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https://gitee.com/mirrors_PX4/PX4-Autopilot.git
synced 2026-10-27 01:53:35 +08:00
Use a single inverse implementation for a single matrix size
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+13
-13
@@ -305,7 +305,7 @@ SquareMatrix<Type, M> expm(const Matrix<Type, M, M> & A, size_t order=5)
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* inverse based on LU factorization with partial pivotting
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*/
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template<typename Type, size_t M>
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bool inv(const SquareMatrix<Type, M> & A, SquareMatrix<Type, M> & inv)
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bool inv(const SquareMatrix<Type, M> & A, SquareMatrix<Type, M> & inv, size_t rank = M)
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{
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SquareMatrix<Type, M> L;
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L.setIdentity();
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@@ -316,12 +316,12 @@ bool inv(const SquareMatrix<Type, M> & A, SquareMatrix<Type, M> & inv)
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//printf("A:\n"); A.print();
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// for all diagonal elements
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for (size_t n = 0; n < M; n++) {
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for (size_t n = 0; n < rank; n++) {
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// if diagonal is zero, swap with row below
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if (fabs(static_cast<float>(U(n, n))) < FLT_EPSILON) {
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//printf("trying pivot for row %d\n",n);
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for (size_t i = n + 1; i < M; i++) {
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for (size_t i = n + 1; i < rank; i++) {
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//printf("\ttrying row %d\n",i);
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if (fabs(static_cast<float>(U(i, n))) > 1e-8f) {
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@@ -349,12 +349,12 @@ bool inv(const SquareMatrix<Type, M> & A, SquareMatrix<Type, M> & inv)
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}
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// for all rows below diagonal
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for (size_t i = (n + 1); i < M; i++) {
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for (size_t i = (n + 1); i < rank; i++) {
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L(i, n) = U(i, n) / U(n, n);
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// add i-th row and n-th row
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// multiplied by: -a(i,n)/a(n,n)
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for (size_t k = n; k < M; k++) {
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for (size_t k = n; k < rank; k++) {
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U(i, k) -= L(i, n) * U(n, k);
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}
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}
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@@ -367,9 +367,9 @@ bool inv(const SquareMatrix<Type, M> & A, SquareMatrix<Type, M> & inv)
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//SquareMatrix<Type, M> Y = P;
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// for all columns of Y
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for (size_t c = 0; c < M; c++) {
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for (size_t c = 0; c < rank; c++) {
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// for all rows of L
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for (size_t i = 0; i < M; i++) {
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for (size_t i = 0; i < rank; i++) {
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// for all columns of L
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for (size_t j = 0; j < i; j++) {
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// for all existing y
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@@ -392,14 +392,14 @@ bool inv(const SquareMatrix<Type, M> & A, SquareMatrix<Type, M> & inv)
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//SquareMatrix<Type, M> X = Y;
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// for all columns of X
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for (size_t c = 0; c < M; c++) {
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for (size_t c = 0; c < rank; c++) {
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// for all rows of U
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for (size_t k = 0; k < M; k++) {
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for (size_t k = 0; k < rank; k++) {
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// have to go in reverse order
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size_t i = M - 1 - k;
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size_t i = rank - 1 - k;
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// for all columns of U
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for (size_t j = i + 1; j < M; j++) {
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for (size_t j = i + 1; j < rank; j++) {
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// for all existing x
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// subtract the component they
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// contribute to the solution
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@@ -416,8 +416,8 @@ bool inv(const SquareMatrix<Type, M> & A, SquareMatrix<Type, M> & inv)
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}
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//check sanity of results
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for (size_t i = 0; i < M; i++) {
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for (size_t j = 0; j < M; j++) {
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for (size_t i = 0; i < rank; i++) {
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for (size_t j = 0; j < rank; j++) {
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if (!is_finite(P(i,j))) {
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return false;
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}
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