This document discusses the metric K-center problem and algorithms for approximating its solution. It shows that metric K-center is NP-hard by reducing it to the dominating set problem. It also proves that metric K-center cannot be approximated within a factor of 2, unless P=NP. The document then describes Gonzalez's algorithm, which achieves an approximation ratio of 2 by greedily selecting centers to maximize the minimum distance from existing centers.

1. Metric K-Center
Ken Liu
CoreTech
Trend Micro Inc.

2. Outline
• Preliminary
• NP-hard
• Unapproximability
• Gonzalez Algorithm
• Approximation ratio
• Conclusion

3. Preliminary
• A metric space is a pair (M, d) where M is a set of nodes
and d is a distance function on M such that:
– d(x, y) >=0 (non-negative),
– d(x, y) = 0 iff x=y (identity of indiscernibles),
– d(x, y) = d(y, x) (symmetry) and
– d(x, y) <= d(x, z) + d(y, z) (triangle inequality)
• Notation overloading
– Given x ∈ M and Y ⊆ M, let d(x, Y) = min {d(x, y): y ∈ Y}

4. Metric K-Center Problem
• Given a metric space (M, d), find k center
C = {c1, c2, ..., ck} to minimize the radius
– max { d(x, C): u ∈ M}
• Geometric view
– Voronoi diagram
– Each node is covered by some center
c1
c2
c3
c4

5. NP-hard
• Dominating set
– Given a graph G=(V, E), a k-dominating set of size k is a set C = {c1, c2, ..., ck}
⊆ V such that each node v ∈ V is adjacet to some node in C.
– To determine if a graph has a k-dominating set is known to be NP-hard
• Theorem: Metrick K-center is NP-hard
Proof:
– Given G = (V, E), let d(x, y) = 2 if (x, y) ∉ E and d(x, y) = 1 otherwise.
– (V, d) is a metric space.
– G has a k-dominating set => radius of optimal k centers for (V, d) is 1
– G has no k-dominating set => radius of optimal k centers for (V, d) is 2
– G = (V, E) has a k-dominating  radius of optimal k centers for (V, d) is 1

6. Unapproximibility
• Theorem: Assuming NP≠P, Metrick K-center cannot be
approximated within 2.
Proof:
– Given G = (V, E), let d(x, y) = 2 if (x, y) ∉ E and d(x, y) = 1 otherwise.
– G has a k-dominating set => radius of optimal k centers for (V, d) is 1
– G has no k-dominating set => radius of optimal k centers for (V, d) is 2
– Assume for contradiction that we have a polynomial (2 –ε)-
approximation algorithm called A
– G has a k-dominating set  A(V, d) < 2 - ε
– So we have a polynomial algorithm for the k-dominating set, -><-

7. Gonzalez’s Algorithm (1985)
• Input: a metric space (M, d) and a positive integer
• Output: a set of k centers
1. C = {}
2. c1 := a randomly chosen node in M
3. C.add(c1)
4. for i in {2, ..., k}:
ci := the node c ∈ M – C with max d(c, C)
C.add(ci)
5. return C

8. Approximation ratio
• Theorem: Gonzalez’s algorithm has an approximation ratio of 2
• proof:
– Let C = {c1, c2, ..., ck} be the output of Gonzalez algorithm and r be its
radius.
– Let ck+1 be the farest node from C. Note that
• d(ci, cj) >= r for all i ≠ j
– Let C’ = {c’1, c’2, ..., c’k} be the optimal solution and r* be its radius
– By Pigenhole theorem we have two nodes ci, cj being covered by the
same center c’ in C’
– r <= d(ci, cj) <= d(ci, c’) + d(cj, c’) <= d(ci, C’) + d(cj, C’) <= 2r*
c’
ci
cj
<=r*
>=r

9. Conclusion
• Gonzalez’s algorithm is simple yet optimal

10. Conclusion
• Gonzalez’s algorithm is simple yet optimal