Research notes on every problem, and a library of the papers behind them. Built from the open erdos repository.
Updated
Statement
Setting as on Theorem 9: a split graph has independent and a clique, cliques are inclusion-maximal complete subgraphs on at least two vertices, and is the least size of a vertex set meeting all of them (pp. 117--118, 124).
Proposition 10 (p. 125, quoted). "For every there exists a split-graph such that every clique of has at least vertices, and ."
With Theorem 9 this shows that the paper's statement holds in split graphs for and for no (p. 124).
The construction (p. 125). and ; the pairs are taken for , and for even the last is reset to ; is joined to every vertex of outside . The proof states that and that each has degree .
For odd this works as printed: the pairs partition , no vertex of meets every clique, and . For even the printed indexing leaves in no pair, so lies in and in every , and that graph has . Taking , with the pairs for and , gives and for , which is the bound the proof displays, so the proposition holds for every . This adjustment is the corpus's reading, not the paper's.
Source. Zsolt Tuza, Covering all cliques of a graph, Discrete Math. 86 (1990), 117--126, doi:10.1016/0012-365X(90)90354-K. Statement and proof p. 125. The edition read is identified on the source card.
Read depth. Claims checked: the statement and the construction were read clause by clause on the printed page, and the construction was checked for odd and even as recorded above. Nothing here is independently reviewed.
Proof pointer
Page 125, as recorded above. The cliques are , of vertices, and the sets , of vertices. A single vertex of misses , and a single vertex of misses the clique of the whose pair contains it, so once the pairs cover ; two vertices of from different pairs meet every clique.
Dependencies
None beyond the definitions.
Bears on
- Problem 611: the examples show that the strongly chordal bound of Theorem 7, with the least clique order, does not extend to split graphs. They have , so they do not bear against the problem's question.