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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

Bridges 2004, Winfield KS

Hamiltonian Cycleson Symmetrical Graphs

Carlo H. Séquin

EECS Computer Science Division

University of California, Berkeley

map of k nigsberg
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.




  • 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.
this is a test closed book
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
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
The Platonic Solids in 3D
  • Hamiltonian Cycles ?
  • Eulerian Cycles ?
the octahedron
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 cycleon polyhedron edges.The Cuboctahedron
  • Flattened net ofcuboctahedronto show symmetry.

Thecyanand theredcycles arecongruent (mirrored)!

larger challenges
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 k 5 k 7 and k 9
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.




Can we make the i Hamiltonian cycles in each graph congruent ?

complete graphs k 2 i 1

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
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
The 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 paths


4D Simplex: 2 Hamiltonian Paths

Two identical paths, complementing each other

4d cross polytope 3 paths
4D Cross Polytope: 3 Paths
  • All vertices have valence 6 !


the most satisfying solution

C4 (C2)

The Most Satisfying Solution ?
  • Each Path has its own C2-symmetry.
  • 90°-rotationaround z-axischanges coloron all edges.
the 24 cell is more challenging

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


The 24-Cell Is More Challenging
  • Valence 8:

-> 4 Hamiltonian pathsare needed !

24 cell shell based approach
This is a mess …hard to deal with !

Exploiting the concentric shells:

5 families of edges.

24-Cell: Shell-based Approach
shell schedule for the 24 cell




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.
24 cell 4 hamiltonian cycles
24-Cell: 4 Hamiltonian Cycles


 4-fold symmetry

why shells make task easier
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.
tetrahedral symmetry on 0cta shell





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
“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




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 !

exploiting c 3 symmetry for the 24 cell





  • 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)


pipe cleaner models for the 24 cell




Pipe-Cleaner Models for the 24-Cell

That is actually how I first found the tetrahedral solution !


rapid prototyping model of the 24 cell
Rapid Prototyping Model of the 24-Cell
  • Noticethe 3-foldpermutationof colorsMade on the Z-corp machine.
the uncolored 120 cell
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
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
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
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
Dodecahedral Double Shell

Colored by two congruent Hamiltonian cycles

the 600 cell
The 600-Cell
  • 120 vertices,valence 12;
  • 720 edges;

 Make 6 cycles, length 120.

search on the 600 cell
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
Shells in the 600-Cell


Number of segments of each type in each Hamiltonian cycle




shells in the 600 cell45
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
Start With a Simpler Model …


  • 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: 32V


the complete itds 4 shells 192 edges
The Complete ITDS: 4 shells, 192 edges






one cycle on the itds
One Cycle on the ITDS






broken part on zcorp machine
Broken 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

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.






when is i o symmetry possible
When Is I/O Symmetry Possible ?


  • 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 edge patterns
All Possible Edge Patterns

( shown on one tetrahedral face )

Constraintsbetween2 edges of one cycle











possible colorings for intra shell edges
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
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
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.


comparison itds 600 cell
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??


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 …
double volution shell
Double Volution Shell

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.
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.