Newsgroups: sci.image.processing
From: RGreason@colours.demon.co.uk (Roy Greason)
Path: cantaloupe.srv.cs.cmu.edu!das-news2.harvard.edu!news2.near.net!news.mathworks.com!udel!gatech!howland.reston.ans.net!pipex!bt!btnet!peernews.demon.co.uk!colours.demon.co.uk!RGreason
Subject: Re: [Question] Generate random dots
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In article: <277260268wnr@dstrip.demon.co.uk>  Steve@dstrip.demon.co.uk 
(Steve Rencontre) writes:
> Xref: news.demon.co.uk sci.image.processing:8187
> Newsgroups: sci.image.processing
> From: Steve@dstrip.demon.co.uk (Steve Rencontre)
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colours.demon.co.uk!news.demon.co.uk!peernews.demon.co.uk!dstrip.demon.co.uk
!Steve
> Subject: Re: [Question] Generate random dots
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> Date: Fri, 21 Apr 1995 11:01:33 +0000
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> 
> Interesting.
> 
> OTTOMH, I can't think of a good way of doing *random* dots without
> a neighbourhood search.
> 
> What *might* possibly work is to use an inverse 2D FFT. Create a 2D
> spectrum that looks like low-frequency noise, then IFFT it and put a dot
> at the centre of each generated black region. Of course, a 200x200 IFFT
> might take longer to calculate than doing a search... I'm also not sure
> it would do what you want :-)
> 
> 
> 
> -----------------------------------------------------------------------
> Steve Rencontre               |  steve@dstrip.demon.co.uk (business) 
> If it works, it's obsolete.   |  steveren@cix.compulink.co.uk (private)
> -----------------------------------------------------------------------
> 
How about producing a temp image where each random pixel is circled 10 
pixels wide and filled.  Then the next random generation could look at the 
pixel position in the temp image and if it is not part of a filled area,  
then add a pixel to the main image and produce another filled circle with 
the same centre in the temp image.  However perhaps filling a circle is no 
quicker, (or is even slower) than a neighbourhood search? <:-)

-- 
-------------------------------------------------------------------------
	Roy Greason.

	RGreason@colours.co.uk
	R.Greason@uclan.ac.uk
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