Problem: wrong solutions?

[steve3141]

Simple example

variable x(2)
parameter b(2)
minimize norm(x-b)
norm(x,1) <= 1

With b=array([0, 0]) we get solution x=[0, 0] as expected. But b=array([0.3, 0.3]), for example, does not give solution x=b as expected; it gives x=[ 0.0005921, 0.0005921] We'd expect solution to be x=b for b=[a,a] with 0<=a<=0.5. Either I'm missing something -- which is possible -- or there's a bug in the problem setup. I haven't examined the transformations in detail yet but will do and let you know if I find anything. Thanks.

[steve3141]

My mistake; I don't know what norm(x,1) actually does -- I'd assumed it was the 1-norm (which is true in matlab CVX but not qcml...) What I meant to use was "norm1(x)".

[echu]

Oh, just in case this confuses anyone in the future: QCML allows single-argument functions (of vectors) to take multiple arguments by applying the function row-wise. In other words if a function f maps a vector to a scalar, then calling f on multiple vectors effectively calls:

f(x,y,z) = map(f, zip(x,y,z))

where x, y, and z are vector expressions. This behavior is only for functions that map vectors to scalars.

The usual example is max. The expression max(x) gives a scalar that is maximum element of the vector x. The expression max(x,y) gives the elementwise maximum between two vectors.

Thus, norm(x,y) produces a vector

[ ||(x_1, y_1)||_2, ||(x_2, y_2)||_2, ... ]

So the constraint norm(x,1) <= 1 is a vector expression that says ||(x_1,1)||_2 <=1 and ||(x_2,1)||_2 <= 1. This is strange, but consistent. As you stated, norm1 is the thing you're looking for. :)

On Wed, Jul 16, 2014 at 8:27 AM, steve3141 notifications@github.com wrote:

Closed #47 https://github.com/cvxgrp/qcml/issues/47.

— Reply to this email directly or view it on GitHub https://github.com/cvxgrp/qcml/issues/47#event-142222391.