Low Weight Perfect Matchings

Answering a question posed by Caro, Hansberg, Lauri, and Zarb, we show that for every positive integer $n$ and every function $\sigma\colon E(K_{4n})\to\{-1,1\}$ with $\sigma\left(E(K_{4n})\right)=0$, there is a perfect matching $M$ in $K_{4n}$ with $\sigma(M)=0$. Strengthening a result of Caro and Yuster, we show that for every positive integer $n$ and every function $\sigma\colon E(K_{4n})\to\{-1,1\}$ with $\left|\sigma\left(E(K_{4n})\right)\right|<n^2+11n+2,$ there is a perfect matching $M$ in $K_{4n}$ with $|\sigma(M)|\leq 2$. Both these results are best possible.

As a variation of this problem, they ask whether, for every positive integer n and every labeling σ: E(K 4n ) → {−1, 1} of the edges of the complete graph K 4n of order 4n with σ (E(K 4n )) = 0, there is a perfect matching M in K 4n with σ(M) = 0. We answer their question in the affirmative. Under the hypothesis of Theorem 1, the existence of a perfect matching M in K 4n with |σ(M)| ≤ 2 already follows from more general results due to Caro and Yuster, cf. Theorem 1.1 in [3]. More precisely, Caro and Yuster showed that the weaker hypothesis |σ (E(K 4n ))| ≤ 2(4n − 1) suffices for the existence of such a perfect matching M with |σ(M)| ≤ 2. As observed in [2], for infinitely many positive integers n, there are functions σ: E(K 4n ) → {−1, 1} with σ (E(K 4n )) = 4 √ n − 2 such that σ(M) = 0 for every perfect matching M in K 4n . Slightly modifying their construction, we obtain the following proposition, which implies that Theorem 1 is best possible for infinitely many values of n.
Considering the construction in the proof of Proposition 2 suggests that zero weight perfect matchings are excluded rather by parity reasons than by the imbalance |σ (E(K 4n ))| of σ. We confirm this with our second main result showing that much weaker conditions on the imbalance imply the existence of low weight perfect matchings.
Theorem 3. For every positive integers n and k with k ≥ 2, and every function σ: For k = 2, Theorem 3 implies the following strengthening of the above-mentioned consequence of the result of Caro and Yuster.
All proofs are given in the next section. For a survey concerning related results, we refer the reader to [1] and the introduction of [2].

Proofs
We start with the proof of our first main result.
Proof of Theorem 1. We suppose, for a contradiction, that σ: E(K 4n ) → {−1, 1} is such that σ (E(K 4n )) = 0 but that σ(M) = 0 for every perfect matching M in K 4n . First, we consider the case n = 1. The edge set of K 4 is the union of three edge-disjoint perfect matchings M 1 , For a matching M, we denote by M + and M − the sets of plus-edges and minus-edges in M, respectively. We choose a perfect matching M in K 4n such that |σ(M)| is as small as possible.
We start with some easy observations. Claim 1. For every two edges e in M + and f in M − , there are no two disjoint minus-edges between e and f . In particular, there are at most two minus-edges between e and f .
Proof of Claim 1. If there are two disjoint minus-edges e ′ and f ′ between two edges e ∈ M + and f ∈ M − , then the perfect matching and, hence, the total number of plus-edges is at least contradicting (1).
Claim 3. For every two edges u 1 u 2 and v 1 v 2 in M + , if u 1 v 1 is a minus-edge, then u 2 v 2 is also a minus-edge. Furthermore, σ(M) = 2.
Proof of Claim 3. If u 1 u 2 and v 1 v 2 are two edges in M + such that u 1 v 1 is a minus-edge and u 2 v 2 is a plus-edge, then the perfect matching Since σ(M) = 2, the matching M contains exactly n + 1 plus-edges and n − 1 minus-edges. We call a perfect matching N in K 4n good if σ(N) = σ(M). If N is a good matching, then We distinguish the following two cases.
Case 1. Every good matching contains a special edge.
Let e + be a special edge in M. Proof of Claim 4. Suppose, for a contradiction, that there is a plus-edge between e + and some edge e in M + \ {e + }. By Claim 3, there are at least two plus-edges between e + and e. Since e + is special, it follows that there are at most 4n − 2 minus-edges in V (M + ). Therefore, by Claim 1, the total number of plus-edges is at least In this case, we may assume that the good matching M chosen at the beginning of the proof contains no special edge.
which is odd. Nevertheless, by Claim 3, the number of minus-edges in V (M + ) is even, which is a contradiction, and completes the proof.
The following is based on a construction given at the end of [2].
Proof of Proposition 2. Let n be such that there exists a positive even integer k with 4n = k 2 +4. Let (A, B) be a partition of the vertex set of K 4n with |A| = 1 2 (k 2 +k)+2 and |B| = 1 2 (k 2 −k)+2. Now, we define a function σ: E(K 4n ) → {−1, 1} such that all edges between A and B receive the value 1 and all remaining edges receive the value −1. Note that and thus, Now, suppose, for a contradiction, that M is a perfect matching in K 4n with σ(M) = 0. Clearly, M contains n plus-edges between A and B. Hence, the number of vertices in A that are not covered by a plus-edge in M is Since this is an odd number, by construction, not all these vertices can be covered by minusedges in M, which is a contradiction, and completes the proof.
For the proof of our second main result, Theorem 3, we need the following extremal result about matchings due to Erdős and Gallai [4].
Theorem 5 (Erdős and Gallai [4]). If G is a graph of order 4n with matching number n − k for some positive integer k, then the number of edges of G is at most 4n 2 − 3n+k 2 , with equality if and only if G is the complement of the disjoint union of a complete graph of order 3n + k and n − k isolated vertices.
We proceed to the proof of our second main result.
Proof of Theorem 3. Let σ: E(K 4n ) → {−1, 1} be such that σ (E(K 4n )) ≥ 0 and |σ(M)| ≥ 2k ≥ 4 for every perfect matching M in K 4n . As observed above, |σ(M)| is even for every perfect matching M in K 4n . Therefore, it remains to show that σ (E(K 4n )) is at least the term stated in the theorem. We distinguish the following two cases.  Note that σ(M) ≥ 2k for every perfect matching M in K 4n . Let ν be the matching number of the graph G = (V (K 4n ), σ −1 (−1)).
Let ν = n − k ′ for some integer k ′ ≥ k. By Theorem 5, we obtain which completes the proof.