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?