To keep the discussion more concrete, suppose we have a variable $x=(x_0,x_1,x_2,x_3,x_4,x_5)$ which consists of 3 groups $(x_0,x_1,x_2)$, $(x_3)$ and $(x_4,x_5)$. Let's say we want to express:

- an upper bound $b_i$ on the total value within each group,
- a joint volatility constraint such as $$\gamma\geq\left\|G\cdot \left[\begin{array}{c} x_0+x_1+x_2-i_0 \\ x_3-i_1 \\ x_4+x_5-i_2\end{array}\right]\right\|_2$$ for some matrix $G$ and constant index weights $i_0,i_1,i_2$
- the constraint that all values within one group have the same sign.

**Pick, slice**

A straightforward solution is to pick (Expr.pick) the content of each group into a separate view and add relevant constraints in a loop over the groups. One could also take slices (Expr.slice) when the groups form contiguous subsequences. This approach is implemented below in Python Fusion.

**Loop-free**

The previous solution requires picking and stacking expressions in a loop over all groups. This can sometimes be slow. A nicer and more efficient solution uses a bit of linear algebra to perform the grouping. We first encode the groups via a sparse

*membership matrix*$\mathrm{Memb}$, where the rows are groups, columns are entries of $x$, and there is a $1$ whenever an entry belongs to a group. In our example $$\mathrm{Memb}=\left[\begin{array}{cccccc}1&1&1&0&0&0\\0&0&0&1&0&0\\0&0&0&0&1&1\end{array}\right] .$$
This matrix is easy to construct in Fusion.

Note that $\mathrm{Memb}\cdot x$ is the vector of all group sums, in our case $$\mathrm{Memb}\cdot x = \left[\begin{array}{c} x_0+x_1+x_2 \\ x_3 \\ x_4+x_5\end{array}\right].$$

That means we can express both of the previous models in a single call without looping. Since $\mathrm{Memb}$ is very sparse, this becomes a very efficient representation. Of course it is important to keep $\mathrm{Memb}$ as a sparse matrix.

**Loop-free same sign**

We can now use the same matrix to model the last problem from the introduction: all entries within each group must have the same sign. We introduce a sequence of binary variables $z=(z_0,\ldots,z_{g-1})$, one for each of the $g$ groups. The $j$-th variable will determine the sign of elements in the $j$-th group. That is imposed by constraints $$-M(1-z_j)\leq x_i\leq Mz_j,$$ whenever $x_i$ belongs to $j$-th group. We can use the pick/slice strategy per group, as before, or observe that $\mathrm{Memb}^T\cdot z$ produces the vector with the correct binary variable for each entry in $x$. I our case, if $z=(z_0,z_1,z_2)$ then $$\mathrm{Memb}^T\cdot z = (z_0,z_0,z_0,z_1,z_2,z_2)^T.$$ Now each inequality in (4) can be written as a single constraint: