Here is the problem by Louis Carrol: make a contur 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 contur 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 the 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, the 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
the graphs on a sphere. Trying to find all nonisomorphic
projections of alternating knots and links with *n*
crossings, we need to find all nonisomorphic 4-regular planar
graphs with *n* vertices and *vice versa* Among them,
we could distinguish graphs with or without digons. If we denote
digons by colored edges, we could imagine the trefoil as a
triangle with all colored edges, the knot 4_{1} as a
tetrahedron with two colored nonadjacent edges, Borromean rings
as an octahedron...

After that, you could 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
"Geometry of links" by S.V.Jablan.