complexity results for three-dimensional orthogonal

graph drawing

maurizio “titto” patrignanithird university of rome

graph drawing 2005

three-dimensional orthogonal drawings• nodes are (distinct) points in 3D space• edges are composed by sequences of axis-

parallel segments

node

bend

edge

what we know• existence

– only graphs of maximum degree six admit such drawings– all graphs of maximum degree six admit such drawings

• volume (n3/2) [rosenberg 1983] (n3/2) [eades, stirk, and whitesides 1996]

• bends– in the optimal O(n3/2) volume we can draw with up to 7 bends per

edge in O(n3/2) time [eades, symvonis, and whitesides 2000]– in O(n3) volume we can draw with up to 3 bends per edges in linear

time [papakostas and tollis 1999][eades, symvonis, and whitesides, 2000]

– in O(n2) volume we can draw with 6 bends per edge in linear time and handle insertions/deletions in O(1) time [closson, gartshore, johansen, and wismath. 2000]

what we do not know

1. is three bends per edge the lower bound?or rather: does every graph admit a 2-bend drawing?

2. can we extend to 3D the topology-shape-metrics approach?

2-bend drawing problem• an algorithm to produce 2-bends drawings

could be particularly effective for information visualization

• is such algorithm does not exists then the algorithms we have are the best possible

• the K7 graph that was thought to require 3 bends turned out to admit a 2-bend drawing [wood ’97]

• problem #46 of the open problem project [demaine, mitchell, and o’rourke]

topology-shape-metrics approach in 2DV={1,2,3,4,5,6}E={(1,4),(1,5),(1,6), (2,4),(2,5),(2,6), (3,4),(3,5),(3,6)}

planarization

orthogonalization

compaction

6

1 25

34

1 25

6

3

4

1 25

6

3

4

topology-shape-metrics approach in 3D

V={1,2,3,4,5}E={(1,2),(1,3),(1,4), (2,3),(2,4),(2,5), (3,4),(3,5)}

orthogonalization

compaction

1

2

3 4

5

1

23 4

5

simple and not simple shape graphs

simple shape graph(admitting non-intersecting metrics)

not simple shape graph(always intersects)

simplicity testing in 3D• known results:

– simplicity test for cycles [di battista, liotta, lubiw, and whitesides, ‘01]

– simplicity test for paths (with additional constraints) [di battista, liotta, lubiw, and whitesides, ‘02]

– the above two characterizations are not easy to extend to simple graphs (theta graphs) [di giacomo, liotta, and patrignani, ‘04]

• simplicity testing is an open problem in the general case – problem #20 of [brandenburg, eppstein, goodrich,

kobourov, liotta, and mutzel. ’03]

two open problems

1. existence of a 2-bend drawing2. simplicity testing

can complexity considerations give us some insight?

what we showgiven a 6-degree graph we prove that:• simplicity testing is NP-complete

if you fix edge shapes (with a maximum of 2 bends per edge) finding the metrics corresponding to a non intersecting drawing is NP-complete

• 2-bend routing is NP-completeif you fix node positions finding a routing without intersections with a maximum of two bends per edge is NP-complete

how we prove the statementsreductions from the 3SAT problem:instance: a set of clauses {c1, c2, …, cm} each

containing three literals from a set of boolean variables {v1, v2, …, vn}

question: can truth values be assigned to the variables so that each clause contains at least one true literal?

example of 3SAT instance: (v1 v3 v4) (v1 v2 v5) (v2 v3 v5)

c1 c2 c3

we consider a generic target problemstructure of the target problem:instance: a graph G and a set of constraints S

expressed with respect to its nodes and edges

question: does G admit a 3D orthogonal drawing satisfying S?

the 3SAT reduction frameworkvariable gadgets

clause gadgets

joint gadgets

use of the 3SAT frameworktheoremif these four statements are true

– there is at least one non intersecting drawing for each truth assignment satisfying the 3SAT instance

– in any non intersecting drawing if a variable gadget is true, the corresponding joint gadget is true and vice versa

– in any non intersecting drawing of a clause gadget one of the literals is true

– the construction can be done in polynomial time

then the target problem is NP-hard

simplicity testing problem

instance: a graph G and a shape for each edgequestion: does G admit a 3D orthogonal

drawing where the edges have the prescribed shape?

variable gadgettrue variable false variable

variable gadget propagating truth valuesfalse variable

joint-gadget

T

T

F

F

joint-gadget

T

T

T

F

F

F

F

T

clause gadget

from the joint gadget

from the joint gadget

from the variable gadget

all literals false intersecting clause gadget

F

F

F

F

F

T F

FF

T

T T

F F T T

T T

F F

T T T T

2-bend routing problem

instance: a graph G and the coordinates for its nodes

question: does G admit a 2-bend orthogonal drawing where the nodes have such coordinates?

variable gadgetfalse variabletrue variable

variable gadget propagating truth valuesvariable gadgetto clause

gadget c1

to clause gadget c2

to clause gadget c3

joint gadgetfrom the

variable gadget

joint gadgetfrom the

variable gadget

joint gadgetfrom the

variable gadgetto the

clause gadget

clause gadget

conclusionssimplicity testing is NP-hard

• any characterization of simple orthogonal shapes involves a hard computation

• even if we were able to find simple orthogonal shapes the compaction step would be NP-hard

• open problems:– are there classes of graphs such that the compaction

step is polynomial?– given a generic graph, are there families of shapes such

that the metrics can always be computed in polynomial time?

conclusions

2-bend routing is NP-hard • yet another problem where two bends per

edge implies NP-hardnesstwo bends per edge + fixed shape NP-hardnesstwo bends per edge + fixed positions NP-hardnesstwo bends per edge + diagonal layout NP-hardness

• [wood, 2004]

• open problem:– what is the problem of finding a 2-bend

drawing of a graph?

thank you!