Geometric and Algebraic Methods for H-Cycles

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Geometric and Algebraic Methods for H-Cycles

• Workshop, Adelaide, 14-15 December, 2012
• David G. Glynn, CSEM

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Outline of Talk

• Cubic graph
• Hamilton Cycle/Edge 3-colouring
• Circuits, Bonds
• Bond Matroid (Dual to Cycle Matroid) of the Graph
• Sylvester Problem in Geometry
• Quantum Sets of Lines
• Coordinates (of points and lines)
• Binary Codes of the Graph
• Weights of Words
• Locally Minimal Words of 3r,r1,d code
• Finite Fields, GF(2), GF(4)
• Algebraic equations for the H-Cycle problem
• Finding H-cycles in bipartite graphs using
  determinants

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Cubic Graphs (simple)

• We shall see how to get all the H-cycles from the
  Tait colourings.
• First we need matroid representation of G.
• There are 2 choices
• Circuit or Bond matroid rep.

• Every H-cycle gives a Tait edge 3-colouring.
• Just colour the edges alternately round the cycle
  a and b, and the other transverse edges c.
• Snarks have no edge 3-colourings and so no
  H-cycles.

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Bond Matroid of Graph

• Every graph has a regular rep over any field,
  but usually choose GF(2).
• A (connected) graph with v vertices and e edges
  has a bond rep in PG(e-v,2) as e points (the
  edges).

• Our case cubic graph, 2e3v, so let v2r (even),
  and e3r.
• Bond matroid is set of 3r points in PG(r,2).
• Basic properties of bond matroid follow.

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Bond Matroid B(G) Graph G

• 3r points in PG(r,2)
• Basis of Matroid (r1 points)
• Circuit of matroid
• Small circuits
• 1 zero point (vector)
• 2 identical points
• 3 points in a line

• 3r edges in graph
• Complement of spanning tree
• Bond of graph
• Small bonds
• Isthmus (bridge)
• 2-bond
• 3-bond e.g. edges through vertex

Can assume no 3-bonds except vertex 3-bonds,
and no isthmuses or 2-bonds. (Just makes it
easier for this talk.)
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What of B(G) H-Cycle?

• A circuit of G corresponds to the complement of a
  hyperplane of B(G), a maximal set of points not
  ctg a basis.
• Therefore an H-cycle (circuit of size 2r)
  corresponds to the complement of a hyperplane
  intersecting B(G) in r points, so that these r
  points are independent in PG(r,2). (r edges are
  perfect matching)

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Sylvesters problem

• For which sets S of points generating a geometry
  of dimension r is there a hyperplane containing
  precisely r independent points?

• For r2, AG(2,R) or Euclidean plane, S lt8,
  there is a line ctg 2 points of S.
• For AG(2,C), this is not true in general.
• The set of 9 inflection points of an elliptic
  curve has no 2-secants.

So H-cycle problem is a special case of
Sylvesters.
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Perfect Matchings

• A perfect matching (PM) is a set r edges of G
  that cover all 2r vertices.
• The complement of an H-cycle (a circuit of size
  2r) is a PM.
• In general, the complement of a PM is an
  edge-disjoint union of k circuits.
• Some may be odd circuits.
• If all are even, then this corresponds in B(G) to
  a secundum skew to B(G).

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Skew Secundums

• A secundum in PG(r,2) is a subspace of dim r-2.
• The dual is a line.
• There are three hyperplanes passing through a
  Skew Sec
• Each intersects B(G) in r points (3 PMs).

• A hyperplane h of PG(r,2) contains 0 or 2k SSs,
  where the complement of h in B(G) is the union of
  k1 even cycles.
• The dual of the set of all SSs of B(G) is a set
  DSS of lines of PG(r,2).

H-cycles correspond to points on one line of DSS.
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A Quantum Interlude

• A quantum set of lines of binary space PG(r,2) is
  a set of n lines such that their Grassmann
  coordinates sum to zero.
• G-coords of a line generated by points a(a_i),
  and b(b_i) in homogeneous coords are the r
  choose 2 subdets of (ab) of size 2. The geometry
  of additive quantum codes. Glynn, Gulliver, Maks,
  Gupta 2004., on internet

The 2r lines of B(G) are a quantum set of lines.
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Coordinates for the Graph

• Each edge of G (i.e. point of B(G)) has r
  hom-coords over GF(2).
• Each vertex of G (i.e. line of B(G)) has r choose
  2 G-coords over GF(2).
• GK_4, M(G) is Pasch configuration of 6 points
  and 4 lines in PG(2,2).
• Edge hom-coords are 1,2,3,12,23,123 vertex
  G-coords are 12, 13,23, 12,13, 23. (For
  hom-coords ij means x_i x_j 1 , for G-coords,
  ij, means ij subdet 1.)

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Result using B(G)

• If a cubic graph has one Hamilton cycle it has at
  least three. (Tutte)
• Pf The symmetric difference of the lines of DSS
  (dual of skew-secundums) corresponds to the set
  of H-cycles.
• The (geometric) binary code generated by the
  characteristic functions of the lines in PG(r,2)
  has minimal distance (or weight) three.
• If there are 3 H-cycles they correspond to a line
  of DSS. If there are 4 H-cycles they correspond
  to four non-collinear points on a plane of DSS.

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Another Result using B(G)

• Any edge of G contains an even number of
  H-cycles.
• Pf The geometric code generated by the
  characteristic functions of the lines of PG(r,2)
  is orthogonally dual to the code of complements
  of hyperplanes.
• Using duality of PG(r,2), this means that any
  point of PG(r,2) is complement to an even number
  of hyperplanes through the SSs. In particular,
  the points of B(G) are complement to an even
  number of H-cycle hyperplanes.

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Summary of a Method

• We can find Tait colourings by coordinatizing the
  matroid B(G) and searching for SSs.
• It starts with solving 2r linear equations with r
  choose 2 variables since a secundum and a line
  are skew iff the inner product of their coords is
  1.
• Then the H-cycles come from the SSs.

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Bipartite Graphs

• The number of H-cycles in a 3-edge connected
  bipartite cubic graph is even. (Bosak, 1967)
• Pf We can obtain a system of r-1 algebraic
  equations E_i 1 over GF(2) having 2r-2
  variables and the equation E Prod_i E_i 1
  has degree 2r-2 in the 2r-2 variables.
• The number (mod 2) of solutions is therefore the
  coefficient of the diagonal monomial x_1
  x_2r-2 in this product. It turns out that it
  is easily calculated to be zero.
• Thus the number of SSs to B(G) is even.
• However, in general, the parities of the number
  of SSs and the number of H-cycles of any cubic
  graph are the same.

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Using GF(4) to Find Tait Colourings

• GF(4) GF(2)w, where w2 w1.
• Given a cubic graph G let the coordinates of the
  3r points Pj of B(G) be a_ij in GF(2), where
  i0,,r, j1,,3r.
• Let vj be variables over GF(4) (thus vj4vj).
• For each point of B(G) (i.e. edge of G) let Pj
  Sigmaj aij.vj (a linear form).
• Prod over j Pj 1 has one solution for each Tait
  colouring
• Corollary G is a snark iff this is identically
  zero.
• The short proof of these facts uses some
  geometry, such as embedding PG(r,2) in PG(r,4),
  and looking at Baer involutions etc. The main
  idea is that the dual of B(G) is a collection of
  3r hyperplanes, and the points not covered by
  these in PG(r,4) correspond to the SSs of B(G).

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Determinants can find them!

• Consider a bipartite cubic graph G
• G has r r vertices and 3r edges.
• G corresponds to r by r binary matrix M.
• Use the cycle code C 3r,r1,d_2
• Cycles are disjoint unions of circuits.
• Closed under symmetric differences
• C has rank r1, distance d ( girth)
• Basis of C from spanning tree T

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M(x1,,xr1) and H-cycles

• Let x1,,xr1 be r1 variables over GF(2) and
  C1,,Cr1 be the r1 circuits generating the
  cycle code.
• Represent Ci by an r by r binary matrix Mi.
• Let M Sigma xi . Mi
• Then rk(M) r-1 iff it is an H-cycle.
• I.e. iff any submatrix M of M has det(M) 1.
• So, the number of H-cycles of G is the number of
  solutions of det(M) 1, for xi in GF(2).
• Cor Det(M) 0 iff no H-cycles.

• NB det(M)1 is of degree r-1 and has r1
  variables.
• Thm (Warning 1936) if a pol equation of degree d
  in mgtd variables over GF(q) has at least one
  solution then it has at least q(m-d).
• Cor if G has an H-cycle then it has at least 22
  4. (Tutte)
• Thm (Ax-Katz 1964) The number of solns of above
  pol equation is divisible by qm/d.
• Cor The number of H-cycles of a cubic bipartite
  graph is even. (Kotzig, 1958).

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Sub H-Cycles

• Suppose there are only a very small number of
  H-cycles in a cubic bipartite graph G. Then
  there will be at least some sub-H-cycles of large
  sizes.
• e.g. for (2r-2)-cycles. The number of solutions
  of a pol equation of degree r1 of degree r-2 is
  divisible by 8 (Ax-Katz).
• If there are between 0 and 6 H-cycles in G then
  for an r-2 by r-2 submatrix of M there are at
  least 8 solutions in some cases. So some of these
  solutions correspond to (2r-2)-cycles.
• For 2k-cycles (k gt r-2) consider r-3 by r-3
  subdets det(M) 1, and there will be at least
  16 solutions, corresponding to such cycles. etc.
• Details are omitted here for example, the number
  of square subdeterminants of size r-2 of an r by
  r H-cycle matrix M that have det 1 is
  r2(r2-1)/12.

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