Download presentation

Presentation is loading. Please wait.

Published byGretchen Edling Modified about 1 year ago

1

2
Geometric and Algebraic Methods for H-Cycles Workshop, Adelaide, December, 2012 David G. Glynn, CSEM

3
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,r+1,d] code Finite Fields, GF(2), GF(4) –Algebraic equations for the H-Cycle problem –Finding H-cycles in bipartite graphs using determinants 3

4
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. 4

5
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, 2e=3v, so let v=2r (even), and e=3r. Bond matroid is set of 3r points in PG(r,2). Basic properties of bond matroid follow. 5

6
Bond Matroid B(G) – Graph G 3r points in PG(r,2) Basis of Matroid (r+1 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.) 6

7
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) 7

8
Sylvester’s problem For which sets S of points generating a geometry of dimension r is there a hyperplane containing precisely r independent points? For r=2, AG(2,R) or Euclidean plane, |S| <∞, 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. 8 So H-cycle problem is a special case of Sylvester’s.

9
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).

10
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 PM’s). A hyperplane h of PG(r,2) contains 0 or 2^k SS’s, where the complement of h in B(G) is the union of k+1 even cycles. The dual of the set of all SS’s of B(G) is a set DSS of lines of PG(r,2). 10 H-cycles correspond to points on one line of DSS.

11
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] 11 The 2r lines of B(G) are a quantum set of lines.

12
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). G=K_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.) 12

13
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. 13

14
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 SS’s. In particular, the points of B(G) are complement to an even number of H-cycle hyperplanes. 14

15
Summary of a Method We can find Tait colourings by coordinatizing the matroid B(G) and searching for SS’s. 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 SS’s. 15

16
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 SS’s to B(G) is even. –However, in general, the parities of the number of SS’s and the number of H-cycles of any cubic graph are the same. 16

17
Using GF(4) to Find Tait Colourings GF(4) := GF(2)[w], where w^2 = w+1. Given a cubic graph G let the coordinates of the 3r points Pj of B(G) be a_ij in GF(2), where i=0,…,r, j=1,…,3r. Let vj be variables over GF(4) (thus vj^4=vj). 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 SS’s of B(G). 17

18
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,r+1,d]_2 –Cycles are disjoint unions of circuits. –Closed under symmetric differences –C has rank r+1, distance d (= girth) –Basis of C from spanning tree T 18

19
M(x1,…,xr+1) and H-cycles Let x1,…,xr+1 be r+1 variables over GF(2) and C1,…,Cr+1 be the r+1 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 r+1 variables. Thm (Warning 1936): if a pol equation of degree d in m>d 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 2^2 = 4. (Tutte) Thm (Ax-Katz 1964): The number of solns of above pol equation is divisible by q^[m/d]. Cor The number of H-cycles of a cubic bipartite graph is even. (Kotzig, 1958). 19

20
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 r+1 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 >= 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 r^2(r^2-1)/12. 20

21

Similar presentations

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google