数学代写|图论作业代写Graph Theory代考|$k$-Flows for small $k$

Doug I. Jones

Doug I. Jones

Lorem ipsum dolor sit amet, cons the all tetur adiscing elit

如果你也在 怎样代写图论Graph Theory 这个学科遇到相关的难题,请随时右上角联系我们的24/7代写客服。图论Graph Theory有趣的部分原因在于,图可以用来对某些问题中的情况进行建模。这些问题可以在图表的帮助下进行研究(并可能得到解决)。因此,图形模型在本书中经常出现。然而,图论是数学的一个领域,因此涉及数学思想的研究-概念和它们之间的联系。我们选择包含的主题和结果是因为我们认为它们有趣、重要和/或代表主题。

图论Graph Theory通过熟悉许多过去和现在对图论的发展负责的人,可以增强对图论的欣赏。因此,我们收录了一些关于“图论人士”的有趣评论。因为我们相信这些人是图论故事的一部分,所以我们在文中讨论了他们,而不仅仅是作为脚注。我们常常没有认识到数学是一门有生命的学科。图论是人类创造的,是一门仍在不断发展的学科。

couryes-lab™ 为您的留学生涯保驾护航 在代写图论Graph Theory方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写图论Graph Theory代写方面经验极为丰富,各种代写图论Graph Theory相关的作业也就用不着说。

数学代写|图论作业代写Graph Theory代考|$k$-Flows for small $k$

数学代写|图论作业代写Graph Theory代考|$k$-Flows for small $k$

Trivially, a graph has a 1-flow (the empty set) if and only if it has no edges. In this section we collect a few simple examples of sufficient conditions under which a graph has a 2-, 3- or 4-flow. More examples can be found in the exercises.

Proposition 6.4.1. A graph has a 2-flow if and only if all its degrees are even.

Proof. By Theorem 6.3.3, a graph $G=(V, E)$ has a 2-flow if and only if it has a $\mathbb{Z}_2$-flow, i.e. if and only if the constant $\operatorname{map} \vec{E} \rightarrow \mathbb{Z}_2$ with value $\overline{1}$ satisfies (F2). This is the case if and only if all degrees are even.

For the remainder of this chapter, let us call a graph even if all its vertex degrees are even.

Proposition 6.4.2. A cubic graph has a 3-flow if and only if it is bipartite.

Proof. Let $G=(V, E)$ be a cubic graph. Let us assume first that $G$ has a 3 -flow, and hence also a $\mathbb{Z}3$-flow $f$. We show that any cycle $C=x_0 \ldots x{\ell} x_0$ in $G$ has even length (cf. Proposition 1.6.1). Consider two consecutive edges on $C$, say $e_{i-1}:=x_{i-1} x_i$ and $e_i:=x_i x_{i+1}$. If $f$ assigned the same value to these edges in the direction of the forward orientation of $C$, i.e. if $f\left(e_{i-1}, x_{i-1}, x_i\right)=f\left(e_i, x_i, x_{i+1}\right)$, then $f$ could not satisfy (F2) at $x_i$ for any non-zero value of the third edge at $x_i$. Therefore $f$ assigns the values $\overline{1}$ and $\overline{2}$ to the edges of $C$ alternately, and in particular $C$ has even length.

Conversely, let $G$ be bipartite, with vertex bipartition ${X, Y}$. Since $G$ is cubic, the map $\vec{E} \rightarrow \mathbb{Z}3$ defined by $f(e, x, y):=\overline{1}$ and $f(e, y, x):=\overline{2}$ for all edges $e=x y$ with $x \in X$ and $y \in Y$ is a $\mathbb{Z}{3^{-}}$ flow on $G$. By Theorem 6.3.3, then, $G$ has a 3-flow.

数学代写|图论作业代写Graph Theory代考|Flow-colouring duality

In this section we shall see a surprising connection between flows and colouring: every $k$-flow on a plane multigraph gives rise to a $k$-vertexcolouring of its dual, and vice versa. In this way, the investigation of $k$-flows appears as a natural generalization of the familiar map colouring problems in the plane.

Let $G=(V, E)$ and $G^=\left(V^, E^\right)$ be dual plane multigraphs. For simplicity, let us assume that $G$ and $G^$ have neither bridges nor loops and are non-trivial. For edge sets $F \subseteq E$, let us write
$$
F^:=\left{e^ \in E^* \mid e \in F\right} .
$$
Conversely, if a subset of $E^$ is given, we shall usually write it immediately in the form $F^$, and thus let $F \subseteq E$ be defined implicitly via the bijection $e \mapsto e^*$.

Suppose we are given a circulation $g$ on $G^$ : how can we employ the duality between $G$ and $G^$ to derive from $g$ some information about $G$ ? The most general property of all circulations is Proposition 6.1.1, which says that $g(X, \bar{X})=0$ for all $X \subseteq V^$. By Proposition 4.6.1, the minimal cuts $E^(X, \bar{X})$ in $G^$ correspond precisely to the cycles in $G$. Thus if we take the composition $f$ of the maps $e \mapsto e^$ and $g$, and sum its values over the edges of a cycle in $G$, then this sum should again be zero.
Of course, there is still a technical hitch: since $g$ takes its arguments not in $E^$ but in $\overrightarrow{E^}$, we cannot simply define $f$ as above: we first have to refine the bijection $e \mapsto e^$ into one from $\vec{E}$ to $\overrightarrow{E^}$, i.e. assign to every $\vec{e} \in \vec{E}$ canonically one of the two directions of $e^*$. This will be the purpose of our first lemma. After that, we shall show that $f$ does indeed sum to zero along any cycle in $G$.

If $C=v_0 \ldots v_{\ell-1} v_0$ is a cycle with edges $e_i=v_i v_{i+1}$ (and $v_{\ell}:=v_0$ ), we shall call
$$
\vec{C}:=\left{\left(e_i, v_i, v_{i+1}\right) \mid i<\ell\right}
$$
a cycle with orientation. Note that this definition of $\vec{C}$ depends on the vertex enumeration chosen to denote $C$ : every cycle has two orientations. Conversely, of course, $C$ can be reconstructed from the set $\vec{C}$. In practice, we shall therefore speak about $C$ freely even when, formally, only $\vec{C}$ has been defined.

数学代写|图论作业代写Graph Theory代考|$k$-Flows for small $k$

图论代考

数学代写|图论作业代写Graph Theory代考|$k$-Flows for small $k$

简单地说,当且仅当图没有边时,它有一个1流(空集)。在本节中,我们收集了几个简单的充分条件示例,在这些条件下,图具有2流、3流或4流。更多的例子可以在练习中找到。

提案6.4.1当且仅当图的所有度都是偶数时,图具有2流。

证明。根据定理6.3.3,图$G=(V, E)$有一个2流当且仅当它有一个$\mathbb{Z}_2$流,即当且仅当值为$\overline{1}$的常数$\operatorname{map} \vec{E} \rightarrow \mathbb{Z}_2$满足(F2)。这是当且仅当所有度都是偶数时的情况。

在本章的剩余部分,我们称一个图为偶数,如果它所有顶点的度数都是偶数。

提案6.4.2。三次图具有3流当且仅当它是二部图。

证明。设$G=(V, E)$为三次图。首先假设$G$有一个3 -流,因此也有一个$\mathbb{Z}3$ -流$f$。我们证明$G$中的任何循环$C=x_0 \ldots x{\ell} x_0$都有偶数长度(参见命题1.6.1)。考虑$C$上的两条连续边,比如$e_{i-1}:=x_{i-1} x_i$和$e_i:=x_i x_{i+1}$。如果$f$在$C$的正向方向上给这些边赋相同的值,即如果$f\left(e_{i-1}, x_{i-1}, x_i\right)=f\left(e_i, x_i, x_{i+1}\right)$,那么对于$x_i$的第三条边的任何非零值,$f$都不能满足$x_i$的(F2)。因此,$f$将值$\overline{1}$和$\overline{2}$交替分配给$C$的边,特别是$C$具有偶数长度。

反过来,设$G$是二部的,顶点为二分${X, Y}$。由于$G$是立方的,由$f(e, x, y):=\overline{1}$和$f(e, y, x):=\overline{2}$对所有边$e=x y$与$x \in X$和$y \in Y$定义的映射$\vec{E} \rightarrow \mathbb{Z}3$是$G$上的$\mathbb{Z}{3^{-}}$流。根据定理6.3.3,$G$有一个3流。

数学代写|图论作业代写Graph Theory代考|Flow-colouring duality

在本节中,我们将看到流和着色之间的惊人联系:平面多图上的每个$k$ -流都会产生其对偶的$k$ -顶点着色,反之亦然。这样,对$k$ -flow的研究似乎是对平面上熟悉的地图着色问题的自然概括。

设$G=(V, E)$和$G^=\left(V^, E^\right)$为双平面多图。为简单起见,让我们假设$G$和$G^$既没有桥也没有循环,并且是非平凡的。对于边集$F \subseteq E$,我们这样写
$$
F^:=\left{e^ \in E^* \mid e \in F\right} .
$$
相反,如果给出了$E^$的一个子集,我们通常会立即将其写成$F^$的形式,因此让$F \subseteq E$通过双射$e \mapsto e^*$隐式地定义。

假设我们有一个循环 $g$ 在 $G^$ :我们如何利用两者之间的二元性 $G$ 和 $G^$ 源于… $g$ 一些关于 $G$ ? 所有循环中最普遍的性质是命题6.1.1,它说 $g(X, \bar{X})=0$ 对所有人 $X \subseteq V^$. 根据提案4.6.1,最低限度的削减 $E^(X, \bar{X})$ 在 $G^$ 精确地对应于 $G$. 因此,如果我们考虑组合 $f$ 在地图上 $e \mapsto e^$ 和 $g$,并将其值在循环的边缘上求和 $G$,那么这个和应该还是0。
当然,还有一个技术难题:since $g$ 不接受它的论点 $E^$ 但是在 $\overrightarrow{E^}$,我们不能简单地定义 $f$ 如上所述:我们首先必须改进双射 $e \mapsto e^$ Into one from $\vec{E}$ 到 $\overrightarrow{E^}$,即分配给每一个 $\vec{e} \in \vec{E}$ 的两个方向之一 $e^*$. 这就是我们第一个引理的目的。在那之后,我们将展示它 $f$ 确实沿着任意循环和为零吗 $G$.

如果$C=v_0 \ldots v_{\ell-1} v_0$是一个边为$e_i=v_i v_{i+1}$(和$v_{\ell}:=v_0$)的循环,我们将调用
$$
\vec{C}:=\left{\left(e_i, v_i, v_{i+1}\right) \mid i<\ell\right}
$$
有方向的循环注意,$\vec{C}$的定义依赖于用来表示$C$的顶点枚举:每个循环有两个方向。当然,反过来,$C$也可以由集合$\vec{C}$重建。因此,在实践中,我们将自由地谈论$C$,即使在形式上只定义了$\vec{C}$。

统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。

Days
Hours
Minutes
Seconds

hurry up

15% OFF

On All Tickets

Don’t hesitate and buy tickets today – All tickets are at a special price until 15.08.2021. Hope to see you there :)