)and hence it is (1 , a ). If F has type (1 , a ) by Lemma 19 (iii) the ∗ -deck has(1 , a − , ∗ ). That is, F has type (1 , a ) if and only if the ∗ -deck has (1 , a − , ∗ ),implying that we can decide the type of F from its ∗ -deck.Now we assume a = 1 and the types in the ∗ -deck with a 0 are (0 , , ∗ ). F has one of three types: (0 , ,
1) and (0 , , , F has type (0 , , , F has type (0 , , ∗ )when deleted. By Lemma 19 (v), the component is K , and the ∗ -deck of F has precisely three (0 , , ∗ ). On the other hand, if F had type (1 ,
1) then byLemma 19 (iii) there are at least four (0 , , ∗ ) in the ∗ -deck of F . If F had type(0 ,
0) then its ∗ -deck also have at least four (0 , , ∗ ) since we assumed none ofthe components are K and every tree has at least two leaves. Thus (0 , , ∗ )appears precisely three times in the ∗ -deck if and only if F has type (0 , ∗ -deck has at least four times (0 , , ∗ ). F has type (0 , , ∗ -deck has only (0 , , ∗ ).(ii) the type-deck consists of precisely (0 , , ∗ ) and (0 , , ∗ -deck is(1 , , ∗ ), in which case F has type (1 , P . Thus a component of type (0) has aninternal edge and by removing an internal edge from (0 ,
0) we get (0 , a, a ). If a = 1 then F has type (0 , F has type (1 , F has type (0 , F , every internal edge produces a forest of type (1 , F have odd degree.15herefore, F has type (1 ,
1) if F has a vertex of degree 2, and F has type(0 ,
0) otherwise. That is, we can determine the type of F given all the assump-tions so far. This completes Case 2. Case 3. F has more than two components. By the assumptions afterLemma 19 the ∗ -deck has a type (0 , a , a , . . . , a k , ∗ ) where a i > i .The question is to decide whether F has a component of type (0) or not. If ithas, then the (0) component is unique and the ∗ -deck has a type without zero.We replace (1 , ∗ ) with a (0) to recover the type of F . If F does not have acomponent of type (0), then F has type (1 , a , a , . . . , a k ).Suppose the ∗ -deck of F has another type (0 , b , b , . . . , b k , ∗ ) where { a i :1 ≤ i ≤ k } 6 = { b i : 1 ≤ i ≤ k } as multisets. Then F has a component of type(0), since otherwise the zeros in the ∗ -deck types come from a (1) of F and allsuch types must be the same up to order of entries.Thus we assume (0 , a , a , . . . , a k , ∗ ) is the only type in the ∗ -deck with a0. If the type of F had a 0 and two distinct nonzero numbers then the ∗ -deckcontains two distinct types with a 0. Hence we may assume that(0 , a , a , a , . . . , a k , ∗ ) = (0 , a − , a, a, . . . , a, ∗ ) for some a ≥ . The type of F is either (0 , a, a, a, . . . , a ) or (1 , a − , a, a, . . . , a ). Lemma 19 (iii)implies that in the latter case the ∗ -deck has (1 , a − , a − , a, . . . , a, ∗ ), whereasthe former case cannot have the same type. Thus we can decide the type of F from the ∗ -deck in Case 3.In all cases we can decide the type of F from the degree sequence and thetype-deck of F . Therefore Lemma 18 holds and Lemma 17 follows. In this section we prove some facts about Euler polynomials, the number ofdominating sets, and sums over spanning forests of a graph. The commontheme of these results is a very simple way of proving by just using differentrepresentations of the bipartition polynomial.We denote by F ( G ) the set of spanning forests of the graph G . Theorem 20
The Euler polynomial of a graph G of order n and size m satisfies E ( G, z ) = (1 + z ) m − n ( − z ) n X H ∈F ( G ) (cid:18) − zz (cid:19) k ( H ) (cid:18) − z z (cid:19) ext( H ) . Proof.
We use the forest representation of the bipartition polynomial that isgiven in Theorem 6 and Equation (14); we obtain E ( G, z ) = (1 + z ) m n B (cid:18) G ; 1 , , − z z (cid:19) = (1 + z ) m n X H ∈F ( G ) iso( H ) (cid:18) − z z (cid:19) n − k ( H ) (cid:18) − z z (cid:19) ext( H ) k ( H ) − iso( H ) = (1 + z ) m − n ( − z ) n X H ∈F ( G ) (cid:18) − zz (cid:19) k ( H ) (cid:18) − z z (cid:19) ext( H ) . | Comp( H ) | +iso( H ) = k ( H ).The next theorem provides a representation of the Euler polynomial as asum ranging over subsets of the vertex set. Theorem 21
The Euler polynomial of a graph G = ( V, E ) satisfies E ( G, z ) = (1 + z ) | E | | V | X W ⊆ V (cid:18) − z z (cid:19) | ∂W | . Proof.
The result can be obtained via the multiplicative representation ofthe bipartition polynomial according to Theorem 2. The substitution of thisrepresentation for the bipartition polynomial in Equation (14) yields E ( G, z ) = (1 + z ) m n B (cid:18) G ; 1 , , − z z (cid:19) = (1 + z ) m n X W ⊆ V Y v ∈ N G ( W ) "(cid:18) − z z (cid:19) | ∂v ∩ ∂W | = (1 + z ) m n X W ⊆ V (cid:18) − z z (cid:19) | ∂W | . The last equality follows from the fact that | N G ( v ) ∩ W | is the number of edgesthat connect v to a vertex of W . Hence when we take the product over allvertices in N G ( W ), then we count each edge in ∂W exactly once.The following statement can be proven also via the principle of inclusion–exclusion. However, our knowledge about the bipartition polynomial offers aneven faster way of proof. Theorem 22
The number d ( G ) of dominating sets of a graph G satisfies d ( G ) = 2 n X W ⊆ V ( − | W | (cid:18) (cid:19) | N G [ W ] | . Proof.
Here we use the product representation of the bipartition polynomial ofa simple graph. The restriction to simple graphs does not change the dominationpolynomial. According to Equation (8), we obtain d ( G ) = D ( G, n B (cid:18) G ; − , , − (cid:19) = 2 n X W ⊆ V (cid:18) − (cid:19) | W | Y v ∈ N G ( W ) (cid:20)(cid:18) (cid:19) h | N G ( v ) ∩ W | − i + 1 (cid:21) = 2 n X W ⊆ V (cid:18) − (cid:19) | W | Y v ∈ N G ( W ) (cid:18) (cid:19) = 2 n X W ⊆ V (cid:18) − (cid:19) | W | (cid:18) (cid:19) | N G ( W ) | , which is equivalent to the statement of the theorem.17 heorem 23 Let G = ( V, E ) be a graph with a linearly ordered edge set and F ( G ) the set of all spanning forests of G with external activity 0. Then X H ∈F ( G ) ( − k ( H ) = ( − n k ( G ) . Proof.
The statement follows immediately from the forest representation ofthe bipartition polynomial that is given in Theorem 6 by substituting x = 1, y = 1, z = − B ( G ; x, y, z ). Theorem 24
Let G = ( V, E ) be a simple undirected graph with n vertices. Thenumber of bicolored subgraphs of G with exactly i isolated vertices and exactly j edges is given by the coefficient of x i z j in the polynomial (2 x − n B (cid:18) G ; 12 x − , x − , z (cid:19) . Proof.
An edge subset F ⊆ E of G induces a subgraph that can be properlycolored with two colors if and only if ( V, F ) is a bipartite graph. The number ofbicolored graphs with edge set F is then given by 2 k ( V,F ) . Substituting x and y by 1 / (2 x −
1) in B ( G ; x, y, z ) and multiplying the resulting expression with(2 x − n yield(2 x − n X F ⊆ E ( V,F ) is bipartite z | F | (cid:18) x x − (cid:19) iso( V,F ) Y ( U,A ) ∈ Comp(
V,F ) (cid:18) x − (cid:19) | U | . Using the equations iso(
V, F ) + | Comp(
V, F ) | = k ( V, F ) andiso(V , F) + X ( U,A ) ∈ Comp(
V,F ) | U | = n, we obtain(2 x − n B (cid:18) G ; 12 x − , x − , z (cid:19) = X F ⊆ E ( V,F ) is bipartite k ( V,F ) x iso( V,F ) z | F | , which proves the statement. Corollary 25
Let G be a simple graph of order n . The number of bicoloredsubgraphs of G without any isolated vertices is ( − n B ( G ; − , − , . Proof.
This follows immediately from the last line of the proof of Theorem 24by substituting x = 0 and z = 1. The bipartition polynomial emerges as a powerful tool for proving equationsin graphical enumeration. It shows nice relations to other graph polynomials,offers a couple of useful representations, and is polynomially reconstructible.However, there are still many open questions for the bipartition polynomial; weconsider the following ones most interesting:18
Equation (15) gives a relation between the bipartition polynomial of aplanar graph and its dual. However, this equation is restricted to theevaluation of B ( G ; x, y, z ) at x = y = 1. Is there a way to generalize thisresult? • The Ising and matching polynomial of a graph G can be derived from thecorresponding polynomials of the complement ¯ G . Can we calculate thebipartition polynomial of a graph from the bipartition polynomial of itscomplement? • There are known pairs of nonisomorphic graphs with the same bipartitionpolynomial. However, despite all efforts by extensive computer search, wecould not find a pair of nonisomorphic trees with coinciding bipartitionpolynomial. We know that no such pair for trees with order less than 19exists. Is the bipartition polynomial able to distinguish all nonisomorphictrees?
We thank Jo Ellis-Monaghan, Andrew Goodall, Johann A. Makowsky, and IainMoffatt – the organizers of the Dagstuhl seminar
Graph Polynomials: Towardsa Comparative Theory (2016). This excellent and fruitful workshop initiatedthe cooperation of the authors of this paper.
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