basegfx/source/workbench/gauss.hxx

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  30
  31 /** This method eliminates elements below main diagonal in the given
  32     matrix by gaussian elimination.
  33
  34     @param matrix
  35     The matrix to operate on. Last column is the result vector (right
  36     hand side of the linear equation). After successful termination,
  37     the matrix is upper triangular. The matrix is expected to be in
  38     row major order.
  39
  40     @param rows
  41     Number of rows in matrix
  42
  43     @param cols
  44     Number of columns in matrix
  45
  46     @param minPivot
  47     If the pivot element gets lesser than minPivot, this method fails,
  48     otherwise, elimination succeeds and true is returned.
  49
  50     @return true, if elimination succeeded.
  51  */
  52 template <class Matrix, typename BaseType>
  53 bool eliminate(         Matrix&                 matrix,
  54                     int                         rows,
  55                     int                         cols,
  56                     const BaseType& minPivot    )
  57 {
  58     BaseType    temp;
  59     int                 max, i, j, k;   /* *must* be signed, when looping like: j>=0 ! */
  60
  61     /* eliminate below main diagonal */
  62     for(i=0; i<cols-1; ++i)
  63     {
  64         /* find best pivot */
  65         max = i;
  66         for(j=i+1; j<rows; ++j)
  67             if( fabs(matrix[ j*cols + i ]) > fabs(matrix[ max*cols + i ]) )
  68                 max = j;
  69
  70         /* check pivot value */
  71         if( fabs(matrix[ max*cols + i ]) < minPivot )
  72             return false;       /* pivot too small! */
  73
  74         /* interchange rows 'max' and 'i' */
  75         for(k=0; k<cols; ++k)
  76         {
  77             temp = matrix[ i*cols + k ];
  78             matrix[ i*cols + k ] = matrix[ max*cols + k ];
  79             matrix[ max*cols + k ] = temp;
  80         }
  81
  82         /* eliminate column */
  83         for(j=i+1; j<rows; ++j)
  84             for(k=cols-1; k>=i; --k)
  85                 matrix[ j*cols + k ] -= matrix[ i*cols + k ] *
  86                     matrix[ j*cols + i ] / matrix[ i*cols + i ];
  87     }
  88
  89     /* everything went well */
  90     return true;
  91 }
  92
  93
  94 /** Retrieve solution vector of linear system by substituting backwards.
  95
  96     This operation _relies_ on the previous successful
  97     application of eliminate()!
  98
  99     @param matrix
 100     Matrix in upper diagonal form, as e.g. generated by eliminate()
 101
 102     @param rows
 103     Number of rows in matrix
 104
 105     @param cols
 106     Number of columns in matrix
 107
 108     @param result
 109     Result vector. Given matrix must have space for one column (rows entries).
 110
 111     @return true, if back substitution was possible (i.e. no division
 112     by zero occured).
 113  */
 114 template <class Matrix, class Vector, typename BaseType>
 115 bool substitute(        const Matrix&   matrix,
 116                     int                         rows,
 117                     int                         cols,
 118                     Vector&                     result  )
 119 {
 120     BaseType    temp;
 121     int                 j,k;    /* *must* be signed, when looping like: j>=0 ! */
 122
 123     /* substitute backwards */
 124     for(j=rows-1; j>=0; --j)
 125     {
 126         temp = 0.0;
 127         for(k=j+1; k<cols-1; ++k)
 128             temp += matrix[ j*cols + k ] * result[k];
 129
 130         if( matrix[ j*cols + j ] == 0.0 )
 131             return false;       /* imminent division by zero! */
 132
 133         result[j] = (matrix[ j*cols + cols-1 ] - temp) / matrix[ j*cols + j ];
 134     }
 135
 136     /* everything went well */
 137     return true;
 138 }
 139
 140
 141 /** This method determines solution of given linear system, if any
 142
 143     This is a wrapper for eliminate and substitute, given matrix must
 144     contain right side of equation as the last column.
 145
 146     @param matrix
 147     The matrix to operate on. Last column is the result vector (right
 148     hand side of the linear equation). After successful termination,
 149     the matrix is upper triangular. The matrix is expected to be in
 150     row major order.
 151
 152     @param rows
 153     Number of rows in matrix
 154
 155     @param cols
 156     Number of columns in matrix
 157
 158     @param minPivot
 159     If the pivot element gets lesser than minPivot, this method fails,
 160     otherwise, elimination succeeds and true is returned.
 161
 162     @return true, if elimination succeeded.
 163  */
 164 template <class Matrix, class Vector, typename BaseType>
 165 bool solve(     Matrix&         matrix,
 166             int                 rows,
 167             int                 cols,
 168             Vector&             result,
 169             BaseType    minPivot        )
 170 {
 171     if( eliminate<Matrix,BaseType>(matrix, rows, cols, minPivot) )
 172         return substitute<Matrix,Vector,BaseType>(matrix, rows, cols, result);
 173
 174     return false;
 175 }