By Alan Albert Bertossi; Alberto Montresor
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What we would like to avoid is writing down every possible level sequence and discarding the ones that aren't regular. Instead, Beyer and Hedetniemi [BH] found a way to hop directly from one of those level sequences to its successor. It hinges on the positions of the 2's in the original sequence. ,ln) be given. If L contains no entry > 2 then the algorithm halts and all trees of n vertices have been generated. Else, let lp be the rightmost entry of L that is > 2. Let lq be the rightmost position preceding p such that lq = lp-\ (vertex q is the parent of vertex p).
1 we show an ordered tree and its level sequence. The vertices of the tree are unlabeled but we show next to each vertex the epoch at which it gets visited. The appearances of the 2's in a level sequence of a tree T are important. , we are at the root of one of the subtrees at the root of T. Hence all entries in the level sequence, beginning with that 2, and ending just before the next 2, comprise the level sequence of a single root subtree of T. An ordered tree of course comes equipped with an ordering of its subtrees.
The level subsequences of all subtrees of T can be ordered lexicograph23 24 CHAPTER 5 FIG. 1. An ordered, rooted tree, with its visitation and level sequences. ically. Then we can say that U < V if L(U) < L(V) in that ordering. In general, if T is some given rooted tree, then there will be many ordered trees T' such that T' is an ordering of T. We will now specify a single ordered tree T* to be the canonical ordered tree that corresponds to T. It is the one ordered tree, among all of those that correspond to T, whose level sequence is the largest, in the lexicographic ordering of level sequences.
Algoritmi e Strutture di Dati by Alan Albert Bertossi; Alberto Montresor