xref: /aoo42x/main/basegfx/source/workbench/gauss.hxx (revision 07a3d7f1)
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21 
22 
23 
24 /** This method eliminates elements below main diagonal in the given
25     matrix by gaussian elimination.
26 
27     @param matrix
28     The matrix to operate on. Last column is the result vector (right
29     hand side of the linear equation). After successful termination,
30     the matrix is upper triangular. The matrix is expected to be in
31     row major order.
32 
33     @param rows
34     Number of rows in matrix
35 
36     @param cols
37     Number of columns in matrix
38 
39     @param minPivot
40     If the pivot element gets lesser than minPivot, this method fails,
41     otherwise, elimination succeeds and true is returned.
42 
43     @return true, if elimination succeeded.
44  */
45 template <class Matrix, typename BaseType>
eliminate(Matrix & matrix,int rows,int cols,const BaseType & minPivot)46 bool eliminate( 	Matrix&			matrix,
47                     int				rows,
48                     int				cols,
49                     const BaseType& minPivot	)
50 {
51 	BaseType	temp;
52 	int			max, i, j, k;	/* *must* be signed, when looping like: j>=0 ! */
53 
54 	/* eliminate below main diagonal */
55 	for(i=0; i<cols-1; ++i)
56 	{
57 		/* find best pivot */
58 		max = i;
59 		for(j=i+1; j<rows; ++j)
60 			if( fabs(matrix[ j*cols + i ]) > fabs(matrix[ max*cols + i ]) )
61 				max = j;
62 
63 		/* check pivot value */
64 		if( fabs(matrix[ max*cols + i ]) < minPivot )
65 			return false;	/* pivot too small! */
66 
67 		/* interchange rows 'max' and 'i' */
68 		for(k=0; k<cols; ++k)
69 		{
70 			temp = matrix[ i*cols + k ];
71 			matrix[ i*cols + k ] = matrix[ max*cols + k ];
72 			matrix[ max*cols + k ] = temp;
73 		}
74 
75 		/* eliminate column */
76 		for(j=i+1; j<rows; ++j)
77 			for(k=cols-1; k>=i; --k)
78 				matrix[ j*cols + k ] -= matrix[ i*cols + k ] *
79 					matrix[ j*cols + i ] / matrix[ i*cols + i ];
80 	}
81 
82 	/* everything went well */
83 	return true;
84 }
85 
86 
87 /** Retrieve solution vector of linear system by substituting backwards.
88 
89 	This operation _relies_ on the previous successful
90     application of eliminate()!
91 
92     @param matrix
93     Matrix in upper diagonal form, as e.g. generated by eliminate()
94 
95     @param rows
96     Number of rows in matrix
97 
98     @param cols
99     Number of columns in matrix
100 
101     @param result
102     Result vector. Given matrix must have space for one column (rows entries).
103 
104     @return true, if back substitution was possible (i.e. no division
105     by zero occurred).
106  */
107 template <class Matrix, class Vector, typename BaseType>
substitute(const Matrix & matrix,int rows,int cols,Vector & result)108 bool substitute(	const Matrix&	matrix,
109                     int				rows,
110                     int				cols,
111                     Vector&			result	)
112 {
113 	BaseType	temp;
114 	int			j,k;	/* *must* be signed, when looping like: j>=0 ! */
115 
116 	/* substitute backwards */
117 	for(j=rows-1; j>=0; --j)
118 	{
119 		temp = 0.0;
120 		for(k=j+1; k<cols-1; ++k)
121 			temp += matrix[ j*cols + k ] * result[k];
122 
123 		if( matrix[ j*cols + j ] == 0.0 )
124 			return false;	/* imminent division by zero! */
125 
126 		result[j] = (matrix[ j*cols + cols-1 ] - temp) / matrix[ j*cols + j ];
127 	}
128 
129 	/* everything went well */
130 	return true;
131 }
132 
133 
134 /** This method determines solution of given linear system, if any
135 
136 	This is a wrapper for eliminate and substitute, given matrix must
137 	contain right side of equation as the last column.
138 
139     @param matrix
140     The matrix to operate on. Last column is the result vector (right
141     hand side of the linear equation). After successful termination,
142     the matrix is upper triangular. The matrix is expected to be in
143     row major order.
144 
145     @param rows
146     Number of rows in matrix
147 
148     @param cols
149     Number of columns in matrix
150 
151     @param minPivot
152     If the pivot element gets lesser than minPivot, this method fails,
153     otherwise, elimination succeeds and true is returned.
154 
155     @return true, if elimination succeeded.
156  */
157 template <class Matrix, class Vector, typename BaseType>
solve(Matrix & matrix,int rows,int cols,Vector & result,BaseType minPivot)158 bool solve(	Matrix&		matrix,
159             int			rows,
160             int			cols,
161             Vector&		result,
162             BaseType	minPivot	)
163 {
164 	if( eliminate<Matrix,BaseType>(matrix, rows, cols, minPivot) )
165 		return substitute<Matrix,Vector,BaseType>(matrix, rows, cols, result);
166 
167 	return false;
168 }
169