Disjoint Compatibility via Graph Classes

Abstract

Two plane drawings of graphs on the same set of points are called disjoint compatible if their union is plane and they do not have an edge in common. Let $S$ be a convex point set of $2n \geq 10$ points and let $\mathcal{H}$ be a family of plane drawings on $S$. Two plane perfect matchings $M_1$ and $M_2$ on $S$ (which do not need to be disjoint nor compatible) are \emph{disjoint $\mathcal{H}$-compatible} if there exists a drawing in $\mathcal{H}$ which is disjoint compatible to both $M_1$ and $M_2$ In this work, we consider the graph which has all plane perfect matchings as vertices and where two vertices are connected by an edge if the matchings are disjoint $\mathcal{H}$-compatible. We study the diameter of this graph when $\mathcal{H}$ is the family of all plane spanning trees, caterpillars or paths. We show that in the first two cases the graph is connected with constant and linear diameter, respectively, while in the third case it is disconnected.

Figures (19)

• Figure 1: Two plane perfect matchings (in blue) on the same set of twelve points in convex position which are disjoint $\mathcal{T}$-compatible. The complying disjoint compatible spanning tree is drawn in green.
• Figure 2: Left: A matching $M$ and two semicycles $X_1$ (red edges) and $X_2$ (blue edges) with their convex hulls. The cycle $\overline{X_1}$ is an inside $4$-cycle, since the boundary of the red shaded area contains at least two (in fact three) diagonals. The cycle $\overline{X_2}$ is a $4$-ear. Right: The matching resulting from rotating the cycle $X_1$.
• Figure 3: Two plane matchings (in blue and red) on $S$ which whose symmetric difference is an inside cycle. The complying disjoint compatible spanning tree is drawn in green.
• Figure 4: Two plane matchings (in blue and red) on $S$ which whose symmetric difference are two inside cycles ($C_1$ and $C_2$). The yellow regions are the convex hulls of the $A_i$s with $1\leq i \leq 5$.
• Figure 5: Rotation of a 6-ear in 3 steps (in each step we rotate the grey inside cycle).

Theorems & Definitions (30)

• Lemma 0
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• Lemma 1
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