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Topic: Mean Value Blurring (Read 1958 times) 

Barukh
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Mean Value Blurring
« on: Apr 8^{th}, 2014, 9:00am » 
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Consider a representation of a visual image as rectangular matrix NxN, where every element represents the color of a pixel – a positive real number in interval [0..R]. This image undergoes an iterative process of blurring according to the following procedure: At iteration i, every pixel color is replaced by an arithmetic mean of colors of itself and all its neighbours (horizontal, vertical, diagonal) at the iteration i1. Assume that all operations are performed mathematically. We are interested in the following question: What happens after infinite number of blurring iterations? Specifically, is it necessarily the case that the final image will be homogeneous (that is, every pixel has the same color)? Consider also the following variants of image representation: A. Colors are positive integers, and integer arithmetic is used. B. Colors are positive integers, but this time floating point arithmetic is used, and the result is round to the nearest integer.

« Last Edit: Apr 8^{th}, 2014, 9:01am by Barukh » 
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Barukh
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Re: Mean Value Blurring
« Reply #1 on: Apr 9^{th}, 2014, 1:37am » 
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Just to be clear: In image processing, the procedure described above is called Mean Filtering with 3x3 square kernel.

« Last Edit: Apr 9^{th}, 2014, 1:38am by Barukh » 
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towr
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Re: Mean Value Blurring
« Reply #2 on: Apr 9^{th}, 2014, 10:23am » 
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How do we treat the edge cases? In the original statement, well, there's just fewer neighbours on the edges; but if you use a 3x3 kernel, then it matter what you do with the nonexistent values beyond the border. (If you set the beyondborder values to zero, or extend to infinity, then all color bleeds off eventually.)


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Barukh
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Re: Mean Value Blurring
« Reply #3 on: Apr 9^{th}, 2014, 10:36am » 
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on Apr 9^{th}, 2014, 10:23am, towr wrote:How do we treat the edge cases? 
 You count only the neighbours in the image (so, for instance, the corner pixel will have 3 neighbours).


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