Efficient 4x4 matrix inverse (affine transform)

问题:

I was hoping someone can point out an efficient formula for 4x4 affine matrix transform. Currently my code uses cofactor expansion and it allocates a temporary array for each cofactor. It's easy to read, but it's slower than it should be.

Note, this isn't homework and I know how to work it out manually using 4x4 co-factor expansion, it's just a pain and not really an interesting problem for me. Also I've googled and came up with a few sites that give you the formula already (http://www.euclideanspace.com/maths/algebra/matrix/functions/inverse/fourD/index.htm). However this one could probably be optimized further by pre-computing some of the products. I'm sure someone came up with the "best" formula for this at one point or another?

回答1:

You should be able to exploit the fact that the matrix is affine to speed things up over a full inverse. Namely, if your matrix looks like this

A = [ M   b  ]
    [ 0   1  ]

where A is 4x4, M is 3x3, b is 3x1, and the bottom row is (0,0,0,1), then

inv(A) = [ inv(M)   -inv(M) * b ]
         [   0            1     ]

Depending on your situation, it may be faster to compute the result of inv(A) * x instead of actually forming inv(A). In that case, things simplify to

inv(A) * [x] = [ inv(M) * (x - b) ]
         [1] = [        1         ] 

where x is a 3x1 vector (usually a 3D point).

Lastly, if M represents a rotation (i.e. its columns are orthonormal), then you can use the fact that inv(M) = transpose(M). Then computing the inverse of A is just a matter of subtracting the translation component, and multiplying by the transpose of the 3x3 part.

Note that whether or not the matrix is orthonormal is something that you should know from the analysis of the problem. Checking it during runtime would be fairly expensive; although you might want to do it in debug builds to check that your assumptions hold.

Hope all of that is clear...

回答2:

Just in case someone would like to save some typing, here's an AS3 version I wrote based on page 9 (more efficient version of Laplace Expansion Theorem) of the link posted above by phkahler:

public function invert() : Matrix4 {
    var m : Matrix4 = new Matrix4();

    var s0 : Number = i00 * i11 - i10 * i01;
    var s1 : Number = i00 * i12 - i10 * i02;
    var s2 : Number = i00 * i13 - i10 * i03;
    var s3 : Number = i01 * i12 - i11 * i02;
    var s4 : Number = i01 * i13 - i11 * i03;
    var s5 : Number = i02 * i13 - i12 * i03;

    var c5 : Number = i22 * i33 - i32 * i23;
    var c4 : Number = i21 * i33 - i31 * i23;
    var c3 : Number = i21 * i32 - i31 * i22;
    var c2 : Number = i20 * i33 - i30 * i23;
    var c1 : Number = i20 * i32 - i30 * i22;
    var c0 : Number = i20 * i31 - i30 * i21;

    // Should check for 0 determinant

    var invdet : Number = 1 / (s0 * c5 - s1 * c4 + s2 * c3 + s3 * c2 - s4 * c1 + s5 * c0);

    m.i00 = (i11 * c5 - i12 * c4 + i13 * c3) * invdet;
    m.i01 = (-i01 * c5 + i02 * c4 - i03 * c3) * invdet;
    m.i02 = (i31 * s5 - i32 * s4 + i33 * s3) * invdet;
    m.i03 = (-i21 * s5 + i22 * s4 - i23 * s3) * invdet;

    m.i10 = (-i10 * c5 + i12 * c2 - i13 * c1) * invdet;
    m.i11 = (i00 * c5 - i02 * c2 + i03 * c1) * invdet;
    m.i12 = (-i30 * s5 + i32 * s2 - i33 * s1) * invdet;
    m.i13 = (i20 * s5 - i22 * s2 + i23 * s1) * invdet;

    m.i20 = (i10 * c4 - i11 * c2 + i13 * c0) * invdet;
    m.i21 = (-i00 * c4 + i01 * c2 - i03 * c0) * invdet;
    m.i22 = (i30 * s4 - i31 * s2 + i33 * s0) * invdet;
    m.i23 = (-i20 * s4 + i21 * s2 - i23 * s0) * invdet;

    m.i30 = (-i10 * c3 + i11 * c1 - i12 * c0) * invdet;
    m.i31 = (i00 * c3 - i01 * c1 + i02 * c0) * invdet;
    m.i32 = (-i30 * s3 + i31 * s1 - i32 * s0) * invdet;
    m.i33 = (i20 * s3 - i21 * s1 + i22 * s0) * invdet;

    return m;
}

This successfully produced an identity matrix when I multiplied various 3D transformation matrices by the inverse returned from this method. I'm sure you can search/replace to get this into whatever language you'd like.

回答3:

To follow-up on pkhaler's and Robin Hilliard's excellent responses above, here is Robin's ActionScript 3 code converted into a C# method. Hopefully this can save some typing for other C# developers, as well as C/C++ and Java developers in need of a 4x4 matrix inversion function:

public static double[,] GetInverse(double[,] a)
{
    var s0 = a[0, 0] * a[1, 1] - a[1, 0] * a[0, 1];
    var s1 = a[0, 0] * a[1, 2] - a[1, 0] * a[0, 2];
    var s2 = a[0, 0] * a[1, 3] - a[1, 0] * a[0, 3];
    var s3 = a[0, 1] * a[1, 2] - a[1, 1] * a[0, 2];
    var s4 = a[0, 1] * a[1, 3] - a[1, 1] * a[0, 3];
    var s5 = a[0, 2] * a[1, 3] - a[1, 2] * a[0, 3];

    var c5 = a[2, 2] * a[3, 3] - a[3, 2] * a[2, 3];
    var c4 = a[2, 1] * a[3, 3] - a[3, 1] * a[2, 3];
    var c3 = a[2, 1] * a[3, 2] - a[3, 1] * a[2, 2];
    var c2 = a[2, 0] * a[3, 3] - a[3, 0] * a[2, 3];
    var c1 = a[2, 0] * a[3, 2] - a[3, 0] * a[2, 2];
    var c0 = a[2, 0] * a[3, 1] - a[3, 0] * a[2, 1];

    // Should check for 0 determinant
    var invdet = 1.0 / (s0 * c5 - s1 * c4 + s2 * c3 + s3 * c2 - s4 * c1 + s5 * c0);

    var b = new double[4, 4];

    b[0, 0] = ( a[1, 1] * c5 - a[1, 2] * c4 + a[1, 3] * c3) * invdet;
    b[0, 1] = (-a[0, 1] * c5 + a[0, 2] * c4 - a[0, 3] * c3) * invdet;
    b[0, 2] = ( a[3, 1] * s5 - a[3, 2] * s4 + a[3, 3] * s3) * invdet;
    b[0, 3] = (-a[2, 1] * s5 + a[2, 2] * s4 - a[2, 3] * s3) * invdet;

    b[1, 0] = (-a[1, 0] * c5 + a[1, 2] * c2 - a[1, 3] * c1) * invdet;
    b[1, 1] = ( a[0, 0] * c5 - a[0, 2] * c2 + a[0, 3] * c1) * invdet;
    b[1, 2] = (-a[3, 0] * s5 + a[3, 2] * s2 - a[3, 3] * s1) * invdet;
    b[1, 3] = ( a[2, 0] * s5 - a[2, 2] * s2 + a[2, 3] * s1) * invdet;

    b[2, 0] = ( a[1, 0] * c4 - a[1, 1] * c2 + a[1, 3] * c0) * invdet;
    b[2, 1] = (-a[0, 0] * c4 + a[0, 1] * c2 - a[0, 3] * c0) * invdet;
    b[2, 2] = ( a[3, 0] * s4 - a[3, 1] * s2 + a[3, 3] * s0) * invdet;
    b[2, 3] = (-a[2, 0] * s4 + a[2, 1] * s2 - a[2, 3] * s0) * invdet;

    b[3, 0] = (-a[1, 0] * c3 + a[1, 1] * c1 - a[1, 2] * c0) * invdet;
    b[3, 1] = ( a[0, 0] * c3 - a[0, 1] * c1 + a[0, 2] * c0) * invdet;
    b[3, 2] = (-a[3, 0] * s3 + a[3, 1] * s1 - a[3, 2] * s0) * invdet;
    b[3, 3] = ( a[2, 0] * s3 - a[2, 1] * s1 + a[2, 2] * s0) * invdet;

    return b;
}

回答4:

IIRC you can greatly shrink the code and time by precomputing a bunch (12?) 2x2 determinants. Split the matrix in half vertically and compute every 2x2 in both the upper and lower half. One of these smaller determinants is used in every term you'll need for the bigger computation and they each get reused.

Also, don't use a separate determinant function - reuse the sub-determinants you computed for the adjoint to get the determinant.

Oh, just found this.

There are some improvements you can make knowing its a certain kind of transform too.

回答5:

I believe the only way to compute an inverse is to solve n times the equation: A x = y, where y spans the unit vectors, i.e., the first one is (1,0,0,0), the second is (0,1,0,0), etc.

(Using the cofactors (Cramer's rule) is a bad idea, unless you want a symbolic formula for the inverse.)

Most linear algebra libraries will allow you to solve those linear systems, and even to compute an inverse. Example in python (using numpy):

from numpy.linalg import inv
inv(A) # here you go