Provable intractability. NP-completeness. Decision problems. P problems. Algorithms. Two causes of intractability:

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1 Provable intractabilit -completeness Selected from Computer and Intractabilit b MR Gare and DS Johnson (979) Two causes of intractabilit: The problem is so difficult that an exponential amount of time is needed to discover a solution The solution itself has a length beond an polnomial function of the input sie Eg, given a weighted graph G and an integer B, list all Hamiltonian circuits having length of B or less Intractabilit of this sort can be easil recognied This tpe of intractabilit can be regarded as a signal that the problem is not defined realisticall, because we are asking for more information than we could ever use 007/0/8 CS,OC 007/0/8 CS,OC Decision problems PRIME Instance: An integer a Z + Question: Is a a prime? P problems REACHABILITY Instance: A graph G=(V, E) and two vertices u and v in V Question: Does G contain a path from u to v? Onl two possible answers: es or no Problem description, two parts: Problem name Generic instance of the problem Yes-no question asked in terms of the generic instance CLIQE Instance: A graph G=(V, E) and a positive integer k Question: Does G contain a clique of sie k or more? -DIMENSIONAL MATCHING (DM) Instance: A set M W X Y, where W, X, and Y are disoint sets having the same number q of elements Question: Does M contain a matching, that is, a subset M M such that M =q and no two elements of M agree in an coordinate? A problem Π belongs to the class of P if for it there is a polnomial algorithm On the right are three known P problems MINIMM TREE Instance: A graph G=(V, E), a weight function W mapping each e E to a integer W(e) Z +, and an integer k Z + Question: Does G contains a tree of sie k or less? -DIMENSIONAL MATCHING (DM) Instance: A set M X Y, where X and Y are disoint sets having the same number q of elements Question: Does M contains a matching? 007/0/8 CS,OC 007/0/8 CS,OC 4

2 problems A problem Π belongs to the class of if for it there exists a nondeterministic polnomial algorithm CLIQE CLIQE Instance: A graph G=(V, E) and a positive integer k Question: Does G contain a clique of sie k or more? Nondetermistic algorithm: At some points, the algorithm faces several choices The algorithm does not specif which choice the next step should take A combination of the choices is called a computation (path) Nondetermistic Alg A for Π : For a es instance, there is at least one computation leading to es ; For a no instance, all the possible combinations lead to no Y/N Deterministic Alg Linear structured, Single outcome Y N Y Y Y N Y N Nondeterministic Alg Tree-like structure, Multiple outcomes Nondetermistic polnomial Alg Ever computation completes within polnomial time Alg for CLIQE =Φ For each v V do Let =+{v} or = EndFor If k and forms a clique return YES Else Return NO EndIf v v v v v v v v v v N Y N N Y N N N Graph G computation on the left graph G 007/0/8 CS,OC 5 007/0/8 CS,OC 6 SAT Alg for CLIQE For each u do Let t(u)=t or F EndFor If t satisfies Ψ return YES Else Return NO EndIf An instance ={u, u, u }, ={C, C } where C = { u, u, u } and That is, C = { u, u, u } ( ) ( ) SATISFIABILITY (SAT) Instance: A set of Boolean variables and a collection C of clauses over such that the form a Boolean formula Ψ in conunction form Question: Does a satisfing assignment for Ψ exist? Y ψ = u u u u u u 007/0/8 CS,OC 7 N u u u u u u u Alg on the instance Y Y Y Y N Y P=? P vs A foremost open question in contemporar mathematics and computer science Needs a maor breakthrough P : All the P problems are problem The deterministic algorithm is a special kind of the nondeterminisctic algorithm Decision Problems Decidable Problems 007/0/8 CS,OC 8 P

3 Another definition for algorithms An algorithm has two phases: Guess a structure Based on the guess, compute and answer YES or NO in polnomial time of the input instance Notations: D Π : the set of all instances for probelem Π ; Y Π : the set of es-instances for Π; N Π : that of no-instances Obviousl we have D Π = Y Π +N Π An algorithm A solves problem Π : For an YES instance I Y Π, there exists a guess, such that in the second phase the algorithm will repl with YES in polnomial time For an NO instance I N Π, ever guess will lead the second phase of the algorithm A to repl NO in polnomial time In other words: For ver I D Π there is a guess that makes algorithm A to answer YES in polnomial time if and onl if I Y Π 007/0/8 CS,OC 9 Problems SAT : Guess an truth assignment t Verif whether or not t satisfies all the clauses (polnomial time doable) CLIQE : Guess a subset V Verif whether k and ever pair of vertices in is connected b an edge We are not sure that PRIME, but its complementar problem COMPOSITE is reall in (COMPOSITE is to ask whether a given number k is a composite number): Guess a number that is no less than and no more than k If k is dividable b answer YES, otherwise NO 007/0/8 CS,OC 0 -complete Problems Intuitivel, -complete problems are the most intractable problems in A formal definition for it needs to introduce the concept of reduction How to prove completeness of Obviousl, once an complete problem has a polnomial time algorithm, all problems will have Turing reduction: Given two problems and, if there there is an algorithm A for that uses an algorithm for as subroutine then we sa that is Turing reducible to Karp reduction: Given two problems and, if there there is transformation (function) T:D D such that I Y if and onl if T(I) Y, then we term T a reduction (transformation) from to Polnomial reduction: A Karp reduction T that transforms to is termed a polnomial reduction (transformation) if T can be computed in polnomial time We use to denote is polnomial reducible to -completeness: C and for all 007/0/8 CS,OC B the definition: Proving that for ever C, there is a polnomial transformation f transforming to B polnomial transformation from a known C problem (This is correct because and ): showing that, Selecting a know -complete problem C, Constructing a transformation f from to,and 4 Proving that f is a polnomial transformation: ) showing that f is computable in polnomial time, and ) Proving that I Y f(i) Y The first method need to do thing in a high abstract wa The second is much easier, provided that we have known C problems To use the second, we need a seed 007/0/8 CS,OC

4 SAT, the seed for -completeness Cook s theorem (97): SAT is -complete Proof: Omitted (SA Cook, Proc rd Ann ACM Smp on Theor of Computing, pp5-58) Take SAT as a seed, we will breed (cultivate) following six basic -complete problems -SATISFIABILITY (SAT) Instance: Collection C={c, c,, c m } of clauses on a set = {u, u,, u n } of Boolean variables such that c i = for i m Question: Is there a truth assignment for that satisfies all the clauses in C? -DIMENSIONAL MATCHING (DM) C Instance: A set M W X Y, where W, X, and Y are disoint sets having the same number q of elements Question: Does M contain a matching, that is, a subset M M such that M =q and no two elements of M agree in an coordinate? 007/0/8 CS,OC SAT P Basic C Problems VERTEX COVER (VC) Instance: A graph G=(V, E) and a positive integer k Question: Is there a vertex cover of sie k or less for G, that is, a subset V V such that V k and for each edge {u,v} E, at least one of u and v belongs to V? CLIQE Instance: A graph G=(V, E) and a positive integer k Question: Does G contain a clique of sie k or more, that is, a subset V V such that V k and ever two vertices in V are oined b an edge in E? HAMILTONIAN CIRCIT (HC) Instance: A graph G=(V, E) Question: Does G contain a Hamiltonian circuit, that is, an ordering v, v,, v n of the vertices of G, where n= V, such that {v n, v } E and {v i, v i+ } E for all i, i < n? 007/0/8 CS,OC 4 Basic C Problems SAT is -complete (I) PARTITION Instance: A finite set A={a, a,, a n } of positive integers Question: Is there a subset A A such that a A a = a A-A a? On the left is the diagram of the sequence of transformations used to prove the -completeness of the six basic problems DM SAT SAT VC PARTITION HC CLIQE SAT because we can guess a truth assignment for and then check in polnomial time whether that truth setting satisfies all the given threeliteral clauses The transformation from SAT: Let ={u, u,, u n } and C={c, c,, c m } be an instance of SAT, then the C and of SAT are = m and C = C Thus we need onl to show how C and can be constructed from c m Let c ={,,, k }, where each i takes either the positive form of u l or the negative form of u l Then the wa to construct C and depends on k Case k=: = { } and C = {, }{,, }{,, }{,, }} Case k=: = { } and C = Case k=: = φ and C = { c } Case 4 k>: {, }{,, }} i = { i k } and C = {, } i i + k {, i + } i k 4} {, k, k } { } 007/0/8 CS,OC 5 007/0/8 CS,OC 6 4

5 SAT is -complete (II) SAT is -complete (III) To prove that the construction is indeed a transformation, we must show that the set C of clauses is satisfiable if and onl if C is IF part: Suppose t: {T, F} is a truth assignment satisfing C We extend t to to get t as follows Case or : The clauses in C are alread satisfied b t, so we can arbitraril extend t to, sa b setting t ()=T for all l= or : Set t ( i )=T for i k-, which satisfies all the clauses of C l=k- or k: Set t ( i )=F for i k-, which satisfies all the clauses of C Otherwise, l k-: Set t ( i )=T for i l-, the first l- clauses of C are satisfied Set t ( i )=F for l- i k-, all of the l-th clause to the (k-)-th clause of C are satisfied { } l l The ( l ) th clause, l of C Case : A trivial case has been satisfied b t( l ) = T Case 4: Since t satisfies C, there These choices for t will insure that C is must be a least number l such satisfied and so C is, b t that t( l )=T 007/0/8 CS,OC 7 ONLY-IF part: Conversel, if t is a satisfing truth assignment for, it is eas to verif that the restriction of t to the variables in must be a satisfing truth assignment for C The transformation can be performed in polnomial time: To see this, it suffices to observe that the number of three-literal clauses in C is bounded b a polnomial in mn The whole proof for the -completeness of SAT is finished 007/0/8 CS,OC 8 5