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  2. Boolean satisfiability problem - Wikipedia

    en.wikipedia.org/wiki/Boolean_satisfiability_problem

    A variant of the 3-satisfiability problem is the one-in-three 3-SAT (also known variously as 1-in-3-SAT and exactly-1 3-SAT). Given a conjunctive normal form with three literals per clause, the problem is to determine whether there exists a truth assignment to the variables so that each clause has exactly one TRUE literal (and thus exactly two ...

  3. Not-all-equal 3-satisfiability - Wikipedia

    en.wikipedia.org/wiki/Not-all-equal_3-satisfiability

    Unlike 3SAT, some variants of NAE3SAT in which graphs representing the structure of variables and clauses are planar graphs can be solved in polynomial time.In particular this is true when there exists a planar graph with one vertex per variable, one vertex per clause, an edge for each variable–clause incidence, and a cycle of edges connecting all the variable vertices.

  4. SAT solver - Wikipedia

    en.wikipedia.org/wiki/SAT_solver

    In computer science and formal methods, a SAT solver is a computer program which aims to solve the Boolean satisfiability problem (SAT). On input a formula over Boolean variables, such as "(x or y) and (x or not y)", a SAT solver outputs whether the formula is satisfiable, meaning that there are possible values of x and y which make the formula true, or unsatisfiable, meaning that there are no ...

  5. MAX-3SAT - Wikipedia

    en.wikipedia.org/wiki/MAX-3SAT

    For every R, add clauses representing f R (x i1,...,x iq) using 2 q SAT clauses. Clauses of length q are converted to length 3 by adding new (auxiliary) variables e.g. x 2 ∨ x 10 ∨ x 11 ∨ x 12 = ( x 2 ∨ x 10 ∨ y R) ∧ ( y R ∨ x 11 ∨ x 12). This requires a maximum of q2 q 3-SAT clauses. If z ∈ L then there is a proof π such ...

  6. Sharp-SAT - Wikipedia

    en.wikipedia.org/wiki/Sharp-SAT

    #SAT is harder than SAT in the sense that, once the total number of solutions to a Boolean formula is known, SAT can be decided in constant time. However, the converse is not true, because knowing a Boolean formula has a solution does not help us to count all the solutions , as there are an exponential number of possibilities.

  7. 1 + 2 + 3 + 4 + ⋯ - ⋯ - Wikipedia

    en.wikipedia.org/wiki/1_%2B_2_%2B_3_%2B_4_%2B_%E...

    where f (2k−1) is the (2k − 1)th derivative of f and B 2k is the (2k)th Bernoulli number: B 2 = ⁠ 1 / 6 ⁠, B 4 = ⁠− + 1 / 30 ⁠, and so on. Setting f ( x ) = x , the first derivative of f is 1, and every other term vanishes, so [ 15 ]

  8. TI-Nspire series - Wikipedia

    en.wikipedia.org/wiki/TI-Nspire_series

    They have a thinner design, with a thickness of 1.57 cm (almost half of the TI-89), a 1,200 mA·h (1,060 mAh before 2013) rechargeable battery (wall adapter is included in the American retail package), a 320 by 240 pixel full color backlit display (3.2" diagonal), and OS 3.0 which includes features such as 3D graphing.

  9. 1 − 2 + 3 − 4 + ⋯ - ⋯ - Wikipedia

    en.wikipedia.org/wiki/1_%E2%88%92_2_%2B_3_%E2%88...

    The idea becomes clearer by considering the general series 1 − 2x + 3x 2 − 4x 3 + 5x 4 − 6x 5 + &c. that arises while expanding the expression 1 ⁄ (1+x) 2, which this series is indeed equal to after we set x = 1.