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 :
 x: x  L1 iff r(x)  L2
L is NP-hard means:
 L’  NP: L’ ≤ L
L is in NPC means:
L  NP and L is NP-hard

2. How to establish NP-hardness
Lemma: If L1 is NP-hard and L1 ≤ L then L2 is NP-hard.

3. Boolean functions f: {false, true}n  {false,true}
Example: XOR(x1, x2) = x1  x2
XOR(true , true) = false
In this course 0=false, 1=true.
XOR(1,1)=0

4. Two special classes of Boolean formulae
f(x1, x2) = x1  x2
DNF: f(x1, x2) = (x1 ∧ ¬x2) ∨ (x2 ∧ ¬x1)
CNF: f(x1, x2) = (¬x1 ∨ ¬x2) ∧ (x1 ∨ x2)
A CNF is any conjunction of clauses (disjunctions of literals). A DNF is any disjunction of terms (conjunctions of literals).
Any function on n variables can be described by a CNF (DNF) formula containing at most 2n clauses (terms), each containing at most n literals.

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 ∨ ¬x2) ∧ (x1 ∨ x2)
Output: Yes.
Input: (¬x1 ∨ ¬x2) ∧ (x1 ∨ x2) ∧ (x1 ∨ ¬x2 ) ∧ (¬x1 ∨ x2)
Output: No.

6. SAT
SAT is in NP.
Cook’s theorem (1972): SAT is NP-hard.

7. Circuits A circuit C is a directed acyclic graph.
Nodes in C are called gates.
Types of gates:
Variable (or input) gates of indegree 0, labeled X1, X2, …Xn
Constant gates of indegree 0, labeled 0,1
AND-, OR-gates of indegree 2,
NOT-, COPY-gates of indegree 1.
m distinguished gates are output gates.
The circuit C computes a Boolean function
C: {0,1}n  {0,1}m

8. A formula can be viewed as a special kind of circuit
f(X1, X2, X3) = ¬X1 ∨(X1 ∧ X2 ∧ X3) ⋁ ⋀ ⋀ X3 X1 X2

9. Are formulas and circuits in fact the same thing?
Not quite!
Given a circuit, we can write down an equivalent formula, but in may become much bigger.
A circuit is allowed to reuse the result of a sub-computation without doing the computation again!

10. Multi-output circuits
Any function f: {0,1}n  {0,1}m is represented by some circuit.
Proof:
Any function f: {0,1}n  {0,1} can be described by a Boolean formula.
A formula can also be viewed as a circuit.
Combine m such circuits.

11. Cooks’ theorem: SAT is NP-hard
Proof of Cook’s theorem:
CIRCUIT SAT is NP-hard.
CIRCUIT SAT reduces to SAT.
Hence, SAT is NP-hard.

12. 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.

13. Example ⋁
Input: ⋀
Output: Yes ⋀ X3 X1 X2

14. Example ⋀
Input: ⋀
Output: No ⋀ X3 X1 X2

15. Cooks’ theorem: SAT is NP-hard
Proof of Cook’s theorem:
CIRCUIT SAT is NP-hard.
CIRCUIT SAT reduces to SAT.
Hence, SAT is NP-hard.
We first prove CIRCUIT SAT reduces to SAT

16. Circuits vs. Turing Machines
We consider Circuits also as computational devices themselves.
Like Turing Machines, circuits C: {0,1}n  {0,1} solve decision problems on {0,1}n.
Unlike Turing machines, circuits takes inputs of a fixed input length n only.

17. Theorem
Given Turing Machine M running in time at most p(n)≥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
∀x ∈ {0,1}n: Cn(x)=1 iff M accepts x.
The map 1n  Cn is polynomial time computable.

18. Intuition behind proof

19. Problem: Cycles!
Flip-Flop, stores one bit.

20. The Tableau Method … Time t Time 1 Can be replaced by acyclic Boolean
circuit of size ≈ s s 2 1
Time 0

21. Cell state vectors
Given a Turing Machine computation, an integer t and an integer i let cti  {0,1}s 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(t+1)i = cti if the computation has already terminated at time t.

22. Crucial Observation If we know the Turing machine and
ct-1,i-1, ct-1,i, ct-1, i+1, 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,i+1).
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.

23. t(n) x1 xn
2 t(n)

24. Cooks’ theorem: SAT is NP-hard
Proof of Cook’s theorem:
CIRCUIT SAT is NP-hard.
CIRCUIT SAT reduces to SAT.
Hence, SAT is NP-hard.

25. 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
∀x: x ∈ L ⇔ r(x) ∈ CIRCUIT SAT
The only thing we know about L is that there is a language L’ in P and a polynomial p, so that: