sparse image representation via combined transforms - Convex ...
sparse image representation via combined transforms - Convex ... sparse image representation via combined transforms - Convex ...
22 CHAPTER 2. SPARSITY IN IMAGE CODING the broadcasting business), the sender (TV station) can usually afford expensive equipment and a long processing time, while the receiver (single television set) must have cheap and fast (even real-time) algorithms. Our objective is to find sparse decompositions. Following the above rule, in our algorithm we will tolerate high complexity in decomposing, but superposition must be fast and cheap. Ideally, the order of complexity of superpositioning must be no higher than the order of complexity of a Fast Fourier Transform (FFT). We choose FFT for comparison because it is a well-known technique and a milestone in the development of signal processing. Also by ignoring the logarithm factor, the order of complexity of FFT is almost equal to the order of complexity of copying a signal or image from one disk to another. In general, we can hardly imagine any processing scheme that can have a lower order of complexity than just copying. We will see that the order of complexity of our superposition algorithm is indeed no higher than the order of complexity of doing an FFT. 2.4 Proof Proof of Theorem 2.1 We only need to prove that N∑ log x 2 i ≥ i=1 N∑ log yi 2 . (2.5) i=1 Let’s define the following function for a sequence x: f(x) = N∑ log x 2 i . i=1 Consider the following procedure: 1. The sequence x (0) is the same sequence as x: x (0) = x. 2. For any integer n ≥ 1, we do the following: (a) Pick an index l such that l =min{i : y i >x (n−1) i }.
2.4. PROOF 23 The index l does not exist if and only if the two sequences {x i , 1 ≤ i ≤ N} and {y i , 1 ≤ i ≤ N} are the same. (b) Pick an index u such that u = max{i : y i f(x (n) ),n ≥ 1. And because f(x) =f(x (0) )andf(y) is equal to the last function value in f(x (n) ),n ≥ 1, we have f(x) ≥ f(y), which is the inequality (2.5). The last item concludes our proof for Theorem 2.1.
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2.4. PROOF 23<br />
The index l does not exist if and only if the two sequences {x i , 1 ≤ i ≤ N} and<br />
{y i , 1 ≤ i ≤ N} are the same.<br />
(b) Pick an index u such that<br />
u = max{i : y i f(x (n) ),n ≥ 1. And because f(x) =f(x (0) )andf(y) is<br />
equal to the last function value in f(x (n) ),n ≥ 1, we have f(x) ≥ f(y), which is the<br />
inequality (2.5).<br />
The last item concludes our proof for Theorem 2.1.
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