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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 ?.

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hamiltonian cycles on symmetrical graphs

Bridges 2004, Winfield KS

Hamiltonian Cycleson Symmetrical Graphs

Carlo H. Séquin

EECS Computer Science Division

University of California, Berkeley

map of k nigsberg
Map 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.

definitions

START

END

  • Hamiltonian Path:Visits all vertices once.
  • Hamiltonian Cycle:A closed Ham. Path.
Definitions
  • Eulerian Path:Uses all edges of a graph.
  • Eulerian Cycle:A closed Eulerian Paththat returns to the start.
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.
  • No Eulerian Paths: >2 odd-valence vertices.
  • Hamiltonian Cycles? – YES!
  • = Projection of a cube (edge frame); Do other Platonic solids have Hamiltonian cycles ?
the platonic solids in 3d
The Platonic Solids in 3D
  • Hamiltonian Cycles ?
  • Eulerian Cycles ?
the octahedron
The 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 cuboctahedron
Hamiltonian cycleon polyhedron edges.The Cuboctahedron
  • Flattened net ofcuboctahedronto show symmetry.

Thecyanand theredcycles arecongruent (mirrored)!

larger challenges
Larger 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 k 5 k 7 and k 9
Complete Graphs K5, K7, and K9
  • 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 k 2 i 1

The common Hamiltonian cycle for all K2i+1

Complete Graphs K2i+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.
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 4d
The 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 paths

C2

4D Simplex: 2 Hamiltonian Paths

Two identical paths, complementing each other

4d cross polytope 3 paths
4D Cross Polytope: 3 Paths
  • All vertices have valence 6 !

C3

the most satisfying solution

C4 (C2)

The Most Satisfying Solution ?
  • Each Path has its own C2-symmetry.
  • 90°-rotationaround z-axischanges coloron all edges.
the 24 cell is more challenging

Natural to consider one C4-axis for replicating the Hamiltonian cycles.

C4

The 24-Cell Is More Challenging
  • Valence 8:

-> 4 Hamiltonian pathsare needed !

24 cell shell based approach
This is a mess …hard to deal with !

Exploiting the concentric shells:

5 families of edges.

24-Cell: Shell-based Approach
shell schedule for the 24 cell

OUTER SHELL

MIDDLE SHELL

INNER SHELL

This is the visitation schedule for one Ham. cycle.

Shell-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.
24 cell 4 hamiltonian cycles
24-Cell: 4 Hamiltonian Cycles

Aligned:

 4-fold symmetry

why shells make task easier
Why 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 shell

C3*

C3*

C3*

C3*

Tetrahedral Symmetry on “0cta”-Shell
  • C3*-rotations that keep one color in place, cyclically exchange the three other ones:
tetrahedral symmetry for the 24 cell
“Tetrahedral” Symmetry for the 24-Cell
  • 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 …

Note that the same-colored edges are disconnected !

another shell schedule for the 24 cell

OUTER SHELL

MIDDLE SHELL

INNER SHELL

Another Shell-Schedule for the 24-Cell
  • Stay on each shell for only one edge at a time:

This is the schedule needed for overall tetrahedral symmetry !

exploiting c 3 symmetry for the 24 cell

OUTER SHELL

I/O-MIRROR

MIDDLE SHELL

INNER SHELL

  • We can also use inside / outside symmetry,so only 1/6 of the cycle needs to be found !
Exploiting C3-Symmetry for the 24-Cell
  • Only 1/3 of the cycle needs to be found.(C3-axis does not go through any edges, vertices)

REPEATED UNIT

pipe cleaner models for the 24 cell

CENTRAL CUBOCTA

OUTER

OCTA

Pipe-Cleaner Models for the 24-Cell

That is actually how I first found the tetrahedral solution !

INNER OCTA

rapid prototyping model of the 24 cell
Rapid Prototyping Model of the 24-Cell
  • Noticethe 3-foldpermutationof colorsMade on the Z-corp machine.
the uncolored 120 cell
The Uncolored 120-Cell
  • 120-Cell:
    • 600 valence 4-vertices, 1200 edges
    • --> May yield 2 Hamiltonian cycles length 600.
brute force approach for the 120 cell
Brute-force approach for the 120-Cell

Thanks to Mike Pao for his programming efforts !

  • 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.
a more promising approach
A 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 shells
Simpler 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 shell
Dodecahedral Double Shell

Colored by two congruent Hamiltonian cycles

the 600 cell
The 600-Cell
  • 120 vertices,valence 12;
  • 720 edges;

 Make 6 cycles, length 120.

search on the 600 cell
Search 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 cell
Shells in the 600-Cell

INNERMOST TETRAHEDRON

Number of segments of each type in each Hamiltonian cycle

INSIDE / OUTSIDE SYMMETRY

OUTERMOST TETRAHEDRON

CONNECTORS SPANNING THE CENTRAL SHELL

shells in the 600 cell1
Shells 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: 32V

CONNECTORS

the complete itds 4 shells 192 edges
The Complete ITDS: 4 shells, 192 edges

TETRA

ICOSA

ICOSA

TETRA

SHELLS CONNECTORS

one cycle on the itds
One Cycle on the ITDS

SHELLS CONNECTORS

TETRA

ICOSA

ICOSA

TETRA

broken part on zcorp machine
Broken 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

Directionality !!

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 (BCM, YRG)

C3 (GCY, MBR)

C3 (RMY, CGB)

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 !

all possible edge patterns
All Possible Edge Patterns

( shown on one tetrahedral face )

Constraintsbetween2 edges of one cycle

INTRA-SHELL

EDGES

12

6

INTER-SHELL

EDGES

Twocompletely

independentsets

12

12+12

possible colorings for intra shell edges
Possible Colorings for Intra-Shell Edges

Basic Tetrahedron (4):

Tetras with Offset Edges (12):

0 1 2 3 4 5

6 7 8 9 10 11

inter shell edge colorings
Inter-shell Edge Colorings

Adding the second half-edge:

0 1 2 3 4 5

6 7 8 9 10 11

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)

Always two options – but only 12 unique solutions!

combinatorics for the itds
Combinatorics 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: 103until success.

32

comparison itds 600 cell
Comparison: ITDS 600-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 …
double volution shell
Double Volution Shell

Resulting from the two complementary Hamiltonian paths on cuboctahedron

conclusions
Conclusions
  • 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.
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