Hamiltonian cycles on symmetrical graphs
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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

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Bridges 2004, Winfield KS

Hamiltonian Cycleson 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 ?

Leonhard Euler (1707-83) says: NO ! (1735)– because there are vertices with odd valence.


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!)

  • What Eulerian / Hamiltonian Path / Cycle(s)does the following graph contain ?


Answer:

  • It admits an Eulerian Cycle !– but no Hamiltonian Path.


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

  • Hamiltonian Cycles ?

  • Eulerian Cycles ?


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 ?


Hamiltonian cycleon polyhedron edges.

The Cuboctahedron

  • Flattened net ofcuboctahedronto show symmetry.

Thecyanand theredcycles arecongruent (mirrored)!


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 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 ?


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 …

  • 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

From BRIDGES’2002 Talk


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


C2

4D Simplex: 2 Hamiltonian Paths

Two identical paths, complementing each other


4D Cross Polytope: 3 Paths

  • All vertices have valence 6 !

C3


There are many different options:

Hypercube: 2 Hamitonian Cycles


C4 (C2)

The Most Satisfying Solution ?

  • Each Path has its own C2-symmetry.

  • 90°-rotationaround z-axischanges coloron all edges.


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

C4

The 24-Cell Is More Challenging

  • Valence 8:

    -> 4 Hamiltonian pathsare needed !


This is a mess …hard to deal with !

Exploiting the concentric shells:

5 families of edges.

24-Cell: Shell-based Approach


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.


One Cycle – Showing Broken Symmetry

Almost C2


That is what I had to inspect for …

C2-Symmetry  More than One Cycle!

C2


Path Composition


24-Cell: 4 Hamiltonian Cycles

Aligned:

 4-fold symmetry


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.


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

  • 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 !


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 !


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


INSIDE-OUTSIDE SYMMETRY

One of the C3-symmetric Cycles


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

  • Noticethe 3-foldpermutationof colorsMade on the Z-corp machine.


3D Color Printer(Z Corporation)


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

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

  • 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

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

Colored by two congruent Hamiltonian cycles


Physical Model of Penta-Double-Shell


The 600-Cell

  • 120 vertices,valence 12;

  • 720 edges;

     Make 6 cycles, length 120.


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

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-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 …

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)

Just 4 nested shells (192 edges):

  • Tetrahedron: 4V, 6E

  • Icosahedron: 12V, 30E

  • Icosahedron: 12V, 30E

  • Tetrahedron: 4V, 6E

    total: 32V

CONNECTORS


ITDS: The 2 Icosahedral Shells


The Complete ITDS: 4 shells, 192 edges

TETRA

ICOSA

ICOSA

TETRA

SHELLS CONNECTORS


One Cycle on the ITDS

SHELLS CONNECTORS

TETRA

ICOSA

ICOSA

TETRA


The Composite ITDS


ITDS: Composite of 6 Ham. Cycles


One Vertex of ITDS (valence 12)


Broken Part on Zcorp machine

Icosi-tetrahedral Double Shell


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 ?


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 ?

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 Positions


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

Basic Tetrahedron (4):

Tetras with Offset Edges (12):

0 1 2 3 4 5

6 7 8 9 10 11


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

  • 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

  • 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” … ?

  • 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-Cell


As a Sculpture


Double Volution Shell

Resulting from the two complementary Hamiltonian paths on cuboctahedron


As a Sculpture


4D Cross Polytope


As a Sculpture


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)

  • 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)

  • 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 ?


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