### 数学代写|组合优化代写Combinatorial optimization代考|Preliminaries

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## 数学代写|组合优化代写Combinatorial optimization代考|Preliminaries

The main result of this paper is the following theorem.
Theorem 2. Let $G=(V, A)$ be a directed graph, then $P C_{p}(G)$ is integral for any integer $p$ if and only if
(C1) it does not contain as a subgraph any of the graphs $H_{1}, H_{2}, H_{3}, H_{4}, H_{5}$, $H_{6}$ of Fig. 1, and
(C2) it does not contain a non-directed $g$-odd $Y$-cycle $C$ with an arc $(u, v)$ with both $u$ and $v$ not in $V(C)$.

Given a directed graph $G=(V, A)$, a subgraph induced by the nodes $v_{1}, \ldots, v_{r}$ of $D$ is called a bidirected cycle if the only arcs in this induced subgraph are $\left(v_{i}, v_{i+1}\right)$ and $\left(v_{i+1}, v_{i}\right)$, for $i=1, \ldots, r$, with $v_{r+1}=v_{1}$. We denote it by $B I C_{r}$.

For a directed graph $G=(V, A)$ and an arc $(u, v) \in A$, define $G(u, v)$ to be the graph obtained by removing $(u, v)$ from $G$, and adding a new arc $\left(u, v^{\prime}\right)\left(v^{\prime}\right.$ is a new pendent node). The rest of this section is devoted to definitions with respect to a feasible point in $P C_{p}(G)$.

Definition 3. A vector $(x, y) \in \mathbb{R}^{|A|+|V|}$ will be denoted by $z$, i.e., $z(u)=y(u)$ for all $u \in V$ and $z(u, v)=x(u, v)$ for all $(u, v) \in A$. Given a vector $z$ and $a$ labeling function $l: V \cup A \rightarrow{-1,0,1}$, we define a new vector $z_{l}$ from $z$ as follows:
$$\begin{gathered} z_{l}(u)=z(u)+l(u) \epsilon, \text { for all } u \in V, \text { and } \ z_{l}(u, v)=z(u, v)+l(u, v) \epsilon, \text { for all }(u, v) \in A \end{gathered}$$
where $\epsilon$ is a sufficiently small positive scalar. When we assign labels to some nodes and arcs without specifying the labels of the remaining nodes and arcs, it means that they are assigned the label zero.

Definition 4. When dealing with a vector $z \in P C_{p}(G)$, we say that the arc $(u, v)$ is tight if $z(u, v)=z(v)$. Also we say that an odd directed cycle $C$ is tight if $z(A(C))=(|A(C)|-1) / 2$.

## 数学代写|组合优化代写Combinatorial optimization代考|The Proof of Theorem 2

We will sketch quickly the necessity part of the proof. With each of the graphs $H_{1}, H_{2}, H_{3}$ and $H_{6}$ in Fig. 1 we show a fractional extreme point of $P C_{3}(G)$ when $G$ is restricted to these graphs. The graphs $H_{4}$ and $H_{5}$ show an extreme point of $P C_{2}(G)$. The numbers near the nodes correspond to $y$ variables. The $x$ variables take the value $\frac{1}{2}$ for $H_{1}, H_{2}$ and $H_{4}$. For the graphs $H_{3}, H_{5}$ and $H_{6}$ the arcs take the value $\frac{1}{3}$, except the arc in the right in $H_{6}$ that takes the value $\frac{2}{3}$.

To prove necessity we just have to notice that the extreme points for the subgraphs $H_{i}$, for $i=1, \ldots, 6$, in Fig. 1 may be extended to extreme points for any graph containing these subgraphs by setting each remaining node variable to one and each remaining arc variable to zero. For condition (C2) when the graph contains a non-directed g-odd $Y$-cycle $C$ with an arc $(u, v)$ having both nodes $u$ and $v$ not in $C$, we construct an extreme point of $P C(G)$ where $p=$ $\frac{|C|+|C|+1}{2}+|V|-|V(C)|-1$ as follows. All the nodes in $C$ take the value 0 , the nodes in $\bar{C}$ and $\hat{C}$ with the node $u$ take the value $\frac{1}{2} ;$ all the arcs in $C$ with $(u, v)$ take the value $\frac{1}{2}$. All other nodes take the value 1 and the other arcs take the value 0, except for each unique arc leaving each node in $\hat{C}$ (see the definition of a $Y$-cycle) they take the value $\frac{1}{2}$. One way to see that these are indeed extreme points is to start adding $\epsilon$ to one of the components and try to keep satisfying as equation the same constraints that the original vector satisfies as equation. First we conclude that we have to add or subtract $\epsilon$ to other components and this leads to the violation of Eq. (1) or to the impossibility of keeping tight the inequality that are satisfied as equation.

The rest of this section is devoted to the sufficiency part. Denotes by Pair $(G)$ the set of pair of nodes ${u, v}$ such that both arcs $(u, v)$ and $(v, u)$ exist. The proof of this theorem will be done by induction on the number of $|P a i r(G)|$. This result has been proved in [3] for oriented graphs, that is when $|P a i r(G)|=0$. This case is the starting point of the induction. Assume that Theorem 2 is true for any directed graph $H$ with $|P a i r(H)| \leq m$ and let us show that it holds for any directed graph $G$ with $|\operatorname{Pair}(G)|=m+1$. Let $G=(V, A)$ be a directed graph with $|P a i r(G)|=m+1 \geq 1$ satisfying conditions (C1) and (C2) of Theorem 2. Suppose the contrary, that is $P C_{p}(G)$ is not integral. Let $\bar{z}$ be a fractional extreme point of $P C_{p}(G)$. Next we will give some useful lemmas and then in Subsect. $3.1$ and $3.2$ the proof is completed. In Subsect. $3.1$ we show the theorem when there is no g-odd $Y$-cycle and in Subsect. $3.2$ we complete the proof when such a cycle exists.

## 数学代写|组合优化代写Combinatorial optimization代考|G Does Not Contain a Non-directed g-Odd Y -Cycle

When we have two arcs $(u, v)$ and $(v, u)$, from Lemma 3, the graph $G(u, v)$ satisfies condition (C1), which is not the case for condition (C2), even when $G$ does not contain a non-directed g-odd $Y$-cycle. For example we may have a g-odd cycle $C$ which is not a Y-cycle and an arc $(s, t)$ with both $s$ and $t$ not in $V(C)$. Now if we have an arc $(u, v)$ in $A(C)$ and $(v, u) \in A \backslash A(C)$, and if $v \in \hat{C}$ and $u \in C$ this same cycle may become a Y-cycle in $G(v, u)$ and so with the arc $(s, t)$ condition $(\mathrm{C} 2)$ is violated.

Next we will show that we may always find a pair of $\operatorname{arcs}(u, v)$ and $(v, u)$ where at least one of the graphs $G(u, v)$ or $G(v, u)$ satisfies (C2).

Lemma 8. Let $P=v_{1}, \ldots, v_{k}$ a maximal bidirected path, different from a bidirected cycle. We have the following

(i) If $G\left(v_{k}, v_{k-1}\right)$ contains a non-directed $g$-odd $Y$-cycle $C$, then the unique arc leaving $v_{k}$ is $\left(v_{k}, v_{k-1}\right)$,
(ii) If $G\left(v_{k-1}, v_{k}\right)$ contains a non-directed $g$-odd $Y$-cycle $C$, then there is at most one another arc leaving $v_{k-1}$ which is $\left(v_{k-1}, v_{k-2}\right)$.

Let $P=v_{1}, \ldots, v_{k}$ be a maximal bidirected path. From Lemma 4 the extremities of $P$ cannot coincide, that is $P$ is not a bidirected cycle. We will treat two cases (1) none of the arcs $\left(v_{k-1}, v_{k}\right)$ and $\left(v_{k}, v_{k-1}\right)$ belong to an odd directed cycle tight for $\bar{z}$ and of size at least five, (2) at least one of these arcs belong to such a cycle.

Case 1. None of the arcs $\left(v_{k-1}, v_{k}\right)$ and $\left(v_{k}, v_{k-1}\right)$ belong to an odd directed cycle of size at least five. From the lemma above it is easy to see that at least one of the graphs $G\left(v_{k}, v_{k-1}\right)$ or $G\left(v_{k-1}, v_{k}\right)$ does not contain a g-odd $Y$-cycle. In fact, assume that both graphs contain a g-odd $Y$-cycle. When $G\left(v_{k-1}, v_{k}\right)$ contains a g-odd $Y$-cycle $C$, we must have $v_{k} \in \dot{C}$. But this is impossible since Lemma 8 (i) implies that $\left(v_{k}, v_{k-1}\right)$ is the unique arc leaving $v_{k}$. Therefore, we only need to treat the following three cases, ordered as follows.

## 数学代写|组合优化代写Combinatorial optimization代考|Preliminaries

(C1) 它不包含任何图作为子图H1,H2,H3,H4,H5, H6图 1 和
(C2) 它不包含非定向G-奇怪的是-循环C带弧线(在,在)既在和在不在在(C).

## 数学代写|组合优化代写Combinatorial optimization代考|G Does Not Contain a Non-directed g-Odd Y -Cycle

(一) 如果G(在ķ,在ķ−1)包含一个非定向G-奇怪的是-循环C，则唯一弧离开在ķ是(在ķ,在ķ−1),
(ii) 如果G(在ķ−1,在ķ)包含一个非定向G-奇怪的是-循环C, 那么最多有一个弧离开在ķ−1这是(在ķ−1,在ķ−2).

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