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

2 3Sat CNF formula: a formula  of n variables (x i ) given by m clauses (C j ), 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 Z 2, 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 Each vertex V is a set of u variables. Each vertex W is a set of u clauses When an assignment  to LC satisfies the edge (V,W)? If  satisfies W, and  (V) is a restriction of  (W). Bipartite graph … … Constraints Functions  

5 type-0 block A set of Tu clauses and u variables. type-1 block A set of (T+1)u clauses. 0 is the family of all type-0 blocks 1 is the family of all type-1 blocks Given W 2 1 (V 2 0 ): M W (M V ) – 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 (2 u 7 Tu ) values.

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

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

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 M W. If V is a random sub-block of W then Pr V [  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 M W and  2 . If V is a random sub block of W then Pr V [ 8  ’ 2 ,  ’  ,  V,W (  )   V,W (  ’)] ¸ 1-|  |/T

10 The Smoothness Lemma: For any set 0   µ M W Proof:

11 Label Cover + with smoothness property. 3Sat CNF formula Our Plan 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  satisfiable.

12 Long Code F V is the set of all functions f:M V ! {-1,1}. F W is the set of all functions f:M W ! {-1,1}. Long code of an assignment x 2 M V is the mapping A:F V ! {-1,1} where A(f)=f(x). Size: 2 2 |V|

13 Building… V W … … AWAW AVAV We will replace each vertex W (V) in a set of boolean variables, a variable for each bit of A W (A V ), the long code of W (V). (W,f) ! X W,f. (V,f) ! X V,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:M V ! {-1,1} with the uniform probability. Pick a function g:M W ! {-1,1} with the uniform probability.

16 Accept unless A(f)=B(g)=B(h)=1. Define a function h:M W ! {-1,1} independently 8 y 2 M W if f(  (y))=1 then h(y)=-g(y) if f(  (y)=-1 then: Equivalent: Accept if the clause X V,f Ç X W,g Ç X W,h is satisfiable  {0,1} ! {1,-1}  x)=(-1) x

17 Completeness 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. How many clauses we got ? Polynomially in n for constant u !!!

18 Fourier Analysis Reminder: F V is the set of all functions f:M V ! {-1,1}. Orthonormal basis to F V 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 -2 v  f 2 F V A(f)B(f)

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

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

21 E g,h, [B(g)B(h)]

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

23 Pr[s x =1] ¸ 1- 

24 e(x) · 1- 

25 E f,g,h, [A(f)B(g)B(h)]

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

27

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

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: F U ! {-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 For the choice Previous theorem on Label Cover+

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