1. Complexity 1 Hardness of Approximation

2. Complexity 2 Introduction Objectives: –To show several approximation problems are NP-hard Overview: –Reminder: How to show inapproximability? –Probabilistic Checkable Proofs –Hardness of approximation for clique

3. Complexity 3 Optimization Problems Consider an optimization problem P: instances: x 1,x 2,x 3,… optimization measure feasible solutions all graphs Example: all cliques in that graph the clique’s size (max)

4. Complexity 4 Each Instance Has an Optimal Solution OPT x1x1 x2x2 x3x3 x4x4

5. Complexity 5 Approximation (Max Version) OPT xixi

6. Complexity 6 How To Show Hardness of Approximation? Hardness of distinguishing far off instances  Hardness of approximation OPT AB gap xixi

7. Complexity 7 Gap Problems (Max Version) Instance: … Problem: to distinguish between the following two cases: The maximal solution  B The maximal solution ≤ A

8. Complexity 8 Formally: Claim: If the [A,B]-gap version of a problem is NP-hard, then that problem is NP-hard to approximate within factor B/A.

9. Complexity 9 Formally: Proof: Suppose there is an approximation algorithm that outputs C so that C/C*≤B/A A proper distinguisher: * If C  B, return ‘YES’ * Otherwise return ‘NO’

10. Complexity 10 Proof Since C*≥AC/B, (1) If C>B (we answer ‘YES’), then necessarily C*>A (the correct answer cannot be ‘NO’). (2) If C*≤A (the correct answer is ‘NO’), then necessarily C≤B (we answer ‘NO’)

11. Complexity 11 Idea We’ve shown “standard” problems are NP-hard by reductions from 3SAT. We want to prove gap-problems are NP- hard, Why won’t we prove some canonical gap- problem is NP-hard and reduce from it? If a reduction reduces one gap-problem to another we refer to it as approximation-preserving

12. Complexity 12 Gap-3SAT[  ] Instance: a set of clauses {c 1,…,c m } over variables v 1,…,v n. Problem: to distinguish between the following two cases: There exists an assignment which satisfies all clauses. No assignment can satisfy more than 7/8+  of the clauses.

13. Complexity 13 Gap-3SAT: Example ( x 1  x 2  x 3 ) ( x 1   x 2   x 2 ) (  x 1  x 2  x 3 ) (  x 1   x 2   x 2 ) (  x 1  x 2   x 3 ) (  x 3   x 3   x 3 )  = { x 1  F ; x 2  T ; x 3  F } satisfies 5/6 of the clauses

14. Complexity 14 Why 7/8? Claim: For any set of clauses with exactly three independent literals, there always exists an assignment which satisfies at least 7/8.

15. Complexity 15 The Probabilistic Method Proof: Consider a random assignment. x1x1 x2x2 x3x3 xnxn...

16. Complexity 16 1. Find the Expectation Let Y i be the random variable indicating the outcome of the i-th clause. For any 1  i  m, E[Y i ]=0·1/8+1·7/8=7/8 E[  Y i ] =  E[Y i ] = 7/8m

17. Complexity 17 2. Conclude Existence Expectedly, the number of clauses satisfied is 7/8m. Thus, there exists an assignment which satisfies at least that many.

18. Complexity 18 PCP (Without Proof) Theorem (PCP): For any  >0, Gap-3SAT[  ] is NP-hard. This is tight! Gap-3SAT[0] is polynomial time decidable

19. Complexity 19 Approximation Preservation AB YES don’t care NO YES don’t care NO

20. Complexity 20 Hardness of Approximation Do the reductions we’ve seen also work for the gap versions? We’ll revisit the CLIQUE example.

21. Complexity 21 CLIQUE Construction...... a part for each clause a vertex for each literal edge indicates consistency

22. Complexity 22 Approximation Preservation If there is an assignment which satisfies all clauses, there is a clique of size m. If there is a clique of size (7/8+  )m, there is an assignment which satisfies more than 7/8+  of the clauses.

23. Complexity 23 Gap-CLIQUE (Ver1) The following problem is NP-hard for any  >0: Instance: a graph G=(V,E) composed of m independent sets of size 3. Problem: to distinguish between: There’s a clique of size m Every clique is of size at most (7/8+  )m

24. Complexity 24 Corollary Theorem: for any  >0, CLIQUE is hard to approximate within a factor of 1/(7/8+  )

25. Complexity 25 Amplification The bigger the gap is, the better the hardness result. We’ll see how a gap can be amplified.

26. Complexity 26......... Amplification A part for every k vertices vertex for each Boolean assignment edge indicates consistency Given an instance of the Gap-CLIQUE problem and a constant k:

27. Complexity 27 Boolean assignments A Boolean assignment over k vertices {v 1,…,v k } is a function A:{v 1,…,v k }  {0,1}. Think about it as if it indicates whether each vertex belongs to the clique.

28. Complexity 28 Good Assignments

29. Complexity 29 Consistency Two assignments are inconsistent, when they give the same vertex different truth-values.... n

30. Complexity 30 Consistency They are also inconsistent, if they both assign 1 to two vertices not connected by an edge. non-edge

31. Complexity 31 Correctness

32. Complexity 32 Chromatic Number Instance: a graph G=(V,E). Problem: To minimize k, so that there exists a function f:V  {1,…,k}, for which (u,v)  E  f(u)  f(v)

33. Complexity 33 Chromatic Number

34. Complexity 34 Chromatic Number Observation: Each color group is an independent set

35. Complexity 35 Clique Cover Number (CCN) Instance: a graph G=(V,E). Problem: To minimize k, so that there exists a function f:V  {1,…,k}, for which (u,v)  E  f(u)=f(v)

36. Complexity 36 Clique Cover Number (CCN)

37. Complexity 37 Reduction Idea...... CLIQUECCN...... q cyclic shift- morphic clique preserving m......

38. Complexity 38 Correctness

39. Complexity 39 Transformation T:V  [q] for any v 1,v 2,v 3,v 4,v 5,v 6, T(v 1 )+T(v 2 )+T(v 3 )  T(v 4 )+T(v 5 )+T(v 6 ) (mod q) {v 1,v 2,v 3 }={v 4,v 5,v 6 } T is unique for triplets

40. Complexity 40 Observations Such T is unique for pairs and for single vertices as well: If T(x)+T(u)=T(v)+T(w), then {x,u}={v,w} If T(x)=T(y) (mod q), then x=y

41. Complexity 41 feasible values Greedy Construction v6v6 v2v2 v1v1 v5v5 v3v3 v4v4 vertices we determined forbidden values

42. Complexity 42 Greedy Construction - Analysis At most values are ruled out totally, so for q=n 5 the greedy construction works. Corollary: There exists a polynomial time algorithm which constructs a triplet unique transformation with q=n 5

43. Complexity 43 Using the Transformation 01234… (q-1) vivi vjvj T(v i )=1 T(v j )=4 CLIQUE CCN

44. Complexity 44 Completing the CCN Graph Construction T(s) T(t) (s,t)  E CLIQUE  (T(s),T(t))  E CCN

45. Complexity 45 Completing the CCN Graph Construction T(s) T(t) Close the set of edges under shift: For every (x,y)  E, if x’-y’=x-y (mod q), then (x’,y’)  E

46. Complexity 46 Max Clique of G-clique and G-ccn Lemma: Max-Clique(G-clique) = Max-Clique(G-CCN) Corollary: –MAX-clique(G-clique) = m  CCN(G-ccn)=q –MAX-clqiue(G-clique)  q

47. Complexity 47 Edge Origin Unique T(s) T(t) First Observation: This edge comes only from (s,t)

48. Complexity 48 Triangle Consistency Second Observation: A triangle only come from a triangle

49. Complexity 49 Clique Preservation Corollary: {c 1,…,c k } is a clique in the CCN graph iff {T(c 1 ),…,T(c k )} is a clique in the CLIQUE graph.

50. Complexity 50 Summary We’ve seen how to show hardness of approximation results in general, and even proven several such using the PCP theorem: –CLIQUE –CHROMATIC NUMBER 