# 数学代写|组合优化代写Combinatorial optimization代考|CAAM574002

## 数学代写|组合优化代写Combinatorial optimization代考|Total Dual Integrality

In this and the next section we focus on integral polyhedra:
Definition 5.12. A polyhedron $P$ is integral if $P=P_I$.
Theorem 5.13. (Hoffman [1974], Edmonds and Giles [1977]) Let P be a rational polyhedron. Then the following statements are equivalent:
(a) $P$ is integral.
(b) Each face of $P$ contains integral vectors.
(c) Each minimal face of $P$ contains integral vectors.
(d) Each supporting hyperplane of $P$ contains integral vectors.
(e) Each rational supporting hyperplane of $P$ contains integral vectors.
(f) $\max {c x: x \in P}$ is attained by an integral vector for each $c$ for which the maximum is finite.
(g) $\max {c x: x \in P}$ is an integer for each integral $c$ for which the maximum is finite.

Proof: We first prove (a) $\Rightarrow$ (b) $\Rightarrow$ (f) $\Rightarrow$ (a), then (b) $\Rightarrow(\mathrm{d}) \Rightarrow(\mathrm{e}) \Rightarrow(\mathrm{c}) \Rightarrow(\mathrm{b})$, and finally (f) $\Rightarrow(\mathrm{g}) \Rightarrow(\mathrm{e})$
(a) $\Rightarrow$ (b): Let $F$ be a face, say $F=P \cap H$, where $H$ is a supporting hyperplane, and let $x \in F$. If $P=P_I$, then $x$ is a convex combination of integral points in $P$, and these must belong to $H$ and thus to $F$.
(b) $\Rightarrow$ (f) follows directly from Proposition 3.4, because ${y \in P: c y=$ $\max {c x: x \in P}}$ is a face of $P$ for each $c$ for which the maximum is finite.
(f) $\Rightarrow$ (a): Suppose there is a vector $y \in P \backslash P_I$. Then (since $P_I$ is a polyhedron by Theorem $5.1$ ) there is an inequality $a x \leq \beta$ valid for $P_I$ for which $a y>\beta$. Then clearly (f) is violated, since $\max {a x: x \in P}$ (which is finite by Proposition 5.2) is not attained by any integral vector.
(b) $\Rightarrow$ (d) is also trivial since the intersection of a supporting hyperplane with $P$ is a face of $P .(\mathrm{d}) \Rightarrow(\mathrm{e})$ and $(\mathrm{c}) \Rightarrow(\mathrm{b})$ are trivial.
(e) $\Rightarrow(\mathrm{c})$ : Let $P={x: A x \leq b}$. We may assume that $A$ and $b$ are integral. Let $F=\left{x: A^{\prime} x=b^{\prime}\right}$ be a minimal face of $P$, where $A^{\prime} x \leq b^{\prime}$ is a subsystem of $A x \leq b$ (we use Proposition 3.9). If $A^{\prime} x=b^{\prime}$ has no integral solution, then by Lemma $5.11$ – there exists a rational vector $y$ such that $c:=y A^{\prime}$ is integral but $\delta:=y b^{\prime}$ is not an integer. Adding integers to components of $y$ does not destroy this property ( $A^{\prime}$ and $b^{\prime}$ are integral), so we may assume that all components of $y$ are positive. Observe that $H:={x: c x=\delta}$ is a rational hyperplane which contains no integral vectors.

## 数学代写|组合优化代写Combinatorial optimization代考|Totally Unimodular Matrices

Definition 5.19. A matrix $A$ is totally unimodular if each subdeterminant of $A$ is $0,+1$, or $-1$.

In particular, each entry of a totally unimodular matrix must be $0,+1$, or $-1$. The main result of this section is:

Theorem 5.20. (Hoffman and Kruskal [1956]) An integral matrix $A$ is totally unimodular if and only if the polyhedron ${x: A x \leq b, x \geq 0}$ is integral for each integral vector $b$.

Proof: Let $A$ be an $m \times n$-matrix and $P:={x: A x \leq b, x \geq 0}$. Observe that the minimal faces of $P$ are vertices.

To prove necessity, suppose that $A$ is totally unimodular. Let $b$ be some integral vector and $x$ a vertex of $P . x$ is the solution of $A^{\prime} x=b^{\prime}$ for some subsystem $A^{\prime} x \leq b^{\prime}$ of $\left(\begin{array}{c}A \ -I\end{array}\right) x \leq\left(\begin{array}{l}b \ 0\end{array}\right)$, with $A^{\prime}$ being a nonsingular $n \times n$-matrix. Since $A$ is totally unimodular, $\mid$ det $A^{\prime} \mid=1$, so by Cramer’s rule $x=\left(A^{\prime}\right)^{-1} b^{\prime}$ is integral.
We now prove sufficiency. Suppose that the vertices of $P$ are integral for each integral vector $b$. Let $A^{\prime}$ be some nonsingular $k \times k$-submatrix of $A$. We have to show $\left|\operatorname{det} A^{\prime}\right|=1$. W.l.o.g., $A^{\prime}$ contains the elements of the first $k$ rows and columns of $A$.
Consider the integral $m \times m$-matrix $B$ consisting of the first $k$ and the last $m-k$ columns of $(A I)$ (see Figure 5.2). Obviously, $|\operatorname{det} B|=\left|\operatorname{det} A^{\prime}\right|$.

To prove $|\operatorname{det} B|=1$, we shall prove that $B^{-1}$ is integral. Since $\operatorname{det} B \operatorname{det} B^{-1}$ $=1$, this implies that $|\operatorname{det} B|=1$, and we are done.

Let $i \in{1, \ldots, m}$; we prove that $B^{-1} e_i$ is integral. Choose an integral vector $y$ such that $z:=y+B^{-1} e_i \geq 0$. Then $b:=B z=B y+e_i$ is integral. We add zero components to $z$ in order to obtain $z^{\prime}$ with
$$\left(\begin{array}{ll} A & I \end{array}\right) z^{\prime}=B z=b \text {. }$$
Now $z^{\prime \prime}$, consisting of the first $n$ components of $z^{\prime}$, belongs to $P$. Furthermore, $n$ linearly independent constraints are satisfied with equality, namely the first $k$ and the last $n-k$ inequalities of
$$\left(\begin{array}{c} A \ -I \end{array}\right) z^{\prime \prime} \leq\left(\begin{array}{l} b \ 0 \end{array}\right)$$

# 组合优化代考

## 数学代写|组合优化代写combinatoroptimization代考|Total Dual Integrality

.

(a) $P$是一个积分。
(b) $P$的每个面都包含积分向量。
(c) $P$的每个最小面都包含积分向量。
(d) $P$的每个支持超平面都包含积分向量。
(e) $P$的每个支持超平面都包含积分向量。
(f) $\max {c x: x \in P}$是由每个$c$的最大值的一个积分向量得到的。
(b) 的每个面都包含积分向量。
(c) 的每个最小面都包含积分向量。
(d) 的每个支持超平面都包含积分向量
(g) $\max {c x: x \in P}$是每个整数$c$的整数，其最大值为有限。

(a) $\Rightarrow$ (b):让 $F$ 要有面子，说 $F=P \cap H$，其中 $H$ 是支持的超平面，让 $x \in F$。如果 $P=P_I$，那么 $x$ 积分点的凸组合在吗 $P$，这些一定属于 $H$ 因此 $F$.
(b) $\Rightarrow$ (f)直接遵循命题3.4，因为 ${y \in P: c y=$ $\max {c x: x \in P}}$ 是一张 $P$ 对于每一个 $c$ 其最大值是有限的
(f) $\Rightarrow$ (a):假设有一个向量 $y \in P \backslash P_I$。Then (since) $P_I$ 根据定理，是一个多面体 $5.1$ )存在不平等 $a x \leq \beta$ 适用于 $P_I$ 因为 $a y>\beta$。那么显然(f)被违反了，因为 $\max {a x: x \in P}$ (根据命题5.2是有限的)不能由任何积分向量得到。
(b) $\Rightarrow$ (d)也很简单，因为支撑超平面与 $P$ 是一张 $P .(\mathrm{d}) \Rightarrow(\mathrm{e})$ 和 $(\mathrm{c}) \Rightarrow(\mathrm{b})$
(e) $\Rightarrow(\mathrm{c})$ :让 $P={x: A x \leq b}$。我们可以假设 $A$ 和 $b$ 都是积分。让 $F=\left{x: A^{\prime} x=b^{\prime}\right}$ 做一个最小的面孔 $P$，其中 $A^{\prime} x \leq b^{\prime}$ 是一个子系统 $A x \leq b$ (我们使用3.9号提案)。如果 $A^{\prime} x=b^{\prime}$ 没有积分解，那么由引理 $5.11$ 存在一个有理数向量 $y$ 如此这般 $c:=y A^{\prime}$ 是积分，但是 $\delta:=y b^{\prime}$ 不是整数。的组件中添加整数 $y$ 不会破坏此属性( $A^{\prime}$ 和 $b^{\prime}$ 是积分)，所以我们可以假设?的所有分量 $y$ 都是积极的。注意 $H:={x: c x=\delta}$ 是一个不包含积分向量的有理超平面。

## 数学代写|组合优化代写组合优化代考|完全单模矩阵

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