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Probably this is well known. There is probabilistic reduction from SAT to Unique SAT (0 or 1 solutions).

According to answer and comments derandomizing the reduction would imply $PH \subseteq \oplus P$.

There is simple reduction from SAT to 0,1 integer linear program with zero or one solutions.

Given SAT formula $\phi$ on variables $x_1,\ldots x_n$, introduce binary variables $b_i \in \{0,1\}$.

Define map literals to linear polynomials $f(x_i)=b_i$ and $f(\lnot x_i)=1-b_i$.

For a clause $l_1 \lor \cdots \lor l_k$ add constraint $\sum_{i=1}^k f(l_i) \ge 1$.

So far the solutions of the constraints are in one to one correspondence with the solutions of $\phi$.

To make the solution unique, add the optimization

$$\text{maximize} \sum_{i=1}^n 2^{(i-1)}b_i$$

Two sets of distinct powers of two have equal sums iff the sets are equal, so the objective function have unique maximum if the constraints are satisfied. The ordering of $x_i$ doesn't matter. So the 0,1 ILP has zero or one solutions and it is not easier than NP-complete and coNP-complete.

In what complexity class is solving the 0,1 ILP?

As far as I know ILP are not a complexity class.

$\log_2(n)$ calls to NP oracle would solve it via binary search, though I don't see a direct reduction to SAT.

This appears close to UP though there is no certificate.

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up vote 10 down vote accepted

OptP-complete. Krentel showed that MAX-SAT, finding the lexicographically maximum satisfying assignment, is OptP-complete and the reduction above reduces Max-SAT to ILP. ILP sits in OptP pretty much by definition.

Note that you need n calls to an NP-oracle to solve ILP via binary search, O(log n) isn't sufficient.

There really isn't much of a connection to UP, though Mulmuley, Vazirani and Vazirani give an isolation lemma that gives an alternate proof of Valiant-Vazirani via a maximization problem.

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Thank you. log_2(n) was indeed wrong. – joro Jul 17 '14 at 14:07

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