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Questions tagged [total-unimodularity]

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Question: are there any graph automorphism that imply total unimodularity of the graph's adjacency matrix, resp. for which automorphisms exist graphs whose adjacency matrix is not totally unimodular
Manfred Weis's user avatar
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Given a matrix $M$, we will refer to the submatrix formed by the first $k$ rows as $M([k], \cdot)$. Let $A$ be a $m\times n$ totally unimodular matrix where $m \leq n$. We define a new $m\times (n+1)$ ...
Kevin S.'s user avatar
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I have a combinatorial problem whose solution has an LP-formulation that contains assignment constraints as well as flow-conservation constraints. Here are the highly simplified constraints, of which ...
Manfred Weis's user avatar
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Question: which, if any, of the criteria for total unimodularity of matrices are easier to check, in the sense of computational complexity, for a circulant matrix $A\in\lbrace -1,0,+1\rbrace^{n\times ...
Manfred Weis's user avatar
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Let $A=[a_{ik}]$ be a matrix with the consecutive ones property in each column, i.e. each column consists of a single consecutive block of $1$'s (with zeros everywhere else). Is there anything at all ...
Tom Solberg's user avatar
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This is not my area of expertise, so forgive me if the question is a bit naive. Given a collection of vectors $v_1,\ldots,v_d$ in $\mathbb{R}^n$ (with $d\geq n$), there is a corresponding set of ...
Yoav Len's user avatar
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Given a polyhedron $$Ax\geq b$$ where every entry of $A,b$ are non-negative and $A\in\{0,1\}^{m\times n}$ and there are $O(1)$ (say $\leq8$) non-negative entries per row of $A$ is it possible to ...
Turbo's user avatar
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Does there exist examples of $m \times n$ matrices with $m > n$ with the property that the determinant of every $n \times n$ submatrix is at least $1$ in absolute value? (The $1$ can be replaced by ...
Sandeep Silwal's user avatar
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Let G be the node-arc incidence matrix of a given directed network (rows of $G$ correspond to nodes and its columns correspond to arcs). Let $B_1,\dots, B_K$ denote a partition of the nodes of the ...
Ozzy's user avatar
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Given a matrix $M\in\{0,1\}^{m\times n}$ is there an algorithm to tell if we can convert some of $1$s to $-1$s and make $M$ Totally Unimodular and output such a Totally Unimodular in polynomial in $mn$...
Turbo's user avatar
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Is it possible to prove this matrix family only contains totally unimodular matrices? The matrix has dimensions $\frac{3n(n-1)}2$ rows and $n+\frac{n(n-1)}2$ columns. To every pair $(i,i')$ with $1\...
Turbo's user avatar
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Given $AX\leq B$ where $A\in\Bbb Z^{m\times n}$,$B\in\Bbb Z^m$ finding $X\in\Bbb Z^n$ where $m\geq n$ is the integer programming problem. If $A$ is totally unimodular then the problem is solvable in ...
Turbo's user avatar
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Consider a directed graph $G=(V,\mathbb{A})$ and a set of flow constraints of the following form: $$ \sum_{(u,v)\in\mathbb{A}}x_{u,v} - \sum_{(v,u)\in \mathbb{A}}x_{v,u} = 0 \forall v \in V$$ ...
Ricardo's user avatar
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A unimodular matrix $M$ is a square integer matrix having determinant $+1$ or $−1$. A totally unimodular matrix (TU matrix) is a matrix for which every square non-singular submatrix is unimodular. A ...
qianchi 's user avatar
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A matrix is totally uni-modular if the determinant of any (square) sub-matrix is {+1, 0, -1}. My question is, "Is there a way to transform(linear or non) a general matrix into a totally uni-modular ...
Clark's user avatar
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I'm working on a set of problems for which I can formulate binary integer programs. When I solve the linear relaxations of these problems, I always get integer solutions. I would like to prove that ...
AaronDefazio's user avatar
4 votes
1 answer
938 views

I have a {0,1}, invertible, triangular matrix, that I would like to show is totally unimodular. Are there any known results on the total unimodularity of classes of triangular matrices?
Michael Biro's user avatar
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16 votes
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One of the consequences of Seymour's characterization of regular matroids is the existence of a polynomial time recognition algorithm for totally unimodular matrices (i.e. matrices for which every ...
Gordon Royle's user avatar
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