Graceful and Strongly Graceful Permutations

TL;DR

This paper introduces new permutations related to graceful labelings of trees, proving that certain lobsters with perfect matchings are strongly graceful and proposing a method to approach Bermond's conjecture on all lobsters being graceful.

Contribution

It defines generalized strongly graceful permutations and uses them to prove strong gracefulness for specific lobster trees, advancing the understanding of Bermond's conjecture.

Findings

New permutations related to graceful labelings are discovered.

Lobsters with perfect matchings are shown to be strongly graceful.

A tractable method for approaching Bermond's conjecture is developed.

Abstract

A graceful labelling of a graph G is an injective function f from the set of vertices of G into the set {0,1,...,|EG|} such that if edge uv is assigned the label |f(u)-f(v)| then all edge labels have distinct values. A strong graceful labelling of a tree T with a perfect matching is a graceful labelling of T with the additional property that the sum of the vertex labels of each odd labelled edge add up to |ET|. A lobster or a 2-distant tree is a tree T that contains a path P such that any vertex of T is a distance at most 2 from a vertex of P. In this paper, we define generalised strongly graceful permutations and discover two new permutations in addition to the known permutation that is obtained by replacing each vertex label f(v) by |ET|- f(v). We use these permutations to prove, by induction, that a lobster with a perfect matching that consists of the set of end edges of the lobster, is strongly graceful. Further, we show that there exist strongly graceful labellings that assign the label 0 to four specific vertices of any tree belonging to this family of lobsters. By using the technique developed in this paper we will, further, present a tractable way for proving an equivalent form of Bermond conjecture which states that all lobsters are graceful. Two out of a total of three cases of the proposed equivalent form of Bermond conjecture are completed leaving the third case open for refutation or completion.