How to: pixel packs in an image to representative point coordinates (x,y) Let's have an image (gray-scale or even binary) as shown on the following figure at the left hand side, the goal is to generate a list of points i.e., coordinates in the form of (x,y) for each pack of the dark pixels in the image.
What are the proper image processing tools to do this and where are they available?  





Updates:
1)
Here you may find some more details about the problem. (Note the variation in the size of packs)  

  

I may suggest having packs detected to compute the convex-hull boundary for each and then find the representative centroid {see this for details}.

  



2)
Here is the result produced by application of Distance Transform (suggested by "Libor"). Note my annotations on the figure. The method does not work as it was promising!  



3)
Erosion eliminates small packs!  

from __future__ import division
from scipy import zeros, ndimage as dsp
from pylab import subplot,plot,matshow,show

img = zeros((30,30))
img[10:14,10:14] = 1
img[16:17,16:17] = 1
img[19:23,19] = 1
img[19,19:23] = 1

subplot(221)
matshow(img,0)

subplot(222)
y = dsp.binary_erosion(img,[[1,1],[1,1]])
matshow(y,0)

subplot(223)
y = dsp.binary_erosion(img,[[0,1,0],[1,1,1],[0,1,0]])
matshow(y,0)

subplot(224)
y = dsp.binary_erosion(img,[[1,1,1],[1,1,1],[1,1,1]])
matshow(y,0)

show()




4)
Well here is a Python (i.e., the language of love :) ) implementation of the labeling idea (also proposed by "Jean-Yves" below):

subplot(221)
l,n = dsp.label(img)
sl = dsp.find_objects(l)
for s in sl:
    x = (s[1].start+s[1].stop-1)/2
    y = (s[0].start+s[0].stop-1)/2
    plot(x,y,'wo')


and the result:  



Note that although it is done in Python so quick due to Scipy performance, the background procedure in label function should be an exhausting iteration. This may be considered as a trade-off. So for a while I keep being eager to seek more efficient algorithms. And also note that in the given code above I found the center of geometry so simply while for complex or asymmetric shapes this may cause positioning to be biased. That is it is a work in progress ;).

5)
Here is a complex case (a real image) captured from here on which the labelling proposal applied and you see the results. Note that it took only 0.015 s for whole procedure including labelling and finding the objects. Scipy guys, did very well job, I think. Wow! {right-click on image, click on view image for full resolution}
Just a naive suggestion: do you know about component labeling? 

The technique is about finding chunks of "touching" pixels and assigning them a label, e.g. an integer number. You can then interrogate each chink separately, looking for the pixel that share the same label.

In MATLAB, here is the function that does it trivially: bwlabel