Nå kan du bestille Complete fra Vitusapotek på nett. Se vårt utvalg The clique decision problem is NP-complete.It was one of Richard Karp's original 21 problems shown NP-complete in his 1972 paper Reducibility Among Combinatorial Problems. This problem was also mentioned in Stephen Cook's paper introducing the theory of NP-complete problems. Because of the hardness of the decision problem, the problem of finding a maximum clique is also NP-hard

The Clique Decision Problem belongs to NP-Hard - A problem L belongs to NP-Hard if every NP problem is reducible to L in polynomial time.Now, let the Clique Decision Problem by C. To prove that C is NP-Hard, we take an already known NP-Hard problem, say S, and reduce it to C for a particular instance * In this article, we will prove that the Clique Detection Problem is NP-Complete by the help of Independent Set problem, which is NP-Complete*. Refer to Proof that Clique Decision problem is NP-Complete, for the proof with the help of Boolean Satisfiability Problem.. Clique Problem is in NP If any problem is in NP, then, given a 'certificate', which is a solution to the problem and an.

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**Clique****problem**is**NP-Complete**The algorithm above does not work. You can find a counterexample, where the algorithm removes a vertex that is part of the largest**clique**, but here is a simple observation - To prove that CLIQUE is NP-complete, we need to reduce SAT to CLIQUE. It's a bit easier to reduce 3-SAT to CLIQUE (although we could do a direct reduction from SAT). Therefore I'm going to do this in two steps SAT -> 3-SAT -> CLIQUE Generally spea..
- In this Wikipedia article about the Clique problem in graph theory it states in the beginning that the problem of finding a clique of size K, in a graph G is NP-complete:. Cliques have also been studied in computer science: finding whether there is a clique of a given size in a graph (the clique problem) is NP-complete, but despite this hardness result many algorithms for finding cliques have.
- In computer science, the clique problem is the computational problem of finding a maximum clique, or all cliques, in a given graph. It is NP-complete, one of Karp's 21 NP-complete problems. It is also fixed-parameter intractable, and hard to approximate.Nevertheless, many algorithms for computing cliques have been developed, either running in exponential time (such as the Bron-Kerbosch.

$\begingroup$ @hardmath NP means that a yes answer can be checked in polynomial time. If a no answer can be checked in polynomial time, the problem is in co-NP. $\endgroup$ - Robert Israel Dec 30 '16 at 2:2 ** Now that you know the problem is in NP, you can use reduction to prove that the problem is NP-complete or NP-hard**. (If the problem was in NP, there would be absolutely no use in going to the reduction step because the problem cannot be NP-complete if it is not in NP.) For the reduction, you should transform the clique problem into your problem Corollary 2 : The CLIQUE problem is NP-complete. A clique in a simple undirected graph G= (V;E) is a subset C V of vertices, each pair of which is connected by an edge in E. In other words, a clique is a complete subgraph of G. The size of a clique is the number of vertices it contains; that is, jCj. The CLIQUE problem is to nd a clique of. Introduction to Complexity Theory: CLIQUE is NP-complete In this lecture, we prove that the CLIQUE problem is NP-complete. A clique is a set of pairwise adjacent vertices; so what's the CLIQUE problem: CLIQUE: Given a graph G(V;E) and a positive integer k, return 1 if and only if there exists a set of vertice

Recall that to show that a decision problem (language) L is NP-complete we need to show: (i) L 2NP. (That is, given an input and an appropriate certi cate, we can guess the solution and verify whether the input is in the language), and (ii) L is NP-hard, which we can show by giving a reduction from some known NP-complete problem L0 to L, that. ** Why is the clique problem NP-complete? [duplicate] Ask Question Asked 7 years, 9 months ago**. Active 1 year ago. Viewed 25k times 8. 4 $\begingroup$ This question already has an answer here: Closed 7 years ago. Possible Duplicate: Is the k-clique problem NP-complete? I've. Prove Clique Problem is NP Complete (English+Hindi) University Academy- Formerly-IP University CSE/IT. P and NP Class Problem and Polynomial Time Solvable (English+Hindi) - Duration: 9:07 NP-Hard Graph Problem - Clique Decision Problem CDP is proved as NP-Hard PATREON : https://www.patreon.com/bePatron?u=20475192 Courses on Udemy =====..

- clique problem also has many applications in bioinformatics and computational chemistry. Most versions of the clique problem are hard. The clique decision problem is NP-complete (one o
- The clique decision problem is NP-complete (one of Karp's 21 NP-complete problems). The problem of finding the maximum clique is both fixed-parameter intractable and hard to approximate. And, listing all maximal cliques may require exponential time as there exist graphs with exponentially many maximal cliques
- The Maximal Clique Problem. Although many theoretical papers have been published on DNA computing since Adleman's first crude demonstration in 1994, it would be over two years, in 1997, until another NP complete problem would be solved
- Show that CLIQUE is NP-complete using a reduction from HALF-CLIQUE. Solution: Clearly Half-CLIQUE NP, since an oracle can guess a certificate consisting of a set of k=|V|/2 vertices, and we can test if these vertices are fully-connected in time O(V2). To show that CLIQUE is NP-Complete, we can restrict the instances of our target problem
- istic algorithm. In this algorithm, first we try to deter
- 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 a clique of size ≥k? Note that Max-Clique is clearly in NP. Theorem 20.2 Max-Clique is NP-Complete

- This problem is a simpler (but still NP-complete) version of the form given in Garey and Johnson. For relevant variations and potential heuristic approaches, the papers P.E. Dunne and P.H. Leng, An algorithm for optimising signal selection in demand-driven circuit simulation, Transactions of the Society for Computer Simulation , vol. 8, no.4, pp. 269-280, 199
- NP-complete problems are the hardest in NP: if any NP-complete problem is p-time solvable, then all problems in NP are p-time solvable How to formally compare easiness/hardness of problems? Reductions Reduce language L 1 to L 2 via function f: 1. CLIQUE NPC CLIQUE =.
- imum 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
- Question: Suppose The Only Known NP-Complete Problem Is INDEPENDENT-SET. Use This Knowledge To Prove That The CLIQUE Problem Is NP-Complete. Described Below Is The CLIQUE Problem In Detail: Clique INSTANCE: An Undirected Graph G(V E) And A Positive Integer K. QUESTION: Does Graph G Have A Clique Of Size K, I.e.
- 2 A Special Class of Problems: NP-Complete NP-Complete is a special class of intractable problems. This class contains a very diverse set of problems with the following intriguing properties: 1. We only know how to solve these problems in exponential time e.g. O(2O(nk)). 2. If we can solve any NP-Complete problem in polynomial time, then we will b
- cannot calculate the solution in polynomial time the problem then becomes the NP-Hard problem given that it includes the version of NP-Complete. The clique problem resides with two well-studied problems: Maximum Clique Problem: The maximum clique is the maximal clique with the maximum cardinality or weight
- 4) Clique ϵ NP:-Proof: - As you know very well, you can get the Clique through 3CNF and to convert the decision-based NP problem into 3CNF you have to first convert into SAT and SAT comes from NP. So, concluded that CLIQUE belongs to NP. Proof of NPC:-Reduction achieved within the polynomial time from 3CNF to Clique

- clique problem是问一个图中是否有大小是k以上的clique。 任意挑出k个点，我们可以简单的判断出这k个点是不是一个clique，所以这个问题属于NP。 证明备巩胶clique problem是NP完备可以很简单的从独立顶点集问题（促射凳Independent set problem）reduce
- Un problème NP-difficile est un problème qui remplit la seconde condition, et donc peut être dans une classe de problème plus large et donc plus difficile que la classe NP. Bien qu'on puisse vérifier rapidement toute solution proposée d'un problème NP-complet, on ne sait pas en trouver efficacement
- NP-Complete problems are in both NP and NP-Hard, and therefore their solutions can be checked in polynomial time, but they cannot be solved in polynomial time. The clique cover problem is an NP-Complete problem and can include findin
- The problem: Clique (I've also seen Max Clique or Clique Decision Problem (CDP)) In it, they reduce 3SAT to Clique, proving Clique is NP-Complete, and then reduce Clique to VC. This works in the exact same way as the reduction from VC to Clique that I'll be doing here next. Leave a comment. Posted in Core Problems
- Problem Set 6 Solutions Problem 1. [Clique] Solution To prove that Half-Clique is NP-complete we have to prove that 1) Half-Clique 2NP 2) Half-Clique is NP-hard 1) To prove that Half-Clique 2NP we consider an instance of the problem (G;jVj=2) and a subset HC V. To prove that that HCis an actual solution to the problem we have t
- Graph clique problem: given a graph and an integer k, is there a subgraph in G that is a complete graph of size k? Theorem: The clique problem is NP-complete. Proof: Reduction from 3-SAT: Literals become nodes; k clauses induce node groups; Connect all inter-group compatible nodes / literals. Example: (x+y+z)(x'+y'+z)(x'+y+z') Z Y X Z' Y X' X' Y'
- Some NP-Complete Problems 10.1 Statements of the Problems In this chapter we will show that certain classical algo-rithmic problems are NP-complete. The problem Clique is obviously in NP. To show that it is NP-complete, we reduce Inde-pendent Setto it. 10.1. STATEMENTS OF THE PROBLEMS 64

Problem Types A clique in an undirect graph G=(V,E) is a subset U of V such that every pair of vertices in U is joined by an edge. E.g., mutual friends on facebook, genes that vary together An optimization problem: How large is the largest clique in G A search problem: Find the/a largest clique in The problem you are trying to solve, I as understand, is the decision version of the Clique problem, which is effectively a NP-Complete problem. You are correct to believe that a polynomial solution for this problem would constitute a proof that [.. Reading time: 35 minutes. In this article, we will go through a simple yet elegant algorithm to find a clique of a given size. Clique is an interesting topic in itself given that the clique decision problem is NP-Complete and clique arises in almost all real-life applications involving graphs

Prove that Exact4SAT is NP-complete. Solution: This problem is in NP as this problem is a special case of SAT, which is in NP (A polytime verification algorithm for SAT would also work as verification here.) We reduce from 3-SAT * Das Cliquenproblem (mit CLIQUE notiert) ist ein Entscheidungsproblem der Graphentheorie*.Das Cliquenproblem ist eines der 21 klassischen NP-vollständigen Probleme, deren Zugehörigkeit zu dieser Klasse Richard M. Karp 1972 bewies

AproblemB is NP-hard if every problem in NP has a polytime reduction to B.If,in 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 NP-Complete Problems. These are what are known as NP-complete problems. NP-complete is a concept in complexity theory used to describe a category of problems for which there is no known correct and fast solution. In other words, the solution to an NP-complete problem can be quickly verified, but there is no known way to quickly find a solution 09:46am April 16, 2003 † We can reduce a clique problem to an independent set problem. Suppose we have a clique problem for general graph. We already know it is NP-Complete. We can create an independent set problem for G0, which is the complement of G in the clique problem

S) is complete. Finding the largest clique in a graph is an NP-hard problem, called the maximum clique problem (MCP). Cliques are intimately related to vertex covers and independent sets. Given a graph G, and defining E* to be the complement of E, S is a maximu Since V 0 is a **clique** of size b in G iff V 0 is an independent set of G 0 of size b and the construction of G 0 and b from G and b can be done in polynomial time, we can conclude that the independent set **problem** is **NP**-hard, too. To sum up, the independent set **problem** is **NP-complete** NP-complete problems are interesting to researchers thinking about this, because any NP problem can be rephrased as an NP-complete problem. This means that if any problem in NP is not in P, then an NP-complete problem must be not in P. So a lot of researchers are focused on taking some NP-complete problem and proving that it can't be solved. P1. Consider the clique problem restricted to graphs in which every vertex has degree at most 3. Call this problem CLIQUE-3. (a) Prove that CLIQUE-3 is in NP. (b) What is wrong with the following proof of NP-completeness for CLIQUE-3? We know that the clique problem in general graphs is NP-complete, so it is enough to presen

problem in NP. Thus it is sufficient to present a polynomial-time algorithm for any NP-complete problem to prove that P=NP, as mentioned in [4] and [5]. The NP-complete problem considered in this paper is the Clique problem which is one of Richard Karp's 21 problems as shown NP-complete in hi Thus, \(X\) is as hard as any problem in NP. A problem \(X\) is defined to be NP-complete if \(X\) is in NP, and \(X\) is NP-hard. The requirement that a problem be NP-hard might seem to be impossible, but in fact there are hundreds of such problems, including TRAVELING SALESMAN. Another such problem is called K-CLIQUE NP-Hard and NP-Complete problems. Today, we discuss NP-Completeness. Recall from 6.006: • P = the set of problems that are solvable in polynomial time. If the problem has size. n, the problem should be solved in. n. O (1). • NP = the set of decision problems solvable in nondeterministic polynomial time. The output of these problems is a YES. CLIQUE = f(G;k) : graph G has a clique of size kg We will show that CLIQUE is an NP complete problem. To do so, we start with the easier direction: Lemma 0.5. CLIQUE 2NP: Proof. The veri er can take an indicator vector for the set of vertices in the clique, y. Checking whether the size of y is at least k can be done in O(n) time, and checkin

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 graphs). In this lecture we will show that the colouring problem on arbitrary graphs becomes NP-complete even for k= 3 ! Crazy! No? Think about it: One easy to rule out a 3-colouring is to check if Ghas a clique of size 4, like in the example below1: How hard is it to check for a clique of size at least k= 4? We just showed that CLIQUE is NP. NP-complete problems Your favorite topic certainly has an NP-complete problem in it Even the other sciences are not safe: biology, chemistry, physics have NP-complete problems too! Theorem (Cook-Levin): SAT and 3-SAT are NP-complete Corollary:SAT ∈∈∈∈P if and only if P = NP Given a favorite problem ΠΠΠΠ∈∈∈∈NP

In 1972, Karp introduced a list of twenty-one NP-complete problems, one of which was the problem of finding a maximum clique in a graph.Given a graph, one must find a largest set of vertices such that any two vertices in the set are connected by an edge. Such a set of vertices is called a maximum clique of the graph and in general can be very difficult to find NP problem: - Suppose a DECISION-BASED problem is provided in which a set of inputs/high inputs you can get high output. Criteria to come either in NP-hard or NP-complete. The point to be noted here, the output is already given, and you can verify the output/solution within the polynomial time but can't produce an output/solution in polynomial time First of all, there is needed to outline some aspects involving problems belonging to NP (NP-Hard) class. As a parenthesis, NP stands for nondeterministic polynomial time algorithm which refers to the fact of discovering a nondeterministic Turing machine able to solve the problem in cause within a polynomial number of discovering nondeterministic movements Cliquen Problem Wir wollen wissen ob es in einem Graphen G eine Clique größergleich k gibt Eine Clique ist eine Menge an Knoten die paarweise adjazent sind Wir wollen zeigen das das Cliquen Problem NP-nollständig ist und führen es einfach auf das Independent set Problem zurüc

Only a decision problem can be NP complete. However, an optimization problem may be the NP hard. Furthermore if L1 is a decision problem and L2 an optimization problem, then it is possible that L1 α L2. One can trivially show that the knapsack decision problem reduces to knapsack optimization problem. For the clique problem one can easily show. Prove the Clique problem is polynomially transformable to the vertex cover problem and thus conclude the vertex cover problem is also NP-complete. Definition: a clique in an undirected graph in a is a subset of its vertices such that every two vertices in the subset are connected by an edge The natural algorithmic problem is, given a graph, nd the largest independent set. To turn this optimisation problem into a decision problem, we de ne IND as: The set of pairs (G;K), where G is a graph, and K is an integer, such that G contains an independent set with K or more vertices. IND is clearly in NP. We now show it is NP-complete

- These problems are all hard: the clique decision problem is NP-complete , the problem of finding the maximum clique is both fixed-parameter intractable and hard to approximate, and listing all maximal cliques may require exponential time as there exist graphs with exponentially many maximal cliques
- As noted in the earlier answers, NP-hard means that any problem in NP can be reduced to it. This means that any complete problem for a class (e.g. PSPACE) which contains NP is also NP-hard. In order to get a problem which is NP-hard but not NP-complete, it suffices to find a computational class which (a) has complete problems, (b) provably contains NP, and (c) is provably different from NP
- 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.
- Thus the vertex cover problem is in the class NP. Proof that vertex cover is NP Hard - To prove that Vertex Cover is NP Hard, we take some problem which has already been proven to be NP Hard, and show that this problem can be reduced to the Vertex Cover problem. For this, we consider the Clique problem, which is NP Complete (and hence NP Hard)

- NP complete problems •problem A is NP-complete if-A is in NP (poly-time to verify proposed solution) -any problem in NP reduces to A •second condition says: if one solves pb A, it solves via polynomial reductions all other problems in NP •CIRCUIT SAT is NP-complete (see book)-and so the other problems discussed here, because they reduce to i
- Longest Path Problem. Given an undirected graph G and two vertices u and v, find a longest simple path from u to v. All of those problems are NP-hard. Each one is closely related to a known NP-complete problem. Let's take the Maximum Clique problem (MAX-CLIQUE) as an example. CP is the Clique problem
- ating set problem, i.e., the do

- NP-complete . 계산 복잡도 이론 (Computational Complexity Theory) 에서, NP-complete 문제는 NP 중에서 가장 어려운 문제들이다 (그것들이 대부분 P 에는 속하지 않는다는 점에서).그 이유는 만일 어떤 NP-complete 문제를 빨리 푸는 방법을 발견할 수 있다면, 모든 NP 문제들을 빨리 푸는데 그 알고리즘을 사용할 수 있기.
- A lo mejor el NP-completo no dura. Ah, una cosa más, se cree que si alguien consigue alguna vez un problema NP-completo en tiempo P, entonces todos los problemas NP-completos se podrían resolver usando el mismo método, así que todo el conjunto de problemas NP-completos dejaría de existir. El problema del viajant
- NP-complete problems AmirHossein Ghamarian December 2, 2008 1. The CLIQUE problem is the decision problem deﬁned in your book: Given an undirected graph G = (V;E) and a natural number k, decide whether there is a clique of size k in G. The MAXIMUM
- Em ciência da computação, o problema do clique refere-se a qualquer problema que possui como objetivo encontrar subgrafos completos (cliques) em um grafo.Como exemplo, o problema de encontrar conjuntos de nós em que todos os elementos estão conectados entre si. Por exemplo, o problema clique surge no cenário seguinte
- Informally, a search problem B is NP-Hard if there exists some NP-Complete problem A that Turing reduces to B. The problem in NP-Hard cannot be solved in polynomial time, until P = NP. If a problem is proved to be NPC, there is no need to waste time on trying to find an efficient algorithm for it

- g Problem Graph isomorphism NP and P NP-Complete NP-Hard Certificates Oracles Deter
- NP-complete problems are the problems that are both NP-hard, and in NP. 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
- problem： Prove that the following problem is NP-complete:given an undirected graph G=(V,E) and an integer k,return a clique of size k as well as an independent set of size k,provided both exist
- 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-complete

Thus, IS ∈ NP. - NP-Hard: In order to show that IS ∈ NP-Hard, we will reduce from Clique to IS. Our reduction consists of demonstrating a polynomial time conversion of an instance of Clique to an instance of IS, and an if-and-only-if proof that a yes instance of Clique maps to a yes instance of IS and vice versa. 1 C - Clique Problem题目重述：The clique problem is one of the most well-known NP-complete problems. Under some simplification it can be formulated as follows. Consider an undirected graph G. It is required to find a subset of vertices C of the maximum size su Theorem: 3-SAT is NP-complete. Next we will prove the NP-completeness of the clique decision problem with a reduction from the 3-SAT problem. Speci cally, we will prove: Theorem: If we have an algorithm Asolving the clique decision problem in polynomial time, we can solve the 3-SAT problem using Ain polynomial time G&J don't even bother to prove Clique is NP-Complete, just stating that it (and Independent Set) is a different version of Vertex Cover. But the problem comes up often enough that it's worth seeing in its own right. The problem: Clique (I've also seen Max Clique or Clique Decision Problem (CDP) NP Complete Problem: the clique and vertex cover !! xdizen asked on 2003-12-02. Math / Science; 10 Comments. 1 Solution. 1,349 Views. Last Modified: 2007-12-19. HI all I am lost in this stuff and i need some help ! THE CLIQUE PROBLEM Given an.

I need to find a solution to the following question: The problem of the Two Clique is in P or NP-complete (assuming P != NP)? The Two Clique problem is the following: Given a graph G = (V, E), can V be partitioned into two non-empty parts V1, V2 such that V = V1 ∪ V2 and both V1 and V2 are cliques in G TheoremSAT is NP-complete Proof idea:The turing machine program for any problem in NP can be veri ed by a polynomial sized SAT instance that encodes that the input is well formed and that each step follows legally from the next. ImplicationWe now have one NP-complete problem. We will now reduce other problems to it NP-Complete Problems, Part I Jim Royer April 1, 2019 All known algorithms for the Hamiltonian Path Search Problem are exponential-time. NP-Completeness April 1, 201913/27. Cuts and bisections max-sized clique for the graph. NP-Completeness The images are from Wikipedia. April 1, 201917/27

Jogger is NP-complete. 1. Jogger is in NP: Given a path P, we can check in O(|P|) whether or not the sum of all edge weights is equal to i. 2. Consider the Subset Sum (SS) problem. 1, which is a known NP-complete problem. Given a set S of positive integers, is there a subset S. 0 ⊆ S such that sum of the elements of S. 0. is t. Example: S. Problem #2 We will denote Vertex Cover by VC. Given a certiﬁcate to VC, a potential subset of vertices S, we can check in polynomial time that each edge in G is adjacent to some vertex in S. Thus, VC is in NP. To show that VC is NP-complete, we reduce the Clique problem to it, which we know is NP-complete. The Clique problem is to determine if 1 De nition of **NP-Complete** Languages 2 naesat is **NP-complete** 3 0-1 integer programming is **NP-complete** 4 independent set is **NP-complete** 5 **clique** is **NP-complete** 6 Reductions and Computational Feasibility John E. Savage (Brown University) CSCI 1590 Intro to Computational Complexity February 2, 2009 2 / 1

28.15.1. Reduction of 3-SAT to Clique¶. The following slideshow shows that an instance of 3-CNF Satisfiability problem can be reduced to an instance of Clique problem in polynomial time Find an already known NP-complete problem R 0, and come up with a transform that reduces R 0 to R. For this strategy to become effective, we need at least one NP-complete problem. This is provided by Cook's Theorem below. Cook's Theorem: SAT is NP-complete. Back to Top VI. NP-Completeness of the k-Clique Problem. The k-clique problem was. In the problem that we study, a partition is performed on the vertices and not on the edges. This problem has also been referred to as the clique cover problem (11). We know that both clique-partitioning and coloring are NP-Complete problems (3) 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., ( Proving NP-Completeness by Reduction To prove a problem is NP-complete, use the ear-lier observation: If Sis NP-complete, T2NP and S P T, then Tis NP-complete

As a consequence, MBP is also NP-complete for general graphs. 2. The reduction. We define MBP as follows: Maximum edge biclique problem (MBP): Given a bipartite graph G=(V 1 ∪V 2,E) and a positive integer K, does G contain a biclique with at least K edges? Theorem 1. MBP is NP-complete. Proof. We shall reduce CLIQUE to MBP Step 2: Pick a known NP-complete problem. State what problem Y you are re-ducing to X. You need to show that Y ≤ p X. You may use any problem Y which we have proved in class to be NP-complete, as well as any problem you have proved to be NP-complete on the homework assignments. Some problems will be far easier to use than others in your proof A problem Y ∈NP with the property that for every problem X in NP, X polynomial transforms to Y. Cook's theorem. CNF-SAT is NP-complete. Recipe to establish NP-completeness of problem Y. Step 1. Show that Y ∈NP. Step 2. Show that CNF-SAT (or any other NP-complete problem) transforms to Y. Example: CLIQUE is NP-complete. Step 1. CLIQUE ∈NP NP HARD: NP-hard (Non-deterministic Polynomial-time hard), in computational complexity theory, is a class of problems that are, informally, at least as hard as the hardest problems in NP. A problem H is NP-hard if and only if there is an NP-complete problem L that is polynomial time Turing-reducible to H (i.e., L ≤ TH). 5 If P= NPthen Shortest-Path is NP-complete. [TRUE] Any problem in Pis polynomial-time reducible to any other problem in P. If P= NPthen the same would hold for NP. Since Shortest-Path 2NP, it follows that for all X 2NP, X P Shortest-Path. Therefore, Shortest-Path is NP-complete. It is possible that Independent-Set 2Pand Ham-cycle 62P [FALSE. Maximum weight clique problem requires a maximum weight clique. Furthermore it differs from the NP-complete problem Subset Product in that it does not contain the product value B in its input