L is in NP means:

1

• L is in NP means
• There is a language L in P and a polynomial p
  so that
• L1 L2 means
• For some polynomial time computable map r
• 8 x x 2 L1 iff r(x) 2 L2
• L is NP-hard means
• 8 L 2 NP L L
• L is in NPC means
• L 2 NP and L is NP-hard

2
Polynomial time computable maps

• f 0,1 ! 0,1 is called polynomial time
  computable if for some polynomial p,
• - For all x, f(x) p(x).
• - Lf 2 P.

3
Equivalent definition

• A map is polynomial time computable if and only
  if there is a Turing machine that on every input
  x accepts after at most a polynomial number of
  steps and leaves f(x) on its tape when
  terminating.

4
How to establish NP-hardness

• Lemma If L1 is NP-hard and L1 L2 then L2 is
  NP-hard.

5
SAT

• SAT Given a Boolean function in CNF
  representation, is there a way to assign truth
  values to the variables so that the function
  evaluates to true?
• SAT Given a CNF, is it true that it does not
  represent the constant-0 function?
• Input ( x1 Ç x2) Æ (x1 Ç x2)
• Output Yes.
• Input ( x1 Ç x2) Æ (x1 Ç x2) Æ (x1 Ç x2 ) Æ
  ( x1 Ç x2)
• Output No.

6
SAT

• SAT is in NP.
• Cooks theorem (1971) SAT is NP-hard.

7
TSP
HAMILTONIAN CYCLE
MIN VERTEX COLORING
SAT
MAX INDEPENDENT SET
SET COVER
ILP
MILP
KNAPSACK
TRIPARTITE MATCHING
BINPACKING
8
Usefulness of NPC

• Languages in NPC are the least likely problems in
  NP to be in P.
• Suppose we would like to find algorithm for L 2
  NPC.
• If we believe that P is not NP, we know that no
  worst case efficient algorithm exists.
• If we have no opinion about P vs. NP, we know
  that if we find an efficient algorithm for L,
  well earn 1,000,000.

9
Cooks theorem SAT is NP-hard

• Proof of Cooks theorem
• CIRCUIT SAT is NP-hard.
• CIRCUIT SAT reduces to SAT.
• Hence, SAT is NP-hard.

10
Boolean Circuits
?
?
?

X3
X2
X1
11
Circuits vs. Turing Machines

• Circuits can be given as inputs to algorithms but
  they can also be seen as computational devices
  themselves!
• Like Turing Machines, circuits
  C 0,1n ! 0,1 solve decision problems
  on 0,1n.
• Unlike Turing machines, circuits takes inputs of
  a fixed input length n only.

12
Theorem

• Given Turing Machine M running in time at most
  p(n) on inputs of length n, where p is a
  polynomial.
• For every n, there is a circuit Cn with at most
  O(p(n)2) gates so that
• 8 x 2 0,1n Cn(x)1 iff M accepts x.
• The map 1n ! Cn is polynomial time computable.

13
Remark

• This is really just like the Polynomial Turings
  Thesis, only in reverse
• We show that a reasonable sequential model of
  computation (computation by uniform families
  of circuits) has at least as much power as Turing
  Machines.

14
Intuition behind proof
15
Problem Cycles!
Flip-Flop, stores one bit.
16
The Tableau Method

Time t

Time 1
Can be replaced by acyclic Boolean circuit of
size s
Time 0
17
Cell state vectors

• Given a Turing Machine computation, an integer t
  and an integer i let cti 2 0,1s be a Boolean
  representation of the following information, a
  cell state vector
• The symbol in cell i at time t
• Whether or not the head is pointing to cell i at
  time t
• If the head is pointing to cell i, what is the
  state of the finite control of the Turing machine
  at time t?
• The integer s depends only on the Turing machine
  (not the input to the computation, nor t,i).
• To make cti defined for all t, we let c(t1)i
  ct i if the computation has already terminated at
  time t.

18
Crucial Observation

• If we know the Turing machine and ct-1,i-1,
  ct-1,i, ct-1, i1, we also can determine ct,i.
• In other words, there is a Boolean function h
  depending only on the Turing machine so that
  ct,i h(ct-1,i-1, ct-1,i, ct-1,i1).
• A circuit D for h is the central building block
  in a circuit computing all cell state vectors for
  all times for a given input.

19
t(n)
x1
xn
2 t(n)
20
Cooks theorem SAT is NP-hard

• Proof of Cooks theorem
• CIRCUIT SAT is NP-hard.
• CIRCUIT SAT reduces to SAT.
• Hence, SAT is NP-hard.

21
CIRCUIT SAT

• CIRCUIT SAT Given a Boolean circuit, is there a
  way to assign truth values to the input gates, so
  that the output gate evaluates to true?
• Generalizes SAT, as CNFs are formulas and
  formulas are circuits.

22
Example
?
Input
?
Output Yes
?

X3
X2
X1
23
Example
?
Input
?
Output No
?

X3
X2
X1
24
CIRCUIT SAT is NP-hard

• Given an arbitrary language L in NP we must show
  that L reduces to CIRCUIT SAT.
• This means We must construct a polynomial time
  computable map r mapping instances of L to
  circuits, so that
• 8 x x 2 L , r(x) 2 CIRCUIT SAT
• The only thing we know about L is that there is a
  language L in P and a polynomial p, so that

25
Theorem

• Given Turing Machine M for L running in time at
  most q(n) on inputs of length n, where q is a
  polynomial.
• For every n, there is a circuit Cn with at most
  O(q(n)2) gates so that
• 8 x 2 0,1n Cn(x)1 iff M accepts x.
• The map 1n ! Cn is polynomial time computable.

26
Cooks theorem SAT is NP-hard

• Proof of Cooks theorem
• CIRCUIT SAT is NP-hard.
• CIRCUIT SAT reduces to SAT.
• Hence, SAT is NP-hard.

27
TSP
HAMILTONIAN CYCLE
MIN VERTEX COLORING
SAT
MAX INDEPENDENT SET
SET COVER
ILP
MILP
KNAPSACK
TRIPARTITE MATCHING
BINPACKING
28
Remarks on Papadimitrious terminology

• When Papadimitriou writes log space reduction,
  just substitute polynomial time reduction.
• When Papadimitriou writes NL, just substitute P.
• Papadimtrious concepts are more restrictive, but
  the more restrictive definitions will play no
  role in this course.