Complexity of Approximation

1. Complexity of Approximation How hard is it to approximate this problem?

2. Traveling Salesman • Given n cities with a distance table, find a minimum total-distance tour to visit each city exactly once.

3. Approximate Traveling Salesman

4. Theorem Proof: Given a graph G=(V,E), define a distance table on V as follows:

5. Contradiction Argument • Suppose r-approximation exists. Then we have a polynomial-time algorithm to solve Hamiltonian Cycle as follow: r-approximation solution <r |V| if and only if G has a Hamiltonian cycle

6. Many-One Reductions with Gap

7. In general: Then, (/)-approx- is NP-hard.

8. For max problem , if a reduction • <satisfies Then, (/)-approx-is as hard as  Proof If opt > then any (/)-approx must be >opt > If opt <  then any approx must be >  So, from the output of a (/)-approx, we can tell whether x is in 

9. Example Dominating Set: Given a graph G = (V, E) and an integer k, determine whether G has a dominating set of size < k. (known as NP-complete) Metric-k-centers: Given a graph with metric distance between vertices and an integer k, select k vertices as centers so that the max distance of any vertex to its nearest center is minimized. Metric-k-centers has no (2-)-approx unless DS in P

10. Reduction with gap from DS to Metric k-Centers Maps (G,k) to (G,k,d) where the distance d is: G has a dominating set D of size k, select D as the k centers, with max distance 1. G does not have a dominating set D of size k any k-center set C has max distance > 2. Gap is 

11. Planar-CVC-4: given a planar graph with max degree 4, and integer k > 0, find a connected vertex cover of G of size k. NP-complete Bottleneck Steiner Tree in Rectilinear Plane: Given n terminals on the rectilinear plane and an integer k, find a Steiner tree of k Steiner points that minimizes the longest edge length. has no (2-)-approx unless P = NP

12. Reduction with gap 2 • Embed Planar-CVC-4 instance G on the rectilinear plane with edge length at least 2k+2. • Create P(G): G P(G)

13. Gap Amplification and Preservation

14. Edge Disjoint Path Given a graph G and k pairs of vertices, find disjoint paths that connect the max number of given pairs. EDP-2 (with k = 2) is NP-hard EDP has an NP-hard gap [1gap

15. Gap-Amplification Reductionfrom EDP-2 to EDP

16. Gap [1+ , 2] of EDP-2 Gap [1+, k] of EDP Choose 2 # of edges = m = (k(k-1)/2) |E| + k We get: EDP does not have a poly-time m -approximation, unless P = NP, (for any  < ¼). ½- 

17. Max-3Lin Given a system of 3-variable linear equations over GF(2), find an assignment of variables satisfying the max number of equations Max-3Sat

18. If an assignment satisfies the lin eq then it satisfies 4 clauses; If it does not satisfy the lin eq then it can satisfy only 3 clauses.

19. So, this is a reduction amplifying the gap [0.5+of Max-3Lin to gap [3.5+–] of Max-3Sat. From PCP work, it is known that Max-3Lin has an NP-hard gap [0.5+ So, (8/7)-approx-Max-3Sat is NP-hard.

20. Theorem

21. Max-3Lin < Min-VC

22. Now, connect two vertices if their labels are conflicting. If an assignment to xi's satisfies k vertices, then these vertices are independent, and so the remaining 4m – k vertices form a vertex cover. A gap [0.5+e, 1-e] of Max-3Lin a gap [3+e, 3.5-e] of Min-VC

23. The class APX • A problem  is in APX if r-approx- is poly-time solvable for some r > 0. • A problem is APX-complete if every problem in APX is E-reducible to it. • An APX-complete problem does not have a PTAS, unless P = NP. • E.g: Bottlenect Steiner Tree, Metric-k- Centers, Max-3Sat are APX-complete.

24. (under certain assumption about complexity classes) c not (n )-approx for some c Clique not c-approx for some c not (clog n)-approx for some c Set Cover APX- complete Max 3Sat APX TSP- MAXSNP- complete PTAS Bin Packing Knapsack FPTAS

25. L-reduction

27. E-Reduction The error ratio is preserved where E-reduction is a little weaker than L-reduction.

28. MAX SNP A subclass of APX. Many problems are proved L-complete for MAX SNP. L-complete MAX SNP E-complete for APX

29. Theorem This is an important result proved using PCP system. Theorem

30. Properties (P1) (P2)

33. (P2) APX PTAS

38. MAX3SAT-3 Theorem

39. VC-b VC-4 is MAX SNP-complete Proof.

41. Assume F has n variables and m clauses. If an assignment satisfies k clauses, then the corresponding vc is of size n + 3m – k. Since opt(SAT) > m/2, we get opt(VC) = n + 3m – opt(SAT) < 4m < 8 opt(SAT) Also, |opt(VC) – appr(VC)| = |opt(SAT) – appr(SAT)| This is an L-reduction.

42. Theorem

43. 1 2 1 2 v 3 3 4 5 4 5 G G’

44. Each vertex can cover at most b edges

45. For a vc S' of G', define _ S = { v | cvS' } Then, S is a vc of G, and Now, opt(G') = opt(G) + 2m – n

46. Theorem Proof.

47. VC-3 < VC-CG(VC in cubic graph) Assume: G has i vertices of degree 1, jvertices of degree 2. deg(x)=2 deg(y)=1

48. (L1) H has a min vc of size 3(2i+j) i + 2j < 6 opt(G) G has > (i+2j)/2 edges but each vertex can cover at most 3 edges So, opt(G') < opt(G) + 3(2i+j) < 37 opt(G). (L2) opt(G') = opt(G) + 3(2i+j) appr(G') = apprx(G) + 3(2i+j) |opt(G) – appr(G)| = |opt(G') – appr(G')|

49. Maj-DS Given a graph G, find a vertex subset D such that for each v not in D, at least half of its neighbors are in D VC-CG < Maj-DS

50. G If e6 incident on v4 G'