Improved algorithms for ranking and unranking (k, m)-ary trees in B-order

Abstract

Du and Liu (Eur J Comb 28:1312–1321, 2007) introduced (k, m)-ary trees as a generalization of k-ary trees. In a (k, m)-ary tree, every node on even level has degree k (i.e., has k children), and every node on odd level has degree m (which is called a crucial node) or is a leaf. In particular, a (k, m)-ary tree of order n has exactly n crucial nodes. Recently, Amani and Nowzari-Dalini (Bull Iranian Math Soc 45(4):1145–1158, 2019) presented a generation algorithm to produce all (k, m)-ary trees of order n in B-order using Zaks’ encoding, and showed that the generated ordering of this encoding results in a reverse-lexicographical ordering. They also proposed the corresponding ranking and unranking algorithms for (k, m)-ary trees according to such a generated ordering. These algorithms take $\mathcal {O}(kmn^2)$ time and space for building a precomputed table in which (k, m)-Catalan numbers (i.e., a kind of generalized Catalan numbers) are stored in advance. Then, each ranking and unranking algorithm can be performed subsequently in $\mathcal {O}(n)$ and $\mathcal {O}(n\log n)$ time, respectively. In this paper, we revisit the ranking and unranking problems. With the help of an encoding scheme called “right-distance” introduced by Wu et al. (Math Comput Model 53:1331–1335, 2011a; IEICE Trans Inf Syst E94–D:226–232, 2011b), we propose new ranking and unranking algorithms for (k, m)-ary trees of order n in B-order using Zaks’ encoding. We show that both algorithms can be improved in $\mathcal {O}(kmn)$ time and $\mathcal {O}(n)$ space without building the precomputed table.

Appendix A: Proof of Theorem 5

The proof is by induction on $\ell +d$. The result is trivially true for $d=0$ (i.e., $B_{\ell ,0}=0$, the case in the amendment of Eq. (3.1)) or $\ell \leqslant 1$ (i.e., $B_{0,d}=1$ for $d\ne 0$ and $B_{1,d}=d$, the first two conditions of Eq. (3.1)). We now consider $\ell \geqslant 2$ and $d\geqslant 1$ and suppose that the theorem holds for $B_{\ell ',d'}$ with $\ell '<\ell$ or $d'<d$. There are two case as follows:

If $d=1$, by the third condition of Eq. (3.1) and the induction hypothesis, we have

$$\begin{aligned} B_{\ell ,1}= & {} B_{\ell -1,km} =\frac{km}{km(\ell -1)+km}{km(\ell -1)+km\atopwithdelims ()\ell -1} =\frac{1}{\ell }{km\ell \atopwithdelims ()\ell -1} \\= & {} \frac{1}{\ell }\cdot \frac{km\ell \,!}{(\ell -1)!\ (km\ell -\ell +1)!} = \frac{1}{km\ell +1}\cdot \frac{(km\ell +1)!}{\ell \,!\,(km\ell +1-\ell )!}\\= & {} \frac{1}{km\ell +1}{km\ell +1\atopwithdelims ()\ell }. \end{aligned}$$

Otherwise (i.e., $\ell ,d>1$), by the last condition of Eq. (3.1) and the induction hypothesis, we have

$$\begin{aligned} B_{\ell ,d}= & {} B_{\ell ,d-1}+B_{\ell -1,km+d-1} \\= & {} \frac{d-1}{km\ell +d-1}{km\ell +d-1\atopwithdelims ()\ell }\\&+\frac{km+d-1}{km(\ell -1)+km+d-1}{km(\ell -1)+km+d-1\atopwithdelims ()\ell -1}\\= & {} \frac{d-1}{km\ell +d-1}{km\ell +d-1\atopwithdelims ()\ell } +\frac{km+d-1}{km\ell +d-1}{km\ell +d-1\atopwithdelims ()\ell -1}\\= & {} \left( \frac{d-1}{km\ell +d-1}\cdot \frac{km\ell +d-\ell }{km\ell +d}+ \frac{km+d-1}{km\ell +d-1}\cdot \frac{\ell }{km\ell +d}\right) {km\ell +d\atopwithdelims ()\ell }\\= & {} \frac{d}{km\ell +d}{km\ell +d\atopwithdelims ()\ell }. \end{aligned}$$

This completes the proof. $\square$