Computing Maximal Layers of Points in $E^{f(n)}$

Abstract

In this paper we present a randomized algorithm for computing the collection of maximal layers for a point set in $E^{k}$ ($k = f(n)$). The input to our algorithm is a point set $P = \{p_1,\ldots ,p_n\}$ with $p_i \in E^{k}$. The proposed algorithm achieves a runtime of $O\left( kn^{2 - {1 \over \log {k}} + \log _k{\left( 1 + {2 \over {k+1}}\right) }}\log {n}\right)$ when P is a random order and a runtime of $O(k^2 n^{3/2 + (\log _{k}{(k-1)})/2}\log {n})$ for an arbitrary P. Both bounds hold in expectation. Additionally, the run time is bounded by $O(kn^2)$ in the worst case. This is the first non-trivial algorithm whose run-time remains polynomial whenever f(n) is bounded by some polynomial in n while remaining sub-quadratic in n for constant k (in expectation). The algorithm is implemented using a new data-structure for storing and answering dominance queries over the set of incomparable points.

Appendix

1.1 Proof of Lemma 3

Proof

Recall that T is a linear extension of P. Since p precedes q in T, $\mu (p) > \mu (q)$. Hence, $\exists$ $j^{'} \in \{1,\ldots ,k\}$ such that $p[j'] > q[j']$. Let $j' = {\text {argmax}}_{1 \le j \le k}{p[j]}$. We compute the probability ${\text {Pr}}[p[j] > q[j]\ |\ \mu (p) > \mu (q)]$ in two parts over the disjoint sets $\{j = j^{'}\}$ and $\{j \ne j^{'}\}$:

$$\begin{aligned} {\text {Pr}}[p[j] > q[j]\ |\ \mu (p) > \mu (q)]&= {\text {Pr}}[p[j] > q[j]\ |\ j = j^{'}, \mu (p) > \mu (q)]{\text {Pr}}[j = j^{'}] \nonumber \\&+ {\text {Pr}}[p[j] > q[j]\ |\ j \not = j^{'}, \mu (p) > \mu (q)]{\text {Pr}}[j \not = j^{'}]\nonumber \\&= 1{1 \over k} + \left( 1 - {1 \over 2}{\mu (q) \over \mu (p)}\right) \left( 1 - {1 \over k}\right) \end{aligned}$$

(3)

Since,

$$\begin{aligned} {\text {Pr}}[p[j] > q[j]\ |\ j \not = j^{'}, \mu (p) > \mu (q)]&= {{(\mu (p) - \mu (q))\mu (q) + {\mu (q)^2 \over 2}} \over {\mu (p)\mu (q)}}=1 - {1 \over 2}{\mu (q) \over \mu (p)} \end{aligned}$$

This follows from the fact that p[j] and q[j] are independent random variables uniformly distributed over $[0, \mu (p)]$ and $[0, \mu (q)]$ (given $\mu (p) > \mu (q)$) respectively. In the set $\{j = j^{'}\}$ clearly $p[j] > q[j]$. However, in the set $\{j \ne j^{'}\}$ the probability that $p[j] > q[j]$ is $\left( 1 - {1 \over 2}{\mu (q) \over \mu (p)}\right)$. We note that $\mu (p), \mu (q)$ are themselves random variables. More importantly they are i.i.d random variables having the following distribution:

$$\begin{aligned} {\text {Pr}}[\mu (p) < t] = t^{k} \end{aligned}$$

on the interval [0, 1]. This follows from how points in P are constructed. We take the expectation of both side of Eq. 3 over the event space generated by $\mu (p), \mu (q)$ on the set $\{\mu (p) > \mu (q)\}$:

$$\begin{aligned} {\text {Pr}}[p[j] > q[j]\ |\ \mu (p) > \mu (q)]&= 1 - {1 \over 2}\mathbb {E}\left[ {\mu (q) \over \mu (p)}\ \mid \ \mu (p) > \mu (q)\right] \left( 1 - {1 \over k}\right) \\&= 1 - {1 \over 2}\left( {k \over {k+1}}\right) \left( 1 - {1 \over k}\right) = 1 - {1 \over 2}\left( {{k - 1} \over {k+1}}\right) \end{aligned}$$

Now, let us compute, $\mathbb {E}\left[ {\mu (q) \over \mu (p)}\ \mid \ \mu (p) > \mu (q)\right]$. Recall that $\mu (p) = max_{1 \le i \le k}{p[j]}$. So the distribution function of $\mu (p)$ is,

$$\begin{aligned} F[\mu (p)] = {\text {Pr}}[\mu (p) < t] = {\text {Pr}}\left[ \bigwedge _{1\le i\le k}{p[i] < t}\right] = \prod _{1\le i\le k}{\text {Pr}}[p[i] < t] = t^k \end{aligned}$$

Where the second equality comes from that fact that each component of p are independent and identically distributed on [0, 1] with uniform probability. Hence,

$$\begin{aligned} \mathbb {E}\left[ {\mu (q) \over \mu (p)}\ \mid \ \mu (p) > \mu (q)\right]&= \int _{\mu (p) > \mu (q)}{{\mu (q) \over \mu (p)}}dF[\mu (p)]dF[\mu (q)] \\&= {{k^2} \over {{\text {Pr}}[\mu (p) > \mu (q)]}}\int _{0}^{1}{\int _{0}^{\mu (p)}{\mu (p)^{k-2}\mu (q)^{k}}d\mu (p)d\mu (p)} \\&= {k \over {k+1}} \end{aligned}$$

A similar argument can be used to prove the second claim.    $\square$

1.2 Solving a(w, d)

To simplify our calculations we modify the recurrence slightly: With $a(w,d) = k^d b(w, d)$, the recurrence equation becomes,

$$\begin{aligned} b(w,d) = {1 \over k^d}b(w-1,d-1) + {(1 - {1 \over k^d})}b(w-1,d) \end{aligned}$$

Let, $G_d(z) = \sum _{w = 0}^{\infty }{b(w,d)z^w}$. We note that $b(w, d) = 0$ when $w \le d$. Then we have,

$$\begin{aligned} G_d(z)&= {z \over k^d}G_{d-1}(z) + z(1 - {1 \over k^d})G_d(z)\\&= {z \over {k^d (1 - (1 - {1 \over k^d})z)}}G_{d-1}(z) \\&\ldots \\&= {{z^d} \over {\prod _{i = 1}^{d}{k^i(1 - (1 - {1 \over k^i})z)}}}G_0(z) \end{aligned}$$

But, $G_0(z) = \sum _{w = 0}^{\infty }{b(w, 0)z^w} = \sum _{w = 1}^{\infty }{z^w} = {z \over {1 - z}}$ as $b(w, 0) = a(w, 0) = 1$ when $w \ge 1$. Hence,

$$\begin{aligned} b(w, d)&= k^{-d(d + 1)/2}[z^{w - d - 1}] G_d(z)\nonumber \\&= k^{-d(d + 1)/2}[z^{w - d - 1}]{{1} \over {(1 - z)\prod _{i = 1}^{d}{(1 - (1 - k^{-i})z)}}} \end{aligned}$$

(4)

where the notation $[z^i]p(z)$ means the coefficient of $z^i$ in the polynomial p(z) as usual. Using partial fractions: Let,

$$\begin{aligned} {{1} \over {(1 - z)\prod _{i = 1}^{d}{(1 - (1 - k^{-1})z)}}} \equiv {\beta _0 \over {1 - z}} + \sum _{i = 1}^{d}{\beta _i \over {(1 - (1 - k^{-i})z)}} \end{aligned}$$

For which we get the following solution,

$$\begin{aligned}&\beta _0 = k^{d(d + 1)/2} \\&\beta _i = {{k^{d(d + 1)/2}(1 - k^{-i})^d} \over {\prod _{j \ne i, j \ge 1}^{d}{(1 - k^{j - i})}}} \end{aligned}$$

Substituting these in Eq. 4 above we get, $b(w,d) = 1 - \sum _{i = 1}^{d}{{(1 - k^{-i})^{w - 1}}\over {\prod _{j = 1, j \ne i}^{d}{(1 - k^{j - i})}}}$, which gives us the desired result for a(w, d).