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 ?