hamiltonian cycles on symmetrical graphs
DESCRIPTION
Bridges 2004, Winfield KS. Hamiltonian Cycles on Symmetrical Graphs. Carlo H. Séquin EECS Computer Science Division University of California, Berkeley. Map of Königsberg. Can you find a path that crosses all seven bridges exactly once – and then returns to the start ?. - PowerPoint PPT PresentationTRANSCRIPT
Hamiltonian CyclesHamiltonian Cycles on Symmetrical Graphson Symmetrical Graphs
Carlo H. Séquin
EECS Computer Science Division
University of California, Berkeley
Bridges 2004, Winfield KS
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.
DefinitionsDefinitions
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.
This is a Test … (closed book!)This is a Test … (closed book!)
What Eulerian / Hamiltonian Path / Cycle(s)does the following graph contain ?
Answer:Answer:
It admits an Eulerian Cycle !– but no Hamiltonian Path.
Another Example … (extra credit!)Another Example … (extra credit!) What paths/cycles exist on this graph?
No Eulerian Cycles: Not all valences are even.
Hamiltonian Cycles? – YES!
No Eulerian Paths: >2 odd-valence vertices.
= Projection of a cube (edge frame); Do other Platonic solids have Hamiltonian cycles ?
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.
Are there other (semi-)regular polyhedra for which we can do that ?
The CuboctahedronThe Cuboctahedron
Hamiltonian cycleon polyhedron edges.
Flattened net ofcuboctahedronto show symmetry.
The cyan and the red cycles are congruent (mirrored)!
Larger ChallengesLarger Challenges
All these graphs have been planar … boring !
Our examples had only two Hamiltonian cycles.
Can we find graphs that are covered by three or more Hamiltonian cycles ?
Graphs need to have vertices of valence ≥ 6.
Can we still make those cycles congruent ?
Graphs need to have all vertices equivalent.
Let’s look at complete graphs,
i.e., N fully connected vertices.
Complete Graphs KComplete Graphs K55, K, K77, , andand K K99
5, 7, 9 vertices – all connected to each other.
Let’s only consider graphs with all even vertices,i.e., only K2i+1.
K5 K7 K9
Can we make the i Hamiltonian cycles in each graph congruent ?
Complete Graphs KComplete Graphs K22ii+1+1
K2i+1 will need i Hamiltonian cycles for coverage.
Arrange nodes with i-fold symmetry: 2i-gon C2i
Last node is placed in center.
The common Hamiltonian cycle for all K2i+1
Make Constructions in 3D …Make Constructions in 3D …
We would like to have highly symmetrical graphs.
All vertices should be of the same even valence.
All vertices should be connected equivalently.
Graph should allow for some symmetrical layout in 3D space.
Where can we obtain such graphs? From 4D!
(But don’t be afraid of the 4D source … This is really just a way of getting interesting 3D wire frames on which we can play Euler’s coloring game.
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
4D Simplex: 2 Hamiltonian Paths4D Simplex: 2 Hamiltonian Paths
Two identical paths, complementing each other
C2
4D Cross Polytope: 3 Paths4D Cross Polytope: 3 Paths
All vertices have valence 6 !
C3
Hypercube: 2 Hamitonian CyclesHypercube: 2 Hamitonian Cycles
There are many different options:
The Most Satisfying Solution ?The Most Satisfying Solution ?
Each Path has its own C2-symmetry.
90°-rotationaround z-axischanges coloron all edges.
C4 (C2)
The 24-Cell Is More ChallengingThe 24-Cell Is More Challenging Valence 8:
-> 4 Hamiltonian pathsare needed !
Natural to consider one C4-axis for replicating the Hamiltonian cycles.
C4
24-Cell: Shell-based Approach24-Cell: Shell-based Approach
This is a mess …hard to deal with !
Exploiting the concentric shells:
5 families of edges.
Shell-Schedule for the 24-CellShell-Schedule for the 24-Cell
The 24 vertices lie on 3 shells.
Pre-color the shells individuallyobeying the desired symmetry.
Rotate shells against each other.
OUTER SHELL
MIDDLE SHELL
INNER SHELL
This is the visitation schedule for one Ham. cycle.
One Cycle – One Cycle – Showing Broken SymmetryShowing Broken Symmetry
Almost C2
CC22-Symmetry -Symmetry More than One Cycle! More than One Cycle!
C2
That is what I had to inspect for …
Path CompositionPath Composition
24-Cell: 4 Hamiltonian Cycles24-Cell: 4 Hamiltonian Cycles
Aligned:
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.
Tetrahedral Symmetry on “0cta”-ShellTetrahedral Symmetry on “0cta”-Shell
C3*-rotations that keep one color in place, cyclically exchange the three other ones:
CC33**
CC33**CC33**
CC33**
““Tetrahedral” Symmetry for the 24-CellTetrahedral” Symmetry for the 24-Cell
Note that the same-colored edges are disconnected !
All shells must have the same symmetry orientation- this reduces the size of the search tree greatly.
Of course such a solution may not exist …
Another Shell-Schedule for the 24-CellAnother Shell-Schedule for the 24-Cell
Stay on each shell for only one edge at a time:
OUTER SHELL
MIDDLE SHELL
INNER SHELL
This is the schedule needed for overall tetrahedral symmetry !
Only 1/3 of the cycle needs to be found.(C3-axis does not go through any edges, vertices)
Exploiting Exploiting CC33-Symmetry-Symmetry for the 24-Cell for the 24-Cell
OUTER SHELL
MIDDLE SHELL
INNER SHELL
REPEATED UNIT
I/O-MIRROR
We can also use inside / outside symmetry,so only 1/6 of the cycle needs to be found !
One of the One of the CC33-symmetric-symmetric Cycles CyclesINSIDE-OUTSIDE SYMMETRY
INNER OCTA
That is actually how I first found the tetrahedral solution !
Pipe-Cleaner Models for the 24-CellPipe-Cleaner Models for the 24-Cell
CENTRAL
CUBOCTA
OUTEROCTA
Rapid Prototyping Model of the 24-CellRapid Prototyping Model of the 24-Cell
Noticethe 3-foldpermutationof colors
Made on the Z-corp machine.
3D Color Printer3D Color Printer (Z Corporation)(Z Corporation)
The Uncolored 120-CellThe Uncolored 120-Cell
120-Cell: 600 valence 4-vertices, 1200 edges
--> May yield 2 Hamiltonian cycles length 600.
Brute-force approach for the 120-CellBrute-force approach for the 120-Cell
Assign opposite edges different colors (i.e., build both cycles simultaneously).
Do path-search with backtracking.
Came to a length of 550/600, but then painted ourselves in a corner !(i.e., could not connect back to the start).
Perhaps we can exploit symmetryto make search tree less deep.
Thanks to Mike Pao for his programming efforts !
A More Promising ApproachA More Promising Approach
Clearly we need to employ the shell-based approach for these monsters!
But what symmetries can we expect ?
These objects belong to the icosahedral symmetry group which has 6 C5-axes and 10 C3-axes.
Can we expect the individual paths to have 3-fold or 5-fold symmetry ? This would dramatically reduce the depth of the search tree !
Simpler Model with Dodecahedral ShellsSimpler Model with Dodecahedral Shells
Just two shells (magenta) and (yellow)Each Ham.-path needs 15 edges on each shell,and 10 connectors (cyan) between the shells.
Dodecahedral Double ShellDodecahedral Double Shell
Colored by two congruent Hamiltonian cycles
Physical Model of Penta-Double-ShellPhysical Model of Penta-Double-Shell
The 600-CellThe 600-Cell
120 vertices,valence 12;
720 edges;
Make 6 cycles, length 120.
Search on the 600-CellSearch on the 600-Cell
Search by “loop expansion”:
Replace an edge in the current pathwith the two other edges of a triangleattached to the chosen edge.
Always keeps path a closed cycle !
This quickly worked for finding a full cycle.
Also worked for finding 3 congruent cycles of length 120.
When we tried to do 4 cycles simultaneously, we got to 54/60 using inside/outside symmetry.
Shells in the 600-CellShells in the 600-Cell
Number of segments of each type in each Hamiltonian cycle
INNERMOST TETRAHEDRON
OU
TE
RM
OS
T T
ET
RA
HE
DR
ON
CONNECTORS SPANNING THE CENTRAL SHELL
INSIDE / OUTSIDE SYMMETRY
Shells in the 600-CellShells in the 600-Cell
Summary of features:
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.
Inside/outside symmetry
What other symmetries are there … ?
Start With a Simpler Model …Start With a Simpler Model …
Specifications:
All vertices of valence 12
Overall symmetry compatible with “tetra-6”
Inner-, outer-most shells = tetrahedra
No edge intersections
As few shells as possible …
…. This is tricky …
Icosi-Tetrahedral Double-Shell (ITDS)Icosi-Tetrahedral Double-Shell (ITDS)
Just 4 nested shells (192 edges):
Tetrahedron: 4V, 6E
Icosahedron: 12V, 30E
Icosahedron: 12V, 30E
Tetrahedron: 4V, 6E
total: 32VCONNECTORS
ITDS: The 2 Icosahedral ShellsITDS: The 2 Icosahedral Shells
The Complete ITDS: 4 shells, 192 edgesThe Complete ITDS: 4 shells, 192 edges
SHELLS CONNECTORS
TETRA
ICOSA
ICOSA
TETRA
One Cycle on the ITDSOne Cycle on the ITDS
SHELLS CONNECTORS
TETRA
ICOSA
ICOSA
TETRA
The Composite ITDSThe Composite ITDS
ITDS: Composite of 6 Ham. CyclesITDS: Composite of 6 Ham. Cycles
One Vertex of ITDS (valence 12)One Vertex of ITDS (valence 12)
Broken Part on Zcorp machineBroken Part on Zcorp machine
Icosi-tetrahedral Double Shell
What Did I Learn from the ITDS ?What Did I Learn from the ITDS ?
A larger, more complex modelto exercise the shell-based approach.
Shells, or subsets of edges cannot just be rotated as in the first version of the 24-Cell.
The 6-fold symmetry, corresponding to six differently colored edges on a tetrahedron,is actually quite tricky !
Not one of the standard symmetry groups.
What are the symmetries we can hope for ?
The Symmetries of the Composite ?The Symmetries of the Composite ? 4 C3 rotational axes (thru tetra vertices)
that permute two sets of 3 colors each.
Inside/outside mirror symmetry.
C3 (RGB, CMY)
C3 (RMY, CGB) C3 (GCY, MBR)
C3 (BCM, YRG)
Directionality !!
???
When Is I/O Symmetry Possible ?When Is I/O Symmetry Possible ?Hypothesis:
When the number of edges in one Ham. cycle that cross the central shell is 4i+2
The 600-Cell cannot accommodate I/O symmetry !
Basic Tetra -- truncated -- or beveled
Dual (mid-face) -- Dual truncated -- Mid-edge .
All Possible Shell Vertex PositionsAll Possible Shell Vertex Positions
All Possible Edge PatternsAll Possible Edge Patterns( shown on one tetrahedral face )
12
6
12+12
12
INTRA-SHELLEDGES
INTER-SHELLEDGES
Two completely independent sets
Constraints between2 edges of one cycle
0 1 2 3 4 5
Possible Colorings for Intra-Shell EdgesPossible Colorings for Intra-Shell Edges
Tetras with Offset Edges (12):
6 7 8 9 10 11
Basic Tetrahedron (4):
0+3(9) 1+7(10) 2+8(11) 3+0(6) 4+7(10) 5+8(11)
6+3(9) 7+1(4) 8+2(5) 9+0(6) 10+1(4) 11+2(5)
Inter-shell Edge ColoringsInter-shell Edge Colorings
Adding the second half-edge:
0 1 2 3 4 5
6 7 8 9 10 11
Always two options – but only 12 unique solutions!
Combinatorics for the ITDSCombinatorics for the ITDS
Total colorings: 6192 10149
Pick 192 / 6 edges: ( 192 ) 1037
Pick one edge at every vertex: 1232 1034
Assuming inside-out symmetry: 1216 1017
All shell combinations: 42 *5762 *126 *124 1017
Combinations in my GUI: 42 *576 *126 *124 1014
Constellations examined: 103 until success.
32
Comparison: Comparison: ITDSITDS 600-Cell600-Cell
Total Colorings: [10149] 6720 10560
Pick 120 edges: [1037] ( 720 ) 10168
Pick one edge at every vertex: [1034] 12120 10130
Hope for inside-out symmetry: [1017] 1260 1065
All shell combinations: [1017] 42 *124 *1254 1063
Shells with I/O symmetry: [1014] 4 *122 *1228 1032
Constellations to examine: [103] 10??
120
Where is the “Art” … ?Where is the “Art” … ?
Can these Math Models lead to something artistic as well ?
Any constructivist sculptures resulting from these efforts ?
Suppose you had to show the flow of the various Hamiltonian cycles without the use of color …
Complementary Bands in the 5-CellComplementary Bands in the 5-Cell
As a SculptureAs a Sculpture
Double Double VolutionVolution Shell Shell
Resulting from the two complementary Hamiltonian paths on cuboctahedron
As a SculptureAs a Sculpture
4D Cross Polytope4D Cross Polytope
As a Sculpture
ConclusionsConclusions
Finding a Hamiltonian path/cycle is an NP-hard computational problem.
Trying to get Eulerian coverage with a set of congruent Hamiltonian paths is obviously even harder.
Taking symmetry into account judiciously can help enormously.
Conclusions (2)Conclusions (2)
The simpler 4D polytopes yielded their solutions relatively quickly.
Those solutions actually do have nice symmetrical paths!
The two monster polytopes presented a much harder problem than first expected-- mostly because I did not understand what symmetries can truly be asked for.
Conclusions (3)Conclusions (3)
The 24-Cell, Double-Penta-Shell, andIcosi-Tetra Double Shell, have given me a much deeper understanding of the symmetry issues involved.
Now it’s just a matter of programming these insights into a procedural search to find the 2 remaining solutions.
The 24-Cell is really unusually symmetrical and the most beautiful of them all.
QUESTIONS ?QUESTIONS ?