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by maweki 2308 days ago
I did not completely understand to what graphs this applies. Certainly not generic ones, that can't be true.

Is it only complete ones? It seems intuitively true that complete graphs can be tiled this way.

2 comments

I was wondering the same thing. It does concern complete graphs only, but note also the additional requirement of edge-disjointedness (that is, every edge in the large graph is covered by exactly one copy of the subgraph).

>A typical decomposition question asks whether the edges of some graph G can be partitioned into disjoint copies of another graph H. One of the oldest and best known conjectures in this area, posed by Ringel in 1963, concerns the decomposition of complete graphs into edge-disjoint copies of a tree. It says that any tree with n edges packs 2n+1 times into the complete graph K2n+1. In this paper, we prove this conjecture for large n.

https://arxiv.org/abs/2001.02665

The edges of complete graphs with 2n + 1 vertices can be covered with the edges of 2n + 1 identical copies of any tree with n + 1 vertices.