What Is the Hardest Problem in NP?
Question:
We do not know if problems in NP have polynomial time
deterministic algorithms. Can we at least point out to some
problem(s) in NP that are the most difficult ones to solve?
Computability and Complexity
Answer:
This is indeed possible and SAT is one example of such a problem.
In fact there are many other problems are in NP and difficult for
the class NP (CLIQUE, HAMPATH, SUBSET-SUM, 3SAT, . . . ).
We call them NP-complete problems.
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NP-completeness
Cook-Levin Theorem (SAT is NP-complete)
given by Jiri Srba
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Consequence:
If just one NP-complete problem can be solved in P then all
other problems in NP will be solvable in P (hence P=NP).
If P 6= NP then none of the NP-complete problems is solvable
in deterministic polynomial time.
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NP-Completeness
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Cook-Levin Theorem
Definition (NP-Completeness)
Cook-Levin Theorem
The language SAT is NP-complete.
A language B is NP-complete iff
1
B ∈ NP (containment in NP), and
2
for every A ∈ NP we have A ≤P B (NP-hardness).
Proof: SAT is clearly in NP. We have to show that SAT is NP-hard:
Every language A ∈ NP is poly-time reducible to SAT.
Theorem
If B is NP-complete and B ∈ P,
then P = NP.
Let us assume any given A ∈ NP, so
there is a nondeterm. decider M for A running in time O(nk ).
Proof: We know that if A ≤P B and B ∈ P then A ∈ P.
Our aim: for any input string w construct in polynomial time a
Boolean formula φ such that
Theorem
If B is NP-complete, B ≤P C , and C ∈ NP,
then C is NP-complete.
M accepts w if and only if φ is satisfiable.
Proof: Because ≤P is transitive (see the tutorial).
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Table of Configurations of M Run on ababa
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Boolean Formula Describing Table of Configurations
Assume a table of configurations when M is run on w = w1 . . . wn .
We will construct a formula φ such that M accepts w iff φ ∈ SAT .
def
φ = φcell ∧ φstart ∧ φaccept ∧ φmove
The set of variables contains:
xi,j,s
where 1 ≤ i, j ≤ nk and s ∈ C is a tape symbol or a control state.
Note: There are only polynomially many variables w.r.t. to n.
Intuition:
Variable xi,j,s is true if and only if cell[i, j] contains the symbol s.
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Definition of φcell
Definition of φstart
Every cell[i, j] contains exactly one symbol s.
def
φcell =
h _
^
1≤i,j≤nk
xi,j,s
^
∧
s∈C
The first row contains the initial configuration q0 w1 . . . wn .
i
(xi,j,s ∨ xi,j,t )
def
φstart = x1,1,# ∧ x1,2,q0 ∧ x1,3,w1 ∧ x1,4,w2 ∧ . . . x1,n+2,wn ∧
s,t∈C ,s6=t
x1,n+3,t ∧ . . . x1,nk −1,t ∧ x1,nk ,#
Note: φstart is of polynomial size w.r.t. to n.
Note: φcell is of polynomial size w.r.t. to n.
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Definition of φaccept
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Definition of φmove
Every window in the table is legal.
Let LW denote the set of all legal windows.
There is an accepting configuration in the table.
def
φmove =
_
def
φaccept =
^
legal-window(i, j)
1≤i<nk ,1<j<nk
xi,j,qaccept
1≤i,j≤nk
def
legal-window(i, j) =
_
xi,j−1,a1 ∧xi,j,a2 ∧xi,j+1,a3 ∧xi+1,j−1,a4 ∧xi+1,j,a5 ∧xi+1,j+1,a6
Note: φaccept is of polynomial size w.r.t. to n.
a1 ,a2 ,a3
∈LW
a4 ,a5 ,a6
Note: φmove is of polynomial size w.r.t. to n.
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Cook-Levin Theorem — Summary
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Exam Questions
Let A ∈ NP be decided by poly-time nondeterministic TM M.
For every w ∈ Σ∗ we constructed in polynomial time a
def
formula φ = φcell ∧ φstart ∧ φaccept ∧ φmove .
We argued that M accepts w if and only if φ is satisfiable.
Definition of NP-completeness.
Hence SAT is NP-hard.
Theorems about NP-completeness.
Clearly SAT is in NP.
SAT and 3SAT are NP-complete.
Conclusion: SAT is NP-complete.
Corollary
The language 3SAT is NP-complete.
Proof: A small modification of the proof for SAT.
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