I am trying to formulate a problem as QUBO problem and am not able to transform the inequality constraint.
$$ \sum_i^N x_i \geq 1 $$ into a suitable penalty function. For N = 2, the penalty term can be written as $$ P(1-x_1-x_2+x_1x_2)$$ as mentioned in A Tutorial on Formulating and Using QUBO Models. Can someone help me in generalising the above constraint to equivalent penalty.
It would help if you used slack variables and squared penalties. One way of doing it is by defining the following penalty $$P\left(\sum_{i=1}^{N}x_i - 1 - Z \right)^2,$$ where $Z$ is an integer variable such that $0 \leq Z \leq N-1$.
If you use the D-Wave API, I believe you can easily define an integer (discrete) variable $Z$. If you use something else, you may need to express $Z$ as an expansion of auxiliary binary variables: $$ Z = \sum_{k=0}^{M-1} 2^ky_k + ry_M, $$ where $M = \lfloor \log_2 Z \rfloor$ and $r$ is the remainder so that $Z \leq N-1$. The remainder can be written as $r = N + 1 - 2^M$. For more details on this, see Section 2.4.
Intuitively, whenever $\sum_i x_i \geq 1$, the values of $Z$ make the entire penalty term disappear. However, when $\sum_i x_i =0$, no matter what $Z$ is, we get the penalty $P(1 + Z)^2$.
The following naive generalization might be useful:
\begin{equation} 1 - \sum_\limits{i=1}^{n}x_i + \sum\limits_{\substack{\{j,k\}\in \{1,...,n\}^2\\ j\not=k}} x_j x_k \end{equation}
If all variables are set to $0$, this expression has a value of $1$. If one or two variables are set to $1$, the expression's value is $0$. If $m>2$ variables are set to $1$, its value is $1-m+$$m \choose 2$. Since $m \choose 2$$\ge m$, the expression's value is at least $1$. Thus, minimizing this expression guarantees that at least one variable is set to $1$. Nevertheless, it also forces your solution to have less than three variables set to $1$, i.e., $\sum\limits_{i=1}^n x_i \le 2$. You could partition your set of variables to avoid the effect of this last extra constraint.
For instance,
\begin{equation} 1-x_1-x_2-x_3+x_1x_2+x_1x_3+x_2x_3 \end{equation}
equals 1 when $x_1=x_2=x_3=1$. Thus, such assignment of values is unfeasible, i.e., it does not correspond to the minimum value of 0.
