Clustering a 2d Pareto Front: P-center Problems Are Solvable in Polynomial Time

Abstract

Having many non dominated solutions in bi-objective optimization problems, this paper aims to cluster the Pareto front using Euclidean distances. The p-center problems, both in the discrete and continuous versions, become solvable with a dynamic programming algorithm. Having N points, the complexity of clustering is $O(KN\log N)$ (resp. $O(KN\log ^2 N)$) time and O(N) memory space for the continuous (resp. discrete) K-center problem for $K\geqslant 3$, and in $O(N\log N)$ time for such 2-center problems. Furthermore, parallel implementations allow quasi-linear speed-up for the practical applications.

Appendix A: Proof of the Lemmas 2 and 4

Proof of Lemma 2: $\forall k \in [\![i,i']\!], \; {\parallel } x_j - x_k {\parallel } \leqslant \max \left( {\parallel }x_j - x_i {\parallel },{\parallel } x_j - x_i {\parallel }\right)$, using Proposition 1. Then:

$f_{ctr}^{{\mathcal {D}}}(P) = \min _{j \in [\![i,i']\!], x_j \in P} \max \big ( \max \left( {\parallel }x_j - x_i {\parallel },{\parallel }x_j - x_i {\parallel } \right) , \max _{ k \in [\![i,i']\!]} {\parallel }x_j - x_k {\parallel } \big )$

$f_{ctr}^{{\mathcal {D}}}(P) = \min _{j \in [\![i,i']\!], x_j \in P} \max \left( {\parallel }x_j - x_i {\parallel }, {\parallel } x_j - x_{i'} {\parallel } \right)$. It proves (16). We prove now a stronger result than (15): the application $x \in {\mathbb {R}}^2 \longmapsto \max _{p \in P} {\parallel }x - p {\parallel } \in {\mathbb {R}}$ as a unique minimum reached for $x = \frac{x_i + x_j}{2}$:

$\displaystyle \forall x \in {\mathbb {R}}^2-\left\{ \frac{x_i + x_j}{2}\right\} , \, \max _{p \in P} {\parallel }x - p{\parallel } > \frac{1}{2} {\parallel } x_i - x_j {\parallel } = \max _{p \in P} {\parallel }\frac{x_i + x_j}{2} - p {\parallel } (19)$

We prove (19) analytically, denoting $\text{ diam }(P) = \frac{1}{2} {\parallel }x_i - x_j {\parallel }$. We firstly use the coordinates defining as origin the point $O = \frac{x_i + x_j}{2}$ and that $x_i = (- \frac{1}{2} \text{ diam }(P),0)$ and $x_j = ( \frac{1}{2} \text{ diam }(P),0)$. Let x a point in ${\mathbb {R}}^2$ and $(x_1,x_2)$ its coordinates. We minimize $\max _{p \in P} d(x,x_p)$ distinguishing the cases:

if $x_1>0$, $d(x,x_i)^2 = (x_1 + \frac{1}{2} \text{ diam }(P))^2 +x_2^2 \geqslant (x_1 + \frac{1}{2} \text{ diam }(P))^2 > (\frac{1}{2} \text{ diam }(P))^2$

if $x_1<0$, $d(x,x_j)^2 = (x_1 - \frac{1}{2} \text{ diam }(P))^2 +x_2^2 \geqslant (x_1 - \frac{1}{2} \text{ diam }(P))^2 > (\frac{1}{2} \text{ diam }(P))^2$

if $x_1=0$ and $x_2 \ne 0$, $d(x,x_i)^2 = (\frac{1}{2} \text{ diam }(P))^2 +x_2^2 > (\frac{1}{2} \text{ diam }(P))^2$

In these three sub-cases, $\max _{p \in P} d(x,x_p) \geqslant d(x,x_i) >\frac{1}{2} \text{ diam }(P)$. The three cases allow to reach any point of ${\mathbb {R}}^2$ except $x_0 = \frac{x_i + x_j}{2}$. To prove the last equality, we use the coordinates such that $x_i = (- \frac{1}{2} \text{ diam }(P);0)$ and $x_j = ( \frac{1}{2} \text{ diam }(P);0)$. The origin $x_0$ has coordinates $(\frac{1}{2 \sqrt{2}} \text{ diam }(P), \frac{1}{2 \sqrt{2}} \text{ diam }(P))$. Let $x=(x^1,x^2) \in P$, such that $x\ne x_i, x_j$. Thus $x_i \prec x \prec x_j$. The Pareto dominance induces $0 \leqslant x_1,x_2\leqslant \frac{1}{\sqrt{2}} \text{ diam }(P)$. $d(x,x_0)^2= (x_1 - \frac{1}{2 \sqrt{2}} \text{ diam }(P))^2 + (x_2 - \frac{1}{2 \sqrt{2}} \text{ diam }(P))^2$

$d(x,x_0)^2 \leqslant (\frac{1}{2 \sqrt{2}} \text{ diam }(P))^2 + (\frac{1}{2 \sqrt{2}} \text{ diam }(P))^2 = 2 \frac{1}{8} \text{ diam }(P))^2$

$d(x,x_0) \leqslant \frac{1}{2} \text{ diam }(P)$, which proves (19) as $d(x_0,x_i) = d(x_0,x_j) = \frac{1}{2} \text{ diam }(P)$. $\square$

Proof of Lemma 4: Let $i<i'$. We define $g_{i,i',j}, h_{i,i',j}$ with:

$g_{i,i'} :j \in [\![i,i']\!] \longmapsto {\parallel } x_j - x_i {\parallel }$ and $h_{i,i'} :j \in [\![i,i']\!] \longmapsto {\parallel } x_j - x_{i'} {\parallel }$

Using Proposition 1, g is strictly decreasing and h is strictly increasing.

Let $A=\{j \in [\![i,i']\!] | \forall m \in [\![i,j]\!] g_{i,i'}(m) < h_{i,i'}(m) \}$. $g_{i,i'}(i) = 0$ and $h_{i,i'}(i) = {\parallel } x_{i'} - x_i {\parallel }>0$ so that $i \in A$, $A\ne \emptyset$. We note $l=\max A$. $h_{i,i'}(i') = 0$ and $g_{i,i'}(i') = {\parallel }x_{i'} - x_i {\parallel }>0$ so that $i' \notin A$ and $l<i'$. Let $j \in [\![i,l-1]\!]$. $g_{i,i'}(j) < g_{i,i'}(j+1)$ and $h_{i,i',j}(j+1) < h_{i,i'}(j)$. $f_{i,i'}(j+1)= \max \left( g_{i,i'}(j+1),h_{i,i'}(j+1)\right) = h_{i,i}(j+1)$ and $f_{i,i'}(j)= \max (g_{i,i'}(j),h_{i,i'}(j)) = h_{i,i}(j)$ as $j,j+1 \in A$. Hence, $f_{i,i'}(j+1)= h_{i,i'}(j+1) < h_{i,i'}(j) = f_{i,i'}(j)$. It proves that $f_{i,i'}$ is strictly decreasing in $[\![i,l]\!]$. $l+1 \notin A$ and $g_{i,i'}(l+1) > h_{i,i'}(l+1)$ to be coherent with $l=\max A$. Let $j \in [\![l+1,i'-1]\!]$. $j+1 > j \geqslant l+1$ so $g_{i,i'}(j+1)> g_{i,i'}(j) \geqslant g_{i,i'}(l+1)> h_{i,i'}(l+1) \geqslant h_{i,i'}(j) > h_{i,i'}(j+1)$. It implies $f_{i,i'}(j+1)= g_{i,i'}(j+1)$ and $f_{i,i'}(j)= g_{i,i'}(j)$ and $f_{i,i'}(j+1) < f_{i,i'}(j)$. $g_{i,i'}(j) < g_{i,i'}(j+1)$ and $h_{i,i',j}(j+1) < h_{i,i'}(j)$. $f_{i,i'}(j+1)= \max (g_{i,i'}(j+1),h_{i,i'}(j+1)) = g_{i,i}(j+1)$ and $f_{i,i'}(j)= \max (g_{i,i'}(j),h_{i,i'}(j)) = h_{i,i}(j)$ as $j,j+1 \in A$. Hence, $f_{i,i'}(j+1)= h_{i,i'}(j+1) > h_{i,i'}(j) = f_{i,i'}(j)$, $f_{i,i'}$ is strictly increasing in $[\![l+1,i']\!]$. $\square$