1. Hamiltonian Cycles on Symmetrical Graphs Carlo H. Séquin EECS Computer Science Division University of California, Berkeley Bridges 2004, Winfield KS

2. Map of Königsberg u 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.

3. Definitions u Eulerian Path: Uses all edges of a graph. u Eulerian Cycle: A closed Eulerian Path that returns to the start. END START u Hamiltonian Path: Visits all vertices once. u Hamiltonian Cycle: A closed Ham. Path.

4. This is a Test … (closed book!) u What Eulerian / Hamiltonian Path / Cycle(s) does the following graph contain ?

5. Answer: u It admits an Eulerian Cycle ! – but no Hamiltonian Path.

6. Another Example … (extra credit!) u What paths/cycles exist on this graph? u No Eulerian Cycles: Not all valences are even. u Hamiltonian Cycles? – YES! u No Eulerian Paths: >2 odd-valence vertices. u = Projection of a cube (edge frame); Do other Platonic solids have Hamiltonian cycles ?

7. The Platonic Solids in 3D u Hamiltonian Cycles ? u Eulerian Cycles ?

8. The Octahedron u All vertices have valence 4. u They admit 2 paths passing through. u Pink edges form Hamiltonian cycle. u Yellow edges form Hamiltonian cycle. u The two paths are congruent ! u All edges are covered. u Together they form a Eulerian cycle. u Are there other (semi-)regular polyhedra for which we can do that ?

9. The Cuboctahedron u Hamiltonian cycle on polyhedron edges. u Flattened net of cuboctahedron to show symmetry. The cyan and the red cycles are congruent (mirrored)!

10. Larger Challenges u All these graphs have been planar … boring ! u Our examples had only two Hamiltonian cycles. u Can we find graphs that are covered by three or more Hamiltonian cycles ?  Graphs need to have vertices of valence ≥ 6. u Can we still make those cycles congruent ?  Graphs need to have all vertices equivalent. u Let’s look at complete graphs, i.e., N fully connected vertices.

11. Complete Graphs K 5, K 7, and K 9 u 5, 7, 9 vertices – all connected to each other. u Let’s only consider graphs with all even vertices, i.e., only K 2i+1. K5K5 K7K7 K9K9 Can we make the i Hamiltonian cycles in each graph congruent ?

12. Complete Graphs K 2i+1 u K 2i+1 will need i Hamiltonian cycles for coverage. u Arrange nodes with i-fold symmetry: 2i-gon  C 2i u Last node is placed in center. The common Hamiltonian cycle for all K 2i+1

13. Make Constructions in 3D … u We would like to have highly symmetrical graphs. u All vertices should be of the same even valence. u All vertices should be connected equivalently. u Graph should allow for some symmetrical layout in 3D space. u 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.

14. The 6 Regular Polytopes in 4D From BRIDGES’2002 Talk

15. Which 4D-to-3D Projection ?? u 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

16. 4D Simplex: 2 Hamiltonian Paths Two identical paths, complementing each other C2C2

17. 4D Cross Polytope: 3 Paths u All vertices have valence 6 ! C3C3

18. Hypercube: 2 Hamitonian Cycles u There are many different options:

19. The Most Satisfying Solution ? u Each Path has its own C 2 -symmetry. u 90°-rotation around z-axis changes color on all edges. C 4 (C 2 )

20. The 24-Cell Is More Challenging u Valence 8: -> 4 Hamiltonian paths are needed ! u Natural to consider one C 4 -axis for replicating the Hamiltonian cycles. C4C4

21. 24-Cell: Shell-based Approach u This is a mess … hard to deal with ! Exploiting the concentric shells: 5 families of edges.

22. Shell-Schedule for the 24-Cell u The 24 vertices lie on 3 shells. u Pre-color the shells individually obeying the desired symmetry. u Rotate shells against each other. OUTER SHELL MIDDLE SHELL INNER SHELL This is the visitation schedule for one Ham. cycle.

23. One Cycle – Showing Broken Symmetry Almost C 2

24. C 2 -Symmetry  More than One Cycle! C2C2 u That is what I had to inspect for …

25. Path Composition

26. 24-Cell: 4 Hamiltonian Cycles Aligned:  4-fold symmetry

27. Why Shells Make Task Easier u Decompose problem into smaller ones: l Find a suitable shell schedule; l Prepare components on shells compatible with schedule; l Find a coloring that fits the schedule and glues components together, by “rotating” the shells and connector edges within the chosen symmetry group. u Fewer combinations to deal with. u Easier to maintain desired symmetry.

28. Tetrahedral Symmetry on “0cta”-Shell u C 3 *-rotations that keep one color in place, cyclically exchange the three other ones: C3*C3*C3*C3* C3*C3*C3*C3* C3*C3*C3*C3* C3*C3*C3*C3*

29. “Tetrahedral” Symmetry for the 24-Cell Note that the same-colored edges are disconnected ! u All shells must have the same symmetry orientation - this reduces the size of the search tree greatly. u Of course such a solution may not exist …

30. Another Shell-Schedule for the 24-Cell u 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 !

31. u Only 1/3 of the cycle needs to be found. (C 3 -axis does not go through any edges, vertices) Exploiting C 3 -Symmetry for the 24-Cell OUTER SHELL MIDDLE SHELL INNER SHELL REPEATED UNIT I/O-MIRROR u We can also use inside / outside symmetry, so only 1/6 of the cycle needs to be found !

32. One of the C 3 -symmetric Cycles INSIDE-OUTSIDE SYMMETRY

33. INNER OCTA That is actually how I first found the tetrahedral solution ! Pipe-Cleaner Models for the 24-Cell CENTRAL CUBOCTA OUTER OCTA

34. Rapid Prototyping Model of the 24-Cell u Notice the 3-fold permutation of colors Made on the Z-corp machine.

35. 3D Color Printer (Z Corporation)

36. The Uncolored 120-Cell u 120-Cell: l 600 valence 4-vertices, 1200 edges l --> May yield 2 Hamiltonian cycles length 600.

37. Brute-force approach for the 120-Cell u Assign opposite edges different colors (i.e., build both cycles simultaneously). u Do path-search with backtracking. u Came to a length of 550/600, but then painted ourselves in a corner ! (i.e., could not connect back to the start). u Perhaps we can exploit symmetry to make search tree less deep. Thanks to Mike Pao for his programming efforts !

38. A More Promising Approach u Clearly we need to employ the shell-based approach for these monsters! u But what symmetries can we expect ? l These objects belong to the icosahedral symmetry group which has 6 C 5 -axes and 10 C 3 -axes. l Can we expect the individual paths to have 3-fold or 5-fold symmetry ? This would dramatically reduce the depth of the search tree !

39. 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.

40. Dodecahedral Double Shell Colored by two congruent Hamiltonian cycles

41. Physical Model of Penta-Double-Shell

42. The 600-Cell l 120 vertices, valence 12; l 720 edges;  Make 6 cycles, length 120.

43. Search on the 600-Cell u Search by “loop expansion”: l Replace an edge in the current path with the two other edges of a triangle attached to the chosen edge. l  Always keeps path a closed cycle ! u This quickly worked for finding a full cycle. l Also worked for finding 3 congruent cycles of length 120. l When we tried to do 4 cycles simultaneously, we got to 54/60 using inside/outside symmetry.

44. Shells in the 600-Cell Number of segments of each type in each Hamiltonian cycle INNERMOST TETRAHEDRON OUTERMOST TETRAHEDRON CONNECTORS SPANNING THE CENTRAL SHELL INSIDE / OUTSIDE SYMMETRY

45. Shells in the 600-Cell Summary of features: u 15 shells of vertices u 49 different types of edges: l 4 intra shells with 6 (tetrahedral) edges, l 4 intra shells with 12 edges, l 28 connector shells with 12 edges, l 13 connector shells with 24 edges. u Inside/outside symmetry u What other symmetries are there … ?

46. Start With a Simpler Model … Specifications: u All vertices of valence 12 u Overall symmetry compatible with “tetra-6” u Inner-, outer-most shells = tetrahedra u No edge intersections u As few shells as possible … …. This is tricky … 

47. Icosi-Tetrahedral Double-Shell (ITDS) Just 4 nested shells (192 edges): u Tetrahedron: 4V, 6E u Icosahedron: 12V, 30E u Tetrahedron: 4V, 6E total: 32V CONNECTORS

48. ITDS: The 2 Icosahedral Shells

49. The Complete ITDS: 4 shells, 192 edges SHELLS CONNECTORS TETRA ICOSA TETRA

50. One Cycle on the ITDS SHELLS CONNECTORS TETRA ICOSA TETRA

51. The Composite ITDS

52. ITDS: Composite of 6 Ham. Cycles

53. One Vertex of ITDS (valence 12)

54. Broken Part on Zcorp machine Icosi-tetrahedral Double Shell

55. What Did I Learn from the ITDS ? A larger, more complex model to exercise the shell-based approach. u Shells, or subsets of edges cannot just be rotated as in the first version of the 24-Cell. u The 6-fold symmetry, corresponding to six differently colored edges on a tetrahedron, is actually quite tricky ! u Not one of the standard symmetry groups. u What are the symmetries we can hope for ?

56. The Symmetries of the Composite ? u 4 C 3 rotational axes (thru tetra vertices) that permute two sets of 3 colors each. u Inside/outside mirror symmetry. C 3 (RGB, CMY) C 3 (RMY, CGB) C 3 (GCY, MBR) C 3 (BCM, YRG) Directionality !! ???

57. When Is I/O Symmetry Possible ? Hypothesis: u 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 !

58. Basic Tetra -- truncated -- or beveled Dual (mid-face) -- Dual truncated -- Mid-edge. All Possible Shell Vertex Positions

59. All Possible Edge Patterns ( shown on one tetrahedral face ) 12 6 12+12 12 INTRA- SHELL EDGES INTER- SHELL EDGES Two completely independent sets Constraints between 2 edges of one cycle

60. 0 1 2 3 4 5 Possible Colorings for Intra-Shell Edges Tetras with Offset Edges (12): 6 7 8 9 10 11 Basic Tetrahedron (4):

61. 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 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!

62. Combinatorics for the ITDS u Total colorings: 6 192  10 149 u Pick 192 / 6 edges: ( 192 )  10 37 u Pick one edge at every vertex: 12 32  10 34 u Assuming inside-out symmetry: 12 16  10 17 u All shell combinations: 4 2 *576 2 *12 6 *12 4  10 17 u Combinations in my GUI: 4 2 *576 *12 6 *12 4  10 14 u Constellations examined:  10 3 until success. 32

63. Comparison: ITDS  600-Cell u Total Colorings: [10 149 ]  6 720  10 560 u Pick 120 edges: [10 37 ]  ( 720 )  10 168 u Pick one edge at every vertex: [10 34 ]  12 120  10 130 u Hope for inside-out symmetry: [10 17 ]  12 60  10 65 u All shell combinations: [10 17 ]  4 2 *12 4 *12 54  10 63 u Shells with I/O symmetry: [10 14 ]  4 *12 2 *12 28  10 32 u Constellations to examine: [10 3 ]   10 ?? 120

64. Where is the “Art” … ? u Can these Math Models lead to something artistic as well ? u Any constructivist sculptures resulting from these efforts ? l Suppose you had to show the flow of the various Hamiltonian cycles without the use of color …

65. Complementary Bands in the 5-Cell

66. As a Sculpture

67. Double Volution Shell Resulting from the two complementary Hamiltonian paths on cuboctahedron

68. As a Sculpture

69. 4D Cross Polytope

70. As a Sculpture

71. Conclusions u Finding a Hamiltonian path/cycle is an NP-hard computational problem. u Trying to get Eulerian coverage with a set of congruent Hamiltonian paths is obviously even harder. u Taking symmetry into account judiciously can help enormously.

72. Conclusions (2) u The simpler 4D polytopes yielded their solutions relatively quickly. u Those solutions actually do have nice symmetrical paths! u 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.

73. Conclusions (3) u The 24-Cell, Double-Penta-Shell, and Icosi-Tetra Double Shell, have given me a much deeper understanding of the symmetry issues involved. u Now it’s just a matter of programming these insights into a procedural search to find the 2 remaining solutions. u The 24-Cell is really unusually symmetrical and the most beautiful of them all.

74. QUESTIONS ?