The quadratic M-convexity testing problem

TL;DR

This paper investigates the quadratic M-convexity testing problem, establishing its computational complexity and providing an efficient algorithm for cases where it is polynomial-time solvable, thus advancing understanding in discrete convex analysis.

Contribution

The paper proves that QMCTP is co-NP-complete generally but polynomial-time solvable under a natural assumption, and introduces an $O(n^2)$ algorithm for such cases.

Findings

QMCTP is co-NP-complete in general.

QMCTP is polynomial-time solvable under a natural assumption.

An $O(n^2)$-time algorithm is proposed for the polynomial-time solvable case.

Abstract

M-convex functions, which are a generalization of valuated matroids, play a central role in discrete convex analysis. Quadratic M-convex functions constitute a basic and important subclass of M-convex functions, which has a close relationship with phylogenetics as well as valued constraint satisfaction problems. In this paper, we consider the quadratic M-convexity testing problem (QMCTP), which is the problem of deciding whether a given quadratic function on $\{0,1\}^n$ is M-convex. We show that QMCTP is co-NP-complete in general, but is polynomial-time solvable under a natural assumption. Furthermore, we propose an $O(n^2)$-time algorithm for solving QMCTP in the polynomial-time solvable case.

Equations

$$f(x)+f(y)\ge f(x-{\chi }_i+{\chi }_j)+f(y+{\chi }_i-{\chi }_j),$$

$$f(x_1,x_2,\dots ,x_n):=\{\begin{array}{ll}\sum _{i\in [n]}{a}_i{x}_i+\sum _{1\le i<j\le n}{a}_{ij}{x}_i{x}_j& \text{if i∈[n]∑xi=r},\\ +\infty & \text{otherwise}\end{array}$$

$$a_{ij}+a_{kl}\ge \min \{a_{ik}+a_{jl},a_{il}+a_{jk}\}$$

$$f(x_1,x_2,x_3,x_4,x_5):=\{\begin{array}{ll}{x}_1{x}_3+2x_1{x}_4+(+\infty )\cdot x_1{x}_5+x_3{x}_5+2x_4{x}_5& \text{if ∑ixi=3},\\ +\infty & \text{otherwise}.\end{array}$$

$$f_G(x_1,x_2,\dots ,x_n):=\{\begin{array}{ll}\sum _{1\le i<j\le n}{a}_{ij}{x}_i{x}_j& \text{if i∈[n]∑xi=r},\\ +\infty & \text{otherwise},\end{array}$$

$$a_{ij}+a_{kl}\ge \min \{a_{ik}+a_{jl},a_{il}+a_{jk}\}$$

$$a_{ij}+a_{kl}=a_{il}+a_{jk}$$

$$f(z+{\chi }_i+{\chi }_j)+f(z+{\chi }_k+{\chi }_l)$$

$$\ge \min \{f(z+{\chi }_i+{\chi }_k)+f(z+{\chi }_j+{\chi }_l),f(z+{\chi }_i+{\chi }_l)+f(z+{\chi }_j+{\chi }_k)\}$$

$$a_{ij}+a_{kl}\ge \min \{a_{ik}+a_{jl},a_{il}+a_{jk}\}.$$

$$f(z+{\chi }_i+{\chi }_j)+f(z+{\chi }_k+{\chi }_l)$$

$$\ge \min \{f(z+{\chi }_i+{\chi }_k)+f(z+{\chi }_j+{\chi }_l),f(z+{\chi }_i+{\chi }_l)+f(z+{\chi }_j+{\chi }_k)\}$$

$$g(x):=\sum _{i\in [n]}{a}_i{x}_i+\sum _{1\le i<j\le n}{a}_{ij}{x}_i{x}_j$$

$$f(z+{\chi }_i+{\chi }_j)=g(z)+a_i+a_j+\sum _{p\in \text{supp}(z)}{a}_{ip}+\sum _{p\in \text{supp}(z)}{a}_{jp}+a_{ij},$$

$$f(z+{\chi }_k+{\chi }_l)=g(z)+a_k+a_l+\sum _{p\in \text{supp}(z)}{a}_{kp}+\sum _{p\in \text{supp}(z)}{a}_{lp}+a_{kl}.$$

$$f(z+{\chi }_i+{\chi }_j)+f(z+{\chi }_k+{\chi }_l)$$

$$\ge \min \{f(z+{\chi }_i+{\chi }_k)+f(z+{\chi }_j+{\chi }_l),f(z+{\chi }_i+{\chi }_l)+f(z+{\chi }_j+{\chi }_k)\}$$

$$\displaystyle\Leftrightarrow\$$

$$g(x)\le {\alpha }^{\ast }+\sum _{i\in [n]}{p}_i^{\ast }{x}_i\le f(x)(x\in \mathrm{dom}f\cap \mathrm{dom}g).$$

$$g(x):=\{\begin{array}{ll}f(x)& \text{if f(x)<+∞},\\ -\infty & \text{if f(x)=+∞}\end{array}$$

$$f(x)=g(x)\le {\alpha }^{\ast }+\sum _{i\in [n]}{p}_i^{\ast }{x}_i\le f(x)(x\in \mathrm{dom}f).$$

$${\hat{a}}_{ij}=\{\begin{array}{ll}{a}_{ij}+b& \text{if i∗∈{i,j}},\\ a_{ij}& \text{if i∗∈{i,j}},\end{array}(i,j\in [n],i\ne j)$$

Full text

**Keywords: ** discrete convex analysis, M-convex, testing problem

1 Introduction

A function $f$ on $\{0,1\}^{n}$ is said to be M-convex [7] if it satisfies the following generalization of matroid exchange axiom:

Exchange Axiom:

For $x,y\in\textrm{dom }f$ and $i\in\textrm{supp}(x)\setminus\textrm{supp}(y)$, there exists $j\in{\rm supp}(y)\setminus\textrm{supp}(x)$ such that

[TABLE]

where \textrm{dom }f:=\{x\in\{0,1\}^{n}\mid\text{f(x) takes a finite value}\} is the effective domain of $f$, $\textrm{supp}(x):=\{i\mid x_{i}=1\}$ for $x=(x_{1},x_{2},\dots,x_{n})\in\{0,1\}^{n}$, and $\chi_{i}$ is the $i$th unit vector. In general, M-convex functions are defined on the integer lattice $\mathbf{Z}^{n}$. In this paper, we restrict ourselves to M-convex functions defined on $\{0,1\}^{n}$, which are equivalent to the negative of valuated matroids introduced by Dress–Wenzel [3, 4]. M-convex functions play a central role in discrete convex analysis [8]. Indeed, M-convex functions appear in many areas such as operations research, economics, and game theory (see e.g., [8, 9, 10]). Quadratic M-convex functions also appear in many areas, and constitute a basic and important class of discrete functions. Quadratic M-convex functions have a close relationship with tree metrics [5], which is an important concept for mathematical analysis in phylogenetics (see e.g., [12]). Recently, Iwamasa–Murota–Živný [6] have revealed hidden quadratic M-convexity in valued constraint satisfaction problems (VCSPs) with joint winner property [2], and presented a perspective to their polynomial-time solvability from discrete convex analysis.

In this paper, we consider the quadratic M-convexity testing problem (QMCTP) defined as follows. Let $\overline{\mathbf{R}}:=\mathbf{R}\cup\{+\infty\}$ and $[n]:=\{1,2,\dots,n\}$ for a positive integer $n$ with $n\geq 4$.

Given:

$a_{i}\in\mathbf{R}$ for $i\in[n]$, $a_{ij}\in\overline{\mathbf{R}}$ for $1\leq i<j\leq n$, and a positive integer $r$ with $2\leq r\leq n-2$.

Question:

Is the quadratic function $f:\{0,1\}^{n}\rightarrow\overline{\mathbf{R}}$ defined by

[TABLE]

M-convex?

Notice that if $r=1$ or $r=n-1$, then $f$ of the form (1) is always a linear function. Here we assume that $a_{ij}=a_{ji}$ for distinct $i,j\in[n]$ and $\textrm{dom }f$ is nonempty. In this paper, functions can take the infinite value $+\infty$, where $a<+\infty$, $a+\infty=+\infty$ for $a\in\mathbf{R}$, and $0\cdot(+\infty)=0$. In the case where $a_{ij}$ takes a finite value for all distinct $i,j\in[n]$, the following theorem is immediate from [8, Theorem 6.4] (see also [8, Proposition 6.8]).

Theorem 1.1** ([8]; see also [11, Theorem 5.2]).**

Suppose that $a_{ij}$ takes a finite value for all distinct $i,j\in[n]$. Then a function of the form (1) is M-convex if and only if

[TABLE]

holds for every distinct $i,j,k,l\in[n]$.

By Theorem 1.1, if $a_{ij}$ is a finite value for all distinct $i,j\in[n]$, then QMCTP is solvable in polynomial time. However, if $a_{ij}$ can take the infinite value $+\infty$ for some distinct $i,j\in[n]$, there exists an example such that the condition (2) does not characterize M-convexity. Indeed, define $f:\{0,1\}^{5}\rightarrow\overline{\mathbf{R}}$ by

[TABLE]

Then $f$ is M-convex; this can be verified by the definition of M-convexity. However, the condition (2) is violated since $a_{12}+a_{34}<\min\{a_{13}+a_{24},a_{14}+a_{23}\}$ with $a_{12}+a_{34}=0$, $a_{13}+a_{24}=1$, and $a_{14}+a_{23}=2$. Thus, in the general case, the complexity of QMCTP is not settled yet.

A quadratic function of the form (1) with an infinite quadratic coefficient arises naturally from a binary VCSP function. A binary VCSP function $F$ satisfying the joint winner property can be transformed to a function represented as the sum two special M-convex functions of the form (1) with an infinite quadratic coefficient. This fact explains the polynomial-time solvability of $F$ (see [6] for details). The class of functions represented as the sum of two general quadratic M-convex functions corresponds to a new tractable class of binary VCSPs. Thus we need to consider QMCTP in the general case for the first step to identify such a new tractable class.

In this paper, we settle QMCTP by showing the following negative result.

Theorem 1.2**.**

QMCTP* is co-NP-complete.*

We also prove a positive result under the following natural condition.

Condition A:

For any $i\in[n]$, there exists $x\in\textrm{dom }f$ with $x_{i}=1$.

Theorem 1.3**.**

If Condition A holds, QMCTP is solvable in $O(n^{2})$ time.

Note that checking Condition A is an NP-complete problem, since it is almost equivalent to checking the existence of a stable set of size $r-1$.

The remainder of this paper is organized as follows. In Section 2, we prove Theorem 1.2 and introduce three types of functions. In Section 3, we present a characterization of M-convexity under Condition A for three types. This characterization implies Theorem 1.1. In Section 4, we propose an $O(n^{2})$-time algorithm for each type of QMCTP, and prove the validity of these algorithms. Thus, we show Theorem 1.3.

2 Co-NP-Completeness of QMCTP

In this section, we show the co-NP-completeness of QMCTP in the general case. In order to show Theorem 1.2, we prepare some lemmas.

In the terminology of discrete convex analysis, a set $X\subseteq\{0,1\}^{n}$ is said to be M-convex if for $x,y\in X$ and $i\in\textrm{supp}(x)\setminus\textrm{supp}(y)$, there exists $j\in{\rm supp}(y)\setminus\textrm{supp}(x)$ such that $x-\chi_{i}+\chi_{j},y+\chi_{i}-\chi_{j}\in X$. That is, an M-convex set $X$ is nothing but the base family of some matroid if we identify a 0-1 vector with a subset of $[n]$. Note that if $f$ is M-convex, then $\textrm{dom }f$ is M-convex.

Lemma 2.1**.**

Suppose that $f$ is a function of the form (1) such that ${\rm dom}\ f$ is M-convex. For some distinct $i,j\in[n]$, assume that there exist $x,y\in{\rm dom}\ f$ with $x_{i}=1$ and $y_{j}=1$. Then, if $a_{ij}<+\infty$, there exists $z\in{\rm dom}\ f$ with $z_{i}=z_{j}=1$.

Proof.

Take $x,y\in\textrm{dom }f$ with $|\textrm{supp}(x)\setminus\textrm{supp}(y)|$ minimum satisfying $x_{i}=y_{j}=1$. It suffices to show $|\textrm{supp}(x)\setminus\textrm{supp}(y)|=0$. Suppose, to the contrary, that $|\textrm{supp}(x)\setminus\textrm{supp}(y)|>0$. First we assume $|\textrm{supp}(x)\setminus\textrm{supp}(y)|\geq 2$. Then there exists $i^{\prime}\neq i$ such that $i^{\prime}\in\textrm{supp}(x)\setminus\textrm{supp}(y)$. By the M-convexity of $\textrm{dom }f$ for $x$, $y$, and $i^{\prime}$, there exists $j^{\prime}\in\textrm{supp}(y)\setminus\textrm{supp}(x)$ such that $x-\chi_{i^{\prime}}+\chi_{j^{\prime}}\in\textrm{dom }f$. If $j^{\prime}=j$, then $x^{\prime}:=x-\chi_{i^{\prime}}+\chi_{j}$ satisfies $x_{i}^{\prime}=x_{j}^{\prime}=1$, a contradiction. If $j^{\prime}\neq j$, then $x^{\prime}:=x-\chi_{i^{\prime}}+\chi_{j^{\prime}}$ satisfies $x^{\prime}_{i}=y_{j}=1$ and $|\textrm{supp}(x^{\prime})\setminus\textrm{supp}(y)|<|\textrm{supp}(x)\setminus\textrm{supp}(y)|$. This is also a contradiction to the choice of $x$ and $y$. Hence we have $|\textrm{supp}(x)\setminus\textrm{supp}(y)|=1=|\textrm{supp}(y)\setminus\textrm{supp}(x)|$.

Since $x,y\in\textrm{dom }f$, it holds that $a_{kl},a_{ik},a_{jk}<+\infty$ for any $k,l\in\textrm{supp}(x-\chi_{i})(=\textrm{supp}(y-\chi_{j}))$. Moreover, we have $a_{ij}<+\infty$ by the assumption. Hence we obtain $z:=x-\chi_{k}+\chi_{j}\in\textrm{dom }f$ for $k\in\textrm{supp}(x-\chi_{i})$. Hence $z$ satisfies $z_{i}=z_{j}=1$, a contradiction. Thus, we have $|\textrm{supp}(x)\setminus\textrm{supp}(y)|=0$. $\Box$

For a function $f$ of the form (1), we define an undirected graph $G_{f}=([n],E_{f})$ by $E_{f}:=\{\{i,j\}\mid i,j\in[n],\ i\neq j,\ a_{ij}=+\infty\}$. Notice that Condition A holds if and only if, for each $i\in[n]$, there is a stable set in $G_{f}$ of size $r$ containing $i$.

Lemma 2.2**.**

Suppose that Condition A holds. Then ${\rm dom}\ f$ is an M-convex set if and only if each connected component of $G_{f}$ is a complete graph.

Proof.

(if part). Let $A_{1},A_{2},\dots,A_{m}$ be the connected components of $G_{f}$. Then $\textrm{dom }f$ is represented by $\textrm{dom }f=\{x\in\{0,1\}^{n}\mid\sum_{i}x_{i}=r,\ |\textrm{supp}(x)\cap A_{p}|\leq 1\text{ for all }p\in[m]\}$. Hence $\textrm{dom }f$ can be regarded as the base family of a partition matroid. This implies that $\textrm{dom }f$ is M-convex.

(only-if part). We prove the contrapositive. Suppose that some connected component of $G_{f}$ is not complete. That is, there exist distinct $i,j,k\in[n]$ such that $\{i,j\},\{j,k\}\in E_{f}$ and $\{i,k\}\not\in E_{f}$. By Condition A, $a_{ik}<+\infty$, and Lemma 2.1, there exists $x\in\textrm{dom }f$ with $x_{i}=x_{k}=1$.

Take any $x,y\in\textrm{dom }f$ with $x_{i}=x_{k}=1$ and $y_{j}=1$. Since $a_{ij}=a_{jk}=+\infty$, we have $\textrm{supp}(x)\setminus\textrm{supp}(y)\supseteq\{i,k\}$. Then for all $j^{\prime}\in\textrm{supp}(y)\setminus\textrm{supp}(x)$, it holds that $x-\chi_{i}+\chi_{j^{\prime}}\not\in\textrm{dom }f$ or $y+\chi_{i}-\chi_{j^{\prime}}\not\in\textrm{dom }f$. Indeed, if $j^{\prime}=j$, then $x-\chi_{i}+\chi_{j}\not\in\textrm{dom }f$ holds from $a_{kj}=+\infty$, and if $j^{\prime}\neq j$, then $y+\chi_{i}-\chi_{j^{\prime}}\not\in\textrm{dom }f$ holds from $a_{ij}=+\infty$. This implies that $\textrm{dom }f$ is not M-convex. $\Box$

Here we consider the following problem (P), which is the problem for testing the M-convexity of $\textrm{dom }f$:

Given:

A graph $G=(V,E)$ having a stable set of cardinality $r$.

Question:

Let $T:=\bigcup\{S\subseteq V\mid S\text{ is a stable set of }G\text{ with }|S|=r\}$. Is each connected component of the subgraph of $G$ induced by $T$ a complete graph?

Lemma 2.3**.**

The problem (P) is co-NP-complete.

Proof.

It is clear that the problem (P) is in co-NP. We show the co-NP-hardness of (P) by reduction from the stable set problem, which is an NP-complete problem: Given $G=(V,E)$ and a positive integer $k\leq|V|$, we determine whether $G$ contains a stable set of size at least $k$. For a given graph $G=(V,E)$ and a positive integer $m$, define $G_{m}:=(V\cup V_{m},E\cup E_{m})$ by $|V_{m}|=m$, $V_{m}\cap V=\emptyset$, and $E_{m}:=\{\{i,j\}\mid i\in V,\ j\in V_{m}\}$. Let $T_{m}:=\bigcup\{S\subseteq V\cup V_{m}\mid S\text{ is a stable set of }G_{m}\text{ with }|S|=m\}$. Since $V_{m}$ is a stable set of $G_{m}$ satisfying $|V_{m}|=m$, we have $T_{m}\supseteq V_{m}$. If $T_{m}\supsetneq V_{m}$, the subgraph of $G_{m}$ induced by $T_{m}$ is not complete by the definition of $E_{m}$. Hence each connected component of the subgraph of $G_{m}$ induced by $T_{m}$ is complete if and only if $G$ does not have a stable set of cardinality at least $m$. Therefore we have the cardinality of a maximum stable set of $G$ by solving (P) for $G_{k}$ ($k=|V|,|V|-1,\dots,1$). Indeed, the first $k$ such that we output “no” by solving (P) for $G_{k}$ is equal to the cardinality of a maximum stable set. Since the maximum stable set problem has a polynomial-time reduction to the complement of (P), (P) is co-NP-hard. $\Box$

We are now ready to prove Theorem 1.2.

Proof of Theorem 1.2.

It is clear that QMCTP is in co-NP. We show the co-NP-hardness of QMCTP by reduction from the problem (P). Let $G=([n],E)$ be a graph having a stable set of cardinality $r$. We define $f_{G}$ by

[TABLE]

where $a_{ij}:=+\infty$ for $\{i,j\}\in E$ and $a_{ij}:=0$ for $\{i,j\}\not\in E$. Note that $x\in\textrm{dom }f$ if and only if $\textrm{supp}(x)$ is a stable set of $G$. We have $\textrm{dom }f_{G}\neq\emptyset$ by the assumption that $G$ has a stable set of cardinality $r$. We define $X$ by $X:=\bigcup\{S\subseteq[n]\mid S\text{ is a stable set of }G\text{ of size }r\}$. Then there exists $x\in\textrm{dom }f_{G}$ with $x_{i}=1$ if and only if $i\in X$. For $x\in\{0,1\}^{X}$, define $\tilde{x}\in\{0,1\}^{n}$ by $\tilde{x}_{i}:=x_{i}$ if $i\in X$ and $\tilde{x}_{i}:=0$ if $i\in[n]\setminus X$. Moreover define $f_{G}|_{X}(x):=f_{G}(\tilde{x})$ for $x\in\{0,1\}^{X}$. By the definition of $X$, $f_{G}$ is M-convex (i.e., $\textrm{dom }f_{G}$ is M-convex) if and only if $f_{G}|_{X}$ is M-convex (i.e., $\textrm{dom }f_{G}|_{X}$ is M-convex). Furthermore, by Lemma 2.2, $f_{G}|_{X}$ is M-convex if and only if each connected component of the subgraph of $G$ induced by $X$ is complete. This means that we can solve (P) by solving QMCTP for $f_{G}$. $\Box$

3 Characterization of Quadratic M-Convexity

In this section, we present a characterization of M-convexity under Condition A, which implies Theorem 1.1. By Lemma 2.2, we see that the following Condition B is necessary for the M-convexity.

Condition B:

Each connected component of $G_{f}$ is a complete graph.

Therefore, in this section, we can assume that a function $f$ of the form (1) satisfies Conditions A and B. Let $A_{1},A_{2},\dots,A_{m}$ be the vertex sets of the connected components of $G_{f}$ of size at least two, and define $A_{0}:=[n]\setminus\bigcup_{p=1}^{m}A_{p}$, which denotes the set of isolated vertices. Then we classify the types of $f$ as follows.

Type I:

$|A_{0}|+m\geq r+2$.

Type II:

$|A_{0}|+m=r+1$.

Type III:

$|A_{0}|+m=r$.

If $|A_{0}|+m<r$, then we have $\textrm{dom }f=\emptyset$. Hence we exclude this case.

Theorem 3.1**.**

Suppose that a function $f$ of the form (1) satisfies Conditions A and B. Then the following hold.

(I):

$f$* of Type I is M-convex if and only if it holds that*

[TABLE]

for every distinct $i,j,k,l\in[n]$.

(II):

$f$* of Type II is M-convex if and only if it holds that*

[TABLE]

for every $p\in[m]$, distinct $i,k\in A_{p}$, and distinct $j,l\in[n]\setminus A_{p}$.

(III):

$f$* of Type III is M-convex if and only if it holds that*

[TABLE]

for every distinct $p,q\in[m]$, distinct $i,k\in A_{p}$, and distinct $j,l\in A_{q}$.

Moreover, if $f$ is an M-convex function of Type II or III, then $f$ is a linear function on ${\rm dom\ }f$, that is, there exist $p_{i}\in\mathbf{R}$ for each $i\in[n]$ and $\alpha\in\mathbf{R}$ satisfying $f(x)=\sum_{i}p_{i}x_{i}+\alpha$ for any $x=(x_{1},x_{2},\dots,x_{n})\in{\rm dom}\ f$.

The function defined in (3) is an example of Type II. If $a_{ij}$ is finite value for all distinct $i,j\in[n]$, the function $f$ is of Type I. Hence Theorem 3.1 implies Theorem 1.1 as the finite case. By Theorem 3.1, we see that QMCTP is solvable in polynomial time under Conditions A and B.

In the proof of Theorem 3.1, we use the following facts about the local exchange axiom characterizing M-convexity, which are immediate corollaries of [8, Theorem 6.4] (see also [8, Proposition 6.8]).

Theorem 3.2** ([8]).**

A function $f:\{0,1\}^{n}\rightarrow\overline{\mathbf{R}}$ with ${\rm dom}\ f\subseteq\{x\in\{0,1\}^{n}\mid\sum_{i}x_{i}=r\}$ is M-convex if and only if ${\rm dom}\ f$ is M-convex and

[TABLE]

holds for all $z\in\{0,1\}^{n}$ and all distinct $i,j,k,l\in[n]$ such that $z+\chi_{i}+\chi_{j},z+\chi_{k}+\chi_{l}\in{\rm dom}\ f$.

Lemma 3.3**.**

A function $f$ of the form (1) is M-convex if and only if for every distinct $i,j,k,l\in[n]$ such that there exists $z\in\{0,1\}^{n}$ with $z+\chi_{i}+\chi_{j},z+\chi_{k}+\chi_{l}\in{\rm dom}\ f$, it holds that

[TABLE]

Recall that, for a function $f$ of the form (1), $\textrm{dom }f$ is always M-convex, since we assume that $f$ satisfies Conditions A and B. By Lemma 3.3, the condition (2) in Theorem 1.1 (or the condition (4) in Theorem 3.1) is sufficient for M-convexity. However, this is not necessary in general.

Proof of Lemma 3.3.

Take any $z\in\{0,1\}^{n}$ and distinct $i,j,k,l\in[n]$ such that $z+\chi_{i}+\chi_{j},z+\chi_{k}+\chi_{l}\in{\rm dom}\ f$. By Theorem 3.2, it suffices to show that for such $i,j,k,l$,

[TABLE]

holds if and only if $a_{ij}+a_{kl}\geq\min\{a_{ik}+a_{jl},a_{il}+a_{jk}\}$ holds (note that the inequality $a_{ij}+a_{kl}\geq\min\{a_{ik}+a_{jl},a_{il}+a_{jk}\}$ is independent of the choice of $z$).

Define $g:\{0,1\}^{n}\rightarrow\overline{\mathbf{R}}$ by

[TABLE]

for $x=(x_{1},x_{2},\dots,x_{n})\in\{0,1\}^{n}$. Then we have

[TABLE]

Since $f(z+\chi_{i}+\chi_{j})$ and $f(z+\chi_{k}+\chi_{l})$ take finite values, each term of (7) and (8), i.e., $g(z)$, $a_{ij}$, $a_{kl}$, and $a_{ip},a_{jp},a_{kp},a_{lp}$ for $p\in\textrm{supp}(z)$, also takes a finite value. Hence we obtain

[TABLE]

$\Box$

For the function $f$ defined in (3), we can see that there is no $z\in\{0,1\}^{5}$ such that $z+\chi_{1}+\chi_{2}$ and $z+\chi_{3}+\chi_{4}$ both belong to $\textrm{dom }f$. This is the reason why the inequality $a_{12}+a_{34}\geq\min\{a_{13}+a_{24},a_{14}+a_{23}\}$ is not necessary for the M-convexity of $f$.

A function $f$ is said to be M-concave if $-f$ is M-convex. The following theorem (M-separation theorem) holds.

Theorem 3.4** ([8, Theorem 8.15]).**

Suppose that $f:\{0,1\}^{n}\rightarrow\mathbf{R}\cup\{+\infty\}$ is M-convex and $g:\{0,1\}^{n}\rightarrow\mathbf{R}\cup\{-\infty\}$ is M-concave satisfying ${\rm dom}\ f\cap{\rm dom}\ g\neq\emptyset$ and $g(x)\leq f(x)$ for any $x\in{\rm dom}\ f\cap{\rm dom}\ g$. Then there exist $\alpha^{*}\in\mathbf{R}$ and $p^{*}\in\mathbf{R}^{n}$ such that

[TABLE]

We are now ready to prove Theorem 3.1.

Proof of Theorem 3.1.

First we show a characterization of M-convexity. For $i\in[n]$ denote by $B_{i}$ the connected component of $G_{f}$ containing $i$. That is, $B_{i}=\{i\}$ for $i\in A_{0}$, and $B_{i}=A_{p}$ for $i\in A_{p}$. Note that $x\in\textrm{dom }f$ if and only if $\sum_{i}x_{i}=r$ and $|\textrm{supp}(x)\cap A_{p}|\leq 1$ for $p\in[m]$. If $a_{ij}=+\infty$ or $a_{kl}=+\infty$, then it holds that $f(z+\chi_{i}+\chi_{j})=+\infty$ or $f(z+\chi_{k}+\chi_{l})=+\infty$ for all $z\in\{0,1\}^{n}$. In the following, we consider each type in turn.

Type I.

We show that for all distinct $i,j,k,l\in[n]$ with $a_{ij}<+\infty$ and $a_{kl}<+\infty$, there exists $z\in\{0,1\}^{n}$ such that $z+\chi_{i}+\chi_{j},z+\chi_{k}+\chi_{l}\in\textrm{dom }f$. $|A_{0}\setminus(B_{i}\cup B_{j}\cup B_{k}\cup B_{l})|+|\{A_{1},A_{2},\dots,A_{m}\}\setminus\{B_{i},B_{j},B_{k},B_{l}\}|\geq r-2$ holds since $|A_{0}|+m\geq r+2$. Therefore we can take $z\in\{0,1\}^{n}$ satisfying $\textrm{supp}(z)\subseteq[n]\setminus(B_{i}\cup B_{j}\cup B_{k}\cup B_{l})$, $|\textrm{supp}(z)\cap A_{p}|\leq 1$ for $p\in[m]$, and $\sum_{i}z_{i}=r-2$. Then $z+\chi_{i}+\chi_{j},z+\chi_{k}+\chi_{l}\in\textrm{dom }f$ holds for such $z$.

By Lemma 3.3, $f$ is M-convex if and only if for every distinct $i,j,k,l\in[n]$ with $a_{ij},a_{kl}<+\infty$, it holds that $a_{ij}+a_{kl}\geq\min\{a_{ik}+a_{jl},a_{il}+a_{jk}\}$. Moreover, if $a_{ij}=+\infty$ or $a_{kl}=+\infty$, then $a_{ij}+a_{kl}\geq\min\{a_{ik}+a_{jl},a_{il}+a_{jk}\}$ automatically holds. Hence $f$ is M-convex if and only if for every distinct $i,j,k,l\in[n]$, it holds that $a_{ij}+a_{kl}\geq\min\{a_{ik}+a_{jl},a_{il}+a_{jk}\}$.

Type II.

We show that for distinct $i,j,k,l\in[n]$ with $a_{ij},a_{kl}<+\infty$, there exists $z\in\{0,1\}^{n}$ such that $z+\chi_{i}+\chi_{j},z+\chi_{k}+\chi_{l}\in\textrm{dom }f$ if and only if $(B_{i}\cup B_{j})\cap(B_{k}\cup B_{l})\neq\emptyset$ holds (note that we have $B_{i}\cap B_{j}=B_{k}\cap B_{l}=\emptyset$ since $a_{ij},a_{kl}<+\infty$).

Suppose $(B_{i}\cup B_{j})\cap(B_{k}\cup B_{l})=\emptyset$ (i.e., $B_{i}$, $B_{j}$, $B_{k}$, and $B_{l}$ are all disjoint). Then $|A_{0}|+m\geq 4$. Hence $r\geq 3$ since $|A_{0}|+m=r+1$. Furthermore we obtain $|A_{0}\setminus(B_{i}\cup B_{j}\cup B_{k}\cup B_{l})|+|\{A_{1},A_{2},\dots,A_{m}\}\setminus\{B_{i},B_{j},B_{k},B_{l}\}|=r-3$. Hence for all $z\in\{0,1\}^{n}$ such that $\sum_{i}z_{i}=r-2$ and $|\textrm{supp}(z)\cap A_{p}|\leq 1$ for $p\in[m]$ with $A_{p}\neq B_{i},B_{j},B_{k},B_{l}$, it holds that $|\textrm{supp}(z)\cap(B_{i}\cup B_{j}\cup B_{k}\cup B_{l})|\neq\emptyset$. This means $z+\chi_{i}+\chi_{j}\not\in\textrm{dom }f$ or $z+\chi_{k}+\chi_{l}\not\in\textrm{dom }f$. Thus, for $i,j,k,l\in[n]$ with $(B_{i}\cup B_{j})\cap(B_{k}\cup B_{l})=\emptyset$, there is no $z$ satisfying $z+\chi_{i}+\chi_{j},z+\chi_{k}+\chi_{l}\in\textrm{dom }f$.

Suppose $(B_{i}\cup B_{j})\cap(B_{k}\cup B_{l})\neq\emptyset$. Without loss of generality, we also suppose $B_{i}\cap B_{k}\neq\emptyset$. Then there exists $p\in[m]$ such that $B_{i}=B_{k}=A_{p}$. Since $|A_{0}|+m=r+1$, we have $|A_{0}\setminus(B_{i}\cup B_{j}\cup B_{k}\cup B_{l})|+|\{A_{1},A_{2},\dots,A_{m}\}\setminus\{B_{i},B_{j},B_{k},B_{l}\}|=|A_{0}\setminus(B_{j}\cup B_{l})|+|\{A_{1},A_{2},\dots,A_{m}\}\setminus\{A_{p},B_{j},B_{l}\}|\geq r-2$. Therefore we can take $z\in\{0,1\}^{n}$ satisfying $\textrm{supp}(z)\subseteq[n]\setminus(B_{i}\cup B_{j}\cup B_{k}\cup B_{l})$, $|\textrm{supp}(z)\cap A_{p}|\leq 1$ for $p\in[m]$, and $\sum_{i}z_{i}=r-2$. Then $z+\chi_{i}+\chi_{j},z+\chi_{k}+\chi_{l}\in\textrm{dom }f$ holds for such $z$.

By Lemma 3.3, $f$ is M-convex if and only if for every $p\in[m]$, distinct $i,k\in A_{p}$, and distinct $j,l\in[n]\setminus A_{p}$, it holds that $a_{ij}+a_{kl}\geq\min\{a_{ik}+a_{jl},a_{il}+a_{jk}\}$. Since $a_{ik}=+\infty$, the above inequality can be represented as $a_{ij}+a_{kl}\geq a_{il}+a_{jk}$. Moreover, by replacing $j$ with $l$, we have $a_{ij}+a_{kl}\leq a_{il}+a_{jk}$. Hence $f$ is M-convex if and only if for every $p\in[m]$, distinct $i,k\in A_{p}$, and distinct $j,l\in[n]\setminus A_{p}$, it holds that $a_{ij}+a_{kl}=a_{il}+a_{jk}$.

Type III.

We show that for distinct $i,j,k,l\in[n]$ with $a_{ij},a_{kl}<+\infty$, there exists $z\in\{0,1\}^{n}$ such that $z+\chi_{i}+\chi_{j},z+\chi_{k}+\chi_{l}\in\textrm{dom }f$ if and only if $B_{i}\cup B_{j}=B_{k}\cup B_{l}$ holds.

Suppose $B_{i}\cup B_{j}\neq B_{k}\cup B_{l}$. Without loss of generality, we also suppose $B_{i}\neq B_{k}$ and $B_{i}\neq B_{l}$. Then $|A_{0}|+m\geq 3$. Hence $r\geq 3$ since $|A_{0}|+m=r$. Furthermore we obtain $|A_{0}\setminus(B_{i}\cup B_{j}\cup B_{k}\cup B_{l})|+|\{A_{1},A_{2},\dots,A_{m}\}\setminus\{B_{i},B_{j},B_{k},B_{l}\}|\leq|A_{0}\setminus(B_{i}\cup B_{k}\cup B_{l})|+|\{A_{1},A_{2},\dots,A_{m}\}\setminus\{B_{i},B_{k},B_{l}\}|=r-3$. Hence for all $z\in\{0,1\}^{n}$ such that $\sum_{i}z_{i}=r-2$ and $|\textrm{supp}(z)\cap A_{p}|\leq 1$ for $p\in[m]$ with $A_{p}\neq B_{i},B_{j},B_{k},B_{l}$, it holds that $|\textrm{supp}(z)\cap(B_{i}\cup B_{j}\cup B_{k}\cup B_{l})|\neq\emptyset$. This means $z+\chi_{i}+\chi_{j}\not\in\textrm{dom }f$ or $z+\chi_{k}+\chi_{l}\not\in\textrm{dom }f$. Thus, for $i,j,k,l\in[n]$ with $(B_{i}\cup B_{j})\cap(B_{k}\cup B_{l})=\emptyset$, there is no $z$ satisfying $z+\chi_{i}+\chi_{j},z+\chi_{k}+\chi_{l}\in\textrm{dom }f$.

Suppose $B_{i}\cup B_{j}=B_{k}\cup B_{l}$. Without loss of generality, we also suppose $B_{i}=B_{k}$ and $B_{j}=B_{l}$. Then there exist distinct $p,q\in[m]$ such that $B_{i}=B_{k}=A_{p}$ and $B_{j}=B_{l}=A_{q}$. Since $|A_{0}|+m=r$, we have $|A_{0}\setminus(B_{i}\cup B_{j}\cup B_{k}\cup B_{l})|+|\{A_{1},A_{2},\dots,A_{m}\}\setminus\{B_{i},B_{j},B_{k},B_{l}\}|=|A_{0}|+|\{A_{1},A_{2},\dots,A_{m}\}\setminus\{A_{p},A_{q}\}|=r-2$. Therefore we can take $z\in\{0,1\}^{n}$ satisfying $\textrm{supp}(z)\subseteq[n]\setminus(B_{i}\cup B_{j}\cup B_{k}\cup B_{l})$, $|\textrm{supp}(z)\cap A_{p}|\leq 1$ for $p\in[m]$, and $\sum_{i}z_{i}=r-2$. Then $z+\chi_{i}+\chi_{j},z+\chi_{k}+\chi_{l}\in\textrm{dom }f$ holds for such $z$.

By Lemma 3.3, $f$ is M-convex if and only if for every distinct $p,q\in[m]$, distinct $i,k\in A_{p}$, and distinct $j,l\in A_{q}$, it holds that $a_{ij}+a_{kl}\geq\min\{a_{ik}+a_{jl},a_{il}+a_{jk}\}$. Since $a_{ik}=a_{jl}=+\infty$, the above inequality can be represented as $a_{ij}+a_{kl}\geq a_{il}+a_{jk}$. Moreover, by replacing $j$ with $l$, we have $a_{ij}+a_{kl}\leq a_{il}+a_{jk}$. Hence $f$ is M-convex if and only if for every distinct $p,q\in[m]$, distinct $i,k\in A_{p}$, and distinct $j,l\in A_{q}$, it holds that $a_{ij}+a_{kl}=a_{il}+a_{jk}$.

Linearity.

Then we show linearity of an M-convex function $f$ of Type II or III. By the characterization of Type II or III, the function $g$ defined by

[TABLE]

is M-concave for an M-convex function $f$ of Type II or III. By Theorem 3.4, there exist $\alpha^{*}\in\mathbf{R}$ and $p^{*}\in\mathbf{R}^{n}$ such that

[TABLE]

This means that $f$ is a linear function on $\textrm{dom }f$. $\Box$

4 Testing Quadratic M-Convexity in Quadratic Time

In this section, we present an $O(n^{2})$-time algorithm for QMCTP under the assumption that a function $f$ of the form (1) satisfies Condition A (and Condition B). By Theorem 3.1, it suffices to give an $O(n^{2})$-time algorithm for checking the condition (4), (5), or (6) in Theorem 3.1 for each type, respectively.

4.1 Algorithms

Our idea used in a proposed algorithm for Type I is that the quadratic coefficients $(a_{ij})_{i,j\in[n]}$ of input $f$ are transformed into another $(\hat{a}_{ij})_{i,j\in[n]}$ which has an easily checkable property if $(a_{ij})_{i,j\in[n]}$ satisfies (4). For Types II and III, we give simpler conditions equivalent to (5) and (6), and check the new one.

We say that $(a_{ij})_{i,j\in[n]}$ satisfies the anti-tree metric property if $(a_{ij})_{i,j\in[n]}$ satisfies (4), that is, $a_{ij}+a_{kl}\geq\min\{a_{ik}+a_{jl},a_{il}+a_{jk}\}$ holds for all distinct $i,j,k,l\in[n]$. We also say that $(a_{ij})_{i,j\in[n]}$ satisfies the anti-ultrametric property if $a_{ij}\geq\min\{a_{ik},a_{jk}\}$ holds for all distinct $i,j,k\in[n]$.

Algorithm I (for Type I).

Step 1:

Define $\alpha:=\min\{a_{ij}\mid i,j\in[n]\}$, $b_{i}:=\min\{a_{ij}\mid j\in[n]\setminus\{i\}\}-\alpha$ for $i\in[n]$, and $\hat{a}_{ij}:=a_{ij}$ for distinct $i,j\in[n]$.

Step 2:

Update $\hat{a}_{ij}\leftarrow\hat{a}_{ij}-b_{i}-b_{j}$ for distinct $i,j\in[n]$.

Step 3:

If $(\hat{a}_{ij})_{i,j\in[n]}$ satisfies the anti-ultrametric property, output that “$f$ is M-convex.” Otherwise, output that “$f$ is not M-convex.” $\Box$

In Algorithms II and III, denote $A_{p}$ by $[n_{p}]$ for each $p\in[r]$, where $n_{p}:=|A_{p}|$.

Algorithm II (for Type II).

Step:

For all $p\in[r]$, if $a_{ij}+a_{i+1,j+1}=a_{i+1,j}+a_{i,j+1}$ holds for every $i\in[n_{p}-1]$ and $j\in\{n_{p}+1,n_{p}+2,\dots,n-1\}$, output that “$f$ is M-convex.” Otherwise, output that “$f$ is not M-convex.” $\Box$

Algorithm III (for Type III).

Step:

For all distinct $p,q\in[r]$, if $a_{ij}+a_{i+1,j+1}=a_{i+1,j}+a_{i,j+1}$ holds for every $i\in[n_{p}-1]$ and $j\in[n_{q}-1]$, output that “$f$ is M-convex.” Otherwise, output that “$f$ is not M-convex.” $\Box$

Theorem 4.1**.**

Algorithms I, II, and III work correctly and run in $O(n^{2})$ time.

We can check whether $f$ satisfies Condition B, i.e., each connected component of $G_{f}$ is a complete graph, in $O(n^{2})$ time. Thus, by Theorem 4.1, we obtain Theorem 1.3. In the rest of this section, we give the proof of Theorem 4.1. It is clear that the running time of Algorithms II and III are $O(n^{2})$. In Section 4.2, we show the validity of Algorithms I, II, and III, and show that we can check the anti-ultrametric property of given $(\hat{a}_{ij})_{i,j\in[n]}$ in $O(n^{2})$ time in Step 3 of Algorithm I.

4.2 Proof of Theorem 4.1

For brevity of notation, we denote $\min\{a_{ij}\mid j\in[n]\setminus\{i\}\}$ by $\min_{j}a_{ij}$.

4.2.1 Validity of Algorithm I.

Observe that if $(a_{ij})_{i,j\in[n]}$ satisfies the anti-tree metric property, so does $(\hat{a}_{ij})_{i,j\in[n]}$ defined by

[TABLE]

for some $i^{*}\in[n]$ and $b\in\mathbf{R}$. This means that the (inverse) operation of Step 2 does not change the anti-tree metric property. Furthermore, It is known that the anti-ultrametric property is stronger than the anti-tree metric property [5, 12]. Thus, if Algorithm I returns the output “$f$ is M-convex,” then $(a_{ij})_{i,j\in[n]}$ satisfies the anti-tree metric property. Therefore, the validity of Algorithm I is established by proving that Algorithm I returns “$f$ is M-convex” whenever $(a_{ij})_{i,j\in[n]}$ satisfies the anti-tree metric property. We need some lemmas to show this statement. In the following, suppose that $(a_{ij})_{i,j\in[n]}$ satisfies the anti-tree metric property. Recall that $\alpha:=\min\{a_{ij}\mid i,j\in[n]\}$.

Lemma 4.2**.**

Suppose that $\min_{j}a_{ij}=\alpha$ holds for all $i\in[n]$. Then $(a_{ij})_{i,j\in[n]}$ satisfies the anti-ultrametric property.

Proof.

Suppose, to the contrary, that there exist distinct $i,j,k\in[n]$ with $a_{ij}<\min\{a_{ik},a_{jk}\}$. Then $a_{ik}>\alpha<a_{jk}$ holds. Hence, by the assumption, there exists $l\in[n]\setminus\{i,j,k\}$ satisfying $a_{kl}=\alpha$. Then we obtain $a_{ik}>a_{ij}<a_{jk}$ and $a_{jl}\geq a_{kl}\leq a_{il}$. Thus, for such $i,j,k,l\in[n]$, it holds that $a_{ik}+a_{jl}>a_{ij}+a_{kl}<a_{il}+a_{jk}$. This contradicts the anti-tree metric property of $(a_{ij})_{i,j\in[n]}$. $\Box$

By Lemma 4.2, it suffices to show that if $(a_{ij})_{i,j\in[n]}$ satisfies the anti-tree metric property, it holds that $\min_{j}\hat{a}_{ij}=\alpha$ for any $i\in[n]$ after Step 2 of Algorithm I. In the following, we prove this.

Lemma 4.3**.**

Suppose that $\min_{j^{\prime}}a_{ij^{\prime}}=a_{ij}>\alpha$ holds for $i,j\in[n]$. Then there exists $k\in[n]$ such that $a_{jk}=\alpha$.

Proof.

We show this by induction on $n$ (the number of variables of $f$).

In the case of $n=4$, it suffices to prove that if $\min\{a_{12},a_{13},a_{14}\}=a_{12}>\alpha$, we have $\min\{a_{23},a_{24}\}=\alpha$. Suppose, to the contrary, that $\min\{a_{23},a_{24}\}>\alpha$. By the assumption and $\min\{a_{12},a_{13},a_{14},a_{23},a_{24},a_{34}\}=\alpha$, we obtain $a_{34}=\alpha$. Then, since $\min\{a_{12},a_{13},a_{14}\}=a_{12}$, we have $a_{14}\geq a_{12}\leq a_{13}$, and since $\min\{a_{23},a_{24}\}>\alpha$, we have $a_{23}>a_{34}<a_{24}$. Therefore we have $a_{14}+a_{23}>a_{12}+a_{34}<a_{13}+a_{24}$. This contradicts the anti-tree metric property of $(a_{ij})_{i,j\in[n]}$.

In the case of $n\geq 5$, it suffices to prove that if $\min\{a_{12},a_{13},\dots,a_{1n}\}=a_{12}>\alpha$, we have $\min\{a_{23},a_{24},\dots,a_{2n}\}=\alpha$. Suppose, to the contrary, that $\min\{a_{23},a_{24},\dots,a_{2n}\}=a_{23}>\alpha$. Since $(a_{ij})_{i,j\in[n]}$ defines an M-convex function, $(a_{ij})_{i,j\in\{2,3,\dots,n\}}$ also defines an M-convex function. Moreover, since $\min\{a_{12},a_{13},\dots,a_{1n}\}>\alpha$, we have $\min\{a_{ij}\mid i,j\in\{2,3,\dots,n\}\}=\alpha$. By $\min\{a_{23},a_{24},\dots,a_{2n}\}=a_{23}>\alpha$ and the induction hypothesis, there exists $k\in\{2,3,\dots,n\}$ such that $a_{3k}=\alpha$ (without loss of generality assume $k=4$). Since $\min\{a_{12},a_{13},\dots,a_{1n}\}=a_{12}$ and $\min\{a_{23},a_{24},\dots,a_{2n}\}>\alpha$, it holds that $a_{14}\geq a_{12}\leq a_{13}$ and $a_{23}>a_{34}<a_{24}$. Therefore we have $a_{14}+a_{23}>a_{12}+a_{34}<a_{13}+a_{24}$. This contradicts the anti-tree metric property of $(a_{ij})_{i,j\in[n]}$. $\Box$

Define $G_{\min}:=(V_{\min},E_{\min})$ by $E_{\min}:=\{\{i,j\}\mid a_{ij}=\alpha\}$ and $V_{\min}:=\{i\in[n]\mid i\in\exists e\in E_{\min}\}$.

Lemma 4.4**.**

$G_{\min}$* is connected. Moreover, for all $i,j\in V_{\min}$, there exists an $i$-$j$ path having at most two edges in $G_{\min}$.*

Proof.

It suffices to prove that there exists an $i$-$j$ path having at most two edges in $G_{\min}$ for all distinct $i,j\in V_{\min}$. Suppose, to the contrary, that $G_{\min}$ does not have an $i$-$j$ path with at most two edges for some $i,j\in V_{\min}$. Hence we have $\{i,j\}\not\in E_{\min}$. Furthermore there exist $k,l\in V_{\min}$ such that $\{i,k\},\{j,l\}\in E_{\min}\not\ni\{i,l\},\{j,k\}$. For such $i,j,k,l$, we obtain $a_{ij}+a_{kl}>a_{ik}+a_{jl}<a_{il}+a_{jk}$, a contradiction. $\Box$

Lemma 4.5**.**

Suppose that $\min_{j^{\prime}}a_{ij^{\prime}}=a_{ij}>\alpha$ holds for $i,j\in[n]$. Define $b_{i}:=a_{ij}-\alpha$. Then $a_{ik}-b_{i}\geq\min_{i^{\prime}}a_{i^{\prime}k}$ holds for all $k\in[n]\setminus\{i\}$.

Proof.

Take any $k\in[n]\setminus\{i\}$. If $\min_{i^{\prime}}a_{i^{\prime}k}=\alpha$, then it holds that $a_{ik}-b_{i}=a_{ik}-a_{ij}+\alpha\geq\alpha=\min_{i^{\prime}}a_{i^{\prime}k}$. This means that if $k=j$, the statement holds by Lemma 4.3. Hence, in the following, suppose $\min_{i^{\prime}}a_{i^{\prime}k}>\alpha$ (note that $k\neq j$ holds). If $a_{ik}=\min_{i^{\prime}}a_{i^{\prime}k}$, we have $\min_{j^{\prime}}a_{ij^{\prime}}=\alpha$ by Lemma 4.3. This contradicts $\min_{j^{\prime}}a_{ij^{\prime}}=a_{ij}>\alpha$. Thus we obtain $a_{ik}>\min_{i^{\prime}}a_{i^{\prime}k}$. We consider the following three cases.

(Case 1: $\min_{i^{\prime}}a_{i^{\prime}k}=a_{jk}$). By Lemma 4.3, $a_{jl}=\alpha$ holds for some $l\in[n]\setminus\{i,j,k\}$. Since $(a_{ij})_{i,j\in[n]}$ satisfies the anti-tree metric property, we have $a_{ik}+a_{jl}\geq\min\{a_{ij}+a_{kl},a_{il}+a_{jk}\}$. Moreover, by $\min_{j^{\prime}}a_{ij^{\prime}}=a_{ij}$ and $\min_{i^{\prime}}a_{i^{\prime}k}=a_{jk}$, it holds that $a_{il}\geq a_{ij}$ and $a_{kl}\geq a_{jk}$, respectively. Hence $a_{ik}+a_{jl}\geq\min\{a_{ij}+a_{kl},a_{il}+a_{jk}\}\geq a_{ij}+a_{jk}$ holds. Since $a_{jl}=\alpha$ and $b_{i}=a_{ij}-\alpha$ hold, we obtain $a_{ik}-b_{i}\geq a_{jk}=\min_{i^{\prime}}a_{i^{\prime}k}$.

(Case 2: $\min_{i^{\prime}}a_{i^{\prime}k}=a_{kl}$ for some $l\neq j$ and $a_{jl}=\alpha$). Since $(a_{ij})_{i,j\in[n]}$ satisfies the anti-tree metric property, we have $a_{ik}+a_{jl}\geq\min\{a_{ij}+a_{kl},a_{il}+a_{jk}\}$. Moreover, by $\min_{j^{\prime}}a_{ij^{\prime}}=a_{ij}$ and $\min_{i^{\prime}}a_{i^{\prime}k}=a_{kl}$, it holds that $a_{il}\geq a_{ij}$ and $a_{jk}\geq a_{kl}$, respectively. Hence $a_{ik}+a_{jl}\geq\min\{a_{ij}+a_{kl},a_{il}+a_{jk}\}\geq a_{ij}+a_{kl}$ holds. Since $a_{jl}=\alpha$ and $b_{i}=a_{ij}-\alpha$ hold, we obtain $a_{ik}-b_{i}\geq a_{kl}=\min_{i^{\prime}}a_{i^{\prime}k}$.

(Case 3: $\min_{i^{\prime}}a_{i^{\prime}k}=a_{kl}$ for some $l\neq j$ and $a_{jl}>\alpha$). By Lemma 4.3, we have $j,l\in V_{\min}$ since $\min_{j^{\prime}}a_{ij^{\prime}}=a_{ij}>\alpha$ and $\min_{i^{\prime}}a_{i^{\prime}k}=a_{kl}>\alpha$. By Lemma 4.4, there exists a $j$-$l$ path having at most two edges. Then the assumption of $a_{jl}>\alpha$ means that there exists $p\in[n]\setminus\{i,j,k,l\}$ such that $a_{jp}=a_{lp}=\alpha$.

Since $(a_{ij})_{i,j\in[n]}$ satisfies the anti-tree metric property, it holds that $a_{ik}+a_{jp}\geq\min\{a_{ip}+a_{jk},a_{ij}+a_{kp}\}$. By $\min_{j^{\prime}}a_{ij^{\prime}}=a_{ij}$ and $\min_{i^{\prime}}a_{i^{\prime}k}=a_{kl}$, it holds that $a_{ip}\geq a_{ij}$ and $a_{kp}\geq a_{kl}$, respectively. Moreover, by $a_{jk}\geq a_{kl}$, we have $a_{ik}+a_{jp}\geq\min\{a_{ip}+a_{jk},a_{ij}+a_{kp}\}\geq a_{ij}+a_{kl}$. Since $a_{jp}=\alpha$ and $b_{i}=a_{ij}-\alpha$ hold, we obtain $a_{ik}-b_{i}\geq a_{kl}=\min_{i^{\prime}}a_{i^{\prime}k}$. $\Box$

Let $(\hat{a}_{ij})_{i,j\in[n]}$ be the quadratic coefficients after Step 2, that is, $\hat{a}_{ij}=a_{ij}-b_{i}-b_{j}$ for distinct $i,j\in[n]$. We show that $\min_{j}\hat{a}_{ij}=\alpha$ for any $i\in[n]$.

Take any $i\in[n]$. Since $b_{i}\geq 0$, it holds that $\min_{j}\hat{a}_{ij}\leq\min_{j}a_{ij}-b_{i}=\alpha$. Next we prove $\min_{j}\hat{a}_{ij}\geq\alpha$. If $\min_{j}a_{ij}=\alpha$, we have $b_{i}=0$. Hence we obtain $\min_{j}\hat{a}_{ij}=\min_{j}\{a_{ij}-b_{j}\}\geq\alpha$. If $\min_{j}a_{ij}>\alpha$, by Lemma 4.5, we obtain $\min_{j}\hat{a}_{ij}\geq\min_{j}\{\min_{k}a_{jk}-b_{j}\}=\alpha$.

4.2.2 Time Complexity of Algorithm I.

It is clear that Steps 1 and 2 in Algorithm I can be done in $O(n^{2})$ time. In the following, we devise an $O(n^{2})$-time algorithm for determining whether $(\hat{a}_{ij})_{i,j\in[n]}$ satisfies the anti-ultrametric property, while a direct verification of checking the anti-ultrametric property in Step 3 takes $O(n^{3})$ time.

First we present a key lemma for designing an $O(n^{2})$-time algorithm. This fact is well known [5, 6] and the point here is to allow the infinite value.

Lemma 4.6** ([6, Lemma 8]).**

$(\hat{a}_{ij})_{i,j\in[n]}$* satisfies the anti-ultrametric property if and only if there exist some laminar family $\mathcal{L}$ on $[n]$ and some $c_{U}\in\overline{\mathbf{R}}$ for $U\in\mathcal{L}$ such that*

• •

$[n]\in\mathcal{L}$,

• •

if $U\subsetneq U^{\prime}$, then $c_{U}>c_{U^{\prime}}$ holds,

• •

$\hat{a}_{ij}=c_{U(i,j)}$* holds for any distinct $i,j\in[n]$, where $U(i,j)$ is the minimal element in $\mathcal{L}$ including $\{i,j\}$.*

By Lemma 4.6, we obtain the following natural procedure Decompose, which updates a laminar family $\mathcal{L}$ and defines $c_{U}\in\overline{\mathbf{R}}$ for $U\in\mathcal{L}$. Suppose that we are given $U\subseteq[n]$ and $w\in\overline{\mathbf{R}}$.

Procedure: $\texttt{Decompose}(U,w)$.

Step 1:

If $|U|\leq 1$ or $w=+\infty$, then stop.

Step 2:

Take any $i\in U$. Define $e:=\min\{\hat{a}_{ij}\mid j\in U\setminus\{i\}\}$ and $X:=\operatornamewithlimits{argmin}\{\hat{a}_{ij}\mid j\in U\setminus\{i\}\}$.

Step 3:

If $e>w$, then $\mathcal{L}\leftarrow\mathcal{L}\cup\{U\}$, $c_{U}:=e$, and $w\leftarrow e$.

Step 4:

Execute $\texttt{Decompose}(X,w)$ and $\texttt{Decompose}(U\setminus X,w)$. $\Box$

For initialization, let $\mathcal{L}:=\{[n]\}$ and $c_{[n]}:=\alpha$. Observe that if $(\hat{a}_{ij})_{i,j\in[n]}$ satisfies the anti-ultrametric property, $\texttt{Decompose}([n],\alpha)$ constructs an appropriate laminar family $\mathcal{L}$ and $c_{U}$ for $U\in\mathcal{L}$ corresponding to $(\hat{a}_{ij})_{i,j\in[n]}$. Moreover $\texttt{Decompose}([n],\alpha)$ runs in $O(n^{2})$ time.

We are ready to describe an algorithm for checking the anti-ultrametric property as follows.

Algorithm A (for checking the anti-ultrametric property).

Step 1:

Define $\mathcal{L}:=\{[n]\}$ and $c_{[n]}:=\alpha$.

Step 2:

Execute $\texttt{Decompose}([n],\alpha)$.

Step 3:

Make a copy of $\mathcal{L}$ and denote it by $\mathcal{L}^{\prime}$, that is, $\mathcal{L}^{\prime}:=\{U^{\prime}\mid U\in\mathcal{L}\}$ (the base set of $\mathcal{L}^{\prime}$ is also $[n]$).

Step 4:

While $\mathcal{L}^{\prime}\neq\emptyset$, do the following:

Step 4-1:

Take any minimal element $U^{\prime}\in\mathcal{L}^{\prime}$. Define $a_{ij}^{\prime}:=c_{U}$ for $\{i,j\}\subseteq U$ and $\{i,j\}\cap U^{\prime}\neq\emptyset$.

Step 4-2:

Let $U_{+}\in\mathcal{L}$ be the minimal element in $\mathcal{L}$ with $U\subsetneq U_{+}$. Update $U_{+}^{\prime}\leftarrow U_{+}^{\prime}\setminus U$.

Step 4-3:

Update $\mathcal{L}^{\prime}\leftarrow\mathcal{L}^{\prime}\setminus U^{\prime}$.

Step 5:

If $(\hat{a}_{ij})_{i,j\in[n]}=(a_{ij}^{\prime})_{i,j\in[n]}$, then output “$(\hat{a}_{ij})_{i,j\in[n]}$ satisfies the anti-ultrametric property.” Otherwise, output “$(\hat{a}_{ij})_{i,j\in[n]}$ does not satisfy the anti-ultrametric property.” $\Box$

In Step 4 of Algorithm A, note that we define the value of $a_{ij}^{\prime}$ exactly once for every distinct $i,j\in[n]$. Hence the time complexity of Step 4 is $O(n^{2})$ time. Thus, we see that Algorithm A runs in $O(n^{2})$ time. By Lemma 4.6, the validity of Algorithm A is clear. Therefore we obtain the following theorem.

Theorem 4.7**.**

Algorithm A works correctly and runs in $O(n^{2})$ time.

By Theorem 4.7, we can determine whether $(\hat{a}_{ij})_{i,j\in[n]}$ satisfies the anti-ultrametric property in $O(n^{2})$ time.

Remark 4.8**.**

The procedure Decompose has already been proposed in the preprint version of [5] in the context of M-convexity, and [1, 13] in the context of ultrametrics. However, these papers deal with a different case where $a_{ij}$ takes a finite value for all distinct $i,j\in[n]$ and the effective domain is subset of general integral vectors.

Remark 4.9**.**

We can devise another simpler $O(n^{2})$-time algorithm for Type I if $a_{ij}$ takes a finite value for all distinct $i,j\in[n]$. Indeed, for the finite case, it is known that $(a_{ij})_{i,j\in[n]}$ satisfies the anti-tree metric property if and only if $(\hat{a}_{ij})_{i,j\in[n-1]}$ satisfies the anti-ultrametric property, where $\hat{a}_{ij}:=a_{ij}-a_{in}-a_{jn}$ for all distinct $i,j\in[n-1]$ (see e.g., [12]). Hence it suffices to check the anti-ultrametric property of $(\hat{a}_{ij})_{i,j\in[n]}$. However, in the general case, the above algorithm does not work, since the relation between anti-tree metric property and anti-ultrametric property fails. For example, consider the case of $a_{15}=+\infty$, $a_{12}=a_{34}=2$, $a_{13}=a_{24}=a_{45}=1$, $a_{14}=a_{23}=a_{25}=a_{35}=0$. Then $(a_{ij})_{i,j\in[5]}$ does not satisfy the anti-tree metric property since $a_{12}+a_{34}(=4)>a_{13}+a_{24}(=2)>a_{14}+a_{23}(=0)$. However, $(\hat{a}_{ij})_{i,j\in[4]}$ satisfies the anti-ultrametric property since $\hat{a}_{12}=\hat{a}_{13}=\hat{a}_{14}=-\infty$, $\hat{a}_{23}=\hat{a}_{24}=0$, and $\hat{a}_{34}=1$.

4.2.3 Validity of Algorithms II and III.

Let $N$ and $M$ be positive integers. It suffices to prove that $a_{ij}+a_{kl}=a_{il}+a_{kj}$ holds for every $i,k\in[N]$ with $i<k$ and $j,l\in[M]$ with $j<l$ if $a_{ij}+a_{i+1,j+1}=a_{i+1,j}+a_{i,j+1}$ holds for every $i\in[N-1]$ and $j\in[M-1]$. We show this by induction on $(k-i)+(l-j)$. For $s\geq 1$ and $t\geq 1$, take any $i,k\in[N]$ with $k=i+s$ and $j,l\in[M]$ with $l=j+t$. The case $s+t=2$ holds by the assumption. Suppose $s+t\geq 3$. Without loss of generality, $s\geq 2$. By the induction hypothesis, we have $a_{ij}+a_{k-1,l}=a_{il}+a_{k-1,j}$ and $a_{k-1,j}+a_{kl}=a_{k-1,l}+a_{kj}$. Hence we obtain $a_{ij}+a_{kl}=a_{il}+a_{kj}$. This completes the induction step.

Acknowledgments

The author thanks Hiroshi Hirai and Kazuo Murota for careful reading and numerous helpful comments. The author also thanks the referees for helpful comments. In particular, Algorithms II and III (for Types II and III, respectively) are suggested by one of the referees. This research was supported by JSPS Research Fellowship for Young Scientists.