drawing graphs on few lines and few planes
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The Complexity of Drawing Graphs
on Few Lines and Few Planes
Steven Chaplick Krzysztof FleszarFabian Lipp Alexander Wolff
Julius-Maximilians-Universitat Wurzburg, Germany
Alexander RavskyNational Academy of Sciences of Ukraine, Lviv, Ukraine
Oleg VerbitskyHumboldt-Universitat zu Berlin, Germany
Our Task
For any planar graph (like K4) there is a crossing-freestraight-line drawing in the plane.
Our Task
For any planar graph (like K4) there is a crossing-freestraight-line drawing in the plane.
But what about beyond-planar graphs?
Our Task
For any planar graph (like K4) there is a crossing-freestraight-line drawing in the plane.
But what about beyond-planar graphs?Use more than one plane to draw it.
Our Task
For any planar graph (like K4) there is a crossing-freestraight-line drawing in the plane.
How many planes in 3-space do we need to cover all verticesand edges of a crossing-free straight-line drawing of K5?
But what about beyond-planar graphs?Use more than one plane to draw it.
Our Task
For any planar graph (like K4) there is a crossing-freestraight-line drawing in the plane.
How many planes in 3-space do we need to cover all verticesand edges of a crossing-free straight-line drawing of K5?
But what about beyond-planar graphs?Use more than one plane to draw it.
Our Task
For any planar graph (like K4) there is a crossing-freestraight-line drawing in the plane.
How many planes in 3-space do we need to cover all verticesand edges of a crossing-free straight-line drawing of K5?
But what about beyond-planar graphs?Use more than one plane to draw it.
Our Task
For any planar graph (like K4) there is a crossing-freestraight-line drawing in the plane.
How many planes in 3-space do we need to cover all verticesand edges of a crossing-free straight-line drawing of K5? 3
But what about beyond-planar graphs?Use more than one plane to draw it.
Our Task
For any planar graph (like K4) there is a crossing-freestraight-line drawing in the plane.
How many planes in 3-space do we need to cover all verticesand edges of a crossing-free straight-line drawing of K5?
We propose the numberof planes needed as aparameter for classifyingbeyond-planar graphs.
3
But what about beyond-planar graphs?Use more than one plane to draw it.
Related Work: Low Visual Complexity
Minimum number of segments[Durocher et al., JGAA 2013]
Related Work: Low Visual Complexity
Minimum number of segments[Durocher et al., JGAA 2013]
9 segments5 lines
Related Work: Low Visual Complexity
Minimum number of segments[Durocher et al., JGAA 2013]
9 segments5 lines
Low number of arcs[Schulz, JGAA 2015]
Related Work: Low Visual Complexity
Minimum number of segments[Durocher et al., JGAA 2013]
9 segments5 lines
Low number of arcs[Schulz, JGAA 2015]
dodecaedron
Related Work: Low Visual Complexity
Minimum number of segments[Durocher et al., JGAA 2013]
9 segments5 lines
Low number of arcs
10 arcs
[Schulz, JGAA 2015]
dodecaedron
Definitions
Let G be a graph and 1 ≤ m < d .
Affine cover number ρmd (G ):
minimum number of m-dimensional hyperplanes in Rd s.t.G has a crossing-free straight-line drawing that is containedin these planes
Definitions
Let G be a graph and 1 ≤ m < d .
Interesting cases:
• Line cover numbers in 2D and 3D: ρ12, ρ13• Plane cover numbers: ρ23
Affine cover number ρmd (G ):
minimum number of m-dimensional hyperplanes in Rd s.t.G has a crossing-free straight-line drawing that is containedin these planes
Definitions
Let G be a graph and 1 ≤ m < d .
Interesting cases:
• Line cover numbers in 2D and 3D: ρ12, ρ13• Plane cover numbers: ρ23
Affine cover number ρmd (G ):
minimum number of m-dimensional hyperplanes in Rd s.t.G has a crossing-free straight-line drawing that is containedin these planes
• Combinatorial bounds for various classes of graphs• Relations to other graph characteristics
[Chaplick et al., GD 2016]
Known results:
Our Results
• Computing line cover numbers is ∃R-hard(ρ12 and ρ13)
• Computing line cover numbers is fixed-parameter tractable(ρ12 and ρ13)
• Computing plane cover numbers is NP-hard(ρ23)
Our Results
• Computing line cover numbers is ∃R-hard(ρ12 and ρ13)
• Computing line cover numbers is fixed-parameter tractable(ρ12 and ρ13)
• Computing plane cover numbers is NP-hard(ρ23)
∃R-hardness
Decision problem for the existential theory of the reals:decide if first-order formula about the reals of the form∃x1 . . . ∃xmΦ(x1, . . . , xm) is true where formula Φ uses:• no quantifiers• constants 0, 1• basic arithmetic operations• order and equality relations
∃R-hardness
Decision problem for the existential theory of the reals:decide if first-order formula about the reals of the form∃x1 . . . ∃xmΦ(x1, . . . , xm) is true where formula Φ uses:• no quantifiers• constants 0, 1• basic arithmetic operations• order and equality relations
∃R: problems that can be reduced to that problem[Schaefer, GD 2009]
∃R-hardness
Decision problem for the existential theory of the reals:decide if first-order formula about the reals of the form∃x1 . . . ∃xmΦ(x1, . . . , xm) is true where formula Φ uses:• no quantifiers• constants 0, 1• basic arithmetic operations• order and equality relations
∃R: problems that can be reduced to that problem[Schaefer, GD 2009]
∃R
PSPACE
NP
∃R-hardness
Decision problem for the existential theory of the reals:decide if first-order formula about the reals of the form∃x1 . . . ∃xmΦ(x1, . . . , xm) is true where formula Φ uses:• no quantifiers• constants 0, 1• basic arithmetic operations• order and equality relations
∃R: problems that can be reduced to that problem
Natural ∃R-complete problems:• rectilinear crossing number• recognition of segment intersection graphs• recognition of unit disk graphs
[Schaefer, GD 2009]
∃R
PSPACE
NP
Arrangement Graph Recognition (AGR)
Simple line arrangement: set of ` lines in R2
Arrangement Graph Recognition (AGR)
Simple line arrangement: set of ` lines in R2
Arrangement Graph Recognition (AGR)
Simple line arrangement: set of ` lines in R2
• Each pair of lines has exactly one intersection• No three lines share a common point
Arrangement Graph Recognition (AGR)
Simple line arrangement: set of ` lines in R2
• Each pair of lines has exactly one intersection• No three lines share a common point
Arrangement graph:`(`− 1)/2 vertices`(`− 2) edges
Arrangement Graph Recognition (AGR)
Simple line arrangement: set of ` lines in R2
• Each pair of lines has exactly one intersection• No three lines share a common point
Arrangement graph:`(`− 1)/2 vertices`(`− 2) edges
Arrangement Graph Recognition (AGR)
Simple line arrangement: set of ` lines in R2
• Each pair of lines has exactly one intersection• No three lines share a common point
Arrangement graph:`(`− 1)/2 vertices`(`− 2) edges
Problem AGR: Given a graph G = (V , E ), decide if it is anarrangement graph of some set of lines.
Arrangement Graph Recognition (AGR)
Simple line arrangement: set of ` lines in R2
• Each pair of lines has exactly one intersection• No three lines share a common point
Arrangement graph:`(`− 1)/2 vertices`(`− 2) edges
Problem AGR: Given a graph G = (V , E ), decide if it is anarrangement graph of some set of lines.
AGR is known to be ∃R-hard
Line Cover Number is ∃R-hard
Reduction from AGRGiven a graph G = (V , E ).
Line Cover Number is ∃R-hard
Add tails ⇒ G ′: all vertices have degree 1 or 4.
Reduction from AGRGiven a graph G = (V , E ).
Line Cover Number is ∃R-hard
Add tails ⇒ G ′: all vertices have degree 1 or 4.
Reduction from AGRGiven a graph G = (V , E ).
G is an arrangement graph iff ρ12(G ′) ≤ `Claim:
[Durocher, Mondal, Nishat, Whitesides, JGAA 2013]Proof similar to hardness proof for segment number.
Theorem: Deciding whether ρ12(G ) ≤ k and ρ13(G ) ≤ k is∃R-complete.
Our Results
• Computing line cover numbers is ∃R-hard(ρ12 and ρ13)
• Computing line cover numbers is fixed-parameter tractable(ρ12 and ρ13)
• Computing plane cover numbers is NP-hard(ρ23)
Fixed-parameter Tractability
Kernelization:
Assume that G has a k-line cover.Remove connected components that are paths.Contract long paths to at most
(k2
)vertices.
Reduce G to an instance G ′
with size bounded by f (k)
Fixed-parameter Tractability
Kernelization:
Assume that G has a k-line cover.Remove connected components that are paths.Contract long paths to at most
(k2
)vertices.
at most(k2
)intersections
Reduce G to an instance G ′
with size bounded by f (k)
Fixed-parameter Tractability
Kernelization:
Assume that G has a k-line cover.Remove connected components that are paths.Contract long paths to at most
(k2
)vertices.
at most(k2
)intersections
Reduce G to an instance G ′
with size bounded by f (k)
Fixed-parameter Tractability
Kernelization:
Assume that G has a k-line cover.Remove connected components that are paths.Contract long paths to at most
(k2
)vertices.
degree 2degree > 2
degree 1
at most(k2
)intersections
Reduce G to an instance G ′
with size bounded by f (k)
Fixed-parameter Tractability
Kernelization:
Assume that G has a k-line cover.Remove connected components that are paths.Contract long paths to at most
(k2
)vertices.
Number of paths: O(k2)Vertices per path: at most
(k2
)
degree 2degree > 2
degree 1
at most(k2
)intersections
Reduce G to an instance G ′
with size bounded by f (k)
Fixed-parameter Tractability
Kernelization:
Assume that G has a k-line cover.Remove connected components that are paths.Contract long paths to at most
(k2
)vertices.
Number of paths: O(k2)Vertices per path: at most
(k2
)
degree 2degree > 2
}O(k4) vertices
O(k4) edges
degree 1
at most(k2
)intersections
Reduce G to an instance G ′
with size bounded by f (k)
Fixed-parameter Tractability
Test line cover number for reduced graph G ′ of size O(k4).Describe as a formula Φ in the existential theory of the realsand use Renegar’s decision algorithm.
Fixed-parameter Tractability
Test line cover number for reduced graph G ′ of size O(k4).Describe as a formula Φ in the existential theory of the realsand use Renegar’s decision algorithm.
For graph G and integer k, in time2O(k3) + O(n + m) we can decide• whether ρ1d(G ) ≤ k and, if so,• give a combinatorial description
of the drawing.
Theorem:
Our Results
• Computing line cover numbers is ∃R-hard(ρ12 and ρ13)
• Computing line cover numbers is fixed-parameter tractable(ρ12 and ρ13)
• Computing plane cover numbers is NP-hard(ρ23)
Detour: 3-SAT Variant
(x1, x2, x6) (x1, x4, x5) (x1, x3, x4) (x3, x5, x6) (x2, x3, x6)
We use a reduction from Positive Planar Cycle 1-in-3-SAT
Detour: 3-SAT Variant
(x1, x2, x6) (x1, x4, x5) (x1, x3, x4) (x3, x5, x6) (x2, x3, x6)
We use a reduction from Positive Planar Cycle 1-in-3-SAT
Detour: 3-SAT Variant
(x1, x2, x6) (x1, x4, x5) (x1, x3, x4) (x3, x5, x6) (x2, x3, x6)
We use a reduction from Positive Planar Cycle 1-in-3-SAT
Detour: 3-SAT Variant
(x1, x2, x6) (x1, x4, x5) (x1, x3, x4) (x3, x5, x6) (x2, x3, x6)
x1
x2x3
x4
x5x6
We use a reduction from Positive Planar Cycle 1-in-3-SAT
Detour: 3-SAT Variant
(x1, x2, x6) (x1, x4, x5) (x1, x3, x4) (x3, x5, x6) (x2, x3, x6)
x1
x2x3
x4
x5x6
We use a reduction from Positive Planar Cycle 1-in-3-SAT
Detour: 3-SAT Variant
(x1, x2, x6) (x1, x4, x5) (x1, x3, x4) (x3, x5, x6) (x2, x3, x6)
x1
x2x3
x4
x5x6
We use a reduction from Positive Planar Cycle 1-in-3-SAT
Detour: 3-SAT Variant
(x1, x2, x6) (x1, x4, x5) (x1, x3, x4) (x3, x5, x6) (x2, x3, x6)
x1
x2x3
x4
x5x6
We use a reduction from Positive Planar Cycle 1-in-3-SAT
NP-hardness proof:
Detour: 3-SAT Variant
(x1, x2, x6) (x1, x4, x5) (x1, x3, x4) (x3, x5, x6) (x2, x3, x6)
x1
x2x3
x4
x5x6
We use a reduction from Positive Planar Cycle 1-in-3-SAT
NP-hardness proof:
⇓Positive Planar Cycle 1-in-3-SAT is NP-hard
combined and modified
Planar Cycle 3-SAT[Kratochvıl, Lubiw, Nesetril, 1991]
Positive Planar 1-in-3-SAT[Mulzer, Rote, 2008]
+
Intersection Line Gadget
intersectionline gadget G
Intersection Line Gadget
Assume that G is embeddedon two planes
intersectionline
intersectionline gadget G
Intersection Line Gadget
Assume that G is embeddedon two planes
intersectionline
⇒ Red vertices are placedon the intersection line
intersectionline gadget G
Reduction to Plane Cover Number
Theorem: Deciding ρ23(G ) = 2 is NP-hard.
Proof:
(x1, x2, x6) (x1, x4, x5) (x1, x3, x4) (x3, x5, x6) (x2, x3, x6)
x1
x2
x3
x4
x5
x6
Reduction to Plane Cover Number
Theorem: Deciding ρ23(G ) = 2 is NP-hard.
Proof:
x1
x2
x3
x4
x5
x6
Reduction to Plane Cover Number
Theorem: Deciding ρ23(G ) = 2 is NP-hard.
Proof:
x1
x2
x3
x4
x5
x6
Reduction to Plane Cover Number
Theorem: Deciding ρ23(G ) = 2 is NP-hard.
Proof:
x1
x2
x3
x4
x5
x6
Reduction to Plane Cover Number
Theorem: Deciding ρ23(G ) = 2 is NP-hard.
Proof:
x1
x2
x3
x4
x5
x6
Reduction to Plane Cover Number
Theorem: Deciding ρ23(G ) = 2 is NP-hard.
Proof:
x1
x2
x3
x4
x5
x6
Reduction to Plane Cover Number
Theorem: Deciding ρ23(G ) = 2 is NP-hard.
Proof:
x1 x3
x4
x5
x6
x2
Reduction to Plane Cover Number
Theorem: Deciding ρ23(G ) = 2 is NP-hard.
Proof:
x3
x4
x5
x6
x1
x2
Reduction to Plane Cover Number
Theorem: Deciding ρ23(G ) = 2 is NP-hard.
Proof:
x3
x4
x5
x6
x2
x1
Reduction to Plane Cover Number
Theorem: Deciding ρ23(G ) = 2 is NP-hard.
Proof:
x4
x5
x6
x2
x1 x3
Reduction to Plane Cover Number
Theorem: Deciding ρ23(G ) = 2 is NP-hard.
Proof:
x5
x6
x2
x1 x3
x4
Reduction to Plane Cover Number
Theorem: Deciding ρ23(G ) = 2 is NP-hard.
Proof:
x6
x2
x1 x3
x4
x5
Reduction to Plane Cover Number
Theorem: Deciding ρ23(G ) = 2 is NP-hard.
Proof:x2
x1 x3
x4
x5
x6
Reduction to Plane Cover Number
Theorem: Deciding ρ23(G ) = 2 is NP-hard.
Proof:x2
x1 x3
x4
x5
x6
clauseslie oninter-sectionline
Reduction to Plane Cover Number
Theorem: Deciding ρ23(G ) = 2 is NP-hard.
Proof:x2
x1 x3
x4
x5
x6
clauseslie oninter-sectionline blue vertices not on
intersection line
Reduction to Plane Cover Number
Theorem: Deciding ρ23(G ) = 2 is NP-hard.
Proof:x2
x1 x3
x4
x5
x6
clauseslie oninter-sectionline blue vertices not on
intersection lineeach variable onlyon one plane
Reduction to Plane Cover Number
Theorem: Deciding ρ23(G ) = 2 is NP-hard.
Proof:x2
x1 x3
x4
x5
x6
clauseslie oninter-sectionline blue vertices not on
intersection lineeach variable onlyon one plane
caterpillar onone plane
Hardness of the Plane Cover Number
Theorem: Deciding ρ23(G ) = 2 is NP-hard.
Proof:
Hardness of the Plane Cover Number
Theorem: Deciding ρ23(G ) = 2 is NP-hard.
Proof:
Hardness of the Plane Cover Number
Theorem: Deciding ρ23(G ) = 2 is NP-hard.
Proof:
Hardness of the Plane Cover Number
Theorem: Deciding ρ23(G ) = 2 is NP-hard.
Proof:
Hardness of the Plane Cover Number
Theorem: Deciding ρ23(G ) = 2 is NP-hard.
Proof:
Hardness of the Plane Cover Number
Theorem: Deciding ρ23(G ) = 2 is NP-hard.
Proof:
Hardness of the Plane Cover Number
Theorem: Deciding ρ23(G ) = 2 is NP-hard.
Proof:
Hardness of the Plane Cover Number
Theorem: Deciding ρ23(G ) = 2 is NP-hard.
Proof:
Hardness of the Plane Cover Number
Theorem: Deciding ρ23(G ) = 2 is NP-hard.
Proof:
false
true
Hardness of the Plane Cover Number
Theorem: Deciding ρ23(G ) = 2 is NP-hard.
Corollary: Deciding ρ23(G ) = k is NP-hard for any k ≥ 2.(add more blocking gadgets)
Plane cover number not FPT in k.
Proof:
false
true
Conclusion
• Computing the line cover numbers is hard, butfixed-parameter tractable.Can it be computed efficiently for trees?
Conclusion
• Positive Planar Cycle1-in-3-SAT is NP-hard.
• Computing the plane covernumber is NP-hard andcontained in ∃R.Is it also ∃R-hard?
• Approximation Algorithms?
• Computing the line cover numbers is hard, butfixed-parameter tractable.Can it be computed efficiently for trees?
Conclusion
• Positive Planar Cycle1-in-3-SAT is NP-hard.
• Computing the plane covernumber is NP-hard andcontained in ∃R.Is it also ∃R-hard?
• Approximation Algorithms?
• Computing the line cover numbers is hard, butfixed-parameter tractable.Can it be computed efficiently for trees?
Conclusion
• Positive Planar Cycle1-in-3-SAT is NP-hard.
• Computing the plane covernumber is NP-hard andcontained in ∃R.Is it also ∃R-hard?
• Approximation Algorithms?
• Computing the line cover numbers is hard, butfixed-parameter tractable.Can it be computed efficiently for trees?
Conclusion
• Positive Planar Cycle1-in-3-SAT is NP-hard.
• Computing the plane covernumber is NP-hard andcontained in ∃R.Is it also ∃R-hard?
• Approximation Algorithms?
• Computing the line cover numbers is hard, butfixed-parameter tractable.Can it be computed efficiently for trees?
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