2022
DOI: 10.1002/jgt.22799
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Graphs with no induced house nor induced hole have the de Bruijn–Erdös property

Abstract: A set of n $n$ points in the plane which are not all collinear defines at least n $n$ distinct lines. Chen and Chvátal conjectured in 2008 that a similar result can be achieved in the broader context of finite metric spaces. This conjecture remains open even for graph metrics. In this article we prove that graphs with no induced house nor induced cycle of length at least 5 verify the desired property. We focus on lines generated by vertices at distance at most 2, define a new notion of 'good … Show more

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Cited by 4 publications
(7 citation statements)
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“…It is not difficult to see that for any tree T (or, more generally, any graph with a bridge) we have gp − (T ) = 2, so it is possible to have equality with the lower bound in Inequality (1). In Section 2 we will take a closer look at the graphs G which satisfy gp − (G) = 2.…”
Section: Lower General Position Sets In Graphsmentioning
confidence: 99%
“…It is not difficult to see that for any tree T (or, more generally, any graph with a bridge) we have gp − (T ) = 2, so it is possible to have equality with the lower bound in Inequality (1). In Section 2 we will take a closer look at the graphs G which satisfy gp − (G) = 2.…”
Section: Lower General Position Sets In Graphsmentioning
confidence: 99%
“…If G contains no triangle, then ( 5) and (6) show that every (j, ℓ) in D 1 ∪ D 3 with j < ℓ has ℓ = j + 1 and (3), (4) show that every (j, ℓ) in D 2 with j < ℓ has j = 0, ℓ = n − 1. Therefore…”
Section: Metric and Non-metric Hypergraphsmentioning
confidence: 99%
“…Since no edge of P extends to a triangle, implications (3), (4) show that none of {0, 2}, {1, 3}, {2, 4} can belong to P and implications ( 5), (6) show that neither of {0, 3}, {1, 4} can belong to P . Hence each of the three edges of P must be one of {0, 1}, {1, 2}, {2, 3}, {3, 4}, {0, 4}.…”
Section: Metric and Non-metric Hypergraphsmentioning
confidence: 99%
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