Who Needs Crossings?: Noncrossing Linkages are Universal, and Deciding (Global) Rigidity is Hard
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2025
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| _version_ | 1866915564419547136 |
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| author | Abel, Zachary Demaine, Erik D. Demaine, Martin L. Eisenstat, Sarah Lynch, Jayson Schardl, Tao B. |
| author_facet | Abel, Zachary Demaine, Erik D. Demaine, Martin L. Eisenstat, Sarah Lynch, Jayson Schardl, Tao B. |
| contents | We exactly settle the complexity of graph realization, graph rigidity, and graph global rigidity as applied to three types of graphs: "globally noncrossing" graphs, which avoid crossings in all of their configurations; matchstick graphs, with unit-length edges and where only noncrossing configurations are considered; and unrestricted graphs (crossings allowed) with unit edge lengths (or in the global rigidity case, edge lengths in $\{1,2\}$). We show that all nine of these questions are complete for the class $\exists\mathbb{R}$, defined by the Existential Theory of the Reals, or its complement $\forall\mathbb{R}$; in particular, each problem is (co)NP-hard.
One of these nine results--that realization of unit-distance graphs is $\exists\mathbb{R}$-complete--was shown previously by Schaefer (2013), but the other eight are new. We strengthen several prior results. Matchstick graph realization was known to be NP-hard (Eades \& Wormald 1990, or Cabello et al.\ 2007), but its membership in NP remained open; we show it is complete for the (possibly) larger class $\exists\mathbb{R}$. Global rigidity of graphs with edge lengths in $\{1,2\}$ was known to be coNP-hard (Saxe 1979); we show it is $\forall\mathbb{R}$-complete.
The majority of the paper is devoted to proving an analog of Kempe's Universality Theorem--informally, "there is a linkage to sign your name"--for globally noncrossing linkages. In particular, we show that any polynomial curve $ϕ(x,y)=0$ can be traced by a noncrossing linkage, settling an open problem from 2004. More generally, we show that the regions in the plane that may be traced by a noncrossing linkage are precisely the compact semialgebraic regions (plus the trivial case of the entire plane). Thus, no drawing power is lost by restricting to noncrossing linkages. We prove analogous results for matchstick linkages and unit-distance linkages as well. |
| format | Preprint |
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arxiv_https___arxiv_org_abs_2510_17737 |
| institution | arXiv |
| publishDate | 2025 |
| record_format | arxiv |
| spellingShingle | Who Needs Crossings?: Noncrossing Linkages are Universal, and Deciding (Global) Rigidity is Hard Abel, Zachary Demaine, Erik D. Demaine, Martin L. Eisenstat, Sarah Lynch, Jayson Schardl, Tao B. Computational Geometry Metric Geometry We exactly settle the complexity of graph realization, graph rigidity, and graph global rigidity as applied to three types of graphs: "globally noncrossing" graphs, which avoid crossings in all of their configurations; matchstick graphs, with unit-length edges and where only noncrossing configurations are considered; and unrestricted graphs (crossings allowed) with unit edge lengths (or in the global rigidity case, edge lengths in $\{1,2\}$). We show that all nine of these questions are complete for the class $\exists\mathbb{R}$, defined by the Existential Theory of the Reals, or its complement $\forall\mathbb{R}$; in particular, each problem is (co)NP-hard. One of these nine results--that realization of unit-distance graphs is $\exists\mathbb{R}$-complete--was shown previously by Schaefer (2013), but the other eight are new. We strengthen several prior results. Matchstick graph realization was known to be NP-hard (Eades \& Wormald 1990, or Cabello et al.\ 2007), but its membership in NP remained open; we show it is complete for the (possibly) larger class $\exists\mathbb{R}$. Global rigidity of graphs with edge lengths in $\{1,2\}$ was known to be coNP-hard (Saxe 1979); we show it is $\forall\mathbb{R}$-complete. The majority of the paper is devoted to proving an analog of Kempe's Universality Theorem--informally, "there is a linkage to sign your name"--for globally noncrossing linkages. In particular, we show that any polynomial curve $ϕ(x,y)=0$ can be traced by a noncrossing linkage, settling an open problem from 2004. More generally, we show that the regions in the plane that may be traced by a noncrossing linkage are precisely the compact semialgebraic regions (plus the trivial case of the entire plane). Thus, no drawing power is lost by restricting to noncrossing linkages. We prove analogous results for matchstick linkages and unit-distance linkages as well. |
| title | Who Needs Crossings?: Noncrossing Linkages are Universal, and Deciding (Global) Rigidity is Hard |
| topic | Computational Geometry Metric Geometry |
| url | https://arxiv.org/abs/2510.17737 |