简单的3x3矩阵逆码(C++)

Question

计算3x3矩阵逆的最简单方法是什么？

我只是在寻找一个简短的代码片段,它可以解决非奇异矩阵,可能使用Cramer的规则.它不需要高度优化.我更喜欢简单而不是速度.我宁愿不链接其他库.

Answer 1

这是batty答案的一个版本,但这会计算正确的逆.batty的版本计算逆的转置.

// computes the inverse of a matrix m
double det = m(0, 0) * (m(1, 1) * m(2, 2) - m(2, 1) * m(1, 2)) -
             m(0, 1) * (m(1, 0) * m(2, 2) - m(1, 2) * m(2, 0)) +
             m(0, 2) * (m(1, 0) * m(2, 1) - m(1, 1) * m(2, 0));

double invdet = 1 / det;

Matrix33d minv; // inverse of matrix m
minv(0, 0) = (m(1, 1) * m(2, 2) - m(2, 1) * m(1, 2)) * invdet;
minv(0, 1) = (m(0, 2) * m(2, 1) - m(0, 1) * m(2, 2)) * invdet;
minv(0, 2) = (m(0, 1) * m(1, 2) - m(0, 2) * m(1, 1)) * invdet;
minv(1, 0) = (m(1, 2) * m(2, 0) - m(1, 0) * m(2, 2)) * invdet;
minv(1, 1) = (m(0, 0) * m(2, 2) - m(0, 2) * m(2, 0)) * invdet;
minv(1, 2) = (m(1, 0) * m(0, 2) - m(0, 0) * m(1, 2)) * invdet;
minv(2, 0) = (m(1, 0) * m(2, 1) - m(2, 0) * m(1, 1)) * invdet;
minv(2, 1) = (m(2, 0) * m(0, 1) - m(0, 0) * m(2, 1)) * invdet;
minv(2, 2) = (m(0, 0) * m(1, 1) - m(1, 0) * m(0, 1)) * invdet;

Answer 2

你为什么不尝试自己编码呢？把它作为挑战.:)

对于3×3矩阵

alt text http://mathworld.wolfram.com/images/equations/MatrixInverse/NumberedEquation3.gif

矩阵逆是

alt text http://mathworld.wolfram.com/images/equations/MatrixInverse/NumberedEquation4.gif

我假设你知道矩阵| A |的决定因素是什么 是.

图像(c)Wolfram | Alpha和 mathworld.wolfram(06-11-09,22.06)

Answer 3

这段代码计算矩阵A 的转置逆:

double determinant =    +A(0,0)*(A(1,1)*A(2,2)-A(2,1)*A(1,2))
                        -A(0,1)*(A(1,0)*A(2,2)-A(1,2)*A(2,0))
                        +A(0,2)*(A(1,0)*A(2,1)-A(1,1)*A(2,0));
double invdet = 1/determinant;
result(0,0) =  (A(1,1)*A(2,2)-A(2,1)*A(1,2))*invdet;
result(1,0) = -(A(0,1)*A(2,2)-A(0,2)*A(2,1))*invdet;
result(2,0) =  (A(0,1)*A(1,2)-A(0,2)*A(1,1))*invdet;
result(0,1) = -(A(1,0)*A(2,2)-A(1,2)*A(2,0))*invdet;
result(1,1) =  (A(0,0)*A(2,2)-A(0,2)*A(2,0))*invdet;
result(2,1) = -(A(0,0)*A(1,2)-A(1,0)*A(0,2))*invdet;
result(0,2) =  (A(1,0)*A(2,1)-A(2,0)*A(1,1))*invdet;
result(1,2) = -(A(0,0)*A(2,1)-A(2,0)*A(0,1))*invdet;
result(2,2) =  (A(0,0)*A(1,1)-A(1,0)*A(0,1))*invdet;

虽然问题规定了非奇异矩阵,但您可能仍想检查行列式是否等于零(或非常接近零)并以某种方式标记它以确保安全.

Answer 4

对我们未知(雅虎)海报的所有应有的尊重,我看着这样的代码,只是在里面死了一点.调试字母汤非常难以调试.任何地方的任何一个错字都可能真的毁了你的一整天.可悲的是,这个特殊的例子缺少带有下划线的变量.当我们有a_b-c_d*e_f-g_h时,它会更有趣.特别是当使用_和 - 具有相同像素长度的字体时.

根据他的建议接受Suvesh Pratapa,我注意到:

Given 3x3 matrix:
       y0x0  y0x1  y0x2
       y1x0  y1x1  y1x2
       y2x0  y2x1  y2x2
Declared as double matrix [/*Y=*/3] [/*X=*/3];

(A)当拿一个3x3阵列的小调时,我们有4个感兴趣的值.较低的X/Y索引始终为0或1.较高的X/Y索引始终为1或2.始终!因此:

double determinantOfMinor( int          theRowHeightY,
                           int          theColumnWidthX,
                           const double theMatrix [/*Y=*/3] [/*X=*/3] )
{
  int x1 = theColumnWidthX == 0 ? 1 : 0;  /* always either 0 or 1 */
  int x2 = theColumnWidthX == 2 ? 1 : 2;  /* always either 1 or 2 */
  int y1 = theRowHeightY   == 0 ? 1 : 0;  /* always either 0 or 1 */
  int y2 = theRowHeightY   == 2 ? 1 : 2;  /* always either 1 or 2 */

  return ( theMatrix [y1] [x1]  *  theMatrix [y2] [x2] )
      -  ( theMatrix [y1] [x2]  *  theMatrix [y2] [x1] );
}

(B)现在是行列式:(注意减号!)

double determinant( const double theMatrix [/*Y=*/3] [/*X=*/3] )
{
  return ( theMatrix [0] [0]  *  determinantOfMinor( 0, 0, theMatrix ) )
      -  ( theMatrix [0] [1]  *  determinantOfMinor( 0, 1, theMatrix ) )
      +  ( theMatrix [0] [2]  *  determinantOfMinor( 0, 2, theMatrix ) );
}

(C)逆现在是:

bool inverse( const double theMatrix [/*Y=*/3] [/*X=*/3],
                    double theOutput [/*Y=*/3] [/*X=*/3] )
{
  double det = determinant( theMatrix );

    /* Arbitrary for now.  This should be something nicer... */
  if ( ABS(det) < 1e-2 )
  {
    memset( theOutput, 0, sizeof theOutput );
    return false;
  }

  double oneOverDeterminant = 1.0 / det;

  for (   int y = 0;  y < 3;  y ++ )
    for ( int x = 0;  x < 3;  x ++   )
    {
        /* Rule is inverse = 1/det * minor of the TRANSPOSE matrix.  *
         * Note (y,x) becomes (x,y) INTENTIONALLY here!              */
      theOutput [y] [x]
        = determinantOfMinor( x, y, theMatrix ) * oneOverDeterminant;

        /* (y0,x1)  (y1,x0)  (y1,x2)  and (y2,x1)  all need to be negated. */
      if( 1 == ((x + y) % 2) )
        theOutput [y] [x] = - theOutput [y] [x];
    }

  return true;
}

并使用一些质量较低的测试代码进行完善:

void printMatrix( const double theMatrix [/*Y=*/3] [/*X=*/3] )
{
  for ( int y = 0;  y < 3;  y ++ )
  {
    cout << "[  ";
    for ( int x = 0;  x < 3;  x ++   )
      cout << theMatrix [y] [x] << "  ";
    cout << "]" << endl;
  }
  cout << endl;
}

void matrixMultiply(  const double theMatrixA [/*Y=*/3] [/*X=*/3],
                      const double theMatrixB [/*Y=*/3] [/*X=*/3],
                            double theOutput  [/*Y=*/3] [/*X=*/3]  )
{
  for (   int y = 0;  y < 3;  y ++ )
    for ( int x = 0;  x < 3;  x ++   )
    {
      theOutput [y] [x] = 0;
      for ( int i = 0;  i < 3;  i ++ )
        theOutput [y] [x] +=  theMatrixA [y] [i] * theMatrixB [i] [x];
    }
}

int
main(int argc, char **argv)
{
  if ( argc > 1 )
    SRANDOM( atoi( argv[1] ) );

  double m[3][3] = { { RANDOM_D(0,1e3), RANDOM_D(0,1e3), RANDOM_D(0,1e3) },
                     { RANDOM_D(0,1e3), RANDOM_D(0,1e3), RANDOM_D(0,1e3) },
                     { RANDOM_D(0,1e3), RANDOM_D(0,1e3), RANDOM_D(0,1e3) } };
  double o[3][3], mm[3][3];

  if ( argc <= 2 )
    cout << fixed << setprecision(3);

  printMatrix(m);
  cout << endl << endl;

  SHOW( determinant(m) );
  cout << endl << endl;

  BOUT( inverse(m, o) );
  printMatrix(m);
  printMatrix(o);
  cout << endl << endl;

  matrixMultiply (m, o, mm );
  printMatrix(m);
  printMatrix(o);
  printMatrix(mm);  
  cout << endl << endl;
}

事后:

您可能还想检测非常大的决定因素,因为舍入误差会影响您的准确性!