# Hamiltonian Cycles on Symmetrical Graphs - PowerPoint PPT Presentation

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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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#### Presentation Transcript

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 ?

• 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:

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

### 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

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.

### 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

### 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 … ?

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

### The Complete ITDS: 4 shells, 192 edges

TETRA

ICOSA

ICOSA

TETRA

SHELLS CONNECTORS

### One Cycle on the ITDS

SHELLS CONNECTORS

TETRA

ICOSA

ICOSA

TETRA

### 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 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

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 …

### Double Volution Shell

Resulting from the two complementary Hamiltonian paths on cuboctahedron

### 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.