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This algorithm takes a finite number of steps to reach a solution and smoothly improves its candidate solution as it goes (so it can find good approximate solutions when cut off at a reasonable number of iterations), but is very slow in practice, owing largely to the computation of the pseudoinverse ((A P) T A P) −1. [1]
In numerical linear algebra, the Jacobi method (a.k.a. the Jacobi iteration method) is an iterative algorithm for determining the solutions of a strictly diagonally dominant system of linear equations. Each diagonal element is solved for, and an approximate value is plugged in. The process is then iterated until it converges.
For example, in the MATLAB or GNU Octave function pinv, the tolerance is taken to be t = ε⋅max(m, n)⋅max(Σ), where ε is the machine epsilon. The computational cost of this method is dominated by the cost of computing the SVD, which is several times higher than matrix–matrix multiplication, even if a state-of-the art implementation ...
The program is solvable in polynomial time if the graph has all undirected or all directed edges. Variants include the rural postman problem. [3]: ND25, ND27 Clique cover problem [2] [3]: GT17 Clique problem [2] [3]: GT19 Complete coloring, a.k.a. achromatic number [3]: GT5 Cycle rank; Degree-constrained spanning tree [3]: ND1
The Boolean satisfiability problem (SAT) asks to determine if a propositional formula (example depicted) can be made true by an appropriate assignment ("solution") of truth values to its variables. While it is easy to verify whether a given assignment renders the formula true , [ 1 ] no essentially faster method to find a satisfying assignment ...
NumPy (pronounced / ˈ n ʌ m p aɪ / NUM-py) is a library for the Python programming language, adding support for large, multi-dimensional arrays and matrices, along with a large collection of high-level mathematical functions to operate on these arrays. [3]
a k is the "contrapoint," i.e., a point such that f(a k) and f(b k) have opposite signs, so the interval [a k, b k] contains the solution. Furthermore, |f(b k)| should be less than or equal to |f(a k)|, so that b k is a better guess for the unknown solution than a k. b k−1 is the previous iterate (for the first iteration, we set b k−1 = a 0).
The solution is obtained iteratively via (+) = (), where the matrix is decomposed into a lower triangular component , and a strictly upper triangular component such that = +. [4] More specifically, the decomposition of A {\displaystyle A} into L ∗ {\displaystyle L_{*}} and U {\displaystyle U} is given by: