GRAPHS Here is the problem by Louis Carrol: make a contour drawing of this figure by a singular closed line, never passing twice through the same point. Find all solutions! The first question is: is there any solution? The answer is - yes, because the contour is an Eulerian graph: in every vertex there is an even number of edges. Try to prove this general statement! How many different solutions exist if a curve may have a finite number of self-intersection points? Every Eulerian graph is a projection of some knot or link and vice versa Such a projection is called regular if the graph is 4-regular, i.e. if the valence of every vertex is 4. Otherwise, a projection is irregular. By slightly changing it, it is always possible to turn some irregular projection of a knot or link into a regular one. Two knot or link projections are isomorphic (or simply, equal or same) if they are isomorphic as graphs on a sphere. Trying to find all non isomorphic projections of alternating knots and links with n crossings, we need to find all non isomorphic 4-regular planar graphs with n vertices and vice versa Among them, we can distinguish graphs with or without digons. If we denote digons by colored edges, we can imagine a trefoil as a triangle with all colored edges, a knot 4_{1} as a tetrahedron with two colored non-adjacent edges, Borromean rings as an octahedron... After that, we can replace every digon by a chain of digons, and obtain different families of knots and links. For example, this is the family generated by the knot 4_{1} If you like that "geometrical" way of thinking about knots and links, see the paper "Ordering Knots" by S.V.Jablan. |