Efficient and Not-So-Efficient Algorithms. NP-Complete Problems. Search Problems. Propositional Logic. DPV Chapter 8: NP-Complete Problems, Part 1

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1 Efficient and Not-So-Efficient Algoritms NP-Complete Problems DPV Capter 8: NP-Complete Problems, Part 1 Jim Royer EECS November 24, 2009 Problem spaces tend to be big: A grap on n vertices can ave up to n n 2 spanning trees. A grap on n vertices can ave Θ(2 n ) many pats between verts s and t. Etc. Te Good News In our previous work, out of wit Θ(2 n ) (or worse) many coices, we ave found te rigt answer in time O(n k ) for some k. Te Bad News Not all are so nice. Uncredited diagrams are from DPV or omemade. NP-Completeness 1 / 1 NP-Completeness 2 / 1 Searc Problems Propositional Logic Searc are tose of te form: Given:... Find:... (wit a large searc space) Satfiability (as a searc problem) Given: A boolean formula in conjunctive normal form (CNF). Find: A satfying assignment for it (if it as one). We need to define some terms. Te formulas of propositional logic are given by te grammar: P ::= Var P P P P P Var ::= te syntactic category of variables A trut assignment a function I : Variables { False, True }. A trut assignment I determines te value of a formula as follows: I[[x]] = True iff I(x) = True (x a variable) I[[ p]] = True iff I[[p]] = False I[[p q]] = True iff I[[p]] = I[[q]] = True. I[[p q]] = True iff I[[p]] = True or I[[q]] = True. A satfying assignment for a formula p an I wit I[[p]] = True. Te only known algoritms for SAT run in exponential time. NP-Completeness 3 / 1 NP-Completeness 4 / 1

2 Conjunctive Normal Form Back to Searc Problems, 1 Instead of writing x we write x. A variable x and te negation of a variable x are called literals. A clause a djunction of literals. E.g.: (x y z). A conjunctive normal form formula a conjunction of clauses. E.g.: (x y z) (x y) (y z) (z x) (x y z) Satfiability (as a searc problem) Given: A boolean formula in conjunctive normal form (CNF). Find: A satfying assignment for it (if it as one). Note te differences wit te boolean circuit evaluation problem. If a CNF formula as n variables, tere are 2 n possible assignments. Elements of a Searc Problem I: an instance of te problem S: a possible solution for I C : (Instances) (Pot. Solutions) { True, False } C(I, S) = te result of cecking if S solves I C sould be computable in O( I k ) for some k. (T implies tat S cannot be too large.) For SAT: An instance : a CNF formula (x y) (y x) A pot. solution : a trut assignment x True, y True poly-time cecker : boolean circuit evaluation NP-Completeness 5 / 1 NP-Completeness 6 / 1 Back to Searc Problems, 2 Elements of a Searc Problem I: an instance of te problem S: a possible solution for I C : (Instances) (Pot. Solutions) { True, False } C(I, S) = te result of cecking if S solves I C sould be computable in O( I k ) for some k. (T implies tat S cannot be too large.) Variations of SAT Horn Clause SAT Easy (Capter 5) 2SAT Easy 3SAT Hard, it seems nsat te restricted version of SAT in wic eac clause as at most n literals. NP-Completeness 7 / 1 Traveling Salesman Traveling Salesman (TS) as a searc problem Given: n vertices and all n (n 1)/2-many dtances between tem b a budget (number) Find: An ordering of 1,..., n: π(1), π(2),..., π(n) (a tour) so tat d π(1),π(2) + d π(2),π(3) + + d π(n),π(1) b. Traveling Salesman (TS) as an optimization problem Given: n vertices and all n (n 1)/2-many dtances between tem. Find: An ordering of 1,..., n: π(1), π(2),..., π(n) so tat te tour s cost d π(1),π(2) + d π(2),π(3) + + d π(n),π(1) minimal. Te two are equivalent: A solution to te optimization problem, solves te searc problem. (Wy?) Given a way to solve te searc problem, you can construct a solution to te optimization problem via binary searc. (How?) Te only known algoritms for tese are exponential time. TS a restriction of te Minimal Spanning Tree problem in wic te MST allowed no brances. NP-Completeness 8 / 1

3 Euler and Hamiltonian Pats, 1 Euler and Hamiltonian Pats, 2 A pat in an undirected grap an Euler pat wen it uses eac edge of te grap exactly once. (Te pat may pass troug a vertex many times.) If te pat a cycle, ten it called an Euler Tour or an Euler Circuit. A pat in an undirected grap a Rudrata pat (or more usually a Hamiltonian pat) wen it uses eac vertex of te grap exactly once. If te pat a cycle, ten it called a Rudrata Cycle or Hamiltonian Cycle. Teorem (Euler) G as an Euler pat G connected and as at most two vertices of odd degree. No Euler Pat Euler Tour: A B... K Images from: ttp://en.wikipedia.org/wiki/eulerian_pat NP-Completeness 9 / 1 Image from: ttp://en.wikipedia.org/wiki/hamiltonian_pat Tere a nice poly-time algoritm for te Euler Pat Searc problem. (See ttp://en.wikipedia.org/wiki/eulerian_pat.) All known algoritms for te Hamiltonian Pat Searc Problem are exponential-time. NP-Completeness 10 / 1 Cuts and bections Integer Linear Programming A cut in a grap a set of edges wic, if removed, dconnect te grap. A minimum cut a cut of smallest size. Minimum Cut Problem Given: An undirected grap G and a budget b (a number), Find: A cut of G of at most b edges. Balanced Cut Problem Given: An undirected grap G = (V, E) and a budget b (a number), Find: A partition of V: S and T wit S, T V /3 suc tat te number of edges between S and T at most b. You can use Ford-Fulkerson to solve Min-Cut in poly-time. Te only known algoritms for Balanced-Cut are exponential time. Balanced-Cuts are important in clustering. (See DPV.) Integer Linear Programming (ILP) Given: constraints A x b and objective function ct x and goal: g Find: A vector of integers x satfying A x b and ct x g. or equivalently Given: constraints A x b Find: A vector of integers x satfying A x b. ( c T x g incorporated into te constraints.) Zero-One Equations (ZOE) Given: constraints A x b Find: A vector of 0 s and 1 s x satfying A x b. ILP and ZOE sow up in lots of optimization work. Te only known algoritms for Balanced-Cut are exponential time. Image from: ttp://en.wikipedia.org/wiki/cut_(grap_teory) NP-Completeness 11 / 1 NP-Completeness 12 / 1

4 Tree-dimensional matcing Independent set, vertex cover, and clique, 1 3D Matcing Given: R A B C were A = B = C = n. Find: A subset M R of n many triples suc tat if (a, b, c) and (a, b, c ) are dtinct elements of M, ten a = a, b = b, and c = c. 2D matcing poly-time (via Ford-Fulkerson). S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 255 Figure 8.4 A more elaborate matcmaking scenario. Eac triple sown as a triangularsaped node joining boy, girl, and pet. Al Bob Cet Armadillo Alice Bobcat Canary Te only known algoritms for 3D Matcing are exponential time. Beatrice Carol complete our story we will introduce a few more ard, wic will play a role later in te capter, wen we relate te computational difficulty of all tese. Te reader invited to skip aead to Section 8.2 and ten return to te definitions of tese as required. Tree-dimensional matcing Recall te BIPARTITE MATCHING problem: given a bipartite grap wit n nodes on eac side (te boys and te girls), find a set of n djoint edges, or decide tat no suc set exts. In Section 7.3, we saw ow to efficiently solve t problem by a reduction to maximum flow. However, tere an interesting generalization, called 3D MATCHING, for wic no polynomial algoritm known. In t new setting, tere are n boys and n girls, but also n pets, NP-Completeness and te compatibilities 13 / 1 among tem are specified by a set of triples, eac containing a boy, a girl, and a pet. Intuitively, a triple (b, g, p) means tat boy b, girl g, and pet p get along well togeter. We want to find n djoint triples and tereby create n armonious ouseolds. Can you spot a solution in Figure 8.4? Independent set, vertex cover, and clique, 2 Suppose G = (V, E) an undirected grap and U V. (a) U independent wen for eac u, v U, (u, v) / E. Independent set, vertex cover, and clique In te INDEPENDENT SET problem (recall Section 6.7) we are given a grap and an integer g, and te aim to find g vertices tat are independent, tat, no two of wic ave an edge between tem. Can you find an independent set of tree vertices in Figure 8.5? How about four vertices? We saw in Section 6.7 tat t problem can be solved efficiently on trees, but for general graps no polynomial algoritm known. Tere are many oter searc about graps. In VERTEX COVER, for example, te input a grap and a budget b, and te idea to find b vertices tat cover (touc) every edge. Can you cover all edges of Figure 8.5 wit seven vertices? Wit six? (And do you see te (b) U a vertex cover wen eac edge of E as at least one endpoint in U. (c) U a clique wen for eac dtinct u, v U, (u, v) E. Independent Set Problem Find: An independent set for G of size b. Vertex Cover Problem Find: A vertex cover for G of size b. Clique Problem Find: A clique in G of size b. Suppose G = (V, E) an undirected grap and U V. (a) U independent wen for eac u, v U, (u, v) / E. (b) U a vertex cover wen eac edge of E as at least one endpoint in U. (c) U a clique wen for eac dtinct u, v U, (u, v) E. Te blue vertices are a max-sized independent set for te grap. Te red vertices are a min-sized vertex cover for te grap. Te images are from Wikipedia. NP-Completeness 14 / 1 Independent set, vertex cover, and clique, 3 Independent Set Problem Find: An independent set for G of size b. Vertex Cover Problem Find: A vertex cover for G of size b. Clique Problem Find: A clique in G of size b. Te red vertices are a max-sized clique for te grap. Te only known algoritms for tese are exponential time. NP-Completeness 15 / 1 NP-Completeness 16 / 1

5 Longest Pat, Knapsack, Subset Sum In sort, te world full of searc, some of wic can be solved efficiently oters seem to Problems: be very ard. Hard T and depicted Easyin te following table. Te Longest Pat Problem Given: A undirected grap G. Find: A longest simple pat in G. (Simple pat a pat wit no repeated vertices.) Knapsack Given: Weigts w 1,..., w n, values v 1,..., v n, Total Capacity: W, and Goal: G (all positive integers). Find: A selection of items wit total weigt W and total value G. Subset Sum Given: A multet of integers M and goal G. Find: An { x 1,..., x k } M suc tat G = x x k. NP-Completeness 17 / 1 Hard (NP-complete) Easy (in P) 3SAT TRAVELING SALESMAN PROBLEM LONGEST PATH 3D MATCHING KNAPSACK INDEPENDENT SET INTEGER LINEAR PROGRAMMING RUDRATA PATH BALANCED CUT 2SAT, HORN SAT MINIMUM SPANNING TREE SHORTEST PATH BIPARTITE MATCHING UNARY KNAPSACK INDEPENDENT SET on trees LINEAR PROGRAMMING EULER PATH MINIMUM CUT All of te ard above are ard for te same reason. T table wort contemplating. On te rigt we ave tat can be Te only known algoritms for tese are exponential time. In fact, tey are all te same problem in dgue. efficiently. On te left, we ave a bunc of ard nuts tat ave escaped efficient soluti?? But didn t we ave an LP solution to Knapsack? Yes, but... many decades or centuries. NP-Completeness 18 / 1 P and NP, 1 Recall: Elements of a Searc Problem I: an instance of te problem S: a possible solution for I C : (Instances) (Pot. Solutions) { True, False } C(I, S) = te result of cecking if S solves I C sould be computable in O( I k ) for some k. (T implies tat S cannot be too large.) (NP and P: As searc ) (a) An efficient cecking algoritm an algoritm tat computes C as above in O( I k )-time for some k. (b) Eac efficient cecking algoritm C determines a searc problem: S a solution for I C(I, S) = True. (c) NP = te class of all searc tat ave eff. cecking algoritms. (d) P = te class of all searc for wic one can find solutions (or determine tere are none) in polynomial time. NP-Completeness 19 / 1 P and NP, 2 A decion problem a problem wit an Yes/No answer. SAT as a decion problem: Does CNF formula F ave satfying assignment? Hamiltonian Pat as a decion problem: Does G ave a Hamiltonian Pat? (NP and P: As decion ) (a) NP = te class of all decion tat ave eff. cecking algoritms. (b) P = te class of all decion tat ave polytime algoritms. In Algoritms: searc are a bit more natural tan decion. In Computational Complexity Teory: te reverse Wic better? Te searc and decion forms of NP are rougly of equivalent ardness. Te $1,000,000,000 Question: P? = NP. Collect your prize ere: ttp:// Q: Wat te bas for believing all tose we just lted are ard? NP-Completeness 20 / 1

6 efficient algoritm (besides, of course, te torical fact tat many clever matematicians and computer scientts ave tried ard and failed to find any)? Suc evidence provided by reductions, Formalizing wic translate Reductions, one searc problem 1 into anoter. Wat tey demonstrate tat te on te left side of te table are all, in some sense, exactly te same problem, except tat tey are stated in different languages. Wat s more, we will also use reductions to sowreducing tat tese A are tote problem ardest B searc reprasing A in NP if B s language even one of tem as a polynomial E.g., Te time reprasing algoritm, te tenboolean every problem CircuitinEval NPProblem as a polynomial as an LPtime problem. algoritm. Tus if we believe tat P NP, ten all tese searc are ard. We defined reductions in Capter 7 and saw many examples of tem. Let s now specialize t definition to searc. A reduction from searc problem A to searc problem B A problem A reducible to problem B (written: A B [DPV] or A B [tese a polynomial-time algoritm f tat transforms any instance I of A into an instance f(i) of B, togeter slides]) wit wen anoter tere polynomial-time are two polytime algoritm computable tat functions maps any f and solution S of f(i) back S. Dasgupta, C.H. Papadimitriou, and U.V. Vazirani 259 into a solution f transforms (S) of I; an seea-instance te following I into diagram. a B-instance If f(i) f as (I). no solution, ten neiter does I. Tese two transforms translationa procedures B-solutionfS and of f (I) imply for into tatan anya-solution algoritm(s) for Bof can I be converted tat into an P algoritm NP te for task A by of bracketing finding a it proof between for a f given and. matematical assertion a searc problem and terefore in NP (after all, wen a formal proof of a matematical statement written out in excruciating detail, Algoritm it can be cecked for A mecanically, line by line, by an efficient algoritm). So if P = NP, tere would be an efficient metod to prove any teorem, tus eliminating te need for matematicians! All insolution all, tere S of f(i) are a variety of Solution reasons wy it Instance Instance f(i) Algoritm widely believed (S) of I I tat f P NP. However, proving t as turned out to be extremely difficult, one of te deepest and most importantfor unsolved B puzzles of matematics. Reductions, again No solution to f(i) No solution to I Even And if now we accept we cantat finally P define NP, te wat class about of te te ardest specific searc. on te left side of te table? On te bas of wat evidence do we believe tat tese particular ave no A searc problem NP-complete NP-Completeness if all oter searc 21 / 1 efficient algoritm (besides, of course, te torical fact tat many reduce clever to it. matematicians and computer scientts ave tried ard and failed to find any)? Suc evidence provided T by reductions, a very strong wic requirement translate one indeed. searc For problem a problem into to anoter. be NP-complete, Wat tey it demonstrate must be useful in tat solving te every searc on te problem left side of in te te table world! are It all, in remarkable some sense, tat exactly suc te same problem, ext. except But tey Formalizing tat do, tey and are te stated firstreductions, column in different of te languages. table 3 we Wat s saw earlier more, we filled will wit also use te reductions most famous to sow examples. tat tese In Section 8.3 we are sall te see ardest ow searc all tese in reduce NP if to even one one anoter, of tem and as also a polynomial wy all oter time searc algoritm, ten reduce every to problem tem. in NP as a polynomial time algoritm. Tus if we believe tat P NP, ten all tese searc are ard. We defined reductions in Capter 7 and saw many examples of tem. Let s now specialize t definition Reductions to searc compose. A reduction from searc problem A to searc problem B a polynomial-time I.e., A Balgoritm and B Cf implies tat transforms tat A any C. instance I of A into an instance f(i) of B, togeter If wit A Banoter and Apolynomial-time NP-ard, tenalgoritm so B. tat maps any solution S of f(i) back into a solution (S) of I; see te following diagram. If f(i) as no solution, ten neiter does I. Tese two If A translation NP-complete procedures and B f and an NP-problem imply tat any wit algoritm A B, for B can be converted into an algoritm ten B for also A by NP-complete. bracketing it between (Wy f and t. andy?) Instance I f Instance f(i) Algoritm for A Algoritm for B Solution S of f(i) No solution to f(i) And now we can finally define te class of te ardest searc. Solution (S) of I No solution to I Even if we accept tat P NP, wat about te specific on te left side of te table? On te bas of wat evidence do we believe tat tese particular ave no efficient algoritm (besides, of course, te torical fact tat many clever matematicians Formalizing Reductions, 2 and computer scientts ave tried ard and failed to find any)? Suc evidence provided by reductions, wic translate one searc problem into anoter. Wat tey demonstrate tat te on te left side of te table are all, in some sense, exactly te same problem, except tat tey are stated in different languages. Wat s more, we will also use reductions to sow(a) tat Atese problem NP-ard are te wen ardest all searc in NP if even one of tem as a polynomial NP- time algoritm, reduceten to it. every problem in NP as a polynomial time algoritm. Tus if we(b) believe A problem tat P NP-complete NP, ten allwen teseit searc are ard. We defined NP and reductions NP-ard. in Capter 7 and saw many examples of tem. Let s now specialize t definition to searc. A reduction from searc problem A to searc problem B a polynomial-time If A B andalgoritm A ard, ften tatsotransforms B. any instance I of A into an instance f(i) of B, togeter (Wy?) wit anoter polynomial-time algoritm tat maps any solution S of f(i) back into a solution (S) of I; see te following diagram. If f(i) as no solution, ten neiter does S. Dasgupta, If AC.H. B and Papadimitriou, B easy, ten and sou.v. I. Tese two translation procedures f and A. Vazirani 259 imply tat any algoritm for B can be converted into an algoritm (Wy?) for A by bracketing it between f and Image. from ttp:// tat P NP te task of finding a proof for a given matematical assertion a searc problem and terefore in NP (after Algoritm all, wen fora Aformal proof of a matematical statement written out in excruciating detail, it can be cecked mecanically, line by line, by an efficient algoritm). So if P = NP, tere would be an efficient Solution Smetod of f(i) to prove Solution any teorem, tus Instance Instance f(i) Algoritm eliminating (S) of I I te need f for matematicians! All in all, tere are a variety of reasons wy it widely believed tat P NP. However, forproving B tno as solution turned to f(i) out to be extremely difficult, No solution to I one of te deepest and most important unsolved puzzles of matematics. Reductions, And now we again can finally define te class of te ardest searc. NP-Completeness 22 / 1 Even if we accept tat P NP, wat about te specific on te left side of te table? A On searc teproblem bas of wat NP-complete evidence do if all weoter believe searc tat tese particular reduce to it. ave no efficient algoritm (besides, of course, te torical fact tat many clever matematicians T a very strong requirement indeed. For a problem to be NP-complete, it must be useful and computer Formalizing scientts ave Reductions, tried ard and4 failed to find any)? Suc evidence provided in solving every searc problem in te world! It remarkable tat suc ext. by reductions, wic translate one searc problem into anoter. Wat tey demonstrate But tey do, and te first column of te table we saw earlier filled wit te most famous tat te on te left side of te table are all, in some sense, exactly te same problem, examples. StepsIninSection proving 8.3 we tat sall asee reduction ow all tese works reduce to one anoter, and also except tat tey are stated in different languages. Wat s more, we will also use reductions to wy all oter searc reduce to tem. sowargue tat tese tat: are te ardest searc in NP if even one of tem as a polynomial time algoritm, ten every problem in NP as a polynomial time algoritm. Tus 1. f and are polytime computable. (Some and waving on t usually OK.) if we believe tat P NP, ten all tese searc are ard. We defined 2. If f (I) reductions as solution in Capter S, ten7 Iand assaw solution many(s). examples of tem. Let s now specialize t definition 3. If f (I) to as searc no solution,. ten A reduction I as nofrom solution. searc problem A to searc problem B a polynomial-time 3 algoritm f tat transforms any instance I of A into an instance f(i) of. If I as a solution, ten f (I) also as solution. B, togeter wit anoter polynomial-time algoritm tat maps any solution S of f(i) back into a solution 3 and (S) 3 are of I; equivalent: see te following (A = diagram. B) If f(i) ( B as = no A). solution, ten neiter does I. Tese two translation However, 3 procedures f and imply tat any algoritm for B can be converted into an algoritm for A byusually bracketing easier it between to prove. f and. Instance I f Instance f(i) Algoritm for A Algoritm for B Solution S of f(i) No solution to f(i) Solution (S) of I No solution to I A searc problem NP-complete if all oter searc reduce to it. NP-Completeness 23 / 1 T a very strong requirement indeed. For a problem to be NP-complete, it must be useful in solving every searc problem in te world! It remarkable tat suc ext. And now we can finally define te class of te ardest searc. NP-Completeness 24 / 1 A searc problem NP-complete if all oter searc reduce to it.

7 Te Plan of 8.3 eductions between searc. All of NP SAT Algoritms Warm up: Rudrata (s, t)-pat Rudrata Cycle Rudrata (s, t)-pat Given: G = (V, E) and s, t V. Find: A pat from s to t in G passing troug eac vertex once. 3SAT INDEPENDENT SET 3D MATCHING VERTEX COVER CLIQUE ZOE SUBSET SUM ILP RUDRATA CYCLE TSP Rudrata Cycle Given: G = (V, E). Find: A cycle in G passing troug eac vertex once. Rudrata (s, t)-pat Rudrata Cycle: Given an instance of Rudrata (s, t)-pat: (G, s, t) construct G by adding an new vertex x and new edges (x, s) and (x, t). If P a Rudrata-cyclc in G, ten leaving x, (x, s) and (x, t) out of P gives an (s, t)-rudrata pat in G. If G as a (s, t)-rudrata pat P, ten adding (x, s) and (x, t) to P yields a Rudrata cycle in G. If G as no Rudrata cycle, ten G as no (s, t)-rudrata pat. NP-Completeness 25 / 1 NP-Completeness 26 / 1 reductions 3SAT Independent Set, 1 see tat te searc of Section 8.1 can be reduced to one anoter as igure 8.7. As a consequence, tey are all NP-complete. e tackle 3SATte specific reductions in te tree, let s warm up by relating two versions Tink circuits, but not too literally. G G 3SAT Independent Set, 2 To build a satfying assignment of an instance of 3SAT, we ave to pick out at least one literal per clause to be TRUE constently! ta problem. Consider: (x y z) (x y z) (x y z) (x y). Given: A CNF formula θ in wic eac clause as at most 3 literals. Find: A satfying assignment for θ. A clause (l 1 l 2 l 3 ) represented by a triangle wit verts labeled: l 1, l 2, & l 3. Idea: If v U an indep. set and v as label l, s, t)-path RUDRATA CYCLE ten te trut assignment corresponding to U sets l to TRUE. Independent Set Problem UDRATA CYCLE problem: given a grap, tere a cycle tat passes troug eac Set g = # of clauses. So any indep. set must ave one vertex from eac triangle. ly once? Given: WeG can = (V, also E) formulate and b. te closely related RUDRATA (s, t)-path problem, (Wy?) Find: An independent set for G of size vertices s and t are specified, and we b. 264 want a pat starting at s and ending at t But we ave to ave constency cecks. We don t want to set x and x to TRUE. Algoritms I.e., Find U V wit U b and ( u, v U)[(u, v) / E]. oug eac vertex exactly once. Is it possible tat RUDRATA CYCLE easier tan So put an edge between eac occurrence of a variable x and eac occurrence of x. Figure 8.8 Te grap corresponding to (x y z) (x y z) (x y z) (x y). T enforces constency. (Wy?), t)-path? We will sow by a reduction tat te answer no. Puzzle: ction maps an instance (G = (V, E), s, t) of RUDRATA (s, t)-path into an instance Tese are very different of RUDRATA CYCLE as follows: G looking. How to we get simply a reduction? G wit an additional vertex x and y y y y es {s, x} and {x, t}. For instance: x z x z x z x s NP-Completeness 27 / 1 s NP-Completeness 28 / 1

8 3SAT Independent Set, 3 Construction: 3SAT instance IS instance Given an instance of 3SAT C 1,..., C k. For eac clause C i = (l 1, l 2, l 3 ), build a triangle wit vertices labeled l 1, l 2, & l 3. Add an edge between eac literal and all of its opposites. Set g = k. Construction: IS solution 3SAT solution Given an independent set U: for eac v U wit label l, set l to TRUE. For eac variable tat does not yet ave a trut value, set it to TRUE. Claim 1: Bot 264 constructions are poly-time. Algoritms Claim 2: Given an indep. set, we can construct a satfying assignment for C 1,..., C k. Claim 3 Figure 8.8 Te grap corresponding to (x y z) (x y z) (x y z) (x y). : If C 1,..., C k as an satfying assignment, ten tere size g indep. set. (x y z) (x y z) (x y z) (x y) y y y y x z x z x z x SAT 3SAT, 1 Construction: SAT instance 3SAT instance Translate eac clause to sequence of clauses. (l 1 l 2 l k ) (l 1 l 2 y 1 ) (y 1 l 3 y 2 ) (y k 3 l k 1 l k ) were k > 3 and y 1,..., y k 3 are new vars. Clauses wit 3 literals translate to temselves. Construction: 3SAT solution SAT solution Take te trut assignment for te 3SAT formula & restrict it to te original vars. Claim 1: Bot constructions are poly-time. Claim 2: If te 3SAT formula as satfying assignment, ten restriction satfies te SAT formula. Claim 3 : If te SAT formula satfiable, so te 3SAT formula. NP-Completeness 29 / 1 NP-Completeness 30 / 1 Let us tink. To form a satfying trut assignment we must pick one literal from eac clause and give it te value true. But our coices must be constent: if we coose x in one clause, we cannot coose x in anoter. Any constent coice of literals, one from eac clause, specifies a trut assignment (variables for wic neiter literal as been cosen can take on eiter value). So, let us represent a clause, say (x y z), by a triangle, wit vertices labeled x, y, z. Wy triangle? Because a triangle as its tree vertices maximally connected, and tus forces us to pick only one of tem for te independent set. Repeat t construction for all clauses a clause wit two literals will be represented simply by an edge joining te literals. (A clause wit one literal silly and can be removed in a preprocessing step, since te value of te variable determined.) In te resulting grap, an independent set as to pick at most one literal from eac group (clause). To force exactly one coice from eac clause, take te goal g to be te number of clauses; in our example, g = 4. All tat msing now a way to prevent us from coosing opposite literals (tat, bot x and x) in different clauses. But t easy: put an edge between any two vertices tat correspond to opposite literals. Te resulting grap for our example sown in Figure 8.8. Let s recap te construction. Given an instance I of 3SAT, we create an instance (G, g) of INDEPENDENT SET as follows. Grap G as a triangle for eac clause (or just an edge, if te clause as two literals), wit vertices labeled by te clause s literals, and as additional edges between any two vertices tat represent opposite literals. Te goal g set to te number of clauses. Clearly, t construction takes polynomial time. However, recall tat for a reduction we do not just need an efficient way to map instances of te first problem to instances of te second (te function f in te diagram on page 259), but also a way to reconstruct a solution to te first instance from any solution of te second (te function ). As always, tere are two tings to sow. 1. Given an independent set S of g vertices in G, it possible to efficiently recover a satfying trut assignment to I.