INHERENT LIMITATIONS OF COMPUTER PROGRAMS

CSci 4011 INHERENT LIMITATIONS OF COMPUTER PROGRAMS

TIMECOMPLEXITY 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(nc) c  ℕ IMPORTANTCOMPLEXITY CLASS P is the class of problems that can be solved in polynomial time: they are efficientlydecidable

NON-DETERMINISTIC PROGRAMS …are just like standard programs, except: 1. There is a special instruction, guess(), that can return 0 or 1. 2. 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(nc) c  ℕ { L | L is decided by a O(t(n))-time non-deterministic Turing machine } Definition: NTIME(t(n)) =

logical operations parentheses variables BOOLEAN FORMULAS A satisfying assignment is a setting of the variables that makes the formula true (x  y)  z  = x = 1, y = 1, z = 1 is a satisfying assignment for  (x  y)  (z  x) 0 0 1 0 A Boolean formula is satisfiable if there exists a satisfying assignment for it SAT = {  |  is a satisfiable Boolean formula }

clauses A 3cnf-formula is of the form: (x1  x2 x3)  (x4  x2  x5) (x3  x2  x1) YES (x1 x̅2 x1) NO (x3  x1) (x3  x̅2 x̅1) NO (x1  x2  x3)  (x̅4 x2  x1) (x3  x1  x̅1) NO (x1  x̅2 x3) (x3  x̅2 x̅1) 3SAT = {  |  is a satisfiable 3cnf-formula }

3SAT = {  |  is a satisfiable 3cnf-formula } Theorem: 3SAT  NP On input : 1. Check if the formula is in 3cnf 2. For each variable xi: a. Set temp = guess(). b. Scan across , replacing xi with temp. 3. Test if the assignment satisfies  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: • If L = { x | y |y| = poly(|x|) and V(x,y) accepts } • then L  NP Because we can guess y and then run V (2) If L  NPthen 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 } V(,y) = “Accept if  is in 3cnf and (y) is true” {  | y such that y is a satisfying assignment to  } SAT = 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

e f b a g d i c h HAMILTONIAN PATHS

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

e f b a g d c K-CLIQUES

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 = { 〈y1,…yn, t〉 | there exist bits b1…bn so that Σi bi yi = t }. FUSION = { 〈SIM,1k〉 | 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 Recognition vs Generation Checking if a song sounds like Mozart Composing a song that sounds like Mozart Checking if a proof is correct Finding a correct proof Checking if a fusion generator is efficient Finding an efficient fusion generator Verifying that p ⨉ q = N Finding prime factors of N Appreciating funny jokes Writing funny jokes If P=NP, especially by a fast algorithm, then the tasks on left and right are all equally difficult!

If P = NP… Mathematicians would be out of a job Cryptography as we know it would not be possible 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 = {〈C1,C2〉| ∀x. C1(x) = C2(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… Generation is as easy as recognition: Being a chef is as easy as eating. Writing symphonies is as easy as listening to them. Writing Shakespeare is as easy as recognizing Shakespeare. 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

A B ƒ ƒ ∀w. w  A  ƒ(w)  B

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

HARDEST PROBLEMS IN NP 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)

NP P B B is NP-Complete

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

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.

INHERENT LIMITATIONS OF COMPUTER PROGRAMS