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 ?.
Bridges 2004, Winfield KS
Hamiltonian Cycleson Symmetrical Graphs
Carlo H. Séquin
EECS Computer Science Division
University of California, Berkeley
Leonhard Euler (1707-83) says: NO ! (1735)– because there are vertices with odd valence.
Hamiltonian cycleon polyhedron edges.
Thecyanand theredcycles arecongruent (mirrored)!
Graphs need to have vertices of valence ≥ 6.
Graphs need to have all vertices equivalent.
i.e., N fully connected vertices.
Can we make the i Hamiltonian cycles in each graph congruent ?
The common Hamiltonian cycle for all K2i+1
From BRIDGES’2002 Talk
Cell-first Face-first Edge-first Vertex-first
Use Cell-first: High symmetry; no coinciding vertices/edges
Two identical paths, complementing each other
There are many different options:
-> 4 Hamiltonian pathsare needed !
This is a mess …hard to deal with !
Exploiting the concentric shells:
5 families of edges.
This is the visitation schedule for one Ham. cycle.
That is what I had to inspect for …
Note that the same-colored edges are disconnected !
This is the schedule needed for overall tetrahedral symmetry !
That is actually how I first found the tetrahedral solution !
Thanks to Mike Pao for his programming efforts !
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
Make 6 cycles, length 120.
Number of segments of each type in each Hamiltonian cycle
INSIDE / OUTSIDE SYMMETRY
CONNECTORS SPANNING THE CENTRAL SHELL
Summary of features:
Just 4 nested shells (192 edges):
Icosi-tetrahedral Double Shell
A larger, more complex modelto exercise the shell-based approach.
C3 (RGB, CMY)
C3 (BCM, YRG)
C3 (GCY, MBR)
C3 (RMY, CGB)
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
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!
Resulting from the two complementary Hamiltonian paths on cuboctahedron