Extreme functions with an arbitrary number of slopes

Abstract

For the one dimensional infinite group relaxation, we construct a sequence of extreme valid functions that are piecewise linear and such that for every natural number $k\ge 2$, there is a function in the sequence with k slopes. This settles an open question in this area regarding a universal bound on the number of slopes for extreme functions. The function which is the pointwise limit of this sequence is an extreme valid function that is continuous and has an infinite number of slopes. This provides a new and more refined counterexample to an old conjecture of Gomory and Johnson stating that all extreme functions are piecewise linear. These constructions are extended to obtain functions for the higher dimensional group problems via the sequential-merge operation of Dey and Richard.

A Appendix

Theorem 8

Let $n\ge 1$ be a natural number. A function ${ \theta }:\mathbb {R}^n\rightarrow \mathbb {R}_+$ is minimal/extreme/facet for $I_b$ when $b\in [0,1/2]^n{\setminus }\{0\}$ if and only if ${ \tilde{\theta }}:\mathbb {R}^n\rightarrow \mathbb {R}_+$ defined by ${ \tilde{\theta }}(x) := { \theta }(-x)$ is minimal/extreme/facet for $I_{\mathbf {1}-b}$, respectively, where $\mathbf {1}\in \mathbb {R}^n$ is the vector of all ones.

Proof

The result essentially follows by applying Theorem 8.2 in [16]. However, since that paper considers the so-called mixed-integer problem, while we are looking at the pure integer problem, we include a proof for completeness.

We show one direction as the other follows from swapping the roles of b and $\mathbf {1}-b$. Suppose that ${ \theta }$ is minimal for $I_b$ with $b\in [0,1/2]^n{\setminus }\{0\}$. Define ${ \tilde{\theta }}(x):= { \theta }(-x)$. We check that ${ \tilde{\theta }}$ is minimal using Theorem 1. If $w\in \mathbb {Z}^n$ then so is $-w$, and therefore ${ \tilde{\theta }}(w) = { \theta }(-w) = 0$ since ${ \theta }$ is minimal. Let $x, y\in \mathbb {R}^n$ and note that

$$\begin{aligned} { \tilde{\theta }}(x+y) = { \theta }(-x-y) \le { \theta }(-x)+{ \theta }(-y) = { \tilde{\theta }}(x)+{ \tilde{\theta }}(y), \end{aligned}$$

where the inequality follows from the subadditivity of ${ \theta }$. Hence ${ \tilde{\theta }}$ is subadditive. Finally, let $r\in \mathbb {R}^n$ and note that

$$\begin{aligned} { \tilde{\theta }}(r)+{ \tilde{\theta }}((\mathbf {1}-b)-r) = { \theta }(-r)+{ \theta }(b-(\mathbf {1}-r)) = { \theta }(-r)+{ \theta }(b-(-r)) = 1, \end{aligned}$$

where the second equation follows from the periodicity of ${ \theta }$ and the third equation from the symmetry of ${ \theta }$. Hence ${ \tilde{\theta }}$ is symmetric about $\mathbf {1}-b$. From Theorem 1, ${ \tilde{\theta }}$ is minimal.

Now assume that ${ \theta }$ is extreme. Let $\theta _1, \theta _2$ be valid for $I_{\mathbf {1}-b}$ such that ${ \tilde{\theta }} = \frac{\theta _1+\theta _2}{2}$. We claim that $\tilde{\theta }_i(r):=\theta _i(-r)$, $i=1,2$, is a valid function for $I_b$. This would imply ${ \tilde{\theta }} = \theta _1=\theta _2$ from the extremality of ${ \theta }$. Let $y\in I_b$. Then $\tilde{y}(r) := y(-r)\in I_{\mathbf {1}-b}$. Note that for $i=1,2$,

$$\begin{aligned} \sum _{r\in \mathbb {R}^n}\tilde{\theta }_i(r)y(r) = \sum _{r\in \mathbb {R}^n}\theta _i(-r)y(r) = \sum _{r\in \mathbb {R}^n}\theta _i(-r)\tilde{y}(-r) \ge 1, \end{aligned}$$

since $\theta _i$ is valid for $I_{\mathbf {1}-b}$.

The proof that ${ \theta }$ is a facet if and only if ${ \tilde{\theta }}$ is a facet is similar. $\square$

Proof

We prove this using induction on k. For $k=2$, the result is easily verified using (3). So let $k\ge 3$ and assume that $\pi _{k-1}$ is well-defined, nonnegative, and positive on $\mathbb {R}{\setminus } \mathbb {Z}$. First, we will show that $\pi _k$ is well-defined at the points $\{b\left( \frac{1}{8}\right) ^{k-2}, 2b(\frac{1}{8})^{k-2}, b-2b(\frac{1}{8})^{k-2}, b-b(\frac{1}{8})^{k-2}, b\}$, that is, we will show $\pi _k$ is well-defined at the boundaries of the intervals on which it is defined. This will show that $\pi _k$ is well-defined on [0, 1), and since $\pi _k$ is periodic by definition, it will follow that $\pi _k$ is well-defined everywhere.

Note that

$$\begin{aligned} \frac{4^{2-k}}{1-b}-\left( \frac{1}{1-b}\right) b\left( \frac{1}{8}\right) ^{k-2} = \left( \frac{2^{k-2}-b}{b-b^2}\right) b\left( \frac{1}{8}\right) ^{k-2} = \left( \frac{2^{k-2}-b}{1-b}\right) \left( \frac{1}{8}\right) ^{k-2}>0,\nonumber \\ \end{aligned}$$

(5)

where the inequality follows since $k\ge 3$ and $b\in (0,1)$. Also, observe that

$$\begin{aligned} \frac{1-4^{2-k}}{1-b}-\left( \frac{1}{1-b}\right) \left( b-b\left( \frac{1}{8}\right) ^{k-2}\right)&= \frac{1-2^{k-2}}{1-b}+\left( \frac{2^{k-2}-b}{b-b^2}\right) \left( b-b\left( \frac{1}{8}\right) ^{k-2}\right) \nonumber \\&= \frac{1+b((\frac{1}{8})^{k-2}-1)-4^{2-k}}{1-b}\nonumber \\&\ge \frac{1+\frac{1}{2}((\frac{1}{8})^{k-2}-1)-4^{2-k}}{1-b}\nonumber \\&> 0, \end{aligned}$$

(6)

where the inequalities follow since $b\in (0,\frac{1}{2}]$ and $k\ge 3$. Equations (5) and (6) show that $\pi _k$ is well-defined and positive at the points $b\left( \frac{1}{8}\right) ^{k-2}$ and $b-b\left( \frac{1}{8}\right) ^{k-2}$.

Observe that $b\in I^j_5\cap I^j_6$. Since $\frac{1-2^{k-2}}{1-b}+\left( \frac{2^{k-2}-b}{b-b^2}\right) b = 1$, it follows that $\pi _k$ is well-defined and positive at b.

Notice that $2b\left( \frac{1}{8}\right) ^{k-2}\in I^k_2\subseteq I^{k-1}_1$ by definition. Similarly, $b-2b\left( \frac{1}{8}\right) ^{k-2}\in I^k_4\subseteq I^{k-1}_5$. Therefore, by induction,

$$\begin{aligned} \frac{4^{2-k}}{1-b}-\left( \frac{1}{1-b}\right) 2b\left( \frac{1}{8}\right) ^{k-2} = \pi _{k-1}\left( 2b\left( \frac{1}{8}\right) ^{k-2}\right) >0, \end{aligned}$$

(7)

and

$$\begin{aligned} \frac{1-4^{2-k}}{1-b}-\left( \frac{1}{1-b}\right) \left( b-2b\left( \frac{1}{8}\right) ^{k-2}\right) = \pi _{k-1}\left( b-2b\left( \frac{1}{8}\right) ^{k-2}\right) >0, \end{aligned}$$

(8)

where the inequalities follows since $\pi _{k-1}$ is nonnegative. Thus, $\pi _k$ is well-defined and positive on $2b(\frac{1}{8})^{k-2}$ and $b-2b(\frac{1}{8})^{k-2}$.

Continuity of $\pi _k$ follows from the recursive piecewise definition and the confirmation above that the values are well-defined on the boundaries of the pieces.

We now show that $\pi _k$ is nonnegative and $\pi _k(x)=0$ if and only if $x\in \mathbb {Z}$. Let $x\in [0,1)$. If $x\in I^k_3\cup I^k_6$, then $\pi _k(x) = \pi _{k-1}(x)> 0$ by the induction hypothesis. Observe that $\pi _k$ is affine on the intervals $I^k_1, I^k_2, I^k_4,$ and $I^k_5$. From Equations (5)-(8), $\pi _k$ is positive on the endpoints of each of the latter intervals, except for when $x=0$. Thus, if $x\in I^k_1{\setminus } \{0\}\cup I^k_2\cup I^k_4\cup I^k_5$, then $\pi _k(x)> 0$ and $\pi _k(0)=0$. Finally, if $x\in \mathbb {R}{\setminus } [0,1)$, then by the periodicity of $\pi _k$, it follows that $\pi _k(x)> 0$ and $\pi _k(x)=0$ if $x\in \mathbb {Z}$.$\square$