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 ifa 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 41 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 41 If you like that "geometrical" way of thinking about knots and links, see the paper "Ordering Knots" by S.V.Jablan.