On integer network synthesis problem with tree-metric cost

TL;DR

This paper extends classical network synthesis results to integer capacities, showing that under certain conditions with tree-metric costs, the integer network synthesis problem can be solved efficiently.

Contribution

It introduces an integer version of the network synthesis problem with tree-metric costs and proves polynomial-time solvability when requirements are at least 2.

Findings

INSP can be solved in polynomial time under specified conditions.

Tree-metric costs enable efficient solutions for integer network design.

The result generalizes previous work on half-integral solutions to integer solutions.

Abstract

Network synthesis problem (NSP) is the problem of designing a minimum-cost network (from the empty network) satisfying a given connectivity requirement. Hau, Hirai, and Tsuchimura showed that if the edge-cost is a tree metric, then a simple greedy-type algorithm solves NSP to obtain a half-integral optimal solution. This is a generalization of the classical result by Gomory and Hu for the uniform edge-cost. In this note, we present an integer version of Hau, Hirai, and Tsuchimura's result for integer network synthesis problem (INSP), where a required network must have an integer capacity. We prove that if each connectivity requirement is at least 2 and the edge-cost is a tree metrc, then INSP can be solved in polynomial time.

Equations

$$c^R(e):=R(X_e)(e\in E(T)).$$

$$c^R({\delta }_T(v))\equiv |{\delta }_T(v)\cap F|\text{mbox}mod2(v\in V(T)\setminus V).$$

$$\sum _{e\in E(T)}l(e)R(X_e)+\underset{F}{\min }\sum _{e\in F}l(e),$$

$$c(e):=\sum \{y(ij)\mid ij\in E(K_V),\{i,j\}\cap X_e=1\}(e\in E(T)).$$

$$z(u,v)$$

$$d(u,v)$$

$$|{\delta }_T(v)\cap F|\equiv \{\begin{array}{ll}& 0\text{mbox}mod2\text{mbox}ifv\in I,\\ & 1\text{mbox}mod2\text{mbox}ifv\in J,\end{array}$$

$$I$$

$$J$$

Full text

In this note, we present an integer version of Hau, Hirai, and Tsuchimura’s result for integer network synthesis problem (INSP), where a required network must have an integer capacity. We prove that if each connectivity requirement is at least 2 and the edge-cost is a tree metrc, then INSP can be solved in polynomial time.

Keywords: Combinatorial optimization, network synthesis problem, tree metric, polynomial time algorithm, splitting-off

1 Introduction

Network synthesis problem(NSP) is one of the most simplest network design problems, which asks to find a minimum-cost network (from the empty network) satisfying a given connectivity requirement. In a classic paper [4] of combinatorial optimization, Gomory and Hu algorithmically proved that NSP admits a half-integral optimal solution provided the edge-cost is uniform. Note that this half-integrality property fails under the general edge-cost. Recently, Hau, Hirai, and Tsuchimura [5] showed that this classical result is naturally generalized for a tree-metric edge-cost. Here a tree metric is a special metric represented by the shortest path distance on a weighted tree.

In this note, we continue this line of research, and address the integer network synthesis problem (INSP), where a required network must have an integer capacity. Since INSP is NP-hard in general, it is interesting to explore a class of edge-costs tractable for INSP. An old result, due to Chou and Frank [1], says that INSP can also be solved in polynomial time provided the edge-cost is uniform. Frank [2] developed an elegant framework, based on splitting-off, for connectivity augmentation problem (which generalizes NSP). As a consequence of his result, INSP can be solved in polynomial time provided the edge-cost is node-induced. Here a node-induced cost is a special tree metric corresponding to a star.

The main result of this note is a partial generalization: If each connectivity requirement is at least 2 and the edge-cost is a tree metric, then INSP can be solved in polynomial time. Our algorithm modifies Frank’s framework to apply splitting-off to the tree-network corresponding to the tree-metric, whereas Frank applied splitting-off to the star-network (corresponding to a node-induced cost).

The rest of this paper is organized as follows. In Section 2, we state our result. In Section 3, we describe our algorithm to solve INSP with tree-metric cost.

Notation

Let $\mathbb{R}$ and $\mathbb{R}_{+}$ be the sets of reals and nonnegative reals, respectively. Also, let $\mathbb{Z}$ and $\mathbb{Z}_{+}$ be the sets of integers and nonnegative integers, respectively. For an undirected graph $G$ and a node subset $X\subseteq V(G)$, let $\delta_{G}(X)$ denote the set of edges connecting $X$ and $V(G)\setminus X$. If the graph $G$ is obvious in the context, then $\delta_{G}(X)$ is written as $\delta(X)$. Also $\delta(\{v\})$ is written as $\delta(v)$. The edge-connectivity between $u$ and $v$ in $G$ is the maximum number of pairwise edge-disjoint paths connecting $u$ and $v$. For a function $x\colon S\rightarrow\mathbb{R}$ on a set $S$ and a subset $X\subseteq S$, we denote $\sum_{u\in X}x(u)$ simply by $x(X)$.

2 Network synthesis problem

Let $K_{V}$ denote a complete undirected graph on node set $V$. Let $r\colon V\times V\rightarrow\mathbb{Z}_{+}$ be a connectivity requirement, and let $a\colon E(K_{V})\rightarrow\mathbb{R}_{+}$ be an edge-cost. A realization (of $r$) is a capacity function $y\colon E(K_{V})\rightarrow\mathbb{R}_{+}$ such that for all distinct $s,t\in V$, it holds $y(\delta(X))\geq r(s,t)$ for any subset $X\subseteq V$, i.e., there is an $(s,t)$-flow of value $r(s,t)$ under capacity $y$. The cost of a realization $y$ is defined as $\sum_{e\in E(K_{V})}a(e)y(e)$. The network synthesis problem (NSP) asks to find a realization of minimum-cost.

Gomory and Hu [4] algorithmically proved that NSP admits a half-integral minimum-cost realization under a uniform edge-cost. Let $R(X)\coloneqq\max\{r_{ij}\mid i\in X\not\ni j\}\ (\phi\not=X\subset V)$, where $R(\{u\})$ is simply written as $R(u)$.

Theorem 2.1** ([4]).**

Suppose that $a(e)=1$ for $e\in E(K_{V})$. Then the following hold:

• (1)

A half-integral minimum-cost realization exists, and can be found in polynomial time.

• (2)

The minimum-cost is equal to $\displaystyle\frac{1}{2}\sum_{u\in V}R(u).$

Recently, Hau, Hirai and Tsuchimura [5] generalized Theorem 2.1 to a tree-metric cost. Here an edge-cost $a\colon E(K_{V})\rightarrow\mathbb{R}_{+}$ is called a tree-metric if there exist a tree $T$ with $V\subseteq V(T)$ and a nonnegative edge-length $l\colon E(T)\rightarrow\mathbb{R}_{+}$ such that for all distinct $i,j\in V$, $a(ij)$ is equal to the length of the unique path between $i$ and $j$ in $T$ with respect to $l$. In this case, we say that $a$ is represented by $T$ and $l$. Fix an arbitrary $v\in V$. For edge $e\in E(T)$, let $X_{e}\subseteq V$ denote the subset of $V$ consisting of nodes reachable from $v$ in $T$ not using edge $e$.

Theorem 2.2** ([5]).**

Suppose that $a$ is a tree metric represented by a tree $T$ and a nonnegative edge-length $l\colon E(T)\rightarrow\mathbb{R}_{+}$. Then the following hold:

• (1)

A half-integral minimum-cost realization exists, and can be found in polynomial time.

• (2)

The minimum-cost is equal to $\displaystyle\sum_{e\in E(T)}l(e)R(X_{e}).$

This theorem is a generalization of Theorem 2.1. Indeed, the uniform edge-cost is represented by $T$ and $l$ as:

• •

$V(T)\coloneqq V\cup\{s\}$ and $E(T)\coloneqq\{sv\mid v\in V\}$.

• •

$l(e)\coloneqq 1/2$ for $e\in E(T)$.

Next, we consider an integer version INSP of NSP, where a realization $y$ must satisfy $y(e)\in\mathbb{Z}_{+}$ for each edge $e\in E(K_{V})$. Chou and Frank [1] gave a polynomial time algorithm for finding a minimum-cost integer realization together with a formula of the minimum-cost:

Theorem 2.3** ([1]).**

Suppose that $a(e)=1$ for $e\in E(K_{V})$. Then the following hold:

• (1)

A minimum-cost integer realization can be found in polynomial time.

• (2)

If $R(u)\not=1$ for all $u\in V$, then the minimum-cost is equal to $\displaystyle\left\lceil\frac{1}{2}\sum_{u\in V}R(u)\right\rceil.$

In the case where there is a node $u$ with $R(u)=1$, the problem easily reduces to that on $K_{V\setminus\{u\}}$ (by choosing one edge connecting $u$) [2, 3, 8].

The main result of this note is a tree-metric version of Theorem 2.3. Suppose that $a$ is a tree metric represented by a tree $T$ and a nonnegative edge-length $l\colon E(T)\rightarrow\mathbb{R}_{+}$. From connectivity requirement $R$, define an edge-capacity $c^{R}\colon E(T)\rightarrow\mathbb{Z}_{+}$ on $T$ by

[TABLE]

An inner-odd join (with respect to ($T,R$)) is an edge subset $F\subseteq E(T)$ such that

[TABLE]

Theorem 2.4**.**

Suppose that $a$ is a tree metric represented by a tree $T$ and a nonnegative edge-length $l\colon E(T)\rightarrow\mathbb{R}_{+}$. Also, suppose that $R(X_{e})>1$ for all $e\in E(T)$. Then the following hold:

• (1)

A minimum-cost integer realization can be found in polynomial time.

• (2)

The minimum-cost is equal to

[TABLE]

where the minimum is taken over all inner odd joins $F$ with respect to $(T,R)$.

This generalizes Theorem 2.3 in the case of $R(u)>1$. Indeed, in the above star representation of the uniform cost, a minimum cost inner-odd join is $\{sv\}$ if $\sum_{u\in V}R(u)$ is odd, and $\emptyset$ otherwise. We do not know whether tree-metric weighted INSP including $R(X)\leq 1$ is solvable in polynomial time; see also Remark 1. The proof of Theorem 2.4 is given in the next section.

Remark 1**.**

In the case where $R(X_{e})=0$ for some $e\in E(T)$, one may naturally expect that the problem is decomposed to that on $K_{X_{e}}$ and on $K_{V\setminus X_{e}}$. This is not true for INSP (but is true for NSP). Indeed, let $V\coloneqq\{u_{1},u_{2},u_{3},v_{1},v_{2},v_{3}\}$. The requirement $r$ is defined by $r(u_{i},u_{j})=r(v_{i},v_{j})\coloneqq 3$ for $1\leq i<j\leq 3$ and $r(u_{i},v_{j})\coloneqq 0$ for $1\leq i,j\leq 3$. Let $a$ be a tree metric on $V$ represented by $T,l$, where $V(T):=V\cup\{u,v\}$, $E(T):=\{uv,uu_{1},uu_{2},uu_{3},vv_{1},vv_{2},vv_{3}\}$, $l(uv)=1$, and $l(uu_{i})=l(vv_{i}):=2$ for $i=1,2,3$. Now $R(X_{uv})=0$. However, the problem cannot be decomposed; a minimum-cost integer realization $y$ is given by $y(u_{i}u_{j})=y(v_{i}v_{j})=1$ for $1\leq i<j\leq 3$ and $y(u_{i}v_{j})=1$ if $i=j$, and 0 if $i\neq j$. **

Remark 2**.**

Suppose that we are further given capacity lower bound $g\colon E(K_{V})\rightarrow\mathbb{Z}_{+}$. The edge-connectivity augmentation problem (ECAP) is to find a minimum-cost increase $z\colon E(K_{V})\rightarrow\mathbb{R}_{+}$ such that $g+z$ is a realization of $r$. Frank [2] shows that ECAP admits a half-integral optimal solution provided that the edge-cost is node-induced. From this result and Theorem 2.2, one can naturally conjecture that ECAP with a tree-metric cost has a half-integral optimal solution. But this conjecture is wrong. Indeed, we found instances having optimal values 638+1/3 and 9572+1/4 [7]. **

3 Algorithm

Let $V,r,a$ be an instance of INSP as above. Suppose that $a$ is a tree metric represented by $T$ and $l$, where $X_{e}$ is defined as above for $e\in E(T)$. An integer edge-capacity $c\colon E(T)\rightarrow\mathbb{Z}_{+}$ is said to be $r$-feasible if $c$ satisfies the following:

• (c1)

$c(\delta_{T}(v))$ is even for all $v\in V(T)\setminus V$.

• (c2)

$c(e)\geq R(X_{e})$ holds for all $e\in E(T)$.

The cost of an $r$-feasible edge-capacity $c$ is defined as $\sum_{e\in E(T)}l(e)c(e)$. The essence of Theorem 2.4 is the following:

Theorem 3.1**.**

Suppose that $R(X_{e})>1$ for all $e\in E(T)$. Then the following hold:

• (1)

For an integral realization $y$ of $r$, define an edge-capacity $c\colon E(T)\rightarrow\mathbb{Z}_{+}$ by:

[TABLE]

Then $c$ is $r$-feasible with $\sum_{e\in E(T)}l(e)c(e)=\sum_{e\in E(K_{V})}a(e)y(e)$.

• (2)

For an $r$-feasible edge-capacity $c\colon E(T)\rightarrow\mathbb{Z}_{+}$, there is an integral realization $y$ of $r$ with $\sum_{e\in E(K_{V})}a(e)y(e)\leq\sum_{e\in E(T)}l(e)c(e)$.

Corollary 3.2**.**

The minimum cost of an integer realization of $r$ is equal to the minimum cost of an $r$-feasible capacity in $T$.

From this corollary, we easily obtain Theorem 2.4 (2) as follows: Observe that a minimum-cost $r$-feasible edge-capacity is obtained from $c^{R}$ by increasing $c^{R}(e)$ by 1 on edge $e$ in an inner-odd join $F$, where the cost is equal to $\sum_{e\in E(T)}l(e)c^{R}(e)+l(F)=\sum_{e\in E(T)}l(e)R(X_{e})+l(F)$. Thus we obtain the formula in Theorem 2.4 (2).

We first show (1) in Theorem 3.1.

Proof of Theorem 3.2 (1).

(c1) Consider $ij\in E(K_{V})$ with $y(ij)>0$. Then $y(ij)$ contributes to $c$ along the unique path $P$ in $T$ connecting $i$ and $j$. Thus $y(ij)$ contributes to $c(\delta(v))$ by $2y(ij)$ if $v\in V(P)\setminus V$ and 0 if $v\not\in V(P)$.

(c2) If there exists an edge $e$ such that $c(e)<R(X_{e})$, then the cut size $y(\delta(X_{e}))$ is less than $R(X_{e})$ in graph $K_{V}$ under the capacity $y$. This contradicts the assumption that $y$ is a realization of $r$. ∎

Next we show (2). Let $c\colon E(T)\rightarrow\mathbb{Z}_{+}$ be an $r$-feasible edge-capacity.

Lemma 3.3**.**

Let $G$ be an undirected graph obtained from $T$ by replacing every edge $e\in E(T)$ by $c(e)$ parallel edges. For $u,v\in V$, the edge-connectivity between $u$ and $v$ in $G$ is at least $r(u,v)$.

Proof.

The edge-connectivity between $u$ and $v$ is the minimum of $c(e)$ over edges $e$ in the unique path in $T$ connecting $u$ and $v$. For all $e$ in the path, it holds that $r(u,v)\leq R(X_{e})\leq c(e)$, where the last inequality follows from (c2). Therefore the edge-connectivity between $u$ and $v$ is at least $r(u,v)$. ∎

In order to construct an integral realization of $r$ from edge-capacity $c$, we utilize Mader’s theorem [6], which is a basis of splitting-off technique. The splitting off a pair of edges $uv,vz$ is to replace the two edges $uv$, $vz$ by a new edge $e=uz$. Note that $e$ is a loop if $u=z$. Call a pair of edges $vs,st$ splittable if they can be split off so as to preserve the edge-connectivity between every two nodes other than $s$.

Theorem 3.4** ([6]).**

Let $G=(V,E)$ be a connected undirected graph and let $s$ be a node of $V$ with $|\delta_{G}(s)|\not\in\{1,3\}$. If there is no cut-edge incident to $s$, then there exists a splittable pair $(su,sv)$ of edges.

We are ready to prove Theorem 3.1 (2). For each node $s\in V(T)\setminus V$, We repeatedly use splitting off so as to preserve edge-connectivity between every two nodes other than $s$. More precisely, we execute the Algorithm 1.

It is clear that $|\delta_{G}(s)|$ is always even in Line 4. The following lemma and Theorem 3.4 guarantee that there always exists a splittable pair of edges incident to $s$ in Line 4.

Lemma 3.5**.**

In Line 4, there is no cut-edge incident to $s$.

Proof.

In the initial $G$, the edge-connectivity between every pair of vertices is at least 2 since $c(e)\geq R(X_{e})>1$ (by (c2)). If $s$ has only one neighbor $v$, then there are at least two edges between $s$ and $v$; the claim of this lemma is obvious. If $s$ has at least two neighbors and the cut-edge $e$, the graph is divided into two components $U_{1}$ and $U_{2}$ by the deletion of $e$. Since $s$ has at least two neighbors, we can choose $u_{1}\in U_{1}\setminus\{s\}$ and $u_{2}\in U_{2}\setminus\{s\}$. Since the edge $e$ is a cut-edge, the edge-connectivity between $u_{1}$ and $u_{2}$ is 1. On the other hand, the edge-connectivity between $u_{1}$ and $u_{2}$ is not changed in the middle of the algorithm. This is a contradiction. ∎

The following lemma shows that the cost of realization of $r$ is at most $\sum_{e\in E(T)}l(e)c(e)$ and we have proved Theorem 3.1 (2).

Lemma 3.6**.**

Suppose that $y$ is the output of Algorithm 1. Then $\sum_{e\in E(K_{V})}a(e)y(e)\leq\sum_{e\in E(T)}l(e)c(e)$ holds.

Proof.

Algorithm 1 repeats splitting-off operations. For each step, we define $z\colon V(G)\times V(G)\rightarrow\mathbb{Z}_{+}$ and $d\colon V(G)\times V(G)\rightarrow\mathbb{R}_{+}$ as follows:

[TABLE]

Then the following hold:

• •

Before the initial splitting-off, $\sum_{u,v\in V(G)}d(u,v)z(u,v)=\sum_{e\in E(T)}l(e)c(e)$

• •

After the last splitting-off, $\sum_{u,v\in V(G)}d(u,v)z(u,v)=\sum_{e\in E(K_{V})}a(e)y(e)$

Thus it suffices to show that $\sum_{u,v\in V(G)}d(u,v)z(u,v)$ does not increase for each splitting-off operation. Suppose that the splitting-off is applied to a pair $uv,vw$. Then the difference of $\sum_{u,v\in V(G)}d(u,v)z(u,v)$ is $d(u,w)-d(u,v)-d(v,w)$ and is nonpositive by the triangle inequality. Hence $\sum_{u,v\in V(G)}d(u,v)z(u,v)$ is non-increasing. ∎

Algorithm 2 finds a minimum-cost integer realization. To complete the proof of Theorem 2.4, we show that Algorithm 2 runs in polynomial time. First we deal with Line 2 to 4 in Algorithm 2. We consider the following generalized problem: Given a tree $T$, disjoint node sets $I,J\subseteq V(T)$ and a nonnegative edge-cost $l\colon E(T)\rightarrow\mathbb{R}_{+}$, find a minimum-cost edge subset $F\subseteq E(T)$ satisfying:

[TABLE]

where the cost of $F$ is defined as $l(F)$. Such an edge subset is called an $(I,J)$-join. Note that an inner-odd join is exactly an $(I,J)$-join for

[TABLE]

Algorithm 3 is a simple dynamic programming to find a minimum-cost $(I,J)$-join, where $\triangle$ represents symmetric difference. The correctness of this algorithm follows from the observation that $F^{1}\cup F^{2}$ is a minimum-cost $(I,J)$-join not using edge $e$ and that $F^{1}_{e}\cup F^{2}_{e}\cup\{e\}$ is a minimum-cost $(I,J)$-join using edge $e$.

Finally we discuss the time complexity of the whole algorithm. Obviously, Algorithm 3 is implemented by $O(n)$ time. A naive implementation of Algorithm 2 does not yield a strongly polynomial time algorithm, since we replaced each $e\in E(T)$ by $c(e)$ parallel edges. Therefore we regard $c(e)$ as a capacity on $e\in E(T)$ and we use a capacitated version of splitting-off. It takes $O(n^{6})$ time to complete splitting-off at node $s$; see [2]. There may be $O(n)$ nodes in $V(T)\setminus V$. Thus the overall complexity of the whole algorithm is $O(n^{7})$.

Acknowledgments

The work was partially supported by JSPS KAKENHI Grant Numbers 25280004, 26330023, 26280004, 17K00029.