1. 1 COMPOSITION PCP proof by Irit Dinur Presentation by Guy Solomon

2. 2 REMINDER PCP:

3. 3 The PCP theorem: Every NP language has a probabilistically Checkable proof with gap = ½ (:=PCP[1/2,1])

4. 4 Reminder - Constraint Graph (CG) u v y r C is unsatisfied x w C is satisfied

5. 5 Degree reduction Expanderizing Gap Amplification COMPOSITION

6. 6 ALPHABET REDUCTION

7. 7 But first of all…a few definitions. Definition: We say that two strings x, y are from each other if they are differ on at least a fraction of coordinates. 000111 010101 Example: X= Y= We can say that X and Y are 1/3 – far (relative) or 2-far (absolute) Hamming distance

8. 8 INPUTOUTPUT Assignment Tester is a reduction

9. 9 q- Assignment Tester ‘s output

10. 10 YES ! AT is…..a PCP verifier

11. 11 INPUTOUTPUT Assignment Tester is a reduction

12. 12 YES ! AT is…..a PCP verifier Circular argument ?? AT on constant size constraint

13. 13 2-query Assignment Tester outputs a constraint graph 2 query AT : Boolean function  system of constraints Each constraint depends on at most 2 variables Output : Constraint graph

14. 14 q - query AT  2 – query AT reduction

15. 15 q-Assignment Tester  2-Assignment Tester

16. 16 COMPOSITION THEOREM

17. 17 Proof – Basic idea

18. 18 Basic Idea – What we want to do ? v uw Stage 1: (Boolean constraint)

19. 19 Stage 2: u v w

20. 20 Stage 3: u v w But how do we do it ?

21. 21 CONSRAINT C  BOOLEAN FUNCTION Encoding the elements of as a binary string Trivial encoding : = {a, b, c, d} a  00 b  01 c  10 d  11 The trivial encoding uses log(| |) bits NO GOOD ! Why ? STAGE 1:

22. 22 ERROR CORRECTING CODES

23. 23 ERROR CORRECTING CODES – CONT.

24. 24 So instead of using the trivial encoding, we will use an error correcting code : e:  where = of relative distance = ¼, i.e., with the following property : x,y, x y  e(x) is -far from e(y) i.e. x y  (e(x), e(y))

25. 25 Now, we can express each of the constraints c C in (G,C, ) as Boolean function ! EXAMPLE: Assume we have the following constraints graph (G,C, ) : = {a,b, c, d} u v w c(u,v) = {(a,d),(b,c)} c(v,w) = {(d,c)}

26. 26 Let’s use the following error correcting code with = ¼ : a  0100 b  1110 c  0000 d  1100 Denote : [u] = [v] = [w] = Example:

27. 27 u v w c(u,v) = {(a,d),(b,c)} c(v,w) = {(d,c)} ENCODING…. u v w c(u,v)={(0100,1100),(1110,0000)} c(v,w)={(1100,0000)}

28. 28 [u] [v] [w] c(u,v)={(0100,1100),(1110,0000)} c(v,w)={(1100,0000)} a d b c d c

29. 29 C is expressed as a Boolean constraint Assignment : u  a v  d w  c Encoding…  Assignment :  0100  1100  0000 C(u,v)  (, ) C(a,d) =1  ) 0, 1, 0, 0, 1, 1, 0, 0 ) = 1

30. 30

31. 31 STAGE 2: u v 2-AT x 1u x 2v x 2u x 5v y2y2 y3y3 y1y1 x 1v x 8u y4y4

32. 32 STAGE 3: v w C (u,v) C (v,w) u x 1u x 1v x 2u x 2v y1y1 x 1v x 2v y1y1 y2y2 x 2w x 1w

33. 33

34. 34 What is the new alphabet size ? Size reduced to a predefined constant – All vertices in G’ take values from

35. 35 Depends on What is the time complexity ? Constant sized constraint  constant time complexity AT time complexity Time complexity linear on number of constraints

36. 36 2-AT ‘s output Constant sized constraint  constant size graph Depends on New graph’s size is linear on number of constraints What is the size of the new graph ?

37. 37 CASE 1 : gap (G) = 0 Claim: gap (G) = 0  gap (G’) = 0 What is the gap of the new graph ? Proof: u v

38. 38 CASE 2 : gap (G) > 0 Proof: Extract….

39. 39 Claim: Extract….

40. 40 Extract…

41. 41 Extract….

42. 42 PROOF OF THE CLAIM Define:

43. 43

44. 44 PROOF OF THE CLAIM-cont. 1/4

45. 45

46. 46 COMPOSITION ERROR CORRECTING CODE 2-ASSIGNMENT TESTER Each constraint in G is a Boolean constraint Paste together all the constraint graphs DEGREE REDUCTION EXPANDERING GAP AMPLIFICATION

47. 47 AT AS A STRONGER PCP REDUCTION REMINDER : PCP THEOREM In our discussion, let fix L to be SAT

48. 48 PCP VERIFIER AS A REDUCTION INPUTOUTPUT

49. 49