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Oct 3, 2010 at 18:57 vote accept Alexander
Oct 3, 2010 at 23:10
Oct 3, 2010 at 18:57 vote accept Alexander
Oct 3, 2010 at 18:57
Oct 1, 2010 at 20:39 comment added Alexander Of course, $G$ and $H$ are not isomorphic.
Oct 1, 2010 at 20:32 comment added Alexander Thank you for the answer, damiano. I know this already) I just have formulated the problem quite incorrectly. Sorry. Indeed, the problem is formulating as follows. Let $G$ and $H$ be the trees with the same degree sequences. Is it true that there always be vertices $q\in V(G)$ and $q′\in V(H)$ such that $(q,p)\in E(G)$ and $(q′,p′)\in E(H)$ for some endvertices $p\in V(G)$ and $p\in V(H)$, and $d(q)=d(q′)$? $d(q)$ - degree of the vertex $q$.
Oct 1, 2010 at 19:32 history answered damiano CC BY-SA 2.5