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P vs. NP Theorem 8.15. If a decision problem P is NP -complete and P∈ P , then P = NP . Proof: See definition of NP -completeness and Lemma 8.6. There are two possible szenarios for the shape of the complexity world: NP P NP -c scenario A P = NP = NP -c scenario B I It is widely believed that P 6= NP , i. e., scenario A holds. I Deciding whether P = NP or P 6= NP is one of the seven millenium prize problems established by the Clay Mathematics Institute in 2000. 227 228

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  • P vs. NP

    Theorem 8.15.

    If a decision problem P is NP-complete and P ∈ P, then P = NP.

    Proof: See definition of NP-completeness and Lemma 8.6.

    There are two possible szenarios for the shape of the complexity world:

    NP

    PNP-c

    scenario A

    P = NP = NP-c

    scenario B

    I It is widely believed that P 6= NP, i. e., scenario A holds.

    I Deciding whether P = NP or P 6= NP is one of the seven milleniumprize problems established by the Clay Mathematics Institute in 2000.

    227

    228

  • Complexity Class coNP and coNP-Complete Problems

    Definition 8.16.

    i The complement of decision problem P = (X ,Y ) is P̄ := (X ,X \ Y ).ii coNP := {P̄ | P ∈ NP}iii A problem P ∈ coNP is coNP-complete if all problems in coNP

    polynomially transform to P.

    Theorem 8.17.

    i P is NP-complete if and only if P̄ is coNP-complete.ii Unless NP = coNP, no coNP-complete problem is in NP.

    iii Unless P = NP, there are problems in NP that are neither in P norNP-complete.

    229

    Complexity Landscape (Widely Believed)

    coNPNP

    PNP-c coNP-c

    Remark.I Almost all problems that are known to be in NP ∩ coNP are also

    known to be in P.I One of few exceptions used to be the problem Primes (given a

    positive integer, decide whether it is prime), before it was shown tobe in P by Agrawal, Kayal, and Saxena in 2002.

    230

  • The Asymmetry of Yes and NoExample:For a given TSP instance, does there exist a tour of cost ≤ 1573084?

    Pictures taken from www.tsp.gatech.edu

    231

    Partition Problem

    Partition Problem

    Given: positive integers a1, . . . , an ∈ Z>0.

    Task: decide whether there exists S ⊆ {1, . . . , n} with∑i∈S

    ai =∑

    i∈{1,...,n}\S

    ai .

    Theorem 8.18.

    i The Partition problem can be solved in pseudopolynomial time.

    ii The Partition Problem is NP-complete.

    Proof of (i): . . .

    232

    www.tsp.gatech.edu

  • Strongly NP-Complete Problems

    Let P = (X ,Y ) be a decision problem, p : Z→ Z a polynomial function.I Let Xp ⊂ X denote the subset of instances x where the absolute value

    of every (integer) number in the input x is at most p(size(x)).

    I Let Pp := (Xp,Y ∩ Xp).

    Definition 8.19.

    A problem P ∈ NP is strongly NP-complete if Pp is NP-complete forsome polynomial function p : Z→ Z.

    Theorem 8.20.

    Unless P = NP, there is no pseudopolynomial algorithm for any stronglyNP-complete problem P.

    Proof: Such an algorithm would imply that the NP-complete problem Ppis in P and thus P = NP.

    233

    Strongly NP-Complete Problems (cont.)Examples:

    I The decision problem corresponding to the Traveling SalespersonProblem (TSP) is strongly NP-complete.

    I The Partition problem is NP-complete but not strongly NP-complete(unless P = NP, by Theorem 8.20).

    I The 3-Partition problem below is strongly NP-complete.

    3-Partition Problem

    Given: positive integers a1, . . . , a3n,B ∈ Z>0 with∑3n

    i=1 ai = nB.

    Task: decide whether the index set {1, . . . , 3n} can be partitioned into ndisjoint subsets S1, . . . ,Sn such that

    ∑i∈Sj ai = B for j = 1, . . . , n.

    Remark.I The 3-Partition problem remains strongly NP-complete if one adds

    the requirement that B/4 < ai < B/2 for all i = 1, . . . , 3n.I Notice that each Sj must contain exactly three indices in this case.

    234

  • NP-Hardness

    Definition 8.21.

    Let P be an optimization or decision problem.i P is NP-hard if all problems in NP polynomially reduce to it.ii P is strongly NP-hard if Pp is NP-hard for some polynomial

    function p.

    Remarks.I A sufficient criterion for an optimization problem to be (strongly)

    NP-hard is the (strong) NP-completeness or (strong) NP-hardness ofthe corresponding decision problem.

    I It is open whether each NP-hard decision problem P ∈ NP isNP-complete (difference between Turing and Karp reduction!).

    I An example of an NP-hard decision problem that does not appear tobe in NP (and thus might not be NP-complete) is the following:

    Given an instance of SAT, decide whether the majority of all truthassignments satisfy all clauses.

    235

    NP-Hardness (cont.)

    Theorem 8.22.

    Unless P = NP, there is no polynomial time algorithm for any NP-harddecision or optimization problem.

    Proof: Clear.

    Remark: It is thus widely believed that problems like the TSP and all otherNP-hard optimization problems cannot be solved in polynomial time.

    236

  • Complexity of Linear Programming

    I As discussed in Chapter 4, so far no variant of the simplex methodhas been shown to have a polynomial running time.

    I Therefore, the complexity of Linear Programming remainedunresolved for a long time.

    I Only in 1979, the Soviet mathematician Leonid Khachiyan provedthat the so-called ellipsoid method earlier developed for nonlinearoptimization can be modified in order to solve LPs in polynomial time.

    I In November 1979, the New York Times featured Khachiyan and hisalgorithm in a front-page story.

    I We will give a sketch of the ellipsoid method and its analysis.

    I More details can, e. g., be found in the book of Bertsimas & Tsitsiklis(Chapter 8) or in the book Geometric Algorithms and CombinatorialOptimization by Grötschel, Lovász & Schrijver (Springer, 1988).

    237

    New York Times, Nov. 27, 1979 238

    IntroductionLinear Programming BasicsThe Geometry of Linear ProgrammingThe Simplex MethodDuality TheoryOptimal Trees and PathsMaximum Flow ProblemsMinimum-Cost Flow ProblemsNP-Completeness