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# 数学代写|优化理论代写Optimization Theory代考|CAAM560

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## 数学代写|优化理论代写Optimization Theory代考|Totally unimodular matrices

Of major importance in relation with subclasses of integer linear programming instances that allow efficient algorithms is the notion of total unimodularity.

Definition 17.1.1 A matrix $A \in \mathbb{Z}^{m \times n}, m, n \in \mathbb{Z}$ is totally unimodular iff for each square submatrix $B$ of $A$ we have $\operatorname{det}(B) \in{-1,0,1}$.

Example 17.1.2 An important class of totally unimodular matrices are incidence matrices of directed graphs. Given a directed graph $G=(V, E), V:=$ $\left{v_1, \ldots, v_m\right}, E:=\left{e_1, \ldots, e_n\right}$, we define its incidence matrix $A=\left[a_{i j}\right]$ by
a_{i j}:=\left{\begin{aligned} -1 & \text { if edge } e_j \text { starts in } v_i \ 1 & \text { if edge } e_j \text { ends in } v_i \ 0 & \text { if } v_i \text { is not incident with } e_j \end{aligned}\right.
By induction on the dimension $k$ of a square submatrix $B$ of $A$ we can show that $A$ is totally unimodular: For $k=1$ it is clear that $\operatorname{det}(B) \in{-1,0,1}$ because all entries of $A$ belong to ${-1,0,1}$. Now for general $k$ note that each column in $B$ has either no, one or two non-zero entries which either are -1 or 1 . If a column in $B$ consists of 0 entries only the determinant is 0 . If a column has precisely one non-zero entry in ${-1,1}$, then Laplace’s expansion rule applied to that column together with the induction hypothesis gives $\operatorname{det}(B)=( \pm 1) \cdot \operatorname{det}(\tilde{B})$ for a submatrix $\tilde{B}$ with $\operatorname{det}(\tilde{B}) \in{-1,0,1}$. Finally, if all columns of $B$ have two non-zero entries, then the structure of $A$ guarantees one to equal 1 and the other to equal -1 . Adding all rows gives the result 0 , thereby showing that the rows are linearly dependent. Thus, $\operatorname{det}(B)=0$.

## 数学代写|优化理论代写Optimization Theory代考|Unimodularity and integer linear programming

Let us now turn to the role totally unimodular matrices play in connection with optimization problems. Recall the Vertex Theorem 6.1.5 from Chapter 6. If a linear programming problem attains its infimum it attains it in some vertex of the feasible set. In general, such a vertex of course is not integral. However, if it were, then it would be as well a solution of the related integer linear programming problem. Therefore, if we could guarantee optimal vertices (or all vertices) to be integral we could use general LP methods as well for integer linear programming. As we shall see next totally unimodular matrices provide such a guarantee.

Let $A \in \mathbb{Z}^{m \times n}$ with row vectors $a_1, \ldots, a_m, b=\left(b_1, \ldots, b_m\right)^T \in \mathbb{Z}^m, M:=$ $\left{x \in \mathbb{R}^n \mid A \cdot x \leq b\right}, \bar{x} \in M$. Recall from Chapters 2 and 6 the definitions $J_0(\bar{x}):=\left{j \mid a_j^T \cdot \bar{x}=b_j\right}$ of active indices in $\bar{x}$ and $\Sigma(\bar{x})=\left{x \in M \mid J_0(x)=\right.$ $J_0(\bar{x})$ of the stratum generated by $\bar{x}$.

Definition 17.2.1 For $A, b, M, \bar{x}$ as above $\Sigma(\bar{x})$ is a minimal stratum if its dimension equals $\operatorname{rank}(A)$.

It is easy to see that if $\Sigma(\bar{a})$ is minimal the rows $\left{a_j \mid j \in J_0(\bar{x})\right}$ span the row-space of $A$. This will be useful in the following

Theorem 17.2.2 Let $A \in \mathbb{Z}^{m \times n}$ be totally unimodular. Then for every $b \in \mathbb{Z}^m$ and for every minimal stratum $\Sigma(x)$ of $M:={x \in \mathbb{R} \mid A \cdot x \leq b}$ there exists an integral point in $M$, i.e. $M \cap \mathbb{Z}^m \neq \emptyset$.

Proof. Let $\bar{x} \in M$ be such that $\Sigma(\bar{x})$ is a minimal stratum. W.1.o.g. assume that the first $r$ many rows $a_1, \ldots, a_r$ are linearly independent and ${1, \ldots, r} \subseteq J_0(\bar{x})$, where $r:=\operatorname{rank}(A)$ (otherwise reorder $A^{\prime} s$ rows). It follows that the $r \times r$ submatrix $A^{\prime}$ of $A$, given by $A^{\prime}=\left[a_{i j}\right]_{1 \leq i, j \leq r}$ is regular. Therefore, the linear system
$$A^{\prime} \cdot y=\left(\begin{array}{c} b_1 \ \vdots \ b_r \end{array}\right)$$
has a unique solution $y^{\prime} \in \mathbb{R}^r$, which actually belongs to $\mathbb{Z}^r$. This follows from integrality of the $b_i$ ‘s, total unimodularity of $A$ (which gives $\operatorname{det}\left(A^{\prime}\right) \in$ ${-1,1})$ and Cramer’s rule.

## Matlab代写

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