Balanced generic circuits without long paths

Csaba Király, Ferenc Péterfalvi
2012 Discrete Mathematics  
We call a graph G = (V , E) a (k, ℓ)-circuit if |E| = k|V | − ℓ + 1 and every X ⊂ V with 2 ≤ |X| ≤ |V | − 1 induces at most k|X | − ℓ edges. A (2, 3)-circuit is also called a generic circuit. We say that a graph is balanced if the difference between its maximum and minimum degrees is at most 1. Graver et al. asked in Graver et al. (1993) [7] whether a balanced generic circuit always admits a decomposition into two disjoint Hamiltonian paths. We show that this does not hold, moreover there are
more » ... lanced (k, k + 1)-circuits for all k ≥ 2 which do not contain any Hamiltonian path nor a path longer than |V | λ for λ > log 8 log 9 ≃ 0, 9464. Proposition 1.1. A graph G is a generic circuit if and only if it can be decomposed into two spanning trees such that no pair of proper subtrees, except single vertices, spans the same vertex set. Having this result one may ask whether a tree decomposition with some special properties exists. Graver et al. posed the following problem. Open Question 1.2 ([7], Exercise 4.69). Does every generic circuit with vertices of degrees 3 and 4 only, have a two tree decomposition into two paths? Note that the smallest (k, k + 1)-circuit is K 2k since a graph with k|V | − k edges cannot be simple if it has less than 2k vertices. Recall the definition of balanced graphs from the abstract. A balanced (k, k + 1)-circuit has vertices with degrees 2k − 1 and 2k only and the number of vertices with degree 2k − 1 is exactly 2k.
doi:10.1016/j.disc.2012.03.031 fatcat:ylplleudtnen5ckbtujhpliasu