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Lecture 5: k-center Problem
1 Introduction
Imagine we have a complete, undirected graph, where each node is a city, and edges represent the shortest distance between these cities. We have funds to build exactly k emergency centers. The k-center problem with triangle inequality is to place our k emergency centers (e.g. marked by Xs in the figure to the right) such that no one has to go too far to get to their closest center.
k-center Problem:
Input: Given G = (V, E), a complete undirected graph whose edges are shortest paths between each pair of nodes. Let Dij denote the path distance between nodes i and j. (Remember, if we start with an incomplete graph, we can make it complete by adding edges where Dij is the length of the shortest existing path between i and j.)
Output: A subset of nodes S ⊆ V with |S| = k, such that the longest distance of a node to its closest node in S is minimized. Specifically, we want to minimize cost(S) = maxj∈V mini∈S Dij.
Our Approximation Algorithm:
We assume that G satisfies the triangle inequality (i.e. Dij + Djk ≥ Dik, ∀i, j, k ∈ V ). So first, we reorder the edges e 1 , e 2 ..., em, in order of cost, such that cost(e 1 ) ≤ cost(e 2 ) ≤ ... ≤ cost(em). Then, we add the lightest edge e 1 and look at that graph, then add the edges e 1 , e 2 , and look at that
Figure 1: G 3 = (V, E 3 ) where E 3 = e 1 , e 2 , e 3
graph, then add edges e 1 , e 2 , e 3 ... Let Gi = (V, Ei), where Ei = {e 1 , e 2 , ..., ei}. Note that our original graph G is now Gm.
Definition 1.0.1 A dominating set of G is a subset S ⊆ V such that every node in V − S is adjacent to a vertex in S. (That is, for each node, either you’re in, or you have a neighbor who is in the dominating set.)
Claim 1.0.2 The optimal solution to a k-center problem is a dominating set in G. (Note that this is a trick question because G is complete. Any one vertex or set of vertices in a complete graph is a dominating set!)
Claim 1.0.3 The optimal solution to a k-center problem in G is a dominat- ing set in Gi for some i ≥ i 0.
This claim is trivially true (when i = m). Look at graphs G 1 , G 2 , G 3 ,..., Gm− 2 , Gm− 1 , Gm. As we move backward from Gm, at which point do we no longer have a dominating set?
Let C∗^ be the cost of an optimal solution to k-center in G. Let ec be the LAST edge of cost c∗. Remember, this edge is not unique! Multiple edges could have the same cost, so ec is the LAST edge, such that for all ec+1...ei, cost(ec+1...) > cost(gc).
Figure 3: Given G = (V, E), E is represented by solid edges, and E^2 is represented by solid and dashed edges.
Notes:
- We can compute the square of a graph by multiplying the adjacency matrices.
- The cube of graph G, denoted G^3 , adds edges between i and j if there exists a path of length 1, 2, or 3 between i and j. This can be extended to create G^4 , G^5 , ...
- This makes no difference for G (which is complete), but we are also going to be looking at Gc, for different c’s, which are certainly not complete graphs.
Definition 1.0.7 An independent set in a graph G = (V, E) is a set S ⊆ V , such that ∀i ∈ S, if (i, j) ∈ E, then j 6 ∈ S.
Definition 1.0.8 A maximal independent set (MIS) in a graph G = (V, E) is an independent set such that ∀v ∈ V , either v ∈ S, or ∃u such that (u, v) ∈ E, and u ∈ S.
Finding a maximum independent set of a graph is NP-hard, but finding a maximal independent set of a graph can be solved in polynomial time using a simple greedy algorithm.
Lemma 1: Let H = (V, E). Let I be an independent set in H^2. Then |I| ≤ |dom(H)|, where dom(H) denotes the size of a minimum cardinality dominating set in H. Note: dom(H) is NP-hard to compute, but any inde- pendent set is smaller than this.
Proof: Let D be a minimum cardinality dominating set in H. For each vertex d ∈ D, its neighborhood forms a clique in H^2. So H^2 contains |D| cliques spanning all the vertices, which implies any independent set in H^2 can pick at most 1 vertex per clique. So |I| ≤ |D|.
If we start with a vertex and its neighbors and square it, we have a clique! Therefore, we can only take 1 vertex from each of these cliques / neighborhoods.
2 2-Approximation Algorithm for k-center
Algorithm A:
- Construct G^21 , G^22 ,... ,G^2 m.
- Compute a maximal independent set (MIS) Li in each graph G^2 i.
- Find the smallest index i, such that |Li| ≤ k, and call that MIS Lj
- Return Lj
Lemma 2: For j as defined in the algorithm above, the cost(ej ) ≤ C∗. Note that cost(ej ) is the most expensive edge in Lj.
Proof: For every i < j, we have |Li| > k since dom(Gi) ≥ |Li| by Lemma
- (That is, if we go through and look at G^21 , G^22 , G^23 , ... , G^2 m, we run out of our k centers before we reach G^2 m. That implies dom(Gi) > k. So the first index for which the k-center problem forms a dominating set > i, so C∗^ > cost(ei).)
Theorem 2.0.9 Algorithm A returns a solution of cost at most 2 ∗ OP T.
Proof: Observe that a maximal independent set in H^2 is also a dominating set in H^2 (any maximal independent set is a dominating set, but not vice
Thus, the the approximation algorithm will output a solution of cost 1 if there is a dominating set in G and output a solution of cost 2 otherwise. So we can use the approximation algorithm to decide whether there is a dominating set in G. Therefore, the approximation algorithm is also NP-hard.
4 Weighted k-center Problem
Idea: We can build as many emergency centers as we like, so long as we don’t exceed our budget of k. It may cost a different amount to build an emergency center at one location versus another.
This is actually the same problem as before, but we are given a weight func- tion on the vertices, w : v → R+, and a bound w ∈ R+. Our goal is to pick S ⊆ V such that w(S) ≤ W , where w is the weight function and W is the specified maximum total weight. Specifically, we want to minimize cost(S) = maxv∈V mins∈S cost(evs). Note that this reduces to the ordinary k-center problem when all vertex weights are 1 and W = k.
Let wdom(G) denote the weight of the min-weight dominating set in G, then with respect to the graph Gi (containing the i lightest edges), we need to find the smallest index i such that wdom(Gi) ≤ W. Notice that if i∗^ is the index, then the cost of the optimal solution is OP T (cost(ei∗^ ))
Given a vertex weighted graph H, let I be an independent set in H^2. For each u ∈ I, let S(u) denote the lightest (i.e. least expensive by weight) neighbor of u in H. We will allow u to be its own neighbor. Let S = {s(u)|u ∈ I}.
Lemma 3: w(S) ≤ wdom(H)
Proof Let D be a min-weight dominating set of H, I claim that I can con- struct a matching between vertices in I and vertices in D who are neighbors that saturates I, because
- Every 2 vertices in I cannot have a common neighbor in D 2) Each vertex in I has a neighbor in D because D is dominating.
Remains to show: each vertex in S weighs less than its corresponding
vertex to I’s matched vertex in D. That’s because we picked the neighbor or I of min-weight (weight at most its neighbor in D).
5 3-Approximation Algorithm for Weighted
k-center:
Algorithm B:
- Construct G^21 , G^22 , G^23 , ..., G^2 m.
- Compute a MIS Li in each graph G^2 i.
- Compute Si = {si(u)|u ∈ Li}.
- Find minimum index i such that w(Si) ≤ W , say j.
- Return Sj.
Theorem 5.0.11 Algorithm B is 3 ∗ OP T for weighted k-center.
Proof Let Si be the solution from the algorithm. Let OP T be the cost of the optimal solution. It holds that w(Si) > W ∀i < j. Therefore, by this and Lemma 3, wdom(Gi) > W ∀i < j. Hence the optimal cost of weighted k-center is at least the cost of the most expensive edge in Gj (i.e. the most expensive edge(s) were 3 hops away). So we know that cost(e) ≤ OP T for every edge in Gj.
Now, notice that Lj is a dominating set in G^2 i. So every vertex in G is on a path of length at most 1 from some vertex in Lj in G^2 i. (Notice that we might have taken a neighbor that was lighter!) Therefore, every vertex is on a path of length at least 3 from the neighbors of the vertices in Lj that we chose in our set.
Since every edge in Gj has weight at most OP T , the triangle inequality guarantees that the edges connecting any vertex in G to some vertex in Sj weigh no more than 3 ∗ OP T.
