bv_cvxbook_extra_exercises

42 56 total variation image interpolation a grayscale

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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(...
View Full Document

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.

Ask a homework question - tutors are online