Inverse Bilinear Interpolation?
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.
p2 --- p3
| |
t | p |
| |
p0 --- p1
s
I'm using bilinear interpolation. S and T are within [0..1] and the interpolated point is given by:
bilerp(s,t) = t*(s*p3+(1-s)*p2) + (1-t)*(s*p1+(1-s)*p0)
Here's the problem.. I have a 2d point p that I know is inside the quad. I want to find the s,t that will give me that point when using bilinear interpolation.
Is there a simple formula to reverse the bilinear interpolation?
Thanks for the solutions. I posted my implementation of Naaff's solution as a wiki.
Answer
I think it's easiest to think of your problem as an intersection problem: what is the parameter location (s,t) where the point p intersects the arbitrary 2D bilinear surface defined by p0, p1, p2 and p3.
The approach I'll take to solving this problem is similar to tspauld's suggestion.
Start with two equations in terms of x and y:
x = (1-s)*( (1-t)*x0 + t*x2 ) + s*( (1-t)*x1 + t*x3 )
y = (1-s)*( (1-t)*y0 + t*y2 ) + s*( (1-t)*y1 + t*y3 )
Solving for t:
t = ( (1-s)*(x0-x) + s*(x1-x) ) / ( (1-s)*(x0-x2) + s*(x1-x3) )
t = ( (1-s)*(y0-y) + s*(y1-y) ) / ( (1-s)*(y0-y2) + s*(y1-y3) )
We can now set these two equations equal to each other to eliminate t. Moving everything to the left-hand side and simplifying we get an equation of the form:
A*(1-s)^2 + B*2s(1-s) + C*s^2 = 0
Where:
A = (p0-p) X (p0-p2)
B = ( (p0-p) X (p1-p3) + (p1-p) X (p0-p2) ) / 2
C = (p1-p) X (p1-p3)
Note that I've used the operator X to denote the 2D cross product (e.g., p0 X p1 = x0*y1 - y0*x1). I've formatted this equation as a quadratic Bernstein polynomial as this makes things more elegant and is more numerically stable. The solutions to s are the roots of this equation. We can find the roots using the quadratic formula for Bernstein polynomials:
s = ( (A-B) +- sqrt(B^2 - A*C) ) / ( A - 2*B + C )
The quadratic formula gives two answers due to the +-. If you're only interested in solutions where p lies within the bilinear surface then you can discard any answer where s is not between 0 and 1. To find t, simply substitute s back into one of the two equations above where we solved for t in terms of s.
I should point out one important special case. If the denominator A - 2*B + C = 0 then your quadratic polynomial is actually linear. In this case, you must use a much simpler equation to find s:
s = A / (A-C)
This will give you exactly one solution, unless A-C = 0. If A = C then you have two cases: A=C=0 means all values for s contain p, otherwise no values for s contain p.