- SAT was the first known NP-complete problem, as proved by Stephen Cook at the University of Toronto in 1971 and independently by Leonid Levin at the National Academy of Sciences in 1973. Until that time, the concept of an NP-complete problem did not even exist. The proof shows how every decision problem in the complexity class NP can be reduced to the SAT problem for CNF formulas, sometimes.
- ism and Turing machines). Here is an intuitive justi cation. SAT is in NP: We nondeter
- istic Turing machine can be modeled as a boolean expression with polynomial size
- NP completeness of SAT Problem explained, P NP NP-HARD NP-COMPLETE, NP completeness of SAT Problem and solution, SAT is NP complete proof, NP completeness of SAT Problem, Satisfiability problem, 3.
- Theorem (Cook-Levin) 3-SAT is NP-complete. Proven in early 1970s by Cook. Slightly di erent proof by Levin independently

- The Cook-Levin theorem asserts that SATISFIABILITY is
**NP-complete**. Although 3-CNF expressions are a subset of the CNF expressions, they are complex enough in the sense that testing for**satis**ability turns out to be**NP-complete**. Theorem : 3SAT is**NP-complete**. Proof : Evidently 3SAT is in**NP**, since**SAT****is**in**NP** - NAE-3-SAT is the special case where each clause has exactly 3 literals. Show that NAE-3-SAT is NP-complete by reducing 3- SAT to it. Solution: NAE-3-SAT, like any variant of SAT, is in NP since the truth assignment is the certificate; we can check every clause in polynomial time to see if it is satisfied
- 4-SAT is a generalization of 3-SAT (k-SAT is SAT where each clause has k or FEWER literals), and so it is automatically complete in NP since 3-SAT is. A more interesting construction is the proof that 3-SAT is NP-Complete. This was originally done by Richard Kar

An instance of Double Sat problem is a boolean formula f. Since an NP-complete problem is a problem which is both NP and NP-Hard, the proof or statement that a problem is NP-Complete consists of two parts: The problem itself is in NP class. All other problems in NP class can be polynomial-time reducible to that 3-SAT is NP-Complete because SAT is - any SAT formula can be rewritten as a conjunctive statement of literal clauses with 3 literals, and the satisifiability of the new statement will be identical to that of the original formula 3-SAT is NP-complete. Because 3-SAT is a restriction of SAT, it is not obvious that 3-SAT is difficult to solve. Maybe the restriction makes it easier. But, in reality, 3-SAT is just as difficult as SAT; the restriction to 3 literals per clause makes no difference. Theorem. 3-SAT is NP-complete

* SAT can be used to prove that other problems are NP complete by showing that the other problem is in NP and that SAT can be reduced to the other problem in polynomial time*. Shortly after Cook published his result, Richard Karp wrote an influential paper that showed many important optimization problems arising in theory and practice were also NP. 3.Conclude that 6=SAT is NP-complete. Solution: 1.In an 6=-assignment each clause has at least one literal assigned 1 and at least one literal assigned 0. The negation of an 6=-assignment preserves this property, so it too is an 6=-assignment. 2.To prove that the given reduction works, we need to show that if the formula ˚is mapped to ˚0, then

To understand why 2SAT isn't NP-hard, you have to consider how easy it is to reduce other problems in NP to it. To get an intuitive understanding of this, look at how SAT can be reduced to 3SAT and try to apply the same techniques to reduce SAT to 2SAT. 2SAT is just not as expressive as 3SAT and other SAT variants What makes a problem harder than another problem? How can we say a problem is the hardest in a complexity class? In this video, we provide a proof sketch o.. In the context of NP-completeness, SAT is the satsifyability problem. And 3-SAT is probably what he's referring to. The satsifyability problem is to find truth values for variables [itex]a_1,a_2...a_n[/itex] i.e. (T or F) so that a particular logical expression is true. 3-SAT restricts this to logical expressions of a particular form Solutions to Assignment#2 Liana Yepremyan 1 4-SAT problem Claim 1.1. 4-SAT is NP-complete. Proof. One way to show this is by reduction from 3-SAT

k-SAT is a special case of SAT. Since SAT is NP-complete, I don't understand why we don't have k-SAT is NP-complete for whatever values of k. In the class, my professor used polynomial reduction fr.. In NP: guess a satisfying assignment and verify that it indeed sat-isﬁes the clauses. NP-hard: We show SAT ≤pm 3SAT. Suppose (U,C) is an instance of satisﬁability. We construct an instance (U′,C′) of 3SAT such that, Cis satisﬁable iﬀ C′ is satisﬁable (and the reduction can be done in poly time) A problem is said to be NP-complete if it, in addition to being in NP, also has the property that any other problem in NP can be reduced to it in polynomial-time. Cook's paper proved SAT to be NP-complete. In fact, since that paper introduced the concept of NP-completeness, SAT was the first problem to be proved NP-complete

In computational complexity theory, the Cook-Levin theorem, also known as Cook's theorem, states that the Boolean satisfiability problem is NP-complete. That is, it is in NP, and any problem in NP can be reduced in polynomial time by a deterministic Turing machine to the Boolean satisfiability problem 8. Show that Monotone NAE-3-SAT, where no variables are negated, is NP-complete. Hint: use two variables in the new formula for each variable of the old. You may repeat variables in a clause. Answer. For membership in NP, as for 3-SAT or NAE-3-SAT the truth assignment is the certiﬁcate, and we can check all the clauses in polynomial time 看懂本文需要提前有哪些知识？知道什么是图灵机，非确定图灵机知道什么是SAT问题什么是NP问题？首先我们要知道，NP一般都是针对决定性问题（Decision problem）的。比如今天我们要讨论的SAT问题，我们要知道一个 This 3-SAT <= NAE 3-SAT reduction is wrong. Suppose x_F is used in place of F. The issue is that the constructed NAE 3-SAT instance is always an yes instance. For instance, the assignment A given by : set all x_i = False, all c_j=True, x_F=True (so first kind (so second kind of clauses satisfied) of clauses satisfied

Thought of as a computational problem, the input to SAT is a clausal formula φ and the problem is to determine whether φ is satisfiable. SAT is NP-complete Cook and Levin independently proved that SAT is NP-complete For any k > 3, k -SAT is obviously NP-complete; as any algorithm for those can be used to solve 3-SAT, as your professor seems to have discussed. So, the fundamental confusion here is that while k -SAT is an instance of the general SAT problem, it does not mean all k -SAT problems are equally difficult

Since $\mathrm{SAT}$ is the first problem proven to be NP-complete, Cook proved that $\mathrm{SAT}$ is NP-complete using the basic definition of NP-completeness which says that to prove that a problem is NP-complete if all NP problems are reducible to it in polynomial time. So, Cook did this using the Turing Machine concept Why is SAT important? • Theoretical importance: - First NP-complete problem (Cook, 1971) • Many practical applications: - Model Checking - Automatic Test Pattern Generation - Combinational Equivalence Checking - Planning in AI - Automated Theorem Proving - Software Verification - 4 An Example • Inputs to SAT solvers are. We now prove that 3SAT is NP-complete, by reduction from SAT. Take a formula F of SAT. We transform it into a formula F0 of 3SAT such that F0 is satis able if and only if F is satis able. Each clause of F is transformed into a sub-expression of F0. Clauses of length 3 are left unchanged Proof that CSATis NP-complete Clearly CSAT2NP: a witness is just a satisfying assignment Need to show that CSATis NP-hard, i.e., L P CSATfor all L2NP Let L2NP We know there is a language L02P and constants ‰,b,csuch that ∀‚2{0,1}∗: ‚2L ()∃'2{0,1}‰j‚jb+c:h‚,'i2L0 Let M0be the polynomial time RAM that recognizes L0 addition, B is in NP, then it is NP-complete. Thus if A is NP-complete, and it has a reduction to another problem B in NP, then B is also NP-complete. 2.3 Examples of Reduction SAT is NP-complete (we will not prove this in class). 1. ILP is NP-complete Let's take the following SAT problem and see if it can be solved by an ILP

NP-Complete: can be solved in Polynomial time only using a Non-deterministic method. NP-Complete may not last. Oh, one more thing, it is believed that if anyone could *ever* solve an NP-Complete problem in P time, then *all* NP-complete problems could also be solved that way by using the same method, and the whole class of NP-Complete. Even though SAT is NP-complete and therefore no known polynomial-time algorithm for it is (yet) known, many improvements over the basic backtracking algorithms have been made over the last few decades. However, here we will look at one of the most basic yet relatively efficient algorithms for solving SAT that 4-SAT is NP-complete. (Don't forget to show that it is in NP.) Ans: To show that 4-SAT is NP-complete, we prove that 4-SAT is in NP and NP-hard. First, 4-SAT is in NP, we can write a nondeterministic polynomial-time al-gorithm which takes a 4-SAT instance and a proposed truth assignment as input. This algorithm evaluates the 4-SAT instance with the truth assign-ment NP-Complete is a class of problems. The class P consists of those problems that are solvable in polynomial time. For example, they could be solved in O (n k) for some constant k, where n is the size of the input. Simply put, you can write a program that will run in reasonable time SAT NPC. Proof. SAT NP since certificate is satisfying assignment of variables. To show SAT is NP-hard, must show every L NP is p-time reducible to it. Idea: Use p-time verifier A(x,y) of L to construct input of SAT s.t. verifier says yes iff satisfiabl

** Introduction to Complexity Theory: 3-Colouring is NP-complete We next show that 3-colouring is NP-complete**. What's the colouring problem on graphs? Given a graph G(V;E), the colouring problem asks for an assignment of kcolours to the vertices c: V ! 3-SAT. Therefore 3-COLOURING is NP-complete NP-complete problems 8.1 Search problems Over the past seven chapters we have developed algorithms for nding shortest paths and minimum spanning trees in graphs, matchings in bipartite graphs, maximum increasing sub-sequences, maximum ows in networks, and so on. All these algorithms are efcient, becaus

Problem 1 (25 points) Show that for n>3, n-sat is NP-complete. (You don't need to show that n-sat is in NP.) Proof: We reduce 3-sat to n-sat as follows. Let ˚be an instance of 3-sat. For any clause (a_b_c) of ˚, replace it with (a_b_c|__{z c} n 2). By repeating this procedure for all clauses of ˚, we derive a new boolean expression ˚0for. 1 point for the overall SAT instance (one 'gadget' per edge). 1 point for conversion of k (if needed). (c)(1 points) Conclude from parts a) and b) that Max-2-SAT is NP-Complete. SOLUTION: Since MaxCut is NP-hard, part b) implies that Max-2-SAT is also NP-hard. Furthermore, Max-2-SAT is in NP because given an assignment of the variables, we ca

- The fact that your problem is NP-complete does not exclude a possibility of a very efficient algorithm that solves your problem in some special restrictive cases. So once again, the fact that the problem is NP-complete means that it is difficult to design an efficient algorithm that works for all possible cases
- A useful example of an NP-complete (in NP and NP-hard) problem is 3-SAT. In 3-SAT, the input is a boolean formula over some number of boolean variables, constrained to be a conjunction (an AND) of some number of disjunctions (ORs) of three literals each, where a literal is a variable or its negation
- Proof that 3SAT is NP-complete Recall 3SAT: Input: ˚a boolean formula in 3CNF Question: is there a satisfying assignment? The language 3SAT is a restriction of SAT, and so 3SAT 2NP. Goddard 19b:
- Many of these problems can be reduced to one of the classical problems called NP-complete problems which either cannot be solved by a polynomial algorithm or solving any one of them would win you a million dollars (see Millenium Prize Problems) and eternal worldwide fame for solving the main problem of computer science called P vs NP.
- Proof that naesat is NP-complete naesat Instance: An instance of 3-sat. Answer: \Yes if each clause is satis able when not all literals have the same value. Theorem naesat is NP-complete. Proof Use the reduction from circuit sat to 3-sat. However, rst convert the circuit from and, or, and not to nand. Replace a step computin
- Theorem 2. SAT is NP-complete. Recall that to show a problem Lis NP-complete, we need to show 2 things: 1. L2NP. 2. Lis NP-hard (i.e. A P Lfor all A2NP). Theorem 1 states that Circuit-SAT is NP-complete, and thus NP-hard, which means for all A2NP: A P Circuit-SAT Therefore there exists a polytime reduction from any problem Ain NP to Circuit-SAT.

NP-Complete . A problem X is NP-Complete if these hold: X is in NP ; Every problem in NP reduces to X ; What are some NP-Complete problems: SAT (proved by Cook) Any problem P such that SAT (or any other NPC problem) reduces to In this post, we will prove that 0-1 integer programming is NP-complete using a reduction from 3-CNF-SAT (which is NP-complete [1]). We will follow the template given in an earlier post. Problem statement. The decision problem for 0-1 integer programming is formulated as follows [2] A language L is NP-complete if and only if. L ∈ NP and; L is NP-hard. The most famous NP-complete language is SAT. It contains all boolean formulas that can be satisfied. For example, (a ∨ b) ∧ (¬a ∨ ¬b) ∈ SAT. A valid witness is {a = 1, b = 0}. The formula (a ∨ b) ∧ (¬a ∨ b) ∧ ¬b ∉ SAT. (How would you prove that?) It is. We will now use the fact that 3-SAT is NP-complete to prove that a natural graph problem called the Max-Clique problem is NP-complete. Deﬁnition 20.2 Max-Clique: Given a graph G, ﬁnd the largest clique (set of nodes such that all pairs in the set are neighbors). Decision problem: Given G and integer k, does G contain

And so NP-complete is a nice answer because this says you're exactly as hard as everything in NP--no harder, no easier. If you draw, in this vague sense, computational difficulty on one axis-- which is not really accurate, but I like to do it anyway-- and you have P is all of these easy problems down here Die Klasse aller NP-vollständigen Probleme wird mit NP-C (complete) bezeichnet. Die Eigenschaften dieser und anderer Klassen werden in der Komplexitätstheorie erforscht, einem Teilgebiet der theoretischen Informatik

NP-complete problem, any of a class of computational problems for which no efficient solution algorithm has been found. Many significant computer-science problems belong to this class—e.g., the traveling salesman problem, satisfiability problems, and graph-covering problems.. So-called easy, or tractable, problems can be solved by computer algorithms that run in polynomial time; i.e., for a. Since any SAT solution also satisfies the 3-SAT instance and any 3-SAT solution sets the variables giving a SAT solution, the transformed problem is equivallent to the original. Note that a slight modification to this construction would serve to prove that 4-SAT, 5-SAT, or any -SAT is also NP-complete Problem 2: (12 points) The proof that SAT is NP-complete appears in Sipser's book, p. 254-259. Part of the main construction involves constructing a formula 13254 6.7, which is expressed as the conjunction of formulas saying that 2 8 3 windows of the tableau are legal. For each of the following, state whether the This is an algorithm for solving SAT, an NP-complete problem. We show first, by reduction from 3-SAT, that this minimum k-cover problem is NP-complete. Then we propose further two greedy.

- ing the {\bf P} versus {\bf NP} problem against the Kleene-Rosser paradox of the $\lambda$-calculus [94], it was found that it represents a counter-example to NP-completeness. We prove that it contradicts the proof of Cook's theorem
- 1. 3SAT is NP-complete. 2. ALL NPC problems can be coded into SAT. (Some directly like 3COL.) Exposition by William Gasarch Algorithms for 3-SAT. Exposition by William Gasarch Algorithms for 3-SAT. Key Idea Behind Recursive 7-ALG KEY1: If F is a 3CNF formula and z is a partial assignment either 1. F(z) = TRUE, or 2. there is a clause C = (L.
- istic Turing machines. The class NP can also be deﬁned using a variant of Turing machines calle
- 3-Coloring is NP-Complete • 3-Coloring is in NP • Certiﬁcate: for each node a color from {1,2,3} • Certiﬁer: Check if for each edge ( u,v), the color of u is diﬀerent from that of v • Hardness: We will show 3-SAT ≤ P 3-Colorin

Proving that a problem is NP is usually trivial, but proving that a problem is NP-hard is not. Boolean satisfiability (SAT) is widely believed to be NP-hard, and thus the usual way of proving that a problem is NP-complete is to prove that there's a polynomial time transformation of the problem to SAT. Why TSP Is Not NP-complete 2-Sat, it follows that Max 2-Sat (as a decision problem) is NP-complete. Notation A CNF formula is represented as a set of clauses. The symbols R and Z denote the sets of reals and integers, respectively. The letter ω denotes the smallest real number such that for all ǫ > 0, n by n matrix multiplication over a ring can be 1 The 3-SAT problem is part of the Karp's 21 NP-complete problems and is used as the starting point to prove that the other problems are also NP-Complete. One example is the independent set problem. The Independent Set Problem can be shown to be NP-Complete by showing that the 3-SAT is polynomially reducible to an independent set problem. 4.4

CIRCUIT-SAT is NP-Complete. Proof: 1. CIRCUIT-SAT NP Proof: Can verify an input assignment satisfies a circuit by computing the output of a finite number of gates, one of which will be the output of the circuit. This can be done in polynomial time. Thus, by definition of NP, CIRCUIT-SAT NP. 2. CIRCUIT-SAT NP-Hard I.e., L CIRCUIT-SAT for every L NP Prove that the Set Intersection problem (defined below) is NP-complete. Two things are required: • Show that Set Intersection is in NP. • Show that CNF-satisfiability is polynomially reducible to Set Intersection. (Note: It is known from Cook's proof that CNF-satisfiabillity is NP-complete.) Definition of the Set Intersection problem Definition: A problem B is NP-complete iff A is NP-hard and A ∈∈∈NP Even though we seem to have lots of hard problems in NP it is not obvious that such super-hard problems even exist! 29 Cook-Levin Theorem Theorem (Cook 1971, Levin 1973): 3-SAT is NP -complete Recal

- 3SAT, or the Boolean satisfiability problem, is a problem that asks what is the fastest algorithm to tell for a given formula in Boolean algebra (with unknown number of variables) whether it is satisfiable, that is, whether there is some combination of the (binary) values of the variables that will give 1. For example, the formula A+1 is satisfiable because, whether A is 0 or 1, the result.
- 2. Next we need to show that CLIQUE is NP-hard; that is we need to show that CLIQUE is at least as hard any other problem in NP. To do so, we give a reduction from 3-SAT (which we've shown is NP-complete) to CLIQUE. Our goal is the following: Given an instance ˚of 3-SAT, we will produce a graph G(V;E) and an integer ksuch that Ghas a cliqu
- Proof that 3SAT is
**NP-complete**. To show that a language is**NP-complete**, all that is required is to show that an existing**NP-complete**problem can be simulated efficiently by the target language. For example, in 3-colorability, I use the fact that 3SAT is an**NP-complete**language to prove that 3-colourability is itself**NP-complete** - istic Turing machine accepts in Polynomial time.
- Lecture 19: SAT is NP-complete 19-4 ˚ startcontains a single clause with nk variables. acceptcontains one variable for each cell of the table, so its size is O(n2k) transitioncontains a formula whose size is xed and depends only on (the transition function of M) for each cell of the table. So, size of ˚ transitionis O(n2k). So, the total size of the formula is O(n2k)
- e if there exists an interpretation that satisfies it. For example, given = ∧ ∨ ∧ This formula is satisfiable because it's true when x is false, y is true and z is false. NP-Complete. A decision problem is NP-complete when it is both in NP and NP-hard. NP means
- istic Turing machine that can solve it in polynomial time. If any problem in NP can be reduced to an SAT problem in Polynomial-time, then it's NP-Complete. We can prove by taking any language and reducing it to SAT in polynomial time

Question: The NP-complete Problem CNF Satisfiability (called SAT) Was Described In Class. 2- SOLN-SAT Is The Problem Where, Given A Boolean Expression In CNF Form, You Are Asked, Does There Exist At Least Two Different Solutions Which Make The Expression True? In This Question You Will Show That 2-SOLN-SAT Is NP-complete. Suppose You Already Know That 2-SOLN-SAT. NP-complete Reductions 1. DOUBLEProve that 3SAT P-SAT, i.e., show DOUBLE SAT is NP complete by reduction from 3SAT. The 3-SAT problem consists of a conjunction of clauses over n Boolean variables, where each clause is a disjunction of 3 literals, e.g., ( To conclude, weve shown that 3-COLOURING is in NP and that it is NP-hard by giving a reduction from 3-SAT. Therefore 3-COLOURING is NP-complete Theorem SAT is NP-complete Proof idea: The turing machine program for any problem in NP can be veriﬁed by a polynomial sized SAT instance that encodes that the input is well formed and that each step follows legally from the next. Implication We now have one NP-complete problem. We will now reduce other problems to it (c)(1 points) Conclude from parts a) and b) that Max-2-SAT is NP-Complete. SOLUTION: Since MaxCut is NP-hard, part b) implies that Max-2-SAT is also NP-hard. Furthermore, Max-2-SAT is in NP because given an assignment of the variables, we can check whether at least k clauses are satis ed. GRADING: 1 point if both the hardness and in NP are there, 0 otherwise.

SAT is NP-Complete SAT ≤ P 3SAT therefore 3SAT is NP-Complete. 3SAT ≤ P CLIQUE therefore CLIQUE is NP-Complete General form: Reduce from known NP-Complete (e.g., SAT) to a new problem. Question: What happens if you reduce to SAT rather than from it 3-SAT is an example of what is called a special case of SAT. Some special cases, like 3 -SAT, are NP complete Others are solvable in polynomial time (chapter 4 Proof that 3SAT is NP-complete. To show that a language is NP-complete, all that is required is to show that an existing NP-complete problem can be simulated efficiently by the target language. For example, in 3-colorability, I use the fact that 3SAT is an NP-complete language to prove that 3-colourability is itself NP-complete Description. 8.3 STINGY SAT is the following problem: given a set of clauses (each a disjunction of literals) and an integer k, find a satisfying assignment in which at most k variables a true, if such an assignment exists. Prove that STINGY SAT is NP-complete.. Proof. 若要证明 STINGY SAT 是 NP-完全问题，就要： 证明 STINGY SAT 是 NP 问题；. Cook [9] has shown that k-SAT is NP-complete for k ≥ 3. LCNF is the class of linear formulas and LSAT is its decision problem. In [3] we present a general method to construct linear minimal unsatisfiable (MU) formulas

** 3-SAT = '2CNF-SAT 'has 3 literals per clause Prop: 3-SAT is NP-complete**. Proof: Show SAT 3-SAT. Example: C (' 1 _' 2 __ ' 7) C0 (' 1 _' 2 _d 1) ^(d 1 _' 3 _d 2) ^(d 2 _' 4 _d 3) ^ (d 3 _' 5 _d 4) ^(d 4 _' 6 _' 7) Claim 7.2 C2SAT , C023-SAT Do this construction for each clause independently SAT is NP-complete Corollary 7.42: 3SAT ≤P HALTTM However HALTTM is not NP-complete. P NP All languages NP-hard NP-complete 21. If P = NP then you can factor any integer in polytime Deﬁne language FACTOR = { ;x, a, b <| x, a, b ∈ ℤ and x has a factor p ∈ ℤ st a ≤ p ≤ b} FACTOR. Strategy 3-SAT Sequencing Problems Partitioning Problems Other Problems Proving Other Problems NP-Complete I Claim: If Y is NP-Complete and X 2NPsuch that Y P X,thenX is NP-Complete. I Given a new problem X, a general strategy for proving it NP-Complete is 1. Prove that X 2NP. 2. Select a problem Y known to be NP-Complete

Strategy 3-SAT Sequencing ProblemsPartitioning ProblemsOther Problems NP vs. co-NP Proving Problems NP-Complete I Claim: If Y is NP-Complete and X 2NPsuch that Y P X, then X is NP-Complete. I Given a new problem X, a general strategy for proving it NP-Complete is 1.Prove that X 2NP Step 2: Reduce SAT as known NP hard problem to 3-Coloring I. Since there are 3 colors, saying a, b and c. There are only 3 situations for each edge to have different colors, the two vertexes an edge can be (a,b)(a,c)(b,c). II. If we can prove there exists assignments of two vertexes x i,y i for edge i (i:1..n)which make the boolean formula true. 3-SAT is NP-Complete Theorem. 3-SAT is NP-complete. Pf. Suffices to show that CIRCUIT-SAT ≤ P 3-SAT since 3-SAT is in NP. Let K be any circuit. Create a 3-SAT variable x i for each circuit element i. Make circuit compute correct values at each node: -x 2 = ¬ x 3 ⇒ add 2 clauses: -x 1 = x 4 ∨ x 5 1 ⇒ add 3 clauses: -x 0 = x 1.

Once one problem (SAT) shown to be NP-Complete, can show many others Example reductions (From CLRS, Ch. 34): Subset-Sum is NP Complete . Subset-Sum Problem: - Given: finite set S of positive integers and integer target t > 0 (That is, input instance is <S, t> Show that DOUBLE-SAT is NP-Complete. (Hint: Reduce 3SAT.) Answer: (1) DOUBLE-SAT 2NP: Simply guess two di erent assignments to all variables and verify that each clause is satis ed in both cases. (2) Reduction of 3SAT to DOUBLE-SAT: Given a 3cnf-function , create a new Boolean function 0by adding a new clause (x[x) to , where xis a new variable. Monotone 3-Sat is the restriction of 3-Sat to monotone formulas, i.e. to formulas in which each clause contains only unnegated variables or only negated variables, respectively. In particular, we show that, for any k≥ 5, Monotone 3-Sat is **NP-complete** even if each variable appears exactly k times unnegated and exactly once negated The advantage of this result over Theorem 2 is that SAT is a natural problem. This completeness result opens up the possibility of showing that several other natural problems in NP are also NP-complete by constructing reductions from SAT. Theorem 3 SAT is NP-complete. Proof. To see that SAT ∈ NP, consider a non-deterministic machine whic This reduction proves that BATTLESHIPS is NP-hard. It is easy to see that a non-deterministic guess is checked to be a solution in polynomial time. Hence BATTLESHIPS is NP-complete. However, although the provided reduction is parsimonious, no such reduction exists from 3-SAT to BIN PACKING. If an instanc SAT is the language of all propositional formulas which are satisfiable. So a particular formula is in SAT, if you can choose assignments (true or false) for each variable causing the whole expression to evaluate true. Now to show a language is NP-Complete it needs to satisfy two constraints. The language needs to be in NP