Linear Programming Using Limited-Precision Oracles

Abstract

Linear programming is a foundational tool for many aspects of integer and combinatorial optimization. This work studies the complexity of solving linear programs exactly over the rational numbers through use of an oracle capable of returning limited-precision LP solutions. Under mild assumptions, it is shown that a polynomial number of calls to such an oracle and a polynomial number of bit operations, is sufficient to compute an exact solution to an LP. Previous work has often considered oracles that provide solutions of an arbitrary specified precision. While this leads to polynomial-time algorithms, the level of precision required is often unrealistic for practical computation. In contrast, our work provides a foundation for understanding and analyzing the behavior of the methods that are currently most effective in practice for solving LPs exactly.

The first author was supported by the Research Campus MODAL funded by the German Ministry of Education and Research under grant number 05M14ZAM.

Appendix

This appendix collects proofs for the main results used in the paper.

Proof of Theorem 2

Proof

Line 10 ensures that at iteration k, $\varDelta _k \leqslant 2^{\lceil \log \alpha \rceil (k-1)}$ and thus $\langle \varDelta _k\rangle \leqslant \lceil \log \alpha \rceil k$. From line 15, the entries in the refined solution vectors $x_k,y_k$ have the form $\sum _{j = 1}^k {\varDelta _j}^{-1} \frac{n_j}{2^p}$ with $n_j \in \mathbb {Z}$, $|n_j| \leqslant 2^{2p}$, for $j = 1,\ldots ,k$. With $D_j := \log (\varDelta _j)$ and $a:= \lceil \log (\alpha )\rceil$ this can be rewritten as

$$\begin{aligned} \sum _{j = 1}^k 2^{-D_j} \frac{n_j}{2^p} = \big ( \sum _{j = 1}^k n_j 2^{a(k-1) - D_j} \big ) / 2^{p+ a(k-1)}. \end{aligned}$$

(6)

The latter is a fraction with integer numerator and denominator. The numerator is bounded by $2^{2p+ ak} - 1$. Hence, $\langle x_k\rangle + \langle y_k\rangle \leqslant (n+m) (2 ak + 3p+ 2)$. The numbers stored in $\hat{ b}, \hat{l}, \hat{ c}$ and $\delta _k$ satisfy the same asymptotic bound. By Lemma 1, the maximum number of iterations is $O(\log (1/\tau )) = O(\langle \tau \rangle )$. All in all, the size of the numbers encountered during the algorithm is $O(\langle A,b,\ell ,c\rangle + (n+m)\langle \tau \rangle )$.

Finally, the total number of elementary operations is polynomially bounded as follows. The initial rounding of the constraint matrix costs $O(nnz)$ elementary operations, where nnz is the number of nonzero entries in A. For each of the $O(\langle \tau \rangle )$ refinements, computing residual errors, maximum violation, and checking termination involves $O(nnz+n+m)$ operations; the computation of the scaling factors takes constant effort, and the update of the transformed problem and the correction of the primal–dual solution vectors costs $O(n+m)$ operations.    $\square$

Proof of Theorem 4

Proof

Let $\varepsilon := \max \{\eta , 1/\alpha \}$, where $\eta$ is the feasibility tolerance of the LP oracle and $\alpha$ is the scaling limit used in Algorithm 1. Conditions (3a–3c) of Theorem 3 follow directly from Lemma 1. We prove conditions (4a) and (4b) by induction. For $k=1$, they follow from (2a) and (2b). Suppose they hold for $k \geqslant 1$. Let $\hat{ x},\hat{ y}, \hat{\mathcal {B}}$ be the last approximate solution returned by the oracle. Then for all $i\not \in \mathcal {B}_{k+1}$

proving (4a). The induction step for (4b) is analogous. Thus, the sequence of basic solutions $x_k,y_k,\mathcal {B}_k$ satisfies the conditions of Theorem 3 and $\mathcal {B}_k$ is optimal after a polynomial number of refinements. According to Theorem 2, this runs in oracle-polynomial time. The linear systems used to compute the basic solutions exactly over the rational numbers can be solved in polynomial time [9].    $\square$

Proof of Lemma 3

Proof

This result inherently relies on the boundedness of the corrector solutions returned by the oracle. Since their entries are in $\mathbb {F}(p)$, $||(\hat{x}_k,\hat{y}_k)||_\infty \leqslant 2^p$. Then $(x_k,y_k) = \sum _{i = 1}^k \frac{1}{\varDelta _{i}} (\hat{x}_i,\hat{y}_i)$ constitutes a Cauchy sequence: for any $k,k' \geqslant K$, $||(x_k,y_k) - (x_{k'},y_{k'})||_\infty \leqslant 2^p\sum _{i = K+1}^\infty \varepsilon ^i = 2^p\varepsilon ^{K+1} / (1 - \varepsilon )$, where $\varepsilon$ is the rate of convergence from Lemma 1. Thus, a unique limit point $(\tilde{x},\tilde{y})$ exists. Using proof techniques as for Theorem 3 one can show that $(\tilde{x},\tilde{y})$ is basic, hence rational.    $\square$

Proof of Theorem 5

Proof

Note that $\varDelta _{k} / \varDelta _{k+1} \leqslant \varepsilon$ for all k implies $\varDelta _{i} \geqslant \varDelta _{j} \varepsilon ^{j-i}$ for all $j \leqslant i$. Then $M_k = \sqrt{\varDelta _{k+1} / (2 \beta ^k)} \geqslant \tilde{q}$ holds if $\sqrt{1/(2 \beta ^k \varepsilon ^k)} \geqslant \tilde{q}$. This holds for all

$$\begin{aligned} k \geqslant K_1 := (2\log \tilde{q} + 1) / \log ( 1/(\beta \varepsilon ) ) \in O(\langle \tilde{q}\rangle ). \end{aligned}$$

(7)

Furthermore, $C\sum _{i=k+1}^{\infty } \varDelta _i^{-1} \leqslant C\sum _{i=k+1}^{\infty } \varepsilon ^{i-k-1} \varDelta _{k+1}^{-1} = C / ((1-\varepsilon ) \varDelta _{k+1})$, which is less than $1 / (2M_k^2) = \beta ^k / \varDelta _{k+1}$ for all

$$\begin{aligned} k > K_2 := (\log C - \log (1-\varepsilon )) / \log \beta \in O(\langle C\rangle ). \end{aligned}$$

(8)

Hence (5) holds for all $k > K := \max \{ K_1, K_2 \}$.    $\square$

Proof of Lemma 4

Proof

From the proof of Theorem 2 we know that at the k-th iteration of the algorithm, the encoding length of the components of $x_k,y_k$ are each bounded by $(2\alpha k +3 p + 2)$, which is O(k) as $\alpha ,p$ are constants. Together with the fact that rational reconstruction can be performed in quadratic time, as discussed above, the first result is established.

To show the second claim we assume that $f>1$ and let K be the final index at which rational reconstruction is attempted. Then if we consider the sequence of indices at which rational reconstruction was applied, that sequence is term-wise bounded above by the following sequence (given in decreasing order): $S = (K, \lfloor K /f\rfloor , \lfloor K /f^2\rfloor ,\ldots , \lfloor K /f^a\rfloor )$ where $a=\lceil \log _f{K}\rceil$. Thus, the cost to perform rational reconstruction at iterations indexed by the sequence S gives an upper bound on the total cost. Again, using the quadratic bound on the cost for rational reconstruction, we arrive at the following cumulative bound, involving a geometric series:

$$\begin{aligned} O\left( (n+m)\sum _{i=0}^a\lfloor K/f^i\rfloor ^2\right)&= O\left( (n+m)K^2 \sum _{i=0}^a f^{-2i}\right) =O((n+m)K^2). \end{aligned}$$

This establishes the result.    $\square$