# Consequences of faster parameterized integer programming

Integer programming in $k$ variables can be done in $k^{O(k)}$ time and $O(k^c)$ space.

Is there any consequence if it can be done in $k^{O(k^\alpha)}$ time and $O(k^c)$ space for some $\alpha\in(0,1)$ under the assumption the integer programming problem we have had at most $1$ solution?

An instance of CNF-SAT with $k$ variables can easily be written as a 0/1 integer linear program over the same variable set, since a clause such as $x_1 \vee x_3 \vee \neg x_4 \vee \neg x_6$ naturally corresponds to a constraint $x_1 + x_3 + (1-x_4) + (1-x_6) \geq 1$, when all variables are forced to take values $0$ and $1$.
Hence if integer programming in $k$ variables can be solved in time $k^{O(k^\alpha)}$ for some $\alpha < 1$, then CNF-SAT with arbitrarily long clauses can be solved in $k^{O(k^\alpha)} = 2^{O(k^\alpha \cdot \log k)}$, which would contradict the Strong Exponential Time Hypothesis since $c \cdot k^\alpha \cdot \log k < k$ for any constant $c$, $\alpha < 1$, and sufficiently large $k$.
• @Ricky Demer Yes, the statement is that for every k and $\epsilon$, there is a k' such a $c^n$ time algorithm for unique k'-SAT leads to a $c^{n(1+\epsilon)}$ time algorithm for k-SAT. Set k=3 and let $\epsilon$ tend to 0 to get the ILP consequence. – daniello Mar 8 '17 at 18:41