The Complexity of Max-Min $k$-Partitioning

TL;DR

This paper establishes the computational complexity of the max-min k-partition problem on weighted graphs, showing it is complete for the class ^P, even in restricted cases, thus highlighting its inherent computational difficulty.

Contribution

It proves the max-min k-partition problem is ^P-complete for various cases, extending known hardness results for related graph partition problems.

Findings

The problem is ^P-complete for k=2 with arbitrary weights.

The problem remains ^P-complete for k=3 with non-negative weights.

This complexity matches known results for MaxCut and Min-3-Cut problems.

Abstract

In this paper we study a max-min $k$-partition problem on a weighted graph, that could model a robust $k$-coalition formation. We settle the computational complexity of this problem as complete for class $\Sigma_2^P$. This hardness holds even for $k=2$ and arbitrary weights, or $k=3$ and non-negative weights, which matches what was known on \textsc{MaxCut} and \textsc{Min-3-Cut} one level higher in the polynomial hierarchy.

Equations

$$W(S)={\sum }_{\{i,j\}\subseteq S}w(i,j).$$

$$W_{-m}(\pi )=\underset{M\subseteq N,|M|\le m}{\min }\{W({\pi }_{-M})\}.$$

$$\forall p:I\to \{0,1\},\exists U\subseteq V^{(p)},|U|\le m,U\text{mbox}isavertexcoverof{G}^{(p)}.$$

$$\forall 2\text{mbox}-partition\pi ,\exists M\subseteq N,|M|\le 2m,W({\pi }_{-M})\le f_{n,m}(m),$$

$$f_{n,m}(x)=2\left(\genfrac{}{}{0ex}{}{n}{2}\right)-\sum _{i=1}^x(n-i)-\sum _{j=1}^{2m-x}(n-j)=g_{n,m}+x(x-2m),$$

$$\forall {\tau }_x:X\to \{0,1\},\exists {\tau }_y:Y\to \{0,1\},F({\tau }_x,{\tau }_y)\text{ is true}.$$

$$\theta -1=\left(\genfrac{}{}{0ex}{}{2m}{2}\right)\Lambda +2\left(\genfrac{}{}{0ex}{}{m+1}{2}\right)\Lambda +{\mu }_1+{\mu }_2,$$

$$\begin{array}[]{cccclr}S^{(p)}&=&K&\cup&\bigcup_{i=1}^{i=\alpha}N_{i,p(X(C_{i}))}&(3m\text{ nodes})\\ S^{1/2}&=&\{v^{1/2}\}&\cup&\bigcup_{i=1}^{i={\alpha}/{2}}N_{i,1-p(X(C_{i}))}&(m+1\text{ nodes})\\ S^{2/2}&=&\{v^{2/2}\}&\cup&\bigcup_{i={\alpha}/{2}+1}^{i=\alpha}N_{i,1-p(X(C_{i}))}&(m+1\text{ nodes})\end{array}$$

$$\overline{M}$$

Full text

1. Preliminaries

A max-min $k$-partition instance is defined by $\langle N,L,w,k,m,\theta\rangle$.

• •

$(N,L,w)$ is a weighted undirected graph. $N=[n]$, where $n\in\mathbb{N}$ is a set of nodes.111Given $n\in\mathbb{N}$, $[n]$ is shorthand of $\{1,\ldots,n\}$. The set of links $L\subseteq{N\choose 2}$ consists of unordered node pairs. Link $\ell=\{i,j\}$ maps to weight $w_{ij}\in\mathbb{Z}$. Equivalently, $w:N^{2}\rightarrow\mathbb{Z}$ satisfies for any $(i,j)\in N^{2}$ that $w(i,i)=0$, $w(i,j)=w(j,i)$ and $w(i,j)\neq 0\Rightarrow\{i,j\}\in L$.

• •

$k$ is the size of a partition, $2\leq k<n$.

• •

$m\in\mathbb{N}$ is the number of nodes that could be removed.

• •

$\theta\in\mathbb{Z}$ is a threshold value.

Let $\pi$ denote a $k$-partition of $N$, which is a collection of node-subsets $\{S_{1},\ldots,S_{k}\}$, such that for each $i\in[k]$, $S_{i}\subseteq N$, and $\forall S_{i},S_{j}\in\pi$, where $i\neq j$, $S_{i}\cap S_{j}=\emptyset$ holds. We say that a $k$-partition $\pi$ is complete when $\bigcup_{i\in[k]}S_{i}=N$ holds (otherwise, it is incomplete). For a complete partition $\pi$ and an incomplete partition $\pi^{\prime}$, we say that $\pi$ subsumes $\pi^{\prime}$ when $S_{i}\supseteq S^{\prime}_{i}$ holds for all $i\in[k]$. For node $i\in N$, $\pi(i)$ is the node-subset to which it belongs. For any $S\subseteq N$, define

[TABLE]

Then, let $W(\pi)$ denote $\sum_{S\in\pi}W(S)$. We require that no node-subset be empty; hence, if some node-subset is empty, we set $W(\pi)=-\infty$.

Given a $k$-partition $\pi=\{S_{1},\ldots,S_{k}\}$ and a set $M\subseteq N$, the remaining incomplete partition $\pi_{-M}$ after removing $M$ is defined as $\{S^{\prime}_{1},\ldots,S^{\prime}_{k}\}$, where $S^{\prime}_{i}=S_{i}\setminus M$. Let $W_{-m}(\pi)$ denote the minimum value after removing at most $m$ nodes, i.e., it is defined as:

[TABLE]

To obtain $W_{-m}(\pi)\neq-\infty$, every $S\in\pi$ needs to contain at least $m+1$ nodes, so that no node-subset of $\pi_{-M}$ is emptied. For partition $\pi=\{S_{1},\ldots,S_{k}\}$, we define its deficit count $\text{df}(\pi)$ as $\sum_{i\in[k]}\max(0,m+1-|S_{i}|)$. Thus, $\text{df}(\pi)=0$ must hold in order to obtain $W_{-m}(\pi)\neq-\infty$.

Definition \thetheorem.

The decision version (1) of our main problem is defined below. It may also be referred to as the defender’s problem.

1. (1)

Max-Min-$k$-Partition: Given a max-min $k$-partition instance, is there any $k$-partition $\pi$ satisfying $W_{-m}(\pi)\geq\theta$? 2. (2)

Max-Min-$k$-Partition/Verif: Given an instance of a max-min $k$-partition and a partition $\pi$, does $W_{-m}(\pi)\geq\theta$ hold?

A key step is to study the natural verification problem (2), to which complement we refer as the attacker’s problem. (Does an attack $M\subseteq N,|M|\leq m$ on $\pi$ exist such that $W(\pi_{-M})\leq\theta-1$?)

2. Complexity of Max-Min-$k$-Partition

In this section, we address the computational complexity of the defender’s problem. The verification (resp. attacker’s) problem itself turns out to be coNP-complete (resp. NP-complete), which intricates one more level in the polynomial hierarchy (PH). We show that Max-Min-$k$-Partition is complete for class $\Sigma_{2}^{P}$, even in two cases:

1. (a)

when $k=2$ for arbitrary link weights $w\lessgtr 0$, or 2. (b)

when $k=3$ for non-negative link weights $w\geq 0$.

These results seem to match what was known on MaxCut Karp (1972) (contained in Min-2-Cut when $w\lessgtr 0$ and NP-complete) and Min-3-Cut Dahlhaus et al. (1992) (NP-complete for $w\geq 0$ when one node is fixed in each node-subset), but one level higher in PH.

Observation 1**.**

Max-Min-$k$-Partition/Verif* is coNP-complete. It holds even for $k=1$, weights $w$ in $\{0,1\}$ and threshold $\theta=1$.*

Proof.

Decision problem Max-Min-$k$-Partition/Verif is in class coNP, since for any no-instance, a failing set $M$ such that $W(\pi_{-M})\leq\theta-1$ is a no-certificate verifiable in polynomial-time.

We show coNP-hardness by reduction from MinVertexCover to the (complement) attacker’s problem. Let graph $G=(V,E)$ and vertex number $m\in\mathbb{N}$ be any instance of MinVertexCover. MinVertexCover asks whether there exists a vertex-subset $U\subseteq V,|U|\leq m$ such that $\forall\{i,j\}\in E,i\in U\mbox{ or }j\in U$, i.e. every edge is covered by a vertex in $U$. We reduce it to an attacker’s instance with nodes $N\equiv V$, weights $w(i,j)\in\{0,1\}$ equal to one if and only if $\{i,j\}\in E$ and threshold $\theta=1$. The verified partition is simply $\pi=\{N\}$. The idea is that constraint $W(\pi_{-M})\leq 0$ is equivalent to damaging every link, hence to finding a vertex-cover $U\equiv M$ with $|M|\leq m$. ∎

We now proceed with the computational complexity of the main defender’s problem under $w\lessgtr 0$ and $w\geq 0$. We show $\Pi_{2}^{P}$-hardness of the $\forall\exists$ complement by reduction from MaxMinVertexCover or $\forall\exists$3SAT. The idea is to (1) enforce that only some proper partitions are meaningful. One possible proper partition corresponds to one choice on $\forall$ in the original problem. Then, (2) within one particular node-subset of a proper partition, we represent the subproblem (e.g. VertexCover or $\textsc{3-SAT}\leq\textsc{IndependentSet}=\textsc{VertexCover}$).

{theorem}

Problem Max-Min-$k$-Partition is $\Sigma_{2}^{P}$-complete, even for $k=2$ node-subsets and $w\in\{-n^{2},1,2\}$.

Proof.

Decision problem Max-Min-$k$-Partition asks whether $\exists\leavevmode\nobreak\ k\mbox{-partition }\pi,\forall\leavevmode\nobreak\ M\subseteq N,|M|\leq m,W(\pi_{-M})\geq\theta$. Therefore, it lies in class $\Sigma_{2}^{P}$, since, for yes-instances, such a $k$-partition $\pi$ is a certificate that can be verified by an NP-oracle on the remaining coNP problem Max-Min-$k$-Partition/Verif. We show $\Sigma_{2}^{P}$-hardness by a (complementary) reduction from $\Pi_{2}^{P}$-complete problem MaxMinVertexCover, defined as follows. Given graph $G=(V,E)$ whose vertices are partitioned by index set $I$ into $V=\bigcup_{i\in I}(V_{i,0}\cup V_{i,1})$, for a function $p:I\rightarrow\{0,1\}$, we define $V^{(p)}=\bigcup_{i\in I}V_{i,p(i)}$ and induced subgraph $G^{(p)}=(V^{(p)},E^{(p)})$. Given $m\in\mathbb{N}$, it asks whether:

[TABLE]

where “$U$ is a vertex cover of $G^{(p)}$” means $\forall\{u,v\}\in E[V^{(p)}]$, $u\in U$ or $v\in U$. Since edges between $V_{i,0}$ and $V_{i,1}$ are never relevant, we can remove them. By (Ko and Lin, 1995, Th. 10, proof), all $V_{i,j}$ sets have the same size, hence set $V^{(p)}$ has a constant size $n$ for any $p$.

The reduction is described in Figure 1. We reduce any instance of MaxMinVertexCover (as described above) to the following complementary instance of Max-Min-$k$-Partition. Nodes $N\equiv V$ are identified with vertices, hence can also be partitioned by $I\times\{0,1\}$ into $N=\bigcup_{i\in I}(N_{i,0}\cup N_{i,1})$ with $N_{i,j}\equiv V_{i,j}$ . We ask for $k=2$ node-subsets and choose a large number $\Lambda$, e.g. $\Lambda=n^{2}$. For every link $\{i,j\}\in{N\choose 2}$, if $\{i,j\}\in E$, we define synergy $w(i,j)=2$; otherwise if $\{i,j\}\notin E$, we define $w(i,j)=1$. However, for every $\ell\in I$ and every $(i,j)\in N_{\ell,0}\times N_{\ell,1}$, we define negative weight $w(i,j)=-\Lambda$. Here, up to $2m$ nodes might fail, and threshold $\theta=f_{n,m}(m)+1$ is defined in the proof. Since we are working on a complementary instance, the question is whether

[TABLE]

where $f_{n,m}:[0,2m]\rightarrow[0,n^{2}]$ is defined later.

This condition is trivially satisfied on 2-partitions $\pi$ where for some $\ell\in[I]$, two nodes $(i,j)\in N_{\ell,0}\times N_{\ell,1}$ are in the same node-subset. Indeed, even with an empty attack $M=\emptyset$, weight $W(\pi_{-\emptyset})$ incurs synergy $w(i,j)=-\Lambda$ and $W(\pi_{-\emptyset})<0\leq f_{n,m}(m)$. Therefore, the interesting part of this condition is on the other 2-partitions: the proper 2-partitions $\pi=\{S_{1},S_{2}\}$, which satisfy $\forall\ell\in[I],\forall(i,j)\in N_{\ell,0}\times N_{\ell,1},\pi(i)\neq\pi(j)$. It’s easy to see that $\pi$ can be characterized by a function $p\!:\!I\!\rightarrow\!\{0,1\}$ such that $S_{1}=\bigcup_{i\in I}N_{i,p(i)}$ and $S_{2}=\bigcup_{i\in I}N_{i,1-p(i)}$, and $|S_{1}|=|S_{2}|=n$. Since the remaining weights inside $S_{1}$ and $S_{2}$ are positive, the largest failures are the most damaging, $|M|=2m$ holds.

We now define function $f_{n,m}$. It maps $x\in[0,2m]$ to the number of in-subset pairs in a proper 2-partition $\pi=\{S_{1},S_{2}\}$ ($|S_{1}|\!=\!|S_{2}|\!=\!n$) after $x$ nodes fail in $S_{1}$ and $2m-x$ in $S_{2}$ (total $2m$ failures). One has:

[TABLE]

where $g_{n,m}$ is constant w.r.t. $x$. Since $f^{\prime}_{n,m}(x)=2(x-m)$ and $f^{\prime\prime}_{n,m}(x)=2$, it is a strictly convex function with minimum point at $x=m$. Therefore, for integers $x\in[2m]$, if $x\neq m$, the inequality $f_{n,m}(x)>f_{n,m}(m)$ holds. By definition, $f_{n,m}(x)$ is a lower bound on $W(\pi_{-M})$ (by assuming that all remaining weights in $\pi_{-M}$ have a value of $1$, instead of $1$ or $2$). Therefore, the main condition can only be satisfied by balanced failures $M=M_{1}\cup M_{2}$ such that $M_{1}\subseteq S_{1}$, $M_{2}\subseteq S_{2}$ and crucially: $|M_{1}|=|M_{2}|=m$.

(yes$\Rightarrow$yes) Any subgraph $G^{(p)}$ admits a vertex cover $U\subseteq V^{(p)}$ with size $|U|\leq m$. Let us show that any proper 2-partition $\pi=\{S_{1},S_{2}\}$ (characterized by a function $p:I\rightarrow\{0,1\}$) can be failed down to $f_{n,m}(m)$. Let $M_{1}\subseteq S_{1}$ correspond to the vertex cover of subgraph $G^{(p)}$ and $M_{2}\subseteq S_{2}$ to the vertex cover of subgraph $G^{(1-p)}$. Then, the failing set $M=M_{1}\cup M_{2}$ has a size of $|M|\leq 2m$, is balanced, and any node pair $\{i,j\}$ of weight two in $\pi$ (edge in $E$) has $i$ or $j$ in $M$, by the vertex covers. All in all, $W(\pi_{-M})=f_{n,m}(m)$.

(yes$\Leftarrow$yes) Any proper 2-partition $\pi=\{S_{1},S_{2}\}$ (characterized by function $p:I\rightarrow\{0,1\}$) admits a well balanced failing set $M=M_{1}\cup M_{2}$ such that $W(\pi_{-M})\!\leq\!f_{n,m}(m)$. Then it must be the case that $M_{1}$ (and $M_{2}$) covers all the node pairs of synergy two in $S_{1}$ (resp. $S_{2}$) that correspond to the edges of $G^{(p)}$ (resp. $G^{(1-p)}$). Then, for any subgraph $G^{(p)}$, attack $U\equiv M_{1}$ is a vertex cover. ∎

Adding a constant to all weights does not preserve optimal solutions. Thus, we cannot modify a problem with negative weights to an equivalent non-negative weight problem. Still, a hardness result for $k=3$ can also be obtained from $\forall\exists$3SAT.

{theorem}

Max-Min-$k$-Partition is $\Sigma_{2}^{P}$-complete, even for $k=3$ node-subsets and weights $w\in\{0,\Lambda,\Lambda+1\}$, where $\Lambda\geq n^{2}$.

Proof.

Let us first recall a classical reduction from 3SAT to IndependentSet, and how the later relates to VertexCover. Let any 3SAT instance be defined by formula $F=C_{1}\wedge\ldots\wedge C_{\alpha}$, where $C_{i}$ is a 3-clause on variables $X$. Every clause $C_{i}=\ell_{i,1}\vee\ell_{i,2}\vee\ell_{i,3}$ is reduced to triangle of vertices $V_{i}=\{v_{i,1},v_{i,2},v_{i,3}\}$ representing the literals of the clause. The set of $3\alpha$ vertices is then $V=\cup_{i=1}^{\alpha}V_{i}$. Between any two subsets $V_{i},V_{j}$, edges exist between two vertices if and only if the corresponding literals are on the same variable and are complementary (hence incompatible). It is easy to see that an independent-set $U\subseteq V$ of size $\alpha$ must have exactly one vertex per triangle $V_{i}$, and will exist (no edges within) if and only if there exists an instantiation of $X$ that makes at least one literal per clause $C_{i}$ true. Given a graph $G=(V,E)$, if $U\subseteq V$ is an independent-set, it means that $i\in U\wedge j\in U\Rightarrow\{i,j\}\notin E$. Hence, contraposition $\{i,j\}\in E\Rightarrow(i\!\in\!V\!\setminus\!U)\vee(j\!\in\!V\!\setminus\!U)$ means that $V\!\setminus\!U$ is a vertex cover. For instance, in the reduction from 3SAT, one can equivalently ask for a vertex cover $V\setminus U$ with size $2\alpha$; that is, two vertices per triangle $V_{i}$: Set $V$ of third vertices shall have no edge left to cover.

Let any instance of $\forall\exists$3SAT be defined by 3CNF formula $F(X,Y)=\bigwedge_{i=1}^{\alpha}C_{i}$ on variables $X=\{x_{1},\ldots,x_{|X|}\}$ and $Y=\{y_{1},\ldots,y_{|Y|}\}$. This problem asks whether:

[TABLE]

Without loss of generality, one can assume there is at most one $X$-literal per clause $C$. Indeed, if there are three $X$-literals, some $\tau_{x}$ can make the clause false, and it is trivially a no-instance. If there are two $X$-literals: $C=x\vee x^{\prime}\vee y$, then by adding a fresh $Y$-variable $z$, one easily obtains $C=(x\vee z\vee y)\wedge(x^{\prime}\vee\neg z\vee y)$. For ease of presentation, we assume exactly one $X$-literal and two $Y$-literals. We extend this proof to including clauses with no $X$-literal, in its final remark. Let $X(C)$ be the $X$-literal in clause $C$.

We build a Max-Min-3-Partition instance on $n\!=\!10\alpha+2$ nodes with $m=2\alpha$ failures . We first describe the nodes. To every clause $C_{i}=\ell^{x}_{i}\vee\ell^{y}_{i}\vee\ell^{y^{\prime}}_{i}$, we associate two node tetrads $N_{i,0}=\{v^{x}_{i,0},v^{y}_{i,0},v^{y^{\prime}}_{i,0},v^{z}_{i,0}\}$ and $N_{i,1}=\{v^{x}_{i,1},v^{y}_{i,1},v^{y^{\prime}}_{i,1},v^{z}_{i,1}\}$ (both depicted in Figure 2) which represent the two scenarios on $X$-literal $\ell^{x}_{i}$: false or true. Hence, there are $2\alpha$ node tetrads and a total of $4m=8\alpha$ nodes in $T=\cup_{i=1}^{\alpha}\cup_{j\in\{0,1\}}N_{i,j}$. There is also a set $K$ of $m=2\alpha$ nodes, and two nodes $v^{1/2},v^{2/2}$. This construct is depicted in Figure 3.

To describe the weights, we define a number $\Lambda\gg 1$, and only three different link weights $0,\Lambda,\Lambda+1$. We call $\Lambda$-link any link with weight $\Lambda$ or $\Lambda+1$. We call 1-link any link with weight $\Lambda+1$. Every pair of nodes in $\cup_{i=1}^{\alpha}\cup_{j\in\{0,1\}}N_{i,j}$ are linked by weight $\Lambda$ or $\Lambda+1$, except ($\star$) we set weights zero (and no link) for every $i,i^{\prime}\in[\alpha]$:

• when

$X(C_{i})=X(C_{i^{\prime}})$ between $N_{i,j}$ and $N_{i^{\prime},1-j}$ for $j\in\{0,1\}$, or

• when

$X(C_{i})=\neg X(C_{i^{\prime}})$ between $N_{i,j}$ and $N_{i^{\prime},j}$ for $j\in\{0,1\}$.

The rationale is to forbid two inconsistent scenarios on a same $X$-variable to coexist in one node-subset.

Whether the $\Lambda$-link is also a 1-link is determined as follows. Inside every node tetrad $N_{i,j}=\{v^{x}_{i,j},v^{y}_{i,j},v^{y^{\prime}}_{i,j},v^{z}_{i,j}\}$, there is a triangle of 1-links: $\{v^{x}_{i,j},v^{y}_{i,j}\}$, $\{v^{y}_{i,j},v^{y^{\prime}}_{i,j}\}$ and $\{v^{y^{\prime}}_{i,j},v^{x}_{i,j}\}$. Only in negative tetrads $N_{i,0}$, there is a 1-link $\{v^{x}_{i,0},v^{z}_{i,0}\}$. Given any tetrad $N_{i,j}$, node $v^{x}_{i,j}$ is not involved in any outgoing 1-link, but only links with weight $\Lambda$. Between any tetrads $N_{i,j}$ and $N_{i^{\prime},j^{\prime}}$ except ($\star$), there is a 1-link between complementary nodes of $Y$-literals; that is, a 1-link exists when the later’s literal is the negation of the former’s.222It is the same idea as in the standard reduction from 3SAT to IndependentSet. Assuming w.l.o.g. that $\alpha$ is even, let $\mu_{1}$ be the number of 1-links in $\bigcup_{i=1}^{i=\alpha/2}N_{i,0}$ and $\mu_{2}$ in $\bigcup_{i=\alpha/2+1}^{i=\alpha}N_{i,0}$.

Inside $K$, every pair of nodes is linked by weight $\Lambda$. Also, every node in $K$ is linked to every node in tetrads $T$ by weight $\Lambda$. Node $v^{1/2}$ is linked to every node in $\bigcup_{i=1}^{i={\alpha}/{2}}\bigcup_{j\in\{0,1\}}N_{i,j}$ by weight $\Lambda$, except for nodes $v^{z}_{i,1}$ by weight $\Lambda+1$; the same holds from node $v^{2/2}$ to every node in $\bigcup_{i={\alpha}/{2}+1}^{i=\alpha}\bigcup_{j\in\{0,1\}}N_{i,j}$. All other weights are zeros. We achieve this construct by defining threshold $\theta$ as:

[TABLE]

and asking whether $\forall 3\text{-part }\pi,\exists M\!\subseteq\!N,|M|\!\leq\!m,W(\pi_{-M})\leq\theta\!-\!1.$

A proper-3-partition $\pi=\{S^{(p)},S^{1/2},S^{2/2}\}$ is characterized by an instantiation $p:X\rightarrow\{0,1\}$ of $X$ variables extended to literals by $p(\neg x)=1-p(x)$, and which defines:

[TABLE]

Note that in $S^{1/2}$ (resp. $S^{2/2}$) the number of 1-links is constant $\mu_{1}$ (resp. $\mu_{2}$) for any $p$, since the formula on $Y$-literals is the same and 1-link $\{v^{1/2},v^{z}_{i,1}\}$ (resp. $\{v^{2/2},v^{z}_{i,1}\}$) compensates for $\{v^{x}_{i,0},v^{z}_{i,0}\}$.

We show that in our construct, any 3-partition which is not a proper-3-partition does trivially satisfy the complement question above. First, let us reason as if all three node-subsets were cliques of $\Lambda$-links. Crucially, in a node-subset of size $\nu$, the number of links ${\nu\choose 2}$ is quadratic. Therefore, the largest node-subsets will be the first attacked, and the only way $\pi_{-M}$ contains as many as ${2m\choose 2}+2{m+1\choose 2}$ $\Lambda$-links is if the node-subsets of $\pi$ had sizes $3m$, $m+1$ and $m+1$. Second, assume $\Lambda$-links are missing in some node-subsets. Then, an attack would focus on more connected subsets and $\pi_{-M}$ cannot contain as many as ${2m\choose 2}+2{m+1\choose 2}$ $\Lambda$-links. Therefore, 3-partition $\pi$ must consist in $\Lambda$-link cliques of size $3m$, $m+1$ and $m+1$. If the largest did not follow consistently some instantiation $p:X\rightarrow\{0,1\}$, then some $\Lambda$-links would be missing (see ($\star$)). Also, the only way to obtain two $\Lambda$-linked cliques of size $m+1$ on $N\setminus S^{(p)}$ is by $S^{1/2}$ and $S^{2/2}$. We also know that $S^{1/2}$ and $S^{2/2}$ contain $\mu_{1}+\mu_{2}$ 1-links.

Crucially, attack $M$ always occurs where it does the largest damage w.r.t. $\Lambda$-links: on node-subset $S^{(p)}$, and the number of remaining $\Lambda$-links is ${2m\choose 2}+2{m+1\choose 2}$. Given a proper-3-partition, what could make the inequality false would be a surviving 1-link in $S^{(p)}\setminus M$. Consequently, condition $\exists M,W(\pi_{-M})\leq\theta-1$ amounts to a $2\alpha$ node attack $M$ that covers every 1-link in $\bigcup_{i=1}^{i=\alpha}N_{i,p(X(C_{i}))}$. A crucial observation is that we necessarily attack/cover exactly two nodes per tetrad $N_{i,j}$, since each tetrad contains a triangle. In negative tetrads $N_{i,0}$, because of 1-link $\{v^{x}_{i,0},v^{z}_{i,0}\}$, one of these nodes has to be $v^{x}_{i,0}\in M$. In positive tetrads $N_{i,1}$, since node $v^{x}_{i,1}$ is not involved in other 1-links than the triangle, choosing both $v^{y}_{i,1}$ and $v^{y^{\prime}}_{i,1}$ in 1-link cover $M$ is the best choice. As in $\textsc{3SAT}\leq\textsc{IndependentSet}$, this amounts to a 1-link-independent-set $\overline{M}=S^{(p)}\setminus(K\cup M)$ with size $2\alpha$ and two nodes per tetrad $N_{i,j}$: first, node $v^{z}_{i,j}$, second if $j=0$ then $v^{y}_{i,0}$ xor $v^{y^{\prime}}_{i,0}$, otherwise if $j=1$ then $v^{x}_{i,1}$.

(yes$\Rightarrow$yes) Assume that for every $\tau_{x}:X\rightarrow\{0,1\}$, there exists $\tau_{y}:Y\rightarrow\{0,1\}$ such that in every clause $C_{i}$ with $\tau_{x}(X(C_{i}))=0$, a $Y$-literal is made true by instantiation $\tau_{y}$. We show that given any proper-3-partition $\{S^{(p)},S^{1/2},S^{2/2}\}$, in $S^{(p)}\setminus K=\bigcup_{i=1}^{i=\alpha}N_{i,p(X(C_{i}))}$, there exists a 1-link-independent-set $\overline{M}$ of size $2\alpha$, as below. Taking $\tau_{x}\equiv p$, let $\tau_{y}:Y\rightarrow\{0,1\}$ be as above mentioned. Then,

[TABLE]

is a 1-link-independent-set of size $2\alpha$: node $v^{y}_{i,0}$ exists since instantiation $\tau_{y}$ gives at least one true literal per clause where $\tau_{x}(X(C_{i}))=0$, and nodes are not 1-linked (no literal contradiction).

(yes$\Leftarrow$yes) Assume that for any $\tau_{x}\equiv p:X\rightarrow\{0,1\}$, a 1-link-independent-set $\overline{M}$ with size $2\alpha$ exists in node-subset $S^{(p)}\setminus K=\bigcup_{i=1}^{i=\alpha}N_{i,p(X(C_{i}))}$. Then, nodes $v^{y}_{i,0}\in\overline{M}$ consistently define $\tau_{y}:Y\rightarrow\{0,1\}$ that makes any clause $C_{i}$ true whenever $\tau_{x}(X(C_{i}))=0$.

Crucially, we also include clauses without any $X$-literal in the same construct. Assume w.l.o.g. that there are less than $\alpha/2$ such $Y$-clauses, within the first indexes in $[\alpha]$. To any $Y$-clause $C=\ell^{y}_{i}\vee\ell^{y^{\prime}}_{i}\vee\ell^{y^{\prime\prime}}_{i}$, one associates two tetrads $N_{i,j}=\{v^{y}_{i,j},v^{y^{\prime}}_{i,j},v^{y^{\prime\prime}}_{i,j},v^{z}_{i,j}\}$, $j\in\{0,1\}$. For $C_{i},C_{i^{\prime}}$ $Y$-clauses, between $N_{i,0}$ and $N_{i^{\prime},1}$ weights are zero. Negative tetrads $N_{i,0}$ are fully $\Lambda$-linked inside, between themselves, with previous tetrads of one $X$-variable and set $K$. Positive tetrads $N_{i,1}$ are fully $\Lambda$-linked inside, between themselves and with $v^{1/2}$. Given a $Y$-clause $C$, we define $X(C)=\emptyset$. For proper-3-partitions, we extend $p(\emptyset)=0$; hence in $\{S^{(p)},S^{1/2},S^{2/2}\}$, for $C_{i}$ a $Y$-clause, one has $N_{i,0}\subseteq S^{(p)}$ and $N_{i,1}\subseteq S^{1/2}$. Similarly, in any $Y$-clause tetrad $N_{i,j}$, there are 1-links $\{\{v^{y}_{i,j},v^{y^{\prime}}_{i,j}\},\{v^{y^{\prime}}_{i,j},v^{y^{\prime\prime}}_{i,j}\},\{v^{y^{\prime\prime}}_{i,j},v^{y}_{i,j}\}\}$, (optional 1-links $\{v^{z}_{i,1},v^{1/2}\}$), and whenever two $Y$-literals are complementary. It follows that the same proof holds. ∎

3. Related Work

Partitioning of a set into (non-empty) subsets may also be referred as coalition structure formation of a set of agents into coalitions. When a number of coalitions $k$ is required and there are synergies between vertices/agents, this problem is referred as $k$-cut, or $k$-way partition, where one minimizes the weight of edges/synergies between the coalitions, or maximizes it inside the coalitions. For positive weights and $k\geq 3$, this problem is NP-complete Dahlhaus et al. (1992), when one vertex is fixed in each coalition. For positive weights and fixed $k$, a polynomial-time $O(n^{k^{2}}T(n,m))$ algorithm exists Goldschmidt and Hochbaum (1994), when no vertex is fixed in coalitions, and where $T(n,m)$ is the time to find a minimum $(s,t)$ cut on a graph with $n$ vertices and $m$ edges. When not too many negative synergies exist (that is, negative edges can be covered by $O(\log(n))$ vertices), an optimal $k$-partition can be computed in polynomial-time SLESS2018217.