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A tree is a connected graph without cycles, with a finite or infinite number of vertices. There are many variants of trees, according to further constraints or decorations.
6
votes
Accepted
Bijective proof of recurrence for rooted unlabeled trees
Pick a canonical ordering of unlabelled rooted trees, say, with lexicographical comparison of tuples $(n, T_1, \ldots, T_k)$, where $n$ is the number of vertices, $T_1, \ldots, T_k$ is the non-descending … Throughout $T, T_1, T_2$ are unlabelled rooted trees in canonical form (that is, subtrees of any vertex are ordered as above). …
0
votes
Accepted
Is there a way to generate a graph of specified treewidth
Here are some classes of graphs that should be interesting:
$k$-trees by definition have treewidth $k$; moreover, adding any edge to a $k$-tree increases treewidth. … This should be an interesting example since $k$-trees are "borderline maximal" with treewidth $k$.
Graphs with treewdith $\leq k$ are famously characterized by a finite set of forbidden minors. …
11
votes
A permutation problem
A simple brute-force of trees yields a counter-example with $n = 7$. Let $\{a_1, \ldots, a_7\} = [7]$, $T$ be a path of four vertices $v_1, v_2, v_3, v_4$, with $v_4$ adjacent to $v_5, v_6, v_7$. …
4
votes
Accepted
Automorphism of the transfinite rooted binary tree
Each connected component of the tree is either rooted at $0$ or at a number which birthday is a limit ordinal, these are exactly the vertices of degree $2$. Let $A$ be the class of all limit ordinals, …
8
votes
Accepted
Spanning $k$-trees
As for the general question about extending results from trees to $k$-trees, the intuition is probably that spanning $k$-trees of $K_n$ can be tamed (in light of results mentioned in OP), while spanning … $k$-trees of a general graph may possess a much more complex structure not exhibited by spanning trees. …