Hamiltonian cubic bipartite graphs

TL;DR

This paper presents a polynomial time algorithm to determine whether a cubic bipartite graph contains a Hamilton cycle, addressing a fundamental problem in graph theory.

Contribution

The paper introduces the first polynomial time algorithm for detecting Hamilton cycles specifically in cubic bipartite graphs.

Findings

Algorithm successfully identifies Hamilton cycles in cubic bipartite graphs

Complexity of the problem is reduced to polynomial time for this class

Provides a new tool for analyzing Hamiltonicity in specific graph classes

Abstract

We provide a polynomial time algorithm to determine a cubic bipartite graph has a hamilton cycle or not.

Full text

keywords : hamilton cycle, cubic bipartite graph, NP-complete problem

MSC : 05C45

1 Notation and property

In this paper, a graph $G:=(V,E)$ is finite, undirected, and simple with the vertex set $V$ and edge set $E$. We follow [1] for basic notation. For a vertex $v\in V(G)$, the open neighborhood, denoted by $N_{G}(v)$, is $\{u\in V(G)\colon\ uv\in E(G)\}$, and the closed neighborhood, denoted by $N_{G}[v]$, is $N_{G}(v)\cup\{v\}$, also for a set $W\subseteq V(G)$, let $N_{G}(W):=\bigcup_{v\in W}N_{G}(v)$ and $N_{G}[W]:=N_{G}(W)\cup W$. For an edge $e\in E(G)$, let $V(e)$ be the set of two ends of $e$.

Proposition A** ([2]).**

Every bridgeless cubic graph has a perfect matching.

In a cubic graph, the complement of a perfect matching is always a 2-factor.

2 Hamiltonian cubic bipartite graphs

We introduce functions ${\bf HC}$ and ${\bf DC}$ as follows.

${\bf HC}$ : Input a cubic bipartite graph $G$.

(1) Obtain a perfect matching $M$ of $G$, and let $F:=G-M$. Note that $F$ is a 2-factor of $G$.

(2) Let $M^{\prime}\subseteq M$ be a maximal subset of $M$ such that for every $m\in M^{\prime}$, both ends of $m$ are not part of the same cycle of $F$. Let $M^{\prime\prime}:=M\setminus M^{\prime}$. Let $J:=G$.

(3) If ${\bf DC}(F,M^{\prime},M^{\prime\prime},J)=\emptyset$ then go to (7).

(4) Let $C:={\bf DC}(F,M^{\prime},M^{\prime\prime},J)$. Let $M(C):=M\cap E(C)$ and $F(C):=E(C)\setminus M(C)$. Let $F:=F-F(C)+M(C)$. If $F$ is a hamilton cycle then go to (7).

(5) For $M:=E(G)\setminus E(F)$, redefine $M^{\prime}$ and $M^{\prime\prime}$ as (2). Let $J:=G$.

(6) Repeat (4) - (5) at most $|V(G)|$ times.

(7) Return $F$.

${\bf DC}$ : Input $F,M^{\prime},M^{\prime\prime}$, and $J$.

(1) Choose $m\in M^{\prime}\cup M^{\prime\prime}$ arbitrarily and let $v_{1},v_{2}\in V(m)$. If $d_{J}(v_{1})=3$ and $v_{1}$ is an end of some marked edge, then for a vertex $w\in N_{J}(v_{1})\setminus\{v_{2}\}$ such that $v_{1}w$ is not marked, let $J:=J-v_{1}w$. If $d_{J}(v_{2})=3$ and $v_{2}$ is an end of some marked edge, then for a vertex $w\in N_{J}(v_{2})\setminus\{v_{1}\}$ such that $v_{2}w$ is not marked, let $J:=J-v_{2}w$.

(2) Repeat (1) until in (1) all $m\in M^{\prime}\cup M^{\prime\prime}$ are chosen.

(3) Choose $m\in M^{\prime}\cup M^{\prime\prime}$ arbitrarily and let $v_{1},v_{2}\in V(m)$. If $d_{J}(v_{1})=2$, then for a vertex $w\in N_{J}(v_{1})\setminus\{v_{2}\}$, mark $v_{1}w$. If $d_{J}(v_{2})=2$, then for a vertex $w\in N_{J}(v_{2})\setminus\{v_{1}\}$, mark $v_{2}w$.

(4) Repeat (1) - (3) until in (3) all $m\in M^{\prime}\cup M^{\prime\prime}$ are chosen.

(5) Choose $m\in M^{\prime}$ arbitrarily and let $v_{1},v_{2}\in V(m)$. If $d_{J}(v_{1})=3$ and $v_{1}$ is not an end of any marked edge, then let $w_{1},w_{2}\in N_{J}(v_{1})\setminus\{v_{2}\}$, $x_{1}\in N_{F}(w_{1})\setminus\{v_{1}\}$, and $x_{2}\in N_{F}(w_{2})\setminus\{v_{1}\}$. If $w_{1}\in V(m^{\prime})$ for some $m^{\prime}\in M^{\prime}$, and $J-v_{1}w_{2}-w_{1}x_{1}$ contains a cycle $C$ such that $v_{1}w_{1}\in E(C)$ and $N_{J-v_{1}w_{2}-w_{1}x_{1}}(V(C))\setminus V(C)=\emptyset$, then mark $v_{1}w_{1}$ and let $J:=J-v_{1}w_{2}$. If $w_{2}\in V(m^{\prime})$ for some $m^{\prime}\in M^{\prime}$, and $J-v_{1}w_{1}-w_{2}x_{2}$ contains a cycle $C$ such that $v_{1}w_{2}\in E(C)$ and $N_{J-v_{1}w_{1}-w_{2}x_{2}}(V(C))\setminus V(C)=\emptyset$, then mark $v_{1}w_{2}$ and let $J:=J-v_{1}w_{1}$. If $d_{J}(v_{2})=3$ and $v_{2}$ is not an end of any marked edge, then let $w_{1},w_{2}\in N_{J}(v_{2})\setminus\{v_{1}\}$, $x_{1}\in N_{F}(w_{1})\setminus\{v_{2}\}$, and $x_{2}\in N_{F}(w_{2})\setminus\{v_{2}\}$. If $w_{1}\in V(m^{\prime})$ for some $m^{\prime}\in M^{\prime}$, and $J-v_{2}w_{2}-w_{1}x_{1}$ contains a cycle $C$ such that $v_{2}w_{1}\in E(C)$ and $N_{J-v_{2}w_{2}-w_{1}x_{1}}(V(C))\setminus V(C)=\emptyset$, then mark $v_{2}w_{1}$ and let $J:=J-v_{2}w_{2}$. If $w_{2}\in V(m^{\prime})$ for some $m^{\prime}\in M^{\prime}$, and $J-v_{2}w_{1}-w_{2}x_{2}$ contains a cycle $C$ such that $v_{2}w_{2}\in E(C)$ and $N_{J-v_{2}w_{1}-w_{2}x_{2}}(V(C))\setminus V(C)=\emptyset$, then mark $v_{2}w_{2}$ and let $J:=J-v_{2}w_{1}$.

(6) Repeat (1) - (5) until in (5) all $m\in M^{\prime}$ are chosen.

(7) Choose $m\in M^{\prime}$ arbitrarily and let $v_{1},v_{2}\in V(m)$. If $d_{J}(v_{1})=3$ and $v_{1}$ is not an end of any marked edge, then let $w_{1},w_{2}\in N_{J}(v_{1})\setminus\{v_{2}\}$. If $w_{1}\in V(m^{\prime})$ for some $m^{\prime}\in M^{\prime}$, and $w_{2}\in V(m^{\prime\prime})$ for some $m^{\prime\prime}\in M^{\prime\prime}$, then mark $v_{1}w_{1}$ and let $J:=J-v_{1}w_{2}$. If $w_{2}\in V(m^{\prime})$ for some $m^{\prime}\in M^{\prime}$, and $w_{1}\in V(m^{\prime\prime})$ for some $m^{\prime\prime}\in M^{\prime\prime}$, then mark $v_{1}w_{2}$ and let $J:=J-v_{1}w_{1}$. If $d_{J}(v_{2})=3$ and $v_{2}$ is not an end of any marked edge, then let $w_{1},w_{2}\in N_{J}(v_{2})\setminus\{v_{1}\}$. If $w_{1}\in V(m^{\prime})$ for some $m^{\prime}\in M^{\prime}$, and $w_{2}\in V(m^{\prime\prime})$ for some $m^{\prime\prime}\in M^{\prime\prime}$, then mark $v_{2}w_{1}$ and let $J:=J-v_{2}w_{2}$. If $w_{2}\in V(m^{\prime})$ for some $m^{\prime}\in M^{\prime}$, and $w_{1}\in V(m^{\prime\prime})$ for some $m^{\prime\prime}\in M^{\prime\prime}$, then mark $v_{2}w_{2}$ and let $J:=J-v_{2}w_{1}$.

(8) Repeat (1) - (7) until in (7) all $m\in M^{\prime}$ are chosen.

(9) Choose $m\in M^{\prime}$ arbitrarily and let $v_{1},v_{2}\in V(m)$. Apply either (*1) or (*2) if both conditions of (*1) and (*2) hold. (*1) If $d_{J}(v_{1})=3$ and $v_{1}$ is not an end of any marked edge, then for two vertices $w_{1},w_{2}\in N_{J}(v_{1})\setminus\{v_{2}\}$, either (a) mark $v_{1}w_{1}$ and let $J:=J-v_{1}w_{2}$, or (b) mark $v_{1}w_{2}$ and let $J:=J-v_{1}w_{1}$. (*2) If $d_{J}(v_{2})=3$ and $v_{2}$ is not an end of any marked edge, then for two vertices $w_{1},w_{2}\in N_{J}(v_{2})\setminus\{v_{1}\}$, either (a) mark $v_{2}w_{1}$ and let $J:=J-v_{2}w_{2}$, or (b) mark $v_{2}w_{2}$ and let $J:=J-v_{2}w_{1}$.

(10) Repeat (1) - (9) until in (9) all $m\in M^{\prime}$ are chosen.

(11) Let $K^{\prime}$ be the union of all disjoint cycles in $J$. Let $k^{\prime}$ be the number of components in $F\cup K^{\prime}$.

(12) Choose $m\in M^{\prime}\cap K^{\prime}$ arbitrarily and let $v_{1},v_{2}\in V(m)$. If $N_{K^{\prime}}(v_{1})\setminus\{v_{2}\}$ or $N_{K^{\prime}}(v_{2})\setminus\{v_{1}\}$ is an end of some edge in $M^{\prime}$, then check $m$.

(13) Repeat (12) until in (12) all $m\in M^{\prime}\cap K^{\prime}$ are chosen.

(14) Choose $m\in M^{\prime}\cap K^{\prime}$ arbitrarily and let $v_{1},v_{2}\in V(m)$. If $N_{K^{\prime}}(v_{1})\setminus\{v_{2}\}$ or $N_{K^{\prime}}(v_{2})\setminus\{v_{1}\}$ is an end of some edge in $M^{\prime\prime}$, then highlight $m$.

(15) Repeat (14) until in (14) all $m\in M^{\prime}\cap K^{\prime}$ are chosen.

(16) Choose a cycle $K\subseteq K^{\prime}$ arbitrarily. Let $k$ be the number of components in $F\cup(K^{\prime}-K)$. If $E(K)$ dose not contain any checked edges and $k=k^{\prime}$, then let $K^{\prime}:=K^{\prime}-K$.

(17) Repeat (16) until in (16) all cycles $K\subseteq K^{\prime}$ are chosen.

(18) Choose a cycle $K\subseteq K^{\prime}$ arbitrarily. Let $k$ be the number of components in $F\cup(K^{\prime}-K)$. If $E(K)$ contains some highlighted edges and $k=k^{\prime}$, then let $K^{\prime}:=K^{\prime}-K$.

(19) Repeat (18) until in (18) all cycles $K\subseteq K^{\prime}$ are chosen.

(20) Choose a cycle $K\subseteq K^{\prime}$ arbitrarily. Let $k$ be the number of components in $F\cup(K^{\prime}-K)$. If $k=k^{\prime}$, then let $K^{\prime}:=K^{\prime}-K$.

(21) Repeat (20) until in (20) all cycles $K\subseteq K^{\prime}$ are chosen.

(22) Return $K^{\prime}$.

Let $G$ be a cubic bipartite graph. ${\bf HC}(G)$ is a 2-factor constructed by applying ${\bf HC}$ to $G$. Note that ${\bf HC}(G)$ is not unique and taken arbitrarily.

Proposition 2.1**.**

If $G$ is hamiltonian then ${\bf HC}(G)$ is a hamilton cycle.

Proof.

Suppose that $G$ has a hamilton cycle $H$ and a 2-factor $F$ of $G$ is not a hamilton cycle. Then all edges of $(E(H)\setminus E(F))\cup(E(F)\setminus E(H))$ and their ends form the union of disjoint cycles $K^{\prime\prime}$ and $F\cup K^{\prime\prime}$ is connected. Note that $E(H)\setminus E(F)\subseteq M$. Also note that every $f\in E(F)\setminus E(H)$ is an edge between two ends of two edges of $E(H)\setminus E(F)$. By the definition of $J$, $K^{\prime}\neq\emptyset$ in ${\bf DC}$ (22). Now, it is impossible for ${\bf DC}$ to return the empty set. Let $F_{n^{\prime}}$ be $F$ led after $n^{\prime}$ times of ${\bf HC}$ (4), let $F_{n^{\prime\prime}}$ be $F$ led after $n^{\prime\prime}$ times of ${\bf HC}$ (4), let $c_{n^{\prime}}$ be the number of cycles in $F_{n^{\prime}}$, and let $c_{n^{\prime\prime}}$ be the number of cycles in $F_{n^{\prime\prime}}$ ($0\leq n^{\prime}<n^{\prime\prime}\leq|V(G)|-1$). Then, by ${\bf DC}$ (5), (7), (16), (18), and (20), for every $n^{\prime}$, there exists $n^{\prime\prime}$ such that $n^{\prime\prime}-n^{\prime}\leq 3$ and $c_{n^{\prime}}>c_{n^{\prime\prime}}$. Therefore, ${\bf HC}$ returns a hamilton cycle at most $|V(G)|$ times of the repetition. ∎

Theorem 2.1**.**

$G$* is hamiltonian if and only if ${\bf HC}(G)$ is a hamilton cycle.*

Proof.

Suppose that $G$ is hamiltonian. By Proposition 2.1, the sufficiency holds. The necessity is obvious from the statement. ∎

Proposition 2.2**.**

All steps of ${\bf HC}$ (including ${\bf DC}$) are computed in polynomial time.

Proof.

It is straightforward from the definitions of ${\bf HC}$ and ${\bf DC}$. ∎