symmetrical hamiltonian manifolds on regular 3d and 4d polytopes
DESCRIPTION
Coxeter Day, Banff, Canada , August 3, 2005. Symmetrical Hamiltonian Manifolds on Regular 3D and 4D Polytopes. Carlo H. Séquin EECS Computer Science Division University of California, Berkeley. START. END. Hamiltonian Path: Visits all vertices once. - PowerPoint PPT PresentationTRANSCRIPT
Symmetrical Hamiltonian ManifoldsSymmetrical Hamiltonian Manifoldson Regular 3D and 4D Polytopeson Regular 3D and 4D Polytopes
Carlo H. Séquin
EECS Computer Science Division
University of California, Berkeley
Coxeter Day, Banff, Canada, August 3, 2005
IntroductionIntroduction
Eulerian Path: Uses all edges of a graph.
Eulerian Cycle: A closed Eulerian Paththat returns to the start. END
START
Hamiltonian Path: Visits all vertices once.
Hamiltonian Cycle: A closed Ham. Path.
Map of KönigsbergMap of Königsberg
Can you find a path that crosses all seven bridges exactly once – and then returns to the start ?
Leonhard Euler (1707-83) says: NO ! (1735)– because there are vertices with odd valence.
The Platonic Solids in 3DThe Platonic Solids in 3D
Hamiltonian Cycles ? Eulerian Cycles ?
The OctahedronThe Octahedron All vertices have valence 4.
They admit 2 paths passing through.
Pink edges form Hamiltonian cycle.
Yellow edges form Hamiltonian cycle.
The two paths are congruent !
All edges are covered.
Together they form a Eulerian cycle.
How many different such Hamiltonian cycles are there ?
Can we do the same for all the other Platonic solids ?
Hamiltonian DissectionsHamiltonian Dissections
Hamiltonian Cycles clearly split genus zero surfaces into two domains.
Are these domains of equal size ?
Are these domains congruent ?
Can they be used to split the solid objectso that it can be taken apart ?
... A nice way to visualize these cycles ...
Dissection of the TetrahedronDissection of the Tetrahedron
Two congruent parts
Dissection of the Hexahedron (Cube)Dissection of the Hexahedron (Cube)
Two congruent parts
Dissection of the OctahedronDissection of the Octahedron
Two congruent parts
The Other Octahedron DissectionThe Other Octahedron Dissection
3-fold symmetry
complement edges are not a Ham. cycle
Dissection of the DodecahedronDissection of the Dodecahedron
¼ + ½ + ¼
Dissection of the IcosahedronDissection of the Icosahedron
based on cycle with S6 - Symmetry
Hamiltonian Cycles on the IcosahedronHamiltonian Cycles on the Icosahedron
... that split the surface into two congruent parts
that transform into each other with a C2-rotation.
Some have even higher symmetry, e.g., D2
*
Another Dissection of the IcosahedronAnother Dissection of the Icosahedron
Not just a conical extrusion from the centroid;
Extra edges in the slide-apart direction.
Multiple Uniform Coverage Multiple Uniform Coverage Can we do what we did for the octahedron
also for the other Platonic solids ?.
The problem is:those have vertices with odd valences.
If we allow to pass every edge twice, this is no longer a problem.
Example: valence_3 vertex:
Try to obtain uniform double edge coverage with multiple copies of one Hamiltonian cycle!
Double Edge Coverage of TetrahedronDouble Edge Coverage of Tetrahedron
3 congruent Hamiltonian cycles
Double Edge Coverage, DodecahedronDouble Edge Coverage, Dodecahedron
3 congruent Hamiltonian cycles
Double Edge Coverage on IcosahedronDouble Edge Coverage on Icosahedron
5 congruent Hamiltonian cycles
Double Edge Coverage on CubeDouble Edge Coverage on Cube
Using 3 Hamiltonian paths – not cycles !
The Different Hamiltonian CyclesThe Different Hamiltonian Cycles
Edges # of H.C. # Dissect. Uniform edge cover
Tetrahedron 4 1 1 yes
Cube 12 1 1 (yes)
Octahedron 12 2 2 yes
Dodecahedron 30 1 0 yes
Icosahedron 30 11 2 yes
Talk OutlineTalk Outline Introduction of the Hamiltonian cycle
The various Ham. cycles on the Platonic solids
Hamiltonian dissections of the Platonic solids
Multiple uniform edge coverage with Ham. cycles
Ham. cycles as constructivist sculptures (art)
Ham. cycles on the 4D regular polytopes
Solutions of the 600-Cell and the 120-Cell
Hamiltonian 2-manifolds on 4D polytopes
Volution surfaces suspended in Ham. cycles (art)
Constructivist SculpturesConstructivist Sculptures
Use Hamiltonian Paths to make constructivist sculptures.
Inspiration by: Peter Verhoeff, Popke Bakker, Rinus Roelofs
Peter VerhoeffPeter Verhoeff
truncatedicosahedron
Hamiltonian CycleHamiltonian Cycle
on the edges of a dodecahedron
CS 184, Fall 2004CS 184, Fall 2004
Student homework
HamCycle_2HamCycle_2
on two stacked dodecahedra
CS 184, F’04CS 184, F’04
““Hamiltonian Path” by Rinus RoelofsHamiltonian Path” by Rinus Roelofs
Space diagonals in a dodecahedron
Dodecahedron with Face DiagonalsDodecahedron with Face Diagonals
Only non-crossing diagonals may be used !
Ham. Cycle with 5-fold SymmetryHam. Cycle with 5-fold Symmetry
on the face diagonals of the dodecahedron
Hamiltonian Cycle with CHamiltonian Cycle with C22-Symmetry-Symmetry
on the face diagonals of the dodecahedron
Sculpture Model of CSculpture Model of C22 Ham. Cycle Ham. Cycle
made on FDM machine
With Prismatic Beams ...With Prismatic Beams ...
... mitring might be tricky !
Sculpture Model of CSculpture Model of C22 Ham. Cycle Ham. Cycle
made on Zcorporation 3D-Printer
““CC22-Symmetrical Hamiltonian Cycle”-Symmetrical Hamiltonian Cycle”
... on face diagonals of the dodecahedron
Count of Different Hamiltonian CyclesCount of Different Hamiltonian Cycles
Edges Face Diag. Space Diag. Diam. Axes
Tetra 4 1HC 0 --- 0 --- 0 ---
Octa 12 2HC 0 --- 0 --- 3 0HC
(three pairs)
Cube 12 1HC 12 0HC
(two tetras)
0 --- 4 0HC
(four pairs)
Icosa 30 11HC
0 --- 30 0HC
( 10 diagonals)
6 0HC
(six pairs)
Dodeca 30 1HC 60 2 ? 60 2 ??
30 0HC
10 0HC
(ten pairs)
Disjoint setsCrossing constraintInteresting !
Paths on the 4D Edge GraphsPaths on the 4D Edge Graphs
The 4D regular polytopes offer several very interesting graphs on which we can study Hamiltonian Eulerian coverage.
Start by finding Hamiltonian cycles.
Then try to obtain uniform edge coverage.
The 6 Regular Polytopes in 4DThe 6 Regular Polytopes in 4D
From BRIDGES’2002 Talk
Which 4D-to-3D Projection ??Which 4D-to-3D Projection ??
There are many possible ways to project the edge frame of the 4D polytopes to 3D.
Example: Tesseract (Hypercube, 8-Cell)
Cell-first Face-first Edge-first Vertex-first
Use Cell-first: High symmetry; no coinciding vertices/edges
Hamiltonian Cycles on the 4D SimplexHamiltonian Cycles on the 4D Simplex
Two identical paths, complementing each other
C2
From BRIDGES’2004 Talk
Ham. Cycles on the 4D Cross PolytopeHam. Cycles on the 4D Cross Polytope
All vertices have valence 6 need 3 paths
C3
Hamiltonian Cycles on the HypercubeHamiltonian Cycles on the Hypercube
Valence-4 vertices requires 2 paths.
There are many different solutions.
24-Cell: 4 Hamiltonian Cycles24-Cell: 4 Hamiltonian Cycles
Aligned to show 4-fold symmetry
Why Shells Make Task EasierWhy Shells Make Task Easier
Decompose problem into smaller ones:
Find a suitable shell schedule;
Prepare components on shells compatible with schedule;
Find a coloring that fits the schedule and glues components together,by “rotating” the shells and connector edges within the chosen symmetry group.
Fewer combinations to deal with.
Easier to maintain desired symmetry.
Rapid Prototyping Model of the 24-CellRapid Prototyping Model of the 24-Cell
Noticethe 3-foldpermutationof colors
Made on the Z-corp machine.
Solutions of the 600- and C120-CellSolutions of the 600- and C120-Cell
600-Cell solution found first:
Paths are “only” 120 edges long.
The 6 congruent copies add many constraints.
Shell-based approach worked well for this.
120-Cell was tougher:
Only 2 colors: Too many possibilities in each shell to enumerate all legal colorings.
Also a daunting challenge for backtracking, because each cycle is 600 edges long.
That is how far I got last year ...
The 600-CellThe 600-Cell
120 vertices,valence 12;
720 edges;
Find 6 cycles, length 120.
Shells in the 600-CellShells in the 600-Cell
Number of segments of each type in each Hamiltonian cycle
OUTERMOST TETRAHEDRON
INN
ER
MO
ST
TE
TR
AH
ED
RO
N
CONNECTORS SPANNING THE CENTRAL SHELL
INSIDE / OUTSIDE SYMMETRY
Shells in the 600-CellShells in the 600-Cell
15 shells of vertices
49 different types of edges:
4 intra shells with 6 (tetrahedral) edges,
4 intra shells with 12 edges,
28 connector shells with 12 edges,
13 connector shells with 24 edges (= two 12-edge shells).
Inside/outside symmetry
Overall tetrahedral symmetry
Shell-Based Search on 600-CellShell-Based Search on 600-Cell Shell Pre-Coloring:
For each (half-)shell of 12 edgesassign two prototype edges of one color, so that five differently colored copies of this pair can be placed without causing any interferences.
We always find exactly 12 different such assignments.
Shell “Rotation”: Add one of the 12 possible shell solutions Check color condition:
each node has 2 edges of all 6 colors Check loop condition:
no cycle shorter than 120 edges allowed. If necessary, backtrack!
One Ham. Cycle on the 600-CellOne Ham. Cycle on the 600-Cell
Thanks to Daniel Chen for programming this.
Hamiltonian Cycles on the 600-CellHamiltonian Cycles on the 600-Cell
1 cycle
Hamiltonian Cycles on the 600-CellHamiltonian Cycles on the 600-Cell
2 cycles
Hamiltonian Cycles on the 600-CellHamiltonian Cycles on the 600-Cell
4 cycles
Hamiltonian Cycles on the 600-CellHamiltonian Cycles on the 600-Cell
6 cycles
The Uncolored 120-CellThe Uncolored 120-Cell
600 vertices of valence 4, 1200 edges.
Find 2 congruent Hamiltonian cycles length 600.
3D Color Printer3D Color Printer (Z Corporation)(Z Corporation)
2004 (Brute-force Approach) for 120-Cell2004 (Brute-force Approach) for 120-Cell
Build both cycles simultaneously: Edges mirrored at 3D centroid get different colors Possible plane-mirror operations or C2 rotations are excluded,
because they all map some edges of the dodecahedron back onto themselves.
Do (single) path search with backtracking:Extend path without closing loop before length 600.
Result: We came to a length of 550/600, but then painted ourselves in a corner !(i.e., could not connect back to the start).
Thanks to Mike Pao for his programming efforts !
Trying to reduce the depth of the search tree, look for symmetries in prototype path itself.
Neither 3-fold nor 5-fold symmetry is possible:
We can also rule out inside/outside (w) symmetry,because of contradiction on intra_shell vs7 (see paper).
Legal coloring,but asymmetrical:
C3-symmetrical,but illegal coloring:
Legal coloring,but asymmetrical:
C5-symmetrical,but illegal cycle:
Symmetry Exploits for the 120-CellSymmetry Exploits for the 120-Cell
Shells in the 120-CellShells in the 120-Cell
Shell-based Approach for 120-Cell ?Shell-based Approach for 120-Cell ?
In the meantime we had solved the 600-Cell.
Shell approach is not practical for 120-Cell
Up to 120 edges per shell, only 2 colors: too many possible shell colorings ! impractical to pre-compute !
Edge-Based Coloring ApproachEdge-Based Coloring Approach
Grow multiple path segments, filling up shells in an orderly manner,avoiding any loop building: over-constrained impasses at the end.
Grow multiple path segments, extending segments in random order,but coloring constrained junctions first: very quick success !
A B
One Ham. Cycle on the 120-CellOne Ham. Cycle on the 120-Cell
Thanks to Daniel Chen for programming this.
Hamiltonian Cycles on the 120-CellHamiltonian Cycles on the 120-Cell
path differentiation with profiles:
120-Cell in De-powder Station120-Cell in De-powder Station
120-Cell with Hamiltonian Cycles120-Cell with Hamiltonian Cycles
Hamiltonian Cycles on 120-CellHamiltonian Cycles on 120-Cell
two paths distinguished by cross sections of the beams (circular / triangular)
Hamiltonian 2-ManifoldsHamiltonian 2-Manifoldswhere: what: connects how: what:
on edge graph:
Ham. Path(1-manifold)
touches all vertices(0-manifolds)
on edge graph:
Ham. Cycle(1-manifold)
passes thru all vertices(0-manifolds)
on polytope
Ham. Surface(2-manifold)
touches all edges(1-manifolds)
on polytope
Ham. Shell(2-manifold)
passes thru all edges(1-manifolds)
Three Levels of ChallengesThree Levels of Challenges
1.) Find a Hamiltonian shell or surface for each 4D polytope.
2.) Find such a 2-manifold of proper geometry, so that multiple copies of it can lead to a uniform coverage of all polytope faces.
3.) Look for maximal symmetry and for other nice properties ...
Hamiltonian Surface on 4D SimplexHamiltonian Surface on 4D Simplex
Moebius strip of 5 triangles: 5 open edges, 5 inner edges;
Inner/outer edges of same color form Hamiltonian cycles !
Two of these will cover all 10 faces of the 4D simplex.
Hamiltonian Closed Shell on HypercubeHamiltonian Closed Shell on Hypercube
Uses 16 out of 24 faces; all inner edges;
3 copies of this 2-manifold yield double coverage.
Hamiltonian Surface on HypercubeHamiltonian Surface on Hypercube
Uses 12 out of 24 faces; 16 inner, 16 outer edges;
This surface is congruent to its complement in 4D !
Two copies (in 4D, not in 3D) yield simple coverage.
Ham. 2-Manifold on 4D Cross PolytopeHam. 2-Manifold on 4D Cross Polytope
16 triangles form a closed Hamiltonian shell (torus);
2 copies of those cover all faces of the Cross Polytope.
What About the 3 Big Ones ??What About the 3 Big Ones ??
Work in progress:
24-Cell: almost there ... ?
120-Cell: first useful results
600-Cell: have not seriously started yet
2-Manifold Coverage of the 24-Cell2-Manifold Coverage of the 24-Cell
Some basic arithmetic:
There are 96 edges of valence 3
Possibility #1: Closed shell of 64 faces, passing through all 96 edges.Euler: 96{#E} – 24{#V} –64{#F} + 2 = 10 Genus 5;should partition 24-Cell into 2 sets of 12 octahedra.
Possibility #2: Open surface of 48 faces, with 48 border edges, and passing through 48 edges. GEP: 1 – 24{V} + 96{E} – 48{F} = 25 Ribbon Loops;might be a single closed band touching itself 24 times(with only 48 border edges, it’s a pretty tangled mess).
Ham. 2-Manifold on 24-CellHam. 2-Manifold on 24-Cell
Found 2 loops of 24 triangles each,-- not yet the desired solution!
2-Manifold Coverage of the 24-Cell2-Manifold Coverage of the 24-Cell
Symmetrical partial solution around z-axis
2-Manifold Coverage of the 120-Cell2-Manifold Coverage of the 120-Cell
Some basic arithmetic:
There are 1200 edges of valence 3.
Looking for: Open surface of 360 pentagons, with 600 border edges, and passing through 600 edges.
GEP: 1 –600{V} +1200{E} –360{F} = 241 Ribbon Loops. Imagine a main loop with 240 side loops;
Needs 480 branch points.
On each pentagon on average 3.333 edges are used by faces of the same color; this is equivalent to 1.333 branches.
360 pentagons * 1.333 branches 480 branch points !
2-Manifold Coverage of 120-Cell2-Manifold Coverage of 120-Cell
Study of the emerging coloring patterns at the core.
2-Manifold Coverage of the 120-Cell2-Manifold Coverage of the 120-Cell
We have found a 2-manifold coverage,with 1-2 pentagons on each edge, and exactly 3 pentagons around each vertex.
This is not congruent to its complement.
Probably does not have maximal possible symmetry.
Can we also have all the pass-thru edges of one color form a Hamiltonian cycle ?
2-Manifold Coverage of the 600-Cell2-Manifold Coverage of the 600-Cell
Some basic arithmetic: There are 720 edges of valence 5,
Valence 5 causes extra conceptual difficulties. 3600 edge uses.
There are several possibilities, e.g.: Open 2-manifold with 400 triangles, with 240 border edges, and passing through 480 edges (aim for coverage with 3 copies of this surface). GEP: 1 – 120{V} + 720{E} – 400{F} = 201 Ribbon Loops.Needs 400 branch points. Every triangle must serve as a branch points – but where do open edges come from ??
Perhaps, try something else ...
2-Manifold Coverage of the 600-Cell2-Manifold Coverage of the 600-Cell
Another attempt:
Open 2-manifold with 480 triangles, with 600 border edges, and passing through 120 edges (aim for double coverage with 5 copies of this surface).
Not enough inner edges to hang everything together ...
Need more thinking ...
Stay tuned ... !
ConclusionsConclusions
Wonderful abstract beauty !
Symmetries, interactions between Ham. cycles and Ham. 2-manifolds.
Mind-bending, headache-creating ...
End on an easier note ...
Make surfaces of a different kind ...
““Volution”Volution” Surfaces Spanning Surfaces Spanning Hamiltonian CyclesHamiltonian Cycles
Back to 3D-space and art ...
VolutionVolution Surfaces (Bridges 2003) Surfaces (Bridges 2003)
“Volution’s Evolution”
Minimal surfaces of different genussuspended in a wire frame composed of
12 quarter-circles on the surface of a cube.
New New VolutionVolution Surfaces Surfaces
Use the Hamiltonian Cycles
found on the Platonic solids
to suspend Volution surfaces.
On the DodecahedronOn the Dodecahedron
2 holes
On the IcosahedronOn the Icosahedron
+ 4 tubes
Many Different Models for IcosahedronMany Different Models for Icosahedron
How I Start Designing these ObjectsHow I Start Designing these Objects
Or with Zome-Tool ModelsOr with Zome-Tool Models
Paper cylinders mark positions of tunnels.
Make a Crude Polyhedral ModelMake a Crude Polyhedral Model
refine with Brakke’s “Surface Evolver”
Make a 3D ObjectMake a 3D Object
Import to SLIDE, apply some surface offset;
export as an STL file, and send to an RP machine.
Icosa_Vol_J9Icosa_Vol_J9
6 tubes
Questions ?Questions ?
QUESTIONS ?QUESTIONS ?