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complexity results for three-dimensional orthogonal graph drawing. maurizio “titto” patrignani third 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. edge. - PowerPoint PPT Presentation
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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!