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 ?
          Answer:

 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
  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
    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



                                Made on
                                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 … ?
     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
                               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
            Adding the second half-edge:



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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