Given a fixed number $r$, such that a vector $v_i \in \{1,0\}^n$ has exactly $r$ ones and $n-r$ zeroes, and a number of inequalities, (say $I$ is this set of inequalities) representing a set $J$ of such vectors; is there way to reduce the number of inequalities $|I|$ representing the set $J$? That is, is there an algorithm to remove redundant inequalities or create a smaller set of inequalities that represents the same set of vectors?
Also, the inequalities are all linear.(That is there is only addition present).
Thanks.
For even $n$, the least number of $0$s in the final product is $|r-\frac{n}{2}|\cdot 2$. If $r$ isn't $\frac{n}{2}$, then no matter how you arrange your ones and zeros, there's going to be $2\cdot|r-\frac{n}{2}|$ zeros and ones facing another zero or one. So, for even $n$, your outcome space is the set of vectors with an even number in $[n, 2\cdot|r-\frac{n}{2}|]$ number of zeros.
For odd $n$, the solution set is vectors such that they have an odd number of zeros in $[n, 2\cdot|r-\lfloor\frac{n}{2}\rfloor|]$.