2009
DOI: 10.1007/s00454-008-9130-6
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Untangling a Planar Graph

Abstract: A straight-line drawing $\delta$ of a planar graph $G$ need not be plane, but can be made so by \emph{untangling} it, that is, by moving some of the vertices of $G$. Let shift$(G,\delta)$ denote the minimum number of vertices that need to be moved to untangle $\delta$. We show that shift$(G,\delta)$ is NP-hard to compute and to approximate. Our hardness results extend to a version of \textsc{1BendPointSetEmbeddability}, a well-known graph-drawing problem. Further we define fix$(G,\delta)=n-shift(G,\delta)$ t… Show more

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Cited by 25 publications
(39 citation statements)
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“…Figure 2 PSGE is related to the notion of planar untangling: Given a straight-line drawing of a planar graph, change the embedding of as few vertices as possible in order to obtain a plane drawing. Goaoc et al [20] described an improvement of a result by Bose et al [4] to show that 4 (n + 1)/2 vertices can always be kept in their original positions. Since we can simply take any plane embedding of G 1 , use the same embedding for G 2 and then untangle G 2 , it immediately follows that every two planar graphs on n vertices admit a 4 (n + 1)/2-PSGE.…”
Section: Introductionmentioning
confidence: 99%
“…Figure 2 PSGE is related to the notion of planar untangling: Given a straight-line drawing of a planar graph, change the embedding of as few vertices as possible in order to obtain a plane drawing. Goaoc et al [20] described an improvement of a result by Bose et al [4] to show that 4 (n + 1)/2 vertices can always be kept in their original positions. Since we can simply take any plane embedding of G 1 , use the same embedding for G 2 and then untangle G 2 , it immediately follows that every two planar graphs on n vertices admit a 4 (n + 1)/2-PSGE.…”
Section: Introductionmentioning
confidence: 99%
“…It is NP-hard to compute fix(G, δ) [5,10] and even to approximate it [5]. At the 5th Czech-Slovak Symposium on Combinatorics in Prague in 1998, Mamoru Watanabe asked whether every polygon on n vertices can be turned into a noncrossing polygon by moving at most εn vertices for some constant ε > 0.…”
mentioning
confidence: 99%
“…Trees √ n/2 [ 2] 3 √ n − 3 [ 2] Outerplanar graphs √ n/2 [ 5,8] 2 √ n − 1 + 1 [ 5] Planar graphs 4 √ (n + 1)/2 [ 2] √ n − 2 + 1 [ 5] In the following, a drawing of a graph G is an injective mapping from V (G) to a set of points in R 2 such that each edge vw ∈ E(G) is represented by the straight-line segment between the images of v and w. A drawing is plane if no two edges cross.…”
mentioning
confidence: 99%
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“…Goaoc et al [4] showed that it is N P-hard to decide whether a given graph can be 1-bend embedded on a given set of points with given vertex-point correspondence. -We show that Geodesic Embeddability on the grid is equivalent to deciding whether the given graph has a rectilinear one-bend drawing on the grid, see Section 2.…”
Section: Introductionmentioning
confidence: 99%