A Randomized Sieving Algorithm for Approximate Integer Programming

Abstract

The Integer Programming Problem (IP) for a polytope $P \subseteq \mathbb{R} ^{n}$ is to find an integer point in P or decide that P is integer free. We give a randomized algorithm for an approximate version of this problem, which correctly decides whether P contains an integer point or whether a (1+ϵ)-scaling of P about its center of gravity is integer free in 2^O(n)(1/ϵ ²)ⁿ-time and 2^O(n)(1/ϵ)ⁿ-space with overwhelming probability. Our algorithm proceeds by reducing the approximate IP problem to an approximate Closest Vector Problem (CVP) under a “near-symmetric” norm. Our main technical contribution is an extension of the AKS randomized sieving technique, first developed by Ajtai et al. (Proceedings of 33rd Symposium on the Theory of Computing (STOC), pp. 601–610, 2001) for lattice problems under the ℓ ₂ norm, to the setting of asymmetric norms. We also present an application of our techniques to exact IP, where we give a nearly optimal algorithmic implementation of the Flatness Theorem, a central ingredient for many IP algorithms. Our results also extend to general convex bodies and lattices.

Appendix

1.1 A.1 Preliminaries

Logconcave Functions

A function $f: \mathbb{R} ^{n} \rightarrow \mathbb{R} _{+}$ is logconcave if for all $\mathbf{x} , \mathbf{y} \in \mathbb{R} ^{n}$, 0≤α≤1, we have that f(x)^α f(y)^1−α≤f(α x+(1−α)y). For a convex body $K \subseteq \mathbb{R} ^{n}$, the indicator function I _K of K is easily seen to be logconcave. A logconcave density $f: \mathbb{R} ^{n} \rightarrow \mathbb{R} _{+}$ is a logconcave function for which $\int_{ \mathbb{R} ^{n}} f( \mathbf{x} )d \mathbf{x} = 1$.

A logconcave random variable $X \in \mathbb{R} ^{n}$ is logconcave if X admits a logconcave density $f: \mathbb{R} ^{n} \rightarrow \mathbb{R} _{+}$. We note that the uniform distribution over K admits a logconcave density, i.e. $\pi _{K}( \mathbf{x} ) = \frac{1}{\mathrm{vol}_{n}(K)}I_{K}[ \mathbf{x} ]$ for $\mathbf{x} \in \mathbb{R} ^{n}$. A classical fact is that for two logconcave random variables $X,Y \in \mathbb{R} ^{n}$, the sum X+Y is also a logconcave random variable. The random variable X (with density f) is isotropic if $\operatorname{E}[X] = \int_{ \mathbb{R} ^{n}} \mathbf {x} f( \mathbf{x} )d \mathbf{x} = \mathbf{0}$ (mean zero), and if $\operatorname{E}[XX^{t}] = (\int_{ \mathbb{R} ^{n}} \mathbf{x} _{i} \mathbf{x} _{j} f( \mathbf{x} )d \mathbf{x} )_{ij} = I_{n}$ (covariance matrix identity), the n×n identity. For any full-dimensional random variable $X \in \mathbb{R} ^{n}$, there exists an affine transformation $T: \mathbb{R} ^{n} \rightarrow \mathbb{R} ^{n}$ (unique up to rotations) such that TX is isotropic. A convex body K is isotropic if the uniform distribution over K is isotropic.

We will need the following two theorems in the proof of Lemma 2. The first theorem due to Kannan, Lovász and Simonovits gives sandwiching estimates for isotropic convex bodies.

Theorem 10

(Isotropic Sandwiching [28])

Let $K \subseteq \mathbb{R} ^{n}$ be an isotropic convex body. Then

$$\sqrt{\frac{n+2}{n}} \cdot B_2^n \subseteq K \subseteq\sqrt{n(n+2)} \cdot B_2^n . $$

The following theorem of Paouris gives strong concentration estimates for logconcave random variables.

Theorem 11

(Measure Concentration [29], Theorem 11)

Let $X \in \mathbb{R} ^{n}$ be an isotropic logconcave random variable. Then for some absolute constant c>0,

$$\Pr[\|X\| \geq c\sqrt{n}t] \leq e^{-\sqrt{n}t} $$

for all t≥1.

Proof of Lemma 2: Approx. Barycenter

Let X ₁,…,X _N denote iid uniform samples over $K \subseteq \mathbb{R} ^{n}$, where $N = (\frac{2c}{\epsilon } )^{2} n$. We will show that for $\mathbf{b} = \frac{1}{N} \sum_{i=1}^{n} X_{i}$, that the following holds

$$ \Pr\bigl[\big\|\pm\bigl( \mathbf{b} - \mathbf{b} (K)\bigr)\big\| _{K- \mathbf{b} (K)} > \epsilon \bigr] \leq4^{-n} $$

(18)

Since the above statement is invariant under affine transformations, we may assume K is isotropic, i.e. $\mathbf{b} (K) = \operatorname{E}[X_{1}] = \mathbf{0}$, the origin, and $\operatorname{E}[X_{1}X_{1}^{t}] = I_{n}$, the n×n identity. Since K is isotropic, from Theorem 10 we have that $B_{2}^{n} \subseteq K$. Therefore to show (18) it suffices to prove that Pr[∥b∥₂>ϵ]≤4⁻ⁿ. Since the X _i ’s are iid isotropic random vectors, we see that $\operatorname{E}[ \mathbf{b} ] = \frac{1}{N} \sum_{i=1}^{N} \operatorname{E}[X_{i}] = 0$ and

$$\operatorname{E}\bigl[ \mathbf{b} \mathbf{b} ^t \bigr] = \frac{1}{N^2} \sum_{i,j \in[N]} E \bigl[X_iX_j^t\bigr] = \frac{1}{N^2} \sum_{i=1}^n E\bigl[X_iX_i^t \bigr] = \frac{1}{N} I_n $$

Now since the X _i s are logconcave, we have that b is also logconcave. Note that the random variable $\sqrt{N} \mathbf{b}$ is logconcave and isotropic, since $\operatorname{E}[\sqrt{N} \mathbf {b} (\sqrt{N} \mathbf{b} )^{t}] = N \operatorname {E}[ \mathbf{b} \mathbf{b} ^{t}] = I_{n}$. Therefore, by the concentration inequality of Paouris Theorem 11 we have that

$$\Pr[\| \mathbf{b} \|_2 > \epsilon ] = \Pr[\|\sqrt {N} \mathbf{b} \|_2 > \sqrt{N}\epsilon ] = \Pr[\|\sqrt{N} \mathbf{b} \|_2 > 2c\sqrt{n}] < e^{-2n} < 4^{-n} $$

as claimed. To prove the theorem, we note that when switching the X _i ’s from truly uniform to 4⁻ⁿ uniform, the above probability changes by at most $n (\frac{2c}{\epsilon ^{2}} ) 4^{-n}$ by Lemma 1. Therefore the total error probability under 4⁻ⁿ-uniform samples is at most 2⁻ⁿ as needed. □

Convexity

Proof of Lemma 3: Estimates for norm recentering

We have $\mathbf{z} \in \mathbb{R} ^{n}$, x,y∈K satisfying (†) ∥±(x−y)∥ _K−y ≤α<1. We prove the statements as follows:

• 1. ∥z−y∥ _K−y ≤τ⇔(z−y)∈τ(K−y)⇔z∈τK+(1−τ)y as needed.

• 2. Let τ=∥z−x∥ _K−x . Then by (1), we have that z∈τK+(1−τ)x. Now note that

  $$(1-\tau) ( \mathbf{x} - \mathbf{y} ) \subseteq|1-\tau| \alpha(K- \mathbf{y} ) $$

  by assumption (†) and (1). Therefore

  $$\begin{aligned} \mathbf{z} \in\tau K + (1-\tau) \mathbf {x} &= \tau K + (1-\tau) \mathbf{y} + (1-\tau ) ( \mathbf{x} - \mathbf{y} ) \\ &\subseteq\tau K + (1-\tau) \mathbf{y} + \alpha|1-\tau |(K- \mathbf{y} ) \\ &= (\tau+ \alpha|1-\tau|) K + (1-\tau-\alpha|1-\tau|) \mathbf{y} \end{aligned}$$

  Hence by (1), we have that

  $$\| \mathbf{z} - \mathbf{y} \|_{K-y} \leq\tau + \alpha|1-\tau| = \| \mathbf{z} - \mathbf {x} \| _{K- \mathbf{x} } + \alpha|1-\| \mathbf {z} - \mathbf{x} \|_{K- \mathbf{x} }| $$

  as needed.

• 3. We first show that

  $$\pm( \mathbf{y} - \mathbf{x} ) \in\frac {\alpha}{1-\alpha}(K- \mathbf{x} ) $$

  By (1) and (†) we have that

  $$\begin{aligned} ( \mathbf{x} - \mathbf{y} ) \in\alpha (K- \mathbf{y} ) &\Leftrightarrow( \mathbf {x} - \mathbf{y} ) - \alpha( \mathbf{x} - \mathbf{y} ) \in \alpha(K- \mathbf{y} ) - \alpha( \mathbf {x} - \mathbf{y} ) \\ &\Leftrightarrow(1-\alpha) ( \mathbf{x} - \mathbf{y} ) \in\alpha(K- \mathbf{x} ) \Leftrightarrow( \mathbf{x} - \mathbf{y} ) \in\frac{\alpha}{1-\alpha}(K- \mathbf{x} ) \end{aligned}$$

  as needed. Next since 0≤α≤1, we have that |1−2α|≤1. Therefore by (†) we have that

  $$(1-2\alpha) ( \mathbf{y} - \mathbf{x} ) \in |1-2\alpha| \alpha(K- \mathbf{y} ) \subseteq\alpha(K- \mathbf{y} ) $$

  since 0∈K−y. Now note that

  $$\begin{aligned} &(1-2\alpha) ( \mathbf{y} - \mathbf{x} ) \in \alpha(K- \mathbf{y} )\\ &\quad\Leftrightarrow (1-2\alpha) ( \mathbf{y} - \mathbf{x} ) + \alpha( \mathbf{y} - \mathbf{x} ) \in\alpha (K- \mathbf{y} ) + \alpha( \mathbf {y} - \mathbf{x} ) \\ &\quad\Leftrightarrow(1-\alpha) ( \mathbf{y} - \mathbf{x} ) \in\alpha(K- \mathbf{x} ) \Leftrightarrow( \mathbf{y} - \mathbf{x} ) \in\frac{\alpha}{1-\alpha}(K- \mathbf{x} ) \end{aligned}$$

  as needed.

  Let τ=∥z−y∥ _K−y . Then by (1), we have that z∈τK+(1−τ)y. Now note that

  $$\begin{aligned} \mathbf{z} \in\tau K + (1-\tau) \mathbf {y} &= \tau K + (1-\tau) \mathbf{x} + (1-\tau ) ( \mathbf{y} - \mathbf{x} ) \\ &\subseteq\tau K + (1-\tau) \mathbf{x} + \frac{\alpha }{1-\alpha}|1-\tau | (K- \mathbf{x} ) \\ &= \biggl(\tau+ \frac{\alpha}{1-\alpha} |1-\tau|\biggr) K + \biggl(1-\tau- \frac {\alpha}{1-\alpha}|1-\tau|\biggr) \mathbf{x} \end{aligned}$$

  Hence by (1), we have that

  $$\| \mathbf{z} - \mathbf{x} \|_{K- \mathbf{x} } \leq\tau+ \frac{\alpha}{1-\alpha} |1-\tau| = \| \mathbf{z} - \mathbf{y} \| _{K- \mathbf{y} } + \frac{\alpha}{1-\alpha} |1-\| \mathbf{z} - \mathbf{y} \| _{K- \mathbf{y} }| $$

  as needed.

 □

Proof of Corollary 1: Stability of symmetry

We claim that (1−∥x∥ _K )(K∩−K)⊆(K−x)∩(x−K). Take z∈K∩−K, then note that

$$\begin{aligned} \| \mathbf{x} + (1-\| \mathbf{x} \| _K) \mathbf{z} \|_K &\leq\| \mathbf{x} \|_K + (1-\| \mathbf{x} \| _K)\| \mathbf{z} \|_K \leq\| \mathbf{x} \|_K + (1- \| \mathbf{x} \|_K)\| \mathbf{z} \|_{K \cap -K} \\ &\leq\| \mathbf{x} \|_K + (1-\| \mathbf{x} \| _K) = 1 \end{aligned}$$

hence x+(1−∥x∥ _K )(K∩−K)⊆K⇔(1−∥x∥ _K )(K∩−K)⊆K−x. Next note that

$$\begin{aligned} &\|- \mathbf{x} + (1-\| \mathbf{x} \| _K) \mathbf{z} \|_{-K} \\&\quad\leq\|- \mathbf{x} \| _{-K} + (1-\| \mathbf{x} \|_K)\| \mathbf{z} \|_{-K} \leq\| \mathbf{x} \|_K + (1-\| \mathbf{x} \|_K)\| \mathbf{z} \|_{K \cap-K} \\ &\quad\leq\| \mathbf{x} \|_K + (1-\| \mathbf{x} \| _K) = 1 \end{aligned}$$

hence −x+(1−∥x∥ _K )(K∩−K)⊆−K⇔(1−∥x∥ _K )(K∩−K)⊆x−K, as needed. Now we see that

$$\mathrm{vol}_n\bigl((K- \mathbf{x} ) \cap( \mathbf{x} -K)\bigr) \geq\mathrm{vol}_n\bigl((1-\| \mathbf{x} \|_K) (K \cap-K)\bigr) = (1-\| \mathbf{x} \|_K)^n \mathrm{vol}_n(K \cap-K) $$

and so the claim follows from Theorem 7. □

1.2 A.2 Integer Programming

Proof of Lemma 4: Well-Centered Bodies

Clearly for any $\mathbf{a} _{0}' \in K_{a,b}$ we have that $K_{a,b} \subseteq \mathbf{a} _{0}' + 2RB_{2}^{n}$, since $K_{a,b} \subseteq K \subseteq \mathbf{a} _{0} + RB_{2}^{n}$. Therefore the we need only worry that K _a,b contains a polynomially sized ball around an easy to compute point in K _a,b . Since a,b correspond to m,u in the while loop (lines 6–13 in Algorithm 1), we have that

$$\begin{aligned} b-a \geq\frac{\delta}{2} \quad\quad \langle \mathbf {x} _l, \mathbf{v} \rangle + \frac {3\delta}{16} \leq b \quad\quad a \leq \langle \mathbf{x} _u, \mathbf{v} \rangle - \frac {7\delta}{16}, \end{aligned}$$

where the last inequality follows since $a+\frac{\delta}{2} \leq b \leq \langle \mathbf{x} _{u}, \mathbf {v} \rangle+\frac{\delta}{16}$. Since $\mathbf{a} _{0}+rB_{2}^{n} \subseteq K$, by assumption $\delta\leq\frac{1}{64} r \| \mathbf{v} \| _{2}$, we also have that

$$\mathrm{width}_K( \mathbf{v} ) \geq\mathrm{width}_{ \mathbf{a} _0+rB_2^n}( \mathbf{v} ) = 2r\| \mathbf{v} \|_2 \geq128 \delta $$

From the above, we additionally conclude that

$$\langle \mathbf{x} _l, \mathbf{v} \rangle \leq \langle \mathbf{a} _0, \mathbf{v} \rangle \leq \langle \mathbf{x} _u, \mathbf{v} \rangle \quad\text{and} \quad \langle \mathbf{x} _u, \mathbf{v} \rangle - \langle \mathbf{x} _l, \mathbf{v} \rangle \geq\mathrm{width}_K( \mathbf{v} ) - \frac{\delta}{8} \geq 127 \delta $$

Let I denote the interval [〈x _l ,v〉,〈x _u ,v〉]∩[a,b]. Combining the above inequalities, it is not hard to check that $\mathrm{length}(I) \geq\frac{3\delta}{16}$ (corresponding to having b shifted as far to the left as possible). Let I _l =[〈x _l ,v〉,〈x,a ₀〉]∩I and I _u =[〈a ₀,v〉,〈x _u ,v〉]∩I. Since I _l and I _u partition I, we must have that either $\mathrm{length}(I_{l}) \geq\frac{3\delta}{32}$ or $\mathrm{length}(I_{u}) \geq\frac{3\delta}{32}$. Assume we are in the former case (the analysis for the latter case is symmetric). Let c denote the midpoint of I _l . Define $\mathbf{a} _{0}'$ as

$$\mathbf{a} _0' = \frac{ \langle \mathbf{a} _0, \mathbf{v} \rangle-c}{ \langle \mathbf{a} _0- \mathbf{x} _l, \mathbf{v} \rangle} \mathbf{x} _l + \frac{c- \langle \mathbf{x} _l, \mathbf{v} \rangle}{ \langle \mathbf{a} _0- \mathbf{x} _l, \mathbf{v} \rangle} \mathbf{a} _0, $$

where we note that $\mathbf{a} _{0}'$ can easily be computed in polynomial time. Since $\mathbf{a} _{0}'$ is a convex combination of x _l and a ₀, and since $\langle \mathbf{a} _{0}', \mathbf {v} \rangle = c \in[a,b]$, we have that $\mathbf{a} _{0}' \in K_{a,b}$. Next by our assumption that $\mathrm{length}(I_{l}) \geq \frac{3\delta}{32}$, that c is the midpoint of I _l , and that ∥x _l −a ₀∥₂≤R, we have that

$$\frac{c- \langle \mathbf{x} _l, \mathbf{v} \rangle}{ \langle \mathbf {a} _0- \mathbf{x} _l, \mathbf{v} \rangle} \geq \frac{3\delta}{64R\| \mathbf{v} \|_2} . $$

Since $\mathbf{a} _{0} + rB_{2}^{n} \subseteq K$, we get that

$$\begin{aligned} K &\supseteq\frac{ \langle \mathbf{a} _0, \mathbf{v} \rangle-c}{ \langle \mathbf{a} _0- \mathbf{x} _l, \mathbf{v} \rangle} \mathbf{x} _l + \frac{c- \langle \mathbf{x} _l, \mathbf{v} \rangle }{ \langle \mathbf{a} _0- \mathbf {x} _l, \mathbf{v} \rangle} \bigl( \mathbf{a} _0 + rB_2^n\bigr) \\ &= \mathbf{a} _0' + r \frac{c- \langle \mathbf{x} _l, \mathbf{v} \rangle}{ \langle \mathbf{a} _0- \mathbf{x} _l, \mathbf{v} \rangle} B_2^n \supseteq \mathbf{a} _0' + \frac{3 r \delta}{64 R \| \mathbf{v} \|_2} B_2^n \end{aligned}$$

Furthermore, note that

$$\max\biggl\{{ \langle \mathbf{x} , \mathbf {v} \rangle: \mathbf{x} \in \mathbf {a} _0' + \frac{3 r \delta}{64 R \| \mathbf{v} \|_2} B_2^n}\biggr\} = c + \frac{3 r \delta}{64 R} \leq c + \frac{3\delta}{64} \leq b $$

since c is the midpoint of I _l ⊆[a,b]. By the symmetric argument, we also have that $\min\{{ \langle \mathbf {x} , \mathbf{v} \rangle: \mathbf {x} \in \mathbf{a} _{0}' + \frac{3 r \delta}{64 R \| \mathbf{v} \|_{2}} B_{2}^{n}}\} \geq c - \frac{3\delta}{64} \geq a$. Therefore, we have that

$$\mathbf{a} _0' + \frac{3 r \delta}{64 R \| \mathbf{v} \|_2} B_2^n \subseteq K_{a,b} $$

as needed. □