GLSL 1.3 inverse 4x4 matrix

Question

I want to calculate the new normal vector and in GLSL 1.4 you can use this formula: normal = mat3(transpose(inverse(modelview))) * in_Normal; But my version of GLSL is 1.3 and the function inverse is not available in this version. Do you know if there is an alternative to this without coding the entire function to inverse a matrix ?

Answer 1

Assuming the last row of your matrix is [ 0 0 0 1 ] (if you use only rotations, scaling and translation, it should), you can simplify your calculations.

Instead of:

normal = mat3(transpose(inverse(modelview))) * in_Normal

use:

normal = transpose(inverse(mat3(modelview))) * in_Normal

Then:

mat3 transpose(mat3 matrix) {
    vec3 row0 = matrix[0];
    vec3 row1 = matrix[1];
    vec3 row2 = matrix[2];
    mat3 result = mat3(
        vec3(row0.x, row1.x, row2.x),
        vec3(row0.y, row1.y, row2.y),
        vec3(row0.z, row1.z, row2.z)
    );
    return result;
}

float det(mat2 matrix) {
    return matrix[0].x * matrix[1].y - matrix[0].y * matrix[1].x;
}

mat3 inverse(mat3 matrix) {
    vec3 row0 = matrix[0];
    vec3 row1 = matrix[1];
    vec3 row2 = matrix[2];

    vec3 minors0 = vec3(
        det(mat2(row1.y, row1.z, row2.y, row2.z)),
        det(mat2(row1.z, row1.x, row2.z, row2.x)),
        det(mat2(row1.x, row1.y, row2.x, row2.y))
    );
    vec3 minors1 = vec3(
        det(mat2(row2.y, row2.z, row0.y, row0.z)),
        det(mat2(row2.z, row2.x, row0.z, row0.x)),
        det(mat2(row2.x, row2.y, row0.x, row0.y))
    );
    vec3 minors2 = vec3(
        det(mat2(row0.y, row0.z, row1.y, row1.z)),
        det(mat2(row0.z, row0.x, row1.z, row1.x)),
        det(mat2(row0.x, row0.y, row1.x, row1.y))
    );

    mat3 adj = transpose(mat3(minors0, minors1, minors2));

    return (1.0 / dot(row0, minors0)) * adj;
}