Inverse Bilinear Interpolation?

Question

I have four 2d points, p0 = (x0,y0), p1 = (x1,y1), etc. that form a quadrilateral. In my case, the quad is not rectangular, but it should at least be convex.

Answer 1

this is my implementation ... I guess it is faster than of tfiniga

void invbilerp( float x, float y, float x00, float x01, float x10, float x11,  float y00, float y01, float y10, float y11, float [] uv ){

// substition 1 ( see. derivation )
float dx0 = x01 - x00;
float dx1 = x11 - x10;
float dy0 = y01 - y00;
float dy1 = y11 - y10;

// substitution 2 ( see. derivation )
float x00x = x00 - x;
float xd   = x10 - x00;
float dxd  = dx1 - dx0; 
float y00y = y00 - y;
float yd   = y10 - y00;
float dyd  = dy1 - dy0;

// solution of quadratic equations
float c =   x00x*yd - y00y*xd;
float b =   dx0*yd  + dyd*x00x - dy0*xd - dxd*y00y;
float a =   dx0*dyd - dy0*dxd;
float D2 = b*b - 4*a*c;
float D  = sqrt( D2 );
float u = (-b - D)/(2*a);

// backsubstitution of "u" to obtain "v"
float v;
float denom_x = xd + u*dxd;
float denom_y = yd + u*dyd;
if( abs(denom_x)>abs(denom_y) ){  v = -( x00x + u*dx0 )/denom_x;  }else{  v = -( y00y + u*dy0 )/denom_y;  }
uv[0]=u;
uv[1]=v;

/* 
// do you really need second solution ? 
u = (-b + D)/(2*a);
denom_x = xd + u*dxd;
denom_y = yd + u*dyd;
if( abs(denom_x)>abs(denom_y) ){  v = -( x00x + u*dx0 )/denom_x;  }else{  v2 = -( y00y + u*dy0 )/denom_y;  }
uv[2]=u;
uv[3]=v;
*/ 
}

and derivation

// starting from bilinear interpolation
(1-v)*(  (1-u)*x00 + u*x01 ) + v*( (1-u)*x10 + u*x11 )     - x
(1-v)*(  (1-u)*y00 + u*y01 ) + v*( (1-u)*y10 + u*y11 )     - y

substition 1:
dx0 = x01 - x00
dx1 = x11 - x10
dy0 = y01 - y00
dy1 = y11 - y10

we get:
(1-v) * ( x00 + u*dx0 )  + v * (  x10 + u*dx1  )  - x   = 0
(1-v) * ( y00 + u*dy0 )  + v * (  y10 + u*dy1  )  - y   = 0

we are trying to extract "v" out
x00 + u*dx0   + v*(  x10 - x00 + u*( dx1 - dx0 ) )  - x = 0
y00 + u*dy0   + v*(  y10 - y00 + u*( dy1 - dy0 ) )  - y = 0

substition 2:
x00x = x00 - x
xd   = x10 - x00
dxd  = dx1 - dx0 
y00y = y00 - y
yd   = y10 - y00
dyd  = dy1 - dy0 

// much nicer
x00x + u*dx0   + v*(  xd + u*dxd )  = 0
y00x + u*dy0   + v*(  yd + u*dyd )  = 0

// this equations for "v" are used for back substition
v = -( x00x + u*dx0 ) / (  xd + u*dxd  )
v = -( y00x + u*dy0 ) / (  yd + u*dyd  )

// but for now, we eliminate "v" to get one eqution for "u"  
( x00x + u*dx0 ) / (  xd + u*dxd )  =  ( y00y + u*dy0 ) / (  yd + u*dyd  )

put denominators to other side

( x00x + u*dx0 ) * (  yd + u*dyd )  =  ( y00y + u*dy0 ) * (  xd + u*dxd  )

x00x*yd + u*( dx0*yd + dyd*x00x ) + u^2* dx0*dyd = y00y*xd + u*( dy0*xd + dxd*y00y ) + u^2* dy0*dxd  

// which is quadratic equation with these coefficients 
c =   x00x*yd - y00y*xd
b =   dx0*yd  + dyd*x00x - dy0*xd - dxd*y00y
a =   dx0*dyd - dy0*dxd