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Jeff
Posts:
107
Registered:
2/27/10
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Re: A BIG problem with simple one
Posted:
Jan 20, 2012 1:00 PM
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"camilla belle" <camilla.belle@poczta.wp.pl> wrote in message <jf9sho$ich$1@newsfeed1.man.lodz.pl>...
> Hi.
>
>
> I have indexed image. Let it be a 10 x 10 matrix , which is filled randomly
> with 1 or 2. Then i put randomly a 3 into the matrix.
> 1 is black, 2 is white and 3 is red, later i use 4 for gray. Imagine it as
> rocks , grass , fire and ash. And now i would like to burn all the grass.
> So we have :
>
> 1 1 1 1 1 1 1 1 1
> 1 1 2 2 2 2 2 2 2
> 2 1 2 2 1 1 2 1 1
> 2 1 2 3 1 1 1 1 1
> 1 1 1 1 1 1 1 3 3
> *
>
> My algorithm is the simplest and only one i could think off.
> 1.Compare all pixels from A to 3
> 2. Save coordinates of pixels==3 in array B.
> 3.For all rows of B do:
> a)get coordinates of pixel P1 and start checking all 8 neighbours
> b)if neigbour is ==2 make it 3
> c)set pixel P1 to 4 which means burned
>
> 4.Repeat from 1.
>
> I am new in programming, matlab , algorithms and image processing . I would
> like to know what is the fastest possible algorithm for this problem or
> similar one( when we have only binary image and we want to set all whites to
> black). Also could anyone write from what domain is this problem ? Is it
> skeletization , dilatation, erozion , maybe i must use morfological
> operation on arrays and so on..... ???
>
> I am asking for help because when i use large matrix like 500x500 , well it
> could take ages to see the result.
>
> Also i wrote algorithm using while , for ... etc. But i know that in matlab
> everything is an array and that's why a man can write code only with arrays
> operation , would it be faster? How i can do that ?
>
> Thanks . Karl.
Could you simplify this problem to just two binary images of "grass" and "fire"? Taking a simple 1x3 array, with fire:
0 1 0
and grass:
1 0 0
dilate the fire to get:
1 1 1
remove the original fire:
1 0 1
find the overlap of the new potential fire with grass to get the new fire:
1 0 0
and remove the new fire from the original grass to get:
0 0 0
If this describes your situation, using imdilate and logical operators should get you what you need.
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