# 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 ?

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

Uses all edges of a graph.
   Eulerian Cycle:
A closed Eulerian Path
END
that returns to the start.

   Hamiltonian Path:
Visits all vertices once.
   Hamiltonian Cycle:
A closed Ham. Path.
This is a Test … (closed book!)

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

 Itadmits 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

 Eulerian   Cycles ?    Hamiltonian   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 ?
The Cuboctahedron

   Hamiltonian cycle             Flattened net of
on polyhedron edges.           cuboctahedron
to show symmetry.

The cyan and the red cycles are congruent (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 ?
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.

The common Hamiltonian cycle for all K2i+1
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
4D Simplex: 2 Hamiltonian Paths

Two identical paths, complementing each other

C2
4D Cross Polytope: 3 Paths
 All   vertices have valence 6 !
C3
Hypercube: 2 Hamitonian Cycles

 There   are many different options:
The Most Satisfying Solution ?

 Each  Path
has its own
C4 (C2)         C2-symmetry.

 90°-rotation
around z-axis
changes color
on all edges.
The 24-Cell Is More Challenging
   Valence 8:                   Natural to consider
C4       one C4-axis
-> 4 Hamiltonian paths            for replicating the
are needed !                   Hamiltonian cycles.
24-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-Cell
 The   24 vertices lie on 3 shells.
 Pre-colorthe shells individually
obeying 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 – Showing Broken Symmetry
Almost C2
C2-Symmetry  More than One Cycle!

C2

   That is what I had to inspect for …
Path Composition
24-Cell: 4 Hamiltonian Cycles

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.
Tetrahedral Symmetry on “0cta”-Shell
   C3*-rotations that keep one color in place,
cyclically exchange the three other ones:

C3*

C3*
C3*
C3*
“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

   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 !
Exploiting C3-Symmetry for the 24-Cell
OUTER SHELL

MIDDLE SHELL
I/O-MIRROR

INNER SHELL

REPEATED UNIT

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

   We can also use inside / outside symmetry,
so only 1/6 of the cycle needs to be found !
One of the C3-symmetric Cycles
INSIDE-OUTSIDE SYMMETRY
Pipe-Cleaner Models for the 24-Cell
That is actually how I first found
the tetrahedral solution !

INNER
OCTA

CENTRAL
CUBOCTA

OUTER
OCTA
Rapid Prototyping Model of the 24-Cell

   Notice
the 3-fold
permutation
of colors

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).
 Perhapswe can exploit symmetry
to 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 path
with the two other edges of a triangle
attached 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

OUTERMOST TETRAHEDRON
INSIDE / OUTSIDE
SYMMETRY

CONNECTORS
SPANNING THE
CENTRAL SHELL

Number of segments of each type in each Hamiltonian cycle
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
CONNECTORS
total: 32V
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 model
to 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 ?
   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)

Directionality !!

C3 (RMY, CGB)                            C3 (GCY, MBR)
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 Shell Vertex Positions
Basic Tetra   --   truncated   --    or beveled

Dual (mid-face) -- Dual truncated --   Mid-edge     .
All Possible Edge Patterns
( shown on one tetrahedral face )

Constraints between
2 edges of one cycle
INTRA-
SHELL
EDGES            6               12

INTER-                                   Two completely
SHELL                                   independent sets
EDGES
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
32
   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.
Comparison: ITDS  600-Cell

   Total Colorings: [10149]  6720  10560

   Pick 120 edges: [1037]  (   720 )
120
 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??
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, and
Icosi-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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