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Nov 29, 2012 at 3:09 vote accept ppyang
Nov 27, 2012 at 16:57 comment added Felix Goldberg What does the dual cone of your cone look like?
Nov 27, 2012 at 16:46 answer added K Audenaert timeline score: 3
Feb 24, 2012 at 5:28 comment added ppyang I think you mean use Bregman successive projection method to find the optimal solution on the intersection of convex sets, however, I am just interested in whether there is any theoretical result about such kind of problems, rather than to get the solution in practical. You have provided a good way to think about it and thank you for your attention!
Feb 24, 2012 at 4:51 comment added Suvrit In that case, just cycle through the constraints enough number of times (if $f$ is quadratic, then you can use Dykstra's method). It seemed to me that you just want a "feasible" solution that is not "too bad". If I get time, I might think more carefully about your question, otherwise, I hope someone else finds the time to handle it. Best,
Feb 24, 2012 at 3:17 comment added ppyang @Suvrit, thank you for your suggestion. I thought about your method and the question seems to be more complicated than expected. I edited the post and gave a more detailed illustration at the end of the original post. How do you think about it? Thank you very much!
Feb 24, 2012 at 3:10 history edited ppyang CC BY-SA 3.0
In response to Suvrit's comment, I added some results of further thinking and more questions.
Feb 23, 2012 at 2:50 comment added Suvrit Using my answer to your previous question, probably this can be found iteratively. First find $X_1 \preceq A_1$ and $X_1 \preceq A_2$, then find $X_2 \preceq X_1$ and $X_2 \preceq A_3$ and so on.
Feb 23, 2012 at 2:23 history edited ppyang CC BY-SA 3.0
added 66 characters in body; edited title
Feb 22, 2012 at 5:37 history edited ppyang CC BY-SA 3.0
added 9 characters in body
Feb 22, 2012 at 4:25 history asked ppyang CC BY-SA 3.0