直接分解算法
2016-08-23
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* To factor the n by n matrix A = (A(I,J)) into the product of the
* lower triangular matrix L = (L(I,J)) and U = (U(I,J)), that is
* A = LU, where the main diagonal of either L or U consists of all ones:
*
* INPUT: dimension n; the entries A(I,J), 1<=I, J<=n, of A;
* the diagonal L(1,1), ..., L(N,N) of L or the diagonal
* U(1,1), ..., U(N,N) of U.
*
* OUTPUT: the entries L(I,J), 1<=J<=I, 1<=I<=n of L and the entries
* lower triangular matrix L = (L(I,J)) and U = (U(I,J)), that is
* A = LU, where the main diagonal of either L or U consists of all ones:
*
* INPUT: dimension n; the entries A(I,J), 1<=I, J<=n, of A;
* the diagonal L(1,1), ..., L(N,N) of L or the diagonal
* U(1,1), ..., U(N,N) of U.
*
* OUTPUT: the entries L(I,J), 1<=J<=I, 1<=I<=n of L and the entries
c++
算法
分解
直接
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