Efficiently finding low-sum copies of spanning forests in zero-sum complete graphs via conditional expectation
aa r X i v : . [ m a t h . C O ] F e b Efficiently finding low-sum copies of spanning forests inzero-sum complete graphs via conditional expectation
Johannes Pardey Dieter Rautenbach
Institute of Optimization and Operations Research, Ulm University,Ulm, Germany, { johannes.pardey,dieter.rautenbach } @uni-ulm.de Abstract
For a fixed positive ǫ , we show the existence of a constant C ǫ with the following property: Given a ± c : E ( K n ) → {− , } of the complete graph K n with c ( E ( K n )) = 0, and a spanning forest F of K n of maximum degree ∆, one can determine in polynomial time an isomorphic copy F ′ of F in K n with | c ( E ( F ′ )) | ≤ (cid:0) + ǫ (cid:1) ∆ + C ǫ . Our approach is based on the method of conditional expectation.
Keywords:
Zero-sum subgraph; zero-sum Ramsey theory; method of conditional expectation
The kind of zero-sum problem that we study here can be traced back to algebraic results such as the well-knownErd˝os-Ginzburg-Ziv theorem [6]. The two survey articles due to Caro [2] as well as Gao and Geroldinger [8]give a detailed account of this area also known as zero-sum Ramsey theory within discrete mathematics andadditive group theory.Several recent papers [3–5,9,10] study (almost) zero-sum spanning forests in edge-labeled complete graphs,and, in the present paper, we contribute an algorithmic approach for finding low-sum spanning forests. Thesetting involves a complete graph K n of order n together with a zero-sum ± -labeling of its edges, that is, afunction c : E ( K n ) → {− , } that satisfies c ( E ( K n )) = X e ∈ E ( K n ) c ( e ) = 0 . For a given spanning forest F of K n , we consider the algorithmic task to efficiently find an isomorphic copy F ′ of F in K n that minimizes | c ( E ( F ′ )) | . The corresponding existence version of this algorithmic task, and, inparticular, the question under which conditions there is a zero-sum copy of F in K n was studied in [3–5, 9, 10],where the arguments are typically non-algorithmic or do not lead to efficient algorithms.The following two simple observations correspond to key existential arguments in this area: • If F is a spanning forest of K n and c is a zero-sum ± K n , then, by symmetry,every edge of K n belongs to the same number of isomorphic copies of F in K n , which implies that theaverage of c ( E ( F ′ )), where F ′ ranges over all isomorphic copies of F in K n , equals 0. In particular, thereare copies F + and F − of F with c ( E ( F + )) ≥ c ( E ( F − )) ≤ • If F , . . . , F k are isomorphic copies of F in K n , c ( E ( F )) ≥ c ( E ( F k )) ≤
0, and each F i +1 arises from F i by removing at most ℓ edges and adding at most ℓ edges, then | c ( E ( F i )) | ≤ ℓ for some i .As observed in [10], these observations yield the following. Proposition 1 (Mohr et al. [10]) . If c : E ( K n ) → {− , } is a zero-sum labeling of K n , and F is a spanningforest of K n of maximum degree ∆ , then there is an isomorphic copy F ′ of F in K n with | c ( E ( F ′ )) | ≤ ∆ + 1 . Conjecture 2 (Mohr et al. [10]) . If c : E ( K n ) → {− , } is a zero-sum labeling of K n , and F is a spanningforest of K n of maximum degree ∆ , then there is an isomorphic copy F ′ of F in K n with | c ( E ( F ′ )) | ≤ ∆ − . In [10] this conjecture was verified for the spanning star K ,n − of K n . For other special spanning forests,in particular, perfect matchings, and under natural divisibility conditions, the existence of zero-sum copieswas shown in [3, 5, 9, 10]. Our main contribution here is the following theorem, which improves Proposition 1in two ways: It strengthens the bound given there almost halfway towards the bound from Conjecture 2, andit provides the existence of an efficient algorithm to find the desired low-sum copy. Theorem 3.
Let ǫ > be fixed. There is a constant C ǫ such that the following holds: Given a zero-sumlabeling c : E ( K n ) → {− , } of K n and a spanning forest F of K n of maximum degree ∆ , one can determinein polynomial time an isomorphic copy F ′ of F in K n with | c ( E ( F ′ )) | ≤ (cid:18)
34 + ǫ (cid:19) ∆ + C ǫ . Our approach is based on the method of conditional expectation [1, 7]. In Section 2 we explain how toimplement this method in the present context, and illustrate it with an algorithmic version of Proposition1. In Section 3 we consider a natural greedy algorithm based on the method of conditional expectation, andprovide the proof of Theorem 3 by analyzing this greedy algorithm. In a conclusion we discuss further possibledevelopments. T via conditional expectation Throughout this section, let K be a complete graph of order n , let c : E ( K ) → {− , } be a zero-sum labelingof K , and let F be a spanning forest of K of maximum degree ∆. Let [ n ] be the set of positive integers atmost n . We may assume that K has vertex set [ n ]. For a subgraph H of K , let c ( H ) abbreviate c ( E ( H )),and let ¯ c ( H ) equal c ( H ) | E ( H ) | . Similarly, for a set E of edges of K , let ¯ c ( E ) equal c ( E ) | E | . As usual, for a set X ofvertices of a graph G , let G [ X ] be the subgraph of G induced by X , and let G − X = G [ V ( G ) \ X ].For a permutation π in S n , let F π be the isomorphic copy of F within K with edge set { π ( u ) π ( v ) : uv ∈ E ( F ) } , that is, within F π , the vertex π ( u ) of K assumes the role of the vertex u of F . The first of the twoobservations mentioned before Proposition 1 can be expressed as follows: Choosing a permutation π from S n uniformly at random, and considering the random variable c ( F π ), linearity of expectation implies that E [ c ( F π )] = 1 n ! X π ∈ S n c ( F π ) = ¯ c ( K ) m ( F ) = 0 . (1)For k in [ n ], and distinct elements i , . . . , i k of [ n ], we consider the expected value of the random variable c ( F π )subject to the condition that π ( j ) = i j for every j in [ k ], that is, E [ c ( F π ) | ( π (1) , . . . , π ( k )) = ( i , . . . , i k )] = 1( n − k )! X π ∈ S n :( π (1) ,...,π ( k ))=( i ,...,i k ) c ( F π ) . (2)By the uniformity of the choice of the random permutation π , we have E [ c ( F π )] = 1 n n X i =1 E [ c ( F π ) | π (1) = i ] . (3)2imilarly, if k < n , then the uniformity of the choice of π implies that the conditional expectation (2) equals1 n − k X i k +1 ∈ [ n ] \{ i ,...,i k } E [ c ( F π ) | ( π (1) , . . . , π ( k + 1)) = ( i , . . . , i k +1 )] . (4)For the approach, it is crucial that the conditional expectation (2) can be calculated efficiently. In fact, linearityof expectation implies that (2) equals c ( F π [ { i , . . . , i k } ]) (5)+ k X j =1 ¯ c ( { i j ℓ : ℓ ∈ [ n ] \ { i , . . . , i k }} ) | N F ( j ) \ [ k ] | (6)+ ¯ c ( K − { i , . . . , i k } ) m ( F − [ k ]) , (7)where • (5) is the weight of the edges uv of F [ k ] whose embedding π ( u ) π ( v ) within K is already completelydetermined by the condition ( π (1) , . . . , π ( k )) = ( i , . . . , i k ), • (6) collects the expected weights of the | N F ( j ) \ [ k ] | edges that each vertex i j of F π , taking the role ofthe vertex j of F , sends into the set [ n ] \ { i , . . . , i k } , and • (7) is the expected weight of the edges of F − [ k ] whose embedding within K − { i , . . . , i k } is still chosenuniformly at random.In view of this representation, we obtain the following. Lemma 4.
Given K , c , F , k , and i , . . . , i k , E [ c ( F π ) | ( π (1) , . . . , π ( k )) = ( i , . . . , i k )] can be computed in polynomial time. Implementing the method of conditional expectation in the present context, we will now explain how todetermine a permutation π in S n , in other words, an isomorphic copy F π of F within K , for which | c ( F π ) | issmall by fixing the values π (1) , . . . , π ( n ) one by one.There are different reasonable ways to do this.Mimicking the proof of Proposition 1 in [10], we obtain the following. Proposition 5.
Given K , c , and F , a permutation π in S n with | c ( F π ) | ≤ ∆ + 1 can be determined inpolynomial time.Proof. By (3) and (4), there is an ordering i , . . . , i n of [ n ] such that E [ c ( F π ) | π (1) = i ] ≥ E [ c ( F π )] = 0 and E [ c ( F π ) | ( π (1) , . . . , π ( k + 1)) = ( i , . . . , i k +1 )] ≥ E [ c ( F π ) | ( π (1) , . . . , π ( k )) = ( i , . . . , i k )]for every k in [ n − i , . . . , i n can be foundin polynomial time. In other words, in polynomial time one can determine a permutation π + in S n with c ( F π + ) ≥
0. Similarly, in polynomial time one can determine a permutation π − in S n with c ( F π − ) ≤ F always involving at least one vertex of degree at most 1,cf. the proof of Proposition 1 in [10], one can determine in polynomial time a sequence π , . . . , π r of permutations3rom S n such that r is polynomially bounded in terms of n , π = π + , π r = π − , and, for every i in [ r ], F π i arises from F π i − by removing at most ∆ + 1 edges and adding at most ∆ + 1 edges. The second of the twoobservations mentioned before Proposition 1 implies min {| c ( F π i ) | : i ∈ { , . . . , r }} ≤ ∆ + 1, and returning apermutation π i minimizing | c ( F π i ) | accomplishes the desired task.Proposition 5 does not really exploit that F is a forest. In fact, it can easily be adapted to the situationin which F is not a forest replacing the bound ∆ + 1 by ∆ + δ , where δ is the minimum degree of F . WhileProposition 5 corresponds to an algorithmic version of the existential argument behind Proposition 1, there isactually a more natural way of implementing the method of conditional expectation for our problem, choosingthe vertices i , . . . , i n one by one in this order in such a way that the absolute value of each conditional expectedvalue (2) is as small as possible. Our proof of Theorem 3 relies on the analysis of this more natural greedyapproach. Note that there is one degree of freedom that we did not exploit so far; we can freely choose theorder in which the vertices of F are embedded one by one into K . Throughout this section, let ǫ >
0, let K be a complete graph of order n , let c : E ( K ) → {− , } be a zero-sumlabeling of K , and let F be a spanning forest of K of maximum degree ∆. In view of the statement of Theorem3 and Proposition 1, we may assume that ǫ < , and that n is sufficiently large in terms of ǫ . Possibly replacing ǫ by a slightly smaller value, we may furthermore assume, for notational simplicity, that ǫn is an integer.Since m ( F ) ≤ n −
1, the forest F has less than ǫn vertices of degree more than ǫ . Hence, since everyinduced subgraph of F is 1-degenerate, we may assume, possibly by reordering/renaming the vertices of K and F , that | N F ( i ) ∩ [ i − | ≤ i in [ ǫn ], and (8) d F ( i ) ≤ ǫ for every i in [ n ] \ [ ǫn ]. (9)Note that the possible reordering/renaming of the vertices of K can be performed in polynomial time.For distinct i , . . . , i k from [ n ], let E [ i , . . . , i k ] = E [ c ( F π ) | ( π (1) , . . . , π ( k )) = ( i , . . . , i k )] . For k = 0, let E [ i , . . . , i k ] = E [ c ( F π )] ( ) = 0.Now, in order to determine F ′ in polynomial time, we consider the following natural greedy algorithm: Choose i , . . . , i n one by one in this order in such a way that in every step | E [ i , . . . , i k ] | is mini-mized, that is, i j = arg min {| E [ i , . . . , i k , p ] | : p ∈ [ n ] \ { i , . . . , i k }} for every j in [ n ] . By (3), (4), and Lemma 4, the values i , . . . , i n , which completely determine π and F ′ , can be determined inpolynomial time. Therefore, in order to complete the proof, it suffices to show that | E [ i , . . . , i k ] | ≤ (cid:18)
34 + 327 ǫ (cid:19) ∆ + (cid:18) ǫ + 4 (cid:19) (10)for every k in [ n ]. We establish (10) using the following two claims. Claim 1.
For every k in { , . . . , n − } , there is some p in [ n ] \ { i , . . . , i k } such that • If E [ i , . . . , i k ] > , then E [ i , . . . , i k , p ] ≤ E [ i , . . . , i k ] , • if E [ i , . . . , i k ] < , then E [ i , . . . , i k , p ] ≥ E [ i , . . . , i k ] , and | E [ i , . . . , i k , p ] − E [ i , . . . , i k ] | ≤ (cid:18) ǫ (cid:19) ∆ + (cid:18) ǫ + 4 (cid:19) . Claim 2.
For every k in { , . . . , n − } , there is some p in [ n ] \ { i , . . . , i k } such that | E [ i , . . . , i k , p ] − E [ i , . . . , i k ] | ≤ (cid:18)
12 + 327 ǫ (cid:19) ∆ + (cid:18) ǫ + 4 (cid:19) . Before we prove these two claims, we explain how they imply (10). Since c is a zero-sum labeling, (10)holds for k = 0. Now, if (10) holds for some k and | E [ i , . . . , i k ] | ≤ ∆, then Claim 2 implies the existence of apossible choice p for i k +1 with | E [ i , . . . , i k +1 ] | bounded as in (10). Therefore, by the selection rule of the greedyalgorithm, (10) holds for k + 1 (instead of k ). Otherwise, if (10) holds for some k but | E [ i , . . . , i k ] | > ∆, thenClaim 1 implies the existence of a possible choice p for i k +1 with | E [ i , . . . , i k +1 ] | bounded as in (10). Again,also in this case, (10) holds for k + 1 (instead of k ). Altogether, a simple inductive argument yields (10) forall k in [ n ].We fix some abbreviating notation.For k ∈ [ n ] and j ∈ [ k ], let c [ k ] = c ( F π [ { i , . . . , i k } ]) ,d k ( j ) = | N F ( j ) \ [ k ] | ,m k = m ( F − [ k ]) , ¯ c k ( j ) = ¯ c ( { i j ℓ : ℓ ∈ [ n ] \ { i , . . . , i k }} ) , and¯ c k = ¯ c ( K − { i , . . . , i k } ) . With these abbreviations, E [ i , . . . , i k ] = c [ k ] + k X j =1 ¯ c k ( j ) d k ( j ) + ¯ c k m k . (11) Claim 3. If p and q are positive integers with q < p and x , . . . , x p ∈ {− , } , then (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) p p X i =1 x i − p − q p − q X i =1 x i (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ qp . Proof. (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) p p X i =1 x i − p − q p − q X i =1 x i (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:18) p − p − q (cid:19) p − q X i =1 x i + 1 p p X i = p − q +1 x i (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ qp ( p − q ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) p − q X i =1 x i (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)| {z } ≤ p − q + 1 p (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) p X i = p − q +1 x i (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)| {z } ≤ q ≤ qp . Below, we shall apply Claim 3 mainly in the following settings: • p = n − k and q = 1, in which case qp = n − k , and • p = (cid:0) n − k (cid:1) and q = n − k −
1, in which case qp = n − k .We proceed to the proof of Claim 1. 5 roof of Claim 1. By symmetry, we may assume that E [ i , . . . , i k ] >
0. By (4), there is some p in [ n ] \{ i , . . . , i k } with E [ i , . . . , i k , p ] ≤ E [ i , . . . , i k ]. We will argue that p already satisfies the desired inequality.Let d = | N F ( k + 1) ∩ [ k ] | and d = d F ( k + 1) − d , that is, d = d k +1 ( k + 1) = | N F ( k + 1) \ [ k + 1] | .Let c = c ( { pi ℓ : ℓ ∈ N F ( k + 1) ∩ [ k ] } ) , ¯ c ′ k ( j ) = ¯ c ( { i j ℓ : ℓ ∈ [ n ] \ { i , . . . , i k , p }} ) for j in [ k ] , ¯ c ′ k ( p ) = ¯ c ( { pℓ : ℓ ∈ [ n ] \ { i , . . . , i k , p }} ) , and¯ c ′ k = ¯ c ( K − { i , . . . , i k , p } ) , that is, going from ¯ c k ( j ) to ¯ c ′ k ( j ) or from ¯ c k to ¯ c ′ k corresponds to the possibly alternative choice of i k +1 as p .Note that m k +1 = m k − d and | c | ≤ d .Furthermore, by Claim 3, | ¯ c ′ k ( j ) − ¯ c k ( j ) | ≤ n − k for j ∈ [ k ] and | ¯ c ′ k − ¯ c k | ≤ n − k .By (11), we have | E [ i , . . . , i k , p ] − E [ i , . . . , i k ] | = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) c + k X j =1 ¯ c ′ k ( j ) d k +1 ( j ) + ¯ c ′ k ( p ) d + ¯ c ′ k m k +1 − k X j =1 ¯ c k ( j ) d k ( j ) − ¯ c k m k (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) c + k X j =1 (cid:16) ¯ c ′ k ( j ) − ¯ c k ( j ) (cid:17) d k +1 ( j ) + k X j =1 ¯ c k ( j ) (cid:16) d k +1 ( j ) − d k ( j ) (cid:17) + ¯ c ′ k ( p ) d + (¯ c ′ k − ¯ c k ) m k − ¯ c ′ k d (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (12) ≤ d + 2 n − k k X j =1 d k +1 ( j ) + k X j =1 | ¯ c k ( j ) | (cid:12)(cid:12)(cid:12) d k +1 ( j ) − d k ( j ) (cid:12)(cid:12)(cid:12) + (cid:16) | ¯ c ′ k | (cid:17) d + 4 n − k m k (13)First, we consider the case that k + 1 ≤ ǫn .Trivially, n − k ≥ (1 − ǫ ) n ǫ< ≥ n, k X j =1 d k +1 ( j ) ≤ m ( F ) < n, and m k ≤ m ( F ) < n. By (8), we have k X j =1 (cid:12)(cid:12)(cid:12) d k +1 ( j ) − d k ( j ) (cid:12)(cid:12)(cid:12) = d ≤ . Since c is a zero-sum labeling, a simple inductive argument based on Claim 3 implies | ¯ c ′ k | ≤ n + 4 n − · · · + 4 n − k ≤ k + 1) n − k ≤ ǫnn − k ≤ ǫ. Now, (13) implies | E [ i , . . . , i k , p ] − E [ i , . . . , i k ] |≤ d |{z} ≤ + 2 n − k | {z } ≤ n k X j =1 d k +1 ( j ) | {z } ≤ n + k X j =1 | ¯ c k ( j ) | | {z } ≤ (cid:12)(cid:12)(cid:12) d k +1 ( j ) − d k ( j ) (cid:12)(cid:12)(cid:12)| {z } ≤ + (cid:16) | ¯ c ′ k | |{z} ≤ ǫ (cid:17) d |{z} ≤ ∆ + 4 n − k | {z } ≤ n m k |{z} ≤ n n n + 1 + (cid:18) ǫ (cid:19) ∆ + 4 n n = (cid:18) ǫ (cid:19) ∆ + 10 ≤ (cid:18) ǫ (cid:19) ∆ + (cid:18) ǫ + 4 (cid:19) . Next, we consider the case that k + 1 > ǫn .By (9), we obtain that d + d = d F ( k + 1) ≤ ǫ , and also that k X j =1 d k +1 ( j ) ≤ n X j = k +1 d F ( j ) ≤ ǫ ( n − k ) . Since F is a forest, we have m k = m ( F − [ k ]) < n − k .Now, (13) implies | E [ i , . . . , i k , p ] − E [ i , . . . , i k ] |≤ d + 2 n − k k X j =1 d k +1 ( j ) | {z } ≤ ǫ ( n − k ) + k X j =1 | ¯ c k ( j ) | | {z } ≤ (cid:12)(cid:12)(cid:12) d k +1 ( j ) − d k ( j ) (cid:12)(cid:12)(cid:12)| {z } ≤ d + (cid:16) | ¯ c ′ k | |{z} ≤ (cid:17) d + 4 n − k m k |{z} ≤ n − k ≤ d + 2 d | {z } ≤ ǫ + 4 ǫ + 4 ≤ ǫ + 4 (14) ≤ (cid:18) ǫ (cid:19) ∆ + (cid:18) ǫ + 4 (cid:19) , which completes the proof.For the proof of Claim 2, we need the following generalization of Theorem 2 from [10]. Claim 4.
Let the positive real ǫ and the integer n be such that ǫn ≥ .If G is a graph of order n and size m such that (cid:12)(cid:12)(cid:12)(cid:12) m − (cid:18) n (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) ≤ ǫ (cid:18) n (cid:19) , then (cid:18) − ǫ (cid:19) n ≤ d G ( u ) ≤ (cid:18)
34 + ǫ (cid:19) n − for some vertex u of G .Proof. Since the statement is trivial for ǫ ≥ , we may assume that ǫ < . For a contradiction, suppose that G is as in the hypothesis but that, for every vertex u of G , either d G ( u ) < (cid:0) − ǫ (cid:1) n or d G ( u ) > (cid:0) + ǫ (cid:1) n − V + = { u ∈ V ( G ) : d G ( u ) > (cid:0) + ǫ (cid:1) n − } and V − = V ( G ) \ V + . Since (cid:0) + ǫ (cid:1) n − > (cid:0) − ǫ (cid:1) n , we have V − = { u ∈ V ( G ) : d G ( u ) < (cid:0) − ǫ (cid:1) n } . We assume that among all counterexamples, the graph G is chosensuch that X u ∈ V + d G ( u ) = (cid:18) n (cid:19) − X u ∈ V − d G ( u )is as large as possible. Note that the desired statement as well as the choice of G are symmetric with respectto forming the complement. 7et n + = | V + | . Clearly, | V − | = n − n + . If n + > n , then2140 n > (cid:18) n (cid:19) > (cid:16) ǫ (cid:17) (cid:18) n (cid:19) ≥ m ≥ X u ∈ V + d G ( u ) > n (cid:18)(cid:18)
34 + ǫ (cid:19) n − (cid:19) ǫn ≥ ≥ n , which is a contradiction. We obtain n + ≤ n . By symmetry with respect to forming the complement, we alsoobtain n − n + ≤ n , and, hence, 14 n ≤ n + ≤ n, (15)which implies, in particular, that every vertex in V + has a neighbor in V − .If V + contains two non-adjacent vertices u and v , and w is a neighbor of u in V − , then G ′ = G − uw + uv isa counterexample contradicting the choice of G . Hence, V + is complete. By symmetry with respect to formingthe complement, the choice of G also implies that V − is independent.Now, let d + be the average degree of the vertices in V + , and let d − be the average degree of the vertices in V − . Clearly, d + > (cid:18)
34 + ǫ (cid:19) n − d − < (cid:18) − ǫ (cid:19) n. (16)Let the real n ′ be such that 2 m = (cid:18) n ′ (cid:19) . If n ′ > (cid:0) ǫ (cid:1) n , then (cid:12)(cid:12)(cid:12)(cid:12) m − (cid:18) n (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12) (cid:18) n ′ (cid:19) − (cid:18) n (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) > (cid:18)(cid:0) ǫ (cid:1) n (cid:19) − (cid:18) n (cid:19) = ǫn (6 n + 4 ǫn − ǫn ≥ > ǫn > ǫ (cid:18) n (cid:19) , which is a contradiction. Conversely, if n ′ < (cid:0) − ǫ (cid:1) n , then (cid:12)(cid:12)(cid:12)(cid:12) m − (cid:18) n (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12) (cid:18) n ′ (cid:19) − (cid:18) n (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) > (cid:18) n (cid:19) − (cid:18)(cid:0) − ǫ (cid:1) n (cid:19) = ǫn (10 n − ǫn − ǫn ≥ > ǫ (10 − ǫ ) n > ǫ (cid:18) n (cid:19) , (cid:18) − ǫ (cid:19) n ≤ n ′ ≤ (cid:18) ǫ (cid:19) n. (17)Note that n ′ − n + n − n + ( ) , ( ) ≥ (cid:0) − ǫ (cid:1) n − (cid:0) − ǫ (cid:1) n + − ǫn + n − n + = (cid:18) − ǫ (cid:19) − ǫ (cid:18) n + n − n + (cid:19) ( ) ≥ − ǫ − ǫ n n ! = 1 − ǫ, which implies 12 n ′ ( n ′ − n + ) ( ) ≥ (cid:18) − ǫ (cid:19) n (cid:18) − ǫ (cid:19) ( n − n + ) >
12 (1 − ǫ ) n ( n − n + )= 2 (cid:18) − ǫ (cid:19) n ( n − n + ) . (18)Since every vertex in V + has exactly n + − V + , and V − is independent, the number of edges in G between V + and V − equals d + n + − n + ( n + −
1) = d − ( n − n + ) , (19)and the sum of all vertex degrees equals d + n + + d − ( n − n + ) = 2 m = (cid:18) n ′ (cid:19) . (20)Adding (19) and (20) implies12 n ′ ( n ′ −
1) + n + ( n + −
1) = 2 d + n + ( ) > (cid:18)(cid:18)
34 + ǫ (cid:19) n − (cid:19) n + ( ) ≥ (cid:18) n ′ − (cid:19) n + or, equivalently, (cid:18) n + − n ′ (cid:19) (cid:0) n + − ( n ′ − (cid:1) > . Since n + ( ) ≤ n ǫ< ,ǫn ≥ < (cid:0) − ǫ (cid:1) n − ( ) ≤ n ′ − , this implies n + < n ′ . Subtracting (19) from (20) implies12 n ′ ( n ′ − − n + ( n + −
1) = 2 d − ( n − n + ) ( ) < (cid:18) − ǫ (cid:19) n ( n − n + ) ( ) ≤ n ′ ( n ′ − n + )9r, equivalently, (cid:18) n + − n ′ (cid:19) ( n + − > . Since ǫ < , ǫn ≥
10, and (15), it follows easily that n + >
1, and, hence, n + > n ′ . The contradiction n ′ < n + < n ′ completes the proof.We are now in the position to complete the proof of Claim 2. Proof of Claim 2. If k + 1 > ǫn , then exactly the same arguments as in the proof of Claim 1 imply that | E [ i , . . . , i k , p ] − E [ i , . . . , i k ] | ( ) ≤ ǫ + 4 ≤ (cid:18)
12 + 327 ǫ (cid:19) ∆ + (cid:18) ǫ + 4 (cid:19) , regardless of the specific choice of p from [ n ] \ { i , . . . , i k } .Hence, we may assume that k + 1 ≤ ǫn .We consider the auxiliary graph G = (cid:18) [ n ] \ [ k ] , c − (1) ∩ (cid:18) [ n ] \ [ k ]2 (cid:19)(cid:19) . Since c is a zero-sum labeling, we have | c − (1) | = (cid:0) n (cid:1) .Note that n (cid:0) n − k (cid:1) ≤ n (cid:0) (1 − ǫ ) n (cid:1) = 2 n (1 − ǫ )((1 − ǫ ) n − ǫ< ≤ n (cid:0) n − (cid:1) ≤ , (21)where the last inequality assumes that n is large enough to ensure n − ≥ n .Since the graph G is obtained from the graph with vertex set [ n ] and edge set c − (1) by removing k vertices,we obtain (cid:12)(cid:12)(cid:12)(cid:12) m ( G ) − (cid:18) n − k (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:12)(cid:12)(cid:12)(cid:12) m ( G ) − (cid:18) n (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:12) (cid:18) n (cid:19) − (cid:18) n − k (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:16) ( n −
1) + ( n −
2) + · · · + ( n − k ) (cid:17) + (cid:12)(cid:12)(cid:12)(cid:12) (cid:18) n (cid:19) − (cid:18) (1 − ǫ ) n (cid:19)(cid:12)(cid:12)(cid:12)(cid:12)| {z } = ǫn ((2 − ǫ ) n − ≤ ǫn ≤ kn + ǫn ≤ ǫn ) ≤ ǫ (cid:18) n − k (cid:19) . Now, applying Claim 4 with ǫ instead of ǫ , implies the existence of a vertex p of G with (cid:18) − ǫ (cid:19) ( n − k ) ≤ d G ( p ) ≤ (cid:18)
34 + 320 ǫ (cid:19) ( n − k ) − . Using the same notation as in the proof of Claim 1, this implies | ¯ c ′ k ( p ) | ≤ n − k − (cid:18)(cid:18)
34 + 320 ǫ (cid:19) ( n − k ) − − (cid:18) − ǫ (cid:19) ( n − k ) (cid:19) +1 ≤ ǫn ≤ − ǫ ) n (cid:18)
12 + 640 ǫ (cid:19) n ǫ< ≤ (1 + 2 ǫ ) (cid:18)
12 + 640 ǫ (cid:19) ǫ< ≤
12 + 963 ǫ . Now, using the notation and some observations from the proof of Claim 1, we have | E [ i , . . . , i k , p ] − E [ i , . . . , i k ] | ( ) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) c + k X j =1 (cid:16) ¯ c ′ k ( j ) − ¯ c k ( j ) (cid:17) d k +1 ( j ) + k X j =1 ¯ c k ( j ) (cid:16) d k +1 ( j ) − d k ( j ) (cid:17) + ¯ c ′ k ( p ) d + (¯ c ′ k − ¯ c k ) m k − ¯ c ′ k d (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ d |{z} ≤ + 2 n − k k X j =1 d k +1 ( j ) | {z } ≤ n n ≤ + k X j =1 | ¯ c k ( j ) | (cid:12)(cid:12)(cid:12) d k +1 ( j ) − d k ( j ) (cid:12)(cid:12)(cid:12)| {z } ≤ + (cid:16) | ¯ c ′ k ( p ) | + | ¯ c ′ k | (cid:17)| {z } ≤ + ǫ + ǫ ≤ +327 ǫ d |{z} ≤ ∆ + 4 n − k m k | {z } ≤ n n ≤ ≤ (cid:18)
12 + 327 ǫ (cid:19) ∆ + (cid:18) ǫ + 4 (cid:19) , which completes the proof.As explained after the statement of Claim 2, this completes the proof of Theorem 3 Unlike the proof of Proposition 5, the proof of Theorem 3 uses that F is a forest. Nevertheless, it is not difficultto generalize the approach to k -degenerate graphs, adapting, in particular, (8) and (9) as well as the estimatesbased thereupon. Another possible and technically straightforward generalization concerns the situation inwhich c is not zero-sum, that is, c ( K ) is not 0. In this case, our approach yields an isomorphic copy F ′ of F for which ¯ c ( F ′ ) is close to ¯ c ( K ). It seems possible to strengthen our approach, or rather the analysis of theconsidered greedy algorithm, in order to get an approximate version of Conjecture 2 with ∆ + replaced by (cid:0) + ǫ (cid:1) ∆ + C ǫ . A key ingredient that needs to be improved for this seems to be Claim 4.The transpositions considered in the proof of Proposition 5 suggest local search as another algorithmicstrategy: Let K be a complete graph with vertex set [ n ], let c : E ( K ) → {− , } be a zero-sum labeling of K ,and let H be a spanning subgraph of K . For two distinct vertives u and v of K , let δ H ( uv ) be the set of edgesof H between { u, v } and ( N H ( u ) ∪ N H ( v )) \ { u, v } , and let H uv arise from H by • removing all edges in δ H ( uv ), and • adding all possible edges between u and N H ( v ) \ { u } as well as between v and N H ( u ) \ { v } ,that is, H uv is an isomorphic copy of H in K in which u and v exchanged their roles.Now, we suppose that c ( H ) > c ( H ) ≤ c ( H uv ) for every edge uv of K , or, equivalently, c ( δ H ( uv )) ≤ c ( δ H uv ( uv )) for every edge uv of K , (22)that is, no local search step replacing H by H uv reduces c ( H ). Standard double-counting arguments imply X uv ∈ ( [ n ]2 ) c ( δ H ( uv )) = X uv ∈ E ( H ) (2 n − c ( uv ) , uv ∈ ( [ n ]2 ) c ( δ H uv ( uv )) = X uv ∈ ( [ n ]2 ) \ E ( H ) ( d H ( u ) + d H ( v )) c ( uv ) + X uv ∈ E ( H ) ( d H ( u ) + d H ( v ) − c ( uv ), and , summing (22) over all edges uv of K , we obtain(2 n − c ( H ) = X uv ∈ E ( H ) (2 n − c ( uv ) ≤ X uv ∈ ( [ n ]2 )( d H ( u ) + d H ( v )) c ( uv ) . The problem now is that the right hand side of this inequality is hard to work with. However, if H is ∆-regular,then, since c is zero-sum, the right hand side evaluates to 0, which implies the contradiction c ( H ) ≤
0. In otherwords, in the case that H is ∆-regular and c ( H ) >
0, there is at least one edge uv of K with c ( H uv ) < c ( H ).Therefore, since | c ( H uv ) − c ( H ) | ≤