is a single constraint, which you can represent as , where are vectors comprised of the variables respectively.

To encode this constraint in the quadratic program you need to set a single row of to be the vector , so that

Off-the-shelf QP-solvers have been used in the past, but for many years now dedicated code is used (much faster and more robust). Those solvers are not (general) QP-solvers anymore and are just build for this one use-case.

sklearn's SVC is a wrapper for libsvm (proof).

As the link says:

Since version 2.8, it implements an SMO-type algorithm proposed in this paper:

R.-E. Fan, P.-H. Chen, and C.-J. Lin. Working set selection using second order information for training SVM. Journal of Machine Learning Research 6, 1889-1918, 2005.

(link to paper)

First difference: cvxopt takes in a minimization problem but you are having a maximumzation problem. This can be resolved by multiplying $-1$ to the maximization objective function.

$P$ describes the quadratic part, Hence let $P_{ij}=y_ik(x_i, x_j)y_j$. The linear objective part is $q\in \mathbb{R}^n$ where $q_i = -1$.

Next let's move on to the inequality constraint, which consists of $c_i \le \frac1{2n\lambda}$ and $-c_i \le 0$. Hence, we can let $G=\begin{bmatrix} I \\ -I\end{bmatrix}$. As an exercise, can you try to write down the corresponding $h$?

$b=0$. As an exercise, can you write down the vector $A$?

The CVXOPT software solves quadratic programs of the form $$ \begin{array}{rl} \min\ & \frac{1}{2}x^TPx+q^Tx \\ \text{s.t.}\ & Gx\leq h \\ & Ax=b \end{array} $$

The problem you wish to solve is $$ \begin{array}{rl} \max\ & \sum_i\alpha_i-\frac{1}{2}\big\|\sum_i\alpha_iy_ix_i\big\|^2 \\ \text{s.t.} & -\alpha_i\leq0\text{ for all }i \\ & \alpha_i\leq{C}\text{ for all }i \\ & \sum_iy_i\alpha_i=0 \end{array} $$

I'm assuming here $x_i\in\mathbb{R}^d$ are vectors and $y_i\in\mathbb{R}$ are scalars for $i=1,\dots,n$.

Objective Function

Let's look at the expression

$$\bigg\|\sum_i\alpha_iy_ix_i\bigg\|^2.$$

Let's define the vector $z_i=y_ix_i$ for all $i=1,\dots,n$. Define the matrix $Z\in\mathbb{R}^{d\times{n}}$ to be the matrix whose $i^\text{th}$ column is $z_i$. Then the expression above can be written as

$$ \|Z\alpha\|^2=\alpha^TZ^TZ\alpha $$

where $\alpha\in\mathbb{R}^n$ is the vector of $\alpha_i$ variables, and we are using the two norm.

Hence our objective function is

$$\max\ \sum_i\alpha_i-\frac{1}{2}\bigg\|\sum_i\alpha_iy_ix_i\bigg\|^2=\max\ -\frac{1}{2}\alpha^TZ^TZ\alpha+\mathbf{1}^T\alpha=\min\ \frac{1}{2}\alpha^TZ^TZ\alpha-\mathbf{1}^T\alpha$$

where $\mathbf{1}$ is the vector of all ones. This is exactly the form required by CVXOPT: take $P=Z^TZ$ and $q=-\mathbf{1}$.

Inequality Constraints

The constraints $-\alpha_i\leq0$ can be written as

$$ -I\alpha\leq\mathbf{0} $$

where $I$ is the $n\times{n}$ identity matrix, and $\mathbf{0}$ is the zero vector. Similarly, the constraints $\alpha_i\leq{C}$ can be written in matrix form as

$$ I\alpha\leq C\mathbf{1}. $$

If we stack these two matrix systems on top of each other, then our inequality constraints can be summarized as $$ \begin{bmatrix}-I\\I\end{bmatrix}\alpha\leq\begin{bmatrix}\mathbf{0}\\C\mathbf{1}\end{bmatrix}. $$

This is the form required by CVXOPT: take

$$ G=\begin{bmatrix}-I\\I\end{bmatrix},\ h=\begin{bmatrix}\mathbf{0}\\C\mathbf{1}\end{bmatrix}. $$

Equality Constraints

I think you can take it from here.

Once you have all these matrices and vectors created, just pass them to solvers.qp() and you should be good to go.