Abstract

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 the consequence of 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.

Highlights

  • In [2] Caro, Hansberg, Lauri, and Zarb considered connected graphs G together with a function σ : E(G) → {−1, 1} labeling the edges of G with −1 or +1, and they studied conditions that imply the existence of different types of spanning trees T with |σ(E(T ))| = σ(e) 1, e∈E(T )where, as usual, for a set E of edges, σ(E) is just the sum of σ(e) over all e in E

  • Answering a question posed by Caro, Hansberg, Lauri, and Zarb, we show that for every positive integer n and every function σ : E(K4n) → {−1, 1} with σ (E(K4n)) = 0, there is a perfect matching M in K4nwith σ(M ) = 0

  • Strengthening the consequence of a result of Caro and Yuster, we show that for every positive integer n and every function σ : E(K4n) → {−1, 1} with |σ (E(K4n))| < n2 +11n+2, there is a perfect matching M in K4n with |σ(M )| 2

Read more

Summary

Introduction

If there are two disjoint minus-edges e and f between two edges e ∈ M + and f ∈ M −, the perfect matching N = (M \ {e, f }) ∪ {e , f } satisfies 0 σ(N ) = σ(M ) − 2 < σ(M ), contradicting the choice of M . If u1u2 and v1v2 are two edges in M + such that u1v1 is a minus-edge and u2v2 is a plus-edge, the perfect matching N = (M \ {u1u2, v1v2}) ∪ {u1v1, u2v2} satisfies 0 σ(N ) = σ(M ) − 2 < σ(M ), contradicting the choice of M .

Results
Conclusion
Full Text
Published version (Free)

Talk to us

Join us for a 30 min session where you can share your feedback and ask us any queries you have

Schedule a call