Springer-Verlag.). (In the context of veri cation, the certi cate consists of the assignment of values to the variables.) 3-SAT is NP-complete. Some clauses My reasoning is that by definition a literal could be 'not a1' which cannot be extended like {a1, a1, a1}. We need to convert that into CNF. By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy. Idea of the proof: encode the workings of a Nondeterministic Turing machine for an instance I of problem X 2NP as a SAT formula so that the formula is satis able if and only if the nondeterministic Turing machine would accept instance I. one SAT solver might be a disadvantage for another. For more details on NPTEL visit http://nptel.iitm.ac.in Use MathJax to format equations. Yep, that works! This is a very challenging course in the specialization. Clause form conversions for Boolean circuits. If you need a reduction from k-SAT to 3-SAT, then ratchet's answer works fine. Here the goal is to reduce an arbitrary SAT problem to 3-SAT in polynomial time using the fewest number of clauses and variables. © 2021 Coursera Inc. All rights reserved. By saying equisatisfiable, we mean that the resulting formula is satisfiable if and only if the initial formula is satisfiable. The 3 variable clause can be copied with no issue. I was reading about NP hardness from here (pages 8, 9) and in the notes the author reduces a problem in 3-SAT form to a graph that can be used to solve the maximum independent set problem.. Planar circuit SAT: This is a variant of circuit SAT in which the circuit, computing the SAT formula, is a planar directed acyclic graph. My question is motivated by curiosity. I want to know in general how can I convert $4-SAT$ to 3-SAT.. And I have a specific case that if you can help me optimize it to 3-SAT it will be greate.. This can be carried out in nondeterministic polynomial time. $\begingroup$ I have been wondering why the extension specifically for k=1 mentioned by ratchet isn't appearing in any book (at least the ones I came across so far). Can a Circle of the Stars Druid roll a natural d3 (or other odd-sided die) to bias their Cosmic Omen roll? Assume for the sake of contradiction that in the current satisfying assignment for F prime, l1 is set to 0, l2 is set to 0, and all the literals from the set A are also set to 0. (I would guess that there are some trade-offs between computation time and the size of the output. Why does the Bible put the evening before the morning at the end of each day that God worked in Genesis chapter one? Reduction of SAT to 3-SAT¶. EEPROM Fatigue - Does it affect only the cells being written excessively, or will it cause global failures? Most students take the SAT in 11th or 12th Grade, some even earlier, in 10th Grade. 3-sat reduces in polynomial time to nae 4-sat. Given a SAT clause of literals x1 or x2 or ... or xn, we can replace out 2 literals a, b as follows. In short, CNF modelling SAT is a globally recognized college admission test, which is conducted by the College Board. I'm leaving this question open for a bit in case someone knows the answer to the more general situation, otherwise I will simply accept ratchet's answer. In this case, the first clause of the formula F prime is satisfied by one of l1 or l2. reduced to solving an instance of 3SAT (or showing it is not satisfiable). For example, suppose we are Can an inverter through a battery charger charge its own batteries? And for sure, we will stop at some point, because at each step, we reduce some long clause with a shorter one. You can also read some recent works and look at the references; for example: Let me please post another solution similar to Ratchel's but somewhat different. Proof : Evidently 3SAT is in NP, since SAT is in NP. In this module you will study the classical NP-complete problems and the reductions between them. This is clearly polynomial in the length of an input formula. Careful. This is jumping ahead a little, but this is a variant of 3SAT that will be helpful in the future (actually, for the next problem) The problem: Not-All-Equal-3SAT … A useful property of Cook's reduction is that it preserves the number of accepting answers. The second clause contains a negation of y and all the other literals which we denote by a set A, okay? The transformation involves taking a boolean formula that would be a "yes" instance to 3-SAT and converting each clause to a set of nodes and edges that are used as an instance of the VC problem. @Mikola Maybe the Tseitin or Plaisted-Greenbaum transformation give you 3CNF? Next we discuss inherently hard problems for which no exact good solutions are known (and not likely to be found) and how to solve them in practice. supports HTML5 video. I mean, both the first two clauses of the formula F prime. Note that the only way that all four of these clauses can be simultaneously satisfied is if z1=T, which also means the original C will be satisfied, If the clause has two literals, C={z1, z2}, then create one new variable v1 and two new clauses: {v1, z1, z2} and {!v1, z1, z2}. If Eturns out to be true, then accept. Although many of the algorithms you've learned so far are applied in practice a lot, it turns out that the world is dominated by real-world problems without a known provably efficient algorithm. The following slideshow shows that an instance of Formula Satisfiability problem can be reduced to an instance of 3 CNF Satisfiability problem in polynomial time. There are usually many ways to model a given problem in CNF, and few guidelines trary SAT instance ˚as input, and transforms it to a 3SAT instance ˚0, such that satisfiabil-ity is preserved, i.e., ˚0 is satisfiable if and only if ˚is satisfiable. 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). Data Structures and Algorithms Specialization, Construction Engineering and Management Certificate, Machine Learning for Analytics Certificate, Innovation Management & Entrepreneurship Certificate, Sustainabaility and Development Certificate, Spatial Data Analysis and Visualization Certificate, Master's of Innovation & Entrepreneurship. This can be carried out in nondeterministic polynomial time. Surprisingly, it can be solved in polynomial time. To view this video please enable JavaScript, and consider upgrading to a web browser that Can anyone see how to do the reduction more directly, skipping the intermediate circuit step and going directly to 3-SAT? Again, the only way to satisfy both of these clauses is to have at least one of z1 and z2 be true, thus satisfying C, If the clause has three literals, C={z1, z2, z3}, just copy C into the 3-SAT instance unchanged, If the clause has more than 3 literals C={z1, z2, ..., zn}, then create n-3 new variables and n-2 new clauses in a chain, where for 2<= j <= n-2, Cij={v1,j-1, zj+1,!vi,j}, Ci1={z1, z2, !vi,1} and Ci,n-2={vi,n-3, zn-1, zn}. In this case, we just set the value of y to 1, right. blocked clauses may be added. The most known algorithm is the Tseitin algorithm (G. Tseitin. I would even be happy with a direct reduction in the special case of n-SAT. This is similar to what will be done for the two art gallery proofs. Here is another 3SAT formula: $(x_1 \lor x_2 \lor x_2)$. We now need to prove that the constructed transformation is correct, that the constructed reduction is correct, and that it takes polynomial time. In NP: guess a satisfying assignment and verify that it indeed sat-isfies the clauses. To construct such a reduction, we need to design a polynomial time algorithm that takes as input a formula in conjunctive normal form, that is, a collection of clauses, and produces an equisatisfiable formula in 3-CNF, that is, a formula in which each clause has at most three literals. So we are okay with it. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. We finish with a soft introduction to streaming algorithms that are heavily used in Big Data processing. 3SAT is the case where each clause has exactly 3 terms. First of all, all the remaining clauses of F prime are satisfied by exactly the same assignment that satisfies the formula F, so our goal is to set y so that both the first two clauses are satisfied. To determine whether a boolean expression Ein CNF is satis able, nondeterministically guess values for all the variables and then evaluate the expression. 28.14.1. uit V’ if ui= T; otherwise uif V’ for 1 i n Does the industry continue to produce outdated architecture CPUs with leading-edge process? Suppose there is a polynomial-time algorithm A for MAX 3-SAT. This video is part of an online course, Intro to Theoretical Computer Science. To view this video please enable JavaScript, and consider upgrading to a web browser that. 3SAT REDUCTION TO CLIQUE (THEOREM 7.32) Proof Idea Polynomial time reduction function which converts Boolean formulas to graphs In the constructed graphs, cliques of a specified size correspond to satisfying assignments of the cnf formula Structures within the graph are designed to mimic the behavior of the variables and clauses Tseitin In this case, the first clause is satisfied by the value of y while the second clause is satisfied by one of the literals from the set A. In, If the clause has only one literal C={z1}, then create two new variables v1 and v2 and four new 3-literal clauses: {v1, v2, z1}, {!v1, v2, z1}, {v1, !v2, z1} and {!v1, !v2, z1}. It's good to know this before trying to solve a problem before the tomorrow's deadline :) Although these problems are very unlikely to be solvable efficiently in the nearest future, people always come up with various workarounds. Here is another: $(x_1) \land (\neg x_1)$. (The reason for going through nae sat is that both max cut and nae sat exhibit a similar kind of symmetry in their solutions.) We do not need to get rid of it. Also trivially convertible. Duplicating variables seems more natural to me, and doesn't violate any definition of 3SAT that I've seen. For a good introduction to CNF encodings read the suggested book Handbook o Satisfiability. 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. The task is to describe a polynomial-time algorithm for: input: a … P. Jackson and D. Sheridan. My reasoning is that by definition a literal could be 'not a1' which cannot be extended like {a1, a1, a1}. the short answer is: since 3SAT is NP-complete, any problem in NP can be p.t. Which governors can flip the Senate as of March 2021? @TayfunPay Can you explain why you consider this solution to be more correct? To do this, consider these two formulas. Is it appropriate to walk out after giving notice before my two weeks are up? I guess I should have thought a bit more carefully before adding that last line, but if I don't get an answer to the more general question I will accept this. Asking for help, clarification, or responding to other answers. 2 Recitation 8 Problem: Reduce SAT to 3-SAT. Auf einen Blick 3sat Livestream, TV-Programm und verpasste Sendungen: Sehen Sie die Videos der 3sat-Mediathek wann und wo sie wollen! To subscribe to this RSS feed, copy and paste this URL into your RSS reader. 3Sat live. Each SAT clause has 1, 2, 3 or more variables. "translated from the Spanish"? Replace a or b in the clause with c. The resulting clause has one less literal. First break that up into these two clauses: You can check that derivation with a truth table. To do this, consider such a long clause and denote by l1 and l2 some two literals of this clause and denote by A the rest of this clause. We will give a polynomial-time algorithm A that given a 3-sat instance constructs an equivalent nae 4-sat … Definition: A Boolean formula is in 3SAT if it in 3CNF form and is also SATisfiable. Then we first of all note an important property. Suppose ˚is satis able and let (x 1;x 2;:::;x n) be the satisfying assignment. Bicycle weigh limit (carrying capacity) increase. Suppose (U,C) is an instance of satisfiability. MathJax reference. Theorem : 3SAT is NP-complete. encodings are compact and mechanisable but in practice do not always lead to Therefore, by adding 4 clauses (a or b or !c) and (a or !b or c) and (!a or b or c) and (!a or !b or c) we can substitute out two variables with one. Right. In The 1 and 2 variable clauses {a1} and {a1,a2} can be expanded to {a1,a1,a1} and {a1,a2,a1} respectively. EDIT: Based on ratchet's answer it is clear now that the reduction to n-SAT is somewhat trivial (and that I really should have thought that one through a bit more carefully before posting). Right? Ultimately any 3SAT formula is either satisfiable (hence can be converted to the 2SAT formula True) or not satisfiable (hence convertible to False). The clause, the first of the two new clauses has length 3. We then proceed to linear programming with applications in optimizing budget allocation, portfolio optimization, finding the cheapest diet satisfying all requirements and many others. Python Programming, Linear Programming (LP), Np-Completeness, Dynamic Programming. Run A on input ’. I don't understand the use of Cook-Levin in (1). We are now going to extend it so that it also satisfies the formula F prime. Is there some technicality that makes this solution better? Okay, for the reverse direction, note that if we have a satisfying assignment that satisfies the formula F prime, then we can just discard the value of the variable y from this assignment, and then what we get is a satisfying assignment for the formula F. Why is that? The bits of P are unit clauses. Recall that a SAT instance is an AND of some clauses, and each clause is OR of some literals. Claim. This particular proof was chosen because it reduces 3SAT to VERTEX COVER and involves the transformation of a boolean formula to something geometrical. © 2013 BMC Software, Inc. All rights reserved. This is the reason (presumably) all authors including Michael R. Garey and David S. Johnson used a different extension presented by 'Carlos Linares López' in his/her post here. is an art and we must often proceed by intuition and experimentation. Goddard 19b: 3. What we need to show is that this clause is satisfied in a formula F, but it must be satisfied, because at least one of l1, l2, and all the literals of A should be satisfied. Is there a historical reason why we say "C sharp" but notate on the staff "sharp C"? Until that time, the concept of an NP-complete problem did not even exist. To get clauses at least size 3, see this answer. Now we are going to do the following, introduce a new, fresh variable y and replace the current clause C with the following two clauses. Die beliebten und populräen Programme zeigen Qualitat in der Prime Time. To construct such a reduction, we need to somehow get rid of all long clauses in our formula. For example, if your formula contains. In the example, the author converts the following 3-SAT problem into a graph. Absence of evidence is not evidence of absence: What does Bayesian probability have to say about it? This is probably beyond the scope of the question, but I wanted to post it anyway. I want to do this so I be able to use sat solvers programs. Proven in early 1970s by Cook. model as variables, and some might take considerable thought to discover. On the other hand, you cannot do {'not a1', 'not a1', 'not a1]} as it needs another logic to identify if the original sat includes a negated literal or not. The channel airs not only in Germany, but also in Austria… The clause with more than 3 variables {a1,a2,a3,a4,a5} can be expanded to {a1,a2,s1}{!s1,a3,s2}{!s2,a4,a5} with s1 and s2 new variables whose value will depend on which variable in the original clause is true. Die Zuschauer, die diesen Kanal ansehen, hat durch Unfragen dem Live TV klar angegeben, was nicht auf dem Kanal sein sollte. If x i is assigned True, we colour v i with Tand v i with F(recall they’re connected to … Making statements based on opinion; back them up with references or personal experience. Such algorithms are usually designed to be able to process huge datasets without being able even to store a dataset. Recall that the only difference between the sets of variables of formulas F and F prime is the variable y. Let formula ’be an instance of 3-SAT. So our goal is to set y so that the resulting assignment satisfies all the clauses of the formula F prime. ∧C k where each C i is an ∨ of three or less literals. Mathematical Logic by Prof.Arindama Singh, Department of Mathematics ,IIT Madras. Videos und Livestreams in der 3sat-Mediathek anschauen! $\endgroup$ – D.W. ♦ Jul 6 '18 at 18:33 The 3-SAT problem is: (a ∨ b ∨ c) ∧ (b ∨ ~c ∨ ~d) ∧ (~a ∨ c ∨ d) ∧ (a ∨ ~b ∨ ~d) There is often a choice of problem features to site design / logo © 2021 Stack Exchange Inc; user contributions licensed under cc by-sa. MAX 3-SAT Theorem (MAX 3-SAT is NP-hard) If MAX 3-SAT can be solved in polynomial time, then so can 3-SAT. Proof. Lot of learning on a complicated subject of NP-Hard problems. To prove that the constructed reduction is correct, we're going to show that the initial formula F with a long clause is satisfiable if and only if the resulting formula where we replaced a long clause with a 3-clause and a shorter clause is also satisfiable. Is it really legal to knowingly lie in public as a public figure? If the assignment returned by A satis es all clauses of ’, then return YES; else return NO. On reducing the hardness of CNF-SAT to k-Clique, ETH-Hardness of $Gap\text-MAX\text-3SAT_{c}$, Satisfiability problems with restricted (not bounded) number of occurrences per variable, One month old puppy pacing in circles and crying. The only way to satisfy this formula is to put X and Y in the right order as the input. That is, from a general version of a problem to its special case. (I'm not quite sure I understand the question fully :) ). $\begingroup$ If you find the last point interesting, you might also be interested to know that #PLANAR-NAE-3SAT (counting solutions) is tractable as well, whereas other seemingly simple SAT variants like PLANAR-MONOTONE-2SAT are tractable (or even trivial) as a decision problem, but #P-hard for counting. The proof is by reduction to planar maximum cut. 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. This means that the total number of iterations is bounded from above by the total number of literals in all the clauses. Is the Pit from a Robe of Useful Items permanent and can it be dispelled? Is there a direct/natural reduction to count non-bipartite perfect matchings using the permanent? Proof. 3Sat Live Stream bietet, die für das Vergnügen des Zuschauers vorbereiteten Programme, in Form von ununterbrochenem fernsehen, an. Exposition by William Gasarch Algorithms for 3-SAT H. H. Hoos and D. G. Mitchell, editors, P. Manolios, D. Vroon, Efficient Circuit to CNF Conversion. Theoretical Computer Science Stack Exchange is a question and answer site for theoretical computer scientists and researchers in related fields. Check out the course here: https://www.udacity.com/course/cs313. Why do translations refer to the original language with a definite article, e.g. Planar NAE 3SAT: This problem is planar equivalent of NAE 3SAT. 3-sat to max cut. While this works, the resulting 3-SAT clauses end up looking almost nothing like the SAT clauses you started with, due to the initial application of the Cook-Levin theorem. ... Post Sales Service So once again, l1 and l2 are two literal of the clause C, which we consider at the moment. Very Very Challenging Course , it test your patience and rewards is extremely satisfying. To get clauses at least size 3 without duplicate literals see this answer. Then you finish the job by the standard reduction of circuit SAT to 3-SAT by replacing gates with clauses. Proof : Evidently 3SAT is in NP, since SAT is in NP. To determine whether a boolean expression Ein CNF is satis able, nondeterministically guess values for all the variables and then evaluate the expression. Realizing no one at my school does quite what I want to do. Video created by University of California San Diego, HSE University for the course "Advanced Algorithms and Complexity". So this is F. It contains a long clause. So we are going to repeat this procedure while there is at least one long clause. Exactly by one literal. This follows from work of Fortnow and Santhanam, see also follow-up work by Dell and van Melkebeek. You've learned the basic algorithms now and are ready to step into the area of more complex problems and algorithms to solve them. Entdecken Sie Dokumentationen, Magazine aus Kultur, Wissenschaft, Gesellschaft und vieles mehr! So roughly speaking, the number of variables in the k-CNF-SAT instance will always depend on the number of clauses in your CNF-SAT formula. Clearly, this can be done in polynomial time. Once again, if the initial formula is F and the resulting formula is F prime, then by saying equisatisfiable, we mean that F is satisfiable if and only if F prime is satisfiable. IUPAC: Would I prioritize low numbering to highest-priority group, OR try to assign lowest numbers overall? SAT appears to be a generalization of 3-SAT and is intuitively more difficult. the best model, and some subformulae might be better expanded. The question said "I would even be happy with a direct reduction in the special case of n-SAT". This assumes duplicate literals are okay. However, it turns out we can reduce SAT to 3-SAT, so 3-SAT is just as hard as SAT. Automation of Reasoning: Classical Papers in Computational Logic, 2:466–483, 1983. I have been wondering why the extension specifically for k=1 mentioned by ratchet isn't appearing in any book (at least the ones I came across so far). NP-hard: We show SAT ≤pm 3SAT. may be omitted by polarity considerations, and implied, symmetry breaking or For a construction of p.t. We then plug the values into the formula and evaluate it. At the same time, this clause is shorter than the following one. The following is the proof that the problem VERTEX COVER is NP-complete. If you want a direct reduction from generic propositional formula to CNF (and to 3-SAT) then - at least from the "SAT solvers perspective" - I think that the answer to your question What is the 'most natural' reduction ...?, is: There is no 'natural' reduction!. Well, for a simple reason. Reduction of 3-SAT to VC The 3-SAT problem is one of the most common NPC problems used on the left side of polynomial time reductions. SAT is in NP: We nondeterministically guess truth values to the variables. And this is F prime that results from F by replacing this long clause with the following two clauses, l1, l2 and y, where y Is a fresh variable and a clause not y or A. I assume now that the formula F is satisfiable and take a satisfying assignment. Then there is just no possibility to satisfy these two clauses because no matter how we assign the value to y either the first clause or the second clause is going to be unsatisfied. So if it satisfies these two literals, we set y to 0. On the other hand, if l1 and l2 are falsified by the current satisfying assignment, then at least one of the literals from the set I should be satisfied. Hence 3COLOR <=p 3SAT. For example… Okay, so to set the variable y, we just check whether the current satisfying assignment of the formula F satisfies one of the literals l1 or l2 or not. Done :) Now we prove that our initial 3-SAT instance ˚is satis able if and only the graph Gas constructed above is 3-colourable. To learn more, see our tips on writing great answers. Isn't boolean-formula-SAT already a special case of circuit-SAT in which the graph structure of the circuit happens to be a tree? Also convertible to 2SAT. Our next reduction is from satisfiability problem to 3-satisfiability problem. Now the reduction that I've always seen in text books goes something like this: First take your instance of SAT and apply the Cook-Levin theorem to reduce it to circuit SAT. Can someone please help me in drawing the below diagram via tikz (for my master thesis). Different encodings may have different advantages and disadvantages such as size or solution density, and what is an advantage for If the 3SAT problem has a solution, then the VC problem has a solution The vertex cover set V’ with exactly n+2m vertices can be obtained as follows : From the truth assignment for {u1, u2, …, un} in 3SAT, we get n vertices from Vu, i.e. The proof shows how every decision problem in the complexity class NP can be reducedto the SAT problem for CNF formulas, sometimes called CNFSAT. BILL- Do examples and counterexamples on the board. Who says the input to SAT has to have "clauses"? By saying long, we mean clauses that contain more that three literals. Reducing 3SAT to SAT We reduce SAT to 3SAT. are known for choosing among them. Solution: We first create a parse tree of the given boolean formula. Convert the multiplication circuit to a 3SAT formula (clauses for each OR,AND,XOR gate, each gate of the circuit should have 3 variables, then the clauses ban the incorrect combinations). And A is the set of all other literals. Thanks for contributing an answer to Theoretical Computer Science Stack Exchange! The second clause, on the other hand, is satisfied by the variable y, right, because we've just assigned the value 0 to y. It only takes a minute to sign up. The language 3SAT is a restriction of SAT, and so 3SAT 2NP. Improving Cook's generic reduction for Clique to SAT? A Habitable Zone Within a Habitable Zone--Would that Make any Difference? reduction of 3COLOR to SAT, you may see section 2 in the following document (the topic is … So the first clause contains literals l1, l2, and y, so it is a disjunction of l1, l2, and y. Thus 3SAT … On the complexity of derivation in propositional calculus. We can iteratively apply this process on each clause until it's of size at most 3. 3Sat is a German TV channel with the focus of providing regional news and other informative programs. SAT is a fairly long exam – 3 hours and 45 minutes in duration, and made up of 10 sections. Concerning time, consider the running time, it is clear because at each iteration we'll replace a clause with a shorter one. We are happy with it. Clearly a degenerate -- though fortunately inadmissible unless P=NP -- solution would be to just solve the SAT problem, then emit a trivial 3-SAT instance...). Less formally, I would like to know: "What is the 'most natural' reduction from SAT to 3-SAT?". We will start with networks flows which are used in more typical applications such as optimal matchings, finding disjoint paths and flight scheduling as well as more surprising ones like image segmentation in computer vision. rev 2021.3.9.38752, The best answers are voted up and rise to the top, Theoretical Computer Science Stack Exchange works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us. EDIT (to include some information on the point of studying 3SAT): If someone gives you an assignment of values to the variables, it is very easy to check to see whether that assignment makes all the clauses … Advanced algorithms build upon basic ones and use new ideas. @crockeea If I had to guess, some algorithms are easier to implement if you assume 3-SAT clauses contain 3 unique variables per clause, To generalize, if we assume all logical operators are binary or unary, then we substitute out binary operators with a single variable (similar to this answer) until we have 3-SAT. Using techniques from parameterized complexity it has been proven that, assuming the polynomial hierarchy doesn't collapse to its third level, there is no polynomial-time algorithm which takes an instance of CNF-SAT on n variables with unbounded clause length, and outputs an instance of k-CNF-SAT (no clauses of length more than k) on n' variables where $n'$ is polynomial in $n$. I learned a lot form going through the programming assignments! This is directly taken from chapter 9 of the 2nd Edition of "The Algorithm Design Manual" by Steven Skiena. You will also practice solving large instances of some of these problems despite their hardness using very efficient specialized software based on tons of research in the area of NP-complete problems. Slightly di erent proof by Levin independently. If Eturns out to be true, then accept.