Hardness Result for MAX-3SAT

1. Hardness Result for MAX-3SAT This lecture is given by: Limor Ben Efraim

2. 3Sat CNF formula: a formula  of n variables (xi) given by m clauses (Cj), each clause contains exactly 3 literals. Max-3Sat: Given: 3Sat CNF formula. Goal: Find an assignment x that maximize the number of satisfied clauses. Hastard (1997), Khot(2002): For any constant  > 0 , it is NP Hard to distinguish whether a MAX-3SAT instance is satisfiable or there is no assignment that satisfies ⅞+ fraction of the clauses. Fact: Any random assignment satisfies 7/8 from the clauses.

3. Max-3Lin-2: Given: a system of linear equations over Z2, exactly 3 variables in each equation. Goal: Find an assignment that maximize the number of satisfied equations. We saw MAX-3Lin-2 Gap(½+,1-) MAX-3Sat Gap(⅞+,1-) 4 gadget

4. Label Cover - Reminder Bipartite graph Each vertex W is a set of u clauses Each vertex V is a set of u variables. … …   Constraints Functions When an assignment  to LC satisfies the edge (V,W)? If  satisfies W, and (V) is a restriction of (W).

5. 0 is the family of all type-0 blocks type-0 block A set of Tu clauses and u variables. type-1 block 1 is the family of all type-1 blocks A set of (T+1)u clauses. Given W 21 (V 20): MW (MV) – the set of all satisfying assignments to W (V). Given W 21 (V 20), how many satisfying assignments there are ? Answer: At most 7(T+1)u (2u7Tu) values.

6. When a type-0 block V is a sub block of type-1 block W ? V,W:MW! MV is the operation of taking a sub assignment. If we can replace u clauses {ci| i=1,2,…u} in W by u variables {xi| i=1,2,…,u} in V such that the variable xi is in the clause ci for 1 · i · u.

7. Label Cover + Bipartite graph Each vertex W is in 1 Each vertex V is in 0 (V,W) 2 E(LG+) if V is a sub-block of W. … … When an assignment  to LC+ satisfies the edge (V,W)? If  satisfies both V and W, and V,W((W))=(V).

8. Theorem: It is NP Hard to distinguish between the following two cases: YES: There is an assignment  that satisfies every edge in the graph NO: No assignment can satisfies more that 2-(u) of the edges

9. Lemma: W 21. Let ,’ 2 MW. If V is a random sub-block of W then PrV [V,W()=V,W(’)] · 1/T Proof:,’ differ at least on one clause. For a choice of a random sub-block V, one replaces at random u clauses out of (T+1)u clauses in W. With probability · 1/T each different clause is replaced. Corollary: W 21. Let 0 µ MW and 2. If V is a random sub block of W then PrV [8’ 2, ’ , V,W() V,W(’)] ¸ 1-||/T

10. The Smoothness Lemma: For any set 0 µ MW Proof:

11. Our Plan Label Cover + with smoothness property. 3Sat CNF formula If there is a satisfiable assignment to the Label Cover+ The 3Sat CNF formula is satisfiable. If the Label Cover+ is 2-(u) satisfiable The 3Sat CNF formula is · ⅞+8>0 satisfiable.

12. Long Code • FV is the set of all functions f:MV! {-1,1}. • FW is the set of all functions f:MW! {-1,1}. • Long code of an assignment x 2 MV is the mapping A:FV! {-1,1} where A(f)=f(x). Size: 22|V|

13. Building… AV AW W V … … We will replace each vertex W (V) in a set of boolean variables, a variable for each bit of AW (AV), the long code of W (V). (W,f) ! XW,f. (V,f) ! XV,f.

14. Building - Continue What are the clauses ? To answer this, we define a test for each (W,V) 2 E(LC+) V is a sub block of W

15. The test • Pick a block W 21 • Pick a random sub-block V of W. • Let  = V,W • Let A,B be the supposed long codes of supposed satisfying assignment to the blocks V,W resp. • Pick a function f:MV! {-1,1} with the uniform probability. • Pick a function g:MW! {-1,1} with the uniform probability.

16. Define a function h:MW! {-1,1} independently 8 y 2 MW • if f((y))=1 then h(y)=-g(y) • if f((y)=-1 then: • Accept unless A(f)=B(g)=B(h)=1. • Equivalent: Accept if the clause XV,fÇ XW,gÇ XW,h is satisfiable :{0,1} !{1,-1} (x)=(-1)x

17. Completeness How many clauses we got ? • This test has perfect completeness. • If f(y|V)=1, by definition one of g(y),h(y) will be -1 B(g)=-1 or B(h)=-1. • If f(y|V)=-1, we have A(f)=-1. Polynomially in n for constant u !!!

18. Fourier Analysis • Reminder: FV is the set of all functions f:MV! {-1,1}. • Orthonormal basis to FV is: (f)= x 2 f(x) 8µ {-1,1}V. v=|V|. • The inner product of 2 functions A,B is (A,B)=2-2vf 2 FV A(f)B(f)

19. Fourier Analysis-Continue Lemma: For any f,g 2 FV and ,µ {-1,1}V: • (fg)=(f)(g) • (f)(f)=M(f). Lemma: 1. Ef[(f)]=0 8µ {-1,1}V , ;. 2. Ef[A(f)]=0 Parseval’s Formula:

20. Soundness (f,g),(f,h) are independent The acceptance criteria can be written as: Ef,g,h[ 1-⅛ (1+A(f)) (1+B(g)) (1+B(h)) ] = ⅞ - ⅛ [ Eg,h(B(g)B(h)) + Ef,g,h ( A(f)B(g)B(h) ) ] We will show that each term · O() For the rest of the proof fix T=

21. Eg,h,[B(g)B(h)]

22. sx is the number of y 2 s.t. y|V=x

23. Pr[sx=1] ¸ 1-

24. e(x) · 1-

25. Ef,g,h,[A(f)B(g)B(h)]

26. Lemma: For any , Proof: the left size is equal to

28. Cauchy-Schwartz inequality E[X]2· E[X2]

29. Cauchy-Schwartz inequality Goal: to see that this is bounded by O()

30. Reminder Label Cover+ 3Sat CNF Formula Given assignment to the 3Sat-CNF formula We can find an assignment to the Label Cover+ Goal: to see that if the assignment satisfies ¸⅞+O() of the clauses Then we can find an assignment that satisfies ¸2-(u) of the edges.

31. The Folding Mechanism Goal: To make sure that A(f)=-A(-f) 8 f. Action: Given A: FU! {-1,1}, define A’: for every pair (f,-f) selecting one of (f,-f). IF: f is selected (A’(f),A’(-f))=(A(f),-A(f)) IF: -f is selected (A’(f),A’(-f))=(-A(-f),A(-f))

32. We will assume all our long codes are of the folding mechanism !!!

33. We will create an assignment to Label Cover+ based on By the folding lemma ,;

34. Previous theorem on Label Cover+ For the choice

35. Summary Label Cover + Gap(2-(u),1) Max-3Sat Gap(⅞+,1) Long Code + Testing FIN