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Given a graph, what are some sufficient (and necessary) conditions to tell if its adjacency matrix has the consecutive ones property?

Similar question for its incidence matrix?

Note that a $\{0,1\}$-valued matrix is said to have the consecutive ones property, if there exists a column permutation such that the ones in each row of the resulting matrix are consecutive. Or the roles of rows and columns can be exchanged in the definition.

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1 Answer 1

Proposition. The adjacency matrix of a graph $\Gamma$ has the a consecutive-ones property if and only if

  1. $\Gamma$ is a tree, or

  2. every induced cycle of $\Gamma$ is a $4$-cycle.

This answer is a work in progress, as I'm stuck on on the 4-cycle case of the converse.

$\Rightarrow$:

An induced $n$-cycle has vertices $0,\ldots, n-1$ with $i$ incident to $j$ if and only if $j=i\pm 1$, taking the indices modulo $n$. So in the adjacency matrix, we've got $$A_{i,i+1}=A_{i,i-1}=A_{i+1,i}=A_{i-1,i}=1$$ for each $i$, and all other entries $0$.

Suppose that $\sigma$ is the permutation of $\{1,\ldots,n\}$ giving rise to a consecutive ones arrangement of $A$. In order to have both $1$'s in a row next to other $1$'s in that column, we need that $\sigma(i)=\sigma(i+2)\mp 1=\sigma(i-2)\pm 1$. However, unless $n=4$, we cannot make such a bijection without leaving some column with $n-2$ zeroes separating a pair of $1$'s. However, if $n=4$, $$A^\sigma=\left(\begin{array}{cccc}1&0&1&0\\1&0&1&0\\0&1&0&1\\0&1&0&1\end{array}\right)$$ is an admissible consecutive ones arrangement.

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