3
$\begingroup$

I can understand just fine how to use 1-dimensional interpolation on data points where one coordinate is a function of the other: y = f(x). However, when we have an image, we generally have points in the image that are not given uniquely by such a function, e.g. two points (x1,y1) and (x1,y2), so how do we describe this in order to interpolate between points in the image?

$\endgroup$
1
  • $\begingroup$ you can assume the function that best represent your image, and use this function to find any value at any location. $\endgroup$
    – HforHesham
    Commented Jan 10, 2014 at 18:59

2 Answers 2

5
$\begingroup$

There are various methods on 2d interpolation (this one, and this one). But most of them considered at least 4 points rather than 2. The simplest 2d interpolation is 3 1d interpolation, in which you interpolate the points between (x1-d1, y1-d2) and (x1+d1, y1-d2) as (x2,y2), then you interpolate the points between (x1-d3, y1+d4) and (x1+d3, y1+d4) as (x3,y3). Finally (x1,y1) is obtained by 1d interpolation of (x2,y2) and (x3,y3).

$\endgroup$
4
$\begingroup$

the simplest interpolation method is nearest neighbour. If you have 4 points with values given by Ms the interpolated value is given by P = f(x,y):

P$\ = f(x,y)$

= M_11 if $\ |x-x_1|<= |x-x_2|$ and $\ |y-y_1|<= |y-y_2|$,

= M_12 if $\ |x-x_1|< |x-x_2|$ and $\ |y-y_2|< |y-y_1|$,

= M_21 if $\ |x-x_2|< |x-x_1|$ and $\ |y-y_1|< |y-y_2|$,

= M_22 if $\ |x-x_2|< |x-x_1|$ and $\ |y-y_2|< |y-y_1|$.

Another method is bilinear interpolation. Given by:enter image description here

In both cases the Ms stand for the values at each of 4 points. The x's and y's are coordinates.

I presume that these methods work better when you have a uniform grid (i.e. your Ms are equally spaced apart). Fancier methods may be able to capture more complex data locations. For example kriging.

You asked about 2 point interpolation in a 2D space. Perhaps this occurs at the edges and part of the reason why you get edge effects. Others will be able to correct me on this. I hope this helps somehwat. enter image description here

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.