Title: Hardness Result for MAX-3SAT
1Hardness 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 ? gt
0 , it is NP Hard to distinguish whether a
MAX-3SAT instance is satisfiable or there is no
assignment that satisfies ?? fraction of the
clauses. - Fact Any random assignment satisfies 7/8 from
the clauses.
3Max-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(??,1-?)
4 gadget
4Label 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 (T1)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(T1)u (2u7Tu) values.
6When a type-0 block V is a sub block of type-1
block W ?
If we can replace u clauses ci i1,2,u in W
by u variables xi i1,2,,u in V such that
the variable xi is in the clause ci for 1 i u.
- ?V,WMW ! MV is the operation of taking a sub
assignment.
7Label 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 (T1)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
11Our 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 ?? 8 ?gt0 satisfiable.
12Long Code
- FV is the set of all functions fMV! -1,1.
- FW is the set of all functions fMW! -1,1.
- Long code of an assignment x 2 MV is the mapping
AFV ! -1,1 where A(f)f(x).
Size 22V
13Building
AV
W
AW
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.
14Building - 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
15The 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 fMV ! -1,1 with the uniform
probability. - Pick a function gMW ! -1,1 with the uniform
probability.
16- Define a function hMW ! -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,f Ç XW,g Ç
XW,h is satisfiable
?0,1 !1,-1 ?(x)(-1)x
17Completeness
How many clauses we got ?
Polynomially in n for constant u !!!
- This test has perfect completeness.
- If f(yV)1, by definition one of g(y),h(y) will
be -1 - B(g)-1 or B(h)-1.
- If f(yV)-1, we have A(f)-1.
18Fourier Analysis
- Reminder FV is the set of all functions fMV!
-1,1. - Orthonormal basis to FV is
- ??(f)? x 2 ? f(x) 8 ? µ -1,1V. vV.
- The inner product of 2 functions A,B is
- (A,B)2-2v ?f 2 FV A(f)B(f)
19Fourier Analysis-Continue
- Lemma For any f,g 2 FV and ?,? µ -1,1V
- ??(fg)??(f)??(g)
- ??(f)??(f)??M ?(f).
- Lemma
- 1. Ef??(f)0 8 ? µ -1,1V , ? ? .
- 2. EfA(f)0
- Parsevals Formula
20Soundness
(f,g),(f,h) are independent
- The acceptance criteria can be written as
- Ef,g,h 1-? (1A(f)) (1B(g)) (1B(h))
- ? - ? 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
21Eg,h,B(g)B(h)
22sx is the number of y 2 ? s.t. yVx
23Prsx1 1-?
24e(x) 1-?
25Ef,g,h,A(f)B(g)B(h)
26Proof the left size is equal to
27(No Transcript)
28Cauchy-Schwartz inequality
EX2 EX2
29Cauchy-Schwartz inequality
Goal to see that this is bounded by O(?)
30Reminder
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
?O(?) of the clauses Then we can find an
assignment that satisfies 2-?(u) of the edges.
31The 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))
32We will assume all our long codes are of the
folding mechanism !!!
33We will create an assignment to Label Cover
based on
By the folding lemma ?,? ?
34For the choice
Previous theorem on Label Cover
35Summary
Label Cover Gap(2-?(u),1)
Max-3Sat Gap(??,1)
Long Code Testing
FIN