How to find an Equivalent point in a Scaled down image?

Question

I would like to calculate the corner points or contours of the star in this in a Larger image. For that I'm scaling down the size to a smaller one & I'm able to get this points clearly. Now How to map this point in original image? I'm using opencv c++.

Answer 1

Consider a trivial example: the image size is reduced exactly by half.

So, the cartesian coordinate (x, y) in the original image becomes coordinate (x/2, y/2) in the reduced image, and coordinate (x', y') in the reduced image corresponds to coordinate (x*2, y*2) in the original image.

Of course, fractional coordinates get typically rounded off, in a reduced scale image, so the exact mapping is only possible for even-numbered coordinates in this example's original image.

Generalizing this, if the image's width is scaled by a factor of w horizontally and h vertically, coordinate (x, y) becomes coordinate(x*w, y*h), rounded off. In the example I gave, both w and h are 1/2, or .5

You should be able to figure out the values of w and h yourself, and be able to map the coordinates trivially. Of course, due to rounding off, you will not be able to compute the exact coordinates in the original image.

Answer 2

I realize this is an old question. I just wanted to add to Sam's answer above, to deal with " rounding off ", in case other readers are wondering the same thing I faced.

This rounding off becomes obvious for even # of pixels across a coordinate axis. For instance, along a 1-D axis, a point demarcating the 2nd quartile gets mapped to an inaccurate value:

axis_prev = [0, 1, 2, 3]
axis_new = [0, 1, 2, 3, 4, 5, 6, 7]

w_prev = len(axis_prev) # This is an axis of length 4
w_new = len(axis_new) # This is an axis of length 8

x_prev = 2
x_new = x_prev * w_new / w_prev

print(x_new)
>>> 4
### x_new should be 5

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In Python, one strategy would be to linearly interpolate values from one axis resolution to another axis resolution. Say for the above, we wish to map a point from the smaller image to its corresponding point of the star in the larger image:

import numpy as np
from scipy.interpolate import interp1d

x_old = np.linspace(0, 640, 641)
x_new = np.linspace(0, 768, 769)

f = interp1d(x_old, x_new)

x = 35
x_prime = f(x)