Given the problem

How could one formulate it in CVX?

I could handle it without the `sqrt`

yet couldn’t do it in this form.

Thank You.

Given the problem

\min_{ x \in {\mathbb{R}}^{n} } \frac{1}{2} {\left\| x - y \right\|}_{2}^{2} + \sum_{i = 1}^{n - 2} \sqrt{{\left( {x}_{i} - {x}_{i + 1} \right)}^{2} + {\left( {x}_{i + 1} - {x}_{i + 2} \right)}^{2}}

How could one formulate it in CVX?

I could handle it without the `sqrt`

yet couldn’t do it in this form.

Thank You.

Hi @Mark_L_Stone, well, I could rewrite the problem as:

\min_{ x \in {\mathbb{R}}^{n} } \frac{1}{2} {\left\| x - y \right\|}_{2}^{2} + \sum_{i = 1}^{n - 2} \sqrt{{\left( {x}_{i} - {x}_{i + 1} \right)}^{2} + {\left( {x}_{i + 1} - {x}_{i + 2} \right)}^{2}} = \min_{ x \in {\mathbb{R}}^{n} } \frac{1}{2} {\left\| x - y \right\|}_{2}^{2} + \sum_{i = 1}^{n - 2} { \left\| {B}_{i} x \right\| }_{2}

Where the matrix {B}_{i} extracts a vector such that

{z}_{i} = {B}_{i} x = \begin{bmatrix}
{x}_{i} - {x}_{i + 1} \\
{x}_{i + 1} - {x}_{i + 2}
\end{bmatrix}

This must be Convex.

Thank You.

Well, there you go. Code that up in a straightforward manner, and CVX should accept it.

Hi @Mark_L_Stone,

That’s the issue, I don’t know how to insert the Loop into CVX.

It can’t be written in one `minimize`

argument.

Thank You.

```
cvx_begin
variable x(n)
Objective = 0.5*sum_square(x-y)
for i=1:n-2
Objective = Objective + norm([x(i)-x(i+1);x(i+1)-x(i+2)])
end
minimize(Objective)
% insert any other constraints
cvx_end
```

Hi @Mark_L_Stone,

It really solved it.

I wasn’t aware it can be done that way (Looping inside CVX Block).

Is there a way to mark question as solved?

Thank You.

There is no marking questions as solved in this forum. Your post stating that it is solved will accomplish that purpose for readers.

Well,

**This question is solved**.

CVX is really awesome.

Just need better support for large size arrays.

Now that you have some experience with CVX, perhaps you should re-read the CVX Users’ Guide, so that you can pick up on things which you didn’t fully absorb the first time.