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# Np complete set

Nå kan du bestille Complete fra Vitusapotek på nett. Se vårt utvalg Informally, NP is set of decision problems which can be solved by a polynomial time via a Lucky Algorithm, a magical algorithm that always makes a right guess among the given set of choices (Source Ref 1). NP-complete problems are the hardest problems in NP set. A decision problem L is NP-complete if Overview. NP-complete problems are in NP, the set of all decision problems whose solutions can be verified in polynomial time; NP may be equivalently defined as the set of decision problems that can be solved in polynomial time on a non-deterministic Turing machine.A problem p in NP is NP-complete if every other problem in NP can be transformed (or reduced) into p in polynomial time Prerequisite: NP Complete Problem: Given a ground set X of elements and also a grouping collection C of subsets available in X and an integer k, the task is to find the smallest subset of X, such that the smallest subset, H hits every set comprised in C.This implies that the intersection of H and S is null for every set S belonging to C, with size ≤ k 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 would cease to exist. Traveling Salesman Proble

Remains NP-complete for 3-sets. Solvable in polynomial time for 2-sets (this is a matching). Feedback vertex set; Feedback arc set; Graph homomorphism problem; Graph coloring; Graph partition into subgraphs of specific types (triangles, isomorphic subgraphs, Hamiltonian subgraphs, forests, perfect matchings) are known NP-complete Independent Set np complete proof + clique, Independent Set problem is np complete, Independent Set problem, Independent Set Problem with example, Independent Set, NP-Completeness

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NP-Complete is the intersection set of NP and NP-Hard. Any NP problem could be transformed into a NP-Complete problem in polynomial time. That means if any of the NP-Complete could have an efficient solution then any NP problem could be solved with same efficiency. Please let me know if I made any mistake NP-complete problems have no known p-time solution, considered intractable. Tractability Difference between tractability and intractability Alphabet = finite set of symbols Language L over is any subset of strings in * Well focus on = {0, 1} L = {10, 11, 101, 111, 1011, } is language of primes Enkel og trygg netthandel hos Komplett med rask levering og bredt sortiment innen elektronikk, datautstyr, PC, TV, mobil, hvitevarer og hjem- og fritidsprodukter

### Complete - i nettapotek - Gratis frakt og rask leverans

• e whether can be partitioned into and , such that: (1
• Some First NP-complete problem We need to nd some rst NP-complete problem. Finding the rst NP-complete problem was the result of the Cook-Levin theorem. We'll deal with this later. For now, trust me that: Independent Set is a packing problem and is NP-complete. Vertex Cover is a covering problem and is NP-complete
• 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.

Problem Set: NP-Complete Reductions 1. [easy] Consider the two problems: DOUBLE-SAT: Given as input a boolean formula , does have at least two satisfying assignments? SAT: Given as input a boolean formula , does have a satisfying assignment? Show that DOUBLE-SAT is NP-complete using a reduction from SAT If there is a problem that is NP (and not P) but not NP Complete, would this be a result of no existing isomorphism between instances of that problem and the NP Complete set? If the two complexity classes are different then by Ladner's theorem there are problems which are $\mathsf{NP}$-intermediate, i.e. they are between $\mathsf{P}$ and $\mathsf{NP\text{-}complete}$ An NP-complete Set The definitions and discussion about P and NP were very interesting. But, of course for any of this discussion to be worthwhile we need to see an NP-complete set. Or at least prove that there is one. The following definitions from the propositional calculus lead to our first NP-complete problem Set Cover ; Vertex Cover; Travelling Salesman Problem; In this context, now we will discuss TSP is NP-Complete. TSP is NP-Complete. The traveling salesman problem consists of a salesman and a set of cities. The salesman has to visit each one of the cities starting from a certain one and returning to the same city 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.

### NP-Completeness Set 1 (Introduction) - GeeksforGeek

1. 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
2. istic polynomial time. The output of these problems is a YES.
3. Theorem: Set Cover is NP-Complete. Proof: First, we argue that Set Cover is in NP, since given a collection of sets C, a certiﬁer can eﬃciently check that C indeed contains at most k elements, and that the union of all sets listed in C does include all elements from the ground set U. We will now show that Vertex Cover ≤P Set Cover
4. NP-complete problems are defined in a precise sense as the hardest problems in P. Even though we don't know whether there is any problem in NP that is not in P, we can point to an NP-complete problem and say that if there are any hard problems in NP, that problems is one of the hard ones
5. Request PDF | Splitting NP-Complete Sets | We show that a set is m-autoreducible if and only if it is m-mitotic. This solves a long standing open question in a surprising way. As a... | Find, read.
6. We show that a set is m-autoreducible if and only if it is m-mitotic. This solves a long-standing open question in a surprising way. As a consequence of this unconditional result and recent work by Glaßer et al., complete sets for all of the following complexity classes are m-mitotic: $\mathrm{NP}$, $\mathrm{coNP}$, $\oplus\mathrm{P}$, $\mathrm{PSPACE}$, and $\mathrm{NEXP}$, as well as all.
7. Claim 1: Set-Cover is in NP. Proof: The following veri er for Set-Cover runs in time polynomial in the length of its rst input: Veri er V(hU;A;ki;hBi) If all the following are all true then accept else reject: { Bis a subset of A { jBj k { 8u 2U 9B 2B[u 2B] The NP-hardness of Set-Cover is implied by the following. Claim 2: Vertex-Cover p Set-Cover

Set-Splitting, i.e., that Set-Splitting is NP-complete by reducing NAE-3-SAT to it. Consider the Set-Splitting problem: given a family F of subsets of a finite set S, decide whether there exists a partition of S into two subsets S 1, S 2 such that all elements of F are split by this partition, i.e., none of the elements of F is completely in S. Proof that SUBSET SUM is NP-complete Recall that input to Subset sum problem is set A= fa1;a2;:::;amgof integers and target t. The question is whether there is A0 Asuch that elements in A0sum to t. We prove this problem is NP-complete. This is again a reduction from 3SAT. The previous ex-ample suggests the approach: deﬁne number A set S ⊂ Σ ∗ is sparse if there is a polynomial p such that the number of strings in S of length at most n is at most p(n).All known NP-complete sets, such as SAT, are polynomial-time isomorphic and are not sparse.The main result of this paper is that if there is a sparse NP-complete set under polynomial-time many-one reductions, then P = NP

NP-complete is a special category of NP problems that have time complexities greater than polynomial time, are verifiable in polynomial time, and belong to a set of problems known as NP-hard. The Undirected Feedback Set Problem asks: Given G. 2) Every problem in NP is reducible to L in polynomial time (Reduction is defined below). 8mm pixel pitches Pick a suitable known NP-complete problem, S (ex: SAT) 3. Show a polynomial algorithm to transform an instance of S into an instance of X SOX RESTOX mnemonic can help. SAT Outside the Box Reduce SAT To X Algorithms NP-Completeness 23 NP-COMPLETENESS PROOF METHOD 24. To show that X is NP-Complete: 1. Show that X is in NP, i.e., a polynomial time. Since the answer here showed that set cover could not be used, I kept searching. I found that the k-coloring problem worked very nicely. Namely the 3-coloring problem is still NP-complete even when constricting k to 3. $\endgroup$ - TheJKFever May 7 '15 at 9:0 NP-complete, so is IS. The following two corollaries are immediate from the above theroem. Corollary 1 : The VERTEX COVER problem is NP-complete. A vertex cover of a simple undirected graph G= (V;E) is a set of vertices such that each edge has at least one of its ends at a vertex of the set. That is, a vertex cover of Gis a subse i ∈ S and set A(x i) = F otherwise. Observe that this is a truth assignment, since all variables are assigned either T or F, but not both. Furthermore, A satisﬁes F, since by our construction, A satisﬁes each clause. Therefore we have shown that 3-SAT ≤ p IS. Thus IS is NP-hard, and since we have shown IS to be in NP, IS is NP-complete.

exhibit an NP-complete set Aand a disjoint set Bin NP such that A[Bis not NP-complete. Since L2NP, there is a constant k 1 such that Lcan be decided in time 2nk. Let t 1 = 2, and t i+1 = tk 2 i. Before we present a formal proof, we provide the main ideas behind the proof. Let us partition into blocks B 1; 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. Proof that INDEPENDENT SET is NP-complete We begin by having an undirected simple graph, G = (V, E), letting a subset I ⊆ V in G be an independent set if no two vertices in set I are connected by an edge in G.1 For every vertex in I, we must check that there is no edge of G that connects this vertex to at least another vertex in I.1 If we run into such an edge, we reject such a set I.

In this post, we will prove that the decision version of the set-covering problem is NP-complete, using a reduction from the vertex covering problem (which is NP-complete). This is Exercise 35.3-2 in CLRS3 . We will follow the template given in an earlier post. Problem statemen It is easy to show that Hitting Set is in NP. A solution just needs to exhibit the set H - one can easily verify in polynomial time whether H is of size k and intersects each of the sets B1, . . . , Bm. We reduce from vertex cover. Consider an instance of the vertex cover problem - graph G = (V, E) and a positive integer k 1 NP-Complete Problems in Graph Theory Bisection Hamilton Path and Circuit Longest Path and Circuit TSP (D) 3-Coloring 2 Sets and Numbers Tripartite Matching Set Covering,Set Packing, and Exact Cover by 3-Sets Integer Programming Knapsack Pseudopolynomial Algorithms and Strong NP-Completeness Williamson NP-Completeness Proof Set Theory & Algebra; Combinatory; Probability; Algorithms; Data Structures; Programming; Digital Logic; CO & Architecture; Operating Systems; Home / Theory of Computation / P, NP, NP-Complete, NP-Hard P, NP, NP-Complete, NP-Hard. July 9, 2016 by arjun Leave a Comment. It might be because of the name but many graduate students find it. National Park Quarter Collector Sets These National Park Quarter Sets make great gifts, and are a excellent way to get your complete collection. Each coin is in brilliant uncirculated condition and are hand selected for quality and eye appeal

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 be able to solve all NP-Complete problems. An Annotated List of Selected NP-complete Problems. The standard textbook on NP-completeness is: . Michael Garey and David Johnson: Computers and Intractability - A Guide to the Theory of NP-completeness; Freeman, 1979.. David Johnson also runs a column in the journal Journal of Algorithms (in the HCL; there is an on-line bibliography of all issues) . On the Web the following sites may be of. 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..

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 证明NP-hard, 需要找到一个已知的NP-complete problem，并且把该问题reduce to Set Cover in polynomial time. （从intuition上讲，这样Set Cover就至少比我们找的NP-complete要难。）因为这个问题和Vertex Cover很像，我们可以证明 Vertex Cover Set Cover. 具体的证明就不赘述了，如果题主需要. ### NP-completeness - Wikipedi

1. 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 I is an independent set of the graph G even though V it's complimentary set of vertices is a vertex cover in this graph
2. ating set, a.k.a. do
3. CiteSeerX - Document Details (Isaac Councill, Lee Giles, Pradeep Teregowda): We study several properties of sets that are complete for NP. We prove that if L is an NP-complete set and S ⊇ L is a p-selective sparse set, then L − S is ≤p m-hard for NP. We demonstrate existence of a sparse set S ∈ DTIME(22n) such that for every L ∈ NP − P, L − S is not ≤p m-hard for NP
4. For example, the construction of a 2-T-complete set for NP that is not 2-tt-complete also separates 2-T-autoreducibility from 2-tt-autoreducibility. Now on home page ad
5. My attempt: To show something is NP Complete, must show it is in NP and a reduction of an NP Hard Problem. Clearly, it is in NP because given a certificate of a scheduling, you can just check there are no conflicts. I want to show this is a reduction of either SAT or Graph Coloring. I'm not sure exactly how to go about that
6. CiteSeerX - Document Details (Isaac Councill, Lee Giles, Pradeep Teregowda): We study several properties of sets that are complete for NP. We prove that if L is an NP-complete set and S ⊇ L is a p-selective sparse set, then L − S is ≤ p m-hard for NP. We demonstrate existence of a sparse set S ∈ DTIME(22n) such that for every L ∈ NP − P, L − S is not ≤ p m-hard for NP
7. istic Turing machine can guess a truth assignment T for E in O(n) time

### Hitting Set problem is NP Complete - GeeksforGeek

NP Certification algorithm intuition. ・Certifier views things from managerial viewpoint. ・Certifier doesn't determine whether s ∈ X on its own; rather, it checks a proposed proof t that s ∈ X. Def. Algorithm C(s, t) is a certifier for problem X if for every string s, s ∈ X iff there exists a string t such that C(s, t) = yes. Def. NP is the set of problems for which there exists a. In this paper, we present several results about collapsing and non-collapsing degrees ofNP-complete sets. The first, a relativized collapsing result, is interesting because it is the strongest known constructive approximation to a relativization o set of size k (the same set of vertices), but the converse is not necessarily true. However, the similarity suggests that if VC in NP-complete, then DS is likely to be NP-complete as well. We will show this fact next. As usual the proof has two parts. First we show that DS 2NP. The certi cate just consists of the subset V0 in the dominating set Problem Set 6 - NP-Complete Reductions 1. Reductions (1a) Hamiltonian Path . = Hamiltonian Circuit . Modify your graph by adding another node that has edges to all the nodes in the original graph

Set: NP-32 B Complete - Set. Set contains. 1x Yamaha NP-32 B; 1x MUSIC STORE Music Store BN1 X bench Adjustable height: 44-58 cm; 1x MUSIC STORE KB-6, Keyboard Stand X Stand, max. 12 kg. 1x MUSIC STORE MS 300 Dynamic Headphons 32 Ohm, 102 dB, 20Hz-20KH easily that NP-complete sets are not p-immune. Assuming only that secure one-way permu-tations exist, we prove that no NP-complete set is DTIME(2 n†)-immune. Also, using these hypotheses we show that no NP-complete set is quasipolynomial-close to P. We introduce a strong but reasonable hypothesis and infer from it that disjoint Turing

### NP-Complete - A Rough Guid

So all the problems we've seen so far have polynomial time algorithms, except a couple in your problem sets, which were actually NP-complete. And the best you could have done was exponential, unless P equals NP. So here's how you can prove this kind of lower bound to say look, I don't need to look for algorithms any more because my problem is. In this paper, we present several results about collapsing and non-collapsing degrees ofNP-complete sets. The first, a relativized collapsing result, is interesting because it is the strongest known constructive approximation to a relativization of Berman and Hartmanis' 1977 conjecture that all ≤ -complete sets forNP arep-isomorphic

3DM Is NP-Complete Theorem Three-dimensional matching (aka 3DM) is NP-complete Proof. 3DM is in NP: a collection of n sets that cover every element exactly once is a certi cate that can be checked in polynomial time. Reduction from 3-SAT. We show that: 3-SAT P 3DM In other words, if we could solve 3DM, we could solve 3-SAT from Airplane Mode and set it to Boat Mode. For security reasons, please leave caps lock on while browsing. This work is licensed under a Creative Commons Attribution-NonCommercial 2.5 License Autoreducibility of NP-Complete Sets John M. Hitchcock and Hadi Shafei Department of Computer Science University of Wyoming Abstract We study the polynomial-time autoreducibility of NP-complete sets and obtain separations under strong hypotheses for NP. Assuming there is a p-generic set in NP, we show the following Start studying P, NP, NP-Complete, NP-Hard. Learn vocabulary, terms, and more with flashcards, games, and other study tools. Search. Create. Log in Sign up. Log in Sign up. 18 terms. jma10blue. P, NP, NP-Complete, NP-Hard. STUDY. PLAY. Decision problem. A question with a yes or no answer. set P. P is the set of all decision problems which can.

We study several properties of sets that are complete for NP. We prove that if L is an NP-complete set and S ⊇ L is a p-selective sparse set, then L − S is ≤ p m-hard for NP. We demonstrate existence of a sparse set S ∈ DTIME(22n) such that for every L ∈ NP − P, L − S is not ≤ p m-hard for NP 28.12.1. NP-Completeness Proofs¶. To start the process of being able to prove problems are NP-complete, we need to prove just one problem $$H$$ is NP-complete. After that, to show that any problem $$X$$ is NP-hard, we just need to reduce $$H$$ to $$X$$.When doing NP-completeness proofs, it is very important not to get this reduction backwards National Park Quarters Complete Date Set America the Beautiful Coins in Deluxe Color Book. 4.2 out of 5 stars 125. $49.95$ 49. 95 $64.99$64.99. Get it as soon as Fri, Nov 6. FREE Shipping by Amazon. Ages: 12 years and up. 2020 P, D National Park Quarter 10 Coin Set Uncirculated. 4.7 out of 5 stars 18 In this module you will study the classical NP-complete problems and the reductions between them. Our last hard set problem that we mentioned here, deals with graphs again. It is called an independent set problem. Here, we're given a graph and the budget b

(2012) Collapsing and Separating Completeness Notions Under Average-Case and Worst-Case Hypotheses. Theory of Computing Systems 51:2, 248-265 National Park Quarters Complete Date Set America the Beautiful Coins in Deluxe Color Book. 4.2 out of 5 stars 125. $49.95$ 49. 95 $64.99$64.99. Get it as soon as Sat, Nov 7. FREE Shipping by Amazon. Ages: 12 years and up. 2020 P, D National Park Quarter 10 Coin Set Uncirculated. 4.9 out of 5 stars 19 Claim: VERTEX COVER is NP-complete Proof: It was proved in 1971, by Cook, that 3SAT is NP-complete. Next, we know that VERTEX COVER is in NP because we could verify any solution in polytime with a simple n 2 examination of all the edges for endpoint inclusion in the given vertex cover

Abstract: All known NP-complete sets are P-isomorphic (i.e. there are polynomial time, one-to-one and onto, polynomial time invertible reductions between any two known NP-complete sets) [BH]. If all NP-complete sets are P-isomorphic, then. P ≠ NP. However it is not known if the existence of more than one P-isomorphism class of NP-complete sets has implications for the P = NP? problem De nition of NP-complete Languages De nition A language L 2 is NP-complete if a) L 2 is in NP and b) for every language L 1 in NP there is polynomial-time reduction from it to L 2 (L 2 is NP-hard.) f: 1 7! 2 is a reduction from L 1 to L 2 if x2L 1 if and only if f(x) 2L 2. fis a polynomial-time reduction if it can also be computed by a DTM in polynomial time in the length of its input Feedback Arc Set in Bipartite Tournaments is NP-Complete Jiong Guo 1Falk Huﬀne¨ r Hannes Moser2 Institut fur¨ Informatik, Friedrich-Schiller-Universitat Jena, Ernst-Abbe-Platz 2, D-07743 Jena, Germany Abstract The Feedback Arc Set problem asks whether it is possible to delete at most k arcs to make a directed graph acyclic National Park Stamps. Customize your Passport with these national park stamps! The Passport To Your National Parks® annual stamp sets, dating back to 1986, feature vibrant images of some of America's most beautiful landscapes

### List of NP-complete problems - Wikipedi

1 Proving NP-completeness In general, proving NP-completeness of a language L by reduction consists of the following steps. 1. Show that the language A is in NP 2. Choose an NP-complete B language from which the reduction will go, that is, B ≤ p A. 3. Describe the reduction function f 4. Argue that if an instance x was in B, then f(x) ∈ A. 5 Abstract. Under the assumption that NP does not have p-measure 0, we investigate reductions to NP-complete sets and prove the following: (1) Adaptive reductions are more powerful than nonadaptive reductions: there is a problem that is Turing-complete for NP but not truth-table-complete Independent set Nun zeigen wir noch das Independent set wirklich in NP-schwer ist indem wir das SAT Problem darauf reduzieren Wir kriegen eine boolsche Formel und bringen sie in KNF wir konstruieren einen Graphen mit den Literalen als Knoten und verbinden alle Knote

### NP-Completeness of Independent Set with Proof By Studies

News updates everyday about in set. Do you want to know the latest news about in set NP-complete problem L 2NP is NP-complete if any language in NP is polynomial-time reducible to L Hardest problem in NP Crescenzi and Kann (UniFi and KTH) Subset Sum October 2011 2 / 8. Basic results Cook-Levin theorem If a set is \good, there exists subset B A such that the sum o The problem you are using is not NP-Complete. Given a set of n integers, there are only $\binom{n}{3}$ possible triplets. So, a brute force algorithm can solve this in polytime. If you apply your idea and reduce from the Partition problem, it should work. $\endgroup$ - user137481 Sep 1 '15 at 21:0 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   NP-complete problems What are the hardest problems in NP? Deﬁnition Problem X is an NP-complete problem if 1 X 2NP, and 2 for all Y 2NP: Y P X. Every problem in NPcan be reduced to an NP-complete problem. Theorem Suppose X is an NP-complete problem. Then X is solvable in polynomial time if and only if P= NP Show that the independent set problem is NP-complete through the following two steps: 1. Show that the problem is in NP. 2. Show that 3SAT is poly-time reducible to the problem Complete Set of 2013 National Park Quarters from the Denver (D) Mint. This National Park Quarter Set contains one of each National Park Quarter released in 2013 from the Denver (D) Mint in Brilliant Uncirculated (BU) Condition. This National Park Continued. not rated \$ 4.9 It's complete and right what Arjun Nayini says, I'll just try to elaborate a bit on the proof that it is so. 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..

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