Complexity of Partitioning Hypergraphs

TL;DR

This paper investigates the computational complexity of partitioning hypergraphs based on a pattern , establishing conditions under which the problem is either polynomial-time solvable or NP-complete for various uniformities.

Contribution

It characterizes the complexity of hypergraph partitioning problems for different uniformities, extending known results to higher dimensions.

Findings

Complexity depends on the pattern  and the uniformity k.

The problem is either polynomial-time solvable or NP-complete for k=3,4.

Results extend to k  for hypergraphs.

Abstract

For a given $\pi=(\pi_0, \pi_1,..., \pi_k) \in \{0, 1, *\}^{k+1}$, we want to determine whether an input $k$-uniform hypergraph $G=(V, E)$ has a partition $(V_1, V_2)$ of the vertex set so that for all $X \subseteq V$ of size $k$, $X \in E$ if $\pi_{|X\cap V_1|}=1$ and $X \notin E$ if $\pi_{|X\cap V_1|}=0$. We prove that this problem is either polynomial-time solvable or NP-complete depending on $\pi$ when $k=3$ or $4$. We also extend this result into $k$-uniform hypergraphs for $k \geq 5$.

Tables

Table 1. Table 1: List of π 𝜋 \pi - partition s with known complexity

$\pi$	$\pi$-partitionable graph	Complexity to decide it
$(0,\ast ,0)$	2-colorable graph	Linear
$(0,1,0)$	Complete bipartite graph	Linear
$(1,\ast ,0)$	Split graph	Linear [3]
$(0,\ast ,\ast ,0)$	$2$-colorable $3$-uniform hypergraph	NP-complete [4]
${\pi }_0=0,{\pi }_k=1$ and ${\pi }_{\ell }=\ast$ otherwise	Split hypergraph	Polynomial-time solvable [1] [5]

Equations

$$V^{\prime }$$

$$E^{\prime }$$

$$\cup \{\{u_1^1,u_1^2,...,u_1^{j_1},u_2^1,...,u_2^{j_2},...,u_k^{j_k}\}\mid \{u_1,u_2,...,u_k\}\in E\}$$

Full text

1 Introduction

For a given $\pi=(\pi_{0},\pi_{1},...,\pi_{k})\in\{0,1,*\}^{k+1}$, the $\pi$-partition is a decision problem with the following input and output.

[TABLE]

We say such partition as a $\pi$-partition and a $k$-uniform hypergraph $G$ is $\pi$-partitionable if it has a $\pi$-partition.

Some special cases of $\pi$-partition have been studied by many people. Table 1 shows a list of some cases of $\pi$-partition which has known time complexity.

The CSP dichotomy conjecture implies that the $\pi$-partition is polynomial-time solvable or NP-complete when $\pi$ does not contains $1$. Recently, Zhuk [6] claims the proof of this conjecture . We will completely classify $\pi$-partition into polynomial-time solvable problems and NP-complete problems for $k=3,4$ and we will prove some partial result for $k\geq 5$.

Before introducing our results, we observe some trivial facts. First, there exist a trivial partition if $\pi_{0}=*$ or $\pi_{k}=*$. For a $\pi\in\{0,1,*\}^{k+1}$, let $\pi^{\prime}$ be a vector in $\{0,1,*\}^{k+1}$ such that every $1$ of $\pi$ changed into [math] and every [math] of $\pi$ changed into $1$. Then by taking complement, it is easy to see that the $\pi$-partition and the $\pi^{\prime}$-partition are polynomial time equivalent. Therefore, we may assume that $\pi_{0}=0$ for all $\pi$. Similarly, by changing label of $V_{1}$ and $V_{2}$, we get the following fact.

Lemma 1.1**.**

The $(\pi_{0},\pi_{1},...,\pi_{k})$-partition and the $(\pi_{k},\pi_{k-1},...,\pi_{0})$-partition are polynomial-time equivalent.

When $k=2$, $(0,*,0)$-partition, $(0,1,0)$-partition and $(0,0,1)$-partition are polynomial-time solvable by using DFS(Depth First Search) and $(0,0,0)$-partition is clearly polynomial-time solvable. $(0,*,1)$-partition is equivalent to deciding whether a graph is a split graph. Hammer and Simeone [3] proved it is polynomial-time solvable. By considering trivial cases and polynomial-time equivalences described above, we get the $\pi$-partition is polynomial-time solvable for all $\pi\in\{0,1,*\}^{3}$.

We proved full dichotomy result when $k=3$ or $4$ and we proved some partial dichotomy result for larger $k$.

Theorem 1.2**.**

For a $\pi\in\{0,1,*\}^{4}$, the $\pi$-partition is NP-complete if $\pi=(0,*,*,0),$ $\pi=(1,*,*,1),$ $\pi=(0,0,*,0),$ $\pi=(0,*,0,0)$, $\pi=(1,*,1,1)$ or $\pi=(1,1,*,1)$ and polynomial-time solvable otherwise.

Theorem 1.3**.**

For a $\pi\in\{0,1,*\}^{5}$, the $\pi$-partition is NP-complete if $\pi=(0,*,*,*,0)$, $\pi=(0,*,*,0,0)$, $\pi=(0,0,*,*,0)$, $\pi=(0,*,0,0,0)$, $\pi=(0,0,0,*,0)$, $\pi=(0,0,*,0,0)$, $\pi=(1,*,*,*,1)$, $\pi=(1,*,*,1,1)$, $\pi=(1,1,*,*,1)$, $\pi=(1,*,1,1,1)$, $\pi=(1,1,1,*,1)$ or $\pi=(1,1,*,1,1)$ and polynomial-time solvable otherwise.

Theorem 1.4**.**

Let $\pi=(\pi_{0},...,\pi_{5})\in\{0,1,*\}^{6}$ be a vector. Suppose $\pi_{0}=0$, $\pi\neq(0,*,0,0,*,0)$, $\pi\neq(0,*,*,0,*,0)$, $\pi\neq(0,*,0,*,*,0)$, $\pi\neq(0,*,0,*,0,0)$ and $\pi\neq(0,0,*,0,*,0)$. Then the $\pi$-partition is NP-complete if $\pi=(0,*,*,*,*,0)$, $\pi=(0,*,*,*,0,0)$, $\pi=(0,0,*,*,*,0)$, $\pi=(0,0,0,*,*,0)$, $\pi=(0,0,*,*,0,0)$, $\pi=(0,*,*,0,0,0)$, $\pi=(0,0,0,0,*,0)$, $\pi=(0,0,0,*,0,0)$, $\pi=(0,0,*,0,0,0)$ or $\pi=(0,*,0,0,0,0)$ and polynomial-time solvable otherwise.

To show these theorems, we first prove the fact that if $\pi$ contains both [math] and $1$, then the $\pi$-partition is polynomial-time solvable in Subsection 3.1. We also prove that if $\pi=(0,*,0,*,...,*,0)$, then the $\pi$-partition is polynomial-time solvable in the same subsection. Then it is enough to check $(0,*,0,0)$-partition is NP-complete or polynomial-time solvable to get full dichotomy result of $k=3$ cases. We will show NP-completeness of this problem in Subsection 3.2. After that, we will look some polynomial-time reductions from a larger $k$ to a smaller $k$ so that we complete the proof of the second and third theorem using the cases of $k=3$ in Subsection 3.3.

2 Preliminary

2.1 Basic notations

For a $k$-uniform hypergraph $G=(V,E)$ and a subset of vertices $V^{\prime}\subseteq V$, a subgraph induced by $V^{\prime}$ be a hypergraph $G^{\prime}$ with the vertex set $V^{\prime}$ and the edge set $E(G^{\prime})=\{e\in E(G)\mid e\subseteq V^{\prime}\}$ and denoted by $G[V^{\prime}]$. If $G[V^{\prime}]$ is an empty graph, we say $V^{\prime}$ is an independent set of $G$. Similarly, if $G[V^{\prime}]$ has edges for every $k$-subset of $V^{\prime}$, we say $V^{\prime}$ is a clique of $G$.

Let $C_{m}^{k}$ be a $k$-uniform hypergraph with a vertex set $\mathds{Z}_{m}$ and an edge set $\{\{i,i+1,...,i+k-1\}\mid 1\leq i\leq m\}$. It is called a k-uniform m-cycle.

2.2 Known results

One important NP-completeness of $\pi$-partition is proved by Lovász [4] which is NP-completeness of 2-coloring problem of hypergraphs.

Theorem 2.1** (Lovász [4]).**

For a given hypergraph $G=(V,E)$, the $2$-coloring of $G$ is a function $f:V\rightarrow\{1,2\}$ such that each edge of $G$ is not monochromatic. Deciding whether a given hypergraph $G$ has $2$-coloring is NP-complete even if $G$ is $3$-uniform hypergraph.

Theorem 2.1 shows the $(0,*,*,0)$-partition is NP-complete.

To prove polynomial solvability results in section 3.1, we will look the theorem by Feder, Hell, Klein and Motwani [1]. The original theorem was for graphs but the proof of the theorem also holds for hypergraphs.

Theorem 2.2** (Feder, Hell, Klein and Motwani [1]).**

Let $\mathcal{S}$ and $\mathcal{D}$ be classes of hypergraphs closed under taking induced subgraph. Suppose there exist a constant $c$ such that every hypergraph $G$ contained in $\mathcal{S}\cap\mathcal{D}$ has at most $c$ vertices. Then, for every $n$-vertex hypergraph $G=(V,E)$, there are at most $n^{2c}$ partitions $(V_{1},V_{2})$ of $V$ such that $G[V_{1}]\in\mathcal{S}$ and $G[V_{2}]\in\mathcal{D}$. Furthermore, we can find all such partitions in $O(n^{2c+2}T(n))$ time where $T(n)$ is the time for recognizing $\mathcal{S}$ and $\mathcal{D}$.

This theorem directly shows the following corollary.

Corollary 2.3**.**

For $\pi=(\pi_{0},\pi_{1},...,\pi_{k})\in\{0,1,*\}^{k+1}$, if $\pi_{0}=0$ and $\pi_{k}=1$, then the $\pi$-partition can be solved in polynomial time. Furthermore, there exist at most $n^{2k}$ $\pi$-partition of an input graph $G$ where $n$ is number of vertices of $G$ and we can find every partition in polynomial time.

Proof.

Take $\mathcal{S}$ be the class of empty hypergraph and $\mathcal{D}$ be the class of complete $k$-uniform hypergraph. In other word, $\mathcal{S}$ is the class of hypergraphs of the form of $G=(V,\emptyset)$ and $\mathcal{D}$ is class of hypergraph of the form of $G=(V,\binom{V}{k})$ where $\binom{V}{k}$ is all $k$-subsets of $V$. Then by applying Theorem 2.2, we can find every $(0,*,*,...,*,1)$-partition of an input $k$-uniform hypergraph $G$ in polynomial time. By finding every $(0,*,*,...,*,1)$-partition of an input $k$-uniform hypergraph $G$ and checking each partition is $\pi$-partition or not, we can find every $\pi$-partition of $G$. ∎

3 Main results

3.1 Polynomial-time solvability

We will use Theorem 2.2 and Corollary 2.3 to prove more general fact.

Proposition 3.1**.**

If $\pi=(\pi_{0},...,\pi_{k})$ contains both $1$ and [math], then the $\pi$-partition is polynomial-time solvable. Furthermore, there exist at most $O(n^{3k})$ $\pi$-partitions for a $n$-vertex hypergraph $G$ and we can find every $\pi$-partition in polynomial-time.

Proof.

Suppose $\pi_{0}=0$ and $\pi_{i}=1$ for some $1\leq i\leq k$. Assume there exists a $\pi$-partition $(V_{1},V_{2})$ of an input hypergraph $G=(V,E)$. If $|V_{1}|\geq k-i$, then for any choice of $U_{1}\subseteq V_{1}$ with $|U_{1}|=k-i$ and every subset $U_{2}\subseteq V_{1}\setminus U_{1}$ of size $i$, $U_{1}\cup U_{2}\notin E(G)$. Similarly, every subset $U_{3}\subseteq V_{2}$ with $|U_{3}|=i$ satisfies $U_{1}\cup U_{3}\in E(G)$. Now, fix $U_{1}\subseteq V_{1}$ and construct an $i$-uniform hypergraph $G_{U_{1}}$ as follows. (Note that $i$ maybe equal to 1.)

$V(G^{\prime})=V(G)\setminus U_{1}$ and $U\in E(G^{\prime})$ if and only if $U\cup U_{1}\in E(G)$.

Then by previous observation, $V_{1}\setminus U_{1}$ is an independent set and $V_{2}$ is a clique.

Now, we will construct an algorithm. For every $U\subseteq V$ of size $k-i$, construct $G_{U}$ with respect to $U$. Use the algorithm from Theorem 2.2 to find every $(1,*,...,*,0)$-partition of $G_{U}$. There are at most $n^{2i}$ $(1,*,...,*,0)$-partitions of $G_{U}$ and we can find every $(1,*,...,*,0)$-partition in polynomial time. For each $(1,*,...,*,0)$-partition $(V_{1},V_{2})$ of $V_{G_{U}}$, check $(V_{1}\cup U,V_{2})$ makes a $\pi$-partition of $G$ or not. After check every $U\subseteq V(G)$ of size $k-i$, check every possibility that $V_{1}$ has size $<k-i$. By previous observation, this algorithm finds all $\pi$-partitions of $G$. ∎

Now, we will see one more polynomial-time solvable class of $\pi$-partitions.

Proposition 3.2**.**

Let $\pi=(\pi_{0},\pi_{1},...,\pi_{k})$. If $\pi_{i}=0$ for all even number $0\leq i\leq k$ and $\pi_{i}=*$ for all odd number $0\leq i\leq k$, then the $\pi$-partition is polynomial-time solvable.

Proof.

For a given $k$-uniform hypergraph $G=(V,E)$, we label the vertices as $u_{1},u_{2},...,u_{|V|}$ and the edges as $e_{1},e_{2},...,e_{|E|}$. Let $M$ be an $|E|\times|V|$ matrix in a Galois field $\mathds{F}_{2}$ where $M_{ij}=1$ if and only if $u_{j}\in e_{i}$. If $G$ has a $\pi$-partition $(V_{1},V_{2})$, then the linear equation $Mx=1$ in $\mathds{F}_{2}$ has a solution which is $x_{i}=1$ if and only if $u_{i}\in V_{1}$. Conversely, if the linear equation $Mx=1$ has a solution, then the vertex partition $(V_{1},V_{2})$ of $G$ where $u_{i}\in V_{1}$ if and only if $x_{i}=1$ is a $\pi$-partition of $G$. Since linear equation can be solved in polynomial time by applying Gaussian elimination, the $\pi$-partition is polynomial-time solvable. ∎

3.2 NP-completeness of the $(0,*,0,0)$-partition

To prove Theorem 1.2, we need one more NP-completeness theorem.

Proposition 3.3**.**

The $(0,*,0,0)$-partition is NP-complete.

Proof.

Let $\pi=(0,*,0,0)$. First observe that the $3$-uniform hypergraph $H=(V,E)$ with $V=\{1,2,3,4\}$ and $E=\{\{1,2,4\},\{1,3,4\},\{2,3,4\}\}$ has the unique $\pi$-partition which is $V_{1}=\{4\}$ and $V_{2}=\{1,2,3\}$. Also, note that $C^{3}_{6}$ has three possible partitions $V_{1}=\{1,4\}$, $V_{1}=\{2,5\}$ and $V_{1}=\{3,6\}$.

We will construct a polynomial-time reduction from the $3$-SAT to the $(0,*,0,0)$-partition. Let $\varphi(x_{1},x_{2},...,x_{n})=\bigwedge_{j=1}^{m}(y_{j}^{1}\vee y_{j}^{2}\vee y_{j}^{3})$ be an input 3-CNF formula where $y_{j}^{k}$ is one of $x_{i}$ or $\neg x_{i}$. If there exists a clause of the form of $(x_{i}\vee x_{i}\vee x_{i})$, by removing clauses containing $x_{i}$ and removing all $\neg x_{i}$, we get a smaller 3-CNF formula which has same satisfiability. Therefore, we may assume that there is no clause consisting of three equal terms.

Let $G_{\varphi}$ be a $3$-uniform hypergraph with the vertex set $\bigcup_{i=1}^{n}\{x_{i}^{1},x_{i}^{2},\neg x_{i}^{1},\neg x_{i}^{2},u_{i}^{1},$ $u_{i}^{2},u_{i}^{3},u_{i}^{4},u_{i}^{5},u_{i}^{6}\}\cup\bigcup_{j=1}^{m}\{w_{i}^{1},w_{i}^{2},w_{i}^{3},w_{i}^{4},w_{i}^{5},w_{i}^{6}\}$. If the $j$-th clause of $\varphi$ is of the form of $x_{j_{1}}\vee x_{j_{1}}\vee x_{j_{2}}$ and $j_{1}\neq j_{2}$, let $E_{j}=\{\{x_{j_{1}}^{1},w_{j}^{1},w_{j}^{2}\},\{{x}_{j_{1}}^{2},w_{j}^{3},w_{j}^{4}\},\{x_{j_{2}}^{1},w_{j}^{5},w_{j}^{6}\},$ $\{w_{j}^{2},w_{j}^{4},w_{j}^{6}\}\}$. Similarly, if the $j$-th clause of $\varphi$ is of the form of $x_{j_{1}}\vee x_{j_{2}}\vee x_{j_{3}}$ and $j_{1},j_{2},j_{3}$ are all distinct, let $E_{j}=\{\{x_{j_{1}}^{1},w_{j}^{1},w_{j}^{2}\},\{{x}_{j_{2}}^{1},w_{j}^{3},w_{j}^{4}\},\{x_{j_{3}}^{1},w_{j}^{5},w_{j}^{6}\},\{w_{j}^{2},w_{j}^{4},w_{j}^{6}\}\}$. If a clause contains $\neg x_{i}$, simply replace $x_{i}^{1},x_{i}^{2}$ into $\neg x_{i}^{1},\neg x_{i}^{2}$. The edge set of $G_{\varphi}$ is $\bigcup_{j=1}^{m}E_{j}\cup\bigcup_{i=1}^{n}\{\{x_{i}^{1},\neg x_{i}^{1},u_{i}^{1}\},\{\neg x_{i}^{1},u_{i}^{1},x_{i}^{2}\},$ $\{u_{i}^{1},x_{i}^{2},\neg x_{i}^{2}\},$ $\{x_{i}^{2},\neg x_{i}^{2},u_{i}^{2}\},$ $\{\neg x_{i}^{2},u_{i}^{2},x_{i}^{1}\},\{u_{i}^{2},x_{i}^{1},\neg x_{i}^{1}\},$ $\{u_{i}^{1},u_{i}^{2},u_{i}^{4}\},$ $\{u_{i}^{1},u_{i}^{3},u_{i}^{4}\},\{u_{i}^{2},u_{i}^{3},u_{i}^{4}\}\}$. Note that the subgraph of $G_{\varphi}$ induced by $\{x_{i}^{1}$, $\neg x_{i}^{1}$, $u_{i}^{1}$, $x_{i}^{2}$, $\neg x_{i}^{2}$, $u_{i}^{2}\}$ is isomorphic to the $C_{6}^{3}$.

We claim that $G_{\varphi}$ has a $\pi$-partition if and only if $\varphi$ is satisfiable. Suppose $G_{\varphi}$ has a $\pi$-partition $(V_{1},V_{2})$. Then by previous observation, $u_{i}^{4}\in V_{2}$ and $u_{i}^{1},u_{i}^{2},u_{i}^{3}\in V_{1}$. Since $\{x_{i}^{1}$, $\neg x_{i}^{1}$, $u_{i}^{1}$, $x_{i}^{2}$, $\neg x_{i}^{2}$, $u_{i}^{2}\}$ makes $C_{6}^{3}$, $x_{i}^{1},x_{i}^{2}\in V_{1}$ and $\neg x_{i}^{1},\neg x_{i}^{2}\in V_{2}$ or $x_{i}^{1},x_{i}^{2}\in V_{2}$ and $\neg x_{i}^{1},\neg x_{i}^{2}\in V_{1}$. For each $i$, assign true to $x_{i}$ if $x_{i}^{1}\in V_{2}$ and assign false to $x_{i}$ if $x_{i}^{2}\in V_{2}$. Suppose the $j$-th clause of $\varphi$ is $x_{j_{1}}\vee x_{j_{2}}\vee x_{j_{3}}$ ($j_{i}$ may equal). For each $E_{j}$, $\{w_{j}^{2},w_{j}^{4},w_{j}^{6}\}\in E_{j}$ so exactly one of $w_{j}^{2},w_{4}^{j},w_{6}^{j}$ is contained in $V_{1}$. This implies at least one of $x_{j_{i}}^{k}$ is contained in $V_{2}$. Therefore, at least one of $x_{j_{i}}$ has true value. If there exists $\neg x_{j_{i}}$ in a clause, by replacing $x_{i}^{1},x_{i}^{2}$ into $\neg x_{i}^{1},\neg x_{i}^{2}$, we get the same result. Therefore, assigned value of $x_{i}$ makes each clause true. So $\varphi$ is satisfiable. Conversely, suppose $\varphi$ is satisfiable. Fix any values of $x_{i}$’s that make $\varphi$ true. We will construct a $\pi$-partition $(V_{1},V_{2})$ of $G_{\varphi}$. If $x_{i}$ is true, make $x_{i}^{1},x_{i}^{2}\in V_{2}$ and $\neg x_{i}^{1},\neg x_{i}^{2}\in V_{1}$. If $x_{i}$ is false, make $x_{i}^{1},x_{i}^{2}\in V_{1}$ and $\neg x_{i}^{1},\neg x_{i}^{2}\in V_{2}$. Suppose the $j$-th clause of $\varphi$ is $x_{j_{1}}\vee x_{j_{2}}\vee x_{j_{3}}$ ($j_{i}$ may equal). Choose one $i$ such that $x_{j_{i}}$ is true. Make $w_{2i}\in V_{1}$, $w_{2i-1}\in V_{2}$ ,$w_{2k}\in V_{2}$ for $k\neq j$ and $w_{2k-1}\in V_{2}$ if $x_{j_{k}}$ is false, $w_{2k-1}\in V_{1}$ if $x_{j_{k}}$ is true for $k\neq j$. Finally, make $u_{i}^{4}\in V_{2}$ and $u_{i}^{1},u_{i}^{2},u_{i}^{3}\in V_{1}$. Then we get it is the $\pi$-partition of $G_{\varphi}$. This process can be done in time polynomial of input size so it gives a polynomial-time reduction from the 3-SAT to the $\pi$-partition. ∎

We can use this theorem to prove the fact that the exact cover problem is NP-complete even if $|\{Y\in S\mid x\in Y\}|=3$ for all $x\in X$. The exact cover problem is a decision problem for a given pair $(X,C)$ where $X$ is a finite set and $C$ is a subset of power set of $X$, deciding there exist a subset $C^{\prime}\subseteq C$ such that each element in $X$ lies in exactly one member of $C^{\prime}$. It is known as NP-complete [2]. For a given $3$-uniform hypergraph $G=(V,E)$, let $X=E$ and $S=\{$set of edges containing $v\mid v\in V\}$, we can easily see that the exact cover problem under the condition and the $(0,*,0,0)$-partition are polynomial-time equivalent.

Now we will prove Theorem 1.2.

Proof of Theorem 1.2.

By Proposition 3.1 and considering trivial cases, we can conclude that $\pi$-partition of $3$-uniform hypergraph is polynomial-time solvable except $\pi=(0,*,*,0)$ ,$\pi=(1,*,*,1)$, $\pi=(0,0,*,0)$, $\pi=(0,*,0,0)$, $\pi=(1,*,1,1)$ or $\pi=(1,1,*,1)$ for $\pi\in\{0,1,*\}^{4}$. The $(0,*,*,0)$-partition is NP-complete by Theorem 2.1 and the $(0,*,0,0)$-partition is NP-complete by Proposition 3.3. By taking complement and by Lemma 1.1, the $(1,*,*,1)$-partition is polynomial-time equivalent to the $(0,*,*,0)$-partition and the $(1,1,*,1)$-partition, the $(1,*,1,1)$-partition and the $(0,0,*,0)$-partition are polynomial-time equivalent to the $(0,*,0,0)$-partition. ∎

3.3 Polynomial-time reductions

In this subsection, we will prove NP-completeness using polynomial-time reduction. Note that by Theorem 3.1 and by taking complement, we may assume that $\pi$ does not contains $1$. We say $\pi$ is $1$-free if $\pi$ does not contains $1$.

First we define a map $\sigma:\{0,*\}^{k+1}\rightarrow\{0,*\}^{k+2}$ as $\sigma(\pi)_{0}=\pi_{0},\sigma(\pi)_{k+1}=\pi_{k}$ and $\sigma(\pi)_{i}=*$ if $\pi_{i}=*$ or $\pi_{i-1}=*$ and $\sigma(\pi)_{i}=0$ otherwise for $1\leq i\leq k$. Then we get the following proposition.

Proposition 3.4**.**

If there are no $i$ such that $\pi_{i}=\pi_{i+2}=*$ and $\pi_{i+1}=0$, there exist a polynomial-time reduction from the $\pi$-partition to the $\sigma(\pi)$-partition.

Proof.

For a given $k$-uniform hypergraph $G=(V,E)$, we construct a $(k+1)$-uniform hypergraph $G^{\prime}=(V^{\prime},E^{\prime})$ with the vertex set $V^{\prime}=V\cup\{u_{1},u_{2},...,u_{k+1}\}$ and the edge set $E^{\prime}=\{e\cup\{u_{i}\}\mid 1\leq i\leq k+1,e\in E\}\cup\{u_{1},...,u_{k+1}\}$. If $G$ has a $\pi$-partition $(V_{1},V_{2})$, then the partition $(V_{1}\cup\{u_{1}\},V_{2}\cup\{u_{2},...,u_{k+1}\})$ is a $\sigma(\pi)$-partition of $G^{\prime}$. Conversely, if $G^{\prime}$ has a $\sigma(\pi)$-partition $(V_{1},V_{2})$, there are at least one pair of integers $(i,j)$ satisfying $u_{i}\in V_{1}$ and $u_{j}\in V_{2}$. Therefore, for each $\{w_{1},w_{2},...,w_{k}\}\in E$, $|\{w_{1},w_{2},...,w_{k}\}\cap V_{1}|=i$ implies that $\sigma(\pi)_{i}=\sigma(\pi)_{i+1}=*$. By assumption, it implies that $\pi_{i}=*$ so $(V_{1}\setminus\{u_{1},...,u_{k+1}\},V_{2}\setminus\{u_{1},...,u_{k+1}\})$ is a $\pi$-partition of $G$. ∎

We can use this proposition and Theorem 2.1 to prove that $2$-coloring problem of $k$-uniform hypergraph is NP-complete for all $k\geq 3$.

Note that number of 0’s in $\sigma(\pi)$ is smaller than $\pi$. On the other hand, the next proposition produce another reduction from the $\pi$-partition to the $\pi^{\prime}$-partition such that number of 0’s in $\pi^{\prime}$ is strictly larger than number of 0’s in $\pi$.

Lemma 3.5**.**

*Suppose $\pi=(\pi_{0},...,\pi_{k})$ and $\pi^{\prime}=({\pi^{\prime}}_{0},...,{\pi^{\prime}}_{k^{\prime}})$ are $1$-free and there are no consecutive ’s in $\pi^{\prime}$. If there exist nonnegative integers $j_{1},j_{2},...,j_{k}$ such that ${\pi^{\prime}}_{\sum_{m\in I}j_{m}}=*$ if $\pi_{|I|}=*$ and ${\pi^{\prime}}_{\sum_{m\in I}j_{m}}=0$ if $\pi_{|I|}=0$ for all $I\subseteq\{1,2,...,k\}$, then there exist a polynomial-time reduction from the $\pi$-partition to the $\pi^{\prime}$-partition.

Proof.

If $\pi^{\prime}$ is a zero vector, then $\pi$ is also a zero vector so clearly there are polynomial-time reduction. Therefore, we may assume that $\pi^{\prime}$ is not a zero vector.

For a given $k$-uniform hypergraph $G=(V,E)$, construct a $k^{\prime}$-uniform hypergraph $G^{\prime}$ as follows.

[TABLE]

Suppose $G^{\prime}$ has a $\pi^{\prime}$-partition $(V^{\prime}_{1},V^{\prime}_{2})$. If $u^{1}\in V^{\prime}_{1}$, then $u^{i}\in V^{\prime}_{1}$ for all $i$ because if not, $|\{w_{u}^{1},w_{u}^{2},...,w_{u}^{k^{\prime}-1},u^{1}\}\cup V^{\prime}_{1}|=|\{w_{u}^{1},w_{u}^{2},...,w_{u}^{k^{\prime}-1},u^{i}\}\cup V^{\prime}_{1}|+1$. It is contradicting to the assumption. By the same argument, if $u^{1}\in V^{\prime}_{2}$, then $u^{i}\in V^{\prime}_{2}$ for all $i$ Let $({V}_{1},{V}_{2})$ be a partition of $V$ where $u\in{V}_{1}$ if and only if $u^{1}\in V^{\prime}_{1}$. For every edge $\{u_{1},u_{2},...,u_{k}\}\in E$, if $|\{u_{1},u_{2},...,u_{k}\}\cap V_{1}|=s$, then there exists $I\subseteq\{1,2,...,k\}$ such that $|I|=s$ and ${\pi^{\prime}}_{\sum_{m\in I}j_{m}}=*$. It implies $\pi_{i}=*$ because if not, then ${\pi^{\prime}}_{\sum_{m\in I}j_{m}}=0$ for all $I\subseteq\{1,2,...,k\}$ with $|I|=s$. Therefore, $(V_{1},V_{2})$ is a $\pi$-partition of $G$. Conversely, assume $G$ has a $\pi$-partition $(V_{1},V_{2})$. We choose any $m$ such that $\pi^{\prime}_{m}=*$. Let $(V^{\prime}_{1},V^{\prime}_{2})$ as $u^{i}\in V^{\prime}_{1}$ if and only if $u\in V_{1}$, $w_{u}^{i}\in V^{\prime}_{1}$ for $1\leq i\leq m-1$, $w_{u}^{i}\in V^{\prime}_{2}$ for $m+1\leq i\leq k^{\prime}-1$ and $w_{u}^{m}\in V^{\prime}_{1}$ if and only if $u\in V_{2}$. Then it is a $\pi^{\prime}$-partition of $G^{\prime}$ since $\pi_{i}=*$ implies ${\pi^{\prime}}_{\sum_{m\in I}j_{m}}=*$ for all $I\subseteq\{1,2,..,k\}$ with $|I|=i$ and $|\{w_{u}^{1},w_{u}^{2},...,w_{u}^{k^{\prime}-1},u^{i}\}\cap V^{\prime}_{1}|=m$ for any $1\leq i\leq k^{\prime},u\in V$. ∎

It is hard to check whether such $j_{1},j_{2},...,j_{k}$ exist or not. However, by taking all $j_{m}=2$, we get the following useful proposition.

Proposition 3.6**.**

Suppose $\pi=(\pi_{0},...,\pi_{k})$ is 1-free. Then there exist a polynomial-time reduction from the $\pi$-partition to the $(\pi_{0},0,\pi_{1},0,\pi_{2},0,...,0,\pi_{k})$-partition.

We observe that no consecutive *’s condition is only needed for making $u^{1},u^{2},...,u^{k^{\prime}}$ belong to the same side. Therefore, for a vector $\pi^{\prime}$, if there exist a hypergraph $H$ and vertices $u_{1},u_{2}\in V(H)$ such that it has at least one $\pi^{\prime}$-partition and for every $\pi^{\prime}$-partition of $H$, $u_{1}$ and $u_{2}$ lie in same side, then the lemma also holds without no consecutive *’s condition. This observation also holds for the next proposition. On the other hand, since $(\pi_{0},0,\pi_{1},0,...,0,\pi_{k})$ has no consecutive *’s, Proposition 3.6 make it easy to apply Lemma 3.4 and the next proposition.

Proposition 3.7**.**

*Suppose $\pi=(\pi_{0},\pi_{1},...,\pi_{k})$ is 1-free and the has no consecutive ’s. Then there exists a polynomial-time reduction from the $\pi$-partition to the $(0,\pi_{0},\pi_{1},...,\pi_{k})$-partition. Furthermore, if $(0,\pi_{0},\pi_{1},...,\pi_{k})=(\pi_{k},\pi_{k-1},...,\pi_{0},0)$, then it also holds without no consecutive * condition.

Proof.

If $(0,\pi_{0},\pi_{1},...,\pi_{k})=(\pi_{k},\pi_{k-1},...,\pi_{0},0)$, for a given $k$-uniform hypergraph $G=(V,E)$, we construct a $k+1$-uniform hypergraph $G^{\prime}=(V^{\prime},E^{\prime})$ with the vertex set $V^{\prime}=V\cup\{u\}$ and the edge set $E^{\prime}=\{e\cup u\mid e\in E\}$. If $G$ has a $\pi$-partition $(V_{1},V_{2})$, then $(V_{1}\cup\{u\},v_{2})$ is a $(0,\pi_{0},\pi_{1},...,\pi_{k})$-partition of $G^{\prime}$. Conversely, suppose $G^{\prime}$ has a $(0,\pi_{0},\pi_{1},...,\pi_{k})$-partition $(V^{\prime}_{1},V^{\prime}_{2})$. If $u\in V^{\prime}_{1}$, then $(V^{\prime}_{1}\setminus\{u\},V^{\prime}_{2})$ is a $\pi$-partition of $G$. If $u\in V^{\prime}_{2}$, then $(V^{\prime}_{1},V^{\prime}_{2}\setminus\{u\})$ is a $(0,\pi_{0},\pi_{1},...,\pi_{k-1})$-partition of $G$ so $(V^{\prime}_{2}\setminus\{u\},V^{\prime}_{1})$ is a $(\pi_{k-1},...,\pi_{0},0)$-partition of $G$ and $(\pi_{k-1},...,\pi_{0},0)=\pi$.

If $(0,\pi_{0},\pi_{1},...,\pi_{k})\neq(\pi_{k},\pi_{k-1},...,\pi_{0},0)$, we choose $m$ such that $\pi_{m}=*$ but $\pi_{k-m-1}=0$. We define a $(k+1)$-uniform hypergraph $G^{\prime}=(V^{\prime},E^{\prime})$ with the vertex set $V^{\prime}=V\cup\{u_{1},...,u_{2k},w_{1},...,w_{k+1}\}$ and the edge set $E^{\prime}=\{\{u_{1},u_{2},...,u_{k},w_{i}\}\mid 1\leq i\leq m+1\}\cup\{\{u_{k+1},u_{k+3},...,u_{2k},w_{i}\}\mid m+2\leq i\leq k+2\}\cup\{e\cup\{w_{1}\}\mid e\in E\}\cup\{\{w_{1},w_{2},...,w_{k+1}\}\}$.

Suppose $G^{\prime}$ has a $(0,\pi_{0},\pi_{1},...,\pi_{k})$-partition $(V^{\prime}_{1},V^{\prime}_{2})$. Then $w_{1},w_{2},...,w_{m+1}$ belongs to the same part because if not, $|\{u_{1},u_{2},...,u_{k},w_{i}\}\cup V_{1}|=|\{u_{1},u_{2},...,u_{k},w_{j}\}|+1$ for some $1\leq i,j\leq m+1$ and it is contradiction to $\pi$ has no consecutive *’s. By the same reason, $w_{m+2},...,w_{k+1}$ belongs to the same part. Since $\pi_{m}=*$ but $\pi_{k-m-1}=0$, $w_{i}\in V^{\prime}_{1}$ for $1\leq i\leq m+1$ and $w_{i}\in V^{\prime}_{2}$ for $m+2\leq i\leq k+1$. Therefore, $|\{v_{1},...,v_{k},w_{1}\}\cap V^{\prime}_{1}|=i$ if and only if $|\{v_{1},...,v_{k}\}\cap V^{\prime}_{1}|=i-1$. It shows $(V^{\prime}_{1}\cap V,V^{\prime}_{2}\cap V)$ is a $\pi$-partition of $G$. Conversely, if $G$ has a $\pi$-partition $(V_{1},V_{2})$, then $(V_{1}\cup\{w_{1},...,w_{m+1}\}\cup\{u_{1},...,u_{m}\}\cup\{u_{k+1},...,u_{k+m+1}\},V_{2}\cup\{w_{m+2},...,w_{k+1}\}\cup\{u_{m+1},...,u_{k}\}\cup\{u_{k+m+2},...,u_{2k}\})$ is a $(0,$ $\pi_{0},\pi_{1},...,\pi_{k})$-partition of $G^{\prime}$.

∎

Corollary 3.8**.**

Let $k\geq 3$ be an integer and $\pi\in\{0,1,*\}^{k+1}$ is $1$-free. If $\pi$ contains exactly one * and $\pi_{0}=\pi_{k}=0$, then the $\pi$-partition is NP-complete.

Proof.

By Proposition 3.3, Proposition 3.7 and Lemma 1.1, it is clear. ∎

Now, we will prove Theorem 1.3 and Theorem 1.4.

Proof of Theorem 1.3.

By Theorem 2.1, Proposition 3.3 and Proposition 3.4, the $(0,*,*,*,0)$-partition, the $(0,0,*,*,0)$-partition and the $(0,*,*,0,0)$-partition are NP-complete. By Corollary 3.8, the $(0,*,0,0,0)$-partition, the $(0,0,*,0,0)$-partition and the $(0,0,$ $0,*,0)$-partition are NP-complete. By Proposition 3.2, the $(0,*,0,*,0)$-partition is polynomial-time solvable. By Proposition 3.1 and considering trivial cases, remaining cases are polynomial-time solvable. By combining these results, we get the proof of Theorem 1.3. ∎

Proof of theorem 1.4.

By applying Proposition 3.3 and Proposition 3.4 to result of Theorem 1.3, we get all NP-completeness of Theorem 1.4. Remaining cases are polynomial-time solvable by Proposition 3.3 and by considering trivial cases. It proves Theorem 1.4. ∎

Acknowledgement

The author would like to thank Jaehyun Koo for pointing out the partitioning problem of a split graph and thank Prof. Sang-il Oum for very helpful advice.