The Complexity of

Optimization Problems

Summary- Complexity of algorithms and problems

- Complexity classes: P and NP

- Reducibility- Karp reducibility- Turing reducibility- NP-complete problems

- Complexity of optimization problems- Classes PO and NPO- NP-hard optimization problems

Uniform and logarithmic cost- Uniform cost: the overall number of instructions

executed by the algorithm before halting

- Logarithmic cost:each instruction has a cost depending on the number of bits of the operands- E.g. product of two n-bit integer costs O(nlogn)

- Same for space measure (but we will talk only of time measure)

Example: xy

- Uniform cost: 2+3y

- Logarithmic cost: aylogy+by2logx(logy+loglogx)+c

begin r:=1;

while y 0 dobegin

r:=r*x; y:=y-1end;

return rend.

Worst case analysis- Instances of the same size may result in different

execution costs (e.g. sorting)

- Cost of applying the algorithm on the worst case instance of a given size

- Gives certainty that the algorithm will perform its task within the established time bound

- It is easier to determine

Input size- Size of input: number of bits needed to present the

specific input

- Existence of encoding scheme which is used to describe any problem instance- For any pair of natural encoding schemes and for any

instance x, the resulting strings are polynomially related- I.e., |ei(x)| pi,j(|ej(x)|) and |ej(x)| pi,j(|ei(x)|)

- Avoid unary base encoding

Asymptotic Analysis- Let t(x) be the running time of algorithm A on input x.

The worst case running time of A is given by t(n)=max(t(x) | x such that |x| n)

- Upper bound: A has complexity O(f(n)) if t(n) is O(f(n)) (that is, we ignore constants)

- Lower bound: A has complexity (f(n)) if t(n) is (f(n))

Complexity of a problem- A problem P

- has a complexity lower bound (f(n)) if any algorithm for P has complexity (f(n))

- has a complexity upper bound O(f(n)) if an algorithm for P exists with complexity O(f(n))

x YP?A

YES

NO

Decision problems- Set of instances partitioned into a YES-subset and a

NO-subset- Given an instance x, decide which subset x belongs to

- A decision problem P is solved by an algorithm A if, for every instance, A halts and returns YES if and only if the instance belongs to the YES-subset

Complexity Classes- For any function f(n), TIME(f(n)) is the set of decision

problems which can be solved with a time complexity O(f(n))- P = the union of TIME(nk) for all k- EXPTIME = the union of TIME (2nk) for all k

- P is contained in EXPTIME

- It is possible to prove (by diagonalization) that EXPTIME is not contained in P

Examples- SATISFYING TRUTH ASSIGNMENT: given a CNF

formula F and a truth assignment f, does f satisfy F?- SATISFYING TRUTH ASSIGNMENT is in P

- SATISFIABILITY (simply, SAT): given a CNF formula F, is F satisfiable?- SAT is in EXPTIME.

- Open problem: SAT is in P?

Class NP- A problem P is in class NP if there exist a polynomial

p and a polynomial-time algorithm A such that, for any instance x, x is a YES-instance if and only if there exists a string y with |y| p(|x|) such that A(x,y) returns YES- y is said to be a certificate- Example: SAT is in NP (the certificate is a truth assignment

that satisfies the formula)

- P is contained in NP (the certificate is the computation of the polynomial-time algorithm)

Non-deterministic algorithms: SAT

TRUE FALSE

v3

TRUE FALSE TRUE FALSETRUE FALSETRUE FALSE

FALSETRUE

f1

YES NO YES YES YES YES YESNO

v3 v3 v3

v2 v2

v1

f2 f3 f4 f5 f6 f7 f8

FALSETRUE(v1 or v2 or (not v3))and

((not v1) or (not v2) or v3)

begin for each variable v

guess Boolean value f(v);if f satisfies F then return YESelse return NO

end.

Non-deterministic algorithms and NP

Every problem in NP admits apolynomial-time non-deterministicalgorithm

begin guess string y with |y| p(|x|);

if A(x,y) returns YES then return YESelse return NO

end.

YES NO YES YES YES YES YESNO

Each computation path, which returns YES, is a certificate of polynomial length that can be checked in polynomial time

Every problem that admits a polynomial-time non-deterministic algorithm is in NP

Karp reducibility

- A decision problem P1 is Karp reducible to a decision problem P2 (in short, P1 P2) if there exists a polynomial-time computable function R such that, for any x, x is a YES-instance of P1 if and only if R(x) is a YES-instance of P2

- If P1 P2 and P2 is in P, then P1 is in P

x IP

R

y IP

A2

answer

1

2

Example: {0,1}-Linear programming- SAT {0,1}-LINEAR PROGRAMMING

- For each Boolean variable v of a CNF Boolean formula F, we introduce a {0,1}-valued variable z

- For each clause l1 or l2 or … or lk of F, we introduce the inequality 1+2 + … + k 1, where

i= z if li= v and i= (1-z) if li= not v

- E.g. (v1 or v2 or (not v3)) and ((not v1) or (not v2) or v3) is transformed into the following two inequalities:

z1+z2+(1-z3) 1 and (1-z1)+(1-z2) +z3 1

- If f is a truth assignment, let g be the natural corresponding {0,1}-value assignment (0=FALSE,1=TRUE)

- f satisfies F if and only g satisfies all inequalities

Turing reducibility

- A decision problem P1 is Turing reducible to a decision problem P2 if there exists a polynomial-time algorithm R solving P1 such that R may access to an oracle algorithm solving P2

R g(x)

oracle for P2

yi IP2y1,y2,... with

x IP1

f(y1),f(y2),...

- If P1 P2 then P1 is Turing reducible to P2

Example: Equivalent formulas- SAT is Turing reducible to EQUIVALENT

FORMULAS- Given a CNF Boolean formula F, query the oracle with

input F and x and (not x)- If the oracle answers YES, then F is not satisfiable,

otherwise F is satisfiable- It is not known whether SAT is Karp reducible to

EQUIVALENT FORMULAS

NP-complete problems- A decision problem P is NP-complete if P is in NP

and, for any decision problem P1 in NP, P1 P

- If P is NP-complete and P is in P, then P=NP- NP-complete problems are the hardest in NP- P versus NP question can be solved by focusing on an NP-

complete problem

- Cook’s Theorem: SAT is NP-complete

Optimization problem- Optimization problem P characterized by

- Set of instances I- Function SOL that associates to any instance the set of

feasible solutions- Measure function m that, for any feasible solution, provides

its positive integer value- Goal, that is, either MAX or MIN

- An optimal solution is a feasible solution y* such thatm(x,y*) = Goal{m(x,y) | y SOL(x)}

- For any instance x, m*(x) denotes optimal measure

MINIMUM VERTEX COVER- INSTANCE: Graph G=(V,E)

- SOLUTION: A subset U of V such that, for any edge (u,v), either u is in U or v is in U

- MEASURE: Cardinality of U

Three problems in one- Constructive problem: given an instance, compute

an optimal solution and its value- We will study these problems

- Evaluation problem: given an instance, compute the optimal value

- Decision problem: given an instance and an integer k, decide whether the optimal value is at least (if Goal=MAX) or at most (if Goal=MIN) k

Class NPO- Optimization problems such that

- I is recognizable in polynomial time- Solutions are polynomially bounded and recognizable in

polynomial time- m is computable in polynomial time

- Example: MINIMUM VERTEX COVER

- Theorem : If P is in NPO, then the corresponding decision problem is in NP

Class PO- NPO problems solvable in polynomial time

- Example: MINIMUM PATH

- An optimization problem P is NP-hard if any problem in NP is Turing reducible to P

- Theorem: If the decision problem corresponding to a NPO problem P is NP-complete, then P is NP-hard- Example: MINIMUM VERTEX COVER

- Corollary: If P NP then PO NPO

Evaluating versus constructing- Decision problem is Turing reducible to evaluation

problem

- Evaluation problem is Turing reducible to constructive problem

- Evaluation problem is Turing reducible to decision problem- Binary search on space of possible measure values

- Is constructive problem Turing reducible to evaluation problem?

MAXIMUM SATISFIABILITY- INSTANCE: CNF Boolean formula, that is, set C of

clauses over set of variables V

- SOLUTION: A truth-assignment f to V

- MEASURE: Number of satisfied clauses

Evaluating versus constructing: MAX SATbegin

for each variable v begin k := MAX SATeval(x);

xTRUE:= formula obtained by setting v to TRUE in x;xFALSE:= formula obtained by setting v to FALSE in x;if MAX SATeval(xTRUE) = k thenbegin

f(v) := TRUE; x := xTRUE

endelse begin

f(v) := FALSE; x := xFALSE

end;return f

end.

Theorem: if the decision problem is NP-complete, then the constructive problem is Turing reducible to the decision problem