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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.

Thanks and regards!

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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.


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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