Scaling, proximity, and optimization of integrally convex functions

Abstract

In discrete convex analysis, the scaling and proximity properties for the class of L$^{\natural }$-convex functions were established more than a decade ago and have been used to design efficient minimization algorithms. For the larger class of integrally convex functions of n variables, we show here that the scaling property only holds when $n \le 2$, while a proximity theorem can be established for any n, but only with a superexponential bound. This is, however, sufficient to extend the classical logarithmic complexity result for minimizing a discrete convex function of one variable to the case of integrally convex functions of any fixed number of variables.

Appendices

Appendix A: An alternative Proof of Theorem 2.4

Here is a proof of Theorem 2.4 (local characterization of integral convexity) that is shorter than the original proof in [2] and valid for functions defined on general integrally convex sets rather than discrete rectangles.

Obviously, (a) implies (b). The proof for the converse, (b) $\Rightarrow$ (a), is given by the following two lemmas, where integral convexity of $\mathrm{dom\,}f$ and condition (b) are assumed.

Lemma A.1

Let $B \subseteq {{\mathbb {R}}}^{n}$ be a box of size two with integer vertices, i.e., $B = [ {a}, {a} + 2 \varvec{1} ]_{{{\mathbb {R}}}}$ for some ${a} \in {{\mathbb {Z}}}^{n}$. Then ${\tilde{f}}$ is convex on $B \cap \overline{\mathrm{dom\,}f}$.

Proof

First, the assumed integral convexity of $\mathrm{dom\,}f$ implies that $B \cap \overline{\mathrm{dom\,}f} = \overline{B \cap \mathrm{dom\,}f}$ and that every point in $B \cap \overline{\mathrm{dom\,}f}$ can be represented as a convex combination of points in $B \cap \mathrm{dom\,}f$. We may assume $B = [ \varvec{0}, 2 \varvec{1} ]_{{{\mathbb {R}}}}$. To prove by contradiction, assume that there exist $x \in B \cap \overline{\mathrm{dom\,}f}$ and $y^{1},\ldots , y^{m} \in B \cap \mathrm{dom\,}f$ such that

$$\begin{aligned} x = \sum _{i=1}^{m} \lambda _{i} y^{i}, \qquad {\tilde{f}}(x) > \sum _{i=1}^{m} \lambda _{i} f(y^{i}), \end{aligned}$$

(A.1)

where $\sum _{i=1}^{m} \lambda _{i} = 1$ and $\lambda _{i} > 0 \ (i=1,\ldots , m)$. We may also assume $x \in [ \varvec{0}, \varvec{1} ]_{{{\mathbb {R}}}}$ without loss of generality. For each $j=1,\ldots , n$, we look at the j-th component of the generating points $y^{i}$ to define

$$\begin{aligned} I_{j}^{0} = \left\{ i \mid y^{i}_{j} = 0 \right\} , \qquad I_{j}^{2} = \left\{ i \mid y^{i}_{j} = 2 \right\} . \end{aligned}$$

Since $x_{j} = \sum _{i=1}^{m} \lambda _{i} y^{i}_{j} \le 1$, if $I_{j}^{2} \not = \emptyset$, then $I_{j}^{0} \not = \emptyset$.

Let $j=n$ and suppose that $I_{n}^{2} \not = \emptyset$. Then $I_{n}^{0} \not = \emptyset$. We may assume $y^{1}_{n} = 0$, $y^{2}_{n} = 2$; $\lambda _{1} > 0$, $\lambda _{2} > 0$. By (2.2) for $(y^{1}, y^{2})$ and the definition of ${\tilde{f}}$ we have

$$\begin{aligned} f(y^{1}) + f(y^{2}) \ge 2 {\tilde{f}}\, \bigg (\frac{y^{1} + y^{2}}{2} \bigg ) = 2 \sum _{k=1}^{l} \mu _{k} f(z^{k}), \end{aligned}$$

where

$$\begin{aligned} \frac{y^{1} + y^{2}}{2} = \sum _{k=1}^{l} \mu _{k} z^{k}, \qquad z^{k} \in N \bigg ( \frac{y^{1} + y^{2}}{2} \bigg ) \cap \mathrm{dom\,}f \quad (k=1,\ldots , l)\qquad \end{aligned}$$

(A.2)

with $\mu _{k} > 0$$(k=1,\ldots , l)$ and $\sum _{k=1}^{l} \mu _{k} = 1$. This implies, with notation $\lambda = \min (\lambda _{1}, \lambda _{2})$, that

$$\begin{aligned} \lambda _{1} f(y^{1}) + \lambda _{2} f(y^{2}) \ge (\lambda _{1} - \lambda ) f(y^{1}) + (\lambda _{2}-\lambda ) f(y^{2}) + 2 \lambda \sum _{k=1}^{l} \mu _{k} f(z^{k}). \end{aligned}$$

Hence

$$\begin{aligned} \sum _{i=1}^{m} \lambda _{i} f(y^{i}) \ge (\lambda _{1} - \lambda ) f(y^{1}) + (\lambda _{2}-\lambda ) f(y^{2}) + 2 \lambda \sum _{k=1}^{l} \mu _{k} f(z^{k}) + \sum _{i=3}^{m} \lambda _{i} f(y^{i}). \end{aligned}$$

Since

$$\begin{aligned} x = (\lambda _{1} - \lambda ) y^{1} + (\lambda _{2}-\lambda ) y^{2} + 2 \lambda \sum _{k=1}^{l} \mu _{k} z^{k} + \sum _{i=3}^{m} \lambda _{i} y^{i}, \end{aligned}$$

we have obtained another representation of the form (A.1). With reference to this new representation define ${\hat{I}}_{n}^{0}$ (resp., ${\hat{I}}_{n}^{2}$) to be the set of indices of the generators whose n-th component is equal to 0 (resp., 2). Since $z^{k}_{n} = 1$ for all k as a consequence of (A.2) with $(y^{1}_{n}+ y^{2}_{n})/2 = (0 + 2)/2 = 1$, we have ${\hat{I}}_{n}^{0} \subseteq I_{n}^{0}$, ${\hat{I}}_{n}^{2} \subseteq I_{n}^{2}$ and $|{\hat{I}}_{n}^{0}| + |{\hat{I}}_{n}^{2}| \le |I_{n}^{0}| + |I_{n}^{2}| - 1$.

By repeating the above process with $j=n$, we eventually arrive at a representation of the form of (A.1) with $I_{n}^{2} = \emptyset$, which means that $y^{i}_{n} \in \{ 0,1 \}$ for all generators $y^{i}$.

Then we repeat the above process for $j=n-1,n-2, \ldots ,1$, to obtain a representation of the form of (A.1) with $y^{i} \in [ \varvec{0}, \varvec{1} ]_{{{\mathbb {Z}}}}$ for all generators $y^{i}$. This contradicts the definition of ${\tilde{f}}$. $\square$

Lemma A.2

For any $x, y \in \overline{\mathrm{dom\,}f}$, ${\tilde{f}}$ is convex on the line segment connecting x and y.

Proof

Let L denote the (closed) line segment connecting x and y, and consider the boxes B, as in Lemma A.1, that intersect L. There exists a finite number of such boxes, say, $B_{1}, \ldots , B_{m}$, and L is covered by the line segments $L_{j} = L \cap B_{j}$$(j=1,\ldots , m)$. That is, $L = \bigcup _{j=1}^{m} L_{j}$. For each point $z \in L {\setminus } \{ x, y \}$, there exists some $L_{j}$ that contains z in its interior. Since $L_{j} \subseteq L \subseteq \overline{\mathrm{dom\,}f}$, ${\tilde{f}}$ is convex on $L_{j}$ by Lemma A.1. Hence⁴${\tilde{f}}$ is convex on L. $\square$

Appendix B: Proof of Proposition 5.3

It is known (cf. [29, proof of Theorem 16.4]) that the set of integer vectors contained in

$$\begin{aligned} F_{A}= & {} \left\{ \sum _{i \in A} \mu _{i}^{+}(\varvec{1}_{A} + \varvec{1}_{i}) + \sum _{i \in A} \mu _{i}^{-} (\varvec{1}_{A} - \varvec{1}_{i}) + \sum _{i \in N {\setminus } A} \mu _{i}^{\circ } (\varvec{1}_{A}+ \varvec{1}_{i})\right. \\&\left. + \lambda \varvec{1}_{A} \, \left| \begin{array}{l} \mu _{i}^{+}, \mu _{i}^{-} \in [0,1]_{{{\mathbb {R}}}}\;(i \in A); \\ \mu _{i}^{\circ } \in [0,1]_{{{\mathbb {R}}}}\; (i \in N {\setminus } A) ; \\ \lambda \in [0,1]_{{{\mathbb {R}}}} \end{array}\right. \right\} \end{aligned}$$

forms a Hilbert basis of ${\tilde{C}}_{A}$. Let z be an integer vector in $F_{A}$. That is, $z \in {{\mathbb {Z}}}^{n}$ and

$$\begin{aligned} z&= \sum _{i \in A} \mu _{i}^{+}\left( \varvec{1}_{A} + \varvec{1}_{i}\right) + \sum _{i \in A} \mu _{i}^{-} \left( \varvec{1}_{A} - \varvec{1}_{i}\right) + \sum _{i \in N {\setminus } A} \mu _{i}^{\circ } \left( \varvec{1}_{A}+ \varvec{1}_{i}\right) + \lambda \varvec{1}_{A} \end{aligned}$$

(B.1)

$$\begin{aligned}&= \sum _{i \in A} \left( \mu _{i}^{+}-\mu _{i}^{-}\right) \varvec{1}_{i} + \sum _{i \in N {\setminus } A} \mu _{i}^{\circ } \left( \varvec{1}_{A} + \varvec{1}_{i}\right) + \left( \lambda + \sum _{i \in A} \left( \mu _{i}^{+} + \mu _{i}^{-}\right) \right) \varvec{1}_{A}\quad \end{aligned}$$

(B.2)

for some $\mu _{i}^{+}, \mu _{i}^{-} \in [0,1]_{{{\mathbb {R}}}}\;(i \in A)$; $\mu _{i}^{\circ } \in [0,1]_{{{\mathbb {R}}}}\; (i \in N {\setminus } A)$; $\lambda \in [0,1]_{{{\mathbb {R}}}}$. Our goal is to show that z can be represented as a nonnegative integer combination of vectors in $B_{A}$.

First note that $\mu _{i}^{\circ } \in \{ 0, 1 \}$ for each $i \in N {\setminus } A$; define $A^{\circ } = \{ i \in N {\setminus } A \mid \mu _{i}^{\circ }=1 \}$. We denote the coefficient of $\varvec{1}_{A}$ in (B.2) as

$$\begin{aligned} \xi = \lambda + \sum _{i \in A} \left( \mu _{i}^{+} + \mu _{i}^{-}\right) \end{aligned}$$

and divide into cases according to whether $\xi$ is an integer or not.

Case 1 ($\xi \in {{\mathbb {Z}}}$): Using $\xi$ we rewrite (B.2) as

$$\begin{aligned} z&= \sum _{i \in A} \left( \mu _{i}^{+}-\mu _{i}^{-}\right) \varvec{1}_{i} + \sum _{i \in N {\setminus } A} \mu _{i}^{\circ } (\varvec{1}_{A} + \varvec{1}_{i}) + \xi \varvec{1}_{A}, \end{aligned}$$

in which $\xi$ is an integer. For each $i \in A$, $\mu _{i}^{+}-\mu _{i}^{-}$ must be an integer, which is equal to 0, 1 or $-1$. Accordingly we define

$$\begin{aligned} A^{=}&= \left\{ i \in A \mid \mu _{i}^{+}-\mu _{i}^{-} = 0 \right\} , \\ A^{>}&= \left\{ i \in A \mid \mu _{i}^{+}-\mu _{i}^{-} = 1 \right\} = \left\{ i \in A \mid \mu _{i}^{+}=1, \ \mu _{i}^{-} = 0 \right\} , \\ A^{<}&= \left\{ i \in A \mid \mu _{i}^{+}-\mu _{i}^{-} = -1 \right\} = \left\{ i \in A \mid \mu _{i}^{+}=0, \ \mu _{i}^{-} = 1 \right\} \end{aligned}$$

to rewrite (B.1) as

$$\begin{aligned} z = \sum _{i \in A^{>}} (\varvec{1}_{A} + \varvec{1}_{i}) + \sum _{i \in A^{<}} (\varvec{1}_{A} - \varvec{1}_{i}) + \sum _{i \in A^{\circ }} (\varvec{1}_{A}+ \varvec{1}_{i}) + \left( \lambda + \sum _{i \in A^{=}} \left( \mu _{i}^{+} + \mu _{i}^{-}\right) \right) \varvec{1}_{A}.\nonumber \\ \end{aligned}$$

(B.3)

Here the coefficient of $\varvec{1}_{A}$ is integral, since

$$\begin{aligned} \lambda + \sum _{i \in A^{=}} (\mu _{i}^{+} + \mu _{i}^{-}) = \xi - \sum _{i \in A^{>} } 1 - \sum _{i \in A^{<}} 1. \end{aligned}$$

Hence (B.3) gives a representation of z as a nonnegative integer combination of vectors in $B_{A}$.

Case 2 ($\xi \not \in {{\mathbb {Z}}}$): Let $\eta$ denote the fractional part of $\xi$, i.e., $\eta = \xi - \lfloor \xi \rfloor$ with $0< \eta < 1$. We rewrite (B.2) as

$$\begin{aligned} z = \sum _{i \in A} (\mu _{i}^{+}-\mu _{i}^{-} + \eta )\varvec{1}_{i} + \sum _{i \in N {\setminus } A} \mu _{i}^{\circ } (\varvec{1}_{A} + \varvec{1}_{i}) + \lfloor \xi \rfloor \varvec{1}_{A}. \end{aligned}$$

(B.4)

For each $i \in A$, $\mu _{i}^{+}-\mu _{i}^{-} + \eta$ must be an integer, which is equal to 1 or 0. Accordingly we define

$$\begin{aligned} A^{+}&= \left\{ i \in A \mid \mu _{i}^{+}-\mu _{i}^{-} + \eta = 1 \right\} , \\ A^{-}&= \left\{ i \in A \mid \mu _{i}^{+}-\mu _{i}^{-} + \eta = 0 \right\} . \end{aligned}$$

Then

$$\begin{aligned} \lfloor \xi \rfloor \ge \min (|A^{+}|, |A^{-}| ), \end{aligned}$$

which follows from

$$\begin{aligned}&\mu _{i}^{+} + \mu _{i}^{-} \left\{ \begin{array}{ll} = 2\mu _{i}^{-} + 1-\eta \ge 1 - \eta &{} (i \in A^{+}) \\ = 2\mu _{i}^{+} + \eta \ge \eta &{} (i \in A^{-}), \end{array} \right. \\&\xi = \lambda + \sum _{i \in A} \left( \mu _{i}^{+} + \mu _{i}^{-}\right) \ge (1 - \eta ) |A^{+}| + \eta |A^{-}| \ge \min (|A^{+}|, |A^{-}| ). \end{aligned}$$

In the case of $|A^{+}| \le |A^{-}|$, we see from (B.4) that

$$\begin{aligned} z&= \sum _{i \in A^{+}} \varvec{1}_{i} + \sum _{i \in A^{\circ }} (\varvec{1}_{A}+ \varvec{1}_{i}) + \lfloor \xi \rfloor \varvec{1}_{A} \\&= \sum _{i \in A^{+}} (\varvec{1}_{A} + \varvec{1}_{i}) + \sum _{i \in A^{\circ }} (\varvec{1}_{A}+ \varvec{1}_{i}) + (\lfloor \xi \rfloor - |A^{+}|) \varvec{1}_{A}, \end{aligned}$$

which is a nonnegative integer combination of vectors in $B_{A}$. In the other case with $|A^{+}| > |A^{-}|$, we have an alternative expression

$$\begin{aligned} z&= - \sum _{i \in A^{-}} \varvec{1}_{i} + \sum _{i \in A^{\circ }} (\varvec{1}_{A}+ \varvec{1}_{i}) + (\lfloor \xi \rfloor +1) \varvec{1}_{A} \\&= \sum _{i \in A^{-}} (\varvec{1}_{A} - \varvec{1}_{i}) + \sum _{i \in A^{\circ }} (\varvec{1}_{A}+ \varvec{1}_{i}) + (\lfloor \xi \rfloor + 1 - |A^{-}|) \varvec{1}_{A}, \end{aligned}$$

which is also a nonnegative integer combination of vectors in $B_{A}$. This completes the proof of Proposition 5.3.