1 COMPOSITION PCP proof by Irit Dinur Presentation by Guy Solomon

2 REMINDER PCP:

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

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

5 Degree reduction Expanderizing Gap Amplification COMPOSITION

6 ALPHABET REDUCTION

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 Example: X= Y= We can say that X and Y are 1/3 – far (relative) or 2-far (absolute) Hamming distance

8 INPUTOUTPUT Assignment Tester is a reduction

9 q- Assignment Tester ‘s output

10 YES ! AT is…..a PCP verifier

11 INPUTOUTPUT Assignment Tester is a reduction

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

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 q - query AT  2 – query AT reduction

15 q-Assignment Tester  2-Assignment Tester

16 COMPOSITION THEOREM

17 Proof – Basic idea

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

19 Stage 2: u v w

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

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 ERROR CORRECTING CODES

23 ERROR CORRECTING CODES – CONT.

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

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

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

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

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

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

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

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

39 Claim: Extract….

40 Extract…

41 Extract….

42 PROOF OF THE CLAIM Define:

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44 PROOF OF THE CLAIM-cont. 1/4

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

48 PCP VERIFIER AS A REDUCTION INPUTOUTPUT

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