mirror of https://github.com/CGAL/cgal
114 lines
4.8 KiB
TeX
114 lines
4.8 KiB
TeX
\begin{ccRefConcept}{LinearAlgebraTraits_d}\ccCreationVariable{LA}
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\ccDefinition
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The data type \ccc{LinearAlgebraTraits_d} encapsulates two classes
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\ccc{Matrix}, \ccc{Vector} and many functions of basic linear algebra.
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An instance of data type \ccc{Matrix} is a matrix of variables of type
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\ccc{NT}. Accordingly, \ccc{Vector} implements vectors of variables of
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type \ccc{NT}. Most functions of linear algebra are \emph{checkable},
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i.e., the programs can be asked for a proof that their output is
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correct. For example, if the linear system solver declares a linear
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system $A x = b$ unsolvable it also returns a vector $c$ such that
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$c^T A = 0$ and $c^T b \neq 0$.
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\ccSetOneOfTwoColumns{5.5cm}
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\ccTypes
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\ccNestedType{NT}{the number type of the components.}
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\ccNestedType{Vector}{the vector type.}
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\ccNestedType{Matrix}{the matrix type.}
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\ccSetTwoOfThreeColumns{2cm}{1cm}
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\ccOperations
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\ccMethod{static Matrix transpose(const Matrix& M);}{returns $M^T$
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(a \ccc{M.column_dimension()} $\times$ \ccc{M.column_dimension()} -
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matrix). }
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\ccMethod{static bool inverse(const Matrix& M, Matrix& I, NT& D,
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Vector& c);}{determines whether \ccc{M} has an inverse. It also
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computes either the inverse as $(1/D) \cdot \ccc{I}$ or when no
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inverse exists, a vector $c$ such that $c^T \cdot M = 0 $.
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\ccPrecond{ $M$ is square.} }
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\ccMethod{static Matrix inverse(const Matrix& M, NT& D) ;}{returns the
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inverse matrix of \ccc{M}. More precisely, $1/D$ times the matrix
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returned is the inverse of \ccc{M}.\\ \ccPrecond{\ccc{determinant(M)
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!= 0}.} \ccPrecond{ $M$ is square.} }
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\ccMethod{static NT determinant (const Matrix& M, Matrix& L, Matrix&
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U, std::vector<int>& q, Vector& c);} {returns the determinant $D$ of
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\ccc{M} and sufficient information to verify that the value of the
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determinant is correct. If the determinant is zero then $c$ is a
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vector such that $c^T \cdot M = 0$. If the determinant is non-zero
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then $L$ and $U$ are lower and upper diagonal matrices respectively
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and $q$ encodes a permutation matrix $Q$ with $Q(i,j) = 1$ iff $i =
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q(j)$ such that $L \cdot M \cdot Q = U$, $L(0,0) = 1$, $L(i,i) = U(i
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- 1,i - 1)$ for all $i$, $1 \le i < n$, and $D = s \cdot U(n - 1,n -
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1)$ where $s$ is the determinant of $Q$. \ccPrecond{\ccc{M} is
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square.}}
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\ccMethod{static bool verify_determinant (const Matrix& M, NT D,
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Matrix& L, Matrix& U, const std::vector<int>& q, Vector&
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c);}{verifies the conditions stated above.}
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\ccMethod{static NT determinant (const Matrix& M);}{returns the
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determinant of \ccc{M}. \ccPrecond{\ccc{M} is square.}}
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\ccMethod{static int sign_of_determinant (const Matrix& M);}{returns
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the sign of the determinant of \ccc{M}. \ccPrecond{\ccc{M} is
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square.}}
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\ccMethod{static bool linear_solver(const Matrix& M, const Vector& b,
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Vector& x, NT& D, Matrix& spanning_vectors, Vector& c);}{ determines
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the complete solution space of the linear system $M\cdot x = b$. If
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the system is unsolvable then $c^T \cdot M = 0$ and $c^T \cdot b
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\not= 0$. If the system is solvable then $(1/D) x$ is a solution,
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and the columns of \ccc{spanning_vectors} are a maximal set of
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linearly independent solutions to the corresponding homogeneous
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system. \ccPrecond{\ccc{M.row_dimension() = b.dimension()}.} }
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\ccMethod{static bool linear_solver(const Matrix& M, const Vector& b,
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Vector& x, NT& D, Vector& c) ;}{ determines whether the linear
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system $M\cdot x = b$ is solvable. If yes, then $(1/D) x$ is a
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solution, if not then $c^T \cdot M = 0$ and $c^T \cdot b \not= 0$.
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\ccPrecond{\ccc{M.row_dimension() = b.dimension()}.} }
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\ccMethod{static bool linear_solver(const Matrix& M, const Vector& b,
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Vector& x, NT& D) ;}{ as above, but without the witness $c$
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\ccPrecond{\ccc{M.row_dimension() = b.dimension()}.} }
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\ccMethod{static bool is_solvable(const Matrix& M, const Vector& b)
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;}{ determines whether the system $M \cdot x = b$ is solvable \\
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\ccPrecond{\ccc{M.row_dimension() = b.dimension()}.} }
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\ccMethod{static bool homogeneous_linear_solver (const Matrix& M,
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Vector& x);}{determines whether the homogeneous linear system
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$M\cdot x = 0$ has a non - trivial solution. If yes, then $x$ is
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such a solution. }
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\ccMethod{static int homogeneous_linear_solver (const Matrix& M,
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Matrix& spanning_vecs);}{determines the solution space of the
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homogeneous linear system $M\cdot x = 0$. It returns the dimension
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of the solution space. Moreover the columns of \ccc{spanning_vecs}
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span the solution space. }
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\ccMethod{static int independent_columns (const Matrix& M,
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std::vector<int>& columns);}{returns the indices of a maximal subset
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of independent columns of \ccc{M}. }
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\ccMethod{static int rank (const Matrix & M);}{returns the rank of
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matrix \ccc{M} }
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\ccHasModels
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\ccRefIdfierPage{CGAL::Linear_algebraHd<RT>} \\
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\ccRefIdfierPage{CGAL::Linear_algebraCd<FT>}
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\end{ccRefConcept}
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