INHERENT LIMITATIONS OF COMPUTER PROGRAMS CSci 4011

TIME COMPLEXITY Definition: Let M be a TM that halts on all inputs. The running time or time-complexity of M is the function f : N  N, where f(n) is the maximum number of steps that M uses on any input of length n. Definition: TIME(t(n)) = { L | L is a language decided by a O(t(n)) time Turing Machine }

P = TIME(n c )  c  ℕ IMPORTANT COMPLEXITY CLASS P is the class of problems that can be solved in polynomial time: they are efficiently decidable

NON-DETERMINISTIC PROGRAMS …are just like standard programs, except: 1. There is a special instruction, guess(), that can return 0 or The program accepts an input if there exists a list of guesses that make it accept. 3. The running time of the program is the maximum number of steps that can be caused by calls to guess().

NP = NTIME(n c )  c  ℕ { L | L is decided by a O(t(n))-time non-deterministic Turing machine } Definition: NTIME(t(n)) =

BOOLEAN FORMULAS (  x  y)  z  = logical operations variables parentheses 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   (x  y)  (z   x) 0010 A Boolean formula is satisfiable if there exists a satisfying assignment for it SAT = {  |  is a satisfiable Boolean formula }

A 3cnf-formula is of the form: (x 1   x 2  x 3 )  (x 4  x 2  x 5 )  (x 3   x 2   x 1 ) clauses (x 1  x̅ 2  x 1 ) (x 3  x 1 )  (x 3  x̅ 2  x̅ 1 ) (x 1  x 2  x 3 )  (x̅ 4  x 2  x 1 )  (x 3  x 1  x̅ 1 ) (x 1  x̅ 2  x 3 )  (x 3  x̅ 2  x̅ 1 ) YES NO 3SAT = {  |  is a satisfiable 3cnf-formula }

Theorem: 3SAT  NP 3SAT = {  |  is a satisfiable 3cnf-formula } 1. Check if the formula is in 3cnf On input  : 2. For each variable x i : 3. Test if the assignment satisfies  a. Set temp = guess(). b. Scan across , replacing x i with temp. Total running time = O(mn)

Theorem: L  NP if and only if there exists a poly-time Turing machine V (for “Verifier”) with L = { x |  y |y| = poly(|x|) and V(x,y) accepts } Proof: (1) If L = { x |  y |y| = poly(|x|) and V(x,y) accepts } then L  NP Because we can guess() y and then run V(x,y) (2) If L  NP then L = { x |  y |y| = poly(|x|) and V(x,y) accepts } Let N be a non-deterministic poly-time TM that decides L and define V(x,y) to accept if y is an accepting computation history of N on x

3SAT = {  |  y such that y is a satisfying assignment to  and  is in 3cnf } SAT = {  |  y such that y is a satisfying assignment to  } V( ,y) = “Accept if  is in 3cnf and  (y) is true” V( ,y) = “Accept if  is a formula and  (y) is true”

A language is in NP if and only if there exist polynomial-length certificates for membership to the language SAT is in NP because a satisfying assignment is a polynomial-length certificate that a formula is satisfiable

HAMILTONIAN PATHS b a e c d f h i g

HAMPATH = { 〈 G,s,t 〉 | G is a directed graph with a Hamiltonian path from s to t } Theorem: HAMPATH  NP The Hamiltonian path itself is a certificate

K- CLIQUES b a e c d f g

CLIQUE = { 〈 G,k 〉 | G is an undirected graph with a k-clique } Theorem: CLIQUE  NP The k-clique itself is a certificate

SEARCHING FOR… A search problem is one where the “answer” is more than a single bit, for example: if 〈 B,G 〉 is a “stable marriage” problem, the “answer” is a stable matching between every boy and some girl. If 〈 C 〉 is a CIRCUIT-SAT problem, the “answer” is an input that makes the circuit output 1. We can think of these problems as “finding a right answer.”

NP = { search problems where it is easy to check if an answer is right. }

EXAMPLES FACTOR = { 〈 N,k 〉 | N has a prime factor ≥ k }. SUBSET-SUM = { 〈 y 1,…y n, t 〉 | there exist bits b 1 …b n so that Σ i b i y i = t }. FUSION = { 〈 SIM,1 k 〉 | SIM is a program that simulates nuclear fusion reactor designs and has an input of length at most k where energy output > energy input }

P = NP? $$$

P VS NP RecognitionvsGeneration Checking if a song sounds like Mozart Composing a song that sounds like Mozart Checking if a proof is correctFinding a correct proof Checking if a fusion generator is efficient Finding an efficient fusion generator Verifying that p q = NFinding prime factors of N Appreciating funny jokesWriting funny jokes If P=NP, especially by a fast algorithm, then the tasks on left and right are all equally difficult!

If P = NP… Cryptography as we know it would not be possible Mathematicians would be out of a job We could determine if cold fusion is possible We could determine if faster-than-light travel is possible.

If P = NP… We could find optimal circuits for any task: EQC = { 〈 C 1,C 2 〉 | ∀ x. C 1 (x) = C 2 (x) } ∈ coNP MIN-CIRCUIT = { 〈 C 〉 | ∀ C’, |C’|≤|C| or (C,C’)  EQC} If P=NP, then EQC ∈ P, so MIN-CIRCUIT ∈ coNP = P.

If P = NP… Writing symphonies is as easy as listening to them. Being a chef is as easy as eating. Writing Shakespeare is as easy as recognizing Shakespeare. Generation is as easy as recognition: In 40+ years, no one has found a proof that P  NP.

POLY-TIME REDUCIBILITY A language A is polynomial time reducible to language B, written A  P B, if there is a polynomial time computable function ƒ : Σ*  Σ*, where for every w, w  A  ƒ(w)  B ƒ is called a polynomial time reduction of A to B

AB ƒ ƒ ∀ w. w  A  ƒ(w)  B

Theorem. If A ≤ P B and B ∈ P, then A ∈ P. (so if A ≤ P B and A  P, then B  P.) Proof. B  P ⇒ TIME COMPLEXITY = O(|w| ab ) |s| b -time program test_b(s) to decide B A ≤ P B ⇒ test_a(w): s = map(w) return test_b(s) |w| a -time function map s.t. map(w)  B iff w  A.

Definition: A language B is NP-complete if: 1. B  NP 2. Every A in NP is poly-time reducible to B (i.e. B is NP-hard) HARDEST PROBLEMS IN NP

B is NP-Complete P NP B

Theorem. If B is NP-Complete, C ∈ NP, and B ≤ P C, then C is NP-Complete. P NP B C

Theorem. If B is NP-Complete and B ∈ P, then P=NP. Corollary. If B is NP-Complete, and P  NP, there is no fast algorithm for B.