bv_cvxbook_extra_exercises

# 42 56 total variation image interpolation a grayscale

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Unformatted text preview: φ(y ) = k=1 where |y |[1] , |y |[2] , |y |[3] , . . . , denote the absolute values of the components of y sorted in decreasing order. (d) A piecewise-linear penalty. 0 |u| ≤ 0.2 |u| − 0.2 0.2 ≤ |u| ≤ 0.3 h(u) = 2|u| − 0.5 |u| ≥ 0.3. m h(yk ), φ(y ) = k=1 (e) Huber penalty. m h(yk ), φ(y ) = h(u) = k=1 u2 | u| ≤ M M (2|u| − M ) |u| ≥ M with M = 0.2. (f) Log-barrier penalty. m h(yk ), φ(y ) = k=1 h(u) = − log(1 − u2 ), dom h = {u | |u| < 1}. Here is the problem. Generate data A and b as follows: m n A b b = = = = = 200; 100; randn(m,n); randn(m,1); b/(1.01*max(abs(b))); (The normalization of b ensures that the domain of φ(Ax − b) is nonempty if we use the log-barrier penalty.) To compare the results, plot a histogram of the vector of residuals y = Ax − b, for each of the solutions x, using the Matlab command hist(A*x-b,m/2); Some additional hints and remarks for the individual problems: 41 (a) This problem can be solved using least-squares (x=A\b). (b) Use the CVX function norm(y,1). (c) Use the CVX function norm_largest(...
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## This note was uploaded on 09/10/2013 for the course C 231 taught by Professor F.borrelli during the Fall '13 term at Berkeley.

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