Discrete (Objective) Space – problem Q D

Given a discrete set of points A ⊆R²+and a rational error tolerance  > 0, we want to compute a minimum cardinality -convex Pareto set of A; the approximate set is allowed to contain only points from A.

We essentially use the same notation here as in the previous subsection. Let P (A) = {p₁, . . . , p_n} denote the Pareto set of A and CP (A) its convex Pareto set. For simplicity, we omit the subscript D and the input set A from some expressions. For example, we denote CPinstead of CP_D(A, ) for a solution to QD(A, ).

Clearly, it suffices to consider the points in P (A) for the approximation, in the sense that there always exists an optimal solution contained in P (A). So, we can assume that A = P (A). We also observe that points in P (A) \ CP (A) (i.e. points dominated by convex combinations of others) - even points of P (A) that do not lie on the lower envelope LE - are actually necessary for the approximation. It is easy to see that if we ignore such points, we lose at most a factor of 2; and as shown in the next section, this is tight.

We first show that a structural lemma very similar to lemma 5.2.5 holds in this setting also. The proof is almost identical, but we sketch it here for completeness.

Lemma 5.2.13. The problem Q_D(A, ) is equivalent to the following: Compute a convex polygonal chainC with minimum number of vertices in P (A) having the following properties: (i) its leftmost (resp. rightmost) vertex(1 + )-covers the leftmost (resp. rightmost) point of P (A) (ii) the curve C lies betweenLE and LE⁰_.

Proof. For necessity, consider an -convex Pareto set CP = {q1, q2, . . . , qr}, with x(qi) < x(qi+1), i ∈ [r − 1], having no redundant points. As observed above, we can assume w.l.o.g. that q_i ∈ P (A)

for all i ∈ [r]. By the definition of P (A), CP defines a polygonal chain C = hq1, . . . , qri in R² such that the function y = C(x) is strictly decreasing. The chain is convex, since otherwise there would exist a redundant point in CP. By definition, for any point in P (A) there exists a point in C that (1 + )-covers it. Similarly to the convex case, we can argue about the necessity of property (i).

Note that, by construction, it suffices to argue that no point of C can be strictly dominated by any point in LE⁰_. For contradiction, suppose that there exists a point c ∈ C strictly dominated by some point in LE⁰_. Then, we claim that there exists a point in CP (A) not (1 + )-covered by any point of C. Sufficiency is also based on arguments parallel to those in Lemma 5.2.5.  In view of this similarity between the problem Q_D and its convex counterpart Q_C, a naive approach to compute CP_D^∗(A, ) would be the following: Given P (A) and , compute its lower envelope LE (the convex polygonal curve having CP (A) as its vertex set) and its scaled counterpart LE⁰_. Select as the leftmost point q1in the approximation the rightmost point of P (A) that (1 + )-covers p₁ and at each iteration select as q_i+1the rightmost point of P (A) -visible from q_i. It is not hard to construct examples for which this approach is suboptimal. It thus turns out that the greedy criterion“pick as next the rightmost point visible from the current one” fails for this variant of the problem.

We describe next a modified criterion that works. It can be assumed that P (A) does not contain any points that are strictly dominated by LE⁰_; such points are redundant and cannot be part of an optimal solution. Let us now define a total order on the points of P (A).

Definition 5.2.14. For p, q ∈ P (A) we say that p is -better than q (we denote this by p _ q) if either (i) the rightmost vertex vp ∈ LE⁰_ -visible from p, lies to the right of the corresponding vertex vqfor q, or (ii) vp = vq = v and the line (defined by the segment) pv lies above the line qv to the right of v.

To simplify the exposition, we extend the definition to sets. Given a set of points S, we say that the point b ∈ S is an -best point in the set if for any other point y ∈ S it holds b  y. We note that there may exist more than one points with this property; in such a case we can arbitrarily pick one of them. The modified algorithm selects a set of points Q ⊆ P (A) as follows: For the computation of the leftmost point q1 ∈ Q, consider the set of eligible points E₁ = {σ ∈ P (A) | x(σ) ≤ (1 + )x(p1)}. If there exists a point in E1 that (1 + )-covers pn, select it and halt.

Otherwise, select an -best point in E1. For each i ≥ 2, select qi from the set (of eligible points) E_i = {σ ∈ P (A) | x(σ) > x(q_i−1) and σ is  − visible from q_i−1}. If one of the points in E_i (1 + )-covers pn, select it and halt. Otherwise, select an -best point in Eiand iterate.

The intuitive justification of the modified criterion is the following: From the analysis of the convex case, we know that a point of the lower envelope is a better choice than all the points to its left (for any  > 0). However, as opposed to the convex case, not all such points can be selected. In any case, a point of CP (A) is a better choice than all the points to its left. So, suppose that we have selected the i-th point qi and consider all the points to its right -visible from it (call this set Ei+1);

the next point qi+1must be one of them. Let v denote the rightmost vertex in Ei+1∩ CP (A). By the analysis of the convex case, we can ignore all the points of Ei+1that lie to the left of v. But how can we “compare” v to the remaining points of Ei+1? These are undominated points that lie to its right, but none of them is in CP (A). As mentioned, it is not always the case that the rightmost point is the correct choice. It turns out that this can be done by comparing the corresponding visibility regions.

Lemma 5.2.15. The set Q output by the above algorithm is an -convex Pareto set of minimum cardinality.

Proof. As in the convex case, it is straightforward to verify that the two properties of Lemma 5.2.13 are satisfied, thus the set Q forms an -convex Pareto set. Let CP_^∗ = {p^∗₁, . . . , p^∗_k} denote the smallest -convex Pareto set. For optimality, we will argue that for all i ∈ [k] it holds p^∗_i ∈ E_i; by construction, this implies the desired result. This is clearly true for i = 1; the leftmost point of any

-convex Pareto set must (1 + )-cover p1, so p^∗₁ ∈ E₁. The proof follows from the next claim:

Claim 5.2.16. For all i ∈ [k − 1], if p^∗_i ∈ E_ithen p^∗_i+1∈ E_i+1.

Proof. The claim holds in vacuum for k = 1. (If k = 1, then a point in E1(1 + )-covers pn; such a point is selected as q₁and forms the minimum cover.) We need to consider the case k ≥ 2. Suppose that for some i ∈ [k − 1] it holds p^∗_i ∈ E_i; either qi ≡ p^∗_i or qi _ p^∗_i. This means that the rightmost vertex v⁰ ∈ CP (A) -visible from qi lies to the right of the rightmost vertex v ∈ CP (A) -visible from p^∗_i. Notice that we can assume w.l.o.g. that the point p^∗_i+1(is equal to or) lies to the right of v. This follows from the analysis of the convex case. The following fact is straightforward from the definition of the relation “” and finishes the proof of the claim:

Algorithm Explicit–Discrete–2D.

Input:P (A) = {p₁, p₂, . . . , p_n} and  > 0.

Output:Q = {q1, . . . , qr} (optimal -convex Pareto set).

Construct the polygonal chain LE⁰_; E₁ = {σ ∈ P (A) | x(σ) ≤ (1 + )x(p₁)};

If ∃q ∈ E1: y(q) ≤ (1 + )y(pn)
then { Q = {q}; halt; } Compute the point q1 =  − best(E1);

Q = {q₁} ; i = 1;

While y(qi) > (1 + )y(pn)
do

{ E_i+1= {σ ∈ P (A) | x(σ) > x(q_i) and σ is  − visible from q_i};

If ∃q ∈ Ei+1: y(q) ≤ (1 + )y(pn)
then { Q = Q ∪ {q}; halt; } Compute the point q_i+1=  − best(E_i+1);

Q = Q ∪ {qi+1};

i = i + 1; }

Table 5.2: Optimal algorithm for explicit two-dimensional discrete case.

Fact 5.2.17. Let x, y ∈ Eiwithx y. Let v be the rightmost vertex v ∈ CP (A) -visible from y.

Thenx -sees v and any point to the right of v -visible from y.

By this fact, p^∗_i+1is -visible from qi ∈ E_i (since it is visible from p^∗_i ∈ E_i and qi _ p^∗_i). Thus,

p^∗_i+1∈ E_i+1. 

 The algorithm is easily seen to run in time O(n²), where n = |P (A)|.