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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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.

Answer: • 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 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.

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

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.

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

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.

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 … ?

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) Just 4 nested shells (192 edges): • Tetrahedron: 4V, 6E • Icosahedron: 12V, 30E • Icosahedron: 12V, 30E • Tetrahedron: 4V, 6E total: 32V CONNECTORS