1. CS154, Lecture 13: P vs NP

2. The EXTENDED Church-Turing Thesis
 Polynomial-Time Turing Machines
Everyone’s Intuitive Notion of Efficient Algorithms
More generally: TM can simulate every “reasonable” model of computation with only polynomial increase in time
A controversial thesis! Potential counterexamples: quantum algorithms

3. Nondeterministic Turing Machines
…are just like standard TMs, except:
1. The machine may proceed according to several possible transitions (like an NFA)
2. The machine accepts an input string if there exists an accepting computation history for the machine on the string

5. Definition: A nondeterministic TM is a 7-tuple T = (Q, Σ, Γ, , q0, qaccept, qreject), where:
Q is a finite set of states
Σ is the input alphabet, where   Σ
Γ is the tape alphabet, where   Γ and Σ  Γ
 : Q  Γ → 2(Q  Γ  {L,R})
q0  Q is the start state
qaccept  Q is the accept state
qreject  Q is the reject state, and qreject  qaccept

6. Defining Acceptance for NTMs
Let N be a nondeterministic Turing machine
An accepting computation history for N on w is a sequence of configurations C0,C1,…,Ct where
1. C0 is the start configuration q0w,
2. Ct is an accepting configuration,
3. Each configuration Ci yields Ci+1
Def. N(w) accepts in t time  Such a history exists
N has time complexity T(n) if for all n, for all inputs of length n and for all histories, N halts in T(n) time

7. TIME(t(n))  NTIME(t(n)) Is TIME(t(n)) = NTIME(t(n)) for all t(n)?
Definition: NTIME(t(n)) =
{ L | L is decided by a O(t(n)) time nondeterministic Turing machine }
TIME(t(n))  NTIME(t(n))
Is TIME(t(n)) = NTIME(t(n)) for all t(n)?

8. What problems can we efficiently solved nondeterministically, but not

9. The Clique Problem k-clique = complete subgraph on k nodes a d f c b e

10. The Clique Problem
Find a clique of 1 million nodes?

11. Assume a reasonable encoding of graphs (example: the adjacency matrix is reasonable)
CLIQUE = { (G,k) | G is an undirected graph with a k-clique }
Theorem: CLIQUE  NTIME(nc) for some c > 1
N((V,E),k): Nondeterministically guess a subset S of V with |S| = k
For all u, v in S, if (u,v) is not in E then reject
Accept

12. The Hamiltonian Path Problemc d f h i g
A Hamiltonian path traverses through each node exactly once

13. HAMPATH = { (G,s,t) | G is a directed graph with
HAMPATH = { (G,s,t) | G is a directed graph with a Hamiltonian path from s to t }
Theorem: HAMPATH  NTIME(nc) for some c > 1
N((V,E),s,t): Nondeterministically guess a sequence v1, …, v|V| of vertices
If vi = vj for some i ≠ j, reject
For all i = 1,…,|V|-1,
if (vi,vi+1) is not in E then reject
If (v1 = s & vn = t) then accept else reject

14. Nondeterministic Polynomial Time
NP = NTIME(nk) 
k  N

15. L = { x |  y ϵ Σ* [|y| ≤ |x|k and V(x,y) accepts ] }
Theorem: L  NP  There is a constant k and polynomial-time TM V such that
L = { x |  y ϵ Σ* [|y| ≤ |x|k and V(x,y) accepts ] }
Proof:
If L = { x | y |y| ≤ |x|k and V(x,y) accepts } then L  NP
Define the NTM N(x): Guess y of length at most |x|k
Run V(x,y) and output answer
Then, L(N) is the set of x s.t. [|y| ≤ |x|k & V(x,y) accepts]
(2) If L  NP then L = { x | y |y| ≤ |x|k and V(x,y) accepts }
Suppose N is a poly-time NTM that decides L. Define V(x,y) to accept iff y encodes an accepting computation history of N on x

16. there are polynomial-length proofs
A language L is in NP
if and only if
there are polynomial-length proofs
(aka. certificates or witnesses) for membership in L
CLIQUE =
{ (G,k) |  subset of nodes S such that S is a k-clique in G }
HAMPATH =
{ (G,s,t) |  Hamiltonian path in graph G
from node s to node t }

17. Boolean Formula Satisfiability
logical
operations
 recedes
 precedes 
parentheses
 =
(x  y)  z
Boolean variables (0 or 1)

18. Boolean Formula Satisfiability
 =
(x  y)  z
A satisfying assignment is a setting of the variables that makes the formula true
x = 1, y = 1, z = 1 is a satisfying assignment for 
(in fact, any assignment with z = 1 is satisfying)
 = (x  y)  (z  x) 1

19. A Boolean formula is satisfiable if there is a true/false setting to the variables that makes the formula true
a  b  c  d
YES
(x  y)  x NO
SAT = {  |  is a satisfiable Boolean formula }

20. A 3cnf-formula has the form:
(x1  x2  x3)  (x4  x2  x5)  (x3  x2  x1)
clauses
literals
3SAT = {  |  is a satisfiable 3cnf-formula }

21. 3SAT = {  |  is a satisfiable 3cnf-formula }
Theorem: 3SAT  NP
We can express 3SAT as
3SAT =
{  |  is in 3cnf and  string y that encodes a satisfying assignment to  }
The number of variables of  is at most ||, so |y| ≤ ||.
Then, argue that the language
3SAT-CHECK = {(,y) |  is in 3cnf and y is a satisfying assignment to }
is in P.
(Similarly, SAT  NP)

22. NP = Problems with the property that, once you have the solution, it is “easy” to verify the solution
When  ϵ SAT,
or (G, k) ϵ CLIQUE,
or (G,s,t) ϵ HAMPATH,
Can prove that with a short proof that can easily been verified
What if  ∉ SAT? (G, k) ∉ CLIQUE? Or (G,s,t) ∉ HAMPATH?

23. P = the problems that can be efficiently solved
NP = the problems where proposed solutions can be efficiently verified
Is P = NP? can problem solving be automated?
Clay Math Institute in the year 2000: “millennium problems”

24. If P = NP:
Mathematicians may be out of a job
Cryptography as we know it may be impossible
In principle, every aspect of life could be efficiently and globally optimized … life as we know it would be different!
Conjecture: P  NP

25. Polynomial Time Reducibility
f : Σ*  Σ* is a polynomial time computable function
if there is a poly-time Turing machine M that on every input w, halts with just f(w) on its tape
Language A is poly-time reducible to language B, written as A P B, if there is a poly-time computable f : Σ*  Σ* so that:
w  A  f(w)  B
f is a polynomial time reduction from A to B
Note there is a k such that for all w, |f(w)| ≤ |w|k

26. f converts any string w into a string f(w) such that w  A  f(w)  Bnc n w
f(w) f n nc
f converts any string w into a string f(w) such that w  A  f(w)  B

27. Theorem: If A P B and B P C, then A P Cg C n nc
ncd

28. Theorem: If A P B and B  P, then A  P
Proof:
Let MB be a poly-time TM that decides B.
Let f be a poly-time reduction from A to B.
We build a machine MA that decides A as follows:
MA = On input w,
1. Compute f(w)
2. Run MB on f(w), output its answer
w  A  f(w)  B

29. Theorem: If A P B and B  NP, then A  NP
Proof:
Analogous…

30. Theorem: If A P B and B  P, then A  P
Theorem: If A P B and B  NP, then A  NP
Corollary: If A P B and A ∉ P, then B ∉ P

31. Definition: A language B is NP-complete if:
1. B  NP
2. Every A in NP is poly-time reducible to B
That is, A ≤P B
When this is true, we say “B is NP-hard”
On homework, you showed
A language L is recognizable iff
L ≤m ATM
ATM is “complete for recognizable languages”:
ATM is recognizable, and for all recognizable L, L ≤m ATM

32. Suppose L is NP-Complete…
If L  P, then P = NP
If L ∉ P, then P ≠ NP

33. Suppose L is NP-Complete…
Then assuming the conjecture P  NP,
L is not decidable in nk time, for every k

34. The Cook-Levin Theorem: SAT and 3SAT are NP-complete
3SAT  NP A satisfying assignment is a “proof” that a 3cnf formula is satisfiable
3SAT is NP-hard Every language in NP can be polynomial-time reduced to 3SAT (complex logical formula)
Corollary: 3SAT  P if and only if P = NP