Random plane increasing trees: Asymptotic enumeration of vertices by distance from leaves

Pub Date : 2022-12-28 DOI:10.1002/rsa.21138
M. Bóna, B. Pittel
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引用次数: 0

Abstract

We prove that for any fixed k$$ k $$ , the probability that a random vertex of a random increasing plane tree is of rank k$$ k $$ , that is, the probability that a random vertex is at distance k$$ k $$ from the leaves, converges to a constant ck$$ {c}_k $$ as the size n$$ n $$ of the tree goes to infinity. We prove that 1−∑j≤kck<22k+3(2k+4)!$$ 1-{\sum}_{j\le k}{c}_k<\frac{2^{2k+3}}{\left(2k+4\right)!} $$ , so that the tail of the limiting rank distribution is super‐exponentially narrow. We prove that the latter property holds uniformly for all finite n$$ n $$ as well. More generally, we prove that the ranks of a finite uniformly random set of vertices are asymptotically independent, each with distribution {ck}$$ \left\{{c}_k\right\} $$ . We compute the exact value of ck$$ {c}_k $$ for 0≤k≤3$$ 0\le k\le 3 $$ , demonstrating that the limiting expected fraction of vertices with rank ≤3$$ \le 3 $$ is 0.9997$$ 0.9997 $$ … We show that with probability 1−n−0.99ε$$ 1-{n}^{-0.99\varepsilon } $$ the highest rank of a vertex in the tree is sandwiched between (1−ε)logn/loglogn$$ \left(1-\varepsilon \right)\log n/\mathrm{loglog}n $$ and (1.5+ε)logn/loglogn$$ \left(1.5+\varepsilon \right)\log n/\mathrm{loglog}n $$ , and that this rank is asymptotic to logn/loglogn$$ \log n/\mathrm{loglog}n $$ with probability 1−o(1)$$ 1-o(1) $$ .
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随机平面递增树:根据离叶子的距离对顶点进行渐近枚举
我们证明了对于任意固定的k $$ k $$,一个随机增长的平面树的随机顶点的秩为k $$ k $$的概率,也就是说,一个随机顶点离叶子的距离为k $$ k $$的概率,随着树的大小n $$ n $$趋近于无穷大,收敛于一个常数ck $$ {c}_k $$。证明了1−∑j≤kck<22k+3(2k+4)!$$ 1-{\sum}_{j\le k}{c}_k<\frac{2^{2k+3}}{\left(2k+4\right)!} $$,使得极限秩分布的尾部是超指数窄的。我们证明后一个性质对所有有限n $$ n $$也一致成立。更一般地,我们证明了一个有限一致随机顶点集合的秩是渐近独立的,每个顶点的分布为{ck}$$ \left\{{c}_k\right\} $$。我们计算了0≤k≤3 $$ 0\le k\le 3 $$时ck $$ {c}_k $$的精确值,证明了秩≤3 $$ \le 3 $$的顶点的极限期望分数为0.9997 $$ 0.9997 $$ .我们证明了以1−n−0.99ε $$ 1-{n}^{-0.99\varepsilon } $$的概率,树中顶点的最高秩夹在(1−ε)logn/ loggn $$ \left(1-\varepsilon \right)\log n/\mathrm{loglog}n $$和(1.5+ε)logn/ loggn $$ \left(1.5+\varepsilon \right)\log n/\mathrm{loglog}n $$之间。这个秩是渐近于logn/loglog $$ \log n/\mathrm{loglog}n $$的概率是1 - 0 (1)$$ 1-o(1) $$。
本文章由计算机程序翻译,如有差异,请以英文原文为准。
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