The Complexity of HCP in Digraps with Degree Bound Two

TL;DR

This paper demonstrates that the Hamiltonian cycle problem in directed graphs with maximum degree two can be solved in polynomial time using novel mappings, challenging the typical NP-complete classification.

Contribution

It introduces two mappings that establish polynomial solutions for HCP in degree-two digraphs and claims to prove P=NP based on these results.

Findings

HCP in degree-two digraphs is polynomially solvable.

A bijection between incidence matrices of digraphs and bipartite graphs.

Finding non-isomorphic Hamiltonian cycles is also polynomial.

Abstract

The Hamiltonian cycle problem (HCP) in digraphs D with degree bound two is solved by two mappings in this paper. The first bijection is between an incidence matrix C_{nm} of simple digraph and an incidence matrix F of balanced bipartite undirected graph G; The second mapping is from a perfect matching of G to a cycle of D. It proves that the complexity of HCP in D is polynomial, and finding a second non-isomorphism Hamiltonian cycle from a given Hamiltonian digraph with degree bound two is also polynomial. Lastly it deduces P=NP base on the results.

Equations

$$\forall x_i\in X\wedge 1\le d(x_i)\le 2$$

$$\forall y_i\in Y\wedge 1\le d(y_i)\le 2$$

$$c_{ij}=\left\{\begin{array}[]{ll}1,&\mbox{if $v_{i}\in e_{j}$;}\\ 0,&\mbox{otherwise.}\end{array}\right.$$

$$c_{ij}=\left\{\begin{array}[]{ll}1,&\mbox{if $<v_{i},v_{i}>\bowtie a_{j}=v_{i}$;}\\ -1,&\mbox{if $a_{j}\bowtie<v_{i},v_{i}>=v_{i}$;}\\ 0,&\mbox{$otherwise$}.\end{array}\right.$$

$$r(C)=n-k$$

$$C^{+}=\left\{c_{ij}|c_{ij}\geq 0\mbox{ otherwise $0$ }\right\}$$

$$C^{-}=\left\{c_{ij}|c_{ij}\leq 0\mbox{ otherwise $0$ }\right\}$$

$$F=\left({\begin{array}[]{c c}C^{+}\\ -C^{-}\end{array}}\right)$$

$$|U|=n$$

$$|V|=n$$

$$|E|=m$$

$$\forall u_i\in U\wedge 1\le d(u_i)\le 2$$

$$\forall v_i\in V\wedge 1\le d(v_i)\le 2$$

$$x=q-k<\frac{n}{4}$$

$$C=\left({\begin{array}[]{llllll}1&0&0&\ldots&0&-1\\ -1&1&0&\ldots&0&0\\ 0&-1&1&\ldots&0&0\\ 0&0&-1&\ldots&0&0\\ 0&0&0&\ldots&0&0\\ 0&0&0&\ldots&-1&1\end{array}}\right).$$

$$F=\left({\begin{array}[]{c c}C^{+}\\ -C^{-}\end{array}}\right)$$

$$G_{i}=\left\{\begin{array}[]{ll}1,&\mbox{if $G_{i}\cap M=\{e_{j},e_{k},\ldots\}$;}\\ 0,&\mbox{if $G_{i}\cap M=\{e_{l},e_{q},\ldots\}$.}\end{array}\right.$$

$$M^{\prime}=\left\{\begin{array}[]{ll}M(t)\otimes G_{t},&\mbox{ $G_{t}$ is a cycle };\\ M(t),&\mbox{otherwise.}\end{array}\right.$$

$$M(t+1)=\left\{\begin{array}[]{ll}M^{\prime},&\mbox{if $r(F^{-1}(M^{\prime}))>r(F^{-1}(M(t)))$ };\\ M(t),&\mbox{otherwise.}\end{array}\right.$$

$$M(t+1)=M^{\prime}\mbox{ if $r(F^{-1}(M^{\prime}))\geq r(F^{-1}(M(t)))$ }$$

Full text

1 Introduction

It is well known that the Hamiltonian cycle problem(HCP) is one of the standard NP-complete problem [1]. As for digraphs, even when the digraphs on this case: planar digraphs with indegree 1 or 2 and outdegree 2 or 1 respectively, it is still $NP-Complete$ which is proved by J.Plesník [2].

Let us named a simple strong connected digraphs with at most indegree 1 or 2 and outdegree 2 or 1 as $\Gamma$ digraphs. This paper solves the HCP of $\Gamma$ digraphs with following main results.

Theorem 1.1

Given an incidence matrix $C_{nm}$ of $\Gamma$ digraph, building a mapping:F=\left({\begin{array}[]{c c}C^{+}\\ -C^{-}\end{array}}\right), then $F$ is a incidence matrix of undirected balanced bipartite graph $G(X,Y;E)$, which obeys the following properties:

1. c1.

$|X|=n$,$|Y|=n$,$|E|=m$** 2. c2.

[TABLE]

[TABLE] 3. c3.

$G$* has at most $\frac{n}{4}$ components which is length of $4$.*

Let us named the undirected balanced bipartite graph $G(X,Y:E)$ of $\Gamma$ digraph as projector graph.

Theorem 1.2

Let $G$ be the projector graph of a $\Gamma$ graph $D(V,A)$, determining a Hamiltonian cycle in $\Gamma$ digraph is equivalent to find a perfect match $M$ in $G$ and $r(C^{\prime})=n-1$, where $C^{\prime}$ is the incidence matrix of $D^{\prime}(V,L)\subseteq D$ and $L=\{a_{i}|a_{i}\in D\wedge e_{i}\in M\}$.

Let the each component of $G$ corresponding to a boolean variable, a monotonic function $f(M)$ is build to represents the number of component in $D$. Based on this function, the maximum number of non-isomorphism perfect matching is linear, thus complexity of $\Gamma$ digraphs has a answer.

Theorem 1.3

Given the incidence matrix $C_{nm}$ of a $\Gamma$ digraph , the complexity of finding a Hamiltonian cycle existing or not is $O(n^{4})$

The concepts of cycle and rank of graph are given in section $2$. Then theorems $1$,$2$,$3$ are proved in sections $3$,$4$,$5$ respectively. The last section discusses the $P$ versus $NP$ in more detail.

2 Definition and properties

Throughout this paper we consider the finite simple (un)directed graph $D=(V,A)$ ($G(V,E)$, respectively), i.e. the graph has no multi-arcs and no self loops. Let $n$ and $m$ denote the number of vertices $V$ and arcs $A$ (edges $E$, respectively), respectively.

As conventional, let $|S|$ denote the number of a set $S$. The set of vertices $V$ and set of arcs of $A$ of a digraph $D(V,A)$ are denoted by $V=\{v_{i}|1\leq i\leq n\}$ and $A=\{a_{j}|(1\leq j\leq m)\wedge a_{j}=<v_{i},v_{k}>,(v_{i}\neq v_{k}\in V)\}$ respectively, where $<v_{i},v_{k}>$ is a arc from $v_{i}$ to $v_{k}$. Let the out degree of vertex $v_{i}$ denoted by $d^{+}(v_{i})$, which has the in degree by denoted as $d^{-}(v_{i})$ and has the degree $d(v_{i})$ which equals $d^{+}(v_{i})+d^{-}(v_{i})$. Let the $N^{+}(v_{i})=\{v_{j}|<v_{i},v_{j}>\in A\}$, and $N^{-}(v_{i})=\{v_{j}|<v_{j},v_{i}>\in A\}$.

Let us define a forward relation $\bowtie$ between two arcs as following, $a_{i}\bowtie a_{j}=v_{k}\>\mbox{iff}\>a_{i}=<v_{i},v_{k}>\wedge a_{j}=<v_{k},v_{j}>$. It is obvious that $|a_{i}\bowtie a_{i}|=0$ .

A *cycle * $L$ is a set of arcs $(a_{1},a_{2},\ldots,a_{l})$ in a digraph $D$, which obeys two conditions:

1. c1.

$\forall a_{i}\in L,\exists a_{j},a_{k}\in L\setminus\{a_{i}\},\;a_{i}\bowtie a_{j}\neq a_{j}\bowtie a_{k}\in V$ 2. c2.

$|\bigcup\limits_{a_{i}\neq a_{j}\in L}{a_{i}\bowtie a_{j}}|=|L|$

If a cycle $L$ obeys the following conditions, it is a simple cycle.

1. c3.

$\forall L^{\prime}\subset L$, $L^{\prime}$ does not satisfy both conditions $c1$ and $c2$.

A Hamiltonian cycle $L$ is also a simple cycle of length $n=|V|\geq 2$ in digraph. As for simplify, this paper given a sufficient condition of Hamiltonian cycle in digraph.

Lemma 1

If a digraph $D(V,A)$ include a sub graph $D^{\prime}(V,L)$ with following two properties, the $D$ is a Hamiltonian graph.

1. c1.

$\forall v_{i}\in D^{\prime}\rightarrow d^{+}(v_{i})=1\wedge d^{-}(v_{i})=1$, 2. c2.

$|L|=|V|\geq 2$* and $D^{\prime}$ is a strong connected digraph.*

A graph that has at least one Hamiltonian cycle is called a Hamiltonian graph. A graph G=$(V;E)$ is bipartite if the vertex set $V$ can be partitioned into two sets $X$ and $Y$ (the bipartition) such that $\exists e_{i}\in E,x_{j}\in X,\forall x_{k}\in X\setminus\{x_{j}\}$, $(e_{i}\bowtie x_{j}\neq\emptyset\rightarrow e_{i}\bowtie x_{k}=\emptyset)$ ($e_{i},Y$, respectively). if $|X|=|Y|$, We call that $G$ is a balanced bipartite graph. A matching $M\subseteq E$ is a collection of edges such that every vertex of $V$ is incident to at most one edge of $M$, a matching of balanced bipartite graph is perfect if $|M|=|X|$. Hopcroft and Karp shows that constructs a perfect matching of bipartite in $O((m+n)\sqrt{n})$ [3]. The matching of bipartite has a relation with neighborhood of $X$.

Theorem 2.1

[4*]**

A bipartite graph $G=(X,Y;E)$ has a matching from $X$ into $Y$ if and only if $|N(S)|\geq S$, for any $S\subseteq X$.*

Lemma 2

A even length of simple cycle consist of two disjoin perfect matching.

Two matrices representation for graphs are defined as follows.

Definition 1

[5] The incidence matrix $C$ of undirected graph $G$ is a two dimensional $n\times m$ table, each row represents one vertex, each column represents one edge, the $c_{ij}$ in $C$ are given by

[TABLE]

It is obvious that every column of an incidence matrix has exactly two $1$ entries.

Definition 2

[5] The incidence matrix $C$ of directed graph $D$ is a two dimensional $n\times m$ table, each row represents one vertex, each column represents one arc the $c_{ij}$ in $C$ are given by

[TABLE]

It is obvious to obtain a corollary of the incidence matrix as following.

Corollary 1

Each column of an incidence matrix of digraph has exactly one $1$ and one $-1$ entries.

Theorem 2.2

[5]** The $C$ is the incidence matrix of a directed graph with $k$ components the rank of $C$ is given by

[TABLE]

In order to convince to describe the graph $D$ properties, in this paper, we denotes the $r(D)=r(C)$.

3 Divided incidence matrix and Projector incidence matrix

Firstly, let us divided the matrix of $C$ into two groups.

[TABLE]

[TABLE]

It is obvious that the matrix of $C^{+}$ represents the forward arc of a digraph and $C^{-}$ matrix represents the backward arc respectively. A corollary is deduced as following.

Corollary 2

A digraph $D=(V,A)$ is strong connected if and only if the rank of divided incidence matrix satisfies $r(C^{+})=r(C^{-})=|V|$.

Secondly, let us combined the the $C^{+}$ and $C^{-}$ as following matrix.

[TABLE]

In more additional, let $F$ represents as an incidence matrix of undirected graph $G(X,Y;E)$. The $F$ is named as *projector incidence matrix * of $C$ and $G$ is named as *projector graph *, where $X$ represents the vertices $V^{+}$ of $D$, $Y$ represents the vertices of $V^{-}$ respectively. In another words we build a mapping $F:D\rightarrow G$ and denotes it as $G=F(D)$. So the $F(D)$ has $2n$ vertices and $m$ edges if $D$ has $n$ vertices and $m$ arcs. We also build up a reverse mapping: $F^{-1}:G\rightarrow D$ When $G$ is a projector graph. To simplify, we also denotes the arcs $a_{i}=F^{-1}(e_{i})$, $v_{i}^{+}=F^{-1}(x_{i})$ and $v_{i}^{-}=F^{-1}(y_{i})$.

3.1 Proof of Theorem 1.1

Firstly, let us prove the theorem 1.1.

Proof

1. c1.

Since $\Gamma$ digraph is strong connected, then each vertices of $\Gamma$ digraph has at least one forward arcs, each row of $C^{+}$ has at least one $1$ entries, and the $U$ represents the $C^{+}$ , so

[TABLE]

the same principle of $C^{-}$, each row of $C^{-}$ has at least one $-1$ entries, and the $V$ represents the $C^{-}$ , so

[TABLE]

Since the columns of $F$ equal to the columns of $C$,

[TABLE] 2. c2.

Since the degree of each $v_{i}$ of $\Gamma$ digraph is $1\leq d^{+}(v_{i})\leq 2$,

[TABLE]

Since the degree of each $v_{i}$ of $\Gamma$ digraph is $1\leq d^{-}(v_{i})\leq 2$,

[TABLE] 3. c3.

Let us prove by contradiction, suppose there are $k>\frac{n}{4}$ components with length of $4$ in $G$. Since $D$ is strong connected, according to the corollary 2, $r(F)=\frac{3n}{2}-q\geq r(C^{+})=n$, where $q\geq k$ is number of components (including $k$ components with length of $4$). Thus $q\leq\frac{n}{2}$, then there are only $x$ components without length $4$, where $x$ is

[TABLE]

Suppose the remind $x$ components with length of $t$ (at least $t$ vertices connected by some edges), then $4k+xt=\frac{3n}{2}$. So $tx=\frac{3n}{2}-4k<\frac{n}{2}$. According to the equation 7, the $t<2$. It is contradict that the $D$ is strong connected.

3.2 The cycle in digraph corresponding matching in projector graph

Secondly, let us given the properties after mapping Hamiltonian cycle $L$ of $D$ into the sub graph $M$ of projector graph $G$.

Lemma 3

If a Hamiltonian cycle $L$ of $D$ mapping into a forest $M$ of projector graph $G$, the forest $M$ consist of $|L|$ number of trees which has only two node and one edge, and $M$ has a unique perfect matching.

Proof

Let the $\Gamma$ digraph $D(V,A)$ has a sub digraph $D^{\prime}(V,L)$ which exists one Hamiltonian cycle and $|L|=n$, the incidence matrix $C$ of $L$ could be permutation as follows.

[TABLE]

Let

[TABLE]

It is obvious that each row of $F$ has only one $1$ entry and each column of $F$ has two $1$ entries.

According to theorem 1.1, $F$ represents a balanced bipartite graph $G(X,Y;E)$ that each vertex has one edge connected, and each edge $e_{i}$ connect on vertex $x_{i}\in X$ , another in $Y$, in another words, $\exists e_{i}\in E\,x_{j}\in X$,$\forall x_{k}\in X\setminus\{x_{j}\}$, $e_{i}\bowtie x_{j}\neq\emptyset\rightarrow e_{i}\bowtie x_{k}=\emptyset$($e_{i},Y$,respectively). According the matching definition, $M$ is a matching, since $|E|=|L|$, $E$ is a perfect matching. and pair of vertices between $X$ and $Y$ only has one edge, so $M$ is a forest, and each tree has only two node with one edge.

4 Proof of Theorem 1.2

Proof

$\Rightarrow$ Let the $\Gamma$ digraph $D(V,A)$ has a sub digraph $D^{\prime}(V,L)$ which is a Hamiltonian cycle and $|L|=n$, let matrix $C^{\prime}$ represents the incidence matrix of $D^{\prime}$, so $r(C^{\prime})=n-1$; According to lemma 3, the projector graph $F(D^{\prime})$ has a perfect matching, thus $F(D)$ also has a perfect matching.

$\Leftarrow$ Let $G(X,Y;E)$ be a projector graph of the $\Gamma$ graph $D(V,A)$,$M$ is a perfect matching in $G$. Let $D^{\prime}(V,L)$ be a sub graph of $D(V,A)$ and $L=\{a_{i}|a_{i}\in D\wedge e_{i}\in M\}$. Since $r(L)=n-1$, $D^{\prime}(V,L)$ is a strong connected digraph. it deduces that $\forall v_{i}\in D^{\prime}$,$d^{+}(v_{i})\geq 1\wedge d^{-}(v_{i})\geq 1$. Suppose $\exists v_{i}\in D^{\prime}$, $d^{+}(v_{i})>1$ ($d^{-}(v_{i})>1$ respectively), Since $|M|=n$, it deduces that $\sum_{i=1}^{n}d(v_{i})>2n+1$, which imply that $|L|>n$. this is contradiction with $L=\{a_{i}|a_{i}\in D\wedge e_{i}\in M\}$ and $|M|=n$. So $\forall v_{i}\in D^{\prime}$, $d^{+}(v_{i})=d^{-}(v_{i})=1$, According the lemma 1, $D^{\prime}$ has a Hamiltonian cycle.

5

Number of perfect matching in projector graph

Let us considering the number of perfect matching in $G$ . Firstly, let us considering a example as shown in figure 1.

\begin{picture}(10.0,4.0)\par \par \put(1.0,0.0){Figure 1. Original Digraph $D$} \put(1.0,1.0){\vector(0,1){0.8}} \put(1.0,1.0){\circle{0.5}} \put(1.2,1.4){$a_{8}$} \par\put(1.0,2.0){\vector(1,0){0.8}} \put(1.0,2.0){\circle{0.5}} \put(1.4,2.1){$a_{1}$} \par\put(2.0,1.0){\vector(-1,0){0.8}} \put(2.0,1.0){\circle{0.5}} \put(1.4,0.7){$a_{22}$} \par\put(2.0,2.0){\vector(0,-1){0.8}} \put(2.0,2.0){\vector(1,0){0.8}} \put(2.0,2.0){\circle{0.5}} \put(2.4,2.1){$a_{2}$} \put(2.2,1.4){$a_{9}$} \par\put(3.0,1.0){\vector(-1,0){0.8}} \put(3.0,1.0){\vector(0,1){0.8}} \put(3.0,1.0){\circle{0.5}} \put(2.4,0.7){$a_{21}$} \put(3.1,1.4){$a_{10}$} \par\put(3.0,2.0){\circle{0.5}} \put(3.0,2.0){\vector(1,0){0.8}} \put(3.4,2.1){$a_{3}$} \par\put(4.0,1.0){\vector(-1,0){0.8}} \put(4.0,1.0){\circle{0.5}} \put(3.4,0.7){$a_{20}$} \par\put(4.0,2.0){\vector(0,-1){0.8}} \put(4.0,2.0){\vector(1,0){0.8}} \put(4.0,2.0){\circle{0.5}} \put(4.4,2.1){$a_{4}$} \put(4.1,1.4){$a_{11}$} \par\put(5.0,1.0){\vector(-1,0){0.8}} \put(5.0,1.0){\vector(0,1){0.8}} \put(5.0,1.0){\circle{0.5}} \put(4.4,0.7){$a_{12}$} \put(5.1,1.4){$a_{19}$} \par\put(5.0,2.0){\circle{0.5}} \put(5.0,2.0){\vector(1,0){0.8}} \put(5.4,2.1){$a_{5}$} \par\put(6.0,1.0){\vector(-1,0){0.8}} \put(6.0,1.0){\circle{0.5}} \put(5.4,0.7){$a_{18}$} \par\put(6.0,2.0){\vector(0,-1){0.8}} \put(6.0,2.0){\vector(1,0){0.8}} \put(6.0,2.0){\circle{0.5}} \put(6.4,2.1){$a_{6}$} \put(6.1,1.4){$a_{13}$} \par\par\put(7.0,1.0){\vector(-1,0){0.8}} \put(7.0,1.0){\vector(0,1){0.8}} \put(7.0,1.0){\circle{0.5}} \put(6.4,0.7){$a_{17}$} \put(7.2,1.4){$a_{14}$} \par\put(7.0,2.0){\circle{0.5}} \put(7.0,2.0){\vector(1,0){0.8}} \put(7.4,2.1){$a_{7}$} \par\put(8.0,1.0){\vector(-1,0){0.8}} \put(8.0,1.0){\circle{0.5}} \put(7.4,0.7){$a_{16}$} \par\put(8.0,2.0){\circle{0.5}} \put(8.0,2.0){\vector(0,-1){0.8}} \put(8.4,1.4){$a_{15}$} \par\par\par\end{picture}

Figure 1. Original Digraph D$$a_{8}

$a_{1}$

$a_{22}$

a_{2}$$a_{9}

a_{21}$$a_{10}

$a_{3}$

$a_{20}$

a_{4}$$a_{11}

a_{12}$$a_{19}

$a_{5}$

$a_{18}$

a_{6}$$a_{13}

a_{17}$$a_{14}

$a_{7}$

$a_{16}$

$a_{15}$

Then the projector graph is shown in figure 2.

\begin{picture}(12.0,4.0)\par \par \put(1.0,0.0){Figure 2. Projector graph $G$} \put(1.0,1.0){\line(0,1){1.8}} \put(1.0,1.0){\circle{0.5}} \put(1.0,3.0){\circle{0.5}} \put(1.1,1.4){$e_{1}$} \par\put(1.5,1.0){\line(0,1){1.8}} \put(1.5,1.0){\circle{0.5}} \put(1.5,3.0){\circle{0.5}} \put(1.6,1.4){$e_{8}$} \par\put(2.0,1.0){\line(0,1){1.8}} \put(2.0,1.0){\circle{0.5}} \put(2.0,3.0){\circle{0.5}} \put(2.1,1.4){$e_{22}$} \par\put(3.0,1.0){\line(0,1){1.8}} \put(3.0,1.0){\line(1,2){1.0}} \put(3.0,1.0){\circle{0.5}} \put(3.0,3.0){\circle{0.5}} \put(2.5,2.0){$e_{9}$} \put(3.2,2.6){$e_{2}$} \par\put(3.3,0.4){$G_{1}$} \par\put(4.0,1.0){\line(0,1){1.8}} \put(4.0,1.0){\line(-1,2){1.0}} \put(4.0,1.0){\circle{0.5}} \put(4.0,3.0){\circle{0.5}} \put(4.1,1.4){$e_{10}$} \put(3.6,2.1){$e_{21}$} \par\par\par\put(4.6,1.0){\line(0,1){1.8}} \put(4.6,1.0){\circle{0.5}} \put(4.6,3.0){\circle{0.5}} \put(4.7,1.4){$e_{3}$} \par\put(5.3,1.0){\line(0,1){1.8}} \put(5.3,1.0){\circle{0.5}} \put(5.3,3.0){\circle{0.5}} \put(5.4,1.4){$e_{20}$} \par\put(6.0,1.0){\line(0,1){1.8}} \put(6.0,1.0){\line(1,2){1.0}} \put(6.0,1.0){\circle{0.5}} \put(6.0,3.0){\circle{0.5}} \put(5.8,2.0){$e_{11}$} \put(6.2,2.6){$e_{4}$} \par\put(6.3,0.4){$G_{2}$} \par\put(7.0,1.0){\line(0,1){1.8}} \put(7.0,1.0){\line(-1,2){1.0}} \put(7.0,1.0){\circle{0.5}} \put(7.0,3.0){\circle{0.5}} \put(7.1,1.4){$e_{19}$} \put(6.6,2.1){$e_{12}$} \par\put(7.7,1.0){\line(0,1){1.8}} \put(7.7,1.0){\circle{0.5}} \put(7.7,3.0){\circle{0.5}} \put(7.7,1.4){$e_{5}$} \par\put(8.3,1.0){\line(0,1){1.8}} \put(8.3,1.0){\circle{0.5}} \put(8.3,3.0){\circle{0.5}} \put(8.3,1.4){$e_{18}$} \par\put(9.0,1.0){\line(0,1){1.8}} \put(9.0,1.0){\line(1,2){1.0}} \put(9.0,1.0){\circle{0.5}} \put(9.0,3.0){\circle{0.5}} \put(8.6,2.0){$e_{6}$} \put(9.2,2.6){$e_{13}$} \par\put(9.3,0.4){$G_{3}$} \par\put(10.0,1.0){\line(0,1){1.8}} \put(10.0,1.0){\line(-1,2){1.0}} \put(10.0,1.0){\circle{0.5}} \put(10.0,3.0){\circle{0.5}} \put(9.6,2.1){$e_{17}$} \put(10.3,1.4){$e_{14}$} \put(10.6,2.0){$\ldots$} \par\end{picture}

Figure 2. Projector graph G$$e_{1}

$e_{8}$

$e_{22}$

e_{9}$$e_{2}

$G_{1}$

e_{10}$$e_{21}

$e_{3}$

$e_{20}$

e_{11}$$e_{4}

$G_{2}$

e_{19}$$e_{12}

$e_{5}$

$e_{18}$

e_{6}$$e_{13}

$G_{3}$

e_{17}$$e_{14}$$\ldots

Given a perfect matching $M$, each component(cycle) in $G$ has two partition edges belong to $M$. Let us code component $G_{i}$ which $|G_{i}|>2$ and matching $M$ to a binary variable.

[TABLE]

Now there are two cases for the number of perfect matching.

1. Label edge.

In that cases, the $Code(M_{1})=\{0,0,1\}$ is different with $Code(M_{2})=\{0,1,0\}$. If there are $k$ number of components(cycles), then there are $2^{k}$ perfect matching. 2. Unlabel edge.

In that cases, the $Code(M_{1})=\{0,0,1\}$ is isomorphic to $Code(M_{2})=\{0,1,0\}$. The same principle that $Code(M_{3})=\{0,1,1\}$ is isomorphic to $Code(M_{4})=\{1,1,0\}$ but is not isomorphic to $Code(M_{1})$.

Then let us summary the maximal number of perfect matching in these two cases.

Lemma 4

The maximal number of labeled perfect matching in a projector graph $G$ is $2^{\frac{n}{4}}$, but the maximal number of unlabeled perfect matching in a projector graph $G$ is $\frac{n}{2}$.

Proof

According to the theorem 1.1, there at most $\frac{n}{4}$ components with a components which is length of $k=4$. When $k$=2, there are only one perfect matching in $G$; When $k=4$, there are $\frac{n}{4}$ components which is $C_{4}$, and so on when $k=6$, there are $\frac{n}{6}$ components which is $C_{6}$, etc, so on. According to the lemma 2, each simple cycle has divided the perfect matching into two class. So maximal number perfect matching in the non isomorphism cycle which is $2^{\frac{n}{4}}$. Since in unlabeled cases, every $C_{4}$ cycle is isomorphism, the maximal number of perfect matching is $2*\frac{n}{4}=\frac{n}{2}$.

Review the example 1 again, it is easy find that follow proposition.

Proposition 1

Given two perfect matching $M1$ and $M2$ in projector graph $G$, if $code(M1)=code(M2)$, then the $r(F^{-1}(M1))=r(F^{-1}(M2))$.

5.1 Proof of Theorem 1.3

Now let us proof the theorem 1.3.

Proof

Let $G$ be a project balanced bipartition of $D$. According theorem 1.1, the $\Gamma$ graph is equivalent to find a perfect match $M$ in a project $G$.

According to the lemma 4, the maximal number non isomorphism perfect matching in $G$ is only $n$.

Thus it is only need exactly enumerate all of non isomorphism perfect matching $M$, then obtain the $value=r(F^{-1}(M))$,if $value=n-1$, then the $e_{i}\in M$ is also $e_{i}\in C$, where $C\subset D$ is a Hamiltonian cycle.

Since the complexity of rank of matrix is $O(n^{3})$, finding a simple cycle in a component with degree $2$ is $O(n^{2})$, and obtaining a perfect matching of a bipartite graph is $O((m+n)\sqrt{n})<O(n^{2})$ [3]. Then all exactly algorithms need to calculate the $n$ time $o(n^{3})$. Thus the complexity is $O(n^{4})$.

Since the non isomorphism perfect matching comes from the coding of edges in the component of $G$, it is not easy implementation.

Let us give two recursive equation to obtain a perfect matching $M$ from $G$. Suppose there are $k$ component $G_{1},G_{2},\ldots G_{k}$ in $G$ where $G_{i}$ is a component with degree $2$ and $|E_{i}|\geq 3$.

[TABLE]

[TABLE]

where $t\leq k-1$, when $t=0$, $M(0)$ is the initial perfect matching from $G$.

When $r(F^{-1}(M(t)))=n-1$, According the theorem 1.1, the $A=F^{-1}(M(t))$ is a Hamiltonian cycle solution. If all of $r(F^{-1}(M(t)))<n-1$, then there has no Hamiltonian cycle in $D$.

Since the non isomorphism perfect matching $M$ in $G$ is poset, the function $r(F^{-1}(M))$ in $G$ is monotonic, so this approach is exactly approach.

Let us give a example to illustrate the approach in detail.

Example 1

Considering the digraph $D$ in figure 1, then the projector graph $G$ in figure 2.

Let $M(0)=\{e_{1},e_{8},e_{22},e_{9},e_{10},e_{3},e_{20},e_{11},e_{19},e_{5},e_{18},e_{6},e_{17},e_{7},e_{15},e_{16}\}$.

Thus the $r(F^{-1}(M(0))=n-3$. Let $M^{\prime}=r(F^{-1}(M(0)\otimes G_{3})$,then $r(F^{-1}(M^{\prime})=n-4$, thus $M(1)=M(0)$ and then turn to $G_{2}$,$G_{1}$. At last it obtain the solution.

Considering the equation 11, let it substituted by following equations when $r(M^{\prime})=n-1$ and $t<k-1$.

[TABLE]

It is obvious that all non-isomorphism Hamiltonian cycle could obtain by the repeat check the equation 12 and the equation $r(M^{\prime})=n-1$.

In conversely, if a Hamiltonian cycle of $\Gamma$ digraphs is given, it represents a perfect matching $M$ in its projector graph $G$. Thus the equation 12 and Theorem 1.3 follows a corollary.

Corollary 3

Given a Hamiltonian $\Gamma$ digraph, the complexity of determining another non-isomorphism Hamiltonian cycle is polynomial time.

5.2 The HCP in digraph with bound two

Let us extend the Theorem 1.3 to digraphs with $d^{+}(v)\leq 2$ and $d^{-}(v)\leq 2$ in this section.

Theorem 5.1

The complexity of finding a Hamiltonian cycle existing or not in digraphs with degree $d^{+}(v)\leq 2$ and $d^{-}(v)\leq 2$ is polynomial time.

Proof

Suppose a digraph $D(V,A)$ having a vertex $v_{i}$ is shown as figure $3$, which is $d^{(}v_{i})=2\wedge d^{-}(v_{i})=2$

\begin{picture}(8.0,4.0)\par \par \par\put(1.0,0.0){Figure 3. A vertex with degree than 2} \put(1.0,1.0){\vector(2,1){1.8}} \par\par\put(1.0,3.0){\vector(2,-1){1.8}} \par\put(3.0,2.0){\circle{0.5}} \par\put(1.2,1.4){$a1$} \put(1.2,3.1){$a2$} \par\put(3.2,2.0){\vector(2,1){1.8}} \put(3.2,2.0){\vector(2,-1){1.8}} \par\put(4.2,1.0){$a3$} \par\put(4.2,2.8){$a4$} \par\par\end{picture}

Figure 3. A vertex with degree than 2

a1$$a2

$a3$

$a4$

Let us spilt this vertex to two vertices that one of vertex has degree with in degree 2 or out degree 1 , another vertex has degree with in degree 1 or out degree 2 as shown in figure $4$. Then the $D$ is derived to a new $\Gamma$ graph $S$.

\begin{picture}(10.0,4.0) \par\put(1.0,0.0){Figrue 4 A vertex in $D$ is mapping to a vertex in $\Gamma$ digraph} \par\put(1.0,1.0){\vector(2,1){1.8}} \par\put(1.0,2.0){\vector(1,0){1.8}} \par\put(3.0,2.0){\circle{0.3}} \par\put(1.2,1.4){$a1$} \put(1.2,2.1){$a2$} \par\put(3.2,2.0){\vector(1,0){1.8}} \par\par\put(5.6,1.4){$a3$} \par\par\par\put(5.0,2.0){\circle{0.3}} \put(6.2,2.8){$a4$} \put(5.2,2.0){\vector(2,1){1.8}} \put(5.2,2.0){\vector(1,0){1.8}} \par\end{picture}

Figrue 4 A vertex in $D$ is mapping to a vertex in $\Gamma$ digraph

a1$$a2

$a3$

$a4$

It is obvious that each vertex in the $\Gamma$ graph $S$ has increase $1$ vertices and $1$ arcs of $D$. Suppose the worst cases is each vertex in $D$ has in degree 2 and out degree 2, the total vertices in $S$ has $2n$ vertices.

According to the theorem 1.3, obtain a Hailtonian cycle $L^{\prime}$ in $S$ is no more then $O(n^{4})$, then the $D$ will has a Hamiltonian cycle $L^{\prime}=L\cap A$.

6

Discussion P versus NP

The $P$ versus $NP$ is a famous open problem in computer science and mathematics, which means to determine whether very language accepted by some nondeterministic algorithm in polynomial time is also accepted by some deterministic algorithm in polynomial time [6]. Cook give a proposition for the $P$ versus $NP$.

Proposition 2

If L is NP-complete and $L\in P$, then $P=NP$.

According above proposition and the result above section, $P$ versus $NP$ problem has a answer.

Theorem 6.1

$P=NP$**

Proof

As the result of [2], the complexity of HCP in digraph with bound two is $NP-complete$. According the theorem 5.1, the complexity of HCP in digraph with bound two is also $P$, thus according to proposition 2, $P=NP$.

In fact, the [2] proves that $3SAT\preceq_{p}HCP\;of\;\Gamma\;digraph$, since $3SAT$ is a $NPC$ problem, which also implies that $P=NP$.

7 Conclusion

According to the theorem 5.1, the complexity of determining a Hamiltonian cycle existence or not in digraph with bound degree two is in polynomial time. And according to the theorem 6.1, $P$ versus $NP$ problem has closed, $P=NP$.

Acknowledgements

The author would like to thank Prof. Kaoru Hirota for valuable suggestions, thank Prof. Jørgen Bang-Jensen who called mine attention to the paper [2], and thank Andrea Moro for useful discussions.