NP-completeness proof via reduction

Question

I'm aware that 0-1 integer programming problem is NP-complete, where the problem is stated as: Given some integer matrix A and some integer vector b, determine whether there exists a vector x consisting of 0's and 1's such that Ax >= b. I've seen that 3-CNF SAT is reducible to this problem.

However, here's a slight variant: Given some integer matrix A and some integer vector b, determine whether there exists a vector x consisting of 0's and 2's such that Ax >= b.

The fact that vector x must consist of 0's and 2's kind of throws me off here. For the original problem, we just could reduce from 3-SAT by adding the inequalities 0 <= xi <= 1 for each boolean variable. But here, I can't just add the following inequalities:

0 <= xi <= 2

because x cannot equal 1.

Am I allowed to add an inequality, such as xi != 1? If not, what reduction can I use to show that the given problem is NP-complete? Thanks for any help.

Answer 1

Suppose you are given a 0-1-Integer-Programming instance. You need to substitute $x_i$'s, $x_i \in \{0,1\}$, with new variables $y_i$'s as G. Bach mentioned. The substitution is $x_i = y_i/2$, $y_i \in \{0,2\}$.

By this substitution, you get
0-1-Integer-Programming $\leq_P$ 0-2-Integer-Programming
and also 0-2-Integer-Programming $\leq_P$ 0-1-Integer-Programming by reverse substitution.
(This is because the polynomial reduction function in this case is invertible).

Therefore 0-2-Integer-Programming is also NP-complete
(from 0-1-Integer-Programming $\leq_P$ 0-2-Integer-Programming).

By the way, you are not allowed to add $x_i \neq 1$ in canonical integer programming problem. Basically it means $x_i \leq 0$ OR $x_i \geq 2$. We can't have OR constraints, all we can have is AND constraints.