Suppose that I have an image with n pixels and the following equation, ∑ r...

myntfalskj4

myntfalskj4

Answered

2022-07-16

Suppose that I have an image with n pixels and the following equation,
r P ( U ( r ) s N ( r ) w r s U ( s ) ) 2 = 0
U = ( U ( 1 ) , U ( 2 ) , . . . , U ( n ) ) are the variables;
P is the set of all the n pixels;
N ( r ) is the set of all neighboring pixels of r. For example, for an image with 3 × 3 pixels, N ( 1 ) = {2, 4, 5}
w r s is a weighting function where s N ( r ) w r s = 1
Note: I just mentioned that it is an image to better explain the role of the set N and the function w r s .
Well, it's easy to show that "one" solution for that equation are the scalar multiples of the all-ones vector. I mean, U = k ( 1 , 1 , . . . , 1 ) is clearly a solution.
My question is: there's a way to proof that this is the "only" solution for the above equation?

Answer & Explanation

Charlize Manning

Charlize Manning

Expert

2022-07-17Added 12 answers

Hint. Suppose you have a solution that is not constant. Then the maximum of the solution occurs at (at least one) location s, with a strictly smaller value in (at least one) neighbor. Can you get a contradiction? You can if the weights are all positive, but not in general.

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