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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

- 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.

- Eulerian Path:Uses all edges of a graph.
- Eulerian Cycle:A closed Eulerian Paththat returns to the start.

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

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

- 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 ?

- Hamiltonian Cycles ?

- Eulerian Cycles ?

- 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.

- Flattened net ofcuboctahedronto show symmetry.

Thecyanand theredcycles arecongruent (mirrored)!

- 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.

- 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

- K2i+1 will need i Hamiltonian cycles for coverage.
- Arrange nodes with i-fold symmetry: 2i-gon C2i
- Last node is placed in center.

- 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.

From BRIDGES’2002 Talk

- 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

Two identical paths, complementing each other

- All vertices have valence 6 !

C3

There are many different options:

C4 (C2)

- 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

- Valence 8:
-> 4 Hamiltonian pathsare needed !

This is a mess …hard to deal with !

Exploiting the concentric shells:

5 families of edges.

OUTER SHELL

MIDDLE SHELL

INNER SHELL

This is the visitation schedule for one Ham. cycle.

- The 24 vertices lie on 3 shells.
- Pre-color the shells individuallyobeying the desired symmetry.
- Rotate shells against each other.

Almost C2

That is what I had to inspect for …

C2

Aligned:

4-fold symmetry

- 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*

- C3*-rotations that keep one color in place, cyclically exchange the three other ones:

- 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

- 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 !

- Only 1/3 of the cycle needs to be found.(C3-axis does not go through any edges, vertices)

REPEATED UNIT

INSIDE-OUTSIDE SYMMETRY

CENTRAL CUBOCTA

OUTER

OCTA

That is actually how I first found the tetrahedral solution !

INNER OCTA

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

- 120-Cell:
- 600 valence 4-vertices, 1200 edges
- --> May yield 2 Hamiltonian cycles length 600.

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.

- 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 !

Just two shells (magenta) and (yellow)

Each Ham.-path needs 15 edges on each shell,and 10 connectors (cyan) between the shells.

Colored by two congruent Hamiltonian cycles

- 120 vertices,valence 12;
- 720 edges;
Make 6 cycles, length 120.

- 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.

INNERMOST TETRAHEDRON

Number of segments of each type in each Hamiltonian cycle

INSIDE / OUTSIDE SYMMETRY

OUTERMOST TETRAHEDRON

CONNECTORS SPANNING THE CENTRAL SHELL

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 … ?

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 …

Just 4 nested shells (192 edges):

- Tetrahedron: 4V, 6E
- Icosahedron: 12V, 30E
- Icosahedron: 12V, 30E
- Tetrahedron: 4V, 6E
total: 32V

CONNECTORS

TETRA

ICOSA

ICOSA

TETRA

SHELLS CONNECTORS

SHELLS CONNECTORS

TETRA

ICOSA

ICOSA

TETRA

Icosi-tetrahedral Double Shell

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 !!

- 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)

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 .

( shown on one tetrahedral face )

Constraintsbetween2 edges of one cycle

INTRA-SHELL

EDGES

12

6

INTER-SHELL

EDGES

Twocompletely

independentsets

12

12+12

Basic Tetrahedron (4):

Tetras with Offset Edges (12):

0 1 2 3 4 5

6 7 8 9 10 11

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!

- 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

- 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

- 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 …

Resulting from the two complementary Hamiltonian paths on cuboctahedron

- 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.

- 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.

- 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.