How does the “icosian calculus” help to find a hamiltonian cycle?

Question

I have read in many places that Hamilton thought of an algebraic structure which he called icosian calculus and used it to find a hamiltonian cycle in the dodecahedron graph and in other platonic solids. But I can't find anywhere an explanation of how it helps. Does anyone know a source to look or can give an explanation?

Answer 1

UPDATED/EDITED: The Icosian calculus used by Hamilton, is a group-theoretic construction, of the different ways to move along edges of the dodecahedron. Hamilton defines 3 different operations for moving along edges, which are represented by elements of a group presentation. Suppose you're moving along a particular edge in the dodecahedron,and moving in a particular direction. We let $i$ denote the action of moving from your current edge, to the edge to your right. Since the dodecahedron has pentagons for faces, 5 right turns brings you back to the edge you began on. So we express this algebraically by writing $i^5=1$ So applying $i$ 5 times is equivalent to an identity element, as it brings us back to the original edge.
Similarly, we let $j$ represent moving around the vertex that you're directed towards while on an edge, in an anti clockwise fashion. So the first application of $j$ also moves you to the edge to the right, but it keeps your "direction" pointing at the same vertex as previously. Since every vertex is incident with 3 edges in the dodecahedron, 3 applications of $j$ bring you back to the original edge so $j^3=1$ Lastly $k$ represents staying on the same edge, and just change direction, thus $k^2=1$

Since Hamiltonian paths of the dodecahedron have length 20, All we need to do know is find all combinations of $i,j,k$ where there are 20 of $i,j$ (representing twenty edge changes, or a path length of 20) such that the $i,j,k$'s multiply together to yield the identity, and no sub string of them yield the identity. This then represents every possible way of starting on one edge, making twenty moves, and returning to the original edge.

If you'd like to investigate this further, try this; https://www.jstor.org/stable/20490184?seq=4#page_scan_tab_contents You have to have a membership to view it, but memberships are free, and even though it costs $10 to download you can view it online for free.